Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Symplectic And Contact Manifolds

  1. Dec 10, 2014 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Hi, we know that every contact manifold has a symplectic submanifold. Is it know whether every symplectic manifold has a contact submanifold?

    A contact manifold is a manifold that admits a (say global) contact form: a nowhere-integrable form/distribution (as in Frobenius' theorem) ## w## so that ##w \wedge dw \neq 0 ##. This form gives rise to a contact distribution as the kernel of ## w##.

    A symplectic manifold is a manifold that admits a symplectic form: a closed non-degenerate 2-form ## \eta##.

    Any refs, comments, etc. appreciated.
     
  2. jcsd
  3. Dec 11, 2014 #2
    I'm not sure that makes sense. By the Whitney embedding theorem, you can always just throw in any submanifold you want if it has low enough dimension. So, there are zillions of contact manifolds that are submanifolds of any symplectic manifold and vice versa (ignoring the lowest dimensions, which are confusing me at the moment). Presumably, you mean that there is some sort of compatibility between the contact form and the symplectic form. It's not clear what that would be. If you restrict a 2-form to a submanifold, it's still a 2-form. If you are the boundary of a symplectic manifold, then you get a contact structure because you can stick in the outward normal into the symplectic form to get a 1-form, which I think you can show is a contact form. Not so clear in general. It's not at all clear how to get from a contact form to a symplectic form on a submanifold.

    Also, locally all contact manifolds and symplectic manifolds are the same, by Darboux's theorem. And I think your question is really a local one, unless you have more requirements on the embedding in mind, so I think you'd just look at R^n with the standard symplectic or contact structure, depending on whether the dimension is even or odd, respectively.

    If you're not talking about submanifolds, there is a construction that will cook up a symplectic manifold out of a contact manifold, called its symplectification. And in the other direction, you can construct the contactification of a symplectic manifold. V I Arnold talks about this in the appendices of his Mathematical Methods of Classical Mechanics book.
     
  4. Dec 11, 2014 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Yes, I just thought about symplectifications and contactifications, but I don't see how it follows that a symplectic manifold admits a contact submanifold. I know how to go in the opposite direction ; if ##w## is a contact form in ## M^{2n+1} ##, then ##dw## is symplectic for a 2n-submanifold of ## M^{2n+1}## by properties of open books and Giroux's correspondence. Do you mean to embed a Darboux ball in a symplectic manifold? And All symplectic are locally alike and All contact are locally alike, but not all contact are locally like symplectic (they don't even compare dimension-wise), unless I can restrict a symplectic structure into a locally contact one; were you talking about contracting the symplectic form into a contact form?. Or maybe you are talking about Lagrangian submanifolds? And I think there is a standard symplectic form in ## \mathbb R^{2n}## , but I don't know of a standard symplectic form.
     
  5. Dec 11, 2014 #4
    When you say submanifold, to me, that contains no reference to the contact or symplectic structure. To me, it means just as a smooth submanifold, so that there may as well be no contact or symplectic structure around at all.
     
  6. Dec 11, 2014 #5

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    No, I mean to find a submanifold of a symplectic manifold so that the submanifold admits a contact structure. I can show that if ## M^{2n+1}## is contact, with contact form ##w## , then ##dw## is symplectic on the fibers of an open book associated to ## M^{2n+1}##; these fibers are symplectic ##(2n+1)## -submanifolds of ## M^{2n}## . But I cannot see how to get a contact form out of a symplectic form; maybe contracting with some vector field? I assume there symplectic form would be related to the symplectic form, but I don't see how.
     
    Last edited: Dec 11, 2014
  7. Dec 11, 2014 #6
    Yes, but the fact that it admits a contact structure is not really relevant. What I'm saying is that locally, your symplectic manifold is just R^2n for some n. So, by the Whitney embedding theorem, as long as your contact manifold has dimension less than n, you can shrink it down so that it fits into that R^2n locally, and embed it in there. Doesn't matter if it's contact or not contact. Unless you mean there's some kind of compatibility condition between the contact form and the symplectic form. But you haven't said what that would be.

    By the way, the open book correspondence, as far as I'm aware is only for 3-manifolds.

    The symplectification and contactification I was talking about work in arbitrary dimensions and are just constructions, not having anything to do directly with submanifolds.

    The symplectification is basically just the cotangent bundle of the contact manifold, with the 0-section removed. The symplectic form is the exterior derivative of the 1-form you get by evaluating the cotangent vectors on the projections of vectors to the manifold.

    I'm not sure how the contactification works. It's some kind of bundle over the symplectic manifold with R or S^1 as fibers.
     
  8. Dec 11, 2014 #7

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    I see, so you suggest using Darboux's local identification/symplectomorphism with the standard symplectic form in ## \mathbb R^{2n}## and then embedding the contact n-manifold in ## \mathbb R^{2n}## ?

    And the Giroux equivalence holds for any odd dimension, i.e., any contact structure ina (2n+1)-manifold is supported by an open book , BUT it is not known if the supporting open book is unique up to stabilization, as it is known to be in 3-dimensions.
     
  9. Dec 11, 2014 #8
    Well, that's the other thing I'm suggesting, except I still have no idea what you mean by "submanifold" in this context. Normally, a submanifold is a submanifold, and having a contact structure doesn't have anything to do with being a submanifold. In my last post, I just meant if you just say submanifold, all the contact and symplectic stuff is not relevant. If you specify how the symplectic form is supposed to relate to the contact form, that's a different story. I suspected you meant there should be some sort of relationship between them, but there's no clear reason to suppose there would be.
     
  10. Dec 11, 2014 #9

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Sorry if I am being unclear, or maybe just dumb.

    But in the contact -symplectic case both structures go together: we use the contact form w in M^{2n+1} and transform it into the symplectic form dw , _and_ this form dw is a symplectic form in the fibers F_i of the open book associated with the contact structure, and these fibers are 2n-submanifolds of M^{2n+1}. So we have both a submanifold given by the fibers F_i of the open book associated to the contact manifold and a symplectic form defined in the fibers. I guess what I am asking is : can we have a 1-form z defined in a symplectic M^2n , so that z satisfies the contact condition ## z \wedge dz \neq 0##in some submanifold of M^2n? If so, what kinds of submanifolds allow for this relationship? Of course, we then need to use some property related to the symplectic form to determine when /how this is possible. Sorry if I am not being too clear.

    And I think you suggested that we can embed a Darboux ball in our symplectic manifold, and, yes, this does not require a symplectic structure, so the condition can be generalized to any manifold having a contact submanifold. But then we could ask: are there other ways of defining submanifolds where the contact condition is satisfied other than by using the Darboux ball?

    So, yet in other words: does a symplectic form present an obstruction to having a 1-form satisfying the contact condition in a given submanifold? Hope I am making sense.
     
    Last edited: Dec 11, 2014
  11. Dec 12, 2014 #10

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Sorry if I was going nowhere with this, thanks for your patience.
     
  12. Dec 13, 2014 #11

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    But I think I can explain what I am aiming for more clearly with by using a separate example: there are Complex manifolds that are not symplectic; so the complex structure somehow presents an obstruction to having a symplectic form defined throughout. My question is whether and how having a global symplectic structure prevents us from having a 1-form satisfying the contact condition on a submanifold, or which submanifolds S of a symplectic manifold allow the existence of a 1-form satisfying the contact condition on S.

    EDIT: Another example: If (X,w) is symplectic and Y is a submanifold, then the restriction of
    (X,w) to Y is not necessarily symplectic.
     
    Last edited: Dec 13, 2014
  13. Dec 14, 2014 #12
    Yes, it's not too clear. I don't see why the symplectic form would be the relevant thing you need to figure out to determine whether a contact manifold can be embedded in such a way that the contact form extends. It would be sufficient to extend to a tubular neighborhood. If the normal bundle is trivial, this wouldn't be hard. I'm not sure what happens if the normal bundle is non-trivial.

    Just to make sure we're on the same page, I actually haven't heard the term Darboux ball, but I just assumed it was the neighborhood in a symplectic manifold that is symplectomorphic to R^2n with the standard symplectic structure, according to Darboux's theorem. Is that right? In that case, it does require a symplectic structure.

    I know what you are saying, but I don't know why you'd expect a symplectic form to be an obstruction to anything like that.
     
  14. Dec 14, 2014 #13
    I don't know why you would think the complex structure is any kind of an obstruction. For example, in a Kahler manifold, the symplectic structure exists quite harmoniously with the complex structure.
     
  15. Dec 15, 2014 #14

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Sorry, re Complex manifolds,I was unclear; I meant that not all Complex manifolds allow a symplectic structure. It seems an obstruction is that the second homology must be non-trivial, and this has somehow to see with the fact that the symplectic form w must generate the fundamental/orientation class, i.e. , the fundamental class must be
    ## w^n=w \cup w \cup w...\cup w =[w \wedge w]^n ##. So, e.g., orientable manifolds M with ##H^2(M) =0## do not allow a symplectic structure.

    And in the case of 4-manifolds, there is an equivalence between being symplectic and allowing a Lefschetz fibration. All I know here is that the Lefschetz pencil , after blowing up finitely-many times allows for a symplectic structure.

    But re the contact submanifold: I don't know the characterization of the set of points where the contact condition is satisfied; of course, if we have a submanifold where the contact condition holds and we have a trivial normal bundle, then we could extend the contact condition. Maybe if the contact condition is an open condition ( so that if it holds in a point, then it holds in a 'hood of the point ), i.e., if ## w \wedge dw(x) \neq 0## , then I think that, from continuity of ## w\wedge dw ## there must be a neighborhood where the condition is satisfied.
     
  16. Dec 15, 2014 #15

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    I think the above isnot a brilliant argument, but clinches it ( if true) : if the contact condition is an open condition, then ## w \wedge dw (x) \neq 0## implies there is an open ball ##B(x,r); r>0 ## where ## w \wedge dw >0 ## holds, and I think this is true by continuity of forms ( these are smooth forms, AFAIK) . So the open ball is itself a contact submanifold. Now, if , as you said, we had a trivial normal bundle, maybe we could extend the contact form along the normal bundle to the points in the open ball.

    But I think there are general methods for creating Contact (2n-1)- and (2n+1)- structures , given a Symplectic 2n-structure; I think that in the (2n-1)-case, we use a vector field to contract the symplectic form; I think the vector field is the Hamiltonian, but I am not sure. And I really don't know at all how to construct the (2n+1)- contact structure.

    As a general comment, I guess we can conclude from the contact condition ## w \wedge (dw)^n \neq 0 ## that ## (dw)^n \neq 0 ## and that ## dw,## is a non-degenerate two-form, i.e., a symplectic form on the pages of the supporting open book.
     
  17. Dec 16, 2014 #16
    I don't know. That sounds right.

    Your normal bundle is always locally trivial, so I think you might be able to patch together the local extensions to get a global one, somehow. There are lots of arguments in topology that go that way. Hirsch's differential topology book probably has some examples. If that ends up working for any manifold, then the conclusion would be that there's not much interest in how the contact manifold is embedded. It's just whether it's contact, I think. I could be wrong, though

    You should probably read the appendices of Arnold's book. It's definitely of interest. He proves Darboux's theorem in a nice way and the proof for contact manifolds is based on symplectifying and applying the symplectic case. The rest of the book is also good motivation, indicating how these concepts relate to physics. If you can understand what he says, you'd have more of a feel for what the relationship between symplectic and contact is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Symplectic And Contact Manifolds
  1. PCA Manifold Submanifold (Replies: 15)

Loading...