# Finding Reeb Vector Fields Associated with Contact Forms

1. Oct 23, 2013

### WWGD

Hi, All:

Let w be a contact form , say in ℝ3, or in some 3-manifold M i.e., a smooth, nowhere-integrable 2-plane subbundle of TM. I'm trying to see how to find the Reeb field Rw associated with w.

My ideas are:

i) Using the actual definition of the Reeb field associated with a contact form w:

Finding Rw as the zero set of dw(Rw, .) , i.e.,

Rw should kill every other vector field in this expression

ii) Using the fact that LRw w =0 ,

i.e., the Lie derivative of w about the Reeb field is zero.

But I have not gotten far using these methods. Anyone have other ideas?

Thanks.

2. Oct 24, 2013

### Ben Niehoff

This is essentially what you do. It should just be a linear algebra problem. Where do you get stuck?

As an example, maybe try the canonical contact form in R^3,

$$\omega = x \, dy + dz$$

3. Oct 24, 2013

### WWGD

Well, I don't know if I'm doing something wrong, but what I'm doing does not get me any useful form for the fields, at least not one that I can recognize as useful; I was trying with the form in ℝ3 (r,θ,z) given by ω:=dz+ r2

Then dω=2rdθdr ; then we want to find a Reeb field Rw=R1∂/∂θ+
R2∂/∂r , where R1(r,θ,z) , R2(r,θ,z) are C , and let V= V1∂/∂θ+ V2∂/∂r be any vector field .

Then I get, setting :
2rdθdr(Rw,.)==0 ,

2rdθdr(R1∂/∂θ + R2∂/∂r, V1∂/∂θ+ V2∂/∂r )==

2r( dθ (Rw )dr(V) -dr(V)dθ(Rw) ) =2r(R1)V2-V1R2 ):==0

Then R1 , R2 are so that R1V2-R2V1=0

What does that tell me about R1 , R2 ?

With the standard form in μ=dz+xdy in ℝ3 , I get:

(Sorry for not being more concise; I'm having trouble working with all these Latex symbols; I'm used to a different version of Latex)

dμ=dydx ; set Rμ=R1∂x+R2∂y, and
V=V1∂x+V2∂y

Then dydx(Rμ,V)= dy(Rμdx(V)-dy(V)dx(Rμ)=

R2V1-R1V2=0

How does that help me figure out R1, R2? Or maybe I'm doing something wrong somewhere?

Last edited: Oct 24, 2013
4. Oct 24, 2013

### WWGD

Also, I was wondering if you, or anyone else, had a suggestion for showing that

for contact w, w/\dw≠ 0 iff dw|Kerw≠0 , i.e., dw restricted to the contact planes is
nowhere zero, so that dw is a symplectic form ( dw, clearly closed, and also non-degenerate).

I have been just cranking out and expanding a generic 1-form w , wedging with dw, and trying to figure out what I need for w/\dw to not be zero, and getting nowhere useful.

Thanks.

5. Oct 24, 2013

### WWGD

I wonder if this too, would work: given a contact form w, we know that the representation of dw as a quadratic form will be odd-dimensional, i.e., as an nxn matrix, where n is odd. The matrix will then be antisymmetric and odd-dimensional, so that its determinant will be zero. Then the representing form Q(a,b) is degenerate, and we then will have some vector v with Q(v,.)=0. But, how do we choose the actual Reeb field from this?

6. Oct 24, 2013

### fzero

You aren't considering the most general possible Reeb field. Try

$$R = R_r \partial_r + R_\theta \partial_\theta + R_z \partial_z,$$

with a similar expression for the general $V$.

7. Oct 24, 2013

### WWGD

Thanks, both for your comments, but my form 2rdθdr does not contain a dz-component, so including a z-component would not

change anything, would it? dθdr( Rz∂z)=0 , right?

8. Oct 24, 2013

### Ben Niehoff

Isn't that exactly what you're trying to find?

9. Oct 24, 2013

### WWGD

Ah, I see, right. But how about the other results with R1,R2 , and A1R2-A2R1=0 ? How does that help me figure out what R1, R2 are?

10. Oct 24, 2013

### Ben Niehoff

1. The solution is not necessarily unique.

2. Your posts would be way easier to read if you used Latex.

11. Oct 24, 2013

### fzero

Can you write down the equations that you get from applying $\omega(R)=1$ and $d\omega(R,V)=0$? I find that they have a unique solution in this case.

12. Oct 25, 2013

### WWGD

fzero:

Do you refer to the Reeb field on (R,θ,z) associated with ω=dz+r2dθ , or the one associated to (x,y,z) , given by ω=xdy+dz ?

Ben:
Please give me some time to learn Latex; I've been using the 'Quick Symbols' drop box in the meantime, how that makes things a bit easier to read.

To All:
Please give me some temporary leeway to ramble-on to see if (where, actually) my knowledge has gaps.

I guess the non-uniqueness is also given by linear algebra, i.e., if we represent dω in some basis (**) as a quadratic form Q(x,y), then Q is odd-dimensional ( since contact forms are only defined for odd-dimensions) and antisymmetric, so that Q(v,. )=0 has non-zero solutions, i.e., Q is degenerate in this case. Right? Still, I'm not sure of why dω ~Q(x,y) is antisymmetric here.

Does the other condition, i.e., ω(Rw)=1 guarantee uniqueness (at least uniqueness up to "something nice") of Rω ? If not, is there some additional condition that would give me uniqueness --up to "something nice (e.g., scalar multiplication)"?

** But then we have the nightmare of deciding if this result depends on the choice of basis.

Last edited: Oct 25, 2013
13. Oct 25, 2013

### Ben Niehoff

I believe so. Consider: We are in $2n + 1$ dimensional space. $d \omega (R, \cdot ) = 0$ gives $2n$ equations, and $\omega(R) = 1$ gives 1 equation. So there are exactly enough equations. The condition $\omega \wedge d\omega \neq 0$ guarantees that these $2n$ equations are linearly-independent.

Right-click on anybody's Latex to see the code written.

14. Oct 25, 2013

### fzero

I computed it for your choice of $\omega$, but it turns out to be the same for the other choice.

I am not familiar with specific uniqueness results. The Reeb field obviously depends on the choice of $\omega$. The condition $\omega(R)=1$ generally fixes $R$ up to scalar multiplication of $\omega$, so a scalar multiple of $R$ would not be Reeb without rescaling $\omega$. Also, since $\omega$ must be completely non-integrable, $d\omega(R,\cdot)=0$ means that most of the coefficients in the expansion of $R$ have to vanish. It seems likely that $d\omega(R,\cdot)=0$ leads to a one-parameter family and $\omega(R)=1$ fixes the parameter.

As I was posting, I see that Ben has given a more refined version of the latter argument.

15. Oct 25, 2013

### WWGD

Ah, actually I think a light bulb went on: I think the condition $\omega(R)=1$ is just saying that $\omega(R)$ is never zero, so that we can normalize it to be 1. And $\omega$ just kills all the points in a contact hyperplane by definition, right? since the contact hyperplanes are precisely the kernel of $\omega$. So I think $\omega(R)(X)$ not being zero is saying that X is never _on_ the contact hyperplanes, so that X is actually transverse to the planes. So a Reeb field must be transverse to the contact planes.

Let me continue working on finding the actual kernel of $d\omega$

16. Oct 25, 2013

### WWGD

I tried again, and I came up with:

1)For $$dw=dz+r^2d\theta$$ , and the Reeb field $$R_r\partial_r +R_\theta\partial_\theta+ R_z\partial_z$$, and a generic $$V= V_r\partial_r + V_\theta \partial_\theta+ V_z \partial_z$$ $$(R_\theta )(V_r)- (R_r)V_\theta =0$$ and, of course, always for any $$R_z$$ since any $$R_z$$ is killed by $$dw=2rdrd\theta$$.
For $$w=xdy+dz$$ , like f_zero said, with $$dw=dydx$$, I get something similar; the Reeb field $$R_x\partial_x+ R_y\partial_y +R_z \partial_Z$$ and a generic vector $$V_x\partial_x +V_y\partial_y + V_z \partial_z$$:

$$d\omega$$ is zero when $$R_yV_x -R_xV_y=0$$ , and any $$R_z$$

I guess there is an "orthogonality" thing here, in that $$<(R_y,R_x),(V_x, -V_y)>=0$$

Last edited: Oct 25, 2013
17. Oct 25, 2013

### WWGD

My apologies; I spent like 25 minutes trying to figure out how to do the correct spacing, and I'm giving up for now.

18. Oct 25, 2013

### Ben Niehoff

Correct spacing with what?

19. Oct 25, 2013

### WWGD

I meant the spacing within and between sentences and expressions, e.g., paragraphs broken in half, etc..

20. Oct 25, 2013

### Ben Niehoff

Use the Quote button to see exactly what someone else typed. There are three commands to insert Latex. "tex" puts things on a separate line. "itex" puts them in-line, and double-# is a shorthand for "itex".