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Let $M$ and $N$ be smooth manifolds with $\dim M<\dim N$.

The spaces $\mathrm{Emb}(M,N)\subset\mathrm{Imm}(M,N)$ of smooth embeddings and immersions $f:M\to N$, respectively, are infinite dimensional Frechet manifolds. They are open subsets of the space $C^{\infty}(M,N)$ of smooth maps $M\to N$, or equivalently, of the space $\Gamma(M,M\times N)$ of smooth sections of the trivial bundle $M\times N\to M$.

The diffeomorphism group $\mathrm{Diff}(M)$ acts naturally on these spaces by precomposition.

My question is what structure the quotient spaces $\mathrm{Emb}(M,N)/\mathrm{Diff}(M)$ and $\mathrm{Imm}(M,N)/\mathrm{Diff}(M)$ have. These spaces seem to arise naturally in geometric problems, where only the image $f(M)\subset N$ of a map $f:M\to N$ is of concern, not the actual map itself.

1) Do $\mathrm{Emb}(M,N)/\mathrm{Diff}(M)$ and $\mathrm{Imm}(M,N)/\mathrm{Diff}(M)$ have a smooth manifold structure?

2) What are their topological properties, e.g. homology/homotopy? Can we compute these via some kind of "Morse" functions?

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  • $\begingroup$ Frechet spaces are hard. For instance the space of invertible bounded linear operators is not open. My guess is you want to see them as Banach manifolds. $\endgroup$ – Charlie Frohman Feb 13 at 17:46
  • $\begingroup$ @CharlieFrohman You most certainly do not want to see them as Banach manifolds, as they are not. Where would the Banach structure come from? (In some applications one works with eg Sobolev spaces of maps for the sake of the analysis to get around the technical issues you mention.) $\endgroup$ – user98602 Feb 13 at 17:50
  • $\begingroup$ Sobolev spaces are Banach. Work in charts. $\endgroup$ – Charlie Frohman Feb 14 at 2:07
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1) See this and its references. The issue in the case of immersions is when an immersion $f: M \to N$ factors through a covering map $g: M \to M'$.

2) For embeddings, this is a ridiculously hard question, analagous to understanding the homology and homotopy of the diffeomorphism group, which is the subject of much (ongoing) research. For instance, the examples of embedded 2-spheres inside $S^3$ are what allowed Hatcher to calculate the homotopy type of $\text{Diff}(S^1 \times S^2)$, and understanding those is essentially equivalent to his very difficult result that $\text{Diff}^+(D^3, \partial D^3)$ is contractible. Similarly, once he has the homotopy type of the space of Haken manifold here one may bootstrap that into an understanding of the topology of diffeomorphism groups.

For immersions this is answered by Hirsch-Smale theory.

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  • $\begingroup$ Thanks for these nice references. I took a look at Hatcher's website, where he has some lecture slides on diffeomorphism groups. He claims that $\pi_0\mathrm{Diff}(M)$ is known for all prime 3-manifolds $M$, but doesn't give a reference. Do you know where I could read about this? $\endgroup$ – srp Feb 13 at 18:14
  • $\begingroup$ He claims this on page 3 of pi.math.cornell.edu/~hatcher/Papers/Diff%28M%292012.pdf $\endgroup$ – srp Feb 13 at 18:15
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    $\begingroup$ @srp I know the slides. This is a long and complicated story, and I doubt I would be able to piece together the references. For Haken manifolds this is a thoerem of Waldhausen which you can read in Hempel's book. For spherical manifolds I would chase through people who have cited Cerf's treatise on the MCG of $S^3$ (I don't see right now what the argument would be). In general, I guess you can take the JSJ decomposition and try to bootstrap from there... I bet there's no harm in emailing the author. If you get detailed references, I'd like to see them! (My email is in my profile.) $\endgroup$ – user98602 Feb 13 at 18:22
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    $\begingroup$ @srp My memory is coming back. You use geometrization to point out that an (oriented, irreducible) manifold is either Haken, or hyperbolic, or Seifert-fibered. So Waldhausen has dealt with the first case, Mostow rigidity is precisely the second case, and then one just has to do the small Seifert-fibered manifolds. That's still going to require some amount of work I'm not familiar enough with to tell you about, but SF's are much more tractable than the general case! $\endgroup$ – user98602 Feb 14 at 20:02

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