# Harmonic polynomials and the spherical Laplacian

Jordan Bell
August 17, 2015

## 1 Topological groups

Let $G$ be a topological group: $(x,y)\mapsto xy$ is continuous $G\times G\to G$ and $x\mapsto x^{-1}$ is continuous $G\to G$. For $g\in G$, the maps $L_{g}(x)=gx$ and $R_{g}(x)=xg$ are homeomorphisms. If $U$ is an open subset of $G$ and $X$ is a subset of $G$, for each $x\in X$ the set $Ux=\{ux:u\in U\}$ is open because $U$ is open and $u\mapsto ux$ is a homeomorphism. Therefore

 $UX=\{ux:u\in U,x\in X\}=\bigcup_{x\in X}Ux$

is open, being a union of open sets.

For a subgroup $H$ of $G$, not necessarily a normal subgroup, define $q:G\to G/H$ by

 $q(g)=gH,\qquad g\in G,$

and assign $G/H$ the final topology for $q$, the finest topology on $G/H$ such that $q:G\to G/H$ 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, $q:G\to G/H$ 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.

###### Proof.

For a topological space $X$, define $\Delta:X\to X\times X$ by $\Delta(x)=(x,x)$. It is a fact that $X$ is Hausdorff if and only if $\Delta(X)$ is a closed subset of $X\times X$. Thus the quotient space $G/H$ is Hausdorff if and only if the image of $\Delta:G/H\to G/H\times G/H$ is closed. The complement of $\Delta(G/H)$ is

 $(G/H\times G/H)-\Delta(G/H)=\{(xH,yH):xH\neq yH\}=\{(q(x),q(y)):x^{-1}y\not\in H\}.$

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

 $p^{-1}(U)=\{(x,y)\in G\times G:x^{-1}y\not\in H\}.$

The map $f:G\times G\to G$ defined by $f(x,y)=x^{-1}y$ is continuous and $G-H$ is open in $G$, so $f^{-1}(G-H)$ is open in $G\times G$. But

 $f^{-1}(G-H)=\{(x,y)\in G\times G:x^{-1}y\not\in H\}=p^{-1}(U),$

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 $\Delta(G\times H)$, 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 $\alpha:G\times K\to K$, denoted

 $\alpha(g,k)=g\cdot k,$

satisfying $e\cdot k=k$ and $(g_{1}g_{2})\cdot k=g_{1}\cdot(g_{2}\cdot k)$. The action is called transitive if for $k_{1},k_{2}\in K$ there is some $g\in G$ such that $g\cdot k_{1}=k_{2}$.

Let $H$ be a closed subgroup of $G$ and let $q:G\to G/H$ 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 $\beta:G\times G/H\to G/H$ by

 $\beta(g,xH)=g\cdot(xH)=(gx)H,\qquad g\in G,\quad xH\in G/H.$

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.

$\beta:G\times G/H\to G/H$ is a transitive left action.

###### Proof.

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

 $(L_{e}\times q)^{-1}(\beta^{-1}(V))=\mu^{-1}(q^{-1}(V)),$

hence $(L_{e}\times q)^{-1}(\beta^{-1}(V))$ is open in $G$. Because $L_{e}:G\to G$ and $q:G\to G/H$ are surjective open maps, the product $L_{e}\times q:G\times G\to G\times G/H$ is a surjective open map, so

 $(L_{e}\times q)((L_{e}\times q)^{-1}(\beta^{-1}(V)))=\beta^{-1}(V)$

is open in $G\times G/H$, showing that $\beta$ is continuous.

For $xH\in G/H$, $e\cdot(xH)=(ex)H=xH$, and for $g_{1},g_{2}\in G$,

 $(g_{1}g_{2})\cdot(xH)=(g_{1}g_{2}xH)=g_{1}\cdot(g_{2}xH)=g_{1}\cdot(g_{2}\cdot% (xH)).$

Therefore $\beta$ is a left action.

For $xH,yH\in G/H$,

 $(yx^{-1})\cdot xH=(yx^{-1}xH)=yH,$

showing that $\beta$ is transitive. ∎

Let $G$ be a compact group and let $\alpha$ be a transitive action of $G$ on a compact Hausdorff space $K$. For any $k_{0}\in K$, let $H=\{g\in G:\alpha(g,k_{0})=k_{0}\}$, the isotropy group of $k_{0}$, 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 $\phi:G/H\to K$ defined by

 $\phi(xH)=\alpha(x,k_{0}),\qquad xH\in G/H$

is a homeomorphism that satisfies

 $\phi(\beta(g,xH))=\alpha(g,\phi(xH)),\qquad g\in G,\quad xH\in G/H,$

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

 $\int_{G}f(x)dm(x)=\int_{G}f(x^{-1})dm(x),\qquad f\in C(G).$

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

Let $H$ be the above isotropy group, and define $m_{G/H}$ on the Borel $\sigma$-algebra of $G/H$ by

 $m_{G/H}=m\circ q^{-1}.$

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

 $m_{G/H}(g\cdot E)=m_{G/H}(E)$

for Borel sets $E$ in $G/H$ and for $g\in G$; we say that $m_{G/H}$ 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 $m_{K}$ on the Borel $\sigma$-algebra of $K$ by

 $m_{K}=m_{G/H}\circ\phi^{-1}=m\circ q^{-1}\circ\phi^{-1}.$

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

## 2 Spherical surface measure

$SO(n)$ is a compact Lie group. $S^{n-1}$ is a topological group, and it is a fact that $\alpha:SO(n)\times S^{n-1}\to S^{n-1}$ defined by

 $\alpha(g,k)=gk,\qquad g\in SO(n),\quad k\in S^{n-1},$

is a transitive left-action. We check that the isotropy group of $e_{n}$ is $SO(n-1)$. Let $q:SO(n)\to SO(n)/SO(n-1)$ be the projection map and define $\phi:SO(n)/SO(n-1)\to S^{n-1}$ by

 $\phi(xSO(n-1))=\alpha(x,e_{n})=xe_{n},\qquad xSO(n-1)\in SO(n)/SO(n-1).$

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 $\mathrm{Vol}(SO(n))=\frac{2^{n-1}\pi^{\frac{(n-1)(n+2)}{4}}}{\prod_{d=2}^{n}% \Gamma(d/2)}.$ 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 $S^{n-1}$ is

 $m_{S^{n-1}}=m\circ q^{-1}\circ\phi^{-1}.$ (1)

It is a fact that the volume of the unit ball in $\mathbb{R}^{n}$ is

 $\omega_{n}=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)},$

and that the surface area of $S^{n-1}$ in $\mathbb{R}^{n}$ is

 $A_{n-1}=n\omega_{n}=n\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}=\frac{% 2\pi^{n/2}}{\Gamma(n/2)}.$

For $E$ a Borel set in $S^{n-1}$, define

 $\sigma(E)=A_{n-1}m_{S^{n-1}}(E).$

Then $\sigma$ is a $SO(n)$-invariant regular Borel measure on $S^{n-1}$, with total measure

 $\sigma(S^{n-1})=A_{n-1}m_{S^{n-1}}(S^{n-1})=A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n% /2)}.$

We call $\sigma$ 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 $\gamma\in SO(n)$ and $f\in C(S^{n-1})$, define

 $(\gamma\cdot f)(x)=f(\gamma^{-1}x)=(f\circ\gamma^{-1})(x),\qquad x\in S^{n-1}.$

Let $\gamma_{n}(x)=(2\pi)^{-n/2}e^{-|x|^{2}/2}$, which satisfies

 $\int_{\mathbb{R}^{n}}\gamma_{n}(x)dx=1,$

and define $I:C(S^{n-1})\to\mathbb{C}$ by

 $I(f)=\int_{\mathbb{R}^{n}}f(x/|x|)\gamma_{n}(x)dx,\qquad f\in C(S^{n-1}),$

which is a positive linear functional. $S^{n-1}$ is a compact Hausdorff space, so by the Riesz representation theorem there is a unique regular Borel measure $\mu$ on $S^{n-1}$ such that

 $I(f)=\int_{S^{n-1}}fd\mu,\qquad f\in C(S^{n-1}).$

Because $I(f)=\int_{\mathbb{R}^{n}}\gamma_{n}(x)dx=1$, $\mu$ is a probability measure. For $\gamma\in SO(n)$, write $g=\gamma\cdot f$, for which $g(x/|x|)=f(\gamma^{-1}(x/|x|))$, and because $|\gamma^{-1}x|=|x|$ for $x\in\mathbb{R}^{n}$ and because Lebesgue measure on $\mathbb{R}^{n}$ is invariant under $SO(n)$, by the change of variables theorem we have

 $I(\gamma\cdot f)=I(g)=\int_{\mathbb{R}^{n}}f\left(\frac{1}{|x|}\gamma^{-1}x% \right)(2\pi)^{-n/2}e^{-|x|^{2}/2}dx=I(f).$

Now define $\nu(E)=\mu(\gamma(E))=((\gamma)^{-1}_{*}\mu)(E)$, the pushforward of $\mu$ by $\gamma^{-1}$. This is a regular Borel probability measure on $S^{n-1}$, and by the change of variables theorem,

 $\int_{S^{n-1}}fd\nu=\int_{S^{n-1}}f\circ\gamma^{-1}d\mu=\int_{S^{n-1}}\gamma% \cdot fd\mu=I(\gamma\cdot f)=I(f).$

Because $I(f)=\int_{S^{n-1}}fd\nu$ for all $f\in C(S^{n-1})$, it follows that $\nu=\mu$. Because $\gamma\in SO(n)$ is arbitrary, this measn that $\mu$ is $SO(n)$-invariant. But $m_{S^{n-1}}$ in (1) is the unique $SO(n)$-invariant regular Borel probability measure on $S^{n-1}$, so $\mu=m_{S^{n-1}}$, so

 $\int_{S^{n-1}}fd\sigma=A_{n-1}\int_{S^{n-1}}fd\mu=A_{n-1}\int_{\mathbb{R}^{n}}% f(x/|x|)(2\pi)^{-n/2}e^{-|x|^{2}/2}dx,$

where $A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$.

## 3 L2(Sn-1) and the spherical Laplacian

For $f,g\in C(S^{n-1})$, let

 $\left\langle f,g\right\rangle=\int_{S^{n-1}}f\overline{g}d\sigma,$

and let $L^{2}(S^{1})$ be the completion of $C(S^{n-1})$ with respect to this inner product.

For $\gamma\in SO(n)$ and $f\in C(S^{n-1})$ we have defined

 $(\gamma\cdot f)(x)=f(\gamma^{-1}x)=(f\circ\gamma^{-1})(x),\qquad x\in S^{n-1}.$

Because $\sigma$ is $SO(n)$-invariant,

 $\displaystyle\left\langle\gamma\cdot f,\gamma\cdot g\right\rangle$ $\displaystyle=\int_{S^{n-1}}f(\gamma^{-1}x)\overline{g}(\gamma^{-1}x)d\sigma(x)$ $\displaystyle=\int_{S^{n-1}}f(x)\overline{g}(x)d((\gamma^{-1})_{*}\sigma)(x)$ $\displaystyle=\int_{S^{n-1}}f(x)\overline{g}(x)d\sigma(x)$ $\displaystyle=\left\langle f,g\right\rangle.$

For $f:S^{n-1}\to\mathbb{C}$, define $F:\mathbb{R}^{n}-\{0\}\to\mathbb{C}$ by

 $F(x)=f(x/|x|).$

We take $f$ to belong to $C^{k}(S^{n-1})$ when $F\in C^{k}(\mathbb{R}^{n}-\{0\})$, $0\leq k\leq\infty$, and we define $\Delta_{S^{n-1}}f$ be the restriction of $\Delta F$ to $S^{n-1}$. We call $\Delta_{S^{n-1}}$ the spherical Laplacian.77 7 cf. N. J. Vilenkin, Special Functions and the Theory of Group Representations, Chapter IX, §1.

###### Theorem 3.

Let $F:\mathbb{R}^{n}\to\mathbb{C}$ be positive-homogeneous of degree $s$ and harmonic and let $f$ be the restriction of $F$ to $S^{n-1}$. Then

 $\Delta_{S^{n-1}}f=-s(n+s-2)f.$
###### Proof.

Let $H(x)=F(x/|x|)=|x|^{-s}F(x)$ and let $r(x)=|x|=(x_{1}^{2}+\cdots+x_{n}^{2})^{1/2}$. We calculate

 $\displaystyle\Delta H$ $\displaystyle=\sum_{i=1}^{n}\partial_{i}^{2}((r^{2})^{-\frac{s}{2}}F)$ $\displaystyle=\sum_{i=1}^{n}\partial_{i}\left(-sx_{i}(r^{2})^{-\frac{s}{2}-1}F% +(r^{2})^{-\frac{s}{2}}\partial_{i}F\right)$ $\displaystyle=\sum_{i=1}^{n}-s(r^{2})^{-\frac{s}{2}-1}F-sx_{i}(2x_{i})\left(-% \frac{s}{2}-1\right)(r^{2})^{-\frac{s}{2}-2}F-sx_{i}(r^{2})^{\frac{s}{2}-1}% \partial_{i}F$ $\displaystyle-sx_{i}(r^{2})^{-\frac{s}{2}-1}\partial_{i}F+(r^{2})^{-\frac{s}{2% }}\partial_{i}^{2}F$ $\displaystyle=-ns(r^{2})^{-\frac{s}{2}-1}F+(r^{2})^{-\frac{s}{2}-2}\sum_{i=1}^% {n}\left(-s(-s-2)x_{i}^{2}F-sx_{i}r^{2}\partial_{i}F-sx_{i}r^{2}\partial_{i}F\right)$ $\displaystyle+(r^{2})^{-\frac{s}{2}}\Delta F$ $\displaystyle=-ns(r^{2})^{-\frac{s}{2}-1}F+(r^{2})^{-\frac{s}{2}-2}\sum_{i=1}^% {n}(s^{2}x_{i}^{2}F+2sx_{i}^{2}F-2sx_{i}r^{2}\partial_{i}F).$

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 $G:\mathbb{R}^{n}-\{0\}\to\mathbb{C}$ is positive-homogeneous of degree $s$ then $x\cdot(\nabla G)(x)=sG(x)$ for all $x$. Therefore

 $\displaystyle\Delta H$ $\displaystyle=-ns(r^{2})^{-\frac{s}{2}-1}F+(r^{2})^{-\frac{s}{2}-2}(s^{2}+2s)|% x|^{2}F-(r^{2})^{-\frac{s}{2}-2}\cdot 2sr^{2}\cdot sF$ $\displaystyle=-ns(r^{2})^{-\frac{s}{2}-1}F+(r^{2})^{-\frac{s}{2}-1}(s^{2}+2s)F% -(r^{2})^{-\frac{s}{2}-1}\cdot 2s^{2}F$ $\displaystyle=-sr^{-s-2}(n+s-2)F.$

For $x\in\mathbb{R}^{n}-\{0\}$,

 $f(x/|x|)=F(x/|x|)=H(x).$

Then $\Delta_{S^{n-1}}f$ is equal to the restriction of $\Delta H$ to $S$, thus for $x\in S$, for which $|r|=1$,

 $(\Delta_{S^{n-1}}f)(x)=-sr^{-s-2}(n+s-2)F(x)=-s(n+s-2)f(x).$

###### Theorem 4.

If $f\in C^{2}(S^{n-1})$ satisfies $\Delta_{S^{n-1}}f=\lambda f$, then $\lambda\leq 0$.

If $g\in C^{2}(S^{n-1})$ satisfies $\Delta_{S^{n-1}}g=\mu g$ with $\lambda\neq\mu$, then $\left\langle f,g\right\rangle=0$.

###### Proof.

Say $\lambda\neq 0$. Then

 $\displaystyle\left\langle f,f\right\rangle$ $\displaystyle=\frac{1}{\lambda}\left\langle\Delta_{S^{n-1}}f,f\right\rangle$ $\displaystyle=\frac{1}{\lambda}\int_{S^{n-1}}(\Delta_{S^{n-1}}f)\overline{f}d\sigma$ $\displaystyle=\frac{1}{\lambda}\int_{S^{n-1}}f\Delta_{S^{n-1}}\overline{f}d\sigma$ $\displaystyle=\frac{1}{\lambda}\int_{S^{n-1}}f\overline{\Delta_{S^{n-1}}f}d\sigma$ $\displaystyle=\frac{1}{\lambda}\int_{S^{n-1}}f\overline{\lambda f}d\sigma$ $\displaystyle=\frac{\overline{\lambda}}{\lambda}\left\langle f,f\right\rangle.$

Because $\lambda\neq 0$, it is not the case that $f=0$, hence $\left\langle f,f\right\rangle>0$. Hence $\frac{\overline{\lambda}}{\lambda}=1$, which means that $\lambda\in\mathbb{R}$. Furthermore,

 $\lambda\left\langle f,f\right\rangle=\left\langle\lambda f,f\right\rangle=% \left\langle\Delta_{S^{n-1}}f,f\right\rangle=\int_{S^{n-1}}(\Delta_{S^{n-1}}f)% \overline{f}d\sigma<0,$

which implies that $\lambda<0$. ∎

We now prove that $\Delta_{S^{n-1}}$ is invariant under the action of $SO(n)$.

###### Theorem 5.

If $f\in C^{2}(S^{n-1})$ and $\gamma\in SO(n)$ then

 $\Delta_{S^{n-1}}(\gamma\cdot f)=\gamma\cdot(\Delta_{S^{n-1}}f).$
###### Proof.

Let $F(x)=f(x/|x|)$, let $g=\gamma\cdot f$, and let $G(x)=g(x/|x|)=f(\gamma^{-1}x/|\gamma^{-1}x|)$. For $x\in\mathbb{R}^{n}-\{0\}$,

 $(\gamma\cdot F)(x)=F(\gamma^{-1}x)=f(\gamma^{-1}x/|\gamma^{-1}x|)=G(x),$

so $\gamma\cdot F=G$. It is a fact that $\Delta(\gamma\cdot F)=\gamma\cdot(\Delta F)$.99 9 Gerald B. Folland, Introduction to Partial Differential Equations, second ed., p. 67, Theorem 2.1. Thus for $x\in S^{n-1}$,

 $(\Delta_{S^{n-1}}g)(x)=(\Delta G)(x)=(\gamma\cdot(\Delta F))(x)=(\Delta F)(% \gamma^{-1}x)=(\Delta_{S^{n-1}}f)(\gamma^{-1}x),$

namely $\Delta_{S^{n-1}}(\gamma\cdot f)=\gamma\cdot(\Delta_{S^{n-1}}f)$. ∎

We now prove that $\Delta_{S^{n-1}}$ 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 $f,g\in C^{2}(S^{n-1})$,

 $\int_{S^{n-1}}(\Delta_{S^{n-1}}f)\cdot gd\sigma=\int_{S^{n-1}}f\cdot\Delta_{S^% {n-1}}gd\sigma.$

$\Delta_{S^{n-1}}$ is negative-definite:

 $\int_{S^{n-1}}(\Delta_{S^{n-1}}f)\cdot\overline{f}\leq 0,$

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

###### Proof.

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

 $(\Delta_{S^{n-1}}f)(x)=(\Delta F)(x),\qquad(\Delta_{S^{n-1}}g)(x)=(\Delta G)(x% ),\qquad x\in S^{n-1}$

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

 $\displaystyle\int_{S^{n-1}}(\Delta_{S^{n-1}}f)(x)\cdot g(x)d\sigma(x)$ $\displaystyle=\int_{S^{n-1}}(\Delta F)(x)\cdot G(x)d\sigma(x)$ $\displaystyle=A_{n-1}\int_{\mathbb{R}^{n}}(\Delta F)(x/|x|)\cdot G(x/|x|)% \gamma_{n}(x)dx$ $\displaystyle=A_{n-1}\int_{\mathbb{R}^{n}}|x|^{2}(\Delta F)(x)\cdot G(x)\gamma% _{n}(x)dx.$

Because

 $\partial_{i}(|x|^{2}G\gamma_{n})=2x_{i}G\gamma_{n}+|x|^{2}\gamma_{n}\partial_{% i}G+|x|^{2}G(-x_{i}\gamma_{n}),$

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

 $\begin{split}&\displaystyle\int_{\mathbb{R}^{n}}(\Delta F)(x)\cdot|x|^{2}G(x)% \gamma_{n}(x)dx\\ \displaystyle=&\displaystyle-\int_{\mathbb{R}^{n}}\sum_{i=1}^{n}(\partial_{i}F% )(x)\partial_{i}(|x|^{2}G(x)\gamma_{n}(x))dx\\ \displaystyle=&\displaystyle-\int_{\mathbb{R}^{n}}\sum_{i=1}^{n}((2G\gamma_{n}% -|x|^{2}G\gamma_{n})\cdot x_{i}\partial_{i}F+|x|^{2}\gamma_{n}\partial_{i}F% \partial_{i}G)dx\\ \displaystyle=&\displaystyle-\int_{\mathbb{R}^{n}}\sum_{i=1}^{n}|x|^{2}\gamma_% {n}\partial_{i}F\cdot\partial_{i}Gdx.\end{split}$

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

 $\int_{S^{n-1}}(\Delta_{S^{n-1}}f)\cdot gd\sigma=\int_{S^{n-1}}f\cdot\Delta_{S^% {n-1}}gd\sigma.$

For $g=\overline{f}$ we have $G=\overline{F}$ and

 $\int_{S^{n-1}}(\Delta_{S^{n-1}}f)\cdot\overline{f}d\sigma=-A_{n-1}\int_{% \mathbb{R}^{n}}\sum_{i=1}^{n}|x|^{2}\gamma_{n}|\partial_{i}F|^{2}dx,$

which is $\leq 0$. If it is equal to $0$ then $(\partial_{i}F)(x)=0$ for all $x\in\mathbb{R}^{n}$, which means that $F$ is constant and hence that $f$ is constant. ∎

## 4 Homogeneous polynomials

For $P(x_{1},\ldots,x_{n})=\sum a_{\alpha}x^{\alpha}\in\mathbb{C}[x_{1},\ldots,x_{n}]$ write

 $P(\partial)=\sum a_{\alpha}\partial^{\alpha},\qquad\overline{P}(x_{1},\ldots,x% _{n})=\sum\overline{a_{\alpha}}x^{\alpha},\qquad\overline{P}(\partial)=\sum% \overline{a_{\alpha}}\partial^{\alpha}.$

For $P,Q\in\mathbb{C}[x_{1},\ldots,x_{n}]$, define1111 11 cf. John E. Gilbert and Margaret A. M. Murray, Clifford Algebras and Dirac Operators in Harmonic Analysis, p. 164, Chapter 3, §3.

 $(P,Q)=(\overline{Q}(\partial P)\Big{|}_{x=0}.$

For $P=\sum a_{\alpha}x^{\alpha}$ and $Q=\sum b_{\beta}x^{\beta}$,

 $(P,Q)=\left(\sum_{\beta}\overline{b_{\beta}}\partial^{\beta}\sum_{\alpha}a_{% \alpha}x^{\alpha}\right)\Big{|}_{x=0}=\sum_{\beta}\overline{b_{\beta}}a_{\beta% }\cdot\beta!.$ (2)
###### Lemma 7.

$(\cdot,\cdot)$ is a positive-definite Hermitian form on $\mathbb{C}[x_{1},\ldots,x_{n}]$.

###### Proof.

It is apparent that $(\cdot,\cdot)$ is $\mathbb{C}$-linear in its first argument and conjugate linear in its second argument. From (2), it satisfies $(P,Q)=\overline{(Q,P)}$, namely, $(\cdot,\cdot)$ is a Hermitian form. For $P\in\mathbb{C}[x_{1},\ldots,x_{n}]$,

 $(P,P)=\sum_{\alpha}a_{\alpha}\overline{a_{\alpha}}\cdot\alpha!=\sum_{\alpha}|a% _{\alpha}|^{2}\cdot\alpha!\geq 0,$

and if $(P,P)=0$ then each $a_{\alpha}$ is equal to $0$, showing that $(\cdot,\cdot)$ is postive-definite. ∎

For $P=\sum_{\alpha}a_{\alpha}x^{\alpha}$ and $Q=\sum_{\beta}b_{\beta}x^{\beta}$,

 $(\Delta P)(x)=\sum_{\alpha}a_{\alpha}\sum_{i=1}^{n}\partial_{i}^{2}x^{\alpha}=% \sum_{\alpha}a_{\alpha}\sum_{i=1}^{n}\frac{\alpha!}{(\alpha-2e_{i})!}x^{\alpha% -2e_{i}},$

and we calculate

 $(\Delta P,Q)=\sum_{\beta}\overline{b_{\beta}}\sum_{i=1}^{n}a_{\beta+2e_{i}}(% \beta+2e_{i})!.$

On the other hand,

 $r^{2}Q(x_{1},\ldots,x_{n})=\sum_{\beta}b_{\beta}x^{\beta}\sum_{i=1}^{n}x_{i}^{% 2}=\sum_{\beta}b_{\beta}\sum_{i=1}^{n}x^{\beta+2e_{i}},$

and we calculate

 $(P,r^{2}Q)=\sum_{\beta}\overline{b_{\beta}}\sum_{i=1}^{n}a_{\beta+2e_{i}}.$
###### Lemma 8.

For $P,Q\in\mathbb{C}[x_{1},\ldots,x_{n}]$,

 $(\Delta P,Q)=(P,r^{2}Q).$

Let $\mathscr{P}_{d}$ be the set of homogeneous polynomials of degree $d$ in $\mathbb{C}[x_{1},\ldots,x_{n}]$, i.e. those $P(x_{1},\ldots,x_{n})\in\mathbb{C}[x_{1},\ldots,x_{n}]$ of the form

 $P(x_{1},\ldots,x_{n})=\sum_{|\alpha|=d}a_{\alpha}x^{\alpha}.$

We include the polynomial $P=0$, and $\mathscr{P}_{d}$ 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.

 $\dim_{\mathbb{C}}\mathscr{P}_{d}=\{\alpha:|\alpha|=d\}=\binom{n+d-1}{d}.$ (3)

Let $\mathscr{A}_{d}$ be the set of those $P\in\mathscr{P}_{d}$ satisfying $\Delta P=0$, i.e. the homogeneous harmonic polynomials of degree $d$.

We prove that $\Delta:\mathscr{P}_{d}\to\mathscr{P}_{d-2}$ 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 $\Delta:\mathscr{P}_{d}\to\mathscr{P}_{d-2}$ is surjective. Its kernel is $\mathscr{A}_{d}$, and

 $\mathscr{A}_{d}^{\perp}=r^{2}\mathscr{P}_{d-2}.$
###### Proof.

By Lemma 8,

 $0=(\Delta P,Q)=(P,r^{2}Q).$

In particular, $(r^{2}Q,r^{2}Q)=0$, and because $(\cdot,\cdot)$ is nondegenerate this means that $r^{2}Q=0$, and therefore $Q=0$. Because $\mathscr{P}_{d-2}$ is a finite-dimensional Hilbert space and the orthogonal complement of the image $\Delta\mathscr{P}_{d}$ is equal to $\{0\}$, it follows that $\Delta\mathscr{P}_{d}=\mathscr{P}_{d-2}$.

If $P\in(r^{2}\mathscr{P}_{d-2})^{\perp}$ then $(P,r^{2}Q)=0$ for all $Q\in\mathscr{P}_{d-2}$, hence $(\Delta P,Q)=0$. In particular $(\Delta P,\Delta P)=0$ and so $\Delta P=0$, which means that $P\in\mathscr{A}_{d}$. On the other hand if $P\in\mathscr{A}_{d}$ then $(P,r^{2}Q)=(\Delta P,Q)=0$, so we get that $(r^{2}\mathscr{P}_{d-2})^{\perp}=\mathscr{A}_{d}$. Because $\mathscr{P}_{d}$ is a finite-dimensional Hilbert space, this implies that $\mathscr{A}_{d}^{\perp}=(r^{2}\mathscr{P}_{d-2})^{\perp\perp}=r^{2}\mathscr{P}% _{d-2}$. ∎

The above theorem tells us that

 $\mathscr{P}_{d}=\mathscr{A}_{d}\oplus\mathscr{A}_{d}^{\perp}=\mathscr{A}_{d}% \oplus r^{2}\mathscr{P}_{d-2}.$

Then,

 $\mathscr{P}_{d-2}=\mathscr{A}_{d-2}\oplus r^{2}\mathscr{P}_{d-4},$

and by induction,

 $\mathscr{P}_{d}=\mathscr{A}_{d}\oplus r^{2}\mathscr{A}_{d-2}\oplus r^{2}% \mathscr{A}_{d-4}\oplus\cdots.$

For $P\in\mathscr{P}_{d}$, there are unique $F_{0}\in\mathscr{A}_{d}$, $F_{2}\in\mathscr{A}_{d-2}$, $F_{4}\in\mathscr{A}_{d-4}$, etc., such that

 $P=F_{0}+r^{2}F_{2}+r^{4}F_{4}+\cdots.$

Let $p$ be the restriction of $P$ to $S^{n-1}$ and let $f_{i}$ be the restriction of $F_{i}$ to $S^{n-1}$. Since $r^{2}=1$ for $x\in S^{n-1}$,

 $p=f_{0}+f_{2}+f_{4}+\cdots.$

We have established the following.

###### Theorem 10.

The restriction of a homogeneous polynomial to $S^{n-1}$ is equal to a sum of the restrictions of homogeneous harmonic polynomials to $S^{n-1}$.

Using $\mathscr{P}_{d}=\mathscr{A}_{d}\oplus r^{2}\mathscr{P}_{d-2}$, we have $\dim_{\mathbb{C}}\mathscr{P}_{d}=\dim_{\mathbb{C}}\mathscr{A}_{d}+\dim_{% \mathbb{C}}\mathscr{P}_{d-2}$, and then using the (3) for $\dim_{\mathbb{C}}\mathscr{P}_{d}$ we get the following.

###### Theorem 11.
 $\dim_{\mathbb{C}}\mathscr{A}_{d}=\binom{n+d-1}{d}-\binom{n+d-3}{d-2}=\binom{n+% d-2}{n-2}+\binom{n+d-3}{n-2}.$

With $n$ fixed, using the asymptotic formula

 $\binom{z+k}{k}=\frac{k^{z}}{\Gamma(z+1)}\left(1+\frac{z(z+1)}{2k}+O(k^{-2})% \right),\qquad k\to\infty,$

we get from the above lemma

 $\dim_{\mathbb{C}}\mathscr{A}_{d}\sim\frac{2}{(n-2)!}d^{n-2}.$

Let $\mathscr{H}_{d}$ be the restrictions of $P\in\mathscr{A}_{d}$ to $S^{n-1}$. We get the following from Theorem 3.

###### Lemma 12.

For $Y\in\mathscr{H}_{d}$,

 $\Delta_{S^{n-1}}Y=\lambda_{d}Y$

where

 $\lambda_{d}=-d(d+n-2)=-\left(d+\frac{n-2}{2}\right)^{2}+\left(\frac{n-2}{2}% \right)^{2}.$

$\lambda_{d}=0$ if and only if $d=0$; if $d_{1} then $\lambda_{d_{2}}<\lambda_{d_{1}}\leq 0$; and $\lambda_{d}\to-\infty$ as $d\to\infty$.

## 5 The Hilbert space L2(Sn-1)

We prove that when $d_{1}\neq d_{2}$, the subspaces $\mathscr{H}_{d_{1}}$ and $\mathscr{H}_{d_{2}}$ of $L^{2}(S^{n-1})$ are mutually orthogonal.

###### Theorem 13.

For $d_{1}\neq d_{2}$, for $Y_{1}\in\mathscr{H}_{d_{1}}$ and for $Y_{2}\in\mathscr{H}_{d_{2}}$,

 $\left\langle Y_{1},Y_{2}\right\rangle=0.$
###### Proof.

From Lemma 12,

 $\Delta_{S^{n-1}}Y_{1}=\lambda_{d_{1}}Y_{1},\qquad\Delta_{S^{n-1}}Y_{2}=\lambda% _{d_{2}}Y_{2}.$

where $\lambda_{d}=-d(d+n-2)$. Because $d_{1}\neq d_{2}$ it follows that $\lambda_{d_{1}}\neq\lambda_{d_{2}}$ and then by Theorem 4, $\left\langle Y_{1},Y_{2}\right\rangle=0$. ∎

For $\phi\in C(S^{n-1})$, write

 $\left\|\phi\right\|_{C^{0}}=\sup_{x\in S^{n-1}}|\phi(x)|.$

Let $A$ be the set of restrictions of all $P\in\mathbb{C}[x_{1},\ldots,x_{n}]$ to $S^{n-1}$. $A$ is a self-adjoint algebra: it is a linear subspace of $C(S^{n-1})$; for $p,q\in A$, with $P,Q\in\mathbb{C}[x_{1},\ldots,x_{n}]$ such that $p$ is the restriction of $P$ to $S^{n-1}$ and $q$ is the restriction of $Q$ to $S^{n-1}$, the product $PQ$ belongs to $\mathbb{C}[x_{1},\ldots,x_{n}]$ and $pq$ is equal to the restriction of $PQ$ to $S^{n-1}$, showing that $A$ is an algebra; and $\overline{p}$ is the restriction of $\overline{P}\in\mathbb{C}[x_{1},\ldots,x_{n}]$ to $S^{n-1}$, showing that $A$ is self-adjoint. For distinct $u=(u_{1},\ldots,u_{n}),v=(v_{1},\ldots,v_{n})$ in $S^{n-1}$, say with $u_{k}\neq v_{k}$, let $P(x_{1},\ldots,x_{n})=x_{k}$ and let $p$ be the restriction of $P$ to $S^{n-1}$. Then $p(u)=u_{k}$ and $p(v)=v_{k}$, showing that $A$ separates points. For $u\in S^{n-1}$, let $P(x_{1},\ldots,x_{n})=1$ and let $p$ be the restriction of $P$ to $S^{n-1}$. Then $p(u)=1$, showing that $A$ is nowhere vanishing. Because $S^{n-1}$ 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 $C(S^{n-1})$: for any $\phi\in C(S^{n-1})$ and for $\epsilon>0$, there is some $p\in A$ such that $\left\|p-\phi\right\|_{C^{0}}\leq\epsilon$.

$L^{2}(S^{n-1})$ is the completion of $C(S^{n-1})$ with respect to the inner product

 $\left\langle f,g\right\rangle=\int_{S^{n-1}}f\cdot\overline{g}d\sigma.$

For $f\in L^{2}(S^{n-1})$ and for $\epsilon>0$, there is some $\phi\in C(S^{n-1})$ with $\left\|\phi-f\right\|_{L^{2}}\leq\epsilon$, and there is some $p\in A$ with $\left\|p-\phi\right\|_{C^{0}}\leq\epsilon$. But for $\psi\in C(S^{n-1})$,

 $\left\|\psi\right\|_{L^{2}}=\left(\int_{S^{n-1}}|\psi|^{2}d\sigma\right)^{1/2}% \leq\left\|\psi\right\|_{C^{0}}\cdot\sqrt{\sigma(S^{n-1})}.$

Then

 $\displaystyle\left\|p-f\right\|_{L^{2}}$ $\displaystyle\leq\left\|p-\phi\right\|_{L^{2}}+\left\|\phi-f\right\|_{L^{2}}$ $\displaystyle\leq\left\|p-\phi\right\|_{C^{0}}\cdot\sqrt{\sigma(S^{n-1})}+\epsilon$ $\displaystyle\leq\epsilon\cdot\sqrt{\sigma(S^{n-1})}+\epsilon.$

This shows that $A$ is dense in $L^{2}(S^{n-1})$ with respect to the norm $\left\|\cdot\right\|_{L^{2}}$.

An element of $\mathbb{C}[x_{1},\ldots,x_{n}]$ can be written as a finite linear combination of homogeneous polynomials. By Theorem 10, the restriction to $S^{n-1}$ of each of these homogeneous polynomials is itself equal to a finite linear combination of homogeneous harmonic polynomials. Thus for $p\in A$ there are $Y_{1}\in\mathscr{H}_{d_{1}},\ldots,Y_{m}\in\mathscr{H}_{d_{m}}$ with $p=Y_{1}+\cdots+Y_{m}$. Therefore, the collection of all finite linear combinations of restrictions to $S^{n-1}$ of homogeneous harmonic polynomials is dense in $L^{2}(S^{n-1})$. Now, Theorem 13 says that for $d_{1}\neq d_{2}$, the subspaces $\mathscr{H}_{d_{1}}$ and $\mathscr{H}_{d_{2}}$ are mutually orthogonal. Putting the above together gives the following.

###### Theorem 14.

$L^{2}(S^{n-1})=\bigoplus_{d\geq 0}\mathscr{H}_{d}$.

For $\phi\in C(S^{n-1})$,

 $\left\|\phi\right\|_{L^{2}}\leq\sqrt{\sigma(S^{n-1})}\cdot\left\|\phi\right\|_% {C^{0}}.$

Similar to Nikolsky’s inequality for the Fourier transform, for $Y\in\mathscr{H}_{d}$, the norm $\left\|Y\right\|_{C^{0}}$ is upper bounded by a multiple of the norm $\left\|Y\right\|_{L^{2}}$ 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 $Y\in\mathscr{H}_{d}$,

 $\left\|Y\right\|_{C^{0}}\leq\sqrt{\frac{\dim_{\mathbb{C}}\mathscr{H}_{d}}{% \sigma(S^{n-1})}}\cdot\left\|Y\right\|_{L^{2}}.$

## 6 Sobolev embedding

Let $P_{d}:L^{2}(S^{n-1})\to\mathscr{H}_{d}$ the projection operator. Thus

 $f=\sum_{d\geq 0}P_{d}f$

in $L^{2}(S^{n-1})$.

We prove the Sobolev embedding for $S^{n-1}$.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 $f\in L^{2}(S^{n-1})$, if $s>n-1$ and

 $\sum_{d\geq 0}(1+d)^{s}\cdot\left\|P_{d}f\right\|_{L^{2}}^{2}<\infty$

then there is some $\phi\in C(S^{n-1})$ such that $\phi=\sum_{d\geq 0}P_{j}f$ in $C(S^{n-1})$, and $f=\phi$ almost everywhere.

###### Proof.

By Theorem 11 there is some $C_{n}$ such that

 $\dim_{\mathbb{C}}\mathscr{H}_{d}\leq C_{n}(1+d)^{n-2}.$

Then by Theorem 15 and the Cauchy-Schwarz inequality,

 $\displaystyle\sum_{d\geq 0}\left\|P_{d}f\right\|_{C^{0}}$ $\displaystyle\leq\sum_{d\geq 0}\sqrt{\frac{\dim_{\mathbb{C}}\mathscr{H}_{d}}{% \sigma(S^{n-1})}}\cdot\left\|P_{d}f\right\|_{L^{2}}$ $\displaystyle\leq\sqrt{\frac{C_{n}}{\sigma(S^{n-1})}}\sum_{d\geq 0}(1+d)^{% \frac{n-2}{2}}\cdot\left\|P_{d}f\right\|_{L^{2}}$ $\displaystyle=\sqrt{\frac{C_{n}}{\sigma(S^{n-1})}}\sum_{d\geq 0}(1+d)^{\frac{s% }{2}}\left\|P_{d}f\right\|_{L^{2}}\cdot(1+d)^{-\frac{s-n+2}{2}}$ $\displaystyle\leq\sqrt{\frac{C_{n}}{\sigma(S^{n-1})}}\left(\sum_{d\geq 0}(1+d)% ^{s}\left\|P_{d}f\right\|_{L^{2}}^{2}\right)\left(\sum_{d\geq 0}(1+d)^{-(s-n+2% )}\right)$ $\displaystyle=\sqrt{\frac{C_{n}}{\sigma(S^{n-1})}}\cdot\zeta(s-n+2)\cdot\sum_{% d\geq 0}(1+d)^{s}\left\|P_{d}\right\|_{L^{2}}^{2}$ $\displaystyle<\infty.$

Therefore $\sum_{d=0}^{m}P_{d}f$ is a Cauchy sequence in the Banach space $C(S^{n-1})$, and hence converges to some $\phi\in C(S^{n-1})$. Because

 $\left\|\sum_{d=0}^{m}P_{d}f-\phi\right\|_{L^{2}}\leq\sqrt{\sigma(S^{n-1})}% \cdot\left\|\sum_{d=0}^{m}P_{d}f-\phi\right\|_{C^{0}},$

the partial sums converge to $\phi$ in $L^{2}(S^{n-1})$, and hence $\phi=f$ in $L^{2}(S^{n-1})$, which implies that $\phi=f$ almost everywhere. ∎

## 7 Hecke’s identity

Hecke’s identity tells us the Fourier transform of a product of an element of $\mathscr{A}_{d}$ 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 $f(u)=e^{-\pi|u|^{2}}P(u)$ with $P\in\mathscr{A}_{d}$,

 $\widehat{f}(v)=(-i)^{d}f(v),\qquad v\in\mathbb{R}^{n}.$
###### Proof.

Let $v\in\mathbb{R}^{n}$. The map $z\mapsto e^{-\pi z\cdot z}P(z-iv)$ is a holomorphic separately in $z_{1},\ldots,z_{n}$, and applying Cauchy’s integral theorem separately for $z_{1},\ldots,z_{n}$,

 $\int_{\mathbb{R}^{n}}e^{-\pi(u+iv)\cdot(u+iv)}P(u)du=\int_{\mathbb{R}^{n}}e^{-% \pi u\cdot u}P(u-iv)du.$

Define $Q:\mathbb{C}^{n}\to\mathbb{C}$ by

 $Q(z)=\int_{\mathbb{R}^{n}}e^{-\pi|u|^{2}}P(z+u)du,\qquad z\in\mathbb{C}^{n},$

and thus

 $\displaystyle Q(-iv)$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-\pi(u+iv)\cdot(u+iv)}P(u)du$ $\displaystyle=\int_{\mathbb{R}^{n}}e^{-\pi|u|^{2}+\pi|v|^{2}-2\pi iu\cdot v}P(% u)du$ $\displaystyle=e^{\pi|v|^{2}}\widehat{f}(v).$

On the other hand, for $t\in\mathbb{R}^{n}$, using spherical coordinates, using the mean value property for the harmonic function $P$, and then using spherical coordinates again,

 $\displaystyle Q(t)$ $\displaystyle=\int_{0}^{\infty}e^{-\pi r^{2}}\left(\int_{S^{n-1}}P(t+w)d\sigma% (w)\right)r^{n-1}dr$ $\displaystyle=\int_{0}^{\infty}e^{-\pi r^{2}}\sigma(S^{n-1})P(t)\cdot r^{n-1}dr$ $\displaystyle=P(t)\int_{0}^{\infty}e^{-\pi r^{2}}\left(\int_{S^{n-1}}d\sigma% \right)r^{n-1}dr$ $\displaystyle=P(t)\int_{\mathbb{R}^{n}}e^{-\pi|x|^{2}}dx$ $\displaystyle=P(t).$

Because $P\in\mathbb{C}[x_{1},\ldots,x_{n}]$, $P$ has an analytic continuation to $\mathbb{C}^{n}$, and then $P(z)=Q(z)$ for all $z\in\mathbb{C}^{n}$. Therefore

 $P(-iv)=Q(-iv)=e^{\pi|v|^{2}}\widehat{f}(v).$

But because $P$ is a homogeneous polynomial of degree $d$, $P(-iv)=(-i)^{d}P(v)$, so

 $(-i)^{d}P(v)=e^{\pi|v|^{2}}\widehat{f}(v),$

i.e.

 $\widehat{f}(v)=(-i)^{d}e^{-\pi|v|^{2}}P(v)=(-i)^{d}f(v),$

proving the claim. ∎

## 8 Representation theory

Let a complex Hilbert space $H$ with $\left\langle\cdot,\cdot\right\rangle$, let $\mathscr{U}(H)$ be the group of unitary operators $H\to H$. For a Lie group $G$, a unitary representation of $G$ on $H$ is a group homomorphism $\pi:G\to\mathscr{U}(H)$ such that for each $f\in H$ the map $\gamma\mapsto\pi(\gamma)(f)$ is continuous $G\to H$.

We have defined $\sigma$ as a unique $SO(n)$-invariant regular Borel measure on $S^{n-1}$. It does not follow a priori that $\sigma$ is $O(n)$-invariant. But in fact, using that $|\gamma x|=|x|$ for $x\in\mathbb{R}^{n}$ and that Lebesgue measure on $\mathbb{R}^{n}$ is $O(n)$-invariant, we check that $\sigma$ is $O(n)$-invariant: for $\gamma\in O(n)$ and a Borel set $E$ in $S^{n-1}$, $\sigma(\gamma E)=\sigma(E)$, i.e. ${\gamma}^{-1}_{*}\sigma=\sigma$.

For $\gamma\in O(n)$ and $f\in L^{2}(S^{n-1})$, define

 $\pi(\gamma)(f)=f\circ\gamma^{-1}.$

$\pi(\gamma)$ is linear. For $f,g\in L^{2}(\gamma)$,

 $\displaystyle\left\langle\pi(\gamma)(f),\pi(\gamma)(g)\right\rangle$ $\displaystyle=\int_{S^{n-1}}f\circ\gamma^{-1}\cdot\overline{g\circ\gamma^{-1}}d\sigma$ $\displaystyle=\int_{S^{n-1}}f\cdot\overline{g}d(\gamma^{-1})_{*}\sigma$ $\displaystyle=\int_{S^{n-1}}f\cdot\overline{g}d\sigma$ $\displaystyle=\left\langle f,g\right\rangle.$

For $f\in L^{2}(S^{n-1})$, let $g=f\circ\gamma$, for which

 $\pi(\gamma)(g)=g\circ\gamma^{-1}=f\circ\gamma\circ\gamma^{-1}=f,$

showing that $\pi(\gamma)$ is surjective. Hence $\pi(\gamma)\in\mathscr{U}(L^{2}(S^{n-1}))$.

For $\gamma_{1},\gamma\in O(n)$ and $f\in L^{2}(S^{n-1})$,

 $\displaystyle\pi(\gamma_{1}\gamma_{2})(f)$ $\displaystyle=f\circ(\gamma_{1}\gamma_{2})^{-1}$ $\displaystyle=f\circ(\gamma_{2}^{-1}\gamma_{1}^{-1})$ $\displaystyle=(f\circ\gamma_{2}^{-1})\circ\gamma_{1}^{-1}$ $\displaystyle=\pi(\gamma_{1})(\pi(\gamma_{2}^{-1}(f))),$

which means that $\pi(\gamma_{1}\gamma_{2})=\pi(\gamma_{1})\pi(\gamma_{2})$, namely $\pi:O(n)\to\mathscr{U}(L^{2}(S^{n-1}))$ is a group homomorphism.

For $\phi\in C(S^{n-1})$ and for $\gamma_{0},\gamma\in O(n)$,

 $\left\|\pi(\gamma)(\phi)-\pi(\gamma_{0})(\phi)\right\|_{L^{2}}^{2}=\left\|\pi(% \gamma_{0}^{-1}\gamma)(\phi)-\phi\right\|_{L^{2}}^{2}\leq\sigma(S^{n-1})\cdot% \left\|\pi(\gamma_{0}^{-1}\gamma)(\phi)-\phi\right\|_{C^{0}}^{2}.$

We take as given that $\left\|\pi(\gamma_{0}^{-1}\gamma)(\phi)-\phi\right\|_{C^{0}}\to 0$ as $\gamma\to\gamma_{0}$ in $O(n)$. Using that $C(S^{n-1})$ is dense in $L^{2}(S^{n-1})$, one then proves that for each $f\in L^{2}(S^{n-1})$, the map $\gamma\mapsto\pi(\gamma)(f)$ is continuous $O(n)\to L^{2}(S^{n-1})$.

###### Lemma 18.

$\pi$ is a unitary representation of the compact Lie group $O(n)$ on the complex Hilbert space $L^{2}(S^{n-1})$.

It is a fact that if $\gamma\in O(n)$ and $P\in\mathscr{P}_{d}$ then $\gamma\cdot P\in\mathscr{P}_{d}$. Furthermore, for $\phi\in C^{2}(S^{n-1})$, $\Delta(\gamma\cdot\phi)=\gamma\cdot(\Delta\phi)$, hence if $P\in\mathscr{A}_{d}$ then $\gamma\cdot P\in\mathscr{A}_{d}$. Then for $Y\in\mathscr{H}_{d}$, $\pi(\gamma)(Y)\in\mathscr{H}_{d}$. This means that each $\mathscr{H}_{d}$ is a $\pi$-invariant subspace.1818 18 cf. Feng Dai and Yuan Xu, Approximation Theory and Harmonic Analysis on Spheres and Balls, Chapter 1.