Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let us define the nth smooth homotopy group of a smooth manifold $M$ to be the group $\pi_n^\infty(S^k)$ of smooth maps $S^n \to S^k$ modulo smooth homotopy. Of course, some care must be taken to define the product, but I don't think this is a serious issue. The key is to construct a smooth map $S^n \to S^n \lor S^n$ (regarded as subspaces of $\mathbb{R}^{n+1}$) which collapses the equator to a point; we then define the product of two (pointed) maps $f, g: S^n \to S^k$ to be the map $S^n \lor S^n \to S^k$ which restricts to $f$ on the left half and $g$ on the right half. To accomplish this, use bump functions to bend $S^n$ into a smooth "dumbell" shape consisting of a cylinder $S^{n-1} \times [0,1]$ with two large orbs attached to the ends, and retract $S^{n-1}$ to a point while preserving smoothness at the ends. Then retract $[0,1]$ to a point, and we're done.

Question: is the natural "forget smoothness" homomorphism $\phi: \pi_n^\infty(S^k) \to \pi_n(S^k)$ an isomorphism? If not, what is known about $\pi_n^\infty(S^k)$ and what tools are used?

In chapter 6 of "From Calculus to Cohomology", Madsen and Tornehave prove that every continuous map between open subsets of Euclidean spaces is homotopic to a smooth map. Thus every continuous map $f$ between smooth manifolds is "locally smooth up to homotopy", meaning that every point in the source has a neighborhood $U$ such that $f|_U$ is homotopic to a smooth map. However it is not clear to me that the local homotopies can be chosen in such a way that they glue together to form a global homotopy between $f$ and a smooth map. This suggests that $\phi$ need not be surjective.

In the same reference as above, it is shown that given any two smooth maps between open subsets of Euclidean spaces which are continuously homotopic, there is a smooth homotopy between them. As above this says that two smooth, continuously homotopic maps between smooth manifolds are locally smoothly homotopic, but I again see no reason why the local smooth homotopies should necessarily glue to form a global smooth homotopy. This suggests that $\phi$ need not be injective.

I am certainly no expert on homotopy theory, but I have read enough to be surprised that this sort of question doesn't seem to be commonly addressed in the basic literature. This leads me to worry that my question is either fatally flawed, trivial, useless, or hopeless. Still, I'm retaining some hope that something interesting can be said.

share|improve this question
add comment

2 Answers

up vote 12 down vote accepted

Yes, the map you mention is an isomorphism. I think the main reason people rarely address your specific question in literature is that the technique of the proof is more important than the theorem. But all the main tools are written up in read-to-use form in Hirsch's Differential Topology textbook.

There are two steps to prove the theorem. Step 1 is that all continuous maps can be approximated by (neccessarily) homotopic smooth maps. The 2nd step is that if you have a continuous map that's smooth on a closed subspace (say, a submanifold) then you can approximate it by a (neccessarily) homotopic smooth map which agrees with the initial map on the closed subspace. So that gives you a well-defined inverse to your map $\phi$.

There are two closely-related proofs of this. Both use embeddings and tubular neighbourhoods to turn this into a problem for continuous maps defined on open subsets of euclidean space. And there you either use partitions of unity or a "smoothing operator", which is almost the same idea -- convolution with a bump function.

share|improve this answer
3  
Thank you for your helpful answer (and reference)! This appears to all be standard fare in differential topology, and perhaps the question would be considered "trivial" to a homotopy theorist as I had feared. Oh well; it was a nice thought. –  Paul Siegel Aug 11 '10 at 16:36
1  
Another good reference for this material is John Lee's book Introduction to Differentiable Manifolds. I sketched Lee's argument in an answer here: mathoverflow.net/questions/23197/…. –  Dan Ramras Aug 11 '10 at 17:20
add comment

Dear Paul, as Ryan says the smooth and continuous homotopy groups of a manifold coincide.

This is stated as Corollary 17.8.1 in Bott and Tu's book Differential Forms in Algebraic Topology (Springer Graduate Texts in Mathematics, #82).The Corollary is to the preceding Proposition 17.8, which says that a continuous map is homotopic to a differentiable one.This is easy but relies on Whitney's embedding theorem for which the authors refer to De Rham's book Variétés différentiables ; you might prefer Hirsch's book in the same Springer series, GTM #33, which is more modern and in English.

As an application Bott and Tu calculate $\pi_q S^n$ for $q\leq n$ by differential methods.

share|improve this answer
1  
The worst part is that I flipped through Bott and Tu before asking this question, thinking "if the answer isn't in here, it probably isn't anywhere". Sigh... –  Paul Siegel Aug 11 '10 at 16:37
    
Note that the embedding theorem is quite easy if you do not care about the dimension in which you embed. –  Lennart Meier Apr 16 '13 at 17:43
add comment

Your Answer

 
discard

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.