Trying to prove one of C.Taubes' theorems gauge-theory-freely One of C.Taubes' theorems says that for a symplectic 4-manifold $X$ with $b^2_+>1$ (where $b^2_+$ denotes the dimension of a maximal positive-definite subspace of $H^2(X;\mathbb R)$ under the intersection form), $\mathrm{Gr}(e)=0$ if $c_1(K)\cdot e-e\cdot e\neq0$. Here $Gr(e)\in\mathbb Z$ is a particular count of $J$-holomorphic curves in $X$ which represent the class $e\in H_2(X;\mathbb Z)$, and $K^{-1}$ denotes the canonical bundle. This theorem is proved using Seiberg-Witten theory. Can one prove this gauge-theory-freely?
 A: The key use of SW theory was to show that $Gr(e)=Gr(c_1(K)-e)$. For the moment, take this equality as granted.
If $Gr(e)\ne0$ then there must exist a $J$-holomorphic curve $C\to X$ such that $[C]=e$, and likewise a $J$-holomorphic curve $C'\to X$ such that $[C']=c_1(K)-e$. To demonstrate the gist of the proof, assume $X$ is not a blow-up, and that $C$ and $C'$ are distinct embedded connected surfaces. By positivity of intersections of holomorphic curves, $e\cdot(c_1(K)-e)=\#(C\cap C')\ge0$, hence $-e\cdot c_1(K)+e\cdot e\le0$. But the dimension of the moduli of $J$-holomorphic curves representing $e$ is $c_1(TX)\cdot e +e\cdot e=-e\cdot c_1(K)+e\cdot e$, which must be nonnegative otherwise the moduli space would be empty. Thus $c_1(K)\cdot e-e\cdot e=0$.
So, we need a way to demonstrate the aforementioned equality of Gromov invariants. I don't know yet (though it's part of my research) how to prove it inherent to $J$-holomorphic curve theory. But you just want a SW-free proof, and that is granted by Donaldson-Smith invariants. These are counts of sections of a certain bundle associated with a given Lefschetz fibration of $X$ (giving pseudolomorphic surfaces in some sense), and were shown to recover Taubes' Gromov invariants. Our desired equality (under a small restrictive assumption*) is a consequence of Serre duality between divisors on Riemann surfaces!
*The restrictive assumption: $b^2_+(X)>b_1(X)+1$. So let's make $X$ simply connected.
