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)?



Legendrian knot in 3-sphere
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](http://arxiv.org/abs/math/0306256) 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*](http://imrn.oxfordjournals.org/content/2009/3/512.abstract), 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.




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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.




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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](http://arxiv.org/abs/math/0611848) by Chantraine on the subject.


