# Symplectic vector fields everywhere transverse to a co-dimension one hypersurface

Usually when speaking about vector fields transverse to a hypersurface in a symplectic manifold, we talk about Liouville vector fields, i.e. vector fields $$X$$ with the property that $$\mathcal{L}_X\omega=\omega$$. This is equivalent (as far I understand) to the hypersurface being of contact type and such vector fields are common. I am interested to see if there are any results about symplectic vector fields everywhere transverse to a codimension one compact hypersurface. It seems that this would be a pretty rare phenomenon, e.g. the sphere in $$\mathbb{R}^{2n}$$ would not possess such a vector field. Are there any results of this nature out there?

I guess my intuition is that as symplectic vector fields are locally Hamiltonian and then locally everywhere tangent to some codimension one hypersurface, you would not expect them to be transverse to another codimension one hypersurface, but of course non-existence of such vector fields should only be clear globally so I don't think I'll be able to translate this into a good argument..

I guess if you are only interested in closed hypersurfaces, the following argument should often work:

A symplectic vector field $$X$$ will preserve the natural volume form $$\omega^n$$. If you take a closed hypersurface $$\Sigma$$ that is separating (e.g. any closed hypersurface of $$\mathbb{R}^{2n}$$), then you can split the symplectic manifold along $$\Sigma$$ into two disconnected domains $$G_0$$ and $$G_1$$. Furthermore $$X$$ will point everywhere along $$\Sigma$$ transversely from $$G_0$$ into $$G_1$$ (or the opposite direction).

As a consequence the flow $$\Phi_t^X$$ maps $$G_0$$ onto a larger domain still containing $$G_0$$ as a proper subset! This contradicts the volume preservation.

Thus it seems to me that the only candidates would be non-separating hypersurfaces, and in particular if your symplectic manifold has trivial homology in degree $$2n-1$$ there is no hope...

• Dear Klaus! Thank you for you answer. If it's alright with you, I will cite your helpful answer in an upcoming paper. I was planning on citing the answer directly, however, if you want acknowledgement in some other form please let me know (I am not yet used to conventions for online forum citations). Apr 19, 2019 at 12:00

This is a great question! (Or rather, I quite liked it. Reasonable people can differ on this, I suppose.) In fact, local considerations (or even semi-local considerations like understanding the symplectic geometry of a Weinstein neighbourhood of the hypersurface in question) can't suffice to rule out the possibility of the existence of a transverse symplectic vector field, as the example of separating and non-separating circles in the torus shows; any two simple closed curves in the torus have neighbourhoods which are symplectomorphic, since they're Lagrangian, but a meridian of the torus obviously admits a transverse symplectic vector field induced by translation in the universal cover in the direction of the equator, while a separating loop admits no transverse symplectic vector field by area considerations.

So global considerations are in order, and with that in mind, we can give a nice account of when a hypersurface admits a transverse symplectic vector field. Recall that any hypersurface $$\iota: N \rightarrow M$$ in a symplectic manifold $$(M,\omega)$$ is coisotropic and so carries a one-dimensional line bundle $$L \rightarrow N$$ whose fiber at $$p$$ is $$(T_pN)^\omega$$. In this setting, we have:

Proposition: There exists a symplectic vector field $$X \in \mathcal{X}(M)$$ which is transverse to $$\iota(N)$$ if and only if there exists a closed one-form $$\sigma \in \Omega^1(N)$$ such that:
(1) $$\sigma$$ is non-vanishing on non-zero vectors in $$L$$ (so in particular, $$\sigma$$ is non-exact), and
(2) $$[\sigma]$$ lies in the image of the induced map on cohomology. ie: $$[\sigma] \in im \big( \iota^*: H^1(M;R) \rightarrow H^1(N;R) \big)$$

The details are easy enough to check and relies simply on the observation that $$X \in \mathcal{X}(M)$$ is a symplectic vector field, transverse to $$N$$ if and only if the one-form $$\tilde{\sigma}:= X\lrcorner\omega$$ is closed and non-vanishing on non-zero vectors in $$L$$ (if $$Y$$ is any non-vanishing section of $$L$$, then $$X(p)$$ lies outside $$^\omega=T_pN$$), along with the fact that the flux map which sends symplectic isotopies to the average of their generating 1-forms is a surjection onto $$H^1(M;R)$$.

As to the preponderance of such objects, in some sense they aren't that uncommon; any odd-dimensional manifold $$N$$ which has a closed $$1-$$form which doesn't vanish on some one-dimensional distribution on $$N$$ embeds in a symplectic manifold which has a symplectic vector field transverse to $$N$$, because we can put a symplectic structure on $$N \times S^1$$ such that the vector field $$\partial_t$$ is symplectic and transverse to $$N \times \lbrace 0 \rbrace$$. But I'll admit that I'm curious to know what more can be said about this aspect of the question.

• Hi Dustin, thanks for your answer! I've been working on this a bit since I asked. In fact $N$ will possess something even more particular - the distribution $ker(\sigma)$ gives a symplectic foliation on $N$. This structure in turn gives further fairly strict requirements on the topology of $N$ (in certain cases). I am actually writing up a paper now that has this as a section (which is of interest to me as it has connection to the existence of certain types of well behaved Poincare sections) so if you're still curious I'll let you when it's on the arxiv! :) Apr 19, 2019 at 11:49
• @RMary You've piqued my interest! Please do let me know whenever you happen to post it! Apr 19, 2019 at 13:15