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We say $G$ is the Zappa-Szep product of two subgroups $K$ and $P$ if $K\cap P = \{e\}$ and the function $K\times P \to G$, $(k,p)\mapsto kp$, is bijective.

The Iwasawa decomposition shows that we can have amenable $K$ and $P$ such that $G=KP$ is non-amenable. However in such examples $K$ is usually not amenable when considered as a discrete group.

Where is a good place for me to look for examples of "naturally occurring" Zappa-Szep products where the two "factors" are amenable as discrete groups, yet the larger group itself is non-amenable? Can this ever happen for e.g. lattices in semisimple Lie groups?

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    $\begingroup$ The first example is $\text{SL}_2(\mathbb{R})$, where $K=\text{SO}(2)$ is abelian and $P$, the upper-right matrices group, is solvable. $\endgroup$ – Uri Bader Jan 9 '18 at 7:59
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    $\begingroup$ Other simple Lie groups will not have this property (unless I am forgetting a very special case). This is true also over the $p$-adics. Lattices in such (including the $\text{SL}_2$ case) will not be either - they have very few amenable subgroups. $\endgroup$ – Uri Bader Jan 9 '18 at 8:08
  • $\begingroup$ @UriBader Thanks - I was going to include SL(2,R) in the question but was typing in haste. I someow think of SL(2,R) with discrete topology as not very natural, so what I was hoping for was something countable and finitely presented. But I admit that this is just wishful thinking $\endgroup$ – Yemon Choi Jan 9 '18 at 8:35
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Here is an observation of Yair Glasner that I previously missed.

Let $k$ be the field of real algebraic numbers and consider $G=\text{PSL}_2(k)$. Let $P=B(\mathcal{k})$ be the standard Borel subgroup and $K=\text{PSO}_2(k)$. Then $K$ and $P$ are solvable, $K\cap P=\{e\}$ and $G=KP$, but $G$ is not amenable (as it contains free groups). All countable examples I am aware of are variants of this example. In particular, I am not aware of any finitely generated example.

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  • $\begingroup$ Thanks a lot for this answer. I'll leave it for a while in case any other answers turn up. $\endgroup$ – Yemon Choi Jan 12 '18 at 20:47
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    $\begingroup$ You need $k$ to consist of real algebraic integers, or at least to have no square root of $-1$. Indeed otherwise the matrix $\begin{pmatrix}1 & 0\\ i & 1\end{pmatrix}$ is not in $KP$. Actually, if you consider $PGL_2$ (to avoid determinant issues), any field $k$ in which $-1$ is not a square works. $\endgroup$ – YCor Jan 14 '18 at 7:39

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