ARCS ON SPHERES INTERSECTING AT MOSTpprzytyc/Thesis.pdf · CHRISTOPHER SMITH AND PIOTR PRZYTYCKIy Abstract. Let pbe a puncture of a punctured sphere, and let Qbe the set of all other

ARCS ON SPHERES INTERSECTING AT MOSTTWICE

CHRISTOPHER SMITH AND PIOTR PRZYTYCKI†

Abstract. Let p be a puncture of a punctured sphere, and letQ be the set of all other punctures. We prove that the maximalcardinality of a set A of arcs pairwise intersecting at most once,which start at p and end in Q, is |χ|(|χ| + 1). We deduce thatthe maximal cardinality of a set of arcs with arbitrary endpointspairwise intersecting at most twice is |χ|(|χ|+ 1)(|χ|+ 2).

1. Introduction

Let S be a punctured sphere with Euler characteristic χ < 0. Weconsider collections of essential simple arcs A between punctures on Ssuch that the arcs in A are pairwise non-homotopic.

Theorem 1.1. The maximal cardinality of a set A of arcs pairwiseintersecting at most twice is

|χ|(|χ|+ 1)(|χ|+ 2).

Remark 1.2. We conjecture that the formula also holds for any con-nected, oriented surfaces with χ < 0. The proof we provide here,however, applies only to the case of spheres.

We will reduce Theorem 1.1 to the following:

Theorem 1.3. Let p be a puncture of S, and Q be the set of all otherpunctures. The maximal cardinality of a set A of arcs pairwise inter-secting at most once, which start at p and end in Q, is

|χ|(|χ|+ 1).

Previous Results. Initially, cardinality questions were asked aboutsets C of essential and nonperipheral simple closed curves on an arbi-trary surface. Juvan, Malnic and Mohar proved that given k boundingthe number of intersections of any two curves in C, there is an upperbound on |C| [JMM96]. An upper bound polynomial in |χ| was ob-tained in [Prz15, Cor 1.6]. Farb and Leininger asked for the preciseasymptotics in the case where k = 1. For S a torus, we have |C| ≤ 3.For S a closed genus 2 surface, Malestein, Rivin, and Theran provedthat the maximal cardinality of C is 12 [MRT14]. They also produced a

† Partially supported by NSERC, National Science Centre DEC-2012/06/A/ST1/00259, and UMO-2015/18/M/ST1/00050.

1

2 CHRISTOPHER SMITH AND PIOTR PRZYTYCKI†

lower quadratic and exponential upper bound on maximal |C| in termsof |χ|, which was obtained independently by Farb and Leininger [Lei11],and later improved to a cubic upper bound in [Prz15, Thm 1.4].

It seems that corresponding questions about arcs are easier to tackle.By [Prz15, Thm 1.2], the maximal cardinality of a setA of arcs pairwiseintersecting at most once is 2|χ|(|χ|+1). By [Prz15, Thm 1.7], given twodistinguished (but not necessarily distinct) punctures p, p′, the maximalcardinality of a set A of arcs pairwise intersecting at most once whichstart at p and end at p′ is 1

2|χ|(|χ|+ 1). Theorem 1.3 is a very useful

addition to this, as it allows us to adapt another proof from [Prz15]to prove the twice intersecting case in Theorem 1.1. Note that ourTheorem 1.3 is much more difficult to prove that any of the theoremsin [Prz15].

Organization. In Section 2 we construct collections of arcs thatshow the bounds in Theorems 1.1 and 1.3 are sharp. We also pro-vide a counter-example showing that Theorem 1.3 does not hold for a3-punctured torus. In Section 3 we show how Theorem 1.3 implies The-orem 1.1. In Section 4 we provide the bulk of the definitions and proofof Theorem 1.3 up to two propositions which are proven in Sections 5and 6.

2. Examples

First, we show an example of a maximal collection of arcs satisfyingthe conditions of Theorem 1.3. We model the n-punctured sphere as adisc whose boundary is the distinguished puncture p, and whose otherpunctures Q lie on a smaller circle on the interior of the disc (seeFigure 1). Note that |Q| = n − 1. From each puncture q ∈ Q, wedraw n − 2 disjoint arcs that are straight line segments dividing thedisc into n − 2 regions, each containing a puncture in Q − {q}. Weobtain (n − 1)(n − 2) arcs which pairwise intersect at most once. As|χ| = n− 2, we have the predicted maximal collection.

Figure 1. On the 5-punctured sphere, for each q ∈ Qwe draw three disjoint arcs, for a total of 12 arcs.

ARCS ON SPHERES INTERSECTING AT MOST TWICE 3

Now we provide a counterexample demonstrating that Theorem 1.3does not hold in the case of a 3-punctured torus. We model the torusas a regular octagon with edges identified as shown in Figure 2. Afterthis identification we have three punctures, {a, b, p} with Q = {a, b},and we consider arcs connecting a or b to p.

After the identifications in Figure 2, two of the edges connect p tothe puncture a; these will be the first two arcs in our collection. Tothese we add the 8 diagonals which run from the vertices labelled a orb to one of the vertices labelled p. Finally we add the three depictedarcs to the collection, for a total of 13 arcs. Since |χ| = 1, Theorem 1.3states that the maximal cardinality of such a collection of arcs shouldonly be 12, showing that this theorem does not hold in general fornon-spheres.

p

a b

a

p

ab

a

Figure 2. The 11th to 13th arcs lying on a 3-puncturedtorus. Notice that no diagonal from either a or b to ptwice intersects any of these three arcs.

Finally, we would like to show that the bound in Theorem 1.1 issharp. To this end consider an n-punctured sphere constructed in thefollowing fashion. Let P be an n-gon, and glue two copies of P togetheralong their edges in the obvious fashion. Remove the vertices to obtainthe n punctures. We will think of the two copies of P as the front andback faces of the sphere.

Our collection A will consist of three types of arcs. First, let E bethe set of all n edges along which the polygons are glued. Second, let

D be the set of the n(n−3)2

diagonals between vertices on each of thetwo faces, for a total of n(n− 3) arcs of this form.

Finally, we form a set C of arcs which cross between faces of thesphere. For each edge e of the polygons, consider the midpoint m of e,and every ordered pair (u, v) of punctures outside e, with u and v notnecessarily distinct. For each such choice of e and (u, v), we includein C the arc whose first half is the straight line segment from u to m onthe front face, and whose second half is the straight line segment fromm to v on the back face. For each edge e, there are (n− 2)2 such arcs,and we have n edges, giving us a total of n(n− 2)2 arcs of this form.


Figure 3. An example of two of the arcs in C on a 6-punctured sphere. The dashed lines indicate where thearc lies on the back face.

u u

a

b

v v

c cm

Figure 4. The disc obtained by cutting the model ofthe 5-punctured sphere along each edge other than e.These arcs necessarily intersect, so cannot be homotopic.

Recalling that n = |χ|+ 2 we obtain that n+n(n− 3) +n(n− 2)2 =|χ|(|χ|+ 1)(|χ|+ 2).

It remains to verify that these arcs satisfy the conditions of Theo-rem 1.1. It is clear that each of the edges in E and the diagonals in Dare homotopically distinct from one another, and as they are straightline segments on their respective faces, they can pairwise intersect atmost once.

Consider the arcs in C. First, we wish to show that arcs in C arehomotopically distinct from each other, and from arcs in D or E . Asarcs in C each necessarily intersect exactly one of the edges, the latteris clear. If we have two arcs in C with the same (u, v) but differingmidpoint m, they must be homotopically distinct, because we knowthat these arcs must intersect the edge e, and do not intersect anyother edge.

Now consider the case where two arcs in C have u, v, and m in com-mon. Consider Figure 4. We cut along the the edges other than e toobtain a space homotopic to a disc. These arcs intersect exactly oncewithout creating any half bigons, and hence cannot be homotopic. Soall arcs in C,D, and E are homotopically distinct.

Finally, we wish to show that arcs in C pairwise intersect at mosttwice with arcs in C ∪ D ∪ E . Note that the arcs in D are diagonals,each lying on exactly one face of S, and the arcs in C each consist oftwo straight line segments, one lying on each face. As such, two arcsin C may pairwise intersect at most twice (at most once on each face),and arcs in D and E pairwise intersect at most once with arcs in C.


3. Reduction to Once Intersecting Case

In this section we adapt a proof from [Prz15] and use Theorem 1.3,whose proof is postponed, in order to obtain the upper bound in The-orem 1.1. We begin with some definitions.

Definition 3.1 ([Prz15, Def 2.1]). A tip τ of A is a pair (α, β) oforiented arcs in A starting at the same puncture and consecutive. Thatis to say that there is no other arc in A issuing from this puncture inthe clockwise oriented cusp sector from α to β.

Let τ = (α, β) be a tip and let Nτ be an open abstract ideal hy-perbolic triangle with vertices a, t, b. The tip τ determines a uniquelocal isometry ντ : Nτ → S sending ta to α and tb to β and mapping aneighbourhood of t to the clockwise oriented cusp sector from α to β.We call ντ the nib of τ .

Proposition 3.2. Suppose that the arcs in A pairwise intersect at mosttwice. Let ν : N = tτNτ → S be the disjoint union of all the nibs ντ .Then for each point p ∈ S, the preimage ν−1(p) consists of at most(|χ|+ 1)(|χ|+ 2) points.

In order to prove Proposition 3.2, we require some further resultsfrom [Prz15].

Definition 3.3 ([Prz15, Def 2.4]). Let n ∈ Nτ be a point in the domainof a nib. The slit at n is the restriction of ντ to the geodesic ray in Nτ

joining t with n.

Lemma 3.4 ([Prz15, Lemma 2.5]). A slit is an embedding.

Lemma 3.5 ([Prz15, Lemma 3.2]). Suppose that the arcs in A pairwiseintersect at most k ≥ 1 times. If for distinct n, n′ ∈ N we have ν(n) =ν(n′), then the images in S of slits at n, n′ intersect at most k−1 timesoutside the endpoint.

Proof of Proposition 3.2. Let p ∈ S be an arbitrary point. Let S ′

be the sphere obtained from S by introducing a puncture at p. ByLemma 3.5 we know that for any two points n, n′ ∈ ν−1(p), the imagesin S of the slits at n and n′ intersect at most once. Therefore on S ′, wehave a collection of simple arcs which start at p and end at puncturesof S, and these arcs pairwise intersect at most once.

If χ is the Euler characteristic of S, then χ− 1 is the Euler charac-teristic of S ′, and so by Theorem 1.3 the maximal size of a collectionof such slits is |χ− 1|(|χ− 1|+ 1) = (|χ|+ 1)(|χ|+ 2). Since each pointin the preimage ν−1(p) contributes an arc to this collection, this givesus the desired bound on the size of ν−1(p). �

Proof of Theorem 1.1. Each arc in A is the first arc of exactly two tips,depending on its orientation, so the area of N is 2|A|π. The area of the


punctured sphere S is 2π|χ|. By Proposition 3.2, the map ν : N → Sis at most (|χ|+ 1)(|χ|+ 2)-to-1, and so we have

2|A|π ≤ 2π|χ|(|χ|+ 1)(|χ|+ 2).

�

4. Minimal Fish, and the Once Intersecting Case

For the remainder of the paper, let p be the distinguished punctureof the punctured sphere S. We will represent S as a closed disc Dwith boundary associated with p, and with a collection of remainingpunctures Q in the interior. Let A be as stated in Theorem 1.3. We willconsider the arcs to be oriented from a puncture in Q to the puncture p.We assume without loss of generality that arcs in A are in minimalposition.

If for each puncture q ∈ Q, the arcs in A beginning at q do notintersect, we see that Theorem 1.3 is easy to prove. There can be atmost |Q|−1 arcs inA from each puncture on D (refer back to Figure 1),and there are |Q| punctures on the disc, giving us |Q| · (|Q| − 1) =|χ| · (|χ| + 1) arcs, recalling |χ| = |Q| − 1. It makes sense then tobegin by considering some properties of arcs beginning at the samepuncture q which intersect.

α

β

pq

Figure 5. A fish (α, β).

Definition 4.1. Let q ∈ Q. A fish F with nose q is an unordered pairof arcs α, β ∈ A from q to p which intersect.

The arcs α and β divide the disc D into three regions. The headof F , denoted by head(F ), is the region adjacent to q, but not to p.The tail of F , denoted by tail(F ), is the region adjacent to p, but notto q. The remaining region is thought of as being outside the fish.

We denote by h(F ) the subset of the punctures Q which are inhead(F ), and by t(F ) the set of punctures which are in tail(F ). Wedo not consider the nose of the fish to be contained in the head. Ifu ∈ t(F ), then we say that F is a u-fish.


α β

γ

r s q

Figure 6. A minimal fish (α, β) and a non-minimal fish(α, γ). In this picture we have r ∼q s.

Definition 4.2. For each puncture q ∈ Q, consider the collection ofarcs in A beginning at q. We assign these arcs a cyclical orderingbased on their intersection points with a sufficiently small circle aboutthe puncture q.

A fish (α, β) with nose q is minimal if there is no arc from q whichbegins in head(F ). Observe that all minimal fish are pairs of consecu-tive arcs, but there are exceptional cases where intersecting consecutivearcs may not be minimal fish. For example, in Figure 6, the arcs α andγ are consecutive arcs which intersect, but due to β they do not forma minimal fish.

Denote by F q the set of minimal fish with nose q, and by Fq the setof minimal q-fish.

Definition 4.3. Let q ∈ Q. The equivalence relation ∼q on Q − {q}is the equivalence relation generated by setting r ∼q s whenever thereis a q-fish F (not necessarily minimal) with nose r such that s ∈ h(F ).(See Figure 6.) Define cq to be the number of equivalence classes of∼q.

Recall our previous discussion that the interesting case for provingTheorem 1.3 was when arcs from the same puncture q intersect. Clearlyif we allow arcs from q to intersect one another, we will be able toinclude more arcs from q in our collection. We can think of this as apositive contribution to |A| at q. What we will show is that this positivecontribution is balanced out by a matching negative contribution.

Each of these extra arcs from q will cause punctures in Q − {q} tolie in the tails of fish with nose q. If u lies in the tail of some fish, thenthis will reduce the number of equivalence classes of ∼u. In turn, thiswill reduce the number of arcs in our collection which may begin at u.

The following lemma, propositions, and corollary can be thoughtof as proving that each positive contribution at one puncture mustbe matched by a negative contribution at some combination of otherpunctures.


We postpone the proofs of Propositions 4.4 and 4.5 to Sections 5and 6 respectively. Here we start by deducing Corollary 4.6 from Propo-sition 4.5. We then show that Proposition 4.4 and Corollary 4.6 implyTheorem 1.3.

Proposition 4.4. Let q ∈ Q. Then

|χ| ≥ |Fq|+ cq.

Proposition 4.5. Let q ∈ Q. If A contains kq arcs from p to q, then

kq ≤ |χ|+∑F∈Fq

|t(F )|.

Corollary 4.6. Under the hypotheses of Proposition 4.5,

kq ≤ cq +∑F∈Fq

|t(F )|.

To prove this corollary, we first need the following lemma.

Lemma 4.7. For each q-fish F , no arc from q to p may pass throughthe region head(F ).

Proof. Let F = (α, β), and let γ ∈ A be an arc from q to p. Observethat q ∈ tail(F ) by hypothesis. For the arc γ to pass through head(F ),it must first exit the tail, then enter the head, and finally exit the head,before arriving at p. This would cause a total of three intersectionsbetween γ and either α or β, which is a contradiction. Note that evenif γ passes directly through the intersection point α ∩ β from tail(F )to head(F ), we still have two intersections there, and one more uponleaving head(F ). �

Proof of Corollary 4.6. By Lemma 4.7, no arcs from q can pass throughhead(F ) for any q-fish F . Consider the modified disc D′ obtained byremoving head(F ) from D for each F ∈ Fq, and let D′ be the connectedcomponent of the resulting surface which contains the boundary p.From the definition of ∼q we see that this D′ will be homeomorphicto an at most (cq + 2)-punctured sphere. One way to see this is toobserve that the punctures of each equivalence class of ∼q become asingle puncture in D′, and the punctures q and p each remain as well,giving at most cq + 2 total punctures. From Proposition 4.5, it followsthat

kq ≤ |χ(D′)|+∑F∈Fq

|t(F )| ≤ cq +∑F∈Fq

|t(F )|.

�

Proof of Theorem 1.3. We begin by summing the inequalities obtainedin Proposition 4.4 and Corollary 4.6 over Q. Recall that |Q| = |χ|+ 1.From Proposition 4.4 we get

|χ|(|χ|+ 1) =∑q∈Q

|χ| ≥∑q∈Q

|Fq|+∑q∈Q

cq.


From Corollary 4.6 we get

|A| ≤∑q∈Q

cq +∑q∈Q

∑F∈Fq

|t(F )|.

Putting these together we obtain

|χ|(|χ|+ 1) ≥∑q∈Q

|Fq| −∑q∈Q

∑F∈Fq

|t(F )|+ |A|.

Now observe that the first two terms on the right hand side are ac-tually counting the same objects. The first term is counting for eachpuncture q, the number of minimal fish with q in their tail. The secondterm is counting for each minimal fish, the number of punctures in thatfish’s tail. These two values therefore cancel out, and we obtain

|χ|(|χ|+ 1) ≥ |A|. �

5. Equivalence Classes

This section is devoted to the proof of Proposition 4.4. We will firstneed one supporting lemma. We keep the notation from Section 4.

Lemma 5.1. Let u ∈ Q. Then there exists a puncture v ∈ Q, v 6= usuch that v is not the nose of a u-fish.

Proof. Suppose toward contradiction that for every puncture qi ∈ Q− {u}there is a fish Fi ∈ F qi with u ∈ t(Fi). Every Fi has at least onehead puncture which, by hypothesis, is also the nose of a u-fish, so wecan choose a finite sequence of such punctures {q1, . . . , qk} such thatqi+1 ∈ h(Fi) and q1 ∈ h(Fk). Furthermore, we may choose that se-quence to be one of minimal length. It can be easily seen that k 6= 2(see Figure 7), so we have k ≥ 3. As we are dealing with cycles, it willbe convenient to write qk+1 = q1.

Figure 7. Two fish cannot produce such a sequence, astheir tails could not intersect.

Since the sequence {qi} is of minimal length, qi /∈ h(Fj) for allj 6= i− 1. Note that if we remove all punctures other than u and the qifrom D, the punctures {qi} and the fish {Fi} still give us a minimal


q1

q2

q3

u

Figure 8. The arcs a1 and a2 drawn as dashed lines.

length sequence of u-fish with the above properties, and so we may as-sume without loss of generality that there are no other punctures on D.Since u cannot be in the head of a u-fish, we have h(Fi) = {qi+1} forall i.

For each i, let ai be an arc lying in head(Fi)−head(Fi+1) connectingqi to qi+1 (see Figure 8), and in minimal position with respect to thearcs of A. Such an arc exists, as head(Fi) − head(Fi+1) is connected.We claim that the ai are disjoint, and therefore form an embeddedpolygon P .

To see this, suppose ai and aj intersect. Since aj lies in head(Fj)−head(Fj+1), we have i 6= j + 1. Similarly, i 6= j − 1. So qi /∈ h(Fj) andqj /∈ h(Fi), as they are not consecutive fish in the sequence {Fi}.

The heads of Fi and Fj intersect in such a way that the nose of eachis not contained in the head of the other (see Figure 9). Let Fi = (α, β)with nose a and Fj = (γ, δ) with nose b. For the heads to intersect, oneof α or β passes through head(Fj). Without loss of generality, supposeβ does so. If α also does, then all four intersections between these twofish occur between the heads, and the tails cannot intersect, so α doesnot intersect head(Fj). If α intersects tail(Fj) before intersecting β,then the tails will be disjoint, so α must intersect β first, then entertail(Fj). However, if this is the case, then b ∈ head(Fi), implyingj = i+ 1, a contradiction. It follows that head(Fi) ∩ head(Fj) = ∅ forall i 6= j − 1, j, j + 1.

Consequently we have an embedded polygon P with sides ai (withthe convention ak+1 = a1) connecting qi to qi+1. For each i, the fishFi consists of the outer arc f out

i which begins outside P , and the innerarc f in

i which begins inside P , neither of which intersect ai. Withoutloss of generality, we may assume that P is oriented as in Figure 10.

Consider the arc f ini for some i, and suppose its first intersection

with P lies on aj. We claim that f ini intersects f in

j prior to its firstpoint of intersection with P .

First, recall that ai is contained in head(Fi), and hence the arcswhich Fi consists of do not intersect ai. Therefore j 6= i. Additionally,


qi

qj

qj+1

qi+1

Figure 9. An example demonstrating why two of thearcs ai cannot intersect; the four intersections caused bythe intersection of the heads prevent the tails of thesefish from intersecting.

f in1 fout1

a1

a2

a3

a4 q1

q2q3

q4

Figure 10. Inner and outer arcs of a fish in the sequence.

ai−1 is contained in head(Fi−1) − head(Fi), and so the arcs which Ficonsists of do not intersect ai−1. Therefore j 6= i− 1.

Because the edge aj is contained entirely within head(Fj), crossing ajrequires crossing both f in

j and f outj . Now refer to Figure 11. For f in

i tocross aj in the indicated direction, it is not possible for it to first crossf outj , then aj, then f in

j . Therefore it must intersect f inj first, then aj,

and finally f outj . Since this intersection with aj is its first intersection

with P , its intersection with f inj occurs before it leaves P . This proves

the claim.Now for each arc f in

i , consider the edge aj of P with which it hasits first intersection, and define a map s(i) = j. Then there existsa minimal length sequence j1, . . . , jn with s(ji) = ji+1 and s(jn) = j1.Define gi = f in

ji. Observe that these gi each intersect the successive gi+1

prior to leaving P . We also note that each arc appears in the sequence{gi} at most once.

Two consecutive arcs gi and gi+1 in the cycle divide D into tworegions. Let Ri be the closed region bounded by gi and gi+1 whichdoes not contain the punctures qji and qji+1

(See Figure 12). DefineR = ∪Ri. The polygon P also divides the disc into two regions. Let Ebe the exterior of P , the region adjacent to p.


foutj

f inj

qj

Figure 11. Example: The dashed line aj, marked withan arrow, cannot be crossed first in the indicated direc-tion by an arc from any other puncture outside head(Fj)which intersects f out

j prior to f inj .

We claim that E ⊂ R. It suffices to show that E ∩ R is both openand closed in E. Each Ri is closed by definition, so certainly E ∩ R isclosed in E. Let x ∈ R. If x does not lie on one of the arcs gi, thenx lies on the interior of Ri for some i, and therefore x has an openneighbourhood in R.

g1

g2

g3

g4R1

R2

R3

R4

Figure 12. The arcs gi, and the regions Ri that theyform. In more complicated examples, these Ri may over-lap.

Suppose x lies on gi for some i. Recall that for any consecutive giand gi+1, their intersection lies inside P . We proved before that giintersects gi+1 prior to leaving P . We will now show that gi intersectsgi−1 prior to leaving P .

If gi leaves P prior to intersecting gi−1, then some arc as intersectsgi before the intersection with gi−1. Consider Figure 13. The arc gi−1divides head(Fi) into two regions, one of which is adjacent to qji anddoes not contain qji+1 or, hence, any other puncture. The arc aji furthersubdivides this into two more regions, one of which lies inside P . LetK be that region (the upper right quarter of head(Fji) in Figure 13).

If as intersects gi before gi intersects gi−1, then it enters the region Kvia the arc gi. It must then proceed to leave K, but it cannot cross ajibecause we have proven that P embeds, and it cannot cross gi−1 prior


giaji

gi−1

qji

Figure 13. If gi leaves P prior to intersecting gi−1, thenthere is an intersection between gi and aji+1 where indi-cated by the arrow.

to gi−1 intersecting aji or it would contradict the definition of gi. Soas must both enter and exit K by crossing gi, forming a bigon betweenas and gi which lies in K. However, there are no punctures in K, andso this is an empty bigon, implying that as is not in minimal positionwith respect to gi, a contradiction.

So as desired, gi intersects both gi−1 and gi+1 prior to leaving Pfor the first time. Since x ∈ E, its position on the arc gi is after theintersections of gi with gi−1 and gi+1. Therefore, a sufficiently smallneighbourhood of x lies in the union Ri−1 ∪ Ri ⊂ R. It follows thatE ∩R is open in R, and hence E ⊂ R.

Now we will show that u ∈ E. Suppose not. Then u lies inside thepolygon P . By Lemma 4.7 no arc from u may pass through the headof any fish with u in its tail. In particular, no arc from u may passthrough Fi for any i. However, ai ∈ head(Fi) for each i, and so no arcfrom u may cross P . It follows that there are no arcs in A from u to p.For the same reason, it would not be possible to add an arc from u to pto the collection A.

Now consider the space obtained by removing every arc in A from D,and consider in particular the component C of that space which con-tains the puncture u. Let α be an embedded arc from u to the boundaryof C. The point x ∈ C where α ends lies on some arc γ ∈ A. Let βbe the subarc of γ running from x to p. Then up to homotopy, the arcobtained by concatenating α and β is an arc from u to p on D whichintersects each arc in A at most once. Therefore, it is possible to addan arc from u to p to the collection A, a contradiction. Therefore,u ∈ E. Consequently, u ∈ Ri for some i.

Without loss of generality, assume u ∈ R1, the region bounded by g1and g2. Recall that qj2 /∈ R1. We claim that qj2+1 ∈ R1. Indeed, thearc g1 by definition intersects aj2 when first leaving P . Since g2 cannotintersect aj2 , because aj2 ∈ head(Fj2), the region R1 contains the halfof aj2 which ends at qj2+1.

Now consider the arc f outj2

. It starts at qj2 , and must be positionedsuch that qj2+1 ∈ head(Fj2). For this to be possible, it must enterR1 by crossing g1, then exit by intersecting g2. After this it is not


possible for f outj2

to reenter R1 without any double intersections, and sotail(Fj2) ∩R1 = ∅. Therefore u /∈ tail(Fj2), a contradiction.

�

Proof of Proposition 4.4. Fix q ∈ Q. We argue by induction. First con-sider a simple base case on the three-punctured sphere whereQ = {q, r}.Then the only equivalence class of ∼q is {r} and there are clearly nofish possible, so |χ| = 1 = |Fq|+ cq.

Now consider the case |χ| ≥ 2. By Lemma 5.1, there exists a punc-ture v other than q which is not the nose of a q-fish. To apply in-

duction, define D to be D with the puncture v removed. By Fq, cq,etc., we refer to those elements in D. It then suffices to show that

|Fq|+ cq ≥ |Fq|+ cq − 1.

Definition 5.2. We say that a fish (α, β) vanishes if after removing v,α and β are no longer in minimal position (they bound an unpuncturedhalf-bigon). Equivalently, v is either the only tail puncture or only headpuncture of (α, β). When we say a minimal fish vanishes, we mean thatit vanishes in this sense, not that it becomes a non-minimal fish.

Note that two distinct minimal fish whose tails intersect cannot beformed with only three arcs (one shared between them); three arcsforming two minimal fish (α, β) and (β, γ) will always have disjointtails (see Figure 14). Hence everywhere that we consider two distinctminimal q-fish, we will only consider the case where four arcs are re-quired to draw the two fish.

Figure 14. If three consecutive arcs form two minimalfish, those fish have disjoint tails.

Before we proceed, it will be useful to prove the following lemmas.

Lemma 5.3. Let (α, β) and (γ, δ) be two non-vanishing minimal q-fishwith common nose u, with all four arcs distinct, and with α < β < γ < δ < α

in the cyclic order about u. Then in D if we consider homotopy equiv-alent arcs α, β, γ, δ in minimal position, then α < β < γ < δ < α.

In other words, while the removal of v may reorder some of the arcsabout q after they are moved to minimal position, the four specific arcsα, β, γ, δ will retain their order with respect to each other under thehypotheses of the lemma.


In order to prove this, we will first prove the following lemma, pro-viding a condition for which the conclusions of Lemma 5.3 hold.

Lemma 5.4. Let α1 < α2 < · · · < αn < α1 be a collection of arcsfrom u to p such that, after removing v, no two arcs αi and αj arehomotopic for i 6= j. If there is no fish F with h(F ) = {v} thenα1 < α2 < · · · < αn < α1.

Proof. We first note that if, after the removal of v, the arcs αi remainin minimal position, then certainly the order of arcs will be unchangedby the removal of v. We may assume, then, that the arcs are no longerin minimal position after the removal of v. This can occur only whenthere is an empty half-bigon. We know by assumption that h(F ) 6= {v}for any fish among the arcs αi, and so any such empty half-bigons mustbe the tails of some fish among the αi.

To show that the order will be unchanged, we will perform a seriesof surgeries on the arcs αi with support away from u, and with eachsurgery reducing the total number of intersections among the collection{αi} by one and preserving the homotopy classes of the arcs. Henceafter finitely many such surgeries, the arcs will be in minimal position,and because every surgery had support away from u, the order of thearcs about u has remained unchanged.

In order to perform these surgeries, we will first show the existence ofa fish (αi, αj) among these arcs with the property that for every otherarc αk with k 6= i, j, if αk intersects tail(αi, αj), it exits the tail beforeproceeding to p.

Let F1 = (αi1 , αi2) be one of the fish with t(F1) = {v}, and supposeF1 is not the desired fish, that is, there is some arc αi3 which enterstail(F1) and does not exit it prior to arriving at p. Without loss of gen-erality, suppose αi2 is the arc which it intersects upon entering tail(F1).Then F2 = (αi2 , αi3) is a fish whose tail is contained in tail(αi1 , αi2),and whose tail is not intersected by αi1 . If this fish is still not the de-sired fish, we repeat the process, at each step obtaining a fish Fk suchthat tail(Fk) ⊂ tail(Fk−1) and αij does not intersect tail(Fk) for anyj < k. As there are only n arcs in the collection, eventually we musthave the desired fish F = (αi, αj).

Let S be a compact set containing tail(F ), not containing any punc-tures, and not intersecting any arc other than the arcs which F consistsof, and arcs which intersect tail(F ). There is a homotopy supportedon S which moves αi to an arc αi and moves αj to an arc αj such thatαi ∩ αj = ∅, and such that for each k 6= i, j, |αk ∩ αi| = |αk ∩ αi| and|αk ∩αj| = |αk ∩ αj|. We note that after applying this homotopy, thereare still no empty half-bigons adjacent to u

So the total number of intersections has been reduced by one by thissurgery. If the arcs are not now in minimal position, we can repeatthis process on the new collection obtained by replacing αi with αi


u

p

α β γ δ

Figure 15. The lifts of α and δ must pass through(β, γ). As a result, tail(α, β) ∩ tail(γ, δ) ⊂ head(β, γ).

and αj with αj, and at each step we will reduce the total number ofintersections by 1 without ever changing the cyclic order of the arcsabout u. �

Proof of Lemma 5.3. It will suffice to show that the four arcs α, β, γ, δsatisfy the conditions of Lemma 5.4. It is clear that no two of thesefour arcs will become homotopic after the removal of v, so in particularwe wish to show that there is no fish F among these four arcs withh(F ) = {v}.

If we think of D as a punctured annulus with u as its inner bound-ary component and p as its outer boundary component, we can thenalso consider the universal cover of D, an infinite strip, as depicted inFigure 15. Let q be one lift of q in the universal cover, and consider thelifts α, β, γ, δ such that q is in both tail(α, β) and tail(γ, δ). Withoutloss of generality, these arcs are ordered α, β, γ, δ in the universal cover.Observe that in D, we must have that (α, δ) is a fish, with β, γ startingin head(α, δ).

Now we wish to show that if there is some fish F among α, β, γ, δ,with h(F ) = {v}, then in particular (β, γ) is such a fish. First, supposewe have any three arcs α1 < α2 < α3 < α1 such that (α1, α3) is a fish,and α2 begins in head(α1, α3). If h(α1, α3) = {v}, then either (α1, α2)is a fish with h(α1, α2) = {v}, or (α2, α3) is a fish with h(α2, α3) = {v}.

Now we can apply this to the configuration in the lemma. If h(α, δ) = {v}then one of (α, β) or (β, δ) is a fish F with h(F ) = {v}. By assumption,(α, β) is a non-vanishing fish, so we must have h(β, δ) = {v}. By thesame process, if this is true, then we must have that (β, γ) is a fish,and h(β, γ) = {v}. In fact, for any fish F among these four arcs tohave h(F ) = {v}, it is necessary for h(β, γ) = {v}. Thus the arcs βand γ intersect as in Figure 15.

In the universal cover, we can observe that tail(α, β) must lie tothe right of β. Similarly, the tail of (γ, δ) must lie to the left of γ.Hence their intersection, and q which lies within it, must lie in theregion which is both right of β and left of γ. This region is head(β, γ).Therefore h(β, γ) 6= {v}, since it must at least contain q. This provesthat no fish F among α, β, γ, δ has h(F ) = {v}, and completes theproof.


q

α β α βεε

Figure 16. On the left, ε intersects only α and thisforms a q-fish ”smaller” than (α, β). On the right, walso intersects β and, depending on which puncture is q,either (α, ε) or (ε, β) is the desired q-fish ϕ(α, β).

�

Lemma 5.5. Let F ′q be the set of non-vanishing minimal q-fish in A.

There is a one-to-one map ϕ : F ′q → Fq.

Proof. In the case that (α, β) remains a minimal fish (α, β) after the

removal of v, we simply define ϕ(α, β) = (α, β).Observe that if a minimal fish does not vanish due to the removal

of v, but does cease to be minimal, it does not cease to be a fish. Nowsuppose F = (α, β) is some minimal fish such that, after the removalof v, F remains a fish but ceases to be minimal. This implies α andβ are no longer consecutive, so there must be at least one arc lyingbetween them which emanates from q in the region head(F ). Let ε besuch an arc. The arc ε must intersect either α or β before proceedingto p. Without loss of generality, suppose it intersects α when leavinghead(F ).

If ε does not intersect β, then the fish (α, ε) is a q-fish. If it doesintersect β then either (α, ε) or (ε, β) is a q-fish. If this resulting fishis minimal, then we are done. If it is not, then we note that there arefewer arcs originating in the head of either (α, ε) or (ε, β) than therewere in (α, β), and therefore we can repeat this process finitely manytimes until we arrive at a fish with no arcs beginning in its head; such afish must be minimal. Define ϕ(α, β) to be the minimal q-fish obtainedby applying this process sufficiently many times.

Let F = (α, β) and F ′ = (γ, δ) be two distinct non-vanishing minimalq-fish. Clearly if they have distinct noses then the above process willpass F and F ′ to distinct minimal fish, so we consider only the casewhere F and F ′ have the same nose s.

Without loss of generality, the four arcs are ordered α < β < γ < δ < αabout the puncture s, with α and δ intersecting so that q lies in the tailof both fish. By Lemma 5.3, we know that α < β < γ < δ < α. Butthis tells us that ϕ(α, β) is a fish consisting of arcs that lie between α

and β, and that ϕ(γ, δ) is a fish consisting of arcs that lie between γ

and δ. Hence ϕ(α, β) 6= ϕ(γ, δ), as required. �


α

β γ

δ q

Figure 17. Two distinct minimal q-fish with the same nose.

We now return to the proof of Proposition 4.4 by considering twocases. Let us first consider the case where {v} is an equivalence classof ∼q, and hence after removing v the number of equivalence classes is

reduced. Observe that if a 6∼q b, then certainly a 6∼q b, as the removalof v does not introduce any new fish. In other words, ∼q is refined bythe removal of the puncture v. As such, cq ≥ cq − 1. It remains to

check that |Fq| ≥ |Fq| in this case.Because {v} is an equivalence class of∼q, we know that v is not in the

head of any q-fish. Furthermore, it cannot be the last puncture in thetail of a q-fish, as every q-fish contains q in its tail by definition. So no q-fish can possibly vanish by the removal of v. Consequently, |Fq| = |F ′q|.There is the possibility that a minimal fish in Fq ceases to be minimal,

but Lemma 5.5 tells us that |F ′q| ≤ |Fq|. Hence, |Fq|+ cq ≥ |Fq|+cq−1as required.

Now we consider the case where {v} was not an equivalence class

of ∼q. We will show that |F ′q| ≥ |Fq| − 1. Since cq ≥ cq as be-fore, this will complete the proof. It will suffice to show that at mostone minimal q-fish vanishes. If this is the case, then by Lemma 5.5,

|Fq| = |F ′q| ≥ |Fq| − 1 as required.Suppose that there exist two distinct vanishing minimal q-fish F

and F ′. Let F = (α, β) and F ′ = (γ, δ) with {v} = h(F ) = h(F ′). Weconsider two possibilities.

Suppose F and F ′ have distinct noses. The nose of one fish cannotbe contained in the head of the other, as each head contains only thepuncture v, which is not the nose of a q fish. So it follows that, whetheror not they share a nose puncture, for the heads of these two fish tointersect, up to homotopy (using the fact that there are no puncturesin the symmetric difference of the heads) each of α and β must bothenter head(F ′) and then exit head(F ′), making four total intersectionswith γ and δ. It follows that no additional intersections can occurbetween these two fish, and so in particular, their tails cannot intersect.Therefore either q /∈ t(F ) or q /∈ t(F ′), a contradiction.


α1

α2

α3

α4

α5

q

Now suppose F and F ′ have a common nose. As above, it is notpossible for both α and β to intersect head(F ′) and still allow F andF ′ to have intersecting tails. If neither α nor β intersects head(F ′), thenthe only possibility is that one head is contained entirely in the other,implying that one of the fish is not minimal. So the only possibility isthat exactly one of α or β intersects head(F ′).

Without loss of generality, suppose β intersects head(F ′). We con-sider two cases here as well. If β first intersects γ and then δ, it ispossible to have h(F ) = h(F ′) = {v}. However, in this case, the arcsγ and δ begin in head(α, β), and so (α, β) is not minimal, a contradic-tion. If β intersects δ first and then γ, then α begins in head(β, δ) andsince α cannot intersect head(γ, δ), it must instead cross β prior to βintersecting head(F ′), and so in this case, v /∈ head(F ).

So we have that there is at most one vanishing minimal q-fish as

required, and |F ′q| ≥ |Fq| − 1. From Lemma 5.5 we know |Fq| ≥ |F ′q|,and we showed cq ≥ cq, and so, as required, we have

|Fq|+ cq ≥ |Fq|+ cq − 1.

�

6. Counting Excess Arcs using Tail Punctures of Fish

Proof of Proposition 4.5. Fix the puncture q ∈ Q and, for simplicity ofnotation, let k = kq. Let {α1, α2, . . . , αk, α1} be the counterclockwisecyclical ordering of the arcs about q.

Consider the discD as a punctured annulus with q as its inner bound-ary, and Q − {q} the collection of punctures. Let D be the universalcover of the annulus: an infinite strip with a boundary edge corre-sponding to p, the boundary of D, and a boundary edge correspondingto q. We lift the arcs in F q to D. Each lift of each arc is an arc betweenthe two edges of D.

Choose α1 to be one of the lifts of α1. Notice that consecutive arcsalong q in D are lifts of consecutive arcs in D about q. Hence we maylabel arcs in D by (. . . , α1, α2, . . . , αk+1, . . . ) where αnk+i is a lift of αifor 1 ≤ i ≤ k and n ∈ Z. In particular, αk+1 is a lift of α1.


q

p

α1

α2 α3 α4 α5 α6

Figure 18. An example lifting of arcs from a puncturedsurface to the infinite strip. Here α6 is a second lift ofα1.

We define a displacement function d(αi, αj) on the arcs in D. If αiand αj are disjoint, then let d(αi, αj) be the number of punctures on

D bounded by those two arcs. If they do intersect, then these twoarcs bound two distinct regions—one adjacent to q and one adjacent top. In this case, let d(αi, αj) be the number of punctures in the regionadjacent to q minus the number of punctures in the region adjacentto p.

This distance function is additive in the following sense: if i < j < l,then d(αi, αj) + d(αj, αl) = d(αi, αl). If the three arcs are disjoint, thisis obvious. If one pair intersects, or if all three intersect, it can easilybe checked that the additivity still holds by individually consideringthe finitely many combinations in which the arcs can intersect. Sincearcs on D do not self intersect, the two lifts α1 and αk+1 of α1 do notintersect, so d(α1, αk+1) = |Q− {q}| = |χ|.

For 1 ≤ i ≤ k, if the arcs αi and αi+1 are disjoint (and hence arenot a minimal fish) then d(αi, αi+1) ≥ 1, as there must be at least onepuncture between them. If they do intersect, then they form a minimalfish Fi, and we have d(αi, αi+1) ≥ 1− |t(F )| as there must be at leastone puncture in h(Fi). Using the additivity of d, and the convention|t(Fi)| = 0 whenever there was no fish Fi, we obtain:

k∑i=1

(1− |t(Fi)|) ≤k∑i=1

d(αi, αi+1) = d(α1, αk+1) = |χ|

Equivalently,

k ≤∑F∈Fq

|t(F )|+ |χ|. �

References

[JMM96] Martin Juvan, Aleksander Malnic, and Bojan Mohar, Systems of curveson surfaces, J. Combin. Theory Ser. B 68 (1996), no. 1, 7–22.

[Lei11] Christopher Leininger, personal communication (2011).[MRT14] Justin Malestein, Igor Rivin, and Louis Theran, Topological designs,

Geom. Dedicata 168 (2014), no. 1, 221–233.[Prz15] Piotr Przytycki, Arcs intersecting at most once, Geom. Funct. Anal. 25

(2015), no. 2, 658–670.


Dept. of Math. & Stats., McGill University, Montreal, Quebec,Canada H3A 0B9

E-mail address: [email protected] address: [email protected]

Documents

ARCS ON SPHERES INTERSECTING AT MOSTpprzytyc/Thesis.pdf · CHRISTOPHER SMITH AND PIOTR PRZYTYCKIy Abstract. Let pbe a puncture of a punctured sphere, and let Qbe the set of all other