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Consider a Coxeter group $W$ with simple generators $S$ and Coxeter matrix $\left( m_{s,t}\right) _{\left( s,t\right) \in S\times S}$.

Let $\mathfrak{M}$ be the set of all pairs $\left(s, t\right) \in S^2$ satisfying $s \neq t$ and $m_{s,t} < \infty$.

For every $\left( s,t\right) \in \mathfrak{M}$, let $c_{s,t}$ be an element of $\left\{ 1,-1\right\} $. Assume that:

  • We have $c_{s,t}=c_{s^{\prime},t^{\prime}}$ for any two elements $\left( s,t\right) $ and $\left( s^{\prime},t^{\prime}\right) $ of $\mathfrak{M}$ for which there exists some $q \in W$ satisfying $s^\prime = qsq^{-1}$ and $t^\prime = qtq^{-1}$.

  • We have $c_{s,t}=c_{t,s}$ for each $\left( s,t\right) \in \mathfrak{M}$.

Let $W^{\prime}$ be the group with the following generators and relations:

Generators: the elements $s\in S$ and an extra generator $q$.

Relations: \begin{align*} s^{2} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\ q^{2} & =1;\\ qs & =sq\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\ \left( st\right) ^{m_{s,t}} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=1;\\ \left( st\right) ^{m_{s,t}} & =q\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=-1. \end{align*}

There is clearly a surjective group homomorphism $\pi:W^{\prime}\rightarrow W$ sending each $s\in S$ to $s$, and sending $q$ to $1$. There is also a group homomorphism $\iota:\mathbb{Z}/2\mathbb{Z} \rightarrow W^{\prime}$ which sends the generator of $\mathbb{Z}/2\mathbb{Z}$ to $q$.

Question. Is $\iota$ injective? Equivalently, is the sequence \begin{equation} 1\longrightarrow\mathbb{Z}/2\mathbb{Z}\overset{\iota}{\longrightarrow}W^{\prime }\overset{\pi}{\longrightarrow}W \longrightarrow1 \end{equation} exact? Equivalently, is $\left\vert \operatorname*{Ker}\pi\right\vert =2$ ?

Background. This would generalize at least one of the two "spin symmetric groups" to the situation of any Coxeter group. It would explain one of the results (Theorem 2.3 (b)) in Alexander Postnikov and Darij Grinberg, Proof of a conjecture of Bergeron, Ceballos and Labbé, and prove a generalization of this result (Conjecture 6.1 (b)).

I have tried generalizing the standard approach to constructing the spin symmetric groups by embedding them in the Hecke-Clifford algebra, but to no avail so far. Nor has the existing literature on central extensions of Coxeter groups been particularly helpful (Howlett's On the Schur multipliers of Coxeter groups counts the extensions abstractly, but doesn't help understanding whether a given one exists; Burichenko's On extensions of Coxeter groups gives a criterion that I don't seem to properly understand, as it gives me wrong answers).

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    $\begingroup$ 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? $\endgroup$
    – Gro-Tsen
    Commented Nov 5, 2017 at 16:14
  • $\begingroup$ @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. $\endgroup$ Commented Nov 5, 2017 at 18:50
  • $\begingroup$ @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.) $\endgroup$ Commented Nov 5, 2017 at 20:56
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Nov 6, 2017 at 4:06
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    $\begingroup$ @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. $\endgroup$ Commented Jun 22, 2019 at 14:57

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