21
$\begingroup$

I'm interested in computing the cohomological dimension of the commutator subgroup $[P_n,P_n]$ of the pure braid group $P_n$. I wasn't able to find a reference in the literature.

Because $[P_n,P_n]$ has an abelian subgroup of rank $\lfloor(n-1)/2\rfloor$ we have$$(n-2)/2\le\text{cd}([P_n,P_n])\le n-2.$$ My guess is that in fact $\text{cd}([P_n,P_n])=n-2$.

There is a right split short exact sequence $$1\to[F_n,F_n]\to[P_{n+1},P_{n+1}]\to[P_n,P_n]\to1,$$ which implies that $[P_n,P_n]$ is an iterated semidirect product of infinitely generated free groups. This suggests an inductive spectral sequence argument but I'm having problems understanding the cohomology groups $H^{n-2}\left([P_n,P_n];H^1([F_n,F_n])\right)$.

EDIT: The commutator subgroup $[B_n,B_n]$ of the full braid group has been studied by Gorin and Lin in "Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids" (1969) Math. USSR Sb. 7 569-596. In particular, it follows from their results that $\text{cd}([B_3,B_3])=1$ and $\text{cd}([B_4,B_4])=2$. However, this is not immediately helpful because $[P_n,P_n]$ is not a finite index subgroup of $[B_n,B_n]$.

$\endgroup$
16
$\begingroup$

Here is the answer: for $n\geq 2$ we have $\mathrm{cd}([P_n,P_n])=n-2$.

https://arxiv.org/abs/1905.05099

$\endgroup$

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.