Relative version of Symplectic Thom conjecture. - MathOverflow

Relative version of Symplectic Thom conjecture.

2011-09-14T06:58:07Z

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?

Answer by Tim Perutz for Relative version of Symplectic Thom conjecture.

2011-09-17T16:39:23Z

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.

Answer by Peter Kronheimer for Relative version of Symplectic Thom conjecture.

2011-10-10T14:06:03Z

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.