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Use the positive Laplace-Beltrami operator on a compact Riemannian manifold without boundary. Its heat kernel is the smooth integral kernel $K(t,x,y)$, $t>0$, of the solution operators for the heat equation:The last condition holds, in particular, in $L^2$ for smooth initial functions. The heat trace isThe heat invariants are the coefficients in the small-time heat kernel expansion of this trace:Equivalently, the local diagonal coefficients $u_j(x,x)$ integrate to $a_j$; $R_g$ is the scalar curvature. Some normalizations absorb $(4\pi {)}^{-d/2}$ into the coefficients. These definitions here use the displayed normalization consistently.

To prove uniqueness, let $K_1,K_2$ satisfy the heat kernel conditions and let $w$ be the difference of their solutions with the same smooth initial function. It solves $w_t=-\Delta w$ and tends to zero in $L^2$ as $t\downarrow 0$. The heat equation energy identity, using integration by parts on the closed manifold, gives for $t>0$Integrating from $s>0$ to $t$ gives $\Vert w(t){\Vert }_{L^2}\le \Vert w(s){\Vert }_{L^2}$; sending $s\downarrow 0$ proves $w(t)=0$. Thus for every smooth test function $f$,For fixed positive $t$ and $x$, smoothness makes this difference a smooth function of $y$. Taking its conjugate as test function shows its squared norm is zero. The heat kernel is unique.

The compact elliptic spectral theorem for the self-adjoint elliptic Laplace-Beltrami operator gives a complete orthonormal basis of smooth eigenfunctions ${\varphi }_j$, with discrete eigenvalues $0\le {\lambda }_0\le {\lambda }_1\le \cdots$ counted with multiplicity. Solving the heat equation for initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$, so the unique heat kernel has its eigenfunction expansion and its heat trace isThe series converges for every $t>0$. This proves that the spectrum with multiplicities determines the heat trace.

Conversely, its large-time behaviour recovers the distinct eigenvalues and their multiplicities successively. The multiplicity of zero is $m_0={\lim }_{t\to \infty }{Z}_M(t)$. After subtracting the already determined terms, letbe the positive remaining sum, and let $\mu$ be its smallest remaining eigenvalue. ThenIndeed, after factoring $e^{-t\mu }$, the next distinct eigenvalue is separated from $\mu$ by a positive gap. The remaining tail tends to zero: for fixed $t_0>0$ its absolute sum is bounded by a constant times $e^{-({\mu }_{\mathrm{next}}-\mu )(t-t_0)}$, using finiteness of $Z_M(t_0)$. Induction recovers every spectral term. Thus the heat trace and the spectrum with multiplicities determine one another.

Finally the leading coefficient among the heat invariants is positive, soThe common heat trace of isospectral manifolds therefore forces them to have the same dimension. Once the dimension is recovered, its leading coefficient also recovers their common volume.

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.

The positive sign printed in the exponential uses the skew-adjoint convention for the Dirac operator: $D^{\ast }=-D$, so $D_{-}=-D_{+}^{\ast }$. Put $A=D_{+}$ and define the nonnegative self-adjoint operatorsFor $t>0$, the assumed compact elliptic spectral theorem for these elliptic differential operators gives discrete eigenvalues, finite multiplicities and trace-class operators $e^{-tP}$, $e^{-tQ}$.

If $Pu=\lambda u$ with $\lambda >0$, then $Q(Au)=A(Pu)=\lambda Au$ and $Au\ne 0$. Conversely $A^{\ast }$ sends the $\lambda$ eigenspace of $Q$ to that of $P$. On these eigenspaces the inverse of $A$ is ${\lambda }^{-1}{A}^{\ast }$. Thus the positive spectra agree with multiplicity, and absolute convergence of the operator traces permits cancellation term by term:The last expression is the Fredholm index of $A$, since its closed range has orthogonal complement $\ker A^{\ast }$. ConsequentlyThis is the supertrace cancellation used in Question 1. If the Dirac operator is instead defined to be self-adjoint, $D_{-}=(D_{+}{)}^{\ast }$ and the heat exponentials must have a minus sign. With that convention the positive exponentials in the printed formula are generally unbounded and have no finite operator trace.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\mathrm{grad}$. A Riemannian submersion is a surjective smooth submersion $\pi :(M,g)\to (N,h)$ for which, at every $p$, the restriction of $d{\pi }_p$ to $H_p=(\ker d{\pi }_p{)}^{\perp }$ is a linear isometry onto $T_{\pi (p)}N$. The spaces $V_p=\ker d{\pi }_p$ and $H_p$ are its vertical and horizontal spaces. Its fibres are totally geodesic submanifolds precisely when ${\nabla }_{U}V$ is vertical for vertical vector fields $U,V$: their second fundamental form vanishes. Equivalently, a geodesic initially tangent to a fibre remains in that fibre while defined.

For a smooth $f:N\to \mathbb{R}$, its basic function $F=f\circ \pi$ is constant along each fibre. The Riemannian gradient of $F$ is the horizontal lift of a vector field through a submersion of ${\mathrm{grad}}_{N}f$, sincefor horizontal $X$, and $dF(U)=0$ for vertical $U$. In particular no derivative of $f$ in a vertical direction occurs.

Here is the needed connection fact, which also follows directly from the Koszul formula: for horizontal lifts $X,Y$ of vector fields $\bar{X},\bar{Y}$ on $N$, the horizontal component of ${\nabla }_{X}Y$ projects to ${\nabla }_{\bar{X}}^N\bar{Y}$. To see this, pair the Koszul formula with a third horizontal lift $Z$. The horizontal inner products are pulled back from $N$, and the horizontal components of their Lie brackets of vector fields project to the brackets on $N$. Thus all six terms are the pullbacks of the corresponding terms on $N$.

Choose an adapted Riemannian orthonormal frame $X_1,\dots ,X_n,U_1,\dots ,U_r$, with the $X_a$ horizontal lifts. Using $\mathrm{Hess}F(A,A)=A(AF)-({\nabla }_{A}A)F$, the connection fact givesFor a vertical $U_b$, both $U_{b}F=0$ and $({\nabla }_{U_b}{U}_b)F=0$, the latter because the fibres are totally geodesic submanifolds. Taking the negative metric trace of the Riemannian Hessian therefore proves the basic-function Laplacian identityThis identity is local and does not require compactness. Vanishing mean curvature of the fibres would already suffice; total geodesicity makes each vertical summand vanish separately.

For the discrete eigenspace assertion, assume the two Riemannian manifolds are closed manifolds. Without a discrete spectral realization, an unrestricted noncompact version need not have an eigenbasis. The projections of the Riemannian product $M\times N$ are Riemannian submersions with totally geodesic submanifolds as fibres. Its Levi-Civita connection splits into the two factor connections. Consequently its positive Laplace-Beltrami operator isThe cross term in the product rule for the positive Laplace-Beltrami operator is zero because the two factor Riemannian gradients are orthogonal.

We use the standard compact elliptic compact elliptic spectral theorem: the positive Laplace-Beltrami operator on a closed manifold is self-adjoint, has compact resolvent, and has a complete orthonormal eigenbasis of smooth eigenfunctions, with finite-dimensional eigenspaces and eigenvalues tending to infinity. Let ${\Delta }_M{u}_i={\mu }_i{u}_i$ and ${\Delta }_N{v}_j={\nu }_j{v}_j$. Fubini's theorem and completeness on each factor show that $u_i(x)v_j(y)$ form a complete orthonormal basis of $L^2(M\times N)$. For example, a function orthogonal to all these products has, for each $i$, zero $u_i$-coefficient as an $L^2(N)$ function, hence is zero.

The displayed operator identity makes $u_i{v}_j$ an eigenfunction with eigenvalue ${\mu }_i+{\nu }_j$. Conversely, if $\Delta w=\lambda w$, self-adjointness of the positive Laplace-Beltrami operator givesAll other coefficients vanish. Only finitely many pairs can have ${\mu }_i+{\nu }_j=\lambda$, since both spectra are nonnegative and have finitely many eigenvalues below any fixed bound. Thus the product Laplacian eigenspace decomposition isThe tensor product summands are mutually orthogonal; their elements are actual smooth eigenfunctions, so this is an equality of eigenspaces, not just a formal expansion.

Let $M$ be a closed manifold with a Riemannian metric and use $\Delta =-\mathrm{div}\mathrm{grad}$, the positive Laplace-Beltrami operator. This is the setting in which the final discrete eigenfunction expansion applies without additional boundary or noncompact spectral hypotheses. A Riemannian heat kernel is a smooth $H(t,x,y)$ for $t>0$ such thatfor every smooth initial function $f$. Here $\nu$ solves the heat equation. The limit is the statement that the initial kernel is the Dirac delta distribution on the diagonal. In the closed setting this uniquely determines the heat kernel; it is symmetric, preserves constants, is nonnegative, and satisfies the semigroup property.

A heat parametrix $P(t,x,y)$ is an approximate version with the same delta initial limit and with errorregular enough at $t=0$ to correct by a convergent integral series. One may construct it by the local Gaussian ansatzwhere $\chi =1$ near the diagonal and is supported in a convex normal neighborhood, and the smooth coefficients solve the usual radial transport equations. The leading coefficient accounts for the Riemannian volume form. Taking $q$ sufficiently large makes the residual extend continuously, with any prescribed finite number of derivatives, to $t=0$. Alternatively a smooth asymptotic summation of all transport coefficients gives a residual vanishing to every order. These local construction, extension and differentiability facts are used here as subsidiary results, as permitted.

For the correction proof take a heat parametrix whose residual is smooth up to $t=0$ on a short interval $[0,T]$. We also use its uniform integral bound ${\sup }_{0<t\le T,x}{\int }_M|P(t,x,y)|d{V}_y<\infty$, its approximate-identity limit, and the following standard differentiability property: convolution with a smooth time-dependent kernel gives a smooth kernel for $t>0$, and differentiating it yields the identity below. These properties follow from the Gaussian estimates and the delta initial limit; no conclusion about the final exact heat kernel is assumed.

Define the Volterra convolution of kernels byIt is associative wherever these integrals converge. The delta term at the upper endpoint givesSeek $H=P+P\ast Q$. Then its error vanishes exactly when $Q+R+R\ast Q=0$. The solution is the Volterra parametrix correctionThe signs matter: the first correction is $-P\ast R$.

To prove convergence, let $|R|\le C$ and $V=\mathrm{vol}(M)$. The time variables in $R^{\ast k}$ range over a simplex of volume $t^{k-1}/(k-1)!$, soThe series for $Q$ therefore converges uniformly on $[0,T]\times M\times M$. Derivatives obey analogous bounds, with polynomial factors in $k$, by the quoted residual extension and differentiation properties. Thus the series can be convolved and differentiated as above. Associativity and absolute convergence give $Q+R+R\ast Q=0$, proving $({\partial }_t+{\Delta }_x)H=0$. The correction $P\ast Q$ has integral norm $O(t)$ by the integral bound on $P$ and boundedness of $Q$. It has zero initial limit, so $H$ has precisely the required delta initial data.

For uniqueness, a smooth solution $w$ with zero initial data satisfies the heat equation energy identityIts initial norm is zero, hence $w=0$. Applying uniqueness to consecutive evolutions proves the semigroup property. Choose $s_0<T$ and, for arbitrary $t>0$, compose enough kernels $H(t/r)$ that $t/r<s_0$. The resulting kernel is independent of the subdivision by uniqueness, is smooth for positive time by the quoted differentiability property, and extends the construction to every $t>0$. This proves a parametrix determines the global heat kernel in the closed setting. The heat equation maximum principle gives nonnegativity and applying uniqueness to the constant initial function gives conservation of total mass. Self-adjointness of the positive Laplace-Beltrami operator $\Delta$ gives symmetry.

Finally let $\{{\varphi }_j\}$ be a complete complex orthonormal eigenbasis of $\Delta$, with eigenvalues ${\lambda }_j\ge 0$, repeated by multiplicity. As subsidiary analytic facts we use the compact elliptic compact elliptic spectral theorem, elliptic regularity bounds making each fixed derivative of ${\varphi }_j$ grow at most polynomially in $1+{\lambda }_j$, and a polynomial eigenvalue-counting bound. The spectral expansion of the Riemannian heat kernel converges because exponential decay makes the following series converge with every derivative when $t\ge \epsilon >0$:Indeed, evolving initial data ${\varphi }_j$ gives $e^{-t{\lambda }_j}{\varphi }_j$ by direct substitution into the heat equation and uniqueness. For general $f\in L^2(M)$, completeness and the heat equation energy identity giveThe smoothly convergent kernel series represents this same operator, and equality for all smooth $f$ identifies it pointwise with the constructed heat kernel. The complex conjugate is required by the inner product; for a real eigenbasis it may be omitted. A noncompact Riemannian manifold may instead require a spectral integral, and the unqualified discrete formula should not be asserted there.