# Homology classes represented by $J$-holomorphic curves

Let $\Sigma$ be a compact Riemann surface with complex structure $j$. Let $(M,J)$ be an almost complex manifold. A map $u: \Sigma \rightarrow M$ is called $J$-holomorphic if $$du \circ j = J \circ du.$$ Now $u_*[\Sigma]$ is an element in $H_2(M, \mathbb{Z})$. Is this element in the free part of $H_2(M, \mathbb{Z})$?

I want to know if I can ignore torsion part when we consider Gromov-Witten invariants. All I could see was that when $M$ is symplectic, pure torsion element cannot represent $J$-holomorphic curves.

[edited] As Sam Lisi pointed out, the 'free part' of $H_2(M, \mathbb{Z})$ doesn't make sanse. I changed my question. Suppose $A$ is represented by $J$-holomorphic curves. If $B$ is a torsion element in $H_2(M, \mathbb{Z})$, can $A+B$ be represented by a $J$-holomorphic curve?

• If the symplectic form on $M$ is exact, then any J-holomorphic curve is not necessarily a nontrivial $\mathbb{R}$-cycle. So i don't think your last sentence is correct. – J. Martel Apr 7 '13 at 20:19
• @J.Martel : If the form on $M$ is exact and $\Sigma$ is closed, then the curve is constant. Perhaps you are thinking of punctured curves? @Hwang: This is probably a stupid comment, but what do you mean by the free part? I thought the splitting wasn't natural. – Sam Lisi Apr 7 '13 at 21:07
• – Mark Gross Apr 7 '13 at 22:53
• For the record, as long as $\Sigma$ is closed and $(M,J,\omega)$ is almost-Kaehler the reason that $u_*[\Sigma]$ can't be torsion is the energy identity. – Nathaniel Bottman Apr 8 '13 at 0:12
• @Sam: I was not thinking clearly. Of course, if $\Sigma$ is closed and $\omega$ is exact, then $\Sigma$ has (by Stokes theorem) zero area (w.r.t. the metric $\omega(J\cdot, \cdot)$). – J. Martel Apr 8 '13 at 1:07