In page 25, Theorem 2.7 Etingof wrote

$A \cong A^*$ as $A$-modules where $A^*:= Hom_k(A,k)$.

I know $A \hookrightarrow A$ by left action, $a \cdot x:= ax$ and $A \hookrightarrow A^*$ by $a \cdot l := la^{-1}$

What is the explicit isomorphism here?

I thought the by making a choice of basis $\{ e_i \}$ the map $e_i \mapsto e'_i$, $\phi:A \rightarrow A'$, is an $A$-module isomoprhism.

But this doesn't seems to be the case, as suppose $a \cdot e_j = \sum t_{ij} e_i$, and $a^{-1} \cdot e_j = \sum t'_{ij} e_i$. Then

$$\phi( a \cdot e_i) (e_j) = t_{ij}, \text{ and } (a \cdot \phi(e_i))(e_j) = e'_i(a^{-1} e_j) = t'_{ij}.$$

So $\phi(a \dot e_i) \not= a \cdot \phi(e_i)$ in general. What's wrong?

If $A$ is a matrix algebra over a ring $R$, then the *trace* defines a linear map $A \to R$ given by $a \mapsto \mathrm{tr}(a)$. This induces a symmetric non-degenerate linear form $$(a,b)=\mathrm{tr}(ab)$$ on $A$, which allows you to identify $A$ with its dual as an $R$-module.

The $A$-action on $A^*$ is defined by the formula $$(af)(b)=f(a^t b)$$ where $a^t$ is the transpose of $a$. In order to make the identification of $A$ with $A^*$ intertwine left multiplication of $A$ on itself with this action, one maps $$A \to A^*, \quad a \mapsto t_a=(b \mapsto \mathrm{tr}(a^t b)).$$

April 16, 2018 12:19 PM

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