Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

During construction of universal bundles one considers (for example) the infinite real projective space $\mathbb{R}\mathbb{P}^\infty$, coming from the sphere $\mathbb{S}^\infty$.

My question is, are there exotic $\mathbb{S}^\infty$'s ?

Edit: Maybe this helps to put things in context: In Husemuller's "Fibre Bundles", you can read in Example 11.3. $G=\mathbb{Z}/2\mathbb{Z}$. The space $E_G(n)$ is just the n-sphere $\mathbb{S}^n$ upto homeomorphism ... The space $E_G$ is $\mathbb{S}^\infty$ and $B_G$ is $\mathbb{R}\mathbb{P}^\infty$. This is a classic example of the Milnor construction of the bundle $(E_G, B_G,\pi)$.

The example below is from the world of functional analysis (a field which I am not exactly familiar with), but the example is nontheless fascinating. In Husemuller (or Milnor) the construction does not requires to take place in a Banach space and is based on the infinite join $G * \ldots *G$ (but maybe you can embed this in a Banach space).

share|cite|improve this question
perhaps this article in "Inventiones mathematicae" helps: – ulead86 Sep 16 '11 at 10:01
@Daniel: that seems rather unrelated... anon: that is also quite unrelated. – Mariano Suárez-Alvarez Sep 16 '11 at 22:27
@Willem: what do you mean by 'exotic'? – Mariano Suárez-Alvarez Sep 16 '11 at 22:28
up vote 9 down vote accepted

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.

share|cite|improve this answer
(My guess is, the question is asking for alternative models of $E_{\mathbb Z/2\mathbb Z}$, not exotic as in exotic sphere) – Mariano Suárez-Alvarez Sep 17 '11 at 2:58
@Mariano: Alas no, my question was about exotic spheres (things that are homeomorphic. but not diffeomorphic to spheres). The infinite join construction is pretty clear... – Willem Noorduin Sep 17 '11 at 12:06
@Ryan, thanks for the inside. Does this mean that $|\Theta_k| \rightarrow 0$ if $k \rightarrow \infty$, or can't you say something like that (I mean, we still don't know what $\Theta_4$ is, allthough the Milnor-Kervaire formulae indicates $|\Theta_4| = 0$) – Willem Noorduin Sep 17 '11 at 12:15
@Willem: I don't believe the number of elements in $\theta_k$ is well-behaved as $k$ increases. And I don't know of any useful stabilization map $\theta_k \to \theta_{k+1}$. One way to put this would be, if there's little connection between $\theta_k$ and $\theta_{k+1}$, why should there be any between $\theta_k$ and $\theta_{\infty}$ ? – Ryan Budney Sep 18 '11 at 1:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.