- This is the exercise problem 21.6: If $V=V(\lambda)$, $\lambda\in\Lambda^+$, prove that $V^*$ is isomorphic (as $L-$module) to $V(-\sigma\lambda)$, where $\sigma\in W$ is the unique element of $W$ sending $\Delta$ to $-\Delta$.
Here $V(\lambda)$ is an irreducible finite dimensional standard cyclic module of weight $\lambda$; $\Lambda^+$ is the set of domiant integral linear functions; $W$ is the Weyl group; $\Delta$ is a base of the root system.
My attempt is to show that both $V^*$ and $V(-\sigma\lambda)$ has the same dimension and so they are isomorphic. I tried to use the theorem in section 21 that saying the set of weights $\Pi(\lambda)$ corresponding to $V(\lambda)$ is permuted by $W$, with $\dim V_\mu=\dim V_{\sigma\mu}$ for $\sigma\in W$. But still struggling how to proceed, any ideas or hints are welcome.
Showing that two representations of a Lie algebra are of the same dimension is not enough to prove that they are equal. But two finite dimensional representations of a finite dimensional semisimple complex Lie algebra which restrict to the same representation of the Cartan subalgebra are isomorphic to one another.
That is, what you should actually check is that the weight space decomposition of $V(\lambda)^*$ with respect to the Cartan is the same as that of $V(-\sigma \lambda)$. To do this, just observe that for a linear form $f \in V(\lambda)^*$ and a vector $v \in V(\lambda)$ of weight $\mu$ with respect to the Cartan, we have $$(hf)(v)=-f(hv)=-\mu(h) f(v)$$ for all $h$ in the Cartan subalgebra. It follows that the set of weights of $V(\lambda)^*$ is the negative of the set of weights of $V(\lambda)$. In particular, the largest dominant weight of $V(\lambda)^*$ is $-\sigma \lambda$, and so since $V(\lambda)^*$ is irreducible it must be equal to $V(-\sigma \lambda)$.