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 ?

2$\begingroup$ On the contrary, if you allow the vector field to be timedependent, then the answer is yes. From the way you worded your question it's not clear if you allow for this or not. $\endgroup$– Ryan BudneyJan 4, 2014 at 0:47

$\begingroup$ 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 timedependent flows of vector fields with restrictions on the time behaviour of the vector field. For example you could look at temporallyanalytic vector fields, and so on. $\endgroup$– Ryan BudneyJan 4, 2014 at 0:57

$\begingroup$ Yes I meant a vector field not depending on time. Still it's good to know the answer in the other case. $\endgroup$– Selim GJan 4, 2014 at 10:51
2 Answers
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 infinitedimensional 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".
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 not 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.