# Low Dimensional Topology

## March 14, 2014

### Train tracks on a torus

Filed under: Curve complexes,Surfaces — Jesse Johnson @ 11:47 am

A little over a year ago, I started writing a series of posts on train tracks and normal loops, then got distracted by other things. In the mean time, I wrote a paper with Yoav Moriah involving train tracks and curve complex distances, which gave me a whole new perspective on what train tracks really mean, more in line with much of Masur and Minsky’s work [1]. So, I want to resuscitate the series of posts on train tracks, but in a slightly different direction than where I was headed before. I’ll start by looking at a very simple case: train tracks on a torus. If you need a review of what train tracks are (the mathematical object, not the literal ones), you can reread my earlier post.

We can form a train track on a torus by taking two essential loops in the torus that intersect once, then smoothing the intersection, as in the Figure below. (I’m drawing the torus as a square with opposite sides identified.) There are two possible ways to smooth the intersection, and for now we’ll just arbitrarily pick one. (Later on, we’ll come back to look at the difference between the two smoothings.) The resutling graph isn’t a train track, bit we can turn it into a train track by taking a regular neighborhood of it, then giving the neighborhood a foliation by intervals perpendicular to the original graph. The original graph (shown in the middle of the Figure) is called a train track diagram.

The question I want to explore in this post is: What loops in the torus are carried by this train track? The answer will be in terms of the slopes of the carried loops. Recall that the universal cover of the torus is the plane. In every isotopy class of essential loops, there is a representative that lifts to a straight line in the universal cover. In fact, there’s an infinite family of such loops that lift to different lines in the plane, but all these lines have the same slope. This slope is what we call the slope of the (isotopy class of the) loop in the torus. In the Figure above, the blue loop has slope 0 and the red loop has slope 1/0 or $\infty$. Note that both of these loops are carried by the train track. (Or, more precisely, they’re isotopic to loops that are carried by the train track.)

In general, we can calculate the absolute value of the slope of a loop by dividing the number of times it intersects the horizontal boundary of the square by the number of times it intersects the vertical boundary. (You can check that this formula holds for the red and blue loops.) For any slope other than 0 and $\infty$, we can figure out the sign as follows: If an arc has one endpoint on the left side of the square and the other endpoint on the top then the loop has positive slope. If an arc has one endpoint on the left and its other endpoint on the bottom then the loop has negative slope. (It’s not too hard to check that a loop that intersects the sides of the square minimally can’t have both types of arcs.)

There are many other loops in the torus, in addition to the red and blue loops above, that are carried by this particular train track. Examples with slopes $1$, $2$ and $\frac{1}{2}$, respectively, are shown in red in the Figure below.

All these loops have positive slopes, and in fact, you can see that no arc from the left side of the square to the bottom of the square can be carried by this train track. So this means that this train track can only carry positive slopes.

On the other hand, we can put in as many copies of either the vertical or the horizontal arc as we want. We can also put in as many arcs as we want from the left side to the top side, and the same number from the bottom to the right side. By choosing the number of such arcs carefully, we can get the intersections between the resulting loops and the sides of the squares to be whatever we want. (If the number of intersections with the top is greater than the number with the bottom, we’ll only use vertical arcs. Otherwise, we’ll only use horizontal arcs.) So, every loop with positive slope will be carried by this train track.

To make it clear, let me summarize what we’ve learned: The train track that we constructed carries all the loops with positive slopes, as well as the loops with slope 0 and $\infty$. Going back to the beginning of the post, note that if we had chosen to smooth the intersection between the original two loops in the opposite way, the resulting train track would have carried all the negative slope loops, as well as 0 and $\infty$. So, we can think of a train track as a way to separate the loops in a surface into two different classes: the loops that are carried and the loops that aren’t.

One way that this gets really interesting is when consider what these two classes look like in the curve complex for the surface. This approach is one of the main tools used in Masur and Minsky’s work on the curve complex [1], particularly their proof that curve complexes of surfaces are Gromov $\delta$-hyperbolic.

Recall that the curve complex for a surface S is the simplicial complex whose vertices represent isotopy classes of essential, simple closed curves in S and whose faces span sets of isotopy classes with pairwise-disjoint representatives. The curve complex for a torus is pretty boring: Any two disjoint essential loops in a torus are parallel (and thus isotopic) to each other, so there are no edges in this curve complex- It’s just an infinite collection of discrete vertices.

So instead, one generally works with the Farey graph for the torus. Much like the curve complex, the vertices of the Farey graph represent isotopy classes of essential loops in the torus. In particular, each vertex represents a rational number (a slope) including $\infty$, and in fact we can arrange the vertices in order by slope along a circle. Since there are no pairs of disjoint loops, we connect any two vertices representing loops that intersect in a single point.

Similarly, we include in the Farey graph all the triangles bounded by loops of three edges. I’ll leave it as an exercise for the reader to check that for every pair of loops in the torus that intersect in exactly one point, there are exactly two other loops such that each of these loops intersects each of the original two loops in a single point. (The two new loops will intersect each other in two points.) So, in other words, each edge in the Farey graph is in the boundary of exactly two triangles. This tells us that the triangles form a surface. In fact, the surface that they form is the disk bounded by the circle along which we placed the vertices in the previous paragraph.

Six of these triangles are shown in the figure on the right, with the slopes corresponding to their vertices indicated as fractions. For each edge in the Farey graph, we can calculate the third vertex representing one of the adjacent triangles as follows: The numerator of the new slope is the sum of the numerators of the original two, and the denominator is the sum of their denominators. Similarly, to get the vertex defining the other triangle, we subtract the numerators and denominators. (To see why this works, you can think about the normal loops and Haken sums that I mentioned in another post from a while back.)

Notice that the triangles in this picture are different sizes, and in fact they get smaller as the numerators and denominators get bigger. But in reality, the edges of the Farey graph should all be the same length. So, you should think about this circle like the boundary of the hyperbolic plane, and the triangles as being ideal triangles. This isn’t exactly right either, since the edges in the Farey graph have finite length, unlike the edges of ideal triangles. But the Farey graph will have the same symmetry group as a tesselation of the hyperbolic plane by ideal triangles.

The Farey graph is closer in structure to a tree. In fact, we can construct a tree by putting a vertex at the center of each triangle and connecting two vertices whenever the corresponding triangles share an edge. The Farey graph will be quasi-isometric to this tree (though if you don’t know what quasi-isometric means, don’t worry about it.) In the same way that each edge in a tree cuts the tree into two separate trees, each edge in the Farey graph cuts the Farey graph (which is really a cell complex) into two disconnected sets of triangles.

Now, lets go back to the train track from the beginning of this post. Recall that the set of loops carried by the train track consisted of all loops with positive slopes, as well as the loops with slopes 0 and $\infty$. These loops make up the right half-circle of the Farey graph. In particular, the subcomplex of the Farey graph spanned by the loops carried by this train track is exactly one of the two components that we get if we cut along the edge spanned by 0 and $\infty$.

Note that when we constructed this train track, we started with any two loops in the torus that intersect in one point, or equivalently, any edge in the Farey graph. We then had a choice of two different ways to smooth the vertex where they intersect into a pair of switches in the train track. If we had made the other choice with our original two loops, we would have gotten a train track that carried all negative slopes, i.e. the other component defined by the edge between 0 and $\infty$. By symmetry, if we had started with a different pair of loops, the two possible train tracks that we could construct from them would similarly define the two different components that we get by cutting the Farey graph along this new edge. (Note that one can also show that every “reasonable” train track in the torus can be constructed from two loops in this way.)

The point of all this is that the different train tracks on the torus can be thought of as defining all the different ways of cutting the Farey graph along single edges. Train tracks in higher genus surfaces play a very similar role, though it’s more complicated because the curve complexes of these surfaces are much less tree-like (though they’re still delta hyperbolic, which is close.) In particular, you can’t separate these complexes by removing a single edge, or indeed any finite collection of simplices. But train tracks still define subsets of loops that are very nice with respect to the curve complex structure.

The reason this turns out to be useful is that it is often possible to prove things about the types of loops that are carried by a given train track, which can then be translated into the language of the curve complex. This is one of the main techniques in Masur and Minsky’s papers on the curve complex [1], and on disk sets of handlebodies [2]. It also proved very useful in my work with Yoav Moriah [3] and his earlier work with Martin Lustig [4]. But a discussion along those lines will have to wait for a future post.

1. “Recall that the set of loops carried by the train track consisted of all loops with positive slopes, as well as the loops with slopes 0 and ∞”

Isn’t the set of loops limited to slopes m/n; m,n ∈ (0,∞) with at least one of m,n = 1 ? For example, I don’t think that the trefoil (3/2) could be carried by that track.

Comment by Jouni Kosonen — March 15, 2014 @ 8:34 am

• It is possible to have both m and n greater than one. If there are vertical arcs (rather than horizontal arcs) in addition to the diagonals, then m will be equal to the number of diagonals in one of the two classes plus the number of verticals. The value n will be the number of diagonals in one of the two classes. If there are horizontal arcs instead of vertical arcs, then it will be the same with n and m reversed. We’re allowed to have as many of the diagonal arcs as we want, as long as there are the same number of each of the two types of diagonals (though I only drew examples with a single diagonal.)

Comment by Jesse Johnson — March 16, 2014 @ 8:18 pm

• Right, something like ╝│╔ would of course work here for 3/2. Sorry about the noise.

Comment by Jouni Kosonen — March 17, 2014 @ 5:24 am

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