AN ALGEBRAIC APPROACH TO HARMONIC POLYNOMIALS ON S3
Kevin Mandira Limanta
A thesis in fulfillment of the requirements for the degree of Master of Science (by Research)
SCHOOL OF MATHEMATICS AND STATISTICS FACULTY OF SCIENCE UNIVERSITY OF NEW SOUTH WALES
June 2017
ORIGINALITY STATEMENT
'I hereby declare that this submission is my own work and to the best of my knowledge it contains no materials previously published or written by another person, or substantial proportions of material which have been accepted for the award of any other degree or diploma at UNSW or any other educational institution, except where due acknowledgement is made in the thesis. Any contribution made to the research by others, with whom I have worked at UNSW or elsewhere, is explicitly acknowledged in the thesis. I also declare that the intellectual content of this thesis is the product of my own work, except to the extent that assistance from others in the project's design and conception or in style, presentation and linguistic expression is acknowledged.'
Signed ....
Date Show me your ways, Lord, teach me your paths. Guide me in your truth and teach me, for you are God my Savior, and my hope is in you all day long.
– Psalm 25:4-5 –
i This is dedicated for you, Papa.
ii Acknowledgement
This thesis is the the result of my two years research in the School of Mathematics and Statistics, University of New South Wales. I learned quite a number of life lessons throughout my entire study here, for which I am very grateful of.
I would like to express my endless gratitude to my supervisor, Associate Professor Norman Wildberger for his constant support and guidance throughout my research study in the School of Mathematics and Statistics, University of New South Wales. His unorthodox view towards mathematics enables me to appreciate mathematics even more. I would also like to thank all of the staff in the school for their help towards my study. I am also indebted to Lembaga Pengelola Dana Pendidikan (Indonesian Endowment Fund for Education) for awarding me the scholarship during my master study in Sydney. Without the aid from them, my dream to pursue this study would have been impossible. Finally, I would like to thank both my parents and my brothers for their constant encouragement while I am studying here.
Sydney, June 2017
Author,
Kevin Mandira Limanta
iii Abstract
In this thesis we are going to study harmonic polynomials on spheres, with the particular attention to the 3-sphere 3. As a Lie group, the algebraic structure of 3 SU (2) enables us to study harmonic polynomials from an algebraic point ' of view. A purely algebraic approach has several advantages, such as the possi- bilities to extend to more general matrix groups and work over general fields such as finite fields, and importantly to rational numbers. This algebraic approach allows us to construct an explicit basis for the space of harmonic polynomials on 3 and describe them from a geometrical point of view. Finally, we give an explicit algebraic formula for zonal harmonic polynomials on 3 which can be identified as characters of Lie group 3 SU (2) ' Contents
0Introduction 1
1 The Algebra of Complex Numbers and Quaternions 5 1.1 The Complex Numbers C ...... 5 1.2 The Quaternions H ...... 8 1.3IdentifyingQuaternionswithMatrices...... 10
2 Spherical Rotation 16 2.1 Rotation in 1 ...... 16 2.2 Describing Rotations by Quaternions ...... 18 2.3SphericalParametrization...... 21 2.3.1 StereographicProjections...... 22 2.3.2 The Standard Theory: When F = R ...... 25 3 Theory of Harmonic Polynomials 27 3.1 Introduction ...... 27 3.2HarmonicPolynomialsontheSphere...... 31 3.3TheKelvinTransform...... 39 3.3.1 InversionintheUnitSphere...... 39 3.3.2 MapsPreservingHarmonicFunctions...... 39 3.4 Spherical Harmonics via Differentiation...... 43 3.5BasesforHarmonicPolynomials:AnalyticApproach...... 49
4 Theory of Harmonic Polynomials: Algebraic Approach 55 4.1 Harmonic Polynomials on 2 ...... 57 4.1.1 ComplexFactorialBasis...... 61 4.1.2 Zonal Harmonic Polynomials on 2 ...... 65 4.1.3 GeneratingFunction...... 72 4.2 Harmonic Polynomials on 3 ...... 73 4.2.1 ComplexFactorialBasis...... 74 4.3ZonalHarmonicsintheFactorialBasis...... 79 4.4ZonalHarmonicsintheComplexFactorialBasis...... 84
i CONTENTS ii
2 1 A Spherical Harmonic Decomposition of ( − ) 90 A.1 Functions on 1 andFourierTheory...... 91
B Integrating Polynomial over a Sphere 93 B.1 Integrating monomials over 1 ...... 94 B.2 Integrating monomials over 2 ...... 96 B.3Integratingmonomialsoverageneralsphere...... 99 Chapter 0
Introduction
In this thesis we are going to study harmonic analysis on spheres, with particular attention to the 3-sphere 3. This sphere is particularly interesting because it possesses a group structure, albeit a non-commutative one. In fact, 1 and 3 are the only two spheres that have a group structure. The algebraic structure of 3 enables us to talk about harmonic analysis on the sphere from a purely algebraic point of view. This thesis will be mainly focused on the study of harmonic polynomials, as we can use the fact that every function defined on a sphere can be decomposed into a series of harmonic polynomials, which serve as the building blocks of spherical functions. The main aim is to introduce the reader to an algebraic way of describing harmonic polynomials on 3. We will be particulary interested in the zonal harmonic polynomials — harmonic polynomials on 3 that are invariant under rotation. These polynomials are compelling because in Lie group theory, zonal harmonic polynomials are characters of the group 3 SU (2). ' This algebraic way of describing harmonic polynomials on 3 connects nat- urally with the theory of quaternions. There are already various studies about spherical harmonic polynomials for a general sphere, using a standard analytical point of view, such as Axler [6] and Müller [22], so one will have options to go with the standard view or the algebraic one. However, since the standard view often incorporates transcendental functions such as the square root function and trigonometricfunctions,itisnotalwayscomputationallyeffective, and does not extend easily to other fields. The algebraic point of view that will be developed here will extend to 2 and also often to orthogonal group of matrices. However, that extension will not be discussed in this thesis and could be a suggestion for
1 0. Introduction 2 future work. Moreover, one may be able to generalize this theory to work over general fields such as finite fields, and importantly to rational numbers. This thesis will be developed as follows. Chapter 0 will be reserved as the introduction to the thesis to give the reader the general idea of where the direc- tionofthethesisisheading.InChapter1,wewillintroducethebasicmaterials needed to explain the algebraic structure of 3 by discussing the algebraic struc- ture of 1 first. In Chapter 2 we see that multiplication of complex numbers or quaternions can be interpreted geometrically as spherical rotations. We will see that incorporating quaternions will be useful to explain rotations on 2,which has no group structure. In Chapter 3 we present the standard theory of har- monic functions. We will introduce first the concept of decomposition of general polynomials into homogeneous polynomials and finally into harmonic polynomi- als. The standard theory constructed later will be a decent tool to explain the decomposition of homogeneous polynomials into harmonic polynomials, as given below.
Theorem 1 Any homogeneous polynomial (x)=(12) (A ) ∈ can be uniquely written in the form
= + (x) 2 + + (x) 2 − ··· − where = and each (A );thatis, is harmonic homogeneous 2 ∈ polynomial of degree . ¥ ¦ At the end of Chapter 3, we will give an explicit basis for harmonic poly- nomials of degree . It is important to note here that although we present the standard theory, we want to extract as much theory as possible to work in a more general algebraic setting. If a certain theory is possible for generalization over a more general field, we will make a distinction. For example, instead of working on Euclidean spaces R,weturnourattentionto-dimensional affine space A over a general field, which might also be the field of rational numbers. There are, however, several steps in which we will resort to an analytical point of view. We will mention it whenever we do so. In the last chapter, we will turn to our purely algebraic setting and extend the ideas developed by Wildberger in the preprint [31]. We will focus on harmonic polynomials on 3 by first discussing harmonic polynomials on 2. We will attempt to approach the study of harmonic polynomials on 2 from an algebraic 0. Introduction 3 point of view and see that the lack of group structure will still be a necessary reason for us to use transcendental functions in describing zonal harmonics. However, that is not the case in 3. The highlight of this thesis will be the derivation of an algebraic version of zonal harmonic polynomials in 3, given in the Theorem below.
Theorem 2 Up to a scalar, there is a unique zonal harmonic polynomial of degree namely
2 b c !(2 2)! (2 2)! (2)! 2 2 2 2 2 2 (x)= ( 1) − − − − − − (2 +1)!( )!( )!! =0 =0 =0 X X X − − With the introduction of the complex factorial basis, we can reformulate zonal harmonic polynomials on 3 into another form, which is given by this Theorem.
Theorem 3 The zonal harmonic polynomials described in factorial basis is equivalent with
( 1) = − ( 1) − − − − =0 Ã=0 ! X X the zonal harmonic polynomials¡ ¢ in complex factorial basis.
In Appendix A, we will present the standard theory of spherical harmonic 2 1 1 decompositions on ( − ), the space of all square-integrable functions on − . In Appendix B, we will also present our own version in the classical problem of integrating polynomials on a sphere, as given in this theorem.
1 2 1 Theorem 4 If (x)=1 2 be any monomials of degree in − 1 ⊂ A ,and is the normalised measure on − defined in a way such that
=1
1 Z− then ( 2)!! (x) = − ( 1)!! ( + 2)!! − Z 1 − =1 − Y precisely when all of the are even integers, and zero otherwise. 0. Introduction 4
This classical problem has been studied extensively from an analytical point of view. See, for example, Baker [7, page 43] and Folland [17], which incorporates the gamma function Γ. While our result is a reformulation of Baker’s and Fol- land’s, the proof is our own work and the result is expressed in terms of rational function while still maintaining the symmetry. Chapter 1
The Algebra of Complex Numbers and Quaternions
In this review chapter we will discuss the tools needed to equip both two spheres 1 and 3 with group structures. We review how the algebra of complex numbers can be used to model the unit circle and how we can associate a matrix to each complex number. Most of what we will do in the case of complex numbers will generalize to the case of quaternions, with the twist that the multiplication is not commutative.
1.1 The Complex Numbers C
Let F be any field for which 2 +1 = 0 has no solution. A complex number system, denoted by C, consists of elements of the form = + —where and belong to F — which then will be called complex numbers.Thisis a quadratic field extension of F.Thenumber which corresponds to the case where =0and =1satisfies 2 = 1. It is important to note that the number − F from our assumption. This set of complex numbers C can be regarded as ∈ F2, with the identification of = + as ( ).Thus,C canbeviewedasa vector space over F, with the addition and scalar multiplication defined as
(11)+(22) (1 + 21 + 2) ≡ (11) (11) ≡
5 1. The Algebra of Complex Numbers and Quaternions 6
for every 1, 2, 1, 2, F.Moreover,C is an algebra with the multiplication ∈ defined as usual as
(11) (22) (12 1212 + 21) (1.1) · ≡ − Note that the multiplication defined above is commutative, since
(22) (11)=(21 2121 + 12) · − =(12 1212 + 21) − =(11) (22) · If = + ,then is called the real part of and is the imaginary part of ,writtenRe ()= and Im ()=, respectively. The conjugate of
= + is = ,andwehavethatforany1, 2 C, − ∈
12 = 12 for if 1 = 1 + 1 and 2 = 2 + 2,then
12 = (12 12)+(12 + 21) − =(12 12) (12 + 21) − − =(1 1)(2 2) − − = 12
Also define the product of any complex number with its conjugate the quad- rance of ,denotedby ().Thus,
()= =( + )( )=2 + 2 (1.2) − We can view this quadrance as a function from C to F. From the commutativity of C it then follows that the quadrance is a multiplicative function, i.e.
(12)= (1) (2) for any 1, 2 C,since ∈
(12)=1212
= 1122
= (1) (2) 1. The Algebra of Complex Numbers and Quaternions 7
Remark 1 Note that this is a departure from the usual approach when the un- derlying field is R, as we do not assume a square root that allows us to define . k k 2 Corollary 1 For any F and C,wehavethat ()= (). ∈ ∈ Define 10 01 1 = and i = 01 10 Ã ! Ã − ! and consider the set of matrices
C = 1 + i F { | ∈ } e = F (Ã !¯ ∈ ) − ¯ ¯ ¯ This is a subalgebra of the algebra 2 (F) ¯over F.Clearly
12 = 1, 1i = i1 = i and i2 = 1 − Since the mapping + (1.3) 7→ à − ! is an algebra isomorphism between C and C, we can identify the complex number + with .Thisidentificatione sometimes helps us in computing the à ! product of several− complex numbers. For example, if we want to compute the product of three complex numbers where the underlying field is the set of rational numbers Q,givenby1 = 5+2, 2 =1 ,and3 =3+2,wecancompute − − the product of these three matrices in 2 (Q):
52 1 1 32 23 15 − − = − 2 5 11 23 15 23 Ã − − !Ã !Ã − ! Ã − − ! giving us 123 = 23 + 15. − A complex number = + is called a unit complex number precisely when its quadrance is 1;thatis, ()=2 + 2 =1. We denote the set of pairs ( ) F2 such that 2 + 2 =1by 1, which will be called the unit circle ∈ 1. The Algebra of Complex Numbers and Quaternions 8 in F2. It is worth to note that the set of all unit complex numbers 1 forms a 1 group under multiplication, for if 1, 2 then (1)= (2)=1so ∈
(12)= (1) (2)=1,
1 hence 12 . The associative property follows from the associativity of ∈ complex numbers, the identity element is 1=1+0;andforevery 1,the ∈ inverse of is since = =1.
2 Example 1 In the finite field F7, it can be checked that +1 = 0 has no solution. The unit circle in this setting is then
1 = (0 1) (1 0) (0 6) (6 0) (2 2) (2 5) (5 2) (5 5) { }
1.2 The Quaternions H
The theory of quaternions was first developed by Sir William Rowan Hamil- ton, one of the most prominent Irish mathematicians in the 19th century. The motivation of the discovery of quaternions started from the desire to generalize complex numbers in a natural way. Hamilton firstcameupwiththeideaofanew algebraic structure whose elements consist of a real part and two distinct imagi- nary parts, which would then be called the Theory of Triplets. Just as complex numbers can be used to represent rotations in two-dimensional space, Hamil- ton hoped that this new structure could represent rotation in three-dimensional space. He worked for years and failed to produce such an algebraic structure. It then occured to Hamilton during a walk in 1843 that instead of having two distinct imaginary parts, three distinct imaginary parts should be considered (see [3, Chapter 12]). Again, let F be any field which has no element that satisfies 2 +1=0. Formally, we have the following definition of quaternions.
Definition 1 The algebra of quaternions, denoted by H,is
1 = 1+ + + : F h i { ∈ } 1. The Algebra of Complex Numbers and Quaternions 9 where the elements satisfy the following multiplication relations:
1 · 1 1 1 (1.4) − − 1 − − 1 − − The multiplication table reads, for example, that = and = . − Compare this with the complex number system. We have, in addition to , two more elements called and which also behave like ;thatis,2 = 2 = 1. Note that from the multiplication table above, we have that = 2 = 1. − − From the table it is also clear that the multiplication is non-commutative. See that, for example, = = = . This is the difference between the complex 6 − numbers and quaternions. The non-commutativity of quaternions makes things considerably more difficult. When performing calculations, we have to be careful with the order. Throughout the text, the zero quaternion is denoted by simply 0. The real part of = ++ +, denoted by Re (),is.Quaternionsofthe form + + are called pure quaternions. We will also write ++ + in short of 1++ +. Similar in the complex number case, we can identify any quaternion = + + + H as a 4-tuple ( ) F4.More ∈ ∈ interestingly, since + + + =( + )+( + ) , we can also identify as ( + + ) C2. ∈ Let 1 = 1 + 1 + 1 +1 and 2 = 2 + 2 + 2 +2 be two quaternions, and take any F.Define the addition and scalar multiplication in a similar ∈ fashion as the case of complex numbers by
1 + 2 (1 + 2)+(1 + 2) +(1 + 2) +(1 + 2) (1.5) ≡ 1 (1)+(1) +(1) +(1) (1.6) ≡ As a consequence of the multiplication table in (1.4), we have that
1 2 (3 + 3 + 3 + 3) · ≡ 1. The Algebra of Complex Numbers and Quaternions 10 where
3 = 12 12 12 12 (1.7) − − − 3 = 12 + 21 + 12 21 − 3 = 12 + 21 12 + 21 − 3 = 12 + 21 + 12 21 −
From now on, we will always write 12 in short of 1 2 for all 12 H. · ∈ Given a quaternion = + + + ,wedefine the conjugate of by
= − − − and the quadrance of by
()= =( + + + )( ) − − − = 2 + 2 + 2 + 2 (1.8)
Also observe that = 2 + 2 + 2 + 2, so any quaternion commutes with its 2 conjugate. Note that we also have ()= () for every H.Fromthe ∈ equation (1.8) above, if we set
1 1 − , ≡ () then it is apparent that 1 1 − = − =1.
1.3 Identifying Quaternions with Matrices
The associative and distributive properties of quaternions are not trivial, though. While these properties can be checked manually by using (1.5) and (1.7), let us introduce another way of viewing quaternions. Let = + + + be any arbitrary quaternion. Let 1, i, j,andk be respectively the following matrices in the algebra 2 (C) over F,thespaceofall2 2 matrices with complex entries: × 10 0 1 i ≡ 01 ≡ 0 Ã ! Ã − ! 01 0 j k ≡ 10 ≡ 0 Ã − ! Ã ! 1. The Algebra of Complex Numbers and Quaternions 11
Note the difference between this i and the previously defined i in the Section 1.1.
Define H = 1 i j k ,thesubalgebraof2 (C) over the field F spanned by the h i matrices 1, i, j,andk. An arbitrary element of this subspace can be written as e + + 1 + i + j + k = + Ã − − ! whichisamatrixoftheform
1 2
2 1 Ã − ! where 1 = + and 2 = + are both complex numbers. This generalizes the identification of complex numbers as matrices, in the special case where the imaginary part of 1 and 2 are both zero. Do note that in the above matrices, we have that i2 = j2 = k2 = ijk = 1 (1.9) − and ij = k jk = i ki = j (1.10) ji = k kj = i ik = j − − − Elements in H correspond to elements in H by + + + + + e (1.11) 7→ + Ã − − ! Onecancheckthatthisidentification respects the relations given in (1.9) and (1.10), and the map (1.11) is an algebra isomorphism, so we can also think of elements in H as simply quaternions as well. This gives a demonstration that the multiplication of quaternions gives an associative and distributive algebra e structure, as can be seen from H.If = + + + , call its corresponding matrix in H by = 1 + i + j + k. This correspondence is especially useful when we are dealing with multiplicatione of quaternions, just like in the case of e complex numbers. From (1.7), we see that if 1 = 1 + 1 + 1 + 1 and
2 = 2 + 2 + 2 + 2,then
3 + 33 + 3 12 = 3 + 33 3 Ã − − ! + + + + = 1 1 1 1 2 2 2 2 1 + 11 1 2 + 22 2 Ã − − !Ã − − !
= 1 2 1. The Algebra of Complex Numbers and Quaternions 12
For example, if 1 =2+3 +4 5 2 = 1+2 3 +4 and 3 =4 2, − − − − then their product can be calculated simply by multiplying
2+3 4 5 1+2 3+4 4 2 = − − − − 1 2 3 4 5 2 3 3+4 1 2 2 4 Ã − − − !Ã − − !Ã − ! 72 + 88 64 + 124 = − − 64 + 124 72 88 Ã − − ! so that 123 = 72 + 88 64 +124. − − By what we have discovered so far, we conclude that the quaternions satisfy the following properties, together make the set of quaternions an algebra.
1. Addition
Commutativity: 1 + 2 = 2 + 1 for all 12 H, • ∈ Associativity: 1 +(2 + 3)=(1 + 2)+3 for all 12 3 H, • ∈ Existence of identity: There exists 0 H such that +0=0+ = • ∈ for all H, ∈ Existence of inverse element: For all H there exists H such • ∈ − ∈ that +( )=( )+ =0. − − 2. Multiplication
Associativity: 1 (23)=(12) 3 for all 123 H, • ∈ Existence of identity: There exists 1 H such that 1=1 = for • ∈ all H, ∈ Existence of inverse element: For all H satisfying () =0,there • 1 1 1 ∈ 6 exists − H such that − = − =1. ∈ 3. Distributive law
Left distributive law: 1 (2 + 3)=12 + 13 for all 123 H, • ∈ Right distributive law: (1 + 2) 3 = 13 + 23 for all 123 H. • ∈ 1. The Algebra of Complex Numbers and Quaternions 13
Another advantage of interpreting quaternions as matrices is that we can retrieve the real part of any quaternion as the trace of the corresponding matrix. Suppose = + + + is any quaternion, then we have 1 = tr ( ) (1.12) 2
For all matrices with complex entries, we define ∗ to be its conjugate transpose or Hermitian transpose; that is, the matrix obtained from the trans- pose by taking the complex conjugate of all entries. If = + + + , then
+ + ∗ ∗ = + Ã − − ! = − − − + Ã − ! =
Also observe that if = + + + ,thenwehavethat
()=2 + 2 + 2 + 2 + + =det + Ã − − ! =det() (1.13)
This is useful when we want to prove that is multiplicative; i.e.
(12)= (1) (2) (1.14) for all 12 H,since ∈
(12)=det(12 )
=det(1 2 )
=det(1 )det(2 )
= (1) (2)
If ()=1,then is a unit quaternion. The case where ()=1 is particularly important because the set H ()=1 is closed under { ∈ | } multiplication, as can be seen from (1.14). In fact, it can be easily verified that 1. The Algebra of Complex Numbers and Quaternions 14 this set is a multiplicative group. The associativity follows from the associativity of general quaternions. The identity element is 1, and the inverse element for is just . We denote this set by 3,the3-sphere of unit quaternions. If the underlying field F is R,wecanthinkof3 as a manifold in R4.Since3 is a manifold and also a group, 3 is a Lie group. Classical theory of Lie groups can be found in Duistermaat [13], Hall [19], or Chevalley [11].
Definition 2 Amatrix (C) is called unitary if ∗ = ,where ∈ is the identity matrix. The unitary group U() is the subgroup of GL ( C) which consists of unitary matrices. The special unitary group SU () is the subgroup of U() that consists of matrices of determinant 1.
Remark 2 Note that if ∗ = then det ()= 1 since determinant is ± multiplicative and det (∗)=det(). Wecouldthenassociatetwospheresthatwehavejustbuiltbythefollowing proposition.
Proposition 1 The group 1 is isomorphic to U(1).Thegroup3 is isomor- phic to SU (2)
Proof. Note that U(1) consists of all matrices of the form () such that
()∗ ()=(1),whichmeansthat =1. This is precisely the set of all unit complex numbers 1, so the said isomorphism is (). For the second 7→ statement, note that there is a bijection between = + + + and + + = .Themap is also a homomorphism since à + ! 7→ − − 3 3 = for all 12 Finally, if then ()=1.Butwe 1 2 1 2 ∈ ∈ know that 1= ()=det(),andthat
∗ = = 1 = 2 proving that is unitary. Hence, SU (2). ∈ Suppose 1 = 1 + 1 + 1 + 1 and 2 = 2 + 2 + 2 + 2,define the quaternion inner product : H H R by h· ·i × →
12 12 + 12 + 12 + 12 h i ≡ From this, it is evident that ()= We also have that h i 1 12 = tr ∗ h i 2 1 2 ¡ ¢ 1. The Algebra of Complex Numbers and Quaternions 15 and the proof is computational. This shows that the theory of quaternions is intimately connected with the algebra of complex matrices. More material about this can be found in Wildberger [32, Part c]. Chapter 2
Spherical Rotation
As mentioned in the Chapter 1, the group of unit complex numbers 1 can be used to explain rotations in the plane. Quaternions were created by Hamilton to help explain rotations in three-dimensional space. In this chapter, we will see how these two algebraic structures are useful for explaining the concept of rotation. Throughout this thesis, let A be the affine space over a field F,which is an algebraic generalization of Euclidean space. When the underlying field is R,thenA is just the usual -dimensional Euclidean space. The material in this chapter is inspired by Wildberger’s quaternion lectures [32, Part b, Part c].
2.1 Rotation in 1
In this section we are going to talk about how complex numbers previously defined can be utilized to describe rotation in two-dimensional affine space A2. Let = + be any complex number. Write [] tobethelinethatpasses through and the origin. The standard theory to measure how far away a particular point from another point, aside from distance, is the notion of angle. As the direction of this thesis suggests, we aim to generalize the work that we develop here to other algebraic settings, so rational functions are preferable to transcendental ones. For this reason, we will find a rational analog of angle. There are several choices and all are developed from rational trigonometry (see Wildberger [30]), but in this section we will stick with only one. 1 Define the turn of = + to be the element ()= := − F,for ∈ =0. It follows that the turn of depends only on [].Define, for =0,the 6 6
16 2. Spherical Rotation 17
linear operator on C 2 ()= C.(2.1) () ∈ From Corollary 1 we observe that 2 =1, () µ ¶ so this is a rotation on 1 determined by the line []. Moreover, we also have that = (2.2) 1 ◦ 2 12 and
= (2.3) 1 for any 1, 2, C and F. Equation (2.3) suggests that rotation on ∈ ∈ can be indexed by lines [] in the two-dimensional affine space A2.Thesetofall one-dimensional subspaces in A2 is called the projective line P1.Itincludes the line [] where = + and =0where the turn is defined; and also [] 6 where = , which is called the pointatinfinity. Hence, the assignment
[] 7→ [] showsusthatrotationson1 can be parametrized by P1.Ifwewrite = + then the above mapping corresponds to the matrix
1 () à − !à − !à − ! 1 (2 2) 2 2 +(2 2) = − − − () 2 (2 2) (2 2) 2 à − − − − − ! so that 1 ()= 2 2 2 + 2 + 2 2 2 + 2 − − − ¡¡ ¢ ¡ ¡ ¢ ¢¢ If we identify by then we can view the linear map as à !
7→ where 1 2 2 2 = − − (2.4) 2 + 2 2 2 2 Ã − ! 2. Spherical Rotation 18
Onecancheckthatthematrix above is orthogonal since the first and the second column are perpendicular with respect to the standard dot product, and 2 2 2 2 2 2 that ( ) +(2) =( + ). The latter also implies that det =1,so − SO (2),thegroupof2 2 orthogonal matrices with determinant 1. ∈ × Note that we can write (2.4) as
1 1 2 2 = − − 1+2 1+2 1 2 Ã − ! where = (),theturnof,forsuch so that Re () =0.Pointatinfinity is 6 10 associated with − . Hence, there is an association 0 1 Ã − !
[] [] 7→ that sends a line [] in P1 to SO (2).
2.2 Describing Rotations by Quaternions
We want to show how quaternions connect to rotations of three-dimensional affine space A3.Let = + + + be a quaternion. Denote by v the imaginary part of ,sothatv = + + and we can write = + v. We can picture as a vector ( ) A4,sowecanthinkofv as a vector ∈ ( ) in the three-dimensional space A3 spanned by , ,and;which ' is perpendicular to the -axis under the usual Euclidean inner product. Now consider the unit sphere 3 in the four-dimensional space A4. It meets the - axis at (1 0 0 0) which corresponds to 1 =1,the-axis at (0 1 0 0) which corresponds to 2 = ,the-axis at (0 0 1 0) which corresponds to 3 = ,and the -axis at (0 0 0 1) which corresponds to 4 = . The equator of this sphere is exactly the intersection between 3 and , which corresponds to the case where =0. In this case, the equation becomes 2 +2 +2 =1which, in the three-dimensional space , is nothing but 2-sphere 2.Now,if = + v is any quaternion then we can think of as the projection of to the -axis and v as the projection to .Alsonotethatifwereflect the point = + v with respect to the -axis then we will get the point v which − is exactly . 2. Spherical Rotation 19
For = + v a nonzero quaternion, define, for any H, ∈ 1 () − = ≡ ()
This is a linear operator on H and has the following properties.
Proposition 2 We have that
1. (1) = 1
2. If w then (w) , ∈ ∈
3. (v)=v
Proof. Clearly, 1 (1) = = =1 () () That proves the first assertion. Next, if w , then we want to show that ∈ Re (w) =0. Note from (1.12), we have that ¡ ¢ 1 Re (w) = tr 2 (w) ³ ´ Since ¡ ¢ 1 = (w) () w and 1 tr = tr ( ) (w) () w ³ ´ 1 = tr ( ) () w
= tr (w) =0 it follows that Re (w) =0,sothat (w) . Here we have used the ∈ fact that tr ()=tr() for any 2 2 complex matrices and ,with ¡ ¢ × = w and = . Now, for the third proposition, note that = + v commutes with v since v commutes with and v itself. Hence, v v (v)= = = v () () 2. Spherical Rotation 20 as desired.
From the Proposition 2 above we conclude that must be a rotation in A3 with v being the axis of rotation. Now, define [] to be the line in ' A4 that passes through and the origin. The set of all lines [] is called the projective space P3.Observe,asinthe1 case, that
= for any H and F. Thusitissensibletomaketheassociation[] [] ∈ ∈ 7→ that tells us that rotations in three-dimensional space can be parametrized by P3.
AsalinearoperatoronthevectorspaceH,themap can be associated to amatrix.Let = + 1 + 2 + 3 be a fixed, arbitrary quaternion and take any = + 1 + 2 + 2 H.Notethat ∈ 1 () − ≡ = () corresponds to the matrix 1 () 1 = () ∗ where
+ 12 + 3 + 12 + 3 = and = 2 + 31 2 + 3 1 Ã− − ! Ã − − ! If we perform the matrix multiplication above and switch back into the form + + + , then we will have that
=
1 2 2 2 2 = + 1 +2(12 3) 2 +2(13 + 2) 3 () 1 − 2 − 3 −
1 ¡¡ ¢ 2 2 2 2 ¢ = 2(12 + 3) 1 + + 2 +2(23 1) 3 () − 1 2 − 3 −
1 ¡ ¡ ¢ 2 2 2 2 ¢ = 2(13 2) 1 +2(23 + 1) 2 + + 3 () − − 1 − 2 3 ¡ ¡ ¢ ¢ 2. Spherical Rotation 21
Wecanencodethatinformationmoreneatly.Ifweidentify with ⎛ 1 ⎞,then ⎜ 2 ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎜ ⎟ we can identify with the linear map ⎝ ⎠
7→ where ()0 0 0 2 2 2 2 1 ⎛ 0 + 1 2 3 2(12 3)2(13 + 2) ⎞ = − − − () 02( + ) 2 2 + 2 2 2( ) ⎜ 1 2 3 1 2 3 2 3 1 ⎟ ⎜ − − 2 2 − 2 2 ⎟ ⎜ 02(13 2)2(23 + 1) 1 2 + 3 ⎟ ⎜ − − − ⎟ ⎝ ⎠ Clearly this leaves the -axis fixed. Denote by the matrix
2 2 2 2 + 2(12 3)2(13 + 2) 1 1 − 2 − 3 − 2( + ) 2 2 + 2 2 2( ) (2.5) () ⎛ 1 2 3 1 2 3 2 3 1 ⎞ − − 2 2 − 2 2 2(13 2)2(23 + 1) + ⎜ − − 1 − 2 3 ⎟ ⎝ ⎠ and observe that = for all H and F. One can compute directly ∈ ∈ that the matrix is orthogonal and see that every 3 3 orthogonal matrix can × be written as for some unit quaternion = + 1 + 2 + 3 (see [15, Chapter 9]). Moreover, det ()=1for all H. We call the group of 3 3 ∈ × orthogonal matrices of determinant 1SO(3), and we close this section by the following result.
Proposition 3 There is an isomorphism between P3 and SO (3). Proof. The mapping
2 2 2 2 + 2(12 3)2(13 + 2) 1 1 − 2 − 3 − [] = 2( + ) 2 2 + 2 2 2( ) [] () ⎛ 1 2 3 1 2 3 2 3 1 ⎞ 7→ − − 2 2 − 2 2 2(13 2)2(23 + 1) + ⎜ − − 1 − 2 3 ⎟ gives the desired isomorphism.⎝ ⎠
2.3 Spherical Parametrization
In this section we are going to present the sphere parametrization from an alge- braic point of view that works well in any field F, provided the equation 2+1 = 0 2. Spherical Rotation 22 doesnothaveasolutioninF. At the end of this section, we will present the standard theory of parametrization when the underlying field is just R.Westart from the simplest sphere, the unit circle 1,beforemovingto2 and finally 3. This is closely related to the rotations we saw earlier.
2.3.1 Stereographic Projections
Stereographic projection is a way to parametrize a sphere by projecting each 1 point in the ( 1)-dimensional sphere − A onto a hyperplane from a − ⊂ fixed point on the sphere, which would be called the pole. We shall illustrate this in the case where =2. Take the unit circle 2 + 2 =1centered at the origin and fixthepoint( 1 0) as our pole. For each F,let[] be the line − ∈ that connects the pole ( 1 0) and the point (0) in P1 A2. This line will − ⊂ intersect the circle exactly once, so that each point on the circle except the pole will correspond to (0) which can be identified as . To put it in another words, each [] P1 will correspond to exactly one point on the unit circle with ( 1 0) ∈ − removed, where F. Thepointatinfinity in P1 will correspond to the pole ∈ ( 1 0). We identify the point at infinity with = . − ∞ For each F, the line that passes through ( 1 0) and (0) in A2 is given ∈ − by the equation = + Substituting it into the equation 2 + 2 =1we get
2 +( + )2 =1 which will give us the quadratic equation
1+2 2 +22 + 2 1 =0 − which factorizes as ¡ ¢ ¡ ¢
( +1) 1+2 1 2 =0 − − We get two values of ,namely¡¡ ¢ ¡ ¢¢
1 2 = 1 or = − − 1+2 Thecasewhere = 1 corresponds to the point ( 1 0) which we know already. − − The other value of will give the value of so that the other intersection point 2. Spherical Rotation 23 is given by
1 2 1 2 ( )= − − + 1+2 1+2 µ µ ¶ ¶ 1 2 2 = − 1+2 1+2 µ ¶ Toputitinanotherwords,
1 2 2 ()= − and ()= 1+2 1+2 where F give us a parametrization of the unit circle 1 with the point ( 1 0) ∈ − removed. Thus if the field F does not have the square root of 1,thenthemap − 1 1 : P 7→ where 1 2 2 − 2 2 if F τ ([]) = 1+ 1+ ∈ ( ³ ( 1 0) ´ if = − ∞ gives the desired parametrization of the whole sphere 1 in terms of the extended, or projective line. We have the following result.
Theorem 5 There is a 1: 1 correspondence between the unit circle 1 and SO (2).
1 2 2 Proof. The point 1+−2 1+2 corresponds to the matrix ³ ´ 1 1 2 2 − − 1+2 1+2 1 2 Ã − ! in SO (2) which is associated to the complex number such that Re () =0. 6 The point ( 1 0) corresponds to − 10 − 0 1 Ã − ! which is associated with the complex number such that Re ()=0.Thisgives us the desired 1: 1correspondence. 2. Spherical Rotation 24
Example 2 If F = F7,then =5corresponds to the point 1 1 1 1 24 26− 10 26− = 4 5− 3 5− =(4 3 3 3) = (5 2) − · · · · · · More precisely,¡ from Example¢ 1, ¡ =0 1, 2, 3,¢4, 5, 6 correspond to the points (1 0), (0 1), (5 5), (2 2), (2 5), (5 2), (0 6), respectively. The point (6 0) = ( 1 0) correspondstothepointofinfinity = . − ∞ To deal with the 2-sphere, it is perhaps better to move our pole to the point (0 0 1) so we can project every point on the sphere with the pole removed − to the -plane, which is the subspace of A3 that consists of points of the form ( 0) where , F. The line connecting the pole and any arbitrary point ∈ ( 0) in the -plane is parametrized by 0 ⎛ ⎞ = ⎛ 0 ⎞ + ⎛ ⎞ 1 1 ⎜ ⎟ ⎜ − ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ where F. Substituting = , = ,and = 1 into the equation ∈ − 2 + 2 + 2 =1,weget ()2 +()2 +( 1)2 =1 − which gives us the quadratic equation 1+2 + 2 2 2 =0 − We get two values of ,namely¡ ¢ 2 =0 or = 1+2 + 2 as long as 2 + 2 = 1. The case where =0corresponds to the pole, and the 6 − other case corresponds to the point 2 2 1 2 2 ( )= − − 1+2 + 2 1+2 + 2 1+2 + 2 µ ¶ This gives us a parametrization for the unit 2-sphere 2 excluding the pole, namely 2 ( )= 1+2 + 2 2 ( )= 1+2 + 2 1 2 2 ( )= − − 1+2 + 2 2. Spherical Rotation 25 for every F such that 2 + 2 = 1. ∈ 6 − Similar to what we have done, the parametrization for the unit 3-sphere excluding the pole, by choosing the pole to be (0 0 0 1) would be − 2 ( )= 1+2 + 2 + 2 2 ( )= 1+2 + 2 + 2 2 ( )= 1+2 + 2 + 2 1 2 2 2 ( )= − − − 1+2 + 2 + 2 where F and 2 +2 +2 +1 =0. This gives us the bijection between P3 ∈ 6 and 3, where the projective space P3 A4 is taken to be the set lines through ⊂ ( 0) and (0 0 0 1). −
2.3.2 The Standard Theory: When F = R
When the underlying field F is real numbers, we can have the usual parametriza- tion of a sphere by using trigonometric functions cos and sin.Equipthefield R with the usual order so that R becomes an ordered field. The equation of a circle in the Cartesian plane R2 centered at ( ) and radius 0 is given by ≥ ( )2 +( )2 = 2, and every point on the circle can be specified with the − − following parametrization:
( )= + cos ( )= + sin where ranges from 0 to 2.Wecall( + cos + sin ) as the polar co- ordinate of the point ( ).Here is called the radial coordinate and the angular coordinate. Obviously, the case =0corresponds to a point ( ). The parametrization of the unit circle is then just a special case of this, with = =0and =1. Hence the parametrizaton of the unit circle with equation 2 + 2 =1is given by
()=cos and ()=sin where 0 2 ≤ Thecasewhere =0corresponds to the point (1 0), = 2 corresponds to the point (0 1), and so on. Combining this and the stereographic projections 2. Spherical Rotation 26 developed in the Subsection 3.3.1, we can conclude that for each [0 2), ∈ there is R such that ∈ 1 2 2 cos = − and sin = 1+2 1+2 In this case, 2 +1is always positive, so () and () are always defined for every R. ∈ If we go one dimension higher to consider the parametrization of general 2- sphere, which is given by ( )2 +( )2 +( )2 =1, it is only natural to − − − parametrize it in the following manners:
( )= + sin cos ( )= + sin sin ( )= + cos where [0 2) and [0]. Herewehavetwoangularcoordinates, ∈ ∈ and . Implicitly, if we combine the parametrization of 2 from stereographic projections above, this means that given any [0] and [0 2),onecan ∈ ∈ always find a pair ( ) R2 so that ∈ 2 sin cos = 1+2 + 2 2 sin sin = 1+2 + 2 1 2 2 cos = − − 1+2 + 2 Note that again, we have 2 + 2 +1 0,so ( ), ( ),and ( ) are always defined for every ( ) R2. ∈ Analogous to the previous cases, the parametrization of 3-sphere where the equation of the sphere is given by ( )2 +( )2 +( )2 +( )2 = 2 − − − − is
( )= + sin sin cos ( )= + sin sin sin ( )= + sin cos ( )= + cos where [0 2) [0] and [0]. ∈ ∈ ∈ Chapter 3
Theory of Harmonic Polynomials
The theory of harmonic functions presented here is the standard analytical ap- proach that works well in any dimension over the field R, but we present it in a more restricted algebraic way so that we can extend the theory to more general fields. It is presented here to give the sense of what the analytical point of view looks like. Since this is usually framed on the field of real numbers R, we will follow that. However, since the purpose of this thesis is to offer an algebraic point of view, certain concepts that have possibilities to work in a more gen- eral algebraic setting will be distinguished whenever necessary. For instance, the usual direction is to define the norm x of x. Here we avoid that and instead k k work with a more general function, namely quadrance. Most of the theory here is developed from Axler [6]. The bilinear form de- fined in Section 4.2 is different from Axler’s approach to decompose the space of homogeneous polynomials, and it is introduced to give a more algebraic ap- proach compared to Axler’s more analytical approach. In particular, the proofs of Lemma 20, Corollary 21, Theorem 22, Corollary 23 and 24 before arriving at the Homogeneous Polynomial Decomposition Theorem are our own version. Every example is also our own, and they are intended to help illustrate the concepts better.
3.1 Introduction
Let A be an affine space over a field F. Here, and throughout this chapter, we will assume that 2, unless stated otherwise. Since we are going to talk about polynomials and rational functions on A, we shall adopt the notion of formal
27 3. Theory of Harmonic Polynomials 28 derivatives, where the derivative is purely algebraic. That means if we have a variable ,weset 1 ( ) − ≡ and we can generalize that to the multivariable case. Recall the Laplacian operator 2 2 2 = 2 + 2 + + 2 (3.1) 4 1 2 ··· which acts on polynomial functions on A. Again, the second derivation here is taken to be the formal derivative. Denote by (A) the space of all polynomial functions on A.
Definition 3 Apolynomialfunction (A) is harmonic precisely when ∈ ∆ 0. ≡ We work mostly with polynomial functions, but the classical theory on R is more general and occasionally we will give examples of other harmonic functions in this case. We will sometimes denote ∆ by
∆ = 2 + 2 + + 2 (3.2) 1 2 ··· 2 where denotes the second partial derivative with respect to the -th variable. Also, denote by x =(12) the typical point in A and define the quadrance of x by (x)=2 + 2 + + 2 . We will give some examples of 1 2 ··· harmonic functions when F = Q.
Example 3 Constant polynomials, linear polynomials are all harmonic func- tions.
2 2 2 Example 4 Let (123)= + 2 .Since 1 2 − 3 2 2 2 ∆ = 2 + 2 + 2 1 2 3 =2+2 4=0 − it follows that is harmonic.
Example 5 Let ( )=22 +52.Overthefield Q, is not harmonic since ∆ =4+10=14=0.ButoverF7, is harmonic since ∆ =14=0. 6 3. Theory of Harmonic Polynomials 29
The above examples are all polynomials which are rational functions. Here is an example, when the underlying field is taken to be the real numbers R,that incorporates transcendental functions exp and sin.
1 Example 6 The function (12)= sin 2 is harmonic since 2 2 ∆ = 2 + 2 1 2 1 1 = sin 2 sin 2 =0 − Throughout this chapter, we shall assume that the underlying field is R, unless stated otherwise. The next example will be important for the rest of this chapter.
1 1 Example 7 For al l 2, the function (x)= (x) − 2 is harmonic. To see this, note that
1 1 1 2 =(2 ) 2 + 2 + + 2 − 2 2 2 + 2 + + 2 − 2 − − 1 2 ··· − 1 2 ··· h 1 2 1 1 i =(2 ) ¡ (x)− 2 (x¢)− 2 − ¡ ¢ − − h i for all 1 2 .Hence ∈ { } 2 ∆ = =1 X 1 1 2 2 2 1 = (2 ) (x)− (x)− − =1 − − X h i 1 2 1 1 =(2 ) (x)− 2 (x)− 2 − − − =1 ³ ´ X 1 1 2 2 1 2 =(2 ) (x)− (x)− − − Ã − =1 ! X 1 1 =(2 ) (x)− 2 (x)− 2 − − =0 ³ ´
Remark 3 Note that in the example above, (x) is a rational function if is even and not rational, otherwise. We can still extract this to a more general algebraic setting if is even, treating the derivative as formal derivative. In both cases, (x) is an algebraic function. 3. Theory of Harmonic Polynomials 30
Below we list several basic properties of any harmonic polynomials.
Proposition 4 Let be a harmonic polynomial defined on A.
1. If y A, then the function (x)= (x y),the-translate of ,defined ∈ − on A + y = w + y : w A is also harmonic. This property is called { ∈ } translation invariance.
2. If 0 = F, then the -dilate of ,defined by (x) on A is also 6 ∈ ≡ harmonic. This property is called dilation invariance.
3. If : A A is an orthogonal linear map, then the rotation of under → , denoted by = , is also harmonic. This property is called rotation ◦ invariance.
Proof. The first proposition is trivial, since
2 2 = so harmonicity of follows from harmonicity of . For the second proposition, note that 2 ∆ ()=∆ (x)= (∆)
on A . It can be inferred that is harmonic if and only if is harmonic. Now we prove the third proposition. Recall that is orthogonal if ( (x)) = (x) for all x A. We can also say that is orthogonal precisely when preserves ∈ the unit sphere. It is well-known that if is orthogonal, then the columns of the matrix corresponding to with respect to the standard basis on A form an orthonormal basis. Let [] be the matrix of relative to the standard basis on A.Foreach 1 2 , ∈ { } ( )= ◦ =1 ◦ X µ ¶ 3. Theory of Harmonic Polynomials 31
Differentiating once more and taking the sum over yields
∆ ( )= ◦ ◦ =1 à =1 ! X X µ ¶ = ◦ =1 =1 X X µ µµ ¶ ¶¶ 2 = ◦ =1 =1 =1 X X X µ ¶ 2 = ◦ =1 =1 Ã=1 ! X X X µ ¶ 2 = 2 =1 ◦ X µ ¶ =(∆) ◦ Thus is harmonic if and only if is harmonic. ◦
3.2 Harmonic Polynomials on the Sphere
Set () to be the subspace of consisting of all polynomials of degree less than or equal to . Clearly
(0) (1) (2) () ⊂ ⊂ ⊂ ···⊂ If we let (A ) to be the space of all homogeneous polynomials of degree on A , 1 2 then any polynomial is a finite linear combination of monomials 1 2 where 1 + 2 + + = .Wehave,forany N, ··· ∈
0 1 2 = () ⊕ ⊕ ⊕ ···⊕ 1 Denote by − the set of unit vectors in A ; that is, the set of all x A ∈ such that (x)=1. This set will be called the ( 1)-sphere inside A.If − 1 is any polynomial on A ,thendenoteby 1 its restriction to − .When | − the context is clear, we will use instead of 1 .Forexample,if =4and | | − 3 2 2 2 4 (x)=1 + 12 + 13 + 14 3 A ∈ then (x)=1 since ¡ ¢ | 3 2 2 2 2 2 2 2 1 + 12 + 13 + 14 = 1(1 + 2 + 3 + 4)=1 3. Theory of Harmonic Polynomials 32
We also denote (A ) and (A ) by the space of all harmonic polynomials on A and the space of all homogeneous harmonic polynomials of degree on A, respectively. It is also evident that (A ) is a vector space over F,and (A ) is a subspace of (A ). It is sometimes convenient to write a homogeneous monomial
1 2 (x)= (12)=1 2
on A by the notion of multi-index =(12).Wedefine to be | | just the sum 1 +2 + +. Inthiscasewewrite as x in short. Since any ··· homogeneous polynomial of degree is just a linear combination of monomials of the form x where = , then every homogeneous polynomial of degree | | can be written as (x)= x (3.3) = |X| Define the bilinear form in the vector space (A ) over F
: (A ) (A ) F h· ·i × → ( ) ( ()) (x) (3.4) 7→ |x=0 where () is the differential operator
()= = |X| associated to (3.3), with
= (1) 1 = ··· 1 1 ··· = 1 1 ··· Hence(3.4)isthedifferential operator () applied to (x) and evaluated at x =0. For example, if =3and (x)= ( )=53 +72 33,then − 3 3 3 ()=5 +7 3 3 2 − 3 We have the following Lemma, which in brief tells us that the bilinear form of two different monomials are always zero. 3. Theory of Harmonic Polynomials 33
Lemma 1 If and are two monomials where = ,then 6 =0 h i
1 2 1 2 Proof. Let (x)=1 2 and (x)=1 2 .Observethat
1 2 1 2 = ··· 1 2 h i 1 2 µ 1 2 ··· ¶ ¯x=0 ¡ ¢¯ Now, if foratleastonevalueof, then the above¯ expression will be ¯ automatically zero. If for at least one value of , then there will still be aterm for some , hence by evaluating at 0 we have the above expression to be zero as well. So if =0itmustbethecasethat = for all h i 6 1 2 ∈ { } Corollary 2 If is defined as above then
= ! h i =1 Y Proof. We have that
1 2 1 2 = ··· (1 2 ) h i 1 2 µ 1 2 ¶ ¯x=0 ··· ¯ =(1 1) (2 1) ¯( 1) ×···× × ×···× ×···×¯ ×···× = ! =1 Y as desired. Now we will see the relation between the Laplacian operator ∆ and the quadrance described by using the Hermitian form defined above. The following shows that the Laplacian operator ∆ and the multiplication by (x) operator are adjoint.
Theorem 6 (Relation of Laplacian and Quadrance) If and are two monomials then, ∆ = (x) h i h i Proof. By the linearity of the bilinear form, we can just assume that both and are monomials. Let (x)=x where =(12), (x)=x where
=(12). From the previous lemma, we must conclude that = | | and = 2 for some 2, otherwise ∆ =0= (x) | | − ≥ h i h i 3. Theory of Harmonic Polynomials 34
We have that
1 2 1 2 ∆ = 1 (1 1) − + + ( 1) − − 1 ··· − 1 so
1 2 − 1 2 ∆ = ( 1) ··· ··· 1 2 1 2 h i − 1 − ¯ =1 µ ··· ··· ¶ ¯x=0 X ¡ ¢¯ (3.5) ¯ Since =2, we have two cases. Either exactly =2for only¯ one | | − | | − and = otherwise; or =1for exactly two values of and = − otherwise. In the first case we can reduce the above equation to get
∆ = ( 1) (1! ( 2)! !) = ! h i − ··· − ··· =1 Y In the second case, (3.5) will reduce to ∆ =0since for two such values of , h i thederivativeistaken 2 times while the power of is = 1 2. − − − Hence ∆ = 1! ! precisely when there is exactly one 1 2 h i ··· ∈ { } such that =2and = for = ; zero otherwise. − 6 Now, note that
1 2 1+2 2 1 2 +2 (x) = ··· 1 2 + + 1 2 h i 1 2 ··· µ 1 2 ··· ¶ ¯x=0 ¡ ¢¯ Again, if exactly =2for only one and = otherwise, then the above¯ − ¯ equation reduces to
1 2 1 +2 (x) = ··· 1 h i 1 2 µ 1 2 ··· ¶ ¯x=0 ¡ ¢¯ = 1! ( +2)! ! ¯ ··· ··· ¯ = 1 ! ! ··· ···
= ! =1 Y If exactly =1for two values of and = otherwise, by the same − reasoning, (x) =0.Since ∆ and (x) agree on every case, h i h i h i we conclude that ∆ = (x) . h i h i Relation of Laplacian and Quadrance Theorem above gives a couple of inter- esting corollaries. 3. Theory of Harmonic Polynomials 35
Corollary 3 For 2, the Laplacian map ≥ ∆ : (A ) 2 (A ) 7→ − is surjective.
Proof. For all (A ) such that ∆ =0,wehavethat ∈ h i 0= ∆ = (x) h i h i More specifically, if we pick = (x) ,wehavethat (x) (x) = h i 0 which implies that (x) =0,forif (x) =0, then by Corollary 2, 6 (x) (x) =0. Since this is true for all x,wemusthavethat =0. h i 6 Hence ∆ maps (A ) onto 2 (A ) − Corollary 4 With respect to the Hermitian form (3.4), all harmonic polynomi- als (A ) are orthogonal to (x) for 2 (A ). ∈ ∈ − Proof. Take any (A ),thenforany 2 (A ) we have ∈ ∈ − (x) = ∆ =0 h i h i where the second equality is justified since ∆ =0. Corollary 4 above is an important ingredient for the following theorem that is the heart of this section:
Theorem 7 (Homogeneous Polynomials Decomposition) If 2,then ≥ (A )= (A ) (x) 2 (A ) ⊕ − The above theorem is the key to the following theorem which states explicitly the decomposition of any homogeneous polynomial of degree .
Theorem 8 Any polynomial (x)=(12) (A ) can be uniquely ∈ written in the form
= + (x) 2 + + (x) 2 (3.6) − ··· − where = and each (A ). 2 ∈ ¥ ¦ 3. Theory of Harmonic Polynomials 36
Proof. The result is obvious for =0or =1, since any homogeneous poly- nomial of degree 0 or 1 is harmonic. Suppose then 2. By the Homogeneous ≥ Polynomial Decomposition Theorem, there is a unique decomposition
= + (x) 2 −
where (A ) and 2 2 (A ). We can proceed by taking the unique ∈ − ∈ − decomposition for 2 as −
2 = 2 + (x) 4 − − −
for 2 2 (A ) and 4 4 (A ) so that − ∈ − − ∈ −
= + (x)( 2 + (x) 4) − − 2 = + (x) 2 + (x) 4 − − Continuing in this fashion, after finitelymanystepswewillarriveat(3.6). As a corollary of the Theorem 8, if we have any polynomial (A ) as ∈ above, then
= + 2 + + 2 (3.7) | | − | ··· − | where and are defined also as above. Now, we turn our attention back to the Homogeneous Polynomials Decom- position Theorem above, and restrict everything to the unit sphere. To be more precise, we have that
(A )= (A ) (x) 2 (A ) ⊕ − 1 fromthetheorem.Ifwejustrestricteverythingin − A , then the left 1 ⊂ 1 hand side becomes ( − ) while the right hand side becomes ( − ) 1 1 1 ⊕1 2 ( − ). We can further decompose 2 ( − ) as 2 ( − ) 4 ( − ) − − − − 1 ⊕ until we are left with 2 ( − ) with = . However note that for this , − 2 1 1 we have 2 ( − )= 2 ( − ) since 2 is either 0 or 1,depending − − ¥ −¦ on the parity of . Hence we have
1 1 1 1 − = − 2 − 2 − − − ⊕ ⊕ ···⊕ ¡ ¢ ¡ ¢ 1 ¡ ¢ ¡ ¢ = 2 − − =0 M ¡ ¢ 3. Theory of Harmonic Polynomials 37
1 Theaboverelationshowsthatwecanregard ( − ) as the finite direct sum of 1 1 the spaces 2 ( − ). Moreover, the spaces ( − ) are pairwise orthogonal − with respect to the inner product
= (3.8) h i 1 Z−
1 where is a surface measure on − . The following theorem uses the analytic approach so the underlying field is R.
Theorem 9 If and are polynomials on A and is harmonic and homoge- neous with degree higher than the degree of ,then
=0 (3.9)
1 Z− Proof. By linearity and the fact that homogeneous polynomials restricted to a sphere can be decomposed as a sum of harmonic polynomials, without loss of generality, we can assume that (A ) and (A ),where. ∈ ∈ Recall Green’s identity
(∆ ∆) = (n n) − − 1 1 Z− Z−
1 where − istheunitopenballcenteredattheorigin, and are both twice- 1 differentiable functions defined on a neighborhood of ,theclosureof − , and are Lebesgue volume measure defined on A and surface area measure 1 on − , respectively. By Green’s theorem, we have
(n n) =0 − 1 Z−
1 Now, for any − , ∈ ( )()= () = () = () n ¯=1 ¯=1 ¯ ¡ ¢¯ ¯ 1 ¯ and by similar reasoning, (n)(¯)= on − ,so¯
0= (n n) = ( ) =( ) − − − 1 1 1 Z− Z− Z− 3. Theory of Harmonic Polynomials 38
Since ,wemusthave =0
1 Z− as desired. Theorem 9 says that a polynomial of degree will be orthogonal to another polynomial of higher degree as long as is harmonic. As an important special case, we have the following Theorem.
Theorem 10 Two homogeneous harmonic polynomials of different degree are 1 1 orthogonal with respect to the inner product (3.8), so that ( − ) and ( − ) are orthogonal as long as = . 6 To close this section, we will give an explicit formula for dim (A ).Note that if =0then 0 (A ) is just the set of constant polynomials, so dim 0 (A )= 1.If =1,then1 (A ) is just the space of all linear polynomials on A ,so dim 1 (A )=.
Theorem 11 (Dimension of (A )) If 2,then ≥
+ 1 + 3 dim (A )= − − (3.10) 1 − 1 µ − ¶ µ − ¶ Proof. First we begin by finding dim (A ). Weknowthatthesetofmono- mials x : =(12) and = is a basis for (A ). Hence, { | | } the dimension of (A ) must be equal to the number of distinct multi-indices such that 1 + 2 + + = for each 0. A basic combinatorial ··· ≥ argument (see Epp [14], for instance) tells us that the number of such solutions is + 1 − 1 µ − ¶ Hence, from the Homogeneous Polynomials Decomposition Theorem, we deduce that
dim (A )=dim (A ) dim 2 (A ) − − + 1 + 3 = − − 1 − 1 µ − ¶ µ − ¶ as desired. 3. Theory of Harmonic Polynomials 39
As a particular example, if =4,wehave
4 +3 +1 2 dim A = =( +1) (3.11) 3 − 3 µ ¶ µ ¶ Now, one may wonder,¡ ¢ given any homogeneous polynomial of degree ,can we find its decomposition represented in (3.6)? If any, is there a finite simple algorithm to find such a decomposition? Although the fact affirms the statements proposed by both questions, we need several basic concepts in hand first.
3.3 The Kelvin Transform
3.3.1 Inversion in the Unit Sphere
When studying functions on unbounded open sets in A, it is often convenient to append a point at infinity to A. From the usual stereographic projection, 1 1 A − is homeomorphic to the unit sphere − in A . Suppose that ∪ {∞} (x)=0precisely when x = 0.ThisworksforQ and real extensions of Q. Consider the map x x∗ where 7→ x if x = 0 (x) 6 ∞ x∗ = 0 if x = ⎧ ∞ ⎨⎪ if x = 0 ∞ 1 This map is called the inversion⎩⎪ of A relative to the unit sphere − .Note that if x = 0 or ,thenx∗ lies on the line through the origin determined by x, 6 ∞ with x 1 1 (x∗)= = (x)= (x)− (x) (x)2 µ ¶ 3.3.2 Maps Preserving Harmonic Functions
The inversion function preserves harmonic functions when =2, as given in the following proposition.
Proposition 5 Let be any harmonic polynomial on some set Ω A2 0 . ⊂ \{ } Then the function : x (x∗) is harmonic on Ω∗ = x∗ : x Ω . 7→ { ∈ } Proof. Let = ( ) be a harmonic polynomial on Ω,then ( )= 2 + 2 2 + 2 µ ¶ 3. Theory of Harmonic Polynomials 40
Now let = and = 2 + 2 2 + 2 so that 2 2 2 2 2 = + +2 2 2 2 µ ¶ µ ¶µ ¶ 2 2 2 + + 2 2 µ ¶ Similarly 2 2 2 2 2 = + +2 2 2 2 µ ¶ µ ¶µ ¶ 2 2 2 + + 2 2 µ ¶ Using the fact that 2 2 2 = − = and = = −(2 + 2)2 − −(2 + 2)2 one may deduce that 2 2 ∆ = + 2 2 2 2 2 2 2 = + + + 2 2 2 µ ¶ õ ¶ µ ¶ ! 2 2 2 +2 + + + 2 2 µ ¶ µ ¶ 2 2 2 + + 2 õ ¶ µ ¶ ! =0
Hence is harmonic. However, the inversion map does not preserve harmonicity for 2.One 1 1 example is (x)= (x) − 2 . This is a rational function when is even. While is harmonic, we see that
1 1 2 x − 1 1 (x)= (x )= = (x) 2 − ∗ (x) µ ¶ which is not in general harmonic. Fortunately there is indeed a transformation which preserves harmonicity for 2,calledtheKelvin transformation. 3. Theory of Harmonic Polynomials 41
Definition 4 Given a polynomial defined on = A 0 ,define the Kelvin \{ } transform () on ∗ by 1 1 ()(x)= (x) − 2 (x∗) Remark 4 Again, note that the transformation above is gives rational function precisely when is even. Note also that is a linear function. One major property of this transformation is that it is its own inverse, as we will see in the following proposition.
Proposition 6 If is a polynomial defined on = A 0 ,then \{ } ( ()) = Proof. Note that
1 1 ( ()(x)) = (x) − 2 (x∗)
³ 1 1 1 ´1 = (x) − 2 (x∗) − 2 ((x∗)∗) 1 1 ³ 1 1 ´ = (x) − 2 (x) 2 − (x) = (x) for every x . Hence ( ()) = . ∈ The Kelvin transform of any harmonic function is harmonic. To see this, we observe the following Lemma.
Lemma 2 If is a homogeneous polynomial of degree on A,then 1 1 1 1 ∆ (x) − 2 − = (x) − 2 − ∆ Proof. TheproofmakesuseoftheproductruleforLaplacians,whichisgiven³ ´ by ∆ ()=∆ +2 + ∆ O · O for any twice continuously differentiable functions and . By letting = 1 1 (x) − 2 − and = ,weget
1 1 1 1 1 1 1 1 ∆ (x) − 2 − = (x) − 2 − ∆ +2 (x) − 2 − + ∆ (x) − 2 − O · O ³ ´ 1 1 ³ ´ 1 ³ ´ = (x) − 2 − ∆ +2(2 2) (x)− 2 − (x ) − − · O 1 + (2 2)( 2) (x)− 2 − − − − 1 1 1 = (x) − 2 − ∆ +2(2 2) (x)− 2 − (x ) − − · O − 1 1 = (x) − 2 − ∆ 3. Theory of Harmonic Polynomials 42
since for any homogeneous polynomial = = x with degree ,wehave | | that P x = · O To see that, note that if is a monomial x where =(12) and = ,wemusthavethat | |
1 1 2 1 1 − 1 2 ··· 1 2 1 ⎛ 2 ⎞ ⎛ 21 2 − ⎞ x = . . ··· ·∇ . · . ⎜ . ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ 1 2 1 ⎟ ⎜ ⎟ ⎜ − ⎟ ⎜ ⎟ ⎜ 1 2 ··· ⎟ 1 2 =(⎝1 +⎠2 +⎝ + ) ⎠ ··· 1 2 ··· =
Hence if is any homogeneous polynomial = = x with degree ,the | | result follows from the linearity of the dot product. P Remark 5 InLemma2above,weusetheanalyticapproachagaintofind the derivative of since might not be a rational function if is odd. We do not have to use analytic approach if is even.
NowwehavecometothemaintheoremabouttheKelvintransform.
Theorem 12 (Preservation of Harmonic Polynomials) If is any poly- nomial in A 0 ,then is harmonic if and only if () is harmonic. \{ } Proof. If is harmonic (i.e. ∆ =0), then ∆ ( ()(x)) = (x)2 ∆ (x) = 0. It then follows that () is harmonic. Now we prove the other direction. ¡ ¢ First suppose is a homogeneous polynomial of degree on A 0 .Then \{ } observe that
1 1 ()(x)= (x) − 2 (x∗)
1 1 x = (x) − 2 (x) µ ¶ 1 1 1 = (x) − 2 (x) (x) 1 1 = (x) − 2 − (x) (3.12) 3. Theory of Harmonic Polynomials 43
With this in mind, we note that
1 1 ∆ ( ()(x)) = ∆ (x) − 2 − (x)
³ 1 1 ´ = (x) − 2 − ∆ (x) = (x)2 ∆ (x) (3.13)
The first equality comes from the previous¡ observation.¢ For the second equality, note that (x)2 ∆ (x) is a homogeneous polynomial of degree +2, hence by applying the reasoning in the previous observation at Equation (3.12), replacing (x) with (x)2 ∆ (x),wefind that
2 1 1 (+2) 2 (x) ∆ (x) = (x) − 2 − (x) ∆ (x) 1 1 ¡ ¢ = (x) − 2 − ∆ (x) which justifies the second equality in Equation (3.13). By this and the linearity of ,weconcludethattheresultistrueforany polynomial. If we have ∆ ()(x) 0 (i.e. if () is harmonic) then it implies ≡ that (x)2 ∆ (x) 0. Applying the Kelvin transform to both sides of the ≡ last equality¡ and using¢ the fact that the Kelvin transform is its own inverse, we may deduce that (x)2 ∆ (x) 0 ≡ Since we have excluded 0 from our domain, then (x) =0, which implies that 6 ∆ =0,provingthat is harmonic.
3.4 Spherical Harmonics via Differentiation
Recall that for any polynomial (x)= x on A ,wecanassociatetoit the differential operator ()= . The following lemma will be useful P to understand the decomposition of homogeneous polynomials. P
1 1 Lemma 3 Suppose 2.If (A ),then () (x) − 2 (A ). ∈ ∈ 1 1³ ´ Proof. We will show first that () (x) − 2 is homogeneous of degree . First we show that the statement³ holds when is´ a monomial by induction. The statement is true when =0,sinceif =0then will just be a constant 3. Theory of Harmonic Polynomials 44 polynomial, say (x)=.Then
1 1 1 1 () (x) − 2 = (x) − 2
³ ´ ³ 1 1 ´ = (x) − 2 ³ 1 1 ´ 1 1 = (x) − 2 (x∗) − 2 = which is homogeneous of degree =0. Now for the inductive part, assume that the proposition is true for some fixed .Let be a multi-index with = , | | and our polynomial is (x)=x. Thisimpliesthat ()= and by our induction hypothesis, there exists (A ) such that ∈ 1 1 (x) − 2 = (x) ³ ´ If we take the Kelvin transform of both sides, we get
1 1 1 1 (x) − 2 = (x) − 2 (x∗) 1 1 = (x) − 2 − (x)
Fix an index 1 2 and differentiate both sides with respect to .We ∈ { } have that
1 1 1 1 1 (x) − 2 =(2 2) (x)− 2 − (x)+ (x) − 2 − − − 1 1 (+1) = (x) − 2 − (2 2) (x)+ (x) − − µ ¶ 1 1 (+1) = (x) − 2 − (x) (3.14) where (x)=(2 2) (x)+ (x) +1 (A ) − − ∈ If we take the Kelvin transform of both sides of the Equation (3.14), we get
1 1 (x) − 2 = (x) µ ¶ 1 1 Since (x) − 2 represents differentiation with any arbitrary multi-index of size +1, this completes the inductive argument. The result for any homogeneous polynomial follows from the linearity of . 3. Theory of Harmonic Polynomials 45
1 1 We still need to prove that () (x) − 2 is harmonic. Note that 1 1 (x) − 2 is harmonic and every partial³ derivative´ of a harmonic function is 1 1 harmonic, hence () (x) − 2 is harmonic. Since the Kelvin transform of any harmonic function is harmonic, this completes the proof of the Lemma. Recall that given any (A ), we have a unique decomposition ∈
= + (x) 2 −
where (A ) and 2 2 (A ). Combined with the Lemma 3, given − − ∈ ∈ any (A ), this polynomial corresponds with two homogeneous harmonic ∈ 1 1 polynomials of degree ,namely and () (x) − 2 .Aswewillseein the next important theorem, these two polynomials³ are related´ to each other, in the sense that they are multiples of each other.
Theorem 13 If 2 and (A ),then ∈ 1 1 () (x) − 2 = ( (x) 2) − − ³ ´ for some 2 2 (A ) and constant which is dependent on and − ∈ − defined by 1 − = (2 2) =0 − − Y By convention, since for =0and =1any polynomial is harmonic, then
= and hence =0.
Proof. The proof is somewhat similar to the proof in the Lemma 3. First, a comment about the constant . In the proof of the Lemma 3, the extra constant (2 2) appears when we move from any monomial of degree to − − +1. Consider the case where isamonomial,sincewecanprovethegeneral result when is any homogeneous polynomial by linearity of . The result is true for =0, since we can choose =0and =1. Suppose that the result is true for some fixed value .Let be a multi-index with size ,and (x)=x. By our induction hypothesis, there exists 2 2 (A ) such that − ∈ − 1 1 (x) − 2 = (x (x) 2) − − ³ ´ Set (x)= (x (x) 2) and note that is a homogeneous harmonic − − polynomial of degree . Applying the Kelvin transform to both sides of the 3. Theory of Harmonic Polynomials 46 equation above, we then have
1 1 1 1 (x) − 2 = (x) − 2 − (x)
Fix an index and differentiate both sides with respect to .Wethenhave,as before,
1 1 (x) − 2
1 1 (+1) 1 = (x) − 2 − (2 2) (x (x) 2)+ (x) − − − − µ ¶ 1 1 (+1) = (x) − 2 − +1 (x (x) ) − for some 1 (A ). Now take the Kelvin transform of both sides of the ∈ − equation above to get
1 1 (x) − 2 = +1 (x (x) ) − µ ¶ Because x represents arbitrary monomial of degree +1, this completes the inductive procedure. In the next theorem we will be able to say something about the algorithm to find the unique decomposition of any homogeneous polynomial mentioned above.
Theorem 14 (Canonical Projection) Suppose 2 and (A ).Then ∈ the canonical projection of onto (A ) is
1 1 () (x) − 2 ³ ´ Proof. From Theorem 13, we can write
1 1 () (x) − 2 = + (x) 2 ³ ´ − We know that the firsttermontherighthandsideisharmonic,hencethisis 1 1 ()(x) − 2 the unique decomposition of mentioned above, so that is the canonical projection of onto (A ). Now let =4and we will shortly see how we can decompose any given homogeneous polynomial of degree . Recall from Equation (3.6) that for 4 ∈ (A ), there is a unique decomposition of the form
2 2 2 2 2 2 2 2 = + + + + 2 + + + + + 2 − ··· − ¡ ¢ ¡ ¢ 3. Theory of Harmonic Polynomials 47
4 where = 2 and (A ). From the Canonical Projection Theorem, we ∈ 4 can compute by using this simple algorithm. First, given any (A ),we ¥ ¦ ∈ can compute explicitly by using the previous theorem. Once we find ,we 4 move to the second step, that is to find 2 2 (A ) in the decomposition − ∈ − = + (x) 2.Wethenfind 2 by using the same procedure described − − above, with 2 in place of this time. After iterating the process more − times, we will arrive at 2 and thus complete the decomposition process. − Let us observe the decomposition process more closely. We now give some examples that illustrate this decomposition. These examples are original.
2 4 Example 8 Consider ( )= 2 (A ). It is not harmonic, since ∈ ∆ =2=0.Since =2,wewouldexpecttofind two harmonic polynomials, 6 namely 2 and 0. Based on Theorem 14,
1 () (x)− = 2 ¡ 24 ¢ 1 1 = 8 2 2 + 2 + 2 + 2 µ µ ¶¶ 1 2 3 = 2 2 + 2 + 2 + 2 − +82 2 + 2 + 2 + 2 − 8 − 1 ³ ¡ 2 2 ¢ 3 ¡ ¢ ´ = (x)− + (x)− −4 2 2 3 1 ¡ 1 2 ¢ x ¡ − ¢ 1 2 x − = (x) − + (x) − −4 (x) (x) (x) µ ¶ µ ¶ µ ¶ 1 = (x)+2 −4 1 = 2 + 2 + 2 + 2 + 2 −4 1 3 1 1 = ¡2 + 2 2 ¢2 −4 4 − 4 − 4
The next step is to find 0. From the decomposition = 2 + (x) 0,wesee that 2 1 2 3 2 1 2 1 2 2 2 2 2 = + + + + + 0 −4 4 − 4 − 4 1 ¡ ¢ which means that 0 = 4 .Fromthis,wecanfind 0 using the algorithm above. 1 But since 0 is already harmonic, then 0 = 4 . Hence, the unique decomposition of in this case is 1 3 1 1 1 (x)=2 = 2 + 2 2 2 + (x) −4 4 − 4 − 4 4 µ ¶ 3. Theory of Harmonic Polynomials 48
Hence, on 3, 1 3 1 1 1 (x)= 2 + 2 2 2 + −4 4 − 4 − 4 4 3 4 Example 9 Consider ( )= 4 (A ).Since =4, we need to find ∈ three harmonic polynomials, namely 4, 2,and0.Now, 1 () (x)− = 4 ¡ 44 ¢ 1 4 1 = 384 3 2 + 2 + 2 + 2 µ µ ¶¶ 1 3 5 4 = 384 (x)− 144 (x)− 384 − 3 ¡ 5 3 4 ¢ = (x)− (x)− − 8 3 5 ¡ 1 ¢ ¡ ¢ x − = (x)− 2 + 2 + 2 + 2 2 + 2 + 2 + 2 (x) µ ¶ µ 4 ¶ µ ¶ 3 1 x − (x)− − 8 (2 + 2 + 2 + 2)2 (x) µ ¶ µ ¶ 3 = 3 (x) − 8 and hence since
= 4 + (x) 2 3 we determine that 2 = 8 . We do the same procedure to find 2.Notethat2 is already a harmonic polynomial, hence 2 = 2 and that 0 0 so that 0 0. ≡ ≡ In other words, 3 3 = 3 (x) + (x) + (x)2 (0) − 8 8 µ ¶ µ ¶ is the desired decomposition.
We will give some more examples without giving the computation process.
1 3 2 2 Example 10 If (x)=( )= 6 ,thenwecanwrite as 3 +( + + 2 2 + )1 where
1 3 1 2 1 2 1 2 1 3 = and 1 = 12 − 12 − 12 − 12 12 Hence, on 3, 1 1 1 1 1 1 3 = 3 2 2 2 + 6 12 − 12 − 12 − 12 12 3. Theory of Harmonic Polynomials 49
1 4 Example 11 If (x)=( )= 24 , then we can write as 2 2 2 2 2 2 2 2 2 = 4 +( + + + )2 +( + + + ) 0 where
5 4 1 4 1 4 1 4 5 2 2 4 = + + + 384 384 384 384 − 192 5 5 1 1 1 22 22 + 22 + 22 + 22, − 192 − 192 192 192 192 3 2 1 2 1 2 1 2 1 2 = + + + and 0 = −64 64 64 64 −64 Thus, restricted to 3, 1 5 1 1 1 5 4 = 4 + 4 + 4 + 4 22 24 384 384 384 384 − 192 5 5 1 1 1 22 22 + 22 + 22 + 22 −192 − 192 192 192 192 3 1 1 1 1 2 + 2 + 2 + −64 64 64 64 − 64
3.5 Bases for Harmonic Polynomials: Analytic Approach
In previous section we have seen that any polynomial of degree can be written as a sum of homogeneous polynomials of degree 0 , and restricted to ≤ ≤ the sphere, any homogeneous polynomial of degree admits a nice decomposi- tion into harmonic polynomials given in (3.7). Thus, one may view harmonic polynomials as building blocks for the space of polynomials on the sphere. It seems reasonable to consider bases for the space of harmonic polynomials on the sphere. We have followed a derivation that works for the field R, although clearly some aspects can be viewed more generally. Recall from the Canonical Projection Theorem, that the canonical projection of (A ) to (A ) is given by 1 1 () (x) − 2 (3.15) 7→ ³ ´ For any homogeneous polynomial of degree ,wecanwrite
1 1 () (x) − 2 = + (x) 2 ³ ´ − 3. Theory of Harmonic Polynomials 50
1 for some 2 (A ). If we restrict both sides to − ,wewillget ∈ − 1 1 () (x) − 2 = + (x) 2 | ⎡ ³ ´ − ⎤¯ ¯ ¯ ⎣ 1 1 ⎦¯ () (x) − 2 ¯ ¯ = + 2 ³ ´¯ − | ¯ ¯ 1 1 ¯ () (x) − 2 ¯ = + ¯2 (3.16) − | Recall from (3.9) that any two polynomials of different degrees will be orthogo- nal, with respect to the inner product (3.8). Combining with what we already 1 developed so far in (3.16), we conclude that 2 is orthogonal to ( − ). − | We have now the following theorem by taking the orthogonal projection of both 1 sides onto ( − ).
1 Theorem 15 The orthogonal projection of onto ( − ) is | 1 1 () (x) − 2 (3.17) Corollary 5 Suppose 2 and (A ),thentheset ∈ 1 1 (x) − 2 : = | | n ³ ´ o spans (A ),andsimilarly,theset
1 1 (x) − 2 : = | | 1 n o spans ( − ). Proof. The proof is straightforward. The first statement follows by choosing the monomial (x)=x of degree . The second statement follows by restricting 1 1 1 1 1 (x) − 2 to − to get (x) − 2 . 1 ³Given that these´ sets are spanning sets for (A ) and ( − ) respec- 1 tively, one may want to find the basis of (A ) and ( − ) in terms of subsets of 1 1 (x) − 2 : = | | and n ³ ´ o 1 1 (x) − 2 : = | | respectively. n o 3. Theory of Harmonic Polynomials 51
1 Theorem 16 (Basis for (A ) and ( − )) If 2, then the set
1 1 (x) − 2 : = and 1 1 | | ≤ n ³ ´ o is a basis for (A ),andtheset
1 1 (x) − 2 : = and 1 1 | | ≤ n o 1 is a basis for ( − )
1 1 Proof. Let = (x) − 2 : = and 1 1 .Wewillshowthat | | ≤ 1 1 n ³ ´ 2 o spans (A ) by showing that (x) − is in the span of for every multi-index of size .If1 =0or ³1 =1, then by´ the definition of ,wehave 1 1 that (x) − 2 . We now proceed by induction on 1. Assume that ∈ 1 1 1 1 and³ that ´ (x) − 2 is in the span of for all multi-indices of 1 1 degree whose first³ components are´ less than 1.Since (x) − 2 is a harmonic 1 1 function, then ∆ (x) − 2 =0,sowehavethat
1 1 1 1 2 2 2 1 (x) − 2 = − (x) − 2 1 2 ··· 1 ³ ´ ³ ³ ´´ 1 1 2 2 2 1 2 = 1 − 2 (x) − − Ã ··· Ã =2 !! X 1 1 2 2 2 1 2 = 1 − 2 (x) − − =2 ··· X ³ ³ ´´ By our induction hypothesis, each of the terms in the last line is in the span of therefore spans (A ). Next it will be shown that
dim ( (A )) | | ≤ where denotes the cardinality of .Since | | 1 1 = (x) − 2 : = and 1 1 | | ≤ n ³ ´ o thenumberofelementsin is at most the number of non-negative integer solutions of
1 + 2 + + = ··· 3. Theory of Harmonic Polynomials 52
where 1 1.If1 =0,then2 + + = so by a basic combinatorial ≤ ··· argument, the number of solution is
+( 1) 1 + 2 − − = − ( 1) 1 2 µ − − ¶ µ − ¶
If 1 =1,then2 + + = 1, so the number of the solutions is ··· − ( 1) + ( 1) 1 + 3 − − − = − ( 1) 1 2 µ − − ¶ µ − ¶ Hence,
+ 2 + 3 − + − | | ≤ 2 2 µ − ¶ µ − ¶ + 1 + 3 = − − 1 − 1 µ − ¶ µ − ¶ =dim( (A ))
wherewehavereferredtoEquation(3.10).Since is a spanning set for (A ), we have that dim (A ), but at the same time dim (A ),sowe | | ≥ ≤ must have that =dim (A ) | | This proves that is a basis for (A ).Now,since is a basis for (A ), 1 by restricting every every element in in − , it follows that
1 1 (x) − 2 : = and 1 1 | | ≤ n o 1 is a basis for ( − ). It is useful to see explicitly what the sets
1 1 (x) − 2 : = and 1 1 | | ≤ n ³ ´ o and 1 1 (x) − 2 : = and 1 1 | | ≤ look like in the case n=4. Below we give one example. o
Example 12 If =2,then =(1234) where 1 + 2 + 3 + 4 =2, so there are nine possibilities: 3. Theory of Harmonic Polynomials 53
1. If =(0 2 0 0),then
1 1 2 2 1 (x) − = 2 (x)− 2 2 3 ³ ´ = ¡ 2 (x)− ¢+8 (x)− − 2 ¡ 1 2 ¢ 3 = (x)− 2 (x∗)− +8 (x∗)− − () Ã µ ¶ ! 2 2 3 1 x − 8 x − = (x)− 2 + − (x) (x)2 (x) Ã µ ¶ µ ¶ ! = 2 (x)+82 − = 22 +62 22 22 − − − 2. If =(0 0 2 0), then similarly,
1 1 2 2 1 (x) − = 3 (x)− ³ ´ = 2¡2 22 +6¢2 22 − − − 3. If =(0 0 0 2),then
1 1 2 2 1 (x) − = 4 (x)− ³ ´ = 2¡2 22 2¢2 +62 − − − 4. If =(0 1 1 0) then
1 1 1 (x) − 2 = 23 (x)− 3 ³ ´ = ¡8 (x)− ¢ 3 ¡ 1 ¢ x − = (x)− 8 (x) (x) (x) à µ ¶µ ¶ µ ¶ ! =8
5. If =(0 1 0 1) then
1 1 1 (x) − 2 = 24 (x)− ³ ´ =8¡ ¢
6. If =(0 0 1 1) then
1 1 1 (x) − 2 = 34 (x)− ³ ´ =8¡ ¢ 3. Theory of Harmonic Polynomials 54
7. If =(1 1 0 0) then
1 1 1 (x) − 2 = 12 (x)− ³ ´ =8¡ ¢
8. If =(1 0 1 0) then
1 1 1 (x) − 2 = 13 (x)− ³ ´ =8¡ ¢
9. If =(1 0 0 1) then
1 1 1 (x) − 2 = 14 (x)− ³ ´ =8¡ ¢
4 Hence a basis for 2 (A ) is
22 +62 22 22, 22 22 +62 22, − − − − − − 22 22 22 +62, 8, 8, 8, 8, 8, 8 ( − − − ) 3 2 1 2 2 3 2 Now since on we have 2 (x)− = 2 (x)− +8 (x)− =8 2 3 − 3 − and 8 (x)− =8, then a basis for 2 ( ) is
82 2, 82 2, 82 2, 8, 8, 8, 8, 8, 8 − − − © ª Chapter 4
Theory of Harmonic Polynomials: Algebraic Approach
The standard theory of spherical functions that leads to the theory of harmonic polynomials developed in the previous chapter has given us a way to compare with the algebraic approach we will shortly see in this chapter. We will first discuss harmonic polynomials on 2 by using the idea of a factorial basis which then can be arranged geometrically using triangular patterns in the plane. This approach was first introduced by Wildberger [31] in his preprint to study spheri- cal harmonic polynomials on a low-dimensional sphere. The core of this thesis is to extend the theory to the case of 3 to explain the associated harmonic polyno- mials purely in an algebraic manner. Finally, we will talk about the rotationally invariant harmonic polynomials on 3 which may be interpreted as characters on the Lie group 3 (2). Most of the results here are more general, and ' work also over the rational numbers, or more general field. To start with, we remind the reader about the concept of factorial and double factorial. They will be used quite often throughout this chapter and in Appendix B.
Definition 5 For any non-negative integer ,thefactorial of ,denotedby!, is given by != ( 1) 1 × − ×···× if is positive, and we define 0! = 1. The double factorial of ,denotedby!!,
55 4. Theory of Harmonic Polynomials: Algebraic Approach 56 is given by ( 2) ( 4) 1 if is odd, !! = × − × − ×···× ( 2) ( 4) 2 if is even ( × − × − ×···× for 0,andwealsodefine 0!! to be 1.
The factorial of can be thought of as the function from the set of non- negative integers Z 0 to itself. Over the field R, there is an analytic continuation ≥ of the factorial function so that at = , it gives the value of !,butwecan define the value at non-integer value of . See Artin [5, Chapter 2] or Rudin [26, Chapter 8] for further explanations.
Definition 6 If = R, there exists a continuous function Γ that satisfies Γ (0) = 1 and Γ ( +1) = Γ () for R. Such a function is called the ∈ gamma function.
We will show that the gamma function Γ is related to the double factorial function.
Proposition 7 If is a positive integer, then 1 (2 1)!! 1 Γ + = − Γ 2 2 2 µ ¶ µ ¶ Proof. The proof is simple and uses the basic property of gamma function. Note that since Γ ( +1)=Γ () for any R,wemusthavethat,for 1, ∈ ≥ 1 1 1 Γ + = Γ 2 − 2 − 2 µ ¶ µ ¶ µ ¶ Recursively, we also have that 1 3 3 Γ = Γ − 2 − 2 − 2 µ ¶ µ ¶ µ ¶ so that 1 1 3 3 Γ + = Γ 2 − 2 − 2 − 2 µ ¶ µ ¶µ ¶ µ ¶ If we follow this process, after a finite number of times we will have that 1 1 3 1 1 Γ + = Γ 2 − 2 − 2 ···2 2 µ ¶ µ ¶µ ¶ µ ¶ (2 1) (2 3) 1 1 = − − Γ 2 × 2 ×···× 2 2 µ ¶ (2 1)!! 1 = − Γ 2 2 µ ¶ 4. Theory of Harmonic Polynomials: Algebraic Approach 57 as desired.
4.1 Harmonic Polynomials on 2
We are now going to use a slightly modified notation for monomials which will turn out to be very useful. This was introduced in Wildberger’s preprint [31].
Definition 7 For any integer , and any variable ,define the notation 1 ≡ ! 1 We define the term ! to be zero if is negative.
Since we are going to investigate polynomials (x) where x 2 A3,it ∈ ⊂ is convenient to let x =( ) instead of (123).Denoteby the monomial !!! and observe that whenever either , ,or is negative, then the above monomial will be automatically zero. Note that : + + = and 0 ≥ 3 3 givesusabasisfor©(A ).Wewillcallthisbasisthefactorialª basis of (A ). We can arrange these basis monomials in a triangle which will be denoted as the level- triangle, as shown in the following examples for =1, 2,and4.Here we set the convention that the distance between each vertex on the side of the triangle is 1. Figure (4.1) shows the level-1 triangle in factorial basis. The next one shows the level-2 triangle and level-4 triangle, as shown in Figure (4.2) and Figure (4.3). Any homogeneous polynomial of degree can be identified by a level- tri- angle of coefficients with respect to the factorial basis of (A ).Forexample look at the triangle in Figure (4.4) below. The triangle in Figure (4.4) corresponds to the polynomial
( )=2 3 +22 2 +4 +42 − − 1 = 2 3 + 2 2 +4 +22 2 − − The action of the Laplacian ∆ is particularly pleasant in the factorial basis. 4. Theory of Harmonic Polynomials: Algebraic Approach 58 z
x y
Figure 4.1: Level 1 triangle in factorial basis z2
xz yz
x2 xy y2
Figure 4.2: Level 2 triangle in factorial basis
Lemma 4 For any integer
2 2 2 ∆ = − + − + − (4.1)
Proof. The proof is computational. The equality in (4.1) is obviously true when either one of , ,or is less than 2. Thus, we may safely assume that all of them are at least 2.Wehave 1 ∆ = ∆ !!! 1 2 2 2 = ( 1) − + ( 1) − + ( 1) − !!! − − − 1 1 1 = ¡ 2 + 2 + ¢2 ( 2)!!! − !( 2)!! − !!( 2)! − − − − 2 2 2 = − + − + − which completes the proof. 4. Theory of Harmonic Polynomials: Algebraic Approach 59
z 4
xz3 yz 3
2 x2z y2z 2 xyz2
x3z x2yz xy2z y3z
x4 x3y x2y2 xy3 y4
Figure 4.3: Level 4 triangle in factorial basis
Now we give a geometrical interpretation of the Lemma 4, which was de- scribed in Wildberger [31]. First imagine stacking the level- triangles of fac- torial basis elements succesively below each other to form a pyramidal packing with =0at the top, =1beneath that, =2beneath that, and so on. If each element is considered to be contained in a sphere, we get a regular spherical packing with each interior sphere touching 6 adjacent spheres on its own level, 3 spheres in the level beneath it and 3 spheres in the level above it. Lemma 4 tells that if we have a homogeneous polynomial of degree , represented in the factorial basis as
( )= ++= X then for each monomial , the Laplacian applied to that monomial sends the coefficient in the level- triangle to the three positions corresponding 2 2 2 to the monomials − , − ,and − in the level-( 2) triangle − 2 levels above it. This leads to the conclusion that is harmonic precisely when in the level- triangle representation of , the sum of all coefficients in every subtriangle of side 2 is zero. For example, in the case of the polynomial presented in Figure (4.4) above, the only subtriangle of side 2 is the triangle with vertices 2, 2,and2,with the corresponding coefficients 1, 2,and4.Since1+2+4 =0,then is not 6 harmonic. Let us give one more example. 4. Theory of Harmonic Polynomials: Algebraic Approach 60
2 4 z
xz -2 4 yz
x2 y2 1 -3xy 2
Figure 4.4: Example of polynomial in level 2 triangle
In the above level-4 triangle, there are 6 subtriangles with side 2 and in every subtriangle, the sum of the coefficients corresponding to each vertex must be zero in order to be harmonic. In this particular example, if
4 3 3 2 2 2 2 2 3 ( )=1 + 2 + 3 + 4 + 5 + 6 + 7 + 2 2 3 4 3 8 + 9 + 10 + 11 + 12 + 2 2 3 4 13 + 14 + 15 then will be harmonic precisely when
1 + 4 + 6 =0 ⎧ 2 + 7 + 9 =0 ⎪ + + =0 ⎪ 3 8 10 ⎪ ⎪ 4 + 11 + 13 =0 ⎨⎪ 5 + 12 + 14 =0 ⎪ ⎪ 6 + 13 + 15 =0 ⎪ ⎪ In this particular example,⎩⎪ we have 15 variables and 6 equations. That gives 3 9 degrees of freedom, so the dimension of 4 (A ) must be 9. In general, in the level triangle, the number of variables is just 1+2+ +( +1)= − ··· 4. Theory of Harmonic Polynomials: Algebraic Approach 61
z 4
a1
xz3 yz 3 a2 a3
2 2 x z a4 a5 a6 y 2z 2 xyz2
x 3z x 2yz 2 y 3z a7 a8 a9 xy z a10
a11 a12 a13 a14 a15 x4 x 3y x 2y2 xy3 y4
Figure 4.5: Level 4 triangle with coefficients in factorial basis
1 ( +1)( +2)and there are 1 ( 1) subtriangles with side 2,so 2 2 −
3 1 1 dim A = ( +1)( +2) ( 1) 2 − 2 − ¡ ¢ =2 +1 (4.2)
This can be justified as well by putting =3to Equation (3.10), so this gives the geometric meaning to the expression dim ( (A )).
4.1.1 Complex Factorial Basis
Now, introduce two variables
+ = and = − = 2 2 With this substitution, since we can express and in terms of and ,itis clear that any homogeneous polynomial in , ,and can be not only expressed in terms of , ,and, but they also have the same degree as . It means that the set : + + = © ª 4. Theory of Harmonic Polynomials: Algebraic Approach 62
z2
uz vz
u2 uv v 2
Figure 4.6: Level 2 triangle in complex factorial basis
3 3 serves as a basis for (A ). Wecallitthecomplex factorial basis of (A ). We can still arrange the elements of this set into a level- triangle as before, as shown in Figure (4.6). The only difference is that and are replaced by and , respectively. As for the action of the Laplacian of any element in the complex factorial basis, it is even simpler.
Lemma 5 For any integer , ,and,wehave
1 1 2 ∆ = − − + − (4.3) Proof. Note that 2 2 = 2 !!! 2 ¡ ¢ (¡ 1)¢ 2 1 1 ( 1) 2 = − − + − − + − − !!! 4 2 4 µ ¶ and by the same principle, 2 ( 1) 2 1 1 ( 1) 2 = − − + − − − − 2 !!! − 4 2 − 4 µ ¶ so that¡ ¢ 2 2 + = 1 1 2 2 !!! − − µ ¶ ¡ ¢ ¡1 1 ¢ = − − ( 1)! ( 1)!! −1 1 − = − − 4. Theory of Harmonic Polynomials: Algebraic Approach 63
Now since 2 2 1 = 2 2 ! µ ¶ ¡ ¢ 2 = − ( 2)! µ 2 − ¶ = − it follows that 2 2 2 ∆ = + + 2 2 2 µ 1 1 ¶2 = − − + − as desired.
Remark 6 Note that the proof above implicitly implies that 2 2 2 + = 2 2 since for every monomial ,wehavethat 2 2 2 + = 1 1 = 2 2 − − µ ¶ ¡ ¢ ¡ ¢ 3 Lemma 5 tells us that if we represent a homogeneous polynomial (A ) ∈ by its triangle of coefficients in the complex factorial basis, then is harmonic if and only if in every vertical column the entries have constant absolute value and 3 are alternating in sign as we move up (or down). The basis elements of (A ) are then represented by vertical columns in the level- triangle, consisting of al- ternating plus and minus ones. A vertical column in called non-degenerate if it consists of more than one entries. A vertical column that is not non-degenerate is called degenerate. Looking back at Figure (4.6), there are five vertical columns, the ones that start with 2, , 2, ,and2. All of the columns other than that which starts with 2 consistofjustonetermwhichisitself.Thecolumnthat starts with 2 represents the polynomial 2 . Consider Figure (4.7). The only − non-degenerate vertical column is marked with an arrow and that corresponds to 2 . − 3 Hence, a basis for 2 (A ) is
22 2 . − © ª 4. Theory of Harmonic Polynomials: Algebraic Approach 64 z2
uz vz
u2 uv v 2
Figure 4.7: Columns in level 2 triangle in complex factorial basis
By the same reasoning, there are seven vertical columns in the level-3 triangle in the complex factorial basis. All of the columns other than which start with 2, 3,and2 consist of just one term which is itself. The column that starts with 2, 3,and2 represents the basis element 2 2 , 3 ,and2 2, 3 − − − respectively. Hence a basis for 3 (A ) is
322 23 2 223 − − − © ª We provide one more example. There are nine vertical columns in the level- 4 triangle in the complex factorial basis. We mark the columns consisting of two and three entries by arrows, so the columns with two entries correspond to polynomials 22 3, 3 2, 3 2,and22 3.Thecolumn − − − − consisting of three entries corresponds to 4 2+22. The other four vertical − columns are degenerate. See Figure (4.8). 3 Hence a basis for 4 (A ) is 4, 3, 22 3, 3 2, 4 2 + 22, − − − 3 2, 22 3, 3, 4 ( − − ) Let be the vertical column in the level-( + + ) triangle in complex factorial¡ basis¢ that has as one of its entries. Thus, is the same column as +1+1 2 .Denoteby the polynomial that is represented − ¡ ¢ by ¡.WehavethefollowingTheorem.¢ £ ¤ ¡ ¢ 4. Theory of Harmonic Polynomials: Algebraic Approach 65
z4
uz3 vz3
u 2z2 uvz2 v 2z2
u 3z u 2vz uv 2z v 3z
u4 u 3v u 2v2 uv 3 v 4
Figure 4.8: Columns in level 4 triangle in complex factorial basis
3 Theorem 17 A basis for ( ) is
, 1 , 22 , ..., 1 , , − − − 1 , ..., 22 , 1 , ( £ ¤ £ − ¤ £ − ¤ £− ¤ £ ¤ ) £ ¤ £ ¤ £ ¤ £ ¤ 4.1.2 Zonal Harmonic Polynomials on 2
In this subsection, we are going to study one special case of harmonic polynomials on 2, called the zonal harmonic polynomials. To do that, we need a few concepts in hand first. Let A3 be the usual three-dimensional affine space over a field F and a particular point x A3 be ( ). In the classical theory when F = R, ∈ the parametrization of the sphere 2 in the Euclidean space R3 distinguishes the direction. Let cos sin 0 − ()=⎛sin cos 0⎞ 001 ⎜ ⎟ ⎝ ⎠ 4. Theory of Harmonic Polynomials: Algebraic Approach 66 be the subgroup of the rotation matrices in three-dimensional space SO (3).We say a polynomial on (R3) transforms by if
x () = (x) ³ ´ for all [0 2) and all relevant x =( ).Here Z. The case =0 ∈ ∈ means that is unchanged by the rotation ().
Definition 8 Apolynomial (R3) is called zonal precisely when is un- ∈ changed by the rotation ();inotherwords,
x () = (x) ³ ´ for every x R3. ∈ Now, define
3 3 R = R : x () = (x) for all [0 2) ∈ ∈ ¡ 3¢ n ¡ ¢ ³ ´ o Since (R ) is finite-dimensional, we must have
3 3 R = R (4.4) Z ¡ ¢ M∈ ¡ ¢ However, in the following Lemma we will see that the infinite direct sum in (4.4) only has finitely many nonzero term.
Lemma 6 For any N, ∈ 3 3 R = R = ¡ ¢ M− ¡ ¢ Proof. Suppose is of degree . Then if we see the polynomial x () | | as a polynomial of ,then is a trigonometric polynomial of degree³ at most ´ in terms of cos and sin . Such a polynomial will be orthogonal to under the usual inner product
2 1 = () 2 Z0 ® so = 0 if . { } | | 4. Theory of Harmonic Polynomials: Algebraic Approach 67
Since the zonal polynomials are unchanged by the rotation along the -axis, then combined with the theory of harmonic polynomials developed in the previ- ous subsection, the unique zonal harmonic homogeneous polynomial of degree must correspond to . Call this zonal harmonic ,thenwehavethat
£ ¤ 2 2 2 4 = − + − (4.5) − − ··· which ends in ( 1) if is odd, or if is even. Here = .We − 2 can rewrite (4.5) as ¥ ¦ 2 b c 2 = ( 1) − (4.6) − =0 X If we expand the relation above and resubstitute and back, we have that
2 b c ( 1) 2 = − () − !!( 2)! =0 − X 2 2 2 b c 1 + 2 = ( 1) − (4.7) − !!( 2)! 4 =0 X − µ ¶ Here are the first few values of :
0 =1,
1 = , 2 1 2 1 2 2 2 = = + , − 2 − 4 3 1 3 1¡ 2 ¢2 3 = = + , − 6 − 4 4 2 2 2 ¡1 4 1¢ 2 2 2 1 2 2 2 4 = + = + + + − 24 − 8 64 ¡ ¢ ¡ ¢ Now, for each , we can express as a polynomial in a single variable because on 2 we have 2 + 2 =1 2. Another way of describing the zonal harmonic − is by rewriting the Equation (4.7) as
2 b c 1 1 2 2 = 1 − (4.8) −4 !!( 2)! − =0 µ ¶ − X ¡ ¢ Hence, the first few values of in terms of only are 4. Theory of Harmonic Polynomials: Algebraic Approach 68
0 =1
1 = 1 2 2 = 3 1 4 − 1¡ 3 ¢ 3 = 5 3 12 − 1 ¡ 4 ¢ 2 4 = 35 30 +3 192 − The polynomials above have a neat¡ description in¢ terms of Legendre poly- nomials, one of many special polynomials. Here we present our own proof to the following important Theorem.
Theorem 18 For each 0, the polynomial above satisfies ≥ 1 ()= () ! where () is the -th Legendre polynomial. TheproofoftheTheoremmakesuseofthefollowingLemma.
Lemma 7 For any non-negative integer and non-negative integer such that ,wehavethat ≤ 2 ¥ ¦ 2 b c ! (2 2)! = − 4!( 2)! ( )! 2 ( )! ( 2)! = X − − − − Proof. The equation is definitely true for =0. Hence, assume that 1.By ≥ shifting the index in the summation in the left hand side, we get that
b 2 c ! b 2 c− ! = 4!( 2)! ( )! 4+ ( + )! ( 2 2)!! = =0 X − − X − − Now we want to follow Zeilberger’s approach in finding a neat formula to the above sum in terms of hypergeometric functions. The algorithm is given in [24, Chapter 3]. Since the first term in the above sum is not 1,wewrite
b 2 c− ! 4+ ( + )! ( 2 2)!! =0 X − − ! b 2 c− !( 2)! = − 4!( 2)! 4 ( + )! ( 2 2)!! =0 − X − − 4. Theory of Harmonic Polynomials: Algebraic Approach 69 so that the firstentryinthelastsumis1. Now we can express the sum
b 2 c− !( 2)! − (4.9) 4 ( + )! ( 2 2)!! =0 X − − in terms of a hypergeometric function. Define !( 2)! = − 4 ( + )! ( 2 2)!! − − so that +1 4 ( + )! ( 2 2)!! = +1 − − 4 ( +1+)! ( 2 2 2)! ( +1)! − − − ( 2 2)( 2 2 1) = − − − − − 4( + +1)( +1) 2 2 1 − − − = − 2 − 2 ( + +1)( +1) ¡ ¢¡ ¢ Hence the sum (4.9) corresponds to the hypergeometric function ! 1 1 2 1 ( 2) ( 2 1) ; +1;1 (4.10) 4!( 2)! −2 − −2 − − − µ ¶ Recall Gauss’ 21 identity given below for rational numbers , ,and. See, for instance Gradshteyn [16, Chapter 9] or Andrews [4, Chapter 2]. Γ ( ) Γ () 21 ( ; ;1)= − − (4.11) Γ ( ) Γ ( ) − − as long as 0. In this case, − − 1 1 = ( 2) = ( 2 1) and = +1 −2 − −2 − − Since = + 1 0,wemusthavethat − − − 2 1 1 1 Γ + 2 Γ ( +1) 21 ( 2) ( 2 1) ; +1;1 = − −2 − −2 − − Γ 1 +1 Γ 1 + 1 µ ¶ ¡ 2 ¢ 2 2 so Equation (4.10) becomes ¡ ¢ ¡ ¢ ! Γ + 1 Γ ( +1) − 2 4!( 2)! × Γ 1 +1 Γ 1 + 1 − ¡ 2 ¢ 2 2 ! Γ + 1 ! = ¡ − ¢ ¡2 ¢ 4!( 2)! × Γ 1 +1 Γ 1 + 1 − 2¡ 2 ¢ 2 ! Γ + 1 = ¡ − ¢ ¡2 ¢ (4.12) 4 ( 2)! × Γ 1 +1 Γ 1 + 1 − 2 ¡ 2 ¢ 2 ¡ ¢ ¡ ¢ 4. Theory of Harmonic Polynomials: Algebraic Approach 70
Proposition 7 states that if isapositiveinteger,wehavethat 1 (2 1)!! 1 Γ + = − Γ 2 2 2 µ ¶ µ ¶ We divide into 2 cases. If is even, write =2 for some Z. Hence we have ∈ that Equation (4.12) reduces to ! (2 2 1)!!Γ 1 1 − − 2 4 ( 2)! × 2 × Γ ( +1)Γ + 1 − − ¡ ¢ 2 (2 2 1)!!Γ 1 ! 2 2 ¡ ¢ = − − 1 4 ( 2)! × 2 − × !(2 1)!!Γ − ¡ ¢ − 2 ! (2 2 1)!! 2 = − − ¡ ¢ 4 ( 2)! × 2 × !( 1)!! − − − 1 ! = (2 2 1)!! ( 2)! × 22+ !( 1)!! × − − − − − − 1 ( 2) 2 = × − ×···× (2 2 1)!! 2 ( 2)! × 2! × − − − 1 2 (2 2) 2 (2 2)! = × − ×···× − 2 ( 2)! × 2! × (2 2) (2 2 2) 2 − − × − − ×···× 1 ( 1) 1 (2 2)! = × − ×···× − 2 ( 2)! × ! × 2 ( ) ( 1) 1 − − − × − − ×···× 1 (2 2)! = − 2 ( 2)! × 2 ( )! − − − (2 2)! = − 2 ( 2)! ( )! − − as desired. If is odd, write =2 +1 for some Z. Hence the Equation (4.12) ∈ simplifies to ! (2 2 1)!!Γ 1 1 − − 2 4 ( 2)! × 2 × Γ + 3 Γ ( +1) − − ¡ ¢ 2 ! (2 2 1)!!Γ 1 2+1 = − − 2 ¡ ¢ 4 ( 2)! × 2 × (2 +1)!!Γ 1 ! − − ¡ ¢ 2 (2 +1)! (2 2 1)!! 2+1 = − − ¡ ¢ 22 ( 2)! × 2 × (2 +1)!!! − − 1 (2 +1)! = (2 2 1)!! ( 2)! × 22+ 1 (2 +1)!!! × − − − − − − 1 (2) (2 2) 2 = × − ×···× (2 2 1)!! 2 ( 2)! × 2! × − − − 4. Theory of Harmonic Polynomials: Algebraic Approach 71 which, as above, reduces to (2 2)! − 2 ( 2)! ( )! − − Sincethestatementistrueforthecase even and odd, we conclude that
b 2 c ! (2 2)! = − 4!( 2)! ( )! 2 ( )! ( 2)! = X − − − − as desired.
ProofofTheorem18. Let ()=! (). Based on (4.8), we have that
2 b c 1 ! 2 2 ()= 1 − −4 !!( 2)! − =0 µ ¶ − X ¡ ¢ 2 b c ! 2 2 = 1 − 4 (!)2 ( 2)! − =0 − X ¡ ¢ 2 b c ! 2 2 2 = ( 1) − − 4 (!)2 ( 2)! − =0 − Ã=0 µ ¶ ! X X 2 b c ( 1) ! 2 = − − 4 (!)2 ( 2)! =0 =0 − µ ¶ X X 2 b c ( 1) ! 2 = − − 4!( 2)! ( )!! =0 =0 X X − − By changing the order of summation, we will have
2 2 b c b c ( 1) ! 2 ()= − − 4!( 2)! ( )!! =0 = X X − − 2 2 2 b c ( 1) − b c ! = − (4.13) ! ⎛ 4!( 2)! ( )!⎞ =0 = − − X ⎜X ⎟ ⎝ ⎠ Incorporating Lemma 7 above, we see that the Equation (4.13) above reduces to
2 2 b c ( 1) − (2 2)! ()= − − ! × 2 ( )! ( 2)! =0 − − X 2 1 b c (2 2)! 2 = ( 1) − − 2 !( )! ( 2)! =0 − X − − 4. Theory of Harmonic Polynomials: Algebraic Approach 72 which is one of the representation of Legendre polynomials. See Silverman [28, Chapter 4] for further references.
4.1.3 Generating Function
2 Now that we have explicitly found our zonal harmonic polynomials on ,it is interesting to consider the generating function of ;thatis,wewouldliketo find an closed, explicit form of
∞ ()= =0 X
Basedontheformulaof given in (4.5), we get
()=1+ + 2 2 + 3 3 + 4 2 + 22 4 + − − − ··· 2 2 3 3 2 2 2 4 = 1+ +¡ + ¢ +¡ 1 ¢ +¡ ¢ ··· − − ··· ∞ ∞ = ¡ ( 1) ()¢¡2 ¢ − Ã=0 !Ã=0 ! X X ∞ ∞ ( 1) = − () 2 ! !! Ã=0 !Ã=0 ! X X ∞ ( 1) = − () 2 !! =0 X The second sum is similar in form with Bessel function of zero order. See Bowman [8, Chapter 1] for explicit formula or Andrews [4, p200] for formula in terms of hypergeometric functions. It is given by
∞ ( 1) 2 0 ()= − !! 2 =0 X ³ ´ where =2√. Hence,
()= 0 2√
2 = 0 ¡ √1 ¢ (4.14) − ³ ´ By Theorem 18, we know that () can be written in terms of Legendre polynomials. Thus, we have the following corollary. 4. Theory of Harmonic Polynomials: Algebraic Approach 73
Corollary 6 The exponential generating function of Legendre polynomial in- volves Bessel function of zero order. More precisely,
∞ () 2 = 0 √1 ! − =0 X ³ ´ where () is the th Legendre polynomial, =0, 1, 2, ....
This adds to the collection of generating functions of Legendre polynomials. The standard generating function of Legendre polynomials is given by
∞ 1 () = √ 2 =0 1 2 + X − More generating functions of Legendre polynomials can be seen in Brafman’s papers in [9] and [10].
4.2 Harmonic Polynomials on 3
Now we move to the case when =4; that is, we consider harmonic polynomials (x) where x 3 A4. Again, for convenience, we write x =( ) instead ∈ ⊂ of (1234).Wedenoteby the factorial monomial
!!!! for any integer . Again, if either one of , , ,or is negative, the above expression will be automatically zero. The following lemma is analogous to the Lemma in our previous section given in (4.1), and the proof is similar.
Lemma 8 For any integer ,
2 2 2 2 = − + − + − + − (4.15) 4 Apart¡ from the¢ standard basis
: + + + = and Z ∈ 4 © ª of (A ),
: + + + = and Z { ∈ } 4. Theory of Harmonic Polynomials: Algebraic Approach 74
4 is also a basis of (A ), which we will call factorial basis of . We can arrange all these factorial basis monomials in a tetrahedron which will be referred as the level- tetrahedron. Analogously to Lemma 4, we can imagine stacking tetrahedrons on four- dimensional space; arranging level- tetrahedron of factorial basis elements suc- cesively below each other, starting from =0,and =1beneath that, =2 beneath that, and so on. Lemma 8 tells us that if we have a homogeneous polynomial of degree , represented in the factorial basis as
( )= +++= X then for each monomial , the Laplacian applied to that monomial sends the coefficient in the level- tetrahedron to the four positions correspond- 2 2 2 2 ingtothemonomials − , − , − ,and − in the level-( 2) tetrahedron 2 levels above it. Similarly, this leads to the conclusion − that is harmonic precisely when in the level- tetrahedron representation of , the sum of all coefficients in every subtetrahedron of side 2 is zero. Since every homogeneous harmonic polynomial is a sum of monomials, we can get a tetrahedral representation for every homogeneous harmonic polynomial. For example, if ( )=2 4 +22 32 +4 22,thenintermsof − − − factorial monomials,
=22 4 +42 62 +4 42 − − − and hence the tetrahedral¡ ¢ representation of would be represented in the fol- lowing two-dimensional representation of a three-dimensional tetrahedron of co- efficients, see Figure (4.9) below.
4.2.1 Complex Factorial Basis
Introduce four new variables that will play a crucial role in understanding zonal harmonic polynomials in A4: + + = = − = = and = − = 2 2 2 2 Since
= + = ( ) = + and = ( ) − − − − 4. Theory of Harmonic Polynomials: Algebraic Approach 75
t 2 -4
y
0 zt
xt 0 0 yt
-6 z 2
xz0 4 yz
2 -4 4 x2 xy y2
Figure 4.9: Example of polynomial in level 2 tetrahedron it is clear that any homogeneous polynomial in and can be rewritten as a homogeneous polynomial of the same degree in and ;andvice versa. That means
: + + + = and Z { ∈ } 4 formsabasisof(A ), and we will call it the complex factorial basis. Here is the crucial Lemma that reveals the usefulness of our new basis. The Laplacian applied to any monomial on 3 is also simple.
Lemma 9 For any integer ,
1 1 1 1 = − − + − − (4.16) 4 Proof. Note that 2 2 = 2 !!!! 2 ¡ ¢ (¡ 1)¢ 2 1 1 ( 1) 2 = − − + − − + − − !!!! 4 2 4 µ ¶ and by the same principle,
2 ( 1) 2 1 1 ( 1) 2 = − − + − − − − 2 !!!! − 4 2 − 4 µ ¶ ¡ ¢ 4. Theory of Harmonic Polynomials: Algebraic Approach 76 so that 2 2 + = 1 1 2 2 !!!! − − µ ¶ 1 1 ¡ ¢ = − −
Similarly, we will have 2 2 + = 1 1 2 2 − − µ ¶ ¡ ¢ so that 2 2 2 2 ∆ = + + + 2 2 2 2 µ 1 1 1 ¶1 ¡ ¢ = − − + − − ¡ ¢ as desired. 4 Supposethatwehaveahomogeneouspolynomial (A ) and we con- ∈ struct its tetrahedron of coefficients in the complex factorial basis. Now, each coefficient is associated with the vector ( ) corresponding to the power of the monomial it is attached to. If we regard the tetrahedron of coefficients as the tetrahedron of the corresponding vectors, then from Lemma 9, is harmonic if andonlyifinthedirectionofthevector(1 1 1 1) the entries have constant − − absolute value and are alternating in sign as we move along the direction of the 4 vector. Thus the basis elements of (A ) correspond to one-dimensional arrays along the direction of vector (1 1 1 1) in the level- tetrahedron and consist − − of alternating plus and minus ones. To see what the harmonic homogeneous polynomials of degree look like, we imagine its tetrahedron of vectors where all the coefficients are initiated to be all 1 first. Starting from the and face, for each point in those two faces we attach the vector (1 1 1 1) and see whether the endpoint is still contained − − in the tetrahedron. If so, then the coefficient is changed to be the different sign from the starting point. If the endpoint of the vector is not in the tetrahedron, the sign of the starting point does not change. Eventually all of the endpoints will have to fall off in the and face.Wenowlookforsomeexamplesfor several values of .
Example 13 In the case =2, the tetrahedron of coefficients in the complex factorial basis will look like Figure (4.10) below. 4. Theory of Harmonic Polynomials: Algebraic Approach 77
v2
y
uv
rv sv
u2
ru su
r2 rs s 2
Figure 4.10: Level 2 tetrahedron in complex factorial basis
4 In this case, the basis for 2 (A ) is given by
2, 2, 2, 2, , , , , − ©4 2 ª Note that dim 2 (A )=9=(2+1)
Example 14 In the case =3, the tetrahedron of coefficients in the complex factorial basis will look like Figure (4.11) below.
4 So the basis for 3 (A ) is given by
3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2 , 2 , 2, 2 ( − − − − ) 4 2 Also note that dim 3 (A )=16=(3+1).
4 Example 15 Belowwegiveabasisfor (A ) for several values of .For 4. Theory of Harmonic Polynomials: Algebraic Approach 78 v3
uv2 rv2 sv2 u 2v y ruv suv r 2v u3 s 2v su2 ru2 2 r 2u rsu s u r 3 r 2s rs2 s3
Figure 4.11: Level 3 tetrahedron in complex factorial basis
4 =4,abasisfor4 (A ) is
4, 3 2, 22 + 22, 3 2, 4, 3, 22, 3, 4, − − − 3, 22, 3, 2 3, 2 2, 2 2, 3, 22, 3, 4, ⎧ − − − ⎫ ⎨⎪ 3, 22, 3, 2 3, 2 2, 2 2 ⎬⎪ − − − ⎪ ⎪ ⎩ Based on these two examples, dim (A ) seems to be a square number given⎭ by ( +1)2, and this is justified from (3.11). To see this geometrically, note that we need to count the number of lines in the direction of vector (1 1 1 1) − − that are contained inside the level- tetrahedron. To count this, we only need to count the number of points in the and face. The number of such points is
(#of points in the face) + (#of points in the face) (#of points in the edge). − Since on each face we have 1+2+ +( +1)= 1 ( +1)( +2)such points ··· 2 4. Theory of Harmonic Polynomials: Algebraic Approach 79 andoneachedgewehave( +1)points, then the number of lines is 1 1 ( +1)( +2)+ ( +1)( +2) ( +1) = ( +1)( +2) ( +1) 2 2 − − =( +1)2 This is well-known, as in fact these harmonics turn out to span the matrix coefficients of the ( +1)irreducible representations of SU (2) 3. '
4.3 Zonal Harmonics in the Factorial Basis
We want to consider zonal homogeneous polynomials in the factorial basis, where zonal in this case means rotationally invariant. For this purpose it will be con- venient to label the entries of level- tetrahedron of basis elements. The rows in the vertical direction will be indexed by =0 1 2 with in row 0 In row we will have a triangle with − at the bottom left of this triangle, − at the bottom right of the triangle, and − at the top of the triangle. Again, we can index the basis elements in this triangle. The rows will be indexed by 1 1 =0 1 2 with − in row 0 and − , − − , , − − , − in row . The elements of the th row will finally be indexed by =0 1 with − − − in the th position. That basis element will be assigned to have coordinates ( ) In other words, the element in the ( ) position is precisely − − − ,where0 . ≤ ≤ ≤ ≤ Now those basis elements whose coordinates ( ) are all even integers form a subtetrahedron of spacing 2 which includes the vertical top term , and we will call those basis elements the even subtetrahedron in the level- tetrahedron.
Theorem 19 (Zonal Harmonic Polynomials on 3 in Factorial Basis) Up to a scalar, there is a unique zonal harmonic polynomial of degree namely
2 b c !(2 2)! (2 2)! (2)! 2 2 2 2 2 2 (x)= ( 1) − − − − − − (2 +1)!( )!( )!! =0 =0 =0 X X X − − Remark 7 Note that the coefficients are independent of . We can say that the coefficient corresponding to the coordinates (2 2 2) of the zonal harmonic polynomial of degree is
! (2 2)! (2 2)! (2)! (222) =( 1) − − − (2 +1)! ( )!( )!! − − 4. Theory of Harmonic Polynomials: Algebraic Approach 80
Proof of Theorem 19. Since the only rotationally invariant homogeneous polynomials in and are (2 + 2 + 2) up to a scalar, then the zonal harmonic polynomial must be of the form
2 b c 2 2 2 2 ( )= 2 ( + + ) − =0 X Now if we expand the above expression, we will get
(2 + 2 + 2) 2 2 ! − Z X∈ (2 + 2) 2 2 = − 2 2 ! − Z =0 µ ¶ X∈ X (2 + 2)2 2 = − 2 2 ! − Z =0 µ ¶ X∈ X 2 2 2 2 2 = − − 2 2 ! − Z =0 =0 µ ¶µ ¶ X∈ X X 2 2 2 2 2 ! ! − − 2 = 2 − ( )!! ( )!! ! Z =0 =0 X∈ X X − − 2 2 2 2 2 = − − 2 2 ( )!( )!! − Z =0 =0 ∈ − − X X X (2 2)!(2 2)!(2)! 2 2 2 2 2 2 = 2 − − − − − ( )! ( )!! Z =0 =0 X∈ X X − − In order for to be harmonic, we must have
(2 2)! (2 2)!(2)! (2 2)! (2 ( +1) 2)! (2)! 0=2 − − + − − ( )! ( )!! 2( +1) ( )! ( +1 )!! − − − − (2 ( +1) 2)! (2 2)! (2)! + − − 2( +1) ( +1 )! ( )!! − − (2 2)! (2 2)! (2( + 1))! + − − 2( +1) ( )! ( )! ( +1)! − − which simplifies to 1 2 = 2( 1) for all integers 1 −2(2 +1) − ≥ 4. Theory of Harmonic Polynomials: Algebraic Approach 81
By normalizing such that 0 =1 we will get
! 2 =( 1) − (2 +1)! which will give us the notation of (x) in Theorem 19.
Here are the first five values of .
0 =1
1 =
2 1 2 1 2 1 2 2 = (4.17) − 3 − 3 − 3 3 1 2 1 2 1 2 3 = − 3 − 3 − 3 4 1 2 2 1 2 2 1 2 2 1 4 1 4 1 4 4 = + + + − 3 − 3 − 3 5 5 5 1 1 1 + 22 + 22 + 22 (4.18) 15 15 15
Note that we can express in terms of alone because is a function of and 2 2 2 2 2 1 2 1 2 1 2 + + =1 . Take for example, 2 = Note that − − 3 − 3 − 3 2 1 2 1 2 1 2 2 = − 3 − 3 − 3 1 1 = 2 2 + 2 + 2 2 − 6 1 1 = 2 ¡1 2 ¢ 2 − 6 − 1 = 42 ¡1 ¢ 6 −
Here are the first five values of ¡in terms¢ of the variable .
0 =1
1 =
1 2 2 = 4 1 (4.19) 6 − 1 ¡ 3 ¢ 3 = 2 6 − 1¡ 4 ¢ 2 4 = 16 12 +1 120 −
It turns out that the polynomials¡ have a nice¢ description as the Cheby- shev polynomials of the second kind, denoted by (),uptoacertain scalar. We have the following Theorem. 4. Theory of Harmonic Polynomials: Algebraic Approach 82
Theorem 20 Let () be the th Chebyshev polynomial of the second kind and 0 is an integer. Then ≥ 1 ()= (); ( +1)! that is, () is proportional to the th Chebyshev polynomial of the second kind.
Proof. One way to show that the claim is indeed true, we can show that for each 0, () takes one of Chebyshev polynomial of the second kind’s direct ≥ form. Now, simple computations show that
=( +1)! ()
2 b c ( 1) ! 2 2 2 2 =( +1)! − + + − (2 +1)! =0 X ¡ ¢ 2 b c ( +1)! 2 2 2 2 = ( 1) + + − − (2 +1)!( 2)! =0 − X ¡ ¢ 2 b c +1 2 2 = ( 1) 1 − − 2 +1 − =0 µ ¶ X ¡ ¢ which is one of the explicit forms of the Chebyshev polynomials of the second kind () (see Weisstein [29, p234], Abramowitz [1, Chapter 22], Oldham [23, Chapter 22], and Rivlin [25]).
Now that we have established that the zonal harmonic polynomials on 3 are nothing other than the Chebyshev polynomials of the second kind, we can find the generating function of the zonal harmonic polynomials in 3.Letus consider the generating function
∞ ∞ 1 () = () (4.20) ( +1)! =0 =0 X X
Theorem 21 The generating function of () is given by
∞ 2 () = sinhc √ 1 − =0 X ³ ´ where 1. | | ≤ 4. Theory of Harmonic Polynomials: Algebraic Approach 83
Proof. It is known that the exponential generating function of () can be given as
∞ () = cosh √2 1 + sinh √2 1 for 1 ! − √2 1 − | | ≤ =0 µ − ¶ X ³ ´ ³ ´ (4.21) (see Rivlin [25]). Integrating both sides of (4.21) gives us
∞ () ∞ () +1 = ( +1)! ⎛ ! ⎞ =0 =0 X X Z0 ⎝ ⎠ ∞ () = ! Ã=0 ! Z0 X = cosh √2 1 + sinh √2 1 − √2 1 − Z0 µ ³ ´ − ³ ´¶ = cosh √2 1 − Z0 ³ ´ + sinh √2 1 √2 1 − − Z0 ³ ´ The first integral could be solved by integration by parts, and so the last equation above reduces to sinh √2 1 sinh √2 1 2 2 √ 1 − 0 − √ 1 − ¯ Z0 − ³ ´¯ − ³ ´ ¯ ¯ + sinh √2 1 √2 1 − − Z0 ³ ´ = sinh √2 1 √2 1 − − ³ ´ Dividing by =0in both sides gives us 6 ∞ ∞ 1 () = () ( +1)! =0 =0 X X = sinh √2 1 √2 1 − − ³ ´ = sinhc √2 1 (4.22) − ³ ´ 4. Theory of Harmonic Polynomials: Algebraic Approach 84 where 1. | | ≤ Note the more compact description of the generating function formula for
(). We treat the function above simply as a formal power series giving ().
4.4 Zonal Harmonics in the Complex Factorial Basis
Now it is interesting to actually express (4.19) in terms of the complex factorial basis. For example,
1 = = ( ) − − which is proportional to . A more complicated attempt would be to substi- − 1 2 1 2 2 2 tute the variables , , ,and in 2 = ( + + ) in terms of , , 2 − 6 ,and. Simple substitution shows that
2 2 2 2 2 2 = + ( ) −3 3 − − 3 4 2 4 = 2 + ( ) 2 −3 3 − − 3 which, up to a scalar, is proportional to 1 2 ( )+2 − 2 −
Another computation shows that 3 is proportional to 1 1 3 2 + 2 3 − 3 − 3 − − ¡ ¢ ¡ ¢ while 4 is proportional to 1 1 1 4 3 2 + 22 + 22 3 2 + 4 − 4 − 6 − − 4 − Ifweseethepatternabove,wespotthetermsinthe¡ ¢ ¡ ¢ ¡Leibniz triangle¢ ,one variant of the Pascal triangle. The th elements in the th row, =0 1 2 3 and =0 1 2,givenby, represent the coefficients − in ( + ) which is . In the Leibniz triangle, the th element in the th row is given by ¡ ¢ 1 = ¡ ¢ 4. Theory of Harmonic Polynomials: Algebraic Approach 85
The terms in are exactly the harmonic polynomials that correspond to the lines with the direction vector (1 1 1 1) that pass through the edge with − − and as the endpoints. For example, consider the case where =4.In the level 4 tetrahedron represented in the complex factorial basis, there are only 5 lines out of 25 lines that pass through the edge with endpoints 4 and 4. Those five are the point 4 itself (the line that passes through this one point in the direction of (1 1 1 1) does not pass any other points), the line that − − passes through 3 which also passes through 2 (so it corresponds to the harmonic polynomial 3 2), the line that passes through 22 which also − passes through and 22 4 (so it corresponds to the harmonic polynomial 22 +22), the line that passes through 3 which also passes through − 2 (so it corresponds to the harmonic polynomial 3 2), and the point 4 − itself (the line that passes through this one point in the direction of (1 1 1 1) − − does not pass any other points). The coefficient of the expression
1 1 2 − − − − + +( 1) − − ··· − is exactly . We have the following Theorem.
Theorem 22 The zonal harmonic polynomials described in factorial basis is a complex multiple of
( 1) = − ( 1) − − − (4.23) − =0 Ã=0 ! X X the zonal harmonic polynomials¡ ¢ in complex factorial basis.
Proof. We need to show that the polynomial given above is a complex multiple of the (), th Chebyshev polynomial of the second kind. First, note that we can expand (4.23) above as
+ ( )!! − − − = ( 1) − − ! !!( )! ( )! =0 =0 − − − X X 1 + ( )!! = ( 1) − () − − − (4.24) ! − !!( )! ( )! =0 =0 X X − − − 2 2 2 2 3 1 Since + + + =1on , it is equivalent with saying that + = 4 . 4. Theory of Harmonic Polynomials: Algebraic Approach 86
Hence, Equation (4.24) can be rewritten as
+ 1 ( 1) ( )!! 1 = − − − − − ! !!( )! ( )! 4 − =0 =0 X X − − − µ ¶ + 1 ( 1) ( )!! 1 = − − ( ) − − − − ! !!( )! ( )! 4 − =0 =0 =0 X X X − − − µ ¶µ ¶ + 1 ( 1) ( )!! = − − − − − (4.25) ! 4!( )! ( )! ( )!! =0 =0 =0 X X X − − − − If we change the order of summation in (4.25) above, we get that
+ 1 ( 1) ( )!! = − − − − − ! 4!( )! ( )! ( )!! =0 =0 = X X X − − − − + 1 ( 1) ( )!! = − − − − − ! 4! =0 =0 X X 1 (4.26) !( )! ( )! ( )! × Ã= ! X − − − − Now, the term in the bracket in Equation (4.26) above is denoted by ( ), then we have that
− 1 ( )= ( + )! ( )! ( )!! =0 − − − − − X 1 − !( )! ( )! = − − − !( )! ( )! ( + )! ( )! ( )!! =0 − − − X − − − − − Here, again we follow the algorithm in Petkovšek [24, Chapter 3] to express ( ) intermsofhypergeometricfunction.Wetakeoutthefirst term in the summation above so that the first term in
− !( )! ( )! − − − ( + )! ( )! ( )!! =0 X − − − − − is 1.Let !( )! ( )! = − − − ( + )! ( )! ( )!! − − − − − 4. Theory of Harmonic Polynomials: Algebraic Approach 87 and note that
+1 ( + )! ( )! ( )!! = − − − − − ( + +1)!( 1)! ( 1)! ( +1)! − − − − − − − ( )( ) = − − − − − ( + +1)( +1) ( ( )) ( ( )) = − − − − − ( + +1)( +1) so 1 ( )= 2 1 ( ( ) ( ); +1;1) !( )! ( )! − − − − − − − − Again, by using Gauss’ identity (see Gradshteyn [16, Chapter 9] or Andrews [4, Chapter 2]) given in (4.11), let
= ( )= + − − − − = ( )= − − − = +1
Since = +1 0,then − − − 1 Γ ( +1)Γ ( +1) ( )= − !( )! ( )! Γ ( +1)Γ ( +1) − − − − 1 ( )!! = − !( )! ( )! ( )!! − − − − ( )! = − (4.27) ( )! ( )! ( )!! − − − − Substituting Equation (4.27) to (4.26) gives us
+ 1 ( 1) ( )!! ( )! = − − − − − − ! 4! ( )! ( )! ( )!! =0 =0 X X µ − − − − ¶ + 1 ( 1) ( )! = − − − − − (4.28) ! 4!( )! ( )! =0 =0 X X − − − If we change the order of summation in (4.28), we have that
+ 1 ( 1) ( )! = − − − − − (4.29) ! 4!( )! ( )! =0 = X X − − − 4. Theory of Harmonic Polynomials: Algebraic Approach 88
Now look at our domain. We can think of that as the pair ( ) where 0 ≤ ≤ ,andall, ,and are integers. Recall that we define 1 to be zero if is ≤ ! negative. Since can be negative if and close enough to (say, for − − example, if = = ), we can remove ( ) from our domain of summation. After removing all such pairs, we can rewrite Equation (4.29) as
2 + 1 b c − ( 1) ( )! = − − − − − ! 4!( )! ( )! =0 = − − − X X 2 1 b c ( 1) ( )! − ( 1) = − − − − − − ! 4! ( )! ( )! =0 Ã = − − − ! X X 2 2 + 1 b c ( 1) ( )! − ( 1) 2 = − − − − − ! 4! ( 2)!! =0 Ã =0 − − ! X X 2 2 1 b c ( )! − ( 2)! 2 = − ( 1) − − − ! 4!( 2)! − ( 2)!! =0 − Ã =0 − − ! X X 2 1 b c ( )! 2 = − ( ) − ! 4!( 2)! − =0 − X 2 1 b c 1 2 = − ( ) − (4.30) ! 22 =0 − X µ ¶ Sincewehavethat + = and = − 2 2 = . Substituting this value to (4.30), we get − 2 1 b c 2 2 = 2 − − () − 2! =0 µ ¶ X 2 1 b c 2 = ( 1) − (2) − 2 ! =0 − µ ¶ X µ ¶ 1 = () .(4.31) 2 ! µ ¶ Here, we use another representation of Chebyshev polynomials of the second 4. Theory of Harmonic Polynomials: Algebraic Approach 89
kind () as 2 b c 2 ()= ( 1) − (2) − =0 − X µ ¶ See Weisstein [29, p234] or Abramowitz [1, Chapter 22].
Equation (4.31) shows that is a complex multiple of ().FromTheorem 20, it has been shown that the zonal harmonic polynomials in factorial is a multiple of (), so our result follows. Note that Theorem 22 gives us another way of viewing the aforementioned Chebyshev polynomials of the second kind.
To close this chapter, we can think of as the character of the Lie group 3. Recall that a character of the group is a trace of an irreducible represen- tation (see Alperin [2, Chapter 6], James [20]). Hence the characters of 3 are Chebyshev polynomials of the second kind. Appendix A
Spherical Harmonic 2 1 Decomposition of −
In this appendix we are going to show that the set¡ of all square-integrable¢ func- 1 tions on − has a natural decomposition over the usual field = R.Define 2 1 the Hilbert space ( − ) which consists of all square-integrable functions on 1 − R equipped with the inner product ⊆ = h i 1 Z− 1 where is the surface area measure on − .Wewanttofind a natural orthog- 2 1 onal decomposition of ( − ). Now, let be any complex Hilbert space. Recall from standard Hilbert theory (see Axler [6, Theorem 5.12]) that we can write
∞ = =0 M precisely when is a closed subspace of for every , is orthogonal to for = ,andforevery there exist such that = 0 + 1 + , 6 ∈ ∈ ··· with the sum converging in the norm of . When all of these conditions are met, we say that is the direct sum of the spaces . Wehavethefollowing 2 1 important theorem about the decomposition of ( − ).
2 1 Theorem 23 (Decomposition of ( − ))
2 1 ∞ 1 − = − =0 ¡ ¢ M ¡ ¢ 90 2 1 A. Spherical Harmonic Decomposition of ( − ) 91
1 Proof. The first condition is met because each ( − ) is finite dimensional 2 1 and hence is a closed subspace of ( − ). The second condition also holds 1 1 since we have already established that ( − ) and ( − ) are orthogonal as long as = . For the third condition, we are going to show that the linear 6 span of
∞ 1 = − H =0 [ 2 1 ¡ ¢ 2 1 is dense in ( − ). From there it can be concluded that for every ( − ) 1 ∈ there exist ( − ) such that ∈
∞ = =0 X To see this, note that if we have a homogeneous polynomial of degree then from (3.7) we have
= + 2 + + 2 | | − | ··· − | 1 so any polynomial ,defined on A , when restricted on − , will be just the sum of finitely many elements in . By Stone-Weierstrass Theorem (see Rudin H 1 [26, Theorem 7.33]) , the set of polynomials defined in A restricted to − 1 1 is dense in ( − ), the space of all continuous functions in − with respect 1 2 1 to the supremum norm. Since we know that ( − ) is dense in ( − ) and 2-norm is never greater than the supremum norm, we conclude that is dense 2 1 H in ( − ), thus implying that the third condition is met.
A.1 Functions on 1 and Fourier Theory
By the decomposition above, we know that every square integrable function 2 1 1 on ( ) can be written as 0 +1 +2 + where each ( ).Weknow 1 1 ··· ∈ that 0 ( )= 1 and ( )= = for 1.Bywriting h i ≥ and in polar coordinates, we have that ® ® + 1 1 = = = 2 2 2 and 1 1 = − = − = − 2 2 2 2 1 A. Spherical Harmonic Decomposition of ( − ) 92
1 Hencewehave ( )= = − so we can write ® ® 2 1 ∞ ∞ = 1 + − = h i =1 = ¡ ¢ M ® M−∞ ® In particular, any 2 (1) canbewrittenas ∈
∞ ()= = X∞ which is nothing other than standard Fourier theory. Due to the decomposition being orthogonal, the coefficients canbecomputedby
= () −
Z1 where, as usual, is the (normalized) measure in 1. The formula above can be found in one of Fourier’s most famous books [18] that is the notation that we use nowadays in the standard Fourier theory. See also Stein [27] or Deitmar [12, Chapter 1] for a more detailed explanation of Fourier theory from the classical point of view. Appendix B
Integrating Polynomial over a Sphere
In this appendix, we would turn our attention to what actually happens when we integrate a polynomial over a sphere. Although this has been studied extensively for decades, see for instance the papers from Baker [7, page 43] and Folland [17], we would like to present a slightly different derivation of a key formula. Both papers by Baker and Folland make use of the gamma function Γ to integrate 1 a homogeneous polynomial (of degree ) (x) over the sphere − .Inthis appendix, we would like to redo the integration and describe the result in terms of a double factorial which we will introduce later. Our general result in (B.15) is aligned with the results of Baker and Folland but the result can be computed efficiently as it is numerically faster to process with computers without having to resort to the gamma function Γ. Alsokeepinmindthatwiththistoolsin hand, we are able to approximate the integral of any square integrable function 1 defined on − ,byAppendixA. Since any polynomial is a sum of homogeneous polynomials, we can assume that is a homogeneous polynomial of degree . We would like to integrate 1 the polynomial over − . To do this, see that we only need to integrate the monomial x where = , since any homogeneous polynomial of degree is | | a linear combination of finitely many monomials of the form x. Let us see what happens in the case where =1and 2 first.
93 B. Integrating Polynomial over a Sphere 94
B.1 Integrating monomials over 1
1 2 Suppose we have ( )= where 1 + 2 = and 0 12 . ≤ ≤ We are interested in finding 1 2
Z1 as a function of 1 and 2, so let’s call it (12).Here is the normalised measure on 1,i.e.measureon1 such that
=1
Z1
To do that, we can parametrize 1 as follows: write =cos and =sin.By choosing 1 = 2 one gets 2 1 ( )= cos1 sin2 (B.1) 1 2 2 Z0
Now, fix nonnegative integers 1 and 2,andlet
1+1 2 1 ()=cos sin −
Note that
1 2 1+2 2 2 = (1 +1)cos sin +(2 1) cos sin − − − 1 2 1 2 2 2 = (1 +1)cos sin +(2 1) cos sin − 1 sin − − − 1 2 1 2 2 = (1 + 2)cos sin +(2 1) cos sin − − − ¡ ¢ By the above equation and the Fundamental Theorem of Calculus, we get that
(2) (0) 0= − = (1 + 2) (12)+(2 1) (12 2) (B.2) 2 − − − which translates into the recursive relation
2 1 (12)= − (12 2) (B.3) 1 + 2 − B. Integrating Polynomial over a Sphere 95
The value (12) will depend on the parity of 2.If2 is odd, then
(12)= (1 1) where is some constant which depends on 1 and
2. However, since
2 1 ( 1) = cos1 sin 1 2 Z0 1 2 = cos1+1 −2 ( +1) 1 ¯0 ¯ =0 ¯ ¯ then (12)=0if 2 is odd. However, when 2 is even, we have
(2 1) (2 3) 1 (12)= − × − ×···× (1 0) (1 + 2) (1 + 2 2) (2 +2) × − ×···× 2 (2 1) (2 3) 1 1 = − × − ×···× cos1 (1 + 2) (1 + 2 2) (2 +2)⎛2 ⎞ × − ×···× Z0 ⎝ ⎠ Let 2 1 = cos 2 Z0 1 andnotethatbywriting =cos− and =cosand doing integration by parts, we have the recurrence relation
=( 1) 2 for 2 − − ≥ and 0 =11 =0. Hence, if is odd, then =0,andif is even, we then have ( 1) ( 3) 1 = − × − ×···× ( 2) 2 × − ×···× B. Integrating Polynomial over a Sphere 96
It then follows that (12) will be nonzero precisely when both 1 and 2 are even, and the value is given by
(2 1) (2 3) 1 (12)= − × − ×···× 1 (1 + 2) (1 + 2 2) (2 +2) × − ×···× (2 1) (2 3) 1 = − × − ×···× (1 + 2) (1 + 2 2) (1 +2) × − ×···× (1 1) (1 3) 1 − × − ×···× (B.4) × 1 (1 2) 2 × − ×···× (1 1)!! (2 1)!! = − − (1 + 2)!! (1 1)!! (2 1)!! = − − (B.5) !! Here we use the notation of double factorial definedinthebeginningof Chapter 4. To sum up, we have the following theorem.
Theorem 24 Let 1 and 2 be any non-negative integers such that 1 + 2 = ,and ( )=1 2 is a monomial defined on 1.If denotes the nor- malized measure on 1 such that
=1,
Z1 then (1 1)!! (2 1)!! 1 2 = − − !! Z1 precisely when both 1 and 2 areeven;andiszerootherwise.
Remark 8 Note that with this statement, all irrationalities have been removed.
B.2 Integrating monomials over 2
In this case, we do the same thing as what we have done previously. Let
1 2 3 ( )= be a monomial of degree ;thatis,1 + 2 + 3 = and 0 1, 2, 3 . We want to establish the following theorem. ≤ ≤
Theorem 25 Let 1, 2,and3 be any non-negative integers such that 1 +
1 2 3 2 2 + 3 = ,and ( )= is a monomial defined on .If B. Integrating Polynomial over a Sphere 97 denotes the normalized measure on 2 such that
=1,
Z2 then (1 1)!! (2 1)!! (3 1)!! 1 2 3 = − − − ( +1)!! Z2 precisely when all 1, 2,and3 are even; and zero otherwise. The usual parametrization of 2 is given by
=sin cos =sin sin =cos where 0 and 0 2 We can choose our measure to be ≤ ≤ ≤ 1 = sin 4 and one can get
2 1 ( )= cos3 sin1+2+1 cos1 sin2 1 2 3 4 Z0 Z0 2 1 = cos3 sin1+2+1 cos1 sin2 4 ⎛ ⎞ ⎛ ⎞ Z0 Z0 ⎝ ⎠ ⎝ ⎠ which will be nonzero precisely when both 1 and 2 are even. In that case, we have
1 (1 1)!! (2 1)!! 3 1+2+1 (123)= − − cos sin 2 (1 + 2)!! Z0 Now, let us focus on what the integral above evaluates to by considering the integral of the form
1 2 (12)= cos sin Z0 Note that the only difference between the above integral and (B.2) is the bound-
+1 2 1 ary. However, in both case, if we also let ()=cos 1 sin − ,wealsohave B. Integrating Polynomial over a Sphere 98
() (0) = 0. Hence, satisfiesthesamerecurrencerelationas.Note − that we are interested in finding
cos3 sin1+2+1
Z0 and since 1 + 2 +1is odd, we must have
cos3 sin1+2+1
Z0 (1 + 2) (1 + 2 2) 2 = × − ×···× (1 + 2 + 3 +1) (1 + 2 + 3 1) (3 +3) × − ×···× cos3 sin × Z0 ( + )!! = 1 2 (1 + 2 + 3 +1) (1 + 2 + 3 1) (3 +3) × − ×···× 1 cos3+1 × − +1 µ 3 ¯0 ¶ (¯ + )!! ¯ 1 2 3+1 = ¯ 1 ( 1) (1 + 2 + 3 +1) (3 +3) (3 +1) − − ×···× × ¡ ¢ which is nonzero precisely when 3 is even. In that case, the value is given by 2( + )!! 1 2 (1 + 2 + 3 +1) (1 + 2 + 3 1) (3 +3) (3 +1) × − ×···× × Hence, if all of 123 are all even, then
(1 1)!! (2 1)!! (123)= − − (1 + 2)!! (1 + 2)!!
× (1 + 2 + 3 +1) (3 +3) (3 +1) ×···× × (1 1)!! (2 1)!! (3 1)!! = − − − (1 + 2 + 3 +1)!! (1 1)!! (2 1)!! (3 1)!! = − − − ( +1)!! and is zero otherwise. B. Integrating Polynomial over a Sphere 99
B.3 Integrating monomials over a general sphere
1 Now, we turn our attention to the case where the underlying surface is − A . Suppose we have a monomial (x)= (12 ) given by ⊆
1 2 (x)=1 2
We would like to find
1 2 1 2
1 Z− where is a measure defined similarly as before. To do this, we could use the spherical coordinates
1 =cos1
2 =sin1 cos 2
3 =sin1 sin 2 cos 3 . .(B.6).
1 =sin1 sin 2 sin 3 sin 2 cos 1 − ··· − − =sin1 sin 2 sin 3 sin 2 sin 1 ··· − − where all [0] for =1 2 2 and 1 [0 2) We then have ∈ − − ∈ 1 = 1 12 1 ( − ) | | ··· − 1 = 1 φ ( − ) | | where is the Jacobian matrix of the above transformation given in (B.6) and we have used the short hand form φ to denote 12 1.Adirect ··· − computation reveals that the Jacobian determinant is given by
2 3 2 =sin − 1 sin − 2 sin 3 sin 2 | | ··· − − 2 − = sin − =1 Y Thus, after rewriting the integrand into spherical coordinates, the integral B. Integrating Polynomial over a Sphere 100 becomes
2 2 1 − sin φ ( 1) − − ··· Ã=1 !Ã=1 ! Z0 Z0 Z0 Y Y 2
1 1 = 1 cos − 1 sin 1 ( − ) ··· − − Z0 Z0 2 − +1+ ++( 1 ) cos sin ··· − − φ =1 Y ¡ ¢ 2 1 − = cos sin+1+ ++( 1 ) ( 1) ··· − − − ⎛=1 ⎞ Y Z0 2 ⎝ ⎠ 1 cos − 1 sin 1 1 × ⎛ − − − ⎞ Z0 ⎝ ⎠ We have the following Theorem.
Theorem 26 For each ,takeany-tuple of non-negative integers (1) 1 2 1 and let (x)=1 2 denote the monomial defined in A .Let − de- 1 note the ( 1)-sphere in A and the normalised measure defined on − − such that =1
1 Z− then 1 (x) = − ( 1)!! (deg ()+ 2)!! − Z 1 − =1 − Y precisely when is even for all 1 2 andiszerootherwise.Here ∈ { } 1 is a constant that depends on . − Proof. The proof goes by induction on . For the base case, consider =2 Previously we have talked about integrating monomials of the form (x)= 1 2 2 2 1 1 2 where (12) Z 0 defined on A over . We have established that ∈ ≥ (1 1)!! (2 1)!! (1 1)!! (2 1)!! (x) = − − = − − (1 + 2)!! (deg ())!! Z1 for even value of 1 and 2 and zero otherwise, so this completes our base case. Now assume that the statement is true for some = . We are going to consider B. Integrating Polynomial over a Sphere 101 thecasewhen = +1. Based on our assumption, if we have the monomial
1 2 (x)=1 2 for any -tuple of nonnegative integers (12) 1 and we integrate it over − ,wewouldhavethat
1 2 1 = − ( 1)!! 1 2 ··· (deg ()+ 2)!! − Z 1 − =1 − Y 1 = − ( 1)!! ( + + + 2)!! 1 =1 − ··· − Y precisely when 1, 2, ..., are even and
1 2 =0 1 2 ··· 1 Z−
1 if at least one of is odd. In case of all are even, if we parametrize − by using the (B.6) described above, we have
2 1 − cos sin+1+ ++( 1 ) ( 1) ··· − − − ⎛=1 ⎞ Y Z0 2 ⎝ ⎠ 1 cos − 1 sin 1 1 (B.7) × − − − Z0 1 = − ( 1)!! (B.8) ( + + + 2)!! 1 =1 − ··· − Y Now, if = +1,takeany( +1)-tuple of non-negative integers (12+1) B. Integrating Polynomial over a Sphere 102 and let (x)=1 2 +1 wherewehave 1 2 ··· +1
1 1 − 1 +1 = cos sin+1+ ++1+( ) 1 +1 () ··· − ⎛=1 ⎞ Z Y Z0 2 ⎝ ⎠ cos sin+1 × ⎛ ⎞ Z0 ⎝ ⎠ 1 1 2+ ++1+ 1 = cos sin ··· − () ⎛ 1 1 1⎞ Z0 1 ⎝ ⎠ − +1+ ++1+( ) cos sin ··· − × ⎛=2 ⎞ Y Z0 ⎝2 ⎠ cos sin+1 (B.9) × Z0 Now, note that based on our induction assumption, the expression
1 2 − +1+ ++1+( ) +1 cos sin ··· − cos sin ⎛=2 ⎞ Y Z0 Z0 ⎝ ⎠ canbethoughtofasthelefthandsideof(B.8)relatedtothepolynomial (x)= 2 3 +1 1 in − , so the expression above is nothing other than 2 3 ··· +1 1 +1 1 ( − ) − ( 1)!! ( + + + 2)!! 2 +1 =2 − ··· − Y if 2, 3, ..., +1 are all even numbers and zero otherwise. If 2, 3, ..., +1 are all even, Equation (B.9) would then become
1 +1 +1 1 1 ( − ) 1 2 = − ( 1)!! 1 2 +1 () ( + + + 2)!! ··· 2 +1 =2 − Z ··· − Y
1 2+ ++1+ 1 cos sin ··· − × ⎛ 1 1 1⎞ Z0 ⎝ ⎠ Now, the integral on the last equation could be simplified based on the parity of B. Integrating Polynomial over a Sphere 103
.If 1 is even, then the last equation reduces to − 1 +1 1 ( − ) − ( 1)!! ()( + + + 2)!! 2 +1 Ã=2 − ! ··· − Y (2 + + +1 + 2) (2 + + +1 + 4) 1 ··· − × ··· − ×···× × (1 + + +1 + 1) (1 + + +1 + 3) (1 +2) ··· − × ··· − ×···× (B.10) cos1 × 1 1 Z0
The value of the last integral on the expression above is zero if 1 is odd, and
(1 1) (1 3) 1 (1 1)!! − × − ×···× 1 = × − 1 (1 2) 2 1!! × − ×···× Z0 if 1 is even. Thus, the expression in (B.10) simplifies to
1 +1 1 ( − ) − ( 1)!! () Ã=2 − ! Y (1 1)!! × − × (1 + + +1 + 1) (1 + + +1 + 3) (1 +2) 1!! ··· − × ··· − ×···× × 1 +1 1 ( − ) 1 = − ( 1)!! () ( + + + 1)!! 1 +1 Ã=1 − ! ··· − Y +1 = ( 1)!! (deg ()+ 1)!! Ã=1 − ! − Y proving the induction step for the case that 1 is even. Now, if 1 is odd, − − B. Integrating Polynomial over a Sphere 104 the equation in (B.9) can be simplified into
1 +1 1 ( − ) − ( 1)!! ()( + + + 2)!! 2 +1 Ã=2 − ! ··· − Y (2 + + +1 + 2) (2 + + +1 + 4) 2 ··· − × ··· − ×···× × (1 + + +1 + 1) (1 + + +1 + 3) (1 +3) ··· − × ··· − ×···× cos1 sin × 1 1 1 Z0 1 +1 1 ( − ) = − ( 1)!! ()( + + + 2)!! 2 +1 Ã=2 − ! ··· − Y (2 + + +1 + 2)!! ··· − × (1 + + +1 + 1) (1 + + + 3) (1 +3) ··· − × ··· − ×···× cos1 sin × 1 1 1 Z0 1 +1 1 ( − ) = − ( 1)!! () Ã=2 − ! Y 1
× (1 + + +1 + 1) (1 + + +1 + 3) (1 +3) ··· − × ··· − ×···× 1 cos1+1 (B.11) × −( +1) 1 µ 1 ¯0 ¶ ¯ ¯ If 1 is odd, then the above equation¯ will vanish. Otherwise, the above expression B. Integrating Polynomial over a Sphere 105 in (B.11) reduces to
1 +1 1 ( − ) − ( 1)!! () Ã=2 − ! Y 1 2
× (1 + + +1 + 1) (1 + + +1 + 3) (1 +3) 1 +1 ··· − × ··· − ×···× µ ¶ 1 +1 2 1 ( − ) 1 − = ( 1)!! ( ) (deg ()+ 1) (deg ()+ 3) (1 +1) − − × − ×···× =2 1 +1 Y 2 1 ( − ) (1 1)!! =2 ( 1)!! − = − − ( ) (deg ()+ 1) (deg ()+ 3) (1 +1) (1 1)!! − × Q− ×···× × − 1 +1 2 1 ( − ) 1 = − ( 1)!! () (deg ()+ 1)!! Ã=1 − ! − Y +1 = ( 1)!! (deg ()+ 1)!! Ã=1 − ! − Y also proving the induction step. Hence, the statement is true for the case = +1,soitistrueforall N. ∈ Note that in the proof previously presented, it implicitly mentions that if 1 is even, then − 1 1 ( − ) = − () and if 1 is odd then 1 − 2 1 ( − ) = − () Thus we have two recurrence relations that are connected to each other. Let be an odd number and write =2 +1. If we combine those two identities together, we have that
2 2 2 1 2 1 2 ( ) ( ) 22 1 ( − ) 2( − ) 2+1 = = − = 2 1 (2+1) (2+1) (2) (2+1) − µ ¶ 1 Now, since we know that the surface area of a ( 1)-sphere − can be ex- − plicitly written as 1 2 2 − = Γ 2 ¡ ¢ ¡ ¢ B. Integrating Polynomial over a Sphere 106 where Γ denotes the gamma function, then we have 2 2 2+1 = 2 1 2+1 Γ () − Γ(+1) µ ¶ ³ 2´ Γ ( +1) =2 2 1 Γ () 2+1 − µ ¶µ ¶ 2Γ ( +1) = 2 1 Γ () −
=22 1 − where in the last equation we have used one of the properties of the gamma function Γ that Γ ( +1)= Γ () for any positive integer .Now,notethat
1 =1based on the result presented in (B.5). We have the following Lemma.
Lemma 10 If is an odd positive integer, then =( 1)!!. − Proof. Write =2 +1 for some . The proof is by induction on .The base case when =0is trivial since 1 =1=0!!=(1 1)!!. Assume that − the statement is true for = 0,thatis,20+1 =(20)!!. Now, for the case
= 0 +1,wehave
20+3 = 2(0+1)+1 =2(0 +1)2(0+1) 1 =(20 +2)20+1 − =(20 +2)(20)!! = (20 + 2)!! = (2 (0 + 1))!! which completes the proof. Based on the above lemma, we can say something as well about the value of
when is even. Let be even and write =2. To see this, note that
2 1 22 1 ( − ) = = − 2 (2) 2 Γ + 1 =2(2 2)!! 2 + 1 − Γ () 2 2 µ ¶ Ã ¡ ¢! 2( 2)!! Γ + 1 = − 2 (B.12) Γ 1 Γ () 2 ¡ ¢ Note another property of gamma¡ ¢ function Γ, called the Duplication Formula (see [4, Theorem 1.5.1]) which states that 1 1 Γ () Γ + =21 2Γ Γ (2) (B.13) 2 − 2 µ ¶ µ ¶ B. Integrating Polynomial over a Sphere 107 foa any positive 0.IfwedividebothsidesbyΓ ()2 we get
Γ + 1 1 Γ (2) 2 =21 2Γ (B.14) Γ () − 2 Γ () Γ () ¡ ¢ µ ¶ Substituting (B.14) to (B.12), we get
2(2 2)!! 1 2 1 Γ (2) = − 2 − Γ Γ 1 2 Γ () Γ () 2 µ µ ¶ ¶ (2 2)!! Γ (2) = −¡ ¢ 22 2 Γ () Γ () − µ ¶ Now, since is an integer, we can simplify the above equation to get (2 2)!! (2 1)! = − − 22 2 ( 1)! ( 1)! − µ − − ¶ (2 2) (2 4) 2 (2 1)! = − × − ×···× − 22 2 ( 1) ( 2) 1 ( 1)! − µ − × − ×···× × − ¶ 1 (2 2) (2 4) 2 (2 1)! = 2 2 − × − ×···× − 2 − ( 1) ( 2) 1 ( 1)! 1 − × − ×···× µ − ¶ 2 − (2 1)! = − 22 2 ( 1)! − µ − ¶ (2 1)! = − 2 1 ( 1)! − − ( 1)! = − ( 2)!! − =( 1)!! − To sum up on what we just did, we have the following theorem.
1 Theorem 27 (Integral of monomials over − ) Let 2 be an integer. ≥ 1 2 1 If (x)=1 2 be any monomials of degree in − A ,and is 1 ⊂ the normalised measure on − defined in a way such that
=1
1 Z− then ( 2)!! (x) = − ( 1)!! (B.15) ( + 2)!! − Z 1 − =1 − Y precisely when all of the are even integers, and zero otherwise. B. Integrating Polynomial over a Sphere 108
Note the simplicity of the result. To compute the integral, one only needs the power of each variables involved in the monomial (hence one can get the degree of the monomial) and the number . All of them are integers so it is pretty easy to compute (B.15) by any computing devices. Even for small value of =deg() one can find the integral manually. Note also that all irrationalities have been removed.
Remark 9 In the main theorem of Folland [17, page 447], it is stated that if
1 2 1 (x)=1 2 and = 2 ( +1) then our theorem above in (B.15) could also be written as
2Γ(1)Γ(2) Γ() Γ( + + ···+ ) if all of are even, (x) = 1 2 ··· ( 0 otherwise. Z 1 It is worth to mention that the definition of − used in [17] is different than what we use, and the measure is not normalized.
The classical integral of a polynomial over a general sphere is presented in terms of the gamma function Γ. While our approach is not as sophisticated as Baker’s or Folland’s, our result is a nice reformulation to the classical result, because the expression ( 2)!! − ( 1)!! ( + 2)!! =1 − − Y is a rational number. Moreover, the symmetry in the above result is also pre- served. Now, it is particularly interesting to see whether the formula will still be neat if one changes the monomial basis to the factorial basis. To see this, suppose we 1 2 +1 1 are now interested in integrating (x)=1 2 +1 over − .Onethen have
1 2 1 Theorem 28 If (x)=1 2 be any monomials of degree in − 1 ⊂ A ,and is the normalised measure on − defined in a way such that
=1
1 Z− then 1 ( 2)!! − (x) = − !! ( + 2)!! Ã ! Z 1 − =1 − Y B. Integrating Polynomial over a Sphere 109
precisely when all of are even integers, and zero otherwise.
Proof. The proof is straightforward. Note that 1 (x) = (x) 1!2! ! 1 1 Z− ··· Z− where (x) is definedasinprevioustheorem.Now,ifallthe are even, the above equation becomes
1 ( 2)!! (x) = − ( 1)!! 1!2! ! ( + 2)!! − Z 1 ··· − =1 − Y ( 2)!! ( 1)!! = − − ( + 2)!! ! =1 − Y ( 2)!! 1 = − ( + 2)!! !! − =1 Y 1 ( 2)!! − = − !! ( + 2)!! Ã=1 ! − Y as desired.
In the particular case where =4, by using two previous theorems, if (x)= ( )= is a monomial of degree ;thatis,0 and ≤ ≤ + + + = ,then 2 = ( 1)!! ( 1)!! ( 1)!! ( 1)!! ( +2)!! − − − − Z3 precisely when are all even numbers, zero otherwise. Also, note that if are all even,
2 1 = ( +2)!! !!!!!!!! Z3 µ ¶ Corollary 7 If are all even then
2 1 ∆ = ( 2)!! !!!!!!!! Z3 − µ ¶ B. Integrating Polynomial over a Sphere 110
Proof. The proof is computational. Observe that from 4.15 and by using our previous result, we have
∆
Z3
2 2 2 2 = − + − + − + −
Z3 2 1 1 1 1 = + + + !! ( 2)!!!!!!!! !! ( 2)!!!!!! !!!! ( 2)!!!! !!!!!! ( 2)!! µ − − − − ¶ 2 = + + + !! !!!!!!!! !!!!!!!! !!!!!!!! !!!!!!!! µ ¶ 2 = !! !!!!!!!! ³2 ´1 = ( 2)!! !!!!!!!! − µ ¶ as desired.
1 2 In general, if we have a monomial (x)= 1 2 of degree defined on A,then
1 2 1 2 2 1 2 2 ∆ = − + + − 1 2 1 2 ··· 1 2 ³ ´ so we have that
1 2 ∆ 1 2
1 Z− ³ ´
1 2 2 1 2 2 = − + + − 1 2 ··· 1 2 1 Z−
1 2 2 1 2 2 = − + + − 1 2 ··· 1 2 1 1 Z− Z− ( 2)!! 1 1 = − + + ( 2+ 2)!! (1 2)!! !! ··· 1!! ( 2)!! − − µ − ··· ··· − ¶ ( 2)!! 1 + 2 + + = − ··· ( + 4)!! 1!!2!! !! − µ 1··· ¶ ( 2)!! − = − !! ( + 4)!! Ã=1 ! − Y precisely when all of are even; zero otherwise. To close this appendix, we have the following Theorem. B. Integrating Polynomial over a Sphere 111
1 2 Theorem 29 Let (x)=1 2 be a monomial of degree on A .If 1 denotes the normalized measure on − so that
=1,
1 Z− then we have that
∆ (x) = ( + 2) (x) − 1 1 Z− Z−
Proof. The proof is simple. Assume all of are zero since if at least one of them is odd, then both sides reduce to zero. Let
1 ( 2)!! − = ∆ (x) = − !! ( + 4)!! Ã ! Z 1 − =1 − Y from our previous observation. Let
1 ( 2)!! − = (x) = − !! ( + 2)!! Ã ! Z 1 − =1 − Y We have that ( + 2)!! = − = ( + 2) ( + 4)!! − − as desired. Bibliography
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