3
$\begingroup$

This question is perhaps a duplicate of this, but I just don't understand the answer and it was written a few years back so I thought I'd try again.

If $\mathfrak{g}$ is simple and we choose a CSA $\mathfrak{h}$, I want to understand why the action of the Weyl group $W$ is irreducible on $\mathfrak{h}^*$.

Fulton and Harris give a proof, which starts with a subrep $\mathfrak{z}\subset\mathfrak{h}^*$, and states that "every root $\alpha\in\mathfrak{h}^*$ of $\mathfrak{g}$ will either lie in $\mathfrak{z}$ or the subspace perpendicular to it (which will be a complementary subrep).

I follow the rest of the proof, but I do not understand why this initial claim is true. Why couldn't a root $\alpha$ be in neither $\mathfrak{z}$ or $\mathfrak{z}^\perp$?

Any/all help/ideas are appreciated!

$\endgroup$

1 Answer 1

4
$\begingroup$

Okay, I found an answer from another source, and it's not that deep. I clearly should have thought more about it!

If we write $B$ for the Killing form, then for any $v\in\mathfrak{z}$, we have $B(v,\check{\alpha})\alpha=v-s_{\alpha}(v)$, where $\check{\alpha}$ is the coroot to $\alpha$ and $s_{\alpha}$ is the orthogonal reflection defined by $\alpha$. Hence if $B(v,\check{\alpha})\neq0$ for some $v\in\mathfrak{z}$ then $\alpha\in\mathfrak{z}$, and otherwise by definition $\alpha\in\mathfrak{z}^{\perp}$.

Thanks to the Lie theorist Prof Fernando for helping me out with my trivial inquiry.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .