S Harmonics

Notes on Spherical Harmonics and
Linear Representations of Lie Groups

Jean Gallier
Department of Computer and Information Science
University of Pennsylvania
Philadelphia, PA 19104, USA
e-mail: jean@cis.upenn.edu
November 13, 2013
Contents
1 Spherical Harmonics
1.1 Introduction, Spherical Harmonics on the Circle . . . . .
1.2 Spherical Harmonics on the 2-Sphere . . . . . . . . . . .
1.3 The Laplace-Beltrami Operator . . . . . . . . . . . . . .
1.4 Harmonic Polynomials, Spherical Harmonics and L2 (S n )
1.5 Spherical Functions and Representations of Lie Groups .
1.6 Reproducing Kernel and Zonal Spherical Functions . . .
1.7 More on the Gegenbauer Polynomials . . . . . . . . . . .
1.8 The Funk-Hecke Formula . . . . . . . . . . . . . . . . . .
1.9 Convolution on G/K, for a Gelfand Pair (G, K) . . . . .
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53
CONTENTS
Chapter 1
Spherical Harmonics and Linear
Representations of Lie Groups
1.1
Introduction, Spherical Harmonics on the Circle
In this chapter, we discuss spherical harmonics and take a glimpse at the linear representation of Lie groups. Spherical harmonics on the sphere, S 2 , have interesting applications in
computer graphics and computer vision so this material is not only important for theoretical
reasons but also for practical reasons.
Joseph Fourier (1768-1830) invented Fourier series in order to solve the heat equation
[12]. Using Fourier series, every square-integrable periodic function, f , (of period 2) can
be expressed uniquely as the sum of a power series of the form
f () = a0 +
(ak cos k + bk cos k),
k=1
where the Fourier coefficients, ak , bk , of f are given by the formulae

Z
Z
Z
1
1
1
f () d, ak =
f () cos k d, bk =
f () sin k d,
a0 =
2

for k 1. The reader will find the above formulae in Fouriers famous book [12] in Chapter
III, Section 233, page 256, essentially using the notation that we use nowdays.
This remarkable discovery has many theoretical and practical applications in physics,
signal processing, engineering, etc. We can describe Fourier series in a more conceptual
manner if we introduce the following inner product on square-integrable functions:
Z
hf, gi =
f ()g() d,
CHAPTER 1. SPHERICAL HARMONICS
which we will also denote by

Z
hf, gi =
f ()g() d,
S1
where S 1 denotes the unit circle. After all, periodic functions of (period 2) can be viewed
as functions on the circle. With this inner product, the space L2 (S 1 ) is a complete normed
vector space, that is, a Hilbert space. Furthermore, if we define the subspaces, Hk (S 1 ),
of L2 (S 1 ), so that H0 (S 1 ) (= R) is the set of constant functions and Hk (S 1 ) is the twodimensional space spanned by the functions cos k and sin k, then it turns out that we have
a Hilbert sum decomposition
M
2
1
L (S ) =
Hk (S 1 )
k=0
into pairwise orthogonal subspaces, where k=0 Hk (S 1 ) is dense in L2 (S 1 ). The functions

cos k and sin k are also orthogonal in Hk (S 1 ).
Now, it turns out that the spaces, Hk (S 1 ), arise naturally when we look for homogeneous solutions of the Laplace equation, f = 0, in R2 (Pierre-Simon Laplace, 1749-1827).
Roughly speaking, a homogeneous function in R2 is a function that can be expressed in polar
coordinates, (r, ), as
f (r, ) = rk g().
Recall that the Laplacian on R2 expressed in cartesian coordinates, (x, y), is given by
f =
2f
2f
+
,
x2
y 2
where f : R2 R is a function which is at least of class C 2 . In polar coordinates, (r, ),

where (x, y) = (r cos , r sin ) and r > 0, the Laplacian is given by

1
f
1 2f
f =
r
+ 2 2.
r r
r
r
If we restrict f to the unit circle, S 1 , then the Laplacian on S 1 is given by
s1 f =
2f
.
2
It turns out that the space Hk (S 1 ) is the eigenspace of S 1 for the eigenvalue k 2 .
To show this, we consider another question, namely, what are the harmonic functions on
R2 , that is, the functions, f , that are solutions of the Laplace equation,
f = 0.
Our ancestors had the idea that the above equation can be solved by separation of variables.
This means that we write f (r, ) = F (r)g() , where F (r) and g() are independent functions.
1.1. INTRODUCTION, SPHERICAL HARMONICS ON THE CIRCLE
To make things easier, let us assume that F (r) = rk , for some integer k 0, which means that
we assume that f is a homogeneous function of degree k. Recall that a function, f : R2 R,
is homogeneous of degree k iff
f (tx, ty) = tk f (x, y)
for all t > 0.
Now, using the Laplacian in polar coordinates, we get

(rk g())
1 2 (rk g())
1
r
+ 2
f =
r r
r
r
2

2g
1
krk g + rk2 2
=
r r
2
g
= rk2 k 2 g + rk2 2
= rk2 (k 2 g + S 1 g).
Thus, we deduce that
f = 0 iff S 1 g = k 2 g,
that is, g is an eigenfunction of S 1 for the eigenvalue k 2 . But, the above equation is
equivalent to the second-order differential equation
d2 g
+ k 2 g = 0,
d2
whose general solution is given by
g() = an cos k + bn sin k.
In summary, we found that the integers, 0, 1, 4, 9, . . . , k 2 , . . . are eigenvalues of S 1
and that the functions cos k and sin k are eigenfunctions for the eigenvalue k 2 , with
k 0. So, it looks like the dimension of the eigenspace corresponding to the eigenvalue k 2
is 1 when k = 0 and 2 when k 1.
It can indeed be shown that S 1 has no other eigenvalues and that the dimensions claimed
for the eigenspaces are correct. Observe that if we go back to our homogeneous harmonic
functions, f (r, ) = rk g(), we see that this space is spanned by the functions
uk = rk cos k,
vk = rk sin k.
Now, (x + iy)k = rk (cos k + i sin k), and since <(x + iy)k and =(x + iy)k are homogeneous
polynomials, we see that uk and vk are homogeneous polynomials called harmonic polynomials. For example, here is a list of a basis for the harmonic polynomials (in two variables)
of degree k = 0, 1, 2, 3, 4:
k
k
k
k
k
=0
=1
=2
=3
=4
1
x, y
x2 y 2 , xy
x3 3xy 2 , 3x2 y y 3
x4 6x2 y 2 + y 4 , x3 y xy 3 .
Therefore, the eigenfunctions of the Laplacian on S 1 are the restrictions of L

the harmonic
2
1
2
1
1
polynomials on R to S and we have a Hilbert sum decomposition, L (S ) =
k=0 Hk (S ).
n
n+1
It turns out that this phenomenon generalizes to the sphere S R
for all n 1.
Let us take a look at next case, n = 2.
1.2
Spherical Harmonics on the 2-Sphere
The material of section is very classical and can be found in many places, for example
Andrews, Askey and Roy [1] (Chapter 9), Sansone [25] (Chapter III), Hochstadt [17] (Chapter
6) and Lebedev [21] (Chapter ). We recommend the exposition in Lebedev [21] because we
find it particularly clear and uncluttered. We have also borrowed heavily from some lecture
notes by Hermann Gluck for a course he offered in 1997-1998.
Our goal is to find the homogeneous solutions of the Laplace equation, f = 0, in R3 ,
and to show that they correspond to spaces, Hk (S 2 ), of eigenfunctions of the Laplacian, S 2 ,
on the 2-sphere,
S 2 = {(x, y, z) R3 | x2 + y 2 + z 2 = 1}.
Then, the spaces Hk (S 2 ) will give us a Hilbert sum decomposition of the Hilbert space,
L2 (S 2 ), of square-integrable functions on S 2 . This is the generalization of Fourier series to
the 2-sphere and the functions in the spaces Hk (S 2 ) are called spherical harmonics.
The Laplacian in R3 is of course given by
f =
2f
2f
2f
+
+
.
x2
y 2
z 2
If we use spherical coordinates

x = r sin cos
y = r sin sin
z = r cos ,
in R3 , where 0 < , 0 < 2 and r > 0 (recall that is the so-called azimuthal angle
in the xy-plane originating at the x-axis and is the so-called polar angle from the z-axis,
angle defined in the plane obtained by rotating the xz-plane around the z-axis by the angle
), then the Laplacian in spherical coordinates is given by

1
1
2 f
r
+ 2 S 2 f,
f = 2
r r
r
r
where
1
S 2 f =
sin

f
1 2f
sin
+
,
sin2 2
1.2. SPHERICAL HARMONICS ON THE 2-SPHERE
is the Laplacian on the sphere, S 2 (for example, see Lebedev [21], Chapter 8 or Section 1.3,
where we derive this formula). Let us look for homogeneous harmonic functions,
f (r, , ) = rk g(, ), on R3 , that is, solutions of the Laplace equation
f = 0.
We get
f =
=
=
=

k
1
1
2 (r g)
r
+
S 2 (rk g)
r2 r
r
r2

1
krk+1 g + rk2 S 2 g
2
r r
rk2 k(k + 1)g + rk2 S 2 g
rk2 (k(k + 1)g + S 2 g).
Therefore,
f = 0 iff S 2 g = k(k + 1)g,
that is, g is an eigenfunction of S 2 for the eigenvalue k(k + 1).
We can look for solutions of the above equation using the separation of variables method.
If we let g(, ) = ()(), then we get the equation

2

sin
+
= k(k + 1),
sin
sin2 2
that is, dividing by and multiplying by sin2 ,

1 2
sin
sin
+ k(k + 1) sin2 =
.

2
Since and are independent functions, the above is possible only if both sides are equal
to a constant, say . This leads to two equations
2
+ = 0
2

sin
sin
+ k(k + 1) sin2 = 0.

However, we want to be a periodic in since we are considering functions on the sphere,

so be must of the form = m2 , for some non-negative integer, m. Then, we know that
the space of solutions of the equation
2
+ m2 = 0
2
10
is two-dimensional and is spanned by the two functions

() = cos m,
() = sin m.
We still have to solve the equation

sin
sin
+ (k(k + 1) sin2 m2 ) = 0,
which is equivalent to
sin2 00 + sin cos 0 + (k(k + 1) sin2 m2 ) = 0.
a variant of Legendres equation. For this, we use the change of variable, t = cos , and we
consider the function, u, given by u(cos ) = () (recall that 0 < ), so we get the
second-order differential equation

m2
2 00
0
(1 t )u 2tu + k(k + 1)
u=0
1 t2
sometimes called the general Legendre equation (Adrien-Marie Legendre, 1752-1833). The
trick to solve this equation is to make the substitution
m
u(t) = (1 t2 ) 2 v(t),
see Lebedev [21], Chapter 7, Section 7.12. Then, we get
(1 t2 )v 00 2(m + 1)tv 0 + (k(k + 1) m(m + 1))v = 0.
When m = 0, we get the Legendre equation:
(1 t2 )v 00 2tv 0 + k(k + 1)v = 0,
see Lebedev [21], Chapter 7, Section 7.3. This equation has two fundamental solution, Pk (t)
and Qk (t), called the Legendre functions of the first and second kinds. The Pk (t) are actually
polynomials and the Qk (t) are given by power series that diverge for t = 1, so we only keep
the Legendre polynomials, Pk (t). The Legendre polynomials can be defined in various ways.
One definition is in terms of Rodrigues formula:
Pn (t) =
1 dn 2
(t 1)n ,
2n n! dtn
see Lebedev [21], Chapter 4, Section 4.2. In this version of the Legendre polynomials they
are normalized so that Pn (1) = 1. There is also the following recurrence relation:
P0 = 1
P1 = t
(n + 1)Pn+1 = (2n + 1)tPn nPn1
n 1,
11
see Lebedev [21], Chapter 4, Section 4.3. For example, the first six Legendre polynomials
are:
1
t
1 2
(3t 1)
2
1 3
(5t 3t)
2
1
(35t4 30t2 + 3)
8
1
(63t5 70t3 + 15t).
8
Let us now return to our differential equation
(1 t2 )v 00 2(m + 1)tv 0 + (k(k + 1) m(m + 1))v = 0.
()
Observe that if we differentiate with respect to t, we get the equation

(1 t2 )v 000 2(m + 2)tv 00 + (k(k + 1) (m + 1)(m + 2))v 0 = 0.
This shows that if v is a solution of our equation () for given k and m, then v 0 is a solution
of the same equation for k and m + 1. Thus, if Pk (t) solves () for given k and m = 0, then
Pk0 (t) solves () for the same k and m = 1, Pk00 (t) solves () for the same k and m = 2, and
in general, dm /dtm (Pk (t)) solves () for k and m. Therefore, our original equation,

m2
(1 t )u 2tu + k(k + 1)
u=0
1 t2
2
00
has the solution

m
u(t) = (1 t2 ) 2
()
dm
(Pk (t)).
dtm
The function u(t) is traditionally denoted Pkm (t) and called an associated Legendre function,
see Lebedev [21], Chapter 7, Section 7.12. The index k is often called the band index .
Obviously, Pkm (t) 0 if m > k and Pk0 (t) = Pk (t), the Legendre polynomial of degree k.
An associated Legendre function is not a polynomial in general and because of the factor
m
(1 t2 ) 2 it is only defined on the closed interval [1, 1].

Certain authors add the factor (1)m in front of the expression for the associated Legendre function Pkm (t), as in Lebedev [21], Chapter 7, Section 7.12, see also footnote 29
on page 193. This seems to be common practice in the quantum mechanics literature where
it is called the Condon Shortley phase factor .
12
The associated Legendre functions satisfy various recurrence relations that allows us to
compute them. For example, for fixed m 0, we have (see Lebedev [21], Chapter 7, Section
7.12) the recurrence
m
m
(t),
(t) = (2k + 1)tPkm (t) (k + m)Pk1
(k m + 1)Pk+1
k1
and for fixed k 2 we have

Pkm+2 (t) =
2(m + 1)t
(t2
1)
1
2
Pkm+1 (t) + (k m)(k + m + 1)Pkm (t),
0mk2
which can also be used to compute Pkm starting from

Pk0 (t) = Pk (t)
kt
k
Pk1 (t) =
1 Pk (t)
1 Pk1 (t).
(t2 1) 2
(t2 1) 2
Observe that the recurrence relation for m fixed yields the following equation for k = m
m
= 0):
(as Pm1
m
Pm+1
(t) = (2m + 1)tPmm (t).
It it also easy to see that
Pmm (t) =
m
(2m)!
(1 t2 ) 2 .
m
2 m!
Observe that
(2m)!
= (2m 1)(2m 3) 5 3 1,
2m m!
an expression that is sometimes denoted (2m 1)!! and called the double factorial .

Beware that some papers in computer graphics adopt the definition of associated Legendre functions with the scale factor (1)m added so this factor is present in these papers,
for example, Green [14].
The equation above allows us to lift Pmm to the higher band m + 1. The computer
graphics community (see Green [14]) uses the following three rules to compute Pkm (t) where
0 m k:
(1) Compute
m
(2m)!
(1 t2 ) 2 .
m
2 m!
If m = k, stop. Otherwise do step 2 once:
Pmm (t) =
m
(2) Compute Pm+1
(t) = (2m + 1)tPmm (t). If k = m + 1, stop. Otherwise, iterate step 3:
13
(3) Starting from i = m + 1, compute

m
m
(i m + 1)Pi+1
(t) = (2i + 1)tPim (t) (i + m)Pi1
(t)
until i + 1 = k.
If we recall that equation () was obtained from the equation
sin2 00 + sin cos 0 + (k(k + 1) sin2 m2 ) = 0
using the substitution u(cos ) = (), we see that
() = Pkm (cos )
is a solution of the above equation. Putting everything together, as f (r, , ) = rk ()(),
we proved that the homogeneous functions,
f (r, , ) = rk cos m Pkm (cos ),
f (r, , ) = rk sin m Pkm (cos ),
are solutions of the Laplacian, , in R3 , and that the functions

cos m Pkm (cos ),
sin m Pkm (cos ),
are eigenfunctions of the Laplacian, S 2 , on the sphere for the eigenvalue k(k + 1). For k
fixed, as 0 m k, we get 2k + 1 linearly independent functions.
The notation for the above functions varies quite a bit essentially because of the choice
of normalization factors used in various fields (such as physics, seismology, geodesy, spectral
analysis, magnetics, quantum mechanics etc.). We will adopt the notation ylm , where l is a
nonnegative integer but m is allowed to be negative, with l m l. Thus, we set
0
Nl Pl (cos )
if m = 0
m
m
m
yl (, ) = 2Nl cos m Pl (cos )
if m > 0
2Nlm sin(m) Plm (cos ) if m < 0

for l = 0, 1, 2, . . ., and where the Nlm are scaling factors. In physics and computer graphics,
Nlm is chosen to be
s
(2l + 1)(l |m|)!
Nlm =
.
4(l + |m|)!
The functions ylm are called the real spherical harmonics of degree l and order m. The index
l is called the band index .
The functions, ylm , have some very nice properties but to explain these we need to recall
the Hilbert space structure of the space, L2 (S 2 ), of square-integrable functions on the sphere.
Recall that we have an inner product on L2 (S 2 ) given by
Z
Z 2 Z
hf, gi =
f g 2 =
f (, )g(, ) sin dd,
S2
14
where f, g L2 (S 2 ) and where 2 is the volume form on S 2 (induced by the metric on

2
2
R3 ). With
space using the norm,
p this inner product, L (S ) is a complete normed 2vector
2
kf k = hf, f i, associated with this inner product, that is, L (S ) is a Hilbert space. Now,
it can be shown that the Laplacian, S 2 , on the sphere is a self-adjoint linear operator with
respect to this inner product. As the functions, ylm1 1 and ylm2 2 with l1 6= l2 are eigenfunctions
corresponding to distinct eigenvalues (l1 (l1 + 1) and l2 (l2 + 1)), they are orthogonal, that
is,
if l1 6= l2 .
hylm1 1 , ylm2 2 i = 0,
It is also not hard to show that for a fixed l,
hylm1 , ylm2 i = m1 ,m2 ,
that is, the functions ylm with l m l form an orthonormal system and we denote
by Hl (S 2 ) the (2l + 1)-dimensional space spanned by these functions. It turns out that
the functions ylm form a basis of the eigenspace, El , of S 2 associated with the eigenvalue
l(l + 1) so that El = Hl (S 2 ) and that S 2 has no other eigenvalues. More is true. It turns
out that L2 (S 2 ) is the orthogonal Hilbert sum of the eigenspaces, Hl (S 2 ). This means that
the Hl (S 2 ) are
(1) mutually orthogonal
(2) closed, and
L
2
(3) The space L2 (S 2 ) is the Hilbert sum,
l=0 Hl (S ), which means that for every function,
2
2
f L (S ), there is a unique sequence of spherical harmonics, fj Hl (S 2 ), so that
f=
fl ,
l=0
P
that is, the sequence lj=0 fj , converges to fP(in the norm on L2 (S 2 )). Observe that
each fl is a unique linear combination, fl = ml aml l ylml .
Therefore, (3) gives us a Fourier decomposition on the sphere generalizing the familiar
Fourier decomposition on the circle. Furthermore, the Fourier coefficients, aml l , can be
computed using the fact that the ylm form an orthonormal Hilbert basis:
aml l = hf, ylml i.
We also have the corresponding homogeneous harmonic functions, Hlm (r, , ), on R3
given by
Hlm (r, , ) = rl ylm (, ).
If one starts computing explicity the Hlm for small values of l and m, one finds that it is
always possible to express these functions in terms of the cartesian coordinates x, y, z as
1.3. THE LAPLACE-BELTRAMI OPERATOR
15
homogeneous polynomials! This remarkable fact holds in general: The eigenfunctions of

the Laplacian, S 2 , and thus, the spherical harmonics, are the restrictions of homogeneous
harmonic polynomials in R3 . Here is a list of bases of the homogeneous harmonic polynomials
of degree k in three variables up to k = 4 (thanks to Herman Gluck):
k
k
k
k
=0
=1
=2
=3
k=4
1
x, y, z
x2 y 2 , x2 z 2 , xy, xz, yz
x3 3xy 2 , 3x2 y y 3 , x3 3xz 2 , 3x2 z z 3 ,
y 3 3yz 2 , 3y 2 z z 3 , xyz
x4 6x2 y 2 + y 4 , x4 6x2 z 2 + z 4 , y 4 6y 2 z 2 + z 4 ,
x3 y xy 3 , x3 z xz 3 , y 3 z yz 3 ,
3x2 yz yz 3 , 3xy 2 z xz 3 , 3xyz 2 x3 y.
Subsequent sections will be devoted to a proof of the important facts stated earlier.
1.3
The Laplace-Beltrami Operator
In order to define rigorously the Laplacian on the sphere, S n Rn+1 , and establish its
relationship with the Laplacian on Rn+1 , we need the definition of the Laplacian on a Riemannian manifold, (M, g), the Laplace-Beltrami operator , as defined in Section ?? (Eugenio
Beltrami, 1835-1900). In that section, the Laplace-Beltrami operator is defined as an operator on differential forms but a more direct definition can be given for the Laplacian-Beltrami
operator on functions (using the covariant derivative, see the paragraph preceding Proposition ??). For the benefit of the reader who may not have read Section ??, we present this
definition of the divergence again.
Recall that a Riemannian metric, g, on a manifold, M , is a smooth family of inner
products, g = (gp ), where gp is an inner product on the tangent space, Tp M , for every
p M . The inner product, gp , on Tp M , establishes a canonical duality between Tp M
and Tp M , namely, we have the isomorphism, [ : Tp M Tp M , defined such that for every
u Tp M , the linear form, u[ Tp M , is given by
u[ (v) = gp (u, v),
v Tp M.
The inverse isomorphism, ] : Tp M Tp M , is defined such that for every Tp M , the

vector, ] , is the unique vector in Tp M so that
gp ( ] , v) = (v),
v Tp M.
The isomorphisms [ and ] induce isomorphisms between vector fields, X X(M ), and oneforms, A1 (M ). In particular, for every smooth function, f C (M ), the vector field
16
corresponding to the one-form, df , is the gradient, grad f , of f . The gradient of f is uniquely

determined by the condition
gp ((grad f )p , v) = dfp (v),
v Tp M, p M.
If X is the covariant derivative associated with the Levi-Civita connection induced by the
metric, g, then the divergence of a vector field, X X(M ), is the function, div X : M R,
defined so that
(div X)(p) = tr(Y (p) 7 (Y X)p ),
namely, for every p, (div X)(p) is the trace of the linear map, Y (p) 7 (Y X)p . Then, the
Laplace-Beltrami operator , for short, Laplacian, is the linear operator,
: C (M ) C (M ), given by
f = div grad f.
Observe that the definition just given differs from the definition given in Section ?? by
a negative sign. We adopted this sign convention to conform with most of the literature on
spherical harmonics (where the negative sign is omitted). A consequence of this choice is
that the eigenvalues of the Laplacian are negative.
For more details on the Laplace-Beltrami operator, we refer the reader to Chapter ?? or
to Gallot, Hulin and Lafontaine [13] (Chapter 4) or ONeill [23] (Chapter 3), Postnikov [24]
(Chapter 13), Helgason [16] (Chapter 2) or Warner [29] (Chapters 4 and 6).
All this being rather abstact, it is useful to know how grad f , div X and f are expressed
in a chart. If (U, ) is a chart of M , with p M and if, as usual,

!

,...,
x1 p
xn p
denotes the basis of Tp M induced by , the local expression of the metric g at p is given by
the n n matrix, (gij )p , with

!

,
(gij )p = gp
.
xi p
xj p
The matrix (gij )p is symmetric, positive definite and its inverse is denoted (g ij )p . We also
let |g|p = det(gij )p . For simplicity of notation we often omit the subscript p. Then, it can be
shown that for every function, f C (M ), in local coordinates given by the chart (U, ),
we have
X
f
grad f =
g ij
,
x
x
j
i
ij
where, as usual
f
(p) =
xj
xj

f=
p
(f 1 )
((p))
uj
17
and (u1 , . . . , un ) are the coordinate functions in Rn . There are formulae for div X and f
involving the Christoffel symbols but the following formulae will be more convenient for our
purposes: For every vector field, X X(M ), expressed in local coordinates as
X=
n
X
i=1
we have
Xi
xi
n

1 X p
div X = p
|g| Xi
|g| i=1 xi
and for every function, f C (M ), the Laplacian, f , is given by

1 X p
ij f
|g| g
.
f = p
xj
|g| i,j xi
The above formula is proved in Proposition ??, assuming M is orientable. A different
derivation is given in Postnikov [24] (Chapter 13, Section 5).
One should check that for M = Rn with its standard coordinates, the Laplacian is given
by the familiar formula
2f
2f
f =
+
+
.
x21
x2n
Remark: A different sign convention is also used in defining the divergence, namely,
n

1 X p
div X = p
|g| Xi .
|g| i=1 xi
With this convention, which is the one used in Section ??, the Laplacian also has a negative
sign. This has the advantage that the eigenvalues of the Laplacian are nonnegative.
As an application, let us derive the formula for the Laplacian in spherical coordinates,
x = r sin cos
y = r sin sin
z = r cos .
We have
= sin cos
+ sin sin
+ cos
= rb
r
x
y
z

= r cos cos
+ cos sin
sin
= rb
x
y
z

= r sin sin
+ sin cos
= r.
b
x
y
18
Observe that rb, b and

b are pairwise orthogonal. Therefore, the matrix (gij ) is given by
1 0
0
0
(gij ) = 0 r2
2
0 0 r sin2
and |g| = r4 sin2 . The inverse of (gij ) is
1 0
0
.
0
(g ij ) = 0 r2
2
2
0 0 r sin
We will let the reader finish the computation to verify that we get

1
f
1
2f
1
2 f
r
+ 2
sin
+ 2 2
.
f = 2
r r
r
r sin
r sin 2
Since (, ) are coordinates on the sphere S 2 via
x = sin cos
y = sin sin
z = cos ,
we see that in these coordinates, the metric, (e
gij ), on S 2 is given by the matrix

1
0
(e
gij ) =
,
0 sin2
that |e
g | = sin2 , and that the inverse of (e
gij ) is

1
0
ij
(e
g )=
.
0 sin2
It follows immediately that
1
S 2 f =
sin
so we have verified that
1
f = 2
r r

f
1 2f
sin
+
,
sin2 2

1
2 f
r
+ 2 S 2 f.
r
r
Let us now generalize the above formula to the Laplacian, , on Rn+1 and the Laplacian,
S n , on S n , where
S n = {(x1 , . . . , xn+1 ) Rn+1 | x21 + + x2n+1 = 1}.
19
Following Morimoto [22] (Chapter 2, Section 2), let us use polar coordinates. The map
from R+ S n to Rn+1 {0} given by
(r, ) 7 r
is clearly a diffeomorphism. Thus, for any system of coordinates, (u1 , . . . , un ), on S n , the
tuple (u1 , . . . , un , r) is a system of coordinates on Rn+1 {0} called polar coordinates. Let
us establish the relationship between the Laplacian, , on Rn+1 {0} in polar coordinates
and the Laplacian, S n , on S n in local coordinates (u1 , . . . , un ).
Proposition 1.1 If is the Laplacian on Rn+1 {0} in polar coordinates (u1 , . . . , un , r)
and S n is the Laplacian on the sphere, S n , in local coordinates (u1 , . . . , un ), then

1
1
n f
r
+ 2 S n f.
f = n
r r
r
r
Proof . Let us compute the (n + 1) (n + 1) matrix, G = (gij ), expressing the metric on Rn+1
e = (e
is polar coordinates and the n n matrix, G
gij ), expressing the metric on S n . Recall
n
that if S , then = 1 and so,
= 0,
ui
as
1 ( )
=
= 0.
ui
2 ui
If x = r with S n , we have
x
=r
,
ui
ui
and
1 i n,
x
= .
r
It follows that
x x

= r2
= r2 geij
ui uj
ui uj
x x
=r
=0
ui r
ui
x x
=
= = 1.
r r
gij =
gin+1
gn+1n+1
Consequently, we get

G=

e 0
r2 G
,
0 1
20
|g| = r2n |e
g | and
1

=

e1 0
r2 G
.
0
1
Using the above equations and

1 X p
ij f
,
|g| g
f = p
xj
|g| i,j xi
we get

n
X
p
p f
1
1 ij f
n
n
p
f =
r |e
+ p
r |e
g | 2 ge
g|
r
xj
r
rn |e
g | i,j=1 xi
rn |e
g | r

n
X
1
p
f
1
f
p
=
|e
g | geij
+ n
rn
2
xj
r r
r
r |e
g | i,j=1 xi

1
1
f
= 2 S n f + n
rn
,
r
r r
r
as claimed.
It is also possible to express S n in terms of S n1 . If en+1 = (0, . . . , 0, 1) Rn+1 , then
we can view S n1 as the intersection of S n with the hyperplane, xn+1 = 0, that is, as the set
S n1 = { S n | en+1 = 0}.
If (u1 , . . . , un1 ) are local coordinates on S n1 , then (u1 , . . . , un1 , ) are local coordinates
on S n , by setting
= sin
e + cos en+1 ,
with
e S n1 and 0 < . Using these local coordinate systems, it is a good exercise to
find the relationship between S n and S n1 , namely

1
1
n1 f
sin
+
S f.
S n f =
n1
sin

sin2 n1
A fundamental property of the divergence is known as Greens Formula. There are
actually two Greens Formulae but we will only need the version for an orientable manifold
without boundary given in Proposition ??. Recall that Greens Formula states that if M is a
compact, orientable, Riemannian manifold without boundary, then, for every smooth vector
field, X X(M ), we have
Z
(div X) M = 0,
M
where M is the volume form on M induced by the metric.
21
If M is a compact, orientable Riemannian manifold, then for any two smooth functions,
f, h C (M ), we define hf, hi by
Z
hf, hi =
f h M .
M
Then, it is not hard to show that h, i is an inner product on C (M ).

An important property of the Laplacian on a compact, orientable Riemannian manifold
is that it is a self-adjoint operator. This fact has already been proved in the more general
case of an inner product on differential forms in Proposition ?? but it might be instructive
to give another proof in the special case of functions using Greens Formula.
For this, we prove the following properties: For any two functions, f, h C (M ), and
any vector field, X C (M ), we have:
div(f X) = f div X + X(f ) = f div X + g(grad f, X)
grad f (h) = g(grad f, grad h) = grad h (f ).
Using these identities, we obtain the following important special case of Proposition ??:
Proposition 1.2 Let M be a compact, orientable, Riemannian manifold without boundary.
The Laplacian on M is self-adjoint, that is, for any two functions, f, h C (M ), we have
hf, hi = hf, hi
or equivalently
Z
hf M .
f h M =
M
Proof . By the two identities before Proposition 1.2,

f h = f div grad h = div(f grad h) g(grad f, grad h)
and
hf = hdiv grad f = div(hgrad f ) g(grad h, grad f ),
so we get
f h hf = div(f grad h hgrad f ).
By Greens Formula,
Z
Z
(f h hf )M =
div(f grad h hgrad f )M = 0,
M
which proves that is self-adjoint.

The importance of Proposition 1.2 lies in the fact that as h, i is an inner product on
C (M ), the eigenspaces of for distinct eigenvalues are pairwise orthogonal. We will make
heavy use of this property in the next section on harmonic polynomials.
22
1.4
Harmonic Polynomials, Spherical Harmonics and

L2(S n)
Harmonic homogeneous polynomials and their restrictions to S n , where

S n = {(x1 , . . . , xn+1 ) Rn+1 | x21 + + x2n+1 = 1},
turn out to play a crucial role in understanding the structure of the eigenspaces of the Laplacian on S n (with n 1). The results in this section appear in one form or another in Stein
and Weiss [26] (Chapter 4), Morimoto [22] (Chapter 2), Helgason [16] (Introduction, Section
3), Dieudonne [6] (Chapter 7), Axler, Bourdon and Ramey [2] (Chapter 5) and Vilenkin
[28] (Chapter IX). Some of these sources assume a fair amount of mathematical background
and consequently, uninitiated readers will probably find the exposition rather condensed,
especially Helgason. We tried hard to make our presentation more user-friendly.
Definition 1.1 Let Pk (n + 1) (resp. PkC (n + 1)) denote the space of homogeneous polynomials of degree k in n + 1 variables with real coefficients (resp. complex coefficients) and let
Pk (S n ) (resp. PkC (S n )) denote the restrictions of homogeneous polynomials in Pk (n + 1) to
S n (resp. the restrictions of homogeneous polynomials in PkC (n + 1) to S n ). Let Hk (n + 1)
(resp. HkC (n + 1)) denote the space of (real) harmonic polynomials (resp. complex harmonic
polynomials), with
Hk (n + 1) = {P Pk (n + 1) | P = 0}
and
HkC (n + 1) = {P PkC (n + 1) | P = 0}.
Harmonic polynomials are sometimes called solid harmonics. Finally, Let Hk (S n ) (resp.
HkC (S n )) denote the space of (real) spherical harmonics (resp. complex spherical harmonics)
be the set of restrictions of harmonic polynomials in Hk (n + 1) to S n (resp. restrictions of
harmonic polynomials in HkC (n + 1) to S n ).
A function, f : Rn R (resp. f : Rn C), is homogeneous of degree k iff
f (tx) = tk f (x),
for all x Rn and t > 0.
The restriction map, : Hk (n + 1) Hk (S n ), is a surjective linear map. In fact, it is a

bijection. Indeed, if P Hk (n + 1), observe that

x
x
k
,
with
S n,
P (x) = kxk P
kxk
kxk
for all x 6= 0. Consequently, if P S n = Q S n , that is, P () = Q() for all S n , then

x
x
k
k
P (x) = kxk P
= kxk Q
= Q(x)
kxk
kxk
1.4. HARMONIC POLYNOMIALS, SPHERICAL HARMONICS AND L2 (S N )
23
for all x 6= 0, which implies P = Q (as P and Q are polynomials). Therefore, we have a
linear isomorphism between Hk (n + 1) and Hk (S n ) (and between HkC (n + 1) and HkC (S n )).
It will be convenient to introduce some notation to deal with homogeneous polynomials.
Given n 1 variables, x1 , . . . , xn , and any n-tuple of nonnegative integers, = (1 , . . . , n ),
let || = 1 + +n , let x = x1 1 xnn and let ! = 1 ! n !. Then, every homogeneous
polynomial, P , of degree k in the variables x1 , . . . , xn can be written uniquely as
X
P =
c x ,
||=k
with c R or c C. It is well known that Pk (n) is a (real) vector space of dimension

dk =

n+k1
k
and PkC (n) is a complex vector space of the same dimension, dk .

We can define an Hermitian inner product on PkC (n) whose restriction to Pk (n) is an
inner product by
polynomial as a differential operator as follows:
P viewinga homogeneous
C
For every P = ||=k c x Pk (n), let
(P ) =
||=k
x1 1
k
.
xnn
Then, for any two polynomials, P, Q PkC (n), let

hP, Qi = (P )Q.
A simple computation shows that
*
+
X
||=k
a x ,
b x
||=k
! a b .
||=k
Therefore, hP, Qi is indeed an inner product. Also observe that

(x21 + + x2n ) =
2
2
+
+
= .
x21
x2n
Another useful property of our inner product is this:

hP, QRi = h(Q)P, Ri.
24
Indeed.
hP, QRi =
=
=
=
=
=
hQR, P i
(QR)P
(Q)((R)P )
(R)((Q)P )
hR, (Q)P i
h(Q)P, Ri.
In particular,
h(x21 + + x2n )P, Qi = hP, (x21 + + x2n )Qi = hP, Qi.
Let us write kxk2 for x21 + + x2n . Using our inner product, we can prove the following
important theorem:
Theorem 1.3 The map, : Pk (n) Pk2 (n), is surjective for all n, k 2 (and simiC
larly for : PkC (n) Pk2
(n)). Furthermore, we have the following orthogonal direct sum
decompositions:
Pk (n) = Hk (n) kxk2 Hk2 (n) kxk2j Hk2j (n) kxk2[k/2] H[k/2] (n)
and
C
C
C
PkC (n) = HkC (n) kxk2 Hk2
(n) kxk2j Hk2j
(n) kxk2[k/2] H[k/2]
(n),
with the understanding that only the first term occurs on the right-hand side when k < 2.
C
Proof . If the map : PkC (n) Pk2
(n) is not surjective, then some nonzero polynomial,
C
Q Pk2
(n), is orthogonal to the image of . In particular, Q must be orthogonal to P
with P = kxk2 Q PkC (n). So, using a fact established earlier,
0 = hQ, P i = hkxk2 Q, P i = hP, P i,

which implies that P = kxk2 Q = 0 and thus, Q = 0, a contradiction. The same proof is
valid in the real case.
We claim that we have an orthogonal direct sum decomposition,
C
PkC (n) = HkC (n) kxk2 Pk2
(n),
and similarly in the real case, with the understanding that the second term is missing if
k < 2. If k = 0, 1, then PkC (n) = HkC (n) so this case is trivial. Assume k 2. Since
C
Ker = HkC (n) and is surjective, dim(PkC (n)) = dim(HkC (n)) + dim(Pk2
(n)), so it is
25
C
sufficient to prove that HkC (n) is orthogonal to kxk2 Pk2
(n). Now, if H HkC (n) and
C
(n), we have
P = kxk2 Q kxk2 Pk2
hkxk2 Q, Hi = hQ, Hi = 0,
C
so HkC (n) and kxk2 Pk2
(n) are indeed orthogonal. Using induction, we immediately get the
orthogonal direct sum decomposition
C
C
C
PkC (n) = HkC (n) kxk2 Hk2
(n) kxk2j Hk2j
(n) kxk2[k/2] H[k/2]
(n)
and the corresponding real version.

Remark: Theorem 1.3 also holds for n = 1.
Theorem 1.3 has some important corollaries. Since every polynomial in n + 1 variables
is the sum of homogeneous polynomials, we get:
Corollary 1.4 The restriction to S n of every polynomial (resp. complex polynomial) in
n + 1 2 variables is a sum of restrictions to S n of harmonic polynomials (resp. complex
harmonic polynomials).
We can also derive a formula for the dimension of Hk (n) (and HkC (n)).
Corollary 1.5 The dimension, ak,n , of the space of harmonic polynomials, Hk (n), is given
by the formula

n+k1
n+k3
ak,n =
k
k2
if n, k 2, with a0,n = 1 and a1,n = n, and similarly for HkC (n). As Hk (n + 1) is isomorphic
to Hk (S n ) (and HkC (n + 1) is isomorphic to HkC (S n )) we have

n+k
n+k2
C
n
n
dim(Hk (S )) = dim(Hk (S )) = ak,n+1 =
.
k
k2
Proof . The cases k = 0 and k = 1 are trivial since in this case Hk (n) = Pk (n). For k 2,
the result follows from the direct sum decomposition
Pk (n) = Hk (n) kxk2 Pk2 (n)
proved earlier. The proof is identical in the complex case.
Observe that when n = 2, we get ak,2 = 2 for k 1 and when n = 3, we get ak,3 = 2k + 1
for all k 0, which we already knew from Section 1.2. The formula even applies for n = 1
and yields ak,1 = 0 for k 2.
26
Remark: It is easy to show that

ak,n+1 =

n+k1
n+k2
+
n1
n1
for k 2, see Morimoto [22] (Chapter 2, Theorem 2.4) or Dieudonne [6] (Chapter 7, formula
99), where a different proof technique is used.
Let L2 (S n ) be the space of (real) square-integrable functions on the sphere, S n . We have
an inner product on L2 (S n ) given by
Z
hf, gi =
f g n ,
Sn
where f, g L2 (S n ) and where n is the volume form on S n (induced by the metric on

2
n
Rn+1 ). With this
using the norm,
p inner product, L (S ) is a complete normed vector2 space
n
kf k = kf k2 = hf, f i, associated with this inner product, that is, L (S ) is a Hilbert space.
In the case of complex-valued functions, we use the Hermitian inner product
Z
hf, gi =
f g n
Sn
and we get the complex Hilbert space, L2C (S n ). We also denote by C(S n ) the space of
continuous (real) functions on S n with the L norm, that is,
kf k = sup{|f (x)|}xS n
and by CC (S n ) the space of continuous complex-valued functions on S n also with the L
norm. Recall that C(S n ) is dense in L2 (S n ) (and CC (S n ) is dense in L2C (S n )). The following
proposition shows why the spherical harmonics play an important role:
Proposition
1.6 The set of all finite linear combinations of elements in
S
C
n
k=0 Hk (S )) is
k=0
Hk (S n ) (resp.
(i) dense in C(S n ) (resp. in CC (S n )) with respect to the L -norm;

(ii) dense in L2 (S n ) (resp. dense in L2C (S n )).
Proof . (i) As S n is compact, by the Stone-Weierstrass approximation theorem (Lang [20],
Chapter III, Corollary 1.3), if g is continuous on S n , then it can be approximated uniformly
by polynomials, Pj , restricted to SSn . By Corollary 1.4, the restriction of each Pj to S n is a
n
linear combination of elements in
k=0 Hk (S ).
(ii) We use the fact that C(S n ) is dense in L2 (S n ). Given f L2 (S n ), for every > 0,
we can choose a continuous function, g, so that kf gk2 < /2. By (i), we can find a linear
27
p
S
n
combination, h, of elements in
vol(S n )), where
k=0 Hk (S ) so that kg hk < /(2
vol(S n ) is the volume of S n (really, area). Thus, we get
p
kf hk2 kf gk2 + kg hk2 < /2 + vol(S n ) kg hk < /2 + /2 = ,
which proves (ii). The proof in the complex case is identical.
We need one more proposition before showing that the spaces Hk (S n ) constitute an
orthogonal Hilbert space decomposition of L2 (S n ).
Proposition 1.7 For every harmonic polynomial, P Hk (n + 1) (resp. P HkC (n + 1)),
the restriction, H Hk (S n ) (resp. H HkC (S n )), of P to S n is an eigenfunction of S n for
the eigenvalue k(n + k 1).
Proof . We have
P (r) = rk H(),
r > 0, S n ,
and by Proposition 1.1, for any f C (Rn+1 ), we have

1
1
n f
f = n
r
+ 2 S n f.
r r
r
r
Consequently,
k
P = (r H) =
=
=
=

k
1
1
n (r H)
S n (rk H)
r
+
rn r
r
r2

1
krn+k1 H + rk2 S n H
n
r r
1
k(n + k 1)rn+k2 H + rk2 S n H
rn
rk2 (k(n + k 1)H + S n H).
Thus,
P = 0 iff S n H = k(n + k 1)H,
as claimed.
From Proposition 1.7, we deduce that the space Hk (S n ) is a subspace of the eigenspace,
Ek , of S n , associated with the eigenvalue k(n + k 1) (and similarly for HkC (S n )). Remarkably, Ek = Hk (S n ) but it will take more work to prove this.
What we can deduce immediately is that Hk (S n ) and Hl (S n ) are pairwise orthogonal
whenever k 6= l. This is because, by Proposition 1.2, the Laplacian is self-adjoint and thus,
any two eigenspaces, Ek and El are pairwise orthogonal whenever k 6= l and as Hk (S n ) Ek
n
and Hl (S n ) El , our claim is indeed true. Furthermore, by Proposition 1.5, each
SHk (S ) nis
finite-dimensional and thus, closed. Finally, we know from Proposition 1.6 that k=0 Hk (S )
is dense in L2 (S n ). But then, we can apply a standard result from Hilbert space theory (for
example, see Lang [20], Chapter V, Proposition 1.9) to deduce the following important result:
28
Theorem 1.8 The family of spaces, Hk (S n ) (resp. HkC (S n )) yields a Hilbert space direct
sum decomposition
2
L (S ) =
Hk (S )
(resp.
L2C (S n )
k=0
HkC (S n )),
k=0
which means that the summands are closed, pairwise orthogonal, and that every f L2 (S n )
(resp. f L2C (S n )) is the sum of a converging series
f=
fk ,
k=0
in the L2 -norm, where the fk Hk (S n ) (resp. fk HkC (S n )) are uniquely determined

a
functions. Furthermore, given any orthonormal basis, (Yk1 , . . . , Yk k,n+1 ), of Hk (S n ), we have
ak,n+1
fk =
ck,mk Ykmk ,
with ck,mk = hf, Ykmk i.
mk =1
The coefficients ck,mk are generalized Fourier coefficients with respect to the Hilbert
basis {Ykmk | 1 mk ak,n+1 , k 0}. We can finally prove the main theorem of this section.
Theorem 1.9
(1) The eigenspaces (resp. complex eigenspaces) of the Laplacian, S n , on S n are the
spaces of spherical harmonics,
Ek = Hk (S n )
(resp.
Ek = HkC (S n ))
and Ek corresponds to the eigenvalue k(n + k 1).

(2) We have the Hilbert space direct sum decompositions
2
L (S ) =
M
k=0
Ek
(resp.
L2C (S n )
Ek ).
k=0
(3) The complex polynomials of the form (c1 x1 + + cn+1 xn+1 )k , with c21 + + c2n+1 = 0,
span the space HkC (n + 1), for k 1.
Proof . We follow essentially the proof in Helgason [16] (Introduction, Theorem 3.1). In (1)
and (2) we only deal with the real case, the proof in the complex case being identical.
(1) We already know that the integers k(n + k 1) are eigenvalues of S n and that
Hk (S n ) Ek . We will prove that S n has no other eigenvalues and no other eigenvectors
29
using the Hilbert basis, {Ykmk | 1 mk ak,n+1 , k 0}, given by Theorem 1.8. Let be
any eigenvalue of S n and let f L2 (S n ) be any eigenfunction associated with so that
f = f.
We have a unique series expansion
f=
k,n+1
aX
X
ck,mk Ykmk ,
k=0 mk =1
with ck,mk = hf, Ykmk i. Now, as S n is self-adjoint and Ykmk = k(n + k 1)Ykmk , the
Fourier coefficients, dk,mk , of f are given by
dk,mk = hf, Ykmk i = hf, Ykmk i = k(n + k 1)hf, Ykmk i = k(n + k 1)ck,mk .
On the other hand, as f = f , the Fourier coefficients of f are given by
dk,mk = ck,mk .
By uniqueness of the Fourier expansion, we must have
ck,mk = k(n + k 1)ck,mk
for all k 0.
Since f 6= 0, there some k such that ck,mk 6= 0 and we must have

= k(n + k 1)
for any such k. However, the function k 7 k(n + k 1) reaches its maximum for k = n1
2
and as n 1, it is strictly decreasing for k 0, which implies that k is unique and that
cj,mj = 0
for all j 6= k.
Therefore, f Hk (S n ) and the eigenvalues of S n are exactly the integers k(n + k 1) so

Ek = Hk (S n ), as claimed.
Since we just proved that Ek = Hk (S n ), (2) follows immediately from the Hilbert decomposition given by Theorem 1.8.
(3) If H = (c1 x1 + + cn+1 xn+1 )k , with c21 + + c2n+1 = 0, then for k 1 is is obvious
that H = 0 and for k 2 we have
H = k(k 1)(c21 + + c2n+1 )(c1 x1 + + cn+1 xn+1 )k2 = 0,
so H HkC (n + 1). A simple computation shows that for every Q PkC (n + 1), if
c = (c1 , . . . , cn+1 ), then we have
(Q)(c1 x1 + + cn+1 xn+1 )m = m(m 1) (m k + 1)Q(c)(c1 x1 + + cn+1 xn+1 )mk ,
30
for all m k 1.
Assume that HkC (n + 1) is not spanned by the complex polynomials of the form (c1 x1 +
+ cn+1 xn+1 )k , with c21 + + c2n+1 = 0, for k 1. Then, some Q HkC (n + 1) is orthogonal
to all polynomials of the form H = (c1 x1 + + cn+1 xn+1 )k , with c21 + + c2n+1 = 0. Recall
that
hP, (Q)Hi = hQP, Hi
and apply this equation to P = Q(c), H and Q. Since
(Q)H = (Q)(c1 x1 + + cn+1 xn+1 )k = k!Q(c),
as Q is orthogonal to H, we get
k!hQ(c), Q(c)i = hQ(c), k!Q(c)i = hQ(c), (Q)Hi = hQ Q(c), Hi = Q(c)hQ, Hi = 0,
which implies Q(c) = 0. Consequently, Q(x1 , . . . , xn+1 ) vanishes on the complex algebraic
variety,
{(x1 , . . . , xn+1 ) Cn+1 | x21 + + x2n+1 = 0}.
By the Hilbert Nullstellensatz , some power, Qm , belongs to the ideal, (x21 + + x2n+1 ),
generated by x21 + +x2n+1 . Now, if n 2, it is well-known that the polynomial x21 + +x2n+1
is irreducible so the ideal (x21 + + x2n+1 ) is a prime ideal and thus, Q is divisible by
x21 + +x2n+1 . However, we know from the proof of Theorem 1.3 that we have an orthogonal
direct sum
C
PkC (n + 1) = HkC (n + 1) kxk2 Pk2
(n + 1).
Since Q HkC (n + 1) and Q is divisible by x21 + + x2n+1 , we must have Q = 0. Therefore,
if n 2, we proved (3). However, when n = 1, we know from Section 1.1 that the complex
harmonic homogeneous polynomials in two variables, P (x, y), are spanned by the real and
imaginary parts, Uk , Vk of the polynomial (x + iy)k = Uk + iVk . Since (x iy)k = Uk iVk
we see that
Uk =

1
(x + iy)k + (x iy)k ,
2
Vk =

1
(x + iy)k (x iy)k ,
2i
and as 1 + i2 = 1 + (i)2 = 0, the space HkC (R2 ) is spanned by (x + iy)k and (x iy)k (for
k 1), so (3) holds for n = 1 as well.
As an illustration of part (3) of Theorem 1.9, the polynomials (x1 + i cos x2 + i sin x3 )k
are harmonic. Of course, the real and imaginary part of a complex harmonic polynomial
(c1 x1 + + cn+1 xn+1 )k are real harmonic polynomials.
In the next section, we try to show how spherical harmonics fit into the broader framework
of linear respresentations of (Lie) groups.
1.5. SPHERICAL FUNCTIONS AND REPRESENTATIONS OF LIE GROUPS
1.5
31
Spherical Functions and Linear Representations of

Lie Groups; A Glimpse
In this section, we indicate briefly how Theorem 1.9 (except part (3)) can be viewed as a
special case of a famous theorem known as the Peter-Weyl Theorem about unitary representations of compact Lie groups (Herman, Klauss, Hugo Weyl, 1885-1955). First, we review
the notion of a linear representation of a group. A good and easy-going introduction to
representations of Lie groups can be found in Hall [15]. We begin with finite-dimensional
representations.
Definition 1.2 Given a Lie group, G, and a vector space, V , of dimension n, a linear
representation of G of dimension (or degree n) is a group homomorphism, U : G GL(V ),
such that the map, g 7 U (g)(u), is continuous for every u V and where GL(V ) denotes
the group of invertible linear maps from V to itself. The space, V , called the representation
space may be a real or a complex vector space. If V has a Hermitian (resp Euclidean) inner
product, h, i, we say that U : G GL(V ) is a unitary representation iff
hU (g)(u), U (g)(v)i = hu, vi,
for all g G and all u, v V.
Thus, a linear representation of G is a map, U : G GL(V ), satisfying the properties:

U (gh) = U (g)U (h)
U (g 1 ) = U (g)1
U (1) = I.
For simplicity of language, we usually abbreviate linear representation as representation. The representation space, V , is also called a G-module since the representation,
U : G GL(V ), is equivalent to the left action, : G V V , with g v = U (g)(v).
The representation such that U (g) = I for all g G is called the trivial representation.
As an example, we describe a class of representations of SL(2, C), the group of complex
matrices with determinant +1,

a b
,
ad bc = 1.
c d
Recall that PkC (2) denotes the vector space of complex homogeneous polynomials of degree
k in two variables, (z1 , z2 ). For every matrix, A SL(2, C), with

a b
A=
c d
for every homogeneous polynomial, Q PkC (2), we define Uk (A)(Q(z1 , z2 )) by
Uk (A)(Q(z1 , z2 )) = Q(dz1 bz2 , cz1 + az2 ).
32
If we think of the homogeneous polynomial, Q(z1 , z2 ), as a function, Q

z1
, then
z2

z1
z1
d b
1 z1
=Q
.
Uk (A) Q
= QA
c a
z2
z2
z2
z1
z2

, of the vector
The expression above makes it clear that

Uk (AB) = Uk (A)Uk (B)
for any two matrices, A, B SL(2, C), so Uk is indeed a representation of SL(2, C) into
PkC (2). It can be shown that the representations, Uk , are irreducible and that every representation of SL(2, C) is equivalent to one of the Uk s (see Brocker and tom Dieck [4], Chapter
2, Section 5). The representations, Uk , are also representations of SU(2). Again, they are
irreducible representations of SU(2) and they constitute all of them (up to equivalence).
The reader should consult Hall [15] for more examples of representations of Lie groups.
One might wonder why we considered SL(2, C) rather than SL(2, R). This is because it
can be shown that SL(2, R) has no nontrivial unitary (finite-dimensional) representations!
For more on representations of SL(2, R), see Dieudonne [6] (Chapter 14).
Given any basis, (e1 , . . . , en ), of V , each U (g) is represented by an n n matrix,
U (g) = (Uij (g)). We may think of the scalar functions, g 7 Uij (g), as special functions on
G. As explained in Dieudonne [6] (see also Vilenkin [28]), essentially all special functions
(Legendre polynomials, ultraspherical polynomials, Bessel functions, etc.) arise in this way
by choosing some suitable G and V . There is a natural and useful notion of equivalence of
representations:
Definition 1.3 Given any two representations, U1 : G GL(V1 ) and U2 : G GL(V2 ), a
G-map (or morphism of representations), : U1 U2 , is a linear map, : V1 V2 , so that
the following diagram commutes for every g G:
V1
U1 (g)
V1
V2
U2 (g)
V2 .
The space of all G-maps between two representations as above is denoted HomG (U1 , U2 ).
Two representations U1 : G GL(V1 ) and U2 : G GL(V2 ) are equivalent iff : V1 V2
is an invertible linear map (which implies that dim V1 = dim V2 ). In terms of matrices, the
representations U1 : G GL(V1 ) and U2 : G GL(V2 ) are equivalent iff there is some
invertible n n matrix, P , so that
U2 (g) = P U1 (g)P 1 ,
g G.
33
If W V is a subspace of V , then in some cases, a representation U : G GL(V ) yields

a representation U : G GL(W ). This is interesting because under certain conditions on
G (e.g., G compact) every representation may be decomposed into a sum of so-called
irreducible representations and thus, the study of all representations of G boils down to the
study of irreducible representations of G (for instance, see Knapp [19] (Chapter 4, Corollary
4.7) or Brocker and tom Dieck [4] (Chapter 2, Proposition 1.9).
Definition 1.4 Let U : G GL(V ) be a representation of G. If W V is a subspace of V ,
then we say that W is invariant (or stable) under U iff U (g)(w) W , for all g G and all
w W . If W is invariant under U , then we have a homomorphism, U : G GL(W ), called
a subrepresentation of G. A representation, U : G GL(V ), with V 6= (0) is irreducible
iff it only has the two subrepresentations, U : G GL(W ), corresponding to W = (0) or
W =V.
An easy but crucial lemma about irreducible representations is Schurs Lemma.
Lemma 1.10 (Schurs Lemma) Let U1 : G GL(V ) and U2 : G GL(W ) be any two real
or complex representations of a group, G. If U1 and U2 are irreducible, then the following
properties hold:
(i) Every G-map, : U1 U2 , is either the zero map or an isomorphism.
(ii) If U1 is a complex representation, then every G-map, : U1 U1 , is of the form,
= id, for some C.
Proof . (i) Observe that the kernel, Ker V , of is invariant under U1 . Indeed, for every
v Ker and every g G, we have
(U1 (g)(v)) = U2 (g)((v)) = U2 (g)(0) = 0,
so U1 (g)(v) Ker . Thus, U1 : G GL(Ker ) is a subrepresentation of U1 and as U1 is
irreducible, either Ker = (0) or Ker = V . In the second case, = 0. If Ker = (0),
then is injective. However, (V ) W is invariant under U2 since for every v V and
every g G,
U2 (g)((v)) = (U1 (g)(v)) (V ),
and as (V ) 6= (0) (as V 6= (0) since U1 is irreducible) and U2 is irreducible, we must have
(V ) = W , that is, is an isomorphism.
(ii) Since V is a complex vector space, the linear map, , has some eigenvalue, C.
Let E V be the eigenspace associated with . The subspace E is invariant under U1
since for every u E and every g G, we have
(U1 (g)(u)) = U1 (g)((u)) = U1 (g)(u) = U1 (g)(u),
34
so U1 : G GL(E ) is a subrepresentation of U1 and as U1 is irreducible and E 6= (0), we

must have E = V .
An interesting corollary of Schurs Lemma is that every complex irreducible representtaion of a commutative group is one-dimensional.
Let us now restrict our attention to compact Lie groups. If G is a compact Lie group,
then it is known that it has a left and right-invariant volume form, G , so we can define the
integral of a (real or complex) continuous function, f , defined on G by
Z
Z
f=
f G ,
G
R
also denoted G f dG or simply G f (t) dt, with G normalized so that G G = 1. (See
Section ??, or Knapp [19], Chapter 8, or Warner [29], Chapters 4 and 6.) Because G is
compact, the Haar measure, G , induced by G is both left and right-invariant (G is a
unimodular group) and our integral has the following invariance properties:
Z
Z
Z
Z
f (t1 ) dt,
f (tu) dt =
f (st) dt =
f (t) dt =
G
for all s, u G (see Section ??).

Since G is a compact Lie group, we can use an averaging trick to show that every (finitedimensional) representation is equivalent to a unitary representation (see Brocker and tom
Dieck [4] (Chapter 2, Theorem 1.7) or Knapp [19] (Chapter 4, Proposition 4.6).
If we define the Hermitian inner product,
Z
hf, gi =
f g G ,
G
then, with this inner product, the space of square-integrable functions, L2C (G), is a Hilbert
space. We can also define the convolution, f g, of two functions, f, g L2C (G), by
Z
Z
1
f (t)g(t1 x)dt
(f g)(x) =
f (xt )g(t)dt =
G
In general, f g 6= g f unless G is commutative. With the convolution product, L2C (G)

becomes an associative algebra (non-commutative in general).
This leads us to consider unitary representations of G into the infinite-dimensional vector
space, L2C (G). The definition is the same as in Definition 1.2, except that GL(L2C (G)) is the
group of automorphisms (unitary operators), Aut(L2C (G)), of the Hilbert space, L2C (G) and
hU (g)(u), U (g)(v)i = hu, vi
with respect to the inner product on L2C (G). Also, in the definition of an irreducible representation, U : G V , we require that the only closed subrepresentations, U : G W , of
the representation, U : G V , correspond to W = (0) or W = V .
35
The Peter Weyl Theorem gives a decomposition of L2C (G) as a Hilbert sum of spaces that
correspond to irreducible unitary representations of G. We present a version of the Peter
Weyl Theorem found in Dieudonne [6] (Chapters 3-8) and Dieudonne [7] (Chapter XXI,
Sections 1-4), which contains complete proofs. Other versions can be found in Brocker and
tom Dieck [4] (Chapter 3), Knapp [19] (Chapter 4) or Duistermaat and Kolk [10] (Chapter
4). A good preparation for these fairly advanced books is Deitmar [5].
Theorem 1.11 (Peter-Weyl (1927)) Given a compact Lie group, G, there is a decomposition of L2C (G) as a Hilbert sum,
M
a ,
L2C (G) =
of countably many two-sided ideals, a , where each a is isomorphic to a finite-dimensional

algebra of n n complex matrices. More precisely, there is a basis of a consisting of
()
smooth pairwise orthogonal functions, mij , satisfying various properties, including
()
()
hmij , mij i = n ,
()
and if we form the matrix, M (g) = ( n1 mij (g)), then the map, g 7 M (g) is an irreducible
unitary representation of G in the vector space Cn . Furthermore, every irreducible
representation of G is equivalent to some M , so the set of indices, , corresponds to the set
of equivalence classes of irreducible unitary representations of G. The function, u , given by
u (g) =
n
X
()
mjj (g) = n tr(M (g))
j=1
is the unit of the algebra a and the orthogonal projection of L2C (G) onto a is the map
f 7 u f,
that is, convolution with u .
Remark: The function, = n1 u = tr(M ), is the character of G associated with the
representation of G into M . The functions, , form an orthogonal system. Beware that
they are not homomorphisms of G into C unless G is commutative. The characters of G are
the key to the definition of the Fourier transform on a (compact) group, G.
A complete proof of Theorem 1.11 is given in Dieudonne [7], Chapter XXI, Section 2,
but see also Sections 3 and 4.
There is more to the Peter Weyl Theorem: It gives a description of all unitary representations of G into a separable Hilbert space (see Dieudonne [7], Chapter XXI, Section 4). If
V : G Aut(E) is such a representation, then for every as above, the map
Z
x 7 V (u )(x) = (V (s)(x))u (s) ds
G
36
is an L
orthogonal projection of E onto a closed subspace, E . Then, E is the Hilbert sum,
E = E , of those E such that E 6= (0) and each such E is itself a (countable) Hilbert
sum of closed spaces invariant under V . The subrepresentations of V corresponding to these
subspaces of E are all equivalent to M = M and hence, irreducible. This is why every
(unitary) representation of G is equivalent to some representation of the form M .
An interesting special case is the case of the so-called regular representation of G in
L2C (G) itself. The (left) regular representation, R, of G in L2C (G) is defined by
(Rs (f ))(t) = s (f )(t) = f (s1 t),
f L2C (G), s, t G.
It turns out that we also get the same Hilbert sum,

M
L2C (G) =
a ,
but this time, the a generally do not correspond to irreducible subrepresentations. However,
()
()
a splits into n left ideals, bj , where bj corresponds to the jth columm of M and all the
()
subrepresentations of G in bj are equivalent to M and thus, are irreducible (see Dieudonne
[6], Chapter 3).
Finally, assume that besides the compact Lie group, G, we also have a closed subgroup, K,
of G. Then, we know that M = G/K is a manifold called a homogeneous space and G acts on
M on the left. For example, if G = SO(n+1) and K = SO(n), then S n = SO(n+1)/SO(n)
(for instance, see Warner [29], Chapter 3). The subspace of L2C (G) consisting of the functions
f L2C (G) that are right-invariant under the action of K, that is, such that
f (su) = f (s)
for all s G and all u K
form a closed subspace of L2C (G) denoted L2C (G/K). For example, if G = SO(n + 1) and
K = SO(n), then L2C (G/K) = L2C (S n ).
It turns out that L2C (G/K) is invariant under the regular representation, R, of G in
L2C (G), so we get a subrepresentation (of the regular representation) of G in L2C (G/K).
Again, the Peter-Weyl gives us a Hilbert sum decomposition of L2C (G/K) of the form
M
L = L2C (G/K) a ,
L2C (G/K) =
for the same s as before. However, these subrepresentations of R in L are not necessarily
irreducible. What happens is that there is some d with 0 d n so that if d 1,
then L is the direct sum of the first d columns of M (see Dieudonne [6], Chapter 6 and
Dieudonne [8], Chapter XXII, Sections 4-5).
We can also consider the subspace of L2C (G) consisting of the functions, f L2C (G), that
are left-invariant under the action of K, that is, such that
f (ts) = f (s)
for all s G and all t K.
37
This is a closed subspace of L2C (G) denoted L2C (K\G). Then, we get a Hilbert sum decomposition of L2C (K\G) of the form
M
L0 = L2C (K\G) a ,
L2C (K\G) =
and for the same d as before, L0 is the direct sum of the first d rows of M . We can also
consider
L2C (K\G/K) = L2C (G/K) L2C (K\G)
= {f L2C (G) | f (tsu) = f (s)}
for all s G and all t, u K.
From our previous discussion, we see that we have a Hilbert sum decomposition
M
L2C (K\G/K) =
L L0
and each L L0 for which d 1 is a matrix algebra of dimension d2 . As a consequence,

the algebra L2C (K\G/K) is commutative iff d 1 for all .
If the algebra L2C (K\G/K) is commutative (for the convolution product), we say that
(G, K) is a Gelfand pair (see Dieudonne [6], Chapter 8 and Dieudonne [8], Chapter XXII,
Sections 6-7). In this case, the L in the Hilbert sum decomposition of L2C (G/K) are nontrivial of dimension n iff d = 1 and the subrepresentation, U, (of the regular representation)
of G into L is irreducible and equivalent to M . The space L is generated by the functions,
()
()
m1,1 , . . . , mn ,1 , but the function
(s) =
1 ()
m (s)
n 1,1
plays a special role. This function called a zonal spherical function has some interesting
properties. First, (e) = 1 (where e is the identity element of the group, G) and
(ust) = (s)
for all s G and all u, t K.
In addition, is of positive type. A function, f : G C, is of positive type iff

n
X
f (s1
j sk )zj z k 0,
j,k=1
for every finite set, {s1 , . . . , sn }, of elements of G and every finite tuple, (z1 , . . . , zn ) Cn .
Because the subrepresentation of G into L is irreducible, the function generates L under
left translation. This means the following: If we recall that for any function, f , on G,
s (f )(t) = f (s1 t),
s, t G,
38
then, L is generated by the functions s ( ), as s varies in G. The function also satisfies

the following property:
(s) = hU(s)( ), i.
The set of zonal spherical functions on G/K is denoted S(G/K). It is a countable set.
The notion of Gelfand pair also applies to locally-compact unimodular groups that are
not necessary compact but we will not discuss this notion here. Curious readers may consult
Dieudonne [6] (Chapters 8 and 9) and Dieudonne [8] (Chapter XXII, Sections 6-9).
It turns out that G = SO(n + 1) and K = SO(n) form a Gelfand pair (see Dieudonne
[6], Chapters 7-8 and Dieudonne [9], Chapter XXIII, Section 38). In this particular case,
= k is any nonnegative integer and L = Ek , the eigenspace of the Laplacian on S n
corresponding to the eigenvalue k(n + k 1). Therefore, the regular representation of
SO(n) into Ek = HkC (S n ) is irreducible. This can be proved more directly, for example,
see Helgason [16] (Introduction, Theorem 3.1) or Brocker and tom Dieck [4] (Chapter 2,
Proposition 5.10).
The zonal spherical harmonics, k , can be expressed in terms of the ultraspherical poly(n1)/2
nomials (also called Gegenbauer polynomials), Pk
(up to a constant factor), see Stein
and Weiss [26] (Chapter 4), Morimoto [22] (Chapter 2) and Dieudonne [6] (Chapter 7). For
1
n = 2, Pk2 is just the ordinary Legendre polynomial (up to a constant factor). We will say
more about the zonal spherical harmonics and the ultraspherical polynomials in the next
two sections.
The material in this section belongs to the overlapping areas of representation theory and
noncommutative harmonic analysis. These are deep and vast areas. Besides the references
cited earlier, for noncommutative harmonic analysis, the reader may consult Folland [11] or
Taylor [27], but they may find the pace rather rapid. Another great survey on both topics
is Kirillov [18], although it is not geared for the beginner.
1.6
Reproducing Kernel, Zonal Spherical Functions

and Gegenbauer Polynomials
We now return to S n and its spherical harmonics. The previous section suggested that
zonal spherical functions play a special role. In this section, we describe the zonal spherical
functions on S n and show that they essentially come from certain polynomials generalizing
the Legendre polyomials known as the Gegenbauer Polynomials. Most proof will be omitted.
We refer the reader to Stein and Weiss [26] (Chapter 4) and Morimoto [22] (Chapter 2) for
a complete exposition with proofs.
Recall that the space of spherical harmonics, HkC (S n ), is the image of the space of homogeneous harmonic poynomials, PkC (n + 1), under the restriction map. It is a finite-dimensional
1.6. REPRODUCING KERNEL AND ZONAL SPHERICAL FUNCTIONS
39
space of dimension

ak,n+1 =

n+k
n+k2
,
k
k2
a
if n 1 and k 2, with a0,n+1 = 1 and a1,n+1 = n + 1. Let (Yk1 , . . . , Yk k,n+1 ) be any

orthonormal basis of HkC (S n ) and define Fk (, ) by
ak,n+1
Fk (, ) =
Yki ()Yki ( ),
, S n .
i=1
The following proposition is easy to prove (see Morimoto [22], Chapter 2, Lemma 1.19 and
Lemma 2.20):
Proposition 1.12 The function Fk is independent of the choice of orthonormal basis. Furthermore, for every orthogonal transformation, R O(n + 1), we have
Fk (R, R ) = Fk (, ),
, S n .
Clearly, Fk is a symmetric function. Since we can pick an orthonormal basis of real

orthogonal functions for HkC (S n ) (pick a basis of Hk (S n )), Proposition 1.12 shows that Fk is
a real-valued function.
The function Fk satisfies the following property which justifies its name, the reproducing
kernel for HkC (S n ):
Proposition 1.13 For every spherical harmonic, H HjC (S n ), we have
Z
H( )Fk (, ) d = j k H(),
, S n ,
Sn
for all j, k 0.
Proof . When j 6= k, since HkC (S n ) and HjC (S n ) are orthogonal and since
Pak,n+1 i
Yk ()Yki ( ), it is clear that the integral in Proposition 1.13 vanishes.
Fk (, ) =
i=1
When j = k, we have
Z
ak,n+1
Z
H( )Fk (, ) d =
Sn
H( )
Sn
X
i=1
ak,n+1
Yki ()
i=1
ak,n+1
Yki ()Yki ( ) d
Z
Sn
H( )Yki ( ) d
Yki () hH, Yki i
i=1
= H(),
40

a
since (Yk1 , . . . , Yk k,n+1 ) is an orthonormal basis.

(k)
In Stein and Weiss [26] (Chapter 4), the function Fk (, ) is denoted by Z ( ) and it is
called the zonal harmonic of degree k with pole .
The value, Fk (, ), of the function Fk depends only on , as stated in Proposition
1.15 which is proved in Morimoto [22] (Chapter 2, Lemma 2.23). The following proposition
also proved in Morimoto [22] (Chapter 2, Lemma 2.21) is needed to prove Proposition 1.15:
Proposition 1.14 For all , , 0 , 0 S n , with n 1, the following two conditions are
equivalent:
(i) There is some orthogonal transformation, R O(n + 1), such that R() = 0 and
R( ) = 0 .
(ii) = 0 0 .
Propositions 1.13 and 1.14 immediately yield
Proposition 1.15 For all , , 0 , 0 S n , if = 0 0 , then Fk (, ) = Fk ( 0 , 0 ).
Consequently, there is some function, : R R, such that Fk (, ) = ( ).
We are now ready to define zonal functions. Remarkably, the function in Proposition
1.15 comes from a real polynomial. We need the following proposition which is of independent
interest:
Proposition 1.16 If P is any (complex) polynomial in n variables such that
P (R(x)) = P (x)
for all rotations, R SO(n), and all x Rn ,
then P is of the form

P (x) =
m
X
cj (x21 + + x2n )j ,
j=0
for some c0 , . . . , cm C.
P
Proof . Write P as the sum of its homogeneous pieces, P = kl=0 Ql , where Ql is homogeneous of degree l. Then, for every > 0 and every rotation, R, we have
k
X
l Ql (x) = P (x) = P (R(x)) = P (R(x)) =
l=0
k
X
l=0
which implies that

Ql (R(x)) = Ql (x),
l = 0, . . . , k.
l Ql (R(x)),
41
If we let Fl (x) = kxkl Ql (x), then Fl is a homogeneous function of degree 0 and Fl is invariant
under all rotations. This is only possible if Fl is a constant function, thus Fl (x) = al for
all x Rn . But then, Ql (x) = al kxkl . Since Ql is a polynomial, l must be even whenever
al 6= 0. It follows that
m
X
P (x) =
cj kxk2j
j=0
with cj = a2j for j = 0, . . . , m and where m is the largest integer k/2.

Proposition 1.16 implies that if a polynomial function on the sphere, S n , in particular,
a spherical harmonic, is invariant under all rotations, then it is a constant. If we relax this
condition to invariance under all rotations leaving some given point, S n , invariant, then
we obtain zonal harmonics.
The following theorem from Morimoto [22] (Chapter 2, Theorem 2.24) gives the relationship between zonal harmonics and the Gegenbauer polynomials:
Theorem 1.17 Fix any S n . For every constant, c C, there is a unique homogeneous
harmonic polynomial, Zk HkC (n + 1), satisfying the following conditions:
(1) Zk ( ) = c;
(2) For every rotation, R SO(n + 1), if R = , then Zk (R(x)) = Zk (x), for all
x Rn+1 .
Furthermore, we have
Zk (x)
= c kxk Pk,n
kxk

,
for some polynomial, Pk,n (t), of degree k.

Remark: The proof given in Morimoto [22] is essentially the same as the proof of Theorem
2.12 in Stein and Weiss [26] (Chapter 4) but Morimoto makes an implicit use of Proposition
1.16 above. Also, Morimoto states Theorem 1.17 only for c = 1 but the proof goes through
for any c C, including c = 0, and we will need this extra generality in the proof of the
Funk-Hecke formula.
Proof . Let en+1 = (0, . . . , 0, 1) Rn+1 and for any S n , let R be some rotation such
that R (en+1 ) = . Assume Z HkC (n + 1) satisfies conditions (1) and (2) and let Z 0 be
given by Z 0 (x) = Z(R (x)). As R (en+1 ) = , we have Z 0 (en+1 ) = Z( ) = c. Furthermore,
for any rotation, S, such that S(en+1 ) = en+1 , observe that
R S R1 ( ) = R S(en+1 ) = R (en+1 ) = ,
and so, as Z satisfies property (2) for the rotation R S R1 , we get
Z 0 (S(x)) = Z(R S(x)) = Z(R S R1 R (x)) = Z(R (x)) = Z 0 (x),
42
which proves that Z 0 is a harmonic polynomial satisfying properties (1) and (2) with respect
to en+1 . Therefore, we may assume that = en+1 .
Write
Z(x) =
k
X
xkj
n+1 Pj (x1 , . . . , xn ),
j=0
where Pj (x1 , . . . , xn ) is a homogeneous polynomial of degree j. Since Z is invariant under

every rotation, R, fixing en+1 and since the monomials xkj
n+1 are clearly invariant under such
a rotation, we deduce that every Pj (x1 , . . . , xn ) is invariant under all rotations of Rn (clearly,
there is a one-two-one correspondence between the rotations of Rn+1 fixing en+1 and the
rotations of Rn ). By Proposition 1.16, we conclude that
j
Pj (x1 , . . . , xn ) = cj (x21 + + x2n ) 2 ,

which implies that Pj = 0 is j is odd. Thus, we can write
[k/2]
Z(x) =
2
2 i
ci xk2i
n+1 (x1 + + xn )
i=0
where [k/2] is the greatest integer, m, such that 2m k. If k < 2, then Z = c0 , so c0 = c

and Z is uniquely determined. If k 2, we know that Z is a harmonic polynomial so we
assert that Z = 0. A simple computation shows that
(x21 + + x2n )i = 2i(n + 2i 2)(x21 + + x2n )i1
and
k2i 2
xn+1
(x1 + + x2n )i = (k 2i)(k 2i 1)xk2i2
(x21 + + x2n )i
n+1
k2i
+ xn+1
(x21 + + x2n )i
= (k 2i)(k 2i 1)xk2i2
(x21 + + x2n )i
n+1
k2i 2
+ 2i(n + 2i 2)xn+1
(x1 + + x2n )i1 ,
so we get
[k/2]1
Z =
(x21 + + x2n )i .
((k 2i)(k 2i 1)ci + 2(i + 1)(n + 2i)ci+1 ) xk2i2
n+1
i=0
Then, Z = 0 yields the relations

2(i + 1)(n + 2i)ci+1 = (k 2i)(k 2i 1)ci ,
i = 0, . . . , [k/2] 1,
43
which shows that Z is uniquely determined up to the constant c0 . Since we are requiring
Z(en+1 ) = c, we get c0 = c and Z is uniquely determined. Now, on S n , we have
x21 + + x2n+1 = 1, so if we let t = xn+1 , for c0 = 1, we get a polynomial in one variable,
[k/2]
Pk,n (t) =
ci tk2i (1 t2 )i .
i=0
Thus, we proved that when Z(en+1 ) = c, we have

x
xn+1
k
k
= c kxk Pk,n
en+1 .
Z(x) = c kxk Pk,n
kxk
kxk
When Z( ) = c, we write Z = Z 0 R1 with Z 0 = Z R and where R is a rotation such
that R (en+1 ) = . Then, as Z 0 (en+1 ) = c, using the formula above for Z 0 , we have

1
1 k
R (x)
0
1

Z(x) = Z (R (x)) = c R (x) Pk,n
en+1
kR1 (x)k

x
k
= c kxk Pk,n
R (en+1 )
kxk

x
k
= c kxk Pk,n
,
kxk
since R is an isometry.
The function, Zk , is called a zonal function and its restriction to S n is a zonal spherical function. The polynomial, Pk,n , is called the Gegenbauer polynomial of degree k and
dimension n + 1 or ultraspherical polynomial . By definition, Pk,n (1) = 1.
The proof of Theorem 1.17 shows that for k even, say k = 2m, the polynomial P2m,n is
of the form
m
X
P2m,n =
cmj t2j (1 t2 )mj
j=0
and for k odd, say k = 2m + 1, the polynomial P2m+1,n is of the form

P2m+1,n =
m
X
cmj t2j+1 (1 t2 )mj .
j=0
Consequently, Pk,n (t) = (1)k Pk,n (t), for all k 0. The proof also shows that the natural
ki
basis for these polynomials consists of the polynomials, ti (1t2 ) 2 , with ki even. Indeed,
with this basis, there are simple recurrence equations for computing the coefficients of Pk,n .
Remark: Morimoto [22] calls the polynomials, Pk,n , Legendre polynomials. For n = 2,
they are indeed the Legendre polynomials. Stein and Weiss denotes our (and Morimotos)
n1
2
Pk,n by Pk
(up to a constant factor) and Dieudonne [6] (Chapter 7) by Gk,n+1 .
44
When n = 2, using the notation of Section 1.2, the zonal functions on S 2 are the spherical
harmonics, yl0 , for which m = 0, that is (up to a constant factor),
r
(2l + 1)
Pl (cos ),
yl0 (, ) =
4
where Pl is the Legendre polynomial of degree l. For example, for l = 2, Pl (t) = 12 (3t2 1).
If we put Z(rk ) = rk Fk (, ) for a fixed , then by the definition of Fk (, ) it is clear that
Z is a homogeneous harmonic polynomial. The value Fk (, ) does not depend of because
by transitivity of the action of SO(n +1) on S n , for any other S n , there is some rotation,
R, so that R = and by Proposition 1.12, we have Fk (, ) = Fk (R, R ) = Fk (, ). To
compute Fk (, ), since
ak,n+1
X

Yki ( ) 2 ,
Fk (, ) =
i=1
and since
a
(Yk1 , . . . , Yk k,n+1 )
is an orthonormal basis of HkC (S n ), observe that

ak,n+1 Z
X
i 2
Yk ( ) d
ak,n+1 =
i=1
(1.1)
Sn
ak,n+1
Z
=
Sn
!
X

2
Yki ( ) d
(1.2)
i=1
Fk (, ) d = Fk (, ) vol(S n ).
(1.3)
Sn
Therefore,
ak,n+1
.
vol(S n )
Beware that Morimoto [22] uses the normalized measure on S n , so the factor involving
vol(S n ) does not appear.
Fk (, ) =
Remark: Recall that

vol(S 2d ) =
2d+1 d
1 3 (2d 1)
if d 1 and vol(S 2d+1 ) =
2 d+1
d!
if d 0.
Now, if R = , then Proposition 1.12 shows that

Z(R(rk )) = Z(rk R()) = rk Fk (R, ) = rk Fk (R, R ) = rk Fk (, ) = Z(rk ).
Therefore, the function Zk satisfies conditions (1) and (2) of Theorem 1.17 with c =
and by uniqueness, we get
ak,n+1
Fk (, ) =
Pk,n ( ).
vol(S n )
Consequently, we have obtained the so-called addition formula:
ak,n+1
vol(S n )
45
Proposition 1.18 (Addition Formula) If (Yk1 , . . . , Yk k,n+1 ) is any orthonormal basis of

HkC (S n ), then
ak,n+1
vol(S n ) X i
Pk,n ( ) =
Y ()Yki ( ).
ak,n+1 i=1 k
Again, beware that Morimoto [22] does not have the factor vol(S n ).
For n = 1, we can write = (cos , sin ) and = (cos , sin ) and it is easy to see that
the addition formula reduces to
Pk,1 (cos( )) = cos k cos k + sin k sin k = cos k( ),
the standard addition formula for trigonometric functions.
Proposition 1.18 implies that Pk,n has real coefficients. Furthermore Proposition 1.13 is
reformulated as
Z
ak,n+1
Pk,n ( )H( ) d = j k H(),
(rk)
vol(S n ) S n
showing that the Gengenbauer polynomials are reproducing kernels. A neat application of
this formula is a formula for obtaining the kth spherical harmonic component of a function,
f L2C (S n ).
P
2
n
Proposition 1.19 For every function,
f
L
C(S
),
if
f
=
C
k=0 fk is the unique decomL
C
k
H
(S
),
then
f
is
given
by
position of f over the Hilbert sum
k
k
k=0
Z
ak,n+1
f ( )Pk,n ( ) d,
fk () =
vol(S n ) S n
for all S n .
Proof . If we recall that HkC (S k ) and HjC (S k ) are orthogonal for all j 6= k, using the formula
(rk), we have
Z
Z X
ak,n+1
ak,n+1
f ( )Pk,n ( ) d =
fj ( )Pk,n ( ) d
vol(S n ) S n
vol(S n ) S n j=0
Z
ak,n+1 X
=
fj ( )Pk,n ( ) d
vol(S n ) j=0 S n
Z
ak,n+1
=
fk ( )Pk,n ( ) d
vol(S n ) S n
= fk (),
as claimed.
We know from the previous section that the kth zonal function generates HkC (S n ). Here
is an explicit way to prove this fact.
46
Proposition 1.20 If H1 , . . . , Hm HkC (S n ) are linearly independent, then there are m

points, 1 , . . . , m , on S n , so that the m m matrix, (Hj (i )), is invertible.
Proof . We proceed by induction on m. The case m = 1 is trivial. For the induction step, we
may assume that we found m points, 1 , . . . , m , on S n , so that the m m matrix, (Hj (i )),
is invertible. Consider the function

H1 () . . . Hm ()

H
()
m+1

H1 (1 ) . . . Hm (1 ) Hm+1 (1 )

7 ..

..
..
.
.
.

.
.
.

H1 (m ) . . . Hm (m ) Hm+1 (m ).
Since H1 , . . . , Hm+1 are linearly independent, the above function does not vanish for all
since otherwise, by expanding this determinant with respect to the first row, we get a linear
dependence among the Hj s where the coefficient of Hm+1 is nonzero. Therefore, we can find
m+1 so that the (m + 1) (m + 1) matrix, (Hj (i )), is invertible.
We say that ak,n+1 points, 1 , . . . , ak,n+1 on S n form a fundamental system iff the
ak,n+1 ak,n+1 matrix, (Pn,k (i j )), is invertible.
Theorem 1.21 The following properties hold:
(i) There is a fundamental system, 1 , . . . , ak,n+1 , for every k 1.
(ii) Every spherical harmonic, H HkC (S n ), can be written as
ak,n+1
H() =
cj Pk,n (j ),
j=1
for some unique cj C.

Proof . (i) By the addition formula,
ak,n+1
vol(S n ) X l
Pk,n (i j ) =
Y (i )Ykl (j )
ak,n+1 l=1 k
a
for any orthonormal basis, (Yk1 , . . . , Yk k,n+1 ). It follows that the matrix (Pk,n (i j )) can be
written as
vol(S n )
(Pk,n (i j )) =
Y Y ,
ak,n+1
where Y = (Ykl (i )), and by Proposition 1.20, we can find 1 , . . . , ak,n+1 S n so that Y and
thus also Y are invertible and so, (Pn,k (i j )) is invertible.
1.7. MORE ON THE GEGENBAUER POLYNOMIALS
47
(ii) Again, by the addition formula,

ak,n+1
vol(S n ) X i
Y ()Yki (j ).
Pk,n ( j ) =
ak,n+1 i=1 k
a
However, as (Yk1 , . . . , Yk k,n+1 ) is an orthonormal basis, (i) proved that the matrix Y is
invertible so the Yki () can be expressed uniquely in terms of the Pk,n ( j ), as claimed.
A neat geometric characterization of the zonal spherical functions is given in Stein and
Weiss [26]. For this, we need to define the notion of a parallel on S n . A parallel of S n
orthogonal to a point S n is the intersection of S n with any (affine) hyperplane orthogonal
to the line through the center of S n and . Clearly, any rotation, R, leaving fixed leaves
every parallel orthogonal to globally invariant and for any two points, 1 and 2 , on such
a parallel there is a rotation leaving fixed that maps 1 to 2 . Consequently, the zonal
function, Zk , defined by is constant on the parallels orthogonal to . In fact, this property
characterizes zonal harmonics, up to a constant.
The theorem below is proved in Stein and Weiss [26] (Chapter 4, Theorem 2.12). The
proof uses Proposition 1.16 and it is very similar to the proof of Theorem 1.17 so, to save
space, it is omitted.
Theorem 1.22 Fix any point, S n . A spherical harmonic, Y HkC (S n ), is constant on
parallels orthogonal to iff Y = cZk , for some constant, c C.
In the next section, we show how the Gegenbauer polynomials can actually be computed.
1.7
More on the Gegenbauer Polynomials
The Gegenbauer polynomials are characterized by a formula generalizing the Rodrigues

formula defining the Legendre polynomials (see Section 1.2). The expression

n2
n2
n2
k1+
1 +
k+
2
2
2
can be expressed in terms of the function as

k + n2
.
n2
Recall that the function is a generalization of factorial that satisfies the equation
(z + 1) = z(z).
48
For z = x + iy with x > 0, (z) is given by

Z
(z) =
tz1 et dt,
where the integral converges absolutely. If n is an integer n 0, then (n + 1) = n!.

It is proved in Morimoto [22] (Chapter 2, Theorem 2.35) that
Proposition 1.23 The Gegenbauer polynomial, Pk,n , is given by Rodrigues formula:

n2
(1)k n2
1
dk

Pk,n (t) =
(1 t2 )k+ 2 ,
n2
n
k
k
2 k + 2 (1 t2 ) 2 dt
with n 2.
The Gegenbauer polynomials satisfy the following orthogonality properties with respect
n2
to the kernel (1 t2 ) 2 (see Morimoto [22] (Chapter 2, Theorem 2.34):
Proposition 1.24 The Gegenbauer polynomial, Pk,n , have the following properties:
Z
n2
2
dt =
Pk,n (t)Pl,n (t)(1 t2 )
n2
2
dt = 0,
vol(S n )
ak,n+1 vol(S n1 )
(Pk,n (t))2 (1 t2 )
k 6= l.
The Gegenbauer polynomials satisfy a second-order differential equation generalizing the

Legendre equation from Section 1.2.
Proposition 1.25 The Gegenbauer polynomial, Pk,n , are solutions of the differential equation
00
0
(1 t2 )Pk,n
(t) ntPk,n
(t) + k(k + n 1)Pk,n (t) = 0.
Proof . For a fixed , the function H given by H() = Pk,n ( ) = Pk,n (cos ), belongs to
HkC (S n ), so
S n H = k(k + n 1)H.
Recall from Section 1.3 that
S n f =
n1
sin

1

1
n1 f
sin
+
S f,
sin2 n1
in the local coordinates where

= sin
e + cos en+1 ,
1.7. MORE ON THE GEGENBAUER POLYNOMIALS
49
with
e S n1 and 0 < . If we make the change of variable t = cos , then it is easy to
see that the above formula becomes
2f
1
f
S n f = (1 t2 ) 2 nt
+
S n1
t
t
1 t2
(see Morimoto [22], Chapter 2, Theorem 2.9.) But, H being zonal, it only depends on ,
that is, on t, so S n1 H = 0 and thus,
k(k + n 1)Pk,n (t) = S n Pk,n (t) = (1 t2 )
Pk,n
2 Pk,n
,
nt
t2
t
which yields our equation.

Note that for n = 2, the differential equation of Proposition 1.25 is the Legendre equation
from Section 1.2.
The Gegenbauer poynomials also appear as coefficients in some simple generating functions. The following proposition is proved in Morimoto [22] (Chapter 2, Theorem 2.53 and
Theorem 2.55):
Proposition 1.26 For all r and t such that 1 < r < 1 and 1 t 1, for all n 1, we
have the following generating formula:
ak,n+1 r Pk,n (t) =
k=0
1 r2
(1 2rt + r2 )
n+1
2
Furthermore, for all r and t such that 0 r < 1 and 1 t 1, if n = 1, then
X
rk
k=1
1
Pk,1 (t) = log(1 2rt + r2 )
k
2
and if n 2, then
X
k=0
1
n1
ak,n+1 rk Pk,n (t) =
n1 .
2k + n 1
(1 2rt + r2 ) 2
In Stein and Weiss [26] (Chapter 4, Section 2), the polynomials, Pk (t), where > 0 are
defined using the following generating formula:
rk Pk (t) =
k=0
1
.
(1 2rt + r2 )
Each polynomial, Pk (t), has degree k and is called an ultraspherical polynomial of degree k
associated with . In view of Proposition 1.26, we see that
n1
ak,n+1 Pk,n (t),
2k + n 1
as we mentionned ealier. There is also an integral formula for the Gegenbauer polynomials
known as Laplace representation, see Morimoto [22] (Chapter 2, Theorem 2.52).
n1
2
Pk
(t) =
50
1.8
The Funk-Hecke Formula
The Funk-Hecke Formula (also known as Hecke-Funk Formula) basically allows one to perform a sort of convolution of a kernel function with a spherical function in a convenient
way. Given a measurable function, K, on [1, 1] such that the integral
Z 1
n2
|K(t)|(1 t2 ) 2 dt
1
makes sense, given a function f L2C (S n ), we can view the expression

Z
K( )f ( ) d
K ? f () =
Sn
as a sort of convolution of K and f . Actually, the use of the term convolution is really
unfortunate because in a true convolution, f g, either the argument of f or the argument
of g should be multiplied by the inverse of the variable of integration, which means that
the integration should really be taking place over the group SO(n + 1). We will come back
to this point later. For the time being, let us call the expression K ? f defined above a
pseudo-convolution. Now, if f is expressed in terms of spherical harmonics as
f=
k,n+1
aX
X
ck,mk Ykmk ,
k=0 mk =1
then the Funk-Hecke Formula states that

K ? Ykmk () = k Ykmk (),
for some fixed constant, k , and so
K ?f =
k,n+1
aX
X
k ck,mk Ykmk .
k=0 mk =1
Thus, if the constants, k are known, then it is cheap to compute the pseudo-convolution
K ? f.
This method was used in a ground-breaking paper by Basri and Jacobs [3] to compute
the reflectance function, r, from the lighting function, `, as a pseudo-convolution K ? ` (over
S 2 ) with the Lambertian kernel , K, given by
K( ) = max( , 0).
Below, we give a proof of the Funk-Hecke formula due to Morimoto [22] (Chapter 2,
Theorem 2.39) but see also Andrews, Askey and Roy [1] (Chapter 9). This formula was first
published by Funk in 1916 and then by Hecke in 1918.
1.8. THE FUNK-HECKE FORMULA
51
Theorem 1.27 (Funk-Hecke Formula) Given any measurable function, K, on [1, 1] such
that the integral
Z 1
n2
|K(t)|(1 t2 ) 2 dt
1
makes sense, for every function, H HkC (S n ), we have

Z 1
Z
2 n2
K( )H() d = vol(Sn1 )
K(t)Pk,n (t)(1 t ) 2 dt H().
Sn
Observe
that when n = 2, the term (1 t2 )
R1
|K(t)| dt makes sense.
1
n2
2
is missing and we are simply requiring that
Proof . We first prove the formula in the case where H is a zonal harmonic and then use the
fact that the Pk,n s are reproducing kernels (formula (rk)).
For all , S n define H by
H() = Pk,n ( )
and F by
Z
K( )Pk,n ( ) d.
F (, ) =
Sn
Since the volume form on the sphere is invariant under orientation-preserving isometries, for
every R SO(n + 1), we have
F (R, R ) = F (, ).
On the other hand, for fixed, it is not hard to see that as a function in , the function
F (, ) is a spherical harmonic, because Pk,n satisfies a differential equation that implies
that S 2 F (, ) = k(k + n 1)F (, ). Now, for every rotation, R, that fixes ,
F (, ) = F (R, R ) = F (, R ),
which means that F (, ) satisfies condition (2) of Theorem 1.17. By Theorem 1.17, we get
F (, ) = F (, )Pk,n ( ).
If we use local coordinates on S n where
= 1 t2
e + t en+1 ,
with
e S n1 and 1 t 1, it is not hard to show that the volume form on S n is given
by
n2
dS n = (1 t2 ) 2 dtdS n1 .
52
Using this, we have

Z
K( )Pk,n ( ) d = vol(S
F (, ) =
Sn
n1
K(t)Pk,n (t)(1 t2 )
n2
2
dt,
and thus,

Z
n1
F (, ) = vol(S )
1
2
K(t)Pk,n (t)(1 t )
n2
2

dt Pk,n ( ),
which is the Funk-Hecke formula when H() = Pk,n ( ).

Let us now consider any function, H HkC (S n ). Recall that by the reproducing kernel
property (rk), we have
Z
ak,n+1
Pk,n ( )H( ) d = H().
vol(S n ) S n
R
Then, we can compute S n K( )H() d using Fubinis Theorem and the Funk-Hecke
formula in the special case where H() = Pk,n ( ), as follows:
Z
K( )H() d
Sn

Z
Z
ak,n+1
Pk,n ( )H( ) d d
=
K( )
vol(S n ) S n
Sn

Z
Z
ak,n+1
=
K( )Pk,n ( ) d d
H( )
vol(S n ) S n
Sn

Z 1
Z
ak,n+1
2 n2
n1
K(t)Pk,n (t)(1 t ) 2 dt Pk,n ( ) d
=
H( )
vol(S )
vol(S n ) S n
1

Z
Z 1
ak,n+1
n1
2 n2
2
dt
Pk,n ( )H( ) d
= vol(S )
K(t)Pk,n (t)(1 t )
vol(S n ) S n
1

Z 1
n1
2 n2
= vol(S )
K(t)Pk,n (t)(1 t ) 2 dt H(),
1
which proves the Funk-Hecke formula in general.

The Funk-Hecke formula can be used to derive an addition theorem for the ultraspherical polynomials (Gegenbauer polynomials). We omit this topic and we refer the interested
reader to Andrews, Askey and Roy [1] (Chapter 9, Section 9.8).
Remark: Oddly, in their computation of K ? `, Basri and Jacobs [3] first expand K in terms
of spherical harmonics as
X
K=
kn Yn0 ,
n=0
1.9. CONVOLUTION ON G/K, FOR A GELFAND PAIR (G, K)
53
and then use the Funk-Hecke formula to compute K ? Ynm and they get (see page 222)
r
4
m
m
kn ,
K ? Yn = n Yn , with n =
2n + 1
for some constant, kn , given on page 230 of their paper (see below). However, there is no
need to expand K as the Funk-Hecke formula yields directly
Z 1

Z
m
m
K( )Yn () d =
K(t)Pn (t) dt Ynm (),
K ? Yn () =
S2
where Pn (t) is the standard Legendre polynomial of degree n since we are in the case of S 2 .
By the definition of K (K(t) = max(t, 0)) and since vol(S 1 ) = 2, we get
Z 1

m
K ? Yn = 2
tPn (t) dt Ynm ,
0
which is equivalent to Basri and Jacobs formula (14) since their n on page 222 is given by
r
4
n =
kn ,
2n + 1
but from page 230,
Z
p
kn = (2n + 1)
tPn (t) dt.
R1
What remains to be done is to compute 0 tPn (t) dt, which is done by using the Rodrigues
Formula and integrating by parts (see Appendix A of Basri and Jacobs [3]).
1.9
Convolution on G/K, for a Gelfand Pair (G, K)
54
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Notes on Spherical Harmonics and

Linear Representations of Lie Groups

Introduction, Spherical Harmonics on the Circle

(ak cos k + bk cos k),

where the Fourier coefficients, ak , bk , of f are given by the formulae

CHAPTER 1. SPHERICAL HARMONICS

which we will also denote by

into pairwise orthogonal subspaces, where k=0 Hk (S 1 ) is dense in L2 (S 1 ). The functions

where f : R2 R is a function which is at least of class C 2 . In polar coordinates, (r, ),

1.1. INTRODUCTION, SPHERICAL HARMONICS ON THE CIRCLE

for all t > 0.

Now, using the Laplacian in polar coordinates, we get

CHAPTER 1. SPHERICAL HARMONICS

Therefore, the eigenfunctions of the Laplacian on S 1 are the restrictions of L

Spherical Harmonics on the 2-Sphere

If we use spherical coordinates

1.2. SPHERICAL HARMONICS ON THE 2-SPHERE

However, we want to be a periodic in since we are considering functions on the sphere,

CHAPTER 1. SPHERICAL HARMONICS

is two-dimensional and is spanned by the two functions

We still have to solve the equation

1.2. SPHERICAL HARMONICS ON THE 2-SPHERE

Observe that if we differentiate with respect to t, we get the equation

has the solution

CHAPTER 1. SPHERICAL HARMONICS

and for fixed k 2 we have

Pkm+1 (t) + (k m)(k + m + 1)Pkm (t),

which can also be used to compute Pkm starting from

1.2. SPHERICAL HARMONICS ON THE 2-SPHERE

(3) Starting from i = m + 1, compute

f (r, , ) = rk sin m Pkm (cos ),

are solutions of the Laplacian, , in R3 , and that the functions

sin m Pkm (cos ),

2Nlm sin(m) Plm (cos ) if m < 0

CHAPTER 1. SPHERICAL HARMONICS

where f, g L2 (S 2 ) and where 2 is the volume form on S 2 (induced by the metric on

1.3. THE LAPLACE-BELTRAMI OPERATOR

homogeneous polynomials! This remarkable fact holds in general: The eigenfunctions of

The Laplace-Beltrami Operator

The inverse isomorphism, ] : Tp M Tp M , is defined such that for every Tp M , the

CHAPTER 1. SPHERICAL HARMONICS

corresponding to the one-form, df , is the gradient, grad f , of f . The gradient of f is uniquely

1.3. THE LAPLACE-BELTRAMI OPERATOR

and for every function, f C (M ), the Laplacian, f , is given by

CHAPTER 1. SPHERICAL HARMONICS

Observe that rb, b and

1.3. THE LAPLACE-BELTRAMI OPERATOR

CHAPTER 1. SPHERICAL HARMONICS

Using the above equations and

where M is the volume form on M induced by the metric.

1.3. THE LAPLACE-BELTRAMI OPERATOR

Then, it is not hard to show that h, i is an inner product on C (M ).

Proof . By the two identities before Proposition 1.2,

which proves that is self-adjoint.

CHAPTER 1. SPHERICAL HARMONICS

Harmonic Polynomials, Spherical Harmonics and

Harmonic homogeneous polynomials and their restrictions to S n , where

for all x Rn and t > 0.

l Ql (x) = P (x) = P (R(x)) = P (R(x)) =