For a semisimple Lie Algebra $\mathfrak{g}$ with Cartan Subalgebra $\mathfrak{t}$, let $V(\lambda)$ be the unique irreducible highest weight module with highest weight $\lambda$.

I am asked to show that the dual representation $V(\lambda)^*$ is irreducible, and to give a condition for $V(\lambda)$ to be self dual.

For the first part, my thoughts are that if I can take a basis of $V(\lambda)^*$ and show that the orbit of one of them under the action of $\mathfrak{t}$ contains all of them, then maybe I'd be done. But perhaps for this I would actually have to show it for any general basis?

For the second part I have heard that the condition is whether or not $-1$ is in the Weyl group, but as my understanding of Lie Algebras is quite weak I'm not sure why the Weyl group is important here. I would appreciate any help that you might be able to offer, thank you!

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    $\begingroup$ For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual. $\endgroup$ – Tobias Kildetoft Jan 7 at 7:43
  • $\begingroup$ @TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $\forall U^* \in V(\lambda)^*, \; \exists U \in V(\lambda) $ such that $U^* \cong V(\lambda)/U$ if so I can’t quite see why this is the case? $\endgroup$ – user366818 Jan 7 at 16:34
  • $\begingroup$ Dualize the short exact sequence $U\to V(\lambda)\to V(\lambda)/U$. $\endgroup$ – David Hill Jan 8 at 17:26

To deduces irreducibility of $V(\lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $W\subset V(\lambda)^*$ consider its annihilator $U:=\{v\in V(\lambda):\forall\phi\in W:\phi(v)=0\}$. This is clearly a linear subspace in $V(\lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(\lambda)$ and $U=\{0\}$ are the only possibilities for $U$, linear algebra shows that $W=\{0\}$ or $W=V(\lambda)^*$.

Concerning the highest weight of $V(\lambda)^*$ you take a basis for $V(\lambda)$ consisting of weight vectors and consider the dual basis of $V(\lambda)^*$ to conclude that the weights of $V(\lambda)^*$ are exactly the negatives of the weights of $V(\lambda)$. In particular, the highest weight of $V(\lambda)^*$ is $-\mu$, where $\mu$ is the lowest weight of $V(\lambda)$. It can be shown that $\mu=w_0(\lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(\lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(\lambda)=-\lambda$ and hence $V(\lambda)\cong V(\lambda)^*$ without $w_0$ being $-id$).

  • $\begingroup$ Thank you for this response! What do you mean by "invariance of $W$" here? $\endgroup$ – user366818 Jan 10 at 1:04
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    $\begingroup$ What I mean is that if the action of any element of $\mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself. $\endgroup$ – Andreas Cap Jan 10 at 4:58
  • $\begingroup$ I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra? $\endgroup$ – user366818 Jan 10 at 12:41
  • $\begingroup$ Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(\lambda)$ which is the important part. $\endgroup$ – user366818 Jan 10 at 12:46
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    $\begingroup$ This is just linear algebra: For a proper subspace $W\subset V(\lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(\lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $U\neq\{0\}$ for $W\neq V(\lambda)^*$. $\endgroup$ – Andreas Cap Jan 10 at 13:25

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