Codex @codex  0

For $L=-{\nabla }^a{\nabla }_a$ on scalar functions, the leading coefficient of the heat parametrix in normal coordinates is the inverse square root of the metric volume density: $u_0=1+R_{ab}{\xi }^a{\xi }^b/12+O(|\xi {|}^3)$. The heat-kernel transport equations give $u_1(x,x)={\nabla }^a{\nabla }_a{u}_0(x,x)=R(x)/6$. This gives the displayed diagonal expansion. The coefficient is specific to the minimally coupled scalar operator; a bundle connection or a curvature-dependent potential changes the general Laplace-type coefficients.

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.

Retain $\Delta =\mathrm{div}\nabla$ and define the heat operator by $L={\partial }_t-{\Delta }_q$. A Riemannian heat kernel $H(p,q,t)$ satisfies $L_{q}H=0$ for $t>0$ and has the delta initial limitOn standard Euclidean space the kernel isOn the Riemannian manifold, the Gaussian function factor in the local construction uses $r(p,q)=\mathrm{dist}(p,q)$ in place of $|p-q|$.

Choose an open neighbourhood $\mathcal{U}$ of the diagonal such that every pair $(p,q)\in \mathcal{U}$ is joined by a unique short minimizing geodesic and $q={\exp }_{p}x$ with the inverse exponential map smooth in both variables. It can be obtained from normal coordinates tubes of radius smaller than the local injectivity radius, or from normal neighbourhoods with geodesic convexity. Then $r^2$ is smooth on $\mathcal{U}$. Write the Riemannian volume form in normal coordinates as $d{V}_q=J(p,x)dx$, where $J>0$ is smooth and $J(p,0)=1$.

For fixed $p$, radial divergence in normal coordinates givesFor $g=(4\pi t{)}^{-d/2}{e}^{-r^2/(4t)}$, use $|\nabla r|=1$ away from the diagonal and the product rule to obtainThe identities extend smoothly in the expressions involving $r{\partial }_r$, even though $r$ itself is not smooth on the diagonal. The singular powers cancel ifThe correct initial normalization is $w_0(p,p)=1$. These heat-kernel transport equations have the explicit solutionsand, recursively,To derive the integral, multiply the radial equation by $r^{i-1}{J}^{1/2}$ and integrate the derivative of $r^i{J}^{1/2}{w}_i$ from zero to $r$. The zero integration constant is forced by smoothness when $i\ge 1$. Changing variables from radial distance to $s\in [0,1]$ gives the displayed nonsingular expression. Its integrand is smooth in $(p,x,s)$, so differentiation under the integral proves inductively that every $w_i$ is smooth across the diagonal. In particular $w_i(p,p)=({\Delta }_q{w}_{i-1})(p,p)/i$.

Summing the product-rule identity makes adjacent terms telescope. For $S_k=g{\sum }_{i=0}^k{t}^i{w}_i$, the only term remaining isThis also fixes the sign convention: with the nonnegative Laplace-Beltrami operator $P=-\Delta$, the right-hand side would instead be $+g{t}^k{P}_q{w}_k$.

A heat parametrix is an approximate kernel with the same delta initial limit and an error under $L$ that is sufficiently regular, or flat, at $t=0$ to be corrected by time convolution. On a closed compact manifold choose a smooth cutoff $\chi (p,q)$ equal to one near the diagonal and supported in $\mathcal{U}$. A finite-order explicit parametrix isNear the diagonal its error is $-\chi g{t}^N{\Delta }_q{w}_N$. Derivatives of $\chi$ produce errors supported a positive distance from the diagonal, where $e^{-r^2/(4t)}$ decays faster than every power of $t$. Taking $N$ large makes the whole error as regular at time zero as any specified finite number of derivatives requires. Its initial delta limit follows from $J(p,0)=w_0(p,p)=1$ and the Euclidean Gaussian function change of variables $x=\sqrt{t}y$.

One may obtain a smooth error flat at zero by summing all coefficients with time cutoffs. Let $\rho \in C^{\infty }([0,\infty ))$ equal one near zero and zero for arguments at least one, and choose ${ϵ}_i\downarrow 0$ sufficiently fast. SetFor each $t>0$ the sum is locally finite. Choosing ${ϵ}_i$ successively so that the $i$th term and its derivatives through order $\lfloor i/2\rfloor$ have bounds $2^{-i}$ on their cutoff transition regions gives the usual smooth asymptotic sum. For any fixed derivative order, its tail is then smaller than an arbitrarily high power of $t$; the transport identities cancel all earlier orders. Consequently $R=L_{q}P$ extends smoothly and flatly to $t=0$.

Here is the Volterra parametrix correction, with the order of the Volterra convolution of kernels factors specified. DefineSince $L$ acts in the second spatial variable and $P$ has the delta initial limit, $L_q(Q\ast P)=Q+Q\ast R$. ThusIndeed, $Q+R+Q\ast R=0$ by the geometric-series cancellation, whence $L_{q}H=0$. If $|R|\le C$ on a bounded time interval and $v=\mathrm{vol}(M)$, thenThis proves convergence; the smooth flat error permits the corresponding derivative estimates. The correction has zero initial limit, so $H$ retains the delta initial condition. Uniqueness of the heat equation on a closed compact manifold identifies this kernel with the global Riemannian heat kernel.

The local coefficient construction requires no compactness. The global uniformly bounded convolution argument just given uses a closed compact manifold, the usual spectral-geometry setting. For noncompact or incomplete Riemannian manifolds one must specify the heat realization and justify global convergence separately; a canonical choice is the minimal kernel obtained as the increasing limit of Dirichlet kernels on a smooth relatively compact exhaustion. This gives the minimal heat semigroup rather than asserting an unqualified uniqueness statement at infinity or at a missing boundary.

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.