Highest weight of a representation of GL_n

Question

Let $G$ be the algebraic group $\mathrm{GL}_n$ (over $\mathbb{C}$ say). We let $T$ be the diagonal torus, $B$ the Borel subgroup of upper triangular matrices and $U$ its unipotent radical. We write $B^{\mathrm{op}}$ and $U^{\mathrm{op}}$ for the opposite groups. We identify $X^\ast(T)$ with $\mathbb{Z}^n$ in the usual way and we fix the set of positive root given by $B$. If $\lambda \in X^\ast(T)$ is a weight of $T$ we extend it to $B$ by $\lambda(U)=1$ (and similarly for $B^{\mathrm{op}}$).

Let $\lambda$ be a dominant weight and consider the representation of $G$ given by $$ V_\lambda := \left\{ f \colon G \to \mathbb{A}^1 \mbox{ such that } f(b^-g)=\lambda(b^-)f(g) \mbox{ for all } (b^-,g) \in B^{\mathrm{op}} \times G \right\}, $$ where $G$ acts (on the left) on $V_\lambda$ via $(g.f)(x)=f(xg)$. Then $V_\lambda$ is (if I understand correctly the theory) the irreducible representation of $G$ with highest (w.r.t. $B$) weight $\lambda$.

Consider now this other representation $$ W_\lambda := \left\{ f \colon G \to \mathbb{A}^1 \mbox{ such that } f(gb)=\lambda(b)f(g) \mbox{ for all } (g,b) \in G \times B \right\}, $$ where $G$ acts (on the left) on $W_\lambda$ via $(g.f)(x)=f(g^{-1}x)$.

What is the highest weight of $W_\lambda$? I guess it is $\lambda'$, where, if $\lambda$ corresponds to $(\lambda_1,\ldots,\lambda_n)$ then $\lambda'$ corresponds to $(-\lambda_n,\ldots,-\lambda_1)$, but I am not sure. Can you provide an explicit isomorphism between $W_\lambda$ and $V_{\lambda'}$? Thank you!

Answer 1

By Peter-Weyl, algebraic functions on $G$ are spanned by the matrix coefficients of simple representations, that is functions of the form $f_{v,w}(g)=\langle gv,w\rangle$ for $v\in V$ and $w\in V^*$ for $V$ an irrep. Your space $V_\lambda$ is the space where $w$ satisfies $(b^-)^{-1}w=\lambda(b^-)w$. That is, $w$ is a lowest weight vector with weight $-\lambda$. Taking dual, we see that $v$ has to live in a simple rep with highest weight $\lambda$.

On the other hand, $W_\lambda$ is the space where $v$ is highest weight of weight $\lambda$, so indeed $W_\lambda\cong V_\lambda^*$, and what you have written is the highest weight of the dual space.

Note that indeed, as suggested in L Spice's comment, taking transpose does give a vector space isomorphism between $W_\lambda$ and $V_\lambda$; this is not an isomorphism of representations. Instead, it twists the action by the automorphism $A\mapsto (A^T)^{-1}$, which switches the a rep of highest weight $\lambda$ to one of lowest weight $-\lambda$, which thus has highest weight $\lambda'$.