Dense subgroup of an extremely amenble group is extremely amenable??

Question

Recall that a topological group $G$ is called extremely amenable if every continuous action of $G$ on a compact space $K$ admit a fixed point. i.e there is a point $\xi\in K$ such that $g.\xi=\xi$ for all $g\in G$.

Now let $H$ be a dense subgroup of an extremely group $G$. My goal is to show that $H$ is also extremely amenable.

More precisly, i don't know if this true that in this case every continuous action of $H$ on a compact space can be extend to an action of $G$ on the same compact space or not? Or may be there is another approach to solve this question.

Thank for any help.

Answer 1

I think you should not try to solve this using the fixed-point definition for extremely amenable groups. This because the "inducing action machinery" outside of the world of locally compact groups is not so useful.

So I would like to suggest you to use the multiplicative left-invariant mean definition. Precisely, a topological group $G$ is extremely amenable if it has a multiplicative left-invariant mean on $\mathcal{C}_{ru}^b(G)$, i.e., a positive functional $m$ on $\mathcal{C}_{ru}^b(G)$ such that $m(1_G)=1$, $m(f\cdot h) = m(f)m(h)$ and $m(gf)=m(f)$ for every $f,h\in \mathcal{C}_{ru}^b(G)$ and $g\in G$. (Here $\mathcal{C}_{ru}^b(G)$ is the set of all bounded right-uniformly continuous functions on $G$).

Let now consider the extension map $ext: \mathcal{C}_{ru}^b(H) \longrightarrow \mathcal{C}_{ru}^b(G)$. Note that this map is well-defined as the extension is unique ( see Bourbaki Topologie générale) and it is linear and positive. Moreover we have that

1. $ext(1_H) = 1_G$
2. $ext(f\cdot h)= ext(f)ext(h)$ for every $f,h \in \mathcal{C}_{ru}^b(H)$
3. $ext(gf)=g\cdot ext(f)$ for every $g\in H$ and $h\in \mathcal{C}_{ru}^b(H)$

Now it is easy to construct a multiplicative left-invariant mean on $\mathcal{C}_{ru}^b(H)$ using the fact that you know that there is one on $\mathcal{C}_{ru}^b(G)$. Hence, $H$ is extremely amenable.

I hope it is clear :-)

Answer 2

The answer by Evian is quite clear. I think there is always a natural translation between the language of invariant means and that of fixed point property. Here is another way to conclude the result by using fixed point properties:

Let $\Gamma$ be a dense subgroup in an extremely amenable group $G$. Let $K$ be a non-empty compact set on which $\Gamma$ acts continuously. To show that $\Gamma$ is also extremely amenable, it suffices to show that the action $\Gamma\times K\to K$ by $(\gamma,x)\mapsto \gamma x$ can be extended continuously to $G$.

Now consider the left-uniform structure of $\Gamma$ and $G$ and the compact uniform structure of $K$. It turns out that every continuous group action on compacta is a uniformly continuous action, i.e. the map $\Gamma\times K\to K$ is uniformly continuous.

Let $(\gamma_\alpha)_\alpha$ be a Cauchy net in $\Gamma$ that converges to $g\in G$. Since uniformly continuous maps preserves Cauchy nets, for any $x\in K$, the net $(\gamma_\alpha x)_\alpha$ is still Cauchy in $K$. As the uniform structure of $K$ is complete (compacta have complete uniform structures), the Cauchy net $(\gamma_\alpha x)_\alpha$ converges to some $y\in K$ and one defines $y=gx$. This action is by definition continuous and by the density of $\Gamma$ in $G$, it is defined on the entire $G$.

Q.E.D.