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The Fischer inner product on polynomials is $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_i$ is adjoint to differentiation by $x_i$. Therefore multiplication by $|x{|}^2$ is adjoint to the ordinary Laplacian, yielding the harmonic decomposition of homogeneous polynomials by finite-dimensional orthogonality.

We use $\Delta =\mathrm{div}\nabla$, whose spectrum on a compact manifold is nonpositive; write $P=-\Delta$ for the positive Laplace-Beltrami operator. This convention will also make the heat operator in Question 3 equal to ${\partial }_t-\Delta$.

Let $e_1,\dots ,e_d$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. Since a geodesic has vanishing acceleration for the Levi-Civita connection, differentiation twice along it gives the Riemannian Hessian:Taking its metric trace yields the geodesic trace formula for the Laplace-Beltrami operatorThe positive-sign convention $P$ puts a minus sign in front of this sum.

For the unit sphere, take $p\in S^d$ and tangent unit vectors $e_i$ orthogonal to $p$. Its great circle geodesics are ${\gamma }_i(t)=p\cos t+e_i\sin t$. For a smooth ambient function $F$ near the sphere, the ordinary chain rule givesThe ambient orthonormal basis is $e_1,\dots ,e_d,p$. Thus, writing $F_r(p)={\partial }_{r}F(rp){|}_{r=1}$ and similarly for $F_{rr}$,This proves the ambient restriction formula for the spherical Laplacian and shows why ambient derivatives normal to the sphere must be subtracted. The equivalent formula in spherical coordinates is

Let ${\mathcal{P}}_{\ell }$ be the vector space of homogeneous polynomials of degree $\ell$ in $d+1$ variables, and ${\mathcal{H}}_{\ell }=\ker \tilde{\Delta }\cap {\mathcal{P}}_{\ell }$ its harmonic polynomials. If $H\in {\mathcal{H}}_{\ell }$, homogeneous polynomial scaling gives $H(r\omega )=r^{\ell }H(\omega )$. Substitution into the spherical coordinates formula givesThe restrictions are the degree-$\ell$ spherical harmonics. A homogeneous polynomial vanishing on the sphere vanishes on every nonzero ray and hence identically, so restriction is injective on ${\mathcal{H}}_{\ell }$.

We prove the harmonic decomposition of homogeneous polynomials that both counts these eigenfunctions and establishes completeness. For an ambient harmonic homogeneous polynomial $H_m$ of degree $m$ and $j\ge 1$, the spherical coordinates formula givesFor $d\ge 1$ its coefficient is nonzero. Inductively assume homogeneous polynomials of degree $\ell -2$ have been decomposed into sums of $r^{2j}{H}_m$. Apply this decomposition to $\tilde{\Delta }Q$ for $Q\in {\mathcal{P}}_{\ell }$, and lift each summand by multiplying by $r^2$ and dividing by the displayed nonzero coefficient. The resulting $R\in r^2{\mathcal{P}}_{\ell -2}$ satisfies $\tilde{\Delta }R=\tilde{\Delta }Q$, so $Q-R\in {\mathcal{H}}_{\ell }$. The same coefficient formula shows that no nonzero element of $r^2{\mathcal{P}}_{\ell -2}$ can be harmonic. Starting at degrees zero and one therefore proves the direct decompositionSince $\dim {\mathcal{P}}_{\ell }=\left(\genfrac{}{}{0ex}{}{\ell +d}{d}\right)$, the multiplicity iswhere the second term is zero for $\ell <2$.

On $r=1$, iterating the decomposition writes every polynomial restriction as a finite sum of spherical harmonics. The Stone-Weierstrass theorem makes these polynomial restrictions dense in $C(S^d)$, since coordinate functions separate points and constants are included; they are consequently dense in $L^2(S^d)$. Integration by parts makes $P$ symmetric on smooth functions and gives orthogonality between distinct eigenvalues. Normalizing within each finite-dimensional eigenspace gives a complete orthonormal eigenbasis $Y_{\ell a}$. To check the full operator spectrum, let $c_{\ell a}=⟨f,Y_{\ell a}⟩$. For smooth $f$, integration by parts gives the coefficients of $Pf$ as ${\lambda }_{\ell }{c}_{\ell a}$. Therefore finite harmonic partial sums approximate both $f$ and $Pf$ in $L^2$, hence approximate in the graph norm. Conversely every coefficient sequence with ${\sum }_{\ell ,a}{\lambda }_{\ell }^2|c_{\ell a}{|}^2<\infty$ is the graph norm limit of its finite harmonic sums. Thus the closure of $P$ has exactly this operator domain and is the real diagonal self-adjoint operator with entries ${\lambda }_{\ell }=\ell (\ell +d-1)$. If $z$ is not one of these entries, the inverse has entries $({\lambda }_{\ell }-z{)}^{-1}$ and is bounded because ${\lambda }_{\ell }\to \infty$. This proves that there are no missing spectral values. The spectrum of the Laplacian on a sphere iswith the above multiplicities; for $\Delta$, negate every eigenvalue. For example $S^1$ has eigenvalues ${\ell }^2$, with multiplicities one at zero and two for $\ell \ge 1$, and $S^2$ has multiplicity $2\ell +1$. The zero-dimensional sphere consists of two isolated points and has only the zero eigenvalue, with multiplicity two.

Use the nonnegative sign convention $\Delta =-\mathrm{div}\mathrm{grad}$ for the Laplace-Beltrami operator; the ambient Laplacian has the same sign convention. Changing both signs changes the signs of the eigenvalues below. Let $e_1,\dots ,e_n$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. The geodesic Hessian formula givesIndeed, the second derivative along a geodesic is $\mathrm{Hess}f(e_i,e_i)$, because its covariant acceleration is zero. Taking the trace of the Hessian matrix in this orthonormal basis gives the displayed Laplace-Beltrami operator.

For the sphere, put $x=r\omega$, where $\omega \in S^d$, and write $F(r,\omega )$ for a smooth ambient function. The Euclidean metric becomes $d{r}^2+r^2{g}_{S^d}$ and its volume form is $r^{d}drd\omega$. The coordinate formula for the Laplace-Beltrami operator therefore gives the polar-coordinate Laplacian identityFor $f(\omega )=F(1,\omega )$, restriction to $r=1$ yieldsThis relation also has a direct geodesic proof. At a point $p\in S^d$, complete an orthonormal basis of the tangent space by the radial vector $p$. Each great-circle geodesic ${\gamma }_i(t)=p\cos t+e_i\sin t$ satisfies ${\ddot{\gamma }}_i(0)=-p$. The chain rule gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-{\partial }_{r}F$. Summing and comparing with the ambient trace gives precisely the radial correction above.

Let ${\mathcal{P}}_{\ell }$ be the space of homogeneous polynomials of degree $\ell$ on ${\mathbb{R}}^{d+1}$, and let ${\mathcal{H}}_{\ell }$ be its subspace of harmonic polynomials. For $H\in {\mathcal{H}}_{\ell }$, homogeneity gives $H(r\omega )=r^{\ell }H(\omega )$. Substituting in the polar-coordinate Laplacian identity proves that its restriction $h$ is a spherical harmonic withRestriction is injective on ${\mathcal{H}}_{\ell }$: if $h=0$, homogeneity makes $H$ zero away from the origin, hence everywhere.

To count and exhaust these eigenfunctions, use the harmonic decomposition of homogeneous polynomialsHere is an algebraic proof rather than an assumption about the spectrum. On polynomials put the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_i$ is adjoint to ${\partial }_i$, so multiplication by $r^2$ is adjoint to $L={\sum }_i{\partial }_i^2$. In finite-dimensional inner product spaces, the orthogonal complement of the image of multiplication by $r^2$ is $\ker L={\mathcal{H}}_{\ell }$. This proves the decomposition. Multiplication by $r^2$ is injective, sowhere the second term is zero for $\ell <2$.

Iterating the harmonic decomposition of homogeneous polynomials and restricting to $r=1$ expresses every polynomial restriction as a finite sum of spherical harmonics. Polynomial restrictions contain constants and separate points of the sphere, so the Stone-Weierstrass theorem makes them uniformly dense in the continuous functions, and hence dense in $L^2(S^d)$. The Laplace-Beltrami operator is self-adjoint, and eigenfunctions with different eigenvalues are orthogonal. If a smooth eigenfunction had an eigenvalue not in the list, it would be orthogonal to a dense subspace and would vanish. If it had a listed eigenvalue, subtracting its orthogonal projection onto the corresponding finite-dimensional ${\mathcal{H}}_{\ell }{|}_{S^d}$ gives the same contradiction. Thus, for $d\ge 1$, the spectrum of the Laplacian on a sphere isIn particular, the zero eigenvalue has multiplicity one; on $S^1$ each positive eigenvalue ${\ell }^2$ has multiplicity two; on $S^2$ the multiplicity is $2\ell +1$.

For the unit $S^d$ and the nonnegative Laplace-Beltrami operator, the eigenvalues are $\ell (\ell +d-1)$, with multiplicities $\left(\genfrac{}{}{0ex}{}{\ell +d}{d}\right)-\left(\genfrac{}{}{0ex}{}{\ell +d-2}{d}\right)$ for $\ell \ge 0$. The eigenfunctions are restrictions of degree-$\ell$ harmonic polynomials. The harmonic decomposition of homogeneous polynomials and the Stone-Weierstrass theorem prove completeness.