# Irreducible Dual Representation

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!

• 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. – Tobias Kildetoft Jan 7 at 7:43
• @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? – user366818 Jan 7 at 16:34
• Dualize the short exact sequence $U\to V(\lambda)\to V(\lambda)/U$. – 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 $$-id$$).
• Thank you for this response! What do you mean by "invariance of $W$" here? – user366818 Jan 10 at 1:04
• 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. – Andreas Cap Jan 10 at 4:58
• 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? – user366818 Jan 10 at 12:41
• Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(\lambda)$ which is the important part. – user366818 Jan 10 at 12:46
• 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 $W\neq V(\lambda)^*$. – Andreas Cap Jan 10 at 13:25