Harmonic polynomials and the spherical Laplacian

Jordan Bell
August 17, 2015

1 Topological groups

Let G be a topological group: (x,y)xy is continuous G×GG and xx-1 is continuous GG. For gG, the maps Lg(x)=gx and Rg(x)=xg are homeomorphisms. If U is an open subset of G and X is a subset of G, for each xX the set Ux={ux:uU} is open because U is open and uux is a homeomorphism. Therefore


is open, being a union of open sets.

For a subgroup H of G, not necessarily a normal subgroup, define q:GG/H by


and assign G/H the final topology for q, the finest topology on G/H such that q:GG/H is continuous (namely, the quotient topology). If U is an open subset of G, then UH is open, and we check that q-1(q(U))=UH. Because G/H has the final topology for q, this means that q(U) is an open set in G/H. Therefore, q:GG/H is an open map.

Theorem 1.

If G is a topological group and H is a closed subgroup, then G/H is a Hausdorff space.


For a topological space X, define Δ:XX×X by Δ(x)=(x,x). It is a fact that X is Hausdorff if and only if Δ(X) is a closed subset of X×X. Thus the quotient space G/H is Hausdorff if and only if the image of Δ:G/HG/H×G/H is closed. The complement of Δ(G/H) is


Call this set U and let p=q×q, which is a product of open maps and thus is itself open G×GG/H×G/H and likewise is surjective. We check that


The map f:G×GG defined by f(x,y)=x-1y is continuous and G-H is open in G, so f-1(G-H) is open in G×G. But


thus p-1(U) is open. As p is surjective, p(p-1(U))=U, and because p is an open map and p-1(U) is an open set, U is an open set. Because U is the complement of Δ(G×H), that set is closed and it follows that G/H is Hausdorff. ∎

Let G be a compact group, let K be a compact Hausdorff space. A left action of G on K is a continuous map α:G×KK, denoted


satisfying ek=k and (g1g2)k=g1(g2k). The action is called transitive if for k1,k2K there is some gG such that gk1=k2.

Let H be a closed subgroup of G and let q:GG/H be the quotient map. We have established that q is open and that G/H is Hausdorff. Because G is compact and q is surjective and continuous, q(G)=G/H is a compact space. We define β:G×G/HG/H by


If xH=yH, then (gx)H=(gy)H, so indeed this makes sense.11 1 cf. Mamoru Mimura and Hiroshi Toda, Topology of Lie Groups, I and II, Chapter I.

Lemma 2.

β:G×G/HG/H is a transitive left action.


Write μ(x,y)=xy. For an open subset V in G/H, we check that


hence (Le×q)-1(β-1(V)) is open in G. Because Le:GG and q:GG/H are surjective open maps, the product Le×q:G×GG×G/H is a surjective open map, so


is open in G×G/H, showing that β is continuous.

For xHG/H, e(xH)=(ex)H=xH, and for g1,g2G,


Therefore β is a left action.

For xH,yHG/H,


showing that β is transitive. ∎

Let G be a compact group and let α be a transitive action of G on a compact Hausdorff space K. For any k0K, let H={gG:α(g,k0)=k0}, the isotropy group of k0, which is a closed subgroup of G. A theorem of Weil22 2 Joe Diestel and Angela Spalsbury, The Joys of Haar Measure, p. 148, Theorem 6.1. states that ϕ:G/HK defined by


is a homeomorphism that satisfies


called an isomorphism of G-spaces.

A Borel measure m on G is called left-invariant if m(gE)=m(E) for all Borel sets E and right-invariant if m(Eg)=m(E) for all Borel sets E. It is proved that there is a unique regular Borel probability measure m on G that is left-invariant.33 3 Walter Rudin, Functional Analysis, second ed., p. 130, Theorem 5.14. This measure is right-invariant, and satisfies


We call m the Haar probability measure on the compact group G.

Let H be the above isotropy group, and define mG/H on the Borel σ-algebra of G/H by


This is a regular Borel probability measure on G/H, and satisfies


for Borel sets E in G/H and for gG; we say that mG/H is G-invariant. A theorem attributed to Weil states that this is the unique G-invariant regular Borel probability measure on G/H.44 4 Joe Diestel and Angela Spalsbury, The Joys of Haar Measure, p. 149, Theorem 6.2. Then define mK on the Borel σ-algebra of K by


This is the unique G-invariant regular Borel probability measure on K.

2 Spherical surface measure

SO(n) is a compact Lie group. Sn-1 is a topological group, and it is a fact that α:SO(n)×Sn-1Sn-1 defined by


is a transitive left-action. We check that the isotropy group of en is SO(n-1). Let q:SO(n)SO(n)/SO(n-1) be the projection map and define ϕ:SO(n)/SO(n-1)Sn-1 by


Then for m the Borel probability measure on SO(n),55 5 SO(n) is a compact Lie group, and more than merely a compact group, it has a natural volume, rather than merely volume 1. It is Vol(SO(n))=2n-1π(n-1)(n+2)4d=2nΓ(d/2). See Luis J. Boya, E. C. G. Sudarshan, and Todd Tilma, Volumes of compact manifolds, http://repository.ias.ac.in/51021/. the unique SO(n)-invariant regular Borel probability measure on Sn-1 is

mSn-1=mq-1ϕ-1. (1)

It is a fact that the volume of the unit ball in n is


and that the surface area of Sn-1 in n is


For E a Borel set in Sn-1, define


Then σ is a SO(n)-invariant regular Borel measure on Sn-1, with total measure


We call σ the spherical surface measure.66 6 cf. Jacques Faraut, Analysis on Lie Groups: An Introduction, p. 186, §9.1 and Claus Müller, Analysis of Spherical Symmetries in Euclidean Spaces, Chapter 1.

For γSO(n) and fC(Sn-1), define


Let γn(x)=(2π)-n/2e-|x|2/2, which satisfies


and define I:C(Sn-1) by


which is a positive linear functional. Sn-1 is a compact Hausdorff space, so by the Riesz representation theorem there is a unique regular Borel measure μ on Sn-1 such that


Because I(f)=nγn(x)𝑑x=1, μ is a probability measure. For γSO(n), write g=γf, for which g(x/|x|)=f(γ-1(x/|x|)), and because |γ-1x|=|x| for xn and because Lebesgue measure on n is invariant under SO(n), by the change of variables theorem we have


Now define ν(E)=μ(γ(E))=((γ)*-1μ)(E), the pushforward of μ by γ-1. This is a regular Borel probability measure on Sn-1, and by the change of variables theorem,


Because I(f)=Sn-1f𝑑ν for all fC(Sn-1), it follows that ν=μ. Because γSO(n) is arbitrary, this measn that μ is SO(n)-invariant. But mSn-1 in (1) is the unique SO(n)-invariant regular Borel probability measure on Sn-1, so μ=mSn-1, so


where An-1=2πn/2Γ(n/2).

3 L2(Sn-1) and the spherical Laplacian

For f,gC(Sn-1), let


and let L2(S1) be the completion of C(Sn-1) with respect to this inner product.

For γSO(n) and fC(Sn-1) we have defined


Because σ is SO(n)-invariant,

γf,γg =Sn-1f(γ-1x)g¯(γ-1x)𝑑σ(x)

For f:Sn-1, define F:n-{0} by


We take f to belong to Ck(Sn-1) when FCk(n-{0}), 0k, and we define ΔSn-1f be the restriction of ΔF to Sn-1. We call ΔSn-1 the spherical Laplacian.77 7 cf. N. J. Vilenkin, Special Functions and the Theory of Group Representations, Chapter IX, §1.

Theorem 3.

Let F:n be positive-homogeneous of degree s and harmonic and let f be the restriction of F to Sn-1. Then


Let H(x)=F(x/|x|)=|x|-sF(x) and let r(x)=|x|=(x12++xn2)1/2. We calculate

ΔH =i=1ni2((r2)-s2F)

Euler’s identity for positive-homogeneous functions88 8 cf. John L. Greenberg, Alexis Fontaine’s ‘Fluxio-differential Method’ and the Origins of the Calculus of Several Variables, Annals of Science 38 (1981), 251–290. states that if G:n-{0} is positive-homogeneous of degree s then x(G)(x)=sG(x) for all x. Therefore

ΔH =-ns(r2)-s2-1F+(r2)-s2-2(s2+2s)|x|2F-(r2)-s2-22sr2sF

For xn-{0},


Then ΔSn-1f is equal to the restriction of ΔH to S, thus for xS, for which |r|=1,


Theorem 4.

If fC2(Sn-1) satisfies ΔSn-1f=λf, then λ0.

If gC2(Sn-1) satisfies ΔSn-1g=μg with λμ, then f,g=0.


Say λ0. Then

f,f =1λΔSn-1f,f

Because λ0, it is not the case that f=0, hence f,f>0. Hence λ¯λ=1, which means that λ. Furthermore,


which implies that λ<0. ∎

We now prove that ΔSn-1 is invariant under the action of SO(n).

Theorem 5.

If fC2(Sn-1) and γSO(n) then


Let F(x)=f(x/|x|), let g=γf, and let G(x)=g(x/|x|)=f(γ-1x/|γ-1x|). For xn-{0},


so γF=G. It is a fact that Δ(γF)=γ(ΔF).99 9 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 67, Theorem 2.1. Thus for xSn-1,


namely ΔSn-1(γf)=γ(ΔSn-1f). ∎

We now prove that ΔSn-1 is symmetric and negative-definite.1010 10 http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf, p. 9, Proposition 4.0.1.

Theorem 6.

For f,gC2(Sn-1),


ΔSn-1 is negative-definite:


and this is equal to 0 only when f is constant.


It is a fact that if F is positive-homogeneous of degree s then ΔF is positive-homogeneous of degree s-2. Let F(x)=f(x/|x|) and G(x)=g(x/|x|), with which


and, because F and G are positive-homogeneous of degree 0,

Sn-1(ΔSn-1f)(x)g(x)𝑑σ(x) =Sn-1(ΔF)(x)G(x)𝑑σ(x)



integrating by parts and using Euler’s identity for positive-homogeneous functions gives us


Because the above expression is the same when F and G are switched, this establishes


For g=f¯ we have G=F¯ and


which is 0. If it is equal to 0 then (iF)(x)=0 for all xn, which means that F is constant and hence that f is constant. ∎

4 Homogeneous polynomials

For P(x1,,xn)=aαxα[x1,,xn] write


For P,Q[x1,,xn], define1111 11 cf. John E. Gilbert and Margaret A. M. Murray, Clifford Algebras and Dirac Operators in Harmonic Analysis, p. 164, Chapter 3, §3.


For P=aαxα and Q=bβxβ,

(P,Q)=(βbβ¯βαaαxα)|x=0=βbβ¯aββ!. (2)
Lemma 7.

(,) is a positive-definite Hermitian form on [x1,,xn].


It is apparent that (,) is -linear in its first argument and conjugate linear in its second argument. From (2), it satisfies (P,Q)=(Q,P)¯, namely, (,) is a Hermitian form. For P[x1,,xn],


and if (P,P)=0 then each aα is equal to 0, showing that (,) is postive-definite. ∎

For P=αaαxα and Q=βbβxβ,


and we calculate


On the other hand,


and we calculate

Lemma 8.

For P,Q[x1,,xn],


Let 𝒫d be the set of homogeneous polynomials of degree d in [x1,,xn], i.e. those P(x1,,xn)[x1,,xn] of the form


We include the polynomial P=0, and 𝒫d is a complex vector space. We calculate1212 12 cf. Arthur T. Benjamin and Jennifer J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, p. 71, Identity 143 and p. 74, Identity 149.

dim𝒫d={α:|α|=d}=(n+d-1d). (3)

Let 𝒜d be the set of those P𝒫d satisfying ΔP=0, i.e. the homogeneous harmonic polynomials of degree d.

We prove that Δ:𝒫d𝒫d-2 is surjective.1313 13 http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf, p. 8, Claim 3.0.3.

Theorem 9.

The map Δ:𝒫d𝒫d-2 is surjective. Its kernel is 𝒜d, and


By Lemma 8,


In particular, (r2Q,r2Q)=0, and because (,) is nondegenerate this means that r2Q=0, and therefore Q=0. Because 𝒫d-2 is a finite-dimensional Hilbert space and the orthogonal complement of the image Δ𝒫d is equal to {0}, it follows that Δ𝒫d=𝒫d-2.

If P(r2𝒫d-2) then (P,r2Q)=0 for all Q𝒫d-2, hence (ΔP,Q)=0. In particular (ΔP,ΔP)=0 and so ΔP=0, which means that P𝒜d. On the other hand if P𝒜d then (P,r2Q)=(ΔP,Q)=0, so we get that (r2𝒫d-2)=𝒜d. Because 𝒫d is a finite-dimensional Hilbert space, this implies that 𝒜d=(r2𝒫d-2)=r2𝒫d-2. ∎

The above theorem tells us that




and by induction,


For P𝒫d, there are unique F0𝒜d, F2𝒜d-2, F4𝒜d-4, etc., such that


Let p be the restriction of P to Sn-1 and let fi be the restriction of Fi to Sn-1. Since r2=1 for xSn-1,


We have established the following.

Theorem 10.

The restriction of a homogeneous polynomial to Sn-1 is equal to a sum of the restrictions of homogeneous harmonic polynomials to Sn-1.

Using 𝒫d=𝒜dr2𝒫d-2, we have dim𝒫d=dim𝒜d+dim𝒫d-2, and then using the (3) for dim𝒫d we get the following.

Theorem 11.

With n fixed, using the asymptotic formula


we get from the above lemma


Let d be the restrictions of P𝒜d to Sn-1. We get the following from Theorem 3.

Lemma 12.

For Yd,




λd=0 if and only if d=0; if d1<d2 then λd2<λd10; and λd- as d.

5 The Hilbert space L2(Sn-1)

We prove that when d1d2, the subspaces d1 and d2 of L2(Sn-1) are mutually orthogonal.

Theorem 13.

For d1d2, for Y1d1 and for Y2d2,


From Lemma 12,


where λd=-d(d+n-2). Because d1d2 it follows that λd1λd2 and then by Theorem 4, Y1,Y2=0. ∎

For ϕC(Sn-1), write


Let A be the set of restrictions of all P[x1,,xn] to Sn-1. A is a self-adjoint algebra: it is a linear subspace of C(Sn-1); for p,qA, with P,Q[x1,,xn] such that p is the restriction of P to Sn-1 and q is the restriction of Q to Sn-1, the product PQ belongs to [x1,,xn] and pq is equal to the restriction of PQ to Sn-1, showing that A is an algebra; and p¯ is the restriction of P¯[x1,,xn] to Sn-1, showing that A is self-adjoint. For distinct u=(u1,,un),v=(v1,,vn) in Sn-1, say with ukvk, let P(x1,,xn)=xk and let p be the restriction of P to Sn-1. Then p(u)=uk and p(v)=vk, showing that A separates points. For uSn-1, let P(x1,,xn)=1 and let p be the restriction of P to Sn-1. Then p(u)=1, showing that A is nowhere vanishing. Because Sn-1 is a compact Hausdorff space, we obtain from the Stone-Weierstrass theorem1414 14 Walter Rudin, Functional Analysis, second ed., p. 122, Theorem 5.7. that A is dense in the Banach space C(Sn-1): for any ϕC(Sn-1) and for ϵ>0, there is some pA such that p-ϕC0ϵ.

L2(Sn-1) is the completion of C(Sn-1) with respect to the inner product


For fL2(Sn-1) and for ϵ>0, there is some ϕC(Sn-1) with ϕ-fL2ϵ, and there is some pA with p-ϕC0ϵ. But for ψC(Sn-1),



p-fL2 p-ϕL2+ϕ-fL2

This shows that A is dense in L2(Sn-1) with respect to the norm L2.

An element of [x1,,xn] can be written as a finite linear combination of homogeneous polynomials. By Theorem 10, the restriction to Sn-1 of each of these homogeneous polynomials is itself equal to a finite linear combination of homogeneous harmonic polynomials. Thus for pA there are Y1d1,,Ymdm with p=Y1++Ym. Therefore, the collection of all finite linear combinations of restrictions to Sn-1 of homogeneous harmonic polynomials is dense in L2(Sn-1). Now, Theorem 13 says that for d1d2, the subspaces d1 and d2 are mutually orthogonal. Putting the above together gives the following.

Theorem 14.


For ϕC(Sn-1),


Similar to Nikolsky’s inequality for the Fourier transform, for Yd, the norm YC0 is upper bounded by a multiple of the norm YL2 that depends on d.1515 15 http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf, p. 12, Proposition 6.0.1.

Theorem 15.

For Yd,


6 Sobolev embedding

Let Pd:L2(Sn-1)d the projection operator. Thus


in L2(Sn-1).

We prove the Sobolev embedding for Sn-1.1616 16 http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf, p. 14, Corollary 7.0.1; cf. Kendall Atkinson and Weimin Han, Spherical Harmonics and Approximations on the Unit Sphere: An Introduction, p. 119, §3.8

Theorem 16 (Sobolev embedding).

For fL2(Sn-1), if s>n-1 and


then there is some ϕC(Sn-1) such that ϕ=d0Pjf in C(Sn-1), and f=ϕ almost everywhere.


By Theorem 11 there is some Cn such that


Then by Theorem 15 and the Cauchy-Schwarz inequality,

d0PdfC0 d0dimdσ(Sn-1)PdfL2

Therefore d=0mPdf is a Cauchy sequence in the Banach space C(Sn-1), and hence converges to some ϕC(Sn-1). Because


the partial sums converge to ϕ in L2(Sn-1), and hence ϕ=f in L2(Sn-1), which implies that ϕ=f almost everywhere. ∎

7 Hecke’s identity

Hecke’s identity tells us the Fourier transform of a product of an element of 𝒜d and a Gaussian.1717 17 Elias M. Stein and Guido Weiss, Introduction to Fourier Analysis on Euclidean Spaces, p. 155, Theorem 3.4; http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf, p. 17, Theorem 9.0.1.

Theorem 17 (Hecke’s identity).

For f(u)=e-π|u|2P(u) with P𝒜d,


Let vn. The map ze-πzzP(z-iv) is a holomorphic separately in z1,,zn, and applying Cauchy’s integral theorem separately for z1,,zn,


Define Q:n by


and thus

Q(-iv) =ne-π(u+iv)(u+iv)P(u)𝑑u

On the other hand, for tn, using spherical coordinates, using the mean value property for the harmonic function P, and then using spherical coordinates again,

Q(t) =0e-πr2(Sn-1P(t+w)𝑑σ(w))rn-1𝑑r

Because P[x1,,xn], P has an analytic continuation to n, and then P(z)=Q(z) for all zn. Therefore


But because P is a homogeneous polynomial of degree d, P(-iv)=(-i)dP(v), so




proving the claim. ∎

8 Representation theory

Let a complex Hilbert space H with ,, let 𝒰(H) be the group of unitary operators HH. For a Lie group G, a unitary representation of G on H is a group homomorphism π:G𝒰(H) such that for each fH the map γπ(γ)(f) is continuous GH.

We have defined σ as a unique SO(n)-invariant regular Borel measure on Sn-1. It does not follow a priori that σ is O(n)-invariant. But in fact, using that |γx|=|x| for xn and that Lebesgue measure on n is O(n)-invariant, we check that σ is O(n)-invariant: for γO(n) and a Borel set E in Sn-1, σ(γE)=σ(E), i.e. γ*-1σ=σ.

For γO(n) and fL2(Sn-1), define


π(γ) is linear. For f,gL2(γ),

π(γ)(f),π(γ)(g) =Sn-1fγ-1gγ-1¯𝑑σ

For fL2(Sn-1), let g=fγ, for which


showing that π(γ) is surjective. Hence π(γ)𝒰(L2(Sn-1)).

For γ1,γO(n) and fL2(Sn-1),

π(γ1γ2)(f) =f(γ1γ2)-1

which means that π(γ1γ2)=π(γ1)π(γ2), namely π:O(n)𝒰(L2(Sn-1)) is a group homomorphism.

For ϕC(Sn-1) and for γ0,γO(n),


We take as given that π(γ0-1γ)(ϕ)-ϕC00 as γγ0 in O(n). Using that C(Sn-1) is dense in L2(Sn-1), one then proves that for each fL2(Sn-1), the map γπ(γ)(f) is continuous O(n)L2(Sn-1).

Lemma 18.

π is a unitary representation of the compact Lie group O(n) on the complex Hilbert space L2(Sn-1).

It is a fact that if γO(n) and P𝒫d then γP𝒫d. Furthermore, for ϕC2(Sn-1), Δ(γϕ)=γ(Δϕ), hence if P𝒜d then γP𝒜d. Then for Yd, π(γ)(Y)d. This means that each d is a π-invariant subspace.1818 18 cf. Feng Dai and Yuan Xu, Approximation Theory and Harmonic Analysis on Spheres and Balls, Chapter 1.