# Centralizer of longest element in a finite irreducible Weyl group: related to folding of ADE graphs?

Say $(W,S)$ is a finite Coxeter group, such as a Weyl group (which satisfies an additional crystallographic condition). Assume also that $W$ is irreducible. Then it has a longest element $w_o$ relative to the given generating set $S$, which can be expressed as the $h/2$-power of a well-chosen Coxeter element when the Coxeter number $h$ is even (or a slightly modified expression if $h$ is odd).

Assume now that $W$ is a Weyl group (the others being a little more complicated to study). A standard fact is that $w_o = -1$ just when the type of $W$ in the classification is not ADE; more precisely, is different from $A_\ell (\ell \geq 2), D_\ell (\ell \geq 5 \text{ odd}), E_6$. When $w_o =-1$, its centralizer is obviously $W$, but otherwise is a proper subgroup.

A recent question here suggests to me a possible uniform treatment of such centralizers in terms of foldings of Coxeter graphs. But I'm not sure how far this is supported by the literature, or exactly how it might work for type $A_\ell$ with $\ell$ even --- then $W$ is the symmetric group $S_{\ell+1}$ and $h=\ell+1$ is odd. (I'm also unsure about the non-crystallographic types).

Note that folding the graph of (say) type $E_6$ yields the graph $F_4$. The Coxeter numbers are always the same when such foldings of ADE graphs occur, e.g., $h=12$ for both $E_6$ and $F_4$. Moreover, $w_o$ is folded into the corresponding longest element $-1$ in accordance with the above expressions as powers of Coxeter elements if $h$ is even. The question cited suggests that the "folded" Weyl group might always embed in $W$ as the precise centralizer of the original $w_o$.

Is this true, and is there a reference? (Further, what can be said in the non-crystallographic case?)

• Just a remark: $w_o=-1$ also holds for type $D_\ell$ with $\ell$ even. Oct 6, 2017 at 9:18
• Thanks for pointing this out (though it had already occurred to me after my hasty post yesterday). I've edited accordingly. Note that section 3.19 of my 1990 book on reflection groups yields the best result. Also, I should have referred to an old post of mine: mathoverflow.net/questions/111469/… Oct 6, 2017 at 11:17
• Folding always comes from a non-trivial group of Dynkin diagram automorphisms, which can be thought of as having an action on the same vector space on which $W$ acts as a reflection group. It seems reasonable to guess that the folded group is the subgroup of $W$ commuting with this action. Maybe it should be obvious but I don't immediately see it. Jan 6, 2018 at 0:32
• @Hugh: Thanks for your comments. I too am unable to see immediately what is going on, though I suspect there is a relevant reference somewhere. (By the way, "group of Dynkin diagram automorphisms" is only meaningful for crystallographic groups, so I'd prefer here to consider Coxeter graphs and graph automorphisms for arbitrary finite reflection groups. Still, Weyl groups are usually of most interest.) Jan 6, 2018 at 18:39

The folding automorphism of a root system is $\tau:\alpha\mapsto -w_0(\alpha)$ (since this preserves the positive roots, hence the simple roots, and is trivial if and only if $w_0=-1$). Since every element of $W$ commutes with $-1$, $\text{Cent}_W(\tau)=\text{Cent}_W(w_0)$.
At least in the case of a root system, $w_0$ is a regular element of $W$ (the sum of the positive roots is a regular eigenvector). Regular Elements of Finite Reflection Groups, by T.A. Springer (Inventiones 1974) especially Theorems 4.2 and 6.4 should go a long way towards answering your question in general.