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?



Is it known that $(F_p^{\times} \ltimes F_p, F_p)$ is not a relative expander family?
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.


