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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.

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  • $\begingroup$ 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_
    Commented Sep 22, 2021 at 1:10
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    $\begingroup$ Not every torsion group is a direct limit of finite groups. Perhaps $\Gamma$ is supposed to be locally finite? $\endgroup$
    – markvs
    Commented 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. $\endgroup$
    – Joel Villatoro
    Commented 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
    Commented 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. $\endgroup$
    – Joel Villatoro
    Commented Sep 22, 2021 at 7:21

1 Answer 1

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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.

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