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.

  • 3
    $\begingroup$ 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 $t > 0$. $\endgroup$ – Phillip Andreae Nov 15 '18 at 1:28
  • $\begingroup$ ... which is the reason that led Dirac to predict the existence of the positron $\endgroup$ – 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.

Also, you should not cross post. I think this is more suitable for Math.SE.

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