# Groups without property (T) but all finite quotients are expanders

What is an example of a group $G$ which

1- is finitely generated by $S$,

2- does not have property (T),

3- admits infinitely many finite quotients which do not factor through an homomorphism $G \to H$ for some [fixed & infinite] property (T) group $H$

4- all the Cayley graphs (w.r.t. $S$) of those finite quotients are $\epsilon$-expanders (for some fixed $\epsilon >0$)

The question is motivated by the false assertion: "a group has property (T) if and only all its finite quotients are $\epsilon$-expanders".

There are "simple" answers if one does not put the factor condition in 3. For example, one could consider $A \times H$ where $A$ is a finitely generated simple amenable group (as proven/constructed by Juschenko-Monod/Matui) and $H$ is some residually finite property (T) group ($SL_3(\mathbb{Z})$ being the canonical choice). Because $A$ is amenable, this does not have (T) and its finite quotients (which all come from $H$) are $\epsilon$ expanders.

Actually, to enlarge the list of examples of this form, a side question would be: what are the finitely generated groups which do not have property (T) and do not have finite non-trivial quotients?

• 3 is impossible because the finite quotient itself has Property T. But in the motivation given by the $A\times H$, all the finite quotient factor through the same property T group, which is of course much stronger. So I thing 3 has to be reformulated. – YCor Oct 5 '15 at 16:32
• Still 3 (as now modified) sounds weird to me. Why not just asking that the group is residually finite? – YCor Oct 5 '15 at 20:05
• You are perfectly right, of course. As I was discussing this with a colleague it seemed like condition 3 would insure more novel (to us) finite quotients. – ARG Oct 6 '15 at 8:54
• Note that the current 3 means: $G/R$ does not have Property T, where $R$ is the intersection of all finite index subgroup. – YCor Jun 24 '16 at 14:52

I think your condition (4) is called "property ($\tau$)". (See Theorem 4.3.2 of Lubotzky's very nice book "Discrete Groups, Expanding Graphs and Invariant Measures".) An example is $G = \mathrm{SL}_2\bigl(\mathbb{Z}[1/p] \bigr)$. (See Example 4.3.3E on page 52 of Lubotzky's book.)

• I think Dave's example qualifies as an answer. Another example would be $SL_2(\mathbb{Z}[\sqrt{p}])$. – Alain Valette Oct 5 '15 at 19:54
• I think it qualifies as a correct answer! I'm not sure about $SL_2(\mathbb{Z}[\sqrt{p}]$ but the expanders from $SL_2(SL_2(\mathbb{Z}[1/p])$ are the same as those obtained from $SL_2(\mathbb{Z})$, which I wanted to avoid. But there does not seem to be a sensible way to put this in the conditions... – ARG Oct 5 '15 at 19:58

A bit late, but let me draw your attention to my first paper:

http://arxiv.org/abs/1005.4566

It shows that property (T) is not determined by the finite quotients, and gives in particular an answer to your question.

Later edit -- Yves' comment calls for more details:

The aim of the above mentioned note is to show that property (T) is not profinite, that is, it is not determined by the set of the finite quotient of the group (a nice exercise is to show that considering the set of isomorphism classes of finite quotient or considering the set of finite quotients counting multiplicity amount to the same - so there is no ambiguity in the above definition). For finitely generated groups, having the same set finite quotients is equivalent to having the same profinite completion.

Now I can explain the simple idea behind the construction. The Congruence subgroup Property (CSP) essentially transforms this problem into a local-to-global problem:

Essentially, under CSP, the profinite completion of $G(\mathbb Z)$ is $$\widehat{G(\mathbb Z)}=\prod_p G(\mathbb Z_p).$$

So in order to have two groups with the same profinite completion we look for two groups $G_1(\mathbb Z)$ and $G_2(\mathbb Z)$ (with CSP) such that they are isomorphic locally. If, moreover, we can arrange that $G_1(\mathbb Z)$ (resp. $G_2(\mathbb Z)$) is a lattice in a product of rank 1 (resp. high rank $\geq 2$) groups, we win, as this should imply that the first does not have $(T)$ while the second has $(T)$.

From this, one is led to the construction naturally. One uses the fact that the quadratic forms $$\sum^n x_i^2-\sum^m y_i^2,\quad \sum^{n-4} x_i^2-\sum^{m+4} y_i^2$$ agrees everywhere locally, but not globally. This leads to the Spin groups construction that appear in Yves' comments.

Another sentence of advertising: This construction is one of three types of constructions that I consider in a different paper, in order to show that the set of isomorphism classes of arithmetic groups in high rank which have the same profinite completion is finite.

• It can be mentioned that one is a lattice in $\mathrm{Spin}(1,n)^2$ (which has (real) rank 2 with simple factors of rank 1) and one is a lattice in $\mathrm{Spin}(5,n-4)^2$ with $n\ge 6$ (the latter has rank $2\min(5,n-4)\ge 4$, with simple factors of rank $\min(5,n-4)\in\{2,3,4,5\}$). – YCor Jun 24 '16 at 11:57