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Is it true that for any surface in a symplectic 4-manifold $X$, representing a given homology class of $H_2(X)$, we can assume it is symplectic? I mean for each second homology class, can we find a symplectic surface representing it?

PVAL-inactive
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user217565
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    If a surface is representing a homology class, then it must already be equipped with an orientation. And for surfaces orientations can be lifted to volume forms, which are the same thing as symplectic forms. – Qiaochu Yuan Mar 06 '15 at 01:23

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No, you cannot. The $\omega$-area of the surface gives an obstruction. For example, if I take $\mathbb{CP}^1 \times \mathbb{CP}^1$, the anti-diagonal (graph of the antipodal map) is Lagrangian. The $\omega$ area of this is therefore $0$. Any other surface that is homologous to this will also have total area $0$ and thus there can't be a symplectic surface representing this class.

Sam Lisi
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  • Off the top of my head, I would guess that if the $\omega$ area of the homology class is positive, you can probably find a symplectic representative, but I haven't thought enough about this. Let me know if this modification of your question is of interest. – Sam Lisi Mar 10 '15 at 00:07
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    If you have an embedded symplectic surface representing some homology class with trivial normal bundle, then by a Moser-type argument you have a tubular neighborhood symplectomorphic to the product. Then by essentially a singular value decomposition argument, there is an almost complex structure on your manifold which restricts to this tubular neighborhood as the standard one which is $\omega$-compatible. So for at least surfaces with trivial normal bundle, one can turn the question into a pseudo-holomorphic existence question. – PVAL-inactive Mar 10 '15 at 00:20