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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:

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


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.

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  • $\begingroup$ Can we suppose that $D$ is self adjoint? $\endgroup$
    – shu
    Aug 20, 2013 at 12:03
  • $\begingroup$ @shu Isn't the first equality in Liviu's answer the self-adjoint-ness? $\endgroup$
    – Student
    Aug 21, 2013 at 13:25
  • $\begingroup$ yes, Liviu's answer is in the self-adjoint case. Such a result also exists for some non self-adjoint operator. But much more complicated. $\endgroup$
    – shu
    Aug 21, 2013 at 13:42
  • $\begingroup$ Interesting question! What's the corresponding operator for the zeta function? Does it has a familiar name? I'd also like to see some applications. $\endgroup$
    – Student
    Nov 9, 2019 at 12:25

1 Answer 1

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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

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

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  • $\begingroup$ Nicolescu Thanks for your efforts. (1) I am not sure I find the last identity "elementary" ;P - I was infact stuck on that before I posted this question! (2) Can you comment on what happens on hyperbolic manifolds? Anything (what if?) changes in the argument of yours? $\endgroup$
    – Student
    Aug 21, 2013 at 13:24
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    $\begingroup$ @Anirbit, Last identity is almost the definition of Gamma function by changing the variable $t\to \lambda t$. $\endgroup$
    – shu
    Aug 21, 2013 at 13:45
  • $\begingroup$ On noncompact manifolds there is a first issue namely the selfadjointness of $D$. If $M$ is a complete manifold, and $D$ is the usual Laplacian, then it has a unique selfadjoint extension to $L^2$. For complete hyperbolic manifolds check this paper and the references therein. archive.numdam.org/ARCHIVE/JEDP/JEDP_1987___/JEDP_1987____A17_0/… $\endgroup$ Aug 21, 2013 at 14:06
  • $\begingroup$ @Liviu Thanks for the reference! Will read that. On a related note I was wondering if this integral representation of the generalized zeta-function related to these kinds of identities like, $\xi(3) = \frac{8\pi^3}{3}\int_0^{\infty}d\lambda \frac{\sqrt{\lambda}}{1+e^{2\pi\sqrt{\lambda}}}$ - could you kindly shed some light? $\endgroup$
    – Student
    Aug 21, 2013 at 21:57
  • $\begingroup$ That integral description of zeta is not that subtle. $\endgroup$ Aug 22, 2013 at 8:11

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