Let $(X, \omega)$ be a closed symplectic 4-manifold. Let $\mathcal{C}=(C_i, mi_i)$ be a holomorphic current in $X$, where $C_i$ is a somewhere injective $J$-holomorphic curve in $X$ and $m_i$ is positive integer. Then we can define the ECH index of $\mathcal{C}$ as follow: $$I(\mathcal{C})= \langle c_1(TX), \mathcal{C}\rangle + \mathcal{C} \cdot \mathcal{C}.$$ This integer comes from the dimension of the moduli space of solutions of he Seiberg Witten equation.

Taubes defines his Gromov invariant by counting $I=0$ holomorphic currents. Fix $A \in H_2(X)$, let $\mathcal{M}(X, \omega)=\{\mathcal{C}=(C_i,m_i) : I(\mathcal{C})=0, [\mathcal{C}]=A\}$ to be moduli space of holomorphic currents with $I=0$. Here $I$ only depends on the homology class $A$. If a multiple cover arises, i.e. $m_i>1$ for some $i$, then $I$ doesn't involve any information about the holomorphic map. In contrast to the usual moduli space of holomorphic curves (considered as holomorphic maps), this moduli space is quite strange; it only consists of currents.

Does make sense to talk about the transversality of $\mathcal{M}(X, \omega)$, or does $\mathcal{M}(X, \omega)$ admit a virtual cycle structure?


This is not Taubes' moduli space. Taubes has many more constraints on the currents, so that his "moduli space" is really a finite set of points (for generic $J$), and he requires special weightings on the multiply-covered curves (which are necessarily unbranched covers of tori due to his constraints) to get a well-defined count from his set of currents. He can also do this when $I>0$, for which he chops down the set of currents by requiring them to pass through a given set of $I/2$ points in $X$.

Transversality concerns the cokernel of the deformation operators of these currents. When the curves are embedded, you have the ordinary deformation operator for which it has no cokernel when $J$ is generic. For multiply-covered curves, you can pull-back the deformation operator of the underlying embedded curve and analyze the (co)kernels. But you need to be careful about branch points, as this will destroy the equality between virtual dimension and actual dimension of the moduli space.

The point is, a moduli of curves (in general) is usually not a manifold when multiply-covered curves exist, so Taubes abandons that and restricts to embedded curves, but he needs some information about multiple covers (namely the multiplicity of covers of certain holomorphic tori). Now, there is a notion of convergence for currents, and a Gromov compactness statement, and Taubes uses this to argue that his "admissible set of currents" is a finite set (for generic $J$).

  • $\begingroup$ Can we see how his constraints arise from the view of Seiberg Witten? The holomorphic currents which counting by Taubes should be limits of Seiberg Witten equations solutions $(A_n, \psi_n)$. The curvature of $A_n$ converges to $\mathcal{C}=\{(C_k, m_k)\}$ as currents and $\psi^{-1}(0)$ converges to the underlying embedding holomorphic curve as point set. By compute the index, we know that $m_k=1$ unless $C_k$ is a torus. But why we can't regard $(C_k, m_k)$ as degree $m_k$-branched cover of torus? $\endgroup$ – trick1234 Apr 24 '16 at 7:57
  • $\begingroup$ It's the index constraint, branched covers have negative index. In terms of (co)homology classes, if the $J$-holomorphic curve is Poincare dual to a multiple of a cohomology class $e\in H^2(X)$ with $e\cdot e=0$ (so $PD(e)$ is represented by a self-intersection zero torus) then the curve is an unbranched cover of that embedded torus. As you mention, the SW solutions mainly recognize the subvarieties (image of the curves). $\endgroup$ – Chris Gerig Apr 24 '16 at 20:10
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    $\begingroup$ The index which you mentioned is Fredholm index of pull-back the deformation operator ? $\endgroup$ – trick1234 Apr 25 '16 at 10:49
  • $\begingroup$ Sorry I meant to say the branched covers will have positive (formal Fredholm) index, which increases by the number of branch points. Already that disagrees with the actual dimension of the moduli (which would increase by twice the number of branch points, due to the 2-dimensional movement of the branch points along the surface). Moral: expect no transversality for branched covers. $\endgroup$ – Chris Gerig Apr 25 '16 at 17:33

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