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Today I came across two possible canonical maps in the context of representation theory from reading stackexchange and a friend's email. However, I don't really know what they are (and have thus far not found satisfactory information about them): 1) $hom_k(V \otimes V,k ) \rightarrow hom_k(V, V^*)$ 2) $(V^*)^* \otimes V \rightarrow hom_k(V, V^*)$

In both cases $V$ is a finite dimensional representation of some (finite) group $G$. In 2) $k$ maybe assumed algebraically closed. Perhaps 1) is a result of exterior derivatives (my guess). Any pointer is welcomed.

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1 Answer 1

up vote 4 down vote accepted

You know there is a natural isomorphism $\hom_k(U\otimes V,W)\to\hom_k(U, \hom_k(V,W))$. For (1), take $U=V$ and $W=k$.

As for the second one, there is no such natural isomorphism: on one hand, $(V^*)^*$ is naturally isomorphic to $V$, so $(V^*)^*\otimes V$ is naturally isomorphic to $V\otimes V$. On the other hand, $\hom(V,V^*)$ is naturally isomorphic to $\hom(V\otimes V,k)$, or, in other words, to $(V\otimes V)^*$. Now there is no natural isomorphism from $V\otimes V$ to $(V\otimes V)^*$.

NB: In fact, $(V\otimes V)^*$ and $V\otimes V$ are in general not even isomorphic, much less naturally isomorphic (naturality is problematic, as the two depend with different variances on $V$ anyways...) One way to see this is to let $G$ be a cyclic group generated by an element $g$ of order $100$, and letting $V$ be the one-dimensional representation of $G$ where $g$ acts by a primitive $100$th root of unity $\omega$. Then $V\otimes V$ is also one-dimensional, with $g$ acting by $\omega^2$, and you can surely describe $(V\otimes V)^*$ and see that it is not isomorphic to $V\otimes V$.

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