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Shalom [edit: originally M. Burger] showed that the pair $(\mathrm{SL}_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$ has Relative Property (T) with respect to standard generating sets.

(The action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{Z}^2$ is the usual one, i.e. the semidirect product can be thought of as a group of affine transformations $x \mapsto A x + b$ where $A \in \mathrm{SL}_2(\mathbb{Z})$ and $b \in \mathbb{Z}^2$.

If we reduce $\mathrm{mod}\ p$, we can think of this as giving an "efficient" way of generating the translations $x \mapsto x + b$ for $b \in F_p^2$.)

A one-dimensional variant in the finite case is whether there exist bounded size subsets $S_p \subset F_p^{\times} \ltimes F_p$ and $\delta > 0$ such that the relative Kazhdan constant:

$\kappa\ (F_p^{\times} \ltimes F_p, F_p, S_p) \ge \delta$

i.e. whether the pairs $(F_p^{\times} \ltimes F_p, F_p)$ can form a relative expander family.

An equivalent formulation: do there exist bounded size sets $S_p$ of affine transformations on $F_p$, such that no non-empty subset $U \subset F_p$, $|U| \leq p/2$ is almost invariant with respect to all of them, i.e.

$\neg \exists U: \forall s \in S: |s(U) \cap U| > \frac{99}{100} |U|$

I believe the answer is no if one uses standard "generating" sets (they needn't actually generate) such as $x \mapsto x + 1,\ x \mapsto ax$, even if $a$ is allowed to vary with $p$. This is very slightly surprising, as these do generate all translations "efficiently" in the weaker sense of logarithmic diameter.

Is there a good argument as to why this should fail in general? Or might there be cunning sets $S_p$ such that relative expansion occurs?

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That (SL$_2(\mathbb Z) \ltimes \mathbb Z^2, \mathbb Z^2$) has relative property (T) was actually first explicitly shown by Margulis (Ergod. Th. & Dynam. Sys., 2:383–396, 1982). –  Jesse Peterson Mar 19 '12 at 20:51
    
Thanks! Since I'm not that familiar with the language that paper is couched in, could you maybe point out which result there is equivalent to relative (T) for $(SL_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$? –  Freddie Manners Mar 19 '12 at 21:30
    
@Freddie: Lemma 1 in his paper shows that (SL$_2(\mathbb R) \ltimes \mathbb R^2, \mathbb R^2$) has relative property (T), (which was already implicit in Kazhdan's original paper). I am mistaken in that he does not seem to explicitly state the case when $\mathbb R$ is replaced by $\mathbb Z$. –  Jesse Peterson Mar 19 '12 at 23:23
    
@Jesse: right, he then says some things about finite covolume lattices, which I thought might have given the result for $\mathbb{Z}$ implicitly, but it's possible it says something subtly different. Looking again, Shalom cites Burger's 1991 paper (Kazhdan constants for $\mathrm{SL}_3(\mathbb{Z})$), who doesn't appear to cite anyone else, so I think I'll give the credit to him. –  Freddie Manners Mar 19 '12 at 23:37
    
Here's an idea: If the result is true, can't we use it to prove that $\text{GL}_2(\mathbb{F}_p)$ is an expander family, in much the same way that the we use the $\text{SL}_2(\mathbb{F}_p)\ltimes\mathbb{F}_p^2$ result to prove that $\text{SL}_3(\mathbb{F}_p)$ is an expander family? Now $\text{GL}_2(\mathbf{F}_p)$ has $\mathbf{F}_p^\times$ as a quotient, so it's definitely not an expander family. –  Sean Eberhard Mar 19 '12 at 23:52

1 Answer 1

up vote 6 down vote accepted

I think that one can show that $(F_p^\times \ltimes F_p,F_p)$ does not have relative property (T), by using the combinatorial proof that semidirect products of amenable groups are again amenable (see Proposition 5 of my notes at http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/ ). A bit more specifically, let $S_p$ be a bounded set of affine transformations on $F_p$, and let $D_p$ be the associated set of dilations on $F_p$, which is then a bounded subset of $F_p^\times$. By the abelian nature of the dilation group, we can then construct a nontrivial subset $F$ of the dilation group $F_p^\times$ which is $99.9\%$-invariant with respect to the dilations of $D_p$ (thus $|dF \Delta F| \leq 10^{-3} |F|$ for all $d \in D_p$), basically by constructing a medium-sized generalised geometric progression using the dilations in $D_p$ as generators. Note that one can make the set $F$ of bounded size (i.e. independent of $p$.) Let $T$ be the set of all translations $t$ such that $SF$ intersects $Ft$ (i.e. the translations in $F^{-1} S_p F$). This is a set whose size is controlled by the size of $S_p$ and of $F$, and in particular is still bounded uniformly in $p$. We can then construct a moderately large (but still of size uniformly bounded in $p$) set $E$ of translations which is $99.9\%$-invariant with respect to any of the translations of $T$. The set $U := EF$ will then (for $p$ large enough) be a non-trivial $99\%$-invariant set with respect to $S_p$, by the argument given in my notes above.

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Thanks! Putting that together with your blog post, I think "set of all group elements $t \in K$" should read ".. $t \in H$" in para 3, proof of Proposition 5. –  Freddie Manners Mar 20 '12 at 11:44
    
So, this nicely encapsulates my approach in special cases. E.g. if our generators are $x \mapsto x + 1$ and $x \mapsto 2 x$, you can say something like "take a bunch of elements in $F_p$ with the same low bits when written in base 2", since it still vaguely makes sense to talk about elements of $F_p$ in base 2; but this fails if you switch 2 for $a \approx p^{1/10}$. So, the generalization here is "take elements of $F_p$ which have distance at most 100 from 0 on the edges: $x \mapsto x + a^r$, for $0 \leq r < 100$" which -- in hindsight -- is clearly the correct generalization. –  Freddie Manners Mar 20 '12 at 11:49
    
Well, maybe "distance at most 100^100" or so. –  Freddie Manners Mar 20 '12 at 11:56

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