Let $\mathfrak{g}$ be a simple Lie algebra. Let $M_{\lambda}$ be the Verma module over $\mathfrak{g}$ of highest weight $\lambda$ and $L_{\lambda}$ be the irreducible $\mathfrak{g}$-module of highest weight $\lambda$.
For a $\mathfrak{g}$-module $U$, the dual $\mathfrak{g}$-module $U^*$ is defined by the rule $x|_{U^*}=(-x|_U)^*$ for any $x\in \mathfrak{g}$. Let $w$ be the longest element of the Weyl group of $\mathfrak{g}$. How to show the following?
(1) If $\lambda$ be a dominant integral weight. Then the $\mathfrak{g}-module$ $L^*_{\lambda}$ dual to $L_{\lambda}$ is isomomorphic to $L_{-w(\lambda)}$.
(2) $Hom_{\mathfrak{g}}(\mathbb{C}, U\otimes L_{\lambda}) = Hom_{\mathfrak{g}}(L^*_{\lambda}, U)$?
I know that $Hom(A, B\otimes C) = Hom(Hom(A, B), C)$. But it seems that we can not use this to show (2).
Thank you very much.
Best Answer
Your statement of the hom-tensor adjunction is incorrect, and we need to use slightly more than the adjunction to prove what you want. In particular, we need that if $V$ is finite dimensional, then $V^**\cong V$ and $\hom(V,W)\cong V^*\otimes W$. Using the fact that $L_{\lambda}$ is finite dimensional, we have $$\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda})\cong\hom_{\mathfrak g}(\mathbb C, U\otimes L_{\lambda}^{**}) \cong \hom_{\mathfrak g}(\mathbb C, \hom_{\mathbb C}(L_{\lambda}^*,U))$$ Applying the adjunction, this is isomorphic to $\hom_{\mathfrak g}(\mathbb C\otimes L_{\lambda}^*,U)\cong \hom_{\mathfrak g}(L_{\lambda}^*,U)$.
For part (1), we need a few ideas (the upshot of which are given by Jyrki). If $V$ is a representation, I will let $V_{\lambda}$ denote the weight space of weight $\lambda$.
Thus, it suffices to show that $L_{\lambda}^*$ has highest weight $-w_0(\lambda)$ where $w_0$ is the longest length word in the Weyl group. As Jyrki said, this follows from the fact that $w_0(\alpha)<0$ for every positive root $\alpha$, as it will be the only dominant weight of the form $-w\lambda$ for $w$ in the Weyl group.