Paletas

4.2.1

Fatgraph realizations

For a free groupFk of rankk, there are (2k−1)! basic fatgraph realizations, because there are that

many distinct cyclic orders on the semigroup generators.

By precomposing with an automorphism of Fk, we can give a fatgraph structure to any surface realization ofFk: suppose (ΣO, fO) is a basic fatgraph realization. Then acting by the automorphism

φ∈Out(F2) gives the surface realization (ΣO, fO◦φ−1), which we will call afatgraph realization.

Again, every surface realization can be represented as a fatgraph realization by choosing a fatgraph structure.

Unfortunately, we need a bit of notation. We will denote by (φ, O) the fatgraph realization ofFk

obtained by acting byφon the basic fatgraph realization induced by the orderOon the semigroup generators ofFk. We denote the rotation quasimorphism obtained onFk by this surface realization

4.2.2

Surface maps into fatgraph realizations

RecallXkis our standardK(Fk,1). Suppose thatg:S→Xk is an incompressible surface map, and (φ, O) is a fatgraph realization (φ might be the identity). Then φ−1◦g :S → Xk has a fatgraph representative, which by abuse of notation we also denote byS. Note there are two fatgraphs in the picture: the target fatgraph surface realization ofFk, and the fatgraph representative of the surface map φ−1_{◦g. Now S is a labeled fatgraph over F}

k, and as such it has a cyclic order on the edges

at each vertex. In addition, the order O induces another cyclic order on the edges at each vertex ofS by simply applyingO to the labels on the edges. We will call this cyclic order on each vertex the pullback ofO. Note this cyclic order might not be well-defined if the labels are not all distinct around a vertex. a A B b b B g fO a b b a

Figure 4.2: Comparing the cyclic orders at each vertex on the edges of the fatgraph coming from the fatgraph structure and from the ordering on labels from the pullback of the orderO on semigroup generators ofF2. See Example 4.2.1.

Example 4.2.1. Figure 4.2 illustrates the different cyclic orders on vertices coming from both the intrinsic fatgraph order and the order induced by the labels. Here the fatgraph realization of F2

is (id,[b, a, B, A]), as shown by the map fO. The surface map g : S → Xk induces a fatgraph structure on the surface S (in the picture, the surface map is implicit from the labeling), and this fatgraph structure includes the cyclic orders on the vertices, which are indicated here by the planar embedding.

The fatgraph realization gives the cyclic order [b, A, B, A] on semigroup generators for F2, and

this induces cyclic orders on the vertices of S. In this case, the vertex on the left has outgoing edges (b, a, B), which is positive in both theS order and the pullback ofO. However, the vertex on the right has outgoing edges (B, b, A), which is positive in the fatgraph order but negative in the pullback ofO.

Theorem 4.2.2. Let g : S → Xk be a surface map. Let (φ, O) be a fatgraph realization of Fk. Give S a labeled fatgraph structure induced by the map φ−1◦g: S →Xk. Then the following are equivalent.

with the orders given by the pullback of O.

2. The induced quasimorphismrot(φ,O)is extremal for ∂S.

3. The mapfO◦φ−1_◦g:S→Σ

O is homotopic to an immersion with geodesic boundary.

Proof. That (2) and (3) are equivalent is the content of Theorem 2.5.5. The equivalence of (1) can be seen intuitively in Figure 4.3. However, the quickest way to see this is to appeal to the fact that the rotation quasimorphism induced by a realization has the area form (divided by−2π) on the hyperbolic surface ΣO as its coboundary (see [7]). Decompose each vertex of the fatgraph

into tripods. There is a pleated representative of S which has one ideal triangle for each tripod. The rotation quasimorphism applied to ∂S will therefore compute the sum of ±1/2 over these triangles, depending on whether they are positive or negative. Since−χ(S) is 1/2 times the number of triangles, we have that rot(φ,O)(∂S) = −χ(S) if and only if all tripods are correctly cyclically

ordered around all vertices.

Condition (1) in Theorem 4.2.2 is a simple combinatorial check, provided that we have a fatgraph representative of φ−1_{◦g : S → X}

k. However, in practice, the map g : S → Xk is given as a

labeled fatgraph, and ifφis not the identity, it appears tricky to obtain a fatgraph representative of

φ−1_◦g:S→X

k, which is the fatgraph of interest in Theorem 4.2.2. In fact, it is possible to push a

fatgraph map forward underφ−1_{with a minimum of effort: simply relabel the edges by applyingφ}−1

and fold (in the sense of Stallings fatgraph folding) the resulting fatgraph. See Figure 4.3, (A)–(C). In addition, ifg:S→Xk is a surface map and (φ, O) a fatgraph realization ofF, it is possible to check condition (1) of Theorem 4.2.2 “all at once,” rather than on each vertex seperately. Consider

S as a labelled fatgraph obtained from φ−1◦g : S → Xk. Take a maximal subgraph T of S — the remaining edges of S determine a set of generators {gi}k

i=1 for π1(S). Since π1(S) ⊆Fk, the

cyclic order O determines a cyclic order on thegi. Furthermore, the cyclic order on each vertex of

S determines a planar embedding ofT and thus a cyclic orderO0 _{on theg}

i. Then clearly

Corollary 4.2.3. Condition (1) of Theorem 4.2.2 is is satisfied if and only ifO0 is compatible with

O, in the above notation.

This idea will be pursued in detail in Section 4.6.