I have three questions actually:


1- is it true that in a sufficiently small neighborhood of Legendrian knot in a 3-manifold we can find another Legendrian knot?


2- If the above is true, suppose we have a Legendrian knot $k$ in a contact 3-manifold, with $tb(k)=n$. Is it true that for any knot $k'$ with $lk(k,k')=n$ and k' Legendrian we have $tb(k')=0$? 


3- Does the Thurston-Bennequin number of a Legendrian knot in an arbitrary 3-manifold tell anything about whether the knot bounds a disk in the 3-manifold? 



question on Thurston-Bennequin number
Let me address questions 2 and 3. Let me point out that everything I'm say is explained in greater detail (and surely in a better way) in Etnyre's survey [Legendrian and transversal knots](http://people.math.gatech.edu/~etnyre/preprints/papers/legsur.pdf).


2- The Thurston-Bennequin number of a component of a link says very little, if nothing at all, about the global topology of a link: in particular, you can fix an arbitrary link, say the unlink with two components in $S^3$, and fix a Legendrian representative (with respect to the standard contact structure, that I will just call $\xi$). That simply means that you have two unlinked unknots that happen to be Legendrian with respect to $\xi$. Now you can play around with each of the two components individually, and make their Thurston-Bennequin number as negative as you want (but not positive - that would violate the Bennequin inequality), and this without changing the topology at all. The operation that I'm hinting at here is called *stabilisation*, which drops the Thurston-Bennequin number by one and is completely *local*, *i.e.* it can be performed inside any given neighbourhood of the knot (in particular, away from the other components if you are working with links).


3- The answer here is yes, in some cases: the simplest case is when the contact structure you are considering is tight. Then the Bennequin inequality tells you that if a Legendrian knot $L$ bounds a surface $S$, then $tb(L) \le -\chi(S)$. In particular, for a knot to bound a disk, you need to have $tb(L) < 0$. Maybe even more interestingly, $tb$ gives bounds on the *slice* genus of a link, by the *slice* Bennequin inequality. Rudolf proved it, and I think he gave the first example of a knot that was topologically slice but not smoothly slice by using it.\*\*


I would like to point out one more thing, at least in the case of knots (the discussion for links is similar, but less studied): usually what's interesting about the Thurston-Bennequin number is the set of values attained by *all* Legendrian representatives of a fixed knot or type. Even more interesting is the set of pairs of values $(tb, rot)$ for all representatives: these are usually plotted on the plane $(tb, rot)$ and they are sometimes referred to as "mountain peaks" (to see why, look at the mountain peaks for [cables of torus knots](http://arxiv.org/abs/1104.0550)).


\*\* Long footnote: one of the knots with this property is the 0-twisted, positively-clasped Whitehead double of the right-handed trefoil, and one can use the existence of such an example to prove the existence of (large) exotic $\mathbb{R}^4$s! This is the content of Exercise 9.4.23 in Gompf and Stipsicz's book *4-manifolds and Kirby calculus*.


