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A topological group $G$ is said to satisfy the Leptin condition if for every compact subset $K\subseteq G$ and for every $\epsilon>0$ there exists a compact subset $L$ such that

$\mu(LK)$ < $ (1+\epsilon)\mu(L)$

where $\mu$ is Haar measure (I'm assuming $G$ locally compact hausdorff). It is known that the Leptin condition is equivalent to amenability, a very different looking property concerning the existence of an invariant mean on the group. How can one prove that nilpotent lie groups satisfy the Leptin condition without passing through amenability? For example $\mathbb{R}^n$ almost trivially satisfies such condition, even if proving amenability is harder. Thanks in advance!

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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 – Yemon Choi Jul 5 '10 at 18:00
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 $\approx$). – Gian Maria Dall'Ara Jul 6 '10 at 10:50
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. – Yemon Choi Jul 6 '10 at 19:43

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