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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?

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  • $\begingroup$ 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. $\endgroup$
    – JHM
    Commented Apr 7, 2013 at 20:19
  • $\begingroup$ @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. $\endgroup$
    – Sam Lisi
    Commented Apr 7, 2013 at 21:07
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    $\begingroup$ See mathoverflow.net/questions/110615/difference-of-curve-classes/… $\endgroup$
    – Mark Gross
    Commented Apr 7, 2013 at 22:53
  • $\begingroup$ 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. $\endgroup$
    – Nathaniel Bottman
    Commented Apr 8, 2013 at 0:12
  • $\begingroup$ @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 $\omega(J\cdot, \cdot)$). $\endgroup$
    – JHM
    Commented Apr 8, 2013 at 1:07

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