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Let $(X,\tau)$ be a homogeneous space, that is for all $x,y \in X$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$. Is there a group operation $*:X\times X\to X$ such that $(X,*,\tau)$ is a topological group?

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  • $\begingroup$ Note that homogeneous space often means that there exists a transitive continuous action of a connected Lie group, which is much more restrictive than your definition: for instance all connected topological manifolds are homogeneous in your sense, but for instance closed surfaces of genus $\ge 2$ are not homogeneous in the stronger sense. Anyway the answers provide counterexamples that are homogeneous in the stronger sense, such as the 2-sphere. $\endgroup$
    – YCor
    Oct 23, 2014 at 18:34

3 Answers 3

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Not necessarily.

It can be shown that every connected topological group has Euler characteristic $0$. Now take the $2$-sphere, for example. It is connected, and the Euler characteristic $\neq0$. It is homogeneous, of course.

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Another way to get negative answers: a topological group does not have the fixed point property (every map from $X$ to itself has a fixed point), as a translation $x \rightarrow a\ast x$ with $a \neq 1$ does not have a fixed point.

But there are homogeneous compact metric space with the fixed point property, e.g. the Hilbert cube $[0,1]^\mathbb{N}$, see this question for reasons why this is so.

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  • $\begingroup$ And a complete proof that the Hilbert cube is homogeneous can be found here, with explicit homeomorphisms. $\endgroup$ Oct 22, 2014 at 19:30
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$S^2$ is a classical counter-example. See Corollary 9.59(iv) on p. 486 in

K.H. Hofmann and S.A. Morris, The Structure of Compact Groups, Berlin, 1998.

Let me give you a method of producing non-separable examples. By Tkachenko's theorem, any $\sigma$-compact group is ccc. So consider homogeneous $\sigma$-compact spaces which are not ccc. You can construct such spaces by taking countable unions of certain (open) long lines.

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