Second Betti number of lattices in $\mathrm{SL}_3(\mathbf{R})$

Question

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

1. $b_1(\Gamma)=0$, and,
2. 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.

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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.

Answer 1

The arithmetic cocompact lattices constructed in (6.7.1) of Witte-Morris' book all have torsion-free finite index subgroups with arbitrarily large second Betti number.

I will briefly recall the construction because this is necessary for the answer. Let $F$ be a totally real number field with elements $a,b,t \in F$ such that $\sigma(a), \sigma(b), \sigma(t) < 0$ for all but one real $\sigma$ embedding of $F$. Let $L = F(\sqrt{t})$ be the degree $2$ extension with Galois group $\mathrm{Gal}(L/F) = \{\mathrm{id}, \tau\}$ and let $\mathcal{O}_L$ denote the ring of integers of $L$. Define $$h = \left(\begin{smallmatrix}a & & \\ & b & \\ & & -1\end{smallmatrix}\right)$$ The arithmetic group $$\Gamma = \{ g \in \mathrm{SL}_3(\mathcal{O}_L) \mid \tau(g)^T h g = h\} \subseteq \mathrm{SU}(h,L/F) $$ embeds as a cocompact lattice in $\mathrm{SL}_3(\mathbb{R})$ (via any one of the two embeddings $L \to \mathbb{R}$).

The non-trivial Galois automorphism $\tau$ of $L/F$ induces an automorphism of the algebraic group $\mathrm{SU}(h,L/F)$ which restricts to an automorphism $\tau$ of $\Gamma$. In particular, we get automorphisms $$\tau^j\colon H^j(\Gamma,\mathbb{C}) \to H^j(\Gamma, \mathbb{C})$$ in the cohomology. It is possible to calculate (or to bound) the Lefschetz number $$ L(\tau) = \sum_{j=0}^5 (-1)^j\mathrm{Tr}(\tau^j) $$ in the cohomology of $\Gamma$. The methods for this have been worked out by Jürgen Rohlfs (and others) in the 80's and 90's. The trick is to use the Lefschetz fixed point theorem on the associated locally symmetric space, which says that the Lefschetz number is the Euler characteristic of the set of fixed points. In the specific example the fixed point set should consist of a bunch of surfaces (and a couple of isolated points). Indeed, on $\mathrm{SL_3}(\mathbb{R})$ the automorphism is $g \mapsto h^{-1}(g^{-1})^Th$ and the group of fixed points is isomorphic to $SO(2,1)$. Since surfaces have non-zero Euler characteristic the method yields a lower bound for the cohomology. Here one can find a decreasing sequence of finite index subgroups $\Gamma_n \leq \Gamma$ such that $$ \sum_{j=0}^5 b_j(\Gamma_n) \gg [\Gamma:\Gamma_n]^{3/8} $$ as $n \to \infty$; see Theorem 4 in my article.

Asymptotically we have the same lower bound for $b_2(\Gamma_n)$. We have Poincare duality and by property (T) we know that $b_1(\Gamma_n) = 0 =b_4(\Gamma_n)$. Since $b_0(\Gamma_n) = b_5(\Gamma_n) = 1$ and we can only have interesting cohomology in degrees $2$ and $3$, and moreover $b_2(\Gamma_n) = b_3(\Gamma_n)$.

(In the case of non-cocompact lattices a similar argument has been carried out by Lee and Schwermer. I did not check whether the other examples of cocompact arithmetic lattices in $\mathrm{SL}_3(\mathbb{R})$ have a useful algebraic finite order automorphism.)

Answer 2

This question came up in a recent research we hold with Lubotzky, Sauer and Weinberger. I will share our findings.

Claim: For every cocompact lattice $\Gamma$ in $G=\text{SL}_3(\mathbb{R})$ there exists a finite index lattice $\Gamma_1\subseteq\Gamma$ with $b_2(\Gamma_1)\neq 0$ (thus also $b_2(\Gamma_2)\neq 0$ for every finite index $\Gamma_2\subseteq \Gamma_1$).

Proof: We assume without loss of the generality that $\Gamma$ is torsion free. By [1, Theorem B], there exists a finite index subgroup $\Gamma_0\subseteq \Gamma$ which surjects on $\mathbb{Z}/2\times \mathbb{Z}/2$. We let $\Gamma_1\lhd \Gamma_0$ be the kernel of this surjection and consider $M=\Gamma_1\backslash G/K$. Note that $\Gamma_0/\Gamma_1$ acts on $M$. By [2, Theorem D] (due to Davis and Weinberger) we get that $M$ is not a rational homology sphere. Since $\Gamma_0$ is a Poincare duality group of dimension 5 and $b_1(\Gamma)=0$ (by property T), we conclude that $b_2(\Gamma_1)\neq 0$. $\square$

The claim also holds for nonuniform lattices by a different method.

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[1]: Lubotzky, Alexander On finite index subgroups of linear groups. Bull. London Math. Soc. 19 (1987), no. 4, 325–328.

[2]: Davis, James F. The surgery semicharacteristic. Proc. London Math. Soc. (3) 47 (1983), no. 3, 411–428.

Answer 3

Disclaimer: I have not had time to check the details yet, but I think the following should work.

Let $f$ be the unique newform of level $30$, weight $2$ and trivial character. Let $\pi(f)$ be the automorphic representation of ${\rm GL}_2$ corresponding to $f$.

Let $\sigma$ be the Gelbart-Jacquet symmetric square lift of $\pi(f)$ to ${\rm GL}_3$. It has regular weight and is therefore cohomological.

Let $D$ be the division algebra over $\mathbb{Q}$ of dimension $9$ with invariants $1/3$ at $2$, $2/3$ at $3$ and $0$ elsewhere. Let $\Gamma$ be the group of elements of reduced norm $1$ in a maximal order of $D$. This is a cocompact lattice in ${\rm SL}_3(\mathbb{R})$. Let $\tau$ be the Jacquet-Langlands transfer of $\sigma$ to $D^\times$, which exists because $\sigma$ is Steinberg at $2$ and $3$ (by work of Badulescu), and is still cohomological.

Then by Matsushima's formula, $\tau$ contributes nontrivially to the $H^2$ and $H^3$ of some torsion-free congruence subgroup of $\Gamma$.