# Finding a metric on a topological space with prescribed isometry group

Let $X$ be a (sufficiently nice) topological space and let $\mathcal{F}$ be a group of homeomorphisms of $X$. Assume that $\mathcal{F}$ is also closed under point-wise convergence. I would like to investigate the possibility of defining a metric $d$ on $X$ such that the the isometries of the metric space $(X,d)$ is precisely the set $\mathcal{F}$. Note that I want the topology induced by $d$ to coincide with the original topology on $X$. What are the conditions on $X$ and $\mathcal{F}$ that allow us to construct such a $d$?

I am interested in this question from the perspective of hyperbolic geometry. On the unit disk the Poincare metric (up to a constant) is the unique metric that has isometry group the holomorphic automorphisms and conjugates. I want to find a metric on $\mathbb{C} \setminus \{0,1\}$ such that the isometries are precisely the holomorphic automorphisms of $\mathbb{C} \setminus \{0,1\}$ and the conjugates of these automorphisms. The Kobayashi pseudodistance on $\mathbb{C} \setminus \{0,1\}$ is certainly one distance that satisfies this condition but the fact that $\mathbb{C} \setminus \{0,1\}$ is Kobayashi hyperbolic is a non-trivial fact. I want to see whether it possible show the existence of such a metric by other means. I also want to see how this new metric is related to the Kobayashi distance

• Note that if $\mathcal{F}$ is a collection of isometries w.r.to some metric, then so is the group it generates. Moreover, the point-wise limit of a sequence of isometries is still an isometry: this in general is not true for just a limit of homeomorphism, so it is a necessary condition. Therefore it would be better to assume from the beginning that $\mathcal{F}$ is a group of homeomorphisms, closed by point-wise convergence. – Pietro Majer Nov 25 '16 at 12:40
• @Pietro Thanks! I have edited the question. – Jaikrishnan Nov 25 '16 at 13:11
• I would consider, for all $(x,y)\in X\times X$, the set $\mathcal{F}(x,y):=\{(f(x),f(y))\,:\, f\in\mathcal{F}\}$. If $\mathcal{F}$ is a group of isometries w.r.to some distance $d$, the sets $D(x,y):=\overline{ \mathcal{F}(x,y)\cup \mathcal{F}(y,x)}$ make a closed partition of $X\times X$ (which is a slightly stronger necessary condition than said above), and $d$ must be constant on each of them. – Pietro Majer Nov 25 '16 at 13:32
• Some related problems were studied by Piotr Niemiec arxiv.org/abs/1201.5675 – Taras Banakh Dec 8 '16 at 19:14