Defining a Contact Structure Globally -- Obstructions?

[WWGD]

Hi,
Let $M^3$ be a 3-manifold embedded in $\mathbb R^3$ and consider a 2-plane field ( i.e. a Contact Structure) assigned at each tangent space $T_p$. I am trying to understand obstructions to defining the plane field as a 1-form ( Whose kernel is the plane field/ Contact Structure) Given a specific point we can define a local form w as a linear map $\mathbb R^3 \rightarrow \mathbb R$ whose kernel is the contact plane. I am curious about the obstructions to defining the contact structure through a global 1-form. I suspect it may have to see with the triviality of either the global (tangent) bundle or the 2-subbundle of the tangent bundle . Is this correct? Can anyone add anything and/or give examples?
Thanks. This seems like @lavinia could know the answer.

 

[lavinia]

Hi WWGD

I don't know anything about contact geometry but it seems if there is a smooth distribution of 2 planes $V$ on a 3 manifold that is the kernel of a global 1 form ,then the tangent bundle splits as $TM=V⊕ξ$ where $ξ$ is a trivial line bundle.

Conversely if with respect to some Riemannian metric $<,>$ , one can choose a globally non-zero vector field $s$ that is orthogonal to $V$, then the 1 form $ω= <s,>$ is a global 1 form whose kernel is $V$.

Notes:

- Every oriented 3 manifold has trivial tangent bundle.
- If the 3 manifold is orientable and the 2 plane distribution is the kernel of a global 1 form. then the 2 plane distribution is also orientable.
- I think a contact distribution is automatically a sub-bundle since locally it is the kernel of a 1 form. Yes?
- It is possible in a oriented 3 manifold to have an unorientable 2 dimensional sub-bundle with non-trivial normal line bundle. It would be interesting to find one that is a contact structure.

Sorry not to be of more help.

 

[WWGD]

Hi WWGD

I don't know anything about contact geometry but it seems if there is a smooth distribution of 2 planes $V$ on a 3 manifold that is the kernel of a global 1 form ,then the tangent bundle splits as $TM=V⊕ξ$ where $ξ$ is a trivial line bundle.

Conversely if with respect to some Riemannian metric $<,>$ , one can choose a globally non-zero vector field $s$ that is orthogonal to $V$, then the 1 form $ω= <s,>$ is a global 1 form whose kernel is $V$.

Notes:

- Every oriented 3 manifold has trivial tangent bundle.
- If the 3 manifold is orientable. then the 2 plane distribution is also orientable.
- I think a contact distribution is automatically a sub-bundle since locally it is the kernel of a 1 form. Yes?
- It is possible in a oriented 3 manifold to have an unorientable 2 dimensional sub-bundle with non-trivial normal line bundle. It would be interesting to find one that is a contact structure.

Sorry not to be of more help.

Actually pretty helpful, Lavinia, thanks, and sorry to put you on the spot :).

 

[lavinia]

It was fun to learn a little about it.

Here is a different idea of obstruction in a special case. This time there will be a global 1 form whose kernel is a distribution of 2 planes but there will be an nice obstruction to this being a contact structure.

For a closed orientable Riemannian 2 manifold such as the sphere, the tangent unit circle bundle is a closed 3 manifold and is also a principal $SO(2)$ bundle. $SO(2)$ acts on a fiber circle by rotation. A connection 1 form is dual to the line bundle along the fiber circles - this is a trivial line bundle - and its kernel is a distribution of 2 planes that is called the horizontal space of the connection. In general this horizontal space not a contact structure. The obstruction is the curvature. The curvature form must be everywhere non-zero in order for the horizontal distribution to be a contact structure. So for instance, the horizontal distribution on the tangent circle bundle to the unit sphere in $R^3$ is a contact structure.

 

[WWGD]

Just to add a few things in the unlikely case you don't know them yet: Contact structures are nowhere-integrable , meaning there is no open set the restriction of whch is the tangent bundle of a manifold. The condition $\theta \wedge d \theta \neq 0$ follows from Frobenius theorem, as the needed condition for a distribution to be somewhere-involutive.

 

[lavinia]

Just to add a few things in the unlikely case you don't know them yet: Contact structures are nowhere-integrable , meaning there is no open set the restriction of whch is the tangent bundle of a manifold. The condition $\theta \wedge d \theta \neq 0$ follows from Frobenius theorem, as the needed condition for a distribution to be somewhere-involutive.

Right. Thank you. I read the proof. It is a nice way to characterize it.

In the curvature case, if $ω$ is the connection 1 form then $dω$ is the curvature 2 form and is equal to $π^{*}-KdV$ where $K$ is the Gauss curvature so $ω∧dω$ is zero at a point only if $K \circ π$ is zero at that point. So if $K$ is never zero, $ω∧dω$ is never zero.