McKean-Singer formula in Heat Kernels and Dirac Operators book

Question

I'm reading "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne. The setting is: $E \to M$ is a $\mathbb{Z}_2$-graded vector bundle on a compact Riemannian manifold $M$ and $D : \Gamma(M, E) \to \Gamma(M, E) $ is a self-adjoint Dirac operator (especially $D^2$ is Laplace). We denote by $D^{\pm}$ the restrictions of $D$ to $\Gamma(M, E^{\pm})$. I have some troubles to understand the last part in the second variant of the proof of Theorem 3.50 McKean-Singer:

It is clamed that the operator $d(e^{-tD^2})/dt$ has a smooth kernel equal to $-D^2e^{-tD^2}$. This don't make any sense to me. Recall if $P: E\to E$ is an integral operator and $s \in E_x, x \in M $ then it's kernel $K: M \times M \to \mathbb{R}$ is defined by equation

$$ Ps_{\vert x}= \int_{\{x \} \times M} K(x,y) s_{\vert y} dy$$

Especially, $K$ maps from $M \times M$, while $-D^2e^{-tD^2}$ (for fixed $t$) maps from $E$. Any idea how $-D^2e^{-tD^2}$ can be interpreted as an integral kernel?

Answer 1

The assertion is supposed to be that $d(e^{-tD^2})/dt$ has the same smooth kernel as $-D^2 e^{-tD^2}$, i.e. they are the same operator. This is because $e^{-tD^2}$ is the solution operator to the heat equation: $\frac{du}{dt} = -D^2 u$. I'd be comfortable saying this is the definition of $e^{-tD^2}$, but you might prefer to construct $e^{-tD^2}$ using the pseudodifferential calculus in which case this fact would follow from the spectral theorem.