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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.

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That isomorphism is wrong. For finite-dimensional vector spaces the LHS is $A^{\ast} \otimes B \otimes C$ while the RHS is $A \otimes B^{\ast} \otimes C$. – Qiaochu Yuan Aug 14 '11 at 22:54
up vote 2 down vote accepted

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 $\lambda$.

  • If $V$ is a finite dimensional representation of $\mathbb g$, $\lambda$ a weight, and $w$ an element of the Weyl group, then $\dim V_{\lambda}=\dim V_{w\lambda}$. We can see this by noting that for each root $\alpha$, we have a copy of $\mathfrak{sl}_2$ which acts on $\bigoplus_i V_{\beta+i\alpha}$ for every weight $\beta$, and standard rep theory for $\mathfrak{sl}_2$ shows that the equality holds when $w=s_{\alpha}$ is a reflection.
  • If $V$ is a finite dimensional representation and $\lambda$ is a weight, then $\dim V_{\lambda}=\dim V^*_{-\lambda}$. This can be shown by taking a basis of weight vectors. If $v$ is a weight vector of weight $\lambda$, the corresponding dual vector $v^*$ has weight $-\lambda$.
  • Irreducible finite dimensional $\mathfrak g$-reps are classified by integral dominant weights via highest weight vectors.
  • If $V$ is irreducible of highest weight $\lambda$, the weights in $V$ will be the convex hull of the weights $w\lambda$, for $w$ in the Weyl group.

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.

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thank you. But it seems that the orders are different in the following: $U\otimes L_{\lambda}** = hom_{\mathfrak{g}}(L_{\lambda}^*, U)$ and $hom(V, W) = V^* \otimes W$. – LJR Aug 15 '11 at 14:28
Yes, because the tensor product is over $\mathbb C$. You will notice that where I use such an isomorphism, the hom is over $\mathbb C$ as well. It is possible to tensor over $\mathfrak g$, but the result will no longer be a representation. – Aaron Aug 15 '11 at 17:40
@user9791: Oops, I interpreted "order" as "order of magnitude", when I think you meant which came first. Tensor products between vector spaces or symmetric modules over commutative rings do not depend on order (and over chain complexes or graded commutative rings, there is a cannonical isomorphism). For a pair of a left and a right module over a noncommutative ring, there is only one way to have the tensor product make sense. Order does matter for bimodules, but this is beyond our scope. – Aaron Aug 15 '11 at 19:50

For (1): You undoubtedly already know that the lowest weight of $L_\lambda$ is $w(\lambda)$: The longest element $w$ maps all the positive roots to negative roots, and hence reverses the partial order. Put this together with the fact that the weights of $V^*$ are the negatives of the weights of $V$, and that negation also reverses the partial order of the weight lattice, and you get that ...

Edit: Oh, and $L_\lambda^*$ is simple, because it's the dual of a simple f.d. module.

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