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Suppose $k$ be a field of characteristic $0$, and $R$ be an unital $k$-algebra. Then, does it hold $$ R \otimes_k \mathrm{End}_R(W) \cong \mathrm{End}_k(W)? $$ for an $R$-module $W$?

This is an attempt to understand the theory of Clifford modules, since there is no proof of this on any literatures. I'm especially interested in the case, $$ R = \mathbb{C}l(V)\\ k = \mathbb{C}\ $$

Might be related : This post, but I couldn't see the direct relation.

ChoMedit
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1 Answers1

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The identity in the title may be too general, but I can show the case you are interested in:

Let $V$ be an even-dimensional Euclidean vector space. The complexification of $C(V)$ (a complex algebra) has the special property that it is isomorphic to $\mathrm{End}(S)$ for some complex vector space $S$, see proposition 3.19 in [1]. So we want to show that $$\mathrm{End}(E)\cong \mathrm{End}(S)\otimes\mathrm{End}_{\mathrm{End}(S)}(E)$$ for an $\mathrm{End}(S)$-module $E$. The key observation is that such an $E$ is isomorphic to the $\mathrm{End}(S)$-module $S\otimes W$ for some complex vector space $W$ (see this answer). Hence $$\mathrm{End}(E)\cong \mathrm{End}(S\otimes W)\cong \mathrm{End}(S)\otimes\mathrm{End}( W)$$ and now the desired result follows from the fact that $$\mathrm{End}( W)\cong \mathrm{End}_{\mathrm{End}(S)}(S\otimes W)\cong \mathrm{End}_{\mathrm{End}(S)}(E).$$

[1] Heat Kernels and Dirac Operators

Filippo
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  • In proposition 3.27, How could we show that this is an isomorphism? Is this because injectivity, and apply dimension counting? Then, how do we know $ \mathrm{dim} \mathrm{End}_{C(V)}(S, E)= \mathrm{dim} E/\mathrm{dim} S$? – ChoMedit Jan 02 '24 at 08:04
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    @ChoMedit You can prove that $S\otimes\mathrm{Hom}_{C(V)}(S, E) \cong E$ by using the fact that there exists a complex vector space $W$ such that $E\cong S\otimes W$ as $\mathrm{End}(S)$-modules. This may seem circular, but apparently the latter result is proven in "Morita theory", see this answer. – Filippo Jan 02 '24 at 08:12
  • @ChoMedit Also I think that the notation $\mathrm{End}{C(V)}(E)$ can be misleading, since commuting with $C(V)$ only implies $\mathbb{R}$-linearity (right?). So it should be emphasized that we actually consider a subspace of $\mathrm{End}(E)=\mathrm{End}\mathbb{C}(E)$. Then the real vector space $C(V)\otimes\mathrm{End}{C(V)}(E)$ has an obvious complex structure and$$\mathrm{End}(E)\cong C(V)\otimes\mathrm{End}{C(V)}(E)$$as complex algebras. – Filippo Jan 02 '24 at 09:53
  • It is important to note that the proposition 3.19 you cite only holds for even dimensional vector spaces. – Nicholas Todoroff Jan 02 '24 at 14:38
  • @NicholasTodoroff Is $\mathrm{dim}(V)=0$ also allowed? What is $C(V)$ in that case? – Filippo Jan 02 '24 at 14:44
  • Define the Clifford algebra as a quotient of the tensor algebra. If $V = {0}$ then $T(V) = \mathbb R$: its universal property guarantees the existence of a homomorphism into any other algebra, the image of $1$ under any homomorphism has to be $1$, and nontrivial algebras exist. The Clifford ideal is ${0}$ because $V$ is, so the quotient of $T(V)$ by this is just $T(V)$, so $C(V) = \mathbb R$. Finally, $\dim(S) = 2^{\dim(V)/2} = 1$ so the proposition is satisfied. – Nicholas Todoroff Jan 02 '24 at 15:03
  • @NicholasTodoroff So that case is also allowed. Interesting, thanks :D – Filippo Jan 02 '24 at 15:33