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Let $S$ be a finite-dimensional complex vector space and $E$ a $\mathrm{End}(S)$-module (you may assume that the complex vector space associated to $E$ is finite-dimensional). In the following $S$ is viewed as an $\mathrm{End}(S)$-module with the obvious action.

Is it correct that the function \begin{align} \mathrm{Hom}_{\mathrm{End}(S)}(S,E)\otimes S&\to E\\ w\otimes s&\mapsto w(s) \end{align} is bijective? AFAIU this is claimed without proof in proposition 3.27 of Heat Kernels and Dirac Operators.

Filippo
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    Related: https://math.stackexchange.com/q/541776/572841 – Filippo Jul 09 '23 at 09:37
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    You should have mentioned that $E$ and $S$ (and therefore $A$) are finite-dimensional vector spaces over a field. Please include that kind of context when asking a question, it makes a huge difference. – Captain Lama Jul 09 '23 at 15:37
  • @CaptainLama Thank you for the comment. In addition, I believe that the problem boils down to the case where $A=\mathrm{End}(S)$ and $A\to\mathrm{End}(S)$ is the identity, i.e. I think that I can erase the $A$ from my question. – Filippo Jul 09 '23 at 18:43
  • Is the tensor product one of modules, or what? – FShrike Jul 09 '23 at 19:44
  • @FShrike Good question. I think that no subscript indicates that we simply consider the tensor product of complex vector spaces...But maybe both works? – Filippo Jul 09 '23 at 20:22

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The result is trivially true if $S=0$ (this case is very weird). Assume $S$ is nonzero.

Morita theory says that the category of $\mathrm{End}(S)$-modules is equivalent to the category of complex vector spaces, and in one direction the equivalence is given by $V \mapsto S \otimes_{\mathbb C} V$. In particular, $E$ is isomorphic as a module to $S \otimes_{\mathbb C} \mathbb C^n \cong S^n$ for some nonnegative integer $n$.

Since $\mathrm{Hom}_{\mathrm{End}(S)}(S,S^n) \cong \bigoplus_n \mathrm{Hom}_{\mathrm{End}(S)}(S,S)$, the problem reduces to the case $E=S$. In this case $\mathrm{Hom}_{\mathrm{End}(S)}(S,S) \cong \mathbb C$, so

  1. The tensor product is over $\mathbb C$
  2. The result is true, since the case of $E=S$ is easy to verify directly.
Ben Williams
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  • "Morita theory says that..." - I asked a question concerning this three weeks ago. In case that you are willing to elaborate, an answer would be much appreciated. Here is the link to the question: https://math.stackexchange.com/q/4733129/572841 – Filippo Jul 27 '23 at 08:58