# Lower Central Series of Pure Braid Groups?

What is the lower central series $\Gamma_k(P_n)$, where $P_n$ is the pure braid group with $n$ strands? We know that $P_n$ is generated by elements $A_{i,j}$; do we know the generators of $\Gamma_k(P_n)$?

• I suppose you're aware that the pure braid group is a semi-direct product of free groups? – Jim Conant Dec 6 '14 at 17:56
• @JimConant, yes. How is this related to the question? – Zuriel Dec 6 '14 at 18:37
• That would be my first line of attack in analyzing the lower central series, since you know what it is for free groups. – Jim Conant Dec 6 '14 at 21:06

As Jim Conant says, the structure of the lower central series of the pure braid group $P_n$ can be deduced from the fact that $P_n$ is an iterated semidirect product of free groups, $P_n=F_{n-1}\rtimes F_{n-2}\cdots F_2\rtimes F_1$, though that's not quite enough: one also needs to know that the action of $P_{n-1}$ on $F_{n-1}$ is by basis-conjugating automorphisms, and thus it's trivial in homology.
Let $1\to A\to B\to C \to 1$ be a split exact sequence of groups, and suppose the induced conjugation action of $C$ on $H_1(A)$ is trivial (that is, $[A, C] \subseteq [A, A]$). Then the sequence of induced maps $1\to \Gamma_k A\to \Gamma_k B\to \Gamma_k C \to 1$ is split exact, for every $k\ge 1$.
• Thanks for the answer! So you mean that there is a split exact sequence $1\to \Gamma_kF_{n-1}\to\Gamma_kP_n\to\Gamma_kP_{n-1}\to1$, right? Hence how to find $\Gamma_kP_n$ exactly? – Zuriel Dec 7 '14 at 5:45
• By induction on $n$, using the known structure of $\Gamma_k F_r$. – Alex Suciu Dec 7 '14 at 5:48
• I suppose that this result about $\Gamma_kP_n$ is known; where can I find a reference? – Zuriel Dec 7 '14 at 5:54