Distance 4 curves on closed surfaces of arbitrary genus
DDISTANCE CURVES ON CLOSED SURFACES OFARBITRARY GENUS
SREEKRISHNA PALAPARTHI AND KUWARI MAHANTA
Abstract.
Let S g denote a closed, orientable surface of genus g ≥ C ( S g )the associated curve complex. The mapping class group of S g acts on C ( S g ) byisometries. Since Dehn twists about certain curves generate the mapping classgroup of S g , one can ask how Dehn twists move specific vertices in C ( S g ) awayfrom themselves. In this article, we answer this question for a specific case whenthe vertices are at a distance 3. We show that if two curves represent vertices ata distance 3 in C ( S g ) then the Dehn twist of one curve about another yields twovertices at distance 4. This produces many examples of curves on S g which areat distance 4 in C ( S g ). We also show that the minimum intersection number ofany two curves at a distance 4 on S g , i min ( g, ≤ (2 g − . Introduction
Let S g denote a closed, orientable surface of genus g ≥
2. On S g , by a curve,we mean an essential simple closed curve throughout this article. For two curves, α, β in S g , i ( α, β ) denotes their geometric intersection number and T α ( β ) denotes theDehn twist of the curve β about the curve α . A simplicial complex, C ( S g ), calledthe complex of curves was introduced by Harvey in ( Harvey, 1981 ). The 0-skeleton, C ( S g ), of this complex is in one-to-one correspondence with isotopy classes of essentialsimple closed curves on S g . Two vertices span an edge in C ( S g ) if and only if thesevertices have mutually disjoint representatives. C ( S g ) can be equipped with a metric, d , by defining the distance between any two vertices to be the minimum number ofedges in any edge path between them. By the distance between two curves on S g , wemean the distance between the corresponding vertices in C ( S g ). i min ( g, n ) denotesthe minimum of the intersection numbers between any two curves on S g which are ata distance n . Section 2 contains definitions and the setup for this work.Masur and Minsky, in their seminal paper ( Masur and Minsky, 2000 ), provedthat the complex of curves is δ -hyperbolic. Later, it was shown that the δ can be cho-sen to be independent of the surface S g , see ( Aougab, 2013 ), (
Bowditch, 2014 ),(
Clay et al., 2014 ), (
Hensel et al., 2015 ), (
Przytycki and Sisto, 2017 ). Thecoarse geometry of the curve complex has various applications to 3-manifolds, Teich-muller theory and mapping class groups. One can see (
Minsky, 2006 ) for many suchapplications.(
Shackleton, 2012 ), (
Webb, 2015 ), (
Watanabe, 2016 ) and (
Birman etal., 2016 ) gave algorithms to compute distances between two vertices in C ( S g ). Inparticular, Birman, Margalit and Menasco ( Birman et al., 2016 ) gave a certainefficient geodesic algorithm to find the distance between any two vertices of C ( S g ).Aougab and Huang ( Aougab and Huang, 2015 ) showed that for genus ≥ i min ( g,
3) = 2 g − g − ≥ Aougab and Taylor, 2014 ) answered a question of DanMargalit and proved that for a surface of genus g ≥ i min ( g, n ) = O ( g n − ). Inparticular, i min ( g,
4) = O ( g ). a r X i v : . [ m a t h . G T ] J a n SREEKRISHNA PALAPARTHI AND KUWARI MAHANTA
In (
Glenn et al., 2017 ), Glenn et. al. gave the mathematics behind the softwarepackage MICC (Metric in the Curve Complex) that partially implements the efficientgeodesic algorithm of (
Birman et al., 2016 ). The authors also gave a test todetermine when two vertices in C ( S g ) are at a distance ≥ Birman et al., 2016 ). Using the MICC software, they showed that i min (2 ,
4) = 12 by giving all minimally intersecting pairs of curves at distance 4. Theyalso gave examples of curves which are at a distance 4 on a surface of genus 3 andconcluded that i min (3 , ≤
29 (refer (
Glenn et al., 2017 ) Theorem 1.8).In the current article, we investigate the effect of certain Dehn twists on distancesin C ( S g ). Remark . If d ( α, γ ) = 1 or , then d ( α, T γ ( α )) = d ( α, γ ) . This observation prompted us to ask how far, in the curve complex, a vertex isdisplaced from itself by a qualified
Dehn twist. As a first attempt, we start with twocurves α and γ on S g which fill S g and ask how far α is displaced from itself by T γ .When α and γ is a filling pair of curves on S g , we show that α and T γ ( α ) is also afilling pair. We further use the distance ≥ Glenn et al., 2017 ) to showthat d ( α, T γ ( α )) = 4 whenever d ( α, γ ) = 3. This gives abundant examples of verticesat distance 4 in C ( S g ). As a consequence, we show that i min ( g, ≤ (2 g − .In particular, using any of the minimally intersecting filling pairs of curves on S ,as described in ( Aougab and Huang, 2015 ), one can get examples of curves atdistance 4 in C ( S ) which intersect at 25 points. This implies that i min (3 , ≤ Setup
For any ordered index in this work, we follow cyclical ordering. For instance, if i ∈ { , , . . . , k } , i = k + 1 will indicate i = 1.Two curves, ν and ν on S g are said to be a filling pair if every component of S g \ { ν , ν } is a disk. A component, D of S g \ { ν , ν } is said to be an 2 n -gon ifits boundary comprises of n arcs of ν and ν . Consider a pair, α, β , of filling curveson S g with geometric intersection number, i ( α, β ) = k . Let their set of intersectionpoints be { w , . . . , w k } . Define a triangulation, G of S as follows:Let { w (cid:48) , . . . , w (cid:48) k } be the 0-skeleton of G . There is an edge between w (cid:48) i and w (cid:48) j in G if and only if there is an arc of α or, β between the two intersection points, w i , w j in S . As α and β are filling pairs, for each disk component, D of S g \ { α, β } attacha disk to the cycle formed by the edges of G corresponding to the arcs of α, β thatform the boundary of D . Using the Euler characteristic of a 2-dimensional complex,the number of faces, f of G is given by f = k + 2 − g .Consider a geodesic, ν , . . . , ν N of length N in C ( S ). An arc, ω in S is a referencearc for the triple ν , ν , ν N if ω and ν are in minimal position and the interior of ω isdisjoint from ν ∪ ν N . The oriented geodesic ν , . . . , ν N is said to be initially efficient if i ( ν , ω ) ≤ N − ω . The authors of ( Birman et al.,2016 ) prove that there exists an initially efficient geodesic between any two verticesof C ( S ).The following theorem from ( Glenn et al., 2017 ) gives a criterion for detectingvertices in C ( S ) at distance at-least 4. Theorem
Glenn et al., 2017 )) . For the filling pair, κ , ω , let Γ ⊂ C ( S ) be the collection of all vertices such that the following hold : (1) for γ ∈ Γ , d ( κ, γ ) = 1 ; and (2) for γ ∈ Γ ; for each segment, b ⊂ ω \ κ , i ( γ, b ) ≤ .Then d ( κ, ω ) ≥ if and only if d ( γ, ω ) ≥ for all γ ∈ Γ . Let λ and µ be two simple closed curves on S g and let R λ and R µ be closedregular neighborhoods of λ and µ respectively. We say that the 4-tuple ( λ, µ, R λ , R µ )is amenable to Dehn twist in special position if the following hold:(1) λ and µ intersect transversely and minimally on S g , ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 3
Figure 1. Q , Q , Q , Q is an example of an initially efficient geo-desic in C ( S ).(2) λ and µ fill S g ,(3) components of R λ ∩ R µ are disjoint disks,(4) components of S \ ( R λ ∪ R µ ) are disks,(5) the number of components of S \ ( R λ ∪ R µ ) is equal to the number of com-ponents of S \ ( λ ∪ µ )(6) each component of S \ ( R λ ∪ R µ ) is contained inside a unique component of S \ ( λ ∪ µ )When λ and µ fill S g , by considering a Euclidean model of S g , it is easy to see thata 4-tuple ( λ, µ, R λ , R µ ) amenable to Dehn twist in special position always exists.Consider a 4-tuple ( λ, µ, R λ , R µ ) which is amenable to Dehn twist in specialposition. Let i ( λ, µ ) = k and K := { , , ..., k } . We construct a curve in the isotopyclass of T λ ( µ ) which we call T λ ( µ ) in special position w.r.t. the -tuple ( λ, µ, R λ , R µ ).Start at any one of the components of R λ ∩ R µ and label it as A . Since µ intersects λ transversely, the arc µ of µ contained in A which has its endpoints X and Y on boundary arcs of R λ is such that X and Y lie on distinct boundary componentof ∂R λ . We call the component of ∂R λ containing X to be ∂ + R λ and the othercomponent containing Y to be ∂ − R λ . Equip A with Euclidean metric so that it is asquare with two opposite sides formed from the arcs of ∂R λ and two remaining sidesformed from arcs of ∂R µ and such that the x -axis lies along µ and the value of the x -coordinate increases from X to Y . This induces an orientation on µ . Next we pick k distinct points { q , q , ..., q k } in the interior of µ such that the x -coordinate of q i is greater than the x coordinate of q j whenever i > j and i, j ∈ K . For each i ∈ K ,let λ i be a curve in R λ which is isotopic to λ and passes through q i . Further for each i, j ∈ K, i (cid:54) = j let λ i and λ j be disjoint.Orient λ such that the y -coordinate on λ increases when following this orienta-tion in the disk A . Starting with A , label the subsequent disk components, R λ ∩ R µ ,as A , A , ..., A k , in the orientation of λ . For each i ∈ K , A i contains a unique arcof µ which we label as µ i . µ i gets an induced orientation from µ . For each i ∈ K ,equip A i with Euclidean metric and assume it to be a square in the xy -plane where µ i lies along the x -axis with the x coordinate increasing along the orientation of µ i .Assume A i to be positioned such that µ i is the line segment joining the mid-pointsof the left and right sides of the square. In this orientation, call the component of ∂R µ which appears above µ i as ∂ + R µ and the the component of ∂R µ below µ i as ∂ − R µ . However, note that the side of A i which is formed of the arcs of ∂ + R λ couldeither be to the right or to the left of this square. Accordingly, the side of A i whichis formed of the arcs of ∂ − R λ could either be to the left or to the right of this square.For i, j ∈ K , by an isotopy inside A i , we can assume that all the arcs of λ j in A i arestraight lines. SREEKRISHNA PALAPARTHI AND KUWARI MAHANTA
For each i, j ∈ K , let u i,j := A i ∩ λ j ∩ ∂ + R µ and v i,j := A i ∩ λ j ∩ ∂ − R µ . Also foreach i ∈ K let the left end point of µ i in the square A i be v i, and the right end pointof µ i in the square A i be u i,k +1 . Construct the Dehn twist of µ about λ as follows:For each j ∈ K ∪ { } draw line segments, θ i,j , connecting v i,j to u i,j +1 . T λ ( µ ) is thecurve (( µ ∪ ( ∪ i ∈ K λ i )) ∩ ( S \ ( ∪ i ∈ K A i )) ∪ ( ∪ i,j ∈ K θ i,j ) . The schematic, figure 2, shows A i before and after this transformation. Figure 3 isthe algorithm to construct T λ ( µ ) in ( Farb and Margalit, 2012 ) using µ and λ i ’s.Figure 2 and 3 together show how the transformed curve in figure 2 is isotopic to T λ ( µ ) in every A i , i ∈ K . In the complement of A i ’s the transformation describedabove does not disturb the curves λ i ’s and µ as prescribed in ( Farb and Margalit,2012 ). When T λ ( µ ) is constructed as above and as shown in figure 2, we say that T λ ( µ ) is in special position w.r.t. λ and µ . We call the k copies of λ , λ i , i ∈ K , and µ to be the scaffolding for T λ ( µ ). We call the Euclidean disks A i , i ∈ K , along withthe line segments θ i,j ’s for j ∈ K to be the disks of transformation for T λ ( µ ). Thepoints u i,j ’s, v i,j ’s, u i,k +1 and v i, for i, j ∈ K shall hold their meaning as defined inthe context of the disks of transformations. So, using these phrases, when T λ ( µ ) is inspecial position w.r.t. λ and µ , the scaffolding of T λ ( µ ) remains unchanged outside itsdisks of transformation. Inside the disks of transformation for T λ ( µ ), the schematicin figure 2 describes the changes to its scaffolding. Figure 2.
Disk of transformation before (figure on the left) andafter (figure on the right) the Dehn twist.
Figure 3.
Surgery of the curves to obtain T λ ( µ ). Distance 4 curves in C ( S g ≥ ) Theorem . Let S be a surface of genus g ≥ . Let α and γ be two curves on S with d ( α, γ ) = 3 . Then, d ( T γ ( α ) , α ) = 4 . ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 5
Proof.
Let ν , ν , ν , ν be a geodesic from the vertex ν corresponding to α to the vertex ν corresponding to γ in C ( S ). Let T γ ( ν ) be the vertex in C ( S )corresponding to T γ ( α ). The existence of the path T γ ( ν ), T γ ( ν ), T γ ( ν ) = ν , ν , ν gives that d ( T γ ( α ) , α ) ≤
4. We prove that d ( T γ ( α ) , α ) ≥ κ = T γ ( α ) and ω = α , hence showing that d ( T γ ( α ) , α ) = 4. Claim . T γ ( α ) and α fill S .Proof of claim 1 : Let i ( γ, α ) = k , K := { , , ..., k } , K − := { , , ..., k − } and K − g := { , , ..., k + 2 − g } . We refer to section 2 for the terminology used here.Since α and γ fill S , there is a 4-tuple ( α, γ, R α , R γ ) which is amenable to Dehn twistin special position. Let T γ ( α ) be in special position w.r.t to α and γ . We denotethe disks of transformation of T γ ( α ) by A i , for i ∈ K . By an isotopy we assume thecurve α to be disjoint from T γ ( α ) \ A i for i ∈ K and in each A i we further assumethe arc α i := α ∩ A i to be a straight line segment below the segment connecting v i, and u i,k +1 . Figure 4.
The scaffolding for T Q ( Q ), where Q and Q are fromexample in figure 1. The shaded region is a rectangle of the scaffold-ing.For i ∈ K , let g i along with α be the scaffolding for T γ ( α ). For j ∈ K − , oneof the components of S \ { g j , g j +1 } is an annulus, G j . Any component of G j \ α isa 4-gon which we call as a rectangle of the scaffolding for T γ ( α ). Figure 4 shows anexample of such a rectangle of the scaffolding. The disks A i , i ∈ K , further divideeach rectangle of the scaffolding into three components. There is a unique i ∈ K suchthat A i and A i +1 intersect a given rectangle of the scaffolding. Denote a rectangleof the scaffolding formed out of G j with its arcs of α lying in A i and A i +1 by B i,j .Denote the sub-rectangles B i,j ∩ A i , by C (cid:48) i,j and B i,j ∩ A i +1 , by C (cid:48)(cid:48) i +1 ,j . Also let B (cid:48) i,j := B i,j \ ( C (cid:48) i,j ∪ C (cid:48)(cid:48) i +1 ,j ). Let B = ∪ ki =1 ∪ k − j =1 B i,j .S \ ( α ∪ γ ) has k + 2 − g disk components by Euler characteristic considerations. If F p is a disk component of S \ ( α ∪ γ ), for some p ∈ K − g , then F (cid:48) p := F p \ B is a singledisk as B intersects any F p only in disks which contain a boundary arc of F p , namelyarcs of γ . The components of S \ ( α ∪ g ∪ · · · ∪ g k ) comprise of k ( k −
1) rectangles ofthe scaffolding for T γ ( α ), namely B i,j where i ∈ K , j ∈ K − , and k +2 − g even sidedpolygonal discs, namely F (cid:48) p , where p ∈ K − g . Let F (cid:48)(cid:48) p denote F (cid:48) p \ R α for p ∈ K − g .For each j ∈ K let w i,j := θ i,j ∩ α i . For each i ∈ K and j ∈ K − , let D (cid:48)(cid:48) i,j be theparallelogram with vertices v i,j , v i,j +1 , w i,j and w i,j +1 and D (cid:48) i,j +1 be the parallelogramwith vertices w i,j , w i,j +1 , u i,j +1 , u i,j +2 . In each disk A i , for i ∈ K , there is a pentagon, P i, , which is above α i and bounded by the lines θ i, , ∂R γ , α i , θ i, and the line segmentof ∂ + R α between u i, and u i, . Likewise, in each disk A i , for i ∈ K , there is a triangle, T i,k +1 , which is bounded by the lines α i , θ i,k and ∂R γ . Figure 5 shows a schematicbefore and after the transformation to the disk A i ; the figure to the left shows therectangles C (cid:48) i, and C (cid:48)(cid:48) i,k and the figure on the right shows P i, and T i,k +1 . SREEKRISHNA PALAPARTHI AND KUWARI MAHANTA
Figure 5.
The disk of transformation for T γ ( α ). The figure onthe left shows the portion of the scaffolding for T γ ( α ). The figureon the right shows the pentagon P i, , the triangle T i,k +1 and theparallelograms formed due to α i and T γ ( α )Figure 6 shows a schematic of R α before and after the transformation to thescaffolding of T γ ( α ). The shaded region in the figure on the left shows C (cid:48) i,j and C (cid:48)(cid:48) i,j − for some indices i, j . The shaded region in the figure on the right shows D (cid:48) i,j and D (cid:48)(cid:48) i,j − for some indices i, j . Figure 6.
A schematic of R α (figure on the left) and after (figureon the right) the Dehn twist.For i ∈ K , note that all the disks A i , occur in some sequence in the annulus R α when moving along α . So, a disk A i is connected to some disk A j on the left and tosome other disk A p on the right by a single arc of α \ R γ , for some distinct indices i, j, p ∈ K . The schematic for two disks A i and A j , for some i, j ∈ K , which areconnected via a single arc of α \ R γ and an arc of T γ ( α ) \ R γ is as shown in the figure7. Note that this schematic is generic since for every j ∈ K , there is a distinct i ∈ K such that A j occurs to the left of A i , in the sense mentioned above. Figure 7.
Two adjacent disks of transformation in R α .Figure 7 is a schematic of a portion of figure 6. In any of the cases, viz. ∂ + R γ and ∂ + R γ face each other, ∂ + R γ and ∂ − R γ face each other or ∂ − R γ and ∂ − R γ face eachother, In this schematic, we see that the pentagon P i, of the disk A i is connected tothe triangle T j,k +1 of A j via an arc of α \ R γ , ω i,j , and an arc of T γ ( α ), η i,j . The disk, ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 7 R i,j outside R γ bounded by ω i,j , η i,j and two arcs of ∂R γ , will be called a conduit .Equip the conduit with the Euclidean metric and assume that R i,j is a rectangle withtwo opposite sides ω i,j and η i,j . Now P i, ∪ R i,j ∪ T j,k +1 is a 4-gon bounded by fourarcs viz. (i) θ i, ∪ η i,j ∪ θ j,k , (ii) α j ∪ ω i,j ∪ α i , (iii) θ i, and (iv) the arc of ∂ + R α between u i, and u i, . This protracted 4-gon will be denoted by D (cid:48) i, .Let S (cid:48) = S \ R α . The components of S \ ( α ∪ T γ ( α )) are the components of S (cid:48) \ T γ ( α ) and the components of R α \ ( α ∪ T γ ( α )) glued at the boundary of R α .Since the changes to the scaffolding of T γ ( α ) is restricted to R α , the components of S (cid:48) \ T γ ( α ) are precisely the disc components of S (cid:48) \ ( g ∪ · · · ∪ g k ).The components of S (cid:48) \ ( g ∪ · · · ∪ g k ) are B (cid:48) i,j , i ∈ { , . . . k } , j ∈ K − , alongwith disks F (cid:48)(cid:48) p , p ∈ K − g , as explained above. The components of R α \ ( α ∪ T γ ( α ))will be examined using the schematic figure 7 of a portion of R α . There are fourkinds of regions in R α . The upper disk regions, like R in the schematic figure 7, thelower disk regions, like R in the schematic figure 7, and the disks D (cid:48) i,j , D (cid:48)(cid:48) i,j , i ∈ K , j ∈ K − . Figure 7 shows how the upper and lower disk regions are glued to disks F (cid:48)(cid:48) p for p ∈ K − g . For each p ∈ K − g , after gluing the lower disk regions and theupper disk regions to the respective disks F (cid:48)(cid:48) p , we get disks which we denote by F (cid:48)(cid:48)(cid:48) p .We know that F (cid:48)(cid:48)(cid:48) p is a disk because the upper and the lower disk regions are disjoint,except for the points w i,j on the boundary and share a single arc of ∂R α with a unique F (cid:48)(cid:48) p . For each p ∈ K − g , we call F (cid:48)(cid:48)(cid:48) p to be the modified disk corresponding to theinitial disk F p .For each i ∈ K and j ∈ K − , the line segment of ∂ + R α between u i,j u i,j +1 isthe common boundary of C (cid:48) i,j and D (cid:48) i,j . Likewise, for each such i, j , the line segmentof ∂ − R α between v i,j v i,j +1 is the common boundary of C (cid:48)(cid:48) i,j and D (cid:48)(cid:48) i,j . So, for such i, j , when considering the components of S \ ( α ∪ g ∪ · · · ∪ g k ) the rectangular core B (cid:48) i,j is connected to C (cid:48) i,j along the boundary segment u i,j u i,j +1 and to C (cid:48)(cid:48) i +1 ,j alongthe boundary segment v i +1 ,j v i +1 ,j +1 , whereas when considering the components of S \ ( α ∪ T γ ( α )), the rectangular core B (cid:48) i,j is connected to D (cid:48) i,j along the boundarysegment u i,j u i,j +1 and D (cid:48)(cid:48) i +1 ,j along the boundary segment v i +1 ,j v i +1 ,j +1 . So therectangles of the scaffolding for T γ ( α ), B i,j , which are components of S \ ( α ∪ g ∪· · · ∪ g k ), after the transformation in the disks of transformation for T γ ( α ) result indisks E i,j := B (cid:48) i,j ∪ D (cid:48) i,j ∪ D (cid:48)(cid:48) i +1 ,j which now are components of S \ ( α ∪ T γ ( α )). Foreach p ∈ K − g , F (cid:48)(cid:48)(cid:48) p is a disk as seen earlier. The components of S \ ( α ∪ T γ ( α )) areprecisely the disks F (cid:48)(cid:48)(cid:48) p and E i,j where p ∈ K − g , i ∈ K and j ∈ K − . This provesthat the components of S \ ( α ∪ T γ ( α )) are all disks and hence proving Claim 1.The components of R γ \ T γ ( α ) are disks and their boundary consists of two arcsegments of T γ ( α ) and one each of ∂ + R γ and ∂ − R γ . We call these disks as rectangulartracks . The word tracks derives its motivation from how these tracks appear in R γ .Figure 8 shows R γ and rectangular tracks inside R γ . Figure 8.
The rectangular tracks shown inside the annulus R γ . SREEKRISHNA PALAPARTHI AND KUWARI MAHANTA
Since i ( α, γ ) = k , there are k components of α ∩ R γ . Every component of α \ T γ ( α )is either contained in R γ or, has a sub-arc which is contained in R γ . For any i ∈ K , α i intersects the rectangular tracks.Let i ∈ K . In the schematic figure 5, A i has exactly k + 1 arcs of T γ ( α ). Call θ i , to be the leftmost arc of A i and θ i ,k to be the rightmost arc of A i . Let usconsider one component of T γ ( α ) ∩ R γ , call it ρ i , which intersects A i in its leftmostarc. This ρ i intersects A i precisely in the arcs θ i , and θ i ,k and it intersects A j forevery j ∈ K \{ i } in the arcs θ j,m where m = ( j − i )( mod k ). This is easily seen fromthe construction of T γ ( α ) in special position w.r.t. α and γ . From this discussion itis clear that ρ i intersects each α j , for j ∈ K , exactly once. It is also clear that, for j ∈ K , the points of ρ i ∩ α j lie on ρ i in the order α i +1 , ..., α k , α , ..., α i − , α i when ρ i is traversed from ∂ + R γ to ∂ − R γ . We now consider two arc components, ρ i and ρ i +1 , of T γ ( α ) ∩ R γ and the rectangular track, T i , which is enclosed by these twocomponents in R γ . We equip this rectangular tracks T i with the Euclidean metricso that the boundary arcs ρ i , ρ i +1 , and the arcs of T i ∩ ∂R γ are all straight linesand so that T i is a rectangle. We refer to T i ∩ ∂ + R γ as the left end of the rectangleand T i ∩ ∂ − R γ as the right end of this rectangular track. We can draw the arcs of α j , for j ∈ K , as straight line segments in the rectangular tracks T i . Figure 9 showsa schematic of T i where i ∈ K . Figure 9.
A rectangular track T i along with arcs of α i in it.From this schematic, at both the left and right end of this rectangular track T i , a i is a common boundary to a triangle and a pentagon. We call α i as the starting arc of this rectangular track T i .Figure 10 shows two possible schematics when A i is pictured in R γ . Figure 10. A i shown inside R γ in the two possible ways. The figureon the left shows α i oriented from top to bottom and the figure onthe right shows α i oriented from bottom to top.For any of the two possible cases observed in figure 10, a portion of one of thetwo pentagons of T i appears in the A i which is between α i and ∂ + R α , where α i is thestarting arc of this track. We call this pentagon the upper pentagon of the rectangulartrack T i , owing to the viewpoint that ∂ + R α is the upper boundary of R α . A portionof the other pentagon of T i appears in A i which is between α i and ∂ − R α . We call this ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 9 pentagon the lower pentagon of the rectangular track . Likewise, we define the uppertriangle and the lower triangle of a rectangular track T i .Let γ ∈ Γ as in the statement of the Theorem 1. We prove that d ( γ, α ) ≥ γ and α fill S . By Theorem 1, this will imply that d ( T γ ( α ) , α ) ≥ i ( γ ∩ α ) (cid:54) = 0 because if γ is disjoint from both α and T γ ( α ) then γ is non-essential as it will lie completely in one of the disc componentsof S \ ( T γ ( α ) ∪ α ). Since d ( γ, γ ) ≥ d ( T γ ( α ) , γ ) − d ( γ, T γ ( α )) = d ( T γ ( α ) , T γ ( γ )) − d ( γ, T γ ( α )) = 3 − i ( γ ∩ γ ) (cid:54) = 0. Since γ intersects γ , itintersects R γ . It cannot be completely contained in R γ because every simple closedcurve contained in an annulus bounds a disk or is isotopic to the core curve of theannulus. Since neither of these is true, it follows that that γ intersects R γ in arcs.Since i ( γ, T γ ( α )) = φ , each component of γ ∩ R γ has to be completely contained inone of the rectangular tracks described by T γ ( α ). Such a component arc of γ couldeither be boundary reducible or essential in R γ .We consider an isotopy I of γ , as follows: In the case that a component arcof γ in R γ is boundary reducible in R γ , we can perform the boundary reduction of γ preserving its minimal intersection position with α and T γ ( α ). This is possiblebecause an arc of γ which is boundary reducible in R γ and is contained in the disk T i will bound a bigon with one boundary arc of R γ in T i . Also, since γ was already inminimal intersection position with α , it does not bound bigons with the arcs α j inside T i . Call the isotopy of γ which reduces all the boundary-reducible arcs of γ ∩ R γ as I . After the isotopy I , we can assume that all the arcs of γ in R γ are essential. Weknow that there is at-least one component of γ ∩ R γ which is an essential arc of R γ as γ cannot be disjoint from R γ . By the hypothesis that i ( γ, b ) ≤ b ⊂ α \ T γ ( α )each rectangular track can contain at-most one component of γ ∩ R γ .Next, we describe an isotopy I of γ such that all the points of γ ∩ α will lie inside R γ and so that no new boundary reducible arc components of γ ∩ R γ are introducedand γ ’s minimal intersection position with α and T γ ( α ) is retained. To this end,suppose that a point of γ ∩ α lies outside R γ .Following the construction of the disk D (cid:48) i, described above using figure 7, we seethat the upper pentagon of the rectangular track T i is connected to the upper triangleof the rectangular track T j via a conduit R i,j where i, j ∈ K are such that A j is tothe left of A i in R α as in schematic 7.If a point of γ ∩ α , x , lies outside R γ , then it has to lie on ω i,j for some i and j such that i, j ∈ K, i (cid:54) = j . We now refer to the dotted line in figure 11. Since theintersection of γ and α is transverse, an arc of γ , call it δ lies on the two sides of theconduit R i,j , one inside and one outside R i,j . The endpoint P of the arc δ inside R i,j is also the endpoint of some other arc of γ as γ is a closed curve. If P connects to anarc of γ lying in the upper triangular region of the track T j , then an essential arc δ of γ ∩ R γ lies in T j with its endpoint Q on ∂R γ in the upper triangle of T j so that δ , thearc P Q and δ together form a bigon with α contradicting the minimal intersectionposition of γ with α . So, P connects to an arc of γ in the upper pentagon in the track T i as is the dotted line in figure 11. Consider an isotopy I which slides the point x onto α i . The image of the arc component of γ ∩ R γ which is in T i , under I has itsendpoint in the lower triangle of T i and the image of x lies in R γ . A schematic forthis isotopy I is shown in figure 11.After finitely many such isotopies, we can now assume that all the points of γ ∩ α lie inside R γ . Now consider an isotopy I of γ as follows: If any of the components of γ ∩ R γ has its endpoint on the boundary of the upper triangle of T j , for some j ∈ K ,then by the above discussion, γ cannot intersect ω i,j or η i,j , for some i ∈ K such thatthe arcs of T i and T j forms the opposite sides of a conduit R i,j . So γ ∩ R i,j is an arc M N which has its endpoints M ∈ T j and N ∈ T i on ∂R γ . Further, since γ is a closedcurve, γ ∩ T i is an arc with its endpoint as N such that N necessarily lies in the upperpentagon of T i . Conversely, if any of the components of γ ∩ R γ has its endpoint, z ,on the boundary of the upper pentagon of T i , then it should be connected to an arc, g , of γ in the conduit R i,j . Note that the endpoints, z , z (cid:48) of g are on ∂R γ . There Figure 11.
The isotopy I moving points of γ ∩ α into R γ .exists an arc component of γ ∩ R γ lying in T j such that z (cid:48) is on the boundary of theupper triangle of T j , as the dotted line in figure 12 shows. If any such arc g of γ exists, consider an isotopy, I , of g such that the image, I ( g ), lies outside R i,j . Aschematic of this is figure 12. Figure 12.
A schematic showing the normalization move, the iso-topy I .The component of γ ∩ R γ in T j now has an endpoint on the boundary of the lowerpentagon of T j and the component of γ ∩ R γ in T i has an endpoint on the boundaryof the lower triangle of T i . Also the image of γ ∩ α under I moves a point of γ ∩ α from the boundary of the upper traingle of T j to the boundary of the lower pentagonof T i . We call I to be a normalization move on γ . After finitely many normalizationmoves performed on γ , wherever applicable, we can assume that every component of γ ∩ R γ is contained in a rectangular track T i for some i ∈ K such that the endpointsof that component lie on the boundary of the lower triangle and the lower pentagonof T i . So a schematic of every component of γ ∩ R γ inside T i is as in figure 13. Figure 13.
The portion of γ in rectified position inside T i .After these isotopies I , I , I of γ , we say that γ is in a rectified position . Wenow prove that γ in rectified position and α fill S . From now on we assume that γ isin a rectified position.For i ∈ K , let H i be the rectangular component of R γ \ ( ∪ i ∈ K α i ) containing thearcs a i and a i +1 on its boundary. Each of these H i contains a unique segment, γ i , ofthe core curve γ . The schematic 14 shows H and γ for instance.We say that an arc, g of γ covers γ i if g ⊂ H i has its end points on α i and α i +1 and g is isotopic in H i to γ i through arcs whose end points stay on α i , α i +1 . Since ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 11
Figure 14.
Schematic showing H and γ in R γ . γ and α form a filling pair, the set of essential arcs, { γ , . . . , γ k } fill S \ α . It followsthat γ fills S along with α if segments of γ \ α cover γ i for all i with i ∈ K .Since γ is in rectified position, each component of γ ∩ R γ already covers all γ i except one as in figure 13. More precisely, if a component of γ ∩ R γ is in a rectangulartrack T i , then γ covers every γ j where j is such that 1 ≤ j ≤ k and j (cid:54) = i −
1. So, if γ ∩ R γ has two distinct components, then each component has to lie in T i for distinct i and hence γ covers γ j for j ∈ { , , . . . k } . We conclude that γ and α fill S in thiscase. Now it remains to show that if there is a single component of γ ∩ R γ , which isan essential arc of R γ and is contained in some rectangular track T i , then γ and α fill S . As in the previous case, γ covers every γ j where j is such that 1 ≤ j ≤ k and j (cid:54) = i −
1. The components of S \ ( α ∪ ≤ j ≤ k,j (cid:54) = i − γ j ) will be disks except possiblyone which could be a cylinder. This can be seen as follows. Since α and γ fill S , thecomponents of S \ α ∪ ≤ j ≤ k γ j are disks. Each segment of γ j \ α for j ∈ { , , ..., k } contributes to two distinct edges of a component J or two separate components J, J (cid:48) of S \ α ∪ ≤ j ≤ k γ j .Let P := γ ∩ α i and P := γ ∩ α i − be points in T i which appear on the uniquecomponent of γ ∩ R γ . Let [ P , P ] represent the arc of γ in R γ with endpoints P and P and γ := γ \ [ P , P ]. γ is contained in all the components of S \ α ∪ ≤ j ≤ k,j (cid:54) = i − γ j which contain the arcs α i − and α i on their boundary. We know that there is at-leastone such component because γ i − is also such an arc which joins α i − to α i . If γ i − is the boundary of J, J (cid:48) , then it would have been an arc which connected α i − on onedisk to α i on another disk. Note that both α i and α i − are also boundary arcs ofboth J and J (cid:48) . So, we would find P on the disk containing α i and P on the diskcontaining α i − . When we join J and J (cid:48) along γ i − we get a disk where γ is an arcfrom P to P intersecting γ i − . Cutting along γ still yields two different disks. Theschematic, figure 15 shows this situation. Figure 15.
The figure on the left shows disks J and J (cid:48) formed bycutting along γ i − . The figure on the right shows the new disksformed when J ∪ J (cid:48) are cut along γ .If γ i − were on the boundary of J representing two edges of J then it would havebeen an arc which connected α i − to α i . When we glue J to itself along γ i − , weget a cylinder, A , where α i and α i − will be arcs on different boundary components of A . So we would find P and P on distinct boundaries of A and hence γ would bean essential arc on A . So cutting A along this arc γ would yield a disk as shown inthe schematic, figure 16. Figure 16.
The disk J glued to itself along γ i − and cut along γ .In any case, we get disks by cutting S \ α along the arcs of γ \ α . So this provesthe theorem. (cid:3) Claim 1 of Theorem 2 shows that α and T γ ( α ) fill S independent of d ( α, γ ) aslong as α and γ is a filling pair i.e. as long as d ( α, γ ) ≥
3. So we have the following.
Corollary . If α and γ are a pair of curves which fill S , then α and T γ ( α ) also fill S . Corollary . For a surface of genus g ≥ , i min ( g, ≤ (2 g − . Proof.
Aougab and Huang (
Aougab and Huang, 2015 ) proved that i min ( g,
3) =2 g − g ≥
3. Now, on S g , for g ≥
3, suppose that α and β are two such minimallyintersecting curves with d ( α, β ) = 3. Then i ( α, T β ( α )) = (2 g − and by Theorem2, d ( α, T β ( α )) = 4. So i min ( g, ≤ (2 g − . (cid:3) An initially efficient geodesicLemma . If α = ν , ν , ν , ν = γ is an initially efficient geodesic then so is T γ ( α ) , T γ ( ν ) , ν , ν , α . Proof.
For p ∈ K − g , let F (cid:48)(cid:48) p be the components of S \{ α, R γ } as in the proof ofTheorem 2. Since the geodesic α, ν , ν , γ is an initially efficient one, each segment of ν intersects every reference arc in E i at most twice. In particular, arcs of ∂ ( R γ ) thatform the edges of E i intersect ν at most twice. It follows from here that there are atthe most two segments of ν in each rectangular track T i as defined in section 3. Aschematic of this is shown in figure 17. Further, since the interior of a reference arc isdisjoint from α ∪ T γ ( α ), it is sufficient to check for the initial efficiency of the geodesic, T γ ( α ) , T γ ( ν ) , ν , ν , α in the modified disks F (cid:48)(cid:48)(cid:48) p , abbreviated F , corresponding to F p , abbreviated E .Since E and F are homeomorphic to a 2 g -gon. Without loss of generality assume E and F to be a regular Euclidean regular polygon with 2 g sides. Starting at anysegment of α in E , we label the edge as α . Label the edges of E in a clockwisedirection, starting at α as γ , α , γ , . . . , γ g . Let S (cid:48) = S \ R γ . Since the componentsof S (cid:48) \ { α, γ } and S (cid:48) \ { α, T γ ( α ) } are the same, it follows that for every edge, a j in F corresponding to α , there exists a unique i ∈ { , . . . , g } such that a j ⊂ α i .Index the edges, a j of F such that j = i . Label the edge of T γ ( α ) in F between a i and a i +1 as t i . Let ω be a reference arc in F with end points on t p and t q for some p, q ∈ { , . . . , g } . Suppose to the contrary that ω ∩ T γ ( ν ) ≥
3. Then there exists threesegments, z , z , z of T γ ( ν ) in F such that z j ∩ ω (cid:54) = φ . For j ∈ { , , } , let the endpoints of z j lie on a j and a j . From our previous discussion on Dehn twist and figure18, there exists arcs of ν in E with end points on γ j and γ j for all j ∈ { , , } .Consider a line segment, ω (cid:48) in E from an interior point of a p to an interior point of ISTANCE 4 CURVES ON CLOSED SURFACES OF ARBITRARY GENUS 13 a q . Then ω (cid:48) is a reference arc for the triple, α , ν , γ and ω (cid:48) ∩ ν ≥
3. This contradictsthat α , ν , ν , γ is an initially efficient geodesic. Hence, ω ∩ T γ ( ν ) ≤ ω for the triple T γ ( α ), T γ ( ν ), α . Figure 17.
There can be at-most two distinct segments of T γ ( ν )in any rectangular component of S \ ( α ∪ T γ ( α )) in R γ Figure 18.
Initial efficiency of T γ ( a ) follows from the initial effi-ciency of a .Since T γ ( α ) , T γ ( ν ) , ν , ν , α is already a geodesic we have that d ( T γ ( ν ) , α ) = 3.This gives that T γ ( ν ) is an initially efficient geodesic of distance 4 from T γ ( α ) to α . (cid:3) References
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Sreekrishna Palaparthi , Department of Mathematics, Indian Institute ofTechnology Guwahati, Assam 781039, India, email: [email protected]