We fix $G=\mathrm{SL}_3(\mathbf{R})$.

Let $\Gamma$ be a torsion-free cocompact lattice in $G$. Is $b_2(\Gamma)=0$?

Here the second Betti number $b_2(\Gamma)$ is both the dimension of the cohomology group $H^2(\Gamma,\mathbf{Q})$ and the dimension of the de Rham cohomology in degree 2 of the locally symmetric space $\Gamma\backslash G/K$, $K=\mathrm{SO}(3)$.

From Kazhdan's Property T, I now that

- $b_1(\Gamma)=0$, and,
- for every finite index subgroup $\Lambda$ of $\Gamma$, the restriction map $H^2(\Gamma,\mathbf{Q})\to H^2(\Lambda,\mathbf{Q})$ is injective. In particular, $b_2(\Lambda)\ge b_2(\Gamma)$.

I do not know the answer to the question for a *single* example of $\Gamma$, so I would accept the answer in a single case.

Actually, for a fixed $\Gamma$, I would consider both a positive or negative answer as remarkable, because:

- If $b_2(\Gamma)=0$, then the (5-dimensional) locally symmetric space $\Gamma\backslash G/K$ is a rational homology sphere (see this MO question);
If $b_2(\Gamma)\neq 0$, then we have a central extension $1\to Z\to\widetilde{\Gamma}\to\Gamma\to 1$, with $Z\simeq\mathbf{Z}$, which does not split (and does not split in restriction to any finite index subgroup, since $\widetilde{\Gamma}$ inherits Property T). Since the fundamental group of $G$ is cyclic of order 2, one can deduce, using superrigidity see that any homomorphism of $\widetilde{\Gamma}$ into any connected Lie group is trivial on $2Z$ (and in particular $\widetilde{\Gamma}$ is linear). If so I'd be very curious about this exotic central extension. For instance, how is $Z$ distorted in $\widetilde{\Gamma}$? It cannot be more than quadratically distorted, because the Dehn function of $\Gamma$ is quadratic.

- [Edit, 2018 Dec 4] In addition, in the latter case case we have another pair of possibilities in which both alternatives appear as surprising. Indeed $H^2_\mathrm{b}(\Gamma,\mathbf{R})=0$ (vanishing of bounded cohomology: Theorem 1.4 of Monod-Shalom 2004). So either (a) in the above central extension, $Z$ is distorted (in contrast to central extensions coming from connected coverings of ambient Lie groups), or (b) $Z$ is undistorted, and this would be a central extension by $Z$ that not represented by a bounded cohomology class. I'm not sure this is known to exist. [/end edit]

In principle my question should be computer-answerable, if in a single case, one can implement a triangulation of the locally symmetric space, of reasonable size.

Additional contextual notes:

as far as I understand, the vanishing results (Matsushima, Zuckerman, Borel-Wallach...) for $b_2$ apply when $G$ is replaced by a simple Lie group of real rank $\ge 3$, hence don't apply here.

by Abert-Bergeron-Biringer-Gelander-Nikolov-Raimbault-Samet (Annals of Math 2017), we have, in $G$ arbitrary simple Lie group of rank $2$ and finite center, and $\Gamma_n$ strictly decreasing sequence of cocompact lattices, $b_2(\Gamma_n)=o([\Gamma:\Gamma_n])$.

for non-cocompact lattices in $G=\mathrm{SL}_3(\mathbf{R})$, the picture is a bit different since the (rational) cohomological dimension is 3 or 4 (rather than 5) and there is no Poincaré duality. For instance, for a finite index subgroup of $\mathrm{SL}_3(\mathbf{Z})$, the rational cohomological dimension is 3, the Euler characteristic is 0, and hence we have $(b_0,b_1,b_2,b_3,b_4,b_5)=(1,0,b_2,1+b_2,0,0)$. We can indeed (typically) have $b_2>0$: it is proved by Ash (Bull AMS, 1977), for $\Gamma=\mathrm{Ker}(\mathrm{SL}_3(\mathbf{Z})\to \mathrm{SL}_3(\mathbf{Z}/7\mathbf{Z}))$, that $b_2\ge 5814$. Is this evidence that cocompact lattices should also have nonzero $b_2$, I don't know.