Neighborhoods of the identity in diffeomorphism groups Finite dimensional Lie groups have the nice property that if $V$ is a small neighborhood of the identity, and $U \subset V$ another neighborhood, then $V$ is covered by $U^k$ (the set of all products of $k$ elements of $U$) for some $k$.  This follows from the fact that V is relatively compact.  
Does the same property hold for small enough neighborhoods of the identity in an infinite dimensional Lie group?  My favorite example is the group of compactly supported diffeomorphisms of $\mathbb{R}^n$ -- is it known whether it has this property?   
Remark: I'm pretty sure this property is not equivalent to local compactness, so the fact that $Diff_c(\mathbb{R}^n)$ is not locally compact doesn't give an answer.  
 A: What helps a little is that for any (second countable connected) smooth manifold $M$, the connected component $Diff_c(M)_0$ of the group $Diff_c(M)$ is perfect (Epstein) and simple (Thurston). On the other hand, $\exp: \mathfrak X_c(M)\to Diff_c(M)$ is far from being locally surjective. Grabowski has shown, that for $\dim(M)\ge 2$ there exists a smooth curve in $Diff_c(M)$ starting at the identity such that the element of these curve (sauf the identity) form a free generating system of a free subgroup of $Diff_c(M)$ such that no element $\ne 1$ of this free subgroup embeds into a flow. This free subgroup of uncountably many generators  is contractible to 1 (slide each generator to 1 along the curve). 
EDIT
Some references: 


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*(1) Martins Bruveris, François-Xavier Vialard: On Completeness of Groups of Diffeomorphisms. arXiv:1403.2089

*(2) S. Haller and T. Rybicki and J. Teichmann: Smooth perfectness for the group of diffeomorphisms. J. Geom. Mech. 5(2013), 281–294. Preprint: arXiv:math/0409605

*(3) David Mumford, Peter W. Michor: On Euler's equation and `EPDiff'. Journal of Geometric Mechanics 5, 3 (2013), 319-344
(pdf)

*(4)  M. Bauer, J. Escher and B. Kolev: Local and Global Well-posedness of the fractional order EPDiff equation on R^d. To appear in Journal of Differential equations.
In [3], Thm 2, see (1) and (4) for more information, it is shown that the right invariant Sobolev metric of integral order $s>\frac{n+3}2$ is geodesically complete on $Diff_{H^\infty}(\mathbb R^n)$ and the corresponding Riemannian exponential mapping is a diffeomorphism from a suitable $H^s$-ball in $\mathfrak X_{H^\infty}(\mathbb R^n)$ into the group. 
This easily implies the result that the OP asked, but only for $H^s$-balls; thus for the Sobolev completion, the topological group (with smooth right translations) 
$$Diff_{H^s}(\mathbb R^n) = \{Id + f: f\in H^s(\mathbb R^n)^n, \det(Id + df) > 0\}.$$
Namely, consider any $H^s$-ball $V$ of radius $r$, Then any open neighborhood in $V$ contains a $H^s$-ball of radius $\epsilon>0$. Papers (1) and (4) show that one can connect any diffeomorphism $\phi\in V$ to the identity by a geodesic $c$ of length $<r$. Cut this geodesic in $\frac{\epsilon}2$-pieces $\{c(t): t_i\le t\le t_{i+1}\}$, $0=t_0<t_1<t_2<\dots<t_N=r$.
Then $c_1(t)=\{c(t): t_1\le t\le r\}. c(t_1)^{-1}$ has length $r-t_1$, and finally 
$\phi=c(r)= c_N(r).c_{N-1}(t_{N-1}).\dots.c_1(t_2).c(t_1)$. 
Moreover, for $Diff_{H^\infty}(\mathbb R^n)$, where $H^\infty$ is the intersection of all Sobolev spaces, a Frechet space, the result also holds, since each neighborhood of the identity in  $Diff_{H^\infty}(\mathbb R^n)$ contains a $H^k$-neigbourhood for sufficiently large  but finite $k$, and we can apply the result with the $H^k$-metric.
But $Diff_c(\mathbb R^n)$ has many more open neighborhoods, and for these nothing follows.     
A: Commenting on Peter Michor's answer, I want to say that we can however use the exponential map.
Is it true that the image of the exponential map is locally dense near the origin? If so, let us take open neighbourhoods $U\subset V$ of the identity such that $\overline{\exp(Vect_c(M))}\supset \overline{V}$. Given any $g\in V$, we can find $h_0\in U$ arbitrarily close to $id$ so that $gh_0^{-1}=f$ is the time $1$ map of a flow $\{f_t\}_{t\in[0,1]}$. Let $k=k(f)$ be the minimum integer such that $f_{1/k}$ belongs to $U$. Then $g$ is the product of $k+1$ diffeomorphisms in $U$:
$$g=f_{1/k}\circ\cdots\circ f_{1/k}\circ h_0.$$
The problem here is that $\overline{U}$ is not compact so it doesn't seem clear that we can have a uniform bound on the choice of $k$ (though we still have some freedom in the choice of $h_0$).
Edit Katie Mann is true that the image of the exponential map can happen to be not dense about the origin. There is one case for which I know that my question has a negative answer and this is a result by Nancy Kopell in Commuting diffeomorphisms, Global Analysis, Proc. Symmpos. Pure Math. vol XIV (1968), 165-184, see also Yoccoz, Petits diviseurs en dimension 1 Astérisque 231, SMF (1995)):
Kopell's Theorem states that the $C^1$ centralizer is trivial for an open dense set of $C^1$ circle diffeomorphisms. In particular there exists an open dense set of diffeomorphisms of the circle which are not the time 1 map of a flow.
