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Let $F$ be a smooth function near the unit sphere $S^d$, and write ${\Delta }_{\mathrm{Euc}}F={\sum }_{a=1}^{d+1}{\partial }_a^{2}F$ for the ordinary ambient second-derivative trace. For the positive Laplace-Beltrami operator $P_{S^d}=-\mathrm{div}\nabla$,evaluated at radius one. The negative of this formula is the restriction identity for the nonpositive Laplace-Beltrami operator convention.

Choose an orthonormal basis $e_i$ of $T_p{S}^d$. The great circles ${\gamma }_i(t)=p\cos t+e_i\sin t$ giveThe negative sum is $P_{S^d}(F|)$; supplementing the tangent basis by the radial vector $p$ gives the displayed formula. If $F$ is a degree-$\ell$ homogeneous polynomial that is a harmonic polynomial, then $F_r=\ell F$, $F_{rr}=\ell (\ell -1)F$ and ${\Delta }_{\mathrm{Euc}}F=0$. Its restriction therefore has positive-convention eigenvalue $\ell (\ell +d-1)$, as in the spectrum of the Laplacian on a sphere.

For a Riemannian submersion with totally geodesic submanifolds as fibers, the positive Laplace-Beltrami operator commutes with pullback of functions. Horizontal terms in the Riemannian Hessian are pulled back from the base, and vertical terms vanish. Minimal fibers suffice because only the trace of the vertical second fundamental form enters.

On a closed manifold with a Riemannian metric the positive Laplace-Beltrami operator is self-adjoint with compact resolvent. It has a complete smooth orthonormal eigenbasis, finite-dimensional eigenspaces, nonnegative eigenvalues tending to infinity, and polynomial eigenvalue-counting bounds. Elliptic regularity bounds derivatives of its eigenfunctions polynomially in their eigenvalues.

For the positive Laplace-Beltrami operator on a closed manifold, the singular support of the wave trace lies in zero together with signed lengths of closed geodesics. Under the clean fixed-set hypotheses, its singularity at a periodic-geodesic length has an expansion determined by the geodesic flow and its linearized return map. In particular, when the relevant leading coefficient does not vanish through cancellation, that length is determined by the spectrum. The inclusion of singular supports alone does not justify an unconditional recovery of every geodesic length.

For a nonnegative self-adjoint operator with compact resolvent, this function counts eigenvalues up to a given level with their multiplicities. For the positive Laplace-Beltrami operator on a closed $n$-dimensional Riemannian manifold, $N(\Lambda )\le C(1+\Lambda {)}^{n/2}$. To prove this bound, multiply each vector in the low-eigenvalue subspace by a finite set of chart cutoffs and record its local Fourier coefficients of frequency at most $R$. The total number of retained coefficients is $O(R^n)$, while the sum of discarded squared norms is at most $C{R}^{-2}(1+\Lambda )\Vert u{\Vert }_2^2$ by Parseval identity and the energy bound. For $R$ a sufficiently large multiple of $\sqrt{1+\Lambda }$, vanishing of all recorded coefficients forces $u=0$. The recording map is injective and proves the dimension bound. Inverting this estimate gives ${\lambda }_j\ge c(j+1{)}^{2/n}-1$.

A densely defined semibounded symmetric operator on a Hilbert space has a canonical semibounded extension that is a self-adjoint operator obtained by closing its quadratic form. After shifting the lower bound to zero, complete the initial form domain in the norm $(\Vert u{\Vert }^2+q[u]{)}^{1/2}$; the associated closed form determines the Friedrichs extension. For the positive Laplace-Beltrami operator on a polygonal surface, this specifies the finite-energy realization at corners. Different choices of the initial form domain impose Dirichlet boundary conditions, Neumann boundary conditions, or mixed boundary conditions; the extension does not choose the boundary condition independently of that domain.

Choose an orthonormal basis $e_i$ of the tangent space at $p$ and the unit-speed geodesics with initial velocities $e_i$. Their covariant accelerations vanish, so $(f\circ {\gamma }_i{)}^{\prime \prime }(0)={\mathrm{Hess}}_{g}f(e_i,e_i)$. Taking the metric trace of the Riemannian Hessian proves the formula for $\Delta =\mathrm{div}\nabla$. The positive Laplace-Beltrami operator has the opposite sign.

On a closed manifold with a Riemannian metric, a smooth solution of ${\partial }_{t}u+\Delta u=0$ for the positive Laplace-Beltrami operator satisfies the displayed identity by integration by parts. Zero initial data force the solution to vanish, proving uniqueness and the semigroup property for heat evolution.

On a closed manifold with a Riemannian metric, the minimum of a smooth heat solution cannot fall below its initial minimum. At a spatial minimum, $\Delta u\le 0$ for the positive Laplace-Beltrami operator. Adding a small increasing function of time and considering the first crossing gives the assertion. Applying the result to nonnegative initial functions shows that the Riemannian heat kernel is nonnegative.

On a closed manifold, the positive Laplace-Beltrami operator has a smooth heat kernel and its heat semigroup is a trace-class operator for positive time. Expanding in a smooth orthonormal eigenbasis gives $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{\varphi }_j(x)\overline{{\varphi }_j(y)}$. Elliptic regularity bounds each derivative of ${\varphi }_j$ polynomially in ${\lambda }_j$, while the exponential dominates those powers, so the expansion converges with all derivatives. Integrating the diagonal and using the unit norm of each eigenfunction proves the formula. For a bundle operator, insert the fibre matrix trace inside the integral.

For the sign convention $\Delta =\mathrm{div}\nabla$, the heat operator is the differential operator $L={\partial }_t-\Delta$. The heat equation is $Lu=0$. If $P=-\Delta$ is the positive Laplace-Beltrami operator, the same operator is ${\partial }_t+P$. A Riemannian heat kernel is a fundamental solution with the initial Dirac delta distribution.

For the positive Laplace-Beltrami operator on a closed compact Riemannian manifold, the heat trace is $Z_M(t)={\sum }_j{e}^{-t{\lambda }_j}={\int }_M{K}_M(t,x,x)dV(x)$. It records every eigenvalue with multiplicity. In a finite free isometric quotient it equals $|U{|}^{-1}{\sum }_{u\in U}{\int }_N{K}_N(t,x,ux)dV(x)$. The orbital integrals form a class function of the ambient finite isometry group, so Gassmann equivalence gives equal quotient heat traces and hence Sunada theorem.

Use the positive Laplace-Beltrami operator ${\Delta }_{+}=-{\mathrm{div}}_g{\mathrm{grad}}_g$. On a closed manifold of positive dimension, its self-adjoint realization has a discrete spectrumcounted with the finite dimensions of the Laplacian eigenspaces. The zero eigenspace consists of functions constant on each connected component. On the unit circle, $-d^2/d{\theta }^2$ has eigenvalue zero once and eigenvalues $k^2$ twice for every integer $k\ge 1$, represented by sine and cosine modes.

The opposite convention $\Delta ={\mathrm{div}}_g{\mathrm{grad}}_g$, used in the coordinate formula for the Laplace-Beltrami operator, negates these eigenvalues. Specifying the sign is therefore part of specifying the spectrum.

On a compact manifold with boundary, Dirichlet boundary conditions or Neumann boundary conditions give corresponding discrete self-adjoint spectra. On a noncompact manifold, continuous spectrum can occur. The operator domain and any boundary conditions belong to the spectral problem, rather than being determined by the local differential formula alone.

For $L=-{\partial }_j(g^{ij}{\partial }_i)$ in geodesic normal coordinates, the Gauss lemma gives $g^{ij}{x}_i=x_j$. Substituting $F=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}{\sum }_{k\ge 0}{t}^k{B}_k$ into $({\partial }_t+L)F=0$ therefore yields $(x\cdot \nabla +k)B_k=-L{B}_{k-1}$, with $B_0=1$. Integration along radial rays gives the displayed formula. Every coefficient is smooth at the centre. Any smooth solution of $(x\cdot \nabla +k)h=0$ is zero for $k>0$, since $s^{k}h(sx)$ is constant and tends to zero at $s=0$. This proves uniqueness. In this special divergence-form case $L1=0$, so every $B_k$ for $k>0$ is zero: the Gaussian itself solves the equation exactly on the coordinate ball. The positive Laplace-Beltrami operator instead has a metric volume-density term and a nonconstant leading amplitude.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.

Choose a smooth cutoff $\eta (x,y)$ that is one near the diagonal and supported where these normal coordinates exist, and form the heat parametrixIt tends to the identity kernel as $t\downarrow 0$: in normal coordinates the Gaussian is an approximate identity, and $u_{0}J=J^{1/2}\to 1$ at the centre. Its residual $R_N=({\partial }_t+{\Delta }_x)P_N$ is $\eta g_t{t}^N{\Delta }_x{u}_N$ plus cutoff-derivative terms supported a positive distance from the diagonal. Those latter terms are smaller than every power of $t$. Hence $R_N=O(t^{N-m/2})$ uniformly. Every prescribed finite number of derivatives has the same kind of estimate after increasing $N$; differentiating a Gaussian costs only finitely many powers of $t$.

For $N$ large enough that the residual is continuous and bounded up to time zero, use the Volterra parametrix correction. For kernels defineSet $Q=-R_N+R_N\ast R_N-R_N\ast R_N\ast R_N+\cdots$. A bounded residual with bound $C$ givesso this series converges and $Q+R_N+R_N\ast Q=0$. Because $P_N$ has identity initial kernel, differentiating the convolution gives $({\partial }_t+{\Delta }_x)(P_N\ast Q)=Q+R_N\ast Q$. Thusis an exact kernel with the correct initial condition. Taking $N$ sufficiently large gives any required finite regularity; uniqueness of the heat equation makes the kernels obtained from different $N$ equal. For uniqueness, the difference applied to smooth initial data has zero initial value, and the energy identity $\frac{d}{dt}\Vert v{\Vert }_2^2=-2\Vert \nabla v{\Vert }_2^2$ forces $v=0$. Parabolic regularity then gives smoothness for positive time.

The Gaussian part of $P_N$ has uniformly bounded spatial $L^1$ norm, so the correction is $O(t^{N+1-m/2})$. This proves, uniformly near the diagonal,with corresponding derivative estimates on increasing the truncation order. In particular on every parabolic neighbourhood $r\le B\sqrt{t}$ it is the stated Gaussian expansion with relative remainder $O(t^{N+1})$, and on the diagonal it yields the complete heat trace expansion. The standard off-diagonal Gaussian version on compact subsets of a sufficiently small convex normal neighbourhood follows from the same convolution estimates with the Gaussian factor retained: the inequalitylocalizes the convolution to the unique joining geodesic; its transverse Hessian is positive, and Gaussian integration gives the successive transport orders. The additive estimate above already specifies an appropriate uniform near-diagonal remainder without asserting a relative estimate far from the diagonal on the basis of a crude bound.

If $M$ is noncompact, use the minimal Friedrichs extension heat semigroup; if there is a boundary, specify boundary conditions. The same construction is local in the interior. One may extend a relatively compact coordinate neighbourhood to a closed auxiliary manifold; localization errors are supported a positive distance away and are exponentially small near the smaller diagonal neighbourhood. Thus the local coefficients remain the same. Finite integrated heat invariants and a discrete spectrum require the global hypotheses used above.

Use the nonnegative convention for the Laplace-Beltrami operator. For a smooth function $f$ on a Riemannian manifold $(M,g)$, three intrinsic definitions areHere $d$ is the exterior derivative, $d^{\ast }$ its formal $L^2$ adjoint for the Riemannian volume density, $\mathrm{grad}f$ the Riemannian gradient, and $\nabla$ the Levi-Civita connection. The bilinear form $\nabla df$ is the Riemannian Hessian. The equally common nonpositive convention negates all three definitions simultaneously.

In local coordinates let $g^{ij}$ be the inverse metric matrix and $\rho =\sqrt{\det (g_{ij})}$. By the defining relation $g(\mathrm{grad}f,V)=df(V)$,The divergence of a Riemannian vector field can be defined by ${\mathcal{L}}_V(d{V}_g)=(\mathrm{div}V)d{V}_g$. Differentiating the coordinate volume density gives $\mathrm{div}V={\rho }^{-1}{\partial }_i(\rho V^i)$. Thus the second definition becomesFor compactly supported smooth $h$, coordinate integration by parts showsSince this holds for every such $h$, the first and second definitions agree. This argument also works on a nonorientable Riemannian manifold, using its volume density; no global orientation is needed.

For the third definition,The Christoffel symbols satisfy ${\Gamma }_{ik}^i={\partial }_k\log \rho$. This follows by tracing their formula and using ${\partial }_k\log \det g=\mathrm{tr}(g^{-1}{\partial }_{k}g)$. Metric compatibility givesSubstitute both identities into the coordinate divergence expression to obtainThis proves all three definitions equivalent. In geodesic normal coordinates at a point $p$, the common value is $-{\sum }_i{\partial }_i^{2}f(p)$. Finally, on a closed manifold,confirming the chosen positive Laplace-Beltrami operator sign convention.

On a closed compact Riemannian manifold, the heat kernel for the positive Laplace-Beltrami operator is the smooth kernel $K_M(t,x,y)$ of $e^{-t{\Delta }_M}$ for $t>0$. It solves $({\partial }_t+{\Delta }_x)K_M=0$ and tends to the delta kernel as $t\downarrow 0$, so thatFor an orthonormal eigenbasis with eigenvalues repeated according to multiplicity,The exponential factors and elliptic regularity ensure smooth convergence for positive time. If a boundary is present, the same construction uses a specified boundary condition preserved by the covering; the closed-manifold case is the one needed below.

Let $p:N\to M=N/U$ be the finite normal locally isometric covering map. The heat equation for $f\circ p$ on $N$ is the lift of the one for $f$ on $M$. Integrate the lifted solution over a fundamental domain for the deck transformation group $U$, splitting the integral over $N$ into its translates. This givesThere is no averaging factor in this kernel formula: each translate accounts for one lift of the integration variable. To verify it directly, the sum solves the lifted heat equation and its integral against $f$ converges to $f(px)$; uniqueness of the heat solution identifies the kernel. Replacing either lift $x$ or $y$ only permutes the sum, since isometries preserve the heat kernel.

Integrating the diagonal and using the covering degree gives the heat trace formulaSuppose $U$ is contained in a finite group $T$ of isometries of $N$. For $s\in T$, change variables $x=sy$ and use invariance of both volume and the heat kernel. ThenThus $F_t$ is a class function on $T$, and for its conjugacy classes $\mathcal{C}$,

Consequently, if $U_1,U_2\le T$ are Gassmann equivalent and act freely, their orders are equal and the displayed class-by-class sums give $Z_{N/U_1}(t)=Z_{N/U_2}(t)$ for every $t>0$. Equal heat traces determine equal eigenvalues with multiplicities: the least eigenvalue at which the multiplicities differed would give a nonzero leading exponential in their trace difference as $t\to \infty$, a contradiction. Hence the two smooth quotient manifolds are isospectral. This proves the Sunada theorem, with the free-action hypothesis ensuring that the quotients are manifolds rather than orbifolds. The theorem alone makes no nonisometry claim.

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.

Interpret $L=-\square$ as the minimally coupled scalar positive Laplace-Beltrami operator on the smooth closed manifold. Writing $dV$ for its metric volume measure (the Riemannian volume form on an oriented chart), integration by parts gives $⟨u,Lu⟩={\int }_M|\nabla u{|}^{2}dV\ge 0$. The compact elliptic spectral theorem supplies an orthonormal basis $u_j$ with $L{u}_j={\lambda }_j{u}_j$, real nonnegative eigenvalues tending to infinity and finite multiplicities. Solve this eigenvalue problem to obtain the spectral zeta functionThe zero eigenvalues are excluded, not assigned inverse powers. Their multiplicity $n_0$ is the number of connected components: $Lu=0$ implies $\int |\nabla u{|}^{2}dV=0$, so $u$ is constant on each component.

An equivalent calculation uses the spectral expansion of the Riemannian heat kernel,Since ${\int }_0^{\infty }{t}^{s-1}{e}^{-t\lambda }dt=\Gamma (s){\lambda }^{-s}$ for $\lambda >0$, the Mellin transform givesThe reduced trace decays exponentially at large $t$. At small $t$, subtract a finite number of terms of the heat kernel expansion, integrate those terms explicitly, and then integrate the remainder. Increasing the number of subtractions extends the formula successively to further left half-planes, producing the meromorphic continuation of ${\zeta }_L$ rather than substituting $s=0$ in its divergent defining series.

For completeness, the needed curvature coefficient comes from the heat-kernel transport equations. In normal coordinates $\xi$ centered at $y$, the Riemannian volume form has density $J(\xi )=1-\frac{1}{6}{R}_{ab}(y){\xi }^a{\xi }^b+O(|\xi {|}^3)$. The leading Gaussian amplitude is thereforeSubstitution of the Gaussian amplitude series into the heat equation gives the first diagonal transport relation $u_1(y,y)={\square }_x{u}_0(x,y){|}_{x=y}=R(y)/6$. The curvature coefficient of the scalar heat kernel is consequentlyIn dimension two this impliesSet $A_0=\mathrm{Area}(M)/(4\pi )$ and $A_1={\int }_{M}RdV/(24\pi )$. Splitting the Mellin integral at $t=1$ and subtracting these two heat terms gives, near $s=0$,where $H$ is holomorphic for $\mathrm{Re}s>-1$. As $1/\Gamma (s)=s+O(s^2)$, only the second term survives at zero. Thus the zero-mode correction to the spectral zeta value isThe Gauss-Bonnet theorem, with $R$ twice the Gaussian curvature, makes this $\boxed{{\zeta }_L(0)=\chi (M)/6-n_0}$. For a connected surface $n_0=1$, so the answer is $\chi (M)/6-1$; a sphere gives $-2/3$, and a flat torus gives $-1$.

The purely local heat coefficient is $A_1=\int RdV/(24\pi )$. Sometimes a formal formula for $\zeta (0)$ quotes this coefficient with zero modes left implicit. For the actual positive-spectrum spectral zeta function defined above, the compact scalar problem necessarily has the additional subtraction $-n_0$. A strictly positive shifted operator is a different problem and has its own heat coefficients.

Use $a^{ij}=g^{ij}$, $r^2=|x{|}^2$, and $E=x\cdot \nabla$. The Gauss lemma from part (a) givesFor the Gaussian $q_t(x)=(4\pi t{)}^{-n/2}{e}^{-r^2/(4t)}$, we have ${\partial }_i{q}_t=-x_i{q}_t/(2t)$. Applying precisely the divergence-form operator in the paper yieldsSince ${\partial }_t{q}_t=(-n/(2t)+r^2/(4t^2))q_t$, the Gaussian itself already satisfies $({\partial }_t+L)q_t=0$.

For completeness, the product rule gives for an arbitrary smooth $G$,Indeed the cross term in $L(q_{t}G)$ is $-2a^{ij}({\partial }_i{q}_t)({\partial }_{j}G)=q_{t}EG/t$. Substituting the formal power series $G={\sum }_{k\ge 0}{t}^k{B}_k$, with $B_0=1$, gives the radial transport equationsAlong a radial segment their unique smooth solution isThe integral is smooth including at the centre, since each derivative adds a nonnegative power of $s$ to the integrable factor $s^{k-1}$. Uniqueness follows because a homogeneous solution $h$ has $d(s^{k}h(sx))/ds=0$; smoothness makes its limit at $s=0$ zero, so $h(x)=0$. This also excludes singular ray-dependent homogeneous solutions.

Here $L1=0$, so the recursion successively gives $\boxed{B_k=0(k\ge 1),G(x,t)=1}$. Thus both existence and uniqueness hold, with no convergence issue because the formal solution is the exact Gaussian. This is the normal-coordinate divergence-form heat parametrix. The positive Laplace-Beltrami operator $-{\rho }^{-1}{\partial }_j(\rho g^{ij}{\partial }_i)$, where $\rho =\sqrt{\det g}$, has an additional first-order term and a different leading amplitude; it is essential to use the operator actually specified here.