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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.

flag	asked Feb 18 at 14:37 Koushik 532●1●10

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? – Stephan Müller Feb 18 at 15:13

i am asking both way – Koushik Feb 18 at 15:39

i don't see why Diff(M) should be a subsheaf of C(M). – Koushik Feb 18 at 15:44

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What is Diff(M) here? The notation is usually used to denote the group of diffeomorphisms, but then Stephan's comment is weird. – Mariano Suárez-Alvarez Feb 18 at 16:29

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$Diff(M)$ is the group of diffeomorphisms, and the question is how properties of $Diff(M)$ reflect in topological properties of $M$. – Alain Valette Feb 18 at 16:53

Andy Putman · Accepted Answer · 2013-02-18 20:16:01Z UTC

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.

Peter Michor · Answer 2 · 2013-02-19 00:08:05Z UTC

$M$ 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.

Diff(M) and connectedness

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