symplectic surfaces in 4-manifolds

Question

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?

Answer 1

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.