Is $\mathbb{S}^\infty$ exotic

Solution 1:

You've got two issues to deal with.

1) What kind of infinite-dimensional manifold do you want to consider $S^\infty$ to be?

and then there's the killer:

2) "Reasonable" Banach, Hilbert, Frechet manifolds have the rather strange property that they're homotopy-equivalent if and only if they're diffeomorphic.

Henderson, David W. (1969). "Infinite-dimensional manifolds are open subsets of Hilbert space". Bull. Amer. Math. Soc. 75:

So as long as you give $S^\infty$ a "reasonable" manifold structure, by-design there's no hope for it to be exotic.

The Henderson theorem has analogues in finite-dimensional manifold theory. The whole story of much classical manifold theory is that the h and s-cobordism theorems tell you how properties of these manifolds reduce to various algebraic properties. But going further, if you "stabilize" a manifold sufficiently, the only "information" contained in that object is the simple homotopy-type of that manifold together with the classifying map of the stable normal bundle. So much of the hard part of manifold theory vanishes when you "crank up" the dimension by stabilization. But when you're dealing with infinite-dimensional manifold, in a certain sense you've already stabilized.

Of course that's more vague and meant to give only partial intuition. But when you look at the details, this story carries-though quite far.