Consider a compact manifold M. For a vector field X on M, let $\phi_X$ denote the diffeomorphism of M given by the time 1 flow of X.

If X and Y are two vector fields, is $\phi_X \circ \phi_Y$ necessarily of the form $\phi_Z$ for some vector field Z?

Since $X\mapsto \phi_X$ can be thought of as the exponential map from the Lie algebra of vector fields to the group of diffeomorphisms, an obvious candidate is that Z should be given by the Baker-Campbell-Hausdorff formula $B(X, Y) = X+Y+\frac{1}{2}[X,Y]+\cdots$. But does this hold in this infinite-dimensional setting? If so, in which sense does the series converge to Z?

Also, I'm interested in the case where M is a symplectic manifold and we consider only symplectic vector fields (ie. vector fields for which the contraction with the symplectic form is a closed 1-form). Locally, X and Y are the Hamiltonian vector fields associated to smooth functions f and g, so I assume that asking whether B(X, Y) makes sense/is symplectic corresponds to asking whether B(f, g) makes sense/defines a smooth function (where, of course, we use the Poisson bracket in the expansion of B(f, g)). The right-hand side of B(f,g) consists of lots of iterated directional derivatives of f and g in the Xf and Xg directions; it is not clear to me that the coefficients in the BCH formula make the series converge (uniformly, say) for any choice of f and g.


To answer your first question, the composition of two time-1 flows won't necessarily be another time-1 flow.

One way to see this is to note that when a time-1 flow $\phi_X$ has a periodic point $P$ (period > 1), then $P$ can't be hyperbolic since it lies on a closed orbit of the flow for $X$. (The eigenvector of $D\phi_X$ tangent to this orbit has corresponding eigenvalue 1.)

Now, take a flow on $S^2$ whose time-1 map rotates the sphere, switching the north and south poles. Take a second flow for which both poles are hyperbolic attracting fixed points. Composing the two time-1 maps gives you a new diffeomorphism with hyperbolic points of period 2.

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    $\begingroup$ This works in the symplectic case as well - you can compose shift-like derivatives to get a hyperbolic one. $\endgroup$ – Sergei Ivanov Mar 20 '10 at 8:31

The answer is no in the symplectic case. I believe there are $X$ and $Y$ such that $\phi_X\circ\phi_Y$ is not generated by any time-independent Hamiltonian flow but I can't produce an example. However I can prove that $\phi_X\circ\phi_Y$ is not generated by a flow defined by a local formula.

Consider the cylinder $S^1\times [0,1]$ and close up its ends to make a sphere. The first field $Y$ rotates the thing along the $S^1$ factor, say, with angular velocity $\sqrt 2$. The second field $X$ is supported is a small round disc $D$ inside the cylinder, and it rotates this disc around its center (with angular velocity decaying to 0 at the boundary).

Let $f=\phi_X\circ\phi_Y$ and suppose it is a time 1 map of a time-independent Hamiltonian flow. Since the sphere is simply connected, there is a globally defined Hamiltonian $H$, and the orbits of the flow are its level curves. Suppose also that the flow field is given by a local formula in terms of $X$ and $Y$. Outside $D$, we have $X=0$, so the formula yields $Y$. This implies that the orbits contain large pieces of the original circles (namely part of the circles outside $D$). Consider such an interval $s$ of the circle through the center of $D$. The image $f(s)$ must be within the same orbit. But $f(s)$ intersects $s$ transversally at some point outside $U$. Moving along $Y$ shows that the orbit contains an open region, a contradiction.


The BCH formula holds in a finite-dimensional Lie algebra for $X,Y$ in some open neighborhood of $0$. Therefore, the coefficients of BCH are bounded by $a^n$ for some $0<a<\infty$. So suppose that $X,Y$ are vector fields for which degree-$n$ bracket monomials (things like $[X,[X,Y]]$, with $n$ many $X$s and $Y$s total) don't grow too fast in your topology — not faster than $b^n$ for some $0 < b < \infty$. Then for $s < b/a$, ${\rm BCH}(sX,sY)$ converges, in your topology, if your topology is complete.

Now, I'm not much of one for infinite-dimensional topological vector spaces, so I can't say which topologies are great to use.


Does it make any sense - if we consider vector fields as derivations of the algebra of smooth functions on M? Then each \phi_X is an automorphism of this algebra and there is an asymptotic formula for its (i.e., \phi_X) action. Product of two such asymptotic representations give as. formula for superposition \phi_X \circ \phi_Y - which leads to BCH for Z. (smth like in universal enveloping algebra of vector fields)... These asymptotic formulas not always converge thou


The answer to the first question, is $X\circ Y$ a vector field if $X$ and $Y$ are, is in general no. The reason is that the operator does not satisfy the Leibniz rule. One of the cool things about the Lie bracket of vector fields is that the problematic terms vanish and it is a vector field.

The question about Baker-Campbell Hausdorff formula is not well stated, you could make it true by being careful about what you mean or make it false.

Finally, $B(X,Y)$ makes sense: you are evaluating a smooth two form on two smooth vector fields.

My guess is that you really want to be asking something about a deformation quantization of a subspace of the smooth functions on the manifold.

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    $\begingroup$ The question is not about $X\circ Y$, it is about their generated time 1 flow maps. $\endgroup$ – Sergei Ivanov Mar 19 '10 at 19:30
  • $\begingroup$ The composition of flows isn't a one parameter group in general so it doesn't define a vector field. $\endgroup$ – Charlie Frohman Mar 19 '10 at 20:57
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    $\begingroup$ You misinterpreted the question. What is asked is whether a single diffeomorphism $\phi_X\circ \phi_Y$ belongs to some one-parameter group (which may have nothing to do with the composition of flows). $\endgroup$ – Sergei Ivanov Mar 19 '10 at 21:14
  • $\begingroup$ One part of the question is about that. And indeed I couldn't see there was a good question in there. $\endgroup$ – Charlie Frohman Mar 20 '10 at 14:47

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