# Does the Weyl group act on its Lie algebra?

I am trying to prove something about the action of a particular Lie algebra on a particular representation (it's the starred claim on page 7 of this paper for those interested). My friend showed me a sketch of a proof that involved acting on an element of the Lie algebra by an element of its Weyl group $W$. However, I was not even aware that the Weyl group had a reasonable action on its Lie algebra.

I am aware that one can use the Killing form to identify a Cartan subalgebra $\mathfrak h$ with its dual $\mathfrak h^*$ and thus obtain an action of $W$ on $\mathfrak h$, but how do we extend this action to all of $\mathfrak g$?

In my case, my Lie algebra is a Kac-Moody algebra, but it behaves so similarly to a complex semisimple Lie algebra that I believe answering this question assuming that $\mathfrak g$ is complex semisimple will be enough.

Additional related question: Is there a reasonable sense in which the Weyl group acts on a representation of $\mathfrak g$?

• The Weyl group acts by automorphisms on both $\mathfrak{g}$ and its integrable representations (this is true in the generality of symmetrizable Kac-Moody algebras). For $\alpha$ a root, $E=E_\alpha$, $F=F_\alpha$ root vectors, I believe the formula for the reflection $s_\alpha$ is $S_\alpha=\mathrm{exp}(F)\mathrm{exp}(-E)\mathrm{exp}(F)$. – David Hill Aug 26 '15 at 20:05
• There are plenty of results that are proved in detail in the complex semisimple case followed by a short justification for it extending pretty much verbatim to the Kac-Moody case. – Matt Samuel Aug 27 '15 at 2:14