We are given a Legendrian knot, fixed up to Legendrian isotopy, in $(S^3,\xi)$ ($\xi$ is the standard contact structure). Does it necessarily bound a symplectic surface in $(B^4,\omega)$ (again $\omega$ is the standard symplectic structure)?
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2 Answers
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No, there are obstructions to this.
Suppose $L$ is a Legendrian knot that bounds a symplectic surface $\Sigma$, and $T$ a (sufficiently close) transverse push-off (see Etnyre's notes or Geiges' Introduction to contact topology for references). Notice that $T$ can be chosen to be $C^\infty$-arbitrarily close to $L$. (This is clear from the construction, although I couldn't find a clean statement in a quick search.)
You can perturb $\Sigma$ along with $K$ while keeping it symplectic, so that it bounds $T$. Now take the double cover of the ball, branched along $T$: this gives a symplectic 4-manifold whose (strongly convex) boundary is the branched cover of the standard contact 3-sphere along $T$. Hence the branched double cover of $\xi$ along $T$ is (strongly) symplectically fillable.
There are plenty of transverse knots whose branched cover is overtwisted, e.g. if $T$ is transverse stabilisation (see, for example, Harvey, Kawamuro, Plamenevskaya, On transverse knots and branched cover, which makes for a very pleasant read). To be even more concrete, take the unknot with Thurston-Bennequin number -3 and rotation number 0: this cannot bound a symplectic surface in the 3-sphere.
I guess that there could be a more elementary argument along the following lines: attach a Weinstein handle along $L$, look for a symplectic approximation of the core disk (relative to its framed boundary). You obtain a symplectic manifold with $H_2=\mathbb{Z}$, whose Chern class is determined by the rotation number (as shown by Gompf), and you can apply the adjunction formula. My guess is that if $tb(L)$ is less than $2g_*(L)-1$ (where $g_*$ is the slice genus) you have no chances.
Finally, one could wonder what happens if you ask for a Lagrangian surface rather than a symplectic one: in this case tb needs to be $2g_*(L)-1$ and rot needs to be 0. There's a nice paper by Chantraine on the subject.
answered Jul 23, 2015 at 7:30
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$\begingroup$ Dear Marco, I do not understand your second comment. $tb(L)<2g_*(L)-1$ would be always true by adjunction, plus I am not sure how this kind of argument will show that $L$ doesn't bound a "symplectic" surface. $\endgroup$– nikitaCommented Jul 23, 2015 at 16:46
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$\begingroup$ as for the first comment, double brached cover of $B^4$ along surface $\Sigma$ is strong convex filling of double branched cover of $S^3$ along $T$. But that's true even if $\Sigma$ is not symplectic, isn't it? $\endgroup$– nikitaCommented Jul 23, 2015 at 17:16
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1$\begingroup$ @nikita: The adjunction inequality says that $tb(L)\le 2g_*(L)-1$, so equality is still possible, but either way if you have a symplectic surface then it actually satisfies an adjunction formula $\langle c_1(\omega), \Sigma\rangle + \Sigma\cdot\Sigma = 2g(\Sigma)-2$. In response to your second comment, if the surface is symplectic then its branched double cover is naturally a symplectic manifold and so the contact structure which comes from taking the branched double cover of the transverse knot in S^3 is in fact symplectically fillable. $\endgroup$ Commented Jul 23, 2015 at 17:42
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$\begingroup$ @StevenSivek: I thought it was obvious at the moment, but now it is quite unclear why this should be true. i.e. why is branched cover of B^4 over a symplectic surface naturally symplectic? $\endgroup$– nikitaCommented Aug 27, 2015 at 20:32
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$\begingroup$ @nikita: this is very similar to what happens in the complex case. Away from the branch set, there is no problem; along the branch set, you work in local coordinates (using the symplectic neighbourhood theorem). $\endgroup$ Commented Aug 27, 2015 at 21:00
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The knot must be quasipositive. You can perturb it to be transverse as in Marco Golla's answer, and then a theorem of Boileau and Orevkov (in Quasi-positivité d'une courbe analytique dans une boule pseudo-convexe, MR1836094) says that transverse knots which bound symplectic surfaces in $B^4$ must be quasipositive.
answered Jul 23, 2015 at 13:37