# Tightness and right-veering monodromies

Update: Andy Wand has posted a preprint of his proof on the arXiv.  Obviously, definitions and content there supercede what was originally written in this series of blog posts.

Update: (12 May 2014) Some minor technical details have been corrected (Thanks to Patrick Massot for pointing these out)

In this second post on the Legendrian surgery conjecture, I want to reinterpret the conjecture in terms of the Giroux correspondence and give an overview of Honda-Kazez-Matic’s criterion for tightness in terms of right-veering monodromies.

Giroux correspondence

The Legendrian surgery conjecture can be reinterpreted via the Giroux correspondence as a statement about the monodromy of an open book supporting a tight contact structure.

An open book decomposition of a 3-manifold M is a pair $(B, \pi)$ of a link $B \subset M$ and a fibration $\pi: M \setminus B \rightarrow S^1$, whose fibers are the interior of compact surface $\Sigma$ with boundary B. Note that flowing once around the base $S^1$ gives a homeomorphism $\phi: \Sigma \rightarrow \Sigma$ that is the identity near the boundary. So, we could instead define an abstract open book, which is a pair $(\Sigma, \phi)$ of a compact surface with nonempty boundary and a monodromy map $\phi$. This defines a closed, oriented 3-manifold
$M = \Sigma \times [0,1] / \Sigma \times \{0\} \cong \phi(\Sigma) \times \{1\} \cup \partial \Sigma \times D^2$
with an obvious open book decomposition.

The same manifold admits many open book decompositions. For instance, take an abstract open book $(\Sigma, \phi)$ and attach a 2-dimensional 1-handle to $\Sigma$ any way you please. Now, choose any essential simple closed curve $\gamma \subset \Sigma' = \Sigma \cup \text{1-handle}$ that intersects the cocore of the 1-handle exactly once. Then the abstract open books $(\Sigma', D_{\gamma}^{\pm} \circ \phi)$ define the same 3-manifold as $(\Sigma, \phi)$, where $D_{\gamma}$ denotes performing a positive Dehn twist along $\gamma$. These are referred to as positive/negative stabilizations of the open book.

Undoing this procedure is called a destabilization. Let $\alpha$ be an arc in $\Sigma$ with boundary points in $\partial \Sigma$ and suppose that $\phi(\alpha), \alpha$ only intersect on the boundary. Let $\gamma$ be the simple closed curve obtained by concatenating $\alpha, \phi(\alpha)$. Then one of the mapping classes $D_{\gamma}^{\pm} \circ \phi$ fixes $\alpha$ up to isotopy and we can choose it to be the identity near $\alpha$. Thus,it restricts to a well-defined monodromy on the surface $\Sigma'$ given by cutting $\Sigma$ along $\alpha$ and a new open book. Choosing $\alpha$ to be the cocore of the 1-handle attached during a stabilization undoes that stabilization

Proposition. All open book decompositions for M are related by a sequence of (de)stabilizations.

There is a deep connection between open books and contact structures. An open book for M supports a contact structure $\xi$ if there is a contact form $\alpha$ for $\xi$ such that (1) B is a positively transverse knot, and (2) $d \alpha$ is a positive area form on all of the pages $\Sigma_{\theta} = \pi^{-1}(\theta)$.

Theorem (Thurston-Winkelnkemper) Every open book supports a unique contact structure.

[Edit: Uniqueness of the contact structure is due to Giroux]

However, contact structures are not supported by a unique open book. Positively (de)stablizing an open book decomposition gives a new OB decomposition supporting the same contact structure. Intuitively, this is because positive stabilizations are essentially given by connecting summing with $(S^3, \xi_{std})$, endowed with some nontrivial open book.

That this is move is enough to classify supporting open books is due to Giroux and the following relationship is called the Giroux correspondence.

Theorem (Giroux) There is a 1-1 correspondence between (abstract) open book decompositions of a closed, oriented 3-manifold M, up to positive (de)stabilization and contact structures on M, up to isotopy isomorphism.

This relies on two facts: (1) every contact structure is supported by some OB decomposition, and (2) all supporting open book decompositions for the same contact structure are related by positive (de)stabilizations.

Thus, all that is needed to specify a contact structure is a compact surface $\Sigma$ and a mapping class $\phi$ for that surface. And this correspondence allows us to study contact geometry algebraically via mapping class groups.

So, to understand tightness in terms of open books, we would like to find some property of mapping classes that is (1) invariant under positive (de)stabilizations, and (2) is equivalent to tightness/overtwistedness. Honda-Kazez-Matic’s non-right-veering condition satisfies (2) but not (1), which is really good but not yet sufficient.

Legendrian surgery also has a nice characterization in terms of open book decompositions.

A curve $\gamma$ embedded on a page $\Sigma_{\theta}$ is really a knot K in M and the page determines a framing of K. Integral Dehn surgery on K can be described via the following modification of the monodromy map:

Lemma. The pair $(\Sigma, D_{\gamma}^k \circ \phi)$ is an abstract open book for $(M_K(-1/k))$.

Proof. (Intuitive) Think of the surgery torus $S^1 \times I \times I$ sitting in the cylinder $\Sigma \times [0,1]$. Now, take a piece of string (thought of as sitting in a page below the solid torus and transverse to $\gamma$) and pull it through the surgery torus.

Geometrically, for any almost all essential simple closed curves sitting in a page, the contact structure supported by that open book can be perturbed to make that curve Legendrian. Moreover, the page framing is exactly the Thurston-Bennequin framing. This is because $d \alpha$ is a positive area form on the page, so by definition the Reeb vector field is always positively transverse to the pages.

Lemma. The contact structure $(M_K(\pm 1), \xi_K(\pm 1))$ is supported by the open book $(\Sigma, D_{\gamma}^{\mp 1})$ (when K lies in a page).

Finally, the following lemma follows easily from Giroux’s proof that every contact structure is supported by some open book.

Lemma. For any Legendrian link L in $(M,\xi)$, there is a supporting open book such that L lies on some page.

This means that Legendrian surgery can be encoded by a tuple $(\Sigma, \phi, \gamma)$ of a compact surface, mapping class and essential simple closed curve. To prove the Legendrian surgery conjecture, we will want our property to satisfy a third condition: (3) it is persistent after positive Dehn twists.

Conjecture. There is some property P of abstract open books $(\Sigma, \phi)$ that is (1) invariant under positive (de)stabilizations, (2) invariant under composition $\phi$ with positive Dehn twists along any essential simple closed curve in $\Sigma$, and is (3) equivalent to the tightness of the supported contact structure.

Right-veering monodromies

One way to study mapping classes $[\phi]$ is by understanding the relation between some collection of arcs or curves $\{\alpha_i\}$ in the surface and their images $\{\phi(\alpha_i)\}$ under a nice representative $\phi$ of the mapping class.
Take a compact surface $\Sigma$, an embedded arc $\alpha \subset \Sigma$ with boundary points on the boundary of $\Sigma$ and a mapping class $[\phi]$. We can also choose a representive $\phi$ of the mapping class that is efficient with respect to $\alpha$, meaning the $\alpha, \phi(\alpha)$ do not form any trivial bigons on the surface that can be isotoped away. Orient $\alpha, \phi(\alpha)$ so that their concatenation is an oriented loop. We say that $latex\alpha$ is right-veering  if the orientation given by $\alpha', \phi(\alpha)'$ at some both boundary point agrees with orientation on $\Sigma$ and that $\alpha$ is left-veering non-right-veering otherwise. See the picture.

[Edit: Non-right veering (i.e. left veering at some endpoint) does not imply left-veering (left-veering at both endpoints)]

Theorem: (Honda-Kazez-Matic) A contact structure $\xi$ is overtwisted if and only if there is an open book $(\Sigma, \phi)$ supporting $\xi$ with a left-veering non-right-veering arc.

So, to know that a contact structure $\xi$ is overtwisted, all we need to know is that there is at least one open book decomposition supporting $\xi$ with at least one single left-veering non-right-veering arc. And, for every overtwisted contact structure, we can find such and open book decomposition and left-veering non-right-veering arc. So this gives an OB decomposition/mapping class group characterization of tightness.

Proof. It’s fairly straightforward to show that every overtwisted contact structure admits some open book with a left-veering non-right-veering arc.

Recall that Eliashberg classified overtwisted contact structures by their homotopy types. Let $(M,\xi)$ be an overtwisted contact structure [Edit: I’ve rewritten the following paragraph -PLC] and $(S^3, \xi_{OT})$ denote the standard overtwisted contact structure . Then since the contact structures $(M, \xi), (M,\xi) \sharp (S^3, \xi_{OT})$ are homotopic, they must be isotopic. [Edit: As Marco points out in the comments below, the overtwisted contact structure on $S^3$ supported by an open book with an annular page and monodromy a single Dehn twist is not homotopic to the standard tight contact structure $\xi_{std}$.  In order to get a contact structure on $M$ homotopic to $\xi$, we need to connect sum with another contact structure $\xi'$ on $S^3$ so that the algebraic topology works out correctly: $(M,\xi)$ and $(M,\xi) \sharp (S^3, \xi_{OT}) \sharp (S^3, \xi')$ are homotopic.]  Now, $(S^3, \xi_{OT})$ has an open book decomposition with an annular page and monodromy given by one negative Dehn twist along the core curve. The connect sum is equivalent to a Murasugi sum of the open books, which in this case is exactly a negative stabilization along some boundary parallel arc. The cocore of the new 1-handle is now a left-veering non-right-veering arc.

There is a unique overtwisted contact structure $(S^3, \xi_{OT})$ homotopic to the standard tight contact structure on $S^3$ .  Since for all contact structures, the connect sum $(M,\xi) \# (S^3, \xi_{std})$ is isomorphic to $(M,\xi)$, this implies that if $\xi$ is overtwisted then $(M,\xi) \# (S^3,\xi_{OT})$ are homotopic, hence isotopic by Eliashberg.

There is another familiar overtwisted contact structure $(S^3, \xi_{Hopf})$ that is supported by an open book with annular pages and monodromy given by one negative Dehn twist along the core curve of the annulus.  We can find a third contact structure $(S^3, \xi')$ such that $(S^3, \xi_{OT})$ and $(S^3, \xi') \# (S^3, \xi_{Hopf})$ are isomorphic.  Thus $(M,\xi)$ is isomorphic to the double connect sum $(M,\xi) \# (S^3, \xi') \# (S^3, \xi_{Hopf})$.  Connect sum is equivalent to a Murasugi sum of the open books, which for $(S^3,\xi_{Hopf})$ is exactly a negative stabilization and the cocore of the new 1-handle is now a non-right-veering arc.

The converse is not too hard and is a straighforward application of convex surface theory:

Recall that a convex surface is a surface S embedded in a contact 3-manifold such that there is a contact vector field $\eta$ transverse to S. The contact structure near S can be completely understood in terms of the isotopy class of the dividing curves, which are given by the points in S where the contact planes contain the contact vector field: $\eta_x \in \xi_x$. Generic surfaces in contact 3-manifolds are convex with transversely cut-out a set of dividing curves.

For example, the horizontal planes in the standard overtwisted contact structure $\xi = \text{ker}(\cos \pi r d z + r \sin \pi r d \theta)$ are convex, because the vertical vector field $\partial_z$ is contact. Note that there are dividing curves when $r = 1/2 + k$ for some nonnegative integer k, which is when $\partial_z \in \xi$.

A bypass D for a convex surface S is a convex disk with Legendrian boundary and a single dividing curve, that intersects the surface S along an arc with boundary on the dividing set and intersecting the dividing set exactly 3 times.

Now, notice that a bypass is essentially half of an overtwisted disk; the idea is to find two bypasses along the same arc on opposite sides of a convex surface, then glue them together to find an overtwisted disk.

Each open book determines a Heegaard splitting of along the surface $\Sigma = - \Sigma_0 \cup \Sigma_{1/2}$ given by gluing together two pages. The contact structure can be isotoped so that this Heegaard surface is convex, with dividing curves exactly given by the binding.  We can assume the compressing disks are convex and it follows that they each have exactly 1 dividing arc, so they are already essentially bypasses. To make it so, cut a little notch in the disk at the binding and push the boundary off a little along the binding.

There are lots of possibly bypasses, one for each compressing disk on each side, but they don’t line up exactly at the Heegaard surface. To achieve this, we need to do what H-K-M call a bypass rotation. Recall that isotoping across a bypass changes the dividing curves as in the picture.

Suppose we have two potential arcs on which to attach bypasses, as in the figure below. Notice that if we first attach a bypass along the right arc, attaching a bypass on the left arc doesn’t change the dividing curves (up to isotopy). It’s trivial. And by what H-K-M refer to as the “Right to Life” principle, there always is a bypass for an trivial arc of attachment.

So, if we see the local picture above and know that a bypass exists for the right arc, we know a bypass exists for the left arc. Note that this is (not) true if we attach a bypass on the left first and then the right; the right arc is not forced to be a trivial arc of attachment.

So, we have a bypass $D_1$ for $\Sigma$ sitting in one of the handlebodies $H_1$, attached along $\alpha'$ and a second bypass $D_2$ sitting in the other handlebody $H_2$ attached along $\phi(\alpha')$. When the arc $\alpha$ is left-veering  non-right-veering, we can perform bypass rotation and find a new bypass $D_1'$ sitting in $H_1$, attached along $\phi(\alpha')$. Thus, $D_1', D_2$ glue up to an overtwisted disk. $\Box$.

However, it is (not) true that every OB decomposition for an overtwisted contact structure has a left-veering non-right-veering arc. By adding some positive stabilizations, we can hide the overtwistedness of the contact structure. Consider the following examples:

(1)Take a left-veering non-right-veering arc $\alpha$ and consider a local picture of one boundary point of this arc that veers left. We can positively stabilize along the green arc in a neighborhood of this point and since the arc intersects the image $\phi(\alpha)$, it gets modified by the Dehn twist and appears to be right-veering now. If we did that at both endpoints, the arc would now be right-veering instead of left-veering.

(2) In a second example, consider a local picture of an entire left-veering arc (left-veering at both ends). Positively stabilize along the green arc, which is a pushoff of $\alpha$ to the left. Again, the Dehn twist drags $\phi(\alpha)$ across the 1-handle and makes a left-veering arc into a right-veering arc.

Note that right/left-veering-ness is localized to the boundary of $\Sigma$. It is completely blind to what happens in the interior of a page. By cleverly applying some positive stabilizations, we can push the overtwistedness or left-veering-ness further and further into the surface.

Intuitively, Andy Wand’s approach is to undo this procedure and bring the negative twisting back to the boundary, where we can apply Honda-Kazez-Matic’s result. He has a characterization of some phenoma in the interior that, after some well-chosen positive (de)stabilizations, result in a left-veering non-right-veering arc at the boundary.

[*In fact, Andy Wand’s method only uses stabilizations, not both stabilizations and destabilizations]

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### 11 responses to “Tightness and right-veering monodromies”

1. Marco Golla

Hey Peter! Nice post!
I have two comments on two things you say:
* In the first lemma: the manifold supported by that open book is actually obtained by (-1/k)-surgery along K (with respect to the page framing). (This is of course irrelevant for the case |k|=1.)
* In the proof of HKM’s right-veering-ness theorem: the contact structure supported by the negative Hopf link is *not* homotopic to the standard one as plane fields (see, for example, Lemma 10 in the paper by Giroux and Goodman “On the stable equivalence of open books in three-manifold”, where in turn they refer to Neumann and Rudolph “The enhanced Milnor number in higher dimensions”). What you can do, instead, is taking another \xi’ in the correct homotopy class (same Euler class, d_3 shifted by +1 or -1), and take the connected sum with \xi_{OT}, and the arguments run in exactly the same way.

2. Thanks
(1) Yup, that’s a typo. Went back and fixed it
(2) I missed that subtlety the first time around; I think the notation in the HKM paper threw me off.

3. Jerry

Sorry to bother with this triviality, but I’m new to OBs and Contact Structures: I’m a bit confused about your notation for the structure on R^3 given by ker( cos(pi*r)dx+ sin(pi* r) d\theta) : is this polar , cylindrical or something else? I can see cos(pi*r)dr +sin(pi*r)(d\theta). If you prefer, after an answer please feel free to delete.

• There was a typo, the contact structure is $\text{ker}(\cos(\pi r) dz + \sin (\pi r) d \theta$ in coordinates $(r,\theta,z)$ on $\mathbb{R}^2 \times \mathbb{R}$; that is, polar coordinates on the $\mathbb{R}^2$-factor and a normal $z$-coordinate on the $\mathbb{R}$-factor.

4. Jerry P.

Hi Peter, a quick one:: So, for the above contract structure, the dividing curves/ the dividing set is given by circles with radius 1/2+k ; k an integer in R^2 x{0}, i.e., in the xy-plane ? It would seem so, since at these points the planes are vertical, coinciding with \Partial/Partial z ?

• Correct

5. 4mflds

Peter: Nice blog: I’m a little lost on the following; could you please give a ref for where this is discussed/proved(I assume one of the HKM papers)? :

1)…Notice that if we first attach a bypass along the right arc, attaching a bypass along the second arc does not change the dividing curves up to isotopy…

2)…Note that this ( a bypass exists for the right, then it exists for the left ) is not true if we attach a bypass on the left first, and then on the right arc, is not forced to be an arc of attachment….

I guess I’m missing some background; could you please give a ref?
Thanks.

• The reference is “Gluing tight contact structures” by Honda.

• Patrick Massot

A slightly more detailed answer could be: point 1) doesn’t need any background, simply look at the picture. The new dividing set is isotopic to the previous one relative to what is not drawn. Point 2) is indeed discussed in “Gluing tight contact structures”. If you decide to read this paper seriously then you should make sure to read the published version and not the arxiv version which hasn’t been kept up to date.

6. Fernando Jackson

So, is it safe to say that any positive Dehn twist D_{\gamma} will preserve
right-veering arcs (because Dehn^+_{\gamma} is a submonoid of
Veer^+ ). And maybe right-veering is preserved also if the monodromy is changed by a negative Dehn twist D^{-} about a handle so that D^{-} can be isotoped to avoid the r.v arcs?

7. Wero.

Re: destabilization (Fourth paragraph below ‘Giroux Correspondence). If you want \alpha to be an arc whose boundary points are in the boundary component, then I don’t think the co-core curve of the attached handle would do, since its endpoints do not lie on the boundary.