# Action of the pure braid group on the commutator subgroup of a free group

Let $P=P_n$ be the pure braid group on $n$ strands and $F=F_n$ the free group on $n$ generators. I'm interested in a nice description of the action of $P$ on the derived subgroup $F'$ which somehow would'nt involve $F$.

More precisely, there is well known action of $P$ on $F$ which has the very particular property of being basis conjugating, i.e. it send every generator of $F$ to a conjugate of itself. In fact, in the following feel free to replace $P$ by the group of all basis conjugating automorphism (which has a somewhat simpler presentation).

This property implies that $P$ acts trivially on the abelianization $F/F'$ (though it is much stronger). It means that $$\forall p \in P, f\in F,\ (p\cdot f)f^{-1}\in F'\ \ (*)$$ where $\cdot$ is the action.

Let $\gamma_k F$ be the $k$th term of the lower central series, so that $\gamma_1F=F$, $\gamma_2F=F'$ and $$\gamma_{k+1}F=[\gamma_kF,F].$$

Then it's a general fact that $(*)$ implies that $\gamma_kP$ acts trivially on $\gamma_lF$ modulo $\gamma_{k+l}F$.

In particular, we have $$\forall p \in P', f\in F',\ (p\cdot f)f^{-1}\in \gamma_4 F$$

but I suspect that it actually belongs to a much smaller group. In fact I was hoping that it belonged to the commutator subgroup of $F'$ $$[F',F']\subset \gamma_4F$$

i.e. that the action of $P'$ on $F'$ would have the same property than the action of $P$ on $F$, but that doesn't seem to be true (I would be happy to be proven wrong here !). So my question can be asked in two different direction:

Is there a nice description/are there nice properties of the action of $P$, or any higher term $\gamma_kP$, on $F'$ given directly in term of $F'$ ?

One can do some pretty explicit computation but it didn't give me any insight yet. In the other but somewhat more concrete direction:

What is the subgroup of $P$ that acts on $F'/F''$ trivially ? Is it non trivial ? Does it contains $\gamma_kP$ for some $k$ ?

Some googling told me that this last question might be related to the Gassner representation of $P_n$, and the so-called magnus representation of the Torelli group of a surface, but I haven't found any answer to this question yet.

See the paper link.springer.com/article/10.1007%2Fs00222-013-0477-9; the action of the pure braid group on the $abelianisation$ of the derived group is described there (see Section(3.4). With some qualification, it is indeed "the" Gassner representation. The trouble is that $F'/F''$ is not a free module over the group algebra ${\mathbb Z}[F/F']$, and hence one embeds the quotient $F'/F''$ into a slightly larger module which is indeed free and which is is exactly the Gassner representation; all this is worked out in the paper I have referred to.

I just now saw the second part of your question: the action of $P$ is indeed highly nontrivial on $F'/F''$. The image is (modulo $d$-th roots of unity, with $n\geq 2d-1$ if you are dealing with the pure braid group on $n$ strands) an arithmetic group in the unitary group associated to the Gassner representation

I could not copy the link so here is the reference: http://www.ams.org/mathscinet-getitem?mr=3219513

• Thanks, that's an interesting ref, though it's mostly computational. For the second part I of course agree that the action is non trivial, I was asking if the kernel was or if there was a description of it. Nov 28, 2014 at 15:30
• in short, you are asking if the kernel of the Gassner representation is non-trivial; I think this isan open problem (though it is known that the Burau rep is not faithful, and Burau and Gassner are related, to the best of my knowledge, it is not known whether Gassner is faithful. You may ask Misha Kapovitch about this Nov 28, 2014 at 15:39
• For a thorough discussion of the status of the faithfulness problem for the Gassner representation, see Joan Birman's review of the paper "Braid groups are linear groups" by Seymour Bachmuth, at ams.org/mathscinet-getitem?mr=1399602 Nov 28, 2014 at 17:10