The curves not carried

Suppose $\tau$ is a train track on a surface $S$. Let $C(\tau)$ be the set of isotopy classes of simple closed curves carried by $\tau$. Masur and Minsky [2004] prove $C(\tau)$ is quasi-convex inside the curve complex $C(S)$. We prove the complement, $C(S) - C(\tau)$, is quasi-convex.


INTRODUCTION
The curve complex C (S), of a surface S, is deeply important in low-dimensional topology. One foundational result, due to Masur and Minsky, states that C (S) is Gromov hyperbolic [3,Theorem 1.1].
Suppose τ is a train track on S. The set C (τ) ⊂ C (S) consists of those curves α carried by τ: we write this as α ≺ τ. Another striking result of Masur and Minsky is that C (τ) is quasi-convex in C (S). This follows from hyperbolicity and their result that splitting sequences of train tracks give rise to quasi-convex subsets in C (S) [4,Theorem 1.3].
We prove a complementary result. This supports the intuition that, for a maximal birecurrent track τ, the carried set C (τ) is like a half-space in a hyperbolic space. Conversely, it is unlike a horoball.
When S is the four-holed sphere or once-holed torus the proof is an exercise in understanding how C (τ) sits inside the Farey graph. So, let S be a connected, compact, oriented surface with χ(S) ≤ −2, and not a four-holed sphere. In these cases, the proof of Theorem 3.1 is roughly as follows. Suppose γ and γ are simple closed curves, not carried by τ. Let [γ, γ ] be a geodesic in C (S). Suppose that α and α are the first and last curves of [γ, γ ] carried by τ. Fix splitting sequences that split τ to α and α , respectively. For each splitting sequence, its vertex set is K 1 -quasi-convex inside of C (S). Since C (S) is Gromov hyperbolic, the geodesic segment [α, α ] is K 1 + δ-close to the union of vertex sets. Proposition 6.1 com-The idea behind Proposition 6.1 is as follows. Since σ = supp(α, τ) is large all components of S − σ are disks or once-holed disks. Since σ is not maximal there is a component Q ⊂ S − σ which is not an ideal triangle or a once-holed ideal monogon. Hence, there is a diagonal δ of Q that is not carried by τ. We then extend δ, in a purely local fashion, to a simple closed curve β. By construction β is in efficient position with respect to τ and meets α at most once. Finally, we appeal to Criteria 4.2 or 4.4 to show that β is not isotopic to a carried curve.

BACKGROUND
We review the basic definitions needed for the rest of the paper.
2.1. The curve complex. Fix S a compact, connected, smooth, oriented surface. Define i (α, β) to be the geometric intersection number between a pair of simple closed curves.
The complex of curves, C (S) is, for us, the following countable graph. Vertices are essential non-peripheral (isotopy classes of ) simple closed curves. Edges are pairs of distinct vertices α and β where i (α, β) = 0. When χ(S) ≤ −2 (and S is not the four-holed sphere) it is an exercise to show C (S) is connected. We may equip C (S) with the usual edge metric, denoted d S . Here is a foundational result due to Masur  A branch B ⊂ τ is either large, mixed, or small as it contains two, one or zero large half-branches. We may split a pre-track τ along a large branch, as shown in Figure 2.4, to obtain a new track τ . Conversely, we fold τ to obtain τ. If a branch is mixed then we may shift along it to obtain τ , as shown in Figure 2.5. Note shifting is symmetric; if τ is a shift of τ then τ is a shift of τ .   Suppose now that τ is a pre-track and N (τ) is a tie neighborhood. Let Q ⊂ S − n(τ) be a component. Let c ± (Q) count the inward and outward corners of Q, respectively. The index of Q is For example a rectangle R has four outward corners and no inward corners. Thus index(R) = 0. Note that index is additive. Since N (τ) is a union of rectangles N (τ) also has index zero.
We may now give our definition of a train track. 2.8. Carried curves and transverse measures. A simple closed curve α ⊂ S is carried by a track τ (written α ≺ τ) if α can be isotoped into the interior of N (τ) and, this done, is transverse to the ties. It is an exercise to show that any carried curve is essential and non-peripheral. We define C (τ) = {α ∈ C (S) | α ≺ τ}. Note that C (τ) is non-empty.
Let B = B τ be the set of branches of τ. Fix a switch s and suppose that the halfbranches A, B , and C are adjacent to s, with A being large. A function µ : B → R ≥0 satisfies the switch condition at s if We call µ a transverse measure if µ satisfies all switch conditions. For example, any carried curve α ≺ τ gives a transverse measure µ α . This permits us to define σ = supp(α, τ), the support of α in τ: a branch B ⊂ τ lies in σ if µ α (B ) > 0.
Since the switch conditions are homogeneous the set of solutions ML(τ) is a rational cone. We projectivize ML(τ) to obtain P (τ), a non-empty convex polytope. All vertices of P (τ) arise from carried curves; we call such curves vertex cycles for τ. Proof. The forward direction is given by Proposition 3.11.3(3) of [5]. The backward direction is an exercise in the definitions.
The usual upper bound on distance in C (S), coming from geometric intersection number, gives the following. Lemma 2.12. For any surface S there is a constant K 0 with the following property.
Suppose τ is a track. Suppose σ is either a split, or a subtrack, of τ. Then the Note that Lemmas 2.11 and 2.12 also hold if τ is a pre-track which is in turn a subtrack of a train track.

Remark 2.15.
In the first statement of their Theorem 1.3 [4, page 310] Masur and Minsky assume their tracks are large and recurrent. However, as they remark after their Lemma 3.1, largeness is not necessary. Also, it is an exercise to eliminate the hypothesis of recurrence, say by using Lemma 2.12 and the subtracks supp(α, τ i ) (for any fixed α ∈ V (τ N )).
A more subtle point is that their Lemmas 3.2, 3.3, and 3.4 use the train track machinery of their earlier paper [3]. Transverse recurrence is used in an essential way in the second paragraph of the proof of Lemma 4.5 of that paper. However the crucial "nesting lemma" [4,Lemma 3.4] can be proved without transverse recurrence. This is done in Lemma 3.2 of [1].
Thus, as stated above, Theorem 2.14 does not require any hypothesis of largeness, recurrence, or transverse recurrence.

PROOF OF THE MAIN THEOREM
We now have enough tools in place to see how Proposition 6.1 implies our main result. Proof. We may assume χ(S) ≤ −2, and that S is not a four-holed sphere. Suppose So, instead, suppose α and α are the first and last curves, along [γ, γ ], carried by τ. Let β be the predecessor of α in [γ, γ ] and let β be the successor of α . Thus, β and β are not carried by τ. The contrapositive of Lemma 2.9 now implies that the tracks σ = supp(α, τ) and σ = supp(α , τ) are not maximal.
For a moment, fix attention on α and σ. We choose a splitting sequence with the following properties: • for all i , the curve α is carried by τ i , and • the support supp(α, τ n ) is a simple closed curve.
From Lemma 2.11 deduce that α is a vertex cycle in τ n . We make similar choices for α and τ . Let V = ∪V (τ i ) be the vertices of the splitting sequence to α and define V similarly. The hyperbolicity of C (S) (Theorem 2.2) and the quasi-convexity of vertex sets (Theorem 2.14) imply the geodesic [α, α ] lies within a K 1 + δ-neighborhood of V ∪ V . To finish the proof we must show that every vertex of V (and of V ) is close to a non-carried curve.
By Lemma 2.11 we may pick vertex cycles α i ∈ V (τ i ) so that: Note that some of the σ i may be pre-tracks, as ∂ h N (σ i ) may contain a component without corners. However, all such pre-tracks are necessarily small. Since σ n = supp(α, τ) is not maximal, likewise none of the σ i are maximal. Let m = max{ | σ is small}. Fix any curve ω ∈ C (S) lying in the complement of σ m . Using ω we deduce d S (α i , α m ) ≤ 2, for any i ≤ m.
If m = n then Lemma 2.12 implies the set V = ∪V (τ i ) lies within a K 0 + 3neighborhood of β, and we are done.
So we may assume that m < n. In this case Lemma 2.12 implies the set is assumed to be a large subtrack of τ. Also, σ i is not maximal. Thus we may apply Proposition 6.1 to obtain a curve β i so that: Applying Lemma 2.12 we deduce, whenever i > m, that V (τ i ) lies within a K 0 + 2neighborhood of β i . Carrying out the same for the splitting sequence from τ to α completes the proof of the theorem.

EFFICIENT POSITION
In order to prove Proposition 6.1, we first give criteria to show that a curve β is not carried by a given track τ. We state these in terms of efficient position, defined previously in [2,Definition 2.3]. See also [6,Definition 3.2].
Suppose τ is a train track and N = N (τ) is a tie neighborhood. A simple arc γ, properly embedded in N , is a carried arc if it is transverse to the ties and disjoint from ∂ h N .  Then no curve isotopic to β is carried by τ.
Proof. Suppose, for a contradiction, that β is isotopic to γ ≺ τ. We now induct on the intersection number |β ∩ γ|.
In the base case β and γ are disjoint; thus β and γ cobound an annulus A ⊂ S. Since β and γ are in efficient position with respect to τ, the intersection A ∩ N (τ) is a union of rectangles, so has index zero. However, one of L or R lies inside of A − N (τ). This contradicts the additivity of index.
In the induction step, β and γ cobound a bigon B ⊂ S. Since γ is carried, the two corners x and y of B lie inside of N (τ). Let β x be the component of β ∩ N that contains x. We call x a carried or dual corner as β x is a carried arc or a tie. We use the same terminology for y.
If x is a carried corner then move along γ ∩ ∂B a small amount, let I x be the resulting tie, and use I x to cut a triangle (containing x) off of B . Do the same at y to obtain B . Now, if both x and y are dual corners then B = B is a bigon. If exactly one of x or y is a dual corner then B is a triangle. In either of these cases index(B ) is positive, contradicting the assumption that β is in efficient position.
So suppose both x and y are carried corners of B ; thus B is a rectangle. Thus B has index zero. Recall that β M is a subarc of β meeting both R and L. Since neither L or R lie in B deduce that β M is disjoint from B . We now define β B = β ∩ B and γ B = γ ∩ B , the two sides of B . We define β to be the curve obtained from β by isotoping β B across B , slightly past γ B . So β is isotopic to β, is in efficient position with respect to τ, has two fewer points of intersection with γ, and contains β M . Thus β M is adjacent to two regions L and R of S − n(β ∪ τ) of negative index, as desired. This completes the induction step and thus the proof of the criterion. Criterion 4.2 is not general enough for our purposes. We also need a criterion that covers a situation where the regions L and R are not immediately adjacent. However the proof of Criterion 4.4 is essentially identical to that of Criterion 4.2 and is accordingly omitted.

EFFICIENT AND CROSSING DIAGONALS
The next tool needed to prove Proposition 6.1 is the existence of crossing diagonals: efficient arcs that cannot be carried.
Let τ be a train track. Suppose σ ⊂ τ is a subtrack. We take N (σ) ⊂ N (τ) to be a tie sub-neighborhood, as follows.
• Every tie of N (σ) is a subarc of a tie of N (τ). To fix ideas, we orient short diagonals and require the component of Q − n(δ), to the right of δ, to be a hexagon. This hexagon meets three (or two) components of ∂ v Q and properly contains one of them, say v. In this situation we say δ cuts v off of Q. In the simplest case a short diagonal δ ⊂ Q is carried by N (τ,Q). We say an efficient diagonal δ is a crossing diagonal if there is • a subarc δ M (or δ I ) and • regions L and R of Q − (δ ∪ n(τ)) satisfying the hypotheses of Criterion 4.2 or Criterion 4.4. Deduce, if β τ is a curve, containing a crossing diagonal δ, then β is not isotopic to a curve carried by τ.

Lemma 5.1. Suppose σ ⊂ τ is a large subtrack. Let Q be a component of S − n(σ) that is not a hexagon or a once-holed bigon. Then for any component
Furthermore δ is properly isotopic, relative to the corners of Q, to a carried or a crossing diagonal.
Proof. Using the orientation of S we orient Q and thus the boundary of Q. Let u and w be the components of ∂ v Q immediately before and after v. (Note that we allow u = w. In this case Q is a once-holed rectangle.) Let h u and h w be the components of ∂ h Q immediately before and after v.
Let N u be the union of the ties of N (τ,Q) meeting h u . As usual, N u is a union of rectangles. (See Figure 5 With I u in hand, let N u be the closure of the component of N u − I u that meets v. We define I w and N w in the same way, with respect to h w . Consider the set X = h u ∪ N u ∪ v ∪ N w ∪ h w . Let N (X ) be a small regular neighborhood of X , taken in S, and set δ = Q ∩ ∂N (X ); see Figure 5.2. We orient δ away from u and towards w. Note that δ cuts a hexagon H off of Q.
We now prove that δ is in efficient position. After an arbitrarily small isotopy, the subarc of δ u ⊂ δ between u and I u is carried; the same holds for the subarc δ w between I w and w. (If I u = u then we take δ u = and similarly for δ w .) All components of δ ∩ N (τ), other than δ u and δ w , are ties. Consider = δ − n(τ). If is connected, then cuts a hexagon R off of Q − n(τ). By additivity of index the region L ⊂ Q − (δ ∪ n(τ)) adjacent to R has index at most zero. If L has index zero, it is a rectangle; we deduce that δ is isotopic to a carried diagonal. If L has negative index then δ is a crossing diagonal, according to Criterion 4.2.
Suppose = δ − n(τ) is not connected. We deduce that the first and last components of cut pentagons off of Q − n(τ); all other components cut off rectangles. When u = w then every region of Q − n(τ) contains at most one component of . In this case an index argument proves that δ is a crossing diagonal, according to Criterion 4.2. If u = w then Q is a once-holed rectangle as shown in Figure 5.3. In this case δ is a crossing diagonal, according to Criterion 4.4.

Lemma 5.4. Suppose σ ⊂ τ is a large subtrack. Let Q be a component of S − n(σ) that is not a hexagon or a once-holed bigon. Then Q has a short crossing diagonal.
Proof. Since σ is large, Q is a disk or a peripheral annulus. Set n = |∂ v Q|. According to Lemma 5.1, for every component v ⊂ ∂ v Q there is a short diagonal δ v cutting a hexagon H v off of Q. Also each δ v is a carried or a crossing diagonal.
Suppose for a contradiction that δ v is carried, for each v ⊂ ∂ v Q. Thus, after an isotopy of each, preserving efficient position, we may assume that the δ v are pairwise disjoint. Now, no hexagon H v can contain another one, because index is additive. We deduce that index(Q) ≤ − n 2 ; this inequality is strict when Q is a peripheral annulus because the component of Q −n(τ) meeting ∂S must also have negative index.
On the other hand, if Q is a disk then index(Q) = 1 − n 2 ; if Q is an annulus then index(Q) = − n 2 . In either case we have a contradiction.

CLOSING UP THE DIAGONAL
After setting some necessary terminology, we give the proof of Proposition 6.1. Suppose τ ⊂ S is a track, and N = N (τ) is a tie neighborhood. Suppose that I ⊂ N is a tie, containing a component u ⊂ ∂ v N . Let R, R , and R be the halfrectangles adjacent to I , where R is large and the others are small. For any unit vector V (x) based at x ∈ interior(I ) we say V (x) is vertical if it is tangent to I , is large if it points into R, and is small otherwise. Suppose α ≺ τ is a carried curve. A point x ∈ α ∩ I is innermost on I if there is a component ⊂ I − (u ∪ α) so that the closure of meets both u and x.
For an oriented curve α, and for any x ∈ α, we write V (x, α) for the unit tangent vector to α at x. If x, y ∈ α then we take [x, y] ⊂ α to be (the closure of) the Also, we take α op to be α equipped with the opposite orientation. We make similar definitions when α is a arc. Proposition 6.1. Suppose τ ⊂ S is a train track. Suppose α ≺ τ and supp(α, τ) is large, but not maximal. Then there is a curve β τ, with i (α, β) ≤ 1, so that any curve isotopic to β is not carried by τ.
Proof. Set σ = supp(α, τ). Fix a component Q of S − n(σ) that is not a hexagon or a once-holed bigon. By Lemma 5.4 there is a short crossing diagonal δ ⊂ Q. Recall that n(δ) cuts a hexagon H off of Q. The orientation on δ is chosen so that H is to the right of δ. Also, H meets three components u, v, w ⊂ ∂ v Q. Note that v is completely contained in H . Let p and q be the initial and terminal points of δ. Thus p ∈ u and q ∈ w; also V (p, δ) is small and V (q, δ) is large. (Equivalently, V (p, δ) points into Q while V (q, δ) points out of Q.) Let J u and J w be the ties of N (σ) containing u and w. Rotate V (p, δ) by π/2, counterclockwise, to get an orientation of J u . We do the same for J w .
Since σ = supp(α, τ), there are pairs of innermost points x R , x L ∈ α ∩ J u and z R , z L ∈ α ∩ J w . We choose names so that x R , p, x L is the order of the points along J u and so that z R , q, z L is the order along J w . Now orient α so that V (x R , α) is small.
One possible configuration of α near the endpoints of δ ⊂ Q.
We now divide the proof into two main cases: either one of V (x L , α), V (z R , α), or V (z L , α) is large, or all three of them are small. In all cases and subcases our goal is to construct a curve β which contains δ and is, after an arbitrarily small isotopy, in efficient position with respect to N (τ). Since β contains δ one of Criterion 4.2 or Criterion 4.4 applies: any curve isotopic to β is not carried by τ.
6.3. A tangent vector to α at x L , z R , or z L is large. This case breaks into subcases depending on whether or not u = w. Suppose first that u = w.
If V (z R , α) is large, then consider the arcs [q, has the desired properties and satisfies i (α, β) = 0. If V (z L , α) is large, then consider the arcs [q, has i (α, β) = 1 because β crosses, once, from the right side to the left side of [z L , x R ]. Suppose now that V (z R , α) and V (z L , α) are small but V (x L , α) is large. Consider the arcs [p, We now turn to the subcase where u = w and V (x L , α) is large. In this case x R = z L , x L = z R , and the points x R , p, q, x L appear, in that order, along J u . Consider the arcs [q, x L ] and [x R , p] ⊂ J u and [x L , x R ] ⊂ α. Then has i (α, β) = 0.
6.4. The tangent vectors to α at x L , z R , and z L are small. Let R ⊂ N (τ) be the biggest rectangle, with embedded interior, where • the components of ∂ v R are subarcs of ties, Since the interior of R is embedded there is a component of ∂ v R disjoint from w. Thus ∂ v R contains some other component, say u , of ∂ v N (σ). Pick a point p in the interior of u . Let ⊂ R be a carried arc starting at q and ending at p .
Let J u be the tie in N (τ) containing u . We orient J u by rotating V (p , ) by π/2, counterclockwise. Let x R and x L be the innermost points of α ∩ J u . Note that V (x L , α) and V (x R , α) are both large. Thus u = u. We have already seen that u = w.
Let Q be the component of S − n(δ ∪ σ) that contains u . As usual we use the orientation on S to orient Q , which in turn induces an orientation on ∂Q . Let v be the component of ∂ v Q immediately before u .
If v = u then Q is a once-holed bigon, contradicting the fact that V (x L , α) and V (x R , α) are both large.
If v ⊂ u −δ then Q = H ⊂ Q is the hexagon to the right of δ. Thus u = v. In this case R ∪Q contains the right side of α. If u = w then the right side of α meets the region Q − (n(δ) ∪Q ), near x L . We deduce that the right side of α is not contained in R ∪Q , a contradiction. If u = w then Q is a once-holed rectangle. In this case R ∪Q, together with a pair of rectangles, is all of S. Thus S is a once-holed torus, contradicting our standing assumption that χ(S) ≤ −2.
If v ⊂ w − δ then Q = Q ∪ N (δ) ∪ H . We deduce that Q is not a once-holed rectangle; so u = w. Also, the left side of α is contained in R ∪Q . However the left side of α meets the hexagon H , near the point x R , giving a contradiction.
To recap: the arc δ ∪ enters Q at p ∈ u ⊂ ∂ v Q . The region Q is not a onceholed bigon; also Q ∩ H = . The component v ⊂ ∂ v Q coming before u is not contained in w − δ.
Let J v be the tie of N (τ) containing v . We orient J v using the orientation of Q . Let y R and y L be the two innermost points of α ∩ J v , where y R comes before y L along J v . Since V (x L , α) is large, the vector V (y L , α) is small.
We now have a final pair of subcases. Either V (y R , α) is large, or it is small. 6.6. The tangent vector to α at y R is small. In this case we consider the component w of ∂ v Q immediately before v . Recall that Q ∩ H = . Now, if w is the left component of w − δ then V (y R , α) being small implies V (z L , α) is large, contrary to assumption. If w is the left component of u − δ then v is contained in w, a contradicton. As usual, let J w be the tie in N (τ) containing w . Since w is not contained in u or w there are a pair of innermost points z R and z L along J w . Applying Lemma 5.1 there is a short diagonal δ in Q that • connects p ∈ u to a point q ∈ w and • cuts v off of Q . Note that v is to the left of δ .
We give J w the orientation coming from ∂ v Q ; this agrees with the orientation given by rotating V (q , δ ) by angle π/2, counterclockwise. We arrange names so the points z R , q , z L come in that order along J w .
Consider the arcs [q , z L ] ⊂ J w , [z L , x R ] ⊂ α, and [x R , p] ⊂ J u . Then has i (α, β) = 1 because β crosses, once, from the right side to the left side of [z L , x R ]. This is the final case, and completes the proof.
.7. One of the four subcases covered by Section 6.6. Here u = w and u = w , so both of Q and Q are once-holed rectangles.