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For a symplectic 4-manifold, it is possible for two symplectic submanifolds to intersect negatively? Actually it is an exercise in the 4-manifold book by Gompf and Stipsicz to find symplectic planes in $\mathbb{R}^4$ with negative intersection. Now I want to know the closed case. Does this happen also for closed symplectic manifolds?

edited question: Let $(M, \omega)$ be a closed symplectic manifold. If $A$ and $B$ are different homology classes represented by symplectic submanifolds of complementary dimension, do they always intersect non-negatively?

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algebraic curves on an algebraic surface may (easily enough) intersect negatively, so yes. –  David Lehavi Dec 27 '10 at 7:34
    
I should have said that self-intersections are not considered. Sorry for stating it carelessly. –  Hwang Dec 27 '10 at 8:25
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Do two different submanifolds in the same class count as a self-intersection? If not, hit a curve with a random symplectomorphism (of which there is always an infinite-dimensional group to choose from) to make the intersection with its unperturbed copy transverse. –  Allen Knutson Dec 27 '10 at 16:22
    
I see. I have to edit the question. –  Hwang Dec 28 '10 at 4:52

1 Answer 1

up vote 2 down vote accepted

The answer to this question is YES. I assume you want $A$ and $B$ to be connected. Already in the case of four manifolds two symplectic surfaces can have negative intersection.

To construct an example, we use that if two symplectic surfaces in a $4$-fold intersect positively, you can always smoothen the neighborhood of their intersection to get a new symplectic surface. (Of course this is not always true in the algebraic case).

Example. Take any symplectic 4-manifold $M$ with a symlectic surface $C$, such that $C^2=-1$ (for example $\mathbb CP^2$ blown up once). Blow up $M$ at a point of $C$ and let $E$ be the exceptional curve, while $C'$ be the proper transform of $C$. Notice that $C'^2=-2$ and $C'$ intersect $E$ positively in one point. So we can smoothen the union $C'\cup E$ (this is the full preimage of $C$ under the blow up, and we can not smooth this union algebraically) and get a surface that we call $B$. Finally we have $B\cdot C'$=$C'^2+C'\cdot E=-2+1<0$.

ADDED. Smoothing. In order to smoothen $C'\cup E$ symplectically we can first chose coordinates in a local chart $U$ and adjust a bit the symplectic form so the $C'\cup E$ is given in $U$ by $zw=0$, and the symplectic form is $-i(dz\wedge d\bar z+dw\wedge d\bar w)$. Then surely the curve given by $zw=\varepsilon$ is symplectic in $U$ (since the smoothen curve is complex, while the symplectic form is Kahler). But the whole curve is now discontinuous on the boundary of $U$. To cure this we consider $zw=\varepsilon (F(|z|^2+|w|^2))$ where $F(0)=1$ in $U/2$ and $F$ is smooth with compact support. It is symplectic again, since inside $U\setminus (U/2)$ it is a little perturbation of the union of lines $z=0$ and $w=0$

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Thank you. Could you briefly explain how we smoothen the intersection? I want to know why the intersection should be positive to do that process. –  Hwang Dec 29 '10 at 11:46
    
I learned a lot. Thank you very much. –  Hwang Dec 29 '10 at 23:46
    
You are welcome! –  Dmitri Dec 29 '10 at 23:58

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