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 $i,j$ or even that $W$ is an affine reflection group). 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 reflection 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 principle 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.

Here are some thougts so far: I can show that elements of $C_P(s_n)$ of length $1$ or $2$ already lie in $Z$ (the case of length $1$ 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 $s_1s_2s_1 \in P$ 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 $s_1,s_3$ do not commute but $s_2$ and $s_1s_3s_1$ do. Any ideas on how to do that?

Edit: I should note that I already posted this question to math.StackExchange (https://math.stackexchange.com/questions/1193740) but did not get any helpful feedback.

Edit 2: Regarding the question whether $s_1s_2s_1$ can centralize $s_3$ (all reflections pairwise distinct; $s_1$ neither centralizing $s_2$ nor $s_3$) in the case of $W$ being an affine reflection group I did a case by case check on the possible Dynkin-diagrams. The cases to consider are $A_3,B_3,\tilde{A_2},\tilde{B_2}$ and $\tilde{G_2}$ and using the standard representation on the root space I found that $s_1s_2s_1$ never centralizes $s_3$.

Edit 3: Here is a proof in the case that $m_{i,n} \neq 3$ for all $1 \leq i \leq n-1$: Assume $C_P(s_n) \neq Z$ and take an element $w \in C_P(s_n) - Z$ of minimal length. Then each reduced expression for $w$ neither starts nor ends with one of the simple reflections in $Z$.

Let $w=s_{i_1}...s_{i_r}$ be such a reduced expression. Since


there is a sequence of braid- and nil-moves that reduces $s_{i_1}...s_{i_r}s_ns_{i_r}...s_{i_1}$ to $s_n$. Since $m_{i_r,n} >3$ we cannot start with a braid- or nil-move involving $s_n$ and since we chose a reduced expression for $w$ there are certainly no nil-moves possible at all. Hence all we can start with is a braid-move in the reduced expression for $w$ (or $w^{-1}$). But after finitely many of such braid-moves the expression we get for $w$ still ends with a simple reflection $s_{i_k}$ which does not commute with $s_n$ (since this would yield an element of shorter length in $C_P(s_n) - Z$). Furthermore $m_{i_k,n} \neq 3$ so we still are unable to perform a braid-move involving $s_n$ and we still have a reduced expression for $w$ so there are no possible nil-moves. In conclusion: After finitely many steps we will never have performed any nil-moves and hence we cannot reduce $ws_nw^{-1}$ to $s_n$ which is a contradiction and hence such a $w$ does not exist.

I hope one can use an analogous argument in the case $m_{i_r,n}=3$.

  • 1
    $\begingroup$ First off, consider that some spherical / finite Coxeter groups, such as those of type $D_n$ have non-trivial center. Hence in there, the centralizer $C_P(s_n)$ is strictly larger than $Z$. This leads to many other examples (including non-spherical ones). Secondly, you may wish to have a look at "On centralizers of parabolic subgroups in Coxeter groups" by Koji Nuida, see arxiv.org/abs/math/0501061 $\endgroup$ – Max Horn Mar 19 '15 at 13:13
  • $\begingroup$ Thanks for your comment. I will have a look at the paper (I actually already found that during my first google search but didn't take much from it, but I will take a closer look). Regarding your first two sentences: The Coxeter-group of type $I_2(4)$ also has non-trivial center but the assertion is true nonetheless so I will need to look at $D_n$ to see what happens there (e.g. in $D_4$ my claim seems to be true). $\endgroup$ – Sebastian Schoennenbeck Mar 19 '15 at 13:25
  • $\begingroup$ I just checked $D_4$ and $D_5$ and in both of them my assertion holds. Certainly my assertion would be wrong if I claimed to get the centralizer of $s_n$ in all of $W$ in the described way (or by adding $s_n$ to $Z$) as for example $I_2(4)$ or the $D_n$ groups show. But since I am only interested in elements which omit $s_n$ in their reduced expressions the situation seems to be at least a little bit more complicated. $\endgroup$ – Sebastian Schoennenbeck Mar 19 '15 at 13:38
  • $\begingroup$ Indeed, I fell for a logical fallacy: In $D_{n+1}$, you have $D_n$ as a subdiagram, and a corresponding special subgroup $P$, say $P=\langle s_2,\ldots s_{n+1}\rangle$, and we are interested in the centralizer of $s_1$. Then both $W$ and $P$ have non-trivial center -- but the central element of $W$ is not contained in $P$, and that of $P$ does not centralize $s_1$. $\endgroup$ – Max Horn Mar 19 '15 at 15:26
  • $\begingroup$ @Sebastian: Though it doesn't directly address your question, a short article by Brink is perhaps helpful ams.org/mathscinet-getitem?mr=1396145 $\endgroup$ – Jim Humphreys Mar 19 '15 at 17:40

I believe I finally found a proof. Surprisingly it does only use 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 occurrences of $s$ (to the left and to the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which happens. 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.


I've tended to lose track of the basic question here, but it may help to focus just on the situation when $W$ is either a finite Coxeter group or an affine Weyl group (assuming in either case that the root system is irreducible). In these cases there is a fairly short list of possible types and considerable detail about the individual root systems in Bourbaki's Chapter VI (along with tables at the end). Case-by-case study is certainly tedious but possible here and so far seems to confirm your expectation, but if that's true for all indicated $W$ it would of course be preferable to have a general proof. (On the other hand, the case of a general Coxeter group seems far more open-ended, in view of Brink's description of centralizers of individual reflections.)

While I haven't looked at all possible cases, observed properties of the allowed root systems here may be helpful. For example, only in affine type $A_n$ for $n \geq 2$ (with $n+1$ vertices forming a circular extended Dynkin diagram) can one remove a single vertex from the graph which is connected to at least two other vertices by edges while leaving a complementary irreducible subsystem. This kind of configuration might be a candidate for a counterexample to your description, though affine $A_n$ can be ruled out directly.

The only general feature of these Coxeter groups which I can cite is the fact that point stabilizers in the closure of a fundamental domain are standard parabolics. But the way in which roots are built up from simple roots in these traditional root systems may also be suggestive. Case-by-case checking of root systems is feasible but lengthy, e.g., for situations in which removing one vertex leaves an irreducible subsystem. Springer's appended tables of positive roots for exceptional types here, arranged by height, facilitate study of those types (finite or affine). Recall that in Bourbaki's definition of affine Weyl groups, the negative of the highest root provides the extra vertex numbered $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.