Cohomological gap in arithmetic groups

Question

$\DeclareMathOperator\SL{SL}$For the sake of this question, let's say that a group $G$ of finite cohomological dimension $n$ has a cohomological gap if for some $0 < i < n$ there is no subgroup $H$ of $G$ with cohomological dimension $\operatorname{cd}(H) = i$.

The question is, can an arithmetic group have a cohomological gap?

I believe the answer is no for torsion-free subgroups of $\SL_n(\mathbb{Z})$, and the same reasoning should apply for torsion-free subgroups of semisimple Chevalley groups over $\mathbb{Z}$. For $\SL_n(\mathbb{Z})$, one just needs to look inside the upper triangular subgroup to find subgroups of all dimensions. My reasoning is this generalizes to Chevalley groups as follows: the Iwasawa decomposition for such a group $G$, viewed as a linear algebraic group, gives us $G(\mathbb{R}) \sim K \times A \times N$, where $K$ is a maximal compact subgroup, $A$ is a maximal torus and $N$ is the subgroup generated by the exponentials of all positive roots. Hence, the dimension of the symmetric space $X$ associated to $G(\mathbb{R})$ is $\dim X = \dim A + \dim N$. Given that this groups split over $\mathbb{Q}$, the $\mathbb{Q}$-rank of $G(\mathbb{Z})$ is equal to $\dim A$. Therefore, as a consequence of the Borel-Serre construction, the virtual cohomological dimension of $G(\mathbb{Z})$ is $\dim N$. However, $N(\mathbb{Z})$ is a nilpotent subgroup of $G(\mathbb{Z})$ of Hirsch length $\dim N$, which gives us the bounds $$\operatorname{vcd}(G(\mathbb{Z})) = \dim N \leq \operatorname{cd}(N(\mathbb{Z}))\,.$$ Since nilpotent groups have no cohomological gap, we can find subgroups of all dimensions in $G(\mathbb{Z})$ by looking inside $N(\mathbb{Z})$.

I don't know how to approach this for general arithmetic or $S$-arithmetic groups. It's quite possible there might be a gap in "relatively small" groups such as torsion-free subgroups of $\SL_3(\mathbb{Z}[\sqrt{2}])$ or cocompact lattices in dimension $\geq 4$. This question is motivated by this related question in the Mathematics StackExchange.

Answer 1

This is to record our current lack of understanding of cohomological dimensions of subgroups of arithmetic groups. I will limit myself to the case of uniform lattices since the situation with nonuniform ones is quite different (as you observed in the $SL(n,\mathbb Z)$ case, one can find unipotent subgroups in nonuniform lattices of all possible intermediate dimensions).

1. Arithmetic lattices of $SO(n,1)$. There are three families of such lattices: (a) The simplest type (associated with quadratic forms over number fields), (b) quaternionic type (associated with division algebras over number fields, they exist for odd $n$), (c) two exotic types in the case $n=7$, associated with rational structures on $SO(7,1)$ coming from the triality (automorphisms of the Dynkin diagram $D_4$). Unfortunately, the only reference I have to the class (c) is page 58 of

J. Tits, Classification of algebraic semisimple groups, Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965), Proc. Sympos. Pure Math., vol. 9, American Mathematical Society, Providence, 1966, pp. 33–62.

namely, ${}^3D^9_{4,1}$ and ${}^6D^9_{4,1}$.

I wish somebody had written a detailed description of this class of arithmetic lattices using a more "concrete" language of something like Jordan algebras...

In all lattices of types (a) and (b) there are subgroups of all intermediate cohomological dimensions (no gaps). This comes from existence of finite index subgroups in such lattices with infinite abelianization (Millson and Li for $n\ge 4$): This produces subgroups of codimension 1 (kernels of epimorphisms to $\mathbb Z$), which suffices. Additionally, one uses the existence of surface subgroups due to Kahn and Markovic. Very little is known in the case of type (c), apart from existence of free subgroups and surface subgroups (and their amalgams which still have cohomological dimensions 1 and 2).

2. Arithmetic lattices of $SU(n,1)$. There are two families of arithmetic lattices: (a) The simplest type (hermitian forms over number fields) and (b) quaternionic type. In the case of type (a), again there are no gaps because of Kazhdan's theorem about existence of finite index subgroups with infinite abelianization.

Again, unfortunately, the only references I know is Kazhdan's original paper and the book by Borel and Wallach:

A. Borel and N. R. Wallach, Continuous cohomology, discrete subgroups, and representations of reductive groups, Ann. of Math. Stud., 94, Princeton Univ. Press, Princeton, NJ, 1980

Both I find unreadable (and this is not for a lack of trying on my part). Ideally, one should be able to write down a Poincare series associated with some holomorphic 1-form on the complex ball that would represent the corresponding cohomology classes (after passage to a sufficiently deep congruence subgroup).

In the case of type (b) all what we know is that there are subgroups of cohomological dimension divisible by 4, and dimensions 0, 1, 2. The first case where we are completely ignorant is $SU(2,1)$ where the existence of subgroups of cohomological dimension 3 is unknown.

3. Lattices in isometry groups of remaining real rank 1 Lie groups $G$ (isometry groups of quaternionic hyperbolic spaces and octonionic hyperbolic plane). Again, very little is known about gaps besides trivialities. For instance if $G$ is the isometry group of the quaternionic hyperbolic $n$-space, $n\ge 2$, then uniform lattices have $vcd=4n$ and there are sublattices of intermediate dimensions $4k, 1\le k< n$. Additionally, there are subgroups of $vcd=0, 1, 2, 3$. But nothing else is known. In the octonionic case, $vcd=8$ and there are subgroups of $vcd=0, 1, 2, 3, 4$. Nothing is known about the remaining dimensions.

4. Irreducible (uniform) lattices Lie groups of rank $n\ge 2$. Again, very little is known. There are free abelian subgroups of ranks $0,...,n$ (Prasad), free subgroups (Tits), as well as (in some cases) thin surface subgroups (Labourie et al) and, again, in some cases, sublattices. But, if you restrict, say, to $G=SL(3,\mathbb R)$, for "generic" lattices, no subgroups of dimensions $3,...,7$ are known. Similarly, for irreducible uniform lattices in $SL(2,\mathbb R)\times SL(2,\mathbb R)$: There are subgroups of $vcd=0,1,2,4$, but the case of $vcd=3$ is wide-open, as far as I know. Basically, it is not even clear where to look for these and maybe they simply do not exist.

Edit 1: In the nonuniform case, if, say, the $\mathbb Q$-rank equals the real rank of the Lie group then the argument in your question goes through. However, in general, the situation with nonuniform lattices is also far from clear.

Edit 2: Consider the case of $SO(n,1)$ and a type (a) uniform arithmetic lattice $\Gamma$ preserving a $K$-rational quadratic form $q$ of real signature $(n,1)$. Let $v\in K^{n+1}$ be a space-like vector. Then $Kv$ is stabilized by a $K$-rational subgroup of $SO(n,1;K)$ isomorphic to $SO(n-1,1;K)$ and its intersection with $\Gamma$ is again an arithmetic lattice $\Gamma'$ of type (a), now in $SO(n-1,1)$. Thus, $\Gamma'$ has $vcd=n-1$. The argument in the case of type (b) lattices is similar: $\Gamma< SO(2m+1,1)$, it contains a sublattice $\Gamma'<SO(2m-1,1)$, again of type (b), etc. Thus, $\Gamma$ contains uniform sublattices $\Gamma_i$ of $vcd=2i+1, i=0, 1,..., m$ (every odd dimension). Every such sublattice $\Gamma_i$ virtually maps onto $\mathbb Z$; with the kernel of $vcd=2i$. Thus, there are no gaps. The proof in the case of lattices of type (a) in $SU(n,1)$ is very similar: One gets sublattices of all even intermediate dimensions and then thin subgroups of all odd intermediate dimensions.