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Hi,

If I have a divergence free vector field defined on a smooth manifold, and I apply some diffeomorphism, what can I say about what happens to the vector field? The example I am using is of an open or closed (pick one) disk embedded in 3 dimensions and we contract the disk to a line. Perhaps the question could be answered by giving first only the most general transformation that could be applied to the vector field (probably general linear transformation) followed by others that may depend increasingly on the metric (which I may want to ignore since I am more interested in diffeomorphisms rather than smooth maps) and yet further depending on the properties of the vector field.

thanks

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    $\begingroup$ Your manifold is probably a smooth manifold? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 7, 2010 at 23:24
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    $\begingroup$ How are you defining divergence? As I recall, to do so you need a volume form on the manifold in question. $\endgroup$
    – drbobmeister
    Commented Dec 7, 2010 at 23:53
  • $\begingroup$ There are a lot of problems with your question. Metric? Is your manifold a Riemannian manifold? To talk about the divergence implies that your manifold is at least orientable (you need a fixed volume form). And since when is a dimension-reducing contraction a diffeomorphism? $\endgroup$
    – Willie Wong
    Commented Dec 7, 2010 at 23:54
  • $\begingroup$ Also, if $\phi: M\to N$ is a diffeomorphism between smooth manifolds, and let $\omega$ be a volume form on $N$ and $X$ a vector field on $M$, then by definition the following two are equivalent: (a) $\mathop{div}_{\phi^*\omega}X = 0$ and (b) $\mathop{div}_\omega \phi_*X$, where $\phi^*\omega$ is the pull-back of $\omega$, and $\phi_*X$ is the push-forward of $X$. So this essentially answers your first question. Please clarify the rest of your post, or the question may risk closure. $\endgroup$
    – Willie Wong
    Commented Dec 7, 2010 at 23:58
  • $\begingroup$ @Ben Sprott: I removed the topological vs. smooth confusion in your post. $\endgroup$
    – Ryan Budney
    Commented Dec 8, 2010 at 0:43

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The divergence of a vector field depends on a volume form, which is a nowhere-zero $n$-form on an oriented $n$-manifold. Given a volume form, most diffeomorphisms change it, but there is a large and rich group of diffeomorphisms that preserve it. For instance, Calabi showed (by an elementary construction) that there is a deformation retraction from the space of all smooth diffeomorphisms on a manifold to the space of volume-preserving diffeormorphisms --- basically because the set of choices for positively-oriented volume forms is contractible. (Note: this retraction cannot be made to preserve the group structure). I.e, you can do basically anything you want with volume preserving vector fields except things that are obviously impossible (like a diffeomorphism of a ball to a proper subset of the ball). Quite a lot is also known about the group structure of volume-preserving diffeomorphisms, as well, if there are questions that involve actual group relations.

There are topological and differentiable consequences of a vector field being divergence free, which are of course preserved by diffeomorphisms. On a closed manifold, the Poincaré recurrence theorem states that for almost every x in $M^n$, the flow line of a divergence free vector field returns infinitely often to every neighborhood of its starting point. This is because for any set of positive measure, the forward flow-tube builds up volume at a steady rate, so if there's an upper bound, it has to come back and intersect itself.

The diffeomorphic image of a divergence free vector field has the property that it can be multiplied by a smooth function --- that is, the speed of the flow can be regulated --- to make it divergence free again. Given a diffeomorphism, just multiply by the ratio of the volume in the domain to the volume in the range. I.e., just like in a stream, the flow slows down when it spreads out and has to speed up to maintain flow through a small gap.

Divergence free vector fields are in 1-1 correspondence with closed $n-1$ forms, by the correspondence $$X \leftrightarrow i_X(\Omega), $$ where $X$ is the vector field, $\Omega$ is the volume form, and $i_X$ means stick $X$ in the first slot of $\Omega$, thought of as a function from an n-tuple of vectors to the real numbers. You can think of $i_X(\Omega)$ as the flux or flow rate of the vector field $X$ through an infinitesimal element of $n-1$-dimensional area. If $X$ is the instantaneous flow of the Mississippi river in time units of second, and you integrate $i_X(dV)$ on a surface that cuts across the entire river, you get the amount of water per second flowing through the Mississippi at that place.

It's easy to work with closed $n-1$ forms, by starting with a basis for cohomology, then modifying by $d(n-2$ form$)$. You can take a diffeomorphism that is the time t flow of a divergence free vector field, and it will preserve divergence-freeness of any other vector field. This does not get all examples even near the identity, but by integrating time-dependent families of divergence free vector fields, you get arbitrary volume-preserving vector field that are isotopic to the identity, and containing a neighborhood of the identity.

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    $\begingroup$ Great awswer; worth re-reading. $\endgroup$
    – drbobmeister
    Commented Dec 8, 2010 at 5:06
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    $\begingroup$ @Bill Thurston : Is there an easy way to see that you cannot deformation retract the group of smooth diffeomorphisms to the group of volume-preserving diffeomorphisms in a way that preserves the group structure? $\endgroup$
    – Andy Putman
    Commented Dec 8, 2010 at 5:07
  • $\begingroup$ @Andy Putnam: Around the time I was getting my Ph.D., and for a few years after, these kinds of things were hot topics. Generalizing a theorem of John Mather, I proved that the identity component of the group of diffeomorphisms of any closed manifold is simple. I.e. it has no group-theoretic retraction whatsoever to anything. Another distinction: there is a characteristic class for foliations, the Godlbillon-Vey invariant, that translates into an element of $H^{n+1)(Diff_\delta(M^n; \mathbb Z)$ that vanishes identically on the group of volume preserving diffeomorphisms. $\endgroup$
    – Bill Thurston
    Commented Dec 8, 2010 at 21:57
  • $\begingroup$ @Andy Putnam: I realized that my comment doesn't actually answer your question about an easy way to see, since it only recites authority. One way to see this when $H^1(M) \ne 0$ is to think about the flux homomorphism on the subgroup that preserves an element of $H^1$. Elements of $H^1(M; \mathbb Z)$ are always induced by maps to $S^1$; take a regular value of such a map, which is a hypersurface $N^{n-1} \subset M$. Map this subgroup of volume-preserving diffeomorphisms to $S^1$ by measuring the signed volume between $N$ and its image. It's easy to see this homomorphism doesn't extend. $\endgroup$
    – Bill Thurston
    Commented Dec 9, 2010 at 0:07
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    $\begingroup$ @Andy Putnam: I might not have published much more than the Bulletin article. I think I was diverted by the classification theory for foliations. I gave lots of talks about it, and there are probably proofs written by others, Mather in particular, perhaps Tsuboi. My memory is hazy about what was written. David Epstein earlier had proved perfect <==> simple, and this is straightforward and nice. Michael Herman's results on KAM theory implied the group for T^n is perfect. The "torus trick" to transfer this fact to other manifolds is cool, but it's likely not been written up in a really nice way. $\endgroup$
    – Bill Thurston
    Commented Dec 10, 2010 at 2:54

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