10
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Isn't $\omega$ being exact a necessary condition for the existence of an embedding? $\endgroup$
    – Deane Yang
    Commented Jan 31, 2016 at 20:42
  • $\begingroup$ @DeaneYang: You mean, because $\omega$ is the pull-back of an exact form? In this case, yes, I see what you mean. $\endgroup$
    – Alex M.
    Commented Jan 31, 2016 at 21:08

1 Answer 1

Reset to default
11
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Apparently, for those embedding results one needs $\omega$ to be integral. I cannot visualize this, so I cannot tell how restrictive this is. $\endgroup$
    – Alex M.
    Commented Jan 31, 2016 at 21:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – Danny Ruberman
    Commented Jan 31, 2016 at 22:28
  • $\begingroup$ If $M$ be a Kahler manifold with $c_1(M)>0$, then you can embed $M$ in $\mathbb CP^N$ $\endgroup$
    – user21574
    Commented Feb 1, 2016 at 2:56
  • $\begingroup$ see also e-collection.library.ethz.ch/eserv/eth:24332/eth-24332-02.pdf $\endgroup$
    – user21574
    Commented Feb 1, 2016 at 2:59
  • 1
    $\begingroup$ If $M^k$ is closed, then the integral of $(f^*\omega)^k$ is zero. But $(g\alpha)^k$ is a volume form, so its integral is positive. So I don't think that version works either. $\endgroup$
    – Danny Ruberman
    Commented Dec 26, 2016 at 13:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .