Natalia M. Pacheco-Tallaj Advisors: Dr. Kevin Schreve and Dr. Nicholas Vlamis. University of Michigan Math Department Summer 2017 Math REU.

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1 THURSTON NORM OF 2 GENERATOR, 1 RELATOR GROUPS Natalia M. Pacheco-Tallaj Avisors: Dr. Kevin Schreve an Dr. Nicholas Vlamis University of Michigan Math Department Summer 2017 Math REU Contents 1 Backgroun Nice way of generating Mo(S) The Thurston Norm Previous relevant results The Algorithm Meaning of P T an eriving the Thurston unit all Sample Outputs = 1 case Results Complicate Thurston Norms Future Directions Enumerating an Detecting Simple Close Curves The Coe Appenix A Interpretation 16 A.1 The Interval Case A.2 1 = 1 example Astract We iscuss an algorithm for rawing the unit all of the Thurston Norm on the secon homology of an oriente 3-manifol. We explore which unit alls are otainale through this algorithm an provie geometric interpretation. We also explore ways to enumerate the otainale unit alls. 1

2 Introuction We are intereste in 3-manifols M with funamental groups that amit (2,1)-presentations, i.e. π 1 (M) = x, y r (1) We usually assume that these manifols have two-imensional first homology groups ( 1 (M) = 2). These properties allow us to apply the results of [4] in orer to provie an algorithm for rawing the unit all of the Thurston norm. The Thurston norm generalizes the genus of a knot K S 3. Recall that every knot K ouns an emee surface in the knot complement S 3 K, the Siefert surface S(K). Of course, if a knot ouns a single surface, it in fact ouns aritrarily many surfaces of ifferent genus. The knot genus recors the simplest possile Siefert surface, i.e. the Siefert surface of minimal genus. For example, the unknot is the only knot with genus 0. The Siefert surface represents a class in the homology group H 2 (S 3 K, (S 3 K)). In fact this same proceure works for any 3-manifol M. Given any element φ in H 2 (M, M), let x M ([φ]) := min{ χ(s) S M an [S] = φ} Thurston showe that x M (φ) is a seminorm on H 2 (M, M, R). By Poincare uality, this also gives a norm on H 1 (M, R). The Thurston Norm measures the simplicity of surfaces representing homology classes, an can tell us a lot aout the geometry of a 3-manifol. Friel an Tillmann [4] provie a way to assign to a group G amitting 2-generator, 1-relator presentation with non-empty, cyclically reuce relator an 1 = 2 a marke polytope in H 1 (G, Z) that contains information aout the group G. Polytopes in the first homology inuce a norm on the first cohomology, which etermines the Thurston norm in the secon homology with Poincare Duality. We provie an algorithm for otaining information aout the Thurston norm of a 3- manifol otaine y gluing a 2-hanle along a separating simple close curve on the surface a genus 2 hanleoy. Given this construction of the manifol, the algorithm thus applies to any manifol with a funamental group as in 1. We aim to etermine which polygons can e otaine through this algorithm, particularly whether all polygons can e otaine an how (i.e. to which manifol they correspon). 1. Backgroun The three-manifols we construct are otaine y gluing a 2-hanle along a separating simple close curve on the surface of a genus 2 hanleoy, i.e. ientifying four consecutive faces of a unit cue along an annular neighorhoo of the separating simple close curve. The property that the curve is separating guarantees that 1 = 2 (see Section 1.2). To generate every simple close curve on T 2 #T 2, the surface of the genus two hanleoy, we use the mapping class group (efine elow) of T 2 #T 2 to otain every separating simple close curve from transformations of the separating simple close curve [x, z] = xzx 1 z 1 epicte on the top-left corner of Figure Nice way of generating Mo(S) Definition Let S e a surface an c a curve on S, an let U e an annular neighorhoo of c on S. Then U is homeomorphic to a cyliner. Let φ : U S 1 I e this homeomorphism. Define a map f : S 1 I S 1 I with f : (e iθ, t) (e i(θ+2πt), t). The Right Dehn Twist aout c, enote T c, is a self-homeomorphism of S given y T c = φ 1 f φ on U an T c = i everywhere else on S. The Left Dehn Twist aout c is Tc 1. 2

3 Figure 1: Right Dehn Twist aout c on a Torus For a surface S g,p with genus g an p punctures, let Hom + (g, p) enote the group of all orientation preserving homeomorphisms of S g,p that fix the set of punctures, an let Hom 0 (g, p) e the path-component of the inentity in Hom + (g, p), i.e. all elements of Hom + (g, p) homotopy equivalent to the ientity map. We can now efine the Mapping Class Group (MCG) of S g,p. Definition The Mapping Class Group of S g,p is the group Mo(S g,p ) = Hom + (g, p)/hom 0 (g, p) of homotopy classes of orientation preserving homeomorphisms of S g,p that fix the set of punctures. Fact (Humphries 1979) For a surface S g,0 of genus g, 2g+1 Dehn Twists generate Mo(S g,0 ). We are concerne with generating the mapping class group Mo(T 2 #T 2 ) of a genus 2 torus with no punctures. The 2g + 1 = 5 Dehn Twists that generate Mo(T 2 #T 2 ) are T a, T, T c, T, T e for curves a,, c,, e as in Figure 2. Figure 2: Dehn Twists aout a,, c,, e generate Mo(T 2 #T 2 ) Figure 3: Funamental group of T 2 #T 2 Fact T a, T, T c, T, T e are given y the following formulae an are the ientity elsewhere, where w, x, y, z are the generators of π 1 (T 2 #T 2 ) as in Figure 3. 3

4 T a : x z 1 x T : z xz T c : x xwz 1 T c : y ywz 1 T : w y 1 w T e : y wy Ta 1 T 1 Tc 1 Tc 1 T 1 Te 1 : x zx : z x 1 x : x xzw 1 : y yzw 1 : w yw : y w 1 y 1.2. The Thurston Norm By an n-cell we will mean a space homeomorphic to the open n-all B n. Definition A cell complex or CW-complex is a Hausorff space X along with a partition P of X into n-cells such that for each n-cell c n there exists a continuous map f : B n X, calle the attaching map, such that B o n maps homeomorphically onto c n an f( B n ) m 1 c for c P a k-cell with k < n. Definition The n th -chain group with real coefficients, enote C n (X), of a CWcomplex X is the real vector space with the n-cells of X as a asis. The n th -ounary map n : C n (X) C n 1 (X) of a CW-complex X is the linear map that takes each n-cell C in the asis of c n to the sum (up to sign) of the asis elements corresponing to the (n-1)-cells in C. Definition The i th -homology group with real coefficients of a CW-complex X is the vector space H i (X) = ker i /im i+1 Fact H i (X) is a topological invariant of the space X. Remark. There are analagously efine homology groups with coefficients in any ring R. In particular, the integral homology groups are enote H i (X, Z). From now on some efinitions will e provie in a specific form that is as general as is neee for the scope of this project. Definition Let S e a surface an P a partition of S as a CW-complex. Let V enote the numer of 0-cells or vertices in P, E enote the numer of 1-cells or eges in P, an F enote the numer of 2-cells or faces in P. The Euler characteristic of S, enote χ(s) is χ(s) := V E + F The Euler characteristic is a topological invariant of S an oes not epen on P. The Euler characteristic of surfaces with equal genus an ounary components is the same an in fact, χ(s g,p ) can e efine in terms of g an p as follows χ(s g,p ) = 2 2g p Definition The n th -cochain group of a space X is efine as C n (X) := Hom(C n (X), R) for a ring R. An element of C n (X, R) is calle a n-cochain. The ounary maps of the chains of X naturally inuce coounary maps etween the cochains of X in the same way that linear maps etween vector spaces inuce ual maps etween ual vector spaces. Thus, if the ounary map i is represente y a matrix A, the corresponing co-ounary map δ i is represente y the transpose A. Just as we efine homology, we can efine cohomology in exactly the way one woul expect 4

5 Definition The ith cohomology of a space X is for some fiel F. H i (X) = ker δ i /im δ i 1 = Hom(H 1 (X), F ) Definition Suppose A X is a CW-sucomplex of X. The relative chain groups C n (X, A) are generate y the n-cells in X A. There are relative homology groups H i (X, A) efine analagously as for homology groups. A relative cycle can e thought of as a chain of cells in X A such that the ounary of this chain is containe in A. Relative cohomology groups are efine similarly. Theorem (Poincarè uality). Suppose M is a finite CW-complex which is also an orientale n-manifol. Then exists an isomorphism ψ i etween C i an C n i (M, M) i n which inuces an isomorphism H i (M) = H n i (M, M). More generally, this is true for any coefficient ring R. Note that any close orientale n-manifol amits a canonical element in H n (M, Z). This is usually calle the funamental class, an can e visualize as the chain which assigns the same constant K to each n-cell. Note that any (n 1)-cell is containe in exactly two n-cells, so y choosing orientations correctly this is an n-cycle. A similar iea works to show that if M has ounary there is a canonical element in H n (M, M). In low imensions it turns out that all the homology groups can e thought of as funamental classes. Fact Let M e a 3-manifol. Then every element in H 2 (M, M) is represente y the funamental class of an immerse surface S. Now, y efinition an y we know that for a 3-manifol M Hom(H 1 (M), R) =H 1 (M) = H 2 (M, M) H 1 (M) = H 2 (M, M) For each loop c i in the asis of H 1 (M), the corresponing asis element φ i in H 1 (M) is such that { 1 i = k φ i (c k ) = 0 i k an each c i can e uniquely assigne an element of H 2 (M, M) represente y a surface S i Previous relevant results We introuce some previous results that are relevant to our work. Definition (Friel an Tillmann). A (2,1)-presentation x, y r for a group G is sai to e a nice presentation if r is a nonempty, cyclically reuce wor an if 1 (G) = 2. Lemma (Friel an Tillmann). Let G e a group amitting a nice (2,1)-presentation an φ : G Z an epimorphism. There exists a nice (2,1)-presentation x, y r for G with φ(x) = 0, φ(y) = 1 that gives rise to the same polytope as the original presentation. A fiere 3-manifol M amits the structure of a fier unle over S 1, i.e. there is a unle S M S 1 where S is some surface. A cohomology class in H 1 (M) is fiere if it is the pullack of the funamental class of S 1. 5

6 Theorem (Thurston). Let N e a 3 manifol. There exists a unique symmetric marke polytope P N in H 1 (N; R) such that for any φ H 1 (N; R) = Hom(π 1 (N), R) we have x N (φ) = max{φ(p) φ(q) p, q P N } Furthermore, φ is fiere if an only if it pairs maximally with a marke vertex. 2. The Algorithm We start with a separating simple close curve on the surface of a genus 2 hanleoy. Particularly, we start with the curve γ 0 = xzx 1 z 1. This curve is chosen so as to guarantee 1 = 2 in the resulting manifol, an etails on why this hols can e foun in Section 2.3. By Fact we know a series of Dehn Twists applie to γ 0 can generate any separating simple close curve, along which we can then paste a 2-hanle to get a 3-manifol. The inputs to the algorithm are then n-wors in T a, T, T c, T, T e for any n, which encoe any possile 3-manifol amitting a nice (2, 1)-presentation for its funamental group. Denoting y γ k the wor in w, x, y, z after the first k Dehn Twist applications to γ 0, the relator for the funamental group of the resulting 3-manifol is otaine y simply ropping the w s an z s from γ n (ecause they are non-trivial in π 1 (T 2 #T 2 ) ut trivial in π 1 of the hanleoy) an cyclically reucing the resulting wor in x, y. We will call r(γ n ) the cyclically reuce relator corresponing to γ n an W T (x, y) (a wor in xy) the equivalent of γ n in the funamental group of M T (i.e. the result of ropping w, z from γ n ). We will let T e the input wor in T a, T, T c, T, T e an their inverses, n e the length of T, so that γ n = T (γ 0 ). M T will enote the manifol resulting from pasting a 2-hanle along γ n, π 1 (M T ) = x, y r with r eing the xy components of γ n. We will use π 1 (M T ) to construct a marke polytope in H 1 (M T ) as follows [4] an as shown in Figure 4: 1. Trace out a walk on the plane y reaing r(γ n ) from left to right, starting at (0, 0) an aing (ɛ, 0) for any x ɛ an (0, ɛ) for any y ɛ. Mark all points in the walk that were crosse only once. 2. Take the convex hull C of these points, preserving the markings of vertex points an let V C enote its set of vertices. 3. For n, m Z, place a point on (n + 1 2, m ) whenever each of the points in S m,n = {(m, n), (m + 1, n), (m, n + 1), (m + 1, n + 1)} is either insie C or in V C. Mark (n + 1 2, m ) if S V C is fully marke. Let the set of all such (n + 1 2, m ) e K. 4. Take the convex hull of K, preserving markings. The resulting polygon P T is the output polygon Meaning of P T an eriving the Thurston unit all The algorithm escrie in the previous susection etermines the Thurston norm y the following theorem. Theorem [3] Let M e an irreucile 3-manifol that amits a cyclically reuce (2, 1)- presentation π = x, y r. Let P T e constructe as in the algorithm aove. Then P N. = PT where P N is as in Theorem an. = means equal up to translation. 6

7 (1) take the path etermine y the relation r marke vertex unmarke vertex an Figure 4: [4] Construction of P T from π 1 (M) Figure 5: The Algorithm: (1) Apply T to γ 0 = [x, z] to get γ n (2) Paste a 2-hanle along γ n to get M T (3) Construct the polytope associate to r(γ n ), the relator of π 1 (M T ). Therefore, P T is the polynomial ual to the unit all of the Thurston norm an it encoes 7

8 exactly the same information. Let φ H 1 (M T ) then ker φ H 1 (M T ) has imension one an is a line on our superposition of H 1 on the plane. Let (ker φ) enote the orthogonal complement of ker φ. The unit norm of φ is then the thickness of P in the irection of (ker φ) as measure y φ. That is, x M (φ) = max{φ(p) φ(q) p, q (ker φ) P } as shown in Figure 6. Figure 6: x M (φ) = φ(p) φ(q) 2.2. Sample Outputs We implemente this algorithm in Python an the pyqtgraph graphics lirary, an elow are some sample inputs an outputs to our software. T = T 3 T 3 T c π 1 (M T ) = x, y y 3 x 3 y 3 x 3 T = T 1 T 1 T ct 1 T c x, y xyxy 2 xyx 1 y 1 x 1 y 2 x 1 y 1 T = T 2 T 2 T 2 c x, y y 2 x 2 y 2 x 2 y 2 x 2 y 2 x = 1 case One can easily tell whether 1 = 1 from the relator fo the funamental group. Let r = x ɛ1 y δ1... x ɛ k y δ k then if k i=1 ɛ 1 = 0 an k i=1 δ 1 = 0, r represents a separating close curve an ouns two genus 1 surfaces, making it trivial in the homology. If on the other han k i=1 ɛ 1 0 or k i=1 δ 1 0 then 1 = 1. 8

9 Assume π = x, y r is the funamental group of M an 1 = 1. We apply Lemma to get a presentationn π = x, y r such that φ(x) = 0, φ(y) = 1 an raw the corresponing polytope as in the 1 = 2 case. When calculating Thurston norms as in Section 2.1, the fact that φ(y) = 0 just means we can ignore the horizontal irection an only care aout the vertical thickness of our polytope. Perhaps this will ecome clearer y example, refer to Appenix A Results We now show the existence of some polygons that can appear as ual polygons to the unit all of the Thurston norm of certain 3-manifols. We say a polygon P is otainale if there is a 3-manifol M with π 1 (M) = x, y r an P M. = P. Lemma Every fully marke rectangle is otainale. Proof. We first prove inuctively that T n T m T c (xzx 1 z 1 ) = xy n wz 1 x m x m zx m zw 1 y n x 1 z 1 x m (2) with T, T, T c as in Fact Equation 2 clearly hols for n = m = 1. Inucting on n, T (n+1) T m T c (xzx 1 z 1 ) = T (xy n wz 1 x m x m zx m zw 1 y n x 1 z 1 x m ) as esire. Next, inucting on m, T n T m+1 c(xzx 1 z 1 ) = T (T n = xy n T (w) z 1 x m x m zx m z T (w 1 ) y n x 1 z 1 x m = xy n+1 wz 1 x m x m zx m zw 1 y (n+1) x 1 z 1 x m T m T c (xzx 1 z 1 )) = xy n w T (z 1 ) x m x m T (z) x m T (z) w 1 y n x 1 T (z 1 ) x m = xy n wz 1 x (m+1) x m+1 zx m+1 zw 1 y n x 1 z 1 x (m+1) as esire, where the first equality is true since an clearly commute. Now, we claim that T = T (h+1) T w+1 T c generates a rectangle of height h an with w with eges parallel to the axes. By 2 we know that the funamental group of M T has relator r = y h+1 x (w+1) x w+1 x w+1 y (h+1) x (w+1) which cyclically reuces to r = y h+1 x w+1 y (h+1) y (w+1) it is easy to see that the walk otaine from this relator never crosses the same point more than one, thus it is fully marke, an that the convex hull C of the walk is a rectangle of with w + 1 an height h + 1 an that the resulting polygon P is thus a fully marke rectangle of with w an height h. Corollary Every fully marke interval on the x or y axes is otainale. Proof. Intervals are rectangles of height or with zero. Vertical intervals of length m can e otaine y T = T (m+1) T T c an horizontal intervals of length m can e otaine y T = T 1 T m+1 T c. Lemma Given a Polygon P, the polygon P otaine y reflecting P aout the x or y axis is otainale. 9

10 Figure 7: In orer to mirror a polygon along the y axis, we want an automorphism of T 2 #T 2 that we can appen to the input wor T which sens x to x 1 ut keeps y fixe. Picture aove is one such map. It keeps the right sie (the y sie) of the genus 2 torus fixe, an rotates the left sie (the x sie) y 180, applying a half Dehn twist in the mile annulus aroun c so that the ounaries of each shape will agree when pasting the torus ack together. Proof. Let T enote a wor in T a, T, T c, T, T e such that the polygon corresponing to M T is P an let c n enote the wor in w, x, y, z such that T (xzx 1 z 1 ) = c n. We efine a mirroring map (Figure 7) as follows (T T a ) 3 :x z 1 x 1 z z 1 [x, 1 ]z 1 Then, y taking (T T a ) 3 (c n ), each z gets replace y z 1 [x 1, z 1 ] which is also trivial in π 1 (M T ) an oes not affect W T. Each x gets replace y z 1 x 1 z which is x 1 in π 1 (M T ) an thus W (T T a) 3 T (x, y) = W T (x 1, y). Then, the polytope P arising from (T T a ) 3 T is the mirror image along the y-axis of P. The proof for mirroring along the x axis is analogous ut uses T, T e instea. Lemma Let n enote the maximum horizontal istance etween two points at any step of the walk an m enote the maximum vertical istance. If n + m 2k + 4 then the marke polygon corresponing to this funamental group cannot e containe within a k y k ox in H 1 Proof. Notice that all the angles in the convex hull C of a walk on the integer lattice will e 90, since the cyclically reuce conition on r implies every x ɛ1 is followe y either an x ɛ1 or a y ɛ2 for ɛ 1, ɛ 2 {0, 1}. Then, (x, y) V C, (x, y ) P T such that x x = 0.5 an y y = 0.5. Then, if there are two points (x 0, y 0 ), (x 1, y 1 ) V C with x 1 x 0 k+2 then there are two points ((x 0, y 0), (x 1, y 1) P T such that x 1 x 0 k + 1. Similarly if y 1 y 0 k + 2 then y 1 y 0 k + 1. Let (x 0, y 0 ), (x 0 + n, y 0) V C maximize horizontal istance an (x 1, y 1 ), (x 1, y 1 + m) V C maximize vertical istance, then if m + n 2k + 4, either n k + 2 or m k + 2 or oth. Lemma For all i Z +, T = T Complicate Thurston Norms (T 1 T c) i+1 gives rise to a polytope P T with 4i vertices. Proof. Denote T i = T 1 (T 1 T c) i+1 an notice that T i+1 = T s T i where T s = T 1 also efine two relate sequences of wors on the funamental group of T 2 #T 2 T 1 T ct. We will A 1 = ywz 1 x A n+1 = B n A n (3) B 1 = y 2 wz 1 x B n+1 = B n B n A n (4) 10

11 A 1 B 1 Figure 8: Each A n, B n lock etermines a walk in the plane roote at the enpoint of the lock that precees it in the wor By Fact 1.1.2, it is easy to see that T s (A 1 ) = y 2 wz 1 x ywz 1 x x 1 zw 1 y 1 x 1 x ywz 1 x = y 2 wz 1 x ywz 1 x = B 1 A 1 = A 2 T s (B 1 ) = y 2 wz 1 x y 2 wz 1 x ywz 1 x x 1 zw 1 y 1 x 1 x ywz 1 x = B 1 B 1 A 1 = B 2 Aitionally, T s (A k ) = A k+1 an T s (B k ) = B k+1 = T s (A k+1 ) = T s (B k A k ) = T s (B k )T s (A k ) = B k+1 A k+1 = A k+2 an T s (B k+1 ) = T s (B k )T s (B k )T s (A k ) = B k+1 B k+1 A k+1 = B k+2 thus y inuction A n+1 = T s (A n ), B n+1 = T s (B n ) (5) Furthermore, each A n, B n is a wor in w, x, y, z which in turn represents a walk on the plane. We can otain the height an with of sai walk y noting w(a 1 ) = 1 = f 0, h(a 1 ) = 1 = f 1, w(b 1 ) = 1 = f 1, h(b 1 ) = 2 = f 2 an w(a i+1 ) = w(b i ) + w(a i ), w(b i+1 ) = w(b i ) + w(b i ) + w(a i ) an thus w(a n ) = f 2n 2 h(a n ) = f 2n 1 w(b n ) = f 2n 1 h(b n ) = f 2n where f k enotes the kth numer in the sequence f 0 = 1, f 1 = 1, f k = f k 1 + f k 2, w(w ) enotes the with of the walk on the plane corresponing to the wor W an h(w ) enotes the height of the walk on the plane corresponing to a the wor W. We claim that T i ([x, z]) = xa 1 A 2... A i B i B i 1... B 1 ywσ i (6) where σ i = T i (x 1 z 1 ) is the walk symmetric to xa 1 A 2... A i B i B i 1... B 1 yw = T i (xz). This is clearly true for i = 1 an follows inuctively from 5 an the fact that T s (x) = xa 1, T s (yw) = B 1 yw. Now let A : (1, 0), (f 0, f 1 ), (f 2, f 3 ),... (f 2i 2, f 2i 1 ) e a sequence of i + 1 elements of Z 2 an B : (f 2i 1, f 2i ), (f 2i 3, f 2i 2 ),... (f 1, f 2 ), (0, 1) e another such sequence, an let (AB), the sequence of length 2i + 2 in Z 2 otaine from appening B to A, efine a sequence of points as follows: p 0 = (AB) 0 = (1, 0) p n = p n 1 + (AB) n (7) an efine a sequence of segments {l n } n y the sequence of line segments etween consecutive points in {p n } n These lines have slopes 0, f1 f 0, f3 f 2,..., f2i 1 l 0 = (0, 0), p 0 l n = p n 1 p n (8) f 2i 2, f 2i f 2i 1, f2i 2 f 2i 3,..., f2 f 1, vertical. We want to prove that {p n } n lie on T i (xz) = xa 1 A 2... A i B i B i 1... B 1 yw an that they oun T i (xz) on the right. Then the point sequence {p n } n where p i is symmetric to p i lies on σ i an ouns it to the left. 11

12 σ i A 1 B 1 A 2 A 1 Figure 9: T i ([x, z]) = xa 1 A 2... A i B i B i 1... B 1 ywσ i It is clear that p 0 lies on T i (xz) = x..., an p 1 = p 0 + (f 0, f 1 ) as well ecause A 1 has with f 0 an height f 1. Similarly, k i, if p k 1 is the last point in xa 1... A k 1 then p k is clearly the last point in xa 1... A k 1 A k ecause (w(a k ), h(a k )) = (f 2n 2, f 2n 1 ) = (AB) k. Thus p 0, p 1,..., p i lie on T i (xz). Now, p i+1 lies at the en of the path xa 1 A 2... A i B i ecause p i lies at the en of the supath xa 1 A 2... A i an p i+1 = p i + (AB) i+1 = p i + (w(b i ), h(b i )). k i + 2 k 2i if p k 1 lies at the en of path xa 1 A 2... A i B i... B 2i+1 (k 1) then p k lies at the en of path xa 1 A 2... A i B i... B 2i+1 (k 1) B 2i+1 k since p k = p k 1 +(AB) k = p k 1 +(w(b 2i+1 k ), h(b 2i+1 k )). Lastly p 2i+1 = p 2i + (0, 1) clearly lies in xa 1 A 2... A i B i B i 1... B 1 yw since p 2i is the enpoint of xa 1 A 2... A i B i B i 1... B 1. On a lock A k, it is easy to see that line connecting the enpoints q 0, q 1 of A k ouns all points in A k to the right. Taking a point q in A k = B k 1 A k 1 ut not at the enpoints, q is either the junction point of A k = B k 1, A k 1 or it is not. If it is then q 0 = (x, y) = q = (x+w(b k 1 ), y+h(b k 1 )) = (x + f 2k 3, y + f 2k 2 ) an q 1 = (x + w(a k ), y + h(a k )) = (x + f 2k 2, y + f 2k 1 ) an you can check that q 0, q, q 1 are oriente clockwise. If q is not the junction point of A k = B k 1, A k 1, let q 2 e this junction point an apply a simiplar argument to q 0, q, q 2 or q 2, q, q 1 as appropiate. A similar argument shows that a line on the enpoints of a B k lock ouns all points in B k to the right. Thus {l n } n ouns T i (xz) to the right. Furthermore, the sequence of lines {l n } n is of increasing slope (this follows from the ientities f k+1 f k+2 f k f k+3 = ( 1) k+1 an fk 2 f k 1f k+1 = ( 1) k which can easily e prove inuctively). Then {p n } n lies on σ i an ouns σ i to the left as esire, thus the set of vertices of the convex hull of T i ([x, ]) is V C = {p n } {p n } (ecause they oun T i ([x, ]) an are convex y our slope argument) which has 4i + 4 elements. Then the resulting polygon P Ti has (4i + 4) 4 vertices, as shown in Figure 10. Lemma For all i Z +, T = T 2 (T 1 T c) i+1 gives rise to a polytope P T with 4i + 2 vertices. Proof. With T i, A k, B k as in the proof of Lemma 3.1.1, y Fact composition of T 1 with T i simply increases the with of A 1, B 1 y 1, thus it oules the with of A k, B k. Exactly the same argument as aove applies to otain the convex hull of 4i + 4 vertices shown in Figure 12 for the walk shown in Figure 11 an as seen in Figure 12, this yiels a polygon of 4i + 2 vertices. 12

13 2i + 4 2i + 1 2i + 3 2i 2i + 2 2i + 1 2i + 5 2i i + 3 4i 4i Figure 10 T 1 (A 1 ) T 1 (B 1 ) T 1 (A 1 ) Figure 11 13

14 2i + 4 2i + 3 2i + 3 2i + 2 2i + 1 2i + 2 2i + 1 2i + 5 2i i + 3 4i i Figure 12 Corollary Given k Z +, a polygon with 2k vertices is otainale. Proof. This follows irectly from Lemma an Lemma For o k, one such polygon P T is given y T = T 2 (T 1 T c) k+1 2. For even k, one such polygon is given y T = T 1 (T 1 T c) k Future Directions 4.1. Enumerating an Detecting Simple Close Curves There are several reasons why preicting or generalizing the relator (an thus the polygon) that will arise from a certain sequence of Dehn Twists applie to [x, z] is quite ifficult. Fact Two Dehn twists T γ1, T γ2 commute if an only if the two curves γ 1, γ 2 amit nonintersecting representatives. Because any consecutive pair of Dehn Twists in our list T a, T, T c, T, T e of generators of Mo(T 2 #T 2 ) on t commute, there oes not seem to e a canonical transformation inuce on P T y composing T γ T for γ = a,, c,, e, since whatever transformation on r an thus on P T epens on the previous Dehn twists present in T. One reasonale approach for trying to prove the otainaility of all polygons following Lemma is confining the search range to polygons within an m n ox an trying to oun the length of T as a wor in T a, T, T c, T, T e. Reucing the wor omain for an n m ox to a finite one coul lea to goo inuctive proofs of the otainaility of polygons (y inucting on its total imension m + n). 14

15 Fact Let H enote the genus 2 hanleoy. Mo(H) < Mo( H) = Mo(T 2 #T 2 ) Some elements of Mo(T 2 #T 2 ) are also in the Mapping Class Group of the two-hanleoy.. In particular, T a, T c, T e Mo(H) an thus P Tγ T = P T for γ = a,, c. That is, within a wor T = T γn... T γ2 T γ1 there coul e any length of trivial suwors. This makes ouning wor size a lot more ifficult. This motivate us to look at other ways to enumerate separating simple close curves without using the Mapping Class Group, an using expressions on the Funamental Group irectly. For instance, [2] provies a way of listing all simple close curves on a once puncutre torus an [1] provies an algorithm for etecting whether a curve on a surface amits a simple representative The Coe We inten to make the coe for rawing the Thurston unit alls availale online. We coul also use this computational tool to a information to we archives such as The Knot Atlas, or maye incorporate the coe to existing topology software. 15

16 A. Interpretation A.1. The Interval Case Let s look at what information we can gather from the following output case (Figure 13). The walk Figure 13 on Z 2 that yiels this interval is encoe y the relator xy m x 1 y m. One way to make a manifol with the funamental group x, y xy m x 1 y m is y ientifying the orange curves an the lue curves in Figure 14. Notice that the orange an lue curves on the surface of the soli torus trace out an annulus, which is homeomorphic to a cylliner. Thus, restricting our attention to the interaction etween this annulus an the cylliner we are ientifying onto it, we can see that x an y m form loops on a torus an thus commute (Figure 15). Now that we have a 3-manifol M with the appropiate funamental group π 1 (M) = x, y xy m x 1 y m, we can erive some information aout it from the polygon in Figure 13. Namely, that if φ H 1 (M, Z) is the element of the first cohomology that sens x 1 an y 0 (first asis element of the cohomology) an ψ H 1 (M, Z) is the element of the first cohomology that sens x 0 an y 1 (secon asis element of the cohomology), then their thurston norms are given y x M (φ) = 0 x M (ψ) = m 1 ecause the iggest thickness of the polygon in the irection of x is 0 an in the irection of y is m 1. Then, the simplest ual surface to the loop x has Euler characteristic 0, so we can fin a surface of Euler characteristic 0 (coul e a torus, an annulus, a Möius strip,... ) emee in M. The simplest ual surface to the loop y has Euler characteristic m 1, so we can fin a surface S with χ(s) = 1 m emee in M. In the case where m = 2, these emee surfaces are not too har to visualize. Our manifol M is forme y pasting the pieces in Figure 14 along like-colore curve, with the orange an lue curves wining aroun the surface of the soli torus only twice. Before pasting, a ual surface to y looks like a isc (Figure 16), an we want to visualize how this isc extens when we attach the cyliner to the torus. The ual surface of y has its ounary on the ounary of M. When attaching the cyliner, the annulus etween the orange an lue curves is remove from the ounary (thus we can remove two pieces from the circle) an the trace parts on the cyliner get ae to the ounary (thus we can a two umps to the circle) as in Figure 17. But the cyliner is not soli an has a hole in the mile, which is also part of the ounary, so we can a two punctures to our umps. 16

17 Figure 14: The orange an lue curves, separate y a istance of ɛ, win aroun y m times on the surface of the soli torus. x is the loop that traverses C I (the thickene cyliner) an closes up in the ɛ-region oune y the lue an orange curves on the surface of the soli torus. Figure 15: y m an x lie on a torus an thus commute 17

18 Figure 16: Dual surface to y in the soli torus Figure 17: Extening to the ual surface of y in x, y xy 2 x 1 y 2 This surface shoul e ual to y in M. The surface is homotopy equivalent to two wege circles, so the Euler characteristic is 1. The fact that the Thurston norm is 1 (which we gathere from the output of the algorithm) assures that we cannot o any etter: this is the simplest surface ual to y. A.2. 1 = 1 example Suppose we have a group G with a presentation π = x, y xy 3 x 2 y 3 x 1 yx 1. This group has 1 = 1. To apply Lemma we construct the following epimorphism φ : G Z φ : x 1 y 1 In general, if the relator of a (2,1) group presentation with 1 = 1 is of the form x ɛ1 y δ1... x ɛ k y δ k an we let p = k i=1 ɛ i an q = k i=1 δ i we can efine an epimorphism from the group to Z y q ψ : x gc(p, q) y p gc(p, q) In their proof of Lemma 1.3.1, Friel an Tillmann escrie an algorithm for converting π to the right presentation, which we apply here irectly. Letting r = xy 3 x 2 y 3 x 1 yx 1 e the relator of π { 1. Pick min φ(x), φ(y) = φ(x) an let c = yx ɛ 1 if φ(x), φ(y) iffer in sign where ɛ = 1 otherwise 2. Replace y with cx ɛ = cx 1 in r an let the result e s s = x(cx 1 ) 3 x 2 (xc 1 ) 3 x 1 cx 1 x 1 sustitute = cx 1 cx 1 cx 2 c 1 xc 1 xc 1 x 1 cx 1 cyclically reuce 3. The esire presentation for G is π = a, c ca 1 ca 1 ca 2 c 1 ac 1 ac 1 a 1 ca 1 Now we raw the polygon associate to this presentation π an erive information aout the Thurston norm exactly the same way as in Section 2.1, ut since φ(c) = 0, x M (φ c ) = max{φ(p) φ(q) p, q (ker φ c ) P } = 0 ecause (ker φ c ) = 0, an any thickness of P in the c irection can thus e ignore. 18

19 References [1] J. Birman an C. Series. An algorithm for simple curves on surfaces. J. Lonon Math. Soc., 29(2): , [2] P. Buser an K.-D. Semmler. The geometry an spectrum of the one hole torus. Comment. Math. Helvetici, 63: , [3] S. Friel, K. Schreve, an S. Tillmann. Thurston norm via Fox calculus. ArXiv e-prints, July [4] Stefan Friel an Stephan Tillmann. Two-generator one-relator groups an marke polytopes. arxiv preprint arxiv: , [5] Martin Kassaov. Generating mapping class groups y involutions. arxiv preprint math/ ,