I hear things that sound like topologists equate $S^\infty$ (defined as the union or "directed colimit" of $n$-spheres) with the actual unit sphere in, say, a nice vector space like $\ell_2$. In what sense is this a rigorous statement? Are they homeomorphic? Or is it weaker?
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1$\begingroup$ Yes, so my question is: in what sense are they topologically the same, despite this difference...? $\endgroup$– ThompsonCommented Oct 17, 2016 at 17:15
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2$\begingroup$ I think the unit sphere in a Hilbert space is also weakly contractible. $\endgroup$– JHFCommented Oct 17, 2016 at 17:55
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1$\begingroup$ Actually, they are not homeomorphic, see mathoverflow.net/questions/103277/…. $\endgroup$– Moishe KohanCommented Oct 17, 2016 at 18:11
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1$\begingroup$ They are both contractible as well, hence are also homotopy equivalent. The limit $S^\infty$ can be embedded into the unit sphere in a Hilbert space. This follows from the embedding of $\mathbb{R}^\infty$ into the seperable infinite dimensional Hilbert space. The other way is not possible I believe, but don't know a proof of the top of my head. You should probably exploit the non-local compactness of the unit sphere in the Hilbert space. It is good to note that the unit sphere is actually homeomorphic (even diffeomorphic in Hilbert manifold sense) to the Hilbert space itself. $\endgroup$– Thomas RotCommented Oct 17, 2016 at 20:55
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1$\begingroup$ See also: math.stackexchange.com/questions/1612091/… This is basically the same question it seems. $\endgroup$– ThompsonCommented Oct 17, 2016 at 22:12
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1 Answer
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They're both contractible, and indeed the inclusion $S^\infty \hookrightarrow S(\ell_2)$ is a homotopy equivalence. Now the main value of this isn't that they're both contractible, it's that for most natural group actions on these spheres, this map is equivariant. (In particular, for $\Bbb Z/2$, $S^1$, $\Bbb Z/n \subset S^1$, $S^3$.) The map between the quotient spaces is also a homotopy equivalence. Quotienting on the left gives you CW complexes modeling $BG$ (respectively, $\Bbb{RP}^\infty$, $\Bbb{CP}^\infty$, the infinite lens spaces $L(n) = \Bbb Z/n$, and $\Bbb{HP}^\infty$), and quotienting on the right gives you Hilbert manifolds modeling these spaces. CW complexes are more useful for the purposes of some parts of algebraic topology, like calculating various cohomology theories on these spaces (the CW structure gives you a nice spectral sequence). On the other hand, the Hilbert manifold structure is more useful for parts of differential topology, including Morse theory or anything where you need the notion of smoothness. So both sides are useful, but since they're naturally homotopy equivalent, when you're doing homotopy-invariant things it doesn't much matter which side you use.