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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!

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  • $\begingroup$ In fact $W_\lambda$ is isomorphic to $V_\lambda$ by the map $(f \in V_\lambda) \mapsto ((g \mapsto f(g^\top)) \in W_\lambda)$. In general, parabolic induction only cares about the Levi component, not the specific parabolic subgroup with that Levi component. $\endgroup$ – LSpice Mar 29 '17 at 17:14
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    $\begingroup$ @LSpice That map doesn't commute with the G-action. $\endgroup$ – Ben Webster Mar 29 '17 at 17:26
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    $\begingroup$ Why this respect the $G$-action? $\endgroup$ – Ricky Mar 29 '17 at 17:30
  • $\begingroup$ @BenWebster (and Ricky), of course you're right. Sorry about that. $\endgroup$ – LSpice Mar 29 '17 at 17:48
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    $\begingroup$ @LSpice You do, of course, get an isomorphism of vector space, and so if you combine with my answer you get the standard fact that if you twist the action of G on a simple representation by inverse transpose, you get the dual representation. $\endgroup$ – Ben Webster Mar 29 '17 at 17:52
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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 $\lambda'$.

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