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Suppose $U,V,W$ are representations of a Lie algebra $\mathfrak g$. I want to show that there is an isomorphism between $Hom_\mathfrak g(V\otimes W, U)$ and $Hom_\mathfrak g(V, U\otimes W^*)$.

I try to put $\phi\mapsto (v\mapsto \phi(v\otimes 1)\otimes g$), but I cannot find a suitable $g\in W^*$. Am I on the right track or it may be better to define an isomorphism on basis?

Also, what is the intuition behind the isomorphism? I cannot see how does $\phi$ being intertwining forces there to be an isomorphism.

1 Answers 1

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Using $\operatorname{Hom}(W,U) \cong U \otimes W^{*}$, you are looking for an isomorphism between $\operatorname{Hom}_{\mathfrak{g}}(V \otimes W, U)$ to $\operatorname{Hom}_{\mathfrak{g}}(V, \operatorname{Hom}(W,U))$. If we were to delete the $\mathfrak{g}$ part, we have a natural isomorphism coming from the universal property of the tensor product. Namely, we have a natural bijection between bilinear maps $V \times W \rightarrow U$ (which can be naturally identified with $\operatorname{Hom}(V,\operatorname{Hom}(W,U))$) and maps from the tensor product $V \otimes W$ to $U$. Hence, we have a natural candidate so you only have to check this indeed gives you an isomorphism of $\mathfrak{g}$-modules.