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

If we denote by $\mathcal{D}$ the Dirac operator whose square coincides with the Laplacian, the trace of an operator $e^{-\mathcal{D}t}$ 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.

I am cross-posting it at here too.