MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $(M, \omega)$ is a symplectic manifold, is it possible to embed (or injectively immerse) it symplectically into a sufficiently large $(\Bbb R ^{2N}, \Omega)$, (with the usual symplectic structure)?

The only thing that I have found is a theorem by Gromov which is conceptually very nice, but which seems to be very difficult to use. First, I should construct some embedding $i_0$ of $M$ - I guess that I could use Whitney here. Next, I should check that $i_0 ^* \Omega$ belongs to the same cohomology class as $\omega$. Finally, I should check that ${\rm d} i_0$ is homotopic through fiberwise-injective bundle maps $: TM \to T \Bbb R ^{2N}$ to some symplectic morphism. This does not look easy at all.

The thing with Gromov's theorem is that it's very general: it embeds in arbitrary symplectic manifolds (not just in $\Bbb R ^{2N}$), and it allows $\omega$ to have arbitrary non-constant rank - hypotheses that are too generous for me.

The closest thing to my needs is a corollary (corollary (a) on page 334 of Gromov's 1986 "Partial Differential Relations") which says that if $\omega$ is an exact symplectic form, then $(M, \omega)$ immerses symplectically into $(\Bbb R ^{2N}, \Omega)$ provided that $\dim M \le N$. Unfortunately, the requirement that $\omega$ be exact is too strong for me.

Does anyone, then, know of more humane conditions (but valid for arbitrary symplectic manifolds) that guarantee what I want?

ADDENDUM: It would be useful too to find symplectic embeddings into some cotangent bundle - but I expect this to be even more complicated.

share|cite|improve this question
2  
Isn't $\omega$ being exact a necessary condition for the existence of an embedding? – Deane Yang Jan 31 at 20:42
    
@DeaneYang: You mean, because $\omega$ is the pull-back of an exact form? In this case, yes, I see what you mean. – Alex M. Jan 31 at 21:08
up vote 10 down vote accepted

Not as stated, there isn't, at least for closed manifolds. For the form $\Omega$ is exact, but on a closed manifold a symplectic form cannot be exact, since a power of it is a volume form. This would apply to immersions as well as embeddings. In particular, this explains the hypothesis of being exact in the Gromov theorem you quoted.

On the other hand, there are embedding results in $\mathbb{C}P^n$ due to Tischler (D. Tischler. Closed 2-forms and an embedding theorem for symplectic manifolds. JDG,(12):229–235, 1977) and Gromov (presumably in the book you quoted). A useful reference may be the thesis of Manuel Araujo which gives a proof; the introduction says it corrects something in Tischler's work.

share|cite|improve this answer
    
Apparently, for those embedding results one needs $\omega$ to be integral. I cannot visualize this, so I cannot tell how restrictive this is. – Alex M. Jan 31 at 21:11
1  
Well, it depends on what you're trying to accomplish by this embedding. I believe that one can approximate an arbitrary form by one with rational periods (integrals over a basis of homology classes) and then take a multiple of the form to make it integral. – Danny Ruberman Jan 31 at 22:28
    
If $M$ be a Kahler manifold with $c_1(M)>0$, then you can embed $M$ in $\mathbb CP^N$ – baba ab dad Feb 1 at 2:56
    

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.