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.



Central extensions of torsion groups by $\mathbb{R}^n$
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](https://arxiv.org/abs/1811.07167)
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.


