Is the Dual of a Reeb field a Contact Form?

[WWGD]

Hi, All:

Let X be a Reeb vector field, and let ω be a 1-form dual to X. Is ω necessarily a contact form?

I know if we have a contact 1-form θ , and Z_θ is the Reeb field associated with θ

, then from the definition of Reeb field, we have θ (Z_θ)=1, which mostly means

that θ is nowhere zero, and we can rescale it to make it be 1 . But is the opposite the case,

i.e., if θ (R_θ)=1 , does it follow that R_θ is the Reeb field associated to

θ ? It seems like, since we're in a 1-dimensional situation, there isn't much room to maneuver

and get different things.

Thanks.

 

[Ben Niehoff]

As far as I remember, $\theta (R) = 1$ is not the only condition for R to be a Reeb field. Don't you also need

d \theta \; (R) = 0 \; ?
That may be sufficient information to answer your question.

 

[WWGD]

Thanks, but the problem is that I don't have an explicit representation for the form, so that I can evaluate whether dw( R,.)=0 ; all I know is that the form w is dual to the Reeb field ( under, say , a choice of Riemannian metric ; I think this is called the Musical isomorphism between tangent and cotangent bundles). Since the form is defined on a 1-manifold, all forms and vector fields are 1-dimensional. I thought this would simplify things, but maybe not.

WWGD: What Would Gauss Do ?

 

[Ben Niehoff]

On a 1-dimensional manifold, isn't every nonvanishing 1-form a contact form? It seems trivial to me...in fact I thought you needed at least 3 dimensions to talk sensibly about contact structures.

 

[WWGD]

Good point; yes; the manifold itself is 3-dimensional (in this case; it must be (2n+1)-dimensional, for finite n), but the contact form is itself a 1-form w so that the 3-form $w \wedge dw ≠ 0$, i.e., w is nowhere-integrable (and, if the form is a global form, then the manifold must be orientable, since then $w \wedge dw ≠ 0$ is a global nowhere-zero form for M). I think this last is a corollary of Frobenius' theorem (the one that gives conditions for the integrability of the manifold). So, at the end of the day, I guess we consider a 1-dimensional subspace of forms in a (graded) space of , dimension, I think, $2^n$ of 1,2, and 3-forms.

 

[Ben Niehoff]

Now I'm confused. Are we talking about a 1-dimensional or a 3-dimensional space?

Where did 1-dimensional come from?

 

[WWGD]

No; sorry, the space itself is 3-dimensional. I'm referring to a contact 1-form on a 3-manifold. The form itself is a 1-form w , satisfying $w \wedge dw ≠ 0$

 

[Ben Niehoff]

Can a 3-dimensional manifold have more than 1 contact structure? That is, where equivalence is modulo exact forms and multiplication by nonzero functions.

I think the answer is yes, since the cone over a contact manifold has a symplectic structure, and in 4 dimensions you can have 3 distinct symplectic structures.

A simpler way to approach your question may be to stick with R^3 and the standard Euclidean metric. If it is true anywhere, it is probably true there. And if it is not true, it should give you a clear idea what prevents it being true.

 

[WWGD]

Well, I was thinking now that there may be just one contact structure on a 3-manifold $M^3$, but I'm pretty sure I'm wrong, but I can't see the flaw in my argument: since the (sub)space of 1-forms in $TM^{*} M^3$ is 1-dimensional, then every 1-form is of the type fdx , for f any $C^ {\infty}$ function... Ouch, I realized just now that the space of 1-forms in $M^3$ is not 1-dimensional; it is 3-dimensional. A basis could be given by {$dx,dy,dz$}. So my argument for why we can have only 1 contact structure (basically, if the space of 1-forms actually was 1-dimensional, then every 1-form would be of the type fdx , and the kernel of fdx equals the kernel of gdx for f,g $C^ {\infty}$) fails.

It seems to come down to linear algebra; having 1-linear maps in a 3-D vector space. If there was just one contact form, that would imply that every single 1-linear map has the same kernel (since the kernel at a point is the contact plane) , which is not the case. ALTHO, we need to mod out by isotopy; the kernels of the 1-forms may be different as assignments of contact planes, but these assignments may become equal up to isotopy. But I think we need the isotopy (done individual-plane-wise) has to be done thru contact structures, to guarantee that the isotope planes are still contact planes.

It would be nice to know of some invariants of contact structures, to help decide if different structures are equivalent. I know only one, but I don't know how to test it: a contact structure can be either tight, or overtwisted; I think it has to see with the orbits of the Reeb field associated with the structure: if the orbits loop, or not. Not very precise; let me look into it. Thanks for your the feedback , it is helping me clear things up.

 

[WWGD]

Hey, Ben, according to Wikipedia ( see the part under "Reeb Field" in http://en.wikipedia.org/wiki/Contact_geometry ), the Reeb field X associated to a contact form $w$ can be defined as the unique element in the kernel of dw satisfying $w(X)=1$. It is interesting to try to tell the difference between a contact vector field -- a vector field whose flow preserves the contact structure ( tho it does not necessarily preserve the form itself; it preserves it only up to product by a function ) , while the Reeb field (the pushforward along the flow) actually preserves the form itself. I'll try to read into that, and into tight and overtwisted structures. Thanks again for the feedback.