Can anything be said about connectedness of a smooth manifold M from some property of Diff(M) in an analogous way Like C(X) has no idempotents iff X is connected.
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$\begingroup$ I guess the question is a bit vague. Is Diff(M) the subsheaf of C(M) of smooth functions on M? Clearly, if Diff(M) has non-trivial idempotents, so has C(M). Are you asking for the converse? $\endgroup$– Stephan MüllerCommented Feb 18, 2013 at 15:13
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$\begingroup$ i am asking both way $\endgroup$– KoushikCommented Feb 18, 2013 at 15:39
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$\begingroup$ i don't see why Diff(M) should be a subsheaf of C(M). $\endgroup$– KoushikCommented Feb 18, 2013 at 15:44
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1$\begingroup$ What is Diff(M) here? The notation is usually used to denote the group of diffeomorphisms, but then Stephan's comment is weird. $\endgroup$– Mariano Suárez-ÁlvarezCommented Feb 18, 2013 at 16:29
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2$\begingroup$ $Diff(M)$ is the group of diffeomorphisms, and the question is how properties of $Diff(M)$ reflect in topological properties of $M$. $\endgroup$– Alain ValetteCommented Feb 18, 2013 at 16:53
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2 Answers
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Let $M$ be a compact oriented manifold. The following hold if and only if $M$ is connected.
1) $\text{Diff}_0(M)$ is simple.
This was proven by Thurston if $M$ is connected; see
MR1445290 (98h:22024) Banyaga, Augustin(1-PAS) The structure of classical diffeomorphism groups. (English summary) Mathematics and its Applications, 400. Kluwer Academic Publishers Group, Dordrecht, 1997. xii+197 pp. ISBN: 0-7923-4475-8
If $M$ is not connected, then $\text{Diff}_0(M)$ contains normal subgroups consisting of elements that fix some connected components and don't fix others.
2) $\text{Diff}_0(M)$ does not decompose as a direct product.
If $M$ is the disjoint union of submanifolds $M_1$ and $M_2$, then it is clear that $\text{Diff}_0(M) = \text{Diff}_0(M_1) \times \text{Diff}_0(M_2)$.
If $M$ is connected, then one can show that $\text{Diff}_0(M)$ does not decompose as a direct product by exhibiting elements $f \in \text{Diff}_0(M)$ whose centralizers consist only of $\langle 1, f, f^2, \ldots \rangle$. There are many such constructions; for instance, see
MR0985855 (90i:58151a) Palis, J.(BR-IMPA); Yoccoz, J.-C.(F-POLY) Rigidity of centralizers of diffeomorphisms. Ann. Sci. École Norm. Sup. (4) 22 (1989), no. 1, 81–98.
A famous conjecture of Smale says that such elements should in fact be generic. This was recently proven by Bonatti-Crovisier-Wilkinson for $C^1$ diffeomorphisms; see
MR2511588 (2010g:37035) Bonatti, Christian(F-DJON-IM); Crovisier, Sylvain(F-PARIS13-AG); Wilkinson, Amie(1-NW) The C1 generic diffeomorphism has trivial centralizer. (English summary) Publ. Math. Inst. Hautes Études Sci. No. 109 (2009), 185–244.
answered Feb 18, 2013 at 17:52
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2$\begingroup$ For part 2, only if $M$ and $N$ are not diffeomorphic. $\endgroup$ Commented Feb 18, 2013 at 18:03
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3$\begingroup$ Part 1, only if all connected components of $M$ are compact, or if you consider the connected component of the group of diffeomorphisms with compact support. $\endgroup$ Commented Feb 18, 2013 at 18:11
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3$\begingroup$ @Will Sawin : I don't think this is a problem since I restricted myself to the connected component of the identity in $\text{Diff}(M)$; such elements cannot permute the components around (which is the problem I think you were referring to). I spent a little time before writing this trying to come up with a clean formulation of what happens to $\text{Diff}(M)$ when some components are diffeomorphic. I couldn't find one, so I took the easy way out. $\endgroup$ Commented Feb 18, 2013 at 20:14
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1$\begingroup$ @Peter Michor : Good point, I was assuming implicitly that everything was compact anyway. I'll add that assumption to the answer. $\endgroup$ Commented Feb 18, 2013 at 20:15
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$M$ is connected if and only if the connected component of $Diff(M)$ (equivalently, of $Diff_c(M)$) acts transitively on $M$.
Edit: I just remembered, that the Lie algebra of compactly supported vector fields determines the base manifold up to diffeomorphism, see: MR0064764 (16,331a) Reviewed Shanks, M. E.; Pursell, Lyle E. The Lie algebra of a smooth manifold. Proc. Amer. Math. Soc. 5, (1954). 468–472.
This is also true for larger Lie algebras, and for complex Stein manifolds, see: MR0516602 (80g:57036) Reviewed Grabowski, J. Isomorphisms and ideals of the Lie algebras of vector fields. Invent. Math. 50 (1978/79), no. 1, 13–33.
Moreover, the group of compactly supported diffeomorphisms determines the base manifold completely, but I cannot find the relevant paper now.
answered Feb 18, 2013 at 18:12