Relation between asymptotic expansion of the heat kernel and wave kernel.

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Question 1. It is known that the heat kernel $e^{-\Delta{t}}$ of a closed and compact manifold of $n$-dimension has an asymptotic expansion around $t=0$: $$ \mathrm{tr}(e^{-\Delta{t}}) \sim (4\pi{t})^{-n/2}\mathrm{vol}(M)(a_0 + a_1 t + \cdots). $$ Then the wave kernel $e^{-\sqrt{\Delta}t}$ has an asymptotic expansion around $t=0$?

Question 2. If the wave kernel $e^{-\sqrt{\Delta}t}$ has an asymptotic expansion around $t=0$, then what is a relation between two asymptotic expansions?


My intuition for the question 1.

The asymptotic expansion of the heat kernel is used to show the analytic continuation of the spectral zeta function: \begin{equation} \begin{split} \zeta_{\Delta}(s) &= \sum_{\lambda>0}\lambda^{-s} \\ &= \frac{1}{\Gamma(s)} \{ \mathcal{M}\sum_{\lambda}e^{-\lambda{t}} \}(s), \end{split} \end{equation} where $\lambda$ runs over the spectrum of the Laplacian.

Also, the wave kernel can be related with the spectral zeta function as follows: \begin{equation} \begin{split} \zeta_{\Delta}(s/2) &= \sum_{\lambda>0}\lambda^{-s} \\ &= \frac{1}{\Gamma(s)} \{ \mathcal{M}\sum_{\lambda}e^{-\sqrt{\lambda}{t}} \}(s). \end{split} \end{equation}

Since the spectral zeta function $\zeta_{\Delta}(s)$ is extended to a meromorphic function over the complex plane, the wave kernel $\mathrm{tr}(e^{-\sqrt{\Delta}{t}})$ has an asymptotic expansion around $t=0$ given by the poles of $\zeta_{\Delta}(s/2)\Gamma(s)$.

But i don't think it is enough for the answer to the Question 1.

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