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Let $G$ denotes an infinite coutable discrete group with Kazhdan's property (T),

My question is:

is it known that the 2nd cohomology group $H^2(G,\mathbb{Z}G)$ is torsion free?

Thanks in advance!


Note that for any finitely presented group $G$, $H^2(G,\mathbb{Z}G)$ is always torsion-free by proposition 13.7.1 in GTM 243(Topological methods in group theory).

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    $\begingroup$ Do you know in the case $SL_3(\mathbf{F}_p[t])$? For this group possibly there is a geometric way of understanding $H^2(G,\mathbf{Z}G)$, and this group is not f.p. (Behr) and has Property T. $\endgroup$
    – YCor
    Nov 18, 2014 at 13:56
  • $\begingroup$ @YCor, thanks for mentioning this group, I learned it from the book on Kazhdan's property (T). And I asked this question because I find a proof that my question has a positive answer but I am not sure whether this is known or not... $\endgroup$
    – Jiang
    Nov 18, 2014 at 15:07
  • $\begingroup$ Where do you use Property T? For instance, the vanishing of $H^1(G,\mathbf{Z}X)$ for every $G$-set $X$ holds for many more groups. $\endgroup$
    – YCor
    Nov 18, 2014 at 21:57
  • $\begingroup$ @YCor, if my proof is correct, then this is a ``direct" consequence of Popa's cocycle superrigidity result for Bernoulli shift of property (T) group (and of course it holds for a more wider class of groups) plus taking advantage of the principal algebraic action setting, although this is not what my primary goal... $\endgroup$
    – Jiang
    Nov 18, 2014 at 22:38
  • $\begingroup$ @YCor, I have decided to write down the proof, you can find it as corollary 4.1 in arxiv.org/abs/1509.08278. Thanks! $\endgroup$
    – Jiang
    Sep 30, 2015 at 10:01

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