7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Could someone (a downvoter perhaps) suggest how this question might be improved? $\endgroup$ – Todd Trimble Jun 11 '17 at 12:44
  • 2
    $\begingroup$ I did not downvote this but as a minimum one should either explain what is $b_2^+$, what is $e$, what is $Gr$, what is $K$, what is $C_1$, and what does gauge-theory-freely mean, or at least give a reference to a text where all this is explained. $\endgroup$ – მამუკა ჯიბლაძე Jun 11 '17 at 13:20
  • 2
    $\begingroup$ I don't understand the downvotes (but I never do). I agree that the question could be improved, in addition to the previous comment by maybe adding some color (for example, are there somewhat similar results where Seiberg-Witten has been supplanted?) $\endgroup$ – Igor Rivin Jun 11 '17 at 13:22
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.