Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Ozsváth and Szabó proved Symplectic Thom conjecture [Annals of Mathematics, 151(2000), 93-124]. It states: An embedded symplectic surface in a closed, symplectic 4-manifold is genus-minimizing in its homology class.

Does a suitable relative version of above hold true ? More specifically, suppose we start with an embedded symplectic surface $\Sigma $ with boundary in symplectic 4-manifold with contact type boundary then is it true that $\Sigma $ is genus-minimizing in its (relative) homology class?

share|improve this question
add comment

2 Answers 2

up vote 12 down vote accepted

I think this must be a consequence of the version of the slice-Bennequin inequality proved by Mrowka and Rollin (but I might be wrong). Perhaps the argument also requires the boundary to have the "strong filling" property (Stein near the boundary).

Given a Legendrian knot $L$ in (to start with) the 3-sphere $S^3$, that inequality says $$ 2 g_*(L) - 1 \ge tb(L) - r(L). $$ ` Every transverse knot $K$ is a push-off of a Legendrian approximation $L$, and the self-linking number of $K$ is related to the invariants of $L$ by $$ sl(K) = tb(L) - r(L). $$ So the slice-Bennequin inequality says $$ 2 g_*(K) -1 \ge sl(K). $$ Unless I'm mistaken, this inequality is an equality for the case of a transverse knot bounding a symplectic surface in the 4-ball.

All this generalizes to the case of a symplectic 4-manifold $X$ that is Stein near its (contact) boundary. Given a Legendrian knot $L$ in the boundary, and a homology class $s$ of surfaces in $X$ with boundary $L$, one has invariants $tb(L,s)$ and $r(L,s)$. Then there is an inequality (as in Mrowka-Rollin), $$ 2 g_*(L,s) - 1 \ge tb(L,s) - r(L,s), $$ where $g_*(L,s)$ is the smallest possible genus of a surface in the class $s$ with boundary $L$. In terms of a transverse push-off $K$, one again has $$ 2 g_*(K,s) - 1 \ge sl(K,s). $$ If $K$ is actually the transverse boundary of a symplectic surface, then one has equality, this being (I think) just the adjunction formula in a relative version.

The Mrowka-Rollin version of the result can today be deduced from the existence of concave fillings (caps). We may assume from the outset that $tb(L,s)$ is positive: if it is not, we may sum in a bunch of Legendrian trefoils until it is. Now enlarge $X$ by first adding a 2-handle along $L$ (standard contact surgery) and then closing it up with a concave filling. The inequality one wants is just the adjunction inequality applied to the homology class formed from $s$ and the 2-disk in the core of the handle.

So with less notation, the answer to the original question is supposed to be: take a Legendrian approximation to the transverse knot and alter things so as to make $tb$ postive; then add a 2-handle and a cap to get a closed symplectic manifold. Then apply the adjuntion inequality to the homology class formed from the symplectic surface and the Lagrangian 2-disk.

share|improve this answer
    
Two other thoughts: (1) You probably don't need the "strong" contact condition at the boundary in order to add the contact handle, so just contact-type boundary may be fine here. (2) Another take on the argument might be that one can apply the symplectic Thom conjecture to surfaces that are only semi-symplectic (the form restricted to the surface is non-negative), provided that the homology class has positive self-intersection. This is because you can alter the symplectic form in the tubular neighborhood, to make it positive on the surface. –  Peter Kronheimer Oct 10 '11 at 15:36
add comment

This is a natural question, and I'm a bit startled to realise that, in this generality, I can't locate a reference for it.

To frame it precisely, let's suppose that $X$ is a compact symplectic 4-manifold with convex contact-type boundary $Y$, and ask whether a compact symplectic surface $\Sigma$ in $X$, transverse to $Y$ and bounding a link $L\subset Y$ transverse to the contact structure, minimises minus the Euler characteristic among surfaces bounding $L$ and homologous to $\Sigma$ relative to $L$.

There's lots in the literature about Bennequin-type inequalities for Legendrian links, notably Mrowka-Rollin's adjunction inequality: http://arxiv.org/abs/math/0410559. But when considering boundaries of symplectic surfaces it seems more natural to take $L$ transverse to the contact structure.

A sufficient condition. Suppose that we can cap $X$ to a closed symplectic manifold $Z$, and cap $\Sigma$ inside $Z$ to a closed symplectic surface $S$. It then follows from the symplectic Thom conjecture in $Z$ that $\Sigma$ is genus-minimizing in the sense I indicated.

A famous example is Kronheimer-Mrowka's proof (see http://www.math.harvard.edu/~kronheim/thom1.pdf) of the Milnor conjecture about the slice genus of algebraic links, in which one completes the (blown up) 4-ball to the (blown up) projective plane and applies the Thom conjecture there. [Experts will spot an anachronism in this summary.]

We do know that any $X$ can be closed up symplectically; see, for instance, Eliashberg's article http://arxiv.org/pdf/math/0311459. It seems plausible that every pair $(X,\Sigma)$, where the boundary of $\Sigma$ is a transverse link, can be closed to a pair $(Z,S)$. Perhaps Eliashberg's argument can be refined to accomplish this.

share|improve this answer
1  
I am not quite sure that every pair $(X,\Sigma) $, where the boundary of $\Sigma $ is a transverse link, can be closed to a pair $(Z,S)$ as you indicate. I seem to have a potential counter-example to it (but I am not sure of it either! :) ), I will post it here if it turns out be, actually, a counter-example. –  Dheeraj Kulkarni Sep 17 '11 at 17:58
1  
Ah, so you're serious! ;) Well, if you have an obstruction to finding such a cap, that would be interesting. –  Tim Perutz Sep 17 '11 at 19:48
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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