A proof of a general slice-Bennequin inequality

In this blog post, I’ll provide a slick proof of a form of the slice-Bennequin inequality (as outlined by Kronheimer in a mathoverflow answer.) The main ingredient is the adjunction inequality for surfaces embedded in closed 4-manifolds. To obtain the slice-Bennequin inequality (which is a statement about surfaces embedded in 4-manifolds with boundary) we use contact surgery and the existence of symplectic caps. This post has four parts: in part I, I explain some background about genus bounds for surfaces in closed 4-manifolds. In part II, we look at the analogous results for surfaces in 4-manifolds with boundary (in which case the surface meets the boundary of the ambient manifold along a knot). In part III I’ll state and explain the results we’ll prove, and part IV is a proof.

Part I: background (closed manifolds)

A fundamental question in 4-dimensional topology is to bound the genera of embedded surfaces in a given second homology class. Gauge theory provided answers to many questions of this flavour:

Adjunction inequality (Kronheimer-Mrowka): Let $X$ be a closed smooth 4-manifold with $b_2^+ \geq 2$ . Let $\Sigma \subset X$ be a smoothly embedded oriented connected surface. If $[\Sigma]^2 \geq 0$ and $[\Sigma] \neq 0$ , then for any basic class $\kappa$ ,
$-\chi(\Sigma) \geq [\Sigma]^2 + |\langle\kappa, [\Sigma]\rangle|.$

Let’s break down some of the definitions. The most elusive is the basic class $\kappa \in H^2(X)$ . This comes from Seiberg-Witten theory: the Seiberg-Witten invariant is a map $SW: Char(X) \to \mathbb{Z}$ , where $Char(X)$ is a subset of $H^2(X)$ consisting of characteristic elements, explicitly elements $\alpha$ such that $\langle \alpha, \beta \rangle = \langle \beta, \beta \rangle$ mod 2 for all $\beta \in H_2(X)$ . A basic class is a characteristic element with non-zero Seiberg-Witten invariant.

Next, we required that $b_2^+ \geq 2$ . Here $b_2$ refers to the second Betti number, and this is further decomposed into $b_2 = b_2^+ + b_2^-$ , where $b_2^+$ is the number of positive eigenvalues of the intersection form, and $b_2^-$ the number of negative eigenvalues.

The above form of the adjunction inequality is difficult to think about, because we haven’t even justified that basic classes exist! Fortunately symplectic manifolds have basic classes – in fact, its first Chern class is a basic class.

Adjunction inequality – symplectic version: Let $X$ be a closed symplectic manifold with $b_2^+ \geq 2$ , and $\Sigma \subset X$ a smoothly embedded oriented connected surface. If $[\Sigma]^2 \geq 0$ and $[\Sigma] \neq 0$ , then
$-\chi(\Sigma) \geq [\Sigma]^2 + |\langle c_1(X), [\Sigma]\rangle|.$

In complex geometry, there is also the adjunction formula which I believe is the namesake of the adjunction inequality. This states that for a complex curve in a complex 4-manifold, the above inequality is in fact an equality. The adjunction formula also generalises to symplectic surfaces in symplectic 4-manifolds. By combining the adjunction inequality and adjunction formula, one obtains a proof of the symplectic Thom conjecture.

Symplectic Thom conjecture (Ozsváth-Szabó): Let $X$ be a symplectic 4-manifold and $\Sigma \subset X$ an embedded symplectic surface. Then $\Sigma$ is genus minimising in its homology class.

Notice that in the Thom conjecture we can actually drop the condition on $b_2^+$ ! By virtue of being symplectic, $b_2^+ \geq 1$ . Therefore there is one case $(b_2^+ = 1)$ that needs to be checked, and the other cases follow immediately from the adjunction inequality and adjunction formula.

Part II: background (manifolds with boundary)

So far we’ve discussed genus bounds for surfaces in closed manifolds. Next we’ll think about manifolds with boundary, in which embedded surfaces meet the boundary along some knot. Historically these problems are knot theoretic: rather than the data of a relative homology class of an embedded surface with boundary, people wish to bound the genus of all embedded surfaces with boundary a given knot. One of the most famous results of this form is the slice-Bennequin inequality.

Slice-Bennequin inequality (Rudolph, Bennequin, …): Let $K$ be a Legendrian knot in $\mathbb{S}^3$ . Let $\Sigma \subset B^4$ be a smoothly embedded surface with $\partial \Sigma = K$ . Then
$-\chi(\Sigma) \geq tb(K) + |r(K)|,$
where $tb(K)$ and $r(K)$ are the Thurston-Bennequin and rotation invariants of $K$ , respectively.

Once again we have a few definitions to work through in order to understand the result. Firstly, what is a Legendrian knot? Usually in topology we consider smoothly embedded knots in smooth manifolds, but when the underlying manifold has more structure we can endow more structure on the knots. $\mathbb{S}^3$ has a standard contact structure – that is, it admits a canonical everywhere non-integrable 2-plane field. A knot in $\mathbb{S}^3$ is Legendrian if it is tangent to the contact distribution everywhere. (One also considers transverse knots, which are knots that are everywhere transverse to the contact distribution.) Legendrian knots have three so-called classical invariants:

The knot type. This is the smooth isotopy class of the underlying topological knot.
The Thurston-Bennequin invariant, $tb(K)$ . Fix a non-vanishing vector field along $K$ which is transverse to the contact distribution, and push $K$ by the vector field to obtain a push-off $K'$ . The linking number $lk(K,K')$ is the Thurston-Bennequin invariant.
The rotation invariant, $r(K)$ . Fix a non-vanishing tangent vector field to $K$ , and a Seifert surface of $K$ . Restricting the contact distribution to the Seifert surface gives a trivial bundle (since all 2-plane bundles over 2-manifolds with boundary are trivial). This induces a trivialisation of the contact distribution restricted to $K$ ; $\xi|_{K} \cong K \times \mathbb{R}^2$ . Viewing $v$ in the latter sense, we can compute its winding number in $\mathbb{R}^2$ as it traverses the knot. This is the rotation number of $K$ .

The statement of the slice-Bennequin inequality is that given a Legendrian knot in $\mathbb{S}^3$ , the genus of any surface in the four-ball bound by the knot is bounded below by its Thurston-Bennequin and rotation invariants. The statement of the theorem depends on the existence of classical invariants – in particular, the rotation number only exists if Seifert surfaces exist. However, in a non-simply connected contact 3-manifold Seifert surfaces need not exist. In order to prove more general form of the slice-Bennequin inequality, we need more general definitions of the invariants.

Part III: general forms of the slice-Bennequin inequality (and symplectic Thom conjecture)

We’ll use the adjunction inequality from part I to prove a form of the slice-Bennequin inequality which holds in more general ambient spaces. Specifically, we’ll prove a symplectic version of the slice-Bennequin inequality (stated below) which can also be derived from the slice-Bennequin inequality obtained in a paper of Mrowka and Rollin. The proof, which I believe isn’t written up anywhere, will follow Kronheimer’s mathoverflow answer.

Slice-Bennequin inequality (Mrowka-Rollin): Let $(Y,\xi)$ be a closed contact 3-manifold, and $K \subset Y$ a Legendrian knot. Let $X$ be a 4-manifold with boundary $Y$ . Suppose there is an orientable smooth surface $\Sigma \subset X$ with boundary $K$ . Then
$-\chi(\Sigma) \geq tb(K,\Sigma) + |r(K,\Sigma,\mathfrak{s},h)|$
for all relative spin $^c$ structures $\mathfrak{s}$ with non-vanishing Seiberg-Witten invariant.

To understand this statement, we must understand a little about what spin $^c$ structures are and their properties. (At the end of the section we’ll also discuss how to redefine the Thurston-Bennequin and rotation invariants in this more general setting.) A spin $^c$ structure is a certain bundle $S \to X$ over a manifold $X$ together with a bundle map $\rho : TX \to \text{End}(S)$ . Here we use a relative version of spin $^c$ structures: $\text{Spin}^c(X,\xi)$ denotes the set of pairs $(\mathfrak s, h)$ where $\mathfrak s$ is a spin $^c$ structure over $X$ , and $h$ is an isomorphism from $\mathfrak s|_{Y}$ to the canonical spin $^c$ structure $\mathfrak s_{\xi}$ on $Y$ induced by $\xi$ .

In part I, we discussed the existence of basic classes, that is, characteristic elements in $H^2(X)$ with non-trivial Seiberg-Witten invariant. A similar result holds here: if $(X,\omega)$ is a symplectic manifold that strongly fills $Y$ , then $\omega$ determines a spin $^c$ structure $\mathfrak{s}_0$ , and the Seiberg-Witten invariant of $\mathfrak{s}_0$ is non-trivial. In fact, spin $^c$ structures are complex vector bundles (with extra structure), so the first Chern class defines a map $c_1: \text{Spin}^c(X) \to H^2(X)$ , and the image of this map is precisely the characteristic elements of $X$ . The spin $^c$ structure $\mathfrak{s}_0$ above is a lift of the first Chern class of the symplectic structure on $X$ . This gives rise to a new statement of the above theorem in the symplectic setting:

Slice-Bennequin inequality – symplectic version: Let $(Y,\xi)$ be a contact 3-manifold strongly filled by a symplectic manifold $X$ . Assume $b_2^+(X) \geq 2$ . Let $K \subset Y$ be a Legendrian knot, and suppose $\Sigma$ is an orientable smooth surface in $X$ bound by $K$ . Then
$-\chi(\Sigma) \geq tb(K,\Sigma) + |r(K,\Sigma)|.$

As in part I, this also gives rise to an answer to a relative Thom conjecture. As was the case in closed manifolds, when we restrict to symplectic surfaces, the inequality becomes an equality. That is, a sort of slice-Bennequin formula holds, so combining it with the slice-Bennequin inequality gives rise to a relative version of the symplectic Thom conjecture:

Relative symplectic Thom conjecture (Gadgil-Kulkarni): Let $(Y,\xi)$ be a contact 3-manifold strongly filled by a symplectic manifold $(X,\omega)$ . Let $K$ be a Legendrian (or transverse) knot in $Y$ , and $\Sigma$ a smoothly embedded symplectic surface in $X$ with boundary $K$ . Then $\Sigma$ is genus minimising in its relative homology class.

Returning to the start of this section – we never actually explained how to define the Thurston-Bennequin and rotation invariants outside of $\mathbb{S}^3$ ! In the sphere we could define them in terms of Seifert surfaces, and moreover the definitions end up being independent of the Seifert surfaces. In the more general setting, we define them in terms of the surface $\Sigma$ appearing in the hypotheses of the theorems:

To define the Thurston-Bennequin invariant, consider a push-off $K'$ of $K$ in $Y$ in the direction of a vector field transverse to the contact distribution $\xi$ (as we did for the usual definition). Now we perturb $\Sigma$ to obtain a new surface $\Sigma'$ in $X$ whose boundary is precisely $K'$ . The intersection number of $\Sigma$ with $\Sigma'$ is the Thurston-Bennequin invariant. It’s straight forward to verify that this agrees with the usual definition, by taking $\Sigma$ to be a Seifert surface pushed from $Y$ into $X$ .

To define the rotation invariant, we start by reformulating the original definition. The initial winding number of the vector field $v$ tangent to $K$ is really the obstruction to extending $v$ to a non-vanishing vector field over the Seifert surface used to define the trivialisation. This can be expressed in terms of characteristic classes. In the symplectic case, given a surface $\Sigma \subset X$ whose boundary is $K$ , we define $r(K,\Sigma)$ to be $\langle c_1(X,v),[\Sigma]\rangle$ where $c_1(X,v)$ is the relative first Chern class of the symplectic manifold $X$ with respect to a trivialisation of $\xi$ along $K$ induced by $v$ . One can show that this agrees with the usual definition of the rotation number in the contact 3-sphere.

In the non-symplectic case, we can further generalise the definition by using spin $^c$ structures. Given $(\mathfrak s, h)$ in $\text{Spin}^c(X,\xi)$ , we define the rotation invariant $r(K,\Sigma, \mathfrak{s},h)$ to be the pairing $\langle c_1(L_{\mathfrak s}, v), [\Sigma]\rangle$ , where $L_{\mathfrak s}$ is the determinant (complex) line bundle of the positive spinors in the spin $^c$ structure $\mathfrak s$ . The isomorphism $h$ determines an isomorphism between this line bundle and $\xi$ over $K$ , so that it makes sense to trivialise on the boundary by $v$ . As mentioned earlier, the symplectic definition above is a special case, so this is indeed a generalisation of the rotation invariant for spheres.

Part IV: A slick proof of the slice-Bennequin invariant using the adjunction inequality

As a reminder, we’ll be proving the following theorem:

Slice-Bennequin inequality – symplectic version: Let $(Y,\xi)$ be a contact 3-manifold strongly filled by a symplectic manifold $X$ . Assume $b_2^+(X) \geq 2$ . Let $K \subset Y$ be a Legendrian knot, and suppose $\Sigma$ is an orientable smooth surface in $X$ bound by $K$ . Then
$-\chi(\Sigma) \geq tb(K,\Sigma) + |r(K,\Sigma)|.$

I’ll start with an overview of the proof, before going into some details. The data we have is that of a surface $\Sigma \subset X$ in a symplectic 4-manifold $X$ , where $X$ is a strong filling of a contact 3-manifold $(Y,\xi)$ , and the boundary of $\Sigma$ is a Legendrian knot in $Y$ . The goal is to bound the genus of $\Sigma$ in terms of the Thurston-Bennequin and rotation invariants of $K$ . Suppose we can embed the symplectic manifold with boundary $X$ into a closed symplectic manifold $\overline X$ in some controlled way (and $\Sigma$ to a closed surface $\overline \Sigma \subset \overline X$ ). From above, we know that the Thurston-Bennequin invariant is somehow related to $[\Sigma]^2 \sim [\overline \Sigma]^2$ , and the rotation invariant to $\langle c_1(X,v), [\Sigma]\rangle \sim \langle c_1(\overline X), [\overline \Sigma]\rangle$ . Therefore the slice-Bennequin inequality should correspond to

$-\chi(\overline{\Sigma}) \geq [\overline \Sigma]^2 + |\langle c_1(\overline X), [\overline \Sigma]\rangle|.$

We have three goals: first, we must construct the manifolds $\overline X$ and $\overline \Sigma$ in a way that hopefully gives the desired control over relevant invariants. Next we must derive the slice-Bennequin inequality from the above adjunction inequality (assuming it holds) by properly relating each term above to the corresponding term in the slice-Bennequin inequality. Finally we’ll verify that all of the conditions for the adjunction inequality for surfaces in closed symplectic manifolds are actually satisfied by $\overline X, \overline \Sigma$ , and so on.

Executing the first goal: as mentioned above, we have a Legendrian knot $K$ in $(Y,\xi)$ , and a surface $\Sigma$ in $(X,\omega)$ where the boundary of $\Sigma$ is $K$ and $X$ is a strong filling of $Y$ . To build an appropriate closed symplectic manifold $\overline X$ , we start by carrying out contact surgery along the knot $K$ . Details on contact surgery are given in a paper by Weinstein:

Contact surgery (Weinstein): Let $(X,\omega,\rho,Y,L)$ be an isotropic setup. If $(Y,L)$ has trivial symplectic normal bundle, and $L$ is a sphere of some dimension, then attaching a handle to $X$ (with attaching sphere $Y$ ) gives rise to a symplectic manifold $X'$ . (Analogously, the handle attachment corresponds to a surgery on $Y$ giving rise to a new contact manifold $Y'$ .) Moreover, there’s a Liouville vector field in $X'$ transverse to $Y'$ .

To use this theorem, let’s break down some definitions and ensure that it applies. An isotropic setup consists of the following data:

$(X,\omega)$ is a symplectic manifold, and $\rho$ is a Liouville vector field. (That is, $\rho$ satisfies $\mathcal L_{\rho} \omega = \omega$ .)
$Y$ is a hypersurface in $X$ , transverse to $\rho$ . (Note that $Y$ is then a contact 3-manifold, with contact form $\iota_{\rho} \omega$ .)
$L$ is an isotropic submanifold of $Y$ . This means its tangent spaces are contained in the contact distribution of $Y$ .
The \emph{symplectic normal bundle} is the quotient bundle $(TL)^\perp/TL$ , where the “orthogonal complement” is the symplectic orthogonal taken in the contact distribution.

In our case, $X$ is the ambient symplectic 4-manifold, $Y$ is the boundary $\partial X$ , and $L$ is the Legendrian knot. $X$ being a strong filling of $Y$ precisely means that there is a Liouville vector field $\rho$ transverse to $Y$ (and pointing outwards) inducing the contact structure on $Y$ . $K$ being Legendrian precisely means that it is isotropic. The final condition is that the symplectic normal bundle must be trivial, but this is automatic because $K$ is half-dimensional isotropic in the contact distribution, so the quotient bundle is the zero bundle. (One might say it’s trivially trivial!) Therefore we can carry out contact surgery. Moreover, the result of the surgery is again a strong filling of $Y'$ by $X'$ (via the last sentence of the contact surgery theorem).

The most helpful aspect of using contact surgery is that the core of the handle being attached is a 2-disk $D$ , and it’s being glued to $K$ . This means we also obtain a closed surface $\overline \Sigma = \Sigma \cup D \in X'$ . Since we’re adding a disk, we’ll be able to control how the Thurston-Bennequin and rotation invariants relate to the homological terms in the adjunction inequality! We now have a closed surface, but still don’t have a closed manifold. For this we make use of a result of Eliashberg:

Existence of symplectic caps (Eliashberg): Let $(X,\omega)$ be a strong filling of a contact manifold. Then $X$ can be capped to produce a closed symplectic manifold – that is, there exists a closed symplectic manifold $\overline X$ in which $X$ embeds symplectically.

Since $X'$ is a strong filling of $Y'$ , we can immediately cap it by the above theorem to obtain a closed symplectic manifold $\overline X$ .

Executing the second goal: Now that we have the data of $\overline \Sigma \subset \overline X$ , the adjunction inequality can be interpreted (although we have yet to verify that it holds in this case). Explicitly, we’ll assume that

$-\chi(\overline \Sigma) \geq [\overline \Sigma]^2 + |\langle c_1(\overline X), [\Sigma]\rangle|.$

The first term relates back to $\chi(\Sigma)$ via the identity

$\chi(\overline \Sigma) = \chi(\Sigma) - 1,$

because we’re adding a single disk to $\Sigma$ to remove a boundary component. Now it suffices to observe that

$tb(K,\Sigma) = [\overline \Sigma]^2 + 1,\quad r(K,\Sigma) = \langle c_1(\overline X), [\overline \Sigma]\rangle$

e.g. as mentioned in Lisca-Matić. Combining these three identities with the adjunction inequality above recovers the slice-Bennequin inequality, as desired.

Executing the third goal: Finally, we need to verify that the data $(\overline X, \overline \Sigma)$ really satisfy the premises of the adjunction inequality. Specifically in our case, we’d like to make use of the symplectic version stated in part 1. Working through the premises one by one, there’s exactly one place where things might go wrong: how do we know that $[\overline \Sigma]^2$ is at least 0 and that $[\overline \Sigma]$ is non-trivial? We can avoid both issues by showing without loss of generality that $[\overline \Sigma]^2$ can be assumed to be at least 1, i.e. $tb(K)$ can be taken to be at least 2.

To achieve this, we’ll replace $K$ with a new Legendrian knot $K'$ by connected-summing a standard Legendrian right-handed trefoil. (Formally to take a connected sum, consider $(X, K)$ and $(B^4, T)$ where $T$ is a Legendrian right-handed trefoil in the boundary of $B^4$ . Then $K \# T$ sits in the connected sum of $\partial X$ and $\mathbb{S}^3$ , i.e. in the boundary of the boundary connected sum of $X$ and $B^4$ .)

The trefoil $T$ bounds a surface of genus 1 in $B^4$ . This surface also gets summed with $\Sigma$ (under the above connected sum) to produce a new surface $\Sigma'$ in $X$ , with boundary $K'$ . Overall the genus of the surface $\Sigma$ has increased by 1, so

$\chi(\Sigma') = \chi(\Sigma) -2.$

On the other hand, the rotation number is additive under connected sums, and the Thurston-Bennequin invariant is “additive plus 1″. Specifically,

$r(K',\Sigma') = r(K,\Sigma) + r(T) = r(K,\Sigma),$

$tb(K',\Sigma') = tb(K,\Sigma) + tb(T) + 1 = tb(K,\Sigma) + 2.$

Plugging all of these into the slice-Bennequin inequality, we find that if $K'$ satisfies the slice-Bennequin inequality, so does $K$ . But we also see that $K'$ has Thurston-Bennequin invariant 2 more than $K$ ! Therefore by adding as many trefoils as needed, eventually the Thurston-Bennequin invariant will be at least 2. It follows that the adjunction inequality for $\overline \Sigma$ in $\overline X$ holds.

This concludes the proof of the slice-Bennequin inequality for Legendrian knots in strongly filled contact 3-manifolds.