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Are there any obstructions known which prevent an even dimensional orientable manifold from being symplectic? I am a novice in this area so I unfortunately I cannot make the question more precise. What I have in mind is raw and is as follows:from the point of handle body theory you can put a symplectic structure on each handle as you are building the manifold, so there must something go wrong on the boundary of the handle which is a three manifold, for example in the case of spheres of dimension >2. I just want to know if there is something related to this question in the literature.

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Maybe this book will help you make precise your question: mathematik.uni-muenchen.de/~kai/research/stein.pdf –  Daniel Pomerleano Feb 18 '13 at 1:24
    
For 4-manifolds, to be symplectic you are required to have $1-b^1(X)+b^2_+(X)=\frac{1}{2}(\chi(X)+\sigma(X))\equiv 0$ mod 2. –  Chris Gerig Feb 18 '13 at 6:12
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Robert's comment is spot on. Describing symplectic manifolds constructively via handle attachment is particularly tricky because you're dealing with a geometric structure. Typically if you figure out a new way to build symplectic manifolds via surgery, it's worth writing a paper about. The most basic ways to build symplectic manifolds via cut and paste are Weinstein surgery and Gompf's fiber sum.

Based on your comments, it sounds like you're interested in thinking about this problem using obstruction theory -- which is designed to address "what goes wrong?"-style questions. For a $2n$-manifold $M$ obstruction theory allows you to figure out if can reduce the structure group of $TM$ from $GL(2n)$ to $Symp(2n)$ which will of course be satisfied if $M$ is symplectic. This approach has the advantage that it turns these types of problems into algebraic ones; deciding whether certain cohomology classes are trivial in the style of Chris Gerig's comment. The classical approach to describing these cohomology classes is to describe them as obstructions to extending a structure (non-vanishing section of some bundle if set up appropriately) over a handle in your manifold: Try carrying out this structure reduction on a neighborhood of the 1-skeleton of $M$. You won't run into trouble, so long as $M$ is orientable. In other words, orientability (which is the same as the vanishing of the first Steifel-Whitney class) is your first obstruction.

The disadvantage however is that obstruction theory only deals with homotopical -- not geometric! -- structures. For example, $S^{1}\times S^{3}\simeq(\mathbb{C}^{2}\setminus \{0\})/(x\sim 2x)$ is a complex manifold, so it's structure group reduces to $U(2)\subset Symp(4)$. However, it can't be symplectic as it's $H^{2}_{dR}$ is zero.

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For a closed manifold of arbitrary dimension, there are two algebraic conditions which must be satisfied. Firstly as Russell states there must be a reduction of the structure group from $GL(2n)$ to $Sp(2n)$, or equivalently from $SO(2n)$ to $U(n)$. This is also necessary for open symplectic manifolds, and is equivalent to the existence of an almost complex structure.

Secondly, for a closed manifold, there must be a homology class $[\omega] \in H^2_{dR}$ so that the $n$-fold wedge product of $[\omega]$ generates $H^{2n}_{dR}$. This only holds for closed manifolds since there are exact symplectic structures on open manifolds.

For 4-manifolds there are further known obstruction due to Taubes, which tell us that the Gromov invariant must be equal to the Seiberg-Witten invariant, in particular the Seiberg-Witten invariant must be non-zero. This shows that the connected sum of three copies of $\mathbb{CP}^2$ admits no symplectic form, even though it satisfies the previous two conditions. As far as I know this is the only known geometric obstruction. For higher dimensional manifolds, it's my understanding that it's an open question whether the two algebraic conditions suffice.

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I should also add that in the case of open manifolds, the reduction of the structure group to $Sp(2n)$ is a necessary and sufficient condition for the existence of a symplectic structure. Furthermore, we can prescribe the homology class $[\omega]$ freely. This result is due to Gromov. However this only works if we don't specify any boundary conditions at infinity. If we require boundary conditions such as symplectic convexity, the question is far more subtle and interesting. –  Emmy Murphy Feb 19 '13 at 17:45
    
"As far as I know this is the only known geometric obstruction": A little pedantically, there are further obstructions in 4 dimensions, beyond the Taubes constraints w.r.t. $[\omega]$ and the a.c. structure: namely, the Taubes constraints on a finite covering space w.r.t to the pullback data! –  Tim Perutz Feb 19 '13 at 20:34
    
(This should have come first, but anyway: welcome to MO!) –  Tim Perutz Feb 19 '13 at 20:47
    
Thank you all. Your answers are very helpful. –  7779052 Feb 21 '13 at 23:01
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Well, already the $4$-sphere, which is even dimensional and orientable, does not carry a symplectic structure. It does not even support a $2$-form that is nowhere degenerate, much less a closed $2$-form that is everywhere nondegenerate.

There are now many known obstructions for compact orientable $4$-manifolds to be symplectic, but the easiest is that there must be an element of the de Rham cohomology group $H^2_{dR}$ whose square in $H^4_{dR}$ is nonzero.

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I am asking if there is known anything from the point of handle body theory, I want to know what does not fit as you are building a,say four manifold, by handle bodies. –  7779052 Feb 18 '13 at 1:02
    
for example for your example S^4, there is a symplectic form on both copies of D^4, then you attach them by the identity map of their boundaries, but the result is not symplectic, I want to know what goes wrong in this process in general. –  7779052 Feb 18 '13 at 1:07
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@kaavek: Generally, you can't perform surgeries preserving the symplectic structure or matching the symplectic structure across boundaries. Asking 'what goes wrong?' is not really the right question. What you should be asking is 'under what conditions can you hope to make symplectic surgery work?', since that is going to be the exception rather than the rule. –  Robert Bryant Feb 18 '13 at 1:34
    
Look at the Mayer-Vietoris sequence when you glue. If the pieces you glue together already have symplectic forms, then Mayer-Vietoris says you might have a problem getting the 2-forms glued together, because the overlap might not have a symplectic form, for cohomological reasons. That puts a general theory behind the $S^4$ example, although of course it is only a necessary condition. –  Ben McKay Feb 18 '13 at 12:31
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