The action of the longest element on weights

Question

I am doing some computations related to some work I do in Lie theory and I need to compute the result of the action

$$w_0 \omega_{i},$$

where $w_0$ denotes the longest element in the Weyl group $W$ of a semisimple simply-connected complex algebraic group $G$ and the $\omega_i$'s denote the fundamental weights. I know I can express $w_0$ as a product of reflections $s_j$ and consider the action of each $s_j$ one by one, but this takes me a quite a lot of time and effort. Is there a shortcut (or a program) to compute this?

Answer 1

Following what @JyrkiLahtonen mentioned, we have $-w_0 \varpi_i = \varpi_{\sigma(i)}$ for some automorphism $\sigma$ of the Dynkin diagram, so all we need to know is which Dynkin diagram automorphism $\sigma$ is. I'll also note that depending on what tools you have on hand, it might be easier to calculate $\sigma$ as the permutation of simple generators induced by conjugation-by-the-longest-element in the Weyl group: $w_0 s_i w_0^{-1} = s_{\sigma(i)}$.

Here I state without proof the automorphism $\sigma$ in each type:

• Type $A_n$: For $n \geq 2$ there is a unique non-identity automorphism (path reversal), which is $\sigma$.
• Type $B_n, C_n$ for $n \geq 2$: The only Dynkin automorphism is the identity, so $\sigma$ is the identity.
• Type $D_n$ for $n \geq 4$: If $n$ is even then $\sigma$ is the identity. If $n$ is odd, then $\sigma$ is the unique non-identity automorphism which interchanges the two interchangable leaves.
• Type $E_6$: $\sigma$ is the unique non-identity automorphism (reversal of the longest path).
• Types $E_7, E_8, F_4, G_2$: The only Dynkin automorphism is the identity.