Question about smoothness of a vector field (Reeb).

[quasar987]

Hello,

I am simply looking for an argument proving the smoothness of the Reeb vector field of a given contact form.

If you don't know the relevant definitions, the problem is simply this: Let M be a manifold of odd dimension 2n+1 and let $\alpha$ be a 1-form on M such that
1) $\alpha$ is nowhere vanishing (Hence, for every p in M, $H_p :=\ker\alpha_p$ has dimension 2n.)
2) The (2n+1)-form $\alpha\wedge d\alpha$ never vanishes. Equivalently, for every p in M, the bilinear form $d\alpha_p|_{H_p\times H_p}$ is nondegenerate.

Since a nondegenerate bilinear form on an odd dimensional vector space does not exist, it must be that for every p in M, $d\alpha_p$ is degenerate. But by 2), the restriction of $d\alpha_p$ to H_p is nondegenerate. So it must be that $\ker d\alpha_p=\{v\in T_pM : d\alpha_p(v,\cdot)=0\in T_p^*M\}$ has dimension 1.

Define a vector field R of M axiomatically by:
(A1) $R_p \in\ker d\alpha_p$ for all p in M,
(A2) $\alpha(R)=1$.

Since $\dim\ker d\alpha_p=1$, axiom (A1) defines uniquely R up to scaling factor, and (A2) fixes that scaling factor. Hence R is well defined (and is called the Reeb vector field associated with the contact form $\alpha$).

I am simply looking for a proof that R is smooth, given that M and $\alpha$ are too! Thanks!

 

[eok20]

Since R is well-defined as you say, you need to check that in a coordinate chart, R varies smoothly. Representing $$d\alpha$$ as a matrix in these coordinates, it is sufficient to show that you can find a smooth non-zero vector field in the kernel (which you can then normalize to be Reeb). To rigorously do this, you can probably argue that row reduction varies smoothly in the coordinates of $$d\alpha$$ so that you can find a smooth basis for its kernel.

 

[lavinia]

Hello,
2) The (2n+1)-form $\alpha\wedge d\alpha$ never vanishes.

The form $\alpha\wedge d\alpha$ is a 3 form - did you mean Hodge dual?

 

[eok20]

The form $\alpha\wedge d\alpha$ is a 3 form - did you mean Hodge dual?

I'm pretty sure he meant $\alpha\wedge (d\alpha)^n.$

 

[quasar987]

I'm pretty sure he meant $\alpha\wedge (d\alpha)^n.$

Correct.

 

[quasar987]

Since R is well-defined as you say, you need to check that in a coordinate chart, R varies smoothly. Representing $$d\alpha$$ as a matrix in these coordinates, it is sufficient to show that you can find a smooth non-zero vector field in the kernel (which you can then normalize to be Reeb). To rigorously do this, you can probably argue that row reduction varies smoothly in the coordinates of $$d\alpha$$ so that you can find a smooth basis for its kernel.

This is exactly what I tried, but I am having trouble making this argument rigourous... :\

 

[lavinia]

I am simply looking for a proof that R is smooth, given that M and $\alpha$ are too!

I don't have all of it but here is a part - I think/I hope

Suppose you have proved that the kernel of the bilinear form is a smooth line bundle. In a coordinate chart it looks like a 2n + 1 ball cross a line and the 1 form $\alpha$ is just a smooth function on it that is linear on each line. Since it is non degenerate, 1 is a regular value and its inverse image is a smooth 2n+1- manifold. This manifold is just a smooth vector field on the ball.

I wonder if the Implicit function theorem can be used for the first part as well.

 

[quasar987]

If we have proved that

$$L:=\bigsqcup_{p\in M}\ker d\alpha_p$$

is a smooth line bundle, then this is equivalent to say that about any point p in M, there is a nowhere vanishing smooth local section of L. Say X is such a smooth local section of L (i.e., a local vector field on M lying in L). Then there is a real valued function f such that locally,

$$R=fX$$

Since $$\alpha$$ and X are nowhere vanishing, we have

$$1=\alpha(R)=\alpha(fX)=f\alpha(X) \Rightarrow f=\frac{1}{\alpha(X)}$$

Thus f is smooth, and hence R=fX is smooth.

So proving the smoothness of R is equivalent to proving that L is a smooth vector bundle.

 

[lavinia]

If we have proved that

$$L:=\bigsqcup_{p\in M}\ker d\alpha_p$$

is a smooth line bundle, then this is equivalent to say that about any point p in M, there is a nowhere vanishing smooth local section of L. Say X is such a smooth local section of L (i.e., a local vector field on M lying in L). Then there is a real valued function f such that locally,

$$R=fX$$

Since $$\alpha$$ and X are nowhere vanishing, we have

$$1=\alpha(R)=\alpha(fX)=f\alpha(X) \Rightarrow f=\frac{1}{\alpha(X)}$$

Thus f is smooth, and hence R=fX is smooth.

So proving the smoothness of R is equivalent to proving that L is a smooth vector bundle.

Right - I also remember thought that the poof of the Implicit function theorem directly supplies local coordinates for a = 1.

 

[quasar987]

What do you mean by that lavinia?

 

[lavinia]

I seem to remember but may be mistaken that the proof that F(x) = constant is a manifold does this by constructing a local parameterization - I will look up the proof to make sure

 

[lavinia]

HI quasarhere is a thought. I wonder what you think. Maybe it could get us somewhere with this.

In the case of a 3 manifold you get that the vector field,R, must satisfy a system of first order linear PDE's. If X and Y are everywhere non-zero linearlly independent vector fields in the kernel of alpha then

alpha([X,R]) = alpha([Y,R]) = 0 I don't know anything about PDE's but wonder if there is an existence theorem here that guarantees smooth solutions.