Harmonic polynomials and the spherical Laplacian

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

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

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

$$(G/H\times G/H)-\mathrm{\Delta }(G/H)=\{(xH,yH):xH\ne yH\}=\{(q(x),q(y)):x^{-1}y\notin H\}.$$

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

$$p^{-1}(U)=\{(x,y)\in G\times G:x^{-1}y\notin H\}.$$

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

$$f^{-1}(G-H)=\{(x,y)\in G\times G:x^{-1}y\notin H\}=p^{-1}(U),$$

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 $\mathrm{\Delta }(G\times H)$, 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

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

If $xH=yH$, then $(gx)H=(gy)H$, so indeed this makes sense.

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 $xH,yH\in G/H$,

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

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 $k_{\mathrm{0}}$, which is a closed subgroup of $G$. A theorem of Weil states that $\varphi :G/H\to K$ defined by

$$\varphi (xH)=\alpha (x,k_0),xH\in G/H$$

is a homeomorphism that satisfies

$$\varphi (\beta (g,xH))=\alpha (g,\varphi (xH)),g\in G,xH\in G/H,$$

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. This measure is right-invariant, and satisfies

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

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 $G/H$. Then define $m_K$ on the Borel $\sigma$-algebra of $K$ by

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

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

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

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 $\varphi :SO(n)/SO(n-1)\to S^{n-1}$ by

$$\varphi (xSO(n-1))=\alpha (x,e_n)=x{e}_n,xSO(n-1)\in SO(n)/SO(n-1).$$

Then for $m$ the Borel probability measure on $SO(n)$, the unique $SO(n)$-invariant regular Borel probability measure on $S^{n-1}$ is

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

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

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

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

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

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

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

We call $\sigma$ the spherical surface measure.

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

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

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

$${\int }_{{ℝ}^n}{\gamma }_n(x)𝑑x=1,$$

and define $I:C(S^{n-1})\to ℂ$ by

$$I(f)={\int }_{{ℝ}^n}f(x/|x|){\gamma }_n(x)𝑑x,f\in C(S^{n-1}),$$

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

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

Because $I(f)={\int }_{{ℝ}^n}{\gamma }_n(x)𝑑x=1$, $\mu$ 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 $x\in {ℝ}^n$ and because Lebesgue measure on ${ℝ}^n$ is invariant under $SO(n)$, by the change of variables theorem we have

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

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

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

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

3 L²(S^n-1) and the spherical Laplacian

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

$$⟨f,g⟩={\int }_{S^{n-1}}f\overline{g}𝑑\sigma ,$$

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

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

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

$⟨\gamma \cdot f,\gamma \cdot g⟩$ $={\displaystyle {\int }_{S^{n-1}}}f({\gamma }^{-1}x)\overline{g}({\gamma }^{-1}x)𝑑\sigma (x)$

$={\displaystyle {\int }_{S^{n-1}}}f(x)\overline{g}(x)d({({\gamma }^{-1})}_{*}\sigma )(x)$

$={\displaystyle {\int }_{S^{n-1}}}f(x)\overline{g}(x)𝑑\sigma (x)$

$=⟨f,g⟩.$

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

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

We take $f$ to belong to $C^k(S^{n-1})$ when $F\in C^k({ℝ}^n-\{0\})$, $0\le k\le \mathrm{\infty }$, and we define ${\mathrm{\Delta }}_{S^{n-1}}f$ be the restriction of $\mathrm{\Delta }F$ to $S^{n-1}$. We call ${\mathrm{\Delta }}_{S^{n-1}}$ the spherical Laplacian.

Theorem 3.

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

$${\mathrm{\Delta }}_{S^{n-1}}f=-s(n+s-2)f.$$

Proof.

Let $H(x)=F(x/|x|)={|x|}^{-s}F(x)$ and let $r(x)=|x|={(x_1^2+\mathrm{\cdots }+x_n^2)}^{1/2}$. We calculate

$\mathrm{\Delta }H$ $={\displaystyle \sum _{i=1}^n}{\partial }_i^2({(r^2)}^{-\frac{s}{2}}F)$

$={\displaystyle \sum _{i=1}^n}{\partial }_i\left(-s{x}_i{(r^2)}^{-\frac{s}{2}-1}F+{(r^2)}^{-\frac{s}{2}}{\partial }_{i}F\right)$

$={\displaystyle \sum _{i=1}^n}-s{(r^2)}^{-\frac{s}{2}-1}F-s{x}_i(2x_i)\left(-{\displaystyle \frac{s}{2}}-1\right){(r^2)}^{-\frac{s}{2}-2}F-s{x}_i{(r^2)}^{\frac{s}{2}-1}{\partial }_{i}F$

$-s{x}_i{(r^2)}^{-\frac{s}{2}-1}{\partial }_{i}F+{(r^2)}^{-\frac{s}{2}}{\partial }_i^{2}F$

$=-ns{(r^2)}^{-\frac{s}{2}-1}F+{(r^2)}^{-\frac{s}{2}-2}{\displaystyle \sum _{i=1}^n}\left(-s(-s-2)x_i^{2}F-s{x}_i{r}^2{\partial }_{i}F-s{x}_i{r}^2{\partial }_{i}F\right)$

$+{(r^2)}^{-\frac{s}{2}}\mathrm{\Delta }F$

$=-ns{(r^2)}^{-\frac{s}{2}-1}F+{(r^2)}^{-\frac{s}{2}-2}{\displaystyle \sum _{i=1}^n}(s^2{x}_i^{2}F+2s{x}_i^{2}F-2s{x}_i{r}^2{\partial }_{i}F).$

Euler’s identity for positive-homogeneous functions states that if $G:{ℝ}^n-\{0\}\to ℂ$ is positive-homogeneous of degree $s$ then $x\cdot (\nabla G)(x)=sG(x)$ for all $x$. Therefore

$\mathrm{\Delta }H$ $=-ns{(r^2)}^{-\frac{s}{2}-1}F+{(r^2)}^{-\frac{s}{2}-2}(s^2+2s){|x|}^{2}F-{(r^2)}^{-\frac{s}{2}-2}\cdot 2s{r}^2\cdot sF$

$=-ns{(r^2)}^{-\frac{s}{2}-1}F+{(r^2)}^{-\frac{s}{2}-1}(s^2+2s)F-{(r^2)}^{-\frac{s}{2}-1}\cdot 2s^{2}F$

$=-s{r}^{-s-2}(n+s-2)F.$

For $x\in {ℝ}^n-\{0\}$,

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

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

$$({\mathrm{\Delta }}_{S^{n-1}}f)(x)=-s{r}^{-s-2}(n+s-2)F(x)=-s(n+s-2)f(x).$$

∎

Theorem 4.

If $f\in C^2(S^{n-1})$ satisfies ${\mathrm{\Delta }}_{S^{n-1}}f=\lambda f$, then $\lambda \le 0$.

If $g\in C^2(S^{n-1})$ satisfies ${\mathrm{\Delta }}_{S^{n-1}}g=\mu g$ with $\lambda \ne \mu$, then $⟨f,g⟩=0$.

Proof.

Say $\lambda \ne 0$. Then

$⟨f,f⟩$ $={\displaystyle \frac{1}{\lambda }}⟨{\mathrm{\Delta }}_{S^{n-1}}f,f⟩$

$={\displaystyle \frac{1}{\lambda }}{\displaystyle {\int }_{S^{n-1}}}({\mathrm{\Delta }}_{S^{n-1}}f)\overline{f}𝑑\sigma$

$={\displaystyle \frac{1}{\lambda }}{\displaystyle {\int }_{S^{n-1}}}f{\mathrm{\Delta }}_{S^{n-1}}\overline{f}𝑑\sigma$

$={\displaystyle \frac{1}{\lambda }}{\displaystyle {\int }_{S^{n-1}}}f\overline{{\mathrm{\Delta }}_{S^{n-1}}f}𝑑\sigma$

$={\displaystyle \frac{1}{\lambda }}{\displaystyle {\int }_{S^{n-1}}}f\overline{\lambda f}𝑑\sigma$

$={\displaystyle \frac{\overline{\lambda }}{\lambda }}⟨f,f⟩.$

Because $\lambda \ne 0$, it is not the case that $f=0$, hence $⟨f,f⟩>0$. Hence $\frac{\overline{\lambda }}{\lambda }=1$, which means that $\lambda \in ℝ$. Furthermore,

which implies that . ∎

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

Theorem 5.

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

$${\mathrm{\Delta }}_{S^{n-1}}(\gamma \cdot f)=\gamma \cdot ({\mathrm{\Delta }}_{S^{n-1}}f).$$

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 $x\in {ℝ}^n-\{0\}$,

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

so $\gamma \cdot F=G$. It is a fact that $\mathrm{\Delta }(\gamma \cdot F)=\gamma \cdot (\mathrm{\Delta }F)$. Thus for $x\in S^{n-1}$,

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

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

We now prove that ${\mathrm{\Delta }}_{S^{n-1}}$ is symmetric and negative-definite.

Theorem 6.

For $f,g\in C^2(S^{n-1})$,

$${\int }_{S^{n-1}}({\mathrm{\Delta }}_{S^{n-1}}f)\cdot g𝑑\sigma ={\int }_{S^{n-1}}f\cdot {\mathrm{\Delta }}_{S^{n-1}}g𝑑\sigma .$$

${\mathrm{\Delta }}_{S^{n-1}}$ is negative-definite:

$${\int }_{S^{n-1}}({\mathrm{\Delta }}_{S^{n-1}}f)\cdot \overline{f}\le 0,$$

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 $\mathrm{\Delta }F$ is positive-homogeneous of degree $s-2$. Let $F(x)=f(x/|x|)$ and $G(x)=g(x/|x|)$, with which

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

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

${\displaystyle {\int }_{S^{n-1}}}({\mathrm{\Delta }}_{S^{n-1}}f)(x)\cdot g(x)𝑑\sigma (x)$ $={\displaystyle {\int }_{S^{n-1}}}(\mathrm{\Delta }F)(x)\cdot G(x)𝑑\sigma (x)$

$=A_{n-1}{\displaystyle {\int }_{{ℝ}^n}}(\mathrm{\Delta }F)(x/|x|)\cdot G(x/|x|){\gamma }_n(x)𝑑x$

$=A_{n-1}{\displaystyle {\int }_{{ℝ}^n}}{|x|}^2(\mathrm{\Delta }F)(x)\cdot G(x){\gamma }_n(x)𝑑x.$

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

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

$${\int }_{S^{n-1}}({\mathrm{\Delta }}_{S^{n-1}}f)\cdot g𝑑\sigma ={\int }_{S^{n-1}}f\cdot {\mathrm{\Delta }}_{S^{n-1}}g𝑑\sigma .$$

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

$${\int }_{S^{n-1}}({\mathrm{\Delta }}_{S^{n-1}}f)\cdot \overline{f}𝑑\sigma =-A_{n-1}{\int }_{{ℝ}^n}\sum _{i=1}^n{|x|}^2{\gamma }_n{|{\partial }_{i}F|}^{2}dx,$$

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

4 Homogeneous polynomials

For $P(x_1,\mathrm{\dots },x_n)=\sum a_{\alpha }{x}^{\alpha }\in ℂ[x_1,\mathrm{\dots },x_n]$ write

$$P(\partial )=\sum a_{\alpha }{\partial }^{\alpha },\overline{P}(x_1,\mathrm{\dots },x_n)=\sum \overline{a_{\alpha }}{x}^{\alpha },\overline{P}(\partial )=\sum \overline{a_{\alpha }}{\partial }^{\alpha }.$$

For $P,Q\in ℂ[x_1,\mathrm{\dots },x_n]$, define

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

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

$$(P,Q)={\left(\sum _{\beta }\overline{b_{\beta }}{\partial }^{\beta }\sum _{\alpha }{a}_{\alpha }{x}^{\alpha }\right)|}_{x=0}=\sum _{\beta }\overline{b_{\beta }}{a}_{\beta }\cdot \beta !.$$ (2)

Lemma 7.

$(\cdot ,\cdot )$ is a positive-definite Hermitian form on $ℂ[x_1,\mathrm{\dots },x_n]$.

Proof.

It is apparent that $(\cdot ,\cdot )$ is $ℂ$-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 $P\in ℂ[x_1,\mathrm{\dots },x_n]$,

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

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 $Q={\sum }_{\beta }{b}_{\beta }{x}^{\beta }$,

$$(\mathrm{\Delta }P)(x)=\sum _{\alpha }{a}_{\alpha }\sum _{i=1}^n{\partial }_i^2{x}^{\alpha }=\sum _{\alpha }{a}_{\alpha }\sum _{i=1}^n\frac{\alpha !}{(\alpha -2e_i)!}{x}^{\alpha -2e_i},$$

and we calculate

$$(\mathrm{\Delta }P,Q)=\sum _{\beta }\overline{b_{\beta }}\sum _{i=1}^n{a}_{\beta +2e_i}(\beta +2e_i)!.$$

On the other hand,

$$r^{2}Q(x_1,\mathrm{\dots },x_n)=\sum _{\beta }{b}_{\beta }{x}^{\beta }\sum _{i=1}^n{x}_i^2=\sum _{\beta }{b}_{\beta }\sum _{i=1}^n{x}^{\beta +2e_i},$$

and we calculate

$$(P,r^{2}Q)=\sum _{\beta }\overline{b_{\beta }}\sum _{i=1}^n{a}_{\beta +2e_i}.$$

Lemma 8.

For $P,Q\in ℂ[x_1,\mathrm{\dots },x_n]$,

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

Let ${𝒫}_d$ be the set of homogeneous polynomials of degree $d$ in $ℂ[x_1,\mathrm{\dots },x_n]$, i.e. those $P(x_1,\mathrm{\dots },x_n)\in ℂ[x_1,\mathrm{\dots },x_n]$ of the form

$$P(x_1,\mathrm{\dots },x_n)=\sum _{|\alpha |=d}{a}_{\alpha }{x}^{\alpha }.$$

We include the polynomial $P=0$, and ${𝒫}_d$ is a complex vector space. We calculate

$${dim}_{ℂ}{𝒫}_d=\{\alpha :|\alpha |=d\}=\left(\genfrac{}{}{0pt}{}{n+d-1}{d}\right).$$ (3)

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

We prove that $\mathrm{\Delta }:{𝒫}_d\to {𝒫}_{d-2}$ is surjective.

Theorem 9.

The map $\mathrm{\Delta }:{𝒫}_d\to {𝒫}_{d-2}$ is surjective. Its kernel is ${𝒜}_d$, and

$${𝒜}_d^{⟂}=r^2{𝒫}_{d-2}.$$

Proof.

By Lemma 8,

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

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 ${𝒫}_{d-2}$ is a finite-dimensional Hilbert space and the orthogonal complement of the image $\mathrm{\Delta }{𝒫}_d$ is equal to $\{0\}$, it follows that $\mathrm{\Delta }{𝒫}_d={𝒫}_{d-2}$.

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

The above theorem tells us that

$${𝒫}_d={𝒜}_d\oplus {𝒜}_d^{⟂}={𝒜}_d\oplus r^2{𝒫}_{d-2}.$$

Then,

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

and by induction,

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

For $P\in {𝒫}_d$, there are unique $F_0\in {𝒜}_d$, $F_2\in {𝒜}_{d-2}$, $F_4\in {𝒜}_{d-4}$, etc., such that

$$P=F_0+r^2{F}_2+r^4{F}_4+\mathrm{\cdots }.$$

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 $x\in S^{n-1}$,

$$p=f_0+f_2+f_4+\mathrm{\cdots }.$$

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

Theorem 11.

$${dim}_{ℂ}{𝒜}_d=\left(\genfrac{}{}{0pt}{}{n+d-1}{d}\right)-\left(\genfrac{}{}{0pt}{}{n+d-3}{d-2}\right)=\left(\genfrac{}{}{0pt}{}{n+d-2}{n-2}\right)+\left(\genfrac{}{}{0pt}{}{n+d-3}{n-2}\right).$$

With $n$ fixed, using the asymptotic formula

$$\left(\genfrac{}{}{0pt}{}{z+k}{k}\right)=\frac{k^z}{\mathrm{\Gamma }(z+1)}\left(1+\frac{z(z+1)}{2k}+O(k^{-2})\right),k\to \mathrm{\infty },$$

we get from the above lemma

$${dim}_{ℂ}{𝒜}_d\sim \frac{2}{(n-2)!}{d}^{n-2}.$$

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

Lemma 12.

For $Y\in {\mathscr{H}}_d$,

$${\mathrm{\Delta }}_{S^{n-1}}Y={\lambda }_{d}Y$$

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 then ; and ${\lambda }_d\to -\mathrm{\infty }$ as $d\to \mathrm{\infty }$.

5 The Hilbert space L²(S^n-1)

We prove that when $d_1\ne d_2$, the subspaces ${\mathscr{H}}_{d_1}$ and ${\mathscr{H}}_{d_2}$ of $L^2(S^{n-1})$ are mutually orthogonal.

Theorem 13.

For $d_1\ne d_2$, for $Y_1\in {\mathscr{H}}_{d_1}$ and for $Y_2\in {\mathscr{H}}_{d_2}$,

$$⟨Y_1,Y_2⟩=0.$$

Proof.

From Lemma 12,

$${\mathrm{\Delta }}_{S^{n-1}}{Y}_1={\lambda }_{d_1}{Y}_1,{\mathrm{\Delta }}_{S^{n-1}}{Y}_2={\lambda }_{d_2}{Y}_2.$$

where ${\lambda }_d=-d(d+n-2)$. Because $d_1\ne d_2$ it follows that ${\lambda }_{d_1}\ne {\lambda }_{d_2}$ and then by Theorem 4, $⟨Y_1,Y_2⟩=0$. ∎

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

$${\parallel \varphi \parallel }_{C^0}=\underset{x\in S^{n-1}}{sup}|\varphi (x)|.$$

Let $A$ be the set of restrictions of all $P\in ℂ[x_1,\mathrm{\dots },x_n]$ 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 $P,Q\in ℂ[x_1,\mathrm{\dots },x_n]$ 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 $ℂ[x_1,\mathrm{\dots },x_n]$ 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 $\overline{P}\in ℂ[x_1,\mathrm{\dots },x_n]$ to $S^{n-1}$, showing that $A$ is self-adjoint. For distinct $u=(u_1,\mathrm{\dots },u_n),v=(v_1,\mathrm{\dots },v_n)$ in $S^{n-1}$, say with $u_k\ne v_k$, let $P(x_1,\mathrm{\dots },x_n)=x_k$ 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 $P(x_1,\mathrm{\dots },x_n)=1$ 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 theorem that $A$ is dense in the Banach space $C(S^{n-1})$: for any $\varphi \in C(S^{n-1})$ and for $ϵ>0$, there is some $p\in A$ such that ${\parallel p-\varphi \parallel }_{C^0}\le ϵ$.

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

$$⟨f,g⟩={\int }_{S^{n-1}}f\cdot \overline{g}𝑑\sigma .$$

For $f\in L^2(S^{n-1})$ and for $ϵ>0$, there is some $\varphi \in C(S^{n-1})$ with ${\parallel \varphi -f\parallel }_{L^2}\le ϵ$, and there is some $p\in A$ with ${\parallel p-\varphi \parallel }_{C^0}\le ϵ$. But for $\psi \in C(S^{n-1})$,

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

Then

${\parallel p-f\parallel }_{L^2}$ $\le {\parallel p-\varphi \parallel }_{L^2}+{\parallel \varphi -f\parallel }_{L^2}$

$\le {\parallel p-\varphi \parallel }_{C^0}\cdot \sqrt{\sigma (S^{n-1})}+ϵ$

$\le ϵ\cdot \sqrt{\sigma (S^{n-1})}+ϵ.$

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

An element of $ℂ[x_1,\mathrm{\dots },x_n]$ 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 $Y_1\in {\mathscr{H}}_{d_1},\mathrm{\dots },Y_m\in {\mathscr{H}}_{d_m}$ with $p=Y_1+\mathrm{\cdots }+Y_m$. 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 $d_1\ne d_2$, the subspaces ${\mathscr{H}}_{d_1}$ and ${\mathscr{H}}_{d_2}$ are mutually orthogonal. Putting the above together gives the following.

Theorem 14.

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

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

$${\parallel \varphi \parallel }_{L^2}\le \sqrt{\sigma (S^{n-1})}\cdot {\parallel \varphi \parallel }_{C^0}.$$

Similar to Nikolsky’s inequality for the Fourier transform, for $Y\in {\mathscr{H}}_d$, the norm ${\parallel Y\parallel }_{C^0}$ is upper bounded by a multiple of the norm ${\parallel Y\parallel }_{L^2}$ that depends on $d$.

Theorem 15.

For $Y\in {\mathscr{H}}_d$,

$${\parallel Y\parallel }_{C^0}\le \sqrt{\frac{{dim}_{ℂ}{\mathscr{H}}_d}{\sigma (S^{n-1})}}\cdot {\parallel Y\parallel }_{L^2}.$$

6 Sobolev embedding

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

$$f=\sum _{d\ge 0}{P}_{d}f$$

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

We prove the Sobolev embedding for $S^{n-1}$.

Theorem 16 (Sobolev embedding).

For $f\in L^2(S^{n-1})$, if $s>n-1$ and

then there is some $\varphi \in C(S^{n-1})$ such that $\varphi ={\sum }_{d\ge 0}{P}_{j}f$ in $C(S^{n-1})$, and $f=\varphi$ almost everywhere.

Proof.

By Theorem 11 there is some $C_n$ such that

$${dim}_{ℂ}{\mathscr{H}}_d\le C_n{(1+d)}^{n-2}.$$

Then by Theorem 15 and the Cauchy-Schwarz inequality,

${\displaystyle \sum _{d\ge 0}}{\parallel P_{d}f\parallel }_{C^0}$ $\le {\displaystyle \sum _{d\ge 0}}\sqrt{{\displaystyle \frac{{dim}_{ℂ}{\mathscr{H}}_d}{\sigma (S^{n-1})}}}\cdot {\parallel P_{d}f\parallel }_{L^2}$

$\le \sqrt{{\displaystyle \frac{C_n}{\sigma (S^{n-1})}}}{\displaystyle \sum _{d\ge 0}}{(1+d)}^{\frac{n-2}{2}}\cdot {\parallel P_{d}f\parallel }_{L^2}$

$=\sqrt{{\displaystyle \frac{C_n}{\sigma (S^{n-1})}}}{\displaystyle \sum _{d\ge 0}}{(1+d)}^{\frac{s}{2}}{\parallel P_{d}f\parallel }_{L^2}\cdot {(1+d)}^{-\frac{s-n+2}{2}}$

$\le \sqrt{{\displaystyle \frac{C_n}{\sigma (S^{n-1})}}}\left({\displaystyle \sum _{d\ge 0}}{(1+d)}^s{\parallel P_{d}f\parallel }_{L^2}^2\right)\left({\displaystyle \sum _{d\ge 0}}{(1+d)}^{-(s-n+2)}\right)$

$=\sqrt{{\displaystyle \frac{C_n}{\sigma (S^{n-1})}}}\cdot \zeta (s-n+2)\cdot {\displaystyle \sum _{d\ge 0}}{(1+d)}^s{\parallel P_d\parallel }_{L^2}^2$

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 $\varphi \in C(S^{n-1})$. Because

$${\parallel \sum _{d=0}^m{P}_{d}f-\varphi \parallel }_{L^2}\le \sqrt{\sigma (S^{n-1})}\cdot {\parallel \sum _{d=0}^m{P}_{d}f-\varphi \parallel }_{C^0},$$

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

7 Hecke’s identity

Hecke’s identity tells us the Fourier transform of a product of an element of ${𝒜}_d$ and a Gaussian.

Theorem 17 (Hecke’s identity).

For $f(u)=e^{-\pi {|u|}^2}P(u)$ with $P\in {𝒜}_d$,

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

Proof.

Let $v\in {ℝ}^n$. The map $z\mapsto e^{-\pi z\cdot z}P(z-iv)$ is a holomorphic separately in $z_1,\mathrm{\dots },z_n$, and applying Cauchy’s integral theorem separately for $z_1,\mathrm{\dots },z_n$,

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

Define $Q:{ℂ}^n\to ℂ$ by

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

and thus

$Q(-iv)$ $={\displaystyle {\int }_{{ℝ}^n}}{e}^{-\pi (u+iv)\cdot (u+iv)}P(u)𝑑u$

$={\displaystyle {\int }_{{ℝ}^n}}{e}^{-\pi {|u|}^2+\pi {|v|}^2-2\pi iu\cdot v}P(u)𝑑u$

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

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

$Q(t)$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\pi r^2}\left({\displaystyle {\int }_{S^{n-1}}}P(t+w)𝑑\sigma (w)\right)r^{n-1}𝑑r$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\pi r^2}\sigma (S^{n-1})P(t)\cdot r^{n-1}𝑑r$

$=P(t){\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-\pi r^2}\left({\displaystyle {\int }_{S^{n-1}}}𝑑\sigma \right)r^{n-1}𝑑r$

$=P(t){\displaystyle {\int }_{{ℝ}^n}}{e}^{-\pi {|x|}^2}𝑑x$

$=P(t).$

Because $P\in ℂ[x_1,\mathrm{\dots },x_n]$, $P$ has an analytic continuation to ${ℂ}^n$, and then $P(z)=Q(z)$ for all $z\in {ℂ}^n$. 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 $⟨\cdot ,\cdot ⟩$, let $𝒰(H)$ 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 $\pi :G\to 𝒰(H)$ 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 $x\in {ℝ}^n$ and that Lebesgue measure on ${ℝ}^n$ 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 $f,g\in L^2(\gamma )$,

$⟨\pi (\gamma )(f),\pi (\gamma )(g)⟩$ $={\displaystyle {\int }_{S^{n-1}}}f\circ {\gamma }^{-1}\cdot \overline{g\circ {\gamma }^{-1}}𝑑\sigma$

$={\displaystyle {\int }_{S^{n-1}}}f\cdot \overline{g}d{({\gamma }^{-1})}_{*}\sigma$

$={\displaystyle {\int }_{S^{n-1}}}f\cdot \overline{g}𝑑\sigma$

$=⟨f,g⟩.$

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 $\pi (\gamma )\in 𝒰(L^2(S^{n-1}))$.

For ${\gamma }_1,\gamma \in O(n)$ and $f\in L^2(S^{n-1})$,

$\pi ({\gamma }_1{\gamma }_2)(f)$ $=f\circ {({\gamma }_1{\gamma }_2)}^{-1}$

$=f\circ ({\gamma }_2^{-1}{\gamma }_1^{-1})$

$=(f\circ {\gamma }_2^{-1})\circ {\gamma }_1^{-1}$

$=\pi ({\gamma }_1)(\pi ({\gamma }_2^{-1}(f))),$

which means that $\pi ({\gamma }_1{\gamma }_2)=\pi ({\gamma }_1)\pi ({\gamma }_2)$, namely $\pi :O(n)\to 𝒰(L^2(S^{n-1}))$ is a group homomorphism.

For $\varphi \in C(S^{n-1})$ and for ${\gamma }_0,\gamma \in O(n)$,

$${\parallel \pi (\gamma )(\varphi )-\pi ({\gamma }_0)(\varphi )\parallel }_{L^2}^2={\parallel \pi ({\gamma }_0^{-1}\gamma )(\varphi )-\varphi \parallel }_{L^2}^2\le \sigma (S^{n-1})\cdot {\parallel \pi ({\gamma }_0^{-1}\gamma )(\varphi )-\varphi \parallel }_{C^0}^2.$$

We take as given that ${\parallel \pi ({\gamma }_0^{-1}\gamma )(\varphi )-\varphi \parallel }_{C^0}\to 0$ 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 $P\in {𝒫}_d$ then $\gamma \cdot P\in {𝒫}_d$. Furthermore, for $\varphi \in C^2(S^{n-1})$, $\mathrm{\Delta }(\gamma \cdot \varphi )=\gamma \cdot (\mathrm{\Delta }\varphi )$, hence if $P\in {𝒜}_d$ then $\gamma \cdot P\in {𝒜}_d$. 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.