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

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.

• You mean "nonempty" compact space, and "extremally amenable" (not "extremely"). And $H$ is endowed with the induced topology.
– YCor
Commented May 1, 2019 at 17:38
• I'm pretty sure that the extension result holds...
– YCor
Commented May 1, 2019 at 17:40

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 :-)

• The definition that you gave is probably for amenable groups but not extremely amenable groups. Commented May 16, 2023 at 14:16
• The definition is for extremely amenable group because for amenability you don't need the fact that the mean is multiplicative Commented May 17, 2023 at 15:32

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.