I am currently reading the book Heat Kernels and Dirac Operators. The definition of the index of a Dirac operator uses the fact that the kernel is finite-dimensional. Unfortunately the proof is rather short:
Proposition $3.37$. If $D$ is a Dirac operator on a compact manifold $M$, then the kernel of $D$ is finite-dimensional.
Proof: The operator $D^2$ has a finite-dimensional kernel, and hence so does the Dirac operator $D$.
Furthermore I found another hint on page $4$ of the book:
Smooth kernels are trace-class, from which we see that the kernel of a Dirac operator is finite-dimensional,
In summary, I believe that the reasoning is as follows: By definition $D^2$ is a generalized Laplacian, and all generalized Laplacians (on a compact manifold) have a finite-dimensional kernel, because they admit a heat kernel and kernels are trace class.
It is not clear to me why the existence of a heat kernel implies a finite-dim. kernel, can someone elaborate please?
