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?


2 Answers 2


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.

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).

** 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.


The answer to your first question is yes by the Legendrian neighbourhood theorem.

More concretely, let $L$ be a closed Legendrian submanifold of a contact manifold $(M, \xi)$. Then a neighbourhood of $L$ in $M$ is contactomorphic to a neighbourhood of the zero section of the 1-jet bundle $J^{1}L \cong T^{*}L\times \mathbb{R}$ so that $L$ is mapped on the zero section.

The contact structure on $J^{1}L$ is defined as $\xi_{\text{jet}} = \text{ker}\,(dz - \lambda)$, where $\lambda$ is the Liouville form on $T^{*}L$ and $z$ is the $\mathbb{R}$-coordinate. For any smooth function $f\colon L \to \mathbb{R}$, the mapping $$j^{1}f\colon L \to J^{1}L, \quad j^{1}f(x) = \bigl(df_{x}, f(x)\bigr)$$ is easily checked to be a Legendrian embedding. The zero section corresponds to the constantly zero function. So if you take a smooth function $f$ which is $C^{1}$-close to the constantly zero function, you will get a Legendrian $C^{1}$-close to your $L$.

For a Legendrian knot $L$ in a contact $3$-manifold, you can take another standard neighbourhood of $L$. Namely $S^{1}\times \mathbb{R}^{2}$ with the contact structure $$\cos\theta\cdot dx - \sin\theta\cdot dy = 0,$$ the knot $L$ corresponds to $S^{1}\times \{0\}$.

Unfortunately I'm not much familiar with the Thurston-Bennequin invariant, but I would recommend Chapter 3 of the book "H.Geiges: An introduction to contact topology, Cambridge University Press, 2008."


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