# Complete length space with discrete action

Fix $x_0\in X$ where $X$ is a complete length space.

Define $$D:=\inf\ \bigg\{d_X(x_0,gx_0) \bigg|g\neq 1\in G\bigg\}$$ where $G \subset {\rm Isom}\ (X)$.

If $X$ is a Riemannian manifold, and if $G$ is not a continuous Lie group, then $D>0$. But if $X$ is not Riemannian manifold and if $X$ is a complete length space, then $D$ may be $0$ ?

Thank you in advance.

[Add]

Def : $G$ is discrete action on a Riemannian manifold $(M,d)$ if for each $x\in M$, there is open neighborhood $U$ s.t. $x\in U$ and $$g\in G,\ U\cap gU\neq \emptyset\Rightarrow g=e$$

Hence if $U$ contains $B_\epsilon (x)$, then $$d(x,gx)>\epsilon$$ for all $g\neq 1$.

Def : $G\subset {\rm Isom}\ X$ where $X$ is a complete length space. If $x\in X$, then $G\cdot x$ is locally path connected if there is $\delta>0$ s.t. $d(x,g\cdot x) <\delta$ implies $$d(x,ax),\ d(ax,gx) <0.51 \delta$$ for some $a \in G$.

So if $X$ is a Riemannian manifold, and $x\in X$, then $G\cdot x$ is locally path connected if $G$-action is discrete or $G$ is Lie group that is a connected Riemannian manifold.

Usually, $G$-action on complete length space $X$ can be thought through the following notion

$$(N_i=S^1\times \mathbb{R},G_i=\mathbb{Z}_i) \rightarrow (X=S^1\times \mathbb{R},G=S^1)$$

• What do you mean by a "continuous Lie group"? For instance, consider ${\mathbb Q}$ acting on the real line by translations. I think, what you meant to say is "if $G$ is a discrete subgroup of $Isom(X)$ equipped with (your favorite) topology". If so, what topology do you have min mind for a general metric space? Lastly, what about the case $g(x_0)=x_0$? – Moishe Kohan Apr 7 '17 at 20:29
• Thank you for your comment. – HK Lee Apr 8 '17 at 0:23
• Your question still makes no sense to me. What does your addition have to do with the question? – Moishe Kohan Apr 8 '17 at 0:58
• I mean : A discrete group $G$ acts on $X$ s.t. $G$ can not be extended to Lie group action and $gx=x\Rightarrow g=e$. Here my question is $\inf\ \{d(x,gx)|g\neq e\}=0$ is possible ? – HK Lee Apr 8 '17 at 1:05

## 1 Answer

Example. $G$ is an uncountable proper subgroup of ${\mathbb R}$ and consider first the isometric action of $G$ on ${\mathbb R}$ (by translations). Next, at each point of $G\cdot 0\subset {\mathbb R}$ attach the unit vertical segment $\subset {\mathbb R}^2$ and equip the reulting "comb" $X$ with the Euclidean path-metric, which easily seen to be complete. The action of $G$ by translations on $X$ preserves this metric and does not extend to a (continuous) Lie group action; at the same time this action is free and every orbit is dense in the real line. In particular, for every $x$ in the real line, $\inf \{d(x,gx): g\ne e\}=0$.

Here is a related question which I do not know how to answer: Suppose that $X$ is a compact geodesic metric space, $G\times X\to X$ is an isometric free action of a compact totally disconnected group. Is it true that $G$ is finite? (This question is mildly related to the Hilbert-Smith conjecture.)

• Thank you for your answer and introducing interesting problem. – HK Lee Apr 10 '17 at 2:04
• Do you have a reference for a "${\it related\ question}$"? – HK Lee May 4 '18 at 4:20