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Let $A$ be a real commutative unital algebra and $M\in A^{n\times n}$ a nilpotent matrix. Is it true that \begin{equation*} (\det(M/\sinh{M}))^{1/2}=\det((M/\sinh{M})^{1/2})? \end{equation*} The square roots are defined in terms of the taylor expansion of $(1+x)^{1/2}$, similarly for $M/\sinh M$.

Motivation: This would give a convenient way to calculate the $\hat{A}$-genus form discussed in this question. In fact this seems to give the correct result for $4$-dim. manifolds: Since \begin{equation*} \sqrt{\frac{x}{\sinh{x}}}=1-\frac{1}{2}\frac{1}{6}x^2+\ldots \end{equation*} we obtain \begin{equation*} \sqrt{\frac{\Omega/2}{\sinh{\Omega/2}}}=1-\frac{1}{6}\frac{1}{8}\Omega^2. \end{equation*} Furthermore $\mathrm{det}(1+A)=1+\mathrm{tr}(A)+\ldots$ and hence \begin{equation*} \hat{A}(\nabla)=1-\frac{1}{6}\frac{1}{8}\mathrm{tr}(\Omega^2) \end{equation*} This is consistent with $(1.36)$ in Heat Kernels and Dirac Operators...

Filippo
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  • If $M$ is nilpotent then $M/\sinh M$ is unipotent so both sides ought to be $1$, no? – coiso Sep 10 '23 at 07:34
  • @coiso You mean that if there is some $k$ such that $M^k=0$ (i.e. $M$ is nilpotent), then $\frac{M}{\sinh{M}}=1$? – Filippo Sep 10 '23 at 07:40
  • I am not assuming that $M^2=0$ if that is the issue. – Filippo Sep 10 '23 at 07:41
  • No, I'm saying the determinant of a unipotent matrix ought to be $1$. Is that wrong? – coiso Sep 10 '23 at 07:46
  • @coiso I see, thank you. Regarding your question: This contradicts the result mentioned in the end of my question, doesnt it? – Filippo Sep 10 '23 at 08:14
  • Do you mean $\det(I+N)=1+{\rm tr}(N)+\cdots$? You seem to have forgotten the "$I+$" in the result. – coiso Sep 10 '23 at 08:20
  • @coiso Fixed, thank you for the remark! – Filippo Sep 10 '23 at 08:24
  • @coiso I required $M$ to be nilpotent because 1) this is the case in the application I mentioned and 2) this allows us to make sense of the RHS (right-hand side) without needing to worry about convergence. But I think that the LHS only makes sense as a limit and this is a subtle issue even if $A$ is a finite-dim. vector space, right? – Filippo Sep 10 '23 at 09:15
  • So I currntly really think that the RHS should be used to the define the genus, whereas the book seems to use the LHS. – Filippo Sep 10 '23 at 09:18
  • I don't know what you mean by genus in this context. But if $M$ is nilpotent, then $B=\sqrt{M/\sinh M}$ is unipotent, and then $\det B=1$, so it's safe to say $\sqrt{\det B}=1$ (assuming we define $\sqrt{1+x}$ by the series you mentioned for nilpotent elements $x\in A$, which applies to $x=0$ in particular). Thus the LHS makes sense without appeal to limits (there is no notion of limits on a generic algebra $A$). – coiso Sep 10 '23 at 09:20
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    @coiso But I showed that $$\det((M/\sinh{M})^{1/2})=1-\frac{1}{2}\frac{1}{6}\mathrm{tr}(M^2)$$if $M^3=0$ in the last part of my question, didnt I? And this is the expected result, so are you really sure that instead$$\det((M/\sinh{M})^{1/2})=\det{B}=1?$$ – Filippo Sep 10 '23 at 09:24
  • You're right: a unipotent matrix over an arbitrary (unital, commutative) algebra needn't have determinant $1$, for instance the $1\times1$ matrix $[1+\varepsilon]$ over $\Bbb R[\varepsilon]/(\varepsilon^k)$. – coiso Sep 10 '23 at 17:53
  • @coiso Thank you for your comment! My last comment was not entirely correct though, the step$$\mathrm{det}(1+R)=1+\mathrm{tr}(R)$$also involved the fact that the entries of $R$ are nilpotent of degree two. – Filippo Sep 11 '23 at 00:50
  • Anyways, I think that it really boils down to the fact that our commutative unital ring has non-zero nilpotent elements. Indeed the $1\times 1$ matrix $1+\epsilon$ is a counterexample whenever $\epsilon$ is such an element, isn't it? – Filippo Sep 11 '23 at 00:54

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