In my last post, I described how a train track on a surface determines a collection of loops in a surface, namely the loops that are carried by the track. Looking at these loops from the perspective of the the Farey graph for the torus, this set consists of the loops corresponding to vertices in one of the components that results from cutting the Farey graph along a certain edge. In the curve complex, train tricks define partitions that are almost as simple, though they are necessarily more complicated because there is no one simplex that separates separates the complex. Still, this type of partition comes in very useful for calculating distances in the curve complex (and was central to my recent preprint with Yoav Moriah) but to see how that works, we need something a bit stronger. In this post, I’ll explain how we can turn the partition defined by a train track into two sets of curves with a buffer between them. By placing these buffers next to each other, we can build larger gaps that imply a lower bound on the distance between certain loops in the curve complex.
Recall that a train track T in a surface S is a subsurface of S endowed with a certain type of singular foliation by intervals. We say that a loop l is carried by T if it is contained in the subsurface and transverse to the intervals in the foliation. Away from the singularities, the intervals in the foliation make up parallel strips, and the singularities define junctions where they come together. If we follow a loop around the train track, the transverse condition implies that each time we enter one of the parallel strips, we have no choice but to follow it to the junction at the other end. However, when the loop enters a junction, it will often have a “choice” of whether to take the branch to the left or right. So as we follow the loop around, depending on the “choices” the loop makes of how to turn, it may end up only crossing some of the parallel bands, and missing others. We say that a carried loop covers the train track T if the loop intersects every fiber, or equivalently if it crosses every band or parallel fibers.
We now have three types of loops in the surface S defined by the train track T: There are the loops that cover T, the loops that are carried by T, but don’t cover it, and the loops that aren’t carried at all. This is where an important observation comes in: It turns out that for a train track T whose complement in S is a collection of triangles, if a loop l covers T and a second loop m is disjoint from l then m must be carried (though not necessarily cover) T. (I don’t know who first noticed this. It’s in Masur and Minsky’s  work, but may go back much earlier.)
To see why this is true, note that the loop m cannot cut across T parallel to the interval fibers because then it would have to cross l. Moreover, any arcs of m outside of T will be contained in the triangular complementary regions and can be pushed into T in a canonical way. If you look carefully at what m can do inside of T without intersecting l, you’ll quickly conclude that it’s possible to isotope m so that it’s transverse to all the fibers and thus carried by T.
What this means in terms of the three classes of loops is that no edge in the curve complex connects a loop that isn’t carried by T to a loop that covers T. In particular, any path from a covering loop to an uncarried loop has to pass through a loop that is carried but doesn’t cover. So, in other words, the set of non-covering carried loops forms a buffer between the covering loops and the uncarried loops.
This is the buffer that I mentioned at the beginning of the post. But now the question is, how can we place these buffers next to each other to make wider ones? The key to this is to construct a second train track U such that T is “carried” by U. By this I mean that the subsurface defined by T is contained in the subsurface defined by U and every interval fiber in T is contained in an interval fiber in U. Note that it follows immediately from definitions that every loop carried by T will be carried by U.
We’ll next add an extra condition that’s slightly more subtle: We want every loops that is carried by T to cover U. Note the difference there: We’re asking for a stronger condition on U. At first, this may seem like too much to ask for, since there will be infinitely many loops carried by T and we don’t want to have to check each one in U. But in fact, we usually only need to check a finite number of things. In particular, we can often arrange so that every band of parallel fibers in T covers U, i.e. intersects every interval fiber in U. Because any loop carried by T must follow at least one of the parallel bands of fibers in T, this guarantees that any loop carried by T will cover U.
In some cases, we may not be able to guarantee that every band in T covers U, but we may still be able to find a subset of the bands in T such that every carried loop must cross one of these bands, and each of these bands covers U.
If we can find a train track U with this condition, then we can compare any loop l that covers T to a loop n that is not carried by U. Since n is also not carried by T, any path from l to n must pass through a loop k that is carried by T. The above condition implies that k covers U, so the path must also contain another loop m that is carried by U, but does not cover U. Thus any path from l to n must pass through at least to other loops, making its length at least three.
We can repeat the process again by constructing a train track V that carries U and has the same filling property. Any loop that is not carried by V will be distance at least four from any loop that covers T. As we build more train tracks in this way, we can find loops that are farther and farther apart. (This is one way to show that the complex of curves has infinite diameter.)
In my paper with Yoav Moriah, we construct a sequence of such train tracks in the bridge surface of a certain type of knot, with the structure of the train tracks determined by a certain type of diagram of the knot. We’re then able to show that every bridge disk below the bridge surface covers one of the train tracks early in the sequence, while no disk above the bridge surface is carried by the last train track in the series. By the above argument, this gives us a lower bound on the distance between the two disk sets. (In practice, we use a slightly different version of the carried condition, which allows the complement of the train track to be any polygon, not just a triangle.) An explicit construction shows a lower bound for the distance. As it turns out, these two bounds are the same, so we’re able to calculate the exact distance for this certain class of knot diagrams.