Consistent 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)

Here is the third and final post on the Legendrian surgery conjecture.

In the previous post, we went through H-K-M’s proof that a contact structure is overtwisted if and only if there is a supporting open book with a left-veering non-right-veering arc. This gives a criterion for establishing tight vs ovetwisted, but its not invariant under positive stabilization. So we can’t just check one supporting open book for \xi, we have to check every single one, and we have to check every possible arc in each open book to see if it is right veering or not.

But the problem might be somewhat manageable:

Let’s start with a simple example.  Suppose our monodromy is not efficient and some arc \phi(\gamma), starts off heading to the right but then immediately crosses back over \gamma, forming a trivial bigon that can be isotoped away. If we remove this bigon by an isotopy, then \phi(\gamma) appears to be left-veering non-right-veering now.

completed_bigon

The arc \phi(\gamma) is right-veering if and only if the curves \gamma, \phi(\gamma) form a second trivial bigon, allowing us to isotope \phi(\gamma) back across \gamma.

completed_bigon2

So, if we have a bigon B_+ formed by the pair \gamma, \phi(\gamma), we want to be able to find a second bigon B_- that completes it, meaning that the two cancel each other out and we preserve right-veering-ness.

Now a second example. Suppose we have a square region, as in the following local picture, and that the monodromy is efficient. Then \alpha_1,\alpha_2 are right-veering. However, suppose that we can destabilize along the right \alpha_2 arc. Remember, this means that \phi( \alpha_2), \alpha_2 do not intersect except at the boundary. The destabilization means that we cut the surface along \alpha_2 and then add a negative Dehn twist along a closed curve isotopic to \alpha_2 \cup \phi (\alpha_2). This is the green curve in the second picture.

positive_square

Notice what happens to \phi'(\alpha_1) after the destabilization: it forms a trivial bigon with \alpha-1. So we can isotope \phi'(\alpha_1) so that it becomes left-veering non-right-veering. If that was it, then this contact structure would be overtwisted.

destab_square

However, suppose that there is a second square region as in the following picture, forming a “singular annulus”

completed_square

Now, when we destabilize, the second square region becomes a trivial bigon that pulls all the way around the diagram to cancel out the bigon formed from the first square region.

destab_completed_square

So, if a contact structure is tight and if we have a square R_+ bounded by \alpha_1, \phi(\alpha_1), \alpha_2, \phi(\alpha_2) such that we can destabilize along one of the arcs, we need to be able to find a second square R_- bounded by the same four arcs and connected at opposite corners. In other words, we need to be able to find a second square R_- that (completes) R_+.

Lets go back to an example from the second post, where we saw how a positive stabilization can hide a left-veering non-right-veering arc. In that example, we can find such a square R_+ bounded by the original arcs and the new arcs added by the stabilization. Furthermore, notice that there is a completing square R_- if and only if \phi(\alpha) crosses back over \alpha.

stabilization_square

Before we continue, lets fix some terminology. A (region) is an immersed 2n-gon in a page whose boundary lies in the curves \{\alpha, \phi(\alpha)\}, alternating between an \alpha and a \phi(\alpha). Furthermore, let’s orient the curves \alpha_i, \phi(\alpha_i), subject to the condition that for each i, \alpha_i \cup \phi(\alpha_i) is a closed, oriented curve. With this orientation, we call a region (positive) if the induced orientation on the boundary agrees with the orientation on the arc basis and a region is (negative) otherwise. Note that an oriented arc basis does not necessarily induce a coherent orientation on the boundary of a region.

Third example: Now let’s take the local picture near a square region and stabilize again. After stabilization, we can find a hexagon region.Note that this region is “completed”, meaning there is a second hexagon region attached at the corners of the first region, if and only if the square region before the stabilization was “completed”.

hexagon_stabilization

We can repeat this process by positively stabilizing as many times as we want. The notion of a “completed” region, of any number of sides, persists after repeated positive stabilizations. And if we can destabilize along the various sides of our 2n-gon, we know we can reduce it back to the case of the canceling bigons in the first example above.

We can also look for uncompleted regions further into the surface: Let’s suppose we have a chain of regions as in the following picture:

support_chain

There are two square regions, R_1,R_2 connected to the boundary, and we can orient the arc basis so these regions are positive. Then implies that R_3,R_4 must be negative regions. Finally, R_5 is a positive region, that is not connected to the boundary at all.  Now let’s do something really tricky: stabilize along the green arc that passes through all five regions.

support_stabilization

Afterwards, we get the following picture:

octogon

In effect, the three positive regions R_1,R_2,R_5 have been amalgamated into a single octagonal region. The regions R_3, R_4 have been amalgamated as well and they have also been joined with the negative regions beyond R_5. The result is that the 8-gon region we find after stabilizing is “completed” if and only if R_5 was “completed” (meaning there is a square negative region forming a singular annulus, as in example 2 above).

Again, “completion” is a property that persists after positive stabilizations. So, if we have an open book for some contact structure, we can look for uncompleted regions in the interior that are connected to the boundary by a chain of regions; then by a stabilization, bring it to the boundary.  Andy refers to any a monodromy with no such uncompleted regions as consistent and it should be a stabilization-invariant characterization of tightness.

These are just some suggestive examples: it seems like a decent amount of work to clean this up into an algorithm that takes an open book with an uncompleted region and, by a sequence of positive (de)stabilizations, find an open book with a left-veering non-right-veering arc.

Finally, let’s get back to the Legendrian surgery conjecture. Andy never elaborated much on this aspect, leaving it as a corollary of the characterization of tightness in terms of open books.

However, the basic idea is that Legendrian surgery can be realized as adding to the monodromy a positive Dehn twist along some essential curve in some open book supporting \xi. The key step now is to prove that positive Dehn twist preserve the consistency of the monodromy or equivalently, that a positive Dehn twist does not create an uncompleted region.

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2 Comments

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2 responses to “Consistent monodromies

  1. lstarkston

    Nice post! It seems like one could choose the basis of arcs {\alpha_i\} and their orientations in different ways, and in some of these ways it may be easier to see the (de)stabilizations needed to get to something left-veering. Do you know if it is necessary to look at all possible arc bases to detect whether your contact structure is tight or overtwisted using this consistency property, or can you just choose a single oriented arc basis arbitrarily and use that to detect tightness/overtwistedness?

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