TITLE: Is the Lisca-Matic bound (aka slice-Bennequin bound) strictly stronger than the Bennequin bound?
QUESTION [10 upvotes]: The Bennequin bound [1] says that, for a transverse knot (or later link) $K$ in $S^3$,
$$\mathrm{sl}(K) \le - \chi(\Sigma)$$
for any Seifert surface $\Sigma$ for $K$, where $\mathrm{sl}$ is the self-linking number.
Lisca and Matic proved [2, Theorem 3.4] that for a Legendrian knot $K$
$$|r(K)| + \mathrm{tb}(K) \le 2g(\Sigma) - 1$$
for any smooth surface $\Sigma \subset B^4$ with boundary $K$, where $\mathrm{tb}$ and $r$ are the Thurston-Bennequin and rotation numbers.
EDIT: This was actually proved by Rudolph [3].
I have several questions here.
(a) Why is the Lisca-Matic bound always stated in terms of Legendrian knots? The combination $|r| + \mathrm{tb}$ is exactly the self-linking number of the transverse push-off, with one of the two orientations. Wouldn't it be simpler to just state it in terms of transverse knots?
(b) Isn't the Lisca-Matic bound strictly stronger than the Bennequin bound? Why isn't it more popular to quote it? Maybe it's about the extension to links (which, in the case of the Bennequin bound, is due to Eliashberg)? It seems that Lisca and Matic only state the result for knots in their paper, but it seems to me that the proof should carry over.
For knots/links in more general 3-manifolds, the Bennequin bound applies as long as you have a tight contact structure, while to state Lisca-Matic you need some sort of filling.
[1] Bennequin, "Entrelacements et équations de Pfaff", In Third Schnepfenried Geometry Conference, Vol. 1, Schnepfenried, 1982, 87–161. Astérisque 107. Paris: Société Mathématique de France, 1983.
[2] Lisca, P., and G. Matić. “Stein 4-manifolds with Boundary and Contact Structures.” 55–66. Topology and Its Applications 88, nos. 1–2, 1998.
[3] Rudolph, Lee, "Quasipositivity as an obstruction to sliceness." Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 51–59.

REPLY [5 votes]: Yes, the slice-Bennequin bound is stronger than Bennequin bound: 
(a) For example, the knot $K \# - K$ is ribbon, so the slice-Bennequin implies that any transverse representative must have negative self-linking number.  But the Bennequin bound only implies it is less than $4 g(K) - 1$, which is positive if $K$ is nontrivial.
(b) Given any transverse representatives $T, -T$ of a knot and its mirror, we can take a transverse connected sum to obtain a knot satisfying $sl(T \# - T) = sl(T) + sl(T) + 1 \leq -1$.  Let $T$ be the positive torus knot $T(p,q)$.  This maximizes the Bennequin bound so we can assume $sl(T) = pq - p - q$.  But this implies that the self-linking number of any transverse representative of the negative torus knot $T(p,q)$ is as most $p + q - pq - 2$.
As Marco points out, one reason to state the result in terms of Legendrian knots is because we have a good local model for attaching a 2-handle along a Legendrian knot.  There is a theory for attaching handles along transverse knots, described in David Gay's thesis, but it is trickier. The approach of Lisca-Matic is to use the genus bound coming from the global adjunction inequality in a Kahler surface of general-type to deduce a slice-genus bound.  
Finally, complex points and Lai indices give another, very satisfying conceptual relation between the Bennequin-Eliashberg bound and the Adjunction Inequality that is more "transverse" than "Legendrian".  A generic real, immersed surface $\Sigma$ in a complex surface $X$ will have finitely many complex points --- i.e. points where the tangent plane is a complex line.  These can be either positive or negative (according to whether the orientation of $\Sigma$ agrees with the complex orientation) and either elliptic or hyperbolic.  The Lai indices of $\Sigma$ count these points with signs:
$I_+(\Sigma) = e_+ - h_+ \qquad I_-(\Sigma) = e_- - h_-$
If $\Sigma$ is embedded, these satisfy the relations
$I_+(\Sigma) + I_-(\Sigma) = \chi(\Sigma) + [\Sigma]^2$
$I_+(\Sigma) - I_-(\Sigma) = \langle c_1(X),[\Sigma] \rangle$
The adjunction inequality is then equivalent to the inequality $I_-(\Sigma) \leq 0$.
There is an analogous story for a convex surface $\Sigma$ in a contact manifold $(M,\xi)$.  The tangencies of $\xi$ along $\Sigma$ are the singularities of the characteristic foliation.  These can be either positive or negative and either elliptic or hyperbolic.  We can then define Lai indices exactly as above.  If $\Sigma$ is a Seifert surface bounded by a transverse knot $K$, then
$sl(K) = I_-(\Sigma) - I_+(\Sigma)$
Moreover, the Bennequin-Eliashberg bound is equivalent to the inequality $I_-(\Sigma) \leq 0$.  Eliashberg proves this directly by showing how to perturb the surface to eliminate all negative elliptic points.
See the books Stein manifolds and holomorphic mappings by Forstneric and Surgery on contact 3-manifolds and Stein surfaces by Ozbagci-Stipsicz.