Shellability and Sphericity of Finite Quasi-arc Complexes

Quasi-triangulations of a non-orientable surface were introduced by Dupont and Palesi (J Algebr Comb 42(2):429–472, 2015). The quasi-arc complex provides an intricate description of the combinatorics of these quasi-triangulations. This is the simplicial complex where vertices correspond to quasi-arcs and maximal simplices to quasi-triangulations. We prove that when the quasi-arc complex is finite then it is shellable and, as a consequence, it is homeomorphic to a sphere.


Introduction
The arc complex Arc(S) of a marked orientable surface S was introduced and studied by Harer [7] whilst investigating the homology of mapping class groups of orientable surfaces. In [4,5], Fomin, Shapiro, and Thurston found there is a strong relation between cluster algebras and these arc complexes. They showed that Arc(S) is a subcomplex of the cluster complex (S) associated to the cluster algebra arising from S. Moreover, it was shown by Fomin and Zelevinsky, almost at the birth of cluster algebras, that the cluster complex of a cluster algebra has a polytopal realisation when the complex is finite, see [6]. These polytopes coming from finite type cluster algebras are known as the generalised associahedra. In particular, Arc(n-gon) is the polytope dual to the (n − 3)-dimensional associahedron.
In [3] Dupont and Palesi consider the quasi-arc complex of unpunctured nonorientable surfaces. Imitating the approach in [5] they describe how the 'lengths' of quasi-arcs are related. In doing so they discover what the analogue of a cluster algebra arising from non-orientable surfaces should be. A natural question is to ask what kind of structure the quasi-arc complex has in this setting. Here, in some sense, the marked Möbius strip M n plays the role of the n-gon-being the only non-orientable surface yielding a finite quasi-arc complex.
For n ∈ {1, 2, 3} it is easy to check that the quasi-arc complex Arc(M n ) of the Möbius strip is a polytope, see Fig. 2. However, in general it is difficult to find a polytopal realisation. The main tool used in the orientable case was the associated root system, which is not available here.
In this paper we prove that the finite type complexes are spherical. Our tool will be shellability. (Theorem 4.44). Arc(M n ) is shellable for n ≥ 1.

Main Theorem
Shellability stems from polytopal theory where it turned out to be the missing piece of the puzzle for obtaining the Euler-Poincaré formula. It has subsequently become a well established idea in combinatorial topology and geometry having some useful implications. For instance, Danaraj and Klee showed in [2] that every shellable pseudo-manifold is a PL-sphere. As a consequence, we obtain that the arc complex of a Möbius strip is homeomorphic to a sphere, or, in other words, that all finite quasi-arc complexes are spherical: Corollary (Corollary 4.47). Let X be a finite quasi-arc complex. Then X is spherical.
The paper is organised as follows. In Sect. 2 we recall the work of Dupont and Palesi in [3]. Here we define the quasi-arc complex of a non-orientable surface and discuss why it is a pseudo-manifold, and when it is finite.
In Sect. 3 we firstly define shellability and recall some fundamental results. Next we restrict our attention to the n-gon and to C n,0 -the cylinder with n marked points on one boundary component, and none on the other. In the interest of introducing key ideas of the paper early on, we present a short proof that both Arc(n-gon) and Arc(C n,0 ) are shellable. As a consequence, applying the result of Danaraj and Klee, we rediscover the classical fact of Harer [7] that Arc(n-gon) and Arc(C n,0 ) are PL-spheres. Section 4 is dedicated to proving the shellability of Arc(M n ) and occupies the bulk of the paper.

Quasi-cluster Algebras
This section recalls the work of Dupont and Palesi in [3]. The flip graph of a bordered surface (S, M) is the graph with vertices corresponding to (quasi-)triangulations and edges corresponding to flips. It is well known that the flip graph of triangulations of (S, M) is connected. Moreover, it can be seen that every one-sided closed curve, in a quasi-triangulation T , is bounded by an arc enclosing a Möbius strip with one marked point on the boundary. Therefore, if we perform a quasi-flip at each one-sided closed curve in T we arrive at a triangulation. As such, we get the following proposition.

Corollary 2.9
The number of quasi-arcs in a triangulation of (S, M) is an invariant of (S, M).

Definition 2.10
The quasi-arc complex Arc(S, M) is the simplicial complex on the ground set A ⊗ (S, M) such that k-simplices correspond to sets of k mutually compatible quasi-arcs. In particular, the vertices in Arc(S, M) are the elements of A ⊗ (S, M) and the maximum simplices are the quasi-triangulations.
Together, Corollary 2.9 and Proposition 2.6 prove the following proposition.

Proposition 2.11
Arc(S, M) is a pseudo-manifold without boundary. That is, each maximal simplex in Arc(S, M) has the same cardinality, and each simplex of codimension 1 is contained in precisely two maximal simplices. Moreover, Arc(M n ) has some seemingly nice properties. Figure 2 shows that for n ∈ {1, 2, 3} it is polytopal.

Shellable
Not The ordering of facets F 1 , F 2 , F 3 , F 4 , F 5 is a shelling for the left complex. However, the right complex is not shellable-the problem here is that v ∈ F 1 ∩ F 5 but there is no 1-simplex which both contains the vertex v and lies in the intersection of two (distinct) facets F i and F j

Shellability
In this section we recall some basic facts about shellability, and introduce the fundamental ideas used throughout this chapter. These well known concepts can be found in [1,2,8].
The following proposition is a simple and well-known result. For instance, see [8].

Proposition 3.4
The simplicial join 1 * 2 is shellable if and only if the simplicial complexes 1 , 2 are both shellable.
In particular, Proposition 3.4 tells us that the cone over a shellable complex is itself shellable.

Proposition 3.5 If
= Arc(S, M) then finding a shelling for is equivalent to ordering the set of triangulations T i of (S, M) so that for all k and j < k there exists i < k such that T i is related to T k by a mutation and T j ∩ T k ⊆ T i ∩ T k . Proof Note that triangulations T i of S correspond to maximal simplices in Arc(S, M) and that partial triangulations T i ∩ T j correspond to simplices of Arc(S). Note that must be pure and (dim(T k ) − 1)-dimensional for all k ≥ 2, it follows that B k is the union of (dim(T k ) − 1)-simplices. So we must have that for all j < k there exists i < k such that T i is a mutation away from T k and the partial triangulation Proposition 3.5 motivates Definition 3.6. Definition 3.6 Given a subcollection of triangulations of a surface S call shellable if it admits an ordering of such that for all k and j < k there exists i < k such that T i is related to T k by a mutation and T j ∩ T k ⊆ T i ∩ T k .  Remark 3.9 Note that taking a cone over can be thought of as disjointly adding one particular arc to every triangulation in .
The following proposition is just a special case of Proposition 3.4. The following theorem provides a very useful application of shellability.  The following proposition will help to prove the shellability of Arc(M n ), and is introduced now to cement key ideas. Proposition 3.13 Arc(C n,0 ) is shellable for n ≥ 1.
Proof Consider the collection of triangulations T (C 1 n,0 ) ⊆ T (C n,0 ) containing a loop at vertex 1. Note that by cutting along the loop we get the (n + 1)-gon (and a copy of C 1,0 ) for n ≥ 2. We will prove by induction on n that T (C 1 n,0 ) is shellable. For n = 1 the set T (C 1 1,0 ) = T (C 1,0 ) is trivially shellable. For n = 2 if we cut along the loop we get the triangle and C 1,0 which are both trivially shellable, so indeed T (C 1 2,0 ) is shellable by Proposition 3.4.
Let Block(i) be the set consisting of all triangulations in T (C 1 n,0 ) containing the triangle with vertices (1, 1, i) for some i ∈ [2, n], see Fig. 5.
Note that Block(i) can be equivalently viewed as the disjoint union of triangulations of the i-gon and the (n − i + 2)-gon. Since T (C 1 k,0 ) can be viewed as triangulations of the (k + 1)-gon, then if we assume T (C 1 k,0 ) is shellable for all k < n, Proposition 3.4 tells us that Block(i) is shellable for all i ∈ [2, n]. Let S(Block(i)) denote a shelling of Block(i).
Claim 3.14 The ordering S(C 1 n,0 ) := list 2 i=n S(Block(i)) is a shelling for T (C 1 n,0 ). Proof of Claim 3.14 Let S precede T in the ordering S(C 1 n,0 ). Then T ∈ Block(k) and S ∈ Block( j) for j ≥ k. If j = k then, because S(Block(k)) is a shelling for Block(k), there exists γ ∈ T such that μ γ (T ) precedes T in the ordering, and T ∩S ⊆ T ∩μ γ (T ). We may therefore assume j > k. In this case, the arc γ = k, 1 ∈ T is not compatible with the arc 1, j ∈ S, so γ / ∈ S. Hence T ∩ S ⊆ T ∩ μ γ (T ). By Proposition 3.5 all that remains to show is that μ γ (T ) occurs before T in the ordering.
Note that we will have a triangle in T with vertices Similarly we can shell T (C i n,0 ) in the same way for all i ∈ [1, n]. Denote a shelling by S(C i n,0 ). Combining these S(C i n,0 ), as described in Claim 3.15 below, we get a shelling for Arc(C n,0 ), which completes the proof of the lemma. Proof Let S precede T in the ordering S(Arc(C n,0 )). Then T ∈ S(C k n,0 ) and S ∈ S(C j n,0 ) for 1 ≤ j ≤ k. Since S(C k n,0 ) is a shelling we may assume j < k. There will be a triangle in T with vertices (k, k, x) for some x ∈ [1, n]\{k}. Fig. 7.
Moreover, the way we constructed the shelling S(C k n,0 ) in Claim 3.14 means that μ γ (T ) precedes T in the ordering. See Fig. 8.
Applying Theorem 3.12 we rediscover the classical result of Harer [7].
then, taking γ to be the loop at k, we get that T := μ γ (T ) precedes T in the ordering S(Arc(C n,0 ))

Shellability of Arc(M n )
In Sect. 3 we achieved shellability of a complex by grouping facets into blocks and finding a 'shelling order' in terms of these blocks. The task was then simplified to finding a shelling of the blocks themselves. Here we essentially follow the same strategy twice. However, on the second iteration of the process we require a specific shelling of the blocks since in general an arbitrary shelling would not suffice.
consist of all triangulations of M n (i.e., no quasitriangulations containing a one-sided curve).

Definition 4.2 Let γ be an arc in
(Informally, a c-arc is an arc that necessarily passes through the cross-cap). Let (i, j) denote a c-arc with endpoints i and j.

Definition 4.4
Let γ be an arc in T ∈ T (M • n ) that is not a c-arc. Call γ a bounding arc (b-arc) if the flip of γ , with respect to T , is a c-arc. Note that, unlike a c-arc, the notion of γ being a b-arc depends on which triangulation γ is in (Fig. 9).
Proof Let S precede T in the ordering. Then S ∈ Block( j) and T ∈ Block(k) where j ≥ k. In particular, T ∈ S( (k) is a shelling we may assume I 1 = I 2 . Suppose that every b-arc in T is also an arc in S. Then I 2 ⊆ I 1 , and since |I 1 | ≤ |I 2 | this implies I 1 = I 2 . So we may assume there is at least one b-arc γ ∈ T that is not Fig. 11 Since x / ∈ I 1 , and the b-arc γ ∈ T flips to a c-arc with endpoint x, then μ γ (T ) precedes T in the ordering an arc in S. Since γ / ∈ S, T ∩ S ⊆ T ∩ μ γ (T ). Moreover, since γ is a b-arc, μ γ (T ) ∈ Block(k + 1). Hence μ γ (T ) precedes T in the ordering, see Fig. 11.
The idea behind Lemma 4.5 is that we are decomposing T (M • n ) into blocks, and ordering these blocks. The ordering is chosen in such a way that if we manage to individually shell the blocks themselves, we will have a shelling of T (M • n ). Figure 12 shows the block structure of T (M • 3 ). In particular, we realise that to shell a block it is sufficient to find a shelling of T (M ⊗ n ). We will split this into two cases: n even and n odd.

Shellability of T (M ⊗ n ) for Even n
In the same vein of our approach thus far, we show T (M ⊗ n ) admits a shelling by ordering smaller blocks of triangulations. In particular, these blocks will be categorised by which diagonal arcs they contain.   Fig. 13.
and j ∈ [1 + n 2 , 1]. Define the length of γ as follows: Remark 4.9 Note that the length of a diagonal arc is n 2 + 1. For a triangulation T ∈ D n {(1, n 2 +1)} we shall see that, excluding the diagonal arc (1, n 2 + 1), the arcs in T will have length all less than n 2 + 1, or all greater than n 2 + 1 (Fig. 14).  . Moreover, for any c-arc γ = (1, n 2 + 1) in T we have the following: )} will contain either the c-arc (2, n 2 + 1) or the c-arc (1, n 2 + 2). Assume the c-arc (2, n 2 + 1) is in T . We will show, by induction on i, the c-arc of maximal length in T with endpoint i ∈ [2, n 2 + 1] must be the c-arc (i, x) where x ∈ [ n 2 + 1, n 2 + i − 1]. Let γ be the c-arc in T of maximal length with endpoint 2. Let j be the other endpoint of γ and suppose for a contradiction j ∈ [ n 2 + 2, 1]. Since (2, n 2 + 1) ∈ T then, as T is a c-triangulation, (2, y) ∈ T for every y ∈ [ n 2 + 1, j]. In particular β := (2, n 2 + 2) ∈ T -which contradicts T ∈ D n {(1, n 2 +1)} . See Fig. 16. By induction, the c-arc α of maximal length in T with endpoint i − 1 is the c-arc . Let γ be the c-arc in T of maximal length with endpoint i. Let j be the other endpoint of γ and for a contradiction suppose j ∈ [ n 2 + i, 1]. But by the maximality of α there will be a c-arc (i, y) ∈ T for all y ∈ [x, j] which contradicts β := (i, n 2 + i) / ∈ T -see Fig. 17. Hence j ∈ [ n 2 + 1, n 2 + i − 1] and l(γ ) < n 2 + 1. In particular, T ∈ T (X n 1 ). If we supposed (1, n 2 + 2) was an arc in T , then an analogous argument shows that T ∈ T (X 2 ) and l(γ ) > n 2 + 1.
The following two definitions are motivated by Lemma 4.11.

Definition 4.12
Consider a c-arc γ in a triangulation in T (X n 1 ). If l(γ ) = n 2 then call γ a max arc.   Proof It follows from the fact that, excluding the c-arc (1, n 2 + 1), the maximal length of any c-arc in S ∈ T (X n 1 ) is strictly less than n 2 + 1, and the minimal length of any c-arc in T ∈ T (X n 2 ) is strictly greater than n 2 + 1.

Corollary 4.15
The triangulation T max in Fig. 18 is the unique triangulation in T (X n 1 ) such that γ ∈T max l(γ ) is maximal. The triangulation T min is the unique triangulation Collapse quadrilateral and relabel marked points Fig. 19 An illustration explaining why triangulating the region between two max arcs, so that no other max arcs are present, is equivalent to T (X 2k 1 ) for some k in T (X n 2 ) such that γ ∈T max l(γ ) is minimal. More explicitly, Proof Consider the partial triangulation P max consisting of all the maximal length c-arcs (i, n 2 +i −1) , i ∈ [2, n 2 +1], which can appear in a triangulation in T (X n 1 ). P max cuts M n into (2 triangles and) quadrilaterals bounded by the two boundary segments [i, i + 1], [ n 2 + i − 1, n 2 + i] and the two c-arcs (i, n 2 + i − 1), (i + 1, n 2 + i), i ∈ [3, n 2 ]. Let T be a triangulation in T (X n 1 ) containing P max . Notice that (i, n 2 + i) / ∈ T by definition of D n {(1, n 2 +1)} , hence (i + 1, n 2 + i − 1) ∈ T for every i ∈ [3, n 2 + 1]} and so T = T max . Moreover, since l(i + 1, n 2 + i − 1) = n 2 − 1 then T max is the unique triangulation in T (X n 1 ) such that γ ∈T l(γ ) is maximal. Analogously we get the result regarding unique minimality of T min .
Consider a partial triangulation of M n consisting of the diagonal arc (1, n 2 + 1) and two max arcs. Cutting along these max arcs we will be left with two regions. Let R be the region that does not contain the diagonal arc (1, n 2 + 1). Note R will contain 2k marked points for some k ∈ {2, . . . , n 2 }.

Lemma 4.16 The set of c-triangulations of R such that no max arcs occur in R is equivalent to T (X
Proof Collapse the quadrilateral (1, 2, n 2 + 1, n) to a c-arc and relabel marked points as shown in Figure 19. Max arcs in R correspond to diagonal arcs in R . Furthermore, up to a relabelling of vertices, triangulating R so that no diagonal arcs occur in the triangulation is precisely triangulating X     Proof If n ∈ {2, 4} then I = ∅ and there is nothing to prove. So assume n ≥ 6.
Suppose m = (i, n 2 + i − 1) ∈ I is not in the triangulation T . We will show there exists a c-arc strictly contained between the endpoints of m.
. Moreover, by maximality of (i, x), (i + 1, x) ∈ T . So indeed there is a c-arc in T strictly contained between the endpoints of m, see Fig. 22.
Of the c-arcs in T that are strictly contained between the endpoints of m, let γ = ( j 1 , j 2 ) be an arc of minimum length. We will show that γ is upper mutable.

Lemma 4.23
There exists an upper shelling for T (X n 1 ). Denote this by S(X n 1 ).
As |J 1 | ≤ |J 2 | and J 1 = J 2 there is a max arc m in S that is not in T . By An analogous argument proves the following lemma.

Lemma 4.25 There exists a lower shelling for T (X n 2 ). Denote this by S(X n 2 ). Lemma 4.26 For any T ∈ T (X n 1 )\{T max }, T max is connected to T by a sequence of lower mutations.
Proof By Lemma 4.22 we can keep performing mutations on upper mutable arcs until we reach a triangulation containing every max arc. By Corollary 4.15 the only triangulation in T (X n 1 ) that contains every max arc is T max . Hence T is connected to T max by a sequence of upper mutations. Equivalently, T max is connected to T by a sequence of lower mutations. • T ∈ T (X n 1 ) and γ is upper mutable. • T ∈ T (X n 2 ) and γ is lower mutable. • γ mutates to a diagonal c-arc.

Lemma 4.28 Let T ∈ D n
{(1, n 2 +1)} and let P T be the partial triangulation of M n consisting of all the special mutable arcs in T . Then any triangulation containing P T cannot contain the diagonal c-arc (i, n 2 + i), where i ∈ {2, . . . , n 2 }. Proof Assume T ∈ T (X n 1 ). An analogous argument works if T ∈ T (X n 2 ). We prove the lemma via induction on the upper shelling order of T (X n 1 ). The first triangulation in the upper shelling ordering is T max . The special mutable arcs in T max are (i, n 2 + i − 2), i ∈ [3, n 2 + 1]. However, the c-arc (i, n 2 + i − 2) is not compatible with the diagonal c-arc (i − 1, n 2 + i − 1). And so ranging i over 3, . . . , n 2 + 1 proves the base inductive case. Let γ be a lower mutable arc in a triangulation T ∈ T (X n 1 ). By Lemma 4.26, to prove the lemma it suffices to show that the special mutable arcs in μ γ (T ) prevent the same diagonal c-arcs as the special mutable arcs in T . Let β 1 , β 2 be the c-arcs containing γ in a quadrilateral. See Fig. 24.
The arcs β 1 and β 2 may be special mutable in T but in μ γ (T ) they definitely won't be. The implication of this is that β 1 and β 2 may be c-arcs in P T , and prevent certain diagonal arcs, but β 1 , β 2 / ∈ P μ γ (T ) so μ γ needs to make up this difference. Indeed, it does make up the difference as the diagonal arcs not compatible with either β 1 or β 2 are precisely the diagonal arcs not compatible with μ γ .

Lemma 4.29
In each c-triangulation T of M n there is at least one diagonal arc.
Proof Let us assume, for a contradiction, that there is no diagonal arc in T . Without loss of generality, we may assume that the c-arc connected to 1, of maximum length, is γ = (1, j 1 ) for some j 1 ∈ [1, n 2 ]. (Otherwise just flip the picture.) Let γ 2 = (2, j 2 ) be the c-arc of maximum length in T that is connected to 2. If j 2 > n 2 then by maximality of γ 1 there is a c-arc (2, n 2 ). Hence, j 2 ∈ [ j 1 , n 2 + 1]. Inductive reasoning shows that the c-arc connected to j 1 − 1 in T , of maximum length, is γ j−1 = ( j − 1, x) for some x ∈ [ j, n 2 + j 1 − 2]. However, then by the maximality of γ j−1 we must have ( j 1 , n 2 + j 1 ) ∈ T . This gives a contradiction, and so the lemma is proved.  Proof Let T, S ∈ T (M ⊗ n ) and suppose S precedes T in the ordering. Then T ∈ O(D n I 1 ) and S ∈ O(D n I 2 ) for some I 1 , I 2 ∈ P(K) where |I 1 | ≤ |I 2 |. If there is a region R in T that contains a special mutable arc γ , such that γ is not an arc in S, then μ γ (T ) precedes T in the ordering and S ∩ T ⊆ μ γ (T ) ∩ T .
So suppose that for every region R of T all special mutable arcs in that region are also arcs in S. Then by Lemma 4.28, I 2 ⊆ I 1 . Since |I 1 | ≤ |I 2 | we must have I 1 = I 2 .
If O(D n I ) was a shelling for D n I then the proof would be finished. However, in general, it is not. To understand how we should proceed let us consider D n Let T be the first triangulation of S(X 2 ) and let S ∈ S(X 1 ). Corollary 4.14 tells us that the only arc T and S share in common is the diagonal c-arc (1, n 2 + 1). If n = 2 then O(D 2 {(1,2)} ) = S, T is a shelling for D n {(1,2)} . However, if n ≥ 4 then there are at least 4 arcs in S and T . Hence, μ γ (T ) / ∈ S(X n 1 ) for any arc γ in T , since μ γ (T ) and S can share at most two arcs in common.
However, as n ≥ 4 the first triangulation of S(X n 2 ) contains (at least one) arc γ that mutates to a diagonal c-arc. And so μ γ (T ) contains more diagonal c-arcs than T . Hence μ γ (T ) precedes T in the overall ordering for T (M ⊗ n ).

Shellability of T (M ⊗ n ) for Odd n
In the even case diagonal arcs were a key ingredient in the shelling of T (M ⊗ n ). We will see 'diagonal triangles' play the same role in the odd case. For the duration of this section we fix n = 2k + 1.

Definition 4.32
A triangle in M n comprising of two c-arcs (i, i + k), (i, i + k + 1) and the boundary segment (i + k, i + k + 1) for some i ∈ [1, n] is called a diagonal triangle (d-triangle). Additionally, call i the special vertex of the d-triangle.
Proof Given T ∈ D n {(k+1,n,1)} we will prove by induction on i ∈ [1, k + 1] that there is no c-arc in T , with endpoint k + i, of length greater than k + 1. For i = 1 this trivially holds. Assume the statement holds true for i and let γ = (x, k + i) and β = (y, k + i + 1) be the c-arcs in T of maximal length which have endpoints k + i and k + i + 1, respectively. Note that x ≤ y and, by induction, we know Proof We will show that if there are two d-triangles there must in fact be a third. Additionally we will show the collection of (legitimate) triangulations in any region cut out in between the three d-triangles is equivalent to D m And applying induction on this we will have the lemma proved. Suppose there are at least two d-triangles in a c-triangulation T . Without loss of generality we may assume the two d-triangles (k +1, 2k +1, 1) and (i, i +k, i +k +1) are in T , for some i ∈ [1, k]. See Fig. 29.
Consider the d-triangles (k + 1, 2k + 1, 1) and (i, i + k, i + k + 1) with special vertices k + 1 and i, respectively. Let R be the region bounded by the c-arcs (1, k + 1), (i, i + k) and the boundary segments   Proof We follow the same idea as in Lemma 4.28. Namely, we will prove the lemma by induction on the shelling order of S(D n {(k+1,n,1)} ).
Let T 1 be the first triangulation in the shelling. Note γ i = (i, k + i − 1) is a special mutable c-arc in T 1 for every i ∈ [2, k + 1]. Moreover γ i is not compatible with the c-arc (i − 1, k + i). Hence there is no d-triangle with special vertex i − 1 or k + i, i ∈ [2, k + 1]. This proves the base inductive case.
Let T ∈ D n {(k+1,n,1)} . To prove the lemma by induction it suffices to show that for any lower mutable arc γ ∈ T , the d-triangles incompatible with P T are precisely the d-triangles incompatible with P μ γ (T ) .
So let γ be a lower mutable arc in T . Let β 1 , β 2 be the c-arcs of the quadrilateral containing γ . See Fig. 32. Note that β 1 and β 2 could be upper mutable in T , but they will definitely not be upper mutable in μ γ (T ). Analogously to the proof of Lemma 4.28, to prove the lemma it suffices to show μ γ is incompatible with all the d-triangles incompatible with either β 1 or β 2 . This follows from the fact that a c-arc α = (x, k + y) of length less than k is incompatible with d-triangles with special vertex z ∈ [y, x − 1] ∪ [k + y + 1, k + x]. See Fig. 33.
An analogous argument to Lemma 4.29 proves the following lemma.  . Proof Let T, S ∈ T (M ⊗ n ) and suppose S precedes T in the ordering. Then T ∈ S(D n I 1 ) and S ∈ S(D n I 2 ) for some I 1 , I 2 ∈ P(K) where |I 1 | ≤ |I 2 |. If there is a region R in T that contains a special arc γ , such that γ is not an arc in S, then μ γ (T ) precedes T in the ordering and S ∩ T ⊆ μ γ (T ) ∩ T . So suppose that for every region R of T all special arcs in that region are also arcs in S. Then by Lemma 4.39 I 2 ⊆ I 1 . Since |I 1 | ≤ |I 2 | we must have I 1 = I 2 . And since S(D n I 1 ) is a shelling for D n I the claim is proved.  Proof Let C consist of all quasi-triangulations of M n containing the one-sided closed curve. Cutting along the one-sided curve in M n we are left with the marked surface C n,0 . Therefore the induced simplicial complex of C is the cone over Arc(C n,0 ). Arc(C n,0 ) is shellable by Proposition 3.13 so Proposition 3.4 tells us C is also shellable. Let S(C) denote a shelling for C. Let S(M • n ) be a shelling of T (M • n ) guaranteed by Lemma 4.43. Coupling these two shellings, as described in Claim 4.45 below, provides us with a shelling of Arc(M n ), and this completes the proof of the theorem. . Proof Suppose S, T ∈ S(M n ) and S precedes T in the ordering. Without loss of generality we may assume S ∈ S(M • n ) and T ∈ S(C). Since T contains the one-sided closed curve γ , and γ / ∈ S then S ∩ T ⊆ μ γ (T ) ∩ T . Moreover, μ γ (T ) ∈ S(M • n ) so precedes T in the ordering.

Remark 4.46
Recall that A denotes the induced simplicial complex of a set of (quasi) triangulations A. Since ∂ T (M • n ) ∼ = Arc(C n,0 ) and C ∼ = Cone(Arc(C n,0 )) then Arc(M n ) is the cone over the boundary of T (M • n ) . Corollary 4.47 Let X be a finite quasi-arc complex. Then X is spherical.
Proof X is a finite shellable pseudo-manifold without boundary by Proposition 2.11, Theorems 2.12 and 4.44. Therefore X is spherical by Theorem 3.12.