Is $H^4(PSL(2,\mathbb{Z}),\mathbb{Z})$ known? I ask this in response to the recent calculation of the same cohomology group for $\mathrm{Co}_0$ and $\mathrm{Co}_1$.
$\begingroup$
$\endgroup$
6
asked Jul 26, 2017 at 10:20
-
8$\begingroup$ Doesn't this follow easily from $PSL(2,\mathbb{Z})\cong C_2\ast C_3$? $\endgroup$– Jeremy RickardCommented Jul 26, 2017 at 10:27
-
5$\begingroup$ So what's the answer then? $\endgroup$– John BaezCommented Jul 26, 2017 at 10:28
-
12$\begingroup$ @JohnBaez $H^n(C_2\ast C_3,\mathbb{Z})\cong H^n(C_2,\mathbb{Z})\oplus H^n(C_3,\mathbb{Z})$, which is $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$ for $n=4$. $\endgroup$– Jeremy RickardCommented Jul 26, 2017 at 10:35
-
4$\begingroup$ The classifying space is the one-point union of those for $C_2$ and $C_3$, hence the homology is the direct sum (except in degree 0). $\endgroup$– ThiKuCommented Jul 26, 2017 at 10:36
-
3$\begingroup$ @JeremyRickard Well that's an anticlimax! $\endgroup$– David Roberts ♦Commented Jul 26, 2017 at 11:18
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
From Jeremy Rickard's comments, the group cohomology (with coefficients a module with trivial action) of a free product of (discrete) groups is sent to direct sum (eg Proposition 1.3.16.3 in C. Löh, Group Cohomology & Bounded Cohomology (pdf)), so $$ H^n(PSL(2,\mathbb{Z}),\mathbb{Z}) \simeq H^n(C_2\ast C_3,\mathbb{Z})\simeq H^n(C_2,\mathbb{Z})\oplus H^n(C_3,\mathbb{Z}), $$ and positive, even-degree cohomology of cyclic groups is $H^{2k}(C_n,\mathbb{Z}) = \mathbb{Z}/n$ ($k\gt 0$) and $0$ in odd degree, hence $H^4(PSL(2,\mathbb{Z}),\mathbb{Z}) = \mathbb{Z}/2 \oplus \mathbb{Z}/3$.