Suppose that $\mathfrak{g}$ is a finite dimensional semisimple Lie algebra and $\eta$ is its Cartanian algebra with the root system $R$ and a basis $S=\{\alpha_1,\cdots,\alpha_n\}$. Suppose that $\lambda\in\eta^*$ is a dominant integral weight(i.e. $\forall H_i,\lambda(H_i)$ is a non-negative integer, where $H_i$ satisfies $\alpha_i(H_i)=2$). Let $L(\lambda)$ be the irreducible finite dimensional representation of $\mathfrak{g}$ and $\tau$ be an element in the Weyl group such that $\tau S=-S$. Show that $L(\lambda)^*$ is isomorphic to $L(-\tau\lambda)$.

My idea is to compare the characters. It is easy to check that $ch(V^*)=\sum\limits_\mu\dim(V_\mu)e^{-\mu}$. But how can I go ahead?


Consider that, the vector space $L(\lambda)$ is the direct sum of the weight spaces $L(\lambda)_\mu$, where $\mu$ is of the form $\lambda-\sum\limits_{i=1}^nk_i\alpha_i$. Suppose $u$ is a primitive vector of $L(-\tau\lambda)$. Then we have the equation $hx_{\tau(\beta_1)}^{k_1}\cdots x_{\tau(\beta_m)}^{k_m}u=(-\tau\lambda+\sum\limits_{i=1}^mk_i\tau(\beta_i))x_{\tau(\beta_1)}^{k_1}\cdots x_{\tau(\beta_m)}^{k_m}u$. It is not difficult to find that for any weight $\mu$ of $L(\lambda)$, $-\tau\mu$ is a weight of $L(-\tau\lambda)$. By the construction of $L(\lambda)$(the quotient of the free module spanned by $x_{-\beta_1}^{k_1}\cdots x_{-\beta_m}^{k_m}v$ divided by the composition of all its proper submodules), we can verify that the map $x_{-\beta_1}^{k_1}\cdots x_{-\beta_m}^{k_m}v\mapsto x_{\tau(\beta_1)}^{k_1}\cdots x_{\tau(\beta_m)}^{k_m}u$ is an isomorphism. Hence $\dim L(\lambda)_\mu=\dim L(-\tau\lambda)_{-\tau\mu}$.

Now we compute the character of $L(-\tau\mu)$.

$ch(L(-\tau\lambda)=\sum\limits_{\mu}\dim L(-\tau\lambda)_{-\tau\mu}e^{-\tau\mu}=\sum\limits_{\mu}\dim L(\lambda)_\mu e^{-\tau\mu}$. Note that $\tau$ preserves the weight and the dimension of eigen spaces of a weight is preserved under the action of Weyl group. So $ch(L(-\tau\lambda))=\sum\limits_\mu\dim L(\lambda)_\mu e^{-\mu}=ch (L(\lambda)^*)$.


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