Notes on Math 321

September 2, 2007 ii

Preface v

1 Curves 1 1.1 Basic Deﬁnitions ...... 1 1.2 Frenet Trihedron ...... 3 1.3 Local canonical form ...... 4 1.4 Exercises ...... 5

2 Regular Surfaces 7 2.1 Digression: Calculus ...... 8 2.1.1 Chain Rule ...... 8 2.1.2 Inverse Function Theorems ...... 8 2.2 Basic Deﬁnitions ...... 9 2.2.1 Proof of theorem 2.2.1 ...... 10 2.2.2 Proof of theorem 2.2.3 ...... 11 2.3 Existence of tangent planes ...... 11 2.4 Local Canonical Form ...... 12 2.5 Exercises ...... 13

3 Gauss Map 15 3.1 Smooth maps between regular surface ...... 15 3.2 Orientation, Gauss map and curvatures ...... 15 3.3 Digression: Linear Algebra ...... 17 3.3.1 One-form ...... 17 3.3.2 Two-form ...... 17 3.3.3 The area form on an oriented Euclidean plane ...... 19 3.4 Fundamental Forms ...... 20 3.4.1 The total diﬀerentials ...... 20 3.4.2 The ﬁrst fundamental two-form ...... 22 3.4.3 The second fundamental two-form ...... 22 3.4.4 Area Form ...... 23 3.5 Exercises ...... 23

iii iv CONTENTS

4 Intrinsic Theory 25 4.1 Vector Fields ...... 25 4.1.1 Lie derivatives of smooth functions ...... 26 4.1.2 Gradient vector fields ...... 27 4.1.3 Lie derivatives of Rm-valued smooth functions ...... 27 4.2 Directional Derivatives ...... 27 4.2.1 Directional derivatives of Rm-valued smooth functions ...... 27 4.2.2 Covariant derivatives of vector fields ...... 28 4.2.3 The covariant differentiation is non-commutative in general ...... 29 4.3 Riemann Curvature ...... 30 4.3.1 Digression: Lie bracket ...... 31 4.3.2 Curvarure Tensor ...... 31 4.4 The covariant differentiation is intrinsic ...... 32 4.5 Riemann curvature and Gauss curvature are all intrinsic ...... 33 4.6 Exercises ...... 33

5 Selected topics 35 5.1 Geodesics ...... 35 5.1.1 Digression ...... 35 Preface

Rough speaking, geometry is the study of interesting shapes and relations among them. (More generally, the core of mathematics is the study of interesting objects and relations among them. The same is true for physics.) Geometry is intuitive, one can quickly get many nice facts by just looking at the geometric shapes — a picture is worth a thousand words. For example, two straight lines on a plane either don’t intersect or intersect exactly at one point. Surely, everybody agrees on this simple fact, but most people cannot give a rigorous proof. To give a proof, one needs algebra, and that lead us to analytic geometry. In a sense algebra is a tool which makes our geometric intuition rigorous. Moreover, algebra extends our understanding of geometry beyond our intuition; for example, the fact that one cannot trisect an angle by using a compass and a ruler is not intuitively clear, but can be proved by using algebra. For the study of more sophisticated shapes, algebra is not enough, we need a great tool, namely calculus, and that leads us to differential geometry. For example, what is the volume of a ball with radius r? Algebra alone cannot give you the answer, but calculus surely can. Roughly speaking, differential geometry is the study of interesting shapes and relations among them by using calculus. But I prefer to view differential geometry as college calculus III — calculus on curved spaces. To me, there is a huge psychological benefit to adopt this view: differential geometry is just a further extension of our regular calculus, so we are just repeating many of the same old calculus ideas most of the time during the learning process; in other words, we are fearless while wandering because we are not far away from our “mother” (i.e., the regular calculus). In this course, we only study curves and surfaces inside R3; these are nice geometric shapes of dimension one and two respectively. Something new such as curvatures and torsion appear naturally, they tell us how the the curves and surfaces sit inside R3 up to the rigid motions of R3 (i.e., the transformations of R3 which preserve both lengths and angles.) These facts are quite intuitive. While the curves are intrinsically flat (i.e., can be straightened locally without stretching); for surface, much less intuitive facts were discovered by Gauss: 1) the Gauss curvature K is intrinsic, i.e., stays the same no matter how you bend the surface without stretching, 2) the total Gauss curvature of an (oriented) closed surface is topological, i.e., stays the same no matter how you stretch the surface. It is worth to mention that the Gauss curvature K also has a meaning in calculus! Recall that, in regular calculus of two variables, for smooth vector field F ,(∂x∂y − ∂y∂x)F is zero. If we extend the calculus to a surface, then (∂x∂y − ∂y∂x)F = KF , so the Gauss curvature measures the the failure of commutativity for mixed partial derivatives. This observation is quite natural and fruitful; as a matter of fact, if we emphasize the view that differential geometry is the calculus on

v vi PREFACE curved spaces, we could rediscover pretty much all mathematics that we need for writing down the fundamental laws of physics: from Riemannian geometry to geometry of vector bundles. Chapter 1

Curves

Prerequisite: Calculus I, i.e., calculus for vector-valued functions of one variable — the kind of calculus needed for the study of the Newtonian mechanics.

Curves are one dimensional nice geometric objects, and it is very useful to view it as the trajectory of a moving particle in R3 or R2.

1.1 Basic Deﬁnitions

Some standard notations:

• I — an open interval, say (−1, 2) or (−∞, ∞) or (2, ∞), etc.;

• t0 — a number in I;

• f — a function on I taking values in R or R2 or R3;

• α — a parameterized smooth curve;

• C — a regular curve.

Some standard terminologies:

• f is smooth — each component function of f is a smooth (i.e., inﬁnitely diﬀerentiable) real-valued function;

Deﬁnition 1.1.1. A parameterized smooth curve in R3 is just a smooth map α: I / R3. The image of α (viewed as a map) is called the trace of α or a curve. We say α is regular if α0 is nonzero everywhere on I.A regular curve in R3 is just the trace of a regular parameterized smooth curve in R3.

In other word, a regular curve in R3 is just a subset of R3 which can be parameterized by a smooth map α with nowhere vanishing ﬁrst derivative. By the way, curves in R2, i.e., plane curves, can be similarly deﬁned.

1 2 CHAPTER 1. CURVES

Example 1.1.1. α(t) = (t, t) is a regular parameterized smooth curve in R2, and its trace ( a straight line) is a regular plane curve.

Example 1.1.2. α(t) = (t3, t3) is a parameterized smooth curve in R2, but not regular at t = 0. However, its trace ( a straight line) is a regular plane curve.

So a regular curve can have a non-regular parametrization.

Example 1.1.3. α(t) = (t3, t2) is a parameterized smooth curve in R2, but it is not a regular parameterized smooth curve. One can show that its trace is not a regular curve1.

So a non-regular curve can have a smooth parametrization.

Example 1.1.4. α(t) = (t, |t|) is not a parameterized smooth curve in R2 because it is not smooth at t = 0. One can show that its trace is not a regular curve either2.

Example 1.1.5. Let f be a real valued smooth function on I. Then the graph of f:

Γ(f) := {(t, f(t) | t ∈ I} is a regular plane curve.

Theorem 1.1.1 (Arclength Parametrization). Let C be a regular curve, then C admits an arclength parametrization, i.e., C is the trace of a regular parameterized smooth curve α such that |α0| = 1 everywhere.

/ 3 Proof. Since C is a regular curve, it must be theR trace of a parameterized smooth curve β: J R such that |β0| 6= 0 everywhere. Let s = g(t) = t |β0(x)| dx. Since3 g0 = |β0| > 0 on J, g is smooth t0 and has a smooth inverse t = f(s) with

1 1 f 0(s) = = . g0(f(s)) |β0(f(s))|

Let α(s) = β(f(s)) for s ∈ I (where I is the image of f). Then |α0(s)| = |β0(f(s))| · f 0(s) = 1.

Example 1.1.6. The unit circle centered at (0, 0) is a regular plane curve. α(s) = (cos t, sin t) is one of its arclength parameterizations.

3 Deﬁnition 1.1.2. Suppose that α: I / R is a regular parameterized smooth curve and t0 ∈ I. The trace of the linear approximation of α around t0 is called the tangent line of α at t0.

Note that the tangent line of α at t0 is the trace of this parameterized curve:

0 t 7→ α(t0) + α (t0)(t − t0).

1The proof is given in example 1.3.1. 2I leave its proof to you. 3I will prove this statement in class. 1.2. FRENET TRIHEDRON 3

1.2 Frenet Trihedron

Suppose that α: I / R3 is a regular parameterized smooth curve. By a re-parametrization if necessary, we can assume that |α0| = 1 on I. Furthermore, we assume that |α00| 6= 0 on I. Let ~t = α0 and we call ~t(s) the tangent vector at s. Since ~t · ~t = 1, ~t · ~t0 = 0 after taking derivatives. In view of the fact that ~t0 = α00 6= ~0, we can introduce unit vector

~t0 α00 ~n = = . |~t0| |α00|

We shall call ~n(s) the normal vector at s, that is because ~t(s) · ~n(s) = 0. The plane spanned by ~t(s) and ~n(s) is called osculating plane at s. Let ~b = ~t × ~n, then ~b(s) is a unit normal vector of the osculating plane at s and is called the binormal vector at s. Note that (~t(s), ~n(s),~b(s)) is an orthornormal frame for R3 and is called the Frenet trihedron at s.

Proposition 1.2.1. There are smooth functions k > 0 and τ on I such that   ~t0 = k~n ~n0 = −k~t −τ~b  ~b0 = τ~n

Proof. The proof is just a bunch of easy computations.

Remark 1.2.1. k(s) is called the curvature at s and τ(s) is called the torsion of s. Roughly speaking, k(s) measures the extent of the bending of tangent line and τ(s) measures the extent of the bending of osculating plane.

We say a map T : R3 / R3 is a rigid motion of R3 if T preserves both distance and angle.

Exercise 1.2.1. Assume T is a rigid motion of R3 such that T (~0) = ~0. Show that 1) T (~x) · T (~y) = x · y. 2) T (~x + ~y) = T (~x) + T (~y). 3) T is continuous, hence T is a linear transformation. 4) A rigid motion of R3 must be an aﬃne linear transformation.

Theorem 1.2.1 (Fundamental Theorem of the Local Theory of Curves). Given smooth functions k(s) > 0 and τ(s), s ∈ I, there exists a regular parameterized smooth curve α: I / R3 such that s is the arc length, k(s) is the curvature and τ(s) is the torsion of α. Moreover, α is unique up to an orientation preserving rigid motion of R3.

Proof. It is a consequence of existence and uniqueness theorem for ordinary diﬀerential equations. 4 CHAPTER 1. CURVES

1.3 Local canonical form

Let α: I / R3 be an arc length parameterized smooth curve, 0 ∈ I. Then 1 α(s) = α(0) + α0(0)s + α00(0)s2 2 1 + α000(0)s3 + ··· 6 1 = α(0) + ~t(0)s + k(0)~n(0)s2 2 1 + α000(0)s3 + ··· (1.1) 6

Since α000 = k0~n + k~n0 = k0~n + k(−k~t − τ~b) = −k2~t + k0~n − kτ~b. Up to a rigid motion, we may assume α(0) = ~0 and (~t(0), ~n(0),~b(0)) = (~i,~j,~k), then 1 α(s) = ~is + k(0)~js2 2 1 + (−k(0)2~i + k0(0)~j − k(0)τ(0)~k)s3 + ··· 6  2  s − k(0) s3  6     0  =  k(0) s2 + k (0) s3 + o(s3) (1.2)  2 6    k(0)τ(0) 3 − 6 s For simplicity we write k(0) as k, k0(0) as k0, and τ(0) as τ. It is not hard to see that s = k2 3 3 x + 6 x + o(x ), it is then clear that the piece of curve near s = 0 is just the graph of map F : I / R2 where k k0 kτ F (x) = ( x2 + x3, − x3) + o(x3). (1.3) 2 6 6 Exercise 1.3.1. Find the curvature and torsion of the graph of F (x) = (cos x, (sin x)3) at x = 0.

Example 1.3.1. The trace of α(t) = (t3, t2) is not a regular curve. Otherwise, there were a regular parameterized curve β(t) = (x(t), y(t)) whose trace is the same as that of α, so we have

y3(t) = x2(t). (1.4)

We may assume that β(0) = (0, 0). Since the tangent line at (0, 0) would be the y-axis by the continuity argument, so x0(0) = 0, then y0(0) = a 6= 0 because β is regular at t = 0. Therefore, we have x(t) = bt2 + o(t2) for some number b and y(t) = at + o(t). In view of eq. (1.4), we have (at + o(t))3 = (bt2 + o(t2))2 = o(t3), i.e., a3t3 = o(t3) as t / 0; then we must have a3 = 0, contradicting to the fact that a 6= 0.

Deﬁnition 1.3.1. A subset of R3 is called an one-manifold inside R3 if, up to a rigid motion of R3, locally it is the graph of a smooth function of the form (y, z) = F (x). 1.4. EXERCISES 5

Here, “locally” means the part of subset inside an open set (could be very small) of R3. So a non-simple (i.e., self-crossing) regular curve is not an one-manifold. Since figure eight is not simple, it is not an one-manifold. A simple regular curve is called a “nice” curve if, for any point p in the curve, the part of curve inside a sufficiently small open ball of R3 centered at p consists of a single connected piece4. A “nice” curve is an one-manifold. An one-manifold may not be a “nice” curve because it might be the disjoint union of two “nice” curves. Example 1.3.2. A line is an one-manifold. The graph of a smooth real-valued function of one- variable is an one-manifold. All conics are one-manifolds, but which one of which is not a curve? In the forthcoming notes, I will talk about the two-manifolds, i.e., the regular (non self-crossing) surfaces defined in chapter two of the textbook. The above definition of one-manifold can be easily generalized as follows: Definition 1.3.2. A subset of R3 is called a two-manifold inside R3 if , up to a rigid motion of R3, locally it is the graph of a smooth function of the form z = F (x, y). Please note that, a regular surface defined in the textbook is nothing but a two-manifold defined as above, so it must be non self-crossing; however, a regular curve defined in the textbook can be self-crossing. Please also note that, a curve must be connected, and a regular surface may not be connected, just like the fact that an one-manifold may not be connected. In summary, we have the following relationships among the terminologies: • regular surface = 2-manifold in R3; • “nice” curve = connected 1-manifold in R2.

1.4 Exercises

Exercise 1.4.1. This exercise tests your understanding of a few basic concepts. 1) Find a non-smooth function whose graph is a regular curve. 2) Find a parameterized smooth curve whose trace is not a regular curve. 3) Find a parameterized smooth curve which is not regular, but its trace is a regular curve. 4) Find a regular curve which is not a 1-manifold. 5) Find a 1-manifold which is not a regular curve. Exercise 1.4.2. Let α(t) = (cos t, sin t, t) (where −∞ < t < ∞) be a parameterized smooth curve, C be the trace of α. 1) Show that C is a simple regular curve. 2) Show that C is a 1-manifold. 3) Find an arc length parametrization for C. 4) Find the Frenet trihedron at t = 0. 5) Find the torsion and curvature at any point of the curve. 6) Find the local canonical form for C at t = 0.

4See examples in class for explanations 6 CHAPTER 1. CURVES Chapter 2

Regular Surfaces

Prerequisite: Calculus II, i.e., multi-variable calculus — the kind of calculus needed for the study of electromagnetism.

Some standard notations:

• U — an open set of Rn;

• ~r0 — a point in U; • f — a smooth map from U or Rm;

n m • Df(~r0) — the diﬀerential of f at ~r0, it is a linear map from R to R ;

• Jf(~r0) — the standard matrix for Df(~r0), called the Jacobian matrix of f at ~r0; • Σ — a regular surface;

• (Σ, p) — a pair such that p ∈ Σ;

• φ:(R2, 0) / (Σ, p) is a local diﬀeomorphism, called the local parametrization of Σ around loc p.

Some standard terminologies:

• An open set of Rn — a subset of Rn such that, for any point of this subset, there is a ball centered at this point which is entirely inside this subset.

• An open set of A ⊂ Rn — the intersection of A with an open set of Rn.

• f: U / Rm is smooth — each component function of f is a smooth (i.e., inﬁnitely diﬀer- entiable) real valued function;

• f: A / B is smooth where A ⊂ Rn and B ⊂ Rm are arbitrary subsets — f has a smooth extension to an open neighborhood of A, i.e., there is an open set U of Rn containing A and a smooth map f˜: U / Rm (not into B in general) such that f˜ = f on A.

7 8 CHAPTER 2. REGULAR SURFACES

• f: A / B is a diffeomorphism — f is one-to-one and onto, both f and its inverse f −1: B / A are smooth; • f:(A, a) / (B, b) is a local diffeomorphism — there is an open set A0 of A and an loc open set B0 of B such that, a ∈ A0, b ∈ B0, and f is a map from A0 to B0 which maps a to b, moreover, f is a diffeomorphism.

2.1 Digression: Calculus

n m Let U be an open set of R , f: U / R a smooth (i.e., inﬁnitely diﬀerentiable) map, ~r0 ∈ U. Then f(~r) = f(~r0) + Df(~r0)(~r − ~r0) + o(|~r − ~r0|) where o(|~r − ~r0|) is a term that goes to zero faster than |~r − ~r0| does as ~r / ~r0. Note that the i i-th component of Df(~r0)(~r − ~r0) is ∇f (~r0) · (~r − ~r0), i.e.,

X3 ∂f i (~r )(xk − xk). ∂xk 0 0 k=1 n m Remark 2.1.1. Df(~r0): R / R is a linear map whose (coordinate) standard matrix is called ∂f i the jacobian matrix, denoted by Jf(~r0) whose (i, j)-entry is ∂xj (~r0).

2.1.1 Chain Rule Theorem 2.1.1 (Chain Rule). Let f: Rn / Rm and g: Rm / Rl be smooth maps. Then ∂(g ◦ f)i X ∂gi ∂f k D(g ◦ f)(x) = Dg(f(x)) ◦ Df(x), i.e., (x) = (f(x)) (x) ∂xj ∂yk ∂xj k for all x ∈ Rn. (Hint for the proof: Try to linearize g ◦ f around x.) In terms of Jacobian matrices, we have J(g ◦ f)(x) = Jg(f(x)) · Jf(x).

2.1.2 Inverse Function Theorems Theorem 2.1.2 (Inverse Function Theorem). Let f be a smooth map from Rn to Rn. Then f is a local diffeomorphism around 0 if and only if Df(0) is an isomorphism. (we write this statement n ∼ n as follows: f: (R , 0) =loc (R , f(0)) ⇐⇒ Df(0) is an isomorphism.) (I.e., f is locally invertible as a smooth map if and only if it is infinitesimally invertible as a linear map. ) Corollary 2.1.3 (Inverse Function Theorem-Injective Form). Let f be a smooth map from Rn n+k n+k ∼ to R . Suppose Df(0) is injective, then there is a local diffeomorphism φ: (R , f(0)) =loc (Rn+k, f(0)) such that for all x closer enough to 0. φ ◦ f(x) = (x, 0). 2.2. BASIC DEFINITIONS 9

(I.e., f is inﬁnitesimally injective if and only if locally it is left equivalent to a standard injection.)

Proof. We may assume the ﬁrst n columns of Jf(0) are linearly independent. Then, if we let F : Rn+k → Rn+k be a smooth map deﬁned as

F (x, y) = f(x) + (0, y), then DF (0, 0) must be invertible. So there is an open set W containing 0 such that F |W is a n+k −1 diﬀeomorphism onto F (W ) ∈ R . Let φ = F |F (W ), then φ ◦ f(x) = φ ◦ F (x, 0) = (x, 0) for (x, 0) in W . The rest is clear.

Corollary 2.1.4 (Inverse Function Theorem-Surjective Form). Let f be a smooth map from Rn+k to Rn with f(0) = 0. Suppose Df(0) is surjective, then there is a local diﬀeomorphism ψ:: n+k ∼ n+k (R , 0) =loc (R , 0) such that for all (x, y) closer enough to 0,

f ◦ ψ(x, y) = x.

(I.e., f is inﬁnitesimally surjective if and only if it is right equivalent to a standard projection.)

Proof. We may assume the ﬁrst n rows of Jf(0) are linearly independent. Then, if we let F : Rn+k → Rn+k be a smooth map deﬁned as

F (x, y) = (f(x, y), y), then DF (0, 0) must be invertible. So there is an open set W containing (0, 0) such that F |W is a n+k −1 diﬀeomorphism onto F (W ) ∈ R . Let ψ = F |F (W ), then f ◦ ψ(x, y) = x for (x, y) in W .

Corollary 2.1.5 (Implicit Function Theorem). Let f: Rn+k /Rn be a smooth map with f(0, 0) = 0. Suppose that Df(0, 0) maps Rn × 0 isomorphically onto Rn. Then there is an open neighborhood of 0 ∈ Rk on which there deﬁnes a unique smooth map g into Rn such that g(0) = 0 and

f(g(y), y) = 0.

Proof. Let F (x, y) = (f(x, y), y), then F (0, 0) = (0, 0) and DF (0, 0) is invertible. Then F is a local diﬀeomorphism around (0, 0). Let ψ be the inverse of F in a product neighborhood of (0, 0), say n+k U × V ⊂ R . Deﬁne g on V by letting g(y) = p1 ◦ ψ(0, y) for any y ∈ V . Then g(0) = 0 and

0 = p1 ◦ F ψ(0, y) = p1 ◦ F (g(y), y) = f(g(y), y).

The uniqueness comes from the uniqueness of the inverse of F .

2.2 Basic Deﬁnitions

Deﬁnition 2.2.1. A subset of R3 is called a regular surface if locally it is the graph of a smooth function of two variables taking value in an one dimensional Euclidean space; i.e., up to a rigid motion of R3, locally it is the graph of a smooth function of the form z = f(x, y). 10 CHAPTER 2. REGULAR SURFACES

Remark 2.2.1. In other word, for Σ ⊂ R3 to be qualiﬁed as a regular surface, for any point p ∈ Σ, we must be able to ﬁnd a 2-dimensional vector subspace V of R3, an open set U of V and a smooth / ⊥ map f: U V (the orthogonal complement of V ) such that Σ ∩ W = Γf where W is an open set of R3 containing p. Of course, up to a rigid motion (which is smooth), we can assume that V is the xy-coordinate plane, so V ⊥ is the z-coordinate axis and then the function is z = f(x, y) for (x, y) in an open set of the xy-coordinate plane.

Remark 2.2.2. Let φ: R3 / R3 be a rigid motion of R3 and Σ ⊂ R3. Then, Σ is a regular surface if and only if φ(Σ) is a regular surface.

Example 2.2.1. Let U be an open set of R2 and f: U / R a smooth function. Then the graph of f is a regular surface.

2 3 2 Example 2.2.2. S := {~r ∈ R | |~r| = 1} is a regular surface. Let ~r0 = (x0, y0, z0) ∈ S . Up to a 2 protation, we may assume that z0 > 0. Then locally around point ~r0,S is the graph of z = f(x, y) = 1 − x2 − y2. In fact, if we take W = {(x, y, z) ∈ R3|z > 0} and U = {(x, y) ∈ R2|x2 + y2 < 1}, 2 then S ∩ W = Γf . Let Σ, W , V , U and f be as in the remark. Let π: R3 / V be the orthogonal projection onto / / V , then π|Γf :Γf U is the inverse of ι: U Γf which maps (x, y) to (x, y, f(x, y)). Note that

ι is smooth and its inverse π|Γf is smooth in the sense that it has a smooth extension to an open 3 set in R (e.g., W ), such an ι is called a diﬀeomorphism of U onto Σ ∩ W ( = Γf ). Therefore, we have shown that, locally, a regular surface must be diﬀeomorphic to an open set of R2. In fact, the following is true.

Theorem 2.2.1. A subset Σ of R3 is a regular surface if and only if locally it is diﬀeomorphic 2 ∼ 2 to an open set of R ; i.e., (Σ, p) =loc (R , 0) for any p ∈ Σ. Corollary 2.2.2. Let Σ ⊂ R3 be a regular surface, φ: R3 / R3 be a diﬀeomorphism. Then the image of Σ under φ is also a regular surface. In particular, the image of a regular surface under a rigid motion of R3 is still a regular surface.

Let f: U / R be a smooth map. We say that y ∈ R is a regular value of f if either f −1(y) is empty or for any x ∈ f −1(y), the rank Df(x) is 1 (=the dimension of R).

3 Theorem 2.2.3. Let U be an open set of R , f : U / R be a smooth map. Suppose that z0 is a −1 regular value of f. Then f (z0) is a regular surface provided that it is non-empty. Example 2.2.3. All but one of the six standard non-degenerate quadrics are regular surfaces.

2.2.1 Proof of theorem 2.2.1 2 ∼ / Proof. Let φ:(R , 0) =loc (Σ, p). I.e., 1) φ: U Σ is a one-to-one map (where U is an open set of R2 containing 0), smooth when viewed as a map into R3 ; 2) φ−1, when viewed as a map into R2, has a smooth extension to Z — an open set of R3 containing the image of φ; 3) φ(0) = p. Since Dφ(0) has rank 2, there exists a coordinate plane V in R3 such that if π: R3 / V is the orthogonal projection onto V , then D(π ◦φ)(0) = Dπ(p)◦Dφ(0) = π ◦Dφ(0) is an isomorphism, so 2.3. EXISTENCE OF TANGENT PLANES 11

π ◦ φ is a local diﬀeomorphism by the inverse function theorem. From the following commutative triangle

(R2, 0) ?? ?? ?? π◦φ ?? φ ?? ? o (V, p¯)π (Σ, p) where φ and π ◦ φ are local diﬀeomorphisms, we conclude that π in the triangle is a local diﬀeo- morphism, too. Since R3 = V ⊕ V ⊥, viewed as a map into R3, π−1 has the following form:

π−1(u, v) = (u, v, f(u, v)). provided that V is the uv-coordinate plane. Here f is a locally deﬁned smooth function. Then (Σ, p) =loc (Γf , p).

2.2.2 Proof of theorem 2.2.3 −1 Proof. Let p ∈ f (z0). We may assume that z0 = 0. Since 0 is a regular value of f, by the inverse function theorem, there is a local diﬀeomorphism φ:(R3, 0) / (R3, p) such that f ◦φ(x, y, z) = z. loc Then, φ−1 −1 ∼ −1 2 (f (0), p) = loc((f ◦ φ) (0), 0) =loc (R , 0). In view of theorem 2.2.1, we are done.

2.3 Existence of tangent planes

3 2 ∼ Let Σ ⊂ R be a regular surface and p ∈ Σ. Then there is a local diﬀeomorphism φ:(R , 0) =loc (Σ, p). Such a φ is called a local parametrization and its inverse is called a local coordinate map. It is customary to write (u, v) for the local parameters, so φ is a R3-valued function of (u, v). The tangent plane of Σ at p, denoted by TpΣ, is deﬁned to be the linear span of

{∂uφ(0), ∂vφ(0)} with the understanding that φ is a R3-valued function of (u, v). One can show that this deﬁnition of tangent plane is actually independent of the choice of the 2 ∼ local parameterizations: suppose that ψ:(R , 0) =loc (Σ, p) is another local parametrization, then 2 ∼ 2 φ = ψ ◦ λ where λ:(R , 0) =loc (R , 0) is a local diﬀeomorphism. By chain rule, we have

(∂uφ(0), ∂vφ(0)) = (∂uψ(0), ∂vψ(0)) · Jλ(0). Since Jλ(0) is nonsingular, we have

span{∂uφ(0), ∂vφ(0)} = span{∂uψ(0), ∂vψ(0)}. 12 CHAPTER 2. REGULAR SURFACES

Remark 2.3.1. Actually, the tangent plane of Σ at p is the parallel translate of span{∂uφ(0), ∂vφ(0)} by p, i.e., TpΣ = p + span{∂uφ(0), ∂vφ(0)}.

Since the identiﬁcation of TpΣ with span{∂uφ(0), ∂vφ(0)} is quite natural, we simply write

TpΣ = span{∂uφ(0), ∂vφ(0)} in the above. Example 2.3.1. The solution set of equation x2 + y2 = z2 is not a regular surface because its tangent plane at (0, 0, 0) does not exist. Let Σ be a regular surface, α: I / R3 be a parameterized smooth curve. We say α is inside 0 Σ if the trace of α is inside Σ. Assume that p ∈ Σ and α(0) = p, then we claim that α (0) ∈ TpΣ. 2 ∼ To show that, we choose a local parametrization φ:(R , 0) =loc (Σ, p). Locally around t = 0, we −1 −1 have α(t) = φ(φ ◦ α(t)) = φ(β(t)) where β = φ ◦ α ≡ (β1, β2); then α0(0) = Dφ(β(0))(β0(0)) = Dφ(0)(β0(0)) 0 0 = β1(0) ∂uφ(0) + β2(0) ∂vφ(0) ∈ TpΣ. (2.1) In fact, more can be proved: Exercise 2.3.1. Let Σ be a regular surface and p ∈ Σ. Show that

2.4 Local Canonical Form

2 ∼ Let Σ be a regular surface and p ∈ Σ, φ:(R , 0) =loc (Σ, p) be a local parametrization. Up to a 3 3 rigid motion of R , we may assume that p is the origin of R , TpΣ is the xy-coordinate plane. Let π be the orthogonal projection of R3 onto the xy-coordinate plane, then D(π ◦ φ)(0) = π ◦ Dφ(0) 2 ∼ is an isomorphism, so π ◦ φ:(R , 0) =loc (Σ, p) is a local diﬀeomorphism. From the following commutative triangle

(R2, 0) ?? ?? ?? π◦φ ?? φ ?? ? 2 o (R , 0)π (Σ, p) where φ and π ◦ φ are local diﬀeomorphisms, we conclude that π in the triangle is a local diﬀeo- −1 morphism with π (u, v) = (u, v, f(u, v)) being a local parametrization. Since p = 0 and TpΣ is the xy-coordinate plane, we have the Taylor expansion of f around (0, 0) as follows:

f(u, v) = Q(u, v) + o(u2 + v2) 2.5. EXERCISES 13 where Q is a quadratic form. We may further assume that Q is in diagonal form. Therefore, we have proved that, up to a rigid motion of R3, locally, a regular surface is the graph of a function of the following form: 1 ¡ ¢ z = f(x, y) = κ x2 + κ y2 + o(x2 + y2) (2.2) 2 1 2 where κ1 and κ2 are some real numbers.

2.5 Exercises

Exercise 2.5.1. Show that quadric ax2 +by2 +cz2 = 1 is diﬀeomorphic to quadric x2 +y2 +z2 = 1 when a, b and c are all positive.

Exercise 2.5.2. Show that the six standard non-degenerate quadrics listed in my notes on quadrics are mutually non-diﬀeomorphic except for a pair.

Exercise 2.5.3. Let A ⊂ R2 be the unit circle centered at the origin, B ⊂ R3 be the intersection of plane x + y + z = 1 with the unit sphere centered at the origin. Show that A is diﬀeomorphic to B.

Exercise 2.5.4. Pick up a quadric, and a point on that quadric. 1) Find the tangent plane of the chosen quadric at the chosen point. 2) Find the local canonical form of the chosen quadric at the chosen point. 14 CHAPTER 2. REGULAR SURFACES Chapter 3

Gauss Map

3.1 Smooth maps between regular surface

Let Σ1 and Σ2 be two regular surfaces and f:Σ1 / Σ2 be a map. We say that f is smooth, if 3 3 as a map into R , it has a smooth extension to an open set of R containing Σ1.

Assume that p1 ∈ Σ1, p2 ∈ Σ2 and f(p1) = p2. We claim that f induces a linear map from Tp1 Σ1 to Tp2 Σ2. To show this claim, we let α be a parameterized smooth curve inside Σ1. Then f ◦ α is a parameterized smooth curve inside Σ2; here the smoothness can be seen this way: f ◦ α = f˜◦ α 3 3 where f˜ is a smooth extension of f to W — an open set of R containing Σ1, i.e., f˜: W / R is ˜ 0 smooth and f = f on Σ1. Now α (0) ∈ Tp1 Σ1 and

0 Tp2 Σ2 3 (f ◦ α) (0) = (f˜◦ α)0(0) 0 = Df˜(p1)(α (0)) (3.1)

In view of exercise 2.3.1, we have shown that f induces a linear map

/ T fp1 : Tp1 Σ1 Tp2 Σ2. (3.2) ˜ Note that T fp1 (~w) = Df(p1)(~w), however, in view of the ﬁrst line of the Eq. (3.1), T fp1 is actually independent of the choice of f˜. ˜ Remark 3.1.1. Actually, what T fp1 really does is to map p1 + ~w to p2 + Df(p1)(~w).

Remark 3.1.2. T fp is well-deﬁned for local map f: (Σ1, p1) / (Σ2, p2) provided that f is 1 loc smooth.

3.2 Orientation, Gauss map and curvatures

Let Σ be a regular surface and p ∈ Σ. We say that Σ is oriented at p if a unit normal vector np to TpΣ is chosen. We say that Σ is oriented if it is smoothly oriented everywhere, i.e., 1) Σ is 2 ∼ oriented at each of its point; 2) for any p ∈ Σ, there is a local parametrization φ:(R , 0) =loc (Σ, p)

15 16 CHAPTER 3. GAUSS MAP

−1 such that (∂uφ × ∂uφ)|φ−1(q) is parallel to nq for all q in the domain of φ . Let us call such a φ an orientation-preserving local parametrization. Let Σ be an oriented regular surface and p ∈ Σ. Then the Gauss map

n :Σ / S2

2 ∼ which maps p to np is a smooth map. To show that, let φ:(R , 0) =loc (Σ, p) be an orientation- preserving local parametrization, then ∂ φ × ∂ φ ∂ φ × ∂ φ n(q) = u u (φ−1(q)) = u u (φg−1(q)) |∂uφ × ∂uφ| |∂uφ × ∂uφ| for all q in the domain of φ−1, hence n is smooth. / 2 The previous section says that we then have a linear map T np: TpΣ Tn(p)S . Note that, 3 2 as a unit vector in R , n(p) is perpendicular to both TpΣ and Tn(p)S , so we have the natural 2 identiﬁcation: Tn(p)S = TpΣ. Therefore, we have shown that T np is an endomorphism of TpΣ. In fact, we have

Theorem 3.2.1. T np is a self-adjoint operator on TpΣ. There is a proof based on direct computation. However, there is more interesting way to prove this theorem. The basic idea is that one just needs to prove this theorem in the simpler case when Σ is represented locally in a canonical form. (that is what canonical form is good for). Note that, to get the canonical form, we may need a rigid transformation R(x) = T (x) + a where a is a vector in R3 and T is an orthogonal transformation of R3. You are invited to do the following exercise.

Exercise 3.2.1. Under the rigid transformation R(x) = T (x) + a, T np is turned into

−1 T ◦ T np ◦ T .

−1 So T np is self adjoint if and only if T ◦ T np ◦ T is. 2 ∼ Now, up to a rigid motion, we have a local a local parametrization φ:(R , 0) =loc (Σ, p) of the form 1 ¡ ¢ φ(x, y) = (x, y, κ x2 + κ y2 + o(x2 + y2)), 2 1 2 then for (x, y, z) ∈ Σ, we have

∂ φ × ∂ φ n(x, y, z) = u u |∂uφ × ∂uφ| p 2 2 = (−κ1x, −κ2y, 1) + o( x + y ). (3.3)

Letn ˜(x, y, z) be equal to the righthand side of the last line in the above equation for any (x, y, z) 3 in a suﬃciently small open neighborhood U of 0 ∈ R , thenn ˜ is a smooth extension of n|U . Then, 2 2 3 for any ~w = (w1, w2) ∈ TpΣ = R = R × 0 ⊂ R , we have

T np(~w) = Dn˜(0)(~w) = (−κ1w1, −κ2w2). 3.3. DIGRESSION: LINEAR ALGEBRA 17

2 So, with the identiﬁcation: TpΣ = R , the standard matrix for T np is µ ¶ −κ 0 1 . 0 −κ2

So T np is self adjoint. In view of exercise 3.2.1, we know that {κ1, κ2} is uniquely determined by 1 (Σ, p) because it is the set of eigenvalues of −T np, and is called the set of principal curvatures of Σ at p.

1 Definition 3.2.1. The mean curvature of Σ at p, denoted by Hp, is defined to be − 2 tr T np. The Gauss curvature of Σ at p, denoted by Kp, is defined to be det T np

κ1+κ2 In terms of the principal curvatures κ1, κ2 at p, it is clear that Hp = 2 and Kp = κ1κ2.

3.3 Digression: Linear Algebra

Let V be a real vector space of dimension n.

3.3.1 One-form An 1-form on V is a linear map T : V / R. The set of all 1-forms on V is called the dual space ∗ of V , denoted by V . Let v = (v1, ··· , vn) be a basis for V ,

φ : V / Rn be the coordinate map with respect to v,

πi : Rn / R be the projection to the i-th component of Rn, and

vˆi = πi ◦ φ.

Then vˆ = (ˆv1, ··· , vˆn) is a basis for V ∗ — the dual basis.

3.3.2 Two-form A 2-form on V is a bilinear map T : V ×V / R. Here “bilinear” means that T is a linear in each variable, i.e., T (x, y) is linear in x for any fixed y and linear in y for any fixed x. We say that the 2-form is symmetric if T (x, y) = T (y, x) for any x, y; we say that the 2-form is antisymmetric if T (x, y) = −T (y, x) for any x, y. An inner product h, i on V is nothing but a symmetric and positive-definite 2-form on V , here positive-definite means that hx, xi ≥ 0 for any x and hx, xi = 0 if and only if x = 0. Remark 3.3.1. Let λ be a two-form on V , then λ defines a linear map λ∗: V / V ∗ in a natural way: λ∗(x)(y) = λ(x, y).

1their geometric meaning will be explained in class. 18 CHAPTER 3. GAUSS MAP

The set of 2-forms, the set of symmetric 2-forms and the set of antisymmetric 2-forms are all vector spaces. They are denoted by T 2V ∗, S2V ∗ and ∧2V ∗ respectively. It is clear that

T 2V ∗ = S2V ∗ ⊕ ∧2V ∗.

Let α and β be two 1-forms on V . For any x, y in V , we deﬁne

α ⊗ β : V × V / R as follows: α ⊗ β(x, y) = α(x)β(y). It is clear that α ⊗ β ( called the tensor product of α with β) is a 2-form on V , i.e., α ⊗ β ∈ T 2V ∗. Let

1 αβ = 2 (α ⊗ β + β ⊗ α) , (3.4) 1 α ∧ β = 2 (α ⊗ β − β ⊗ α) .

Then αβ is a symmetric 2-form on V and α ∧ β is an antisymmetric 2-form on V .

∗ Exercise 3.3.1. Let V be a real vector space of dimension n, θ = (θ1, ··· , θn) be a basis for V . Show that 2 ∗ 1) {θi ⊗ θj | 1 ≤ i, j ≤ n} is a minimal spanning set for T V . 2 ∗ 2) {θiθj | 1 ≤ i ≤ j ≤ n} is a minimal spanning set for S V . 2 ∗ 3) {θi ∧ θj | 1 ≤ i < j ≤ n} is a minimal spanning set for ∧ V . 2 ∗ 2 2 ∗ n(n+1) 2 ∗ n(n−1) 4) dim T V = n , dim S V = 2 and dim ∧ V = 2 .

Exercise 3.3.2. Let λ be a symmetric 2-form on V . Let θ = (θ1, . . . , θn) be a basis for V ∗ and P i j v = (v1, . . . , vn) be the corresponding dual basis of V . Then λ = λijθ θ for some (unique) real ∗ ∗ symmetric matrix [λij]. Show that [λij] is the coordinate matrix for λ : V / V with respect to (v, θ).

Exercise 3.3.3. Let λ be an anti-symmetric 2-form on V . Let θ = (θ1, . . . , θn) be a basis for V ∗ P i j and v = (v1, . . . , vn) be the corresponding dual basis of V . Then λ = λijθ ∧θ for some (unique) ∗ ∗ anti-symmetric real matrix [λij]. Show that [−λij] is the coordinate matrix for λ : V / V with ∗ Pn j respect to (v, θ), i.e., λ (vi) = − j=1 θ λji.

Let V , W be two-dimensional vector spaces, T : V / W a linear map. Let T ∗: W ∗ / V ∗ be the dual of T , and ∧2T ∗ be the resulting endomorphism on ∧2V ∗: ∧2T ∗(α ∧ β) = T ∗(α) ∧ T ∗(β). Suppose that W = V , since dim ∧2V ∗ = 1, then ∧2T ∗ must be the scalar multiplication by a number.

Exercise 3.3.4. Show that this number is just det T . 3.3. DIGRESSION: LINEAR ALGEBRA 19

3.3.3 The area form on an oriented Euclidean plane

By definition, a Euclidean plane is just a pair (V, h, i) where V is a real two-dimensional vector space and h, i is an inner product on V . Observe that ∧2V ∗ is an one dimensional real vector space. By definition, an orientation on V is just a nonzero element of ∧2V ∗ with the understanding that two nonzero elements in ∧2V ∗ represent the same orientation if and only if they differ by the scalar multiplication of a POSITIVE number. It is clear that there are exactly two orientations on V .

Remark 3.3.2. This definition of orientation is intrinsic. The definition given in section 3.2 is extrinsic. The correspondence between these two definitions are fixed in this book as follows: Let V be a 2-dimensional vector subspace of R3, ω be a nonzero element in ∧2V ∗, n be a normal vector of V in R3; then, ω and n represent the same orientation of V if there is a POSITIVE number λ such that ω(v1, v2) = λ(v1 × v2) · n for any two vectors v1 and v2 in V .

An oriented Euclidean plane is just a triple (V, h, i, o) where V is a real two-dimensional vector space, h, i is an inner product on V and o is an orientation. Note that there is an inner product on V ∗ which is naturally induced from the inner product on V : It is the one such that h, i∗: V / V ∗ is an isometry. It turns out that a basis is orthonormal if and only if its dual basis is orthonormal. Then there is a natural inner product on T 2V ∗, too: it is θ ⊗θ the (unique) inner product such that, when θ is an orthonormal basis for V ∗, { i√ j | 1 ≤ i, j ≤ n} 2 is an orthonormal minimal spanning set for T 2V ∗

Deﬁnition 3.3.1. Let (V, h, i, o) be an oriented Euclidean plane. The area form on V is the unique anti-symmetric unit two-form vol which represents orientation o.

Example 3.3.1. Let V = R2(= R2 × 0 ⊂ R3), with the standard inner product and standard orientation given. If we view V as a subset of R3, its standard orientation is more frequently represented by k as in section 3.2. Note that (i, j) is its standard orthonormal basis. Let (ˆi,ˆj) be the corresponding dual basis for V ∗, then vol = ˆi ∧ ˆj . Note that vol is uniquely determined by 1 1 vol(i, j) = 2! = 2! (i × j) · k .

Exercise 3.3.5. Let T be an endomorphism on R2, show that

1 vol(T (i),T (j)) = (T (i) × T (j)) · k. 2!

Exercise 3.3.6. Let a and b be vectors in R3. Show that p |a × b| = (a · a)(a · a) − (a · b)2.

Please be aware of our choice of the correspondence between the two diﬀerent descriptions of orientation: one given in this subsection (the intrinsic one) and one given in section 3.2 (the extrinsic one). 20 CHAPTER 3. GAUSS MAP

3.4 Fundamental Forms

3.4.1 The total diﬀerentials Let Σ be a regular surface, p ∈ Σ, and f:Σ / R be a smooth map. Suppose that α: I / R3 is a parameterized smooth curve inside Σ with α(0) = p. Then f ◦ α: I / R is a smooth function of one variable because f ◦ α = f˜ ◦ α is a composition of two smooth maps, here f˜ is a smooth extension of f. Therefore,

(f ◦ α)0(0) = (f˜◦ α)0(0) = ∇f˜(p) · α0(0).

In view of exercise 2.3.1, we have shown that f induces a linear map

dfp : TpΣ / R (3.5)

∗ which maps ~w to ∇f˜(p) · ~w. It is clear that dfp ∈ (TpΣ) — the cotangent space of Σ at p. We use df to denote the map: p 7→ dfp. In the case f is a locally defined smooth function, df is also defined because dfp only depends on f locally. In any case, we call df the total differential of f. Note that df is in dependent of the choice of f˜ because

0 0 dfp(α (0)) = (f ◦ α) (0) (3.6) for any parameterized smooth curve inside Σ with α(0) = p. 2 ∼ Let Σ be a regular surface, p ∈ Σ and φ:(R , 0) =loc (Σ, p) be a local parametrization. Write φ−1 = (u, v), then u and v are locally deﬁned smooth functions on Σ, so we have du and dv locally deﬁned on Σ. Note that (∂uφ(0), ∂vφ(0)) is a basis for TpΣ, and µ ¯ ¶ ¯ ¯ ¯ ¯ ¯ d ¯ d ¯ d ¯ dup(∂uφ(0)) = dup ¯ φ(t, 0) = ¯ (u(φ(t, 0)) = ¯ t = 1. dt t=0 dt t=0 dt t=0 ¯ d ¯ Similarly, dup(∂vφ(0)) = dt t=0 0 = 0. Likewise, dvp(∂vφ(0)) = 1 and dvp(∂uφ(0)) = 0. Therefore, we have shown that

∗ Proposition 3.4.1. (dup, dvp) is a basis for (TpΣ) ; in fact, it is the dual of (∂uφ(0), ∂vφ(0)).

Some Notations.

1. Let (Σ, p, φ) be as before, and f be a local smooth function or R3-valued local smooth function −1 −1 deﬁned on the domain of φ. We write ∂uf(φ (q)) as fu(q) for any q in the domain of φ . In other word,

∂f −1 fu := ∂u ◦ φ . (3.7)

∂2f −1 ∂3f −1 We also write fuu as ∂u2 ◦ φ , fuvv as ∂v∂v∂u ◦ φ , etc.. 3.4. FUNDAMENTAL FORMS 21

2. Let F = (F 1,F 2,F 3) and G = (G1,G2,G3) be two locally deﬁned smooth maps from Σ to R3. We write dF := (dF 1, dF 2, dF 3) and dG := (dG1, dG2, dG3) and

dF · dG := dF 1dG1 + dF 2dG2 + dF 3dG3. (3.8)

Therefore, if p is in the common domain of F and G, then (dF ·dG)|p is a symmetric two-form on TpΣ. Similarly, we write

dF ∧ dG := dF 1 ∧ dG1 + dF 2 ∧ dG2 + dF 3 ∧ dG3. (3.9)

Proposition 3.4.2. Let f: Σ / R be a locally deﬁned smooth map, φ: (R2, 0) / (Σ, p) be a loc loc local diﬀeomorphism. Show that

df = (f ◦ φ)u du + (f ◦ φ)v dv (3.10) on the common domain of f and φ.

Proof. Just need to verify that df(φu) = (f ◦ φ)u and df(φv) = (f ◦ φ)v. But the veriﬁcation is just a computation. For example, ¯ ¯ ¡ ¢ d ¯ −1 −1 df(φu)|q = ¯ f ◦ φ φ (q) + (t, 0) = ∂u(f ◦ φ)(φ (q)) = (f ◦ φ)u|q. dt t=0

3 Exercise 3.4.1. Let F :Σ / R be a locally deﬁned smooth map. Then dF |p naturally deﬁnes loc 3 1 2 3 a linear map from TpΣ into R : it maps X ∈ TpΣ to (dFp (X), dFp (X), dFp (X)). In other word, 3 dF |p is a R -valued 1-form on TpΣ. Show that

0 0 dFp(α (0)) = (F ◦ α) (0) (3.11) for a parameterized smooth curve inside Σ with α(0) = p; consequently,

dF = (F ◦ φ)u du + (F ◦ φ)v dv. (3.12)

Let (Σ, p, φ) be as before, and ι:Σ / R3 be the inclusion map. Then ι is smooth because we can take the identity map 1: R3 / R3 to be the smooth extension of ι. Since ι ◦ φ = φ, in view of eq. (3.12) in the preceding exercise, locally we have

dι = φu du + φv dv. (3.13) 22 CHAPTER 3. GAUSS MAP

3.4.2 The ﬁrst fundamental two-form

Now if we take V = TpΣ where Σ is a regular surface and p ∈ Σ. Then the dot product for vectors 3 in R deﬁnes a symmetric, positive-deﬁnite 2-form on TpΣ, denoted by Ip. I.e., for any two tangent 3 vectors ~u, ~v in TpΣ ⊂ R , we have Ip(~u,~v) = ~u · ~v. 3 Clearly, Ip is just the inner product on TpΣ induced from the dot product of R . Exercise 3.4.2. In terms of notations introduced in the preceding subsection, show that 1) I = dι · dι. 2 2 2) Locally, we have I = E du + 2F du dv + G dv where E = φu · φu, F = φu · φv, G = φv · φv.

3.4.3 The second fundamental two-form

Let n be the Gauss map. We have shown that T np is self adjoint with respect to inner product Ip, i.e., Ip(x, T np(y)) = Ip(T np(x), y) for any x, y ∈ TpΣ. Deﬁne IIp(x, y) = −Ip(x, T np(y)) for any x, y ∈ TpΣ, then IIp is a symmetric 2-form on TpΣ, called the second fundamental form of Σ at p. Why is IIp is symmetric? well,

IIp(x, y) = −Ip(x, T np(y)) = −Ip(T np(x), y) = −Ip(y, T np(x)) = IIp(y, x). Exercise 3.4.3. In terms of notations introduced in the preceding subsection, show that 1) II = −dn · dι. 2 2 2) locally, we have II = e du + 2f du dv + g dv where e = n · φuu, f = n · φuv, g = n · φvv. 3) (remembering the notation introduced in remark 3.3.1) the following triangle

TpΣ ?? ?? ∗ ?? −IIp ?? T np ?? ? ∗ o (TpΣ) ∗ TpΣ Ip is commutative. Consequently, we have ¯ eg−f 2 ¯ Kp = det(T np) = 2 ¯ , EG−F p ¯ (3.14) 1 1 eG+gE−2fF ¯ Hp = − tr(T np) = 2 ¯ . 2 2 EG−F p Exercise 3.4.4. Let (Σ, p) be as before. Suppose that (Σ, p) is turned into (Σ0, p0) under the rigid / 0 transformation R(x) = T (x) + a. show that TRp = T : TpΣ Tp0 Σ is compatible with the two fundamental forms. I.e., show that, for any x, y in TpΣ, we have 0 0 Ip(x, y) = Ip0 (T (x),T (y)), IIp(x, y) = IIp0 (T (x),T (y)). (3.15) 2 2 3 Exercise 3.4.5. In the local canonical form, we have TpΣ = R = R × 0 ⊂ R . Compute Ip and IIp in this case. 3.5. EXERCISES 23

3.4.4 Area Form

Let Σ be an oriented regular surface and p a point on Σ. Note that TpΣ is an oriented Euclidean plane, so it has an area form volp. Then we have a map vol: p 7→ volp. The basic objective here is to write down a local formula for vol. To do that, we let φ: 2 (R , 0) / (Σ, p) be an orientation-preserving local parametrization, so (∂uφ, ∂vφ, n)|q form an loc right-hand rule for all q in the domain of φ−1. In view of the example 3.3.1 and the subsequent exercises, we know that vol is uniquely determined by

1 vol(∂ φ, ∂ φ) = (∂ φ × ∂ φ) · n, (3.16) u v 2 u v so

vol = (∂uφ × ∂vφ) · n du ∧ dv = |∂uφ × ∂vφ| du ∧ dv, (3.17) i.e, √ vol = EG − F 2 du ∧ dv. (3.18)

Note that, if we wish to indicate the dependence on Σ, we write vol as volΣ. / 2 / 2 Let n:Σ S be the Gauss map. Then T np: TpΣ Tnp S is a linear map, then a linear map 2 ∗ 2 2 ∗ / 2 ∗ ∧ T np : ∧ (Tnp S ) ∧ (TpΣ)

S2 Σ 2 which maps volnp to a scalar multiple of volp . Note that, in the natural identiﬁcation of Tnp S S2 Σ with TpΣ, volnp gets identiﬁed with volp , therefore, in view of exercise 3.3.4 and the fact that det T np = Kp, we have 2 n∗(volS ) = K volΣ

∗ ∗ S2 where K is the Gauss curvature, and notation n is introduced: by deﬁnition, n (vol )|p := 2 ∗ S2 ∧ T np(volnp ).

3.5 Exercises

Exercise 3.5.1. Compute the mean curvature and the Gauss curvature of the sphere with radius r at any point. How about a general non-degenerate quadric?

Let Σ1 and Σ2 be two regular surfaces, p1 ∈ Σ1 and p2 ∈ Σ2. Let f: (Σ1, p1) / (Σ2, p2) be loc a local diﬀeomorphism. We say that f is an local isometry from (Σ1, p1) to (Σ2, p2) if T fq is a (linear) isometry for each q in the domain of f. 24 CHAPTER 3. GAUSS MAP

Exercise 3.5.2. Let Σ be the cylindrical surface with base the unit circle, i.e.,

Σ = {(x, y, z) ∈ R3 | x2 + y2 = 1}.

Show that, 1) there is a local isometry from (Σ, p) to (R2, 0) for any point p ∈ Σ; 2) Σ and R2 have diﬀerent principal curvatures and diﬀerent mean curvature, but the same Gauss curvature.

Exercise 3.5.3 (The third fundamental form). Let III = dn · dn. Show that IIIp(x, y) = 2 Ip(dnp(x), dnp(y)) for any x, y in TpΣ. (Optional question 1: Locally we write III = Edu + Fdudv + Gdv2, can you work out the formulae for E, F and G in terms of E, F , G, e, f, g?) (Optional question 2: Show that dι ∧ dι = dn ∧ dι = dn ∧ dn = 0.) Chapter 4

Intrinsic Theory

4.1 Vector Fields

Let Σ be a regular surface, X:Σ / R3 a smooth map. We say that X is a (smooth) vector field on Σ if X(p) ∈ TpΣ for any p ∈ Σ. In the case that X is only locally defined, X will be called a local vector field. Example 4.1.1. Let Σ be a regular surface and p ∈ Σ. Suppose that φ:(R2, 0) / (Σ, p) is a loc local parametrization, then 3 ∂u :Σ / R loc ¯ ¯ ∂φ¯ q 7→ ¯ = φu(q). (4.1) ∂u φ−1(q) is a local vector field on Σ. Similarly, we have the local vector field ∂v. Note that (∂u, ∂v) is called a local coordinate frame of Σ because, at each point where they are defined, it is a basis for the tangent plane of Σ at that point. Similarly, (du, dv) is called a local coordinate coframe of Σ; in fact, it is the dual of (∂u, ∂v). Note that ∂u = φu and ∂v = φv. Remark 4.1.1. Let X be a local vector field and f be a local smooth function, both are locally defined in an open set U. Then fX: p 7→ f(p)X(p) is a local vector field.

Exercise 4.1.1. Let φ,(∂u, ∂v) be as in the preceding example. Let X be a vector field, show that, locally, 1 2 X = X ∂u + X ∂v for some smooth local functions X1 and X2. The following proposition is very basic and useful: Proposition 4.1.1 (local existence of flow line). Let X be a local vector field defined on open set U of Σ. Then there is an open set W of U × R containing U × 0 and a smooth map Φ: W / Σ such that, 1) Φ(q, 0) = q for any q ∈ U; 2) ∂tΦ(q, t) = X(Φ(q, t)) for any (q, t) ∈ W . Moreover, 1) any two such Φ’s must agree on the common domain of definition; 2) Φ(Φ(q, t), s) = Φ(q, t + s) when both sides make sense.

25 26 CHAPTER 4. INTRINSIC THEORY

The proof is just a direct consequence of local existence and uniqueness theorem for ODE, so it is omitted.

Remark 4.1.2. Map Φ in the proposition is called a ﬂow map for X.

Example 4.1.2. Let Σ be a regular surface and p ∈ Σ. Suppose that φ:(R2, 0) / (Σ, p) is a loc local parametrization. Let Φ(q, t) = φ(φ−1(q) + (t, 0)).

Then Φ is a ﬂow map for ∂u.

Example 4.1.3. Let S1 be the unit circle centered at the origin of R2, and X:S1 / R2 be the vector ﬁeld whose value at (x, y) is (−y, x). Then

Φ((x, y), t) = (x cos t − y sin t, x sin t + y cos t) is a ﬂow map for X. Here W = S1 × R.

4.1.1 Lie derivatives of smooth functions Let Σ be a regular surface, p ∈ Σ. Let X be a local vector ﬁeld and f be a local smooth function, both are locally deﬁned in an open neighborhood of p.

Definition 4.1.1. The Lie derivative of f with respect to X, denoted by LX f, is defined as follows: for any p in the common domain of f and X, ¯ ¯ d ¯ LX f(p) = ¯ f(Φ(p, t)) ( = dfp(X(p)) ) dt t=0 where Φ is a flow map for X.

Note that LX f is a local smooth function, and LX f = 0 if either X = 0 or f is a constant function. Not also that LX f = df(X): p 7→ dfp(X(p)).

Exercise 4.1.2. Let X, Y be local vector ﬁelds, f and g be local smooth functions. Show that, on the common domain of X, f and g, we have

LX (f · g) = LX f · g + f ·LX g.

LgX f = g ·LX f.

LX (f + g) = LX f + LX g.

LX+Y f = LX f + LY f. (4.2)

Here the · between functions means the multiplication of functions. 4.2. DIRECTIONAL DERIVATIVES 27

4.1.2 Gradient vector fields ∗ / ∗ Let Σ be a regular surface and f a local smooth function on Σ. Note that Ip: TpΣ (TpΣ) is an isomorphism. By definition, the gradient vector of f at p, denoted by ∇f|p, is the unique tangent ∗ vector which is mapped to dfp under Ip. Then map ∇f: p 7→ ∇f|p is a local vector field. (Why is ∇f smooth?)

Exercise 4.1.3. Let X be a local vector ﬁeld, f is a local smooth function. Show that

I(X, ∇f) = LX f = df(X).

4.1.3 Lie derivatives of Rm-valued smooth functions Let Σ be a regular surface, p ∈ Σ. Let X be a local vector ﬁeld and F be a local Rm-valued smooth function, both are locally deﬁned in an open neighborhood of p.

Definition 4.1.2. The Lie derivative of F with respect to X, denoted by LX F , is defined as follows: for any p in the common domain of F and X, ¯ ¯ d ¯ LX F (p) = ¯ F (Φ(p, t)) ( = dFp(X(p)) ) dt t=0 where Φ is a flow map for X.

m Note that LX F is a local R -valued smooth function, and LX F = 0 if either X = 0 or F is a constant function.

Exercise 4.1.4. Let X, Y be local vector ﬁelds, g be local smooth function, and F , G be local Rm-valued smooth functions. Show that, on the common domain of X, Y , g, F and G, we have

LX (g · F ) = LX g · F + g ·LX F.

LgX F = g ·LX F.

LX (F + G) = LX F + LX G.

LX+Y F = LX F + LY F. (4.3)

Here the · means the scalar multiplication.

4.2 Directional Derivatives

4.2.1 Directional derivatives of Rm-valued smooth functions

m Let Σ be a regular surface, p ∈ Σ. Let V ∈ TpΣ, F be a local R -valued smooth function. Let X be a local vector ﬁled such that X(p) = V . 28 CHAPTER 4. INTRINSIC THEORY

Definition 4.2.1. The directional derivative of F (as a Rm-valued smooth function) with respect to V , denoted by V (F ), is defined to be (LX F )(p). Exercise 4.2.1. Use the results established in Ex. 4.1.4 to show that this definition makes sense, i.e., independent of choice of X. Moreover, show that, for local smooth function g and local smooth Rm-valued smooth functions F and G, we have

V (g · F ) = V (g) · F (p) + g(p) · V (F ).

V (F + G) = V (F ) + V (G). (4.4)

Exercise 4.2.2. Let g be a smooth function on Σ, (∂u, ∂v) be a local coordinate frame. Then, locally ∂u(g) = (g ◦ φ)u, ∂v(g) = (g ◦ φ)v.

Exercise 4.2.3. Let V1 and V2 be two tangent vectors of Σ at p. Show that V1 = V2 ⇐⇒ V1(g) = V2(g) for any smooth function g locally deﬁned around p.

4.2.2 Covariant derivatives of vector fields Let Σ be a regular surface, X, Y be local vector fields with a non-empty common domain. Since 3 3 Y is a smooth map into R , LX Y is a well-defined R -valued smooth functions on the common domain of X and Y . However, LX Y may not be a vector field anymore, but if we project out the the component in the normal space, we get a vector field. More precisely, for any p ∈ Σ, if we 3 denote by πp the orthogonal projection of R onto TpΣ, and let ∇X Y |p = πp (LX Y (p)), then ∇X Y : p 7→ ∇X Y |p for p in the common domain of X and Y is a vector field — the covariant derivative of Y with respect to X. Exercise 4.2.4. 1) Show that

∇X Y = LX Y − (n ·LX Y )n.

2) Let X, Y , Z be local vector ﬁelds, g be local smooth function. Show that, on the common domain of g, X, Y and Z, we have

∇X (g · Y ) = LX g · Y + g · ∇X Y.

∇gX Y = g · ∇X Y.

∇X (Y + Z) = ∇X Y + ∇X Z.

∇X+Y Z = ∇X Z + ∇Y Z. (4.5)

Here the · means the scalar multiplication. This implies that, locally, ∇ is completely determined by ∇∂u ∂u, ∇∂u ∂v, ∇∂v ∂u,∇∂v ∂v. We are now ready to make a deﬁnition. 4.2. DIRECTIONAL DERIVATIVES 29

Definition 4.2.2. Let Σ be a regular surface, p ∈ Σ. Let V ∈ TpΣ, Y be a local vector field. Let X be a local vector field such that X(p) = V . The directional derivative of Y (as a vector field) with respect to V , denoted by V (Y ), is defined to be (∇X Y )|p. Exercise 4.2.5. Use the results established in Ex. 4.2.4 to show that this definition makes sense, i.e., independent of choice of X. Moreover, show that, for local smooth function g and local vector field Y and Z, we have

V (g · Y ) = V (g) · Y (p) + g(p) · V (Y ).

V (Y + Z) = V (Y ) + V (Z). (4.6)

Let Σ be a regular surface and p ∈ Σ. Suppose that φ:(R2, 0) /(Σ, p) is a local parametriza- loc tion. Then ∂u and ∂v are local vector ﬁelds around p. Suppose that X is a local vector ﬁeld around p, we wish to calculate ∇∂u X and ∇∂v X. First of all, ¯ ¯ d ¯ −1 ∂ L∂u X|q = ¯ X(φ(φ (q) + (t, 0))) = (X ◦ φ)|φ−1(q) = (X ◦ φ)u(q). (4.7) dt t=0 ∂u In other word,

L∂u X = (X ◦ φ)u. (4.8)

Since ∂u ◦ φ = ∂uφ and ∂v ◦ φ = ∂vφ, we have

L∂u ∂u = φuu,

L∂u ∂v = L∂v ∂u = φuv,

L∂v ∂v = φvv. (4.9) Therefore,

∇∂u ∂u = φuu − (n · φuu)n,

∇∂u ∂v = ∇∂v ∂u = φuv − (n · φuv)n,

∇∂v ∂v = φvv − (n · φvv)n. (4.10)

4.2.3 The covariant differentiation is non-commutative in general Let Σ be a regular surface and p ∈ Σ. Suppose that φ:(R2, 0) /(Σ, p) is a local parametrization. loc The purpose here is to point out that, in general, the covariant differentiation is non-commutative. I.e., for vector field Z on regular surface Σ, in general,

R(∂u, ∂v)Z := ∇∂u (∇∂v Z) − ∇∂v (∇∂u Z) 6= 0. In fact, the failure of commutativity is measured by the Gauss curvature. First of all, we would like to mention that, in view of Ex. 4.2.4, by direct computations, we can see that

R(∂u, ∂v)(g · Z) = g · R(∂u, ∂v)Z (4.11) 30 CHAPTER 4. INTRINSIC THEORY for any local smooth function g and local vector ﬁeld Z. Therefore,

R(∂u, ∂v)Z|p depends only on the value of Z at p, so we have a map

F12|p : TpΣ / TpΣ V 7→ R(∂u, ∂v)X|p (4.12) where X is a local vector ﬁeld whose value at p is V . It is not hard to see that F12|p is a linear map, to determine it, we just need to compute F12(∂u) and F12(∂v) at p. To simplify the computation, we may use the local canonical form, i.e., we may assume that, locally around p, Σ is the graph of 1 z = f(x, y) = (κ x2 + κ y2) + o(x2 + y2). 2 1 2 Then φ(u, v) = (u, v, f(u, v)). is a local parametrization. I will leave the rest to you as an exercise.

Exercise 4.2.6. Show that, at point p, we have

F12(∂u) = −K∂v F12(∂v) = K∂u. (4.13) where K = κ1κ2 is the Gauss curvature at p.

4.3 Riemann Curvature

Here we take a closer look at what we have done in the previous section, and that leads to the “discovery” of the Riemann curvature. Let X, Y and Z be local vector ﬁelds, g be a local smooth function. On their common domain, do we have something like Eq. (4.11), i.e.,

(∇X ∇Y − ∇Y ∇X )(gZ) = g(∇X ∇Y − ∇Y ∇X )Z ?

Well, let us compute. Since

∇X ∇Y (gZ) = ∇X (g∇Y Z + Y (g)Z) = g∇X ∇Y Z + X(g)∇Y Z + Y (g)∇X Z + X(Y (g))Z, (4.14) we have

(∇X ∇Y − ∇Y ∇X )(gZ) = g(∇X ∇Y − ∇Y ∇X )Z + (X(Y (g)) − Y (X(g)))Z. (4.15)

Sorry, there is an extra term here because X(Y (g)) − Y (X(g)) is not zero in general. 4.3. RIEMANN CURVATURE 31

4.3.1 Digression: Lie bracket 1 2 1 Locally, with a local parametrization φ chosen, we can write X = X ∂u + X ∂v and Y = Y ∂u + 2 Y ∂v. Then

1 2 X(Y (g)) = X(Y ∂u(g) + Y ∂v(g)) 1 1 2 2 1 2 = X ∂u(Y ∂u(g) + Y ∂v(g)) + X ∂v(Y ∂u(g) + Y ∂v(g)) 1 1 1 2 1 1 1 2 = X Y ∂u(∂u(g)) + X Y ∂u(∂v(g)) + X ∂u(Y )∂u(g) + X ∂u(Y )∂v(g) 2 1 2 2 2 1 2 2 + X Y ∂v(∂u(g)) + X Y ∂v(∂v(g)) + X ∂v(Y )∂u(g) + X ∂v(Y )∂v(g) 1 1 2 2 1 2 2 1 = X Y ∂u(∂u(g)) + X Y ∂v(∂v(g)) + X Y ∂u(∂v(g)) + X Y ∂v(∂u(g)) 1 2 +X(Y )∂u(g) + X(Y )∂v(g) (4.16)

Note that ∂u(∂v(g)) = (g ◦ φ)vu = (g ◦ φ)uv = ∂v(∂u(g)), then we have

£ 1 1 2 2 ¤ X(Y (g)) − Y (X(g)) = (X(Y ) − Y (X ))∂u + (X(Y ) − Y (X ))∂v (g). (4.17)

1 1 2 2 Note that (X(Y ) − Y (X ))∂u + (X(Y ) − Y (X ))∂v is a local vector field, but its value at each point of its domain is independent of φ because of Ex. 4.2.3 and the preceding equation. In other word, there is a (global) vector field on Σ, denoted by [X,Y ], and is uniquely determined by the following defining equation:

[X,Y ](g) := X(Y (g)) − Y (X(g)). (4.18)

Locally, we have

1 1 2 2 [X,Y ] = (X(Y ) − Y (X ))∂u + (X(Y ) − Y (X ))∂v. (4.19)

Remark that [X,Y ] is called the Lie bracket of X and Y .

Exercise 4.3.1. Let X, Y and Z be local vector ﬁelds. On the common domain, show that

[X,Y ] = −[Y,X] skew symmetry [X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0 Jacobi identity. (4.20)

Exercise 4.3.2. Let X, Y be local vector ﬁelds. On the common domain, show that

LX Y − LY X = [X,Y ]. (4.21)

End of digression.

4.3.2 Curvarure Tensor After the digression, we come back to Eq. 4.15 and rewrite it as follows:

(∇X ∇Y − ∇Y ∇X )(gZ) = g(∇X ∇Y − ∇Y ∇X )Z + [X,Y ](g)Z. (4.22)

Since ∇[X,Y ](gZ) = [X,Y ](g)Z + g∇[X,Y ]Z, 32 CHAPTER 4. INTRINSIC THEORY we can rewrite the above equation as

(∇X ∇Y − ∇Y ∇X − ∇[X,Y ])(gZ) = g(∇X ∇Y − ∇Y ∇X − ∇[X,Y ])Z. (4.23) In summary, if we let

R(X,Y ) = ∇X ∇Y − ∇Y ∇X − ∇[X,Y ], (4.24) then

R(X,Y )(gZ) = gR(X,Y )Z. (4.25)

In fact, one can show that R(fX, gY )(hZ) = (fgh)R(X,Y )Z; consequently, at each point p, we have a multi-linear map

Rp : TpΣ × TpΣ × TpΣ / TpΣ (X,Y,Z) 7→ R(X,˜ Y˜ )Z˜|p (4.26) where X˜, Y˜ and Z˜ are local vector ﬁelds whose values at p are X, Y and Z respectively. The map R that assigns Rp to p ∈ Σ is called the Riemann curvature tensor. 1 Exercise 4.3.3. Let (X,Y ) be a basis of TpΣ. Show that hR (X,Y,Y ),Xi K = p . p hX × Y,X × Y i Therefore, if (X,Y ) is a local (tangent) frame, we have hR(X,Y )Y,Xi K = . (4.27) hX × Y,X × Y i

4.4 The covariant diﬀerentiation is intrinsic

The goal of these section is show that the covariant differentiation of vector fields on regular surface Σ is completely determined by the first fundamental form I on Σ. To begin, we need to introduce some notations. Let

1 2 ∇∂u ∂u = Γ11∂u + Γ11∂v. 1 2 ∇∂u ∂v = Γ12∂u + Γ12∂v. 1 2 ∇∂v ∂u = Γ21∂u + Γ21∂v. 1 2 ∇∂v ∂v = Γ22∂u + Γ22∂v. (4.28)

Let g11 = E, g12 = F = g21 and g22 = G. In view of Eq. (4.10), after taking the inner product of the above equations with ∂u = φu and ∂v = φv, we have 1 E = φ · φ = Γ1 g + Γ2 g ≡ Γ . 2 u uu u 11 11 11 12 111 1Hint: use Ex. 4.2.6. 4.5. RIEMANN CURVATURE AND GAUSS CURVATURE ARE ALL INTRINSIC 33

1 F − E = φ · φ = Γ1 g + Γ2 g ≡ Γ . u 2 v uu v 11 12 11 22 211 1 E = φ · φ = Γ1 g + Γ2 g ≡ Γ . 2 v uv u 12 11 12 12 112 1 G = φ · φ = Γ1 g + Γ2 g ≡ Γ . 2 u uv v 12 12 12 22 212

1 2 φvu · φu = Γ21g11 + Γ21g12 ≡ Γ121. 1 2 φvu · φv = Γ21g12 + Γ21g22 ≡ Γ221.

1 F − G = φ · φ = Γ1 g + Γ2 g ≡ Γ . v 2 u vv u 22 11 22 12 122 1 G = φ · φ = Γ1 g + Γ2 g ≡ Γ . 2 v vv v 22 12 22 22 222

It is clear then that Γabc’s are completely determined by the ﬁrst fundamental form. i i Note that φuv = φvu, we have Γjk = Γkj and Γijk = Γikj. Since µ ¶ µ ¶ g g EF 11 12 = g21 g22 FG

i i is a non-singular matrix, we know that Γjk’s and Γabc’s determine each other. Consequently, Γjk’s and the covariant diﬀerentiation are completely determined by the ﬁrst fundamental form.

4.5 Riemann curvature and Gauss curvature are all intrinsic

The last two sections imply that Riemann curvature and Gauss curvature are all intrinsic. That is because the covariant differentiation is intrinsic, Riemann curvature tensor is defined in terms of covariant differentiation and the Gauss curvature can be defined in terms of Riemann curvature tensor and the first fundamental form (see Ex 4.3.3).

4.6 Exercises

Exercise 4.6.1. Let Σ be a regular surface, X, Y , Z be vector ﬁelds on Σ and ∇X Y denote the covariant derivative of Y with respect to X as deﬁned before. Show that

1. ∇X Y − ∇Y X − [X,Y ] = 0 (torsion free);

2. LX (hY,Zi) = h∇X Y,Zi + hY, ∇X Zi (metric compatible ). Show also that the two properties above plus the properties listed in part 2) of Exercise 4.2.4 determine the covariant diﬀerentiation uniquely.

Exercise 4.6.2. Let Σ be a regular surface, X, Y , Z and W be vector ﬁelds on Σ and ∇X Y denote the covariant derivative of Y with respect to X as deﬁned before. Show that 1. hR(X,Y )Z,W i = −hR(Y,X)Z,W i = −hR(X,Y )W, Zi; 34 CHAPTER 4. INTRINSIC THEORY

2. R(X,Y )Z + R(Y,Z)X + R(Z,X)Y = 0;

3. hR(X,Y )Z,W i = hR(Z,W )X,Y i;

4. Let ∇W R be deﬁnes as follows:

(∇W R)(X,Y )Z := R(∇W X,Y )Z + R(X, ∇W Y )Z + R(X,Y )(∇W Z) − ∇W (R(X,Y )Z),

show that 1) (∇W R)(fX, gY )(hZ) = fgh(∇W R)(X,Y )Z and 2)

(∇W R)(X,Y )Z + (∇W R)(Y,Z)X + (∇W R)(Z,X)Y = 0. Chapter 5

Selected topics

5.1 Geodesics

Let Σ be a regular surface, P , Q be two points of Σ, α:[a, b] / Σ be a regular smooth map with α(a) = P and α(b) = Q, then the trace of α is a smooth path in Σ from P to Q. The length of this path is

Z b L[α] = |α0(t)| dt. (5.1) a We wish to ﬁnd a length-minimizing path from P to Q. Note that for a smooth function f(x) to achieve minimal value at x0 we must have ¯ ∂f ¯ ¯ = 0. ∂xi ¯ x0

The same argument implies that, for L to achieve minimal value at α0 we must have ¯ ∂L ¯ ¯ = 0. ∂α(t)¯ α0 The only diﬀerence is that while the index i in xi is discrete, the index t in α(t) is continuous. Note that in the case the index is continuous, people prefer to write ∂ as δ; thus, for L to achieve minimal value at α0 we must have ¯ δL ¯ ¯ = 0. (5.2) δα(t)¯ α0

5.1.1 Digression δL ∂f What is δα(t) then? Well, it is an inﬁnite dimensional analogue of ∂xi , and is called the variational derivative of L with respect to α. Note that X ∂f ∆f = ∆xi + ··· . ∂xi i

35 36 CHAPTER 5. SELECTED TOPICS

Here ··· means terms in higher order of ∆xi’s. So, the inﬁnite dimensional analogue will be Z δL ∆L = dt ∆α(t) + ··· . δα(t)

3 3 Example 5.1.1. Let P , Q be points in R . For each smooth map x:[t1, t2] /R with x(t1) = P and x(t ) = Q, we let 2 Z t2 1 S[x] = |x0(t)|2 dt. t1 2 What is δS δx(t) then? Well, we can compute it this way: Z t2 1 ¡ ¢ ∆S = |(x + ∆x)0(t)|2 − |x0(t)|2 dt 2 Zt1 t2 1 ¡ ¢ = 2x0(t) · (∆x)0(t) + |∆x(t)|2 dt 2 Zt1 t2 = x0(t) · (∆x)0(t) dt + ··· Zt1 t2 = [(x0 · ∆x)0(t) − x00(t) · ∆x(t)] dt + ··· t1 Z t2 = x0(t) · ∆x(t)|t2 + −x00(t) · ∆x(t) dt + ··· t1 Z t1 t2 = −x00(t) · ∆x(t) dt + ··· (5.3) t1 Therefore, δS = −x00(t). δx(t)

δS In reality, people compute δx(t) in a slightly more eﬃcient way as follows: Z t2 δS = x0(t) · δx0(t) dt Zt1 t2 £ ¤ = (x0 · δx)0(t) − x00(t) · δx(t) dt t1 Z t2 = x0(t) · δx(t)|t2 + −x00(t) · δx(t) dt t1 Z t1 t2 = −x00(t) · δx(t) dt (5.4) t1

End of digression. 5.1. GEODESICS 37

After the above digression, we return to our question about length-minimizing path. Let φ be a local parametrization of Σ and write φ−1 ◦ α(t) as x(t) = (x1(t), x2(t)), then

0 −1 0 1 2 α (t) = (φ ◦ φ ◦ α) (t) =x ˙ (t)φu(α(t)) +x ˙ (t)φv(α(t)), therefore, locally we have X 0 2 i j |α | = hij(x)x ˙ x˙ i,j where

h11(x) = E(α) = (∂uφ · ∂uφ)(x) h12(x) = F (α) = (∂uφ · ∂vφ)(x) h22(x) = G(α) = (∂vφ · ∂vφ)(x). (5.5) Instead of computing the variational derivative for L[α], let us ﬁrst compute the variational derivative for Z 1 b S[α] = |α0(t)|2dt. (5.6) 2 a Here it is: Z 1 b δS[α] = δ(|α0(t)|2)dt 2 a Z b 1 i j = δ(hij(x)x ˙ x˙ )dt 2 a Z b 1 k i j i j = (∂khij(x) δx x˙ x˙ + 2hij(x)x ˙ δx˙ )dt 2 a Z b 1 k i j d i j = (∂khij(x) δx x˙ x˙ − [2hij(x)x ˙ ]δx )dt 2 a dt Z b 1 i j d i k = [∂khij(x)x ˙ x˙ − (2hik(x)x ˙ )]δx dt 2 a dt Z b i 1 i j k = −[hik(x)¨x + (∂jhik(x) + ∂ihjk(x) − ∂khij(x))x ˙ x˙ ]δx dt (5.7) a 2 Therefore, we have δS − = h (x)¨xi + Γ (x)x ˙ ix˙ j = h (x)(¨xk + Γk (x)x ˙ ix˙ j). δxl(t) il lij kl ij So, for S to be minimal at x(t), x(t) must satisfy the geodesic equation:

k k i j x¨ + Γij(x)x ˙ x˙ = 0. (5.8)

But how about the variational derivative of Z b L[α] = dt |α0(t)|. a 38 CHAPTER 5. SELECTED TOPICS

Well, you may do the computation directly, but it is bit com-plicated because of the appearance of the square root. So I will do it diﬀerently. Let Z · ¸ 1 b 1 S˜[α, e] = |α0(t)|2 + e(t) dt. (5.9) 2 a e(t)

I claim that if L achieves a minimal at α(t) if and only if S˜ achieves a minimal at (α(t), e(t)) for some e(t). Index

plane curve, 1 rigid motion, 3 smooth, 1 tangent line, 2