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Let $S$ and $W$ be complex and finite-dimensional vector spaces. In the following we consider the obvious action of $\mathrm{End}_{\mathbb C}(S)$ on $S$ and $S\otimes W$. Since $$\forall A\in\mathrm{End}_{\mathbb C}(W):1\otimes A\in\mathrm{End}_{\mathrm{End}_{\mathbb C}(S)}(S\otimes W)$$ we can ask whether the function \begin{align} \mathrm{End}_{\mathbb C}(W)&\to\mathrm{End}_{\mathrm{End}_{\mathbb C}(S)}(S\otimes W)\\ A&\mapsto 1\otimes A \end{align} is bijective. Injectivity should be clear, so the main issue should be surjectivity.

Motivation: AFAIU this is claimed without proof in proposition 3.27 of Heat Kernels and Dirac Operators.

Related: Spinor representation and Clifford modules

Filippo
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    Did you try a dimension argument? – Maximilian Simon Jul 10 '23 at 09:10
  • @MaximilianSimon You mean showing that $V:=\mathrm{End}{\mathbb C}(W)$ and $W:=\mathrm{End}{\mathrm{End}{\mathbb C}(S)}(S\otimes W)$ viewed as $\mathbb C$-vector spaces have the same dimension and then using that an injective function $f\in L{\mathbb C}(V,W)$ is automatically surjective as well? – Filippo Jul 10 '23 at 09:15
  • Yes, I think that should work. – Maximilian Simon Jul 10 '23 at 09:25
  • @MaximilianSimon My naive approach would be to assume a basis $A=(A_1,\ldots,A_n)$ of $\mathrm{End}{\mathbb C}(W)$ and trying to show that $B=(1\otimes A_1,\ldots,1\otimes A_n)$ is a basis of $\mathrm{End}{\mathrm{End}_{\mathbb C}(S)}(S\otimes W)$ - but showing that $B$ spans the entire space boils down to the initial problem, doesn't it? – Filippo Jul 10 '23 at 09:30
  • Wouldn't it suffice to show that if $T=\sum_iA_i\otimes B_i$ is in $\mathrm{End}{\mathrm{End}{\mathbb C}(S)}(S\otimes W)$, then $\forall i:A_i=1$ (maybe by contradiction)? – Filippo Jul 10 '23 at 09:41

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