3
$\begingroup$

So, I have a codimension 2 contact submanifold of a closed contact manifold which has (topologically) trivial normal bundle. The question is, can I find a non-vanishing contact vector field which doesnt vanish on the submanifold, and it is transverse to it?

I want to be able to push it off to get a family of such submanifolds, and hence the question.

Since the space of such vector fields is given by the space of functions on the ambient contact manifold (contact Hamiltonians), my intuition tells me that it is big enough for this question to be reasonable.

Maybe even more wildly, can I approximate a given vector field by a contact vector field? Say C^0 or whatever...

This way I could approximate any non-vanishing section of the normal bundle by a contact vector field... Maybe this is a matter of estimating the size of the corresponding Lie group of contactomorphisms inside the group of diffeomorphisms of the contact manifold...

But now my intuition tells me, since rigidity holds (at least in the symplectic world) and the symplectic group is C^0 closed in the group of volume-preserving diffeomorphisms, that this indeed a stupid question...

$\endgroup$
4
$\begingroup$

For the first question the answer is YES. In a neighborhood $X\times D^2$ choose coordinates and a contact form such that $\alpha=\alpha|_X+xdy$, where $D^2=(x,y)$, $\alpha|_X$ - the contact form on $X$ (page 75, theorem 2.5.15 of Geiges). Consider a family $X_y=X\times(0,y)$. The contact structures on all $X_y$ are the same. Therefore this family can be obtained by contact isotopy (page 90, theorem 2.6.12 of Geiges An introduction to Contact Topology).

For the second question the answer is NO. The group of contactomorphisms is $C^0$-closed in all diffeomorphisms, hence the answer. Here is the counterexample: $(\mathbb{R}^3,\alpha=dz+xdy)$, a vector field $\frac{\partial}{\partial x}$. Suppose that a contact vector field $v$ approximates $\frac{\partial}{\partial x}$. Then by contact condition and Cartan formula on the straight line $x=0,z=0$:

$$0=(\mathcal{L}_v\alpha )(\frac{\partial}{\partial y})=d\alpha(v,\frac{\partial}{\partial y})+d(\alpha(v))(\frac{\partial}{\partial y})$$

So

$$\frac{\partial \alpha(v)}{\partial y}=-d\alpha(v,\frac{\partial}{\partial y}).$$

Since $\alpha(\frac{\partial}{\partial x})=0$ and $d\alpha(\frac{\partial}{\partial x},\frac{\partial}{\partial y})=1$, we can choose $v$ such that $|\alpha(v)|<\frac13$ and $d\alpha(v,\frac{\partial}{\partial y})>\frac23$. The integration of the formula above on the segment $\{(0,y,0),0\le y\le1\}$ gives a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.