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Question: What are the connected components of the familiar spaces of functions between two (let's say compact and smooth, for simplicity) manifolds $M$ and $N$?

Specifically, I'm thinking of the Hölder spaces $\mathcal{C}^{k,\alpha}(M, N)$ and the Sobolev spaces $\mathcal{W}^{k,p}(M, N)$.

Some comments:

  1. For a smooth function $f:M\to N$, it seems clear that, at least, all continuous functions homotopic to $f$ will be connected to it.

  2. This question is inspired by the discussion of $\mathcal{W}^{k,p}(M, N)$ in McDuff-Salamon's book on $J$-holomorphic curves. There it's stated as an offhand remark that the connected components of $\mathcal{W}^{k,p}(M, N)$ (in the case of $M$ oriented & two-dimensional; I'm not sure if this makes a difference) are the completions of the sets {$f:M\to N \text{ smooth}: f_*[M]=A$}, for $A\in H_{\dim M}(N)$.

  3. If the McD-S factoid is true, there should exist sequences of smooth not-all-mutually-homotopic functions which converge in $\mathcal{W}^{k,p}(M, N)$. (This isn't too counterintuitive, since $\mathcal{W}^{k,p}(M, N)$ presumably contains functions which aren't continuous, & so don't themselves have a homotopy class). Can someone give me an example of this phenomenon?

Please feel free to re-tag -- I can't think of anything really appropriate.

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I'm going to guess that $k$ and $p$ were chosen large enough so that $W^{k,p}\subset C^r$ for some $r\ge 0$, (by Sobolev), so that $W^{k,p}$ maps in the same path component are continuous and homotopic. – Paul May 1 '10 at 21:04
Ah, of course you're right -- I forgot about the Sobolev estimates. In that case, should the McD-S remark about "connected components" be interpreted as meaning "unions of connected components"? – macbeth May 1 '10 at 21:32
I think if you use the methods Dylan and Dan mention below you get that $\pi_0(W^{k,p})=\pi_0(C^r)=\pi_0(C^\infy)$. I don't know a good reference to this, but you could check Palais' book called "foundations of global non-linear analysis". Hirsch's book has results that relate $C^r$ and $C^\infty$ for all $r$, so maybe that and Sobolev embedding does the trick. – Paul May 2 '10 at 14:53
Two remarks: first, homology classes of smooth mappings $M\to N$ do not usually determine homotopy classes, though they do, by the Hurewicz theorem, when $M=S^2$ and $N$ is simply connected (maybe that's what you mean about "unions of connected components"?). Second, life on the not-quite-continuous Sobolev borderline (e.g. $W^2_1$ on a surface) is very precarious, both for analysts and topologists. – Tim Perutz May 2 '10 at 14:57
Thanks to all for useful comments. @Tim, yes, that is what I mean: since intuition (supported by the discussion here) is that the connected components are homotopy equivalence classes, and since homology doesn't determine homotopy, it should be false that the connected components are the homology classes. (But perfectly true that homology classes are unions of connected components.) I am feeling happy that here there be no pathology-dragons. – macbeth May 2 '10 at 15:46
up vote 5 down vote accepted

Any continuous map from M to N is homotopic to a smooth map, and if two smooth maps are homotopic, then they are also smoothly homotopic. (More generally, two homotopic functions are homotopic through a homotopy that is smooth except at the endpoints.) The proof involves convolving with Gaussians, and is standard; I think you can find it in Milnor's Topology from a Differentiable Viewpoint, for instance. (It's also appeard on mathoverflow before, but I couldn't find it just now.) The hard issues for smooth vs. continuous functions arise only once you start demanding the maps be injective.

I can't say more about McDuff and Salomon without more context for the quote.

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Thanks. This certainly serves as a proof of my Comment 1: if two functions are homotopic, then in any sensible topology they are in the same connected component. Paul's remark deals with my questions about the converse: if two functions are in the same connected component of $\mathcal{W}^{k,p}(M,N)$, then there is a continuous function $F:[0,1]\to \mathcal{W}^{k,p}(M,N)$ connecting them, and by the Sobolev inequality $||f||_{\mathcal{C}^{0,\mu}}\leq ||f||_{\mathcal{W}^{k,p}}$ (for $kp>\dim M$ and $\mu:=k-np$) this function yields a homotopy. – macbeth May 2 '10 at 8:13
The McDuff-Salamon remark is in J-holomorphic curves and symplectic topology, top of page 45 in the 2004 edition. They are interested in the Banach manifold $\mathcal{W}^{k,p}(M,N)$ (for $M$ a Riemann surface) because they are about to show that a subset of it -- those maps which are J-holomorphic -- is a finite-dimensional submanifold. – macbeth May 2 '10 at 8:22
Now I see it, and I agree with you: it should say "$\mathcal{B}^{k,p}$ is a union of components of $W^{k,p}". – Dylan Thurston May 2 '10 at 15:16

This got too long for a comment to Dylan's answer.

I like the discussion of these ideas in John Lee's book Introduction to Differentiable manifolds (the relevant part isn't in the google preview). He refers to these approximation results as the Whitney Approximation Theorem, and deduces them from the tubular neighborhood theorem and the Whitney Embedding Theorem. Interestingly, he doesn't use convolution.

Here's a sketch: start with a continuous map from $f:M\to N$, and embed N in R^n. First Lee proves that there's smooth map $g: M\to R^n$ close to f (inside a tubular neighborhood of N, say) and then he uses the projection from the tubular neighborhood back to N to get his approximation to f. Note that since balls in $R^n$ are convex, once g is sufficiently close to f there's a linear homotopy linking them, which lies entirely inside the tubular neighborhood.

To produce g, the rough idea is that near any point x in M, the constant function with value f(x) is a "good enough" approximation to f. This gives an open cover of M, and there's a finite subcover, with a subordinate partition of unity $\phi_i$. The approximation to f is now gotten by averaging these constant functions using the partition of unity: $g(x) = \sum \phi_i (x) f(x_i)$.

Lee explains how to modify this in the case where f is already smooth on some closed subset, and you want to leave it unchanged there. (That allows you to approximate homotopies.)

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Nice, I didn't know that proof. – Dylan Thurston May 2 '10 at 15:18

For the Sobolev case, I think may be helpful [together with the references therein].

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Awesome! (Short summary: for $kp\leq \dim M$, the connected components can be very strange.) – macbeth May 4 '10 at 22:10

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