Timeline for How to Prove that nilpotent Lie groups satisfy the Leptin condition?
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| Jul 6, 2010 at 19:43 | comment | added | Yemon Choi | Perhaps there is some discussion of this in Paterson's book on "Amenabillity", if your library has a copy? Unfortunately I am "between libraries" at the moment owing to a change of institution, so I can't go and look up precise references as easily as usual. | |
| Jul 6, 2010 at 10:50 | comment | added | Gian Maria Dall'Ara | Let's say we are able to find a compact neighborhood $V$ of the identity in $G$ such that $V^n$ has measure $\approx n^D$. If $K = V^n$ then by taking $L = V^m$ we have $\frac{\mu(V^{n+m})}{\mu(V^m)}\approx (1+\frac{n}{m})^D$. Maybe I'm missing something evident, but this is less then Leptin's condition (due to the $\approx$). | |
| Jul 5, 2010 at 18:00 | comment | added | Yemon Choi | I'm away from my references so can't give a proper answer, but I think the key point is that nilpotency implies your group has polynomial growth (that is, given some subset A, the measure of $A^n$ is polynomial in n not exponential). I think there has also been some work on constructing explicit Folner-type sets (which wuld seem to correspond to your "Leptin's property") in various semidirect products, but I'd have to check this later | |
| Jul 5, 2010 at 17:22 | history | asked | Gian Maria Dall'Ara | CC BY-SA 2.5 |