Chern character of a super-connection (Heat kernels and Dirac operators)

Question

Let $A$ be a super-connection on a super-bundle $E\to M$, then the differential form \begin{equation} \mathrm{ch}(A)=\mathrm{Str}(e^{-A^2}) \end{equation} is called the chern character of $A$ on page 50 of Heat kernels and Dirac operators. First of all I think that we should replace $A^2$ with the curvature \begin{equation} F\in\Omega(M,\mathrm{End}(E))=\Gamma(M,\Lambda(TM^*)\otimes\mathrm{End}(E)), \end{equation} associated to $A$ through proposition 1.38 in order to make sense of the RHS . But then there is a still a problem: I think that the authors assumed that $F$ is nilpotent (see below), but unlike for the curvature of a covariant derivative there is actually no guarantee for this, is there?

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1. I guess that we could still make sense of $e^{-F}$ as a limit (right?), but I have a bad feeling about this and indeed I think that I found a solid argument:
2. The proof that the chern character is independent of the super-connection uses proposition 1.41, but there we only consider the characteristic form $\mathrm{Str}(f(A^2))$ associated to a polynomial $f$.

Answer 1

After further reading I found that the issue is actually addressed on page 49:

Short answer:

The chern character is indeed a limit and for now I assume that the authors checked that the transgression formula from proposition 1.41 still holds true.

Long answer: The definition of $f(F)$ when $f$ is not a polynomial is kind of hidden on page 49:

• If the grading on the bundle $E$ is the trivial one - i.e. $E^0=E$ and $E^1=\{0\}$ - and the super-connection is a connection, then the curvature $F$ is a section of $$\Lambda^2(TM^*)\otimes\mathrm{End}(E)\subset\Lambda(TM^*)\otimes\mathrm{End}(E)$$ (and the super-trace equals the trace). In particular $F$ is a nilpotent and we can allow $f(z)$ to be any power series in $z$.
• If $A$ is a super-connection, we can allow $f$ to be a power series with infinite radius of convergence.