# Asymptotic expansion of heat operator $e^{-\Delta{t}}$ and $e^{-\mathcal{D}t}$ of Dirac operator

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.

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.
• 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) – reuns Dec 8 '18 at 20:24