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Let $\Sigma_b$ be a closed orientable surface of genus $b \geq 2$, and denote by $\mathsf{P}_2(\Sigma_b)$ the pure braid group with two strands on $\Sigma_b$.

There is a braid $A_{12} \in \Sigma_b$ corresponding to the classical braid generator of the disk, together with $2b$ further generators coming from the usual representation of $\Sigma_b$ as an identification space of a polygon with $4b$ sides, see for instance [B04].

Question. How can we construct finite quotients $\pi \colon \mathsf{P}_2(\Sigma_b) \to G$ such that $\pi(A_{12})$ is non-trivial?

In [CP19] it is remarked that some finite Heisenberg groups do the job, and similar proofs work for other extra-special $p$-groups.

Are there different examples? Is it possible to classify them in some way? For instance, what is the smallest order of such a quotient in terms of $b$?

Note that $A_{12}$ is a commutator in $\Sigma_b$, so our condition $\pi(A_{12}) \neq 1$ implies that $G$ must be non-abelian. For $b=0$, some results are contained in [CKLP19].

Edit. Notation changed in order to be consistent with the one in [CP19].

References.

[B04] P. Bellingeri: On presentations of surface braid groups, J. Algebra 274, No. 2, 543-563 (2004).

[CP19] A. Causin, F. Polizzi: Surface braid groups, finite Heisenberg covers and double Kodaira fibrations, arXiv:1905.03170.

[CKLP19] A. Chudnovsky, K. Kordek, Q. Li, C. Partin: Finite quotients of braid groups, arXiv:1910.07177.

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    $\begingroup$ Just as an aside, the smallest finite quotient for the full surface braid group on any number of strands (for $g>0$) is $\mathbb{Z}/(2)$; the “classical” generators collapse to an order two element in the abelianization. $\endgroup$
    – Santana Afton
    Jan 9, 2020 at 13:57
  • $\begingroup$ Just to be sure: the group in question is $\mathbb{Z} \rtimes \pi_1(\Sigma_g)$ where the action is by monodromy, right? So shouldn't this be about determining the kernel of the mondromy action in $\pi_1(\Sigma_g)$? $\endgroup$
    – Stefan Witzel
    Jan 20, 2020 at 8:20
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    $\begingroup$ @StefanWitzel: I am not sure of understanding your comment. The subgroup $\langle A_{12} \rangle$ is not normal in $\mathsf{P}_2(\Sigma_b)$. In fact, from the presentation (that you can find in my paper with Causin, see p. 10) we have for instance, when $j > k$, $$\tau_{2k} A_{12} \tau_{2k}^{-1} = A_{12} [\rho_{1j}, \, \tau_{2k}].$$ In order to be consistent with the notation in [CP19], I am denoting by $A_{12}$ (instead of $\sigma$) the braid generator of the disk and by $\rho_{ij}$, $\tau_{ij}$ the geometric generators coming from the polygon. $\endgroup$
    – Francesco Polizzi
    Jan 20, 2020 at 9:21
  • $\begingroup$ Mapping class groups are residually finite, by Grossman's theorem, so there are lots of finite quotients like you want. Grossman proves her theorem in a way that shows that you can take the quotients to be congruence quotients, meaning that you can find finite quotients by taking characteristic subgroups $C$ of the fundamental group $\pi_1(S)$ and considering the map from the Mapping class group to the finite group $\mathrm{Out}(\pi_1(S)/C)$. Conjecturally these are essentially the only kind of quotients of the mapping class group. $\endgroup$
    – Autumn Kent
    Jan 20, 2020 at 19:32

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