2 act freely on an infinite dimensional sphere?

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Jun 1, 2015 at 17:30	comment	added	Todd Trimble		Another point of view is that $S^\infty$ carries a structure of topological abelian group of exponent 2 (i.e., a topological vector space over $\mathbb{Z}/(2)$) as in this answer: mathoverflow.net/a/43047/2926. Translation by any two linearly independent elements would also answer your question affirmatively.
Jun 1, 2015 at 12:53	comment	added	Eric Wofsey		Alternatively, $S^\infty$ is homeomorphic to $\mathbb{R}^\infty$ (see this answer, for instance), and clearly $\mathbb{R}^\infty$ is homeomorphic to its square.
Jun 1, 2015 at 6:59	comment	added	Ryan Budney		Isn't $S^\infty$ homeomorphic to $S^\infty \times S^\infty$? That would answer your question. I think you can construct the map fairly explicitly, $S^n \times S^n \to S^{2n}$. You think of this map as crushing the two factors $S^n \times \{1\} \cup \{1\} \times S^n$ to a common point, with the map a homeomorphism otherwise. If you do this fairly naturally, this induces a map of the colimits $S^\infty \times S^\infty \to S^\infty$. This map is not a homeomorphism itself but it looks like it can be fixed as it fails to be a homeomorphism only by crushing two contractible subspaces.
Jun 1, 2015 at 5:40	history	asked	Jens Reinhold	CC BY-SA 3.0