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$\DeclareMathOperator\GL{GL}$It is known that every braid group $B_n$ embeds as a subgroup of $\GL_m(\mathbb{Z}[q^{\pm 1},t^{\pm 1}])$, where $m=n(n-1)/2$ (see Krammer - Braid groups are linear). This is via the Lawrence–Krammer representation, proved to be faithful for all $n$ independently by Bigelow and Krammer. Thus, braid groups are linear.

My question is whether it is known that they are not linear over $\mathbb{Z}$. That is, is it known that, for large enough $n$, $B_n$ does not embed as a subgroup into any $\GL_m(\mathbb{Z})$? I haven't found any references to this being an open problem, so I'm wondering if there's just some established reason it can't happen.

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    $\begingroup$ I guess it's an open question (I wondered about it long ago), for $n\ge 4$. I don't remember there's a known representation that would be a candidate to be faithful. $\endgroup$
    – YCor
    Commented Feb 5, 2022 at 13:22
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    $\begingroup$ Braid groups are subgroups of mapping class groups. Kontsevich gave a candidate representation of mapping class groups into $GL_m(\mathbb Z)$, with $m$ is about $g^g$. $\endgroup$
    – Ben Wieland
    Commented Feb 5, 2022 at 14:45
  • $\begingroup$ Yeah, I guess it's just open, and perhaps so open that very few people even talk about it in the literature. But I didn't know about this proposed Kontsevich representation, good to know. $\endgroup$
    – Matt Zaremsky
    Commented Feb 6, 2022 at 18:28
  • $\begingroup$ Is it known if $B_n$ is not virtually special? I think there was a question whether it is even CAT(0) and I do not know if it is resolved. $\endgroup$
    – markvs
    Commented Feb 7, 2022 at 2:57
  • $\begingroup$ see this question, answers and comments mathoverflow.net/questions/42836/… $\endgroup$
    – markvs
    Commented Feb 7, 2022 at 3:12

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