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I would like to prove that if $H\subset G$ is a normal amenable subgroup such that $G/H$ is amenable, then $G$ is amenable. The definition of amenability that I'm using is the following:

A group $G$ is amenable if every action of $G$ by homeomorphisms of a compact metric space admits an invariant probability measure.

This definition can be found on Navas's "Groups of Circle Diffeomorphisms". I've tried a lot of different ways but i couldn't prove it, I know there are many equivallent definitions for amenability but I would like (if possible) a proof that only uses this definition.

Here is what I've done so far: If $G$ acts on $(M,d)$ then $G/H$ acts on $M/H$ (the quotient of $M$ by the orbits of $H$), the problem is that this group is not necessarily metric, whe could endow the quotient group with the pseudometric $d'$ given in wikipedia https://en.wikipedia.org/wiki/Metric_space#Quotient_metric_spaces (the topology could be weaker than the quotient topology), and then do another quotient $X=(M/H)/\sim$ where $[x]\sim [y]$ if $d'([x],[y])=0$. Here $X$ is a compact metric space and we could take the action of $G/H$ on $X$ given by ${[g]}({[[x]]})=[[y]]$ if $[[g(x)]]=[[y]]$, since $G/H$ is amenable there exists an invariant probability measure, namely $\nu$. Now the sets $A_{[[x]]}=\lbrace y\in M:[[y]]=[[x]]\rbrace$ are compact and invariant under the action of $H$, so each one has an invariant probability measure namely $\mu_{[[x]]}$ and we could define the probability measure on $M$ as $$\mu(B)=\int_X \mu_{[[x]]}(B\cap A_{[[x]]})d\nu([[x]]).$$

I don't know if this works in general, i couldn't prove or disprove it, i suppose this doesn't work since there could be some internal shifting of the orbits of $H$ in the sets $A_{[[x]]}$, but I hope this gives you some insight of what I'm trying so far.

I hope I was clear, many thanks in advance.

Something that might help: The space of probability measures on a metric space is compact, so you could use convergence of probability neasures.

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    $\begingroup$ Have you seen this direct proof here, using the Følner property? $\endgroup$ Dec 14 '20 at 15:08
  • $\begingroup$ Yes I’ve seen many direct proofs using the følner property, the problem is that i’ve not found any proof of the equivalence of the definitions. I can proove følner implies amenable but not the other way around. Yet I would prefer a prrof involving this definition. $\endgroup$ Dec 14 '20 at 21:57
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I think, the equivalence of Navas' definition and the standard notion of amenablity is called Bogolyubov-Dey theorem. You can find it in many places, see for instance Proposition 3.6 in

Grigorchuk, Rostislav; de la Harpe, Pierre, Amenability and ergodic properties of topological groups: from Bogolyubov onwards, Ceccherini-Silberstein, Tullio (ed.) et al., Groups, graphs and random walks. Selected papers of the workshop, Cortona, Italy, June 2–6, 2014 on the occasion of the 60th birthday of Wolfgang Woess. Cambridge: Cambridge University Press (ISBN 978-1-316-60440-3/pbk; 978-1-316-57657-1/ebook). London Mathematical Society Lecture Note Series 436, 215-249 (2017). ZBL1397.43001.

(Read here for a free version.) Given this result, you can use many of the available proofs of the fact that the class of amenable groups is closed under extensions, e.g. here or one of many other books dealing with amenable groups.

Edit. It is clear from the context of the book that Navas defines amenability (and, for instance, property T) only for groups equipped with discrete topology. It is unfortunate that he never mentions amenability in the context of topological groups (equipped with nondiscrete topology), uses a nonstandard definition of amenability and provides no references (as far as I can tell) for a general textbook treatment of amenable groups (and there are several of these, see references here, at least in the case of locally compact groups which includes discrete groups).

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  • $\begingroup$ Thanks, I’ll take a look and see if I can use it. $\endgroup$ Dec 15 '20 at 0:01
  • $\begingroup$ Okay, I've read it, it is very interesting but sadly I can't use it because it uses the topology of $G$, but I'm not using the topology of the group in my proof (I would have to introduce the topology of the group and only use it for this demonstration). many thanks anyway, this article may still be useful. $\endgroup$ Dec 15 '20 at 2:37
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    $\begingroup$ @MatíasUres: I do not understand what you mean: amenability is a property of a topological group. If your group is not given a topology, you are working with discrete topology. Their proof, of course, will apply in this setting too. $\endgroup$ Dec 15 '20 at 4:08
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Fix a compact metric space $M.$ Let $W(M)$ denote the Wasserstein space for $M$: the space of probability measures on $M,$ with the Wasserstein metric. The important property is that this metric gives the topology of weak convergence, making $W(M)$ a compact metric space.

Let $W(M)^H$ denote the subspace of $H$-invariant measures. This is closed, so it's also a compact metric space.

An action of $G$ on $M$ gives an action $(gp)(A)=p(g^{-1}A)$ on $W(M).$ Since $H$ is normal, $G$ preserves $W(M)^H$: if $p$ is $H$ invariant then $p(g^{-1}hA)=p((g^{-1}hg)g^{-1}A)=p(g^{-1}A).$ But $H$ acts trivially on $W(M)^H,$ so in fact $G/H$ acts on $W(M)^H.$ Since $G/H$ is amenable there's a $G$-invariant measure $\xi$ on $W(M)^H.$

This is a probability measure on a space of probability measures. To get a measure on the original space $M,$ we need integration of measures. Or in other words the multiplication of the Kantorovich monad. Define $E\xi\in W(M)$ by $(E\xi)(A)=\int p(A)d\xi(p)$ for each Borel $A.$ The $G$-invariance of $\xi$ implies the $G$-invariance of $E\xi$: $$(gE\xi)(A)=\int (gp)(A)d\xi(p)=(E\xi)(A).$$

Finally I would like to mention that the same argument works if you drop the metrizability condition everywhere. The existence of an invariant probability measure for every $G$-action on a compact Hausdorff space is one of the few definitions of amenability that generalizes usefully to non-locally compact groups.

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  • $\begingroup$ Thank you very much, I didn't know of the Wasserstein metric, this is perfect. $\endgroup$ Dec 22 '20 at 23:37

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