# 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

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.

• I suspect that the Dirac operator $\mathcal{D}$ generally has a spectrum that is unbounded in both directions, i.e., it has arbitrarily positive and arbitrarily negative eigenvalues. The arbitrarily negative eigenvalues would mean that the "heat trace" $\sum_\mu e^{-t \mu}$ (sum over the eigenvalues of $\mathcal{D}$) does not converge for any $t > 0$. – Phillip Andreae Nov 15 '18 at 1:28
• ... which is the reason that led Dirac to predict the existence of the positron – lcv Dec 17 '18 at 7:52

The comment by Philip is correct. Let $$M= \mathbb{S}^{1}$$ and let $$D=i\frac{\partial}{\partial x}$$, then $$D^{2}=-\Delta$$ has spectrum in $$\mathbb{N}$$ with eigenfunctions $$e^{inx}$$. The Dirac operator $$D$$ have spectrum in $$\mathbb{Z}$$, since $$in*i=-n\in \mathbb{Z}$$. So the heat trace cannot be defined in general unless you use some regularization procedure.