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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.


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.

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  • $\begingroup$ It seems plausible that surface subgroups represented by sublattices in $O(2,1;\mathbb{R})$ might be homologically non-trivial. If one could find a sublattice of complementary dimension intersecting this generically, then one ought to be able to prove that it becomes homologically non-trivial in a finite-index subgroup using double coset separability arguments. $\endgroup$ – Ian Agol Oct 27 '18 at 18:19
  • $\begingroup$ Maybe one can compute $b_2$ for the following non-division algebra example. Can't remember where I saw it, but it's a standard kind of thing. Take $F = Q(\sqrt{2})$ and $L = Q(\sqrt[4]{2})$. Take the unitary group $\Gamma = SU_3(A)$ , for the trivial Hermitian form and nontrivial element of $Gal(L/F)$, and $A$ a suitable ring of integers. Then $\Gamma$ will embed into $SL_3(R)$ (at one place of $F$) as a cocompact lattice... not sure about torsion. $\endgroup$ – Marty Oct 28 '18 at 16:38
  • $\begingroup$ @Marty cocompact lattices in $\mathrm{SL}_3(\mathbf{R})$ are reasonably classified (they're arithmetic and the ways to produce arithmetic lattices are classified, for instance in Witte-Morris' book arxiv.org/abs/math/0106063, §6.7, at least up to commensurability). $\endgroup$ – YCor Oct 28 '18 at 16:49
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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.)

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  • $\begingroup$ Great, thank you. Your $\Gamma_n$ are, in addition, torsion-free by construction. A side question: in your article, beginning of page 13 (published, p721) you write "So every arithmetic subgroup of $\mathrm{SL}_3$ contains a cofinal $p$-tower (for every odd prime p) with some Betti number growing faster than the square root of the index." Here you write the exponent $3/8$ (and not $1/2$) and you say that you "did not check the other examples of cocompact lattices in $\mathrm{SL}_3(\mathbf{R})$". How can I conciliate these statements? $\endgroup$ – YCor Oct 29 '18 at 16:03
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    $\begingroup$ @YCor: The example on p.721 refers only to the split algebraic group $\mathrm{SL}_3$ defined over $\mathbb{Q}$, i.e. to certain non-cocompact lattices. Cocompact lattices are defined using other algebraic groups (e.g. $\mathrm{SU}(h,L/F)$ as above) and one needs to find a suitable automorphism for each of these groups. The exponent $3/8$ is the ratio of the dimension of the fixed point group $\mathrm{SO}(2,1)$ and the dimension of $\mathrm{SL}_3(\mathbb{R})$. In the non-cocompact cases above one can do slightly better (and find a $4$-dim fixed point group). $\endgroup$ – Steffen Kionke Oct 29 '18 at 16:24
  • $\begingroup$ Thanks. If I understand correctly, the cocompact lattices commensurability-conjugation correspond to $\mathbf{Q}$-anisotropic $\mathbf{R}$-split $\mathbf{Q}$-forms of $\mathrm{SL}_3$. These come into two families: $\mathrm{SU}(h,L/F)$ as you consider above, and the other family are the $\mathrm{SL}_1(D)$ where $D$ is a central division algebra of degree 3, itself $\mathbf{Q}$-anisotropic and $\mathbf{R}$-split. So one has to find a non-identity finite order $\mathbf{Q}$-defined automorphism of $\mathrm{SL}_1(D)$? (Having positive-dimensional set of fixed points is not an issue.) $\endgroup$ – YCor Oct 29 '18 at 17:14
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    $\begingroup$ Yes, you need a finite order automorphism of $\mathrm{SL}_1(D)$ defined over $\mathbb{Q}$ such that the group of fixed points $G$ has positive dimension and its arithmetic subgroups have non-zero Euler characteristic. (As a remark: all automorphisms of $D$ are inner by Skolem-Noether, but I don't see how one can write down useful elements in $D$.) $\endgroup$ – Steffen Kionke Oct 29 '18 at 17:45
  • $\begingroup$ Actually, I have checked that the only finite order, non-identity automorphisms of $\mathrm{SL}_1(D)$ have order 3, and for each such automorphism, the connected component of the group of fixed points is a 2-dimensional $\mathbf{Q}$-anisotropic torus— so the arithmetic subgroups have $\chi=0$. (The group of algebraic automorphisms of $\mathrm{SL}_1(D)$, for $D$ central division algebra of degree 3, is reduced to $\mathrm{PGL}_1(D)$, i.e., is not twice bigger.) $\endgroup$ – YCor Nov 5 '18 at 17:26
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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 $\Gamma$.

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  • $\begingroup$ @Venkataramana Thanks, that is probably why I should have checked the details... :-) There is something I don't understand in your commment : why is the representation contributing to $H^2$ not tempered? Are you saying that the $H^2_{\rm cusp}$ is trivial for congruence subgroups of ${\rm SL}_3(\mathbb{Z})$? Many examples have been computed and these cohomology spaces are non-trivial. $\endgroup$ – Aurel Oct 28 '18 at 11:04
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    $\begingroup$ @Venkataramana: I think, it is an old result of Lee-Schwermer that there is a lot of cuspidal cohomology for principal congruence subgroups of $SL_3(\mathbb{Z})$. See link.springer.com/article/10.1007%2FBF01394023 $\endgroup$ – Steffen Kionke Oct 29 '18 at 13:33
  • $\begingroup$ @Kionke: Sorry. Representations contributing to $H^2$ for SL(3) is indeed tempered, so my comments were wrong (and deleted). $\endgroup$ – Venkataramana Oct 29 '18 at 15:24
  • $\begingroup$ Steffen's method does not apply to arithmetic subgroups of $\mathrm{SL}_1(D)$, if I'm correct. So it would be interesting to check this case. Note: $\mathrm{SL}_1(D)$ is a torsion-free group (for any central cubic division algebra $D$ over $\mathbf{Q}$). $\endgroup$ – YCor Nov 5 '18 at 17:36

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