4
$\begingroup$

In Braid Groups of Kassel, Turaev, it mentions that $\mathcal{B}_n$ is a residually finite group. The definition that they give as a residually finite group is a group $G$ such that for each $g\in G-\{e_G\}$ ($e_G$ the identity of $G$), there exists an homomorphism $f$ to a finite group $H$ such that $f(g)\neq e_H$. My question is:

How can I obtain the group $H$ for a given element $g\in \mathcal{B}_n$ and the homomorphism that fulfills this?

I hope you can help me. Nice Holidays.

$\endgroup$
4
  • $\begingroup$ You need $H$ to be a finite group. $\endgroup$
    – Derek Holt
    Dec 15 '15 at 19:08
  • $\begingroup$ Someone explained this to me once. Something along the lines of: "$B_n$ is a subgroup of $\text{Aut}(F_n)$, which is much easier to show is residually finite." I don't remember how to do either half of that argument. $\endgroup$
    – user98602
    Dec 15 '15 at 19:18
  • 1
    $\begingroup$ Well if $g$ is not a pure braid, the obvious quotient map to the symmetric group will do. For pure braids, I assume you'll want to comb the braid. So use the fact that $P_n=F_n\rtimes P_{n-1}$ where $F_n$ is the free group on $n$ generators (which is residually finite) and then use induction $\endgroup$
    – Dan Rust
    Dec 15 '15 at 19:18
  • 1
    $\begingroup$ @DanRust I just posted the same argument. :) $\endgroup$ Dec 15 '15 at 19:19
2
$\begingroup$

If the braid is not pure, then you can detect it by a homomrphism to the symmetric group. So this reduces to showing that the pure braid group $P_n$ is residually finite. $P_n$ is an iterated semi-direct product of free groups. (This is called combing the braid.) The result now follows because free groups are residually finite. https://mathoverflow.net/questions/20471/why-are-free-groups-residually-finite

$\endgroup$
4
  • $\begingroup$ By the Symmetric Group which one you mean? $S_n$ for the group $\mathcal{B}_n$? I'm looking for an explicit example of this homomorphisms between a $\mathcal{B}_n$ to that $H$ group mentioned in the question. Can be this possible? $\endgroup$
    – iam_agf
    Dec 15 '15 at 19:55
  • $\begingroup$ @MonsieurGalois The group H depends on the element. So, yes, for many elements you can use the homomorphism to $S_n$, but it won't work on the pure braid subgroup, when you have to start using different homomorphisms to other finite groups. $\endgroup$ Dec 16 '15 at 0:37
  • $\begingroup$ For the group $P_n$ which group(s) can be used? $\endgroup$
    – iam_agf
    Dec 16 '15 at 0:51
  • $\begingroup$ @MonsieurGalois: $P_n$ is an iterated semidirect product of free groups. In the link I gave, the groups $H$ are constructed for free groups. $\endgroup$ Dec 16 '15 at 0:53
2
$\begingroup$

Another possibility is to embed $\mathcal{B}_n$ into the automorphism group $\mathrm{Aut}(\mathbb{F}_n)$ of the free group $\mathbb{F}_n$. Now, Baumslag gave a very short prove of the fact that, for any finitely generated residually finite group $G$, $\mathrm{Aut}(G)$ is also residually finite. The conclusion follows from the residual finiteness of finitely generated free groups (see for example the beautiful proof of Stallings in Topology of finite graphs), since a subgroup of a residually finite group is clearly residually finite itself.

For more details, see Basic results on braid groups and the references therein.

$\endgroup$

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.