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Assume that X is a compact metrizable topological space for which the action of homeomorphism group is transitive.

Is there a compatible metric d on X such that the action of group of isometries of X is a transitive action?

What about if we restrict X to be a manifold?

Jim Belk
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1 Answers1

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This is a follow-up on the comment by Normal:

Suppose that $(X,d)$ is a compact metric space. Then the isometry group $G$ of $(X,d)$ is a compact topological group (when equipped with the topology of uniform convergence): This is immediate from the Arzela-Ascoli Theorem. Now, one can ask, which compact topological manifolds admit a faithful transitive action of a compact topological group $G$. Such questions are addressed by the theory of Topological Transformation Groups developed by Gleason, Montgomery, von Neumann, Yamabe, Zippin and others, motivated by Hilbert's 5th Problem. For the details of this theory see the book "Transformation Groups" by Montgomery and Zippin. See for instance, this wikipedia article and this one. The conclusion is that if $G$ is a compact group acting transitively on a compact topological manifold $M$ (these assumptions can be relaxed), then $G$ is a Lie group; denoting by $G_x< G$ the stabilized of a point $x\in M$, we obtain that $G_x< G$ is a Lie subgroup (since it is closed) and $M$ is homeomorphic to $G/G_x$. From this, with a bit of the Lie theory (since fundamental groups of connected topological groups are abelian), one concludes that, say, among 2-dimensional compact oriented manifolds $M$, only $S^2$ and $T^2$ admit a transitive topological group action. Or, you can use the fact that $G$ admits a biinvariant Riemannian metric, hence, $M$ admits a $G$-invariant Riemannian metric, hence, $\chi(M)\ge 0$, since the group of conformal automorphisms of a Riemann surface of genus $\ge 2$ has to be finite (Hurwitz theorem).

There is, most likely, a faster way to prove this avoiding the fully-developed machinery of the theory of the Transformation Group theory (maybe chasing old papers by Montgomery and Zippin from 1930s dealing with topological group actions on surfaces).

Moishe Kohan
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  • +1 for your very interesting answer. i need to more background to understand the details. BTW what do you mean by 'faithfull" action? – Ali Taghavi Jan 16 '16 at 15:48
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    @AliTaghavi: Faithful means that every non-identity element acts nontrivially. Such actions are also known as "effective". – Moishe Kohan Jan 16 '16 at 15:59
  • thank you.I think the terminology effective is more custom. if i am not mistaken an effective action of a Lie group G on a manifold gives an embedding of the lie algebra on $\chi^{\infty}(M)$? – Ali Taghavi Jan 16 '16 at 20:15
  • @AliTaghavi: Yes, it does. – Moishe Kohan Jan 16 '16 at 20:37