Mellin transform between heat kernel and zeta-function

Question

For some notion of a "positive operator" $D$ of "Laplacian type" one seems to be able to define a notion of a zeta-function as $\xi(s,f,D) = Tr_{L^2}(f D^{-s})$ where $f \in L^2$ (the space of square-integrable functions on the chosen manifold). Also one now defines the generalized heat-kernel corresponding to this as $K(t,f,D) = Tr_{L^2}(f\, exp(-tD))$.

Now I am faced with the following identity for which I would like to know the proof:

$$ \xi(s,f,D) = \frac{1}{\Gamma(s)}\int_0^{\infty} t^{s-1} K(t,f,D) \, dt $$

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It would be helpful if someone could also elaborate on the notion of what is a "positive operator of Laplacian type" (..I have some rough idea based on specific examples..) and if someone could specify peculiarities in the above equations that are likely to come up if one goes to non-compact manifolds like hyperbolic spaces.

I am mostly interested in doing this on hyperbolic spaces.

Answer 1

Being of Laplacian type means that $D$ is a second order p.d.o. whose principal symbol coincides with that of a Laplacian of a metric $g$ on the manifold. Being positive signifies that that for any smooth functions with compact support $\newcommand{\bR}{\mathbb{R}}$ $f_0, f_1: M\to \bR$, $f_0\neq 0$, you have

$$ \int_M (Df_0)\cdot f_1 dV_g =\int_M f_0 \cdot (D f_1) dV_g,\;\; \int_M (Df_0) f_0 > 0. $$

In case the manifold $M$ is compact, $\dim M=m$, then $D$ has a discrete spectrum

$$ 0<\lambda_0\leq \lambda_1\leq \cdots. $$

Let $(\Psi_k)_{k\geq 0}$ denote an orthonormal basis of $L^2(M)$ consisting of eigenfunctions, $D\Psi_k=\lambda_k\Psi_k$.

For $s>-m$, then $fD^{-s}$ is an integral operator with integral kernel $\newcommand{\eK}{\mathscr{K}}$

$$ \eK_{f D^{-s}}(x,y) =\sum_{k\geq 0} \lambda_k^{-s} f(x)\Psi_k(x) \Psi_k(y), $$

while $fe^{-tD}$ is an integral operator with kernel

$$\eK_{fe^{-tD}}(x,y)= \sum_{k\geq 0} e^{-t\lambda_k} f(x)\Psi_k(x) \Psi_k(y). $$

Then

$$ \xi(s,f,d)=\int_M \eK_{fD^{-s}} (x,x) dV_g(x)=\sum_{k\geq 0} \lambda_k^{-s} \int_M f(x) |\Psi_k(x)^2| dV_g (x), $$

$$ K(t,f, D)=\int_M \eK_{fe^{-tD}} (x,x) dV_g(x)=\sum_{k\geq 0} e^{-\lambda_k t} \int_M f(x) |\Psi_k(x)^2| dV_g (x). $$

Now use the elementary identity

$$ \lambda^{-s}=\frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-\lambda t} dt. $$