Let us consider a (not necessarily finite) Coxeter group $W$ generated by a finite set of involutions $S=\{s_1,...,s_n\}$ subject (as usual) to the relations $(s_is_j)^{m_{i,j}}$ with $m_{i,j}=m_{j,i}$ and $m_{i,j}=1$ if and only if $i=j$ (if necessary you may also assume that $m_{i,j}<\infty$ for all $i,j$). Let $P\leq W$ be a subgroup generated by all but one of the $s_i$, say wlog $P=\langle s_1,...,s_{n-1}\rangle$.

I am interested in the centralizer of $s_n$ in $P$. In particular I would like to know if $C_P(s_n)=C_W(s_n) \cap P=\langle s_i~|~ 1\leq i\leq n-1, m_{i,n}=2\rangle=:Z$ always holds.

Obviously this is true if $n=2$ and I believe (though I have not written it down rigorously) I can prove it for Coxeter groups of type $A_n$ by using the standard isomorphism to $S_n$. On the other hand the centralizer of $s_n$ in $W$ is not necessarily a standard parabolic subgroup (look at the dihedral group of order $8$ for example).

There are some results on centralizers of reflections in Coxeter groups and on normalizers/centralizers of parabolic subgroups (which is the same in this special case) to be found in the literature but most deal with the centralizer in $W$. In principal it should be possible to obtain the centralizer in $P$ from these results by simply taking the intersection but the results I found so far are not explicit/ simple enough for this to be a feasible solution.

Edit: Some thoughts so far: I can show that elements of $C_P(s_n)$ of length $1$ or $2$ already lie in $Z$ (the case $1$ being trivial) and that elements of $C_P(s_n)$ of length $3$ where all three occurring simple reflections are pairwise distinct already belong to $Z$.

On the other hand look at $s_1s_2s_1 \in C_P(s_n)$ which centralizes $s_n$ if and only if $s_2$ centralizes $s_1s_ns_1$. I don't see any reason why this should not be the case so I tried constructing a counterexample consisting of $s_1,s_2$ and $s_3$ such that $s_1,s_2$ do not commute and $s_1s_3$ do not commute but $s_2$ and $s_1s_3s_1$ do. Any ideas on how to do that?

  • $\begingroup$ If it is a finite reflection group coming from symmetries of a root system (IOW types $A,B,C,D,E,F,G$) then it is easy to calculate the order of the centralizer, because $s_n$ is the reflection w.r.t. to the hyperplane having the related simple root $\alpha$ as its normal. The conjugacy class of $s_n$ consists of similar reflection related to roots of the same length. So we can calculate the size of the conjugacy class, and hence the order of the centralizer. This leads to a case-by-case check of whether you get equality or a strict inclusion. $\endgroup$ – Jyrki Lahtonen Mar 18 '15 at 20:57
  • $\begingroup$ @JyrkiLahtonen: Certainly true but this only calculates the (order of the) centralizer in $W$ not in $P$. Plus I am (more) interested in the case where $W$ is infinite (though I think I can assume that $W$ is (possibly affine) reflection group). $\endgroup$ – Sebastian Schoennenbeck Mar 19 '15 at 7:20
  • $\begingroup$ Ok. I missed a point or three. An interesting question anyway! $\endgroup$ – Jyrki Lahtonen Mar 19 '15 at 7:28

I posted the question over at MathOverflow and just posted a proof. So to not have the question without an answer on this site, here it goes:

Surprisingly the proof only uses standard facts on Coxeter-groups (exchange condition, solving the word problem via braid-moves,...).

Let me first make the notation a bit easier:

Claim: Let $P\leq W$ be a special subgroup of $W$ generated by some subset $S'\subsetneq S$ of $S$ and $s \in S-S'$. Then the centralizer of $s$ in $P$ is generated by those involutions in $S'$ which commute with $s$, $C_P(s)=\langle s' \in S'~|~ s's=ss' \rangle$.

Proof: Let $w \in C_P(s)$ and $w=s_1...s_r$ be a reduced expression. By induction it is enough to prove that $s_rs=ss_r$ since the elements of length $1$ in the centralizer are precisely the simple involutions commuting with $s$.

We have $\ell(ws)=\ell(w)+1$ and since $\ell(wsw^{-1})=\ell(s)=1$ we conclude that $\ell(wss_r)<\ell(ws)$, so $\ell(wss_r)=\ell(w)$. By the exchange condition there is a reduced expression for $ws$ ending in $s_r$ and since $s_1...s_rs$ is already a reduced expression for $ws$ there exists a finite series of braid-moves connecting these two expressions.

The expression $s_1...s_rs$ contains $s$ only once and no simple involution that does not commute with $s$ shows up to the right of $s$. Consider now any braid-move in this situation. If $s$ is not involved in the move the two conditions obviously still hold afterwards. If $s$ is involved the other simple involution involved must commute with $s$ since any braid-move involving $s$ and a non-commuting $s'$ requires either at least two occurrence of $s$ (two the left and two the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which there is. Hence any braid-move fixes our two conditions and after finitely many braid moves there is still no simple involution to the right of $s$ which does not commute with $s$.

On the other hand there is, as noted above, a finite series of braid-moves after which the expression ends in $s_r$ so $s_r$ has to commute with $s$ as asserted.


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