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For a closed Riemannian manifold $M$ of $n$-dimension, we consider the Laplace-Beltrami operator $\Delta$.

It is known that we have an asymptotic expansion for the trace of heat operator $e^{-\Delta{t}}$ as follows $$ \mathrm{tr}(e^{-\Delta{t}})=\sum_{\lambda}e^{-\lambda{t}}\overset{t\downarrow0}{\sim}t^{-\frac{n}{2}}\sum_{n} \alpha_{n}t^{n}, $$ where $\lambda$ runs over the set of spectrum of Laplacian $\Delta$.

My question is that

Denote by $\mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, $i.e.$ $\mathcal{D}^{2}=\Delta$. Then the sum of positive eigenvalues of an operator $e^{-\mathcal{D}t}$ $$\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}{t}}$$ has an asymptotic expansion around $t=0$? If it exists, then is it possible to induce a relation between coefficients?

I know that the proof for the case of heat operator follows from the construction of heat kernel. But I wonder that the same construction can be applied to the Dirac operator.

Thank you for your time and effort.

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We define the spectral zeta function of the Laplacian $\Delta$ by $$ \zeta_{\Delta}(s) = \sum_{\lambda\in\mathrm{Sp}(\Delta)}\lambda^{-s}. $$ It is known that $\zeta_{\Delta}(s)$ converges absolutely over $\mathrm{Re}(s)>\frac{n}{2}$, and has an analytic continuation to a meromorphic function over all complex plane. It can be expressed by $$\zeta_{\Delta}(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\lambda{t}}dt,$$ where $\Gamma(s):=\int_{0}^{\infty}t^{s-1}e^{-t}dt$ is the Gamma function.

From the identities $$ \Gamma(s)\lambda^{-s}=\int_{0}^{\infty}t^{s-1}e^{-\lambda{t}}dt $$ and $$ \Gamma(2s)\lambda^{-s}=\int_{0}^{\infty}t^{2s-1}e^{-\sqrt{\lambda}t}dt, $$ we deduce the following relation $$ \zeta_{\Delta}(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\lambda{t}}dt \\ =\frac{1}{\Gamma(2s)}\int_{0}^{\infty}t^{2s-1}\sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}t}dt. $$ Since the spectral zeta function $\zeta_{\Delta}(s)$ is meromorphic over a complex plane, $i.e.$ is analytic except for discrete singularities, the series $$ \sum_{\lambda\in\mathrm{Sp}(\Delta)}e^{-\sqrt{\lambda}t} $$ has an asymptotic expansion.

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  • $\begingroup$ If the spectral zeta function is meromorphic over a complex plane and $\zeta_\Delta(s) \Gamma(2s)$ satisfies the conditions of some tauberian theorem then the series has an asymptotic expansion given by the poles of $\zeta_\Delta(s) \Gamma(2s)$. (try with $-1 + \sum_n e^{-n^2 x}$ whose Mellin transform $\Gamma(s) \zeta(2s)$ has only one pole) $\endgroup$
    – reuns
    Commented Dec 8, 2018 at 20:24

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