# 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".

21. Oct 30, 2013

### WWGD

Thanks all; I know this solution gives me a basis for the solution space, but I'm still trying to show uniqueness with the additional condition that $$w(R_w)=1$$

22. Oct 31, 2013

### fzero

No, if you want

$$R_yV_x -R_xV_y=0$$

for all $V_x,V_y$ then you must have $R_x=R_y=0$ (just try $V_x=0$, then $V_y=0$ as a separate case).

Then the condition $\omega(R)=1$ fixes $R_z$.

23. Oct 31, 2013

### WWGD

Ah, thanks. Just curious; I found out that the Reeb field associated to $S^2$ with the form:

$$x_1dy_1 -y_1dx_1+x_2dy_2-y_2dx_2$$ is given as $V(z)= i.z$ , where $i,z$ are Complex.

I'm trying to verify this; can I see $V_z$ as the Real vector field $$y\partial-x\partial y$$ ?

(think of $$iz=i(x+iy)=-y+ix$$ , i.e., as a vector field normal to the 2-sphere?

24. Oct 31, 2013

### fzero

First of all, this is $S^3$, not the 2-sphere. Second, since we are treating the $S^3$ as embedded in $\mathbb{R}^4$, there are restrictions on the coordinates $(x_i,y_i)$ in order that we are dealing with $S^3$. Points on the $S^3$ are those that satisfy $x_1^2+y_1^2+x_2^2 + y_2^2 =1$. This condition also defines a normal form

$$\nu = x_1 dx_1 + y_1 dy_1 + x_2 dy_2 + y_2 dy_2,$$

and vector fields tangent to the $S^3$ have to satisfy $\nu(V)=0$. You would need to use this condition to derive the Reeb field from the definitions.

With complex coordinates, we can use the chain rule to determine the relation

$$\partial_{z_i} = \frac{1}{2} \left(\partial_{x_i} - i \partial_{y_i} \right)$$

or any similar ones that we need. The Reeb field is

$$R = \sum_i \left( -y_i \partial_{x_i} + x_i \partial_{y_i} \right),~~~~(*)$$

so if we write this in complex coordinates, we'll find

$$R = \sum_i \left( \frac{i}{2} ( z_i - \bar{z}_i) (\partial_{z_i} + \partial_{\bar{z}_i}) +\frac{i}{2} ( z_i + \bar{z}_i) (\partial_{z_i} - \partial_{\bar{z}_i} \right) = \sum_i i z_i \partial_{z_i}.$$

So the expression that you have for the Reeb field in complex coordinates is correct. You should do the computation the other way now. Start with that expression, obtain the appropriate relations between ordinary and complex derivatives from the chain rule, and use them to obtain (*).

25. Nov 3, 2013

### WWGD

Thanks for your reply, $f_0$. Hope this will be the last question in this post.
I get your first comment; dumb of me; contact structures are defined only in odd dimensions. And I guess you're looking at $S^3$ as $f^{-1}(1)$ , so that the tangent space of a level set is the kernel of df (since the level set is of the form ${x: f(x)=c}$ and $d/dx(c)=0$

And I think I get the last part as using the transforms:

$Rez=\frac{z+z^-}{2}$ and $Imz= \frac{z-z^-}{2i}$ to go back-and-forth between the Real and the Complex versions (didn't you mean for $2i$ to go in the denominator?.)

But I don't get the $$\partial_{z_k}= \frac{ \partial_{x_k}-i\partial_{y_k}}{2}$$

(Could we please use k to index $z_k$ to avoid confusion with the complex $i$? )

Don't we just start at $$z_k=x_k+iy_k$$ and use linearity of $\partial$ so that

$$\partial z_k = \partial x_k+ i\partial y_k$$ ?

EDIT: How about this idea for finding $V(z)=iz$ as a complex vector field as a Real vector field:

While vector fields do not necessarily push forward along a map, we use the fact that $\mathbb R^2$ and $\mathbb C$ are diffeomorphic , thru, e.g., $$(x,y) \rightarrow x+iy$$ , so that in this case, vector fields from $\mathbb R^2$ do pushforward into $\mathbb C$ ? I'm trying to avoid some other things I'm not that familiar with, like complexifying (sections) of the tangent bundle of $\mathbb R^2$ , tho if this is simpler, I'm game.

Last edited: Nov 3, 2013