4
$\begingroup$

Let $\Gamma$ be a torsion group (i.e. every element has finite order). I am interested in understanding central extensions of the form: $\require{AMScd}$ \begin{CD} 0 @>>> \mathbb{R}^n @>\exp>> G @>\pi>> \Gamma @>>> 1\\ \end{CD} Equivalently, I want examples of groups $\Gamma$ with non-trivial classes in $H^2(\Gamma,\mathbb{R}^n)$. When $\Gamma$ is finite I'm aware that $H^2(\Gamma,\mathbb{R}^n) = 0$ but I have little intuition or examples for infinite torsion groups.

Interestingly, any central extension such a $\Gamma$ by $\mathbb{R}^n$ has a canonical set-theoretic splitting $\sigma \colon \Gamma \to G$ with the property that:

$$ \sigma(\gamma) = g \quad \Leftrightarrow \quad \pi(g) = \gamma \text{ and } g \text{ has finite order}$$ where $\pi \colon G \to \Gamma$ is the projection. It is not too hard to show that if any group-theoretic splitting exists then it must be $\sigma$.

Using usual group cohomology arguments, this gives rise to a cocycle: $$ \alpha \colon G \times G \to \mathbb{R}^n $$ I've managed to prove a few interesting properties of $\alpha$ (for example, $\alpha$ must be symmetric) but have not been able to show it is zero and I suspect a counterexample probably exists.

$\endgroup$
6
  • $\begingroup$ Can't you write $\Gamma$ as direct limit of finite groups and check if $H^2(\lim\limits_{\rightarrow}\Gamma_i,\mathbb{R})=\lim\limits_{\leftarrow}H^2(\Gamma_i,\mathbb{R})=0$? $\endgroup$
    – tj_
    Sep 22, 2021 at 1:10
  • 1
    $\begingroup$ Not every torsion group is a direct limit of finite groups. Perhaps $\Gamma$ is supposed to be locally finite? $\endgroup$
    – markvs
    Sep 22, 2021 at 1:16
  • $\begingroup$ @MarkSapir I don't mean to say $\Gamma$ is locally finite, I think this is why the search for a counterexample has been difficult. If every element of $\Gamma$ is contained in a finite subgroup then $H^2(\Gamma,\mathbb{R}^n) = 0$. This is due to the fact that $\alpha$ will vanish when restricted to any finite subgroup. $\endgroup$ Sep 22, 2021 at 7:06
  • $\begingroup$ The canonical splitting you describe is characterized by the property that the associated 2-cocycle $\Gamma\times\Gamma\to\mathbf{R}^n$ vanishes on $H\times H$ for every finite cyclic subgroup $H$ of $\Gamma$, and satisfies the property that it vanishes on $H\times H$ for every finite subgroup $H$ of $\Gamma$. $\endgroup$
    – YCor
    Sep 22, 2021 at 7:11
  • $\begingroup$ In my previous comment, I mean to say "every pair of elements of" $\Gamma$ not "every element". Every element of $\Gamma$ is contained in a finite cyclic subgroup, of course. $\endgroup$ Sep 22, 2021 at 7:21

1 Answer 1

2
$\begingroup$

The paper S. I. Adyan and V. S. Atabekyan, V. S. Central extensions of free periodic groups, Mat. Sb. 209 (2018), no. 12, 3–16; translation in Sb. Math. 209 (2018), no. 12, 1677–1689 proves that if $n\geq 665$ is odd and $m\geq 2$, then the Schur multiplier $H^2(B(m,n),\mathbb Z)$ for the free Burnside group $B(m,n)$ of exponent $n$ on $m$-generators is free abelian of countable rank. In the arXiv version this is Corollary 3 and so using the universal coefficients theorem, $H^2(B(m,n),\mathbb R^n)$ is quite huge. If you want an explicit class, then that is beyond my competency range.

$\endgroup$

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.