Tasty Bits of Several Complex Variables

Home , Antiholomorphic function, Pluriharmonic function, Removable singularity

A whirlwind tour of the subject

Jiří Lebl

December 23, 2020 (version 3.4) 2

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Acknowledgments: I would like to thank Debraj Chakrabarti, Anirban Dawn, Alekzander Malcom, John Treuer, Jianou Zhang, Liz Vivas, Trevor Fancher, Nicholas Lawson McLean, Alan Sola, Achinta Nandi, Sivaguru Ravisankar, Richard Lärkäng, Elizabeth Wulcan, Tomas Rodriguez, and students in my classes for pointing out typos/errors and helpful suggestions.

During the writing of this book, the author was in part supported by NSF grant DMS-1362337.

More information:

See https://www.jirka.org/scv/ for more information (including contact information). The LATEX source for the book is available for possible modiﬁcation and customization at github:

https://github.com/jirilebl/scv Contents

Introduction 5

0.1 Motivation, single variable, and Cauchy’s formula ...... 5

1 Holomorphic functions in several variables 11

1.1 Onto several variables ...... 11

1.2 Power series representation ...... 16

1.3 Derivatives ...... 24

1.4 Inequivalence of ball and polydisc ...... 28

1.5 Cartan’s uniqueness theorem ...... 32

1.6 Riemann extension theorem, zero sets, and injective maps ...... 35

2 Convexity and pseudoconvexity 40

2.1 Domains of holomorphy and holomorphic extensions ...... 40

2.2 Tangent vectors, the Hessian, and convexity ...... 44

2.3 Holomorphic vectors, the Levi form, and pseudoconvexity ...... 51

2.4 Harmonic, subharmonic, and plurisubharmonic functions ...... 63

2.5 Hartogs pseudoconvexity ...... 73

2.6 Holomorphic convexity ...... 80

3 CR functions 86

3.1 Real-analytic functions and complexiﬁcation ...... 86

3.2 CR functions ...... 92

3.3 Approximation of CR functions ...... 98

3.4 Extension of CR functions ...... 105

¯ 4 The 휕-problem 109

4.1 The generalized Cauchy integral formula ...... 109 ¯ 4.2 Simple case of the 휕-problem ...... 111

4.3 The general Hartogs phenomenon ...... 113

5 Integral kernels 116

5.1 The Bochner–Martinelli kernel ...... 116

5.2 The Bergman kernel ...... 119

5.3 The Szegö kernel ...... 123 4 CONTENTS

6 Complex analytic varieties 125

6.1 The ring of germs ...... 125

6.2 Weierstrass preparation and division theorems ...... 127

6.3 The dependence of zeros on parameters ...... 132

6.4 Properties of the ring of germs ...... 135

6.5 Varieties ...... 137

6.6 Hypervarieties ...... 140

6.7 Irreducibility, local parametrization, and Puiseux theorem ...... 143

6.8 Segre varieties and CR geometry ...... 146

A Basic notation and terminology 151

B Results from one complex variable 152

C Diﬀerential forms and Stokes’ theorem 162

D Basic terminology and results from algebra 170

Further reading 173

Index 174

List of Notation 179 Introduction

This book is a polished version of my course notes for Math 6283, Several Complex Variables, given in Spring 2014, Spring 2016, and Spring 2019 semesters at Oklahoma State University. It is meant for a semester-long course. Quite a few exercises of various diﬃculty are sprinkled throughout the text, and I hope a reader is at least attempting all of them. Many are required later in the text. The reader should attempt exercises in sequence as earlier exercises can help or even be required to solve later exercises. The prerequisites are a decent knowledge of vector calculus, basic real analysis, and a working knowledge of complex analysis in one variable. Measure theory (Lebesgue integral and its convergence theorems) is useful, but it is not essential except in a couple of places later in the book. The ﬁrst two chapters and most of the third is accessible to beginning graduate students after one semester of a standard single-variable complex analysis graduate course. From time to time (e.g.

proof of Baouendi–Trèves in chapter 3 , and most of chapter 4 , and chapter 5 ), basic knowledge of

diﬀerential forms is useful, and in chapter 6 we use some basic ring theory from algebra. By design, it can replace the second semester of complex analysis. This book is not meant as an exhaustive reference. It is simply a whirlwind tour of several complex variables with slightly more material than can be covered within a semester. See the end of

the book for a list of books for reference and further reading. There are also appendices for a list of

one-variable results, an overview of diﬀerential forms, and some basic algebra. See appendix B ,

appendix C , and appendix D .

0.1 Motivation, single variable, and Cauchy’s formula

Let us start with some standard notation. We use ℂ for the complex√ numbers, ℝ for real numbers, ℤ for integers, ℕ = {1, 2, 3,...} for natural numbers, 푖 = −1. Throughout this book, we use the standard terminology of domain to mean connected open set. We will try to avoid using it if connectedness is not needed, but sometimes we use it just for simplicity. √ As complex analysis deals with the complex√ numbers, perhaps we should begin with −1. We start with the real numbers, ℝ, and we add −1 into our ﬁeld. We call this square root 푖, and we write the complex numbers, ℂ, by identifying ℂ with ℝ2 using 푧 = 푥 + 푖푦, where 푧 ∈ ℂ, and (푥, 푦) ∈ ℝ2. A subtle philosophical issue is that there are two square roots of −1. Two chickens are running around in our yard, and because we like to know which is which, we catch one and write “푖” on it. If we happened to have caught the other chicken, we would have gotten an exactly equivalent theory, which we could not tell apart from the original. 6 INTRODUCTION

Given a complex number 푧, its “opposite” is the complex conjugate of 푧 and is deﬁned as

def 푧¯ = 푥 − 푖푦. The size of 푧 is measured by the so-called modulus, which is just the Euclidean distance:

def √ √︃ |푧| = 푧푧¯ = 푥2 + 푦2. If 푧 = 푥 + 푖푦 ∈ ℂ for 푥, 푦 ∈ ℝ, then 푥 is called the real part and 푦 is called the imaginary part. We write 푧 + 푧¯ 푧 − 푧¯ Re 푧 = Re(푥 + 푖푦) = = 푥, Im 푧 = Im(푥 + 푖푦) = = 푦. 2 2푖 A function 푓 : 푈 ⊂ ℝ푛 → ℂ for an open set 푈 is said to be continuously differentiable, or 퐶1 if the first (real) partial derivatives exist and are continuous. Similarly, it is 퐶푘 or 퐶푘 -smooth if the first 푘 partial derivatives all exist and are differentiable. Finally, a function is said to be 퐶∞ or ∗ 푘 simply smooth if it is infinitely differentiable, or in other words, if it is 퐶 for all 푘 ∈ ℕ. Complex analysis is the study of holomorphic (or complex-analytic) functions. Holomorphic functions are a generalization of polynomials, and to get there one leaves the land of algebra to arrive in the realm of analysis. One can do an awful lot with polynomials, but sometimes they are just not enough. For example, there is no nonzero polynomial function that solves the simplest of differential equations, 푓 0 = 푓 . We need the exponential function, which is holomorphic. Let us start with polynomials. In one variable, a polynomial in 푧 is an expression of the form

푑 ∑︁ 푗 푃(푧) = 푐푗 푧 , 푗=0 where 푐푗 ∈ ℂ and 푐푑 ≠ 0. The number 푑 is called the degree of the polynomial 푃. We can plug in some number 푧 and simply compute 푃(푧), so we have a function 푃 : ℂ → ℂ. We try to write ∞ ∑︁ 푗 푓 (푧) = 푐푗 푧 푗=0 and all is very ﬁne until we wish to know what 푓 (푧) is for some number 푧 ∈ ℂ. We usually mean

∞ 푑 ∑︁ 푗 ∑︁ 푗 푐푗 푧 = lim 푐푗 푧 . 푑→∞ 푗=0 푗=0 As long as the limit exists, we have a function. You know all this; it is your one-variable complex analysis. We usually start with the functions and prove that we can expand into series. Let 푈 ⊂ ℂ be open. A function 푓 : 푈 → ℂ is holomorphic (or complex-analytic) if it is complex-diﬀerentiable at every point, that is, if 푓 (푧 + 휉) − 푓 (푧) 푓 0(푧) = lim exists for all 푧 ∈ 푈. 휉∈ℂ→0 휉

∗While 퐶∞ is the common deﬁnition of smooth, not everyone always means the same thing by the word smooth.I have seen it mean diﬀerentiable, 퐶1, piecewise-퐶1, 퐶∞, holomorphic, . . . 0.1. MOTIVATION, SINGLE VARIABLE, AND CAUCHY’S FORMULA 7

Importantly, the limit is taken with respect to complex 휉. Another vantage point is to start with a ∗ continuously diﬀerentiable 푓 , and say 푓 = 푢 +푖 푣 is holomorphic if it satisﬁes the Cauchy–Riemann equations: 휕푢 휕푣 휕푢 휕푣 = , = − . 휕푥 휕푦 휕푦 휕푥 The so-called Wirtinger operators,

휕 1 휕 휕 휕 1 휕 휕 def= − 푖 , def= + 푖 , 휕푧 2 휕푥 휕푦 휕푧¯ 2 휕푥 휕푦

provide an easier way to understand the Cauchy–Riemann equations. These operators are determined by insisting 휕 휕 휕 휕 푧 = 1, 푧¯ = 0, 푧 = 0, 푧¯ = 1. 휕푧 휕푧 휕푧¯ 휕푧¯ The function 푓 is holomorphic if and only if 휕 푓 = 0. 휕푧¯ That seems a far nicer statement of the Cauchy–Riemann equations, and it is just one complex equation. It says a function is holomorphic if and only if it depends on 푧 but not on 푧¯ (perhaps that might not make a whole lot of sense at ﬁrst glance). Let us check:

휕 푓 1 휕 푓 휕 푓 1 휕푢 휕푣 휕푢 휕푣 1 휕푢 휕푣 푖 휕푣 휕푢 = + 푖 = + 푖 + 푖 − = − + + . 휕푧¯ 2 휕푥 휕푦 2 휕푥 휕푥 휕푦 휕푦 2 휕푥 휕푦 2 휕푥 휕푦

This expression is zero if and only if the real parts and the imaginary parts are zero. In other words 휕푢 휕푣 휕푣 휕푢 − = , + = . 휕푥 휕푦 0 and 휕푥 휕푦 0

That is, the Cauchy–Riemann equations are satisﬁed. If 푓 is holomorphic, the derivative in 푧 is the standard complex derivative you know and love:

휕 푓 0 푓 (푧0 + 휉) − 푓 (푧0) (푧0) = 푓 (푧0) = lim . 휕푧 휉→0 휉

That is because 휕 푓 1 휕푢 휕푣 푖 휕푣 휕푢 휕푢 휕푣 휕 푓 = + + − = + 푖 = 휕푧 2 휕푥 휕푦 2 휕푥 휕푦 휕푥 휕푥 휕푥 1 휕푢 휕푣 휕 푓 = + 푖 = . 푖 휕푦 휕푦 휕(푖푦)

A function on ℂ is really a function defined on ℝ2 as identified above and hence it is a function 푥 푦 푥 푧+푧¯ 푦 푧−푧¯ 푧 of and . Writing = 2 and = 2푖 , we think of it as a function of two complex variables and ∗Holomorphic functions end up being infinitely differentiable anyway, so this hypothesis is not overly restrictive. 8 INTRODUCTION

푧¯. Pretend for a moment as if 푧¯ did not depend on 푧. The Wirtinger operators work as if 푧 and 푧¯ really were independent variables. For instance: 휕 휕 푧2푧¯3 + 푧10 = 2푧푧¯3 + 10푧9 and 푧2푧¯3 + 푧10 = 푧2(3¯푧2) + 0. 휕푧 휕푧¯ So a holomorphic function is a function not depending on 푧¯. The most important theorem in one variable is the Cauchy integral formula. Theorem 0.1.1 (Cauchy integral formula). Let 푈 ⊂ ℂ be a bounded domain where the boundary 휕푈 is a piecewise smooth simple closed path (a Jordan curve). Let 푓 : 푈 → ℂ be a continuous function, holomorphic in 푈. Orient 휕푈 positively (going around counter-clockwise). Then 1 ∫ 푓 (휁) 푓 (푧) = 푑휁 for all 푧 ∈ 푈. 2휋푖 휕푈 휁 − 푧 The Cauchy formula is the essential ingredient we need from one complex variable. It follows ∗ from Green’s theorem (Stokes’ theorem in two dimensions). You can look forward to Theorem 4.1.1 for a proof of a more general formula, the Cauchy–Pompeiu integral formula. As a differential form, 푑푧 = 푑푥 + 푖 푑푦. If you are uneasy about differential forms, you probably defined the path integral above directly using the Riemann–Stieltjes integral in your one-complex- variable class. Let us write down the formula in terms of the standard Riemann integral in a special case. Take the unit disc def 픻 = 푧 ∈ ℂ : |푧| < 1 . The boundary is the unit circle 휕픻 = 푧 ∈ ℂ : |푧| = 1 oriented positively, that is, counter-clockwise. Parametrize 휕픻 by 푒푖푡, where 푡 goes from 0 to 2휋. If 휁 = 푒푖푡, then 푑휁 = 푖푒푖푡 푑푡, and 1 ∫ 푓 (휁) 1 ∫ 2휋 푓 (푒푖푡)푒푖푡 푓 (푧) = 푑휁 = 푑푡. 푖푡 2휋푖 휕픻 휁 − 푧 2휋 0 푒 − 푧 If you are not completely comfortable with path integrals, try to think about how you would parametrize the path, and write the integral as an integral any calculus student would recognize. I venture a guess that 90% of what you learned in a one-variable complex analysis course (depending on who taught it) is more or less a straightforward consequence of the Cauchy integral formula. An important theorem from one variable that follows from the Cauchy formula is the maximum modulus principle (or just the maximum principle). Let us give its simplest version. Theorem 0.1.2 (Maximum modulus principle). Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℂ is holomorphic. If for some 푧0 ∈ 푈 sup | 푓 (푧)| = | 푓 (푧0)|, 푧∈푈

then 푓 is constant, that is, 푓 ≡ 푓 (푧0). That is, if the supremum is attained in the interior of the domain, then the function must be constant. Another way to state the maximum principle is to say: If 푓 extends continuously to the boundary of a domain, then the supremum of | 푓 (푧)| is attained on the boundary. In one variable you learned that the maximum principle is really a property of harmonic functions.

∗If you wish to feel inadequate, note that this theorem, on which all of complex analysis (and all of physics) rests, was proved by George Green, who was the son of a miller and had one year of formal schooling. 0.1. MOTIVATION, SINGLE VARIABLE, AND CAUCHY’S FORMULA 9

Theorem 0.1.3 (Maximum principle). Let 푈 ⊂ ℂ be a domain and ℎ : 푈 → ℝ harmonic, that is, 휕2ℎ 휕2ℎ ∇2ℎ = + = 0. 휕푥2 휕푦2

If for some 푧0 ∈ 푈 sup ℎ(푧) = ℎ(푧0) or inf ℎ(푧) = ℎ(푧0), 푧∈푈 푧∈푈 then ℎ is constant, that is, ℎ ≡ ℎ(푧0). In one variable, if 푓 = 푢 + 푖푣 is holomorphic for real valued 푢 and 푣, then 푢 and 푣 are harmonic. Locally, any harmonic function is the real (or imaginary) part of a holomorphic function, so in one complex variable, studying harmonic functions is almost equivalent to studying holomorphic functions. Things are decidedly diﬀerent in two or more variables. Holomorphic functions admit a power series representation in 푧 at each point 푎: ∞ ∑︁ 푗 푓 (푧) = 푐푗 (푧 − 푎) . 푗=0 푧 휕 푓 No ¯ is necessary, since 휕푧¯ = 0. Let us see the proof using the Cauchy integral formula as we will require this computation in several variables as well. Given 푎 ∈ ℂ and 휌 > 0, deﬁne the disc of radius 휌 around 푎 def Δ휌 (푎) = 푧 ∈ ℂ : |푧 − 푎| < 휌 .

Suppose 푈 ⊂ ℂ is open, 푓 : 푈 → ℂ is holomorphic, 푎 ∈ 푈, and Δ휌 (푎) ⊂ 푈 (that is, the closure of the disc is in 푈, and so its boundary 휕Δ휌 (푎) is also in 푈). For 푧 ∈ Δ휌 (푎) and 휁 ∈ 휕Δ휌 (푎),

푧 − 푎 |푧 − 푎| = < 1. 휁 − 푎 휌 0 |푧 − 푎| ≤ 휌0 < 휌 푧−푎 ≤ 휌 < In fact, if , then 휁−푎 휌 1. Therefore, the geometric series ∞ ∑︁ 푧 − 푎 푗 1 휁 − 푎 = = 휁 − 푎 1 − 푧−푎 휁 − 푧 푗=0 휁−푎 − 푗 푧, 휁 0 푎 휕 푎 Í 푧 푎 converges uniformly absolutely for ( ) ∈ Δ휌 ( ) × Δ휌 ( ) (that is, 푗 휁−푎 converges uniformly). Let 훾 be the path going around 휕Δ휌 (푎) once in the positive direction. Compute 1 ∫ 푓 (휁) 푓 (푧) = 푑휁 2휋푖 훾 휁 − 푧 1 ∫ 푓 (휁) 휁 − 푎 = 푑휁 2휋푖 훾 휁 − 푎 휁 − 푧 ∞ 1 ∫ 푓 (휁) ∑︁ 푧 − 푎 푗 = 푑휁 2휋푖 휁 − 푎 휁 − 푎 훾 푗=0 ∞ ∑︁ 1 ∫ 푓 (휁) = 푑휁 (푧 − 푎) 푗 . 2휋푖 휁 푎 푗+1 푗=0 훾 ( − ) 10 INTRODUCTION

In the last equality, we may interchange the limit on the sum with the integral either via Fubini’s

푧 푀 푓 (휁) = | 푓 (휁)| theorem or via uniform convergence: is ﬁxed and if is the supremum of 휁−푎 휌 on 휕Δ휌 (푎), then

푓 (휁) 푧 − 푎 푗 |푧 − 푎| 푗 |푧 − 푎| ≤ 푀 , < . 휁 푎 휁 푎 휌 and 휌 1 − − 1 The key point is writing the Cauchy kernel 휁−푧 as 1 1 휁 − 푎 = , 휁 − 푧 휁 − 푎 휁 − 푧

and then using the geometric series. Not only have we proved that 푓 has a power series, but we computed that the radius of convergence is at least 푅, where 푅 is the maximum 푅 such that Δ푅 (푎) ⊂ 푈. We also obtained a formula for the coeﬃcients 1 ∫ 푓 (휁) 푐 = 푑휁. 푗 푗+1 2휋푖 훾 (휁 − 푎) For a set 퐾, denote the supremum norm:

def k 푓 k퐾 = sup| 푓 (푧)|. 푧∈퐾 By a brute force estimation, we obtain the very useful Cauchy estimates:

1 ∫ 푓 (휁) 1 ∫ k 푓 k k 푓 k |푐 | = 푑휁 ≤ 훾 |푑휁 | = 훾 . 푗 푗+1 푗+1 푗 2휋푖 훾 (휁 − 푎) 2휋 훾 휌 휌

푗 휕 We diﬀerentiate Cauchy’s formula times (using the Wirtinger 휕푧 operator),

휕 푗 푓 1 ∫ 푗! 푓 (휁) 푓 ( 푗) (푧) = (푧) = 푑휁, 푗 푗+1 휕푧 2휋푖 훾 (휁 − 푧)

and therefore 휕 푗 푓 푗! 푐 = 푓 ( 푗) (푎) = (푎). 푗 휕푧 푗 Hence, we can control derivatives of 푓 by the size of the function:

휕 푗 푓 푗 푓 ( 푗) !k k훾 푓 (푎) = (푎) ≤ . 휕푧 푗 휌 푗 This estimate is one of the key properties of holomorphic functions, and the reason why the correct topology for the set of holomorphic functions is the same as the topology for continuous functions. Consequently, obstructions to solving problems in complex analysis are often topological in character.

For further review of one-variable results, see appendix B . Chapter 1

Holomorphic functions in several variables

1.1 Onto several variables

푛 times z }| { Let ℂ푛 = ℂ × ℂ × · · · × ℂ denote the 푛-dimensional complex Euclidean space. Denote by 푧 = 푛 (푧1, 푧2, . . . , 푧푛) the coordinates of ℂ . Let 푥 = (푥1, 푥2, . . . , 푥푛) and 푦 = (푦1, 푦2, . . . , 푦푛) denote 푛 푛 2푛 coordinates in ℝ . We identify ℂ with ℝ by letting 푧 = 푥 + 푖푦, that is, 푧푗 = 푥푗 + 푖푦푗 for every 푗. As in one complex variable, we write 푧¯ = 푥 − 푖푦. We call 푧 the holomorphic coordinates and 푧¯ the antiholomorphic coordinates. 푛 Deﬁnition 1.1.1. For 휌 = (휌1, 휌2, . . . , 휌푛) where 휌푗 > 0 and 푎 ∈ ℂ , deﬁne a polydisc

def 푛 Δ휌 (푎) = 푧 ∈ ℂ : |푧푗 − 푎푗 | < 휌푗 for 푗 = 1, 2, . . . , 푛 .

We call 푎 the center and 휌 the polyradius or simply the radius of the polydisc Δ휌 (푎). If 휌 > 0 is a number, then def 푛 Δ휌 (푎) = 푧 ∈ ℂ : |푧푗 − 푎푗 | < 휌 for 푗 = 1, 2, . . . , 푛 . In two variables, a polydisc is sometimes called a bidisc. As there is the unit disc 픻 in one variable, so is there the unit polydisc in several variables: 푛 푛 픻 = 픻 × 픻 × · · · × 픻 = Δ1(0) = 푧 ∈ ℂ : |푧푗 | < 1 for 푗 = 1, 2, . . . , 푛 . In more than one complex dimension, it is diﬃcult to draw exact pictures for lack of real dimensions on our paper. We visualize the unit polydisc in two variables by drawing the following picture by plotting just against the modulus of the variables:

|푧2|

휕픻2 픻2

|푧1| 12 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Recall the Euclidean inner product on ℂ푛:

def h푧, 푤i = 푧1푤¯ 1 + 푧2푤¯ 2 + · · · + 푧푛푤¯ 푛.

The inner product gives us the standard Euclidean norm on ℂ푛:

def √︁ √︁ 2 2 2 k푧k = h푧, 푧i = |푧1| + |푧2| + · · · + |푧푛| .

This norm agrees with the standard Euclidean norm on ℝ2푛. Deﬁne balls as in ℝ2푛:

def 푛 퐵휌 (푎) = 푧 ∈ ℂ : k푧 − 푎k < 휌 ,

And deﬁne the unit ball as

def 푛 픹푛 = 퐵1(0) = 푧 ∈ ℂ : k푧k < 1 .

A ball centered at the origin can also be pictured by plotting against the modulus of the variables since the inequality deﬁning the ball only depends on the moduli of the variables. Not every domain can be drawn like this, but if it can, it is called a Reinhardt domain, more on this later. Here is a picture of 픹2:

|푧2|

휕픹2

픹2

|푧1|

Deﬁnition 1.1.2. Let 푈 ⊂ ℂ푛 be an open set. A function 푓 : 푈 → ℂ is holomorphic if it is locally ∗ bounded and holomorphic in each variable separately. In other words, 푓 is holomorphic if it is locally bounded and complex-diﬀerentiable in each variable separately: 푓 (푧 , . . . , 푧 + 휉, . . . , 푧 ) − 푓 (푧) lim 1 푘 푛 exists for all 푧 ∈ 푈 and all 푘 = 1, 2, . . . , 푛. 휉∈ℂ→0 휉

In this book, however, the words “diﬀerentiable” and “derivative” (without the “complex-”) refer to plain vanilla real diﬀerentiability.

As in one variable, we deﬁne the Wirtinger operators

휕 1 휕 휕 휕 1 휕 휕 def= − 푖 , def= + 푖 . 휕푧푗 2 휕푥푗 휕푦푗 휕푧¯푗 2 휕푥푗 휕푦푗

∗ For every 푝 ∈ 푈, there is a neighborhood 푁 of 푝 such that 푓 |푁 is bounded. It is a deep result of Hartogs that we might in fact just drop “locally bounded” from the deﬁnition and obtain the same set of functions. 1.1. ONTO SEVERAL VARIABLES 13

An alternative definition is to say that a continuously differentiable function 푓 : 푈 → ℂ is holomorphic if it satisfies the Cauchy–Riemann equations 휕 푓 = 0 for 푗 = 1, 2, . . . , 푛. 휕푧¯푗 As in one variable, if you defined partial derivatives of holomorphic functions as the limits of the definition above, then you would get the Wirtinger 휕 for holomorphic functions. That is, if 푓 is 휕푧푘 holomorphic, then 휕 푓 푓 (푧 , . . . , 푧 + 휉, . . . , 푧 ) − 푓 (푧) (푧) = lim 1 푘 푛 . 휕푧푘 휉∈ℂ→0 휉 Due to the following proposition, the alternative definition using the Cauchy–Riemann equations is just as good as the definition we gave. Proposition 1.1.3. Let 푈 ⊂ ℂ푛 be an open set and suppose 푓 : 푈 → ℂ is holomorphic. Then 푓 is infinitely differentiable.

Proof. Suppose Δ = Δ휌 (푎) = Δ1 × · · · × Δ푛 is a polydisc centered at 푎, where each Δ푗 is a disc and suppose Δ ⊂ 푈, that is, 푓 is holomorphic on a neighborhood of the closure of Δ. Let 푧 be in Δ. Orient 휕Δ1 positively and apply the Cauchy formula (after all 푓 is holomorphic in 푧1): 1 ∫ 푓 (휁 , 푧 , . . . , 푧 ) 푓 (푧) = 1 2 푛 푑휁 . 휋푖 휁 − 푧 1 2 휕Δ1 1 1

Apply it again on the second variable, again orienting 휕Δ2 positively: 1 ∫ ∫ 푓 (휁 , 휁 , 푧 , . . . , 푧 ) 푓 (푧) = 1 2 3 푛 푑휁 푑휁 . 2 (휁 − 푧 )(휁 − 푧 ) 2 1 (2휋푖) 휕Δ1 휕Δ2 1 1 2 2 Applying the formula 푛 times we obtain 1 ∫ ∫ ∫ 푓 (휁 , 휁 , . . . , 휁 ) 푓 (푧) = ··· 1 2 푛 푑휁 ··· 푑휁 푑휁 . (1.1) ( 휋푖)푛 (휁 − 푧 )(휁 − 푧 )···(휁 − 푧 ) 푛 2 1 2 휕Δ1 휕Δ2 휕Δ푛 1 1 2 2 푛 푛

As 푓 is bounded on the compact set 휕Δ1 × · · · × 휕Δ푛, we find that 푓 is continuous in Δ. We may differentiate underneath the integral, because 푓 is bounded on 휕Δ1 × · · · × 휕Δ푛 and so are the derivatives of the integrand with respect to 푥푗 and 푦푗 , where 푧푗 = 푥푗 + 푖푦푗 , as long as 푧 is a positive distance away from 휕Δ1 × · · · × 휕Δ푛. We may differentiate as many times as we wish.

In ( 1.1 ) above, we derived the Cauchy integral formula in several variables. To write the formula more concisely we apply the Fubini’s theorem to write it as a single integral. We will write it down using diﬀerential forms. If you are unfamiliar with diﬀerential forms, think of the integral as the iterated integral above, and you can read the next few paragraphs a little lightly. It is enough to

understand real diﬀerential forms; we simply allow complex coeﬃcients here. See appendix C for

an overview of diﬀerential forms, or Rudin [ R2 ] for an introduction with all the details. Given real coordinates 푥 = (푥1, . . . , 푥푛), a one-form 푑푥푗 is a linear functional on tangent vectors such that 푑푥 휕 = 1 and 푑푥 휕 = 0 if 푗 ≠ 푘. Because 푧 = 푥 + 푖푦 and 푧¯ = 푥 − 푖푦 , 푗 휕푥푗 푗 휕푥푘 푗 푗 푗 푗 푗 푗

푑푧푗 = 푑푥푗 + 푖 푑푦푗 , 푑푧¯푗 = 푑푥푗 − 푖 푑푦푗 . 14 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

훿푘 훿 푗 훿푘 푗 푘 Let 푗 be the Kronecker delta, that is, 푗 = 1, and 푗 = 0 if ≠ . Then as expected, 휕 휕 휕 휕 푑푧 훿푘 , 푑푧 , 푑푧 , 푑푧 훿푘 . 푗 = 푗 푗 = 0 ¯푗 = 0 ¯푗 = 푗 휕푧푘 휕푧¯푘 휕푧푘 휕푧¯푘 One-forms are objects of the form 푛 ∑︁ 훼푗 푑푧푗 + 훽푗 푑푧¯푗 , 푗=1 where 훼푗 and 훽푗 are functions (of 푧). Two-forms are combinations of wedge products, 휔 ∧ 휂, of one-forms. A wedge of a two-form and a one-form is a three-form, etc. A 푘-form is an object that can be integrated on a so-called 푘-chain, for example, a 푘-dimensional surface. The wedge product takes care of the orientation as it is anticommutative on one-forms: For one-forms 휔 and 휂, we have 휔 ∧ 휂 = −휂 ∧ 휔. At this point, we need to talk about orientation in ℂ푛, that is, ordering of the real coordinates. There are two natural real-linear isomorphisms of ℂ푛 and ℝ2푛. We identify 푧 = 푥 + 푖푦 as either

(푥, 푦) = (푥1, . . . , 푥푛, 푦1, . . . , 푦푛) or (푥1, 푦1, 푥2, 푦2, . . . , 푥푛, 푦푛).

If we take the natural orientation of ℝ2푛, it is possible (if 푛 is even) that we obtain two opposite orientations on ℂ푛 (if 푛 is even, the real linear map that takes one ordering to the other has determinant −1). The orientation we take as the natural orientation of ℂ푛 (in this book) corresponds to the second ordering above, that is, (푥1, 푦1, . . . , 푥푛, 푦푛). Either isomorphism may be used in computation as long as it is used consistently, and the underlying orientation is kept in mind. Theorem 1.1.4 (Cauchy integral formula). Let Δ ⊂ ℂ푛 be a polydisc. Suppose 푓 : Δ → ℂ is a continuous function holomorphic in Δ. Write Γ = 휕Δ1 × · · · × 휕Δ푛 oriented appropriately (each 휕Δ푗 has positive orientation). Then for 푧 ∈ Δ ∫ 푓 (휁 , 휁 , . . . , 휁 ) 푓 푧 1 1 2 푛 푑휁 푑휁 푑휁 . ( ) = 푛 1 ∧ 2 ∧ · · · ∧ 푛 (2휋푖) Γ (휁1 − 푧1)(휁2 − 푧2)···(휁푛 − 푧푛)

We stated a more general result where 푓 is only continuous on Δ and holomorphic in Δ. The proof of this slight generalization is contained within the next two exercises.

Exercise 1.1.1: Suppose 푓 : 픻2 → ℂ is continuous and holomorphic on 픻2. For every 휃 ∈ ℝ, prove 푖휃 푖휃 푔1(휉) = 푓 (휉, 푒 ) and 푔2(휉) = 푓 (푒 , 휉) are holomorphic in 픻.

Exercise 1.1.2: Prove the theorem above, that is, the slightly more general Cauchy integral formula where 푓 is only continuous on Δ and holomorphic in Δ.

The Cauchy integral formula shows an important and subtle point about holomorphic functions in several variables: The value of the function 푓 on Δ is completely determined by the values of 푓 1.1. ONTO SEVERAL VARIABLES 15 on the set Γ, which is much smaller than the boundary of the polydisc 휕Δ. In fact, the Γ is of real dimension 푛, while the boundary of the polydisc has real dimension 2푛 − 1. We call the set Γ = 휕Δ1 × · · · × 휕Δ푛 the distinguished boundary. Let us draw the unit bidisc: Γ = 휕픻 × 휕픻 |푧2|

휕픻2 픻2

|푧1|

The set Γ is a 2-dimensional torus, like the surface of a donut. Whereas the set 휕픻2 = (휕픻 × 픻) ∪ (픻 × 휕픻) is the union of two ﬁlled donuts, or more precisely it is both the inside and the outside of the donut put together, and these two things meet on the surface of the donut. So the set Γ is quite small in comparison to the entire boundary 휕픻2.

Exercise 1.1.3: Suppose Δ is a polydisc, Γ its distinguished boundary, and 푓 : Δ → ℂ is continuous on Δ and holomorphic on Δ. Prove | 푓 (푧)| achieves its maximum on Γ.

Exercise 1.1.4: A ball is different from a polydisc. Prove that for every 푝 ∈ 휕픹푛 there exists a continuous 푓 : 픹푛 → ℂ, holomorphic on 픹푛 such that | 푓 (푧)| achieves a strict maximum at 푝. Exercise 1.1.5: Show that in the real setting, differentiable in each variable separately does not 푓 (푥, 푦) = 푥푦 imply differentiable even if the function is locally bounded. Let 푥2+푦2 outside the origin and 푓 (0, 0) = 0. Prove that 푓 is a locally bounded function in ℝ2, which is differentiable in each variable separately (all partial derivatives exist at every point), but 푓 is not even continuous. There is something very special about the holomorphic category.

Exercise 1.1.6: Suppose 푈 ⊂ ℂ푛 is open. Prove that 푓 : 푈 → ℂ is holomorphic if and only if 푓 is locally bounded and for every 푎, 푏 ∈ ℂ푛, the function 휁 ↦→ 푓 (휁푎 + 푏) is holomorphic on the open set {휁 ∈ ℂ : 휁푎 + 푏 ∈ 푈}.

Exercise 1.1.7: Prove a several complex variables version of Morera’s theorem (see Theorem B.4 ). A triangle 푇 ⊂ ℂ푛 is the closed convex hull of three points, so including the inside. Orient 푇 in some way and orient 휕푇 accordingly. A triangle 푇 lies in a complex line if its vertices 푎, 푏, 푐 satisfy 휁 (푏 − 푎) = 푐 − 푎 for some 휁 ∈ ℂ. Suppose 푈 ⊂ ℂ푛 is open and 푓 : 푈 → ℂ is continuous. Prove that 푓 is holomorphic if and only if ∫ 푓 (푧) 푑푧푘 = 0 휕푇 for every triangle 푇 ⊂ 푈 that lies in a complex line, and every 푘 = 1, 2, . . . , 푛. Hint: The previous exercise may be useful. 16 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES 1.2 Power series representation

As you noticed, writing out all the components can be a pain. Just as we write vectors as 푧 instead of (푧1, 푧2, . . . , 푧푛), we similarly deﬁne the so-called multi-index notation to deal with more complicated formulas such as the ones above. 훼 ∈ 푛 ∪ { } Let ℕ0 be a vector of nonnegative integers (where ℕ0 = ℕ 0 ). We write def 1 def 1 푧 def 푧 푧 푧푛 푧훼 푧훼1 푧훼2 ··· 푧훼푛 , , 1 , 2 ,..., , = 1 2 푛 = = 푧 푧1푧2 ··· 푧푛 푤 푤1 푤2 푤푛 |훼| 훼 훼 훼 def 휕 def 휕 1 휕 2 휕 푛 def |푧|훼 = |푧 |훼1 |푧 |훼2 · · · |푧 |훼푛 , = ··· , 푑푧 = 푑푧 ∧ 푑푧 ∧ · · · ∧ 푑푧 , 1 2 푛 휕푧훼 휕푧훼1 휕푧훼2 휕푧훼푛 1 2 푛 1 2 푛 def def |훼| = 훼1 + 훼2 + · · · + 훼푛, 훼! = 훼1!훼2! ··· 훼푛!.

We can also make sense of this notation, especially the notation 푧훼, if 훼 ∈ ℤ푛, that is, if it includes 훼 푛 negative integers. Although usually, is assumed to be in ℕ0. Furthermore, when we use 1 as a vector it means (1, 1,..., 1). If 푧 ∈ ℂ푛, then

− 푧 ( − 푧 , − 푧 ,..., − 푧 ), 푧훼+1 푧훼1+1푧훼2+1 ··· 푧훼푛+1. 1 = 1 1 1 2 1 푛 or = 1 2 푛 It goes without saying that when using this notation it is important to be careful to always realize which symbol lives where, and most of all, to not get carried away. For instance, one could interpret 1 푧 in two diﬀerent ways depending on if we interpret 1 as a vector or not, and if we are expecting a vector or a number. Best to just keep to the limited set of cases as given above. In this notation, the Cauchy formula becomes the perhaps deceptively simple ∫ 푓 (휁) 푓 푧 1 푑휁. ( ) = 푛 (2휋푖) Γ 휁 − 푧 Let us move to power series. For simplicity, we start with power series at the origin. Using the multinomial notation, we write such a series as

∑︁ 훼 푐훼푧 . ∈ 푛 훼 ℕ0 You have to admit that the above is far nicer to write than, for example, for 3 variables writing

∞ ∞ ∞ ∑︁ ∑︁ ∑︁ 푐 푧 푗 푧푘 푧ℓ , 푗 푘ℓ 1 2 3 (1.2) 푗=0 푘=0 ℓ=0 which is not even exactly the deﬁnition of the series sum (see below). When it is clear from context that we are talking about a power series and all the powers are nonnegative, we often write just

∑︁ 훼 푐훼푧 . 훼 It is important to note what this means. The sum does not have a natural ordering. We are 훼 ∈ 푛 푛 summing over ℕ0, and there is no natural ordering of ℕ0. It makes no sense to talk about 1.2. POWER SERIES REPRESENTATION 17

conditional convergence. When we say the series converges, we mean absolutely. Fortunately, power series converge absolutely, and so the ordering does not matter. If you want to write the limit in terms of partial sums, you pick some ordering of the multi-indices, 훼(1), 훼(2),..., and then

푚 ∑︁ 훼 ∑︁ 훼(푘) 푐훼푧 = lim 푐훼(푘) 푧 . 푚→∞ 훼 푘=1

By the Fubini theorem (for sums) this limit is also equal to the iterated sum such as ( 1.2 ). Í 푐 푧훼 푧 푋 Í 푐 푧훼 A power series 훼 훼 converges uniformly absolutely for ∈ when 훼 | 훼 | converges uniformly for 푧 ∈ 푋. As in one variable, we need the geometric series in several variables. For 푧 ∈ 픻푛 (unit polydisc),

∞ ∞ ∞ 1 1 ∑︁ ∑︁ ∑︁ = = © 푧 푗 ª © 푧 푗 ª ··· © 푧 푗 ª 1 − 푧 (1 − 푧 )(1 − 푧 )···(1 − 푧 ) 1 ® 2 ® 푛 ® 1 2 푛 푗=0 푗=0 푗=0 ∞ ∞ ∞ « ¬ « ¬ « ¬ ∑︁ ∑︁ ∑︁ 푗1 푗2 푗푛 ∑︁ 훼 = ··· 푧1 푧푛 ··· 푧푛 = 푧 . 푗1=0 푗2=0 푗푛=0 훼

The series converges uniformly absolutely on all compact subsets of the unit polydisc: Any compact set in the unit polydisc is contained in a closed polydisc Δ centered at 0 of radius 1 − 휖 for some 휖 > 0. The convergence is uniformly absolute on Δ. This claim follows by simply noting the same fact for each factor is true in one dimension. We now prove that holomorphic functions are precisely those having a power series expansion.

푛 Theorem 1.2.1. Let Δ = Δ휌 (푎) ⊂ ℂ be a polydisc. Suppose 푓 : Δ → ℂ is a continuous function holomorphic in Δ. Then on Δ, 푓 is equal to a power series converging uniformly absolutely on compact subsets of Δ: ∑︁ 훼 푓 (푧) = 푐훼 (푧 − 푎) . (1.3) 훼

Conversely, if 푓 : Δ → ℂ is deﬁned by ( 1.3 ) converging uniformly absolutely on compact subsets of Δ, then 푓 is holomorphic on Δ.

Proof. First assume a continuous 푓 : Δ → ℂ is holomorphic on Δ. Write Γ = 휕Δ1 × · · · × 휕Δ푛 and orient it positively. Take 푧 ∈ Δ and 휁 ∈ Γ. As in one variable, write the Cauchy kernel as

1 1 1 1 ∑︁ 푧 − 푎 훼 = = . 휁 − 푧 휁 − 푎 푧−푎 휁 − 푎 휁 − 푎 1 − 휁−푎 훼

1 1 1 1 Here, make sure you interpret the formulas as 휁−푧 = (휁 −푧 )···(휁 −푧 ) , 휁−푎 = (휁 −푎 )···(휁 −푎 ) and 1 1 푛 푛 1 1 푛 푛 푧−푎 = 푧1−푎1 ,..., 푧푛−푎푛 . 휁−푎 휁1−푎1 휁푛−푎푛 The multivariable geometric series is a product of geometric series in one variable, and geometric series in one variable are uniformly absolutely convergent on compact subsets of the unit disc. So the series above converges uniformly absolutely for (푧, 휁) ∈ 퐾 × Γ for every compact subset 퐾 of Δ. 18 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Compute

∫ 푓 (휁) 푓 푧 1 푑휁 ( ) = 푛 (2휋푖) Γ 휁 − 푧 1 ∫ 푓 (휁) ∑︁ 푧 − 푎 훼 = 푑휁 ( 휋푖)푛 휁 − 푎 휁 − 푎 2 Γ 훼 ∑︁ 1 ∫ 푓 (휁) = 푑휁 (푧 − 푎)훼. ( 휋푖)푛 휁 푎 훼+1 훼 2 Γ ( − )

The last equality follows by Fubini or uniform convergence just as it does in one variable. Uniform absolute convergence (as 푧 moves) on compact subsets of the ﬁnal series follows from the uniform absolute convergence of the geometric series. It is also a direct consequence of the Cauchy estimates below. We have shown that ∑︁ 훼 푓 (푧) = 푐훼 (푧 − 푎) , 훼 where 1 ∫ 푓 (휁) 푐 = 푑휁. 훼 푛 훼+1 (2휋푖) Γ (휁 − 푎) Notice how strikingly similar the computation is to one variable. Let us prove the converse statement. The limit of the series is continuous as it is a uniform-on- compact-sets limit of continuous functions, and hence it is locally bounded in Δ. Next, we restrict to each variable in turn, ∑︁ 훼 푧푗 ↦→ 푐훼 (푧 − 푎) . 훼 This function is holomorphic via the corresponding one-variable argument.

The converse statement also follows by applying the Cauchy–Riemann equations to the series termwise. We leave that as an exercise. First, one must show that the term-by-term derivative series also converges uniformly absolutely on compact subsets. Then one applies the theorem from real analysis about derivatives of limits: If a sequence of functions and its derivatives converges uniformly, then the derivatives converge to the derivative of the limit.

Exercise 1.2.1: Prove the claim above that if a power series converges uniformly absolutely on compact subsets of a polydisc Δ, then the term-by-term derivative converges. Do the proof without using the analogous result for single variable series.

A third way to prove the converse statement of the theorem is to note that partial sums are holomorphic and write them using the Cauchy formula. Uniform convergence shows that the limit also satisﬁes the Cauchy formula, and diﬀerentiating under the integral obtains the result. 1.2. POWER SERIES REPRESENTATION 19

Exercise 1.2.2: Follow the logic above to prove the converse of the theorem. Do the proof without using the analogous result for single variable series. Hint: Let Δ00 ⊂ Δ0 ⊂ Δ be two polydiscs with the same center 푎 such that Δ00 ⊂ Δ0 and Δ0 ⊂ Δ. Apply Cauchy formula on Δ0 for 푧 ∈ Δ00.

Next we organize some immediate consequences of the theorem and the calculation in the proof. 푛 Proposition 1.2.2. Let Δ = Δ휌 (푎) ⊂ ℂ be a polydisc, and Γ its distinguished boundary. Suppose 푓 : Δ → ℂ is a continuous function holomorphic in Δ. Then, for 푧 ∈ Δ, 휕|훼| 푓 1 ∫ 훼! 푓 (휁) (푧) = 푑휁. 훼 푛 훼+1 휕푧 (2휋푖) Γ (휁 − 푧)

In particular, if 푓 is given by ( 1.3 ), 1 휕|훼| 푓 푐 = (푎), 훼 훼! 휕푧훼 and we have the Cauchy estimates: k 푓 k |푐 | ≤ Γ . 훼 휌훼 In particular, the coefficients of the power series depend only on the derivatives of 푓 at 푎 (and so on the values of 푓 in an arbitrarily small neighborhood of 푎) and not the specific polydisc used in the theorem. Proof. Using the Leibniz rule (taking derivatives under the integral), as long as 푧 ∈ Δ (not on the boundary), we can differentiate under the integral. We are talking regular real partial differentiation, and we use it to apply the Wirtinger operator. The point is that " # 휕 1 푘 = . 휕푧 푘 푘+1 푗 (휁푗 − 푧푗 ) (휁푗 − 푧푗 ) Let us do a single derivative to get the idea: 휕 푓 휕 ∫ 푓 (휁 , 휁 , . . . , 휁 ) 푧 1 1 2 푛 푑휁 푑휁 푑휁 ( ) = 푛 1 ∧ 2 ∧ · · · ∧ 푛 휕푧1 휕푧1 (2휋푖) Γ (휁1 − 푧1)(휁2 − 푧2)···(휁푛 − 푧푛) 1 ∫ 푓 (휁 , 휁 , . . . , 휁 ) = 1 2 푛 푑휁 ∧ 푑휁 ∧ · · · ∧ 푑휁 . 휋푖 푛 2 1 2 푛 (2 ) Γ (휁1 − 푧1) (휁2 − 푧2)···(휁푛 − 푧푛) How about we do it a second time: 휕2 푓 1 ∫ 2 푓 (휁 , 휁 , . . . , 휁 ) (푧) = 1 2 푛 푑휁 ∧ 푑휁 ∧ · · · ∧ 푑휁 . 휕푧2 (2휋푖)푛 휁 푧 3 휁 푧 휁 푧 1 2 푛 1 Γ ( 1 − 1) ( 2 − 2)···( 푛 − 푛)

Notice the 2 before the 푓 . Next derivative, a 3 is coming out. After 푗 derivatives in 푧1 you get the constant 푗!. It is exactly the same thing that happens in one variable. A moment’s thought will 훼 ∈ 푛 convince you that the following formula is correct for ℕ0: 휕|훼| 푓 1 ∫ 훼! 푓 (휁) (푧) = 푑휁. 훼 푛 훼+1 휕푧 (2휋푖) Γ (휁 − 푧) 20 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Therefore, 휕|훼| 푓 훼! 푐 = (푎). 훼 휕푧훼 We obtain the Cauchy estimates as before:

휕|훼| 푓 1 ∫ 훼! 푓 (휁) 1 ∫ 훼!| 푓 (휁)| 훼! (푎) = 푑휁 ≤ |푑휁 | ≤ k 푓 k . 훼 푛 훼+1 푛 훼+1 훼 Γ 휕푧 (2휋푖) Γ (휁 − 푎) (2휋) Γ 휌 휌

As in one-variable theory, the Cauchy estimates prove the following proposition.

푛 Proposition 1.2.3. Let 푈 ⊂ ℂ be an open set. Suppose the sequence 푓푗 : 푈 → ℂ converges uniformly on compact subsets to 푓 : 푈 → ℂ. If every 푓푗 is holomorphic, then 푓 is holomorphic and |훼| |훼| n 휕 푓푗 o 휕 푓 the sequence 휕푧훼 converges to 휕푧훼 uniformly on compact subsets.

Exercise 1.2.3: Prove the proposition above.

Let 푊 ⊂ ℂ푛 be the set where a power series converges such that it diverges on the complement. The interior of 푊 is called the domain of convergence. In one variable, every domain of convergence is a disc, and hence it is described with a single number (the radius). In several variables, the domain of convergence is not as easy to describe. For the multivariable geometric series the domain of convergence is the unit polydisc, but more complicated examples are easy to ﬁnd.

Example 1.2.4: In ℂ2, the power series

∞ ∑︁ 푧 푧푘 1 2 푘=0

converges absolutely on the set

2 2 푧 ∈ ℂ : |푧2| < 1 ∪ 푧 ∈ ℂ : 푧1 = 0 , and nowhere else. This set is not quite a polydisc. It is neither an open set nor a closed set, and its 2 closure is not the closure of the domain of convergence, which is the set 푧 ∈ ℂ : |푧2| < 1 .

Example 1.2.5: The power series ∞ ∑︁ 푧푘 푧푘 1 2 푘=0 converges absolutely exactly on the set

2 푧 ∈ ℂ : |푧1푧2| < 1 . 1.2. POWER SERIES REPRESENTATION 21

The picture is deﬁnitely more complicated than a polydisc: . |푧2| .

···

|푧1|

Í 1 푧 푗 푧푘 Exercise 1.2.4: Find the domain of convergence of 푗,푘 푘! 1 2 and draw the corresponding picture. Í 푐 푧 푗 푧푘 Exercise 1.2.5: Find the domain of convergence of 푗,푘 푗,푘 1 2 and draw the corresponding 푘 picture if 푐푘,푘 = 2 , 푐0,푘 = 푐푗,0 = 1 and 푐푗,푘 = 0 otherwise. Exercise 1.2.6: Suppose a power series in two variables can be written as a sum of a power series in 푧1 and a power series in 푧2. Show that the domain of convergence is a polydisc.

푛 Suppose 푈 ⊂ ℂ is a domain such that if 푧 ∈ 푈 and |푧푗 | = |푤푗 | for all 푗, then 푤 ∈ 푈. Such a 푈 is called a Reinhardt domain. The domains we were drawing so far are Reinhardt domains. They are exactly the domains that you can draw by plotting what happens for the moduli of the variables. A domain is called a complete Reinhardt domain if whenever 푧 ∈ 푈, then for 푟 = (푟1, . . . , 푟푛) where 푟푗 = |푧푗 | for all 푗, we have that the whole polydisc Δ푟 (0) ⊂ 푈. So a complete Reinhardt domain is a union (possibly inﬁnite) of polydiscs centered at the origin.

Exercise 1.2.7: Let 푊 ⊂ ℂ푛 be the set where a certain power series at the origin converges. Show that the interior of 푊 is a complete Reinhardt domain.

Theorem 1.2.6 (Identity theorem). Let 푈 ⊂ ℂ푛 be a domain (connected open set) and let 푓 : 푈 → ℂ be holomorphic. Suppose 푓 |푁 ≡ 0 for a nonempty open subset 푁 ⊂ 푈. Then 푓 ≡ 0. Proof. Let 푍 be the set where all derivatives of 푓 are zero; then 푁 ⊂ 푍, so 푍 is nonempty. The set 푍 is closed in 푈 as all derivatives are continuous. Take an arbitrary 푎 ∈ 푍. Expand 푓 in a power series around 푎 converging to 푓 in a polydisc Δ휌 (푎) ⊂ 푈. As the coeﬃcients are given by derivatives of 푓 , the power series is the zero series. Hence, 푓 is identically zero in Δ휌 (푎). Therefore, 푍 is open. As 푍 is also closed and nonempty, and 푈 is connected, we have 푍 = 푈. The theorem is often used to show that if two holomorphic functions 푓 and 푔 are equal on a small open set, then 푓 ≡ 푔. 22 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Theorem 1.2.7 (Maximum principle). Let 푈 ⊂ ℂ푛 be a domain (connected open set). Let 푓 : 푈 → ℂ be holomorphic and suppose | 푓 (푧)| attains a local maximum at some 푎 ∈ 푈. Then 푓 ≡ 푓 (푎).

Proof. Suppose | 푓 (푧)| attains a local maximum at 푎 ∈ 푈. Consider a polydisc Δ = Δ1×· · ·×Δ푛 ⊂ 푈 centered at 푎. The function 푧1 ↦→ 푓 (푧1, 푎2, . . . , 푎푛)

is holomorphic on the disc Δ1 and its modulus attains the maximum at the center. Therefore, it is constant by maximum principle in one variable, that is, 푓 (푧1, 푎2, . . . , 푎푛) = 푓 (푎) for all 푧1 ∈ Δ1. For any ﬁxed 푧1 ∈ Δ1, consider the function

푧2 ↦→ 푓 (푧1, 푧2, 푎3, . . . , 푎푛).

This function, holomorphic on the disc Δ2, again attains its maximum modulus at the center of Δ2 and hence is constant on Δ2. Iterating this procedure we obtain that 푓 (푧) = 푓 (푎) for all 푧 ∈ Δ. By the identity theorem we have that 푓 (푧) = 푓 (푎) for all 푧 ∈ 푈.

Exercise 1.2.8: Let 푉 be the volume measure on ℝ2푛 and hence on ℂ푛. Suppose Δ centered at 푎 ∈ ℂ푛, and 푓 is a function holomorphic on a neighborhood of Δ. Prove 1 ∫ 푓 (푎) = 푓 (휁) 푑푉 (휁), 푉 (Δ) Δ where 푉 (Δ) is the volume of Δ and 푑푉 is the volume measure. That is, 푓 (푎) is an average of the values on a polydisc centered at 푎.

Exercise 1.2.9: Prove the maximum principle by using the Cauchy formula instead. (Hint: Use the previous exercise)

Exercise 1.2.10: Prove a several variables analogue of the Schwarz’s lemma: Suppose 푓 is |훼| 푛 푓 푘 휕 푓 holomorphic in a neighborhood of 픻 , (0) = 0, and for some ∈ ℕ we have 휕푧훼 (0) = 0 whenever |훼| < 푘. Further suppose for all 푧 ∈ 픻푛, | 푓 (푧)| ≤ 푀 for some 푀. Show that

| 푓 (푧)| ≤ 푀k푧k푘 for all 푧 ∈ 픻푛.

Exercise 1.2.11: Apply the one-variable Liouville’s theorem to prove it for several variables. That is, suppose 푓 : ℂ푛 → ℂ is holomorphic and bounded. Prove 푓 is constant.

Exercise 1.2.12: Improve Liouville’s theorem slightly in ℂ2. A complex line though the origin is the image of a linear map 퐿 : ℂ → ℂ푛. a) Prove that for any collection of ﬁnitely many complex lines through the origin, there exists an entire nonconstant holomorphic function (푛 ≥ 2) bounded (hence constant) on these complex lines. b) Prove that if an entire holomorphic function in ℂ2 is bounded on countably many distinct complex lines through the origin, then it is constant. c) Find a nonconstant entire holomorphic function in ℂ3 that is bounded on countably many distinct complex lines through the origin. 1.2. POWER SERIES REPRESENTATION 23

Exercise 1.2.13: Prove the several variables version of Montel’s theorem: Suppose { 푓푘 } is a uniformly bounded sequence of holomorphic functions on an open set 푈 ⊂ ℂ푛. Show that there 푓 exists a subsequence { 푘푗 } that converges uniformly on compact subsets to some holomorphic function 푓 . Hint: Mimic the one-variable proof.

Exercise 1.2.14: Prove a several variables version of Hurwitz’s theorem: Suppose { 푓푘 } is a sequence of nowhere zero holomorphic functions on a domain 푈 ⊂ ℂ푛 converging uniformly on compact subsets to a function 푓 . Show that either 푓 is identically zero, or that 푓 is nowhere zero.

Hint: Feel free to use the one-variable result .

Exercise 1.2.15: Suppose 푝 ∈ ℂ푛 is a point and 퐷 ⊂ ℂ푛 is a ball centered at 푝 ∈ 퐷.A holomorphic function 푓 : 퐷 → ℂ can be analytically continued along a path 훾 : [0, 1] → ℂ푛, 훾(0) = 푝, if for every 푡 ∈ [0, 1] there exists a ball 퐷푡 centered at 훾(푡), where 퐷0 = 퐷, and a holomorphic function 푓푡 : 퐷푡 → ℂ, where 푓0 = 푓 , and for each 푡0 ∈ [0, 1] there is an 휖 > 0 such 푡 푡 < 휖 푓 푓 퐷 퐷 that if | − 0| , then 푡 = 푡0 in 푡 ∩ 푡0 . Prove a several variables version of the Monodromy theorem: If 푈 ⊂ ℂ푛 is a simply connected domain, 퐷 ⊂ 푈 a ball and 푓 : 퐷 → ℂ a holomorphic function that can be analytically continued from 푝 ∈ 퐷 to every 푞 ∈ 푈, then there exists a unique holomorphic function 퐹 : 푈 → ℂ such that 퐹|퐷 = 푓 .

Let us define notation for the set of holomorphic functions. At the same time, we notice that the set of holomorphic functions is a commutative ring under pointwise addition and multiplication. Definition 1.2.8. Let 푈 ⊂ ℂ푛 be an open set. Define O(푈) to be the ring of holomorphic functions. The letter O is used to recognize the fundamental contribution to several complex variables by ∗ Kiyoshi Oka .

Exercise 1.2.16: Prove that O(푈) is actually a commutative ring with the operations

( 푓 + 푔)(푧) = 푓 (푧) + 푔(푧), ( 푓 푔)(푧) = 푓 (푧)푔(푧).

Exercise 1.2.17: Show that O(푈) is an integral domain (has no zero divisors) if and only if 푈 is connected. That is, show that 푈 being connected is equivalent to showing that if ℎ(푧) = 푓 (푧)푔(푧) is identically zero for 푓 , 푔 ∈ O(푈), then either 푓 or 푔 is identically zero.

A function 퐹 deﬁned on a dense open subset of 푈 is meromorphic if locally near every 푝 ∈ 푈, 퐹 = 푓/푔 for 푓 and 푔 holomorphic in some neighborhood of 푝. We remark that it follows from a deep result of Oka that for domains 푈 ⊂ ℂ푛, every meromorphic function can be represented as 푓/푔 globally. That is, the ring of meromorphic functions is the ﬁeld of fractions of O(푈). This problem is the so-called Poincaré problem, and its solution is no longer positive once we generalize 푈 to complex manifolds. The points of 푈 through which 퐹 does not extend holomorphically are called the poles of 퐹. Namely, poles are the points where 푔 = 0 for every possible representation 푓/푔. Unlike in one variable, in several variables poles are never isolated points. There is also a new type of singular point for meromorphic functions in more than one variable:

∗ See https://en.wikipedia.org/wiki/Kiyoshi_Oka 24 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Exercise 1.2.18: In two variables one can no longer think of a meromorphic function 퐹 simply taking on the value of ∞, when the denominator vanishes. Show that 퐹(푧, 푤) = 푧/푤 achieves all values of ℂ in every neighborhood of the origin. The origin is called a point of indeterminacy.

Exercise 1.2.19: Prove that zeros (and so poles as we will see later, though this is a bit harder 푛 to see) are never isolated in ℂ for 푛 ≥ 2. Hint: Consider 푧1 ↦→ 푓 (푧1, 푧2, . . . , 푧푛) as you move

푧2, . . . , 푧푛 around, and use perhaps Hurwitz .

1.3 Derivatives

Given a function 푓 = 푢 + 푖푣, the complex conjugate is 푓¯ = 푢 − 푖푣, defined simply by 푧 ↦→ 푓 (푧). When 푓 is holomorphic, then 푓¯ is called an antiholomorphic function. An antiholomorphic function is a function that depends on 푧¯ but not on 푧. So if we write the variable, we write 푓¯ as 푓¯(푧¯). Let us see why this makes sense. Using the definitions of the Wirtinger operators, 휕 푓¯ 휕 푓 휕 푓¯ 휕 푓 = = 0, = , for all 푗 = 1, . . . , 푛. 휕푧푗 휕푧¯푗 휕푧¯푗 휕푧푗 For functions that are neither holomorphic or antiholomorphic, we pretend they depend on both 푧 and 푧¯. Since we want to write functions in terms of 푧 and 푧¯, let us figure out how the chain rule works for Wirtinger derivatives, rather than writing derivatives in terms of 푥 and 푦. Proposition 1.3.1 (Complex chain rule). Suppose 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 are open sets and suppose 푓 : 푈 → 푉, and 푔 : 푉 → ℂ are (real) differentiable functions (mappings). Write the variables as 푛 푚 푧 = (푧1, . . . , 푧푛) ∈ 푈 ⊂ ℂ and 푤 = (푤1, . . . , 푤푚) ∈ 푉 ⊂ ℂ . Then for 푗 = 1, . . . , 푛, 푚 푚 휕 ∑︁ 휕푔 휕 푓 휕푔 휕 푓¯ 휕 ∑︁ 휕푔 휕 푓 휕푔 휕 푓¯ [푔 ◦ 푓 ] = ℓ + ℓ , [푔 ◦ 푓 ] = ℓ + ℓ . (1.4) 휕푧 휕푤 휕푧 휕푤¯ 휕푧 휕푧¯ 휕푤 휕푧¯ 휕푤¯ 휕푧¯ 푗 ℓ=1 ℓ 푗 ℓ 푗 푗 ℓ=1 ℓ 푗 ℓ 푗 Proof. Write 푓 = 푢 + 푖푣, 푧 = 푥 + 푖푦, 푤 = 푠 + 푖푡, 푓 is a function of 푧, and 푔 is a function of 푤. The composition plugs in 푓 for 푤, and so it plugs in 푢 for 푠, and 푣 for 푡. Using the standard chain rule, 휕 1 휕 휕 [푔 ◦ 푓 ] = − 푖 [푔 ◦ 푓 ] 휕푧푗 2 휕푥푗 휕푦푗 푚 1 ∑︁ 휕푔 휕푢 휕푔 휕푣 휕푔 휕푢 휕푔 휕푣 = ℓ + ℓ − 푖 ℓ + ℓ 2 휕푠 휕푥 휕푡 휕푥 휕푠 휕푦 휕푡 휕푦 ℓ=1 ℓ 푗 ℓ 푗 ℓ 푗 ℓ 푗 푚 ∑︁ 휕푔 1 휕푢 휕푢 휕푔 1 휕푣 휕푣 = ℓ − 푖 ℓ + ℓ − 푖 ℓ 휕푠 2 휕푥 휕푦 휕푡 2 휕푥 휕푦 ℓ=1 ℓ 푗 푗 ℓ 푗 푗 푚 ∑︁ 휕푔 휕푢 휕푔 휕푣 = ℓ + ℓ . 휕푠 휕푧 휕푡 휕푧 ℓ=1 ℓ 푗 ℓ 푗 For ℓ = 1, . . . , 푚, 휕 휕 휕 휕 휕 휕 = + , = 푖 − . 휕푠ℓ 휕푤ℓ 휕푤¯ ℓ 휕푡ℓ 휕푤ℓ 휕푤¯ ℓ 1.3. DERIVATIVES 25

Continuing:

푚 휕 ∑︁ 휕푔 휕푢 휕푔 휕푣 [푔 ◦ 푓 ] = ℓ + ℓ 휕푧 휕푠 휕푧 휕푡 휕푧 푗 ℓ=1 ℓ 푗 ℓ 푗 푚 ∑︁ 휕푔 휕푢 휕푔 휕푢 휕푔 휕푣 휕푔 휕푣 = ℓ + ℓ + 푖 ℓ − ℓ 휕푤 휕푧 휕푤¯ 휕푧 휕푤 휕푧 휕푤¯ 휕푧 ℓ=1 ℓ 푗 ℓ 푗 ℓ 푗 ℓ 푗 푚 ∑︁ 휕푔 휕푢 휕푣 휕푔 휕푢 휕푣 = ℓ + 푖 ℓ + ℓ − 푖 ℓ 휕푤 휕푧 휕푧 휕푤¯ 휕푧 휕푧 ℓ=1 ℓ 푗 푗 ℓ 푗 푗 푚 ∑︁ 휕푔 휕 푓 휕푔 휕 푓¯ = ℓ + ℓ . 휕푤 휕푧 휕푤¯ 휕푧 ℓ=1 ℓ 푗 ℓ 푗 The 푧¯ derivative works similarly. Because of the proposition, when we deal with an arbitrary, possibly nonholomorphic, function 푓 , we often write 푓 (푧, 푧¯) and treat 푓 as a function of 푧 and 푧¯. Remark 1.3.2. It is good to notice the subtlety of what we just said. Formally it seems as if we are treating 푧 and 푧¯ as independent variables when taking derivatives, but in reality they are not independent if we actually wish to evaluate the function. Under the hood, a smooth function that is not necessarily holomorphic is really a function of the real variables 푥 and 푦, where 푧 = 푥 + 푖푦. Remark 1.3.3. Another remark is that we could have swapped 푧 and 푧¯, by flipping the bars everywhere. There is no difference between the two, they are twins in effect. We just need to know which one is which. After all, it all starts with taking the two square roots of −1 and deciding which one is 푖 (remember the chickens?). There is no “natural choice” for that, but once we make that choice we must be consistent. And once we picked which root is 푖, we also picked what is holomorphic and what is antiholomorphic. This is a subtle philosophical as much as a mathematical point. Definition 1.3.4. Let 푈 ⊂ ℂ푛 be open. A mapping 푓 : 푈 → ℂ푚 is said to be holomorphic if each component is holomorphic. That is, if 푓 = ( 푓1, . . . , 푓푚), then each 푓푗 is a holomorphic function. As in one variable, the composition of holomorphic functions (mappings) is holomorphic. Theorem 1.3.5. Let 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 be open sets and suppose 푓 : 푈 → 푉 and 푔 : 푉 → ℂ푘 are both holomorphic. Then the composition 푔 ◦ 푓 is holomorphic.

Proof. The proof is almost trivial by chain rule. Again let 푔 be a function of 푤 ∈ 푉 and 푓 be a function of 푧 ∈ 푈. For any 푗 = 1, . . . , 푛 and any 푝 = 1, . . . , 푘, we compute 0 0 휕 푚 휕푔 휕 푓 7 휕푔 7휕 푓¯ ∑︁ © 푝 ℓ 푝 ℓ ª 푔푝 ◦ 푓 = + ® = 0. 휕푧¯ 휕푤 휕푧¯ 휕푤¯ 휕푧¯ ® 푗 ℓ=1 ℓ 푗 ℓ 푗 « ¬ For holomorphic functions the chain rule simpliﬁes, and it formally looks like the familiar vector calculus rule. Suppose again 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 are open, and 푓 : 푈 → 푉 and 푔 : 푉 → ℂ are 푛 푚 holomorphic. Name the variables 푧 = (푧1, . . . , 푧푛) ∈ 푈 ⊂ ℂ and 푤 = (푤1, . . . , 푤푚) ∈ 푉 ⊂ ℂ . In 26 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

¯ formula ( 1.4 ) for the 푧푗 derivative, the 푤¯ 푗 derivative of 푔 is zero and the 푧푗 derivative of 푓ℓ is also zero because 푓 and 푔 are holomorphic. Therefore, for any 푗 = 1, . . . , 푛,

푚 휕 ∑︁ 휕푔 휕 푓 [푔 ◦ 푓 ] = ℓ . 휕푧 휕푤 휕푧 푗 ℓ=1 ℓ 푗

Exercise 1.3.1: Prove using only the Wirtinger derivatives that a holomorphic function that is real-valued must be constant.

Exercise 1.3.2: Let 푓 be a holomorphic function on ℂ푛. When we write 푓¯ we mean the function 푧 ↦→ 푓 (푧), and we usually write 푓¯(푧¯) as the function is antiholomorphic. However, if we write 푓¯(푧) we really mean 푧 ↦→ 푓 (푧¯), that is, composing both the function and the argument with conjugation. Prove 푧 ↦→ 푓¯(푧) is holomorphic, and prove 푓 is real-valued on ℝ푛 (when 푦 = 0) if and only if 푓 (푧) = 푓¯(푧) for all 푧 ∈ ℂ.

For a 푈 ⊂ ℂ푛, a holomorphic mapping 푓 : 푈 → ℂ푚, and a point 푝 ∈ 푈, deﬁne the holomorphic derivative, sometimes called the Jacobian matrix, 휕 푓 퐷 푓 (푝) def= 푗 (푝) . 휕푧 푘 푗 푘 The notation 푓 0(푝) = 퐷 푓 (푝) is also used.

Exercise 1.3.3: Suppose 푈 ⊂ ℂ푛 is open, ℝ푛 is naturally embedded in ℂ푛. Consider a holo- 푚 푚 푚 morphic mapping 푓 : 푈 → ℂ and suppose that 푓 |푈∩ℝ푛 maps into ℝ ⊂ ℂ . Prove that given 푛 푛 푚 푝 ∈ 푈 ∩ ℝ , the real Jacobian matrix at 푝 of the map 푓 |푈∩ℝ푛 : 푈 ∩ ℝ → ℝ is equal to the holomorphic Jacobian matrix of the map 푓 at 푝. In particular, 퐷 푓 (푝) is a matrix with real entries.

Using the holomorphic chain rule above, as in the theory of real functions, the derivative of the composition is the composition of derivatives (multiplied as matrices). Proposition 1.3.6 (Chain rule for holomorphic mappings). Let 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 be open sets. Suppose 푓 : 푈 → 푉 and 푔 : 푉 → ℂ푘 are both holomorphic, and 푝 ∈ 푈. Then

퐷(푔 ◦ 푓 )(푝) = 퐷푔 푓 (푝) 퐷 푓 (푝).

In shorthand, we often simply write 퐷(푔 ◦ 푓 ) = 퐷푔퐷 푓 .

Exercise 1.3.4: Prove the proposition.

Suppose 푈 ⊂ ℂ푛, 푝 ∈ 푈, and 푓 : 푈 → ℂ푚 is a diﬀerentiable function at 푝. Since ℂ푛 is identiﬁed with ℝ2푛, the mapping 푓 takes 푈 ⊂ ℝ2푛 to ℝ2푚. The normal vector calculus Jacobian at 푝 of this mapping (a 2푚 × 2푛 real matrix) is called the real Jacobian, and we write it as 퐷ℝ 푓 (푝). 1.3. DERIVATIVES 27

Proposition 1.3.7. Let 푈 ⊂ ℂ푛 be an open set, 푝 ∈ 푈, and 푓 : 푈 → ℂ푛 be holomorphic. Then 2 |det 퐷 푓 (푝)| = det 퐷ℝ 푓 (푝). The expression det 퐷 푓 (푝) is called the Jacobian determinant and clearly it is important to know if we are talking about the holomorphic Jacobian determinant or the standard real Jacobian determinant det 퐷ℝ 푓 (푝). Recall from vector calculus that if the real Jacobian determinant det 퐷ℝ 푓 (푝) of a smooth mapping is positive, then the mapping preserves orientation. In particular, the proposition says that holomorphic mappings preserve orientation.

Proof. The real mapping using our identiﬁcation is (Re 푓1, Im 푓1,..., Re 푓푛, Im 푓푛) as a function of (푥1, 푦1, . . . , 푥푛, 푦푛). The statement is about the two Jacobians at 푝, that is, the derivatives at 푝. Hence, we can assume that 푝 = 0 and 푓 is complex linear, 푓 (푧) = 퐴푧 for some 푛 × 푛 matrix 퐴. It is just a statement about matrices. The matrix 퐴 is the (holomorphic) Jacobian matrix of 푓 . Let 퐵 be the real Jacobian matrix of 푓 . Let us change the basis of 퐵 to be (푧, 푧¯) using 푧 = 푥 + 푖푦 and 푧¯ = 푥 − 푖푦 on both the target and the source. The change of basis is some invertible complex matrix 푀 such that 푀−1퐵푀 (the real Jacobian matrix 퐵 in these new coordinates) is a matrix of the derivatives of ( 푓1, . . . , 푓푛, 푓¯1,..., 푓¯푛) in terms of (푧1, . . . , 푧푛, 푧¯1,..., 푧¯푛). In other words, 퐴 0 푀−1퐵푀 = . 0 퐴 Thus det(퐵) = det(푀−1푀퐵) = det(푀−1퐵푀) = det(퐴) det(퐴) = det(퐴) det(퐴) = |det(퐴)|2 . The regular implicit function theorem and the chain rule give that the implicit function theorem holds in the holomorphic setting. The main thing to check is to check that the solution given by the standard implicit function theorem is holomorphic, which follows by the chain rule. Theorem 1.3.8 (Implicit function theorem). Let 푈 ⊂ ℂ푛 × ℂ푚 be an open set, let (푧, 푤) ∈ ℂ푛 × ℂ푚 be our coordinates, and let 푓 : 푈 → ℂ푚 be a holomorphic mapping. Let (푧0, 푤0) ∈ 푈 be a point such that 푓 (푧0, 푤0) = 0 and such that the 푚 × 푚 matrix 휕 푓 푗 (푧0, 푤0) 휕푤 푘 푗 푘 is invertible. Then there exists an open set 푉 ⊂ ℂ푛 with 푧0 ∈ 푉, open set 푊 ⊂ ℂ푚 with 푤0 ∈ 푊, 푉 × 푊 ⊂ 푈, and a holomorphic mapping 푔 : 푉 → 푊, with 푔(푧0) = 푤0 such that for every 푧 ∈ 푉, the point 푔(푧) is the unique point in 푊 such that 푓 푧, 푔(푧) = 0.

Exercise 1.3.5: Prove the holomorphic implicit function theorem above. Hint: Check that the normal implicit function theorem for 퐶1 functions applies, and then show that the 푔 you obtain is holomorphic.

Exercise 1.3.6: State and prove a holomorphic version of the inverse function theorem. 28 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES 1.4 Inequivalence of ball and polydisc

Deﬁnition 1.4.1. Two domains 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푛 are said to be biholomorphic or biholomorphically equivalent if there exists a one-to-one and onto holomorphic map 푓 : 푈 → 푉 such that the inverse 푓 −1 : 푉 → 푈 is holomorphic. The mapping 푓 is said to be a biholomorphic map or a biholomorphism. As function theory on two biholomorphic domains is the same, one of the main questions in complex analysis is to classify domains up to biholomorphic transformations. In one variable, there is the rather striking theorem due to Riemann: Theorem 1.4.2 (Riemann mapping theorem). If 푈 ⊂ ℂ is a nonempty simply connected domain such that 푈 ≠ ℂ, then 푈 is biholomorphic to 픻. In one variable, a topological property on 푈 is enough to classify a whole class of domains. It is one of the reasons why studying the disc is so important in one variable, and why many theorems are stated for the disc only. There is simply no such theorem in several variables. We will show momentarily that the unit ball and the polydisc,

푛 푛 푛 픹푛 = 푧 ∈ ℂ : k푧k < 1 and 픻 = 푧 ∈ ℂ : |푧푗 | < 1 for 푗 = 1, . . . , 푛 , are not biholomorphically equivalent. Both are simply connected (have no holes), and they are the two most obvious generalizations of the disc to several variables. They are homeomorphic, that is, topology does not see any diﬀerence.

푛 Exercise 1.4.1: Prove that there exists a homeomorphism 푓 : 픹푛 → 픻 , that is, 푓 is a bĳection, and both 푓 and 푓 −1 are continuous.

2 Let us stick with 푛 = 2. Instead of proving that 픹2 and 픻 are biholomorphically inequivalent we will prove a stronger theorem. First a definition. Definition 1.4.3. Suppose 푓 : 푋 → 푌 is a continuous map between two topological spaces. Then 푓 is a proper map if for every compact 퐾 ⊂⊂ 푌, the set 푓 −1(퐾) is compact. The notation “⊂⊂” is a common notation for relatively compact subset, that is, the closure is compact in the relative (subspace) topology. Often the distinction between compact and relatively compact is not important. For instance, in the definition above we can replace compact with relatively compact. So the notation is sometimes used if “compact” is meant. Vaguely, “proper” means that “boundary goes to the boundary.” As a continuous map, 푓 pushes compacts to compacts; a proper map is one where the inverse does so too. If the inverse is a continuous function, then clearly 푓 is proper, but not every proper map is invertible. For example, the map 푓 : 픻 → 픻 given by 푓 (푧) = 푧2 is proper, but not invertible. The codomain of 푓 is important. If we replace 푓 by 푔 : 픻 → ℂ, still given by 푔(푧) = 푧2, then the map is no longer proper. Let us state the main result of this section. Theorem 1.4.4 (Rothstein 1935). There exists no proper holomorphic mapping of the unit bidisc 2 2 2 픻 = 픻 × 픻 ⊂ ℂ to the unit ball 픹2 ⊂ ℂ . 1.4. INEQUIVALENCE OF BALL AND POLYDISC 29

As a biholomorphic mapping is proper, the unit bidisc is not biholomorphically equivalent to the unit ball in ℂ2. This fact was first proved by Poincaré by computing the automorphism groups 2 of 픻 and 픹2, although his proof assumed the maps extended past the boundary. The first complete proof was by Henri Cartan in 1931, though popularly the theorem is attributed to Poincaré. It seems standard practice that any general audience talk about several complex variables contains a mention of Poincaré, and often the reference is to this exact theorem. We need some lemmas before we get to the proof of the result. First, a certain one-dimensional object plays an important role in the geometry of several complex variables. It allows us to apply one-variable results in several variables. It is especially important in understanding the boundary behavior of holomorphic functions. It also prominently appears in complex geometry. Definition 1.4.5. A nonconstant holomorphic mapping 휑 : 픻 → ℂ푛 is called an analytic disc. If the mapping 휑 extends continuously to the closed unit disc 픻, then the mapping 휑 : 픻 → ℂ푛 is called a closed analytic disc. Often we call the image Δ = 휑(픻) the analytic disc rather than the mapping. For a closed analytic disc we write 휕Δ = 휑(휕픻) and call it the boundary of the analytic disc. In some sense, analytic discs play the role of line segments in ℂ푛. It is important to always have in mind that there is a mapping defining the disc, even if we are more interested in the set. Obviously for a given image, the mapping 휑 is not unique. 2 2 Let us consider the boundaries of the unit bidisc 픻 × 픻 ⊂ ℂ and the unit ball 픹2 ⊂ ℂ . We notice the boundary of the unit bidisc contains analytic discs {푝} × 픻 and 픻 × {푝} for 푝 ∈ 휕픻. That is, through every point in the boundary, except for the distinguished boundary 휕픻 × 휕픻, there exists an analytic disc lying entirely inside the boundary. On the other hand, the ball contains no analytic discs in its boundary. 2푛−1 푛 Proposition 1.4.6. The unit sphere 푆 = 휕픹푛 ⊂ ℂ contains no analytic discs. Proof. Suppose there is a holomorphic function 푔 : 픻 → ℂ푛 such that the image 푔(픻) is inside the unit sphere. In other words 2 2 2 2 k푔(푧)k = |푔1(푧)| + |푔2(푧)| + · · · + |푔푛 (푧)| = 1 for all 푧 ∈ 픻. Without loss of generality (after composing with a unitary matrix) assume that 푔(0) = (1, 0, 0,..., 0). Consider the first component and notice that 푔1(0) = 1. If a sum of positive numbers is less than or equal to 1, then they all are, and hence |푔1(푧)| ≤ 1. Maximum principle says that 푔1(푧) = 1 for all 푧 ∈ 픻. But then 푔푗 (푧) = 0 for all 푗 = 2, . . . , 푛 and all 푧 ∈ 픻. Therefore, 푔 is constant and thus not an analytic disc. The fact that the sphere contains no analytic discs is the most important geometric distinction between the boundary of the polydisc and the sphere.

Exercise 1.4.2: Modify the proof to show some stronger results. a) Let Δ be an analytic disc and Δ ∩ 휕픹푛 ≠ ∅. Prove Δ contains points not in 픹푛. b) Let Δ be an analytic disc. Prove that Δ ∩ 휕픹푛 is nowhere dense in Δ. 2 c) Find an analytic disc in ℂ , such that (1, 0) ∈ Δ, Δ ∩ 픹2 = ∅, and locally near (1, 0) ∈ 휕픹2, 2 2 the set Δ ∩ 휕픹2 is the curve deﬁned by Im 푧1 = 0, Im 푧2 = 0, (Re 푧1) + (Re 푧2) = 1. 30 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Before we prove the theorem, let us prove a lemma making the statement about proper maps taking boundary to boundary precise.

Lemma 1.4.7. Let 푈 ⊂ ℝ푛 and 푉 ⊂ ℝ푚 be bounded domains and let 푓 : 푈 → 푉 be continuous. Then 푓 is proper if and only if for every sequence {푝푘 } in 푈 such that 푝푘 → 푝 ∈ 휕푈, the set of limit points of 푓 (푝푘 ) lies in 휕푉.

Proof. Suppose 푓 is proper. Take a sequence {푝푘 } in 푈 such that 푝푘 → 푝 ∈ 휕푈. Then take any 푓 푝 푓 푝 푞 푉 퐸 푓 푝 convergent subsequence ( 푘푗 ) of ( 푘 ) converging to some ∈ . Consider = ( 푘푗 ) as a set. Let 퐸 be the closure of 퐸 in 푉 (subspace topology). If 푞 ∈ 푉, then 퐸 = 퐸 ∪ {푞} and 퐸 is compact. Otherwise, if 푞 ∉ 푉, then 퐸 = 퐸. The inverse image 푓 −1(퐸) is not compact (it contains a sequence going to 푝 ∈ 휕푈) and hence 퐸 is not compact either as 푓 is proper. Thus 푞 ∉ 푉, and hence 푞 ∈ 휕푉. As we took an arbitrary subsequence of 푓 (푝푘 ) , 푞 was an arbitrary limit point. Therefore, all limit points are in 휕푉. Let us prove the converse. Suppose that for every sequence {푝푘 } in 푈 such that 푝푘 → 푝 ∈ 휕푈, the set of limit points of 푓 (푝푘 ) lies in 휕푉. Take a closed set 퐸 ⊂ 푉 (subspace topology) and −1 −1 −1 look at 푓 (퐸). If 푓 (퐸) is not compact, then there exists a sequence {푝푘 } in 푓 (퐸) such that −1 푝푘 → 푝 ∈ 휕푈. That is because 푓 (퐸) is closed (in 푈) but not compact. The hypothesis then says that the limit points of 푓 (푝푘 ) are in 휕푉. Hence 퐸 has limit points in 휕푉 and is not compact.

Exercise 1.4.3: Let 푈 ⊂ ℝ푛 and 푉 ⊂ ℝ푚 be bounded domains and let 푓 : 푈 → 푉 be continuous. Suppose 푓 (푈) ⊂ 푉, and 푔 : 푈 → 푉 is deﬁned by 푔(푥) = 푓 (푥) for all 푥 ∈ 푈. Prove that 푔 is proper if and only if 푓 (휕푈) ⊂ 휕푉.

Exercise 1.4.4: Let 푓 : 푋 → 푌 be a continuous function of locally compact Hausdorﬀ topological spaces. Let 푋∞ and 푌∞ be the one-point compactiﬁcations of 푋 and 푌. Then 푓 is a proper map if and only if it extends as a continuous map 푓∞ : 푋∞ → 푌∞ by letting 푓∞|푋 = 푓 and 푓∞(∞) = ∞.

We now have all the lemmas needed to prove the theorem of Rothstein.

2 푖휃 Proof of Theorem 1.4.4 . Suppose there is a proper holomorphic map 푓 : 픻 → 픹2. Fix some 푒 푖휃 in the boundary of the disc 픻. Take a sequence 푤푘 ∈ 픻 such that 푤푘 → 푒 . The functions

푔푘 (휁) = 푓 (휁, 푤푘 ) map the unit disc into 픹2. By the standard Montel’s theorem , by passing to a subsequence we assume that the sequence of functions converges (uniformly on compact subsets) to 푖휃 2 a limit 푔 : 픻 → 픹2. As (휁, 푤푘 ) → (휁, 푒 ) ∈ 휕픻 , then by Lemma 1.4.7 we have that 푔(픻) ⊂ 휕픹2 and hence 푔 must be constant. 푔0 푔0 Let 푘 denote the derivative (we differentiate each component). The functions 푘 converge to 푔0 = 0. So for an arbitrary fixed 휁 ∈ 픻, 휕 푓 (휁, 푤 ) → 0. This limit holds for all 푒푖휃 and 휕푧1 푘 푖휃 some subsequence of an arbitrary sequence {푤푘 } where 푤푘 → 푒 . The holomorphic mapping 푤 ↦→ 휕 푓 (휁, 푤), therefore, extends continuously to the closure 픻 and is zero on 휕픻. We apply the 휕푧1 maximum principle or the Cauchy formula and the fact that 휁 was arbitrary to find 휕 푓 ≡ 0. By 휕푧1 symmetry 휕 푓 ≡ 0. Therefore, 푓 is constant, which is a contradiction as 푓 was proper. 휕푧2 1.4. INEQUIVALENCE OF BALL AND POLYDISC 31

The proof is illustrated in the following picture:

휁, 푒푖휃 |푧2| ( ) (휁, 푤푘 ) 푤2 푒푖휃 (휁, 푤 ) 2 푤1 푤푘 (휁, 푤1)

|푧1|

In the picture, on the left-hand side is the bidisc, and we restrict 푓 to the horizontal gray lines (where the second component is fixed to be 푤푘 ) and take a limit to produce an analytic disc in the boundary of 픹 . We then show that 휕 푓 = 0 on the vertical gray line (where the first component is 2 휕푧1 fixed to be 휁). The right-hand side shows the disc where 푧1 = 휁 is fixed, which corresponds to the vertical gray line on the left.

The proof says that the reason why there is not even a proper mapping is the fact that the boundary of the polydisc contains analytic discs, while the sphere does not. The proof extends easily to higher dimensions as well, and the proof of the generalization is left as an exercise.

Theorem 1.4.8. Let 푈 = 푈0 × 푈00 ⊂ ℂ푛 × ℂ푘 , 푛, 푘 ≥ 1, and 푉 ⊂ ℂ푚, 푚 ≥ 1, be bounded domains such that 휕푉 contains no analytic discs. Then there exist no proper holomorphic mapping 푓 : 푈 → 푉.

Exercise 1.4.5: Prove Theorem 1.4.8 .

The key takeaway from this section is that in several variables, to see if two domains are equivalent, the geometry of the boundaries makes a diﬀerence, not just the topology of the domains. The following is a fun exercise in one dimension about proper maps of discs:

Exercise 1.4.6: Let 푓 : 픻 → 픻 be a proper holomorphic map. Then

푘 Ö 푧 − 푎푗 푓 (푧) = 푒푖휃 , 1 − 푎¯ 푧 푗=1 푗

−1 for some real 휃 and some 푎푗 ∈ 픻 (that is, 푓 is a ﬁnite Blaschke product). Hint: Consider 푓 (0).

In several variables, when 픻 is replaced by a ball, this question (what are the proper maps) becomes far more involved, and if the dimensions of the balls are diﬀerent, it is not solved in general. 32 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Exercise 1.4.7: Suppose 푓 : 푈 → 픻 be a proper holomorphic map where 푈 ⊂ ℂ푛 is a nonempty

domain. Prove that 푛 = 1. Hint: Consider the same idea as in Exercise 1.2.19 . 푓 푚 푓 Exercise 1.4.8: Suppose : 픹푛 → ℂ is a nonconstant continuous map such that |픹푛 is 푓 푧 푧 푓 holomorphic and k ( )k = 1 whenever k k = 1. Prove that |픹푛 maps into 픹푚 and furthermore that this map is proper.

1.5 Cartan’s uniqueness theorem

The following theorem is another analogue of Schwarz’s lemma to several variables. It says that for a bounded domain, it is enough to know that a self mapping is the identity at a single point to show that it is the identity everywhere. As there are quite a few theorems named for Cartan, this one is often referred to as the Cartan’s uniqueness theorem. It is useful in computing the automorphism groups of certain domains. An automorphism of 푈 is a biholomorphic map from 푈 onto 푈. Automorphisms form a group under composition, called the automorphism group. As 푛 exercises, you will use the theorem to compute the automorphism groups of 픹푛 and 픻 . Theorem 1.5.1 (Cartan). Suppose 푈 ⊂ ℂ푛 is a bounded domain, 푎 ∈ 푈, 푓 : 푈 → 푈 is a holomorphic mapping, 푓 (푎) = 푎, and 퐷 푓 (푎) is the identity. Then 푓 (푧) = 푧 for all 푧 ∈ 푈.

Exercise 1.5.1: Find a counterexample to the theorem if 푈 is unbounded. Hint: For simplicity take 푎 = 0 and 푈 = ℂ푛.

Before we get into the proof, let us write down the Taylor series of a function in a nicer way, splitting it up into parts of diﬀerent degree. A polynomial 푃 : ℂ푛 → ℂ is homogeneous of degree 푑 if

푃(푠푧) = 푠푑 푃(푧)

for all 푠 ∈ ℂ and 푧 ∈ ℂ푛. A homogeneous polynomial of degree 푑 is a polynomial whose every monomial is of total degree 푑. For instance, 푧2푤 − 푖푧3 + 9푧푤2 is homogeneous of degree 3 in the variables (푧, 푤) ∈ ℂ2. A polynomial vector-valued mapping is homogeneous of degree 푑, if each component is. If 푓 is holomorphic near 푎 ∈ ℂ푛, then write the power series of 푓 at 푎 as

∞ ∑︁ 푓푗 (푧 − 푎), 푗=0 where 푓푗 is a homogeneous polynomial of degree 푗. The 푓푗 is called the degree 푗 homogeneous part of 푓 at 푎. The 푓푗 would be vector-valued if 푓 is vector-valued, such as in the statement of the ∗ theorem. In the proof, we will require the vector-valued Cauchy estimates (exercise below) .

∗The normal Cauchy estimates could also be used in the proof of Cartan by applying them componentwise. 1.5. CARTAN’S UNIQUENESS THEOREM 33

푚 Exercise 1.5.2: Prove a vector-valued version of the Cauchy estimates. Suppose 푓 : Δ푟 (푎) → ℂ 푛 is continuous function holomorphic on a polydisc Δ푟 (푎) ⊂ ℂ . Let Γ denote the distinguished boundary of Δ. Show that for any multi-index 훼 we get

휕|훼| 푓 훼! (푎) ≤ k 푓 (푧)k . 훼 훼 sup 휕푧 푟 푧∈Γ

Proof of Cartan’s uniqueness theorem. Without loss of generality, assume 푎 = 0. Write 푓 as a power series at the origin, written in homogeneous parts: ∞ ∑︁ 푓 (푧) = 푧 + 푓푘 (푧) + 푓푗 (푧), 푗=푘+1 where 푘 ≥ 2 is an integer such that 푓2(푧), 푓3(푧), . . . , 푓푘−1(푧) is zero. The degree-one homogeneous part is simply the vector 푧, because the derivative of 푓 at the origin is the identity. Compose 푓 with itself ℓ times: 푓 ℓ (푧) = 푓 ◦ 푓 ◦ · · · ◦ 푓 (푧). | {z } ℓ times As 푓 (푈) ⊂ 푈, then 푓 ℓ is a holomorphic map of 푈 to 푈. As 푈 is bounded, there is an 푀 such that k푧k ≤ 푀 for all 푧 ∈ 푈. Therefore, k 푓 (푧)k ≤ 푀 for all 푧 ∈ 푈, and k 푓 ℓ (푧)k ≤ 푀 for all 푧 ∈ 푈. If we plug in 푧 + 푓푘 (푧) + higher order terms into 푓푘 , we get 푓푘 (푧) + some other higher order 2 terms. Therefore, 푓 (푧) = 푧 + 2 푓푘 (푧) + higher order terms. Continuing this procedure, ∞ ℓ ∑︁ 푓 (푧) = 푧 + ℓ 푓푘 (푧) + 푓˜푗 (푧), 푗=푘+1

for some other degree 푗 homogeneous polynomials 푓˜푗 . Suppose Δ푟 (0) is a polydisc whose closure is in 푈. Via Cauchy estimates, for any multinomial 훼 with |훼| = 푘,

|훼| ℓ |훼| 훼! 휕 푓 휕 푓 푀 ≥ (0) = ℓ (0) . 푟훼 휕푧훼 휕푧훼

ℓ 휕 |훼| 푓 푓 The inequality holds for all ∈ ℕ, and so 휕푧훼 (0) = 0. Therefore, 푘 ≡ 0. On the domain of convergence of the expansion, we get 푓 (푧) = 푧, as there is no other nonzero homogeneous part in the expansion of 푓 . As 푈 is connected, then the identity theorem says 푓 (푧) = 푧 for all 푧 ∈ 푈. As an application, let us classify all biholomorphisms of all bounded circular domains that ﬁx a point. A circular domain is a domain 푈 ⊂ ℂ푛 such that if 푧 ∈ 푈, then 푒푖휃 푧 ∈ 푈 for all 휃 ∈ ℝ. Corollary 1.5.2. Suppose 푈, 푉 ⊂ ℂ푛 are bounded circular domains with 0 ∈ 푈, 0 ∈ 푉, and 푓 : 푈 → 푉 is a biholomorphic map such that 푓 (0) = 0. Then 푓 is linear.

For example, 픹푛 is circular and bounded. So a biholomorphism of 픹푛 (an automorphism) that ﬁxes the origin is linear. Similarly, a polydisc centered at zero is also circular and bounded. 34 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Proof. The map 푔(푧) = 푓 −1 푒−푖휃 푓 (푒푖휃 푧) is an automorphism of 푈 and via the chain rule, 푔0(0) = 퐼. Therefore, 푓 −1 푒−푖휃 푓 (푒푖휃 푧) = 푧, or in other words 푓 (푒푖휃 푧) = 푒푖휃 푓 (푧). 푓 푓 푧 Í∞ 푓 푧 푓 푗 Write near zero as ( ) = 푗=1 푗 ( ) where 푗 are homogeneous polynomials of degree (notice 푓0 = 0). Then ∞ ∞ ∞ ∞ ∑︁ 푖휃 푖휃 ∑︁ ∑︁ 푖휃 ∑︁ 푖 푗휃 푒 푓푗 (푧) = 푒 푓푗 (푧) = 푓푗 (푒 푧) = 푒 푓푗 (푧). 푗=1 푗=1 푗=1 푗=1 푖휃 푖 푗휃 푖( 푗−1)휃 By the uniqueness of the Taylor expansion, 푒 푓푗 (푧) = 푒 푓푗 (푧), or 푓푗 (푧) = 푒 푓푗 (푧), for all 푗, all 푧, and all 휃. If 푗 ≠ 1, we obtain that 푓푗 ≡ 0, which proves the claim.

Exercise 1.5.3: Show that every automorphism 푓 of 픻푛 (that is, a biholomorphism 푓 : 픻푛 → 픻푛) is given as 푧1 − 푎1 푧2 − 푎2 푧푛 − 푎푛 푓 (푧) = 푃 푒푖휃1 , 푒푖휃2 , . . . , 푒푖휃푛 1 − 푎¯1푧1 1 − 푎¯2푧2 1 − 푎¯푛푧푛 for 휃 ∈ ℝ푛, 푎 ∈ 픻푛, and a permutation matrix 푃. 푎 푃 푧 h푧,푎i 푎 푎 푃 푧 Exercise 1.5.4: Given ∈ 픹푛, define the linear map 푎 = h푎,푎i if ≠ 0 and 0 = 0. Let √︁ 2 푠푎 = 1 − k푎k . Show that every automorphism 푓 of 픹푛 (that is, a biholomorphism 푓 : 픹푛 → 픹푛) can be written as 푎 − 푃 푧 − 푠 (퐼 − 푃 )푧 푓 (푧) = 푈 푎 푎 푎 1 − h푧, 푎i for a unitary matrix 푈 and some 푎 ∈ 픹푛. 푛 Exercise 1.5.5: Using the previous two exercises, show that 픻 and 픹푛, 푛 ≥ 2, are not biholomorphic via a method more in the spirit of what Poincaré used: Show that the groups of automorphisms of the two domains are different groups when 푛 ≥ 2. Exercise 1.5.6: Suppose 푈 ⊂ ℂ푛 is a bounded open set, 푎 ∈ 푈, and 푓 : 푈 → 푈 is a holomorphic mapping such that 푓 (푎) = 푎. Show that every eigenvalue 휆 of the matrix 퐷 푓 (푎) satisfies |휆| ≤ 1. Exercise 1.5.7 (Tricky): Find a domain 푈 ⊂ ℂ푛 such that the only biholomorphism 푓 : 푈 → 푈 is the identity 푓 (푧) = 푧. Hint: Take the polydisc (or the ball) and remove some number of points (be careful in how you choose them). Then show that 푓 extends to a biholomorphism of the polydisc. Then see what happens to those points you took out. Exercise 1.5.8: a) Show that Cartan’s uniqueness theorem is not true in the real case, even for rational functions. That is, find a rational function 푅(푡) of a real variable 푡, such that 푅 that takes (−1, 1) to (−1, 1), 푅0(0) = 1, and 푅(푡) is not the identity. You can even make 푅 bĳective.

b) Show that also Exercise 1.5.6 is not true in the real case. For any 훼 ∈ ℝ ﬁnd a rational function 푅(푡) of a real variable 푡, such that 푅 takes (−1, 1) to (−1, 1) and 푅0(0) = 훼. Exercise 1.5.9: Suppose 푈 ⊂ ℂ푛 is an open set, 푎 ∈ 푈, 푓 : 푈 → 푈 is a holomorphic mapping, 푓 (푎) = 푎, and suppose that |휆| < 1 for every eigenvalue 휆 of 퐷 푓 (푎). Prove that there exists a ℓ neighborhood 푊 of 푎, such that limℓ→∞ 푓 (푧) = 푎 for all 푧 ∈ 푊. 1.6. RIEMANN EXTENSION THEOREM, ZERO SETS, AND INJECTIVE MAPS 35 1.6 Riemann extension theorem, zero sets, and injective maps

∗ In one dimension if a function is holomorphic in 푈 \{푝} and locally bounded in 푈, in particular

bounded near 푝, then the function extends holomorphically to 푈 (see Proposition B.22 (i) ). In several variables the same theorem holds, and the analogue of a single point is the zero set of a holomorphic function. Theorem 1.6.1 (Riemann extension theorem). Let 푈 ⊂ ℂ푛 be a domain, 푔 ∈ O(푈), and 푔 is not identically zero. Let 푁 = 푔−1(0) be the zero set of 푔. If 푓 ∈ O(푈 \ 푁) is locally bounded in 푈, then there exists a unique 퐹 ∈ O(푈) such that 퐹|푈\푁 = 푓 . The proof is an application of the Riemann extension theorem from one dimension. Proof. Take any 푝 ∈ 푁, and let 퐿 be a complex line through 푝. That is, 퐿 is an image of an affine mapping 휑 : ℂ → ℂ푛 defined by 휑(휉) = 푎휉 + 푝, for a vector 푎 ∈ ℂ푛. The composition 푔 ◦ 휑 is a holomorphic function of one variable, and it is either identically zero, or the zero at 휉 = 0 is isolated. The function 푔 is not identically zero in any neighborhood of 푝. So there is some line 퐿 such that 푔 ◦ 휑 is not identically zero, or in other words, 푝 is an isolated point of 퐿 ∩ 푁. 0 0 Write 푧 = (푧1, . . . , 푧푛−1) and 푧 = (푧 , 푧푛). Without loss of generality 푝 = 0, and 퐿 is the line obtained by 푧0 = 0. So 푔 ◦ 휑 is 휉 ↦→ 푔(0, 휉). There is some small 푟 > 0 such that 푔 is nonzero on 0 the set given by |푧푛| = 푟 and 푧 = 0. By continuity, 푔 is nonzero on the set given by |푧푛| = 푟 and k푧0k < 휖 for some 휖 > 0. In particular, for any fixed small 푠 ∈ ℂ푛−1, with k푠k < 휖, setting 푧0 = 푠, the zeros of 휉 ↦→ 푔(푠, 휉) are isolated.

푧푛 휖

푔(푧) = 0

푧0 푟 푔(푧) = 0

푧0 = 푠

0 For k푧 k < 휖 and |푧푛| < 푟, write ∫ 0 0 1 푓 (푧 , 휉) 퐹(푧 , 푧푛) = 푑휉. 2휋푖 |휉|=푟 휉 − 푧푛 The function 휉 → 푓 (푧0, 휉) extends holomorphically to the entire disc of radius 푟 by the Riemann extension from one dimension. By Cauchy integral formula, 퐹 is equal to 푓 at the points where they are both deﬁned. By diﬀerentiating under the integral, the function 퐹 is holomorphic in all variables. In a neighborhood of each point of 푁, 푓 extends to a continuous (holomorphic in fact) function. A continuous extension of 푓 must be unique on the closure of 푈 \ 푁 in the subspace topology,

∗ 푓 : 푈 \ 푋 → ℂ is locally bounded in 푈 if for every 푝 ∈ 푈, there is a neighborhood 푊 of 푝 such that 푓 is bounded on 푊 ∩ (푈 \ 푋). 36 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

(푈 \ 푁) ∩푈. The set 푁 has empty interior, so (푈 \ 푁) ∩푈 = 푈. Hence, 퐹 is the unique continuous extension of 푓 to 푈. The set of zeros of a holomorphic function has a nice structure at most points. Theorem 1.6.2. Let 푈 ⊂ ℂ푛 be a domain and 푓 ∈ O(푈) and 푓 is not identically zero. Let −1 푁 = 푓 (0). Then there exists a open and dense (subspace topology) subset 푁reg ⊂ 푁 such that near each 푝 ∈ 푁reg, after possibly reordering variables, 푁 can be locally written as

푧푛 = 푔(푧1, . . . , 푧푛−1) for a holomorphic function 푔.

Proof. If 푁 is locally a graph at 푝, then it is a graph for every point of 푁 near 푝. So 푁reg is open. If for every point 푝0 ∈ 푁 and every neighborhood 푊 of 푝0, we show that 푁 ∩ 푊 has a regular point, then 푁reg is dense. Replacing 푁 with 푁 ∩ 푊, it thus suffices to show 푁reg is nonempty. Since 푓 is not identically zero, then not all derivatives (of arbitrary order) of 푓 vanish identically on 푁. If some first order derivative of 푓 does not vanish identically on 푁, let ℎ = 푓 . Otherwise, suppose 푘 is such that a derivative of 푓 of order 푘 does not vanish identically on 푁, and all derivatives of 푓 order less than 푘 vanish identically on 푁. Let ℎ be one of the derivatives of order 푘 − 1. We obtain a function ℎ : 푈 → ℂ, holomorphic, vanishing on 푁, and such that without loss of generality the 푧푛 derivative does not vanish identically on 푁. Then there is some point 푝 ∈ 푁 such that 휕ℎ (푝) ≠ 0. We apply the implicit function theorem at 푝 to find 푔 such that 휕푧푛 ℎ 푧1, . . . , 푧푛−1, 푔(푧1, . . . , 푧푛−1) = 0,

and 푧푛 = 푔(푧1, . . . , 푧푛−1) is the unique solution to ℎ = 0 near 푝. The zero set of ℎ contains 푁, the zero set of 푓 . We must show equality near 푝. That is, we need 0 to show that near 푝, every zero of ℎ is also a zero of 푓 . Write 푝 = (푝 , 푝푛). Then the function 휉 ↦→ 푓 (푝0, 휉)

has an isolated zero in a small disc Δ around 푝푛 and is nonzero on the circle 휕Δ. By Rouché’s 0 0 0

theorem , 휉 ↦→ 푓 (푧 , 휉) must have a zero for all 푧 suﬃciently close to 푝 (close enough to make 0 0 0 0 0 | 푓 (푞 , 휉) − 푓 (푧 , 휉)| < | 푓 (푞 , 휉)| for all 휉 ∈ 휕Δ). Since 푔(푧 ) is the unique solution 푧푛 to ℎ(푧 , 푧푛) = 0 near 푝 and the zero set of 푓 is contained in the zero set of ℎ, we are done. The zero set 푁 of a holomorphic function is a so-called subvariety or an analytic set although

the general deﬁnition of a subvariety is more complicated, and includes more sets. See chapter 6 . Points where 푁 is a graph of a holomorphic mapping are called regular points, and we write them as 푁reg as above. In particular, since 푁 is a graph of a single holomorphic function, they are called regular points of (complex) dimension 푛 − 1, or (complex) codimension 1. The set of regular points is what is called an (푛 − 1)-dimensional complex submanifold. It is also a real submanifold of real dimension 2푛 − 2. The points on a subvariety that are not regular are called singular points. 푈 2 푓 푧 푧2 푧2 푋 푓 −1 푓 푧 , 푧 Example 1.6.3: For = ℂ , let ( ) = 1 − 2 and consider = (0). As ∇ = (2 1 2 2), outside of the origin, we can solve for 푧1 or 푧2 and so all points of 푋 \ 0 are regular. In fact, 푧1 = 푧2 and 푧1 = −푧2 are the two possibilities. In no neighborhood of the origin, however, is there a way to 1.6. RIEMANN EXTENSION THEOREM, ZERO SETS, AND INJECTIVE MAPS 37

solve for either 푧1 or 푧2, since you always get two possible solutions: If you could solve 푧1 = 푔(푧2), then both 푧2 = 푔(푧2) and −푧2 = 푔(푧2) must be true, a contradiction for any nonzero 푧2. Similarly, we cannot solve for 푧1. So the origin is a singular point. To see that you may have need to use derivatives of the function, notice that the function 휑(푧) = (푧2 − 푧2)2 has the same zero set 푋, but both 휕휑 and 휕휑 vanish on 푋. Using ℎ = 휕휑 or 1 2 휕푧1 휕푧2 휕푧1 ℎ = 휕휑 in the proof will work. 휕푧2 푓 푧 푧2푧 It may be also good to consider a function such as ( ) = 1 2 to see that we do not always get the entire set of regular functions with one neighborhood 푊. The zero set is where 푧1 = 0 or where 푧2 = 0, and all points except the origin are regular. We use ℎ = 푓 to show that points outside the origin where 푧2 = 0 are regular. To show that points outside the origin where 푧1 = 0 are regular, we must restrict to some neighborhood 푊 of those points and use ℎ = 휕 푓 . 휕푧1

푥2 푥2 Example 1.6.4: The theorem is not true in the nonholomorphic setting. The set where 1 + 2 = 0 in ℝ2 is only the origin, clearly not a graph of any function of one variable. The ﬁrst part of the theorem works, but the ℎ you ﬁnd is either 2푥1 or 2푥2, and its zero set is too big.

푋 푧 ∈ 2 푧2 푧3 Exercise 1.6.1: Find all the regular points of the subvariety = ℂ : 1 = 2 . Hint: The trick is showing that you’ve found all of them.

Exercise 1.6.2: Suppose 푈 ⊂ ℂ푛 is a domain and 푓 ∈ O(푈). Show that the complement of the zero set, 푈 \ 푓 −1(0), is connected.

Let us now prove that a one-to-one holomorphic mapping is biholomorphic, a result deﬁnitely not true in the smooth setting: 푥 ↦→ 푥3 is smooth, one-to-one, onto map of ℝ to ℝ, but the inverse is not diﬀerentiable.

Theorem 1.6.5. Suppose 푈 ⊂ ℂ푛 is an open set and 푓 : 푈 → ℂ푛 is holomorphic and one-to-one. Then the Jacobian determinant is never equal to zero on 푈. In particular, if a holomorphic map 푓 : 푈 → 푉 is one-to-one and onto for two open sets 푈, 푉 ⊂ ℂ푛, then 푓 is biholomorphic.

The function 푓 is locally biholomorphic, in particular 푓 −1 is holomorphic, on the set where the Jacobian determinant 휕 푓 퐽 (푧) = 퐷 푓 (푧) = 푗 (푧) 푓 det det 휕푧 푘 푗 푘 is not zero. This follows from the inverse function theorem, which is just a special case of the implicit function theorem. The trick is to show that 퐽푓 happens to be nonzero everywhere. In one complex dimension, every holomorphic function 푓 can, in the proper local holomorphic 푑 coordinates (and up to adding a constant), be written as 푧 for 푑 = 0, 1, 2,...: Near a 푧0 ∈ ℂ, there 푑 exists a constant 푐 and a local biholomorphic 푔 with 푔(푧0) = 0 such that 푓 (푧) = 푐 + 푔(푧) . Such a simple result does not hold in several variables in general, but if the mapping is locally one-to-one, then the present theorem says that such a mapping can be locally written as the identity. 38 CHAPTER 1. HOLOMORPHIC FUNCTIONS IN SEVERAL VARIABLES

Proof of the theorem. We proceed by induction. We know the theorem for 푛 = 1. Suppose 푛 > 1 and suppose we know the theorem is true for dimension 푛 − 1. Suppose for contradiction that 퐽푓 = 0 somewhere. First suppose that 퐽푓 is not identically zero. Find a regular point 푞 on the zero set of 퐽푓 . Write the zero set of 퐽푓 near 푞 as

푧푛 = 푔(푧1, . . . , 푧푛−1) for some holomorphic 푔. If we prove the theorem near 푞, we are done. Without loss of generality assume 푞 = 0. The biholomorphic (near the origin) map Ψ(푧1, . . . , 푧푛) = 푧1, 푧2, . . . , 푧푛−1, 푧푛 − 푔(푧1, . . . , 푧푛−1) −1 takes the zero set of 퐽푓 to the set given by 푧푛 = 0. By considering 푓 ◦ Ψ instead of 푓 , we may assume that 퐽푓 = 0 on the set given by 푧푛 = 0. We may also assume that 푓 (0) = 0. If 퐽푓 vanishes identically, then there is no need to do anything other than a translation. In either case, we may assume that 0 ∈ 푈, 푓 (0) = 0, and 퐽푓 = 0 when 푧푛 = 0. We wish to show that all the derivatives of 푓 in the 푧1, . . . , 푧푛−1 variables vanish whenever 푧푛 = 0. This would clearly contradict 푓 being one-to-one, as 푓 (푧1, . . . , 푧푛−1, 0) would be constant. So for any point on 푧푛 = 0 we consider one of the components of 푓 and one of the derivatives of that component. Without loss of generality, suppose the point is 0, and for contradiction suppose 휕 푓1 휕푧 (0) ≠ 0. The map 1 퐺(푧1, . . . , 푧푛) = 푓1(푧), 푧2, . . . , 푧푛 is biholomorphic on a small neighborhood of the origin. The function 푓 ◦ 퐺−1 is holomorphic and one-to-one on a small neighborhood. By the deﬁnition of 퐺, −1 푓 ◦ 퐺 (푤1, . . . , 푤푛) = 푤1, ℎ(푤) , where ℎ is a holomorphic mapping taking a neighborhood of the origin in ℂ푛 to ℂ푛−1. The mapping

휑(푤2, . . . , 푤푛) = ℎ(0, 푤2, . . . , 푤푛) is a one-to-one holomorphic mapping of a neighborhood of the origin in ℂ푛−1 to ℂ푛−1. By the induction hypothesis, the Jacobian determinant of 휑 is nowhere zero. If we diﬀerentiate 푓 ◦ 퐺−1, we notice 퐷( 푓 ◦ 퐺−1) = 퐷 푓 ◦ 퐷(퐺−1). So at the origin det 퐷( 푓 ◦ 퐺−1) = det 퐷 푓 det 퐷(퐺−1) = 0. We obtain a contradiction, as at the origin det 퐷( 푓 ◦ 퐺−1) = det 퐷휑 ≠ 0. The theorem is no longer true if the domain and range dimensions of the mapping are not equal.

푋 푧 ∈ 2 푧2 푧3 Exercise 1.6.3: Take the subvariety = ℂ : 1 = 2 . Find a one-to-one holomorphic mapping 푓 : ℂ → 푋. Then note that the derivative of 푓 vanishes at a certain point. So

Theorem 1.6.5 has no analogue when the domain and range have diﬀerent dimension.

Exercise 1.6.4: Find a continuous function 푓 : ℝ → ℝ2 that is one-to-one but such that the inverse 푓 −1 : 푓 (ℝ) → ℝ is not continuous. 1.6. RIEMANN EXTENSION THEOREM, ZERO SETS, AND INJECTIVE MAPS 39

This is an appropriate place to state a well-known and as yet unsolved conjecture (and most likely ridiculously hard to solve): the Jacobian conjecture. This conjecture is a converse to the theorem above in a special case: Suppose 퐹 : ℂ푛 → ℂ푛 is a polynomial map (each component is a polynomial) and the Jacobian derivative 퐽퐹 is never zero, then 퐹 is invertible with a polynomial inverse. Clearly 퐹 would be locally one-to-one, but proving (or disproving) the existence of a global polynomial inverse is the content of the conjecture.

Exercise 1.6.5: Prove the Jacobian conjecture for 푛 = 1. That is, prove that if 퐹 : ℂ → ℂ is a polynomial such that 퐹0 is never zero, then 퐹 has an inverse, which is a polynomial.

푛 푛 Exercise 1.6.6: Let 퐹 : ℂ → ℂ be an injective polynomial map. Prove 퐽퐹 is a nonzero constant. Exercise 1.6.7: Prove that the Jacobian conjecture is false if “polynomial” is replaced with “entire holomorphic,” even for 푛 = 1.

Exercise 1.6.8: Prove that if a holomorphic 푓 : ℂ → ℂ is injective, then it is onto, and therefore 푓 (푧) = 푎푧 + 푏 for 푎 ≠ 0.

We remark that while every injective holomorphic map of 푓 : ℂ → ℂ is onto, the same is not true in higher dimensions. In ℂ푛, 푛 ≥ 2, there exist so-called Fatou–Bieberbach domains, that is, proper subsets of ℂ푛 that are biholomorphic to ℂ푛. Chapter 2

Convexity and pseudoconvexity

2.1 Domains of holomorphy and holomorphic extensions

It turns out that not every domain in ℂ푛 is a natural domain for holomorphic functions.

푛 ∗ Deﬁnition 2.1.1. Let 푈 ⊂ ℂ be a domain (connected open set). The set 푈 is a domain of holomorphy if there do not exist nonempty open sets 푉 and 푊, with 푉 ⊂ 푈 ∩ 푊, 푊 ⊄ 푈, and 푊 connected, such that for every 푓 ∈ O(푈) there exists an 퐹 ∈ O(푊) with 푓 (푧) = 퐹(푧) for all 푧 ∈ 푉.

푈 푉 푊

The idea is that if a domain 푈 is not a domain of holomorphy and 푉, 푊 exist as in the deﬁnition, then 푓 “extends across the boundary” somewhere.

푛 Example 2.1.2: The unit ball 픹푛 ⊂ ℂ is a domain of holomorphy. Proof: Consider 푈 = 픹푛, and suppose 푉, 푊 as in the deﬁnition exist. As 푊 is connected and open, it is path connected. There exist points in 푊 that are not in 픹푛, so there is a path 훾 in 푊 that goes from a point 푞 ∈ 푉 to some 푝 ∈ 휕픹푛 ∩ 푊. Without loss of generality (after composing with rotations, that is, unitary matrices), assume 푝 = (1, 0, 0,..., 0). Take the function 푓 (푧) = 1 . The function 퐹 must agree with 푓 on 1−푧1 the component of 픹푛 ∩ 푊 that contains 푞. But that component also contains 푝 and so 퐹 must blow up (in particular it cannot be holomorphic) at 푝. The contradiction shows that no 푉 and 푊 exist. In one dimension this notion has no real content: Every domain is a domain of holomorphy.

Exercise 2.1.1 (Easy): In ℂ, every domain is a domain of holomorphy.

푛 Exercise 2.1.2: If 푈푗 ⊂ ℂ are domains of holomorphy (possibly an inﬁnite set of domains), then Ñ 푈 the interior of 푗 푗 is either empty or every connected component is a domain of holomorphy. ∗Domain of holomorphy can make sense for disconnected sets (not domains), and some authors do deﬁne it so. 2.1. DOMAINS OF HOLOMORPHY AND HOLOMORPHIC EXTENSIONS 41

Exercise 2.1.3 (Easy): Show that a polydisc in ℂ푛 is a domain of holomorphy.

Exercise 2.1.4: ∞ a) Given 푝 ∈ 휕픹푛, ﬁnd a function 푓 holomorphic on 픹푛, 퐶 -smooth on 픹푛 (all real partial 푝 derivatives of all orders extend continuously to 픹푛), that√ does not extend past √as a holomorphic function. Hint: For the principal branch of · the function 휉 ↦→ 푒−1/ 휉 is holomorphic for Re 휉 > 0 and extends to be continuous (even smooth) on all of Re 휉 ≥ 0. b) Find a function 푓 holomorphic on 픹푛 that does not extend past any point of 휕픹푛.

Various notions of convexity will play a big role later on. A set 푆 is geometrically convex if 푡푥 + (1 − 푡)푦 ∈ 푆 for all 푥, 푦 ∈ 푆 and 푡 ∈ [0, 1]. The exercise below says that every geometrically convex domain is a domain of holomorphy. Domains of holomorphy are often not geometrically convex (e.g. any domain in ℂ is a domain of holomorphy), so classical convexity is not the correct notion, but it is in the right direction.

Exercise 2.1.5: Show that a geometrically convex domain in ℂ푛 is a domain of holomorphy.

In the following when we say 푓 ∈ O(푈) extends holomorphically to 푉 where 푈 ⊂ 푉, we mean that there exists a function 퐹 ∈ O(푉) such that 푓 = 퐹 on 푈. Remark 2.1.3. The subtlety of the definition of a domain of holomorphy is that it does not necessarily talk about functions extending to a larger set, since we must take into account single-valuedness. For instance, let 푓 be the principal branch of the logarithm defined on the slit plane 푈 = ℂ \{푧 ∈ ℂ : Im 푧 = 0, Re 푧 ≤ 0}. We can locally define an extension from one side through the boundary of the domain, but we cannot define an extension on a open set that contains 푈. This example should be motivation for why we let 푉 be a proper subset of 푈 ∩ 푊, and why 푊 need not include all of 푈. In dimension two or more, not every domain is a domain of holomorphy. We have the following theorem. The domain 퐻 in the theorem is called the Hartogs figure. 푚 푘 Theorem 2.1.4. Let (푧, 푤) = (푧1, . . . , 푧푚, 푤1, . . . , 푤푘 ) ∈ ℂ × ℂ be the coordinates. For two numbers 0 < 푎, 푏 < 1, define the set 퐻 ⊂ 픻푚+푘 by 푚+푘 푚+푘 퐻 = (푧, 푤) ∈ 픻 : |푧푗 | > 푎 for 푗 = 1, . . . , 푚 ∪ (푧, 푤) ∈ 픻 : |푤푗 | < 푏 for 푗 = 1, . . . , 푘 . If 푓 ∈ O(퐻), then 푓 extends holomorphically to 픻푚+푘 . In ℂ2 if 푚 = 1 and 푘 = 1, the figure looks like (the 푐 will come up in the proof): |푤| In diagrams, the Hartogs figure is 1 often drawn as:

푏 퐻

|푧| 푎 푐 1 42 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Proof. Pick a 푐 ∈ (푎, 1). Let

푚 Γ = 푧 ∈ 픻 : |푧푗 | = 푐 for 푗 = 1, . . . , 푚 .

That is, Γ is the distinguished boundary of 푐픻푚, a polydisc centered at 0 of radius 푐 in ℂ푚. Define the function ∫ 푓 (휉, 푤) 퐹 푧, 푤 1 푑휉. ( ) = 푚 (2휋푖) Γ 휉 − 푧 Clearly, 퐹 is well-defined on 푐픻푚 × 픻푘 as 휉 only ranges through Γ and so as long as 푤 ∈ 픻푘 then (휉, 푤) ∈ 퐻. The function 퐹 is holomorphic in 푤 as we can differentiate underneath the integral and 푓 is 푤 퐻 퐹 푧 1 푧 holomorphic in on . Furthermore, is holomorphic in as the kernel 휉−푧 is holomorphic in as long as 푧 ∈ 푐픻푚. For any fixed 푤 with |푤푗 | < 푏 for all 푗, the Cauchy integral formula says 퐹(푧, 푤) = 푓 (푧, 푤) for all 푧 ∈ 푐픻푚. Hence, 퐹 = 푓 on the open set 푐픻푚 × 푏픻푘 , and so they are equal on (푐픻푚 × 픻푘 ) ∩ 퐻. Combining 퐹 and 푓 we obtain a holomorphic function on 픻푚+푘 that extends 푓 .

The theorem is used in many situations to extend holomorphic functions. We usually need to translate, scale, rotate (apply a unitary matrix), and even take more general biholomorphic mappings of 퐻, to place it wherever we need it. The corresponding polydisc—or the image of the polydisc under the appropriate biholomorphic mapping if one was used—to which all holomorphic functions on 퐻 extend is denoted by 퐻b and is called the hull of 퐻. Let us state a simple but useful case of the so-called Hartogs phenomenon.

Corollary 2.1.5. Let 푈 ⊂ ℂ푛, 푛 ≥ 2, be an open set and 푝 ∈ 푈. Then every 푓 ∈ O(푈 \{푝}) extends holomorphically to 푈.

Proof. Without loss of generality, by translating and scaling (those operations are after all holomor- 푝 ,..., , 3 푛 푈 phic), we assume that = 0 0 4 and the unit polydisc 픻 is contained in . We ﬁt a Hartogs 퐻 푈 푚 푛 푘 푛 푛−1 1 푎 푏 1 ﬁgure in by letting = − 1 and = 1, writing ℂ = ℂ × ℂ , and taking = = 2 . Then 푛 퐻 ⊂ 푈, and 푝 ∈ 픻 \ 퐻. Theorem 2.1.4 says that 푓 extends to be holomorphic at 푝.

This result provides another reason why holomorphic functions in several variables have no isolated zeros (or poles). Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, and 푓 ∈ O(푈) with 푓 being zero only at 푝, that 푓 −1 푝 1 푈 푝 푓 is (0) = { }. Then 푓 would be holomorphic in \{ }. It would not be possible to extend through 푝 (not even continuously let alone holomorphically), and we obtain a contradiction. The extension works in an even more surprising fashion. We could take out a very large set, for example, any geometrically convex subset:

Exercise 2.1.6: Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, be an open set and 퐾 ⊂⊂ 푈 is a compact geometrically convex subset. If 푓 ∈ O(푈 \ 퐾), then 푓 extends to be holomorphic in 푈. Hint: Find a nice point on 휕퐾 and try extending a little bit. Then make sure your extension is single-valued. 2.1. DOMAINS OF HOLOMORPHY AND HOLOMORPHIC EXTENSIONS 43

Convexity of 퐾 is not needed; we only need that 푈 \ 퐾 is connected, however, the proof is much harder. The single-valuedness of the extension is the key point that makes the general proof harder. Notice the surprising fact that every holomorphic function on the shell 푛 픹푛 \ 퐵1−휖 (0) = 푧 ∈ ℂ : 1 − 휖 < k푧k < 1 for any 휖 > 0 automatically extends to a holomorphic function of 픹푛. We need 푛 > 1. The extension result decisively does not work in one dimension; for example take 1/푧. There is a related fact about zero sets. If 푛 ≥ 2 and 푓 ∈ O(픹푛) the set of its zeros must “touch the boundary” (in other words −1 푓 (0) is not compact) or be empty: If the set of zeros were compact in 픹푛, then we could try to extend the function 1/푓 .

Exercise 2.1.7 (Hartogs triangle): Let 2 푇 = (푧1, 푧2) ∈ 픻 : |푧2| < |푧1| . Show that 푇 is a domain of holomorphy. Then show that if

푇e = 푇 ∪ 퐵휖 (0) for an arbitrarily small 휖 > 0, then 푇e is not a domain of holomorphy. In fact, every function holomorphic on 푇e extends to a holomorphic function of 픻2. Exercise 2.1.8: Take the natural embedding of ℝ2 ⊂ ℂ2. Suppose 푓 ∈ O(ℂ2 \ ℝ2). Show that 푓 extends to be holomorphic in all of ℂ2. Hint: Change coordinates before using Hartogs. Exercise 2.1.9: Suppose 2 푈 = (푧, 푤) ∈ 픻 : 1/2 < |푧| . Draw 푈. Let 훾 = 푧 ∈ ℂ : |푧| = 3/4 oriented positively. If 푓 ∈ O(푈), then show that the function 1 ∫ 푓 (휉, 푤) 퐹(푧, 푤) = 푑휉 2휋푖 훾 휉 − 푧

is well-deﬁned in (3/4)픻 × 픻, holomorphic where deﬁned, yet it is not necessarily true that 퐹 = 푓 on the intersections of their domains. Exercise 2.1.10: Suppose 푈 ⊂ ℂ푛 is an open set such that for every 푧 ∈ ℂ푛 \{0}, there is a 휆 ∈ ℂ such that 휆푧 ∈ 푈. Let 푓 : 푈 → ℂ be holomorphic with 푓 (휆푧) = 푓 (푧) whenever 푧 ∈ 푈, 휆 ∈ ℂ and 휆푧 ∈ 푈. a) (easy) Prove that 푓 is constant. b) (hard) Relax the requirement on 푓 to being meromorphic: 푓 = 푔/ℎ for holomorphic 푔 and ℎ. Find a nonconstant example, and prove that such an 푓 must be rational (that is, 푔 and ℎ must be polynomials). Exercise 2.1.11: Suppose 3 푈 = 푧 ∈ 픻 : 1/2 < |푧1| or 1/2 < |푧2| . 3 Prove that every function 푓 ∈ O(푈) extends to 픻 . Compare to Exercise 2.1.9 . 푛 푛 Exercise 2.1.12: Suppose 푈 = ℂ \{푧 ∈ ℂ : 푧1 = 푧2 = 0}, 푛 ≥ 2. Show that every 푓 ∈ O(푈) extends holomorphically to ℂ푛. 44 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

2 2 Example 2.1.6: By Exercise 2.1.8 , 푈1 = ℂ \ ℝ is not a domain of holomorphy. On the other hand, 푈 = ℂ2 \{푧 ∈ ℂ2 : 푧 = 0} is a domain of holomorphy; simply use 푓 (푧) = 1 as the 2 2 푧2 function that cannot extend. Therefore, 푈1 and 푈2 are rather diﬀerent as far as complex variables are concerned, yet they are the same set if we ignore the complex structure. They are both simply a 4-dimensional real vector space minus a 2-dimensional real vector subspace. That is, 푈1 is the set where either Im 푧1 ≠ 0 or Im 푧2 ≠ 0, while 푈2 is the set where either Re 푧2 ≠ 0 or Im 푧2 ≠ 0. The condition of being a domain of holomorphy, requires something more than just some real geometric condition on the set. Namely, we have shown that the image of a domain of holomorphy via an orthonormal real-linear mapping (so preserving distances, angles, straight lines, etc.) need not be a domain of holomorphy. Therefore, when we want to “rotate” in complex analysis we need to use a complex linear mapping, so a unitary matrix.

2.2 Tangent vectors, the Hessian, and convexity

An exercise in the previous section showed that every convex domain is a domain of holomorphy. However, classical convexity is too strong.

Exercise 2.2.1: Show that if 푈 ⊂ ℂ푚 and 푉 ⊂ ℂ푘 are both domains of holomorphy, then 푈 × 푉 is a domain of holomorphy.

In particular, the exercise says that for any domains 푈 ⊂ ℂ and 푉 ⊂ ℂ, the set 푈 ×푉 is a domain of holomorphy in ℂ2. The domains 푈 and 푉, and hence 푈 ×푉, can be spectacularly nonconvex. But we should not discard convexity completely, there is a notion of pseudoconvexity, which vaguely means “convexity in the complex directions” and is the correct notion to distinguish domains of holomorphy. Let us ﬁgure out what classical convexity means locally for a smooth boundary.

Definition 2.2.1. A set 푀 ⊂ ℝ푛 is a 퐶푘 -smooth hypersurface if at each point 푝 ∈ 푀, there exists a 푘-times continuously differentiable function 푟 : 푉 → ℝ with nonvanishing derivative, defined in a neighborhood 푉 of 푝 such that 푀 ∩ 푉 = 푥 ∈ 푉 : 푟(푥) = 0 . The function 푟 is called the defining function of 푀 (at 푝). An open set (or domain) 푈 ⊂ ℝ푛 with 퐶푘 -smooth boundary is a set where 휕푈 is a 퐶푘 -smooth hypersurface, and at every 푝 ∈ 휕푈 there is a defining function 푟 such that 푟 < 0 for points in 푈 and 푟 > 0 for points not in 푈. If we say simply smooth, we mean 퐶∞-smooth, that is, the 푟 above is infinitely differentiable.

푈 푟 < 0 푟 = 0 푝 푉 푟 > 0 2.2. TANGENT VECTORS, THE HESSIAN, AND CONVEXITY 45

What we really defined is an embedded hypersurface. In particular, in this book the topology on the set 푀 will be the subset topology. Furthermore, in this book we generally deal with smooth (that is, 퐶∞) functions and hypersurfaces. Dealing with 퐶푘 -smooth functions for finite 푘 introduces technicalities that make certain theorems and arguments unnecessarily difficult. As the derivative of 푟 is nonvanishing, a hypersurface 푀 is locally the graph of one variable over the rest using the implicit function theorem. That is, 푀 is a smooth hypersurface if it is locally a set defined by 푥푗 = 휑(푥1, . . . , 푥푗−1, 푥푗+1, . . . , 푥푛) for some 푗 and some smooth function 휑. The definition of an open set with smooth boundary is not just that the boundary is a smooth hypersurface, that is not enough. It also says that one side of that hypersurface is in 푈 and one side is not in 푈. That is because if the derivative of 푟 never vanishes, then 푟 must have different signs on different sides of 푥 ∈ 푉 : 푟(푥) = 0 . The verification of this fact is left to the reader (Hint: Look at where the gradient points to). 푛 푛 2푛 Same definition works for ℂ , where we treat ℂ as ℝ . For example, the ball 픹푛 is a domain with smooth boundary with defining function 푟(푧, 푧¯) = k푧k2 − 1. We can, in fact, find a single global defining function for every open set with smooth boundary, but we have no need of this. In ℂ푛 a hypersurface defined as above is a real hypersurface, to distinguish it from a complex hypersurface that would be the zero set of a holomorphic function, although we may at times leave out the word “real” if it is clear from context. 푛 푛 Definition 2.2.2. For a point 푝 ∈ ℝ , the set of tangent vectors 푇푝ℝ is given by 휕 휕 푇 푛 ,..., . 푝ℝ = spanℝ 휕푥1 푝 휕푥푛 푝 푛 That is, a vector 푋푝 ∈ 푇푝ℝ is an object of the form 푛 ∑︁ 휕 푋푝 = 푎푗 , 휕푥 푝 푗=1 푗

for real numbers 푎푗 . For computations, 푋푝 could be represented by an 푛-vector 푎 = (푎1, . . . , 푎푛). However, if 푝 ≠ 푞, then 푇 ℝ푛 and 푇 ℝ푛 are distinct spaces. An object 휕 is a linear functional 푝 푞 휕푥푗 푝 on the space of smooth functions: When applied to a smooth function 푔, it gives 휕푔 . Therefore, 휕푥푗 푝 푋푝 is also such a functional. It is the directional derivative from calculus; it is computed as 푋푝 푓 = ∇ 푓 |푝 ·(푎1, . . . , 푎푛). Deﬁnition 2.2.3. Let 푀 ⊂ ℝ푛 be a smooth hypersurface, 푝 ∈ 푀, and 푟 is a deﬁning function at 푝, 푛 then a vector 푋푝 ∈ 푇푝ℝ is tangent to 푀 at 푝 if 푛 ∑︁ 휕푟 푋푝푟 = 0, or in other words 푎푗 = 0. 휕푥 푝 푗=1 푗

The space of tangent vectors to 푀 is denoted by 푇푝 푀, and is called the tangent space to 푀 at 푝.

The space 푇푝 푀 is an (푛 −1)-dimensional real vector space—it is a subspace of an 푛-dimensional 푛 푇푝ℝ given by a single linear equation. Recall from calculus that the gradient ∇푟|푝 is “normal” to 푀 at 푝, and the tangent space is given by all the 푛-vectors 푎 that are orthogonal to the normal, that is ∇푟|푝 · 푎 = 0. 46 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

We cheated in the terminology, and assumed without justiﬁcation that 푇푝 푀 depends only on 푀, not on 푟. Fortunately, the deﬁnition of 푇푝 푀 is independent of the choice of 푟 by the next two exercises.

Exercise 2.2.2: Suppose 푀 ⊂ ℝ푛 is a smooth hypersurface and 푟 is a smooth defining function for 푀 at 푝. a) Suppose 휑 is another smooth defining function of 푀 on a neighborhood of 푝. Show that there exists a smooth nonvanishing function 푔 such that 휑 = 푔푟 (in a neighborhood of 푝). b) Now suppose 휑 is an arbitrary smooth function that vanishes on 푀 (not necessarily a defining function). Again show that 휑 = 푔푟, but now 푔 may possibly vanish. Hint: First suppose 푟 = 푥푛, then find a 푔 such that 휑 = 푥푛푔. Then find a local change of variables to make 푀 into the set given by 푥푛 = 0. A useful one variable calculus fact: If 푓 (0) = 0 and 푓 is 푠 ∫ 1 푓 0(푡푠) 푑푡 = 푓 (푠) ∫ 1 푓 0(푡푠) 푑푡 푠 smooth, then 0 , and 0 is a smooth function of .

Exercise 2.2.3: Show that 푇푝 푀 is independent of which deﬁning function we take. That is, prove that if 푟 and 푟˜ are deﬁning functions for 푀 at 푝, then Í 푎 휕푟 = 0 if and only if Í 푎 휕푟˜ = 0. 푗 푗 휕푥푗 푝 푗 푗 휕푥푗 푝

The tangent space 푇푝 푀 is the set of derivatives along 푀 at 푝. If 푟 is a deﬁning function of 푀,

and 푓 and ℎ are two smooth functions such that 푓 = ℎ on 푀, then Exercise 2.2.2 says that

푓 − ℎ = 푔푟, or 푓 = ℎ + 푔푟,

for some smooth 푔. Applying 푋푝 we ﬁnd

푋푝 푓 = 푋푝 ℎ + 푋푝 (푔푟) = 푋푝 ℎ + (푋푝푔)푟 + 푔(푋푝푟) = 푋푝 ℎ + (푋푝푔)푟.

So 푋푝 푓 = 푋푝 ℎ on 푀 (where 푟 = 0). In other words, 푋푝 푓 only depends on the values of 푓 on 푀.

푛 Example 2.2.4: If 푀 ⊂ ℝ is given by 푥푛 = 0, then 푇푝 푀 is given by derivatives of the form

푛−1 ∑︁ 휕 푋푝 = 푎푗 . 휕푥 푝 푗=1 푗

That is, derivatives along the ﬁrst 푛 − 1 variables only.

Deﬁnition 2.2.5. The disjoint union

푛 Ø 푛 푇ℝ = 푇푝ℝ 푝∈ℝ푛 is called the tangent bundle. There is a natural identiﬁcation ℝ푛 × ℝ푛 푇ℝ푛: 푛 휕 푛 푛 ∑︁ 푛 (푝, 푎) ∈ ℝ × ℝ ↦→ 푎푗 ∈ 푇ℝ . 휕푥 푝 푗=1 푗 2.2. TANGENT VECTORS, THE HESSIAN, AND CONVEXITY 47

The topology and smooth structure on 푇ℝ푛 comes from this identification. The wording “bundle” (a 푛 푛 −1 푛 bundle of fibers) comes from the natural projection 휋 : 푇ℝ → ℝ , where fibers are 휋 (푝) = 푇푝ℝ . A smooth vector field in 푇ℝ푛 is an object of the form 푛 ∑︁ 휕 푋 = 푎 , 푗 휕푥 푗=1 푗 푛 푛 where 푎푗 are smooth functions. That is, 푋 is a smooth function 푋 : 푉 ⊂ ℝ → 푇ℝ such that 푛 푛 푋(푝) ∈ 푇푝ℝ . Usually we write 푋푝 rather than 푋(푝). To be more fancy, say 푋 is a section of 푇ℝ . Similarly, the tangent bundle of 푀 is Ø 푇 푀 = 푇푝 푀. 푝∈푀

A vector field 푋 in 푇 푀 is a vector field such that 푋푝 ∈ 푇푝 푀 for all 푝 ∈ 푀. Before we move on, let us note how smooth maps transform tangent spaces. Given a smooth mapping 푓 : 푈 ⊂ ℝ푛 → ℝ푚, the derivative at 푝 is a linear mapping of the tangent spaces: 푛 푚 푛 푚 퐷 푓 (푝) : 푇푝ℝ → 푇푓 (푝)ℝ . If 푋푝 ∈ 푇푝ℝ , then 퐷 푓 (푝)푋푝 should be in 푇푓 (푝)ℝ . The vector 푚 퐷 푓 (푝)푋푝 is defined by how it acts on smooth functions 휑 of a neighborhood of 푓 (푝) in ℝ :

퐷 푓 (푝)푋푝 휑 = 푋푝 (휑 ◦ 푓 ). It is the only reasonable way to put those three objects together. When the spaces are ℂ푛 and ℂ푚, we denote this derivative as 퐷ℝ 푓 to distinguish it from the holomorphic derivative. As far as calculus computations are concerned, the linear mapping 퐷 푓 (푝) is the Jacobian matrix acting on vectors in the standard basis of the tangent space as given above. This is why we use the same notation for the Jacobian matrix and the derivative acting on tangent spaces. To verify this claim, it is enough to see where the basis element 휕 goes, and the form of 퐷 푓 (푝) as a matrix follows by the chain 휕푥푗 푝 푓 푥 , 푥 푥 푥 푥2, 푥 푥 푥 푥 rule. For example, the derivative of the mapping ( 1 2) = ( 1 + 2 2 + 1 3 1 + 4 2 + 1 2) at the origin is given by the matrix 1 2 , and so the vector 푋 = 푎 휕 + 푏 휕 gets taken to 3 4 푝 휕푥1 0 휕푥2 0 퐷 푓 (0)푋 = (푎 + 2푏) 휕 + (3푎 + 4푏) 휕 , where we let (푦 , 푦 ) be the coordinates on the target. 0 휕푦1 0 휕푦2 0 1 2 You should check on some test function, such as 휑(푦1, 푦2) = 훼푦1 + 훽푦2, that the definition above is satisfied. Now that we know what tangent vectors are and how they transform, let us define convexity for domains with smooth boundary. Definition 2.2.6. Suppose 푈 ⊂ ℝ푛 is an open set with smooth boundary, and 푟 is a defining function for 휕푈 at 푝 ∈ 휕푈 such that 푟 < 0 on 푈. If 푛 푛 ∑︁ 휕2푟 ∑︁ 휕 푎푗 푎ℓ ≥ 0, for all 푋푝 = 푎푗 ∈ 푇푝휕푈, 휕푥 휕푥 푝 휕푥 푝 푗=1,ℓ=1 푗 ℓ 푗=1 푗

then 푈 is said to be convex at 푝. If the inequality above is strict for all nonzero 푋푝 ∈ 푇푝휕푈, then 푈 is said to be strongly convex at 푝. ∗ A domain 푈 is convex if it is convex at all 푝 ∈ 휕푈. If 푈 is bounded , we say 푈 is strongly convex if it is strongly convex at all 푝 ∈ 휕푈. ∗Matters are a little more complicated with the “strong” terminology if 푈 is unbounded. 48 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

The matrix 휕2푟

휕푥 휕푥 푗 ℓ 푝 푗ℓ is the Hessian of 푟 at 푝. So, 푈 is convex at 푝 ∈ 휕푈 if the Hessian of 푟 at 푝 as a bilinear form is positive semidefinite when restricted to 푇푝휕푈. In the language of vector calculus, let 퐻 be the Hessian of 푟 at 푝, and treat 푎 ∈ ℝ푛 as a column vector. Then 휕푈 is convex at 푝 whenever 푡 푛 푎 퐻푎 ≥ 0, for all 푎 ∈ ℝ such that ∇푟|푝 · 푎 = 0. This bilinear form given by the Hessian is the second fundamental form from Riemannian geometry in mild disguise (or perhaps it is the other way around). We cheated a little bit, since we have not proved that the notion of convexity is well-defined. In particular, there are many possible defining functions.

Exercise 2.2.4: Show that the definition of convexity is independent of the defining function. Hint: If 푟˜ is another defining function near 푝, then there is a smooth function 푔 > 0 such that 푟˜ = 푔푟.

Example 2.2.7: Let us prove that the unit disc in ℝ2 is convex (actually strongly convex). Let (푥, 푦) be the coordinates and let 푟(푥, 푦) = 푥2 + 푦2 − 1 be the deﬁning function. The tangent space of the circle is one-dimensional, so we simply need to ﬁnd a single nonzero tangent vector at each point. Consider the gradient ∇푟 = (2푥, 2푦) to check that 휕 휕 푋 = 푦 − 푥 휕푥 휕푦

is tangent to the circle, that is, 푋푟 = 푋(푥2 + 푦2 − 1) = (2푥, 2푦)·(푦, −푥) = 0 on the circle (by chance 푋푟 = 0 everywhere). The vector ﬁeld 푋 is nonzero on the circle, so at each point it gives a basis of the tangent space.

푟 > 0 ∇푟 푟 < 0 푋 푦 휕 푥 휕 = 휕푥 − 휕푦 The Hessian matrix of 푟 is " 휕2푟 휕2푟 # 2 2 0 휕푥 휕푥휕푦 = . 휕2푟 휕2푟 0 2 휕푦휕푥 휕푦2 Applying the vector (푦, −푥) gets us 2 0 푦 푦 −푥 = 2푦2 + 2푥2 = 2 > 0. 0 2 −푥 So the domain given by 푟 < 0 is strongly convex at all points. 2 푟 휕 푟 휕 In general, to construct a tangent vector ﬁeld for a curve in ℝ , consider 푦 휕푥 − 푥 휕푦 . In higher dimensions we can just run through enough pairs of variables to get a basis of 푇 푀. 2.2. TANGENT VECTORS, THE HESSIAN, AND CONVEXITY 49

Exercise 2.2.5: Show that if an open set with smooth boundary is strongly convex at a point, then it is strongly convex at all nearby points. On the other hand ﬁnd an example of an open set with smooth boundary that is convex at one point 푝, but not convex at points arbitrarily near 푝.

Exercise 2.2.6: Show that the domain in ℝ2 deﬁned by 푥4 + 푦4 < 1 is convex, but not strongly convex. Find all the points where the domain is not strongly convex.

3 푥2 푥2 2 < 푥 Exercise 2.2.7: Show that the domain in ℝ deﬁned by ( 1 + 2) 3 is strongly convex at all points except the origin, where it is just convex (but not strongly).

∗ In the following, we use the big-oh notation, although we use a perhaps less standard shorthand . A smooth function is 푂(ℓ) at a point 푝 (usually the origin), if all its derivatives of order 0, 1, . . . , ℓ−1 vanish at 푝. For example, if 푓 is 푂(3) at the origin, then 푓 (0) = 0, and its ﬁrst and second derivatives vanish at the origin. For computations it is often useful to use a more convenient deﬁning function, that is, it is convenient to write 푀 as a graph. Lemma 2.2.8. Suppose 푀 ⊂ ℝ푛 is a smooth hypersurface, and 푝 ∈ 푀. Then after a rotation and translation, 푝 is the origin, and near the origin 푀 is given by

푦 = 휑(푥),

where (푥, 푦) ∈ ℝ푛−1 × ℝ are our coordinates and 휑 is a smooth function that is 푂(2) at the origin, namely, 휑(0) = 0 and 푑휑(0) = 0. If 푀 is the boundary of an open set 푈 with smooth boundary and 푟 < 0 on 푈, then the rotation can be chosen such that 푦 > 휑(푥) for points in 푈. 푦

푦 > 휑(푥) 푈 푦 = 휑(푥)

푥

Proof. Let 푟 be a deﬁning function at 푝. Take 푣 = ∇푟|푝. By translating 푝 to zero, and applying a rotation (an orthogonal matrix), we assume 푣 = (0, 0,..., 0, 푣푛), where 푣푛 < 0. Denote our 푥, 푦 푛−1 푟 푣 휕푟 coordinates by ( ) ∈ ℝ × ℝ. As ∇ |0 = , then 휕푦 (0) ≠ 0. We apply the implicit function theorem to ﬁnd a smooth function 휑 such that 푟 푥, 휑(푥) = 0 for all 푥 in a neighborhood of the origin, and (푥, 푦) : 푦 = 휑(푥) are all the solutions to 푟 = 0 near the origin.

∗The standard notation for 푂(ℓ) is 푂(k푥kℓ ) and is usually deﬁned to mean that 푓 (푥) is bounded as 푥 → 푝. k푥 kℓ 50 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

What is left is to show that the derivative at 0 of 휑 vanishes. As 푟 푥, 휑(푥) = 0 for all 푥 in a neighborhood of the origin, we differentiate. For every 푗 = 1, . . . , 푛 − 1, 푛−1 ! 휕 h i ∑︁ 휕푟 휕푥 휕푟 휕휑 휕푟 휕푟 휕휑 = 푟 푥, 휑(푥) = ℓ + = + . 0 휕푥 휕푥 휕푥 휕푦 휕푥 휕푥 휕푦 휕푥 푗 ℓ=1 ℓ 푗 푗 푗 푗 At the origin, 휕푟 (0, 0) = 0 and 휕푟 (0, 0) = 푣 ≠ 0, and therefore 휕휑 (0) = 0. 휕푥푗 휕푦 푛 휕푥푗 To prove the final statement, note that 푟 < 0 on 푈. It is enough to check that 푟 is negative for , 푦 푦 > 휕푟 , 푣 < (0 ) if 0 is small, which follows as 휕푦 (0 0) = 푛 0. The advantage of this representation is that the tangent space at 푝 can be identified with the 푥 coordinates for the purposes of computation. Considering 푥 as a column vector, the Taylor expansion of a smooth function 휑 at the origin is 1 휑(푥) = 휑(0) + ∇휑| · 푥 + 푥푡 퐻푥 + 퐸 (푥), 0 2 h 휕2휑 i where 퐻 = is the Hessian matrix of 휑 at the origin, and 퐸 is 푂(3), namely, 퐸 (0) = 0, 휕푥푗 휕푥푘 0 푗 푘 and all first and second order derivatives of 퐸 vanish at 0. For the situation of the lemma above, the 휑 is 푂(2) at the origin, i.e. 휑(0) = 0 and ∇휑|0 = 0. So we write the hypersurface 푀 as 1 푦 = 푥푡 퐻푥 + 퐸 (푥). 2 푀 휕푈 푦 > 1 푥푡 퐻푥 퐸 푥 If is the boundary of an open set, then we pick the rotation so that 2 + ( ) on 푈. It is an easy exercise to show that 푈 is convex at 푝 if 퐻 positive semidefinite, and 푈 is strongly convex at 푝 if 퐻 is positive definite.

Exercise 2.2.8: Prove the statement above about 퐻 and convexity at 푝.

Exercise 2.2.9: Let 푟 be a deﬁning function at 푝 for a smooth hypersurface 푀 ⊂ ℝ푛. We say 푀 is convex from both sides at 푝 if both the set given by 푟 > 0 and the set given by 푟 < 0 are convex at 푝. Prove that if a hypersurface 푀 ⊂ ℝ푛 is convex from both sides at all points, then it is locally just a hyperplane (the zero set of a real aﬃne function).

Recall that 푈 is geometrically convex if for every 푝, 푞 ∈ 푈 the line between 푝 and 푞 is in 푈, or in other words 푡 푝 + (1 − 푡)푞 ∈ 푈 for all 푡 ∈ [0, 1]. In particular, geometric convexity is a global condition. You need to know all of 푈. On the other hand, the notion of convex for a smooth boundary is local in that you only need to know 휕푈 in a small neighborhood. For domains with smooth boundaries the two notions are equivalent. Proving one direction is easy.

Exercise 2.2.10: Suppose a domain 푈 ⊂ ℝ푛 with smooth boundary is geometrically convex. Show that it is convex.

The other direction is considerably more complicated, and we will not worry about it here. Proving a global condition from a local one is often trickier, but also often more interesting. Similar diﬃculties will be present once we move back to several complex variables and try to relate pseudoconvexity with domains of holomorphy. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 51 2.3 Holomorphic vectors, the Levi form, and pseudoconvexity

푛 2푛 푛 2푛 As ℂ is identified with ℝ using 푧 = 푥 + 푖푦, we have 푇푝ℂ = 푇푝ℝ . If we take the complex span instead of the real span we get the complexified tangent space 휕 휕 휕 휕 푇 푛 , ,..., , . ℂ ⊗ 푝ℂ = spanℂ 휕푥1 푝 휕푦1 푝 휕푥푛 푝 휕푦푛 푝 푛 We simply replace all the real coefficients with complex ones. The space ℂ ⊗ 푇푝ℂ is a 2푛- dimensional complex vector space. Both 휕 and 휕 are in ℂ ⊗ 푇 ℂ푛, and in fact: 휕푧푗 푝 휕푧¯푗 푝 푝 휕 휕 휕 휕 푇 푛 , ,..., , . ℂ ⊗ 푝ℂ = spanℂ 휕푧1 푝 휕푧¯1 푝 휕푧푛 푝 휕푧¯푛 푝 Define 휕 휕 휕 휕 푇 (1,0) 푛 def ,..., 푇 (0,1) 푛 def ,..., . 푝 ℂ = spanℂ and 푝 ℂ = spanℂ 휕푧1 푝 휕푧푛 푝 휕푧¯1 푝 휕푧¯푛 푝

(1,0) 푛 (0,1) 푛 The vectors in 푇푝 ℂ are the holomorphic vectors and vectors in 푇푝 ℂ are the antiholomorphic vectors. We decompose the full tangent space as the direct sum

푛 (1,0) 푛 (0,1) 푛 ℂ ⊗ 푇푝ℂ = 푇푝 ℂ ⊕ 푇푝 ℂ .

(0,1) 푛 A holomorphic function is one that vanishes on 푇푝 ℂ . Let us note what holomorphic functions do to these spaces. Given a smooth mapping 푓 from 푛 푚 푛 푛 푚 ℂ to ℂ , its derivative at 푝 ∈ ℂ is a real-linear mapping 퐷ℝ 푓 (푝) : 푇푝ℂ → 푇푓 (푝)ℂ . Given the basis above, this mapping is represented by the standard real Jacobian matrix, that is, a real 2푚 × 2푛 matrix that we wrote before as 퐷ℝ 푓 (푝). Proposition 2.3.1. Let 푓 : 푈 ⊂ ℂ푛 → ℂ푚 be a holomorphic mapping with 푝 ∈ 푈. Suppose 푛 푚 퐷ℝ 푓 (푝) : 푇푝ℂ → 푇푓 (푝)ℂ is the real derivative of 푓 at 푝. We naturally extend the derivative to 푛 푚 퐷ℂ 푓 (푝) : ℂ ⊗ 푇푝ℂ → ℂ ⊗ 푇푓 (푝)ℂ . Then 퐷 푓 (푝) 푇 (1,0) 푛 ⊂ 푇 (1,0) 푚 퐷 푓 (푝) 푇 (0,1) 푛 ⊂ 푇 (0,1) 푚. ℂ 푝 ℂ 푓 (푝) ℂ and ℂ 푝 ℂ 푓 (푝) ℂ

(1,0) 푛 If 푓 is a biholomorphism, then 퐷ℂ 푓 (푝) restricted to 푇푝 ℂ is a vector space isomorphism. (0,1) 푛 Similarly for 푇푝 ℂ .

Exercise 2.3.1: Prove the proposition. Hint: Start with 퐷ℝ 푓 (푝) as a real 2푚 × 2푛 matrix to show it extends (it is the same matrix if you think of it as a matrix and use the same basis vectors). Think of ℂ푛 and ℂ푚 in terms of the 푧s and the 푧¯s and think of 푓 as a mapping

(푧, 푧¯) ↦→ 푓 (푧), 푓¯(푧¯).

Write the derivative as a matrix in terms of the 푧s and the 푧¯s and 푓 s and 푓¯s and the result will follow. That is just changing the basis. 52 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

When talking about only holomorphic functions and holomorphic vectors, when we say derivative of 푓 , we mean the holomorphic part of the derivative, which we write as 퐷 푓 (푝) 푇 (1,0) 푛 → 푇 (1,0) 푚. : 푝 ℂ 푓 (푝) ℂ

(1,0) 푛 That is, 퐷 푓 (푝) is the restriction of 퐷ℂ 푓 (푝) to 푇푝 ℂ . In other words, let 푧 be the coor- n o dinates on ℂ푛 and 푤 be coordinates on ℂ푚. Using the bases 휕 ,..., 휕 on ℂ푛 and 휕푧1 푝 휕푧푛 푝 n o 휕 ,..., 휕 on ℂ푚, the holomorphic derivative of 푓 : ℂ푛 → ℂ푚 is represented as 휕푤1 푓 (푝) 휕푤푚 푓 (푝) the 푚 × 푛 Jacobian matrix 휕 푓 푗 , 휕푧 푘 푝 푗 푘 which we have seen before and for which we also used the notation 퐷 푓 (푝). As before, define the tangent bundles ℂ ⊗ 푇ℂ푛, 푇 (1,0)ℂ푛, and 푇 (0,1)ℂ푛, by taking the disjoint unions. One can also define vector fields in these bundles. 푛 Let us describe ℂ ⊗푇푝 푀 for a smooth real hypersurface 푀 ⊂ ℂ . Let 푟 be a real-valued defining 푛 function of 푀 at 푝. A vector 푋푝 ∈ ℂ ⊗ 푇푝ℂ is in ℂ ⊗ 푇푝 푀 whenever 푋푝푟 = 0. That is, 푛 푛 ∑︁ 휕 휕 ∑︁ 휕푟 휕푟 푋푝 = 푎푗 + 푏푗 ∈ ℂ ⊗ 푇푝 푀 whenever 푎푗 + 푏푗 = 0. 휕푧 푝 휕푧¯ 푝 휕푧 푝 휕푧¯ 푝 푗=1 푗 푗 푗=1 푗 푗

Therefore, ℂ ⊗ 푇푝 푀 is a (2푛 − 1)-dimensional complex vector space. We decompose ℂ ⊗ 푇푝 푀 as

(1,0) (0,1) ℂ ⊗ 푇푝 푀 = 푇푝 푀 ⊕ 푇푝 푀 ⊕ 퐵푝, where (1,0) def (1,0) 푛 (0,1) def (0,1) 푛 푇푝 푀 = ℂ ⊗ 푇푝 푀 ∩ 푇푝 ℂ , and 푇푝 푀 = ℂ ⊗ 푇푝 푀 ∩ 푇푝 ℂ .

The 퐵푝 is just the “leftover” and must be included, otherwise the dimensions will not work out. Make sure that you understand what all the objects are. The space 푇푝 푀 is a real vector space; (1,0) (0,1) ℂ⊗푇푝 푀, 푇푝 푀, 푇푝 푀, and 퐵푝 are complex vector spaces. To see that these give vector bundles, we must first show that their dimensions do not vary from point to point. The easiest way to see this fact is to write down convenient local coordinates. First, let us see what a biholomorphic map does to the holomorphic and antiholomorphic vectors. A biholomorphic map 푓 is a diffeomorphism. And if a real hypersurface 푀 is defined by a function 푟 near 푝, then the image 푓 (푀) is also a real hypersurface is given by the defining function 푟 ◦ 푓 −1 near 푓 (푝). Proposition 2.3.2. Suppose 푀 ⊂ ℂ푛 is a smooth real hypersurface, 푝 ∈ 푀, and 푈 ⊂ ℂ푛 is an open set such that 푀 ⊂ 푈. Let 푓 : 푈 → ℂ푛 be a holomorphic mapping such that 퐷 푓 (푝) is invertible (a biholomorphism near 푝). Let 퐷ℂ 푓 (푝) be the complexified real derivative as before. Then 퐷 푓 (푝) 푇 (1,0) 푀 = 푇 (1,0) 푓 (푀), 퐷 푓 (푝) 푇 (0,1) 푀 = 푇 (0,1) 푓 (푀). ℂ 푝 푓 (푝) ℂ 푝 푓 (푝) That is, the spaces are isomorphic as complex vector spaces. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 53

The proposition is local, if 푈 is only a neighborhood of 푝, replace 푀 with 푀 ∩ 푈.

Proof. The proof is an application of Proposition 2.3.1 . As the map is a biholomorphism at 푝,

퐷 푓 (푝) 푇 (1,0) 푛 = 푇 (1,0) 푛, 퐷 푓 (푝) 푇 (0,1) 푛 = 푇 (0,1) 푛, ℂ 푝 ℂ 푓 (푝) ℂ ℂ 푝 ℂ 푓 (푝) ℂ and 퐷ℂ 푓 (푝) ℂ ⊗ 푇푝 푀 = ℂ ⊗ 푇푓 (푝) 푓 (푀).

퐷 푓 (푝) 푇 (1,0) 푀 푇 (1,0) 푓 (푀) 푇 (0,1) 푀 푇 (0,1) 푓 (푀) Then it is clear that ℂ must take 푝 to 푓 (푝) and 푝 to 푓 (푝) . In the next proposition it is important to note that a translation and applying a unitary matrix are biholomorphic changes of coordinates.

Proposition 2.3.3. Let 푀 ⊂ ℂ푛 be a smooth real hypersurface, 푝 ∈ 푀. After a translation and a rotation by a unitary matrix, 푝 is the origin, and near the origin, 푀 is written in variables (푧, 푤) ∈ ℂ푛−1 × ℂ as Im 푤 = 휑(푧, 푧,¯ Re 푤), with the 휑(0) and 푑휑(0) = 0. Consequently

휕 휕 휕 휕 푇 (1,0) 푀 ,..., , 푇 (0,1) 푀 ,..., , 0 = spanℂ 0 = spanℂ 휕푧1 0 휕푧푛−1 0 휕푧¯1 0 휕푧¯푛−1 0 휕 퐵 . 0 = spanℂ 휕(Re 푤) 0

(1,0) (0,1) In particular, dimℂ 푇푝 푀 = dimℂ 푇푝 푀 = 푛 − 1 and dimℂ 퐵푝 = 1. If 푀 is the boundary of a open set 푈 with smooth boundary, the rotation can be chosen so that Im 푤 > 휑(푧, 푧,¯ Re 푤) on 푈.

Proof. We apply a translation to put 푝 = 0 and in the same manner as in Lemma 2.2.8 apply a ∇푟 − 휕 휑( ) 푑휑( ) unitary matrix to make sure that is in the direction 휕(Im 푤) 0. That 0 = 0 and 0 = 0 follows as before. As a translation and a unitary matrix are holomorphic and in fact biholomorphic,

then via Proposition 2.3.1 we obtain that the tangent spaces are all transformed correctly. 휕 푀 The rest of the proposition follows at once as 휕(Im 푤) 0 is the normal vector to at 0. Remark 2.3.4. When 푀 is of smaller dimension than 2푛 − 1 (no longer a hypersurface, but a submanifold of higher codimension), then the proposition above does not hold. That is, we would (1,0) (0,1) still have dimℂ 푇푝 푀 = dimℂ 푇푝 푀, but this number need not be constant from point to point. Fortunately, when talking about domains with smooth boundaries, the boundaries are hypersurfaces, and this complication does not arise.

Deﬁnition 2.3.5. Suppose 푈 ⊂ ℂ푛 is an open set with smooth boundary, and 푟 is a deﬁning function for 휕푈 at 푝 ∈ 휕푈 such that 푟 < 0 on 푈. If

푛 2 푛 ∑︁ 휕 푟 ∑︁ 휕 (1,0) 푎¯푗 푎ℓ ≥ 0 for all 푋푝 = 푎푗 ∈ 푇푝 휕푈, 휕푧¯ 휕푧 푝 휕푧 푝 푗=1,ℓ=1 푗 ℓ 푗=1 푗 54 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

then 푈 is said to be pseudoconvex at 푝 (or Levi pseudoconvex). If the inequality above is strict for (1,0) all nonzero 푋푝 ∈ 푇푝 휕푈, then 푈 is said to be strongly pseudoconvex. If 푈 is pseudoconvex, but not strongly pseudoconvex at 푝, then we say that 푈 is weakly pseudoconvex. ∗ A domain 푈 is pseudoconvex if it is pseudoconvex at all 푝 ∈ 휕푈. For a bounded 푈, we say 푈 is strongly pseudoconvex if it is strongly pseudoconvex at all 푝 ∈ 휕푈. (1,0) For 푋푝 ∈ 푇푝 휕푈, the sesquilinear form

푛 ∑︁ 휕2푟 L(푋푝, 푋푝) = 푎¯푗 푎ℓ 휕푧¯ 휕푧 푝 푗=1,ℓ=1 푗 ℓ

is called the Levi form at 푝. So 푈 is pseudoconvex (resp. strongly pseudoconvex) at 푝 ∈ 휕푈 if the Levi form is positive semidefinite (resp. positive definite) at 푝. The Levi form can be defined for any real hypersurface 푀, although one has to decide which side of 푀 is “the inside.”

The matrix 휕2푟

휕푧 휕푧 ¯푗 ℓ 푝 푗ℓ is called the complex Hessian of 푟 at 푝. So, 푈 is pseudoconvex at 푝 ∈ 휕푈 if the complex Hessian of (1,0) 푟 at 푝 as a sesquilinear form is positive (semi)deﬁnite when restricted to tangent vectors in 푇푝 휕푈. For example, the unit ball 픹푛 is strongly pseudoconvex as can be seen by computing the Levi form directly from 푟(푧, 푧¯) = k푧k2 − 1, that is, the complex Hessian of 푟 is the identity matrix. We remark that the complex Hessian is not the full Hessian. Let us write down the full Hessian, 휕 휕 using the basis of 휕푧 s and 휕푧¯ s. It is the symmetric matrix

2 2 2 2  휕 푟 ··· 휕 푟 휕 푟 ··· 휕 푟   휕푧1휕푧1 휕푧1휕푧푛 휕푧1휕푧¯1 휕푧1휕푧¯푛   ......   ......   2 2 2 2   휕 푟 ··· 휕 푟 휕 푟 ··· 휕 푟   휕푧푛휕푧1 휕푧푛휕푧푛 휕푧푛휕푧¯1 휕푧푛휕푧¯푛  .  휕2푟 휕2푟 휕2푟 휕2푟   휕푧¯ 휕푧 ··· 휕푧¯ 휕푧 휕푧¯ 휕푧¯ ··· 휕푧¯ 휕푧¯   1. 1 . 1. 푛 1. 1 . 1. 푛   ......     2 2 2 2   휕 푟 ··· 휕 푟 휕 푟 ··· 휕 푟   휕푧¯푛휕푧1 휕푧¯푛휕푧푛 휕푧¯푛휕푧¯1 휕푧¯푛휕푧¯푛  The complex Hessian is the lower left, or the transpose of the upper right, block. In particular, it is a smaller matrix. And we also apply it only to a subspace of the complexiﬁed tangent space. Let us illustrate the change of basis on one dimension. The change of variables is left to student for higher dimensions. Let 푧 = 푥 + 푖푦 be in ℂ, and denote by 푇 the change of basis matrix:

" 2 2 # " 2 2 # 1 1 휕 푟 휕 푟 휕 푟 휕 푟 /2 /2 푡 휕푥휕푥 휕푥휕푦 휕푧휕푧 휕푧휕푧¯ 푇 = , 푇 2 2 푇 = 2 2 . −푖/2 푖/2 휕 푟 휕 푟 휕 푟 휕 푟 휕푦휕푥 휕푦휕푦 휕푧휕푧¯ 휕푧휕¯ 푧¯ Let us also see how a complex linear change of variables acts on the Hessian matrix. A complex 푛 푛 휕 linear change of variables is not an arbitrary 2 × 2 matrix. If the Hessian is in the basis of 휕푧 s

∗The deﬁnition for unbounded domains is not consistent in the literature. Sometimes strictly pseudoconvex is used. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 55

h i 휕 푛 × 푛 퐴 퐴 ⊕ 퐴 퐴 0 and 휕푧¯ s, an complex linear matrix acts on the Hessian as , that is 0 퐴 . Write the h 푡 i full Hessian as 푋 퐿 , where 퐿 is the complex Hessian. The complex linear change of variables 퐴 퐿 푋 transforms the full Hessian as 퐴 0푡 푋 퐿푡 퐴 0 퐴푡 푋 퐴 (퐴∗퐿퐴)푡 = , 0 퐴 퐿 푋 0 퐴 퐴∗퐿퐴 퐴푡 푋 퐴 푡 where 퐴∗ = 퐴 is the conjugate transpose of 퐴. So 퐴 transforms the complex Hessian 퐿 as 퐴∗퐿퐴, that is, by ∗-congruence. Star-congruence preserves the inertia (the number of positive, negative, and zero eigenvalues) of a Hermitian matrix by the Sylvester’s law of inertia from linear algebra. The Levi form itself does depend on the defining function, but the signs of the eigenvalues do not. It is common to say “the Levi form” without mentioning a specific defining function even though that is not completely correct. The proof of the following proposition is left as an exercise. Proposition 2.3.6. Let 푈 ⊂ ℂ푛 be an open set with smooth boundary and 푝 ∈ 휕푈. The inertia of the Levi form at 푝 does not depend on the defining function at 푝. In particular, the definition of pseudoconvexity and strong pseudoconvexity is independent of the defining function.

Exercise 2.3.2: If 푟 is real-valued, then the complex Hessian of 푟 is Hermitian, that is, the matrix is equal to its conjugate transpose.

Exercise 2.3.3: Prove Proposition 2.3.6 .

Exercise 2.3.4: Show that a convex domain with smooth boundary is pseudoconvex, and show that (a bounded) strongly convex domain with smooth boundary is strongly pseudoconvex.

Exercise 2.3.5: Show that if an open set with smooth boundary is strongly pseudoconvex at a point, it is strongly pseudoconvex at all nearby points.

We are generally interested what happens under a holomorphic change of coordinates, that is, a biholomorphic mapping. And as far as pseudoconvexity is concerned we are interested in local changes of coordinates as pseudoconvexity is a local property. Before proving that pseudoconvexity is a biholomorphic invariant, let us note where the Levi form appears in the graph coordinates from

Proposition 2.3.3 , that is, when our boundary (the hypersurface) is given near the origin by Im 푤 = 휑(푧, 푧,¯ Re 푤), where 휑 is 푂(2). Let 푟(푧, 푧,¯ 푤, 푤¯ ) = 휑(푧, 푧,¯ Re 푤) − Im 푤 be our deﬁning function. The complex Hessian of 푟 has the form 퐿 0 휕2휑 where 퐿 = . 0 0 휕푧 휕푧 ¯푗 ℓ 0 푗ℓ n o Note that 퐿 is an (푛−1) × (푛−1) matrix. The vectors in 푇 (1,0) 휕푈 are the span of 휕 ,..., 휕 . 0 휕푧1 0 휕푧푛−1 0 푛 푇 (1,0) 휕푈 (푎, ) ∈ 푛 푎 ∈ 푛−1 That is, as an -vector, a vector in 0 is represented by 0 ℂ for some ℂ . The Levi form at the origin is then 푎∗퐿푎, in other words, it is given by the (푛 − 1) × (푛 − 1) matrix 퐿. If this matrix 퐿 is positive semideﬁnite, then 휕푈 is pseudoconvex at 0. 56 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Example 2.3.7: Let us change variables to write the ball 픹푛 in diﬀerent local holomorphic coordinates where the Levi form is displayed nicely. The sphere 휕픹푛 is deﬁned in the variables 푛 푍 = (푍1, . . . , 푍푛) ∈ ℂ by k푍 k = 1. Let us change variables to (푧1, . . . , 푧푛−1, 푤) where 푍 푗 1 + 푍푛 푧푗 = for all 푗 = 1, . . . , 푛 − 1, 푤 = 푖 . 1 − 푍푛 1 − 푍푛

This change of variables is a biholomorphic mapping from the set where 푍푛 ≠ 1 to the set where 푤 ≠ −푖 (exercise). For us it is suﬃcient to notice that the map is invertible near (0,..., 0, −1), which follows by simply computing the derivative. Notice that the last component is the inverse of the Cayley transform (that takes the disc to the upper half-plane). We claim that the mapping takes the unit sphere given by k푍 k = 1 (without the point (0,..., 0, 1)), to the set deﬁned by

2 2 Im 푤 = |푧1| + · · · + |푧푛−1| ,

and that it takes (0,..., 0, −1) to the origin (this part is trivial). Let us check:

푍 2 푍 2 푖 1+푍푛 − 푖 1+푍푛 2 2 1 푛−1 1−푍푛 1−푍푛 |푧1| + · · · + |푧푛−1| − Im 푤 = + · · · + − 1 − 푍푛 1 − 푍푛 2푖 |푍 |2 |푍 |2 1 + 푍 1 + 푍¯ = 1 + · · · + 푛−1 − 푛 − 푛 2 2 푍 |1 − 푍푛| |1 − 푍푛| 2(1 − 푛) 2(1 − 푍¯푛) |푍 |2 + · · · + |푍 |2 + |푍 |2 − 1 = 1 푛−1 푛 . 2 |1 − 푍푛|

2 2 2 2 Therefore, |푍1| + · · · + |푍푛| = 1 if and only if Im 푤 = |푧1| + · · · + |푧푛−1| . As the map takes the point (0,..., 0, −1) to the origin, we can think of the set given by

2 2 Im 푤 = |푧1| + · · · + |푧푛−1|

as the sphere in local holomorphic coordinates at (0,..., 0, −1) (by symmetry of the sphere we could have done this at any point by rotation). In the coordinates (푧, 푤), the ball (the inside of the sphere) is the set given by 2 2 Im 푤 > |푧1| + · · · + |푧푛−1| . In these new coordinates, the Levi form is just the identity matrix at the origin. In particular, the domain is strongly pseudoconvex. We have not yet proved that pseudoconvexity is a biholomorphic invariant, but when we do, it will also mean that the ball is strongly pseudoconvex. Not the entire sphere gets transformed, the points where 푍푛 = 1 get “sent to inﬁnity.” The 2 2 hypersurface Im 푤 = |푧1| + · · · + |푧푛−1| is sometimes called the Lewy hypersurface, and in the ∗ literature some even say it is the sphere . Pretending 푧 is just one real direction, it looks like this:

∗That is not, in fact, completely incorrect. If we think of the sphere in the complex projective space, we are simply looking at the sphere in a diﬀerent coordinate patch. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 57

Im 푤

푧 Re 푤

2 2 As an aside, the hypersurface Im 푤 = |푧1| + · · · + |푧푛−1| is also called the Heisenberg group. The group in this case is the group deﬁned on the parameters (푧, Re 푤) of this hypersurface with the group law deﬁned by (푧, Re 푤)(푧0, Re 푤0) = (푧 + 푧0, Re 푤 + Re 푤0 + 2 Im 푧 · 푧0).

Exercise 2.3.6: Prove the assertion in the example about the mapping being biholomorphic on the sets described above.

Let us see how the Hessian of 푟 changes under a biholomorphic change of coordinates. That is, let 푓 : 푉 → 푉0 be a biholomorphic map between two domains in ℂ푛, and let 푟 : 푉0 → ℝ be a smooth function with nonvanishing derivative. Let us compute the Hessian of 푟 ◦ 푓 : 푉 → ℝ. We ﬁrst compute what happens to the nonmixed derivatives. As we have to apply chain rule twice, to keep track better track of things, we write the derivatives as functions. For clarity, let 푧 be the coordinates in 푉 and 휁 the coordinates in 푉0. That is, 푟 is a function of 휁 and 휁¯, 푓 is a function of 푧, and 푓¯ is a function of 푧¯. Therefore, 푟 ◦ 푓 is a function of 푧 and 푧¯.

0 푛 휕2(푟 ◦ 푓 ) 휕 ∑︁ 휕푟 휕 푓 휕푟 휕 푓¯ (푧, 푧¯) = 푓 (푧), 푓¯(푧¯) ℓ (푧) + 푓 (푧), 푓¯(푧¯) ℓ (푧¯) 휕푧 휕푧 휕푧 휕휁 휕푧 휕휁¯ 휕푧 푗 푘 푗 ℓ=1 ℓ 푘 ℓ 푘 0 푛 ∑︁ 휕2푟 휕 푓 휕 푓 휕2푟 휕 푓¯ 휕 푓 = 푓 (푧), 푓¯(푧¯) 푚 (푧) ℓ (푧) + 푓 (푧), 푓¯(푧¯) 푚 (푧¯) ℓ (푧) 휕휁 휕휁 휕푧 휕푧 휕휁¯ 휕휁 휕푧 휕푧 ℓ,푚=1 푚 ℓ 푗 푘 푚 ℓ 푗 푘 푛 ∑︁ 휕푟 휕2 푓 + 푓 (푧), 푓¯(푧) ℓ (푧) 휕휁 ¯ 휕푧 휕푧 ℓ=1 ℓ 푗 푘 푛 푛 ∑︁ 휕2푟 휕 푓 휕 푓 ∑︁ 휕푟 휕2 푓 = 푚 ℓ + ℓ . 휕휁 휕휁 휕푧 휕푧 휕휁 휕푧 휕푧 ℓ,푚=1 푚 ℓ 푗 푘 ℓ=1 ℓ 푗 푘

h 2 i h 2 i The matrix 휕 (푟◦ 푓 ) can have diﬀerent eigenvalues than the matrix 휕 푟 . If 푟 has nonvanishing 휕푧푗 휕푧푘 휕휁푗 휕휁푘 gradient, then using the second term, we can (locally) choose 푓 in such a way as to make the matrix h 2 i 휕 (푟◦ 푓 ) be the zero matrix (or anything else) at a certain point by choosing the second derivatives 휕푧푗 휕푧푘 h 2 i of 푓 arbitrarily at that point. See the exercise below. Nothing about the matrix 휕 푟 is preserved 휕휁푗 휕휁푘 under a biholomorphic map. And that is precisely why it does not appear in the deﬁnition of h 2 i h 2 i pseudoconvexity. The story for 휕 (푟◦ 푓 ) and 휕 푟 is exactly the same. 휕푧¯푗 휕푧¯푘 휕휁¯푗 휕휁¯푘 58 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Exercise 2.3.7: Given a real function 푟 with nonvanishing gradient at 푝 ∈ ℂ푛. Find a local change of coordinates 푓 at 푝 (so 푓 ought to be a holomorphic mapping with an invertible 2 2 derivative at 푝) such that 휕 (푟◦ 푓 ) and 휕 (푟◦ 푓 ) are just the zero matrices. 휕푧푗 휕푧푘 푝 휕푧¯푗 휕푧¯푘 푝

Let us look at the mixed derivatives: 푛 휕2(푟 ◦ 푓 ) 휕 ∑︁ 휕푟 휕 푓 (푧, 푧¯) = 푓 (푧), 푓¯(푧¯) ℓ (푧) 휕푧¯ 휕푧 휕푧¯ 휕휁 휕푧 푗 푘 푗 ℓ=1 ℓ 푘 0 푛 휕2푟 휕 푓¯ 휕 푓 푛 휕푟 휕2 푓 ¨¨* ∑︁ 푓 푧 , 푓¯ 푧 푚 푧 ℓ 푧 ∑︁ 푓 푧 , 푓¯ 푧 ℓ¨ 푧 = ( ) ( ¯) ( ¯) ( ) + ( ) ( ¯) ¨¨ ( ) 휕휁¯ 휕휁 휕푧¯ 휕푧 휕휁 ¨휕푧¯ 휕푧 ℓ,푚=1 푚 ℓ 푗 푘 ℓ=1 ℓ 푗 푘 푛 ∑︁ 휕2푟 휕 푓¯ 휕 푓 = 푚 ℓ . 휕휁¯ 휕휁 휕푧¯ 휕푧 ℓ,푚=1 푚 ℓ 푗 푘 The complex Hessian of 푟 ◦ 푓 is the complex Hessian 퐻 of 푟 conjugated as 퐷∗퐻퐷, where 퐷 is the holomorphic derivative matrix of 푓 at 푧 and 퐷∗ is the conjugate transpose. Sylvester’s law of inertia says that the number of positive, negative, and zero eigenvalues of 퐷∗퐻퐷 is the same as that for 퐻. The eigenvalues might have changed, but their sign did not. We are only considering 퐻 and 퐷∗퐻퐷 on a subspace. In linear algebra language, consider an invertible 퐷, a subspace 푇, and its image 퐷푇. Then the inertia of 퐻 restricted to 퐷푇 is the same as the inertia of 퐷∗퐻퐷 restricted to 푇. Let 푀 be a smooth real hypersurface given by 푟 = 0, then 푓 −1(푀) is a smooth real hypersurface (1,0) −1 given by 푟 ◦ 푓 = 0. The holomorphic derivative 퐷 = 퐷 푓 (푝) takes 푇푝 푓 (푀) isomorphically to 푇 (1,0) 푀 퐻 푇 (1,0) 푀 퐷∗퐻퐷 푓 (푝) . So is positive (semi)deﬁnite on 푓 (푝) if and only if is positive (semi)deﬁnite (1,0) −1 on 푇푝 푓 (푀). We have almost proved the following theorem. In short, pseudoconvexity is a biholomorphic invariant. Theorem 2.3.8. Suppose 푈, 푈0 ⊂ ℂ푛 are open sets with smooth boundary, 푝 ∈ 휕푈, 푉 ⊂ ℂ푛 a neighborhood of 푝, 푞 ∈ 휕푈0, 푉0 ⊂ ℂ푛 a neighborhood of 푞, and 푓 : 푉 → 푉0 a biholomorphic map with 푓 (푝) = 푞, such that 푓 (푈 ∩ 푉) = 푈0 ∩ 푉0. Then the inertia of the Levi form of 푈 at 푝 is the same as the inertia of the Levi form of 푈0 at 푞. In particular, 푈 is pseudoconvex at 푝 if and only if 푈0 is pseudoconvex at 푞. Similarly, 푈 is strongly pseudoconvex at 푝 if and only if 푈0 is strongly pseudoconvex at 푞.

푈 푈0 푓 푝 푞 푉0 푉

To finish proving the theorem, the only thing left is to observe that if 푓 (푈 ∩ 푉) = 푈0 ∩ 푉0, then 푓 (휕푈 ∩푉) = 휕푈0 ∩푉0, and to note that if 푟 is a defining function for 푈0 at 푞, then 푓 ◦ 푟 is a defining function for 푈 at 푝. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 59

Exercise 2.3.8: Find an example of a bounded domain with smooth boundary that is not convex, but that is pseudoconvex.

While the Levi form is not invariant under holomorphic changes of coordinates, its inertia is. Putting this together with the other observations we made above, we will ﬁnd the normal form for the quadratic part of the deﬁning equation for a smooth real hypersurface under biholomorphic transformations. It is possible to do better than the following lemma, but it is not possible to always get rid of the dependence on Re 푤 in the higher order terms. Recall that 푂(푘) at the origin means a function that together with its derivatives up to order 푘 − 1 vanish at the origin. Lemma 2.3.9. Let 푀 be a smooth real hypersurface in ℂ푛 and 푝 ∈ 푀. Then there exists a local biholomorphic change of coordinates taking 푝 to the origin and 푀 to the hypersurface given by

훼 훼+훽 ∑︁ 2 ∑︁ 2 Im 푤 = |푧푗 | − |푧푗 | + 퐸 (푧, 푧,¯ Re 푤), 푗=1 푗=훼+1

where 퐸 is 푂(3) at the origin. Here 훼 is the number of positive eigenvalues of the Levi form at 푝, 훽 is the number of negative eigenvalues, and 훼 + 훽 ≤ 푛 − 1. Proof. Change coordinates so that 푀 is given by Im 푤 = 휑(푧, 푧,¯ Re 푤), where 휑 is 푂(2). Apply Taylor’s theorem to 휑 up to the second order:

휑(푧, 푧,¯ Re 푤) = 푞(푧, 푧¯) + (Re 푤)(퐿푧 + 퐿푧) + 푎(Re 푤)2 + 푂(3), where 푞 is quadratic, 퐿 : ℂ푛−1 → ℂ is linear, and 푎 ∈ ℝ. If 퐿 ≠ 0, do a linear change of coordinates in the 푧 only to make 퐿푧 = 푧1. So assume 퐿푧 = 휖푧1 where 휖 = 0 or 휖 = 1. 0 02 0 Change coordinates by leaving 푧 unchanged and letting 푤 = 푤 + 푏푤 + 푐푤 푧1. Ignore 푞(푧, 푧¯) for a moment, as this change of coordinates does not aﬀect it. Also, only work up to second order.

푤 − 푤¯ 푤 + 푤¯ 푤 + 푤¯ 2 − Im 푤 + 휖 (Re 푤)(푧 + 푧¯ ) + 푎(Re 푤)2 = − + 휖 (푧 + 푧¯ ) + 푎 1 1 2푖 2 1 1 2 푤0 + 푏푤02 + 푐푤0푧 − 푤¯ 0 − 푏¯푤¯ 02 − 푐¯푤¯ 0푧¯ = − 1 1 2푖 푤0 + 푏푤02 + 푐푤0푧 + 푤¯ 0 + 푏¯푤¯ 02 + 푐¯푤¯ 0푧¯ + 휖 1 1 (푧 + 푧¯ ) 2 1 1 (푤0 + 푏푤02 + 푐푤0푧 + 푤¯ 0 + 푏¯푤¯ 02 + 푐¯푤¯ 0푧¯ )2 + 푎 1 1 4 푤0 − 푤¯ 0 = − 2푖 (휖푖 − 푐)푤0 + 휖푖푤¯ 0푧 + (휖푖 + 푐¯)푤¯ 0 + 휖푖푤0푧¯ + 1 1 2푖 (푖푎 − 2푏)푤02 + (푖푎 + 2푏¯)푤¯ 02 + 2푖푎푤0푤¯ 0 + + 푂(3). 4푖 60 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

We cannot quite get rid of all the quadratic terms in 휑, but we choose 푏 and 푐 to make the second order terms not depend on Re 푤0. Set 푏 = 푖푎 and 푐 = 2푖휖, and add 푞(푧, 푧¯) + 푂(3) into the mix to get 2 − Im 푤 + 휑(푧, 푧,¯ Re 푤) = − Im 푤 + 푞(푧, 푧¯) + 휖 (Re 푤)(푧1 + 푧¯1) + 푎(Re 푤) + 푂(3) 푤0 − 푤¯ 0 푤0 − 푤¯ 0 푤0 − 푤¯ 0 2 = − + 푞(푧, 푧¯) − 휖푖 (푧 − 푧¯ ) + 푎 + 푂(3) 2푖 2푖 1 1 2푖 0 0 0 2 = − Im 푤 + 푞(푧, 푧¯) − 휖푖(Im 푤 )(푧1 − 푧¯1) + 푎(Im 푤 ) + 푂(3). The right-hand side is the deﬁning equation in the (푧, 푤0) coordinates. However, it is no longer written as a graph of Im 푤0 over the rest, so we apply the implicit function theorem to solve for Im 푤0 and write the hypersurface as a graph again. The expression for Im 푤0 is 푂(2), and therefore 0 0 2 −푖휖 (Im 푤 )(푧1 − 푧¯1) + 푎(Im 푤 ) is 푂(3). So if we write 푀 as a graph, Im 푤0 = 푞(푧, 푧¯) + 퐸 (푧, 푧,¯ Re 푤0), then 퐸 is 푂(3). We write the quadratic polynomial 푞 as

푛−1 ∑︁ 푞(푧, 푧¯) = 푎푗 푘 푧푗 푧푘 + 푏푗 푘 푧¯푗 푧¯푘 + 푐푗 푘 푧¯푗 푧푘 . (2.1) 푗,푘=1

As 푞 is real-valued, it is left as an exercise to show that 푎푗 푘 = 푏푗 푘 and 푐푗 푘 = 푐푘 푗 . That is, the matrix [푏푗 푘 ] is the complex conjugate of [푎푗 푘 ] and [푐푗 푘 ] is Hermitian. We make another change of coordinates. We ﬁx the 푧s again, and we set

푛−1 0 00 ∑︁ 푤 = 푤 + 푖 푎푗 푘 푧푗 푧푘 . (2.2) 푗,푘=1 In particular, 푛−1 푛−1 0 00 ∑︁ 00 ∑︁ Im 푤 = Im 푤 + Im 푖 푎푗 푘 푧푗 푧푘 = Im 푤 + 푎푗 푘 푧푗 푧푘 + 푏푗 푘 푧¯푗 푧¯푘 , 푗,푘=1 푗,푘=1 0 0 00 as 푎푗 푘 = 푏푗 푘 . Plugging ( 2.2 ) into Im 푤 = 푞(푧, 푧¯) + 퐸 (푧, 푧,¯ Re 푤 ) and solving for Im 푤 cancels the holomorphic and antiholomorphic terms in 푞, and leaves 퐸 as 푂(3). After this change of coordinates we may assume 푛−1 ∑︁ 푞(푧, 푧¯) = 푐푗 푘 푧푗 푧¯푘 . 푗,푘=1

That is, 푞 is a sesquilinear form. Since 푞 is real-valued, the matrix 퐶 = [푐푗 푘 ] is Hermitian. In linear algebra notation, 푞(푧, 푧¯) = 푧∗퐶푧, where we think of 푧 as a column vector. If 푇 is a linear transformation on the 푧 variables, say 푧0 = 푇푧, we obtain 푧0∗퐶푧0 = (푇푧)∗퐶푇푧 = 푧∗(푇 ∗퐶푇)푧. Thus, we normalize 퐶 up to ∗-congruence. A Hermitian matrix is ∗-congruent to a diagonal matrix with only 1s, −1s, and 0s on the diagonal, again by Sylvester’s law of inertia. Writing out what that means is precisely the conclusion of the proposition. 2.3. HOLOMORPHIC VECTORS, THE LEVI FORM, AND PSEUDOCONVEXITY 61

Exercise 2.3.9: Prove the assertion in the proof, that is, if 푞 is a quadratic as in ( 2.1 ) that is real-valued, then 푎푗 푘 = 푏푗 푘 and 푐푗 푘 = 푐푘 푗 .

∗ 푛 Lemma 2.3.10 (Narasimhan’s lemma ). Let 푈 ⊂ ℂ be an open set with smooth boundary that is strongly pseudoconvex at 푝 ∈ 휕푈. Then there exists a local biholomorphic change of coordinates ﬁxing 푝 such that in these new coordinates, 푈 is strongly convex at 푝 and hence strongly convex at all points near 푝.

Exercise 2.3.10: Prove Narasimhan’s lemma. Hint: See the proof of Lemma 2.3.9 .

Exercise 2.3.11: Prove that an open 푈 ⊂ ℂ푛 with smooth boundary is pseudoconvex at 푝 if and only if there exist local holomorphic coordinates at 푝 such that 푈 is convex at 푝.

Narasimhan’s lemma only works at points of strong pseudoconvexity. For weakly pseudoconvex points the situation is far more complicated. The diﬃculty is that we cannot make a 푈 that is weakly pseudoconvex at all points near 푝 to be convex at all points near 푝. Let us prove the easy direction of the famous Levi problem. The Levi problem was a long- † 푛 standing problem in several complex variables to classify domains of holomorphy in ℂ . The answer is that a domain is a domain of holomorphy if and only if it is pseudoconvex. Just as the problem of trying to show that the classical geometric convexity is the same as convexity as we have deﬁned it, the Levi problem has an easier direction and a harder direction. The easier direction is to show that a domain of holomorphy is pseudoconvex, and the harder direction is to show that a

pseudoconvex domain is a domain of holomorphy. See Hörmander’s book [ H ] for the proof of the hard direction. Theorem 2.3.11 (Tomato can principle). Suppose 푈 ⊂ ℂ푛 is an open set with smooth boundary and at some point 푝 ∈ 휕푈 the Levi form has a negative eigenvalue. Then every holomorphic function on 푈 extends to a neighborhood of 푝. In particular, 푈 is not a domain of holomorphy. Pseudoconvex at 푝 means that all eigenvalues of the Levi form are nonnegative. The theorem says that a domain of holomorphy must be pseudoconvex. The theorem’s name comes from the proof, and sometimes other theorems using a similar proof of a “tomato can” of analytic discs are called tomato can principles. The general statement of proof of the principle is that “an analytic function holomorphic in a neighborhood of the sides and the bottom of a tomato can extends to the inside.” And the theorem we named after the principle states that “if the Levi form at 푝 has a negative eigenvalue, we can ﬁt a tomato can from inside the domain over 푝.” Proof. We change variables so that 푝 = 0, and near 푝, 푈 is given by

푛−1 2 ∑︁ 2 0 0 Im 푤 > −|푧1| + 휖푗 |푧푗 | + 퐸 (푧1, 푧 , 푧¯1, 푧¯ , Re 푤), 푗=2

∗A statement essentially of Narasimhan’s lemma was already used by Helmut Knesser in 1936. †E. E. Levi stated the problem in 1911, but it was not completely solved until the 1950s, by Oka and others. 62 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

0 where 푧 = (푧2, . . . , 푧푛−1), 휖푗 = −1, 0, 1, and 퐸 is 푂(3). We embed an analytic disc via the map 휑 휉 ↦→ (휆휉, 0, 0,..., 0) for some small 휆 > 0. Clearly 휑(0) = 0 ∈ 휕푈. For 휉 ≠ 0 near the origin

푛−1 2 2 ∑︁ 2 2 2 −휆 |휉| + 휖푗 |0| + 퐸 (휆휉, 0, 휆휉,¯ 0, 0) = −휆 |휉| + 퐸 (휆휉, 0, 휆휉,¯ 0, 0) < 0. 푗=2

That is because the function above has a strict maximum at 휉 = 0 by the second derivative test. Therefore, for 휉 ≠ 0 near the origin, 휑(휉) ∈ 푈. By picking 휆 small enough, 휑(픻 \{0}) ⊂ 푈. As 휑(휕픻) is compact we can “wiggle it a little” and ﬁnd discs in 푈. In particular, for all small enough 푠 > 0, the closed disc given by

휑 휉 ↦→푠 (휆휉, 0, 0,..., 0, 푖푠)

is entirely inside 푈 (that is, for slightly positive Im 푤). Fix such a small 푠 > 0. Suppose 휖 > 0 is small and 휖 < 푠. Deﬁne the Hartogs ﬁgure 퐻 = (푧, 푤) : 휆 − 휖 < |푧1| < 휆 + 휖 and |푧푗 | < 휖 for 푗 = 2, . . . , 푛 − 1, and |푤 − 푖푠| < 푠 + 휖 ∪ (푧, 푤) : |푧1| < 휆 + 휖, and |푧푗 | < 휖 for 푗 = 2, . . . , 푛 − 1, and |푤 − 푖푠| < 휖 .

0 The set where |푧1| = 휆, 푧 = 0 and |푤| ≤ 푠, is inside 푈 for all small enough 푠, so an 휖-neighborhood of that is in 푈. For 푤 = 푖푠 the whole disc where |푧1| ≤ 휆 is in 푈, so an 휖-neighborhood of that is in 푈. Thus, for small enough 휖 > 0, 퐻 ⊂ 푈. We are really just taking a Hartogs ﬁgure in the 푧1, 푤 0 variables, and then “fattening it up” to the 푧 variables. We picture the Hartogs ﬁgure in the |푧1| and |푤 − 푖푠| variables. The boundary 휕푈 and 푈 are only pictured diagrammatically. Also, we make a “picture” the analytic discs giving the “tomato can.” In the picture, the 푈 is below its boundary 휕푈, unlike usually.

휕푈 휕푈 |푤 − 푖푠| 푠 + 휖 푈푈 푠 휑 (푧, 푤) = (0, 0) 퐻 Tomato can of 휖 analytic discs

휑푠 휖휖 휆 |푧1|

The origin is in the hull of 퐻, and so every function holomorphic in 푈, and so in 퐻, extends through the origin. Hence 푈 is not a domain of holomorphy. 2.4. HARMONIC, SUBHARMONIC, AND PLURISUBHARMONIC FUNCTIONS 63

Exercise 2.3.12: For the following domains in 푈 ⊂ ℂ2, ﬁnd all the points in 휕푈 where 푈 is weakly pseudoconvex, all the points where it is strongly pseudoconvex, and all the points where it is not pseudoconvex. Is 푈 pseudoconvex? a) Im 푤 > |푧|4 b) Im 푤 > |푧|2(Re 푤) c) Im 푤 > (Re 푧)(Re 푤)

Exercise 2.3.13: Let 푈 ⊂ ℂ푛 be an open set with smooth boundary that is strongly pseudoconvex at 푝 ∈ 휕푈. Show that 푝 is a so-called peak point: There exists a neighborhood 푊 of 푝 and a holomorphic 푓 : 푊 → ℂ such that 푓 (푝) = 1 and | 푓 (푧)| < 1 for all 푧 ∈ 푊 ∩ 푈 \{푝}.

Exercise 2.3.14: Suppose 푈 ⊂ ℂ푛 is an open set with smooth boundary. Suppose for 푝 ∈ 휕푈, there is a neighborhood 푊 of 푝 and a holomorphic function 푓 : 푊 → ℂ such that 푑푓 (푝) ≠ 0, 푓 (푝) = 0, but 푓 is never zero on 푊 ∩ 푈. Show that 푈 is pseudoconvex at 푝. Hint: You may need

the holomorphic implicit function theorem, see Theorem 1.3.8 . Note: The result does not require the derivative of 푓 to not vanish, but is much harder to prove without that hypothesis.

A hyperplane is the “degenerate” case of normal convexity. There is also a flat case of pseudoconvexity. A smooth real hypersurface 푀 ⊂ ℂ푛 is Levi-flat if the Levi form vanishes at every point of 푀. The zero matrix is positive semidefinite and negative semidefinite, so both sides of 푀 are pseudoconvex. Conversely, the only hypersurface pseudoconvex from both sides is a Levi-flat one.

Exercise 2.3.15: Suppose 푈 = 푉 ×ℂ푛−1 ⊂ ℂ푛, where 푉 ⊂ ℂ is an open set with smooth boundary. Show that 푈 is has a smooth Levi-ﬂat boundary.

Exercise 2.3.16: Prove that a real hyperplane is Levi-ﬂat.

Exercise 2.3.17: Consider 푓 ∈ O(푈) for an open 푈 ⊂ ℂ푛. Let 푀 = 푧 ∈ 푈 : Im 푓 (푧) = 0 . Show that if 푑푓 (푝) ≠ 0 for some 푝 ∈ 푀, then near 푝, 푀 is a Levi-ﬂat hypersurface.

Exercise 2.3.18: Suppose 푀 ⊂ ℂ푛 is a smooth Levi-ﬂat hypersurface, 푝 ∈ 푀, and a complex line 퐿 is tangent to 푀 at 푝. Prove that 푝 is not an isolated point of 퐿 ∩ 푀.

Exercise 2.3.19: Suppose 푈 ⊂ ℂ푛 is an open set with smooth boundary and 휕푈 is Levi-ﬂat. Show that 푈 is unbounded. Hint: If 푈 were bounded, consider the point on 휕푈 farthest from the origin.

2.4 Harmonic, subharmonic, and plurisubharmonic functions

푛 2 ∗ Deﬁnition 2.4.1. Let 푈 ⊂ ℝ be an open set. A 퐶 -smooth function 푓 : 푈 → ℝ is harmonic if 휕2 푓 휕2 푓 ∇2 푓 = + · · · + = 0 on 푈. 휕푥2 휕푥2 1 푛 ∗Recall the operator ∇2, sometimes also written Δ, is the Laplacian. It is the trace of the Hessian matrix. 64 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

∗ A function 푓 : 푈 → ℝ ∪ {−∞} is subharmonic if it is upper-semicontinuous and for every ball 퐵푟 (푎) with 퐵푟 (푎) ⊂ 푈, and every function 푔 continuous on 퐵푟 (푎) and harmonic on 퐵푟 (푎), such that 푓 (푥) ≤ 푔(푥) for 푥 ∈ 휕퐵푟 (푎), we have

푓 (푥) ≤ 푔(푥), for all 푥 ∈ 퐵푟 (푎).

In other words, a subharmonic function is a function that is less than every harmonic function on every ball. We remark that when 푛 = 1 in the definition of a subharmonic function, it is the same as the standard definition of a convex function of one real variable, where affine linear functions play the role of harmonic functions: A function of one real variable is convex if for every interval it is less than the affine linear function with the same end points. After all, a function of one real variable is harmonic if the second derivative vanishes, and it is therefore affine linear. In one real dimension it is also easier to picture. The function 푓 is convex if on every interval [훼, 훽], 푓 ≤ 푔 for every affine linear 푔 bigger than 푓 at the endpoints 훼 and 훽. In particular, we can take the 푔 that is equal to 푓 at the endpoints. The picture is analogous for subharmonic functions for 푛 > 1, but it is harder to draw.

푦 = 푔(푥)

푦 = 푓 (푥) = 훼 푥 = 훽푥

We will consider harmonic and subharmonic functions in ℂ ℝ2. Let us go through some basic results on harmonic and subharmonic functions in ℂ that you have seen in detail in your one-variable class. Consequently, we leave some of these results as exercises. In this section (and not just here) we often write 푓 (푧) for a function even if it is not holomorphic.

Exercise 2.4.1: An upper-semicontinuous function achieves a maximum on compact sets.

Exercise 2.4.2: Let 푈 ⊂ ℂ be open. Show that for a 퐶2 function 푓 : 푈 → ℝ,

휕2 1 푓 = ∇2 푓 . 휕푧휕푧¯ 4 Use this to show that 푓 is harmonic if and only if it is (locally) the real or imaginary part of a holomorphic function. Hint: The key is ﬁnding an antiderivative of a holomorphic function.

Exercise 2.4.3: Prove the identity theorem. Let 푈 ⊂ ℂ be a domain and 푓 : 푈 → ℝ harmonic such that 푓 = 0 on a nonempty open subset of 푈. Then 푓 ≡ 0.

It follows from the exercise that a harmonic function is infinitely differentiable. It is useful to find a harmonic function given boundary values. This problem is called the Dirichlet problem, and

∗ Recall 푓 is upper-semicontinuous if lim sup푡→푥 푓 (푡) ≤ 푓 (푥) for all 푥 ∈ 푈. 2.4. HARMONIC, SUBHARMONIC, AND PLURISUBHARMONIC FUNCTIONS 65

it is solvable for many (though not all) domains. The proof of the following special case is contained in the exercises following the theorem. The Poisson kernel for the unit disc 픻 ⊂ ℂ is 1 1 − 푟2 1 1 + 푟푒푖휃 푃푟 (휃) = = Re , for 0 ≤ 푟 < 1. 2휋 1 + 푟2 − 2푟 cos 휃 2휋 1 − 푟푒푖휃 Theorem 2.4.2. Let 푢 : 휕픻 → ℝ be a continuous function. The function 푃푢 : 픻 → ℝ, deﬁned by ∫ 휋 푖휃 푖푡 푖휃 푖휃 푃푢(푟푒 ) = 푢(푒 )푃푟 (휃 − 푡) 푑푡 if 푟 < 1 and 푃푢(푒 ) = 푢(푒 ), −휋 is harmonic in 픻 and continuous on 픻.

In the proof, it is useful to consider how the graph of 푃푟 as a function of 휃 looks for a ﬁxed 푟. The following are the graphs of 푃푟 for 푟 = 0.5, 푟 = 0.7, and 푟 = 0.85 on [−휋, 휋]:

-3 -2 -1 0 1 2 3 2.0 2.0

1.5 1.5

1.0 1.0

0.5 0.5

0.0 0.0

-3 -2 -1 0 1 2 3

Exercise 2.4.4: a) Prove 푃푟 (휃) > 0 for all 0 ≤ 푟 < 1 and all 휃. ∫ 휋 푃 (휃) 푑휃 = b) Prove −휋 푟 1. c) Prove for any given 훿 > 0, sup{푃푟 (휃) : 훿 ≤ |휃| ≤ 휋} → 0 as 푟 → 1.

Exercise 2.4.5: Prove Theorem 2.4.2 using the following guideline: a) Poisson kernel is harmonic as a function of 푧 = 푟푒푖휃 ∈ 픻, and hence 푃푢 is harmonic. b) 푃 acts like an approximate identity: Prove that 푃푢(푟푒푖휃) → 푢(푒푖휃) uniformly as 푟 → 1 (Hint: Split the integral to [−훿, 훿] and the rest and use the previous exercise). c) Prove that 푃푢(푧) tends to 푢(푧0) as 푧 ∈ 픻 → 푧0 ∈ 휕픻.

Exercise 2.4.6: State and prove a version of Theorem 2.4.2 for an arbitrary disc Δ푟 (푎). Exercise 2.4.7: Prove that the Dirichlet problem is not solvable in the punctured disc 픻 \{0}. Hint: Let 푢 = 0 on 휕픻 and 푢(0) = 1. The solution would be less than −휖 log|푧| for every 휖 > 0.

The Poisson kernel is a reproducing kernel for holomorphic functions, as (the real and imaginary parts of) holomorphic functions are harmonic. Poisson kernel exists for higher dimensions as well. Solving the Dirichlet problem using the Poisson kernel leads to the following result. 66 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Proposition 2.4.3 (Mean-value property and sub-mean-value property). Let 푈 ⊂ ℂ be an open set.

(i) A continuous function 푓 : 푈 → ℝ is harmonic if and only if

∫ 2휋 1 푖휃 푓 (푎) = 푓 (푎 + 푟푒 ) 푑휃 whenever Δ푟 (푎) ⊂ 푈. 2휋 0

(ii) An upper-semicontinuous function 푓 : 푈 → ℝ ∪ {−∞} is subharmonic if and only if

∫ 2휋 1 푖휃 푓 (푎) ≤ 푓 (푎 + 푟푒 ) 푑휃 whenever Δ푟 (푎) ⊂ 푈. 2휋 0

For the sub-mean-value property you may have to use the Lebesgue integral to integrate an upper-semicontinuous function, and to use the version of the Poisson integral above, you need to approximate by continuous functions on the boundary in the right way. On ﬁrst reading, feel free to think of continuous subharmonic functions and not too much will be lost.

Exercise 2.4.8: Fill in the details of the proof of Proposition 2.4.3 .

Exercise 2.4.9: Let 푈 ⊂ ℂ be open. Show that if 푓 : 푈 → ℝ ∪ {−∞} is subharmonic, then

lim sup 푓 (푤) = 푓 (푧) for all 푧 ∈ 푈. 푤→푧

Exercise 2.4.10: Suppose 푈 ⊂ ℂ is open and 푔 : 푈 → ℝ is harmonic. Then 푓 : 푈 → ℝ ∪ {−∞} is subharmonic if and only if 푓 − 푔 is subharmonic.

Proposition 2.4.4 (Maximum principle). Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℝ ∪ {−∞} is subharmonic. If 푓 attains a maximum in 푈, then 푓 is constant.

Proof. Suppose 푓 attains a maximum at 푎 ∈ 푈. If Δ푟 (푎) ⊂ 푈, then

1 ∫ 2휋 푓 (푎) ≤ 푓 (푎 + 푟푒푖휃) 푑휃 ≤ 푓 (푎). 2휋 0

Hence, 푓 = 푓 (푎) almost everywhere on 휕Δ푟 (푎). By upper-semicontinuity, 푓 = 푓 (푎) everywhere on 휕Δ푟 (푎). This was true for all 푟 with Δ푟 (푎) ⊂ 푈, so 푓 = 푓 (푎) on Δ푟 (푎), and so the set where 푓 = 푓 (푎) is open. The set where an upper-semicontinuous function attains a maximum is closed. So 푓 = 푓 (푎) on 푈 as 푈 is connected.

Exercise 2.4.11: Prove that subharmonicity is a local property. That is, given an open set 푈 ⊂ ℂ, a function 푓 : 푈 → ℝ ∪ {−∞} is subharmonic if and only if for every 푝 ∈ 푈 there exists a neighborhood 푊 of 푝, 푊 ⊂ 푈, such that 푓 |푊 is subharmonic. Hint: Perhaps try to use the

maximum principle and Exercise 2.4.10 . 2.4. HARMONIC, SUBHARMONIC, AND PLURISUBHARMONIC FUNCTIONS 67

Exercise 2.4.12: Suppose 푈 ⊂ ℂ is a bounded open set, 푓 : 푈 → ℝ ∪ {−∞} is an upper- semicontinuous function, such that 푓 |푈 is subharmonic, 푔 : 푈 → ℝ is a continuous function such that 푔|푈 is harmonic and 푓 (푧) ≤ 푔(푧) for all 푧 ∈ 휕푈. Prove that 푓 (푧) ≤ 푔(푧) for all 푧 ∈ 푈. Exercise 2.4.13: Let 푔 be a function harmonic on a disc Δ ⊂ ℂ and continuous on Δ. Prove that for every 휖 > 0 there exists a function 푔휖 , harmonic in a neighborhood of Δ, such that 푔(푧) ≤ 푔휖 (푧) ≤ 푔(푧) + 휖 for all 푧 ∈ Δ. In particular, to test subharmonicity, we only need to consider those 푔 that are harmonic a bit past the boundary of the disc.

Proposition 2.4.5. Suppose 푈 ⊂ ℂ is an open set and 푓 : 푈 → ℝ is a 퐶2 function. The function 푓 is subharmonic if and only if ∇2 푓 ≥ 0.

Proof. Suppose 푓 is a 퐶2-smooth function on a subset of ℂ ℝ2 with ∇2 푓 ≥ 0. We wish to show that 푓 is subharmonic. Take a disc Δ such that Δ ⊂ 푈. Consider a function 푔 continuous on Δ, harmonic on Δ, and such that 푓 ≤ 푔 on the boundary 휕Δ. Because ∇2( 푓 − 푔) = ∇2 푓 ≥ 0, we assume 푔 = 0 and 푓 ≤ 0 on the boundary 휕Δ. Suppose ∇2 푓 > 0 at all points on Δ. Suppose 푓 attains a maximum in Δ, call this point 푝. The Laplacian ∇2 푓 is the trace of the Hessian matrix, but for 푓 to have a maximum, the Hessian must have only nonpositive eigenvalues at the critical points, which is a contradiction as the trace is the sum of the eigenvalues. So 푓 has no maximum inside, and therefore 푓 ≤ 0 on all of Δ. 2 2 2 Next suppose ∇ 푓 ≥ 0. Let 푀 be the maximum of 푥 + 푦 on Δ. Take 푓푛 (푥, 푦) = 푓 (푥, 푦) + 1 푥2 푦2 1 푀 2 푓 > 푓 푓 푛 ( + ) − 푛 . Clearly ∇ 푛 0 everywhere on Δ and 푛 ≤ 0 on the boundary, so 푛 ≤ 0 on all of Δ. As 푓푛 → 푓 , we obtain that 푓 ≤ 0 on all of Δ. The other direction is left as an exercise.

Exercise 2.4.14: Finish the proof of the proposition above.

In analogy to convex functions, a 퐶2-smooth function 푓 of one real variable is convex if and only if 푓 00(푥) ≥ 0 for all 푥.

Proposition 2.4.6. Suppose 푈 ⊂ ℂ is an open set and 푓훼 : 푈 → ℝ ∪ {−∞} is a family of subharmonic functions. Let 휑(푧) = sup 푓훼 (푧). 훼 If the family is ﬁnite, then 휑 is subharmonic. If the family is inﬁnite, 휑(푧) ≠ ∞ for all 푧, and 휑 is upper-semicontinuous, then 휑 is subharmonic.

Proof. Suppose Δ푟 (푎) ⊂ 푈. For any 훼,

∫ 2휋 ∫ 2휋 1 푖휃 1 푖휃 휑(푎 + 푟푒 ) 푑휃 ≥ 푓훼 (푎 + 푟푒 ) 푑휃 ≥ 푓훼 (푎). 2휋 0 2휋 0 Taking the supremum on the right over 훼 obtains the results. 68 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Exercise 2.4.15: Prove that if 휑 : ℝ → ℝ is a monotonically increasing convex function, 푈 ⊂ ℂ is an open set, and 푓 : 푈 → ℝ is subharmonic, then 휑 ◦ 푓 is subharmonic.

Exercise 2.4.16: Let 푈 ⊂ ℂ be open, { 푓푛} a sequence of subharmonic functions uniformly bounded above on compact subsets, and {푐푛} a sequence of positive real numbers such that Í∞ 푐 < 푓 Í∞ 푐 푓 푛=1 푛 ∞. Prove that = 푛=1 푛 푛 is subharmonic. Make sure to prove the function is upper-semicontinuous.

Exercise 2.4.17: Suppose 푈 ⊂ ℂ is a bounded open set, and {푝푛} a sequence of points in 푈. For 푧 푈 푓 푧 Í∞ −푛 푧 푝 ∈ , deﬁne ( ) = 푛=1 2 log| − 푛|, possibly taking on the value −∞. a) Show that 푓 is a subharmonic function in 푈. 푈 푝 1 푓 b) If = 픻 and 푛 = 푛 , show that is discontinuous at 0 (the natural topology on ℝ ∪ {−∞}). c) If {푝푛} is dense in 푈, show that 푓 is discontinuous on a dense set. Hint: Prove that 푓 −1(−∞) is a small (but dense) set. Another hint: Integrate the partial sums, and use polar coordinates.

There are too many harmonic functions in ℂ푛 ℝ2푛. To get the real and imaginary parts of holomorphic functions in ℂ푛, we require a smaller class of functions than all harmonic functions. Deﬁnition 2.4.7. Let 푈 ⊂ ℂ푛 be open. A 퐶2-smooth 푓 : 푈 → ℝ is pluriharmonic if for every 푎, 푏 ∈ ℂ푛, the function of one variable 휉 ↦→ 푓 (푎 + 푏휉)

is harmonic (on the set of 휉 ∈ ℂ where 푎 + 푏휉 ∈ 푈). That is, 푓 is harmonic on every complex line. A function 푓 : 푈 → ℝ ∪ {−∞} is plurisubharmonic, sometimes plush or psh for short, if it is upper-semicontinuous and for every 푎, 푏 ∈ ℂ푛, the function of one variable 휉 ↦→ 푓 (푎 + 푏휉) is subharmonic (whenever 푎 + 푏휉 ∈ 푈). A harmonic function of one complex variable is in some sense a generalization of an aﬃne linear function of one real variable. Similarly, as far as several complex variables are concerned, a pluriharmonic function is the right generalization to ℂ푛 of an aﬃne linear function on ℝ푛. In the same way plurisubharmonic functions are the correct complex variable generalizations of convex functions. A convex function of one real variable is like a subharmonic function, and a convex function of several real variables is a function that is convex when restricted to any real line.

Exercise 2.4.18: Let 푈 ⊂ ℂ푛 be open. Prove that a 퐶2-smooth 푓 : 푈 → ℝ is pluriharmonic if and only if 휕2 푓 = 0 on 푈 for all 푗, 푘 = 1, . . . , 푛. 휕푧¯푗 휕푧푘 Exercise 2.4.19: Show that a pluriharmonic function is harmonic. On the other hand, ﬁnd an example of a harmonic function that is not pluriharmonic. 2.4. HARMONIC, SUBHARMONIC, AND PLURISUBHARMONIC FUNCTIONS 69

Exercise 2.4.20: Let 푈 ⊂ ℂ푛 be open. Show that 푓 : 푈 → ℝ is pluriharmonic if and only if it is locally the real or imaginary part of a holomorphic function. Hint: Using a previous exercise 휕 푓 is holomorphic for all 푘. Assume that 푈 is simply connected, 푝 ∈ 푈, and 푓 (푝) = 0. Consider 휕푧푘 the line integral from 푝 to a nearby 푞 ∈ 푈:

푛 ∫ 푞 ∑︁ 휕 푓 퐹(푞) = (휁) 푑휁 . 휕푧 푘 푝 푘=1 푘

Prove that it is path independent, compute derivatives of 퐹, and ﬁnd out what is 퐹 + 퐹¯ − 푓 .

Exercise 2.4.21: Prove the maximum principle: If 푈 ⊂ ℂ푛 is a domain and 푓 : 푈 → ℝ ∪ {−∞} is plurisubharmonic and achieves a maximum at 푝 ∈ 푈, then 푓 is constant.

Proposition 2.4.8. Let 푈 ⊂ ℂ푛 be open. A 퐶2-smooth 푓 : 푈 → ℝ is plurisubharmonic if and only if the complex Hessian matrix 휕2 푓 휕푧 휕푧 ¯푗 푘 푗 푘 is positive semideﬁnite at every point.

Proof. First let us suppose that the complex Hessian has a negative eigenvalue at a point 푝 ∈ 푈. h 2 i 푝 푓 휕 푓 After a translation assume = 0. As is real-valued, the complex Hessian 휕푧¯ 휕푧 is Hermitian. 푗 푘 0 푗 푘 A complex linear change of coordinates acts on the complex Hessian by ∗-congruence, and therefore h 2 i 휕 푓 we can diagonalize, using Sylvester’s Law of Inertia again. So assume that 휕푧¯ 휕푧 is diagonal. 푗 푘 0 푗 푘 If the complex Hessian has a negative eigenvalue, then one of the diagonal entries is negative. 휕2 푓 Without loss of generality suppose < 0. The function 푧1 ↦→ 푓 (푧1, 0,..., 0) has a negative 휕푧¯1휕푧1 0 Laplacian and therefore is not subharmonic, and thus 푓 itself is not plurisubharmonic. For the other direction, suppose the complex Hessian is positive semidefinite at all points. Let 푝 ∈ 푈. After an affine change of coordinates assume that the line 휉 ↦→ 푎 + 푏휉 is simply setting all but the first variable to zero, that is, 푎 = 0 and 푏 = (1, 0,..., 0). As the complex Hessian is positive 2 semidefinite, 휕 푓 ≥ 0 for all points (푧 , 0,..., 0). We proved above that ∇2푔 ≥ 0 implies 푔 is 휕푧¯1휕푧1 1 subharmonic, and we are done.

Exercise 2.4.22: Suppose 푈 ⊂ ℂ푛 is open and 푓 : 푈 → ℂ is holomorphic. a) Show log| 푓 (푧)| is plurisubharmonic. In fact, it is pluriharmonic away from the zeros of 푓 . b) Show | 푓 (푧)|휂 is plurisubharmonic for all 휂 > 0.

Exercise 2.4.23: Show that the set of plurisubharmonic functions on an open set 푈 ⊂ ℂ푛 is a cone in the sense that if 푎, 푏 > 0 are constants and 푓 , 푔 : 푈 → ℝ ∪ {−∞} are plurisubharmonic, then 푎 푓 + 푏푔 is plurisubharmonic. 70 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Theorem 2.4.9. Suppose 푈 ⊂ ℂ푛 is an open set and 푓 : 푈 → ℝ ∪ {−∞} is plurisubharmonic. For every 휖 > 0, let 푈휖 ⊂ 푈 be the set of points further than 휖 away from 휕푈. Then there exists a smooth plurisubharmonic function 푓휖 : 푈휖 → ℝ such that 푓휖 (푧) ≥ 푓 (푧), and

푓 (푧) = lim 푓휖 (푧) for all 푧 ∈ 푈. 휖→0 That is, 푓 is a limit of smooth plurisubharmonic functions. The idea of the proof is important and useful in many other contexts. Proof. We smooth 푓 out by convolving with so-called mollifiers, or approximate delta functions. Many different mollifiers work, but let us use a specific one for concreteness. For 휖 > 0, define

( 2 퐶푒−1/(1−k푧k ) if k푧k < 1, 1 푔(푧) = and 푔휖 (푧) = 푔(푧/휖). 0 if k푧k ≥ 1, 휖2푛

It is left as an exercise that 푔, and so 푔휖 , is smooth. The function 푔 has compact support as it is only nonzero inside the unit ball. The support of 푔휖 is the 휖-ball. Both are nonnegative. Choose 퐶 so that ∫ ∫ 푔 푑푉 = 1, and therefore 푔휖 푑푉 = 1. ℂ푛 ℂ푛 Here 푑푉 is the volume measure. The function 푔 only depends on k푧k. To get an idea of how these 2 2 푒−1/(1−푥2) 1 푒−1/ 1−(푥/0.5) 1 푒−1/ 1−(푥/0.25) functions work, consider the graphs of , 0.5 , and 0.25 .

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 1.50 1.50

1.25 1.25

1.00 1.00

0.75 0.75

0.50 0.50

0.25 0.25

0.00 0.00

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

Compare the graphs to the graphs of the Poisson kernel as a function of 휃, which is also a type of molliﬁer. In fact, the idea of integrating against the right approximate delta function with the desired properties is similar to the solution of the Dirichlet problem using the Poisson kernel. The function 푓 is bounded above on compact sets as it is upper semicontinuous. If 푓 is not 푓 푓 , −1 bounded below, we replace with max 휖 , which is still plurisubharmonic. Therefore, without loss of generality we assume that 푓 is locally bounded. For 푧 ∈ 푈휖 , we deﬁne 푓휖 as the convolution with 푔휖 : ∫ ∫ 푓휖 (푧) = ( 푓 ∗ 푔휖 )(푧) = 푓 (푤)푔휖 (푧 − 푤) 푑푉 (푤) = 푓 (푧 − 푤)푔휖 (푤) 푑푉 (푤). ℂ푛 ℂ푛 2.4. HARMONIC, SUBHARMONIC, AND PLURISUBHARMONIC FUNCTIONS 71

The two forms of the integral follow easily via change of variables. We are perhaps abusing notation a bit as 푓 is only defined on 푈, but it is not a problem as long as 푧 ∈ 푈휖 , because 푔휖 is then zero when 푓 is undefined. By differentiating the first form under the integral, we find that 푓휖 is smooth. Let us show that 푓휖 is plurisubharmonic. We restrict to a line 휉 ↦→ 푎 + 푏휉. We wish to test subharmonicity by the sub-mean-value property using a circle of radius 푟 around 휉 = 0: ∫ 2휋 ∫ 2휋 ∫ 1 푖휃 1 푖휃 푓휖 (푎 + 푏푟푒 ) 푑휃 = 푓 푎 + 푏푟푒 − 푤 푔휖 (푤) 푑푉 (푤) 푑휃 2휋 0 2휋 0 ℂ푛 ∫ ∫ 2휋 1 푖휃 = 푓 푎 − 푤 + 푏푟푒 푑휃 푔휖 (푤) 푑푉 (푤) ℂ푛 2휋 0 ∫ ≥ 푓 (푎 − 푤)푔휖 (푤) 푑푉 (푤) = 푓휖 (푎). ℂ푛

For the inequality we used 푔휖 ≥ 0. So 푓휖 is plurisubharmonic. Let us show that 푓휖 (푧) ≥ 푓 (푧) for all 푧 ∈ 푈휖 . As 푔휖 (푤) only depends on |푤1|,..., |푤푛|, we notice that 푔휖 (푤1, . . . , 푤푛) = 푔휖 (|푤1|,..., |푤푛|). Without loss of generality we consider 푧 = 0, and we use polar coordinates for the integral. ∫ 푓휖 (0) = 푓 (−푤)푔휖 (|푤1|,..., |푤푛|) 푑푉 (푤) ℂ푛 ∫ 휖 ∫ 휖 ∫ 2휋 ∫ 2휋 푖휃1 푖휃푛 = ··· ··· 푓 (−푟1푒 ,..., −푟푛푒 ) 푑휃1 ··· 푑휃푛 0 0 0 0 푔휖 (푟1, . . . , 푟푛) 푟1 ··· 푟푛 푑푟1 ··· 푑푟푛 ∫ 휖 ∫ 휖 ∫ 2휋 ∫ 2휋 푖휃2 푖휃푛 ≥ ··· ··· (2휋) 푓 (0, −푟2푒 ,..., −푟푛푒 ) 푑휃2 ··· 푑휃푛 0 0 0 0 푔휖 (푟1, . . . , 푟푛) 푟1 ··· 푟푛 푑푟1 ··· 푑푟푛 ∫ 휖 ∫ 휖 푛 ≥ 푓 (0) ··· (2휋) 푔휖 (푟1, . . . , 푟푛) 푟1 ··· 푟푛 푑푟1 ··· 푑푟푛 0 0 ∫ = 푓 (0) 푔휖 (푤) 푑푉 (푤) = 푓 (0). ℂ푛

The second equality above follows as 푔휖 is zero outside the polydisc of radius 휖. For the inequalities, 푔 ≥ 휋 = ∫ 2휋 푑휃 we again needed that 휖 0. The penultimate equality follows from the fact that 2 0 . Finally, we show lim휖→0 푓휖 (푧) = 푓 (푧). For subharmonic, and so for plurisubharmonic, functions, 푓 휁 푓 푧 훿 > 휖 > 푓 휁 푓 푧 훿 휁 퐵 푧 lim sup휁→푧 ( ) = ( ). So given 0 ﬁnd an 0 such that ( ) − ( ) ≤ for all ∈ 휖 ( ). ∫ ∫ 푓휖 (푧) − 푓 (푧) = 푓 (푧 − 푤)푔휖 (푤) 푑푉 (푤) − 푓 (푧) 푔휖 (푤) 푑푉 (푤) 퐵휖 (0) 퐵휖 (0) ∫ = 푓 (푧 − 푤) − 푓 (푧) 푔휖 (푤) 푑푉 (푤) 퐵휖 (0) ∫ ≤ 훿 푔휖 (푤) 푑푉 (푤) = 훿. 퐵휖 (0)

Again we used that 푔휖 is nonnegative. We ﬁnd 0 ≤ 푓휖 − 푓 (푧) ≤ 훿, and so 푓휖 (푧) → 푓 (푧). 72 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Exercise 2.4.24: Show that 푔 in the proof above is smooth on all of ℂ푛.

Exercise 2.4.25: ∫ 2휋 푓 (푎 + 푟푒푖휃) 푑휃 푟 a) Show that for a subharmonic function, 0 is a monotone function of (Hint: Try a 퐶2 function ﬁrst and use Green’s theorem).

b) Use this fact to show that the 푓휖 (푧) from Theorem 2.4.9 is monotone decreasing in 휖. Exercise 2.4.26: Let 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 be open. Prove that if 푔 : 푈 → 푉 is holomorphic and 푓 : 푉 → ℝ is a 퐶2 plurisubharmonic function, then 푓 ◦ 푔 is plurisubharmonic. Then use it to prove the same fact for all plurisubharmonic functions (Hint: monotone convergence).

Exercise 2.4.27: Show that plurisubharmonicity is a local property, that is, 푓 is plurisubharmonic if and only if 푓 is plurisubharmonic in some neighborhood of each point.

Exercise 2.4.28: Using the computation from Theorem 2.4.9 show that if 푓 is pluriharmonic, ∞ then 푓휖 = 푓 (where it makes sense), obtaining another proof that a pluriharmonic function is 퐶 .

Exercise 2.4.29: Let the 푓 in Theorem 2.4.9 be continuous and suppose 퐾 ⊂⊂ 푈. For small enough 휖 > 0, 퐾 ⊂ 푈휖 . Show that 푓휖 converges uniformly to 푓 on 퐾.

푘 Exercise 2.4.30: Let the 푓 in Theorem 2.4.9 be 퐶 -smooth for some 푘 ≥ 0. Show that all derivatives of 푓휖 up to order 푘 converge uniformly on compact sets to the corresponding derivatives of 푓 . See also previous exercise.

Let us prove the theorem of Radó, which is a complementary result to the Riemann extension theorem. Here on the one hand the function is continuous and vanishes on the set you wish to extend across, but on the other hand you know nothing about this set. It is sometimes covered in a one-variable course, and in several variables it follows directly from the one-variable result. Theorem 2.4.10 (Radó). Let 푈 ⊂ ℂ푛 be open and 푓 : 푈 → ℂ a continuous function that is holomorphic on the set 푈0 = 푧 ∈ 푈 : 푓 (푧) ≠ 0 . Then 푓 ∈ O(푈). Proof. First assume 푛 = 1. As the theorem is local, it is enough to prove it for a small disc Δ such that 푓 is continuous on the closure Δ, let Δ0 be the part of the disc where 푓 is nonzero. If Δ0 is empty, then we are done as 푓 is just identically zero and hence holomorphic. Let 푢 be the real part of 푓 . On Δ0, 푢 is a harmonic function. Let 푃푢 be the Poisson integral of 푢 on Δ. Hence 푃푢 equals 푢 on 휕Δ, and 푃푢 is harmonic in all of Δ. Consider the function 푃푢(푧) −푢(푧) on Δ. The function is zero on 휕Δ and it is harmonic on Δ0. By rescaling 푓 we can without loss of generality assume that | 푓 (푧)| < 1 for all 푧 ∈ Δ. For any 푡 > 0, the function 푧 ↦→ 푡 log | 푓 (푧)| is subharmonic on Δ0 and upper-semicontinuous on Δ0. Further, it is negative on 휕Δ. The function 푧 ↦→ −푡 log | 푓 (푧)| is superharmonic (minus a subharmonic function) on Δ0, lower-semicontinuous on Δ0 and positive on 휕Δ. On the set Δ \ Δ0 where 푓 is zero, the two functions are −∞ and ∞ respectively. Therefore, for all 푡 > 0 and 푧 ∈ 휕Δ ∪ (Δ \ Δ0), we have 푡 log | 푓 (푧)| ≤ 푃푢(푧) − 푢(푧) ≤ −푡 log | 푓 (푧)| . (2.3) 2.5. HARTOGS PSEUDOCONVEXITY 73

Applying the maximum principle to the subharmonic functions 푧 ↦→ 푡 log | 푓 (푧)| − 푃푢(푧) − 푢(푧) 0 and 푧 ↦→ 푡 log | 푓 (푧)| − 푢(푧) − 푃푢(푧) shows that ( 2.3 ) holds for all 푧 ∈ Δ and all 푡 > 0.

푡 log| 푓 | < 0 Δ0 푓 ≠ 0 푃푢 − 푢 = 0 푊 Δ \ Δ0 푓 = 0 푡 log| 푓 | = −∞

Taking the limit 푡 → 0 shows that 푃푢 = 푢 on Δ0. Let 푊 = Δ \ Δ0. On 푊, 푢 = 0 and so 푃푢 − 푢 is harmonic on 푊 and continuous on 푊. Furthermore, 푃푢 − 푢 = 0 on Δ0 ∪ 휕Δ, and so 푃푢 − 푢 = 0 on 휕푊. By the maximum principle, 푃푢 = 푢 on 푊 and therefore on all of Δ. Similarly, if 푣 is the imaginary part of 푓 , then 푃푣 = 푣 on Δ. In other words, 푢 and 푣 are harmonic on Δ. As Δ is simply connected, let 푣˜ be the harmonic conjugate of 푢 that equals 푣 at some point of Δ0. As 푓 is holomorphic on Δ0, the harmonic functions 푣˜ and 푣 are equal the nonempty open subset Δ0 of Δ and so they are equal everywhere. Consequently, 푓 = 푢 + 푖푣 is holomorphic on Δ. The extension of the proof to several variables is left as an exercise.

Exercise 2.4.31: Use the one-variable result to extend the theorem to several variables.

2.5 Hartogs pseudoconvexity

It is worth mentioning explicitly that by the exercises of the previous section, plurisubharmonicity is preserved under holomorphic mappings. That is, if 푔 is holomorphic and 푓 is plurisubharmonic, then 푓 ◦ 푔 is plurisubharmonic. In particular, if 휑 : 픻 → ℂ푛 is an analytic disc and 푓 is plurisubharmonic in a neighborhood of 휑(픻), then 푓 ◦ 휑 is subharmonic.

∗ 푛 Definition 2.5.1. Let F be a class of (extended )-real-valued functions defined on an open 푈 ⊂ ℝ . If 퐾 ⊂ 푈, define 퐾b, the hull of 퐾 with respect to F, as the set

def n o 퐾b = 푥 ∈ 푈 : 푓 (푥) ≤ sup 푓 (푦) for all 푓 ∈ F . 푦∈퐾

† An open set 푈 is said to be convex with respect to F if for every 퐾 ⊂⊂ 푈, the hull 퐾b ⊂⊂ 푈.

Clearly 퐾 ⊂ 퐾b. The key is to show that 퐾b is not “too large” for 푈. Keep in mind that the functions in F are deﬁned on 푈, so 퐾b depends on 푈 not just on 퐾. An easy mistake is to consider ∗By extended reals we mean ℝ ∪ {−∞, ∞}. †Recall that ⊂⊂ means relatively compact, that is, the closure in the relative (subspace) topology is compact. 74 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

functions defined on a larger set, obtaining a smaller F and hence a larger 퐾b. Sometimes it is useful to write 퐾bF to denote the dependence on F, especially when talking about several different hulls. For example, if 푈 = ℝ and F is the set of real-valued smooth 푓 : ℝ → ℝ with 푓 00(푥) ≥ 0, then for any 푎, 푏 ∈ ℝ we have {푎, 푏} = [푎, 푏]. In general, if F was the set of convex functions, then a domain 푈 ⊂ ℝ푛 is geometrically convex if and only if it is convex with respect to convex functions, although let us not define what that means except for smooth functions in exercises below.

Exercise 2.5.1: Suppose 푈 ⊂ ℝ푛 is a domain. a) Show that 푈 is geometrically convex if and only if it is convex with respect to the aﬃne linear functions. b) Suppose 푈 has smooth boundary. Show that 푈 is convex if and only if it is convex with respect to the smooth convex functions on 푈, that is, with respect to smooth functions with positive semideﬁnite Hessian.

Exercise 2.5.2: Show that every open set 푈 ⊂ ℝ푛 is convex with respect to real polynomials.

Theorem 2.5.2 (Kontinuitätssatz—Continuity principle). Suppose an open set 푈 ⊂ ℂ푛 is convex with respect to plurisubharmonic functions, then given any collection of closed analytic discs 푈 Ð 휕 푈 Ð 푈 Δ훼 ⊂ such that 훼 Δ훼 ⊂⊂ , we have 훼 Δ훼 ⊂⊂ .

Various similar theorems are named the continuity principle. Generally what they have in common is the family of analytic discs whose boundaries stay inside a domain, and whose conclusion has to do with extension of holomorphic functions, or with domains of holomorphy.

Proof. Let 푓 be a plurisubharmonic function on 푈. If 휑훼 : 픻 → 푈 is the holomorphic (in 픻) mapping giving the closed analytic disc, then 푓 ◦ 휑훼 is subharmonic. By the maximum principle, 푓 on Δ훼 must be less than or equal to the supremum of 푓 on 휕Δ훼, so Δ훼 is in the hull of 휕Δ훼. In other Ð Ð 휕 Ð 푈 words 훼 Δ훼 is in the hull of 훼 Δ훼 and therefore 훼 Δ훼 ⊂⊂ by convexity.

Let us illustrate the failure of the continuity principle. If you have discs (denoted by straight line segments) that approach the boundary as in the ﬁgure below, then the domain is not not convex with respect to plurisubharmonic functions. In the diagram, the boundaries of the discs are denoted by the dark dots at the end of the segments. In fact, if we replace discs with line segments, this is the standard convexity, see the exercises below.

Exercise 2.5.3: Suppose 푈 ⊂ ℂ푛 is a domain and 퐾 ⊂⊂ 푈 is a nonempty compact subset. Prove that 푈 \ 퐾 is not convex with respect to plurisubharmonic functions. 2.5. HARTOGS PSEUDOCONVEXITY 75

Exercise 2.5.4: Suppose 푈 ⊂ ℂ푛 is a domain with smooth boundary, 푝 ∈ 휕푈, and Δ is an aﬃne linear analytic disc with 푝 ∈ Δ, but Δ \{푝} ⊂ 푈. Prove that 푈 is not convex with respect to the plurisubharmonic functions.

Exercise 2.5.5: Prove the corresponding Kontinuitätssatz, and its converse, for geometric con- 푛 vexity: Prove that a domain 푈 ⊂ ℝ is geometrically convex if and only if whenever [푥훼, 푦훼] ⊂ 푈 Ð 푥 , 푦 푈 Ð 푥 , 푦 푈 is a collection of straight line segments such that 훼{ 훼 훼} ⊂⊂ implies 훼 [ 훼 훼] ⊂⊂ .

Deﬁnition 2.5.3. Let 푈 ⊂ ℂ푛 be open. An 푓 : 푈 → ℝ is an exhaustion function for 푈 if

푧 ∈ 푈 : 푓 (푧) < 푟 ⊂⊂ 푈 for every 푟 ∈ ℝ.

A domain 푈 ⊂ ℂ푛 is Hartogs pseudoconvex if there exists a continuous plurisubharmonic exhaustion function. The set {푧 ∈ 푈 : 푓 (푧) < 푟} is called the sublevel set of 푓 , or the 푟-sublevel set.

Example 2.5.4: The unit ball 픹푛 is Hartogs pseudoconvex. The continuous function 푧 ↦→ − log1 − k푧k

is an exhaustion function, and it is easy to check directly that it is plurisubharmonic. Example 2.5.5: The entire ℂ푛 is Hartogs pseudoconvex as k푧k2 is a continuous plurisubharmonic exhaustion function. Also, because k푧k2 is plurisubharmonic, then given any 퐾 ⊂⊂ ℂ푛, the hull 퐾b with respect to plurisubharmonic functions must be bounded. In other words, ℂ푛 is convex with respect to plurisubharmonic functions. Theorem 2.5.6. Suppose 푈 ( ℂ푛 is a domain. The following are equivalent: (i) − log 휌(푧) is plurisubharmonic, where 휌(푧) is the distance from 푧 to 휕푈. (ii) 푈 has a continuous plurisubharmonic exhaustion function, that is, 푈 is Hartogs pseudoconvex. (iii) 푈 is convex with respect to plurisubharmonic functions deﬁned on 푈.

Proof. (i) ⇒ (ii) : If 푈 is bounded, the function − log 휌(푧) is clearly a continuous exhaustion function. If 푈 is unbounded, take 푧 ↦→ max{− log 휌(푧), k푧k2}.

(ii) ⇒ (iii) : Suppose 푓 is a continuous plurisubharmonic exhaustion function. If 퐾 ⊂⊂ 푈, then for some 푟 we have 퐾 ⊂ {푧 ∈ 푈 : 푓 (푧) < 푟} ⊂⊂ 푈. But then by deﬁnition of the hull 퐾b we have 퐾b ⊂ {푧 ∈ 푈 : 푓 (푧) < 푟} ⊂⊂ 푈. 푛 (iii) ⇒ (i) : For 푐 ∈ ℂ with k푐k = 1, let 휌푐 (푧) be the radius of the largest aﬃne disc centered at 푧 in the direction 푐 that still lies in 푈. That is, 휌푐 (푧) = sup 휆 > 0 : 푧 + 휁푐 ∈ 푈 for all 휁 ∈ 휆픻 .

As 휌(푧) = inf푐 휌푐 (푧), − log 휌(푧) = sup − log 휌푐 (푧) . k푐k=1 푛 If we prove that for any 푎, 푏 ∈ ℂ the function 휉 ↦→ − log 휌푐 (푎 + 푏휉) is subharmonic, then 휉 ↦→ − log 휌(푎 + 푏휉) is subharmonic, and we are done. Here is the setup, the disc is drawn as a line: 76 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

휌 푧 푧 푧 푐 ( )

휌(푧) 푐

푈푈

Suppose Δ ⊂ ℂ is a disc such that 푎 + 푏휉 ∈ 푈 for all 휉 ∈ Δ. If 푢 is a harmonic function on Δ continuous on Δ such that − log 휌푐 (푎 + 푏휉) ≤ 푢(휉) on 휕Δ, we must show that the inequality holds

on Δ. By Exercise 2.4.13 we may assume 푢 is harmonic on a neighborhood of Δ and so let 푢 = Re 푓 for a holomorphic function 푓 . Fix a 휉 ∈ 휕Δ. We have − log 휌푐 (푎 + 푏휉) ≤ Re 푓 (휉), or in other words − Re 푓 (휉) − 푓 (휉) 휌푐 (푎 + 푏휉) ≥ 푒 = 푒 . − 푓 (휉) Using 휁 = 푡푒 in the deﬁnition of 휌푐 (푎 + 푏휉), the statement above is equivalent to saying that

(푎 + 푏휉) + 푡푒− 푓 (휉)푐 ∈ 푈 for all 푡 ∈ 픻.

This statement holds whenever 휉 ∈ 휕Δ. We must prove that it also holds for all 휉 ∈ Δ. − 푓 (휉) The function 휑푡 (휉) = (푎 + 푏휉) + 푡푒 푐 gives a closed analytic disc with boundary inside 푈. We have a family of analytic discs, parametrized by 푡, whose boundaries are in 푈 for all 푡 with |푡| < 1. For 푡 = 0 the entire disc is inside 푈. As 휑푡 (휉) is continuous in both 푡 and 휉 and Δ is compact, 휑푡 (Δ) ⊂ 푈 for 푡 in some neighborhood of 0. Take 0 < 푡0 < 1 such that 휑푡 (Δ) ⊂ 푈 for all 푡 with |푡| < 푡 . Then 0 Ø Ø 휑푡 (휕Δ) ⊂ 휑푡 (휕Δ) ⊂⊂ 푈,

|푡|<푡0 |푡|≤푡0

because continuous functions take compact sets to compact sets. Kontinuitätssatz implies Ø 휑푡 (Δ) ⊂⊂ 푈.

|푡|<푡0 Ð 휑 (Δ) ⊂⊂ 푈 Ð 휑 (Δ) ⊂⊂ 푈 휖 > By continuity again |푡|≤푡0 푡 , and so |푡|<푡0+휖 푡 for some 0. Consequently − 푓 (휉) 휑푡 (Δ) ⊂ 푈 for all 푡 with |푡| < 1. Thus (푎 + 푏휉) + 푡푒 푐 ∈ 푈 for all 휉 ∈ Δ and all |푡| < 1. This − Re 푓 (휉) implies 휌푐 (푎 + 푏휉) ≥ 푒 for all 휉 ∈ Δ, which in turn implies − log 휌푐 (푎 + 푏휉) ≤ Re 푓 (휉) = 푢(휉) for all 휉 ∈ Δ. Therefore, − log 휌푐 (푎 + 푏휉) is subharmonic.

푛 푛 Exercise 2.5.6: Show that if 푈1 ⊂ ℂ and 푈2 ⊂ ℂ are Hartogs pseudoconvex domains, then so are all the topological components of 푈1 ∩ 푈2. Exercise 2.5.7: Show that if 푈 ⊂ ℂ푛 and 푉 ⊂ ℂ푚 are Hartogs pseudoconvex domains, then so is 푈 × 푉.

Exercise 2.5.8: Show that every domain 푈 ⊂ ℂ is Hartogs pseudoconvex. 2.5. HARTOGS PSEUDOCONVEXITY 77

Ð 푈 Exercise 2.5.9: Show that the union 푗 푗 of a nested sequence of Hartogs pseudoconvex domains 푛 푈푗−1 ⊂ 푈푗 ⊂ ℂ is Hartogs pseudoconvex.

2 2 Exercise 2.5.10: Let ℝ ⊂ ℂ be naturally embedded (that is, it is the set where 푧1 and 푧2 are real). Show that the set ℂ2 \ ℝ2 is not Hartogs pseudoconvex.

Exercise 2.5.11: Let 푈 ⊂ ℂ푛 be a domain and 푓 ∈ O(푈). Prove that 푈0 = 푧 ∈ 푈 : 푓 (푧) ≠ 0

is a Hartogs pseudoconvex domain. Hint: See also Exercise 1.6.2 .

Exercise 2.5.12: Suppose 푈, 푉 ⊂ ℂ푛 are biholomorphic domains. Prove that 푈 is Hartogs pseudoconvex if and only if 푉 is Hartogs pseudoconvex.

2 Exercise 2.5.13: Let 푈 = 푧 ∈ ℂ : |푧1| > |푧2| . a) Prove that 푈 is a Hartogs pseudoconvex domain. b) Find a closed analytic disc Δ in ℂ2 such that 0 ∈ Δ (0 ∉ 푈) and 휕Δ ⊂ 푈. c) What do you think would happen if you tried to move Δ a little bit to avoid the intersection

with the complement? Think about the continuity principle . Compare with Exercise 2.5.4 .

Exercise 2.5.14: Let 푈 ⊂ ℂ푛 be a domain, and 푓 : 푈 → ℝ be a continuous function, plurisubharmonic and negative on 푈, and 푓 = 0 on 휕푈. Prove that 푈 is Hartogs pseudoconvex.

The statement corresponding to Exercise 2.5.9 on nested unions for domains of holomorphy is the Behnke–Stein theorem, which follows using this exercise and the solution of the Levi problem. Although historically Behnke–Stein was proved independently and used to solve the Levi problem.

Exercise 2.5.12 says that (Hartogs) pseudoconvexity is a biholomorphic invariant. That is a good indication that we are looking at a correct notion. It also allows us to change variables to more convenient ones when proving a speciﬁc domain is (Hartogs) pseudoconvex. It is not immediately clear from the deﬁnition, but Hartogs pseudoconvexity is a local property.

Lemma 2.5.7. A domain 푈 ⊂ ℂ푛 is Hartogs pseudoconvex if and only if for every point 푝 ∈ 휕푈 there exists a neighborhood 푊 of 푝 such that 푊 ∩ 푈 is Hartogs pseudoconvex.

Proof. One direction is trivial, so consider the other. Suppose 푝 ∈ 휕푈, and let 푊 be such that 푈 ∩푊 is Hartogs pseudoconvex. Intersection of Hartogs pseudoconvex domains is Hartogs pseudoconvex, so assume 푊 = 퐵푟 (푝). Let 퐵 = 퐵푟/2(푝). If 푞 ∈ 퐵 ∩ 푈, the distance from 푞 to the boundary of 푊 ∩ 푈 is the same as the distance to 휕푈. The setup is illustrated in the following ﬁgure.

푊 푈

푟/2 푞 푝 퐵 78 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

The part of the boundary 휕푈 in 푊 is marked by a thick black line, the part of the boundary of 휕(푊 ∩ 푈) that arises as the boundary of 푊 is marked by a thick gray line. A point 푞 ∈ 퐵 is marked and a ball of radius 푟/2 around 푞 is dotted. No point of distance 푟/2 from 푞 is in 휕푊, and the distance of 푞 to 휕푈 is at most 푟/2 as 푝 ∈ 휕푈 and 푝 is the center of 퐵. Let dist(푥, 푦) denote the euclidean ∗ distance function . Then for 푧 ∈ 퐵 ∩ 푈 − log dist(푧, 휕푈) = − log dist푧, 휕(푈 ∩ 푊). The right-hand side is plurisubharmonic as 푈 ∩ 푊 is Hartogs pseudoconvex. Such a ball 퐵 exists around every 푝 ∈ 휕푈, so near the boundary, − log dist(푧, 휕푈) is plurisubharmonic. If 푈 is bounded, then 휕푈 is compact. So there is some 휖 > 0 such that − log dist(푧, 휕푈) is plurisubharmonic if dist(푧, 휕푈) < 2휖. The function 휑(푧) = max− log dist(푧, 휕푈), − log 휖 is a continuous plurisubharmonic exhaustion function. The proof for unbounded 푈 requires some function of k푧k2 rather than a constant 휖, and is left as an exercise.

Exercise 2.5.15: Finish the proof of the lemma for unbounded domains.

It may seem that we deﬁned a totally diﬀerent concept, but it turns out that Levi and Har- togs pseudoconvexity are one and the same on domains where both concepts make sense. As a consequence of the following theorem we say simply “pseudoconvex” and there is no ambiguity. Theorem 2.5.8. Let 푈 ⊂ ℂ푛 be a domain with smooth boundary. Then 푈 is Hartogs pseudoconvex if and only if 푈 is Levi pseudoconvex. Proof. Suppose 푈 ⊂ ℂ푛 is a domain with smooth boundary that is not Levi pseudoconvex at

푝 ∈ 휕푈. As in Theorem 2.3.11 , change coordinates so that 푝 = 0 and 푈 is deﬁned by

푛−1 2 ∑︁ 2 Im 푧푛 > −|푧1| + 휖푗 |푧푗 | + 푂(3). 푗=2

For a small ﬁxed 휆 > 0, the closed analytic discs deﬁned by 휉 ∈ 픻 ↦→ (휆휉, 0, ··· , 0, 푖푠) are in 푈 for all small enough 푠 > 0. The origin is a limit point of the insides, but not a limit point

of their boundaries. Kontinuitätssatz is not satisﬁed, and 푈 is not convex with respect to the plurisubharmonic functions. Therefore, 푈 is not Hartogs pseudoconvex. Next suppose 푈 is Levi pseudoconvex. Take any 푝 ∈ 휕푈. After translation and rotation by a unitary, assume 푝 = 0 and write a deﬁning function 푟 as 0 0 푟(푧, 푧¯) = 휑(푧 , 푧¯ , Re 푧푛) − Im 푧푛, 0 where 푧 = (푧1, . . . , 푧푛−1). Levi pseudoconvexity says 푛 푛 ∑︁ 휕2푟 ∑︁ 휕푟 푎¯푗 푎ℓ ≥ 0 whenever 푎푗 = 0, (2.4) 휕푧¯ 휕푧 푞 휕푧 푞 푗=1,ℓ=1 푗 ℓ 푗=1 푗

∗If either 푥 and/or 푦 are sets of points, we take the inﬁmum of the euclidean distance over all the points. 2.5. HARTOGS PSEUDOCONVEXITY 79

푞 휕푈 푠 푞 푞 , . . . , 푞 , 푞 푖푠 for all ∈ near 0. Let be a small real constant, and let e = ( 1 푛−1 푛 + ). None of the 휕푟 휕푟 휕2푟 휕2푟 derivatives of 푟 depends on Im 푧푛, and therefore = and = for all 푗 and 휕푧ℓ 푞e 휕푧ℓ 푞 휕푧¯푗 휕푧ℓ 푞e 휕푧¯푗 휕푧ℓ 푞

ℓ. So condition ( 2.4 ) holds for all 푞 ∈ 푈 near 0. We will use 푟 to manufacture a plurisubharmonic exhaustion function, that is one with a semideﬁnite Hessian. Starting with 푟, we already have what we need in all but one direction. Let ∇ 푟| = 휕푟 ,..., 휕푟 denote the gradient of 푟 in the holomorphic directions only. Given 푧 푞 휕푧1 푞 휕푧푛 푞 푛 푞 ∈ 푈 near 0, decompose an arbitrary 푐 ∈ ℂ as 푐 = 푎 + 푏, where 푎 = (푎1, . . . , 푎푛) satisﬁes 푛 ∑︁ 휕푟 푎푗 = 푎, ∇푧푟|푞 = 0. 휕푧 푞 푗=1 푗

Taking the orthogonal decomposition, 푏 is a scalar multiple of ∇푧푟|푞. By Cauchy–Schwarz,

푛 푛 ∑︁ 휕푟 ∑︁ 휕푟 푐푗 = 푏푗 = 푏, ∇푧푟|푞 = k푏kk∇푧푟|푞 k. 휕푧 푞 휕푧 푞 푗=1 푗 푗=1 푗

As ∇푧푟|0 = (0,..., 0, −1/2푖), then for 푞 suﬃciently near 0 we have that k∇푧푟|푞 k ≥ 1/3, and

푛 푛 1 ∑︁ 휕푟 ∑︁ 휕푟 k푏k = 푐푗 ≤ 3 푐푗 . k∇ 푟| k 휕푧 푞 휕푧 푞 푧 푞 푗=1 푗 푗=1 푗 As 푐 = 푎 + 푏 is the orthogonal decomposition, k푐k ≥ k푏k. The complex Hessian matrix of 푟 is continuous, and so let 푀 ≥ 0 be an upper bound on its operator norm for 푞 near the origin. Again using Cauchy–Schwarz 푛 푛 ∑︁ 휕2푟 ∑︁ 휕2푟 푐¯푗 푐ℓ = (푎¯푗 + 푏¯푗 )(푎ℓ + 푏ℓ) 휕푧¯ 휕푧 푞 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ 푗=1,ℓ=1 푗 ℓ 푛 ∑︁ 휕2푟 = 푎¯푗 푎ℓ 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ 푛 푛 푛 ∑︁ 휕2푟 ∑︁ 휕2푟 ∑︁ 휕2푟 + 푏¯푗 푐ℓ + 푐¯푗 푏ℓ + 푏¯푗 푏ℓ 휕푧¯ 휕푧 푞 휕푧¯ 휕푧 푞 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ 푗=1,ℓ=1 푗 ℓ 푗=1,ℓ=1 푗 ℓ 푛 휕2푟 ∑︁ 2 ≥ 푎¯푗 푎ℓ − 푀k푏kk푐k − 푀k푐kk푏k − 푀k푏k 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ ≥ −3푀k푐kk푏k. Together with what we know about k푏k, for 푞 ∈ 푈 near the origin

푛 휕2푟 푛 휕푟 ∑︁ 2 ∑︁ 푐¯푗 푐ℓ ≥ −3푀k푐kk푏k ≥ −3 푀k푐k 푐푗 . 휕푧¯ 휕푧 푞 휕푧 푞 푗=1,ℓ=1 푗 ℓ 푗=1 푗 For 푧 ∈ 푈 sufficiently close to 0 define 푓 (푧) = − log−푟(푧) + 퐴k푧k2, 80 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY where 퐴 > 0 is some constant we will choose later. The log is there to make 푓 blow up as we approach the boundary. The 퐴k푧k2 is there to add a constant diagonal matrix to the complex Hessian of 푓 , which we hope is enough to make it positive semidefinite at all 푧 near 0. Compute: 휕2 푓 1 휕푟 휕푟 1 휕2푟 = − + 퐴훿ℓ, 2 푗 휕푧¯푗 휕푧ℓ 푟 휕푧¯푗 휕푧ℓ 푟 휕푧¯푗 휕푧ℓ ℓ ∗

훿 푓 푐 푞 푈 where 푗 is the Kronecker delta . Apply the complex Hessian of to at ∈ near the origin (recall that 푟 is negative on 푈 and so for 푞 ∈ 푈, −푟 = |푟|): 2 푛 휕2 푓 푛 휕푟 푛 휕2푟 ∑︁ 1 ∑︁ 1 ∑︁ 2 푐¯푗 푐ℓ = 푐ℓ + 푐¯푗 푐ℓ + 퐴k푐k 휕푧¯ 휕푧 푞 푟2 휕푧 푞 |푟| 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ ℓ=1 ℓ 푗=1,ℓ=1 푗 ℓ 2 푛 휕푟 2푀 푛 휕푟 1 ∑︁ 3 ∑︁ 2 ≥ 푐ℓ − k푐k 푐푗 + 퐴k푐k . 푟2 휕푧 푞 |푟| 휕푧 푞 ℓ=1 ℓ 푗=1 푗 Now comes a somewhat funky trick. As a quadratic polynomial in k푐k, the right-hand side of the inequality is always nonnegative if 퐴 > 0 and if the discriminant is negative or zero. Let us set the discriminant to zero: 푛 !2 푛 2 32푀 ∑︁ 휕푟 1 ∑︁ 휕푟 0 = 푐푗 − 4퐴 푐ℓ . |푟| 휕푧 푞 푟2 휕푧 푞 푗=1 푗 ℓ=1 ℓ 퐴 34 푀2 퐴 All the nonconstant terms go away and = 4 makes the discriminant zero. Thus for that , 푛 ∑︁ 휕2 푓 푐¯푗 푐ℓ ≥ 0. 휕푧¯ 휕푧 푞 푗=1,ℓ=1 푗 ℓ In other words, the complex Hessian of 푓 is positive semideﬁnite at all points 푞 ∈ 푈 near 0. The function 푓 (푧) goes to inﬁnity as 푧 approaches 휕푈. So for every 푡 ∈ ℝ, the 푡-sublevel set (the set where 푓 (푧) < 푡) must be a positive distance away from 휕푈 near 0. We have constructed a local continuous plurisubharmonic exhaustion function for 푈 near 푝. If we intersect with a small ball 퐵 centered at 푝, then we get that 푈 ∩ 퐵 is Hartogs pseudoconvex. This is true at all 푝 ∈ 휕푈, so 푈 is Hartogs pseudoconvex.

2.6 Holomorphic convexity

Deﬁnition 2.6.1. Let 푈 ⊂ ℂ푛 be a domain. For a set 퐾 ⊂ 푈, deﬁne the holomorphic hull

def n o 퐾b푈 = 푧 ∈ 푈 : | 푓 (푧)| ≤ sup| 푓 (푤)| for all 푓 ∈ O(푈) . 푤∈퐾

A domain 푈 is holomorphically convex if whenever 퐾 ⊂⊂ 푈, then 퐾b푈 ⊂⊂ 푈. In other words, 푈 is † holomorphically convex if it is convex with respect to moduli of holomorphic functions on 푈.

∗ ℓ ℓ Recall 훿푗 = 0 if 푗 ≠ ℓ and 훿푗 = 1 if 푗 = ℓ. † Sometimes simply 퐾b is used, but we use 퐾b푈 to emphasize the dependence on 푈. 2.6. HOLOMORPHIC CONVEXITY 81

It is a simple exercise (see below) to show that a holomorphically convex domain is Hartogs pseudoconvex. We will prove that holomorphic convexity is equivalent to being a domain of holomorphy. That a Hartogs pseudoconvex domain is holomorphically convex is the Levi problem for Hartogs pseudoconvex domains and is considerably more diﬃcult. The thing is, there are lots of plurisubharmonic functions, and they are easy to construct; we can even construct them locally, and then piece them together by taking maxima. There are far fewer holomorphic functions, and we cannot just construct them locally and expect the pieces to somehow ﬁt together. As it is so fundamental, let us state it as a theorem.

Theorem 2.6.2 (Solution of the Levi problem). A domain 푈 ⊂ ℂ푛 is holomorphically convex if and only it is Hartogs pseudoconvex.

Proof. The forward direction follows from an exercise below. We skip the proof of the backward

direction in order to save some hundred pages or so. See Hörmander’s book [ H ] for the proof.

Exercise 2.6.1: Prove that a holomorphically convex domain is Hartogs pseudoconvex. See

Exercise 2.4.22 .

Exercise 2.6.2: Prove that every domain 푈 ⊂ ℂ is holomorphically convex by giving a topological description of 퐾b푈 for any compact 퐾 ⊂⊂ 푈. Hint: Runge may be useful. Exercise 2.6.3: Suppose 푓 : ℂ푛 → ℂ is holomorphic and 푈 is a topological component of 푧 ∈ ℂ푛 : | 푓 (푧)| < 1 . Prove that 푈 is a holomorphically convex domain.

푛 Exercise 2.6.4: Compute the hull 퐾b픻푛 of the set 퐾 = 푧 ∈ 픻 : |푧푗 | = 휆푗 for 푗 = 1, . . . , 푛 , where 0 ≤ 휆푗 < 1. Prove that the unit polydisc is holomorphically convex. Exercise 2.6.5: Prove that a geometrically convex domain 푈 ⊂ ℂ푛 is holomorphically convex.

Exercise 2.6.6: Prove that the Hartogs ﬁgure (see Theorem 2.1.4 ) is not holomorphically convex.

Exercise 2.6.7: Let 푈 ⊂ ℂ푛 be a domain and 푓 ∈ O(푈). Show that if 푈 is holomorphically convex, then 푈e = 푧 ∈ 푈 : 푓 (푧) ≠ 0 is holomorphically convex. Hint: First see Exercise 1.6.2 . Exercise 2.6.8: Suppose 푈, 푉 ⊂ ℂ푛 are biholomorphic domains. Prove that 푈 is holomorphically convex if and only if 푉 is holomorphically convex.

Exercise 2.6.9: In the deﬁnition of holomorphic hull of 퐾, replace 푈 with ℂ푛 and O(푈) with holomorphic polynomials on ℂ푛, to get the polynomial hull of 퐾. Prove that the polynomial hull 푛 of 퐾 ⊂⊂ ℂ is the same as the holomorphic hull 퐾bℂ푛 . Exercise 2.6.10:

a) Prove the Hartogs triangle 푇 (see Exercise 2.1.7 ) is holomorphically convex. b) Prove 푇 ∪ 퐵휖 (0) (for a small enough 휖 > 0) is not holomorphically convex.

푛 푛 Exercise 2.6.11: Show that if domains 푈1 ⊂ ℂ and 푈2 ⊂ ℂ are holomorphically convex, then so are all the topological components of 푈1 ∩ 푈2. 82 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Exercise 2.6.12: Let 푛 ≥ 2. a) Let 푈 ⊂ ℂ푛 be a domain and 퐾 ⊂⊂ 푈 a nonempty compact subset. Show that 푈 \ 퐾 is not holomorphically convex. b) Let 푈 ⊂ ℂ푛 be a bounded holomorphically convex domain. Prove that ℂ푛 \ 푈 is connected. c) Find an unbounded holomorphically convex domain 푈 ⊂ ℂ푛 where ℂ푛 \ 푈 is disconnected. Ð 푈 Exercise 2.6.13 (Behnke–Stein again): Show that the union 푗 푗 of a nested sequence of holo- 푛 morphically convex domains 푈푗−1 ⊂ 푈푗 ⊂ ℂ is holomorphically convex.

Using an exercise above we see that ℂ푛 is both holomorphically convex and a domain of holomorphy. In fact, these two notions are equivalent for all other domains in ℂ푛.

Theorem 2.6.3 (Cartan–Thullen). Let 푈 ( ℂ푛 be a domain. The following are equivalent:

(i) 푈 is a domain of holomorphy.

(ii) For all 퐾 ⊂⊂ 푈, dist(퐾, 휕푈) = dist(퐾b푈, 휕푈). (iii) 푈 is holomorphically convex.

Proof. Let us start with (i) ⇒ (ii) . Suppose there is a 퐾 ⊂⊂ 푈 with dist(퐾, 휕푈) > dist(퐾b푈, 휕푈). After possibly a rotation by a unitary, there exists a point 푝 ∈ 퐾b푈 and a polydisc Δ = Δ푟 (0) with polyradius 푟 = (푟1, . . . , 푟푛) such that 푝 + Δ = Δ푟 (푝) contains a point of 휕푈, but Ø 퐾 + Δ = Δ푟 (푞) ⊂⊂ 푈. 푞∈퐾

휕푈

푝

퐾 Δ푟 (푝)

퐾 + Δ

If 푓 ∈ O(푈), then there is an 푀 > 0 such that | 푓 | ≤ 푀 on 퐾 + Δ as that is a relatively compact set. By the Cauchy estimates for each 푞 ∈ 퐾 we get

훼 휕 푓 푀훼! (푞) ≤ . 휕푧훼 푟훼 2.6. HOLOMORPHIC CONVEXITY 83

This inequality therefore holds on 퐾b푈 and hence at 푝. The series

∑︁ 1 휕훼 푓 (푝)(푧 − 푝)훼 훼 휕푧훼 훼 !

converges in Δ푟 (푝). Hence 푓 extends to all of Δ푟 (푝) and Δ푟 (푝) contains points outside of 푈, in other words, 푈 is not a domain of holomorphy.

The implication (ii) ⇒ (iii) is immediate.

Finally, we prove (iii) ⇒ (i) . Suppose 푈 is holomorphically convex. Let 푝 ∈ 휕푈. By convexity 퐾 퐾 푈 Ð 퐾 푈 퐾 퐾 choose nested compact sets 푗−1 ( 푗 ⊂⊂ such that 푗 푗 = , and (d푗 )푈 = 푗 . As the sets exhaust 푈, we can perhaps pass to a subsequence to ensure that there exists a sequence of points 푝푗 ∈ 퐾푗 \ 퐾푗−1 such that lim푗→∞ 푝푗 = 푝. − 푗 As 푝푗 is not in the hull of 퐾푗−1, there is a function 푓푗 ∈ O(푈) such that | 푓푗 | < 2 on 퐾푗−1, but

푗−1 ∑︁ | 푓 (푝 )| > 푗 + 푓 (푝 ) . 푗 푗 푘 푗 푘=1 푗 Í∞ 푓 푧 Finding such a function is left as an exercise below. For any , the series 푘=1 푘 ( ) converges −푘 uniformly on 퐾푗 as for all 푘 > 푗, | 푓푘 | < 2 on 퐾푗 . As the 퐾푗 exhaust 푈, the series converges uniformly on compact subsets of 푈. Consequently,

∞ ∑︁ 푓 (푧) = 푓푘 (푧) 푘=1 is a holomorphic function on 푈. We bound

푗−1 ∞ ∞ ∑︁ ∑︁ ∑︁ −푘 | 푓 (푝푗 )| ≥ | 푓푗 (푝푗 )| − 푓푘 (푝푗 ) − 푓푘 (푝푗 ) ≥ 푗 − 2 ≥ 푗 − 1.

푘=1 푘= 푗+1 푘= 푗+1

푛 So lim푗→∞ 푓 (푝푗 ) = ∞. Clearly there cannot be any open 푊 ⊂ ℂ containing 푝 to which 푓 extends

(see deﬁnition of domain of holomorphy ). As any connected open 푊 such that 푊 \ 푈 ≠ ∅ must contain a point of 휕푈, we are done.

By Exercise 2.6.8 , holomorphic convexity is a biholomorphic invariant. Thus, being a domain of holomorphy is also a biholomorphic invariant. This fact is not easy to prove from the deﬁnition of a domain of holomorphy, as the biholomorphism is deﬁned only on the interior of our domains. Holomorphic convexity is an intrinsic notion; it does not require knowing anything about points outside of 푈. It is a much better way to think about domains of holomorphy. Holomorphic ∗ convexity generalizes easily to more complicated complex manifolds , while the notion of a domain of holomorphy only makes sense for domains in ℂ푛.

∗Manifolds with complex structure, that is, “manifolds with multiplication by 푖 on the tangent space.” 84 CHAPTER 2. CONVEXITY AND PSEUDOCONVEXITY

Exercise 2.6.14: Prove the existence of the function 푓푗 ∈ O(푈) as indicated in the proof of Cartan–Thullen above.

Exercise 2.6.15: Show that if 푈 ⊂ ℂ푛 is holomorphically convex, then there exists a single function 푓 ∈ O(푈) that does not extend through any point 푝 ∈ 휕푈.

2 2 Exercise 2.6.16: We know 푈 = ℂ \{푧 ∈ ℂ : 푧1 = 0} is a domain of holomorphy. Use part (ii) of the theorem to show that if 푊 ⊂ ℂ2 is a domain of holomorphy and 푈 ⊂ 푊, then either 푊 = 푈 2 or 푊 = ℂ . Hint: Suppose 퐿 ⊂ 푊 is a complex line and 퐾 is a circle in 퐿. What is 퐾b푊 ?

In the following series of exercises, which you should most deﬁnitely do in order, you will solve the Levi problem (and more) for complete Reinhardt domains. Recall that a domain 푈 is a complete Reinhardt domain if whenever (푧1, . . . , 푧푛) is in 푈 and 푟푗 = |푧푗 |, then the entire closed polydisc Δ푟 (0) ⊂ 푈. We say a complete Reinhardt domain 푈 is logarithmically convex if there exists a 푛 (geometrically) convex 퐶 ⊂ ℝ such that 푧 ∈ 푈 if and only if (log|푧1|,..., log|푧푛|) ∈ 퐶.

Exercise 2.6.17: Prove that a logarithmically convex complete Reinhardt domain is the intersection of sets of the form

푛 푛 훼1 훼푛 훽 푧 ∈ ℂ : 훼1 log|푧1| + · · · + 훼푛 log|푧푛| < 훽 = 푧 ∈ ℂ : |푧1| · · · |푧푛| < 푒

for some nonnegative 훼1, . . . , 훼푛, and 훽 ∈ ℝ. Exercise 2.6.18: Prove that a complete Reinhardt domain that is Hartogs pseudoconvex is logarithmically convex. 푘 ℓ푘 Exercise 2.6.19: For each ∈ ℕ0, let 푚 ∈ ℕ0 be the smallest nonnegative integer such that ℓ푘 푘훼 푚 ≥ 푚. Prove that the domain of convergence of the power series

∞ ∑︁ ℓ푘 푘 푒−푘 훽푧 1 ··· 푧ℓ푛 1 푛 푘=0

푛 훼 훼 훽 is precisely the set 푧 ∈ ℂ : |푧1| 1 · · · |푧푛| 푛 < 푒 . Hint: That it diverges outside is easy, what is 푘 ℓ푚 훼 1 hard is that it converges inside. Perhaps useful is to notice 푘 − 푚 ≤ 푘 , and furthermore notice 훼 훼 훽 that if 푧 is in the set, there is some 휖 > 0 such that (1 + 휖)|푧1| 1 · · · |푧푛| 푛 = 푒 . Exercise 2.6.20: Prove that a logarithmically convex Reinhardt domain is holomorphically convex and therefore it is a domain of holomorphy.

Exercise 2.6.21: Prove that a complete Reinhardt domain is a domain of holomorphy if and only if it is the domain of convergence of some power series at the origin. Hint: There is a function that does not extend past any boundary point of a holomorphically convex domain.

We (you) have proved the following proposition. 2.6. HOLOMORPHIC CONVEXITY 85

Proposition 2.6.4. Let 푈 ⊂ ℂ푛 be a complete Reinhardt domain. Then the following are equivalent:

(i) 푈 is logarithmically convex.

(ii) 푈 is a domain of holomorphy.

(iii) 푈 is a domain of convergence of some power series at the origin.

(iv) 푈 is Hartogs pseudoconvex. Chapter 3

CR functions

3.1 Real-analytic functions and complexiﬁcation

Deﬁnition 3.1.1. Let 푈 ⊂ ℝ푛 be open. A function 푓 : 푈 → ℂ is real-analytic (or simply analytic if clear from context) if at each point 푝 ∈ 푈, the function 푓 has a convergent power series that converges (absolutely) to 푓 in some neighborhood of 푝. A common notation for real-analytic is 퐶휔. Before we discuss the connection to holomorphic functions, we prove a simple lemma. Lemma 3.1.2. Let ℝ푛 ⊂ ℂ푛 be the natural inclusion and 푉 ⊂ ℂ푛 a domain such that 푉 ∩ ℝ푛 ≠ ∅. Suppose 푓 , 푔 : 푉 → ℂ are holomorphic functions such that 푓 = 푔 on 푉 ∩ ℝ푛. Then 푓 = 푔 on 푉. Proof. Considering 푓 − 푔 we may assume that 푔 = 0. Let 푧 = 푥 + 푖푦 as usual so that ℝ푛 is given by 푦 = 0. Our assumption is that 푓 = 0 when 푦 = 0, so the derivative of 푓 with respect to 푥푗 is zero. When 푦 = 0, the Cauchy–Riemann equations say 휕 푓 휕 푓 0 = = −푖 . 휕푥푗 휕푦푗 Therefore, on 푦 = 0, 휕 푓 = 0. 휕푧푗 The derivative 휕 푓 is holomorphic and 휕 푓 = 0 on 푦 = 0. By induction all holomorphic derivatives of 휕푧푗 휕푧푗 푓 at 푝 ∈ ℝ푛 ∩푉 vanish, and 푓 has a zero power series. Hence 푓 is identically zero in a neighborhood of 푝 in ℂ푛 and by the identity theorem it is zero on all of 푉. We return to ℝ푛 for a moment. We write a power series in ℝ푛 in multinomial notation as usual. 푛 Suppose that for some 푎 ∈ ℝ and some polyradius 푟 = (푟1, . . . , 푟푛), the series

∑︁ 훼 푐훼 (푥 − 푎) 훼 converges whenever |푥푗 − 푎푗 | ≤ 푟푗 for all 푗. Here convergence is absolute convergence. That is,

∑︁ 훼 |푐훼 | |푥 − 푎| 훼 3.1. REAL-ANALYTIC FUNCTIONS AND COMPLEXIFICATION 87

converges. If we replace 푥푗 ∈ ℝ with 푧푗 ∈ ℂ such that |푧푗 − 푎푗 | ≤ |푥푗 − 푎푗 |, then the series still converges. Hence the series ∑︁ 훼 푐훼 (푧 − 푎) 훼 푛 converges absolutely in Δ푟 (푎) ⊂ ℂ . Proposition 3.1.3 (Complexification part I). Suppose 푈 ⊂ ℝ푛 is a domain and 푓 : 푈 → ℂ is real-analytic. Let ℝ푛 ⊂ ℂ푛 be the natural inclusion. Then there exists a domain 푉 ⊂ ℂ푛 such that 푈 ⊂ 푉 and a unique holomorphic function 퐹 : 푉 → ℂ such that 퐹|푈 = 푓 . Among many other things that follow from this proposition, we can now conclude that a real- analytic function is 퐶∞ smooth. Be careful and notice that 푈 is a domain in ℝ푛, but it is not an open set when considered as a subset of ℂ푛. Furthermore, 푉 may be a very “thin” neighborhood around 푈. There is no way of finding 푉 just from knowing 푈. You need to also know 푓 . As an 푓 (푥) = 1 휖 > example, consider 휖2+푥2 for 0, which is real-analytic on ℝ, but the complexification is not holomorphic at ±휖푖.

Proof. We proved the local version already. But we must prove that if we extend our 푓 near every point, we always get the same function. That follows from the lemma above; any two such functions are equal on ℝ푛, and hence equal. There is a subtle topological technical point in this, so let us elaborate. A key topological fact is that we define 푉 as a union of the polydiscs where the series converges. If a point 푝 is in two different such polydiscs, we need to show that the two definitions of 퐹 are the same at 푝. But the intersection of two polydiscs is always connected, and in this case contains a piece of ℝ푛 as well, and we may apply the lemma above.

Exercise 3.1.1: Prove the identity theorem for real-analytic functions. That is, if 푈 ⊂ ℝ푛 is a domain, 푓 : 푈 → ℝ a real-analytic function and 푓 is zero on a nonempty open subset of 푈, then 푓 is identically zero.

Exercise 3.1.2: Suppose 푈 ⊂ ℝ푛 is a domain and 푓 : 푈 → ℝ a real-analytic function. Suppose that 푊 ⊂ 푈 is a nonempty open subset and 푓 |푊 is harmonic. Prove that 푓 is harmonic. Exercise 3.1.3: Let (0, 1) ⊂ ℝ. Construct a real-analytic function on (0, 1) that does not complexify to the rectangle (0, 1) + 푖(−휖, 휖) ⊂ ℂ for any 휖 > 0. Why does this not contradict the proposition?

A polynomial 푃(푥) in 푛 real variables (푥1, . . . , 푥푛) is homogeneous of degree 푑 if 푃(푠푥) = 푠푑 푃(푥) for all 푠 ∈ ℝ and 푥 ∈ ℝ푛. A homogeneous polynomial of degree 푑 is a polynomial whose every monomial is of total degree 푑. If 푓 is real-analytic near 푎 ∈ ℝ푛, then write the power series of 푓 at 푎 as ∞ ∑︁ 푓푗 (푥 − 푎), 푗=0 where 푓푗 is a homogeneous polynomial of degree 푗. The 푓푗 is called the degree 푗 homogeneous part of 푓 at 푎. 88 CHAPTER 3. CR FUNCTIONS

There is usually a better way to complexify real-analytic functions in ℂ푛. Suppose 푈 ⊂ ℂ푛 ℝ2푛, and 푓 : 푈 → ℂ is real-analytic. Assume 푎 = 0 ∈ 푈 for simplicity. Writing 푧 = 푥 + 푖푦, near 0,

∞ ∞ ∑︁ ∑︁ 푧 + 푧¯ 푧 − 푧¯ 푓 (푥, 푦) = 푓 (푥, 푦) = 푓 , . 푗 푗 2 2푖 푗=0 푗=0

The polynomial 푓푗 becomes a homogeneous polynomial of degree 푗 in the variables 푧 and 푧¯. The series becomes a power series in 푧 and 푧¯. As mentioned before, we simply write the function as 푓 (푧, 푧¯), and we consider the power series representation in 푧 and 푧¯ rather than in 푥 and 푦. In multinomial notation, we write a power series at 푎 ∈ ℂ푛 as

∑︁ 훼 훽 푐훼,훽 (푧 − 푎) (푧¯ − 푎¯) . 훼,훽

A holomorphic function is real-analytic, but not vice versa. A holomorphic function is a real-analytic function that does not depend on 푧¯. Before we discuss complexiﬁcation in terms of 푧 and 푧¯, we need the following lemma.

Lemma 3.1.4. Let 푉 ⊂ ℂ푛 × ℂ푛 be a domain, let the coordinates be (푧, 휁) ∈ ℂ푛 × ℂ푛, let

퐷 = (푧, 휁) ∈ ℂ푛 × ℂ푛 : 휁 = 푧¯ ,

and suppose 퐷 ∩ 푉 ≠ ∅. Suppose 푓 , 푔 : 푉 → ℂ are holomorphic functions such that 푓 = 푔 on 퐷 ∩ 푉. Then 푓 = 푔 on all of 푉.

The set 퐷 is sometimes called the diagonal.

Proof. Again assume without loss of generality that 푔 = 0. Whenever (푧, 푧¯) ∈ 푉, we have 푓 (푧, 푧¯) = 0, which is really 푓 composed with the map that takes 푧 to (푧, 푧¯). Using the chain rule

휕 h i 휕 푓 0 = 푓 (푧, 푧¯) = (푧, 푧¯). 휕푧¯푗 휕휁푗

Let us do this again with the 푧푗

휕 h i 휕 푓 0 = 푓 (푧, 푧¯) = (푧, 푧¯). 휕푧푗 휕푧푗

Either way, we get another holomorphic function in 푧 and 휁 that is zero on 퐷. By induction, for all 훼 and 훽 we get 휕|훼|+|훽| h i 휕|훼|+|훽| 푓 0 = 푓 (푧, 푧¯) = (푧, 푧¯). 휕푧훼휕푧¯훽 휕푧훼휕휁 훽 All holomorphic derivatives in 푧 and 휁 of 푓 are zero on every point (푧, 푧¯), so the power series is zero at every point (푧, 푧¯), and so 푓 is identically zero in a neighborhood of any point (푧, 푧¯). The lemma follows by the identity theorem. 3.1. REAL-ANALYTIC FUNCTIONS AND COMPLEXIFICATION 89

Let 푓 be a real-analytic function. Suppose the series (in multinomial notation)

∑︁ 훼 훽 푓 (푧, 푧¯) = 푐훼,훽 (푧 − 푎) (푧¯ − 푎¯) 훼,훽

푛 converges in a polydisc Δ푟 (푎) ⊂ ℂ . By convergence we mean absolute convergence as we discussed before. That is, ∑︁ 훼 훽 |푐훼,훽 | |푧 − 푎| |푧¯ − 푎¯| 훼,훽

converges. The series still converges if we replace 푧¯푗 with 휁푗 where |휁푗 − 푎¯| ≤ |푧¯푗 − 푎¯|. So the series

∑︁ 훼 훽 퐹(푧, 휁) = 푐훼,훽 (푧 − 푎) (휁 − 푎¯) 훼,훽

converges for all (푧, 휁) ∈ Δ푟 (푎) × Δ푟 (푎¯). Putting together the discussion above with the lemma we obtain:

Proposition 3.1.5 (Complexiﬁcation part II). Suppose 푈 ⊂ ℂ푛 is a domain and 푓 : 푈 → ℂ is real-analytic. Then there exists a domain 푉 ⊂ ℂ푛 × ℂ푛 such that

(푧, 휁) : 휁 = 푧¯ and 푧 ∈ 푈 ⊂ 푉,

and a unique holomorphic function 퐹 : 푉 → ℂ such that 퐹(푧, 푧¯) = 푓 (푧, 푧¯) for all 푧 ∈ 푈.

The function 푓 can be thought of as the restriction of 퐹 to the set where 휁 = 푧¯. We will abuse notation and write simply 푓 (푧, 휁) both for 푓 and its extension. The reason for this abuse is evident from the computations above. What we are calling 푓 is a function of (푧, 푧¯) if thinking of it as a function on the diagonal where 휁 = 푧¯, or it is a function of 푧 if thinking of it as just the function 푧 ↦→ 푓 (푧, 푧¯), or it is the function (푧, 휁) ↦→ 푓 (푧, 휁). We have the following commutative diagram:

푧 ↦→ (푧,푧¯) 푈 ⊂ ℂ푛 푉 ⊂ ℂ푛 × ℂ푛

푓 푓 (=퐹) ℂ

All three ways of going from one place to another in the diagram we are calling 푓 . The arrow from 푉 was called 퐹 in the proposition. The notation plays well with differentiation and the Wirtinger operators. Differentiating 푓 (really the 퐹 in the proposition) in 휁푗 and evaluating at (푧, 푧¯) is the same thing as evaluating at (푧, 푧¯) and then differentiating in 푧¯푗 using the Wirtinger operator:

휕퐹 휕 푓 휕 h i 휕 푓 (푧, 푧¯) = (푧, 푧¯) = 푓 (푧, 푧¯) = (푧, 푧¯). 휕휁푗 휕휁푗 휕푧¯푗 휕푧¯푗

If we squint our mind’s eye, we can’t quite see the diﬀerence between 푧¯ and 휁. We have already used this idea for smooth functions, but for real-analytic functions we can treat 푧 and 푧¯ as truly independent variables. The abuse of notation is entirely justiﬁed, at least once it is understood well. 90 CHAPTER 3. CR FUNCTIONS

Remark 3.1.6. The domain 푉 in the proposition is not simply 푈 times the conjugate of 푈. In general, it is much smaller. For example, a real-analytic 푓 : ℂ푛 → ℂ does not necessarily complexify to all of ℂ푛 × ℂ푛. That is because the domain of convergence for a real-analytic function on ℂ푛 is not necessarily all of ℂ푛. In one dimension, the function 1 푓 (푧, 푧¯) = 1 + |푧|2 is real-analytic on ℂ, but it is not a restriction to the diagonal of a holomorphic function on all of ℂ2. The problem is that the complexified function 1 푓 (푧, 휁) = 1 + 푧휁 is undefined on the set where 푧휁 = −1, which by a fluke never happens when 휁 = 푧¯. Remark 3.1.7. This form of complexification is sometimes called polarization due to its relation to ∗ the polarization identities . That is, suppose 퐴 is a Hermitian matrix, we recover 퐴 and therefore the sesquilinear form h퐴푧, 푤i for 푧, 푤 ∈ ℂ푛, by simply knowing the values of 푛 ∗ ∑︁ h퐴푧, 푧i = 푧 퐴푧 = 푎푗 푘 푧¯푗 푧푘 푗,푘=1 푛 for all 푧 ∈ ℂ . In fact, under the hood Proposition 3.1.5 is polarization in an infinite-dimensional Hilbert space, but we digress. The idea of treating 푧¯ as a separate variable is very powerful, and as we have just seen it is completely natural when speaking about real-analytic functions. This is one of the reasons why real-analytic functions play a special role in several complex variables.

Exercise 3.1.4: Let 푈 ⊂ ℂ푛 be an open set and 휑 : 푈 → ℝ a pluriharmonic function. Prove that 휑 is real-analytic.

푛 Exercise 3.1.5: Let 푈 ⊂ ℂ be an open set, 푧0 ∈ 푈. Suppose 휑 : 푈 → ℝ is a pluriharmonic function. You know that 휑 is real-analytic. Using complexiﬁcation, write down a formula for a holomorphic function near 푧0 whose real part is 휑. Exercise 3.1.6: Let 푈 ⊂ ℂ푛 be a domain, and suppose 푓 , 푔 ∈ O(푈). Suppose that 푓 = 푔¯ on 푈. Use complexiﬁcation (complexify 푓 − 푔¯) to show that both 푓 and 푔 are constant.

Example 3.1.8: Not every 퐶∞ smooth function is real-analytic. For 푥 ∈ ℝ, deﬁne ( 2 푒−1/푥 if 푥 > 0, 푓 (푥) = 0 if 푥 ≤ 0.

The function 푓 : ℝ → ℝ is 퐶∞ and 푓 (푘) (0) = 0 for all 푘. The Taylor series of 푓 at the origin does not converge to 푓 in any neighborhood of the origin; it converges to the zero function but not to 푓 . Because of this, there is no neighborhood 푉 of the origin in ℂ such that 푓 is the restriction to 푉 ∩ ℝ of a holomorphic function in 푉. ∗Such as 4h푧, 푤i = k푧 + 푤k2 − k푧 − 푤k2 + 푖k푧 + 푖푤k2 − k푧 − 푖푤k2. 3.1. REAL-ANALYTIC FUNCTIONS AND COMPLEXIFICATION 91

Exercise 3.1.7: Prove the statements of the example above.

Deﬁnition 3.1.9. A real hypersurface 푀 ⊂ ℝ푛 is said to be real-analytic if locally at every point it is the graph of a real-analytic function. That is near every point (locally), after perhaps relabeling coordinates 푀 can be written as a graph 푦 = 휑(푥), where 휑 is real-analytic, (푥, 푦) ∈ ℝ푛−1 × ℝ = ℝ푛.

Compare this definition to Definition 2.2.1 . We could define a real-analytic hypersurface as in

Definition 2.2.1 and then prove an analogue of Lemma 2.2.8 to show that this would be identical to the definition above. The definition we gave is sufficient, and so we avoid the complication and leave it to the interested reader.

Exercise 3.1.8: Show that the definition above is equivalent to an analogue of Definition 2.2.1 . That is, state the alternative definition of real-analytic hypersurface and then prove the analogue

of Lemma 2.2.8 .

A mapping to ℝ푚 is real-analytic if all the components are real-analytic functions. Via complexification we give a simple proof of the following result. Proposition 3.1.10. Let 푈 ⊂ ℝ푛, 푉 ⊂ ℝ푘 be open and let 푓 : 푈 → 푉 and 푔 : 푉 → ℝ푚 be real-analytic. Then 푔 ◦ 푓 is real-analytic. Proof. Let 푥 ∈ ℝ푛 be our coordinates in 푈 and 푦 ∈ ℝ푘 be our coordinates in 푉. We complexify 푓 (푥) and 푔(푦) by allowing 푥 to be a complex vector in a small neighborhood of 푈 in ℂ푛 and 푦 to be a complex vector in a small neighborhood of 푉 in ℂ푘 . So we treat 푓 and 푔 as holomorphic functions. On a certain neighborhood of 푈 in ℂ푛, the composition 푓 ◦ 푔 makes sense and it is holomorphic, as composition of holomorphic mappings is holomorphic. Restricting the complexified 푓 ◦ 푔 back to ℝ푛 we obtain a real-analytic function. The proof demonstrates a simple application of complexification. Many properties of holomorphic functions are easy to prove because holomorphic functions are solutions to certain PDE (the Cauchy–Riemann equations). There is no PDE that defines real-analytic functions, so complexification provides a useful tool to transfer certain properties of holomorphic functions to real-analytic functions. We must be careful, however. Hypotheses on real-analytic functions only give us hypotheses on certain points of the complexified holomorphic functions.

Exercise 3.1.9: Demonstrate the point about complexiﬁcation we made just above. Find a nonconstant bounded real-analytic 푓 : ℝ푛 → ℝ that happens to complexify to ℂ푛.

Exercise 3.1.10: Let 푈 ⊂ ℝ푛 be open. Let 휑 : (0, 1) → 푈 be a real-analytic function (curve), and let 푓 : 푈 → ℝ be real-analytic. Suppose that ( 푓 ◦ 휑)(푡) = 0 for all 푡 ∈ (0, 휖) for some 휖 > 0. Prove that 푓 is zero on the image 휑(0, 1). 92 CHAPTER 3. CR FUNCTIONS 3.2 CR functions

We ﬁrst need to know what it means for a function 푓 : 푋 → ℂ to be smooth if 푋 is not an open set, for example, if 푋 is a hypersurface.

Deﬁnition 3.2.1. Let 푋 ⊂ ℝ푛 be a set. The function 푓 : 푋 → ℂ is smooth (resp. real-analytic) if for each point 푝 ∈ 푋 there is a neighborhood 푈 ⊂ ℝ푛 of 푝 and a smooth (resp. real-analytic) 퐹 : 푈 → ℂ such that 퐹(푞) = 푓 (푞) for 푞 ∈ 푋 ∩ 푈.

For an arbitrary set 푋, issues surrounding this deﬁnition can be rather subtle. The deﬁnition is easy to work with, however, if 푋 is nice, such as a hypersurface, or if 푋 is a closure of a domain with smooth boundary.

Proposition 3.2.2. Suppose 푀 ⊂ ℝ푛 is a smooth (resp. real-analytic) real hypersurface. A function 푓 : 푀 → ℂ is smooth (resp. real-analytic) if and only if whenever near any point we write 푀 in coordinates (푥, 푦) ∈ ℝ푛−1 × ℝ as 푦 = 휑(푥) for a smooth (resp. real-analytic) function 휑, then the function 푓 푥, 휑(푥) is a smooth (resp. real- analytic) function of 푥.

Exercise 3.2.1: Prove the proposition.

Exercise 3.2.2: Prove that if 푀 is a smooth or real-analytic hypersurface, and 푓 : 푀 → ℂ is smooth or real-analytic, then the function 퐹 from the deﬁnition is never unique, even for a ﬁxed neighborhood 푈.

Exercise 3.2.3: Suppose 푀 ⊂ ℝ푛 is a smooth hypersurface, 푓 : 푀 → ℂ is a smooth function, 푝 ∈ 푀, and 푋푝 ∈ 푇푝 푀. Prove that 푋푝 푓 is well-deﬁned. That is, suppose 푈 is a neighborhood of 푝, 퐹 : 푈 → ℂ and 퐺 : 푈 → ℂ are smooth functions that both equal 푓 on 푈 ∩ 푀. Prove that 푋푝 퐹 = 푋푝퐺.

Due to the last exercise, we can apply vectors of 푇푝 푀 to a smooth function on a hypersurface by simply applying them to any smooth extension. We can similarly apply vectors of ℂ ⊗ 푇푝 푀 to smooth functions on 푀, as ℂ ⊗ 푇푝 푀 is simply the complex span of vectors in 푇푝 푀.

Definition 3.2.3. Let 푀 ⊂ ℂ푛 be a smooth real hypersurface. Then a smooth function 푓 : 푀 → ℂ is a smooth CR function if 푋푝 푓 = 0 (0,1) for all 푝 ∈ 푀 and all vectors 푋푝 ∈ 푇푝 푀. Remark 3.2.4. One only needs one derivative (rather than 퐶∞) in the definition above. One can even define a continuous CR function if the derivative is taken in the distribution sense, but we digress. (0,1) Remark 3.2.5. When 푛 = 1, a real hypersurface 푀 ⊂ ℂ is a curve and 푇푝 푀 is trivial. Therefore, all functions 푓 : 푀 → ℂ are CR functions. 3.2. CR FUNCTIONS 93

Proposition 3.2.6. Let 푀 ⊂ 푈 be a smooth (resp. real-analytic) real hypersurface in an open 푛 푈 ⊂ ℂ . Suppose 퐹 : 푈 → ℂ is a holomorphic function, then the restriction 푓 = 퐹|푀 is a smooth (resp. real-analytic) CR function.

Proof. First let us prove that 푓 is smooth. The function 퐹 is smooth and defined on a neighborhood of any point, and so it can be used in the definition. Similarly for real-analytic. Let us show 푓 is CR at some 푝 ∈ 푀. Differentiating 푓 with vectors in ℂ ⊗ 푇푝 푀 is the same as (0,1) (0,1) 푛 differentiating 퐹. As 푇푝 푀 ⊂ 푇푝 ℂ , we have

(0,1) 푋푝 푓 = 푋푝 퐹 = 0 for all 푋푝 ∈ 푇푝 푀.

On the other hand, not every smooth CR function is a restriction of a holomorphic function.

Example 3.2.7: Take the smooth function 푓 : ℝ → ℝ we defined before that is not real-analytic at 2 the origin. Take 푀 ⊂ ℂ be the set defined by Im 푧2 = 0. 푀 is a real-analytic real hypersurface. Clearly 푇 (0,1) 푀 is one-complex-dimensional, and at each 푝 ∈ 푀, 휕 is tangent and spans 푇 (0,1) 푀. 푝 휕푧¯1 푝 푝 Define 푔 : 푀 → ℂ by 푔(푧1, 푧2, 푧¯1, 푧¯2) = 푓 (Re 푧2). 2 Then 푔 is CR as it is independent of 푧¯1. If 퐺 : 푈 ⊂ ℂ → ℂ is a holomorphic function where 푈 is some open set containing the origin, then 퐺 restricted to 푀 must be real-analytic (a power series in Re 푧1, Im 푧1, and Re 푧2) and therefore 퐺 cannot equal to 푔 on 푀.

Exercise 3.2.4: Suppose 푀 ⊂ ℂ푛 is a smooth real hypersurface and 푓 : 푀 → ℂ is a CR function that is a restriction of a holomorphic function 퐹 : 푈 → ℂ deﬁned in some neighborhood 푈 ⊂ ℂ푛 of 푀. Show that 퐹 is unique, that is if 퐺 : 푈 → ℂ is another holomorphic function such that 퐺|푀 = 푓 = 퐹|푀 , then 퐺 = 퐹. Exercise 3.2.5: Show that there is no maximum principle of CR functions. In fact, ﬁnd a smooth real hypersurface 푀 ⊂ ℂ푛, 푛 ≥ 2, and a smooth CR function 푓 on 푀 such that | 푓 | attains a strict maximum at a point.

푛 Exercise 3.2.6: Suppose 푀 ⊂ ℂ , 푛 ≥ 2, is the hypersurface given by Im 푧푛 = 0. Show that every smooth CR function on 푀 is holomorphic in the variables 푧1, . . . , 푧푛−1. Use this to show that for no smooth CR function 푓 on 푀 can | 푓 | attain a strict maximum on 푀. But show that there do exist nonconstant functions such that | 푓 | attains a (nonstrict) maximum 푀.

Real-analytic CR functions on a real-analytic hypersurface 푀 always extend to holomorphic functions of a neighborhood of 푀. To prove this we wish to complexify everything, that is treat the 푧s and 푧¯s as separate variables. The standard way of writing a hypersurface as a graph is not as convenient for this setting, so let us prove that for a real-analytic hypersurface, we can write it as a graph of a holomorphic function in the complexiﬁed variables. That is, using variables (푧, 푤), we will write 푀 as a graph of 푤¯ over 푧, 푧¯, and 푤. This allows us to easily eliminate 푤¯ in any real-analytic expression. 94 CHAPTER 3. CR FUNCTIONS

Proposition 3.2.8. Suppose 푀 ⊂ ℂ푛 is a real-analytic hypersurface and 푝 ∈ 푀. Then after a translation and rotation by a unitary matrix, 푝 = 0, and near the origin in coordinates (푧, 푤) ∈ ℂ푛−1 × ℂ, the hypersurface 푀 is given by

푤¯ = Φ(푧, 푧,¯ 푤),

where Φ(푧, 휁, 푤) is a holomorphic function deﬁned on a neighborhood of the origin in ℂ푛−1 × ℂ푛−1 × ℂ, such that Φ, 휕Φ , 휕Φ vanish at the origin for all 푗, and 푤 = Φ¯ 휁, 푧, Φ(푧, 휁, 푤) for all 푧, 휕푧푗 휕휁푗 휁, and 푤. A local basis for 푇 (0,1) 푀 vector ﬁelds is given by

휕 휕Φ 휕 휕 휕Φ 휕 + = + , 푗 = 1, . . . , 푛 − 1. 휕푧¯푗 휕푧¯푗 휕푤¯ 휕푧¯푗 휕휁푗 휕푤¯

Finally, let M be the set in (푧, 휁, 푤, 휔) ∈ ℂ푛−1 × ℂ푛−1 × ℂ × ℂ coordinates given near the origin by 휔 = Φ(푧, 휁, 푤). Then M is the unique complexiﬁcation of 푀 near the origin in the sense that if 푓 (푧, 푧,¯ 푤, 푤¯ ) is a real-analytic function vanishing on 푀 near the origin, then 푓 (푧, 휁, 푤, 휔) vanishes on M near the origin.

Again as a slight abuse of notation Φ refers to both the function Φ(푧, 휁, 푤) and Φ(푧, 푧,¯ 푤).

Proof. Translate and rotate so that 푀 is given by

Im 푤 = 휑(푧, 푧,¯ Re 푤),

휑 푂 푟 푧, 푧, 푤, 푤 푤−푤¯ 휑푧, 푧, 푤+푤¯ where is (2). Write the deﬁning function as ( ¯ ¯ ) = − 2푖 + ¯ 2 . Complexifying, consider 푟(푧, 휁, 푤, 휔) as a holomorphic function of 2푛 variables, and let M be the set deﬁned by 푟(푧, 휁, 푤, 휔) = 0. The derivative of 푟 in 휔 (that is 푤¯ ) does not vanish near the origin. Use the implicit function theorem for holomorphic functions to write M near the origin as

휔 = Φ(푧, 휁, 푤).

Restrict to the diagonal, 푤¯ = 휔 and 푧¯ = 휁, to get 푤¯ = Φ(푧, 푧,¯ 푤). This is order 2 in the 푧 and the 푧¯ since 휑 is 푂(2). Because 푟 is real-valued, then 푟(푧, 푧,¯ 푤, 푤¯ ) = 푟(푧, 푧,¯ 푤, 푤¯ ) = 푟¯(푧,¯ 푧, 푤,¯ 푤). Complexify to obtain 푟(푧, 휁, 푤, 휔) = 푟¯(휁, 푧, 휔, 푤) for all (푧, 휁, 푤, 휔) near the origin. If 푟(푧, 휁, 푤, 휔) = 0, then

0 = 푟(푧, 휁, 푤, 휔) = 푟¯(휁, 푧, 휔, 푤) = 푟(휁,¯ 푧,¯ 휔,¯ 푤¯ ) = 0.

So, (푧, 휁, 푤, 휔) ∈ M if and only if (휁,¯ 푧,¯ 휔,¯ 푤¯ ) ∈ M. Near the origin, (푧, 휁, 푤, 휔) ∈ M if and only if 휔 = Φ(푧, 휁, 푤), and hence if and only if 푤¯ = Φ(휁,¯ 푧,¯ 휔¯). Conjugating, we get that M is also given by 푤 = Φ¯ (휁, 푧, 휔). As 푧, 휁, 푤, Φ(푧, 휁, 푤) ∈ M, then for all 푧, 휁, and 푤,

푤 = Φ¯ 휁, 푧, Φ(푧, 휁, 푤). 3.2. CR FUNCTIONS 95

The vector field 푋 = 휕 + 휕Φ 휕 annihilates the function Φ(푧, 푧,¯ 푤) − 푤¯ , but that is not enough. 푗 휕푧¯푗 휕푧¯푗 휕휔 The vector field must annihilate a real defining function such as the real part of Φ(푧, 푧,¯ 푤) − 푤¯ . So 푋푗 must also annihilate the conjugate Φ¯ (푧,¯ 푧, 푤¯ ) − 푤, at least on 푀. Compute, for (푧, 푤) ∈ 푀,

휕Φ¯ 휕Φ 휕Φ¯ 푋푗 Φ¯ (푧,¯ 푧, 푤¯ ) − 푤 = (푧,¯ 푧, 푤¯ ) + (푧, 푧,¯ 푤) (푧,¯ 푧, 푤¯ ) 휕푧¯푗 휕푧¯푗 휕푤¯ 휕Φ¯ 휕Φ 휕Φ¯ = 푧,¯ 푧, Φ(푧, 푧,¯ 푤¯ ) + (푧, 푧,¯ 푤) 푧,¯ 푧, Φ(푧, 푧,¯ 푤¯ ) 휕푧¯푗 휕푧¯푗 휕푤¯ 휕 h i 휕 h i = Φ¯ 푧,¯ 푧, Φ(푧, 푧,¯ 푤) = 푤 = 0. 휕푧¯푗 휕푧¯푗

The last claim of the proposition is left as an exercise.

Why do we say the last claim in the proposition proves the “uniqueness” of the complexification? Suppose we defined a complexification M0 by another holomorphic equation 푓 = 0. By the claim, M ⊂ M0, at least near the origin. If the derivative 푑푓 is nonzero at the origin, then 푓 푧, 휁, 푤, 푧, 휁, 푤 휕 푓 Φ( ) = 0 implies that 휕휔 is nonzero at the origin. Using the holomorphic implicit function theorem we can uniquely solve 푓 = 0 for 휔 near the origin, that unique solution is Φ, and hence M0 = M near the origin. As an example, recall that the sphere (minus a point) in ℂ2 is biholomorphic to the hypersurface 푤 푧 2 푤−푤¯ 푧푧 푤 휁 휔 given by Im = | | . That is, 2푖 = ¯. Solving for ¯ and using and obtains the equation for the complexification 휔 = −2푖푧휁 + 푤. Then Φ(푧, 휁, 푤) = −2푖푧휁 + 푤, and Φ¯ (휁, 푧, 휔) = 2푖휁푧 + 휔. Let us check that Φ is the right sort of function: Φ¯ 푧, 휁, Φ(푧, 휁, 푤) = 2푖휁푧 + (−2푖푧휁 + 푤) = 푤. The 휕 푖푧 휕 CR vector field is given by 휕푧¯ + 2 휕푤¯ .

Exercise 3.2.7: Finish the proof of the proposition: Let 푀 ⊂ ℂ푛 be a real-analytic hypersurface given by 푤¯ = Φ(푧, 푧,¯ 푤) near the origin, as in the proposition. Let 푓 (푧, 푧,¯ 푤, 푤¯ ) be a real-analytic function such that 푓 = 0 on 푀. Prove that the complexiﬁed 푓 (푧, 휁, 푤, 휔) vanishes on M.

Exercise 3.2.8: In the proposition we only rotated and translated. Sometimes the following change of coordinates is also done. Prove that one can change coordinates (no longer linear) so that the Φ in the proposition is such that Φ(푧, 0, 푤) = Φ(0, 휁, 푤) = 푤 for all 푧, 휁, and 푤. These coordinates are called normal coordinates.

Exercise 3.2.9: Suppose Φ is a holomorphic function defined on a neighborhood of the origin in ℂ푛−1 × ℂ푛−1 × ℂ. a) Show that 푤¯ = Φ(푧, 푧,¯ 푤) defines a real-analytic hypersurface near the origin if and only 푤 = Φ¯ 휁, 푧, Φ(푧, 휁, 푤) for all 푧, 휁, and 푤. Hint: One direction was proved already. b) As an example, show that 푤¯ = 푧푧¯ does not satisfy the condition above, nor does it define a real hypersurface.

Let us prove that real-analytic CR functions on real-analytic hypersurfaces are restrictions of holomorphic functions. To motivate the proof, consider a real-analytic function 푓 on the circle |푧|2 = 푧푧¯ = 1 ( 푓 is vacuously CR). This 푓 is a restriction of a real-analytic function on a 96 CHAPTER 3. CR FUNCTIONS

neighborhood of the circle, that we write 푓 (푧, 푧¯). On the circle 푧¯ = 1/푧. Thus, 퐹(푧) = 푓 푧, 1/푧 is a holomorphic function deﬁned on a neighborhood of the circle and equal to 푓 on the circle. Our

strategy then is to solve for one of the barred variables via Proposition 3.2.8 , and hope the CR conditions take care of the rest of the barred variables in more than one dimension.

Theorem 3.2.9 (Severi). Suppose 푀 ⊂ ℂ푛 is a real-analytic hypersurface and 푝 ∈ 푀. For every real-analytic CR function 푓 : 푀 → ℂ, there exists a holomorphic function 퐹 ∈ O(푈) for a neighborhood 푈 of 푝 such that 퐹(푞) = 푓 (푞) for all 푞 ∈ 푀 ∩ 푈.

Proof. Write 푀 near 푝 as 푤¯ = Φ(푧, 푧,¯ 푤). Let M be the set in the 2푛 variables (푧, 푤, 휁, 휔) given by 휔 = Φ(푧, 휁, 푤). Take 푓 and consider any real-analytic extension of 푓 to a neighborhood of 푝 ∗ and write it 푓 (푧, 푤, 푧,¯ 푤¯ ). Complexify as before to 푓 (푧, 푤, 휁, 휔). On M we have 푓 (푧, 푤, 휁, 휔) = 푓 푧, 푤, 휁, Φ(푧, 휁, 푤). Let 퐹(푧, 푤, 휁) = 푓 푧, 푤, 휁, Φ(푧, 휁, 푤). Clearly 퐹(푧, 푤, 푧¯) equals 푓 on 푀. As 푓 is a CR function, it is annihilated by 휕 + 휕Φ 휕 on 푀. So 휕푧¯푗 휕푧¯푗 휕푤¯ 휕퐹 휕Φ 휕퐹 휕퐹 + = = 0 휕휁푗 휕휁푗 휕휔 휕휁푗

on 푀 ⊂ M. We have a real analytic function 휕퐹 (푧, 푤, 푧¯) that is zero on 푀, so 휕퐹 (푧, 푤, 휁) = 0 on 휕휁푗 휕휁푗 휕퐹 휕퐹

M ( Proposition 3.2.8 again). As is a function only of 푧, 푤, and 휁 (and not of 휔), = 0 for all 휕휁푗 휕휁푗 (푧, 푤, 휁) in a neighborhood of the origin. Consequently, 퐹 does not depend on 휁, and 퐹 is actually a holomorphic function of 푧 and 푤 only and 퐹 = 푓 on 푀. The most important place where we ﬁnd CR functions that aren’t necessarily real-analytic is as boundary values of holomorphic functions.

Proposition 3.2.10. Suppose 푈 ⊂ ℂ푛 is an open set with smooth boundary. Suppose 푓 : 푈 → ℂ is a smooth function, holomorphic on 푈. Then 푓 |휕푈 is a smooth CR function.

Proof. The function 푓 |휕푈 is clearly smooth. (0,1) Suppose 푝 ∈ 휕푈. If 푋푝 ∈ 푇푝 휕푈 is such that

푛 ∑︁ 휕 푋푝 = 푎푗 , 휕푧¯ 푝 푗=1 푗

take {푞푘 } in 푈 that approaches 푝, then take

푛 ∑︁ 휕 푋푞 = 푎푗 . 푘 휕푧¯ 푞 푗=1 푗 푘 푋 푓 푘 푋 푓 Then 푞푘 = 0 for all and by continuity 푝 = 0.

∗At this point 푓 stands for three distinct objects: the function on 푀, its real-analytic extension to a neighborhood in ℂ푛, and its complexiﬁcation to a neighborhood of (푝, 푝¯) in ℂ푛 × ℂ푛. 3.2. CR FUNCTIONS 97

The boundary values of a holomorphic function deﬁne the function uniquely. That is, if two holomorphic functions continuous up to the (smooth) boundary are equal on an open set of the boundary, then they are equal in the domain: Proposition 3.2.11. Suppose 푈 ⊂ ℂ푛 is a domain with smooth boundary and 푓 : 푈 → ℂ is a continuous function, holomorphic on 푈. If 푓 = 0 on a nonempty open subset of 휕푈, then 푓 = 0 on all of 푈.

Proof. Take 푝 ∈ 휕푈 such that 푓 = 0 on a neighborhood of 푝 in 휕푈. Consider a small neighborhood Δ of 푝 such that 푓 is zero on 휕푈 ∩ Δ. Deﬁne 푔 : Δ → ℂ by setting 푔(푧) = 푓 (푧) if 푧 ∈ 푈 and 푔(푧) = 0 otherwise. It is not hard to see that 푔 is continuous, and it is clearly holomorphic where

it is not zero. Radó’s theorem ( Theorem 2.4.10 ) says that 푔 is holomorphic, and as it is zero on a nonempty open subset of Δ, it is identically zero on Δ, meaning 푓 is zero on a nonempty open subset of 푈, and we are done by identity.

푈 푔 = 푓 푝 Δ 푓 = 0 푔 = 0

Exercise 3.2.10: Find a domain 푈 ⊂ ℂ푛, 푛 ≥ 2, with smooth boundary and a smooth CR function 푓 : 휕푈 → ℂ such that there is no holomorphic function on 푈 or ℂ푛 \ 푈 continuous up to the boundary and whose boundary values are 푓 .

Exercise 3.2.11: a) Suppose 푈 ⊂ ℂ푛 is a bounded open set with smooth boundary, 푓 : 푈 → ℂ is a continuous function, holomorphic in 푈, and 푓 |휕푈 is real-valued. Show that 푓 is constant. b) Find a counterexample to the statement if you allow 푈 to be unbounded.

Exercise 3.2.12: Find a smooth CR function on the sphere 푆2푛−1 ⊂ ℂ푛 that is not a restriction of a holomorphic function of a neighborhood of 푆2푛−1.

Exercise 3.2.13: Show a global version of Severi. Given a real-analytic hypersurface 푀 ⊂ ℂ푛 and a real-analytic CR function 푓 : 푀 → ℂ, show that there exists a neighborhood 푈 of 푀, and an 퐹 ∈ O(푈) such that 퐹|푈 = 푓 .

A problem we tackle next is to try to extend a smooth CR function from the boundary of a domain into a holomorphic function inside. This is a PDE problem where the PDE are the Cauchy– Riemann equations, and the function on the boundary is the boundary condition. Cauchy–Riemann equations are overdetermined, that is, there are too many equations. Not every data on the boundary

gives a solution. Proposition 3.2.10 says that the data being CR is a necessary condition for a

solution (it is not suﬃcient in general). Proposition 3.2.11 says that a solution is unique if it exists. 98 CHAPTER 3. CR FUNCTIONS 3.3 Approximation of CR functions

The following theorem (proved circa 1980) holds in much more generality, but we state its simplest version. One of the simpliﬁcations we make is that we consider only smooth CR functions here, although the theorem holds even for continuous CR functions where the CR conditions are interpreted in the sense of distributions.

Theorem 3.3.1 (Baouendi–Trèves). Suppose 푀 ⊂ ℂ푛 is a smooth real hypersurface, 푝 ∈ 푀 is a 푛 point, and 푧 = (푧1, . . . , 푧푛) are the holomorphic coordinates of ℂ . Then there exists a compact neighborhood 퐾 ⊂ 푀 of 푝, such that for every smooth CR function 푓 : 푀 → ℂ, there exists a sequence {푝푗 } of polynomials in 푧 such that

푝푗 (푧) → 푓 (푧) uniformly in 퐾.

A key point is that 퐾 cannot be chosen arbitrarily, it depends on 푝 and 푀. On the other hand it does not depend on 푓 . Given 푀 and 푝 ∈ 푀 there is a 퐾 such that every CR function on 푀 is approximated uniformly on 퐾 by holomorphic polynomials. The theorem applies in one dimension,

although in that case the theorem of Mergelyan (see Theorem B.31 ) is much more general.

Example 3.3.2: Let us show that 퐾 cannot possibly be arbitrary. For simplicity 푛 = 1. Let 푆1 ⊂ ℂ be the unit circle (boundary of the disc), then every smooth function on 푆1 is a smooth CR function. Let 푓 be a nonconstant real function such as Re 푧. Suppose for contradiction that we could take 퐾 = 푆1 in the theorem. Then 푓 (푧) = Re 푧 could be uniformly approximated on 푆1 by holomorphic polynomials. By the maximum principle, the polynomials would converge on 픻 to a holomorphic function on 픻 continuous on 픻. This function would have nonconstant real boundary values, which is impossible. Clearly 퐾 cannot be the entire circle. 푛 1 푛−1 The example is easily extended to ℂ by considering 푀 = 푆 × ℂ , then Re 푧1 is a smooth CR function on 푀 that cannot be approximated uniformly on 푆1 × {0} by holomorphic polynomials.

The technique of the example above will be used later in a more general situation, to extend CR functions using Baouendi–Trèves. Remark 3.3.3. It is important to note the diﬀerence between Baouendi–Trèves (and similar theorems in complex analysis) and the Weierstrass approximation theorem. In Baouendi–Trèves we obtain an approximation by holomorphic polynomials, while Weierstrass gives us polynomials in the real variables, or in 푧 and 푧¯. For example, via Weierstrass, every continuous function is uniformly approximable on 푆1 via polynomials in Re 푧 and Im 푧, and therefore by polynomials in 푧 and 푧¯. These polynomials do not in general converge anywhere but on 푆1.

Exercise 3.3.1: Let 푧 = 푥 + 푖푦 as usual in ℂ. Find a sequence of polynomials in 푥 and 푦 that converge uniformly to 푒푥−푦 on 푆1, but diverge everywhere else.

The proof is an ingenious use of the standard technique used to prove the Weierstrass approximation theorem. Also, as we have seen molliﬁers before, the technique will not be completely foreign even to the reader who does not know the Weierstrass approximation theorem. Basically what we do is use the standard convolution argument, this time against a holomorphic function. 3.3. APPROXIMATION OF CR FUNCTIONS 99

Letting 푧 = 푥 + 푖푦 we only do the convolution in the 푥 variables keeping 푦 = 0. Then we use the fact that the function is CR to show that we get an approximation even for other 푦. In the formulas below, given a vector 푣 = (푣1, . . . , 푣푛), it will be useful to write 푣 2 def 푣2 푣2. [ ] = 1 + · · · + 푛 The following lemma is a neat application of ideas from several complex variables to solve a problem that does not at ﬁrst seems to involve holomorphic functions. Lemma 3.3.4. Let 푊 be the set of 푛 × 푛 complex matrices 퐴 such that

k(Im 퐴)푥k < k(Re 퐴)푥k for all nonzero 푥 ∈ ℝ푛 and Re 퐴 is positive deﬁnite. Then for all 퐴 ∈ 푊,

∫ 2 푒−[퐴푥] det 퐴 푑푥 = 휋푛/2. ℝ푛 Proof. Suppose 퐴 has real entries and 퐴 is positive deﬁnite (so 퐴 is also invertible). By a change of coordinates ∫ ∫ ∫ ∫ −[퐴푥]2 −[푥]2 −푥2 −푥2 √ 푛 푒 det 퐴 푑푥 = 푒 푑푥 = 푒 1 푑푥1 ··· 푒 푛 푑푥푛 = ( 휋) . ℝ푛 ℝ푛 ℝ ℝ Next suppose 퐴 is any matrix in 푊. There is some 휖 > 0 such that k(Im 퐴)푥k2 ≤ (1−휖2)k(Re 퐴)푥k2 for all 푥 ∈ ℝ푛. That is because we only need to check this for 푥 in the unit sphere, which is compact (exercise). By reality of Re 퐴, Im 퐴, and 푥 we get [(Re 퐴)푥]2 = k(Re 퐴)푥k2 and [(Im 퐴)푥]2 = k(Im 퐴)푥k2. So

2 2 2 2 2 2 푒−[퐴푥] 푒− Re [퐴푥] ≤ 푒−[(Re 퐴)푥] +[(Im 퐴)푥] ≤ 푒−휖 [(Re 퐴)푥] . = Therefore, the integral exists for all 퐴 in 푊 by a similar computation as above. The expression ∫ 2 푒−[퐴푥] det 퐴 푑푥 ℝ푛 is a well-deﬁned holomorphic function in the entries of 퐴, thinking of 푊 as a domain (see exercises 2 2 below) in ℂ푛 . We have a holomorphic function that is constantly equal to 휋푛/2 on 푊 ∩ ℝ푛 and hence it is equal to 휋푛/2 everywhere on 푊.

Exercise 3.3.2: Prove the existence of 휖 > 0 in the proof above.

2 Exercise 3.3.3: Show that 푊 ⊂ ℂ푛 in the proof above is a domain (open and connected).

Exercise 3.3.4: Prove that we can really diﬀerentiate under the integral to show that the integral is holomorphic in the entries of 퐴.

Exercise 3.3.5: Show that some hypotheses are needed for the lemma. In particular, take 푛 = 1 and ﬁnd the exact set of 퐴 (now a complex number) for which the conclusion of the lemma is true. 100 CHAPTER 3. CR FUNCTIONS

Given an 푛 × 푛 matrix 퐴, let k퐴k denote the operator norm, k퐴푣k k퐴k = k퐴푣k = . sup sup 푣 k푣k=1 푣∈ℂ푛,푣≠0 k k

Exercise 3.3.6: Let 푊 be as in Lemma 3.3.4 . Let 퐵 be an 푛 × 푛 real matrix such that k퐵k < 1. Show that 퐼 + 푖퐵 ∈ 푊.

We will be using diﬀerential forms, and the following lemma says that as far as the exterior derivative is concerned, all CR functions behave as restrictions of holomorphic functions.

Lemma 3.3.5. Let 푀 ⊂ ℂ푛 be a smooth real hypersurface, 푓 : 푀 → ℂ be a smooth CR function, 푛 and (푧1, . . . , 푧푛) be the holomorphic coordinates of ℂ . Then at each point 푝 ∈ 푀, the exterior derivative 푑푓 is a linear combination of 푑푧1, . . . , 푑푧푛, thinking of 푧1, . . . , 푧푛 as functions on 푀. In particular, 푑( 푓 푑푧) = 푑푓 ∧ 푑푧 = 0.

Recall the notation 푑푧 = 푑푧1 ∧ 푑푧2 ∧ · · · ∧ 푑푧푛. Proof. After a complex aﬃne change of coordinates, we simply need to prove the result at the 휉 , . . . , 휉 푇 (1,0) 푀 origin. Let 1 푛 be the new holomorphic coordinates and suppose the 0 tangent space is spanned by 휕 ,..., 휕 , and such that 휕 is tangent and 휕 is normal. At the origin, 휕휉1 0 휕휉푛−1 0 휕 Re 휉푛 0 휕 Im 휉푛 0 the CR conditions are 휕 푓 (0) = 0 for all 푗, so 휕휉¯푗 휕 푓 휕 푓 휕 푓 푑푓 (0) = (0) 푑휉1(0) + · · · + (0) 푑휉푛−1(0) + (0) 푑(Re 휉푛)(0). 휕휉1 휕휉푛−1 휕 Re 휉푛

Also, at the origin 푑휉푛 (0) = 푑(Re 휉푛)(0) + 푖푑(Im 휉푛)(0) = 푑(Re 휉푛)(0). So 푑푓 (0) is a linear combination of 푑휉1(0), . . . , 푑휉푛 (0). As 휉 is a complex aﬃne function of 푧, then each 푑휉푗 is a linear combination of 푑푧1 through 푑푧푛, and the claim follows. So if 푓 is a CR function, then 푑( 푓 푑푧) = 푑푓 ∧ 푑푧 = 0 since 푑푧푗 ∧ 푑푧푗 = 0. Proof of the theorem of Baouendi–Trèves. Suppose 푀 ⊂ ℂ푛 is a smooth real hypersurface, and without loss of generality suppose 푝 = 0 ∈ 푀. Let 푧 = (푧1, . . . , 푧푛) be the holomorphic coordinates, 0 write 푧 = 푥 + 푖푦, 푦 = (푦 , 푦푛), and suppose 푀 is given by

0 푦푛 = 휓(푥, 푦 ), where 휓 is 푂(2). The variables (푥, 푦0) parametrize 푀 near 0:

0 푧푗 = 푥푗 + 푖푦푗 , for 푗 = 1, . . . , 푛 − 1, and 푧푛 = 푥푛 + 푖휓(푥, 푦 ).

Deﬁne 0 0 휑(푥, 푦 ) = 푦1, . . . , 푦푛−1, 휓(푥, 푦 ) . Write (푥, 푦0) ↦→ 푧 = 푥 + 푖휑(푥, 푦0) as the parametrization. That is, think of 푧 as a function of (푥, 푦0). 3.3. APPROXIMATION OF CR FUNCTIONS 101

Let 푟 > 0 and 푑 > 0 be small numbers to be determined later. Assume they are small enough so that 푓 and 휑 are defined and smooth on some neighborhood of the set where k푥k ≤ 푟 and k푦0k ≤ 푑. 푛 There exists a smooth 푔 : ℝ → [0, 1] such that 푔 ≡ 1 on 퐵푟/2(0) and 푔 ≡ 0 outside of 퐵푟 (0). Explicit formula can be given. Alternatively we obtain such a 푔 by use of mollifiers on a function that is identically one on 퐵3푟/4(0) and zero elsewhere. Such a 푔 is commonly called a cutoff function.

0 ≤ 푔 ≤ 1 푔 ≡ 0 푟/2 푟 푔 ≡ 1

Exercise 3.3.7: Find an explicit formula for 푔 without using molliﬁers.

Let 0 0 0 퐾 = (푥, 푦 ) : k푥k ≤ 푟/4, k푦 k ≤ 푑 . Let 퐾 = 푧(퐾0), that is the image of 퐾0 under the mapping 푧(푥, 푦0). 0 0 Consider the CR function 푓 to be a function of (푥, 푦 ) and write 푓 (푥, 푦 ). For ℓ ∈ ℕ, let 훼ℓ be a diﬀerential 푛-form deﬁned (thinking of 푤 ∈ ℂ푛 as a constant parameter) by

푛/2 ℓ 2 훼 (푥, 푦0) = 푒−ℓ[푤−푧] 푔(푥) 푓 (푥, 푦0) 푑푧 ℓ 휋 푛/2 ℓ 0 2 = 푒−ℓ[푤−푥−푖휑(푥,푦 )] 푔(푥) 푓 (푥, 푦0) 휋 0 (푑푥1 + 푖푑푦1) ∧ · · · ∧ (푑푥푛−1 + 푖푑푦푛−1) ∧ 푑푥푛 + 푖푑휓(푥, 푦 ) .

The key is the exponential, which looks like the bump function molliﬁer, except that now we have 푤 and 푧 possibly complex. The exponential is also holomorphic in 푤, and that will give us entire holomorphic approximating functions. Fix 푦0 with 0 < k푦0k < 푑 and let 퐷 be deﬁned by

퐷 = (푥, 푠) ∈ ℝ푛 × ℝ푛−1 : k푥k < 푟 and 푠 = 푡푦0 for 푡 ∈ (0, 1) .

퐷 is an (푛 + 1)-dimensional “cylinder.” That is, we take a ball in the 푥 directions and then take a single ﬁxed point 푦0 in the 푠 variables and make a cylinder. Here is a rough diagram: 푠 푦0

푡푦0 퐷 푟 푥 102 CHAPTER 3. CR FUNCTIONS

Orient 퐷 in the standard way as if it sat in the (푥, 푡) variables in ℝ푛 × ℝ. Stokes’ theorem says ∫ ∫ 푑훼ℓ (푥, 푠) = 훼ℓ (푥, 푠). 퐷 휕퐷

Since 푔(푥) = 0 if k푥k ≥ 푟, 훼ℓ is zero on the sides of the cylinder 퐷, so the integral over 휕퐷 only needs to consider the top and bottom of the cylinder. And because of 푔, the integral over the top and bottom can be taken over ℝ푛. As is usual in these sorts of arguments, we do the slight abuse of notation where we ignore that 푓 and 휑 are undeﬁned where 푔 is identically zero: ∫ ℓ 푛/2 ∫ −ℓ[푤−푥−푖휑(푥,푦0)]2 0 0 훼ℓ (푥, 푠) = 푒 푔(푥) 푓 (푥, 푦 ) 푑푥1 ∧ · · · ∧ 푑푥푛−1 ∧ 푑푥푛 + 푖푑푥휓(푥, 푦 ) 휕퐷 휋 푥∈ℝ푛 ℓ 푛/2 ∫ −ℓ[푤−푥−푖휑(푥,0)]2 − 푒 푔(푥) 푓 (푥, 0) 푑푥1 ∧ · · · ∧ 푑푥푛−1 ∧ 푑푥푛 + 푖푑푥휓(푥, 0) , 휋 푥∈ℝ푛 (3.1) where 푑 means the derivative in the 푥 directions only. That is, 푑 휓 = 휕휓 푑푥 + · · · + 휕휓 푑푥 . 푥 푥 휕푥1 1 휕푥푛 푛

We will show that as ℓ → ∞, the left-hand side of ( 3.1 ) goes to zero uniformly for 푤 ∈ 퐾 and the first term on the right-hand side goes to 푓 (푥,˜ 푦0) if 푤 = 푧(푥,˜ 푦0) is in 푀. Hence, we define entire functions that we will show approximate 푓 : ℓ 푛/2 ∫ −ℓ[푤−푥−푖휑(푥,0)]2 푓ℓ (푤) = 푒 푔(푥) 푓 (푥, 0) 푑푥1 ∧ · · · ∧ 푑푥푛−1 ∧ 푑푥푛 + 푖푑푥휓(푥, 0) . 휋 푥∈ℝ푛 푛 Clearly each 푓ℓ is holomorphic and defined for all 푤 ∈ ℂ . In the next claim it is important that 푓 is a CR function. Claim 3.3.6. We have 푛/2 ℓ 2 푑훼 (푥, 푠) = 푒−ℓ[푤−푧(푥,푠)] 푓 (푥, 푠) 푑푔(푥) ∧ 푑푧(푥, 푠), ℓ 휋 and for sufficiently small 푟 > 0 and 푑 > 0,

푛/2 ℓ ∫ 2 lim 푒−ℓ[푤−푧(푥,푠)] 푓 (푥, 푠) 푑푔(푥) ∧ 푑푧(푥, 푠) = 0 ℓ→∞ 휋 (푥,푠)∈퐷 0 0 uniformly as a function of 푤 ∈ 퐾 and 푦 ∈ 퐵푑 (0) (recall that 퐷 depends on 푦 ). 2 Proof. The function (푥, 푠) ↦→ 푒−ℓ[푤−푧(푥,푠)] is CR (as a function on 푀), and so is 푓 (푥, 푠). Therefore,

using Lemma 3.3.5 ,

푛/2 ℓ 2 푑훼 (푥, 푠) = 푒−ℓ[푤−푧(푥,푠)] 푓 (푥, 푠) 푑푔(푥) ∧ 푑푧(푥, 푠). ℓ 휋

Since 푑푔 is zero for k푥k ≤ 푟/2, the integral ∫ ℓ 푛/2 ∫ −ℓ[푤−푧(푥,푠)]2 푑훼ℓ (푥, 푠) = 푒 푓 (푥, 푠) 푑푔(푥) ∧ 푑푧(푥, 푠) 퐷 휋 퐷 3.3. APPROXIMATION OF CR FUNCTIONS 103

is only evaluated for the subset of 퐷 where k푥k > 푟/2. Suppose 푤 ∈ 퐾 and (푥, 푠) ∈ 퐷 with k푥k > 푟/2. Let 푤 = 푧(푥,˜ 푠˜). We need to estimate

−ℓ[푤−푧(푥,푠)]2 −ℓ Re [푤−푧(푥,푠)]2 푒 = 푒 . Then − Re [푤 − 푧]2 = −k푥˜ − 푥k2 + k휑(푥,˜ 푠˜) − 휑(푥, 푠)k2. By the mean value theorem k휑(푥,˜ 푠˜) − 휑(푥, 푠)k ≤ k휑(푥,˜ 푠˜) − 휑(푥, 푠˜)k + k휑(푥, 푠˜) − 휑(푥, 푠)k ≤ 푎k푥˜ − 푥k + 퐴k푠˜ − 푠k, where 푎 and 퐴 are 휕휑 휕휑 푎 = sup (푥,ˆ 푦ˆ0) , 퐴 = sup (푥,ˆ 푦ˆ0)) . 휕푥 휕푦0 k푥ˆk≤푟,k푦ˆ0k≤푑 k푥ˆk≤푟,k푦ˆ0k≤푑 휕휑 휕휑 휑 푥 푦0 Here 휕푥 and 휕푦0 are the derivatives (matrices) of with respect to and respectively, and 휕휑 푟 푑 the norm we are taking is the operator norm. Because 휕푥 is zero at the origin, we pick and small enough (and hence 퐾 small enough) so that 푎 ≤ 1/4. We furthermore pick 푑 possibly even 푑 푟 푟 푥 푟 푥 푟 푤 퐾 smaller to ensure that ≤ 32퐴 . We have that /2 ≤ k k ≤ , but k ˜k ≤ /4 (recall ∈ ), so 푟 5푟 ≤ k푥˜ − 푥k ≤ . 4 4 Also, k푠˜ − 푠k ≤ 2푑 by triangle inequality. Therefore, − Re [푤 − 푧(푥, 푠)]2 ≤ −k푥˜ − 푥k2 + 푎2k푥˜ − 푥k2 + 퐴2k푠˜ − 푠k2 + 2푎퐴k푥˜ − 푥kk푠˜ − 푠k −15 퐴 ≤ k푥˜ − 푥k2 + 퐴2k푠˜ − 푠k2 + k푥˜ − 푥kk푠˜ − 푠k 16 2 −푟2 ≤ . 64 In other words, −ℓ[푤−푧(푥,푠)]2 −ℓ푟2/64 푒 ≤ 푒 , or 푛/2 ℓ ∫ 2 2 푒−ℓ[푤−푧(푥,푠)] 푓 (푥, 푠) 푑푔(푥) ∧ 푑푧(푥, 푠) ≤ 퐶ℓ푛/2푒−ℓ푟 /64, 휋 (푥,푠)∈퐷 for some constant 퐶. Note that 퐷 depends on 푦0. The set of all 푦0 with k푦0k ≤ 푑, is a compact set, so we can make 퐶 large enough to not depend on the 푦0 that was chosen. The claim follows. Claim 3.3.7. For the given 푟 > 0 and 푑 > 0,

ℓ 푛/2 ∫ −ℓ[푥˜+푖휑(푥,푦˜ 0)−푥−푖휑(푥,푦0)]2 0 0 lim 푒 푔(푥) 푓 (푥, 푦 )푑푥1 ∧ · · · ∧ 푑푥푛−1 ∧ 푑푥푛 + 푖푑푥휓(푥, 푦 ) ℓ→∞ 휋 푥∈ℝ푛 = 푓 (푥,˜ 푦0) uniformly in (푥,˜ 푦0) ∈ 퐾0. 104 CHAPTER 3. CR FUNCTIONS

That is, we look at ( 3.1 ) and we plug in 푤 = 푧(푥,˜ 푦 ) ∈ 퐾. The 푔 (as usual) makes sure we never evaluate 푓 , 휓, or 휑 at points where they are not deﬁned. Proof. The change of variables formula implies 휕푧 푑푥 ∧ · · · ∧ 푑푥 ∧ 푑푥 + 푖푑 휓(푥, 푦0) = 푑 푧(푥, 푦0) = (푥, 푦0) 푑푥, 1 푛−1 푛 푥 푥 det 휕푥

휕푧 푥, 푦0 푧 where 휕푥 ( ) is the matrix corresponding to the derivative of the mapping with respect to the 푥 푥, 푦0 variables evaluated at ( ). √ Let us change variables of integration via 휉 = ℓ(푥 − 푥˜):

푛/2 ℓ ∫ 0 0 2 휕푧 푒−ℓ[푥˜+푖휑(푥,푦˜ )−푥−푖휑(푥,푦 )] 푔(푥) 푓 (푥, 푦0) det (푥, 푦0) 푑푥 = 휋 푥∈ℝ푛 휕푥 / √ 2 푛 2 ∫ h 휉 0 0 i 1 − 휉+푖 ℓ 휑 푥˜+ √ ,푦 −휑(푥,푦˜ ) 휉 휉 0 휕푧 휉 0 푒 ℓ 푔 푥˜ + √ 푓 푥˜ + √ , 푦 det 푥˜ + √ , 푦 푑휉. 휋 휉∈ℝ푛 ℓ ℓ 휕푥 ℓ We now wish to take a limit as ℓ → ∞ and for this we apply the dominated convergence theorem. So we need to dominate the integrand. The second half of the integrand is uniformly bounded independent of ℓ as 휕푧 푥 ↦→ 푔(푥) 푓 (푥, 푦0) (푥, 푦0) det 휕푥 is a continuous function with compact support (because of 푔). Hence it is enough to worry about the exponential term. We also only consider those 휉 where the integrand is not zero. Recall that 푟 and 푑 are small enough that 휕휑 1 (푥, 푦0) ≤ , sup 휕푥 ˆ ˆ k푥ˆk≤푟,k푦ˆ0k≤푑 4

0 휉 and as k푥˜k ≤ 푟/4 (as (푥,˜ 푦 ) ∈ 퐾) and 푥˜ + √ ≤ 푟 (because 푔 is zero otherwise), then ℓ 휉 휉 휉 0 0 1 k k 휑 푥˜ + √ , 푦 − 휑(푥,˜ 푦 ) ≤ 푥˜ + √ − 푥˜ = √ . ℓ 4 ℓ 4 ℓ So under the same conditions we have

h √ i2 h √ i2 − 휉+푖 ℓ 휑 푥˜+ √휉 ,푦0 −휑(푥,푦˜ 0) − Re 휉+푖 ℓ 휑 푥˜+ √휉 ,푦0 −휑(푥,푦˜ 0) 푒 ℓ = 푒 ℓ

2 −k휉k2+ℓ 휑 푥˜+ √휉 ,푦0 −휑(푥,푦˜ 0) = 푒 ℓ ≤ 푒−(15/16)k휉k2 .

And that is integrable. Therefore, we take the pointwise limit under the integral to obtain

푛/2 h h 휕휑 i i2 1 ∫ − 휉+푖 (푥,푦˜ 0) 휉 휕푧 푒 휕푥 푔(푥˜) 푓 (푥,˜ 푦0) det (푥,˜ 푦0) 푑휉. 휋 휉∈ℝ푛 휕푥 3.4. EXTENSION OF CR FUNCTIONS 105

In the exponent, we have an expression for the derivative in the 휉 direction with 푦0 ﬁxed. If (푥,˜ 푦0) ∈ 퐾0, then 푔(푥˜) = 1, and so we can ignore 푔. 휕휑 0

퐴 퐼 푖 푥, 푦 Let = + 휕푥 ( ˜ ) . Lemma 3.3.4 says

푛/2 h h 휕휑 i i2 1 ∫ − 휉+푖 (푥,푦˜ 0) 휉 휕푧 푒 휕푥 푓 (푥,˜ 푦0) det (푥,˜ 푦0) 푑휉 = 푓 (푥,˜ 푦0). 휋 휉∈ℝ푛 휕푥 That the convergence is uniform in (푥,˜ 푦0) ∈ 퐾0 is left as an exercise.

Exercise 3.3.8: In the claim above, ﬁnish the proof that the convergence is uniform in (푥,˜ 푦0) ∈ 퐾0. Hint: It may be easier to use the form of the integral before the change of variables and prove that the sequence is uniformly Cauchy.

We are essentially done with the proof of the theorem. The two claims together with ( 3.1 ) show that 푓ℓ are entire holomorphic functions that approximate 푓 uniformly on 퐾. Entire holomorphic functions can be approximated by polynomials uniformly on compact subsets; simply take the partial sums of Taylor series at the origin.

Exercise 3.3.9: Explain why being approximable on 퐾 by (holomorphic) polynomials does not necessarily mean that 푓 is real-analytic.

푛 Exercise 3.3.10: Suppose 푀 ⊂ ℂ is given by Im 푧푛 = 0. Use the standard Weierstrass approximation theorem to show that given a 퐾 ⊂⊂ 푀, and a smooth CR function 푓 : 푀 → ℂ, then 푓 can be uniformly approximated by holomorphic polynomials on 퐾.

3.4 Extension of CR functions

We will now apply the so-called “technique of analytic discs” together with Baouendi–Trèves to prove the Lewy extension theorem. Lewy’s original proof was different and predates Baouendi– Trèves. A local extension theorem of this type was first proved by Helmut Knesser in 1936. Theorem 3.4.1 (Lewy). Suppose 푀 ⊂ ℂ푛 is a smooth real hypersurface and 푝 ∈ 푀. There exists a neighborhood 푈 of 푝 with the following property. Suppose 푟 : 푈 → ℝ is a smooth defining function for 푀 ∩ 푈, denote by 푈− ⊂ 푈 the set where 푟 is negative and 푈+ ⊂ 푈 the set where 푟 is positive. Let 푓 : 푀 → ℝ be a smooth CR function. Then:

(i) If the Levi form with respect to 푟 has a positive eigenvalue at 푝, then 푓 extends to a holomorphic function on 푈− continuous up to 푀 (that is, continuous on {푧 ∈ 푈 : 푟(푧) ≤ 0}). (ii) If the Levi form with respect to 푟 has a negative eigenvalue at 푝, then 푓 extends to a holomorphic function on 푈+ continuous up to 푀 (that is, continuous on {푧 ∈ 푈 : 푟(푧) ≥ 0}). (iii) If the Levi form with respect to 푟 has eigenvalues of both signs at 푝, then 푓 extends to a function holomorphic on 푈. 106 CHAPTER 3. CR FUNCTIONS

So if the Levi form has eigenvalues of both signs, then near 푝 all CR functions are restrictions of holomorphic functions. The function 푟 can be any defining function for 푀. Either we can extend it to all of 푈 or we could take a smaller 푈 such that 푟 is defined on 푈. As we noticed before, once we pick sides (where 푟 is positive and where it is negative), then the number of positive eigenvalues and the number of negative eigenvalues of the Levi form is fixed. A different 푟 may flip 푈− and 푈+, but the conclusion of the theorem is exactly the same. Proof. We prove the first item, and the second item follows by considering −푟. Suppose 푝 = 0 and 푀 is given in some neighborhood Ω of the origin as 푛−1 2 ∑︁ 2 0 0 Im 푤 = |푧1| + 휖푗 |푧푗 | + 퐸 (푧1, 푧 , 푧¯1, 푧¯ , Re 푤), 푗=2 0 where 푧 = (푧2, . . . , 푧푛−1), 휖푗 = −1, 0, 1, and 퐸 is 푂(3). Let Ω− be given by 푛−1 2 ∑︁ 2 0 0 0 > 푟 = |푧1| + 휖푗 |푧푗 | + 퐸 (푧1, 푧 , 푧¯1, 푧¯ , Re 푤) − Im 푤. 푗=2 The (real) Hessian of the function 2 푧1 ↦→ |푧1| + 퐸 (푧1, 0, 푧¯1, 0, 0) is positive definite in a neighborhood of the origin and the function has a strict minimum at 0. There is some small disc 퐷 ⊂ ℂ such that this function is strictly positive on 휕퐷. Therefore, for (푧0, 푤) ∈ 푊 in some small neighborhood 푊 ⊂ ℂ푛−1 of the origin, the function 푛 2 ∑︁ 2 0 0 푧1 ↦→ |푧1| + 휖푗 |푧푗 | + 퐸 (푧1, 푧 , 푧¯1, 푧¯ , Re 푤) − Im 푤 푗=2 is still strictly positive on 휕퐷. We wish to apply Baouendi–Trèves and so let 퐾 be the compact neighborhood of the origin from the theorem. Take 퐷 and 푊 small enough such that (퐷 × 푊) ∩ 푀 ⊂ 퐾. Find the polynomials 0 0 푝푗 that approximate 푓 uniformly on 퐾. Consider 푧1 ∈ 퐷 and (푧 , 푤) ∈ 푊 such that (푧1, 푧 , 푤) ∈ Ω−. 0 Let Δ = 퐷 × {(푧 , 푤)} ∩ Ω−. Denote by 휕Δ the boundary of Δ in the subspace topology of ℂ × {(푧0, 푤)}. 0 The set Ω+ where 푟 > 0 is open and it contains (휕퐷) × {(푧 , 푤)}. Therefore, 휕Δ contains no points of (휕퐷) × {(푧0, 푤)}. Consequently, 휕Δ contains only points where 푟 = 0, that is 휕Δ ⊂ 푀, and also 휕Δ ⊂ 퐷 × 푊. As (퐷 × 푊) ∩ 푀 ⊂ 퐾, we have 휕Δ ⊂ 퐾. 푀 Δ 퐾 푟 < 0 퐷 × 푊 푟 > 0 (푧0, 푤)

푧1 휕Δ 3.4. EXTENSION OF CR FUNCTIONS 107

As 푝푗 → 푓 uniformly on 퐾, then 푝푗 → 푓 uniformly on 휕Δ. As 푝푗 are holomorphic, then by 0 the maximum principle, 푝푗 converge uniformly on all of Δ. In fact, as (푧1, 푧 , 푤) was an arbitrary point in (퐷 × 푊) ∩ Ω−, the polynomials 푝푗 converge uniformly on (퐷 × 푊) ∩ Ω−. Let 푈 = 퐷 × 푊, then 푈− = (퐷 × 푊) ∩ Ω−. Notice 푈 depends on 퐾, but not on 푓 . So 푝푗 converge to a continuous function 퐹 on 푈− ∩ 푈 and 퐹 is holomorphic on 푈−. Clearly 퐹 equals 푓 on 푀 ∩ 푈. To prove the last item, pick a side, and then use one of the ﬁrst two items to extend the function

to that side. Via the tomato can principle ( Theorem 2.3.11 ) the function also extends across 푀 and therefore to a whole neighborhood of 푝. If you were wondering what happened to the analytic discs we promised, the Δ in the above is an analytic disc (simply connected) for a small enough 푈, but it was not necessary to prove that fact. We state the next corollary for a strongly convex domain, even though it holds with far more ∗ generality. It is a simpler version of the Hartogs–Bochner . Later, in Exercise 4.3.4 , you will prove it for strongly pseudoconvex domains. However, the theorem is true for every bounded domain with connected smooth boundary with no assumptions on the Levi form, but a diﬀerent approach would have to be taken.

Corollary 3.4.2. Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, is a bounded domain with smooth boundary that is strongly convex and 푓 : 휕푈 → ℂ is a smooth CR function, then there exists a continuous function 퐹 : 푈 → ℂ holomorphic in 푈 such that 퐹|휕푈 = 푓 . Proof. A strongly convex domain is strongly pseudoconvex, so 푓 must extend to the inside locally near every point. The extension is locally unique as any two extensions have the same boundary values. Therefore, there exists a set 퐾 ⊂⊂ 푈 such that 푓 extends to 푈 \ 퐾. Via an exercise below we can assume that 퐾 is strongly convex and therefore we can apply the special case of Hartogs

phenomenon that you proved in Exercise 2.1.6 to ﬁnd an extension holomorphic in 푈.

Exercise 3.4.1: Prove the existence of the strongly convex 퐾 in the proof of Corollary 3.4.2 above.

Exercise 3.4.2: Show by example that the corollary is not true when 푛 = 1. Explain where in the proof have we used that 푛 ≥ 2.

Exercise 3.4.3: Suppose 푓 : 휕픹2 → ℂ is a smooth CR function. Write down an explicit formula for the extension 퐹.

3 2 2 Exercise 3.4.4: A smooth real hypersurface 푀 ⊂ ℂ is deﬁned by Im 푤 = |푧1| − |푧2| + 푂(3) and 푓 is a real-valued smooth CR function on 푀. Show that | 푓 | does not attain a maximum at the origin.

Exercise 3.4.5: A real-analytic hypersurface 푀 ⊂ ℂ푛, 푛 ≥ 3, is such that the Levi form at 푝 ∈ 푀 has eigenvalues of both signs. Show that every smooth CR function 푓 on 푀 is, in fact, real-analytic in a neighborhood of 푝.

∗What is called Hartogs–Bochner is the 퐶1 version of this theorem where the domain is only assumed to be bounded and the boundary connected, and it was proved by neither Hartogs nor Bochner, but by Martinelli in 1961. 108 CHAPTER 3. CR FUNCTIONS

3 2 2 Exercise 3.4.6: Let 푀 ⊂ ℂ be deﬁned by Im 푤 = |푧1| − |푧2| . a) Show that for this 푀, the conclusion of Baouendi–Trèves holds with an arbitrary compact subset 퐾 ⊂⊂ 푀. b) Use this to show that every smooth CR function 푓 : 푀 → ℂ is a restriction of an entire holomorphic function 퐹 : ℂ3 → ℂ.

Exercise 3.4.7: Find an 푀 ⊂ ℂ푛, 푛 ≥ 2, such that near some 푝 ∈ 푀, for every neighborhood 푊 of 푝 in 푀, there is a CR function 푓 : 푊 → ℂ that does not extend holomorphically to either side of 푀 at 푝.

Exercise 3.4.8: Suppose 푓 : 휕픹푛 → ℂ is a smooth function and 푛 ≥ 2. Prove that 푓 is a CR function if and only if

∫ 2휋 푖휃 푖푘휃 푓 (푒 푣) 푒 푑휃 = 0 for all 푣 ∈ 휕픹푛 and all 푘 ∈ ℕ. 0 Exercise 3.4.9: Prove the third item in the Lewy extension theorem without the use of the tomato can principle. That is, prove in a more elementary way that if 푀 ⊂ 푈 ⊂ ℂ푛 is a smooth real hypersurface in an open set 푈 and 푓 : 푈 → ℂ is continuous and holomorphic in 푈 \ 푀, then 푓 is holomorphic.

Remark 3.4.3. Studying solutions to nonhomogeneous CR equations of the form 푋 푓 = 휓 for a CR vector field 푋, and the fact that such conditions can guarantee that a function must be real-analytic, led Lewy to a famous, very surprising, and rather simple example of a linear partial differential ∗ equation with smooth coefficients that has no solution on any open set . The example is surprising because when a linear PDE has real-analytic coefficients, a solution always exists by the theorem of Cauchy–Kowalevski. Furthermore, if 푋 is a real vector field (푋 is in 푇 푀 not in ℂ ⊗ 푇 푀), then a solution to 푋 푓 = 휓 exists by the method of characteristics, even if 푋 and 휓 are only smooth.

∗Lewy, Hans, An example of a smooth linear partial diﬀerential equation without solution, Annals of Mathematics, 66 (1957), 155–158. Chapter 4

The 휕¯-problem

4.1 The generalized Cauchy integral formula

Before we get into the 휕¯-problem, let us prove a more general version of Cauchy’s formula using Stokes’ theorem (really Green’s theorem). This version is called the Cauchy–Pompeiu integral formula. We only need the theorem for smooth functions, but as it is often applied in less regular contexts and it is just an application of Stokes’ theorem, let us state it more generally. In applications, the boundary is often only piecewise smooth, and again that is all we need for Stokes. Theorem 4.1.1 (Cauchy–Pompeiu). Let 푈 ⊂ ℂ be a bounded open set with piecewise-퐶1 boundary

휕푈 oriented positively (see appendix B ), and let 푓 : 푈 → ℂ be continuous with bounded continuous partial derivatives in 푈. Then for 푧 ∈ 푈:

휕 푓 휁 1 ∫ 푓 (휁) 1 ∫ ¯ ( ) 푓 (푧) = 푑휁 + 휕휁 푑휁 ∧ 푑휁.¯ 2휋푖 휕푈 휁 − 푧 2휋푖 푈 휁 − 푧 If 푓 is holomorphic, then the second term is zero, and we obtain the standard Cauchy formula. If 휁 = 푥 +푖푦, then the standard orientation on ℂ is the one corresponding to the area form 푑퐴 = 푑푥 ∧ 푑푦. The form 푑휁 ∧ 푑휁¯ is the area form up to a scalar. That is,

푑휁 ∧ 푑휁¯ = (푑푥 + 푖 푑푦) ∧ (푑푥 − 푖 푑푦) = (−2푖)푑푥 ∧ 푑푦 = (−2푖)푑퐴.

As we want to use Stokes, we need to write the standard exterior derivative in terms of 푧 and 푧¯. For 푧 = 푥 + 푖푦, we compute: 휕휓 휕휓 휕휓 휕휓 푑휓 = 푑푥 + 푑푦 = 푑푧 + 푑푧.¯ 휕푥 휕푦 휕푧 휕푧¯

Exercise 4.1.1: Observe the singularity in the second term of the Cauchy–Pompeiu formula, and prove that the integral still makes sense (the function is integrable). Hint: polar coordinates.

Exercise 4.1.2: Why can we not diﬀerentiate in 푧¯ under the integral in the second term of the Cauchy–Pompeiu formula? Notice that it would lead to an impossible result. 110 CHAPTER 4. THE 휕¯-PROBLEM

∗ Proof. Fix 푧 ∈ 푈. We wish to apply Stokes’ theorem , but the integrand is not smooth at 푧. Let Δ푟 (푧) be a small disc such that Δ푟 (푧) ⊂⊂ 푈. Stokes now applies on 푈 \ Δ푟 (푧).

푈 푧 Δ푟 (푧)

Via Stokes, we get

휕 푓 휁 ∫ 푓 (휁) ∫ 푓 (휁) ∫ 푓 (휁) ∫ ¯ ( ) 푑휁 − 푑휁 = 푑 푑휁 = 휕휁 푑휁¯ ∧ 푑휁. 휁 푧 휁 푧 휁 푧 휁 푧 휕푈 − 휕Δ푟 (푧) − 푈\Δ푟 (푧) − 푈\Δ푟 (푧) − The second equality follows because holomorphic derivatives in 휁 have a 푑휁 and when we wedge them with 푑휁 we just get zero. We now wish to let the radius 푟 go to zero. Via the exercise above, 휕 푓 (휁) 휕휁¯ 푑휁¯ 푑휁 푈 휁−푧 ∧ is integrable over all of . Therefore,

휕 푓 (휁) 휕 푓 (휁) 휕 푓 (휁) ∫ 휕휁¯ ∫ 휕휁¯ ∫ 휕휁¯ lim 푑휁¯ ∧ 푑휁 = 푑휁¯ ∧ 푑휁 = − 푑휁 ∧ 푑휁.¯ 푟→0 휁 푧 휁 푧 휁 푧 푈\Δ푟 (푧) − 푈 − 푈 −

The second equality is simply swapping the order of the 푑휁 and 푑휁¯. By continuity of 푓 ,

1 ∫ 푓 (휁) 1 ∫ 2휋 lim 푑휁 = lim 푓 (푧 + 푟푒푖휃) 푑휃 = 푓 (푧). 푟→0 휋푖 휁 푧 푟→0 휋 2 휕Δ푟 (푧) − 2 0 The theorem follows.

Exercise 4.1.3: a) Let 푈 ⊂ ℂ be a bounded open set with piecewise-퐶1 boundary and suppose 푓 : 푈 → ℂ 휕 푓 (휁) 퐶1 ∫ 휕푧¯ 푑퐴(휁) = 푧 ∈ 휕푈 푓 | is a function such that 푈 휁−푧 0 for every . Prove that 휕푈 are the boundary values of a holomorphic function in 푈. 휖 > 퐶1 푓 휕 푓 b) Given arbitrary 0, ﬁnd a function on the closed unit disc 픻, such that 휕푧¯ is identically zero outside an 휖-neighborhood of the origin, yet 푓 |휕픻 are not the boundary values of a holomorphic function.

Exercise 4.1.4: Let 푈 ⊂ ℂ and 푓 be as in the theorem, but let 푧 ∉ 푈. Show that

휕 푓 (휁) 1 ∫ 푓 (휁) 1 ∫ 휕휁¯ 푑휁 + 푑휁 ∧ 푑휁¯ = 0. 2휋푖 휕푈 휁 − 푧 2휋푖 푈 휁 − 푧

∗ We are really using Green’s theorem, which is the generalized Stokes’ theorem in 2 dimensions, see Theorem B.2 . 4.2. SIMPLE CASE OF THE 휕¯-PROBLEM 111

4.2 Simple case of the 휕¯-problem

For a smooth function 휓, consider the exterior derivative in terms of 푧 and 푧¯, 휕휓 휕휓 휕휓 휕휓 푑휓 = 푑푧1 + · · · + 푑푧푛 + 푑푧¯1 + · · · + 푑푧¯푛. 휕푧1 휕푧푛 휕푧¯1 휕푧¯푛 Let us give a name to the two parts of the derivative:

def 휕휓 휕휓 def 휕휓 휕휓 휕휓 = 푑푧1 + · · · + 푑푧푛, 휕휓¯ = 푑푧¯1 + · · · + 푑푧¯푛. 휕푧1 휕푧푛 휕푧¯1 휕푧¯푛

Then 푑휓 = 휕휓 + 휕휓¯ . Notice 휓 is holomorphic if and only if 휕휓¯ = 0. The so-called inhomogeneous 휕¯-problem (휕¯ is pronounced “dee bar”) is to solve the equation

휕휓¯ = 푔,

for 휓, given a one-form 푔 = 푔1푑푧¯1 + · · · + 푔푛푑푧¯푛. Such a 푔 is called a (0, 1)-form. The fact that the partial derivatives of 휓 commute, forces certain compatibility conditions on 푔 for us to have any hope of getting a solution (see below).

Exercise 4.2.1: Find an explicit example of a 푔 in ℂ2 such that no corresponding 휓 can exist.

On any open set where 푔 = 0, 휓 is holomorphic. So for a general 푔, what we are doing is ﬁnding a function that is not holomorphic in a speciﬁc way.

Theorem 4.2.1. Suppose 푔 is a (0, 1)-form on ℂ푛, 푛 ≥ 2, given by

푔 = 푔1푑푧¯1 + · · · + 푔푛푑푧¯푛,

푛 where 푔푗 : ℂ → ℂ are compactly supported smooth functions satisfying the compatibility conditions 휕푔 휕푔 푘 = ℓ for all 푘, ℓ = 1, 2, . . . , 푛. (4.1) 휕푧¯ℓ 휕푧¯푘 Then there exists a unique compactly supported smooth function 휓 : ℂ푛 → ℂ such that

휕휓¯ = 푔.

The compatibility conditions on 푔 are necessary, but the compactness is not. Without compactness, the boundary of the set where the equation lives would come into play. Let us not worry about this, and prove that this simple compactly supported version always has a solution. The compactly supported solution is unique: Given any holomorphic 푓 , 휕¯(휓 + 푓 ) = 푔. But since the diﬀerence of any two solutions 휓1 and 휓2 is holomorphic, and the only holomorphic compactly supported function is 0, then the compactly supported solution 휓 is unique. 112 CHAPTER 4. THE 휕¯-PROBLEM

Proof. We really have 푛 smooth functions, 푔1, . . . , 푔푛, so the equation 휕휓¯ = 푔 is the 푛 equations 휕휓 = 푔푘 , 휕푧¯푘

where the functions 푔푘 satisfy the compatibility conditions ( 4.1 ). We claim that the following is an explicit solution:

1 ∫ 푔 (휁, 푧 , . . . , 푧 ) 휓(푧) = 1 2 푛 푑휁 ∧ 푑휁¯ 2휋푖 ℂ 휁 − 푧1 1 ∫ 푔 (휁 + 푧 , 푧 , . . . , 푧 ) = 1 1 2 푛 푑휁 ∧ 푑휁.¯ 2휋푖 ℂ 휁

To show that the singularity does not matter for integrability is the same idea as for the generalized Cauchy formula. Let us check we have the solution. We use the generalized Cauchy formula on the 푧1 variable. Take 푅 large enough so that 푔푗 (휁, 푧2, . . . , 푧푛) is zero when |휁 | ≥ 푅 for all 푗. For every 푗,

∫ 푔 휁, 푧 , . . . , 푧 ∫ 휕푔푗 (휁, 푧 , . . . , 푧 ) 1 푗 ( 2 푛) 1 휕푧¯1 2 푛 푔푗 (푧1, . . . , 푧푛) = 푑휁 + 푑휁 ∧ 푑휁¯ 2휋푖 |휁 |=푅 휁 − 푧1 2휋푖 |휁 |≤푅 휁 − 푧1 휕푔푗 휁, 푧 , . . . , 푧 1 ∫ ( 2 푛) = 휕푧¯1 푑휁 ∧ 푑휁.¯ 2휋푖 ℂ 휁 − 푧1

Using the second form of the deﬁnition of 휓, the compatibility conditions ( 4.1 ), and the computation above we get

휕푔1 (휁 + 푧 , 푧 , . . . , 푧 ) 휕휓 1 ∫ 휕푧¯ 1 2 푛 (푧) = 푗 푑휁 ∧ 푑휁¯ 휕푧¯푗 2휋푖 ℂ 휁 휕푔푗 휁 푧 , 푧 , . . . , 푧 1 ∫ ( + 1 2 푛) = 휕푧¯1 푑휁 ∧ 푑휁¯ 2휋푖 ℂ 휁 ∫ 휕푔푗 (푧 , 푧 , . . . , 푧 ) 1 휕푧¯1 1 2 푛 = 푑휁 ∧ 푑휁¯ = 푔푗 (푧). 2휋푖 ℂ 휁 − 푧1

Exercise 4.2.2: Show that we were allowed to diﬀerentiate under the integral in the computation above.

That 휓 has compact support follows because 푔1 has compact support together with the identity theorem. In particular, 휓 is holomorphic for large 푧 since 휕휓¯ = 푔 = 0 when 푧 is large. When at least one of 푧2, . . . , 푧푛 is large, then 휓 is identically zero simply from its deﬁnition. See the following diagram: 4.3. THE GENERAL HARTOGS PHENOMENON 113

푧2, . . . , 푧푛 large so 휓 = 0 휕휓¯ = 0

푧2, . . . , 푧푛 푔 ≠ 0 휕휓¯ = 0

푧1

푧2, . . . , 푧푛 large so 휓 = 0 휕휓¯ = 0

As 휕휓¯ = 0 on the light gray and white areas in the diagram, 휓 is holomorphic there. As 휓 is zero on the light gray region, it is zero also on the white region by the identity theorem. That is, 휓 is zero on the unbounded component of the set where 푔 = 0, and so 휓 has compact support.

The ﬁrst part of the proof still works when 푛 = 1, we get a solution 휓. However, the last bit of the proof does not work in one dimension, so 휓 does not have compact support.

Exercise 4.2.3: a) Show that if 푔 is supported in 퐾 ⊂⊂ ℂ푛, 푛 ≥ 2, then 휓 is supported in the complement of the unbounded component of ℂ푛 \ 퐾. In particular, show that if 퐾 is the support of 푔 and ℂ푛 \ 퐾 is connected, then the support of 휓 is 퐾. b) Find an explicit example where the support of 휓 is strictly larger than the support of 푔.

Exercise 4.2.4: Find an example of a smooth function 푔 : ℂ → ℂ with compact support, such 휓 휕휓 푔 that no solution : ℂ → ℂ to 휕푧¯ = (at least one of which always exists) is of compact support.

4.3 The general Hartogs phenomenon

We can now prove the general Hartogs phenomenon as an application of the solution of the compactly supported inhomogeneous 휕¯-problem. We proved special versions of this phenomenon using Hartogs ﬁgures before. The proof of the theorem has a complicated history as Hartogs’ original proof from 1906 contained gaps. A fully working proof was ﬁnally supplied by Fueter in 1939 for 푛 = 2 and independently by Bochner and Martinelli for higher 푛 in the early 40s. The proof we give is the standard one given nowadays due to Leon Ehrenpreis from 1961.

Theorem 4.3.1 (Hartogs phenomenon). Let 푈 ⊂ ℂ푛 be a domain, 푛 ≥ 2, and let 퐾 ⊂⊂ 푈 be a compact set such that 푈 \ 퐾 is connected. Every holomorphic 푓 : 푈 \ 퐾 → ℂ extends uniquely to a holomorphic function on 푈. 114 CHAPTER 4. THE 휕¯-PROBLEM

푈퐾

The idea of the proof is extending in any way whatsoever and then using the solution to the 휕¯-problem to correct the result to make it holomorphic. Proof. First find a smooth function 휑 that is 1 in a neighborhood of 퐾 and is compactly supported in 푈 (exercise below). Let 푓0 = (1 − 휑) 푓 on 푈 \ 퐾 and 푓0 = 0 on 퐾. The function 푓0 is smooth on 푈 and it is holomorphic and equal to 푓 near the boundary of 푈, where 휑 is 0. We let 푔 = 휕¯ 푓0 on 푈 푔 휕 푓0 푔 푈 푔 휕푈 , that is 푘 = 휕푧¯ , and we let = 0 outside . As 푘 are identically zero near , we find that 푘 푛 each 푔푘 is smooth on ℂ . The compatibility conditions ( 4.1 ) are satisfied because partial derivatives commute. Let us see why 푔푘 is compactly supported. The only place to check is on 푈 \ 퐾 as elsewhere we have 푔푘 = 0 automatically. Note that 푓 is holomorphic on 푈 \ 퐾 and compute 휕 푓 휕 휕 푓 휕 푓 휕휑 휕휑 0 = (1 − 휑) 푓 = − 휑 − 푓 = − 푓 . 휕푧¯푘 휕푧¯푘 휕푧¯푘 휕푧¯푘 휕푧¯푘 휕푧¯푘

And 휕휑 is compactly supported in 푈 \ 퐾 by construction. Now apply the solution of the compactly 휕푧¯푘 supported 휕¯-problem to ﬁnd a compactly supported function 휓 such that 휕휓¯ = 푔. Set 퐹 = 푓0 − 휓. Let us check that 퐹 is the desired extension. It is holomorphic:

휕퐹 휕 푓0 휕휓 = − = 푔푘 − 푔푘 = 0. 휕푧¯푘 휕푧¯푘 휕푧¯푘

Next, Exercise 4.2.3 and the fact that 푈 \ 퐾 is connected reveals that 휓 must be compactly supported in 푈. This means that 퐹 agrees with 푓 near the boundary (in particular on an open set) and thus everywhere in 푈 \ 퐾 since 푈 \ 퐾 is connected. The hypotheses on dimension and on connectedness of 푈 \ 퐾 are necessary. No such theorem is true in one dimension. If 푈 \ 퐾 is disconnected, a simple counterexample can be constructed. See the exercise below.

Exercise 4.3.1: Show that 휑 exists. Hint: Use molliﬁers.

Exercise 4.3.2: Suppose 푈 ⊂ ℂ푛 is a domain and 퐾 ⊂ 푈 is any compact set (perhaps 푈 \ 퐾 is disconnected). Prove that given 푓 ∈ O(푈 \ 퐾) there exists an 퐹 ∈ O(푈) that equals to 푓 on the intersection of 푈 and the unbounded component of ℂ푛 \ 퐾.

Exercise 4.3.3: Suppose 푈 ⊂ ℂ푛 is a domain and 퐾 ⊂ 푈 is a compact set such that 푈 \ 퐾 is disconnected. Find a counterexample to the conclusion to Hartogs. 4.3. THE GENERAL HARTOGS PHENOMENON 115

One of many consequences of the Hartogs phenomenon is that the zero set of a holomorphic 푓 1 function is never compact in dimension 2 or higher. If it were compact, 푓 would provide a

contradiction, see also Exercise 1.6.2 . Corollary 4.3.2. Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, is a domain and 푓 : 푈 → ℂ is holomorphic. If the zero set 푓 −1(0) is not empty, then it is not compact. Replacing 푈 \ 퐾 with a hypersurface is usually called the Hartogs–Bochner theorem (when the hypersurface is 퐶1 or smooth). The real-analytic case was stated ﬁrst by Severi in 1931. Corollary 4.3.3 (Severi). Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, is a bounded domain with connected real- analytic boundary and 푓 : 휕푈 → ℂ is a real-analytic CR function. Then there exists some 0 푛 0 neighborhood 푈 ⊂ ℂ of 푈 and a holomorphic function 퐹 : 푈 → ℂ for which 퐹|휕푈 = 푓 .

Proof. By Severi’s result ( Theorem 3.2.9 ), for every 푝 ∈ 휕푈, there is a small ball 퐵푝, such that 푓 extends to 퐵푝. Cover 휕푈 by ﬁnitely many such balls so that if 퐵푝 intersects 퐵푞, then the (connected) intersection 퐵푝 ∩ 퐵푞 contains points of 휕푈. The extension in 퐵푝 and in 퐵푞 then agree on a piece of a hypersurface 휕푈, and hence agree. Taking a union of the 퐵푝 we ﬁnd a unique extension in single neighborhood of 휕푈. We write this neighborhood as 푈0 \ 퐾 for some compact 퐾 and a connected 푈0 such that 푈 ⊂ 푈0. Consider the topological components of ℂ푛 \ 퐾. As 휕푈 is connected and 푈 is bounded, the unbounded component of ℂ푛 \ 퐾 must contain all of 휕푈. By boundedness of 푈, all the other components are relatively compact in 푈. If we add them to 퐾, then 퐾 is still compact and 푈0 \ 퐾 is connected. We apply the Hartogs phenomenon.

Exercise 4.3.4 (Hartogs–Bochner again): Let 푈 ⊂ ℂ푛, 푛 ≥ 2, be a bounded domain with connected strongly pseudoconvex smooth boundary and let 푓 : 휕푈 → ℂ be a smooth CR function. Prove that there exists a continuous function 퐹 : 푈 → ℂ holomorphic in 푈 such that 퐹|휕푈 = 푓 . Note: Strong pseudoconvexity is not needed (“bounded with smooth boundary” will do), but that is much more difficult to prove. Exercise 4.3.5: Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, is a bounded domain of holomorphy. Show that ℂ푛 \ 푈 is connected using the Hartogs phenomenon. 푊 ⊂ 푈 ⊂ 푛 푛 ≥ 푧0, 푧0, . . . , 푧0 Exercise 4.3.6: Suppose ℂ , 3, are domains such that for each fixed 3 4 푛, 푧 , 푧 2 푧 , 푧 , 푧0, . . . , 푧0 푈 푊 푧 , 푧 2 푧 , 푧 , 푧0, . . . , 푧0 푈 . ( 1 2) ∈ ℂ : ( 1 2 3 푛) ∈ \ ⊂⊂ ( 1 2) ∈ ℂ : ( 1 2 3 푛) ∈ Prove that every 푓 ∈ O(푊) extends to a holomorphic function on 푈. Note: The fact that 푊 is connected is important. Exercise 4.3.7: a) Prove that if 푛 ≥ 2, no domain of the form 푈 = ℂ푛 \ 퐾 for a compact 퐾 is biholomorphic to a bounded domain. b) Prove that every domain of the form 푈 = ℂ \ 퐾 for a compact 퐾 with nonempty interior is biholomorphic to a bounded domain. Exercise 4.3.8: Suppose 푈 ⊂ ℂ푛, 푛 ≥ 2, is a domain such that for some affine 퐴: ℂ2 → ℂ푛 the set 퐴−1 ℂ푛 \ 푈 has a bounded topological component. Prove that 푈 is not a domain of holomorphy. Chapter 5

Integral kernels

5.1 The Bochner–Martinelli kernel

A generalization of Cauchy’s formula to several variables is called the Bochner–Martinelli integral formula, which in fact reduces to Cauchy’s (Cauchy–Pompeiu) formula when 푛 = 1. As for Cauchy’s formula, we will prove the formula for all smooth functions via Stokes’ theorem. First, let us deﬁne the Bochner–Martinelli kernel: 푛 def (푛 − 1)! ∑︁ 휁¯푗 − 푧¯푗 휔(휁, 푧) = 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑c휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 . (2휋푖)푛 휁 푧 2푛 1 1 푗 푗 푛 푛 푗=1 k − k

The notation 푑c휁¯푗 means that this term is simply left out. Theorem 5.1.1 (Bochner–Martinelli). Let 푈 ⊂ ℂ푛 be a bounded open set with smooth boundary and let 푓 : 푈 → ℂ be a smooth function. Then for 푧 ∈ 푈, ∫ ∫ 푓 (푧) = 푓 (휁)휔(휁, 푧) − 휕¯ 푓 (휁) ∧ 휔(휁, 푧). 휕푈 푈 In particular, if 푓 ∈ O(푈), then ∫ 푓 (푧) = 푓 (휁)휔(휁, 푧). 휕푈 Recall that if 휁 = 푥 + 푖푦 are the coordinates in ℂ푛, the orientation that we assigned to ℂ푛 in this ∗ book is the one corresponding to the volume form

푑푉 = 푑푥1 ∧ 푑푦1 ∧ 푑푥2 ∧ 푑푦2 ∧ · · · ∧ 푑푥푛 ∧ 푑푦푛. With this orientation, 푛 푑휁1 ∧ 푑휁¯1 ∧ 푑휁2 ∧ 푑휁¯2 ∧ · · · ∧ 푑휁푛 ∧ 푑휁¯푛 = (−2푖) 푑푉, and hence 푛 푑휁¯1 ∧ 푑휁1 ∧ 푑휁¯2 ∧ 푑휁2 ∧ · · · ∧ 푑휁¯푛 ∧ 푑휁푛 = (2푖) 푑푉.

∗Again, there is no canonical orientation of ℂ푛, and not all authors follow this (perhaps more prevalent) convention. 5.1. THE BOCHNER–MARTINELLI KERNEL 117

Exercise 5.1.1: Similarly to the Cauchy–Pompeiu formula, note the singularity in the second term of the Bochner–Martinelli formula, and prove that the integral still makes sense (the function is integrable).

Exercise 5.1.2: Check that for 푛 = 1, the Bochner–Martinelli formula reduces to the standard Cauchy–Pompeiu formula.

As usual, we split the derivatives into the holomorphic and antiholomorphic parts. We work with multi-indices. For 훼 and 훽 with |훼| = 푝 and |훽| = 푞, a diﬀerential form

∑︁ 훼 훽 휂 = 휂훼훽 푑푧 ∧ 푑푧¯ |훼|=푝 |훽|=푞

is called a (푝, 푞)-form or a diﬀerential form of bidegree (푝, 푞). Deﬁne

푛 푛 def ∑︁ ∑︁ 휕휂훼훽 def ∑︁ ∑︁ 휕휂훼훽 휕휂 = 푑푧 ∧ 푑푧훼 ∧ 푑푧¯훽, and 휕휂¯ = 푑푧¯ ∧ 푑푧훼 ∧ 푑푧¯훽. 휕푧 푗 휕푧¯ 푗 |훼|=푝 푗=1 푗 |훼|=푝 푗=1 푗 |훽|=푞 |훽|=푞

It is not diﬃcult to see that 푑휂 = 휕휂 + 휕휂¯ as before.

Proof of Bochner–Martinelli. The structure of the proof is essentially the same as that of the Cauchy–Pompeiu theorem for 푛 = 1, although some of the formulas are somewhat more involved. Let 푧 ∈ 푈 be ﬁxed. Suppose 푟 > 0 is small enough so that 퐵푟 (푧) ⊂ 푈. Orient both 휕푈 and 휕퐵푟 (푧) positively. Notice that 푓 (휁)휔(휁, 푧) contains all the holomorphic 푑휁푗 . Therefore,

푑 푓 (휁)휔(휁, 푧) = 휕¯ 푓 (휁)휔(휁, 푧) = 휕¯ 푓 (휁) ∧ 휔(휁, 푧) 푛 (푛 − 1)! ∑︁ 휕 휁¯푗 − 푧¯푗 + 푓 (휁) 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 . (2휋푖)푛 휕휁¯ 휁 푧 2푛 1 1 푛 푛 푗=1 푗 k − k

We compute 푛 푛 2 ! ∑︁ 휕 휁¯푗 − 푧¯푗 ∑︁ 1 휁푗 − 푧푗 = − 푛 = 0. 휕휁¯ 휁 푧 2푛 휁 푧 2푛 휁 푧 2푛+2 푗=1 푗 k − k 푗=1 k − k k − k Therefore, 푑 푓 (휁)휔(휁, 푧) = 휕¯ 푓 (휁) ∧ 휔(휁, 푧). We apply Stokes: ∫ ∫ ∫ ∫ 푓 (휁)휔(휁, 푧) − 푓 (휁)휔(휁, 푧) = 푑 푓 (휁)휔(휁, 푧) = 휕¯ 푓 (휁) ∧ 휔(휁, 푧). 휕푈 휕퐵푟 (푧) 푈\퐵푟 (푧) 푈\퐵푟 (푧) Again, due to the integrability, which you showed in an exercise above, the right-hand side converges to the integral over 푈 as 푟 → 0. Just as for the Cauchy–Pompeiu formula, we now need to show that the integral over 휕퐵푟 (푧) goes to 푓 (푧) as 푟 → 0. 118 CHAPTER 5. INTEGRAL KERNELS

So ∫ ∫ ∫ 푓 (휁)휔(휁, 푧) = 푓 (푧) 휔(휁, 푧) + 푓 (휁) − 푓 (푧)휔(휁, 푧). 휕퐵푟 (푧) 휕퐵푟 (푧) 휕퐵푟 (푧) To ﬁnish the proof, we will show that ∫ 휔(휁, 푧) = 1, and that the second term goes to zero. We 휕퐵푟 (푧) 퐵 푧 휋푛 푟2푛 apply Stokes again and note that the volume of 푟 ( ) is 푛! .

푛 ∫ ∫ (푛 − 1)! ∑︁ 휁¯푗 − 푧¯푗 휔(휁, 푧) = 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑c휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 (2휋푖)푛 휁 푧 2푛 1 1 푗 푗 푛 푛 휕퐵푟 (푧) 휕퐵푟 (푧) 푗=1 k − k 푛 (푛 − 1)! 1 ∫ ∑︁ = (휁¯ − 푧¯ )푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑c휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 (2휋푖)푛 푟2푛 푗 푗 1 1 푗 푗 푛 푛 휕퐵푟 (푧) 푗=1 푛 (푛 − 1)! 1 ∫ ∑︁ = 푑 © (휁¯ − 푧¯ )푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑c휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 ª (2휋푖)푛 푟2푛 푗 푗 1 1 푗 푗 푛 푛® 퐵푟 (푧) 푗=1 « ¬ (푛 − 1)! 1 ∫ = 푛 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 휋푖 푛 푟2푛 1 1 푛 푛 (2 ) 퐵푟 (푧) (푛 − 1)! 1 ∫ = 푛(2푖)푛푑푉 = 1. 휋푖 푛 푟2푛 (2 ) 퐵푟 (푧)

Next, we tackle the second term. Via the same computation as above we ﬁnd

∫ (푛 − 1)! 1 ∫ 푓 (휁) − 푓 (푧)휔(휁, 푧) = 푓 (휁) − 푓 (푧)푛 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 휋푖 푛 푟2푛 1 1 푛 푛 휕퐵푟 (푧) (2 ) 퐵푟 (푧) 푛 ! ∫ ∑︁ 휕 푓 + (휁)(휁¯ − 푧¯ ) 푑휁¯ ∧ 푑휁 ∧ · · · ∧ 푑휁¯ ∧ 푑휁 . 휕휁¯ 푗 푗 1 1 푛 푛 퐵푟 (푧) 푗=1 푗

휕 푓 As 푈 is bounded, | 푓 (휁) − 푓 (푧)| ≤ 푀k휁 − 푧k and (휁)(휁¯푗 − 푧¯푗 ) ≤ 푀k휁 − 푧k for some 푀. So 휕휁¯푗 휕 푓 for all 휁 ∈ 휕퐵푟 (푧), we have | 푓 (휁) − 푓 (푧)| ≤ 푀푟 and (휁)(휁¯푗 − 푧¯푗 ) ≤ 푀푟. Hence 휕휁¯푗

∫ (푛 − 1)! 1 ∫ ∫ 푓 (휁) − 푓 (푧)휔(휁, 푧) ≤ 푛2푛 푀푟 푑푉 + 푛2푛 푀푟 푑푉 = 2푀푟. 휋 푛 푟2푛 휕퐵푟 (푧) (2 ) 퐵푟 (푧) 퐵푟 (푧)

Therefore, this term goes to zero as 푟 → 0.

∫ 푓 (휁)휔(휁, 푧) One drawback of the Bochner–Martinelli formula 휕푈 is that the kernel is not holomorphic in 푧 unless 푛 = 1. It does not simply produce holomorphic functions. If we diﬀerentiate in 푧¯ underneath the 휕푈 integral, we do not necessarily obtain zero. On the other hand, we have an explicit formula and this formula does not depend on 푈. This is not the case for the Bergman and Szegö kernels, which we will see next, although those are holomorphic in the right way. 5.2. THE BERGMAN KERNEL 119

Exercise 5.1.3: Prove that if 푧 ∉ 푈, then rather than 푓 (푧) in the formula you obtain ∫ ∫ 푓 (휁)휔(휁, 푧) − 휕¯ 푓 (휁) ∧ 휔(휁, 푧) = 0. 휕푈 푈

Exercise 5.1.4: Suppose 푓 is holomorphic on a neighborhood of 퐵푟 (푧). a) Using the Bochner–Martinelli formula, prove that 1 ∫ 푓 (푧) = 푓 (휁) 푑푉 (휁), 푉 퐵 푧 푟 ( ) 퐵푟 (푧) where 푉 퐵푟 (푧) is the volume of 퐵푟 (푧). b) Use part a) to prove the maximum principle for holomorphic functions.

Exercise 5.1.5: Use Bochner–Martinelli for the solution of 휕¯ with compact support. That is, 푛 suppose 푔 = 푔1푑푧¯1 + · · · + 푔푛푑푧¯푛 is a smooth compactly supported (0, 1)-form on ℂ , 푛 ≥ 2, and 휕푔푘 = 휕푔ℓ for all 푘, ℓ. Prove that 휕푧¯ℓ 휕푧¯푘 ∫ 휓(푧) = − 푔(휁) ∧ 휔(휁, 푧) ℂ푛 is a compactly supported smooth solution to 휕휓¯ = 푔. Hint: Look at the previous proof.

5.2 The Bergman kernel

Let 푈 ⊂ ℂ푛 be a domain. Deﬁne Bergman space of 푈:

퐴2(푈) def= O(푈) ∩ 퐿2(푈). That is, 퐴2(푈) denotes the space of holomorphic functions 푓 ∈ O(푈) such that ∫ k 푓 k2 def= k 푓 k2 = | 푓 (푧)|2푑푉 < ∞. 퐴2 (푈) 퐿2 (푈) 푈 퐴2(푈) is an inner product space with the 퐿2(푈) inner product ∫ h 푓 , 푔i def= 푓 (푧)푔(푧) 푑푉. 푈 We will prove that 퐴2(푈) is complete, in other words, it is a Hilbert space. We ﬁrst prove that the 퐴2(푈) norm bounds the uniform norm on compact sets. 푛 Lemma 5.2.1. Let 푈 ⊂ ℂ be a domain and 퐾 ⊂⊂ 푈 compact. Then there exists a constant 퐶퐾 , such that 푓 푓 푧 퐶 푓 푓 퐴2 푈 . k k퐾 = sup| ( )| ≤ 퐾 k k퐴2 (푈) for all ∈ ( ) 푧∈퐾 Consequently, 퐴2(푈) is complete. 120 CHAPTER 5. INTEGRAL KERNELS

Proof. As 퐾 is compact there exists an 푟 > 0 such that Δ푟 (푧) ⊂ 푈 for all 푧 ∈ 퐾. Take any 푧 ∈ 퐾,

and apply Exercise 1.2.8 and Cauchy–Schwarz:

1 ∫ | 푓 (푧)| = 푓 (휉) 푑푉 (휉) 푉 푧 Δ푟 ( ) Δ푟 (푧) √︄ √︄ 1 ∫ ∫ ≤ 12 푑푉 (휉) | 푓 (휉)|2 푑푉 (휉) 휋푛푟2푛 Δ푟 (푧) Δ푟 (푧) 1 1 = k 푓 k 2 ≤ k 푓 k 2 . 휋푛/2푟푛 퐴 (Δ푟 (푧)) 휋푛/2푟푛 퐴 (푈)

Taking supremum over 푧 ∈ 퐾 proves the estimate. Therefore, if { 푓푗 } is a sequence of functions in 퐴2(푈) converging in 퐿2(푈) to some 푓 ∈ 퐿2(푈), then it converges uniformly on compact sets, and so 푓 ∈ O(푈). Consequently, 퐴2(푈) is a closed subspace of 퐿2(푈), and hence complete.

For a bounded domain, 퐴2(푈) is always infinite-dimensional, see exercise below. There exist unbounded domains for which either 퐴2(푈) is trivial (just the zero function) or even finite- dimensional. When 푛 = 1, 퐴2(푈) is either trivial, or infinite-dimensional.

Exercise 5.2.1: Show that if a domain 푈 ⊂ ℂ푛 is bounded, then 퐴2(푈) is inﬁnite-dimensional.

Exercise 5.2.2: a) Show that 퐴2(ℂ푛) is trivial (it is just the zero function). b) Show that 퐴2(픻 × ℂ) is trivial. c) Find an example of an unbounded domain 푈 for which 퐴2(푈) is inﬁnite-dimensional. Hint: Think in one dimension for simplicity.

Exercise 5.2.3: Show that 퐴2(픻) can be identiﬁed with 퐴2(픻 \{0}), that is, every function in the latter can be extended to a function in the former.

The lemma says that point evaluation is a bounded linear functional. That is, ﬁx 푧 ∈ 푈 and take 퐾 = {푧}, then the linear operator 푓 ↦→ 푓 (푧)

2 is a bounded linear functional. By the Riesz–Fisher theorem, there exists a 푘푧 ∈ 퐴 (푈), such that

푓 (푧) = h 푓 , 푘푧i.

Deﬁne the Bergman kernel for 푈 as

def 퐾푈 (푧, 휁¯) = 푘푧 (휁).

∗ The function 퐾푈 is deﬁned as (푧, 휁¯) vary over 푈 × 푈 , where we write

푈∗ = {휁 ∈ ℂ푛 : 휁¯ ∈ 푈}. 5.2. THE BERGMAN KERNEL 121

Then for all 푓 ∈ 퐴2(푈) we have ∫ 푓 (푧) = 푓 (휁)퐾푈 (푧, 휁¯) 푑푉 (휁). (5.1) 푈 This last equation is sometimes called the reproducing property of the kernel. It should be noted that the Bergman kernel depends on 푈, which is why we write it as 퐾푈 (푧, 휁¯).

Proposition 5.2.2. The Bergman kernel 퐾푈 (푧, 휁¯) is holomorphic in 푧, antiholomorphic in 휁, and

퐾푈 (푧, 휁¯) = 퐾푈 (휁, 푧¯). 2 Proof. As each 푘푧 is in 퐴 (푈), it is holomorphic in 휁. Hence, 퐾푈 is antiholomorphic in 휁. If we prove 퐾푈 (푧, 휁¯) = 퐾푈 (휁, 푧¯), then we prove 퐾푈 is holomorphic in 푧. 2 As 퐾푈 (푧, 휁¯) = 푘푧 (휁) is in 퐴 (푈), then ∫ 퐾푈 (푧, 휁¯) = 퐾푈 (푧, 푤¯ )퐾푈 (휁, 푤¯ )푑푉 (푤) 푈 ∫ = 퐾푈 (휁, 푤¯ )퐾푈 (푧, 푤¯ )푑푉 (푤) = 퐾푈 (휁, 푧¯) = 퐾푈 (휁, 푧¯). 푈

Therefore, thinking of 휁¯ as the variable, then 퐾푈 is a holomorphic function of 2푛 variables. Example 5.2.3: Let us compute the Bergman kernel (and the Szegö kernel of the next section while we’re at it) explicitly for the unit disc 픻 ⊂ ℂ. Let 푓 ∈ O(픻) ∩ 퐶(퐷), that is, 푓 is holomorphic in 픻 and continuous up to the boundary. Let 푧 ∈ 픻. Then 1 ∫ 푓 (휁) 푓 (푧) = 푑휁. 2휋푖 휕픻 휁 − 푧 On the unit circle 휁휁¯ = 1. Let 푑푠 be the arc-length measure on the circle, parametrized as 휁 = 푒푖푠. Then 푑휁 = 푖푒푖푠푑푠, and 1 ∫ 푓 (휁) 1 ∫ 푓 (휁) 1 ∫ 푓 (휁) 푓 (푧) = 푑휁 = 휁¯ 푑휁 = 푑푠. 2휋푖 휕픻 휁 − 푧 2휋푖 휕픻 1 − 푧휁¯ 2휋 휕픻 1 − 푧휁¯ The integral is now a regular line integral of a function whose singularity, which used to be inside the unit disc, disappeared (we “reﬂected it” to the outside). The kernel 1 1 is called the Szegö 2휋 1−푧휁¯ kernel, which we will brieﬂy mention next. We apply Stokes to the second integral above: 1 ∫ 푓 (휁) 1 ∫ 휕 휁¯ 휁¯ 푑휁 = 푓 (휁) 푑휁¯ ∧ 푑휁 2휋푖 휕픻 1 − 푧휁¯ 2휋푖 픻 휕휁¯ 1 − 푧휁¯ 1 ∫ 푓 (휁) = 푑퐴(휁). 휋 2 픻 (1 − 푧휁¯) The Bergman kernel in the unit disc is, therefore, 1 1 퐾픻 (푧, 휁¯) = . 휋 (1 − 푧휁¯)2

It follows from the exercises below that this function really is the Bergman kernel. That is, 퐾픻 is the unique conjugate symmetric reproducing function that is in 퐴2(픻) for a ﬁxed 휁. We have only shown the formula for functions continuous up to the boundary, but those are dense in 퐴2(픻). 122 CHAPTER 5. INTEGRAL KERNELS

2 푛 Example 5.2.4: In an exercise you found that 퐴 (ℂ ) = {0}. Therefore, 퐾ℂ푛 (푧, 휁¯) ≡ 0.

The Bergman kernel for a more general domain is diﬀcult (usually impossible) to compute explicitly. We do have the following formula however.

푛 Proposition 5.2.5. Suppose 푈 ⊂ ℂ is a domain, and {휑푗 (푧)}푗∈퐼 is a complete orthonormal system for 퐴2(푈). Then ∑︁ 퐾푈 (푧, 휁¯) = 휑푗 (푧)휑푗 (휁), 푗∈퐼 with uniform convergence on compact subsets of 푈 × 푈∗.

2 Proof. For any ﬁxed 휁 ∈ 푈, the function 푧 ↦→ 퐾푈 (푧, 휁¯) is in 퐴 (푈) and so expand this function in terms of the basis and use the reproducing property of 퐾푈

∑︁ ∫ ∑︁ 퐾푈 (푧, 휁¯) = 퐾푈 (푤, 휁¯)휑푗 (푤) 푑푉 (푤) 휑푗 (푧) = 휑푗 (휁)휑푗 (푧). 푗∈퐼 푈 푗∈퐼

The convergence is in 퐿2 as a function of 푧, for a ﬁxed 휁. Let 퐾 ⊂⊂ 푈 be a compact set. Via 2 2 Lemma 5.2.1 , 퐿 convergence in 퐴 (푈) is uniform convergence on compact sets. Therefore, for a ﬁxed 휁 the convergence is uniform in 푧 ∈ 퐾. In particular, we get pointwise convergence. So,

∑︁ 2 ∑︁ 휑푗 (푧) = 휑푗 (푧)휑푗 (푧) = 퐾푈 (푧, 푧¯) ≤ 퐶퐾 < ∞, 푗∈퐼 푗∈퐼

∗ ∗ where 퐶퐾 is the supremum of 퐾푈 (푧, 휁¯) on 퐾 × 퐾 . Hence for (푧, 휁¯) ∈ 퐾 × 퐾 ,

∑︁ √︄∑︁ 2√︄∑︁ 2 휑 (푧)휑 (휁) ≤ 휑 (푧) 휑 (휁) ≤ 퐶 < ∞. 푗 푗 푗 푗 퐾 푗∈퐼 푗∈퐼 푗∈퐼

And so the convergence is uniform on 퐾 × 퐾∗.

Exercise 5.2.4: 푛 a) Show that if 푈 ⊂ ℂ is bounded, then 퐾푈 (푧, 푧¯) > 0 for all 푧 ∈ 푈. b) Why can this fail if 푈 is unbounded? Find a (trivial) counterexample.

Exercise 5.2.5: Show that given a domain 푈 ⊂ ℂ푛, the Bergman kernel is the unique function 퐾푈 (푧, 휁¯) such that 2 1) for a ﬁxed 휁, 퐾푈 (푧, 휁¯) is in 퐴 (푈), 2) 퐾푈 (푧, 휁¯) = 퐾푈 (휁, 푧¯),

3) the reproducing property ( 5.1 ) holds.

Exercise 5.2.6: Let 푈 ⊂ ℂ푛 be either the unit ball or the unit polydisc. Show that 퐴2(푈) ∩ 퐶(푈) is dense in 퐴2(푈). In particular, this exercise says we only need to check the reproducing property on functions continuous up to the boundary to show we have the Bergman kernel. 5.3. THE SZEGÖ KERNEL 123

Exercise 5.2.7: Let 푈, 푉 ⊂ ℂ푛 be two domains and 푓 : 푈 → 푉 a biholomorphism. Prove 퐾푈 (푧, 휁¯) = det 퐷 푓 (푧) det 퐷 푓 (휁) 퐾푉 푓 (푧), 푓 (휁) .

Exercise 5.2.8: Show that the Bergman kernel for the polydisc is

푛 1 Ö 1 퐾 푛 (푧, 휁¯) = . 픻 휋푛 2 푗=1 (1 − 푧푗 휁¯푗 )

훼 Exercise 5.2.9 (Hard): Show that for some constants 푐 , the set of all monomials 푧 gives a 훼 푐 훼 2 complete orthonormal system of 퐴 (픹푛). Hint: To show orthonormality compute the integral 푖휃 푛 using polar coordinates in each variable separately, that is, let 푧푗 = 푟푗 푒 푗 where 휃 ∈ [0, 2휋] and Í 푟2 < 푓 퐴2 푧훼 푗 푗 1. Then show completeness by showing that if ∈ (픹푛) is orthogonal to all , then √︃ 푓 푐 휋푛훼! 훽 = 0. Finding 훼 = (푛+|훼|)! requires the classical function of special function theory. Exercise 5.2.10: Using the previous exercise, show that the Bergman kernel for the unit ball is 푛! 1 퐾픹 (푧, 휁¯) = , 푛 휋푛 (1 − h푧, 휁i)푛+1 where h푧, 휁i is the standard inner product on ℂ푛.

5.3 The Szegö kernel

We use the same technique to create a reproducing kernel on the boundary by starting with 퐿2(휕푈, 푑휎) instead of 퐿2(푈). We obtain a kernel where we integrate over the boundary rather than the domain itself. Let us give a quick overview, but let us not get into the details. Let 푈 ⊂ ℂ푛 be a bounded domain with smooth boundary. Let 퐶(푈) ∩ O(푈) be the holomorphic functions in 푈 continuous up to the boundary. The restriction of 푓 ∈ 퐶(푈) ∩ O(푈) to 휕푈 is a 2 continuous function, and hence 푓 |휕푈 is in 퐿 (휕푈, 푑휎), where 푑휎 is the surface measure on 휕푈. Taking a closure of these restrictions in 퐿2(휕푈) obtains the Hilbert space 퐻2(휕푈), which is called the Hardy space. The inner product on 퐻2(휕푈) is the 퐿2(휕푈, 푑휎) inner product: ∫ h 푓 , 푔i def= 푓 (푧)푔(푧) 푑휎(푧). 휕푈

훼 2 Exercise 5.3.1: Show that monomials 푧 are a complete orthonormal system in 퐻 (휕픹푛). Exercise 5.3.2: Let 푈 ⊂ ℂ푛 be a bounded domain with smooth boundary. Prove that 퐻2(휕푈) is inﬁnite-dimensional.

Given an 푓 ∈ 퐻2(휕푈), write the Poisson integral ∫ 푃 푓 (푧) = 푓 (휁) 푃(푧, 휁) 푑휎(휁), 휕푈 124 CHAPTER 5. INTEGRAL KERNELS where 푃(푧, 휁) is the Poisson kernel. The Poisson integral reproduces harmonic functions. As holomorphic functions are harmonic, we ﬁnd that if 푓 ∈ 퐶(푈) ∩ O(푈), then 푃 푓 = 푓 . Although 푓 ∈ 퐻2(휕푈) is only deﬁned on the boundary, through the Poisson integral, we have the values 푃 푓 (푧) for 푧 ∈ 푈. For each 푧 ∈ 푈,

푓 ↦→ 푃 푓 (푧)

2 deﬁnes a continuous linear functional. Again we ﬁnd a 푠푧 ∈ 퐻 (휕푈) such that

푃 푓 (푧) = h 푓 , 푠푧i.

For 푧 ∈ 푈 and 휁 ∈ 휕푈, we define def 푆푈 (푧, 휁¯) = 푠푧 (휁), although for a fixed 푧 this is a function only defined almost everywhere as it is an element of 2 2 퐿 (휕푈, 푑휎). The function 푆푈 is the Szegö kernel. If 푓 ∈ 퐻 (휕푈), then ∫ 푃 푓 (푧) = 푓 (휁) 푆푈 (푧, 휁¯) 푑휎(휁). 휕푈

As functions in 퐻2(휕푈) extend to 푈, then 푓 ∈ 퐻2(휕푈) may be considered a function on 푈, ∗ where values in 푈 are given by 푃 푓 . Similarly, we extend 푆(푧, 휁¯) to a function on 푈 × 푈 (where the values on the boundary are deﬁned only almost everywhere). We state without proof that if {휑푗 }푗∈퐼 is a complete orthonormal system for 퐻2(휕푈), then ∑︁ 푆푈 (푧, 휁¯) = 휑푗 (푧)휑푗 (휁) (5.2) 푗∈퐼

for (푧, 휁¯) ∈ 푈 × 푈∗, converging uniformly on compact subsets. As before, this formula shows that 푆 ∗ is conjugate symmetric, and so it extends to (푈 × 푈 ) ∪ (푈 × 푈∗).

Example 5.3.1: In Exercise 5.2.3 , we computed that if 푓 ∈ 퐶(픻) ∩ O(픻), then

1 ∫ 푓 (휁) 푓 (푧) = 푑푠. 2휋 휕픻 1 − 푧휁¯

In other words, 푆 (푧, 휁) = 1 1 . 픻 휋 1−푧휁¯

푆 Exercise 5.3.3: Using the formula (5.2) compute 픹푛 . Chapter 6

Complex analytic varieties

6.1 The ring of germs

Definition 6.1.1. Let 푝 be a point in a topological space 푋. Let 푌 be a set and 푈, 푉 ⊂ 푋 be open neighborhoods of 푝. We say that two functions 푓 : 푈 → 푌 and 푔 : 푉 → 푌 are equivalent if there exists a neighborhood 푊 of 푝 such that 푓 |푊 = 푔|푊 . An equivalence class of functions defined in a neighborhood of 푝 is called a germ of a function. Usually it is denoted by ( 푓 , 푝), but we simply say 푓 when the context is clear. The set of germs of complex-valued functions forms a commutative ring, see exercise below to check the details. For example, to multiply ( 푓 , 푝) and (푔, 푝), take two representatives 푓 and 푔 defined on a common neighborhood multiply them and then consider the germ ( 푓 푔, 푝). Similarly, ( 푓 , 푝) + (푔, 푝) is defined as ( 푓 + 푔, 푝). It is easy to check that these operations are well-defined.

Exercise 6.1.1: Let 푋 be a topological space and 푝 ∈ 푋. Let F be a class of complex-valued functions defined on open subsets of such that whenever 푓 : 푈 → ℂ is in F and 푊 ⊂ 푈 is open, then 푓 |푊 ∈ F, and such that whenever 푓 and 푔 are two functions in F, and 푊 is an open set where both are defined, then 푓 푔|푊 and ( 푓 + 푔)|푊 are also in F. Assume that all constant functions are in F. Show that the ring operations defined above on a set of germs at 푝 of functions from F are well-defined, and that the set of germs at 푝 of functions from F is a commutative ring.

Exercise 6.1.2: Let 푋 = 푌 = ℝ and 푝 = 0. Consider the ring of germs of continuous functions (or smooth functions if you wish). Show that for every continuous 푓 : ℝ → ℝ and every neighborhood 푊 of 0, there exists a 푔 : ℝ → ℝ such that ( 푓 , 0) = (푔, 0), but 푔|푊 ≠ 푓 |푊 .

Germs are particularly useful for holomorphic functions because of the identity theorem. In particular, the behavior of the exercise above does not happen for holomorphic functions. Furthermore, for holomorphic functions, the ring of germs is the same as the ring of convergent power series, see exercise below. No similar result is true for only smooth functions. 푛 Deﬁnition 6.1.2. Let 푝 ∈ ℂ . Write 푛O푝 = O푝 as the ring of germs at 푝 of holomorphic functions.

The ring of germs O푝 has many nice properties, and it is generally a “nicer” ring than the ring O(푈) for some open 푈, and so it is easier to work with if we are interested in local properties. 126 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Exercise 6.1.3: a) Show that O푝 is an integral domain (has no zero divisors). b) Prove the ring of germs at 0 ∈ ℝ of smooth real-valued functions is not an integral domain.

Exercise 6.1.4: Show that the units (elements with multiplicative inverse) of O푝 are the germs of functions which do not vanish at 푝.

Exercise 6.1.5: a) (easy) Show that given a germ ( 푓 , 푝) ∈ O푝, there exists a fixed open neighborhood 푈 of 푝 and a representative 푓 : 푈 → ℂ such that any other representative 푔 can be analytically continued from 푝 to a holomorphic function 푈. b) (easy) Given two representatives 푓 : 푈 → ℂ and 푔 : 푉 → ℂ of a germ ( 푓 , 푝) ∈ O푝, let 푊 be the connected component of 푈 ∩ 푉 that contains 푝. Then 푓 |푊 = 푔|푊 . c) There exists a germ ( 푓 , 푝) ∈ O푝, such that for any open neighborhood 푈 of 푝, and any representative 푓 : 푈 → ℂ we can find another representative of 푔 : 푉 → ℂ of that same germ such that 푔|푈∩푉 ≠ 푓 |푈∩푉 . Hint: 푛 = 1 is sufficient.

Exercise 6.1.6: Show that O푝 is isomorphic to the ring of convergent power series.

Deﬁnition 6.1.3. Let 푝 be a point in a topological space 푋. We say that sets 퐴, 퐵 ⊂ 푋 are equivalent if there exists a neighborhood 푊 of 푝 such that 퐴 ∩ 푊 = 퐵 ∩ 푊. An equivalence class of sets is called a germ of a set at 푝. It is denoted by (퐴, 푝), but we may write 퐴 when the context is clear.

The concept of (푋, 푝) ⊂ (푌, 푝) is deﬁned in an obvious manner, that is, there exist representatives 푋 and 푌, and a neighborhood 푊 of 푝 such that 푋 ∩ 푊 ⊂ 푌 ∩ 푊. Similarly, if (푋, 푝), (푌, 푝) are germs and 푋, 푌 are any representatives of these germs, then the intersection (푋, 푝) ∩ (푌, 푝), is the germ (푋 ∩ 푌, 푝) and the union (푋, 푝) ∪ (푌, 푝) is the germ (푋 ∪ 푌, 푝).

Exercise 6.1.7: Check that the deﬁnition of subset, union, and intersection of germs of sets is well-deﬁned.

Let 푅 be some ring of germs of complex-valued functions at 푝 ∈ 푋 for some topological space −1 푋. If 푓 is a complex-valued function, let 푍푓 be the zero set of 푓 , that is 푓 (0). When ( 푓 , 푝) ∈ 푅 is a germ of a function it makes sense to talk about the germ (푍푓 , 푝). We take the zero set of some representative and look at its germ at 푝.

Exercise 6.1.8: Suppose 푓 and 푔 are two representatives of a germ ( 푓 , 푝) show that the germs (푍푓 , 푝) and (푍푔, 푝) are the same. Exercise 6.1.9: Show that if ( 푓 , 푝) and (푔, 푝) are in 푅 and 푓 and 푔 are some representatives, then (푍푓 , 푝) ∪ (푍푔, 푝) = (푍푓 푔, 푝). 6.2. WEIERSTRASS PREPARATION AND DIVISION THEOREMS 127 6.2 Weierstrass preparation and division theorems

Suppose 푓 is (a germ of) a holomorphic function at a point 푝 ∈ ℂ푛. Write

∞ ∑︁ 푓 (푧) = 푓푘 (푧 − 푝), 푘=0

푘 where 푓푘 is a homogeneous polynomial of degree 푘, that is, 푓푘 (푡푧) = 푡 푓푘 (푧). Deﬁnition 6.2.1. Let 푝 ∈ ℂ푛 and 푓 be a function holomorphic in a neighborhood of 푝. If 푓 is not identically zero, deﬁne def ord푝 푓 = min 푘 ∈ ℕ0 : 푓푘 . 0 .

If 푓 ≡ 0, then deﬁne ord푝 푓 = ∞. The number ord푝 푓 is called the order of vanishing of 푓 at 푝. In other words, the order of vanishing of 푓 at 푝 is 푘, whenever all partial derivatives of order less than 푘 vanish at 푝, and there exists at least one derivative of order 푘 that does not vanish at 푝. In one complex variable, a holomorphic function 푓 with ord0 푓 = 푘 can be written (locally) as 푓 (푧) = 푧푘 푢(푧) for a nonvanishing 푢. In several variables, there is a similar theorem, or in fact a pair of theorems, the so-called Weierstrass preparation and division theorems. Deﬁnition 6.2.2. Let 푈 ⊂ ℂ푛−1 be open, and let 푧0 ∈ ℂ푛−1 denote the coordinates. Suppose a polynomial 푃 ∈ O(푈)[푧푛] is monic of degree 푘 ≥ 0, that is,

푘−1 푃 푧0, 푧 푧푘 ∑︁ 푐 푧0 푧 푗 , ( 푛) = 푛 + 푗 ( ) 푛 푗=0 where 푐푗 are holomorphic functions defined on 푈, such that 푐푗 (0) = 0 for all 푗. Then 푃 is called a Weierstrass polynomial of degree 푘. If the 푐푗 are germs in O0 = 푛−1O0, then 푃 ∈ O0 [푧푛] and 푃 is a germ of a Weierstrass polynomial. The definition (and the theorem that follows) still holds for 푛 = 1. If you read the definition carefully, you will find that if 푛 = 1, then the only Weierstrass polynomial of degree 푘 is 푧푘 . Note that for any 푛, if 푘 = 0, then 푃 = 1. The purpose of this section is to show that every holomorphic function in O0 is up to a unit and a possible small rotation a Weierstrass polynomial, which carries the zeros of 푓 . Consequently the algebraic and geometric properties of 푛O0 can be understood via algebraic and geometric properties of 푛−1O0 [푧푛]. Theorem 6.2.3 (Weierstrass preparation theorem). Suppose 푓 ∈ O(푈) for an open 푈 ⊂ ℂ푛−1 × ℂ, where 0 ∈ 푈, and 푓 (0) = 0. Suppose 푧푛 ↦→ 푓 (0, 푧푛) is not identically zero near the origin and its order of vanishing at the origin is 푘 ≥ 1. Then there exists an open polydisc 푉 = 푉0 × 퐷 ⊂ ℂ푛−1 × ℂ with 0 ∈ 푉 ⊂ 푈, a unique 푢 ∈ O(푉), 푢(푧) ≠ 0 for all 푧 ∈ 푉, and a unique Weierstrass polynomial 푃 of degree 푘 with coefficients holomorphic in 푉0 such that 0 0 0 푓 (푧 , 푧푛) = 푢(푧 , 푧푛) 푃(푧 , 푧푛), 0 0 0 and such that all 푘 zeros (counting multiplicity) of 푧푛 ↦→ 푃(푧 , 푧푛) lie in 퐷 for all 푧 ∈ 푉 128 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Proof. There exists a small disc 퐷 ⊂ ℂ centered at zero such that {0} × 퐷 ⊂ 푈 and such that 0 푓 (0, 푧푛) ≠ 0 for 푧푛 ∈ 퐷 \{0}. By continuity of 푓 we ﬁnd a small polydisc 푉 = 푉 × 퐷 such that 푉 ⊂ 푈 and 푓 is not zero on 푉0 × 휕퐷.

zero of 푧푛 ↦→ 푓 (0, 푧푛) the 푘 (푘 = 4) zeros of 0 of order 푘 푧푛 ↦→ 푓 (푧 , 푧푛)

퐷퐷 0 훼1(푧 ) 0 훼2(푧 ) 푧푛 = 0 0 훼3(푧 ) 0 훼4(푧 )

0 0 0 푧푛-plane where 푧 = 0 푧푛-plane where 푧 ∈ 푉

By the one-variable argument principle ( Theorem B.25 ) the number of zeros (with multiplicity) 0 of 푧푛 ↦→ 푓 (푧 , 푧푛) in 퐷 is 휕 푓 0 1 ∫ (푧 , 휁) 휕푧푛 푑휁. 0 2휋푖 휕퐷 푓 (푧 , 휁) As 푓 (푧0, 휁) does not vanish when 푧0 ∈ 푉0 and 휁 ∈ 휕퐷, the expression above is a continuous integer-valued function of 푧0 ∈ 푉0. The expression is equal to 푘 when 푧0 = 0, and so it is equal to 푘 0 0 0 0 0 for all 푧 ∈ 푉 . Write the zeros of 푧푛 ↦→ 푓 (푧 , 푧푛) as 훼1(푧 ), . . . , 훼푘 (푧 ), including multiplicity. The zeros are not ordered in any particular way. Pick some ordering for every 푧0. Write

푘 푃 푧0, 푧 Ö푧 훼 푧0 푧푘 푐 푧0 푧푘−1 푐 푧0 . ( 푛) = 푛 − 푗 ( ) = 푛 + 푘−1( ) 푛 + · · · + 0( ) 푗=1 For a fixed 푧0, 푃 is uniquely defined as the ordering of zeros does not matter in its definition (see exercise below). It is clear that 푢 and 푃 are unique if they exist (that is, if they exist as holomorphic functions).

The functions 훼푗 are not even continuous in general (see Example 6.2.4 ). However, we will prove that the functions 푐푗 are holomorphic. The functions 푐푗 are (up to sign) the elementary symmetric

functions of 훼1, . . . , 훼푘 (see below). It is a standard theorem in algebra (see Exercise 6.2.1 ) that the elementary symmetric functions are polynomials in the so-called power sum functions in the 훼푗 s:

푘 0 ∑︁ 0 푚 푠푚 (푧 ) = 훼푗 (푧 ) , 푚 = 1, . . . , 푘. 푗=1

Therefore, if we show that the power sums 푠푚 are holomorphic, then 푐ℓ are holomorphic.

A reﬁnement of the argument principle (see also Theorem B.25 ) says: If ℎ and 푔 are holomorphic functions on a disc 퐷, continuous on 퐷, such that 푔 has no zeros on 휕퐷, and 훼1, . . . , 훼푘 are the zeros of 푔 in 퐷, then 푘 1 ∫ 푔0(휁) ∑︁ ℎ(휁) 푑휁 = ℎ(훼 ). 2휋푖 푔(휁) 푗 휕퐷 푗=1 6.2. WEIERSTRASS PREPARATION AND DIVISION THEOREMS 129

The formula above with ℎ(휁) = 휁 푚 and 푔(휁) = 푓 (푧0, 휁) says that 푘 휕 푓 (푧0, 휁) ∑︁ 1 ∫ 휕휁 푠 (푧0) = 훼 (푧0)푚 = 휁 푚 푑휁. 푚 푗 2휋푖 푓 (푧0, 휁) 푗=1 휕퐷

The function 푠 is clearly continuous, and if we differentiate under the integral with 휕 for 푚 휕푧¯ℓ ℓ = 1, . . . , 푛 − 1, we find that 푠푚 is holomorphic. Finally, we wish to show that 푃 divides 푓 as claimed. For each fixed 푧0, one variable theory says 0 푓 (푧 ,푧푛) that 푧 ↦→ 0 has only removable singularities, and in fact, it has no zeros as we defined 푃 to 푛 푃(푧 ,푧푛) exactly cancel them all out. The Cauchy formula on 푓/푃 then says that the function ∫ 푓 (푧0, 휁) 푢 푧0, 푧 1 푑휁 ( 푛) = 0 2휋푖 휕퐷 푃(푧 , 휁)(휁 − 푧푛) 0 푓 (푧 ,푧푛) 0 is equal to 0 . The function 푢 is clearly continuous and holomorphic in 푧 for each fixed 푧 . By 푃(푧 ,푧푛) 푛 differentiating under the integral, we find that it is also holomorphic in 푧0. 푓 푧 , 푧 푧2 푧 Example 6.2.4: A useful example to keep in mind is ( 1 2) = 2 − 1, a Weierstrass polynomial 0 √ in 푧2 of degree 푘 = 2. So 푧 = 푧1. For all 푧1 except the origin there are two zeros, ± 푧1. Call one of them 훼1(푧1) and one of them 훼2(푧1). Recall there is no continuous choice of a square root that works for all 푧1, so no matter how you choose, 훼1 and 훼2 will not be continuous. At the origin there is only one zero of order two, so 훼1(0) = 훼2(0) = 0. On the other hand the symmetric functions 푐 푧 훼 푧 훼 푧 푐 푧0 훼 푧 훼 푧 푧 1( 1) = − 1( 1) − 2( 1) = 0 and 0( 1) = 1( 1) 2( 1) = − 1 are holomorphic. The 푘 depends on the coordinates chosen. If we do a linear change of coordinates and consider 푔 푧 , 푧 푓 푧 , 푧 푔 푧 , 푧 푧2 푧 푧 ( 1 2) = − ( 2 1), then ( 1 2) = 1 − 2, which is a Weierstrass polynomial in 2 of degree 푘 훼 푧 푧2 푐 푧 푧2 = 1. After the change, there is only one zero, 1( 1) = 1, and so 0( 1) = − 1.

A function 푓 (푧1, . . . , 푧푛) is symmetric if 푓 = 푓 ◦ 푝 for all permutations of the variables 푝. The elementary symmetric functions of 훼1, . . . , 훼푘 are the coeﬃcients 휎푗 of the polynomial

푘 Ö 푘 푘−1 2 푡 + 훼푗 = 푡 + 휎1 푡 + · · · + 휎푘−2 푡 + 휎푘−1 푡 + 휎푘 . 푗=1 In other words:

휎1 = 훼1 + 훼2 + · · · + 훼푘 , 휎 = 훼 훼 + 훼 훼 + · · · + 훼 훼 , 2 . 1 2 1 3 푘−1 푘 .

휎푘−1 = 훼2훼3 ··· 훼푘 + 훼1훼3훼4 ··· 훼푘 + · · · + 훼1훼2 ··· 훼푘−1, 휎푘 = 훼1훼2 ··· 훼푘 .

So for example when 푘 = 2, then 휎2 = 훼1훼2 and 휎1 = 훼1 + 훼2. The function 휎1 happens to already be a power sum. We can write 휎2 as a polynomial in the power sums: 1 휎 = 훼 + 훼 2 − 훼2 + 훼2 . 2 2 1 2 1 2 130 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Exercise 6.2.1: Show that elementary symmetric functions are polynomials in the power sums.

Exercise 6.2.2: Prove the fundamental theorem of symmetric polynomials: Every symmetric polynomial can be written as a polynomial in the elementary symmetric functions. Use the following procedure. Using double induction, suppose the theorem is true if the number of variables is less than 푘, and the theorem is true in 푘 variables for degree less than 푑. Consider a symmetric 푃(푧1, . . . , 푧푘 ) of degree 푑. Write 푃(푧1, . . . , 푧푘−1, 0) by induction hypothesis as a polynomial in the elementary symmetric functions of one less variable. Use the same coeﬃcients, but plug in the elementary symmetric functions of 푘 variables except the symmetric polynomial in 푘 variables of degree 푘, that is except the 푧1 ··· 푧푘 . You will obtain a symmetric function 퐿(푧1, . . . , 푧푘 ) and you need to show 퐿(푧1, . . . , 푧푘−1, 0) = 푃(푧1, . . . , 푧푘−1, 0). Now use symmetry to prove that 푃(푧1, . . . , 푧푘 ) = 퐿(푧1, . . . , 푧푘 ) + 푧1 ··· 푧푘푄(푧1, . . . , 푧푘 ). Then note that 푄 has lower degree and ﬁnish by induction.

Exercise 6.2.3: Extend the previous exercise to power series. Suppose 푓 (푧1, . . . , 푧푘 ) is a convergent symmetric power series at 0, show that 푓 can be written as a convergent power series in the elementary symmetric functions.

0 Exercise 6.2.4: Suppose 푃(푧 , 푧푛) is a Weierstrass polynomial of degree 푘, and write the zeros 0 0 as 훼1(푧 ), . . . , 훼푘 (푧 ). These are not holomorphic functions, but suppose that 푓 is a symmetric 0 0 convergent power series at the origin in 푘 variables. Show that 푓 훼1(푧 ), . . . , 훼푘 (푧 ) is a holomorphic function of 푧0 near the origin.

The hypotheses of the preparation theorem are not an obstacle. If a holomorphic function 푓 is such that 푧푛 ↦→ 푓 (0, 푧푛) vanishes identically, then we can make a small linear change of coordinates 퐿 (퐿 can be a matrix arbitrarily close to the identity) such that 푓 ◦ 퐿 satisﬁes the hypotheses of the theorem. For example, 푓 (푧1, 푧2, 푧3) = 푧1푧3 + 푧2푧3 does not satisfy the hypotheses of the theorem as 푓 (0, 0, 푧3) ≡ 0. But for an arbitrarily small 휖 ≠ 0, replacing 푧2 with 푧2 + 휖푧3 leads to 푓˜ 푧 , 푧 , 푧 푓 푧 , 푧 휖푧 , 푧 푧 푧 푧 푧 휖푧2 푓˜ , , 푧 휖푧2 푓˜ ( 1 2 3) = ( 1 2 + 3 3) = 1 3 + 2 3 + 3, and (0 0 3) = 3. Thence satisﬁes the hypotheses of the theorem.

Exercise 6.2.5: Prove the fact above about the existence of 퐿 arbitrarily close to the identity. Exercise 6.2.6: Prove that a monic polynomial 푃(휁) of one variable is uniquely determined by its zeros up to multiplicity. That is, suppose 푃 and 푄 are two monic polynomials with the same zeros up to multiplicity, then 푃 = 푄. That proves the uniqueness of the Weierstrass polynomial. Exercise 6.2.7: Suppose 퐷 ⊂ ℂ is a bounded domain, 0 ∈ 퐷, 푈0 ⊂ ℂ푛−1 is a domain, 0 ∈ 푈0, 0 0 0 and 푃 ∈ O(푈 )[푧푛] is a Weierstrass polynomial such that 푃(푧 , 푧푛) is not zero on 푈 × 휕퐷. Then 0 0 for any 푧 ∈ 푈, all zeros of 푧푛 ↦→ 푃(푧 , 푧푛) are in 퐷. Exercise 6.2.8: Let 퐷 ⊂ ℂ be a bounded domain, and 푈0 ⊂ ℂ푛−1 a domain. Suppose 푓 is a continuous function on 푈0 × 퐷 holomorphic on 푈0 × 퐷, where 푓 is zero on at least one point of 0 0 0 푈 × 퐷, and 푓 is never zero on 푈 × 휕퐷. Prove that 푧푛 ↦→ 푓 (푧 , 푧푛) has at least one zero in 퐷 for every 푧0 ∈ 푈0. 6.2. WEIERSTRASS PREPARATION AND DIVISION THEOREMS 131

The order of vanishing of 푓 at the origin is a lower bound on the number 푘 in the theorem. The 푓 (푧 , 푧 ) 푧2 + 푧3 order of vanishing for a certain variable may be larger than this lower bound. If 1 2 = 1 2, then the 푘 we get is 3, but ord0 푓 = 2. We can make a small linear change of coordinates to ensure 푘 = ord0 푓 . With the 푓 as above, 푓 (푧1 + 휖푧2, 푧2) gives 푘 = 2 as expected. The Weierstrass preparation theorem is a generalization of the implicit function theorem. When 0 푘 = 1 in the theorem, then we obtain the Weierstrass polynomial 푧푛 + 푐0(푧 ). That is, the zero set of 푓 is a graph of the holomorphic function −푐0. Therefore, the Weierstrass theorem is a generalization of the implicit function theorem to the case when 휕 푓 is zero. We can still “solve” for 푧 , but we 휕푧푛 푛 ﬁnd 푘 solutions given as the zeros of the obtained Weierstrass polynomial. There is an obvious statement of the preparation theorem for germs.

Exercise 6.2.9: State and prove a germ version of the preparation theorem.

Theorem 6.2.5 (Weierstrass division theorem). Suppose 푓 is holomorphic near the origin, and suppose 푃 is a Weierstrass polynomial of degree 푘 ≥ 1 in 푧푛. Then there exists a neighborhood 푉 of the origin and unique 푞, 푟 ∈ O(푉), where 푟 is a polynomial in 푧푛 of degree less than 푘, and on 푉,

푓 = 푞푃 + 푟.

Note that 푟 need not be a Weierstrass polynomial; it need not be monic nor do the coefficients need to vanish at the origin. It is simply a polynomial in 푧푛 with coefficients that are holomorphic functions of the first 푛 − 1 variables.

Proof. Uniqueness is left as an exercise. Consider a connected neighborhood 푉 = 푉0 × 퐷 of the origin, where 퐷 is a disc, 푓 and 푃 are continuous in 푉0 × 퐷, and 푃 is not zero on 푉0 × 휕퐷. Let ∫ 푓 (푧0, 휁) 푞 푧0, 푧 1 푑휁. ( 푛) = 0 2휋푖 휕퐷 푃(푧 , 휁)(휁 − 푧푛) As 푃 is not zero on 푉0 × 휕퐷, the function 푞 is holomorphic in 푉 (diﬀerentiate under the integral). If 푃 did divide 푓 , then 푞 would really be 푓/푃. But if 푃 does not divide 푓 , then the Cauchy integral formula does not apply and 푞 is not equal to 푓/푃. Interestingly the expression does give the quotient in the division with remainder. Write 푓 using the Cauchy integral formula in 푧푛 and subtract 푞푃 to obtain 푟: ∫ 푓 (푧0, 휁)푃(푧0, 휁) − 푓 (푧0, 휁)푃(푧0, 푧 ) 푟 푧0, 푧 푓 푧0, 푧 푞 푧0, 푧 푃 푧0, 푧 1 푛 푑휁. ( 푛) = ( 푛) − ( 푛) ( 푛) = 0 2휋푖 휕퐷 푃(푧 , 휁)(휁 − 푧푛)

We need to show 푟 is a polynomial in 푧푛 of degree less than 푘. In the expression inside the integral, Í ℎ 푧0, 휁 휁 푗 푧 푗 휁 푧 the numerator is of the form 푗 푗 ( )( − 푛) and is therefore divisible by ( − 푛). The numerator is a polynomial of degree 푘 in 푧푛. After dividing by (휁 − 푧푛), the integrand becomes a polynomial in 푧푛 of degree 푘 − 1. Use linearity of the integral to integrate the coefficients of the polynomial. Each coefficient is a holomorphic function in 푉0 and the proof is finished. Some coefficients may have integrated to zero, so we can only say that 푟 is a polynomial of degree 푘 − 1 or less. 132 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

For example, let 푓 (푧, 푤) = 푒푧 + 푧4푒푤 + 푧푤2푒푤 + 푧푤 and 푃(푧, 푤) = 푤2 + 푧3. Then 푃 is a Weierstrass polynomial in 푤 of degree 푘 = 2. A bit of computation shows 1 ∫ 푒푧 + 푧4푒휁 + 푧휁 2푒휁 + 푧휁 푑휁 = 푧푒푤, so 푓 (푧, 푤) = 푧푒푤 푤2 + 푧3 + 푧푤 + 푒푧 . 2휋푖 (휁 2 + 푧3)(휁 − 푤) 휕픻 |{z} | {z } | {z } 푞 푃 푟 Notice that 푟 is a polynomial of degree 1 in 푤, but it is neither monic, nor do coeﬃcients vanish at 0.

Exercise 6.2.10: Prove the uniqueness part of the theorem.

Exercise 6.2.11: State and prove a germ version of the division theorem.

The Weierstrass division theorem is a generalization of the division algorithm for polynomials with coefficients in a field, such as the complex numbers: If 푓 (휁) is a polynomial, and 푃(휁) is a nonzero polynomial of degree 푘, then there exist polynomials 푞(휁) and 푟(휁) with degree of 푟 less than 푘 such that 푓 = 푞푃 + 푟. If the coefficients are in a commutative ring, we can divide as long as 푃 is monic. The Weierstrass division theorem says that in the case of the ring O푝, we can divide by a monic 푃 ∈ 푛−1O푝 [푧푛], even if 푓 is a holomorphic function (a “polynomial of infinite degree”) as long as 푓 (0, 푧푛) has finite order.

6.3 The dependence of zeros on parameters

Let us prove that the zeros change holomorphically as long as they do not come together. We will prove shortly that the zeros come together only on a small set; it is a zero set of a certain holomorphic function called the discriminant. A set of zeros are said to be geometrically distinct if they are distinct points of ℂ. A zero is called 2 geometrically unique if it is a unique complex number. For example, (푧푛 − 1) has a geometrically 2 unique zero at 1, and (푧푛 − 1) (푧푛 + 1) has two geometrically distinct zeros, 1 and −1. Proposition 6.3.1. Let 퐷 ⊂ ℂ and 푈0 ⊂ ℂ푛−1 be domains, and 푓 ∈ O(푈0 × 퐷). Suppose that for 0 0 0 0 each fixed 푧 ∈ 푈 the function 푧푛 ↦→ 푓 (푧 , 푧푛) has a geometrically unique zero 훼(푧 ) ∈ 퐷. Then 훼 is holomorphic in 푈0. The proposition shows that the regularity conclusion of the implicit function theorem holds under the hypothesis that there exists some local solution for 푧푛. This result holds only for holomorphic functions and not for real-analytic functions. For example, 푥2 − 푦3 = 0 has a unique real solution 푦 = 푥2/3, but that function is not even differentiable. Proof. We must show that 훼 is holomorphic near any point, which, without loss of generality, is the origin. Apply the preparation theorem to find 푓 = 푢푃, where 푃 is a Weierstrass polynomial 0 0 0 0 in O(푉 )[푧푛] for some 푉 ⊂ 푈 and all zeros of 푧푛 ↦→ 푃(푧 , 푧푛) are in 퐷. As 훼 is a geometrically unique zero in 퐷, 푃 푧0, 푧 푧 훼 푧0 푘 푧푘 푘훼 푧0 푧푘−1 ( 푛) = 푛 − ( ) = 푛 − ( ) 푛 + · · · The coefficients of 푃 are holomorphic, so 훼 is holomorphic. 6.3. THE DEPENDENCE OF ZEROS ON PARAMETERS 133

Proposition 6.3.2. Let 퐷 ⊂ ℂ and 푈0 ⊂ ℂ푛−1 be domains, and 푓 ∈ O(푈0 × 퐷). Let 푚 ∈ ℕ be 0 0 0 such that for each 푧 ∈ 푈 , the function 푧푛 ↦→ 푓 (푧 , 푧푛) has precisely 푚 geometrically distinct zeros. 0 0 0 Then locally near each point in 푈 there exist 푚 holomorphic functions 훼1(푧 ), . . . , 훼푚 (푧 ), positive integers 푘1, . . . , 푘푚, and a nonvanishing holomorphic function 푢 such that 푚 0 0 Ö 0 푘푗 푓 (푧 , 푧푛) = 푢(푧 , 푧푛) 푧푛 − 훼푗 (푧 ) . 푗=1

We can only deﬁne 훼1 through 훼푚 locally (on a smaller domain) as we do not know how 훼1 0 through 훼푚 are ordered, and the order could change as we move around 푈 if it is not simply connected. If 푈0 is simply connected, then the functions can be deﬁned globally by analytic 0

continuation. For an example where 푈 is not simply connected, recall Example 6.2.4 . Consider 푈0 = ℂ \{0} and think 퐷 = ℂ rather than a disc for simplicity. Then 푈0 is not simply connected, and there do not exist continuous functions 훼1(푧1) and 훼2(푧1) that are zeros of the Weierstrass 푧2 푧 푧 훼 푧 푧 훼 푧 푧 polynomial, that is 2 − 1 = 2 − 1( 1) 2 − 2( 1) . These would be the two square roots of 1, and there is no continuous (let alone holomorphic) square root deﬁned in ℂ \{0}. Such roots can be chosen to be holomorphic on any smaller simply connected open subset of 푈0, for example, on any disc Δ ⊂ 푈0.

Exercise 6.3.1: Let 퐷 ⊂ ℂ be a bounded domain, 푈0 ⊂ ℂ푛−1 a domain, 푓 a continuous function on 푈0 × 퐷 holomorphic on 푈0 × 퐷, where 푓 is zero on at least one point of 푈0 × 퐷, and 푓 is 0 0 0 0 never zero on 푈 × 휕퐷. Suppose that for each ﬁxed 푧 ∈ 푈 the function 푧푛 ↦→ 푓 (푧 , 푧푛) has at 0 0 0 most one zero in 퐷. Prove that for each 푧 ∈ 푈 푧푛 ↦→ 푓 (푧 , 푧푛) has exactly one zero in 퐷. Note:

And therefore by Proposition 6.3.1 , that zero is a holomorphic function.

Exercise 6.3.2: Prove Proposition 6.3.2 . See the exercise above and Proposition 6.3.1 .

Theorem 6.3.3. Let 퐷 ⊂ ℂ be a bounded domain, 푈0 ⊂ ℂ푛−1 a domain, and 푓 ∈ O(푈0 × 퐷). Suppose the zero set 푓 −1(0) has no limit points on 푈0 × 휕퐷. Then there exists an 푚 ∈ ℕ and a holomorphic function Δ: 푈0 → ℂ, not identically zero, such that for every 푧0 ∈ 푈0 \ 퐸, where −1 0 0 퐸 = Δ (0), 푧푛 ↦→ 푓 (푧 , 푧푛) has exactly 푚 geometrically distinct zeros in 퐷, and 푧푛 ↦→ 푓 (푧 , 푧푛) has strictly less than 푚 geometrically distinct zeros for 푧0 ∈ 퐸. The complement of a zero set of a holomorphic function is connected, open and dense. The function Δ is called the discriminant function and its zero set 퐸 is called the discriminant set. For the quadratic equation, Δ is the discriminant we learned about in high school. 0 0 Proof. The zeros of 푧푛 ↦→ 푓 (푧 , 푧푛) are isolated, and there are ﬁnitely many for every 푧 as 퐷 is bounded and 푓 −1(0) has no limit points on 푈0 ×휕퐷. For any 푝0 ∈ 푈0 we deﬁne two useful paths. Let 훾 be the union of nonintersecting small simple closed curves in 퐷, one around each geometrically 0 distinct zero of 푧푛 ↦→ 푓 (푝 , 푧푛). Let 휆 be a large closed path in 퐷 going exactly once around all the zeros and such that the interior of 휆 is in 퐷. Suppose 훾 and 휆 intersect no zeros. By continuity, the curves 훾 and 휆 do not intersect any zeros for 푧0 near 푝0. Since the set 푓 −1(0) is closed and the zeros do not accumulate on 푈0 × 휕퐷, then for 푧0 near 푝0 the zeros stay a positive distance away from the boundary. So 휆 can be picked to go around all the zeros exactly once for 푧0 near 푝0. 134 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

0 roots of 푧푛 ↦→ 푓 (푧 , 푧푛) 훾 휆 훾 훼 (푧0) 훾 훾 2 훼 푧0 훼 푧0 0 1( ) 4( ) 0 훼3(푧 ) 푧푛-plane for ﬁxed 푧

0 0 0 The argument principle applied to 푧푛 ↦→ 푓 (푧 , 푧푛) using 휆 for 푧 near 푝 shows that the number of zeros (with multiplicity) is bounded by some 푀 ∈ ℕ near 푝0. The 푀 is locally well-defined (it does not depend on 휆 as long as it contains all the zeros), and it is a locally constant function of 푧0 ∈ 푈0. As 푈0 is connected, and it is a fixed number. The maximum number of geometrically distinct zeros must be bounded by 푀. Let 푚 be the maximal number of geometrically distinct zeros and suppose that at some point in 푈0, there are exactly 푚 geometrically distinct zeros. 푈0 푈0 푧0 푈0 푧 푓 푧0, 푧 푚 푈0 Let 푚 ⊂ be the set of ∈ for which 푛 ↦→ ( 푛) has exactly zeros. Write as 푈0 푈0 퐸 퐸 푈0 푈0 푚 푈0 a union of disjoint sets = 푚 ∪ , where = \ 푚. By definition of , 푚 is nonempty. 푝0 푈0 훾 훾0 Suppose ∈ 푚 and goes around the zeros as above. Let be a single component curve of the path 훾 going around one of the zeros. The argument principle with respect to 훾0 says that 훾0 must contain at least one zero for all 푧0 near 푝0. There are only finitely many zeros, and so for 푧0 in 0 0 some neighborhood of 푝 , 푧푛 ↦→ 푓 (푧 , 푧푛) has at least 푚 zeros in 훾, and as 푚 is the maximum, it has 푚 푈0 exactly zeros. In other words, 푚 is open. 푈0 푚 훼 , . . . , 훼 Locally on 푚, there exist holomorphic functions 1 푚 giving the zeros by the previous 푈0 proposition. We cannot define these on all of 푚 as we do not know how they are ordered. The function 0 Ö 0 0 Δ(푧 ) = 훼푗 (푧 ) − 훼푘 (푧 ) 푗≠푘 푧0 푈0 defined for ∈ 푚 does not depend on the order. That means Δ is well-defined as a function on the 푈0 훼 open set 푚, and since 푗 are locally holomorphic, Δ is holomorphic. 0 0 0 Let 푝 ∈ 퐸 ∩ 푈푚, so there are fewer than 푚 zeros at 푝 . Suppose 훾 and 휆 are as above. In each particular component 훾0 of 훾, there must be at least one zero for all 푧0 near 푝0 by the same argument as above. There exist 푧0 arbitrarily near 푝0 where there are 푚 zeros. The region between 휆 and 훾 0 (including the curves) is compact, and so by continuity, if 푧푛 ↦→ 푓 (푝 , 푧푛) was nonzero on it, so is 0 0 0 0 푧푛 ↦→ 푓 (푧 , 푧푛) for 푧 near 푝 . As no zeros of 푧푛 ↦→ 푓 (푧 , 푧푛) lie outside 휆, we have that all zeros lie 훾 푧0 푈0 푝0 훾0 in one of the components of . So if ∈ 푚 near , there must be one component that contains 푧0 푈0 푝0 at least two zeros. Let { ℓ} be an arbitrary sequence of points in 푚 going to . As the number of 푧0 푧 푓 푧0 , 푧 zeros is finite, { ℓ} has a subsequence such that 푛 ↦→ ( ℓ 푛) has at least two zeros in some fixed 훾0 훾 푧0 푧0 푧0 component of for ℓ. Assume { ℓ} is this subsequence. Label the two distinct zeros at ℓ as 훼 푧0 훼 푧0 푝0 훾0 훼 푝0 1( ℓ) and 2( ℓ). At there is only a single (geometrically) zero in , let us name it 1( ). As 푓 −1 훼 푧0 훼 푧0 훼 푝0 ℓ (0) is closed it must be that 1( ℓ) and 2( ℓ) both approach 1( ) as → ∞. As all zeros 푧0 are necessarily bounded, limℓ→∞ Δ( ℓ) = 0. As the limit is zero for a subsequence of an arbitrary sequence, lim Δ(푧0) = 0. 0 0 0 푧 ∈푈푚→푝 6.4. PROPERTIES OF THE RING OF GERMS 135

푈0 푧0 푧0 퐸 We have already deﬁned Δ on 푚, so set Δ( ) = 0 if ∈ . The function Δ is a continuous 0 0

푈 퐸 푈 function on that is zero precisely on and holomorphic on 푚. Radó’s theorem (Theorem 2.4.10) says that Δ is holomorphic in 푈0. The discriminant given above is really the discriminant of the set 푓 −1(0) rather than of the corresponding Weierstrass polynomial. Often for Weierstrass polynomials the discriminant is Î 훼 푧0 훼 푧0 deﬁned as 푗≠푘 푗 ( ) − 푘 ( ) taking multiple zeros into account, and therefore the “discriminant” could be identically zero. It will be clear from upcoming exercises that if the Weierstrass polynomial is irreducible, then the two notions do in fact coincide.

Exercise 6.3.3: Prove that if 푓 ∈ O(푈), then 푈 \ 푓 −1(0) is not simply connected if 푓 −1(0) is nonempty. In particular, in the theorem, 푈0 \ 퐸 is not simply connected if 퐸 ≠ ∅.

Exercise 6.3.4: Let 퐷 ⊂ ℂ be a bounded domain, and 푈0 ⊂ ℂ푛−1 a domain. Suppose 푓 is a continuous function on 푈0 × 퐷 holomorphic on 푈0 × 퐷, and 푓 is never zero on 푈0 × 휕퐷. Suppose 훾 : [0, 1] → 푈0 is a continuous function such that 푓 훾(0), 푐 = 0 for some 푐 ∈ 퐷. Prove that there exists a continuous function 훼 : [0, 1] → ℂ such that 훼(0) = 푐 and 푓 훾(푡), 훼(푡) = 0 for all 푡 ∈ [0, 1]. Hint: Show it is possible for a path arbitrarily close to 훾, but one that stays away from the discriminant.

6.4 Properties of the ring of germs

Let us prove some basic properties of the ring of germs of holomorphic functions. First some algebra terminology. Given a commutative ring 푅, an ideal 퐼 ⊂ 푅 is a subset such that 푓 푔 ∈ 퐼 whenever 푓 ∈ 푅 and 푔 ∈ 퐼 and 푔 + ℎ ∈ 퐼 whenever 푔, ℎ ∈ 퐼. An intersection of ideals is again an ideal, and hence it makes sense to talk about the smallest ideal containing a set of elements. An ideal 퐼 is generated by 푓1, . . . , 푓푘 if 퐼 is the smallest ideal containing { 푓1, . . . , 푓푘 }. We then write 퐼 = ( 푓1, . . . , 푓푘 ). Every element in 퐼 can be written as 푐1 푓1 + · · · + 푐푘 푓푘 where 푐1, . . . , 푐푘 ∈ 푅.A principal ideal is an ideal generated by a single element, that is, ( 푓 ). For convenience, when talking about germs of functions we often identify a representative with the germ when the context is clear. So by abuse of notation, we often write 푓 ∈ O푝 instead of ( 푓 , 푝) ∈ O푝 and ( 푓1, . . . , 푓푘 ) instead of ( 푓1, 푝),..., ( 푓푘 , 푝) . As in the following exercises.

Exercise 6.4.1: a) Suppose 푓 ∈ O푝 is such that 푓 (푝) ≠ 0, and ( 푓 ) is the ideal generated by 푓 . Prove ( 푓 ) = O푝. b) Let 픪푝 = (푧1 − 푝1, . . . , 푧푛 − 푝푛) ⊂ O푝 be the ideal generated by the coordinate functions. Show that if 푓 (푝) = 0, then 푓 ∈ 픪푝. c) Show that if 퐼 ( O푝 is a proper ideal (ideal such that 퐼 ≠ O0), then 퐼 ⊂ 픪푝, that is 픪푝 is a maximal ideal.

Exercise 6.4.2: Suppose 푛 = 1. Show that 1O푝 is a principal ideal domain (PID), that is every ideal is a principal ideal. More precisely, show that given an ideal 퐼 ⊂ 1O푝, then there exists a 푘 = 0, 1, 2,..., such that 퐼 = (푧 − 푝)푘 . 136 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Exercise 6.4.3: If 푈, 푉 ⊂ ℂ푛 are two neighborhoods of 푝 and ℎ : 푈 → 푉 is a biholomorphism. First prove that it makes sense to talk about 푓 ◦ ℎ for any ( 푓 , 푝) ∈ O푝. Then prove that 푓 ↦→ 푓 ◦ ℎ is a ring isomorphism.

A commutative ring 푅 is Noetherian if every ideal in 푅 is ﬁnitely generated. That is, for every ideal 퐼 ⊂ 푅 there exist ﬁnitely many generators 푓1, . . . , 푓푘 ∈ 퐼: Every 푔 ∈ 퐼 can be written as 푔 = 푐1 푓1 + · · · + 푐푘 푓푘 , for some 푐1, . . . , 푐푘 ∈ 푅. In an exercise, you proved 1O푝 is a PID. So 1O푝 is Noetherian. In higher dimensions, the ring of germs may not be a PID, but it is Noetherian.

Theorem 6.4.1. O푝 is Noetherian. Proof. Without loss of generality 푝 = 0. The proof is by induction on dimension. By an exercise above, 1O0 is Noetherian. By another exercise, we are allowed to do a change of coordinates at zero. For induction suppose 푛−1O0 is Noetherian, and let 퐼 ⊂ 푛O0 be an ideal. If 퐼 = {0} or 퐼 = 푛O0, then the assertion is obvious. Therefore, assume that all elements of 퐼 vanish at the origin (퐼 ≠ 푛O0), and that there exist elements that are not identically zero (퐼 ≠ {0}). Let 푔 be such an element. After perhaps a linear change of coordinates, assume 푔 is a Weierstrass polynomial in 푧푛 by the preparation theorem. The ring 푛−1O0 [푧푛] is a subring of 푛O0. The set 퐽 = 푛−1O0 [푧푛] ∩ 퐼 is an ideal in the ring

푛−1O0 [푧푛]. By the Hilbert basis theorem (see Theorem D.4 in the appendix for a proof), as 푛−1O0 is Noetherian, the ring 푛−1O0 [푧푛] is also Noetherian. Thus 퐽 has ﬁnitely many generators, that is 퐽 = (ℎ1, . . . , ℎ푘 ), in the ring 푛−1O0 [푧푛]. By the division theorem, every 푓 ∈ 퐼 is of the form 푓 = 푞푔 + 푟, where 푟 ∈ 푛−1O0 [푧푛] and 푞 ∈ 푛O0. As 푓 and 푔 are in 퐼, so is 푟. As 푔 and 푟 are in 푛−1O0 [푧푛], they are both in 퐽. Write 푔 = 푐1ℎ1 + · · · + 푐푘 ℎ푘 and 푟 = 푑1ℎ1 + · · · + 푑푘 ℎ푘 . Then 푓 = (푞푐1 + 푑1)ℎ1 + · · · + (푞푐푘 + 푑푘 )ℎ푘 . So ℎ1, . . . , ℎ푘 also generate 퐼 in 푛O0.

Exercise 6.4.4: Prove that every proper ideal 퐼 ⊂ O0 where 퐼 ≠ {0} is generated by Weierstrass polynomials. As a technicality, note that a Weierstrass polynomial of degree 0 is just 1, so it works for 퐼 = O0.

Exercise 6.4.5: We saw above that 1O푝 is a PID. Prove that if 푛 > 1, then 푛O푝 is not a PID.

Theorem 6.4.2. O푝 is a unique factorization domain (UFD). That is, up to a multiplication by a unit, every element has a unique factorization into irreducible elements of O푝. Proof. Again assume 푝 = 0 and induct on the dimension. The one-dimensional statement is an

exercise below. If 푛−1O0 is a UFD, then 푛−1O0 [푧푛] is a UFD by the Gauss lemma (see Theorem D.6 ). Take 푓 ∈ 푛O0. After perhaps a linear change of coordinates 푓 = 푞푃, for 푞 a unit in 푛O0, and 푃 a Weierstrass polynomial in 푧푛. As 푛−1O0 [푧푛] is a UFD, 푃 has a unique factorization in 푛−1O0 [푧푛] into 푃 = 푃1푃2 ··· 푃푘 . So 푓 = 푞푃1푃2 ··· 푃푘 . That 푃푗 are irreducible in 푛O0 is left as an exercise. Suppose 푓 = 푞푔˜ 1푔2 ··· 푔푚 is another factorization. The preparation theorem applies to each 푔푗 . Therefore, write 푔푗 = 푢푗 푃e푗 for a unit 푢푗 and a Weierstrass polynomial 푃e푗 . We obtain 푓 = 푢푃e1푃e2 ··· 푃e푚 for a unit 푢. By uniqueness part of the preparation theorem we obtain 푃 = 푃e1푃e2 ··· 푃e푚. Conclusion is obtained by noting that 푛−1O0 [푧푛] is a UFD. 6.5. VARIETIES 137

Exercise 6.4.6: Finish the proof of the theorem by proving 1O푝 is a unique factorization domain.

Exercise 6.4.7: Show that if an element is irreducible in 푛−1O0 [푧푛], then it is irreducible in 푛O0.

6.5 Varieties

−1 As before, if 푓 : 푈 → ℂ is a function, let 푍푓 = 푓 (0) ⊂ 푈 denote the zero set of 푓 . Definition 6.5.1. Let 푈 ⊂ ℂ푛 be an open set. Let 푋 ⊂ 푈 be a set such that near each point 푝 ∈ 푈, there exists a neighborhood 푊 of 푝 and a family of holomorphic functions F defined on 푊 such that Ù 푊 ∩ 푋 = 푧 ∈ 푊 : 푓 (푧) = 0 for all 푓 ∈ F = 푍푓 . 푓 ∈F Then 푋 is called a (complex or complex-analytic) variety or a subvariety of 푈. Sometimes 푋 is called an analytic set. We say 푋 ⊂ 푈 is a proper subvariety if ∅ ≠ 푋 ( 푈. We generally leave out the “complex” from “complex subvariety” as it is clear from context. But you should know that there are other types of subvarieties, namely real subvarieties given by real-analytic functions. We will not cover those in this book. Example 6.5.2: The set 푋 = {0} ⊂ ℂ푛 is a subvariety as it is the only common vanishing point of 푛 푛 functions F = {푧1, . . . , 푧푛}. Similarly, 푋 = ℂ is a subvariety of ℂ , where we let F = ∅. 1/푧 2 Example 6.5.3: The set defined by 푧2 = 푒 1 is a subvariety of 푈 = 푧 ∈ ℂ : 푧1 ≠ 0 . It is not a subvariety of any open set larger than 푈. It is useful to note what happens when we replace “near each point 푝 ∈ 푈” with “near each point 푝 ∈ 푋.” We get a slightly different concept, and 푋 is said to be a local variety. A local variety 푋 is a subvariety of some neighborhood of 푋, but it is not necessarily closed in 푈. As a simple 2 2 example, the set 푋 = 푧 ∈ ℂ : 푧1 = 0, |푧2| < 1 is a local variety, but not a subvariety of ℂ . On the other hand 푋 is a subvariety of the unit ball 푧 ∈ ℂ2 : k푧k < 1 . Note that F depends on 푝 and near each point may have a different set of functions. Clearly the family F is not unique. We will prove below that we would obtain the same definition if we restricted to finite families F. We work with germs of functions. Recall, that when ( 푓 , 푝) is a germ of a function the germ (푍푓 , 푝) is the germ of the zero set of some representative. Let def 퐼푝 (푋) = ( 푓 , 푝) ∈ O푝 : (푋, 푝) ⊂ (푍푓 , 푝) .

That is, 퐼푝 (푋) is the set of germs of holomorphic functions vanishing on 푋 near 푝. If a function vanishes on 푋, then any multiple of it also vanishes on 푋, so 퐼푝 (푋) is an ideal. Really 퐼푝 (푋) depends only on the germ of 푋 at 푝, so deﬁne 퐼푝 (푋, 푝) = 퐼푝 (푋). Every ideal in O푝 is ﬁnitely generated. Let 퐼 ⊂ O푝 be an ideal generated by 푓1, 푓2, . . . , 푓푘 . Write

푉 퐼 def 푍 , 푝 푍 , 푝 푍 , 푝 . ( ) = ( 푓1 ) ∩ ( 푓2 ) ∩ · · · ∩ ( 푓푘 ) 138 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

That is, 푉 (퐼) is the germ of the subvariety “cut out” by the elements of 퐼, since every element of 퐼 vanishes on the points where all the generators vanish. Suppose representatives 푓1, . . . , 푓푘 of the generators are defined in some neighborhood 푊 of 푝, and a germ (푔, 푝) ∈ 퐼 has a representative 푔 defined in 푊 such that 푔 = 푐1 푓1 + · · · + 푐푘 푓푘 , where 푐푘 are also holomorphic functions on 푊. If 푞 푍 푍 푔 푞 푍 푍 푍 푉 퐼 푍 , 푝 ∈ 푓1 ∩ · · · ∩ 푓푘 , then ( ) = 0. Thus, 푓1 ∩ · · · ∩ 푓푘 ⊂ 푔, or in terms of germs, ( ) ⊂ ( 푔 ). The reason why we did not define 푉 (퐼) to be the intersection of zero sets of all germs in 퐼 is that this would be an infinite intersection, and we did not define such an object for germs.

Exercise 6.5.1: Show that 푉 (퐼) is independent of the choice of generators.

Exercise 6.5.2: Suppose 퐼푝 (푋) is generated by the functions 푓1, 푓2, . . . , 푓푘 . Prove 푋, 푝 푍 , 푝 푍 , 푝 푍 , 푝 . ( ) = ( 푓1 ) ∩ ( 푓2 ) ∩ · · · ∩ ( 푓푘 ) Exercise 6.5.3: Given a germ (푋, 푝) of a subvariety at 푝, show 푉 퐼푝 (푋) = (푋, 푝) (see above). Then given an ideal 퐼 ⊂ O푝, show 퐼푝 푉 (퐼) ⊃ 퐼.

As O푝 is Noetherian, 퐼푝 (푋) is finitely generated. Near each point 푝 only finitely many functions are necessary to define a subvariety, that is, by an exercise above, those functions “cut out” the subvariety. When one says defining functions for a germ of a subvariety, one generally means that those functions generate the ideal, not just that their common zero set happens to be the subvariety. A theorem that we will not prove here in full generality, the Nullstellensatz, says that if we take the germ of a subvariety defined by functions in an ideal 퐼 ⊂ O푝, and look at the ideal given by ∗

퐼 that subvariety,√ we obtain the radical of . In more concise language, the Nullstellensatz says 퐼푝 푉 (퐼) = 퐼. Germs of subvarieties are in one-to-one correspondence with radical ideals of O푝.

푋 2 푧2, 푧2 퐼 푧2, 푧2 Example 6.5.4: The subvariety = {0} ⊂ ℂ can be given by F = 1 2 . If = 1 2 ⊂ O0 퐼 푋 퐼 is the ideal of germs generated by these two functions, then 0( ) ≠ . We have seen that the ideal√ 퐼 푋 푧 , 푧 퐼 0( √) is the maximal ideal 픪0 = ( 1 2). If we prove that all the nonconstant monomials are in , 퐼 푧 , 푧 퐼 푧 푧 푧 푧 then = ( 1 2) = 픪0. The only nonconstant√ monomials that are not in are 1, 2, and 1 2, but the square of each of these is in 퐼, so 퐼 = 픪0.

The local properties of a subvariety at 푝 are encoded in the properties of the ideal 퐼푝 (푋). Therefore, the study of subvarieties often involves the study of the various algebraic properties of the ideals of O푝. Let us also mention in passing that the other object that is studied is the so-called coordinate ring O푝/퐼푝 (푋), which represents the functions on (푋, 푝). That is, we identify two functions if they diﬀer by something in the ideal, since then they are equal on 푋. At most points a subvariety behaves like a piece of ℂ푘 , more precisely like a graph over ℂ푘 .A 0 푘 푛−푘 0 푛−푘 푘 푛−푘 graph of a mapping 푓 : 푈 ⊂ ℂ → ℂ is the set Γ푓 ⊂ 푈 × ℂ ⊂ ℂ × ℂ deﬁned by

def 0 푛−푘 Γ푓 = (푧, 푤) ∈ 푈 × ℂ : 푤 = 푓 (푧) .

√ def ∗The radical of 퐼 is deﬁned as 퐼 = { 푓 : 푓 푚 ∈ 퐼, for some 푚}. 6.5. VARIETIES 139

Definition 6.5.5. Let 푋 ⊂ 푈 ⊂ ℂ푛 be a subvariety of an open set 푈. Let 푝 ∈ 푋 be a point. Suppose that after a permutation of coordinates, near 푝 the set 푋 is a graph of a holomorphic mapping. That 0 00 푘 푛−푘 is, after relabeling coordinates, there is a neighborhood 푈 × 푈 ⊂ ℂ × ℂ of 푝, where 푘 ∈ ℕ0, such that 0 00 푋 ∩ (푈 × 푈 ) = Γ푓 for a holomorphic mapping 푓 : 푈0 → ℂ푛−푘 . Then 푝 is a regular point (or simple point) of 푋 and the dimension of 푋 at 푝 is 푘. We write dim푝 푋 = 푘. If all points of 푋 are regular points of dimension 푘, then 푋 is called a complex manifold, or complex submanifold, of (complex) dimension 푘. ∗ As the ambient dimension is 푛, we say 푋 is of codimension 푛 − 푘 at 푝. The set of regular points of 푋 is denoted by 푋reg. Any point that is not regular is singular. The set of singular points of 푋 is denoted by 푋sing. A couple of remarks are in order. A subvariety 푋 can have regular points of several different dimensions, although if a point is a regular point of dimension 푘, then all nearby points are regular points of dimension 푘 as the same 푈0 and 푈00 works. Any isolated point of 푋 is automatically a regular point of dimension 0. Finally remark is that dimension is well-defined. We leave it as an exercise. Sometimes the empty set is considered a complex manifold of dimension −1 (or −∞). Example 6.5.6: The set 푋 = ℂ푛 is a complex submanifold of dimension 푛 (codimension 0). In particular, 푋reg = 푋 and 푋sing = ∅. 푌 푧 3 푧 푧2 푧2 The set = ∈ ℂ : 3 = 1 − 2 is a complex submanifold of dimension 2 (codimension 1). Again, 푌reg = 푌 and 푌sing = ∅. 퐶 푧 ∈ 2 푧3 − 푧2 On the other hand, the so-called cusp, = ℂ : 1 2 = 0 is not a complex submanifold. 퐶 푧 ±푧3/2 The origin is a singular point of (see exercise below). At every other point we can write 2 = 1 , so 퐶reg = 퐶 \{0}, and so 퐶sing = {0}. The dimension at every regular point is 1. Here is a plot of 퐶 in two real dimensions:

−1 −0.5 0 0.5 1 1.5

Exercise 6.5.4: Prove that if 푝 is a regular point of a subvariety 푋 ⊂ 푈 ⊂ ℂ푛 of a domain 푈, then the dimension at 푝 is well-defined. Hint: If there were two possible 푈0 of different dimension (possibly different affine coordinates), construct a map from one such 푈0 to another such 푈0 with nonvanishing derivative. 퐶 푧 ∈ 2 푧3 − 푧2 Exercise 6.5.5: Consider the cusp = ℂ : 1 2 = 0 . Prove that the origin is not a regular point of 퐶. ∗The word ambient is used often to mean the set that contains whatever object we are talking about. 140 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Exercise 6.5.6: Show that 푝 is a regular point of dimension 푘 of a subvariety 푋 if and only if there exists a local biholomorphic change of coordinates that puts 푝 to the origin and near 0, 푋 is given by 푤 = 0, where (푧, 푤) ∈ ℂ푘 × ℂ푛−푘 . In other words, if we allow a biholomorphic change of coordinates, we can let 푓 = 0 in the deﬁnition.

We also define dimension at a singular point. A fact that we will not prove in general is that the set of regular points of a subvariety is open and dense in the subvariety; a subvariety is regular at most points. Therefore, the following definition makes sense without resorting to the convention that max ∅ = −∞. Definition 6.5.7. Let 푋 ⊂ 푈 ⊂ ℂ푛 be a (complex) subvariety of 푈. Let 푝 ∈ 푋 be a point. We define the dimension of 푋 at 푝 to be def dim푝 푋 = max 푘 ∈ ℕ0 : ∀ neighborhoods 푊 of 푝, ∃ 푞 ∈ 푊 ∩ 푋reg with dim푞 푋 = 푘 . If (푋, 푝) is a germ and 푋 a representative, the dimension of (푋, 푝) is the dimension of 푋 at 푝. The dimension of the entire subvariety 푋 is defined to be

def dim 푋 = max dim푝 푋. 푝∈푋 We say that 푋 is of pure dimension 푘 if at all points 푝, dimension of 푋 at 푝 is 푘. We say a germ (푋, 푝) is of pure dimension 푘 if there exists a representative of 푋 that is of pure dimension 푘. We deﬁne the word codimension as before, that is, the ambient dimension minus the dimension of 푋. 퐶 푧 ∈ 2 푧3 − 푧2 Example 6.5.8: We saw that = ℂ : 1 2 = 0 is of dimension 1 at all the regular points, and the only singular point is the origin. Hence dim0 퐶 = 1, and so dim 퐶 = 1. The subvariety 퐶 is of pure dimension 1. We have the following theorem, which we state without proof, at least in the general setting. Theorem 6.5.9. Let 푈 ⊂ ℂ푛 be open and connected and let 푋 ⊂ 푈 be a subvariety, then the set of regular points 푋reg is open and dense in 푋. In fact, 푋sing ⊂ 푋 is a subvariety.

푛 Exercise 6.5.7: Suppose that 푋 ⊂ 푈 ⊂ ℂ is a subvariety of a domain 푈, such that 푋reg is connected. Show that 푋 is of pure dimension. Feel free to assume 푋reg is dense in 푋.

6.6 Hypervarieties

Pure codimension 1 subvarieties are particularly nice. Sometimes pure codimension 1 subvarieties

are called hypervarieties. Back in section 1.6 we proved the following theorem ( Theorem 1.6.2 ). We restate it in the language of varieties. 푛 Theorem 6.6.1. Let 푈 ⊂ ℂ be a domain and 푓 ∈ O(푈). Then 푍푓 is empty, a subvariety of pure codimension 1, or 푍푓 = 푈. Furthermore, if 푍푓 is a pure codimension 1 subvariety, then (푍푓 )reg is open and dense in 푍푓 . 6.6. HYPERVARIETIES 141

We can improve on that slightly. Let us prove that (푍푓 )sing is (locally) contained in the zero set of some holomorphic function that is not zero on any nonempty open subset of 푍푓 . It suﬃces to assume 0 ∈ 푍푓 and to prove this locally near the origin. As usual, after possibly a linear change of 0 푛−1 coordinates, we assume that near the origin 푍푓 is contained in some 푈 × 퐷 ⊂ ℂ × ℂ for a disc 0 퐷 and 푍푓 does not intersect 푈 × 휕퐷. Apply Theorem 6.3.3 to ﬁnd the discriminant function Δ and −1 0 the discriminant 퐸 = Δ (0). Above a neighborhood of each point in 푈 \ 퐸, the set 푍푓 is union

of 푚 distinct graphs of holomorphic functions (see Proposition 6.3.2 ), so those points are regular. 0 Above each point of 퐸 there are only finitely many points of 푍푓 , and 퐸 is nowhere dense in 푈 . So 0 0 the function (푧 , 푧푛) ↦→ Δ(푧 ) does not vanish on any nonempty open subset of 푍푓 . Theorem 6.6.2. If (푋, 푝) is a germ of a pure codimension 1 subvariety, then there is a germ of a holomorphic function 푓 at 푝 such that (푍푓 , 푝) = (푋, 푝). Further, 퐼푝 (푋) is generated by ( 푓 , 푝). Proof. We already proved all the relevant pieces of the first part of this theorem. For the second part there has to exist a germ of a function that vanishes on (푋, 푝). Assume 푝 = 0, and after a linear change of coordinates assume we can apply the Weierstrass preparation theorem to the function. Taking representatives of the germs, we assume 푋 is a pure codimension 1 subvariety of a small enough neighborhood 푈0 × 퐷 ⊂ ℂ푛−1 × ℂ of the origin, where 퐷 is a disc, 0 0 0 and the function that vanishes on 푋 is a Weierstrass polynomial 푃(푧 , 푧푛) defined for 푧 ∈ 푈 , and 0 0 all zeros of 푧푛 ↦→ 푃(푧 , 푧푛) are in 퐷 for 푧 ∈ 푈. 0

Theorem 6.3.3 applies. Let 퐸 ⊂ 푈 be the discriminant set, a zero set of a holomorphic function. 0 0 On 푈 \ 퐸, there are a certain number of geometrically distinct zeros of 푧푛 ↦→ 푃(푧 , 푧푛). Let 푋0 be a topological component of 푋 \(퐸 × 퐷). Above each point 푧0 ∈ 푈0 \ 퐸, let 0 0 0 0 0 0 훼1(푧 ), . . . , 훼푘 (푧 ) denote the distinct zeros that are in 푋 , that is 푧 , 훼푗 (푧 ) ∈ 푋 . If 훼푗 is a holomor- 0 0 0 0 0 0 phic function in some small neighborhood and 푧 , 훼푗 (푧 ) ∈ 푋 at one point, then 푧 , 훼푗 (푧 ) ∈ 푋 for all nearby points too. Furthermore, this means that the set 푋0 contains only regular points of 푋 of dimension 푛 − 1. The number of such geometrically distinct zeros above each point in 푈0 \ 퐸 is locally constant, and as 푈0 \ 퐸 is connected there exists a unique 푘. Take

푘 푘−1 퐹 푧0, 푧 Ö푧 훼 푧0 푧푘 ∑︁ 푔 푧0 푧 푗 . ( 푛) = 푛 − 푗 ( ) = 푛 + 푗 ( ) 푛 푗=1 푗=0 0 The coefficients 푔푗 are well-defined for 푧 ∈ 푈 \ 퐸 as they are independent of how 훼1, . . . , 훼푘 are 0 ordered. The 푔푗 are holomorphic for 푧 ∈ 푈 \ 퐸 as locally we can choose the order so that each 훼푗 is 0 holomorphic. The coefficients 푔푗 are bounded on 푈 and therefore extend to holomorphic functions 0 0 0 0 of 푈 . Hence, the polynomial 퐹 is a polynomial in O(푈 )[푧푛]. The zeros of 퐹 above 푧 ∈ 푈 \ 퐸 are simple and give precisely 푋0. By using the argument principle again, we find that all zeros above points of 퐸 are limits of zeros above points in 푈0 \ 퐸. Consequently, the zero set of 퐹 is the closure 0 0 of 푋 in 푈 × 퐷 by continuity. It is left to the reader to check that all the functions 푔푗 vanish at the origin and 퐹 is a Weierstrass polynomial, a fact that will be useful in the exercises below. 0 0 0 If the polynomial 푃(푧 , 푧푛) is of degree 푚, then 푧 ↦→ 푃(푧 , 푧푛) has at most 푚 zeros. Together with the fact that 푈0 \ 퐸 is connected, this means that 푋 \(퐸 × 퐷) has at most finitely many components (at most 푚). So we can find an 퐹 for every topological component of 푋 \(퐸 × 퐷). Then we multiply those functions together to get 푓 . The fact that this 푓 will generate 퐼푝 (푋) is left as an exercise below. 142 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

In other words, local properties of a codimension 1 subvariety can be studied by studying the zero set of a single Weierstrass polynomial. Example 6.6.3: It is not true that if a dimension of a subvariety in ℂ푛 is 푛 − 푘 (codimension 푘), there are 푘 holomorphic functions that “cut it out.” That only works for 푘 = 1. The set defined by 푧 푧 푧 rank 1 2 3 < 2 푧4 푧5 푧6 is a pure 4-dimensional subvariety of ℂ6, so of codimension 2, and the defining equations are 푧1푧5 − 푧2푧4 = 0, 푧1푧6 − 푧3푧4 = 0, and 푧2푧6 − 푧3푧5 = 0. Let us state without proof that the unique singular point is the origin and there exist no 2 holomorphic functions near the origin that define this subvariety. In more technical language, the subvariety is not a complete intersection. Example 6.6.4: If 푋 is a hypervariety and 퐸 the corresponding discriminant set, it is tempting to say that the singular set of 푋 is the set 푋 ∩ (퐸 × ℂ), which is a codimension 2 subvariety. It is true that 푋 ∩ (퐸 × ℂ) will contain the singular set, but in general the singular set is smaller. 푧2 푧 A simple example of this behavior is the set defined by 2 − 1 = 0. The defining function is a Weierstrass polynomial in 푧2 and the discriminant set is given by 푧1 = 0. However, the subvariety has no singular points. A less trivial example is given in an exercise below. Interestingly we also proved the following theorem. Same theorem is true for higher codimension, but it is harder to prove. Corollary 6.6.5. Let (푋, 푝) is a germ of a subvariety of pure codimension 1. Then there exists a neighborhood 푈 of 푝, a representative 푋 ⊂ 푈 of (푋, 푝) and subvarieties 푋1, . . . , 푋푘 ⊂ 푈 of pure codimension 1 such that (푋푗 )reg is connected for every 푗, and 푋 = 푋1 ∪ · · · ∪ 푋푘 .

Proof. A particular 푋푗 is deﬁned by considering a topological component of 푋 \(퐸 × 퐷) as in the

proof of Theorem 6.6.2 , getting the 퐹, and setting 푋푗 = 푍퐹. The topological component is of course a connected set and it is dense in (푋푗 )reg, which proves the corollary.

Exercise 6.6.1: 푛 푛 푧2 푧2 푧2 a) Prove that the hypervariety in ℂ , ≥ 2, given by 1 + 2 + · · · + 푛 = 0 has an isolated singularity at the origin (that is, the origin is the only singular point). b) For any 0 ≤ 푘 ≤ 푛 − 2, ﬁnd a hypervariety 푋 of ℂ푛 whose set of singular points is a subvariety of dimension 푘. 0 Exercise 6.6.2: Suppose 푝(푧 , 푧푛) is a Weierstrass polynomial of degree 푘 such that for an open 0 0 dense set of 푧 near the origin 푧푛 ↦→ 푝(푧 , 푧푛) has geometrically 푘 zeros, and such that the regular points of 푍푝 are connected. Show that 푝 is irreducible in the sense that if 푝 = 푟푠 for two Weierstrass polynomials 푟 and 푠, then either 푟 = 1 or 푠 = 1. Exercise 6.6.3: Suppose 푓 is a function holomorphic in a neighborhood of the origin with 푧푛 ↦→ 푓 (0, 푧푛) being of ﬁnite order. Show that 푓 푢푝푑1 푝푑2 ··· 푝푑ℓ , = 1 2 ℓ where 푝푗 are Weierstrass polynomials of degree 푘푗 that have generically (that is, on an open 푘 푍 푢 dense set) 푗 distinct zeros (no multiple zeros), the regular points of 푝푗 are connected, and is a nonzero holomorphic function in a neighborhood of the origin. See also the next section, these polynomials will be the irreducible factors in the factorization of 푓 . 6.7. IRREDUCIBILITY, LOCAL PARAMETRIZATION, AND PUISEUX THEOREM 143

Exercise 6.6.4: Suppose (푋, 푝) is a germ of a pure codimension 1 subvariety. Show that the ideal 퐼푝 (푋) is a principal ideal (has a single generator).

Exercise 6.6.5: Suppose 퐼 ⊂ O푝 is an ideal such that 푉 (퐼) is a germ of a pure codimension 1 subvariety. Show that the ideal 퐼 is principal. 퐼 Exercise 6.6.6√ : Let ⊂ O푝 be a principal ideal. Prove the Nullstellensatz for hypervarieties: 푘 퐼푝 푉 (퐼) = 퐼. That is, show that if ( 푓 , 푝) ∈ 퐼푝 푉 (퐼) , then ( 푓 , 푝) ∈ 퐼 for some integer 푘. Exercise 6.6.7: Suppose 푋 ⊂ 푈 is a subvariety of pure codimension 1 for an open set 푈 ⊂ ℂ푛. 0 0 Let 푋 be a topological component of 푋reg. Prove that the closure 푋 is a subvariety of 푈 of pure codimension 1.

6.7 Irreducibility, local parametrization, and Puiseux theorem

Deﬁnition 6.7.1. A germ of a subvariety (푋, 푝) ⊂ (ℂ푛, 푝) is reducible at 푝 if there exist two germs (푋1, 푝) and (푋2, 푝) with (푋1, 푝) ⊄ (푋2, 푝) and (푋2, 푝) ⊄ (푋1, 푝) such that (푋, 푝) = (푋1, 푝) ∪ (푋2, 푝). Otherwise, the germ (푋, 푝) is irreducible at 푝. Similarly globally, a subvariety 푋 ⊂ 푈 is reducible in 푈 if there exist two subvarieties 푋1 and 푋2 of 푈 with 푋1 ⊄ 푋2 and 푋2 ⊄ 푋1 such that 푋 = 푋1 ∪ 푋2. Otherwise, the subvariety 푋 is irreducible in 푈. Example 6.7.2: Local and global reducibility are diﬀerent. The subvariety given by 푧2 푧 푧 2 2 = 1( 1 − 1) 2 ( , ) is irreducible in ℂ (the regular points are connected), but√ locally at the point 1 0 it is reducible. There, the subvariety is a union of two graphs: 푧2 = ± 푧1(푧1 − 1). Here is a plot in two real dimensions: 1

−1 −1 0 1 2

Exercise 6.7.1: Prove a germ of a subvariety (푋, 푝) is irreducible if and only if 퐼푝 (푋) is a prime ideal. Recall an ideal 퐼 is prime if 푎푏 ∈ 퐼 implies either 푎 ∈ 퐼 or 푏 ∈ 퐼. Exercise 6.7.2: Suppose a germ of a subvariety (푋, 푝) is of pure codimension 1. Prove (푋, 푝) is irreducible if and only if there exists a representative of 푋, such that 푋reg is connected. Exercise 6.7.3: Let 푋 ⊂ 푈 be a subvariety of pure codimension 1 of a domain 푈 ⊂ ℂ푛. Prove 푋 is irreducible if and only if the set of regular points is connected. Hint: See previous exercise. 144 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

For complex subvarieties, a subvariety is irreducible if and only if the set of regular points is connected. We omit the proof in the general case, and for hypervarieties it is an exercise above. It then makes sense that we can split a subvariety into its irreducible parts.

Proposition 6.7.3. Let (푋, 푝) ⊂ (ℂ푛, 푝) be a germ of a subvariety. Then there exist ﬁnitely many irreducible subvarieties (푋1, 푝),..., (푋푘 , 푝) such that (푋1, 푝) ∪ ... ∪ (푋푘 , 푝) = (푋, 푝) and such that (푋푗 , 푝) ⊄ (푋ℓ, 푝) for all 푗 and ℓ.

Proof. Suppose (푋, 푝) is reducible: Find (푌1, 푝) ⊄ (푌2, 푝) and (푌2, 푝) ⊄ (푌1, 푝), such that (푌1, 푝)∪ (푌2, 푝) = (푋, 푝). As (푌푗 , 푝) ( (푋, 푝), then 퐼푝 (푌푗 ) ) 퐼푝 (푋) for both 푗. If both (푌1, 푝) and (푌2, 푝) are irreducible, then stop, we are done. Otherwise apply the same reasoning to whichever (or both) (푌푗 , 푝) that was reducible. After ﬁnitely many steps you must come to a stop as you cannot have an inﬁnite ascending chain of ideals since O푝 is Noetherian.

These (푋1, 푝),..., (푋푘 , 푝) are the irreducible components. We omit the proof in general that they are unique. For a germ of a hypervariety, the UFD property of 푛O푝 gives the irreducible components. You found this factorization in an exercise above, and so this factorization is unique. Each irreducible component has the following structure. We give the theorem without proof in the general case, although we have essentially proved it already for pure codimension 1 (to put it together is left as an exercise).

Theorem 6.7.4 (Local parametrization theorem). Let (푋, 0) be an irreducible germ of a subvariety of dimension 푘 in ℂ푛. Let 푋 denote a representative of the germ. Then after a linear change of coordinates, we let 휋 : ℂ푛 → ℂ푘 be the projection onto the ﬁrst 푘 components, and obtain that there exists a neighborhood 푈 ⊂ ℂ푛 of the origin, and a proper subvariety 퐸 ⊂ 휋(푈) such that

(i) 푋0 = 푋 ∩ 푈 \ 휋−1(퐸) is a connected 푘-dimensional complex manifold that is dense in 푋 ∩ 푈.

(ii) 휋 : 푋0 → 휋(푈)\ 퐸 is an 푚-sheeted covering map for some integer 푚.

(iii) 휋 : 푋 ∩ 푈 → 휋(푈) is a proper mapping.

The 푚-sheeted covering map in this case is a local biholomorphism that is an 푚-to-1 map.

Exercise 6.7.4: Use Theorem 6.3.3 to prove the parametrization theorem if (푋, 0) is of pure codimension 1.

Let (푧1, . . . , 푧푛) be the coordinates. The linear change of coordinates needed in the theorem is to ensure that the set deﬁned by 푧1 = 푧2 = ··· = 푧푘 = 0 intersected with 푋 is an isolated point at the origin. This is precisely the same condition needed to apply Weierstrass preparation theorem in the case when 푋 is the zero set of a single function. We saw hypersurfaces are the simpler cases of complex-analytic subvarieties. At the other end of the spectrum, complex-analytic subvarieties of dimension 1 are also reasonably simple for diﬀerent reasons. Locally, complex-analytic subvarieties of dimension 1 are analytic discs. Moreover, they are locally the one-to-one holomorphic images of discs, and so they have a natural topological manifold structure even at singular points. 6.7. IRREDUCIBILITY, LOCAL PARAMETRIZATION, AND PUISEUX THEOREM 145

Example 6.7.5: The image of the analytic disc 휉 ↦→ (휉2, 휉3) is the cusp subvariety defined by 푧3 − 푧2 2 1 2 = 0 in ℂ . The following theorem is often stated only in ℂ2 for zero sets of a single function although it follows in the same way from the local parametrization theorem in higher-dimensional spaces. Of course, we only proved that theorem (or in fact you the reader did so in an exercise), for codimension 1 subvarieties, and therefore, we also only have a complete proof of the following in ℂ2. Theorem 6.7.6 (Puiseux). Let (푧, 푤) ∈ ℂ × ℂ푛−1 be coordinates. Suppose 푓 : 푈 ⊂ ℂ × ℂ푛−1 → ℂℓ is a holomorphic map such that 푓 (푧, 푤) = 0 defines a dimension 1 subvariety 푋 of 푈, 0 ∈ 푋, and 푤 ↦→ 푓 (0, 푤) has an isolated zero at the origin. Then there exists an integer 푘 and a holomorphic function 푔 defined near the origin in ℂ such that for all 휉 near the origin 푓 휉 푘 , 푔(휉) = 0. Proof. Without loss of generality assume (푋, 0) is irreducible, so that the local parametrization theorem applies. We work in a small disc 퐷 ⊂ ℂ centered at the origin, such that the origin is the unique point of the discriminant set (the subvariety 퐸). Let 푁 = {푧 ∈ 퐷 : Im 푧 = 0, Re 푧 ≤ 0}. As 퐷 \ 푁 is simply connected we have the well-defined functions 훼1(푧), . . . , 훼푚 (푧) holomorphic on 퐷 \ 푁 that are solutions to 푓 푧, 훼푗 (푧) = 0. These functions continue analytically across 푁 away from the origin. The continuation equals one of the zeros, e.g. 훼푗 (푧) becomes 훼ℓ (푧) (and by continuity it is the same zero along the entire 푁). So there is a permutation 휎 on 푚 elements such that as 푧 moves counter-clockwise around the origin from the upper half-plane across 푁 to the lower half-plane, 훼푗 (푧) is continued as 훼휎( 푗) (푧). There exists some number 푘 (e.g. 푘 = 푚!) such that 휎푘 is the identity. As 휉 goes around a circle around the origin, 휉 푘 goes around the origin 푘 푘 times. Start at a positive real 휉 and start defining a function 푔(휉) as 훼1(휉 ). Move 휉 around the origin counter-clockwise continuing 푔 analytically. Divide the disc into sectors of angle 2휋/푘, whose 푘 푘 boundaries are where 휉 ∈ 푁. Transition to 훼휎(1) (휉 ) after we reach the boundary of the first sector, 푘 then to 훼휎(휎(1)) (휉 ) after we reach the boundary of the next sector, and so on. After 푘 steps, that is 푘 푘 as 휉 moved all the way around the circle, we are back at 훼1(휉 ), because 휎 is the identity. So 푔(휉) is a well-defined holomorphic function outside the origin. Let 푔(0) = 0, and 푔 is holomorphic at 0 by the Riemann extension theorem. See the following figure for an example:

4 훼2(휉 ) 휉4 휉 4 훼3(휉 ) 푁 4 훼1(휉 )

4 퐷 훼4(휉 )

In the ﬁgure, 푚 = 푘 = 4, and the permutation 휎 takes 1 to 2, 2 to 3, 3 to 4, and 4 to 1. As 휉 moves along the short circular arrow on the right, 휉4 moves along the long circular arrow on the left. The deﬁnition of 푔 is then given in the right-hand diagram. 146 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Exercise 6.7.5: Consider an irreducible germ (푋, 0) ⊂ (ℂ2, 0) deﬁned by an irreducible Weier- strass polynomial 푓 (푧, 푤) = 0 (polynomial in 푤) of degree 푘. Prove there exists a holomorphic 푔 such that 푓 푧푘 , 푔(푧) = 0 and 푧 ↦→ 푧푘 , 푔(푧) is one-to-one and onto a neighborhood of 0 in 푋.

Exercise 6.7.6: Suppose (푋, 0) ⊂ (ℂ2, 0) is a germ of a dimension 1 subvariety. Show that after perhaps a linear change of coordinates, there are natural numbers 푑1, . . . , 푑푘 and holomorphic functions 푐1(푧), . . . , 푐푘 (푧) vanishing at 0, such that 푋 can be deﬁned near 0 by

푘 Ö 푑푗 푤 − 푐푗 (푧) = 0. 푗=1

Exercise 6.7.7: Using the local parametrization theorem, prove that if (푋, 푝) is an irreducible germ of a subvariety of dimension greater than 1, then there exists a neighborhood 푈 of 푝 and a closed subvariety 푋 ⊂ 푈 (whose germ at 푝 is (푋, 푝)), such that for every 푞 ∈ 푋 there exists an irreducible subvariety 푌 ⊂ 푋 of dimension 1 such that 푝 ∈ 푌 and 푞 ∈ 푌.

Exercise 6.7.8: Prove a stronger version of the exercise above. Show that not only is there a 푌, but an analytic disc 휑 : 픻 → 푈 such that 휑(픻) ⊂ 푋, 휑(0) = 푝 and 휑(1/2) = 푞.

Exercise 6.7.9: Suppose 푋 ⊂ 푈 is a subvariety of a domain 푈 ⊂ ℂ푛. Suppose that for any two points 푝 and 푞 on 푋 there exists a ﬁnite sequence of points 푝0 = 푝, 푝1, . . . , 푝푘 = 푞 in 푋, and a sequence of analytic discs Δ푗 ⊂ 푋 such that 푝푗 and 푝푗−1 are in Δ푗 . Exercise 6.7.10: Prove a maximum principle for subvarieties using the exercises above: Suppose 푋 ⊂ 푈 is an irreducible subvariety of an open set 푈, and suppose 푓 : 푈 → ℝ ∪ {−∞} is a plurisubharmonic function. If the modulus of the restriction 푓 |푋 achieves a maximum at some point 푝 ∈ 푋, then the restriction 푓 |푋 is constant. Exercise 6.7.11: Prove that an analytic disc (namely the image of the disc) in ℂ2 is a one- dimensional local variety (that is, a subvariety of some open subset of ℂ2).

Using the Puiseux theorem, we often simply parametrize germs of complex one-dimensional subvarieties. And for larger-dimensional varieties, we can ﬁnd enough one-dimensional curves through any point and parametrize those.

6.8 Segre varieties and CR geometry

The existence of analytic discs (or subvarieties) in boundaries of domains says a lot about the geometry of the boundary.

Example 6.8.1: Let 푀 ⊂ ℂ푛 be a smooth real hypersurface containing a complex hypersurface 푋 (zero set of a holomorphic function with nonzero derivative), at 푝 ∈ 푋 ⊂ 푀. Apply a local biholomorphic change of coordinates at 푝, so that in the new coordinates (푧, 푤) ∈ ℂ푛−1 × ℂ, 푋 is given by 푤 = 0, and 푝 is the origin. The tangent hyperplane to 푀 at 0 contains {푤 = 0}. By rotating 6.8. SEGRE VARIETIES AND CR GEOMETRY 147

the 푤 coordinate (multiplying it by 푒푖휃), we assume 푀 is tangent to the set (푧, 푤) : Im 푤 = 0 . In other words, 푀 is given by Im 푤 = 휌(푧, 푧,¯ Re 푤), where 푑휌 = 0. As 푤 = 0 on 푀, then 휌 = 0 when Re 푤 = 0. That is, 휌 is divisible by Re 푤. So 푀 is deﬁned by 푤 푤 휌 푧, 푧, 푤 , Im = (Re )e( ¯ Re ) 휌 푝 for a smooth function e. The Levi form at the origin vanishes. As = 0 was an arbitrary point on 푀 ∩ 푋, the Levi form of 푀 vanishes on 푀 ∩ 푋. Example 6.8.2: The vanishing of the Levi form is not necessary if the complex varieties in 푀 are small. Consider 푀 ⊂ ℂ3 with a nondegenerate (but not deﬁnite) Levi form:

2 2 Im 푤 = |푧1| − |푧2| . 휃 푀 퐿 푧 푒푖휃 푧 푤 Ð 퐿 For every ∈ ℝ, contains the complex line 휃, given by 1 = 2 and = 0. The union 휃 휃 of those complex lines is not contained in some single unique complex subvariety inside 푀. Any complex subvariety that contains all 퐿휃 must contain the entire complex hypersurface given by 푤 = 0, which is not contained in 푀.

Exercise 6.8.1: Let 푀 ⊂ ℂ푛 be a smooth real hypersurface. Show that if 푀 at 푝 contains 푛−1 a complex submanifold of (complex) dimension more than 2 , then the Levi form must be degenerate, that is, it must have at least one zero eigenvalue.

Exercise 6.8.2: Let 푀 ⊂ ℂ푛 be a smooth pseudoconvex real hypersurface (one side of 푀 is pseudoconvex). Suppose 푀 at 푝 contains a dimension 푘 complex submanifold 푋. Show that the Levi form has at least 푘 zero eigenvalues.

Exercise 6.8.3: Find an example of a smooth real hypersurface 푀 ⊂ ℂ푛 that contains a germ of a singular complex-analytic subvariety (푋, 푝) through a point 푝, which is unique in the sense that if (푌, 푝) is another germ of a complex analytic subvariety in 푀 then (푌, 푝) ⊂ (푋, 푝).

Let us discuss a tool, the Segre variety, that allows us to find such complex subvarieties inside 푀, and much more. Segre varieties only work in the real-analytic setting and rely on complexification. Let 푀 ⊂ ℂ푛 be a real-analytic hypersurface and 푝 ∈ 푀. Suppose 푀 ⊂ 푈, where 푈 ⊂ ℂ푛 is a small domain with a defining function 푟 : 푈 → ℝ for 푀. That is, 푟 is a real-analytic function in 푈 such that 푀 = 푟−1(0), but 푑푟 ≠ 0 on 푀. Define

푈∗ = 푧 ∈ ℂ푛 :푧 ¯ ∈ 푈 .

Suppose 푈 is small enough so that the Taylor series for 푟 converges in 푈 × 푈∗ when treating 푧 and 푧¯ as separate variables. That is, 푟(푧, 휁) is a well-defined function on 푈 × 푈∗, and 푟(푧, 휁) = 0 defines a complexification M in 푈 × 푈∗. Assume also that 푈 is small enough that the complexified 푑푟 does

not vanish on M and that M is connected. See also Proposition 3.2.8 . Given 푞 ∈ 푈, deﬁne the Segre variety Σ푞 (푈, 푟) = 푧 ∈ 푈 : 푟(푧, 푞¯) = 0 = 푧 ∈ 푈 : (푧, 푞¯) ∈ M . 148 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

푈 푟 푟 A priory, the subvariety Σ푝 depends on and . However, if e is a real-analytic function that 푈 푈∗ 푀 푟 complexifies to × and vanishes on , it must also vanish on the complexification M. If e 푑푟 is a defining function as above, that is, edoes not vanish on its zero set and the zero set of the 푟 푈 푈∗ 푟 푧, 휁 푟 complexified eis connected in × , then e( ) = 0 also defines M. Hence the actual does not matter. As long as 푞 ∈ 푀, then 푞 ∈ Σ푞 (푈, 푟), and furthermore the Segre variety is a complex hypersurface for every 푞. It is not hard to see that if 푈e is a small neighborhood of 푞, the same 푟 is a ∗ defining function in 푈e, and we get the same complexification in 푈e × 푈e . So the germ at 푞 ∈ 푈 is well-defined, and we write Σ푞 = Σ푞 (푈, 푟), 푞 .

The Segre variety is well-deﬁned as a germ, and so often when one talks about Σ푞 without mentioning the 푈 or 푟, then one means some small enough representative of a Segre variety or the germ itself.

Exercise 6.8.4: Let 푟 : 푈 → ℝ be a real-valued real-analytic function that complexiﬁes to 푈 ×푈∗. Show that 푟(푧, 휁¯) = 0 if and only if 푟(푤, 휁¯) = 0. In other words, 푧 ∈ Σ휁 (푈, 푟) if and only if 휁 ∈ Σ푧 (푈, 푟).

Example 6.8.3: Suppose we start with the real-analytic hypersurface 푀 given by

Im 푤 = (Re 푤)휌(푧, 푧,¯ Re 푤), with 휌 vanishing at the origin. Rewriting in terms of 푤 and 푤¯ , we find 푤 − 푤¯ 푤 + 푤¯ 푤 + 푤¯ = 휌 푧, 푧,¯ . 2푖 2 2 Setting 푧¯ = 푤¯ = 0, we obtain 푤 푤 푤 = 휌 푧, 0, . 2푖 2 2 As 휌 vanishes at the origin, then near the origin the equation defines the complex hypersurface given by 푤 = 0. So Σ0 is defined by 푤 = 0. This is precisely the complex hypersurface that lies inside 푀. The last example is not a fluke. The most important property of Segre varieties is that it locates complex subvarieties in a real-analytic submanifold. We will phrase it in terms of analytic discs, which is enough as complex subvarieties can be filled with analytic discs, as we have seen. Proposition 6.8.4. Let 푀 ⊂ ℂ푛 be a real-analytic hypersurface and 푝 ∈ 푀. Suppose Δ ⊂ 푀 is an analytic disc through 푝. Then as germs (Δ, 푝) ⊂ Σ푝.

Proof. Let 푈 be a neighborhood of 푝 where a representative of Σ푝 is defined, that is, we assume that Σ푝 is a closed subset of 푈, and suppose 푟(푧, 푧¯) is the corresponding defining function. Let 휑 : 픻 → ℂ푛 be the parametrization of Δ with 휑(0) = 푝. We can restrict 휑 to a smaller disc around the origin, and since we are only interested in the germ of Δ at 푝 this is sufficient (if there are multiple points of 픻 that go to 푝, we repeat the argument for each one). So let us assume without loss of generality that 휑(픻) = Δ ⊂ 푈. Since Δ ⊂ 푀 we have

푟 휑(휉), 휑(휉) = 푟 휑(휉), 휑¯(휉¯) = 0. 6.8. SEGRE VARIETIES AND CR GEOMETRY 149

The function 휉 ↦→ 푟 휑(휉), 휑¯(휉¯) is a real-analytic function of 휉, and therefore for some small neighborhood of the origin, it complexifies. In fact, it complexifies to 픻 × 픻 as 휑(휉) ∈ 푈 for all 휉 ∈ 픻. So we can treat 휉 and 휉¯ as separate variables. By complexification, the equation holds for all such independent 휉 and 휉¯. Set 휉¯ = 0 to obtain

0 = 푟 휑(휉), 휑¯(0) = 푟 휑(휉), 푝¯ for all 휉 ∈ 픻.

In particular, 휑(픻) ⊂ Σ푝 and the result follows.

Exercise 6.8.5: Show that if a real-analytic real hypersurface 푀 ⊂ ℂ푛 is strongly pseudoconvex at 푝 ∈ 푀 (one side of 푀 is strongly pseudoconvex at 푝), then Σ푝 ∩ (푀, 푝) = {푝} (as germs). Exercise 6.8.6: Use the proposition and the exercise above to show that if a real-analytic real hypersurface 푀 is strongly pseudoconvex, then 푀 contains no analytic discs.

We end our discussion of Segre varieties by its perhaps most well-known application, the so- called Diederich–Fornæss lemma. Although we state and prove it only for real-analytic hypersurfaces it works in greater generality. There are two parts to it, although it is generally the corollary that is called the Diederich–Fornæss lemma. First, for real-analytic hypersurfaces each point has a fixed neighborhood such that germs of complex subvarieties contained in the hypersurface extend to said fixed neighborhood. Theorem 6.8.5 (Diederich–Fornæss). Suppose 푀 ⊂ ℂ푛 is a real-analytic hypersurface. For every 푝 ∈ 푀 there exists a neighborhood 푈 of 푝 with the following property: If 푞 ∈ 푀 ∩ 푈 and (푋, 푞) is a germ of a complex subvariety such that (푋, 푞) ⊂ (푀, 푞), then there exists a complex subvariety 푌 ⊂ 푈 (in particular a closed subset of 푈) such that 푌 ⊂ 푀 and (푋, 푞) ⊂ (푌, 푞). Proof. Suppose 푈 is a polydisc centered at 푝, small enough so that the defining function 푟 of 푀 complexifies to 푈 × 푈∗ as above. Suppose 푞 ∈ 푀 ∩ 푈 is a point such that (푋, 푞) is a germ of a positive-dimensional complex subvariety with (푋, 푞) ⊂ (푀, 푞). Most points of a subvariety are regular, so without loss of generality assume 푞 is a regular point, that is, (푋, 푞) is a germ of a complex submanifold. Let 푋 be a representative of the germ (푋, 푞) such that 푋 ⊂ 푀, and 푋 ⊂ 푈, although we do not assume it is closed. Assume 푋 is an image of an open subset 푉 ⊂ ℂ푘 via a holomorphic surjective mapping 휑 : 푉 → 푋. Since 푟 휑(휉), 휑(휉) = 0 for all 휉 ∈ 푉, then we may treat 휉 and 휉¯ separately. In particular, 푟(푧, 휁¯) = 0 for all 푧, 휁 ∈ 푋. Define complex subvarieties 푌0,푌 ⊂ 푈 (closed in 푈) by

0 Ù Ù 푌 = Σ푎 (푈, 푟) and 푌 = Σ푎 (푈, 푟). 푎∈푋 푎∈푌 0 If 푎 ∈ 푌0 and 푏 ∈ 푋, then 푟(푎, 푏¯) = 0. Because 푟 is real-valued, 푟(푏, 푎¯) = 0. Therefore, 푋 ⊂ 푌 ⊂ 푌0. Furthermore, 푟(푧, 푧¯) = 0 for all 푧 ∈ 푌, and so 푌 ⊂ 푀. Corollary 6.8.6 (Diederich–Fornæss). Suppose 푀 ⊂ ℂ푛 is a compact real-analytic hypersurface. Then there does not exist any point 푞 ∈ 푀 such that there exists a germ of a positive-dimensional complex subvariety (푋, 푞) such that (푋, 푞) ⊂ (푀, 푞). 150 CHAPTER 6. COMPLEX ANALYTIC VARIETIES

Proof. Let 푆 ⊂ 푀 be the set of points through which there exists a germ of a positive-dimensional complex subvariety contained in 푀. As 푀, and hence 푆, is compact, there must exist a point 푝 ∈ 푆 that is furthest from the origin. After a rotation by a unitary and rescaling assume 푝 = (1, 0,..., 0). Let 푈 be the neighborhood from the previous theorem around 푝. There exist germs of varieties in 푀 through points arbitrarily close to 푝. So for any distance 휖 > 0, there exists a subvariety 푌 ⊂ 푈 (in particular, 푌 closed in 푈) of positive dimension with 푌 ⊂ 푀 that contains points 휖 close to 푝. Consider the function Re 푧1, whose modulus attains a strict maximum on 푆 at 푝. Because Re 푧1 achieves a maximum strictly smaller than 1 on 휕푈 ∩ 푆, for a small enough 휖, we would obtain a pluriharmonic function with a strict maximum on 푌, which is impossible by the maximum principle

for varieties that you proved in Exercise 6.7.10 . The picture would look as follows:

Re 푧1 = 1

푝

Example 6.8.7: The results above do not work in the smooth setting. Let us disprove the theorem in the smooth setting. Disproving the corollary is an exercise. Let 푔 : ℝ → ℝ be a smooth function that is strictly positive for |푡| > 1, and 푔(푡) = 0 for all |푡| ≤ 1. Deﬁne 푀 in (푧, 푤) ∈ ℂ푛−1 × ℂ by

Im 푤 = 푔k푧k2.

푀 is a smooth real hypersurface. Consider 푝 = (1, 0,..., 0) ∈ 푀. For every 0 < 푠 < 1, let 푞푠 = (푠, 0,..., 0) ∈ 푀 and 푋푠 = (푧, 푤) ∈ 푀 : 푤 = 푠 . Each 푋푠 is a local complex subvariety of dimension 푛 − 1 and (푋푠, 푞푠) ⊂ (푀, 푞푠). The size of 푋푠 goes to zero as 푠 → 1 and 푋푠 cannot extend to a larger complex subvariety inside 푀. So, no neighborhood 푈 at 푝 (as in the theorem) exists.

Exercise 6.8.7: Find a compact smooth real hypersurface 푀 ⊂ ℂ푛 that contains a germ of a positive dimensional complex subvariety.

. . . and that is how using sheep’s bladders can prevent earthquakes! Appendix A

Basic notation and terminology

Let us just quickly review some of the basic notation that we use in this book, that is perhaps not described elsewhere. We use ℂ, ℝ for complex and real numbers, and 푖 for imaginary unit (a square root of −1). We use ℕ = {1, 2, 3,...} for the natural numbers, ℕ0 = {0, 1, 2, 3,...} for the zero-based natural numbers, and ℤ for all integers. When we write ℂ푛 or ℝ푛 we implicitly mean that 푛 ≥ 1, unless otherwise stated. We denote the set subtraction by 퐴 \ 퐵, meaning all elements of 퐴 that are not in 퐵. We denote the complement of a set by 푋푐. The ambient set should be clear. So, for example, if 푋 ⊂ ℂ naturally, then 푋푐 = ℂ \ 푋. The topological closure of a set 푆 is denoted by 푆, its boundary is denoted by 휕푆. If 푆 is relatively compact subset of 푋 (its closure in 푋 is compact) or compact, we write 푆 ⊂⊂ 푋. The notation 푓 : 푋 → 푌 means a function with domain 푋 and codomain 푌. By 푓 (푆) we mean the direct image of 푆 by 푓 . We use the notation 푓 −1 for the inverse image of sets and single points. When 푓 is bĳective (one-to-one and onto), we use 푓 −1 for the inverse mapping. So 푓 −1(푇) for a set 푇 ⊂ 푌 denotes all the points of 푋 that 푓 maps to 푇. Similarly, 푓 −1(푞) could denote all the points that map to 푞, but if the mapping is bĳective, then it means just the unique point mapping to 푞. If we wish to deﬁne a function without necessarily giving it a name, we use the notation 푥 ↦→ 퐹(푥), where 퐹(푥) would generally be some formula giving the output. The notation

푓 |푆

means the restriction of 푓 to 푆: it is a function 푓 |푆 : 푆 → 푌 such that 푓 |푆 (푥) = 푓 (푥) for all 푥 ∈ 푆. A function 푓 : 푈 → ℂ is said to be compactly supported if the support, that is the set {푝 ∈ 푈 : 푓 (푝) ≠ 0}, is compact. To say that two functions are identically equal, we write 푓 ≡ 푔. This means that 푓 (푥) = 푔(푥) for all 푥 in the domain. The notation 푓 ◦ 푔 denotes the composition deﬁned by 푥 ↦→ 푓 푔(푥). We write 푋 def= 푌 to deﬁne 푋 to be 푌 rather than just show equality. Appendix B

Results from one complex variable

We review some results from one complex variable that may be useful for reading this book. The

reader should first look through section 0.1 for basic notation and motivation, although we will review some of the results again here. Let 푈 ⊂ ℂ be open. A function 푓 : 푈 → ℂ is holomorphic if it is complex differentiable at every point, that is, 푓 (푧 + ℎ) − 푓 (푧) 푓 0(푧) = lim ℎ∈ℂ→0 ℎ exists for all 푧 ∈ 푈. For example, polynomials and rational functions in 푧 are holomorphic. Perhaps the most important holomorphic function is the solution to the differential equation 푓 0(푧) = 푓 (푧), 푓 (0) = 1, that is the complex exponential, defined in the entire complex plane: 푓 (푧) = 푒푧 = 푒푥+푖푦 = 푒푥 cos 푦 + 푖 sin(푦). A piecewise-퐶1 path (or curve) in ℂ is a continuous function 훾 : [푎, 푏] → ℂ, continuously differentiable except at finitely many points, such that one-sided limits of 훾0 exist at all points of [푎, 푏] and such that 훾0 (and its one sided limits) is never zero. By abuse of notation, when 훾 is used as a set, we mean the image 훾[푎, 푏]. For a continuous function 푓 : 훾 → ℂ, define ∫ ∫ 푏 푓 (푧) 푑푧 def= 푓 훾(푡)훾0(푡) 푑푡. 훾 푎 As 훾0 is defined at all but finitely many points and is otherwise continuous, the integral is well- defined. Similarly, one defines the more general path integral in 푑푧 = 푑푥 + 푖 푑푦 and 푑푧¯ = 푑푥 − 푖 푑푦. Let 푧 = 훾(푡) = 훾1(푡) + 푖 훾2(푡) = 푥 + 푖 푦 parametrize the path. Then ∫ ∫ 푓 (푧) 푑푧 + 푔(푧) 푑푧¯ = 푓 푥 + 푖 푦 + 푔푥 + 푖 푦 푑푥 + 푖 푓 푥 + 푖 푦 − 푔푥 + 푖 푦 푑푦 훾 훾 ∫ 푏 = 푓 훾(푡)훾0(푡) + 푓 훾(푡)훾0(푡) 푑푡 푎 ∫ 푏 푓 훾 푡 푔훾 푡 훾0 푡 푖 푓 훾 푡 푔훾 푡 훾0 푡 푑푡. = ( ) + ( ) 1( ) + ( ) − ( ) 2( ) 푎

A path is closed if 훾(푎) = 훾(푏), and a path is simple if 훾|(푎,푏] is one-to-one with the possible exception of 훾(푎) = 훾(푏). 153

An open 푈 ⊂ ℂ has piecewise-퐶1 boundary if for each 푝 ∈ 휕푈 there is an open neighborhood 푊 of 푝 such that 휕푈 ∩ 푊 = 훾(푎, 푏) where 훾 : [푎, 푏] → ℂ is an injective piecewise-퐶1 path, and such that each 푝 ∈ 휕푈 is in the closure of ℂ \ 푈. Intuitively the boundary is a piecewise-퐶1 curve that locally cuts the plane into two open pieces. If at each point where the parametrization of 휕푈 훾0 푡 휋 is diﬀerentiable the domain is on the left ( ( ) rotated by 2 points into the domain), then the boundary is positively oriented. As in the introduction, we have the following version of Cauchy integral formula. Theorem B.1 (Cauchy integral formula). Let 푈 ⊂ ℂ be a bounded open set with piecewise-퐶1 boundary 휕푈 oriented positively, and let 푓 : 푈 → ℂ be a continuous function holomorphic in 푈. Then for 푧 ∈ 푈, 1 ∫ 푓 (휁) 푓 (푧) = 푑휁. 2휋푖 휕푈 휁 − 푧 The theorem follows from Green’s theorem, which is the Stokes’ theorem in two dimensions. In the versions we state, one needs to approximate the open set by smaller open sets from the inside to

insure the partial derivatives are bounded. See Theorem 4.1.1 . Let us state Green’s theorem using

the 푑푧 and 푑푧¯ for completeness. See appendix C for an overview of diﬀerential forms. Theorem B.2 (Green’s theorem). Let 푈 ⊂ ℂ be a bounded open set with piecewise-퐶1 boundary 휕푈 oriented positively, and let 푓 : 푈 → ℂ be continuous with bounded continuous partial derivatives in 푈. Then

∫ ∫ ∫ 휕푔 휕 푓 푓 (푧) 푑푧 + 푔(푧) 푑푧¯ = 푑 푓 (푧) 푑푧 + 푔(푧) 푑푧¯ = − 푑푧 ∧ 푑푧¯ 휕푈 푈 푈 휕푧 휕푧¯ ∫ 휕푔 휕 푓 ∫ 휕푔 휕 푓 = (−2푖) − 푑푥 ∧ 푑푦 = (−2푖) − 푑퐴. 푈 휕푧 휕푧¯ 푈 휕푧 휕푧¯ The Cauchy integral formula is equivalent to what is usually called just Cauchy’s theorem: Theorem B.3 (Cauchy). Let 푈 ⊂ ℂ be a bounded open set with piecewise-퐶1 boundary 휕푈 oriented positively, and let 푓 : 푈 → ℂ be a continuous function holomorphic in 푈. Then ∫ 푓 (푧) 푑푧 = 0. 휕푈 There is a converse to Cauchy as well. A triangle 푇 ⊂ ℂ is the convex hull of the three vertices (we include the inside of the triangle), and 휕푇 is the boundary of the triangle oriented counter- clockwise. Let us state the following theorem as an “if and only if,” even though, usually it is only the reverse direction that is called Morera’s theorem. Theorem B.4 (Morera). Suppose 푈 ⊂ ℂ is an open set, and 푓 : 푈 → ℂ is continuous. Then 푓 is holomorphic if and only if ∫ 푓 (푧) 푑푧 = 0 휕푇 for all triangles 푇 ⊂ 푈. As we saw in the introduction, a holomorphic function has a power series. 154 APPENDIX B. RESULTS FROM ONE COMPLEX VARIABLE

Proposition B.5. If 푈 ⊂ ℂ is open and 푓 : 푈 → ℂ is holomorphic, then 푓 is inﬁnitely diﬀerentiable, and if Δ휌 (푝) ⊂ ℂ is a disc, then 푓 has a power series that converges absolutely uniformly on compact subsets of Δ휌 (푝): ∞ ∑︁ 푘 푓 (푧) = 푐푘 (푧 − 푝) , 푘=0 where given a simple closed (piecewise-퐶1) path 훾 going once counter-clockwise around 푝 inside Δ휌 (푝), 푓 (푘) (푝) 1 ∫ 푓 (휁) 푐 = = 푑휁. 푘 푘+1 푘! 2휋푖 훾 (휁 − 푧)

Cauchy estimates follow: If 푀 is the maximum of | 푓 | on the circle 휕Δ푟 (푝), then 푀 |푐 | ≤ . 푘 푟 푘

Conversely, if a power series satisﬁes such estimates, it converges on Δ푟 (푝). A holomorphic 푓 : ℂ → ℂ that is entire. An immediate application of Cauchy estimates is Liouville’s theorem:

Theorem B.6 (Liouville). If 푓 is entire and bounded, then 푓 is constant.

And as a holomorphic function has a power series it satisﬁes the identity theorem:

Theorem B.7 (Identity). Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℂ is holomorphic. If the zero set 푓 −1(0) has a limit point in 푈, then 푓 ≡ 0.

Another consequence of the Cauchy integral formula is that there is a diﬀerential equation characterizing holomorphic functions.

Proposition B.8 (Cauchy–Riemann equations). Let 푈 ⊂ ℂ be open. A function 푓 : 푈 → ℂ is holomorphic if and only if 푓 is continuously diﬀerentiable and

휕 푓 1 휕 푓 휕 푓 = + 푖 = 0 on 푈. 휕푧¯ 2 휕푥 휕푦 Yet another consequence of the Cauchy formula (and one can make an argument that everything in this appendix is a consequence of the Cauchy formula) is the open mapping theorem.

Theorem B.9 (Open mapping theorem). Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℂ is holomorphic and not constant. Then 푓 is an open mapping, that is, 푓 (푉) is open whenever 푉 is open.

The real and imaginary parts 푢 and 푣 of a holomorphic function 푓 = 푢 + 푖푣 are harmonic, that is ∇2푢 = ∇2푣 = 0, where ∇2 is the Laplacian. A domain 푈 is simply connected if every simple closed path is homotopic in 푈 to a constant, in other words, if the domain has no holes. For example a disc is simply connected.

Proposition B.10. If 푈 ⊂ ℂ is a simply connected domain and 푢 : 푈 → ℝ is harmonic, then there exists a harmonic function 푣 : 푈 → ℝ such that 푓 = 푢 + 푖푣 is holomorphic. 155

The function 푣 is called the harmonic conjugate of 푢. For further review of harmonic functions

see section 2.4 on harmonic functions. We have the following versions of the maximum principle. Theorem B.11 (Maximum principles). Suppose 푈 ⊂ ℂ is a domain. (i) If 푓 : 푈 → ℂ is holomorphic and | 푓 | achieves a local maximum in 푈, then 푓 is constant. (ii) If 푈 is bounded and 푓 : 푈 → ℂ is holomorphic in 푈 and continuous, then | 푓 | achieves its maximum on 휕푈. (iii) If 푓 : 푈 → ℝ is harmonic achieves a local maximum or a minimum in 푈, then 푓 is constant. (iv) If 푈 is bounded and 푓 : 푈 → ℝ is harmonic in 푈 and continuous, then 푓 achieves its maximum and minimum on 휕푈. The ﬁrst two items are sometimes called the maximum modulus principle. The maximum principle immediately implies the following lemma. Lemma B.12 (Schwarz’s lemma). Suppose 푓 : 픻 → 픻 is holomorphic and 푓 (0) = 0, then (i) | 푓 (푧)| ≤ |푧|, and (ii) | 푓 0(0)| ≤ 1. 0 Furthermore, if | 푓 (푧0)| = |푧0| for some 푧0 ∈ 픻 \{0} or | 푓 (0)| = 1, then for some 휃 ∈ ℝ we have 푓 (푧) = 푒푖휃 푧 for all 푧 ∈ 픻. The theorem above is actually quite general. Theorem B.13 (Riemann mapping theorem). If 푈 ⊂ ℂ is a nonempty simply connected domain such that 푈 ≠ ℂ, then 푈 is biholomorphic to 픻. Given 푧0 ∈ 푈 there exists a unique biholomorphic 0 0 푓 : 푈 → 픻 such that 푓 (푧0) = 0, 푓 (푧0) > 0, and 푓 maximizes | 푓 (푧0)| among all injective holomorphic maps to 픻 such that 푓 (푧0) = 0. Schwarz’s lemma can also be used to classify the automorphisms of the disc (and hence any simply connected domain). Let Aut(픻) denote the group of biholomorphic (both 푓 and 푓 −1 are holomorphic) self maps of the disc to itself. Proposition B.14. If 푓 ∈ Aut(픻), then there exists an 푎 ∈ 픻 and 휃 ∈ ℝ such that 푧 − 푎 푓 (푧) = 푒푖휃 . 1 − 푎푧¯ Speaking of automorphisms. We have the following version of inverse function theorem. Theorem B.15. Suppose 푈 and 푉 are open subsets of ℂ. (i) If 푓 : 푈 → 푉 is holomorphic and bĳective (one-to-one and onto), then 푓 0(푧) ≠ 0 for all 푧 ∈ 푉, and 푓 −1 : 푉 → 푈 is holomorphic. If 푓 (푝) = 푞, then 1 푓 −1 (푞) = . 푓 0(푝)

(ii) If 푓 : 푈 → 푉 is holomorphic, 푓 (푝) = 푞, and 푓 0(푝) ≠ 0, then there exists a neighborhood 푊 of 푞 and a holomorphic function 푔 : 푊 → 푈 that is one-to-one and 푓 푔(푧) = 푧 for all 푧 ∈ 푊. 156 APPENDIX B. RESULTS FROM ONE COMPLEX VARIABLE

The Riemann mapping theorem actually follows from the following theorem about existence of branches of the logarithm. Theorem B.16. Suppose 푈 ⊂ ℂ is a simply connected domain, and 푓 : 푈 → ℂ is a holomorphic function without zeros in 푈. Then there exists a holomorphic function 퐿 : 푈 → ℂ such that 푒퐿 = 푓 . In particular, we can take roots: For every 푘 ∈ ℕ, there exists a holomorphic function 푔 : 푈 → ℂ such that 푔푘 = 푓 . In one complex variable, zeros of holomorphic functions can be divided out. Moreover, zeros of holomorphic functions are of ﬁnite order unless the function is identically zero. Proposition B.17. Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℂ is holomorphic, not identically zero, and 푓 (푝) = 0 for some 푝 ∈ 푈. There exists a 푘 ∈ ℕ and a holomorphic function 푔 : 푈 → ℂ, such that 푔(푝) ≠ 0 and 푓 (푧) = (푧 − 푝)푘 푔(푧) for all 푧 ∈ 푈. The number 푘 above is called the order or multiplicity of the zero at 푝. We can use this fact and the existence of roots to show that every holomorphic function is locally like 푧푘 . The function 휑 below can be thought of as a local change of coordinates. Proposition B.18. Suppose 푈 ⊂ ℂ is a domain and 푓 : 푈 → ℂ is holomorphic, not identically zero, and 푝 ∈ 푈. Then there exists a 푘 ∈ ℕ, a neighborhood 푉 ⊂ 푈 of 푝, and a holomorphic function 휑 : 푉 → ℂ with 휑0(푝) ≠ 0, such that 휑(푧) 푘 = 푓 (푧) − 푓 (푝) for all 푧 ∈ 푉. Convergence of holomorphic functions is the same as for continuous functions: uniform convergence on compact subsets. Sometimes this is called normal convergence.

Proposition B.19. Suppose 푈 ⊂ ℂ is open and 푓푘 : 푈 → ℂ is a sequence of holomorphic functions which converge uniformly on compact subsets of 푈 to 푓 : 푈 → ℂ. Then 푓 is holomorphic, and 푓 (ℓ) 푓 (ℓ) every derivative 푘 converges uniformly on compact subsets to the derivative . Holomorphic functions satisfy a Heine–Borel-like property:

Theorem B.20 (Montel). Suppose 푈 ⊂ ℂ is open and 푓푛 ⊂ 푈 → ℂ is a sequence of holomorphic functions. If { 푓푛} is uniformly bounded on compact subsets of 푈, then there exists a subsequence converging uniformly on compact subsets of 푈. A sequence of holomorphic functions cannot create or delete zeros out of thin air:

Theorem B.21 (Hurwitz). Suppose 푈 ⊂ ℂ is a domain and 푓푛 ⊂ 푈 → ℂ is a sequence of holomorphic functions converging uniformly on compact subsets of 푈 to 푓 : 푈 → ℂ. If 푓 is not identically zero and 푧0 is a zero of 푓 , then there exists a disc Δ푟 (푧0) and an 푁, such that for all 푛 ≥ 푁, 푓푛 has the same number of zeros (counting multiplicity) in Δ푟 (푧0) as 푓 (counting multiplicity).

A common application, and sometimes the way the theorem is stated, is that if 푓푛 have no zeros in 푈, then either the limit 푓 is identically zero, or it also has no zeros. 157

If 푈 ⊂ ℂ is open, 푝 ∈ 푈, and 푓 : 푈 \{푝} → ℂ is holomorphic, we say that 푓 has an isolated singularity at 푝. An isolated singularity is removable if there exists a holomorphic function 퐹 : 푈 → ℂ such that 푓 (푧) = 퐹(푧) for all 푧 ∈ 푈 \{푝}. An isolated singularity is a pole if lim 푓 (푧) = ∞ (that is | 푓 (푧)| → ∞ as |푧 − 푝| → 0). 푧→푝 An isolated singularity that is neither removable nor a pole is said to be essential. At nonessential isolated singularities the function blows up to a ﬁnite integral order. The ﬁrst part of the following proposition is usually called the Riemann extension theorem. Proposition B.22. Suppose 푈 ⊂ ℂ is an open set, 푝 ∈ 푈, and 푓 : 푈 \{푝} → ℂ holomorphic. (i) If 푓 is bounded (near 푝 is enough), then 푝 is a removable singularity. (ii) If 푝 is a pole, there exists a 푘 ∈ ℕ such that 푔(푧) = (푧 − 푝)푘 푓 (푧) is bounded near 푝 and hence 푔 has a removable singularity at 푝. The number 푘 above is called the order of the pole. There is a symmetry between zeros and 푓 푘 1 푘 푓 푘 1 poles: If has a zero of order , then 푓 has a pole of order . If has a pole of order , then 푓 has a removable singularity, and the extended function has a zero of order 푘. Let ℙ1 = ℂ ∪ {∞} be the Riemann sphere. The topology on ℙ1 is given by insisting that the 1 1 1 1 푓 푈 1 function 푧 is a homeomorphism of ℙ to itself, where ∞ = 0 and 0 = ∞. A function : → ℙ is called meromorphic, if it is not identically ∞, is holomorphic on 푈 \ 푓 −1(∞), and has poles at 푓 −1(∞). A holomorphic function with poles is meromorphic by setting the value to be ∞ at the poles. A meromorphic function is one that can locally be written as a quotient of holomorphic functions. At an isolated singularity we can expand a holomorphic function via the so-called Laurent series by adding all negative powers. The Laurent series also characterizes the type of the singularity. Proposition B.23. If Δ ⊂ ℂ is a disc centered at 푝 ∈ ℂ, and 푓 : Δ \{푝} → ℂ holomorphic, then 푐 ∞ there exists a double sequence { 푘 }푘=−∞ such that ∞ ∑︁ 푘 푓 (푧) = 푐푘 (푧 − 푝) , 푘=−∞ converges absolutely uniformly on compact subsets of Δ. If 훾 is a simple closed piecewise-퐶1 path going once counter-clockwise around 푝 in Δ, then 1 ∫ 푓 (휁) 푐 = 푑휁. 푘 푘+1 2휋푖 훾 (휁 − 푧) The singularity at 푝 is

(i) removable if 푐푘 = 0 for all 푘 < 0.

(ii) pole of order ℓ ∈ ℕ if 푐푘 = 0 for all 푘 < −ℓ and 푐−ℓ ≠ 0.

(iii) essential if for every exist inﬁnitely negative 푘 such that 푐푘 ≠ 0. 158 APPENDIX B. RESULTS FROM ONE COMPLEX VARIABLE

If 푝 is an isolated singularity of 푓 , then call the corresponding 푐−1 the residue of 푓 at 푝, and write it as Res( 푓 , 푝). The proposition says that for a small 훾 around 푝 in the positive direction, 1 ∫ Res( 푓 , 푝) = 푐−1 = 푓 (푧) 푑푧. 2휋푖 훾 Combining this equation with Cauchy’s theorem tells us that to compute integrals of functions with isolated singularities we simply need to ﬁnd the residues, which tend to be simpler to compute. For example, if 푝 is a simple pole (of order 1), then

Res( 푓 , 푝) = lim (푧 − 푝) 푓 (푧). 푧→푝

Theorem B.24 (Residue theorem). Suppose 푈 ⊂ ℂ is an open set, and 훾 is a piecewise-퐶1 simple closed path in 푈 such that the interior of 훾 is in 푈. Suppose that 푓 : 푈 \ 푆 → ℂ is a holomorphic function with isolated singularities in a ﬁnite set 푆, and suppose 푆 lies in the interior of 훾. Then ∫ ∑︁ 푓 (푧) 푑푧 = 2휋푖 Res( 푓 , 푝). 훾 푝∈푆

The identity theorem says that zeros of a nonconstant holomorphic 푓 have no limit points, and 1 푓 so are isolated points. Since 푓 is a meromorphic function with zeros at the poles of , poles are also isolated. Zeros and poles of can be counted fairly easily. Theorem B.25 (Argument principle). Suppose 푈 ⊂ ℂ is an open set, and 훾 is a piecewise-퐶1 simple closed path in 푈 such that the interior of 훾 is in 푈. Suppose that 푓 : 푈 → ℙ1 is a meromorphic function with no zeros or poles on 훾. Then 1 ∫ 푓 0(푧) 푑푧 = 푁 − 푃, 2휋푖 훾 푓 (푧) where 푁 is the number of zeros of 푓 inside 훾 and 푃 is the number of poles inside 훾, both counted with multiplicity. Furthermore, suppose ℎ : 푈 → ℂ is holomorphic. Let 푧1, . . . , 푧푁 be the zeros of 푓 inside 훾 and 푤1, . . . , 푤푃 be the poles of 푓 inside 훾. Then

푁 푃 1 ∫ 푓 0(푧) ∑︁ ∑︁ ℎ(푧) 푑푧 = ℎ(푧 ) − ℎ(푤 ). 2휋푖 푓 (푧) 푗 푗 훾 푗=1 푗=1

The proof is an immediate application of the residue theorem. Simply compute the residues at 푓 푓 푝 푘 ℎ 푧 푓 0(푧) the zeros and poles of . In particular, if has a zero at or multiplicity , then ( ) 푓 (푧) has a 푝 푘 ℎ 푝 푓 푝 푘 ℎ 푧 푓 0(푧) simple pole at with residue ( ). Similarly, if has a pole at of order , then ( ) 푓 (푧) has a simple pole with residue −푘 ℎ(푝) at 푝. In the couple of theorems above, we avoided introducing winding numbers by making 훾 a simple closed curve, so the statements above may be slightly diﬀerent from what you have seen in your one-variable course. Another useful way to count zeros is Rouché’s theorem. 159

Theorem B.26 (Rouché). Suppose 푈 ⊂ ℂ is an open set, and 훾 is a piecewise-퐶1 simple closed path in 푈 such that the interior of 훾 is in 푈. Suppose that 푓 : 푈 → ℂ and 푔 : 푈 → ℂ are holomorphic functions such that | 푓 (푧) − 푔(푧)| < | 푓 (푧)| + |푔(푧)| for all 푧 ∈ 훾. Then 푓 and 푔 have the same number of zeros inside 훾 (up to multiplicity).

In the classical statement of the theorem the weaker inequality | 푓 (푧) − 푔(푧)| < | 푓 (푧)| is used. Notice that either inequality precludes any zeros on 훾 itself. A holomorphic function with an essential singularity achieves essentially every value. A weak version of this result (and an easy to prove one) is the Casorati–Weierstrass theorem: If a holomorphic 푓 has an essential singularity at 푝, then for every neighborhood 푊 of 푝, 푓 푊 \{푝} is dense in ℂ. Let us state the much stronger theorem of Picard: A function with an essential singularity is very wild. It achieves every value (except possibly one) inﬁnitely often.

Theorem B.27 (Picard’s big theorem). Suppose 푈 ⊂ ℂ is open, 푓 : 푈 \{푝} → ℂ is holomorphic, and 푓 has an essential singularity at 푝. Then for every neighborhood 푊 of 푝, 푓 푊 \{푝} is either ℂ or ℂ minus a point.

For example, 푒1/푧 has an essential singularity at the origin and the function is never 0. Since we stated the big theorem, let us also state the little theorem.

Theorem B.28 (Picard’s little theorem). If 푓 : ℂ → ℂ is holomorphic, then 푓 (ℂ) is either ℂ or ℂ minus a point.

One theorem from algebra that is important in complex analysis, and becomes perhaps even more important in several variables is the fundamental theorem of algebra. It really is a theorem of complex analysis and its standard proof is via the maximum principle.

Theorem B.29 (Fundamental theorem of algebra). If 푃 : ℂ → ℂ is a nonzero polynomial of degree 푘, then 푃 has exactly 푘 zeros (roots) in ℂ counted with multiplicity.

The set of rational functions is dense in the space of holomorphic functions, and we even have control over where the poles need to be. Note that a nonconstant polynomial has a “pole at inﬁnity” meaning 푃(푧) → ∞ as 푧 → ∞. Letting ℙ1 again be the Riemann sphere, we have Runge’s approximation theorem.

Theorem B.30 (Runge). Suppose 푈 ⊂ ℂ is an open set and 퐴 ⊂ ℙ1 \ 푈 is a set containing at least one point from each component of ℙ1 \ 푈. Suppose 푓 : 푈 → ℂ is holomorphic. Then for any 휖 > 0 and any compact 퐾 ⊂⊂ 푈, there exists a rational function 푅 with poles in 퐴 such that

|푅(푧) − 푓 (푧)| < 휖 for all 푧 ∈ 퐾.

Perhaps a surprising generalization of the classical Weierstrass approximation theorem, and one of my favorite one-variable theorems, is Mergelyan’s theorem. It may be good to note that Mergelyan does not follow from Runge. 160 APPENDIX B. RESULTS FROM ONE COMPLEX VARIABLE

Theorem B.31 (Mergelyan). Suppose 퐾 ⊂⊂ ℂ is a compact set such that ℂ \ 퐾 is connected and 푓 : 퐾 → ℂ is a continuous function that is holomorphic in the interior 퐾◦. Then for any 휖 > 0 and any compact 퐾 ⊂⊂ 푈, there exists a polynomial 푃 such that |푃(푧) − 푓 (푧)| < 휖 for all 푧 ∈ 퐾. The reason why the theorem is perhaps surprising is that 퐾 may have only a small or no interior. Using a closed interval 퐾 = [푎, 푏] of the real line we recover the Weierstrass approximation theorem. Given an open set 푈 ⊂ ℂ, we say 푈 is symmetric with respect to the real axis if 푧 ∈ 푈 implies 푧¯ ∈ 푈. We divide 푈 into three parts

푈+ = {푧 ∈ 푈 : Im 푧 > 0}, 푈0 = {푧 ∈ 푈 : Im 푧 = 0}, 푈− = {푧 ∈ 푈 : Im 푧 < 0}. We have the following theorem for extending (reflecting) holomorphic functions past boundaries. Theorem B.32 (Schwarz Reflection Principle). Suppose 푈 ⊂ ℂ is a domain symmetric with respect to the real axis, 푓 : 푈+ ∪ 푈0 → ℂ a continuous function holomorphic on 푈+ and real valued on 푈0. Then the function 푔 : 푈 → ℂ defined by

푔(푧) = 푓 (푧) if 푧 ∈ 푈+ ∪ 푈0, 푔(푧) = 푓 (푧¯) if 푧 ∈ 푈−, is holomorphic on 푈. In fact, the reflection is really about harmonic functions. Theorem B.33 (Schwarz reflection principle for harmonic functions). Suppose 푈 ⊂ ℂ is a domain symmetric with respect to the real axis, 푓 : 푈+ ∪ 푈0 → ℝ a continuous function harmonic on 푈+ and zero on 푈0. Then the function 푔 : 푈 → ℝ defined by

푔(푧) = 푓 (푧) if 푧 ∈ 푈+ ∪ 푈0, 푔(푧) = − 푓 (푧¯) if 푧 ∈ 푈−, is harmonic on 푈. Functions may be deﬁned locally, and continued along paths. Suppose 푝 is a point and 퐷 is a disc centered at 푝 ∈ 퐷. A holomorphic function 푓 : 퐷 → ℂ can be analytically continued along a path 훾 : [0, 1] → ℂ, 훾(0) = 푝, if for every 푡 ∈ [0, 1] there exists a disc 퐷푡 centered at 훾(푡), where 퐷0 = 퐷, and a holomorphic function 푓푡 : 퐷푡 → ℂ, where 푓0 = 푓 , and for each 푡0 ∈ [0, 1] there is 휖 > 푡 푡 < 휖 푓 푓 퐷 퐷 an 0 such that if | − 0| , then 푡 = 푡0 in 푡 ∩ 푡0 . The monodromy theorem says that as long as there are no holes, analytic continuation deﬁnes a function uniquely. Theorem B.34 (Monodromy theorem). If 푈 ⊂ ℂ is a simply connected domain, 퐷 ⊂ 푈 a disc and 푓 : 퐷 → ℂ a holomorphic function that can be analytically continued from 푝 ∈ 퐷 to every 푞 ∈ 푈 along any path from 푝 to 푞, then there exists a unique holomorphic function 퐹 : 푈 → ℂ such that 퐹|퐷 = 푓 .

An interesting and useful theorem getting an inequality in the opposite direction from Schwarz’s 1 lemma, and one which is often not covered in a one-variable course is the Koebe 4 -theorem. Think of why no such theorem could possibly hold for just smooth functions. At ﬁrst glance the theorem should seem quite counterintuitive, and at second glance, it should seem outright outrageous. 161

Theorem B.35 (Koebe quarter theorem). Suppose 푓 : 픻 → ℂ is holomorphic and injective. Then 푓 푓 | 푓 0(0)| (픻) contains a disc centered at (0) and radius 4 . 1 The 4 is sharp, that is, it is the best it can be. Finally, it is useful to factor out all the zeros of a holomorphic function, not just ﬁnitely many. Similarly, we can work with poles.

Theorem B.36 (Weierstrass product theorem). Suppose 푈 ⊂ ℂ is a domain, {푎푘 }, {푏푘 } are countable sets in 푈 with no limit points in 푈, and {푛푘 }, {푚푘 } any countable sets of natural numbers. Then there exists a meromorphic function 푓 of 푈 whose zeros are exactly at 푎푘 , with orders given by 푛푘 , and poles are exactly at 푏푘 , with orders given by 푚푘 . Î∞ 푎 For a more explicit statement, we need inﬁnite products. The product 푗=1(1 + 푗 ) converges Î푛 푎 if the sequence of partial products 푗=1(1 + 푗 ) converges. We say that the product converges Î∞ 푎 Í∞ 푎 absolutely if 푗=1(1 + | 푗 |) converges, which is equivalent to 푗=1| 푗 | converging. Deﬁne 푧2 푧푚 퐸 (푧) = (1 − 푧), 퐸 (푧) = (1 − 푧) exp 푧 + + · · · + . 0 푚 2 푚 The function 퐸푚 푧/푎 has a zero of order 1 at 푎. Theorem B.37 (Weierstrass factorization theorem). Let 푓 be an entire function with zeros (repeated according to multiplicity) at points of the sequence {푎푘 } except the zero at the origin, whose order is 푚 (possibly 푚 = 0). Then there exists an entire function 푔 and a sequence {푝푘 } such that

∞ Ö 푧 푓 (푧) = 푧푚푒푔(푧) 퐸 , 푝푘 푎 푘=1 푘 converges uniformly absolutely on compact subsets.

The 푝푘 are chosen such that ∞ 1+푝푘 ∑︁ 푟

푎 푗=1 푘 converges for all 푟 > 0.

***

There are many other useful theorems in one complex variable, and we could spend a lot of time listing them all. However, hopefully the listing above is useful as a refresher for the reader of the most common results, some of which are used in this book, some of which are useful in the exercises, and some of which are just too interesting not to mention. Appendix C

Diﬀerential forms and Stokes’ theorem

Diﬀerential forms come up quite a bit in this book, especially in chapter 4 and chapter 5 . Let us overview their deﬁnition and state the general Stokes’ theorem. No proofs are given, this appendix

is just a bare bones guide. For a more complete introduction to differential forms, see Rudin [ R2 ]. The short story about differential forms is that a 푘-form is an object that can be integrated (summed) over a 푘-dimensional object, taking orientation into account. For simplicity, as in most of this book, everything in this appendix is stated for smooth (퐶∞) objects to avoid worrying about how much regularity is needed. The main point of differential forms is to find the proper context for the Fundamental Theorem of Calculus, ∫ 푏 푓 0(푥) 푑푥 = 푓 (푏) − 푓 (푎). 푎 We interpret both sides as integration. The left-hand side is an integral of the 1-form 푓 0 푑푥 over the 1-dimensional interval [푎, 푏] and the right-hand side is an integral of the 0-form, that is a function, 푓 over the 0-dimensional (two-point) set {푎, 푏}. Both sides consider orientation, [푎, 푏] is integrated from 푎 to 푏, {푎} is oriented negatively and {푏} is oriented positively. The two-point set {푎, 푏} is the boundary of [푎, 푏], and the orientation of {푎, 푏} is induced by [푎, 푏]. Let us define the objects over which we integrate, that is, smooth submanifolds of ℝ푛. Our model for a 푘-dimensional submanifold-with-boundary is the upper-half-space and its boundary:

푘 def 푘 푘 def 푘 ℍ = {푥 ∈ ℝ : 푥푘 ≥ 0}, 휕ℍ = {푥 ∈ ℝ : 푥푘 = 0},

푛 Deﬁnition C.1. Let 푀 ⊂ ℝ have the induced subspace topology. Let 푘 ∈ ℕ0. Let 푀 have the property that for each 푝 ∈ 푀, there exists a neighborhood 푊 ⊂ ℝ푛 of 푝, a point 푞 ∈ ℍ푘 , 푘 ∗ a neighborhood 푈 ⊂ ℍ of 푞, and a smooth one-to-one open mapping 휑 : 푈 → 푀 such that 휑(푞) = 푝, the derivative 퐷휑 has rank 푘 at all points, and 휑(푈) = 푀 ∩ 푊. Then 푀 is an embedded submanifold-with-boundary of dimension 푘. The map 휑 is called a local parametrization. If 푞 is such that 푞푘 = 0 (the last component is zero), then 푝 = 휑(푞) is a boundary point. Let 휕푀 denote the set of boundary points. If 휕푀 = ∅, then we say 푀 is simply an embedded submanifold. The situation for a boundary point and an interior point is depicted in the following diagram. Note that 푊 is a bigger neighborhood in ℝ푛 than the image 휑(푈).

∗By open, we mean that 휑(푉) is a relatively open set of 푀 for every open set 푉 ⊂ 푈. 163

ℝ푛 푀푀 ℝ푛 푊 푝 푊 푝 휕푀 휕푀 휑 휑

ℍ푘 ℍ푘 푈 푞 푈

휕ℍ푘 푞 휕ℍ푘 Boundary pointInterior point

Completely correctly, we should say submanifold of ℝ푘 . Sometimes people (including me) say manifold when they mean submanifold. A manifold is a more abstract concept, but all submanifolds are manifolds. The word embedded has to do with the topology on 푀, and this has to do with the condition 휑(푈) = 푀 ∩ 푊 and 휑 being open. The condition means that 휑 is a homeomorphism onto 푀 ∩ 푊. It is important that 푊 is an open set in ℝ푛. For our purposes here, all submanifolds will be embedded. We have also made some economy in the definition. If 푞 is not on the boundary of ℍ푘 , then we might as well have used ℝ푘 instead of ℍ푘 . A submanifold is something that is locally like ℝ푘 , and if it has a boundary, then near the boundary it is locally like ℍ푘 near a point of 휕ℍ푘 . We also remark that submanifolds are often defined in reverse rather than by parametrizations, that is, by starting with the (relatively) open sets 푀 ∩ 푊, and the maps 휑−1, calling those charts, and calling the entire set of charts an atlas. The particular version of the definition we have chosen makes it easy to evaluate integrals in the same way that parametrizing curves makes it easy to evaluate integrals.

Examples of such submanifolds are domains with smooth boundaries as in Deﬁnition 2.2.1 , we can take the inclusion map 푥 ↦→ 푥 as our parametrization. The domain is then the submanifold 푀 and 휕푀 is the boundary of the domain. Domains are the key application for our purposes. Another example are smooth curves. If 푀 is an embedded submanifold-with-boundary of dimension 푘, then 휕푀 is also an embedded submanifold of dimension 푘 − 1. Simply restrict the parametrizations to the boundary of ℍ푘 . We also need to deﬁne an orientation.

Deﬁnition C.2. Let 푀 ⊂ ℝ푛 be an embedded submanifold-with-boundary of dimension 푘 ≥ 2. Suppose a set of parametrizations can be chosen such that each point of 푀 is in the image of one 휑 푈 푀 휑 푈 푀 of the parametrizations, and if : → and e: e → are two parametrizations such that 휑 푈 휑 푈 ( ) ∩ e( e) ≠ ∅, then the transition map (automatically smooth) deﬁned by 휑 −1 휑 e ◦ 휑−1 휑 푈 휑 푈 on ( ) ∩ e( e) (in other words, wherever it makes sense) is orientation preserving, that is 퐷 휑 −1 휑 > det e ◦ 0 164 APPENDIX C. DIFFERENTIAL FORMS AND STOKES’ THEOREM

at all points. The set of such parametrizations is the orientation on 푀, and we usually take the maximal set of such parametrizations. If 푀 is oriented, then the restrictions of the parametrizations to 휕ℍ푘 give an orientation on 휕푀. We say this is the induced orientation on 휕푀.

For dimensions 푘 = 0 (isolated points) and 푘 = 1 (curves) we must define orientation differently. For 푘 = 0, we simply give each point an orientation of + or −. For 푘 = 1, we need to allow parametrization by open subsets not only of ℍ1 = [0, ∞), but also −ℍ1 = (−∞, 0]. The definition is the same otherwise. To define the orientation of the boundary, if the boundary point corresponds to the 0 in [0, ∞) we give this boundary point the orientation −, and if it corresponds to the 0 in (−∞, 0], then we give this point the orientation +. The reason for this complication is that unlike in ℝ푘 for 푘 ≥ 2, the set ℍ1 = [0, ∞) cannot be “rotated” (in ℝ1) or mapped via an orientation preserving map onto −ℍ1 = (−∞, 0], but in ℝ2 the upper-half-plane ℍ2 can be rotated to the 2 2 lower-half-plane −ℍ = {푥 ∈ ℝ : 푥2 ≤ 0}. For computations, it is often useful for compact curves with endpoints (boundary) to just give one parametrization from [0, 1] or perhaps [푎, 푏], then 푎 corresponds to the − and 푏 corresponds to the +. The fact that the transition map is smooth does require a proof, which is a good exercise in basic analysis. It requires a bit of care at boundary points. An orientation allows us to have a well-defined integral on 푀, just like a curve needs to be oriented in order to define a line integral. However, unlike for curves, not every submanifold of dimension more than one is orientable, that is, admits an orientation. A classical nonorientable example is the Möbius strip. Now that we know “on” what we integrate, let us figure out what “it” is that we integrate. Let us start with 0-forms. We define 0-forms as smooth functions (possibly complex-valued). Sometimes we need a function defined just on a submanifold. A function 푓 defined on a submanifold 푀 is smooth when 푓 ◦ 휑 is smooth on 푈 for every parametrization 휑 : 푈 → 푀. Equivalently, one can prove that 푓 is the restriction of some smooth function defined on some neighborhood of 푀 in ℝ푛. A 0-form 휔 defined on a 0-dimensional oriented submanifold 푀 is integrated as

∫ def ∑︁ 휔 = 휖푝휔(푝), 푀 푝∈푀 where 휖푝 is the orientation of 푝 given as +1 or −1. To avoid problems of integrability, one can assume that 휔 is compactly supported (it is nonzero on at most finitely many points of 푀) or that 푀 is compact (it is a finite set). The correct definition of a 1-form is that it is a “smooth section” of the dual of the vector bundle 푛 푇ℝ . That is, it is something that eats a vector field, and spits out a function. The 1-form 푑푥푘 is supposed to be the object that does 휕 휕 푑푥푘 = 1, 푑푥푘 = 0 if 푗 ≠ 푘. 휕푥푘 휕푥푗

For our purposes here, just suppose that a 1-form in ℝ푛 is an object of the form

휔 = 푔1푑푥1 + 푔2푑푥2 + · · · + 푔푛푑푥푛, 165 where 푔1, 푔2, . . . , 푔푛 are smooth functions. That is, a 1-form is at each point a linear combination of 푑푥1, 푑푥2, . . . , 푑푥푛 that varies smoothly from point to point. Suppose 푀 is a one-dimensional submanifold (possibly with boundary), 휑 : 푈 → 푀 is a parametrization compatible with the orientation of 푀, and 푔푗 is supported in 휑(푈). Deﬁne ∫ 푛 ∫ 휔 def ∑︁ 푔 휑 푡 휑0 푡 푑푡, = 푗 ( ) 푗 ( ) 푀 푗=1 푈

∫ ··· 푑푡 푈 ⊂ where the integral 푈 is evaluated with the usual positive orientation (left to right) as ℝ, and 휑푗 is the 푗th component of 휑. Generally, a 1-form has support bigger than just 휑(푈). In this case, one needs to use a so-called partition of unity to write 휔 as a locally finite sum ∑︁ 휔 = 휔ℓ, ℓ where each 휔ℓ has support in the image of a single parametrization. By locally finite, we mean that on each compact neighborhood only finitely many 휔ℓ are nonzero. Define

∫ def ∑︁ ∫ 휔 = 휔ℓ. 푀 ℓ 푀 The deﬁnition makes sense only if this sum actually exists. For example, if 휔 is compactly supported, then this sum is only ﬁnite, and so it exists. Higher degree forms are constructed out of 1-forms and 0-forms by the so-called wedge product. Given a 푘-form 휔 and an ℓ-form 휂, 휔 ∧ 휂 is a (푘 + ℓ)-form. We require the wedge product to be bilinear at each point: If 푓 and 푔 are smooth functions, then

( 푓 휔 + 푔휂) ∧ 휉 = 푓 (휔 ∧ 휉) + 푔(휂 ∧ 휉), and 휔 ∧ ( 푓 휂 + 푔휉) = 푓 (휔 ∧ 휂) + 푔(휔 ∧ 휉).

The wedge product is not commutative; we require it to be anticommutative on 1-forms. If 휔 and 휂 are 1-forms, then 휔 ∧ 휂 = −휂 ∧ 휔. The negative keeps track of orientation. When 휔 is a 푘-form and 휂 is an ℓ-form,

휔 ∧ 휂 = (−1)푘ℓ휂 ∧ 휔.

We wedge together the basis 1-forms to get all 푘-forms. A 푘-form is then an expression

푛 푛 푛 휔 ∑︁ ∑︁ ∑︁ 푔 푑푥 푑푥 푑푥 , = ··· 푗1,..., 푗푘 푗1 ∧ 푗2 ∧ · · · ∧ 푗푘 푗1=1 푗2=1 푗푘 =1 푔 where 푗1,..., 푗푘 are smooth functions. We can simplify even more. Since the wedge is anticommutative on 1-forms, 푑푥푗 ∧ 푑푥푚 = −푑푥푚 ∧ 푑푥푗 , and 푑푥푗 ∧ 푑푥푗 = 0. 166 APPENDIX C. DIFFERENTIAL FORMS AND STOKES’ THEOREM

푑푥 푑푥 푑푥 푗 , . . . , 푗 In other words, every form 푗1 ∧ 푗2 ∧ · · · ∧ 푗푘 is either zero, if any two indices from 1 푘 푑푥 푑푥 푑푥 푗 < 푗 < < 푗 are equal, or can be put into the form ± 푗1 ∧ 푗2 ∧ · · · ∧ 푗푘 , where 1 2 ··· 푘 . Thus, a 푘-form can always be written as

휔 ∑︁ 푔 푑푥 푑푥 푑푥 . = 푗1,..., 푗푘 푗1 ∧ 푗2 ∧ · · · ∧ 푗푘 1≤ 푗1< 푗2<···< 푗푘 ≤푛 Consider an oriented 푘-dimensional submanifold 푀 (possibly with boundary), a parametrization 휑 푈 푀 푘 휔 휑 푈 푔 : → from the orientation, and a -form supported in ( ) (that is each 푗1,..., 푗푘 is supported in 휑(푈)). Denote by 푡 ∈ 푈 ⊂ ℝ푘 the coordinates on 푈. Deﬁne ∫ ∫ 휔 def ∑︁ 푔 푡 퐷 휑 , 휑 , . . . , 휑 푑푡 = 푗1,..., 푗푘 ( ) det ( 푗1 푗2 푗푘 ) 푀 푈 1≤ 푗1< 푗2<···< 푗푘 ≤푛

∫ ··· 푑푡 푘 푑푡 where the integral 푈 is evaluated in the usual orientation on ℝ with the standard measure 푘 푑푡 푑푡 푑푡 푑푡 퐷 휑 , 휑 , . . . , 휑 on ℝ (think = 1 2 ··· 푛), and ( 푗1 푗2 푗푘 ) denotes the derivative of the mapping ℓ 휑 whose th component is 푗ℓ . Similarly as before, if 휔 is not supported in the image of a single parametrization, write ∑︁ 휔 = 휔ℓ ℓ

as a locally finite sum, where each 휔ℓ has support in the image of a single parametrization of the orientation. Then ∫ def ∑︁ ∫ 휔 = 휔ℓ. 푀 ℓ 푀 Again, the sum has to exist, such as when 휔 is compactly supported and the sum is finite. The only nontrivial differential forms on ℝ푛 are 0, 1, 2, . . . , 푛 forms. The only 푛-forms are object of the form 푓 (푥) 푑푥1 ∧ 푑푥2 ∧ · · · ∧ 푑푥푛.

The form 푑푥1 ∧ 푑푥2 ∧ · · · ∧ 푑푥푛 is called the volume form. Integrating it over a domain (an 푛-dimensional submanifold) gives the standard volume integral. More generally one deﬁnes integration of 푘-forms over 푘-chains, which are just linear combinations of smooth submanifolds, but we do not need that level of generality. In computations, we can avoid sets of zero measure (푘-dimensional), so we can ignore the boundary of the submanifold. Similarly, if we parametrize several subsets we can leave out a measure zero subset. Let us give a couple of examples of computations. Example C.3: Consider the circle 푆1 ⊂ ℝ2. We use a parametrization 휑 : (−휋, 휋) → 푆1 where 휑(푡) = cos(푡), sin(푡) , so the circle is oriented counter-clockwise. Let 휔(푥1, 푥2) = 푃(푥1, 푥2) 푑푥1 + 푄(푥1, 푥2) 푑푥2, then ∫ ∫ 휋 휔 = 푃cos(푡), sin(푡) − sin(푡) + 푄cos(푡), sin(푡) cos(푡) 푑푡. 푆1 −휋 We can ignore the point (−1, 0) as a single point is of 1-dimensional measure zero. 167

Example C.4: Consider a domain 푈 ⊂ ℝ푛, then 푈 is an oriented submanifold. We use the parametrization 휑 : 푈 → 푈, where 휑(푥) = 푥. Then ∫ ∫ ∫ 푓 (푥) 푑푥1 ∧ 푑푥2 ∧ · · · ∧ 푑푥푛 = 푓 (푥) 푑푥1 푑푥2 ··· 푑푥푛 = 푓 (푥) 푑푉, 푈 푈 푈 where 푑푉 is the standard volume measure. Example C.5: Finally, consider 푀 the upper hemisphere of the unit sphere 푆2 ⊂ ℝ3 as a submanifold with boundary. That is consider 푀 푥 3 푥2 푥2 푥2 , 푥 . = ∈ ℝ : 1 + 2 + 3 = 1 3 ≥ 0

The boundary is the circle in the (푥1, 푥2)-plane: 휕푀 푥 3 푥2 푥2 , 푥 . = ∈ ℝ : 1 + 2 = 1 3 = 0 Consider the parametrization of 푀 using the spherical coordinates

휑(휃, 휓) = cos(휃) sin(휓), sin(휃) sin(휓), cos(휓)

for 푈 given by −휋 < 휃 < 휋, 0 < 휓 ≤ 휋/2. After a rotation this is a subset of a half-plane with the points corresponding to 휓 = 휋/2 corresponding to boundary points. We miss the points where 휃 = 휋, including the point (0, 0, 1), but the set of those points is a 1-dimensional curve, and so a set of 2-dimensional measure zero. For the purposes of integration we can ignore it. Let

휔(푥1, 푥2, 푥3) = 푃(푥1, 푥2, 푥3) 푑푥1 ∧ 푑푥2 + 푄(푥1, 푥2, 푥3) 푑푥1 ∧ 푑푥3 + 푅(푥1, 푥2, 푥3) 푑푥2 ∧ 푑푥3.

Then ∫ ∫ 휋 ∫ 휋/2 휕휑 휕휑 휕휑 휕휑 휕휑 휕휑 휕휑 휕휑 휔 = 푃휑(휃, 휓) 1 2 − 2 1 + 푄휑(휃, 휓) 1 3 − 3 1 푀 −휋 0 휕휃 휕휓 휕휃 휕휓 휕휃 휕휓 휕휃 휕휓 휕휑 휕휑 휕휑 휕휑 + 푅휑(휃, 휓) 2 3 − 3 2 푑휃 푑휓 휕휃 휕휓 휕휃 휕휓 ∫ 휋 ∫ 휋/2 h = 푃cos(휃) sin(휓), sin(휃) sin(휓), cos(휓) − cos(휓) sin(휓) −휋 0 + 푄cos(휃) sin(휓), sin(휃) sin(휓), cos(휓) sin(휃) sin2(휓) i + 푅cos(휃) sin(휓), sin(휃) sin(휓), cos(휓) − cos(휃) sin2(휓) 푑휃 푑휓.

The induced orientation on the boundary 휕푀 is the counter-clockwise orientation used in 휋 Example C.3 , because that is the parametrization we get when we restrict to the boundary, 휑(휃, /2) = cos(휃), sin(휃), 0.

The derivative on 푘-forms is the exterior derivative, which is a linear operator that eats 푘-forms and spits out (푘 + 1)-forms. For a 푘-form 휔 푔 푑푥 푑푥 푑푥 , = 푗1,..., 푗푘 푗1 ∧ 푗2 ∧ · · · ∧ 푗푘 168 APPENDIX C. DIFFERENTIAL FORMS AND STOKES’ THEOREM

deﬁne the exterior derivative 푑휔 as

푛 def ∑︁ 휕푔푗 ,..., 푗 푑휔 = 푑푔 ∧ 푑푥 ∧ 푑푥 ∧ · · · ∧ 푑푥 = 1 푘 푑푥 ∧ 푑푥 ∧ 푑푥 ∧ · · · ∧ 푑푥 . 푗1,..., 푗푘 푗1 푗2 푗푘 휕푥 ℓ 푗1 푗2 푗푘 ℓ=1 ℓ

Then deﬁne 푑 on every 푘-form by extending it linearly. For example,

푑 (푃 푑푥2 ∧ 푑푥3 + 푄 푑푥3 ∧ 푑푥1 + 푅 푑푥1 ∧ 푑푥2) 휕푃 휕푄 휕푅 = 푑푥1 ∧ 푑푥2 ∧ 푑푥3 + 푑푥2 ∧ 푑푥3 ∧ 푑푥1 + 푑푥3 ∧ 푑푥1 ∧ 푑푥2 휕푥1 휕푥2 휕푥3 휕푃 휕푄 휕푅 = + + 푑푥1 ∧ 푑푥2 ∧ 푑푥3. 휕푥1 휕푥2 휕푥3

You should recognize the divergence of the vector ﬁeld (푃, 푄, 푅) from vector calculus. All the various derivative operations in ℝ3 from vector calculus make an appearance. If 휔 is a 0-form in ℝ3, then 푑휔 is like the gradient. If 휔 is a 1-form in ℝ3, then 푑휔 is like the curl. If 휔 is a 2-form in ℝ3, then 푑휔 is like the divergence. Something to notice is that 푑(푑휔) = 0 for every 휔, which follows because partial derivatives commute. In particular, we get a so-called complex: If Λ푘 (푀) denotes the 푘-forms on an 푛-dimensional submanifold 푀, then we get the complex 푑 푑 푑 푑 푑 Λ0(푀) → Λ1(푀) → Λ2(푀) → · · · → Λ푛 (푀) → 0. We remark that one can study the topology of 푀 by computing from this complex the cohomology 푘 푘+1 ker(푑 : Λ →Λ ) 푑휔 = 휂 groups, im(푑 : Λ푘−1→Λ푘 ) , which is really about global solvability of the diﬀerential equation

for an unknown 휔. There are variations on this idea and one appears in chapter 4 , but we digress. Let us now state Stokes’ theorem, sometimes called the generalized Stokes’ theorem to distinguish it from the classical Stokes’ theorem you know from vector calculus, which is a special case.

Theorem C.6 (Stokes). Suppose 푀 ⊂ ℝ푛 is an embedded compact smooth oriented (푘 + 1)- dimensional submanifold-with-boundary, 휕푀 has the induced orientation, and 휔 is a smooth 푘-form deﬁned on 푀. Then ∫ ∫ 휔 = 푑휔. 휕푀 푀

One can get away with less regularity, both on 휔 and 푀 (and 휕푀) including “corners.” In ℝ2,

it is easy to state in more generality, see Theorem B.2 . A final note is that the classical Stokes’ theorem is just the generalized Stokes’ theorem with 푛 = 3, 푘 = 2. Classically instead of using differential forms, the line integral is an integral of a vector field instead of a 1-form 휔, and its derivative 푑휔 is the curl operator. 169

As to at least get a ﬂavor of the theorem, let us prove it in a simpler setting, which however is often almost good enough, and it is the key idea in the proof. Suppose 푈 ⊂ ℝ푛 is a domain such that for any 푘 = 1, . . . , 푛 there exist two smooth functions 훼푘 and 훽푘 and 푈 as a set is given by

(푥1, . . . , 푥푘−1, 푥푘+1, . . . , 푥푛) ∈ 휋푘 (푈), 훼푘 (푥1, . . . , 푥푘−1, 푥푘+1, . . . , 푥푛) ≤ 푥푘 ≤ 훽푘 (푥1, . . . , 푥푘−1, 푥푘+1, . . . , 푥푛), where 휋푘 (푈) is the projection of 푈 onto the (푥1, . . . , 푥푘−1, 푥푘+1, . . . , 푥푛) components. Orient 휕푈 as usual. 0 푛−1 Write 푥 = (푥1, . . . , 푥푘−1, 푥푘+1, . . . , 푥푛), and let 푑푉푛−1 be the volume form for ℝ . Consider the (푛 − 1)-form 휔 = 푓 푑푥1 ∧ · · · ∧ 푑푥푘−1 ∧ 푑푥푘+1 ∧ · · · ∧ 푑푥푛. Then 푑휔 = 휕 푓 푑푥 ∧ · · · ∧ 푑푥 . By the fundamental theorem of calculus, 휕푥푘 1 푛 ∫ ∫ 휕 푓 푑휔 = 푑푉푛 푈 푈 휕푥푘 0 ∫ ∫ 훽푘 (푥 ) 휕 푓 = 푑푥푘 푑푉푛−1 0 휕푥 휋푘 (푈) 훼푘 (푥 ) 푘 ∫ 0 = 푓 (푥1, . . . , 푥푘−1, 훽푘 (푥 ), 푥푘+1, . . . , 푥푛) 푑푉푛−1 휋푘 (푈) ∫ 0 − 푓 (푥1, . . . , 푥푘−1, 훼푘 (푥 ), 푥푘+1, . . . , 푥푛) 푑푉푛−1 휋푘 (푈) ∫ = 휔. 휕푈 Any (푛 − 1)-form can be written as a sum of forms like 휔 for various 푘. Integrating each one of them in the correct direction provides the result. Appendix D

Basic terminology and results from algebra

Let us quickly review some basic deﬁnitions and a result or two from commutative that we need in

chapter 6 . See a book such as Zariski–Samuel [ ZS ] for a full reference. Definition D.1. A set 퐺 is called a group if it has an operation 푥 ∗ 푦 defined on it and it satisfies the following axioms: (G1) If 푥 ∈ 퐺 and 푦 ∈ 퐺, then 푥 ∗ 푦 ∈ 퐺.

(G2) (associativity) (푥 ∗ 푦) ∗ 푧 = 푥 ∗ (푦 ∗ 푧) for all 푥, 푦, 푧 ∈ 퐺.

(G3) (identity) There exists an element 1 ∈ 퐺 such that 1 ∗ 푥 = 푥 for all 푥 ∈ 퐺.

(G4) (inverse) For every element 푥 ∈ 퐺 there exists an element 푥−1 ∈ 퐺 such that 푥 ∗ 푥−1 = 0. A group 퐺 is called abelian if it also satisfies: (G5) (commutativity) 푥 ∗ 푦 = 푦 ∗ 푥 for all 푥, 푦 ∈ 퐺. A subset 퐾 ⊂ 퐺 is called a subgroup if 퐾 is a group with the same operation as the group 퐺. If 퐺 and 퐻 are groups, a function 푓 : 퐺 → 퐻 is a group homomorphism if it respects the group law, that is, 푓 (푎 ∗ 푏) = 푓 (푎) ∗ 푓 (푏). If 푓 is bĳective, then it is a group isomorphism. An example of a group is a group of automorphisms. For example, let 푈 ⊂ ℂ be open and suppose 퐺 is the set of biholomorphisms 푓 : 푈 → 푈. Then 퐺 is a group under composition, but 퐺 is not necessarily abelian: If 푈 = ℂ, then 푓 (푧) = 푧 + 1 and 푔(푧) = −푧 are members of 퐺, but 푓 ◦ 푔(푧) = −푧 + 1 and 푔 ◦ 푓 (푧) = −푧 − 1. Definition D.2. A set 푅 is called a commutative ring if it has two operations defined on it, addition 푥 + 푦 and multiplication 푥푦, and if it satisfies the following axioms: (A1) If 푥 ∈ 푅 and 푦 ∈ 푅, then 푥 + 푦 ∈ 푅.

(A2) (commutativity of addition) 푥 + 푦 = 푦 + 푥 for all 푥, 푦 ∈ 푅.

(A3) (associativity of addition) (푥 + 푦) + 푧 = 푥 + (푦 + 푧) for all 푥, 푦, 푧 ∈ 푅.

(A4) There exists an element 0 ∈ 푅 such that 0 + 푥 = 푥 for all 푥 ∈ 푅. 171

(A5) For every element 푥 ∈ 푅 there exists an element −푥 ∈ 푅 such that 푥 + (−푥) = 0. (M1) If 푥 ∈ 푅 and 푦 ∈ 푅, then 푥푦 ∈ 푅. (M2) (commutativity of multiplication) 푥푦 = 푦푥 for all 푥, 푦 ∈ 푅. (M3) (associativity of multiplication) (푥푦)푧 = 푥(푦푧) for all 푥, 푦, 푧 ∈ 푅. (M4) There exists an element 1 ∈ 푅 (and 1 ≠ 0) such that 1푥 = 푥 for all 푥 ∈ 푅. (D) (distributive law) 푥(푦 + 푧) = 푥푦 + 푥푧 for all 푥, 푦, 푧 ∈ 푅. The ring 푅 is called a ﬁeld if furthermore:

(F) For every 푥 ∈ 푅 such that 푥 ≠ 0 there exists an element 1/푥 ∈ 푅 such that 푥(1/푥) = 1.

In a commutative ring 푅, the elements 푢 ∈ 푅 for which there exists an inverse 1/푢 as above are called units. If 푅 and 푆 are rings, a function 푓 : 푅 → 푆 is a ring homomorphism if it respects the ring operations, that is, 푓 (푎 + 푏) = 푓 (푎) + 푓 (푏) and 푓 (푎푏) = 푓 (푎) 푓 (푏), and such that 푓 (1) = 1. If 푓 is bĳective, then it is called a ring isomorphism. Namely, a commutative ring is an abelian additive group (by additive group we just mean we use + for the operation and 0 for the respective identity), with multiplication thrown in. If the multiplication also defines a group on the set of nonzero elements, then the ring is a field. A ring that is not commutative is one that does not satisfy commutativity of multiplication. Some authors define ring without asking for the existence of 1. A ring that often comes up in this book is the ring of holomorphic functions. Let O(푈) be the set of holomorphic functions defined on an open set 푈. Pointwise addition and multiplication give a ring structure on O(푈). The set of units is the set of functions that never vanish in 푈. The set of units is a multiplicative group. Given a commutative ring 푅, let 푅[푥] be the set of polynomials 푘 푘−1 푃(푥) = 푐푘 푥 + 푐푘−1푥 + · · · + 푐1푥 + 푐0, where 푐0, . . . , 푐푘 ∈ 푅. The integer 푘 is the degree of the polynomial and 푐푘 is the leading coefficient of 푃(푥). If the leading coefficient is 1, then 푃 is monic. If 푅 is a commutative ring, then so is 푅[푥]. Similarly, we define the commutative ring 푅[푥1, . . . , 푥푛] of polynomials in 푛 indeterminates.

The most basic result about polynomials, Theorem B.29 the fundamental theorem of algebra, which states that every nonconstant polynomial over 푅 = ℂ has a root, is really a theorem in one complex variable. Definition D.3. Let 푅 be a commutative ring. A subset 퐼 ⊂ 푅 is an ideal if 푓 ∈ 푅 and 푔, ℎ ∈ 퐼 implies that 푓 푔 ∈ 퐼 and 푔 + ℎ ∈ 퐼. In short, 퐼 ⊂ 푅 is an additive subgroup such that 푅퐼 = 퐼. Given a set of elements 푆 ⊂ 푅, the ideal generated by 푆 is the intersection 퐼 of all ideals containing 푆. If 푆 = { 푓1, . . . , 푓푘 } is a finite set, we say 퐼 is finitely generated, and we write 퐼 = ( 푓1, . . . , 푓푘 ). A principal ideal is an ideal generated by a single element. A commutative ring where every ideal is a principal ideal is called a principal ideal domain or a PID. A commutative ring 푅 is Noetherian if every ideal in 푅 is finitely generated. 172 APPENDIX D. BASIC TERMINOLOGY AND RESULTS FROM ALGEBRA

It is not diﬃcult to prove that “an ideal generated by 푆” really is an ideal, that is, the intersection of ideals is an ideal. If an ideal 퐼 is generated by 푓1, . . . , 푓푘 , then every 푔 ∈ 퐼 can be written as

푔 = 푐1 푓1 + · · · + 푐푘 푓푘 ,

for some 푐1, . . . , 푐푘 ∈ 푅. Clearly the set of such elements is the smallest ideal containing 푓1, . . . , 푓푘 . Theorem D.4 (Hilbert basis theorem). If 푅 is a Noetherian commutative ring, then 푅[푥] is Noethe- rian.

As the proof is rather short, we include it here.

Proof. Suppose 푅 is Noetherian, and 퐼 ⊂ 푅[푥] is an ideal. Starting with the polynomial 푓1 of minimal degree in 퐼, construct a (possibly finite) sequence of polynomials 푓1, 푓2,... such that 푓푘 is the polynomial of minimal degree from the set 퐼 \( 푓1, . . . , 푓푘−1). The sequence of degrees deg( 푓1), deg( 푓2),... is by construction nondecreasing. Let 푐푘 be the leading coefficient of 푓푘 . As 푅 is Noetherian, then there exists a finite 푘 such that (푐1, 푐2, . . . , 푐푚) ⊂ (푐1, 푐2, . . . , 푐푘 ) for all 푚. Suppose for contradiction there exists a 푓푘+1, that is, the sequence of polynomials did not end at 푘. In particular, (푐1, . . . , 푐푘+1) ⊂ (푐1, . . . , 푐푘 ) or

푐푘+1 = 푎1푐1 + · · · 푎푘 푐푘 .

As degree of 푓푘+1 is at least the degree of 푓1, through 푓푘 , we can deﬁne the polynomial

deg( 푓푘+1)−deg( 푓1) deg( 푓푘+1)−deg( 푓2) deg( 푓푘+1)−deg( 푓푘 ) 푔 = 푎1푥 푓1 + 푎2푥 푓2 + · · · + 푎푘 푥 푓푘 .

The polynomial 푔 has the same degree as 푓푘+1, and in fact it also has the same leading term, 푐푘+1. On the other hand 푔 ∈ ( 푓1, . . . , 푓푘 ) while 푓푘+1 ∉ ( 푓1, . . . , 푓푘 ) by construction. The polynomial 푔 − 푓푘+1 is also not in ( 푓1, . . . , 푓푘 ), but as the leading terms canceled, deg(푔 − 푓푘+1) < deg( 푓푘+1), but that is a contradiction, so 푓푘+1 does not exist and 퐼 = ( 푓1, . . . , 푓푘 ). Deﬁnition D.5. An element 푓 ∈ 푅 is irreducible if whenever 푓 = 푔ℎ for two elements 푔, ℎ ∈ 푅, then either 푔 or ℎ is a unit. A commutative ring 푅 is a unique factorization domain (UFD) if up to multiplication by a unit, every element has a unique factorization into irreducible elements of 푅.

One version of a result called the Gauss lemma says that just like the property of being Noethe- rian, the property of being a UFD is retained when we take polynomials.

Theorem D.6 (Gauss lemma). If 푅 is a commutative ring that is a UFD, then 푅[푥] is a UFD.

The proof is not diﬃcult, but it is perhaps beyond the scope of this book. Further Reading

[BER] M. Salah Baouendi, Peter Ebenfelt, and Linda Preiss Rothschild, Real submanifolds in complex space and their mappings, Princeton Mathematical Series, vol. 47, Princeton

University Press, Princeton, NJ, 1999. MR 1668103 [B] Albert Boggess, CR manifolds and the tangential Cauchy-Riemann complex, Studies in

Advanced Mathematics, CRC Press, Boca Raton, FL, 1991. MR 1211412 [C] E. M. Chirka, Complex analytic sets, Mathematics and its Applications (Soviet Series),

vol. 46, Kluwer Academic Publishers Group, Dordrecht, 1989. MR 1111477 [D] John P. D’Angelo, Several complex variables and the geometry of real hypersurfaces, Studies

in Advanced Mathematics, CRC Press, Boca Raton, FL, 1993. MR 1224231 [GR] Robert C. Gunning and Hugo Rossi, Analytic functions of several complex variables,

Prentice-Hall Inc., Englewood Cliﬀs, N.J., 1965. MR 0180696 [H] Lars Hörmander, An introduction to complex analysis in several variables, 3rd ed., North- Holland Mathematical Library, vol. 7, North-Holland Publishing Co., Amsterdam, 1990.

MR 1045639 [K] Steven G. Krantz, Function theory of several complex variables, 2nd ed., The Wadsworth & Brooks/Cole Mathematics Series, Wadsworth & Brooks/Cole Advanced Books & Software,

Paciﬁc Grove, CA, 1992. MR 1162310 [R1] Walter Rudin, Function theory in the unit ball of C푛, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Science], vol. 241, Springer-

Verlag, New York, 1980. MR 601594 [R2] , Principles of mathematical analysis, 3rd ed., McGraw-Hill Book Co., New York-Auckland-Düsseldorf, 1976. International Series in Pure and Applied Mathematics.

MR 0385023 [W] Hassler Whitney, Complex analytic varieties, Addison-Wesley Publishing Co., Reading,

Mass.-London-Don Mills, Ont., 1972. MR 0387634 [ZS] Oscar Zariski and Pierre Samuel, Commutative algebra, Volume I, The University Series in Higher Mathematics, D. Van Nostrand Company, Inc., Princeton, New Jersey, 1958. With

the cooperation of I. S. Cohen. MR 0090581 Index

(0, 1)-form, 111 Cartan–Thullen theorem, 82

푝, 푞 ( )-form, 117 Casorati–Weierstrass theorem, 159 ¯

휕 -problem, 111 Cauchy estimates, 10 , 19 , 154

Cauchy formula, 8 , 14

abelian, 170 Cauchy integral formula, 8 ambient, 139 Cauchy integral formula in several variables,

analytic continuation, 23 , 160 14

analytic disc, 29 Cauchy kernel, 10

analytic set, 36 , 137 Cauchy–Pompeiu integral formula, 109

antiholomorphic coordinates, 11 Cauchy–Riemann equations, 7 , 13 antiholomorphic function, 24 center of a polydisc, 11 antiholomorphic vectors, 51 chain rule for holomorphic mappings, 26 approximate delta function, 70 charts, 163 argument principle, 158 chicken atlas, 163 imaginary, 5 automorphism, 32 ∞

퐶 -smooth, 44 automorphism group, 32

circular domain, 33 푘 퐶 -smooth boundary, 44 ball, 12 푘 퐶 -smooth function, 6 Baouendi–Trèves approximation theorem, 98 푘 퐶 -smooth hypersurface, 44

Behnke–Stein theorem, 77 , 82

closed analytic disc, 29

Bergman kernel, 120

closed path, 152 unit disc, 121

codimension, 139 Bergman space, 119

commutative ring, 170 bidegree, 117

compactly supported, 151 bidisc, 11

compatibility conditions, 111 big-oh notation, 49

complete intersection, 142 biholomorphic, 28

complete Reinhardt domain, 21 , 84 biholomorphically equivalent, 28

complex chain rule, 24 biholomorphism, 28

complex conjugate, 6 Bochner–Martinelli integral formula, 116

complex Euclidean space, 11 Bochner–Martinelli kernel, 116

complex Hessian, 54 boundary of a submanifold, 162

complex manifold, 139

Cartan’s uniqueness theorem, 32 complex submanifold, 36 , 139 INDEX 175

complex subvariety, 137 Euclidean distance, 6

complex variety, 137 Euclidean inner product, 12

complex-analytic, 6 Euclidean norm, 12

complex-analytic subvariety, 137 Euclidean space, 11

complex-analytic variety, 137 exhaustion function, 75

complex-diﬀerentiable, 6 exterior derivative, 167

complexiﬁcation, 87 , 89

complexiﬁcation of a real hypersurface, 94 Fatou–Bieberbach domain, 39

complexiﬁed tangent space, 51 ﬁeld, 171

continuity principle, 74 ﬁnitely generated ideal, 171

convergence of power series, 17 fundamental theorem of algebra, 159

converges uniformly absolutely, 17 fundamental theorem of symmetric

convex, 41 , 47 polynomials, 130

with respect to F, 73

Gauss lemma, 172 convex function, 64

generalized Cauchy integral formula, 109 coordinate ring, 138

generalized Stokes’ theorem, 168 CR function, 92

geometric series in several variables, 17 curve, 152

geometrically convex, 41 , 50 cusp, 139

geometrically distinct, 132 cutoﬀ function, 101

geometrically unique, 132

deﬁning function, 44 germ of a function, 125

deﬁning functions, 138 germ of a set, 126

degree, 171 germ of a Weierstrass polynomial, 127

degree 푗 homogeneous part, 32 , 87 Green’s theorem, 153

degree of a polynomial, 6 group, 170

diagonal, 88 group homomorphism, 170

Diederich–Fornæss lemma, 149 group isomorphism, 170

dimension, 140

Hardy space, 123 dimension at a regular point, 139

harmonic, 63 Dirichlet problem, 64

harmonic conjugate, 155 discriminant function, 133

Hartogs ﬁgure, 41 discriminant set, 133

Hartogs phenomenon, 42 , 113 distinguished boundary, 15

Hartogs pseudoconvex, 75 domain, 5

Hartogs triangle, 43 domain of convergence, 20

Hartogs–Bochner, 107 domain of holomorphy, 40

Heisenberg group, 57 domain with smooth boundary, 44

Hessian, 48

elementary symmetric functions, 129 complex, 54

embedded hypersurface, 45 Hilbert basis theorem, 172

embedded submanifold, 162 holomorphic, 6, 12

embedded submanifold-with-boundary, 162 holomorphic coordinates, 11

entire, 154 holomorphic hull, 80

essential singularity, 157 holomorphic implicit function theorem, 27 176 INDEX

holomorphic mapping, 25 Liouville’s theorem, 154

holomorphic vectors, 51 local parametrization, 162

holomorphically convex, 80 Local parametrization theorem, 144

homeomorphism, 28 local variety, 137

homogeneous, 32 locally bounded function, 12

homogeneous part, 32 , 87 locally bounded in 푈, 35

homomorphism, 170 , 171 logarithmically convex, 84

hull, 73

manifold, 163 hull of a Hartogs ﬁgure, 42 map Hurwitz’s theorem, 23 , 156

biholomorphic, 28 hypersurface, 44

proper, 28 hypervariety, 140

maximal ideal, 135

ideal, 135 , 171 maximum modulus principle, 8 , 155

ideal generated by 푆, 171 maximum principle, 8 , 22 , 155

identity theorem, 21 , 154 plurisubharmonic functions, 69

imaginary part, 6 subharmonic functions, 66

implicit function theorem, 27 subvarieties, 146

induced orientation, 164 mean-value property, 66

inertia, 55 Mergelyan’s theorem, 160

inﬁnitely diﬀerentiable, 6 meromorphic, 23 , 157 ¯ inhomogeneous 휕-problem, 111 modulus, 6

integral domain, 23 , 126 molliﬁer, 70

irreducible, 143 , 172 monic, 171

irreducible components, 144 Monodromy theorem, 23 , 160

isolated singularity, 157 Montel’s theorem, 23 , 156

isomorphism, 170 , 171 multi-index notation, 16

multiplicity of a zero, 156

Jacobian conjecture, 39

Jacobian determinant, 27 Narasimhan’s lemma, 61

Jacobian matrix, 26 Noetherian, 136 , 171

normal convergence, 156 1

Koebe -theorem, 161 4 normal coordinates, 95

Koebe quarter theorem, 161 Nullstellensatz, 138 , 143

Kontinuitätssatz, 74

open mapping theorem, 154

Laplacian, 63 open set with smooth boundary, 44

Laurent series, 157 order of a pole, 157 leading coeﬃcient, 171 order of a zero, 156

Levi form, 54 order of vanishing, 127

Levi problem, 61 , 81 orientation, 164

for complete Reinhardt domains, 84 overdetermined, 97

Levi pseudoconvex, 54

Levi-ﬂat, 63 parametrization, 162

Lewy extension theorem, 105 path, 152

Lewy hypersurface, 56 peak point, 63 INDEX 177

Picard’s big theorem, 159 Riemann sphere, 157

Picard’s little theorem, 159 ring, 170

PID, 135 , 171 ring homomorphism, 171 1

piecewise-퐶 boundary, 153 ring isomorphism, 171

pluriharmonic, 68 ring of holomorphic functions, 23

plurisubharmonic, 68 Rothstein’s theorem, 28

plush, 68 Rouché’s theorem, 159

Poincaré problem, 23 Runge’s approximation theorem, 159

point of indeterminacy, 24

Schwarz reﬂection principle, 160

Poisson kernel, 65

Schwarz’s lemma, 22 , 155 polarization, 90

section, 47 pole, 157

Segre variety, 147 poles, 23

Severi’s theorem, 96 , 115 polydisc, 11

simple path, 152 polynomial hull, 81

simple point, 139 polyradius of a polydisc, 11

simply connected, 154 positively oriented, 153

singular point, 36 , 139 power series convergence, 17

smooth boundary, 44 prime ideal, 143

smooth CR function, 92 principal ideal, 135 , 171

smooth function, 6 principal ideal domain, 135 , 171

Stokes’ theorem, 168 proper ideal, 135

strictly pseudoconvex, 54 proper map, 28

strongly convex, 47 pseudoconvex, 54 , 75

strongly pseudoconvex, 54 psh, 68

sub-mean-value property, 66

Puiseux, 145

subgroup, 170 pure dimension, 140

subharmonic, 64

Radó’s theorem, 72 sublevel set, 75

radius of a polydisc, 11 submanifold, 162

real hypersurface, 45 submanifold-with-boundary, 162

real Jacobian, 26 subvariety, 36, 137

real part, 6 support, 151

real-analytic, 86 supremum norm, 10

reducible, 143 symmetric function, 129

regular point, 36 , 139 symmetric with respect to the real axis, 160

Reinhardt domain, 12 , 21 Szegö kernel, 124

complete, 21 , 84 unit disc, 121

logarithmically convex, 84

tangent bundle, 46 removable singularity, 157

tangent space, 45 reproducing property, 121

tangent vector, 45 residue, 158

tomato can principle, 61 residue theorem, 158

transition map, 163

Riemann extension theorem, 35 , 157

Riemann mapping theorem, 28 , 155 UFD, 136 , 172 178 INDEX

uniform absolute convergence, 17 vector ﬁeld, 47

unique factorization domain, 136 , 172

unit, 171 weakly pseudoconvex, 54 unit ball, 12 Weierstrass division theorem, 131 unit disc, 8

Weierstrass factorization theorem, 161 unit polydisc, 11

Weierstrass polynomial, 127 upper-semicontinuous, 64

Weierstrass preparation theorem, 127

variety, 137 Weierstrass product theorem, 161

vector, 45 Wirtinger operators, 7 , 12 List of Notation

Notation Description Page

ℂ complex numbers 5

ℝ real numbers numbers 5

ℤ integers 5

ℕ natural numbers {1, 2, 3,...} 5 √

푖 −1 5

푧¯ complex conjugate 6

|푧| modulus 6

Re 푧 real part 6

Im 푧 imaginary part 6 푘 퐶 푘 times continuously diﬀerentiable 6 , 44 ∞

퐶 inﬁnitely diﬀerentiable 6 , 44 휕 휕 휕 휕

, , , Wirtinger operators 7 , 12 휕푧 휕푧¯ 휕푧 푗 휕푧¯ 푗

푑푧, 푑푧 푗 푑푧 = 푑푥 + 푖 푑푦, 푑푧 푗 = 푑푥 푗 + 푖 푑푦 푗 8 , 13

픻 unit disc 8

≡ identically equal, equal at all points 8 , 151 2 ∇ Laplacian 9

Δ휌 (푎) disc, polydisc 9 , 11

k 푓 k퐾 supremum norm over 퐾 10

h푧, 푤i Euclidean inner product 12

k푧k Euclidean norm 12

퐵휌 (푎) ball 12

픹, 픹푛 unit ball 12

푑푧¯, 푑푧¯ 푗 푑푧¯ = 푑푥 − 푖 푑푦, 푑푧¯ 푗 = 푑푥 푗 − 푖 푑푦 푗 13 푘 푗 푘

훿 훿 훿 푗 푘 푗 Kronecker delta, 푗 = 1, 푗 = 0 if ≠ 14

ℕ0 nonnegative integers {0, 1, 2,...} 16 180 LIST OF NOTATION

Notation Description Page

훼 훼1 훼2 훼푛

푧 푧 푧 ··· 푧 1 2 푛 16

푑푧 푑푧1 ∧ 푑푧2 ∧ · · · ∧ 푑푧푛 16

|훼| 훼1 + 훼2 + · · · + 훼푛 16

훼! 훼1!훼2! ··· 훼푛! 16

휕|훼| 휕훼1 휕훼2 휕훼푛

··· 16 휕푧훼 휕푧훼1 휕푧훼2 휕푧훼푛 1 2 푛 ∑︁ 훼 푐훼푧 multiindex power series 16 훼

O(푈) the set/ring of holomorphic functions on 푈 23 푓 ◦ 푔 composition, 푝 ↦→ 푓 푔(푝) 25 , 151

퐷 푓 , 퐷 푓 (푎) Jacobian matrix / derivative (at 푎) 26 , 47 , 52

퐷ℝ 푓 , 퐷ℝ 푓 (푎) real Jacobian matrix / real derivative (at 푎) 26 , 47 −1 푓 function inverse 28

⊂⊂ compact or relatively compact subset 28 −1 푓 (푆) pullback of a set, {푞 : 푓 (푞) ∈ 푆} 28 2푛−1 푛 2푛−1 푆 unit sphere in ℂ , 푆 = 휕픹푛 29 푛 푇푝ℝ , 푇푝 푀 tangent space 45 휕

tangent vector 45 휕푥 푗 푝 푛 푇ℝ , 푇 푀 tangent bundle 46 , 47

∇푟 gradient of 푟 48

푂(ℓ) big-oh notation 49 푛 ℂ ⊗ 푇푝ℝ , ℂ ⊗ 푇푝 푀 complexiﬁed tangent space 51

(1,0) 푛 (1,0) 푇푝 ℝ , 푇푝 푀 holomorphic tangent vectors 51

(0,1) 푛 (0,1) 푇푝 ℝ , 푇푝 푀 antiholomorphic tangent vectors 51

퐷ℂ 푓 , 퐷ℂ 푓 (푎) complexiﬁed real derivative 51

L(푋푝, 푋푝) Levi form 54 ∗ ∗

푣 , 퐴 conjugate transpose 55

푃푟 (휃) Poisson kernel for the unit disc 65

푑푉, 푑푉 (푤) volume measure, euclidean volume form 70 , 116

푓 ∗ 푔 convolution 70

퐾b, 퐾bF hull of 퐾 with respect to F 73 LIST OF NOTATION 181

Notation Description Page

dist(푥, 푦) distance between points, point and a set, or two sets 78

퐾b푈 holomorphic hull of 퐾 80 휔 퐶 real-analytic 86 2 2 2

푣 푣 푣 [ ] 1 + · · · + 푛 99

푑퐴, 푑퐴(푤) area form, 푑퐴 = 푑푥 ∧ 푑푦 109

푑휓 exterior derivative 111 , 168

휕휓 holomorphic part of exterior derivative 111 , 117 ¯ 휕휓 antiholomorphic part of exterior derivative 111 , 117

푑푥c푗 removed one-form, 푑푥1 ∧ 푑푥c2 ∧ 푑푥3 = 푑푥1 ∧ 푑푥2 116 2 퐴 (푈) Bergman space, square integrable holomorphic functions 119 2 퐿 (푈) square integrable functions 119 2

푓 푓 퐿 k k 퐴2 (푈), k k퐿2 (푈) norm 119 2 h 푓 , 푔i 퐿 inner product 119

퐾푈 (푧, 휁¯) Bergman kernel of 푈 120 ∗

푈 complex conjugate of a domain 120 , 147 2 퐻 (휕푈) Hardy space 123

푆푈 (푧, 휁¯) Szegö kernel of 푈 124

( 푓 , 푝) germ of 푓 at 푝 125

푛O푝, O푝 ring of germs of holomorphic functions at 푝 125

(퐴, 푝) germ of a set 퐴 at 푝 126 −1 푍 푓 zero set of 푓 , 푓 (0) 126 , 137

ord푎 푓 order of vanishing of 푓 at 푎 127

O(푈)[푧푛] polynomial in 푧푛 with coeﬃcients in O(푈) 127

O0 [푧푛] polynomial in 푧푛 with coeﬃcients in O0 127

( 푓 ), ( 푓1, . . . , 푓푛) ideal generated by 푓 or by 푓1, . . . , 푓푛 135 , 171

퐼푝 (푋) ideal of germs at 푝 vanishing on 푋 137

푉 (퐼) the common zero set of germs in 퐼 137

Γ 푓 graph of 푓 138

푋reg regular points of 푋 139

푋sing singular points of 푋 139

dim푝 푋 dimension of 푋 at 푝 140 182 LIST OF NOTATION

Notation Description Page

dim 푋 dimension of 푋 140

Σ푞 (푈, 푟), Σ푞 Segre variety 147

퐴 \ 퐵 set subraction 151

푆 topological closure 151

휕푋 boundary of 푋 (topological, or manifold) 151 , 162

푓 : 푋 → 푌 a function from 푋 to 푌 151

푥 ↦→ 퐹(푥) a function of 푥 151

푓 |푆 restriction of 푓 to 푆 151 def

푋 = 푌 deﬁne 푋 to be 푌 151 ∫

푓 (푧) 푑푧 path integral 152 훾

휔 ∧ 휂 wedge product of diﬀerential forms 165