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From page $50$ of Heat Kernels and Dirac Operators:

If $E$ is a real vector bundle with covariant derivative $\nabla$ and curvature $F$, we associate to it its $\hat{A}$-genus form \begin{equation} \hat{A}(\nabla):=\mathrm{det}^{1/2}\bigg(\frac{F/2}{\sinh{F/2}}\bigg)\in\Gamma(M,\Lambda(TM^*)). \end{equation}

I feel a bit betrayed, because the $\hat{A}$-genus form seems to be defined in terms of another undefined expression.

  • Am I right in assuming that the definition is meant as follows: Fix $x\in M$ and choose a basis $e_1,\ldots,e_n$ of $E_x$, consider the matrix $\Omega\in \Lambda(T_pM^*)^{n\times n}$ with $\Omega^{i}{}_{j}\in \Lambda^2(T_pM^*)\subset\Lambda(T_pM^*)$ given by $$\Omega^{i}{}_{j}(X,Y)=e^iF(X,Y)e_j$$ and set $$\hat{A}(\nabla)_x:=\mathrm{det}^{1/2}\bigg(\frac{\Omega/2}{\sinh{\Omega/2}}\bigg)$$ (of course we have to check that the definition is independent of the basis). But even the definition of the last expression is not entirely clear:
  • First of all, note that the entries of $\Omega$ are in the even subalgebra of $\Lambda(T_pM^*)$, i.e. the determinant makes sense. I would have expected that$$\mathrm{det}^{1/2}\bigg(\frac{\Omega/2}{\sinh{\Omega/2}}\bigg):=\bigg(\mathrm{det}\bigg(\frac{\Omega/2}{\sinh{\Omega/2}}\bigg)\bigg)^{1/2}$$ but the determinant is an element of $\Lambda(T_pM^*)$ and defining the square root is a subtle issue. Setting$$\hat{A}(\nabla)_x:=\mathrm{det}\bigg(\bigg(\frac{\Omega/2}{\sinh{\Omega/2}}\bigg)^{1/2}\bigg)$$seems much more reasonable, as this solves the issue: On the one hand, the function \begin{align} \mathbb R&\to\mathbb R\\ x&\mapsto\sqrt{\frac{x}{\sinh{x}}} \end{align} has a nice taylor series and the other hand we dont even have to worry about convergence as $\Omega$ is nilpotent. (To do: Check consistency with $(1.36)$ for $8$-dim. manifolds.)
Filippo
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    When it comes to characteristic classes, such expression mean that you have a formal power series in some variable $x$, and you substitute $\Omega$ which, as you note, takes away any doubts about convergence because $\Omega$ is nilpotent. – Quaere Verum Sep 11 '23 at 11:34
  • @QuaereVerum So you agree that the definition should be$$\hat{A}(\nabla):=\mathrm{det}\bigg(\bigg(\frac{F/2}{\sinh{F/2}}\bigg)^{1/2}\bigg)$$instead of$$\hat{A}(\nabla):=\bigg(\mathrm{det}\bigg(\frac{F/2}{\sinh{F/2}}\bigg)\bigg)^{1/2}?$$ – Filippo Sep 12 '23 at 04:24
  • No, I do not agree with that. See here if you are confused https://www3.nd.edu/~lnicolae/ind-thm.pdf (page 34 and the preceding discussion). – Quaere Verum Sep 12 '23 at 10:21
  • @QuaereVerum Does he address my objection, namely that it is not clear how (and whether) the square root of the determinant is well defined? – Filippo Sep 12 '23 at 11:06
  • Also, I seem to obtain the correct result by first taking the square root and then the determinant. – Filippo Sep 12 '23 at 11:10
  • I guess that we may to try to define it by considering the taylor expansion of $\sqrt{1+x}$, but then we would either need to prove that the determinant is unipotent, or otherwise that we have convergence w.r.t. the standard topology, but this is a subtle issue since the taylor expansion only has a finite convergence radius (right?). – Filippo Sep 12 '23 at 11:14
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    Yes it is addressed, see lemma 1.2.10 and the discussion right below it. – Quaere Verum Sep 12 '23 at 13:14
  • @QuaereVerum Thank you! – Filippo Sep 12 '23 at 14:27

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