In what sense is $S^\infty$ the same as $\{x \in \ell_2 : \|x\| = 1 \}$?

Question

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?

Answer 1

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.