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Let $M$ be a smooth manifold. Consider the set of diffeomorphisms which are induced by flows of vector fields. (which are not time-dependent)

Is this set a subgroup of $\text{Diff}(M)$?

(Note that not every diffeomorphism which is isotopic to the identity is induced by a flow of a vector field, see here for details).

"A naive attempt":

Maybe it's possible to construct a counter-example when taking $M=\mathbb{S}^2$. Every vector field on $\mathbb{S}^2$ vanishes at some point, hence every flow-diffeomorphism has a fixed point. Maybe we can find two vector fields, such that the composition of their flows is a diffeomorphism without fixed points.

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Asaf Shachar
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    Your attempt won't work. Any diffeo $S^2 \to S^2$ without fixed points is homotopic to the antipodal map, but a composition of flows is clearly homotopic to the identity. – PVAL-inactive Mar 16 '16 at 22:59

1 Answers1

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The relevant theorem is

Theorem (W.Thurston). Let $M$ be a smooth compact manifold, $Diff_o(M)$ is the identity component of the diffeomorphism group of $M$. Then the group $Diff_o(M)$ is simple (as an abstract group).

You can find a proof in

A. Banyaga, The structure of classical diffeomorphism groups. Mathematics and its Applications, 400. Kluwer Academic Publishers Group, Dordrecht, 1997.

Given this, let $G< Diff_o(M)$ denote the subgroup generated by the set $F_M$ of diffeomorphisms given by flows of time-independent vector fields on $M$. It is clear that this subgroup is normal and nontrivial (provided that $dim(M)>0$). Hence, $G= Diff_o(M)$ by Thurston's theorem. It follows that the set $F_M$ cannot form a subgroup unless $F_M= Diff_o(M)$. But, as you already know (see also Lee Mosher's answer here in the case of surfaces), $F_M\ne Diff_o(M)$. Therefore, $F_M$ does not form a subgroup.

Edit. For a direct proof that $F_M$ is not a subgroup see the answer by Martin M.W. to this MO question.

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Moishe Kohan
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