# Flows of vector fields and diffeomorphisms isotopic to the identity

Let $M$ be a compact manifold and $\varphi : M \longrightarrow M$ be a diffeomorphism which is isotopic to the identity. Does there exist a vector field $X$ on $M$ such that $\varphi$ is the flow at time $1$ of $X$? If that is not always the case, where does the obstruction for such a $\varphi$ to be a flow lives in ?

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On the contrary, if you allow the vector field to be time-dependent, then the answer is yes. From the way you worded your question it's not clear if you allow for this or not. – Ryan Budney Jan 4 '14 at 0:47
I suppose if you wanted to quantify the difference between diffeomorphisms isotopic to the identity and flows of vector fields, you could create the intermediate notion of time-dependent flows of vector fields with restrictions on the time behaviour of the vector field. For example you could look at temporally-analytic vector fields, and so on. – Ryan Budney Jan 4 '14 at 0:57
Yes I meant a vector field not depending on time. Still it's good to know the answer in the other case. – Selim G Jan 4 '14 at 10:51

The answer is no. In fact, there are diffeomorphisms arbitrarily close to the identity which are not contained in flows (which are also often called $1$-parameter subgroups; here one is thinking of the set of vector fields on $M$ as a sort of "Lie algebra" for the diffeomorphism group).

Here's an example which I learned about from Warning 1.6 of Milnor's paper "Remarks on infinite-dimensional Lie groups" (which I highly recommend reading if you are interested in things like this). Regard $S^1$ as $\mathbb{R}/2\pi$. Fix some integer $n$ and some $\epsilon$ such that $0<\epsilon<1/n$. Define $$f : S^1 \longrightarrow S^1$$ $$f(x) = x + \pi/n + \epsilon \sin^2(n x).$$ Choosing $n$ large enough and $\epsilon$ small enough, we can make $f$ arbitrarly close to the identity in the $C^{\infty}$ topology. It is a fun exercise to show that $f$ does not lie in any $1$-parameter subgroup of $\text{Diff}(S^1)$.

It is not hard to generalize this kind of behavior to any manifold.

I don't think there is any easy description of the obstruction here; it seems to be a delicate problem in dynamics.

However, it is the case that every element of $\text{Diff}^0(M)$ can be written as a product of finitely many elements that are contained in $1$-parameter subgroups. Indeed, the set of elements of $\text{Diff}^0(M)$ that are contained in $1$-parameter subgroups generate a subgroup $\Gamma$ that is easily seen to be normal. But Thurston proved that $\text{Diff}^0(M)$ is simple for any compact manifold $M$, so we must have $\Gamma = \text{Diff}^0(M)$. For a discussion and proof of Thurston's theorem, I recommend Banyaga's book "The Structure of Classical Diffeomorphism Groups".

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In fact, for every manifold $M$ with $\dim(M)\ge 2$, the group $Diff_c(M)$ of diffeomorphisms with compact support contains a smooth curve $c:\mathbb R\to Diff_c(M)$ (smooth in the sense that the associated mapping $\hat c: \mathbb R\times M\to M$ is smooth) with $c(0)=Id$ such that the $\{c(t): t\ne 0\}$ are a free set of generators for a free subgroup of the diffeomorphism group such that no diffeomorphism in this free subgroup with the exception of the identity embeds into a flow. This is proved in

• Grabowski, J., Free subgroups of diffeomorphism groups, Fundam. Math. 131 (1988), 103–121.

The proof uses clever constructions with Nancy Kopell's diffeomorphisms on $S^1$ (those described in Andy Putman answer) which do embed into a flow. This free subgroup is contractible to the identity: slide the generators $c(t)$ to $c(0)$.

So the image of exponential mapping of $Diff_c(M)$ does not meet very many diffeomorphisms near the identity.

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