Timeline for A spin extension of a Coxeter group?
Current License: CC BY-SA 3.0
13 events
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| Jun 22, 2019 at 14:57 | comment | added | darij grinberg | @DavidESpeyer: Thanks -- I've answered the Question positively in the meantime (a year ago), but only in a month or so will probably be able to write up my proof. Meanwhile, I did look into Howlett's and others' Schur-multiplier papers, but never found myself able to get something out of it that wasn't obviously wrong; it's too much of a foreign language to me. | |
| Jun 22, 2019 at 14:01 | comment | added | David E Speyer | Morris, "Projective representations of finite reflection groups. III.", Comm. Alg. (2004), is doing something very like this, although I haven't unwound his notation to see whether it exactly matches your formulation or not. The title of that paper has the word finite in it, but most of the results don't seem to need it. See also Howlett, "On the Schur multipliers of Coxeter groups", JLMS (1988), which is less concrete but is better about not including unneeded finiteness hypotheses. | |
| Nov 6, 2017 at 4:06 | comment | added | LSpice | This is a quotient of the Tits group (MSN) in a natural way. It seems like maybe Théorème 2.5 there will give you what you want, although I can't see it right away. | |
| Nov 6, 2017 at 0:59 | comment | added | darij grinberg | @VictorProtsak: Thanks. I've reverted to the $\mathfrak{M}$ notation from the paper. | |
| Nov 6, 2017 at 0:59 | history | edited | darij grinberg | CC BY-SA 3.0 |
thanks Victor
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| Nov 6, 2017 at 0:53 | comment | added | Victor Protsak | Sure, $s\ne t$. Actually, your last relation involving $q$ should also have restriction $s\ne t$. | |
| Nov 6, 2017 at 0:37 | comment | added | darij grinberg | @VictorProtsak: I take it you want $s \neq t$, since otherwise picking $s=t$ and $s' = t'$ to be two non-conjugate simple reflections would break it. Anyway, nice question! Perhaps what we did in the proof of Claim 1 in the proof of Lemma 4.0.3 in Alex's and my paper could be of use. | |
| Nov 6, 2017 at 0:24 | comment | added | Victor Protsak | Darij, very interesting question! Clearly your two conditions are necessary: conjugate elements have the same order. But are they sufficient to guarantee that if $st$ and $s't'$ have the same order in $W$, then $c_{s,t}=c_{s',t'}$ ? More precisely, do you know whether the following statement is true for general $W$? $$ $$ If $st$ and $s't'$ are conjugate in $W$ then there is $q\in W$ simultateously conjugating either $s$ to $s'$ and $t$ to $t'$ or $s$ to $t'$ and $t$ to $s'$. | |
| Nov 6, 2017 at 0:02 | comment | added | Victor Protsak | Jim, for any finite dihedral group (finite Coxeter group of rank 2) $\langle s,t\rangle\simeq D_m$ of type $I_2(m)$, in the nontrivial case $c_{s,t}=-1$ this construction produces the dihedral group $D_{2m}$ of type $I_2(2m)$ with double the size. So starting from $W(G_2)=D_6$ you would get $D_{12}$. | |
| Nov 5, 2017 at 20:56 | comment | added | darij grinberg | @Gro-Tsen: Interesting idea; but this would only give two "spin extensions" (and I have no idea how to precisely find their generators and relations), whereas my $c_{s,t}$ in general allow for much more freedom. (Not saying that the approach is broken; could be my extensions are all equivalent.) | |
| Nov 5, 2017 at 18:50 | comment | added | Jim Humphreys | @darij: This is an intriguing area which I haven't gone into, but I wonder for example what would happen in the frequent case that -1 is in a given finite irreducible Coxeter group $W$? (This is a special case of the longest element of $W$, which has been well studied using the classification.) A small example is the $G_2$ case, though symmetric groups of rank > 1 aren't examples. | |
| Nov 5, 2017 at 16:14 | comment | added | Gro-Tsen | I had in mind that every Coxeter group can be realized as isometries of a certain "canonical" symmetric bilinear form (not necessarily definite), so that it embeds in $\mathit{O}(p,q)$: wouldn't its inverse image in $\mathit{Pin}^+(p,q)$ or $\mathit{Pin}^-(p,q)$ answer your question? Or am I talking nonsense? | |
| Nov 4, 2017 at 19:46 | history | asked | darij grinberg | CC BY-SA 3.0 |