Elements of order 2 in a Weyl group.

Question

I would like to prove that any element of order 2 in a Weyl group is the product of commuting root reflections.

I am told that the proof should be by induction on the dimension of the -1 eigenspace.

Any clue would help, thanks !

Answer 1

The fact you are after is an easy consequence of the following generally useful description of involutions (elements whose square is the identity) in the Weyl group.

Proposition. Any involution $w\in W$ can be obtained starting from the involution $e$ by a sequence of length-increasing operations that are either multiplication of an involution by a simple reflection $s_\alpha$ with which it commutes, or conjugation by a simple reflection $s_\alpha$ with which it does not commute.

Proof by induction on the length $\ell(w)$. If $\ell(w)=0$ then $w=e$. So suppose that $\ell(w)>0$, and let $\alpha$ be a simple root such that $\ell(ws_\alpha)<\ell(w)$. Distinguish the cases on whether or not $s_\alpha$ commutes with $w$. If it commutes, then $ws_\alpha$ is an involution, and $w$ is obtained from it by commuting multiplication by $s_\alpha$. If they don't commute, then we can see as follows that $\ell(s_\alpha ws_\alpha)=\ell(w)-2$. From $\ell(ws_\alpha)<\ell(w)$ there is a reduced expression for $w$ that ends with $s_\alpha$, and reversing it we obtain a reduced expression for $w^{-1}=w$ starting with $s_\alpha$, say $w=s_\alpha s_2\ldots s_{\ell(w)}$. Now the exchange condition says that an expression for $ws_\alpha$ can be obtained by striking out one of the generators in the latter reduced expression, and it cannot be the initial $s_\alpha$ as that would give $s_\alpha w$ which is supposed to differ from $ws_\alpha$. Now left-multiplying by $s_\alpha$ gives an expression for $s_\alpha ws_\alpha$ obtained by striking out a generator in $s_2\ldots s_{\ell(w)}$, and therefore of length $\ell(w)-2$. This $s_\alpha ws_\alpha$ is an involution of shorter length than $w$, from which $w$ can be obtained by conjugation by $s_\alpha$ with which it does not commute. QED

If you prefer, the proof can be formulated in terms of the standard reflection representation $\rho$ of $W$. Here $\ell(ws_\alpha)<\ell(w)$ means that $\rho_w(\alpha)$ is a negative root, and the distinction is on whether or not $\rho_w(\alpha)=-\alpha$. If this holds, then the root reflection $\rho_\alpha=\rho(s_\alpha)$ commutes with $\rho(w)$, so $\rho(ws_\alpha)=\rho(w)\rho_\alpha$ is an involution, and so is $ws_\alpha\in W$ since $\rho$ is faithful. If $\rho_w(\alpha)=-\beta$ for some positive root $\beta\neq\alpha$, then from $w^2=e$ we also get $\rho_w(\beta)=-\alpha$. The length of $w$ equals the number of positive roots that $\rho_w$ sends to a negative root, or equivalently the number of pairs $\{\gamma,-\gamma\}$ of opposite roots for which $\rho_w$ "flips signs" (positive goes to negative and vice versa); now with $w'=s_\alpha ws_\alpha$ one has $\rho_w$ sign-flips $\{\gamma,-\gamma\}$ if and only if $\rho(w')$ sign-flips $\{\rho_\alpha(\gamma),-\rho_\alpha(\gamma)\}$, with the two exceptions $\{\gamma,-\gamma\}=\{\alpha,-\alpha\}$ and $\{\gamma,-\gamma\}=\{\beta,-\beta\}$, both of which are sign-flipped by $\rho(w)$ but not by $\rho(w')$, so that $\ell(w')=\ell(w)-2$.

Now back to the question you actually asked of writing an involution $\rho_w$ as product of commuting reflections. First observe that commuting root reflections involve orthogonal roots, so these will necessarily all be in the eigenspace for $-1$ of their product $\rho_w$. The hard part is to show that there are always roots in that eigenspace when it has positive dimension. Here the proposition is useful, as it allows doing recursion on $\ell(w)$ for our involution $w$. If $\alpha$ is such that $\ell(ws_\alpha)<\ell(w)$, then $s_\alpha$ commutes with $w$ if and only if $\rho_w(\alpha)=-\alpha$. If so then $ws_\alpha$ is an involution that we can apply induction to; the roots for it are all contained in the $-1$ eigenspace of $\rho_{ws_\alpha}$, which is the orthogonal complement of $\langle\alpha\rangle$ in the $-1$ eigenspace of $\rho_w$ and we can add $\alpha$ to the collection of roots. If $\rho_w(\alpha)\neq-\alpha$ then $w'=s_\alpha ws_\alpha$ is a shorter conjugate of $w$; by induction we get a set of roots for $w'$, and applying $\rho_\alpha$ to each of them we get a set of roots for $w$.