Grid graphs and Lattice surfaces - Cornell Universitypi.math.cornell.edu/~hrbaik/math711_nov2010.pdf · Veech groups Grid graphs and Lattice surfaces Based on Patrick Hooper’s paper

PreliminariesA construction of translation surfaces

Semi-regular DecompositionVeech groups

Grid graphs and Lattice surfacesBased on Patrick Hooper’s paper

Hyungryul Baik

Department of MathematicsCornell University

Seminar in Analysis, MATH 7110, Nov. 2010

Hyungryul Baik Grid graphs and Lattice surfaces



Outline

1 Preliminaries

2 A construction of translation surfacesFrom bipartite ribbon graphsFrom grid graphs

3 Semi-regular DecompositionSemi-regular polygonsSemi-regular Decomposition

4 Veech groups




Translation surface

A translation surface (X , ω) is a Riemann surface X equippedwith a non-zero holomorphic 1-form ω. The 1-form ω provideslocal charts from X to C defined up to translation, away fromthe zeros of ω. At a zero, we have a chart to the Riemannsurface w = zk+1, where k is the order of the zero. So, atranslation surface inherits a singular Euclidean metric bypulling back the Euclidean metric via the charts. A zero of orderk of ω yields a cone singularity with cone angle 2π(k + 1).Equivalently, we can see the translation surfaces as a finiteunion of polygonal subsets of C as we did in the class.




Affine homeomorphisms and Derivatives

Definition #1

Let (X , ω) and (X0, ω0) be two translation surfaces. Ahomeomorphism f : X → X0 is called affine if it preserves

underlying affine structures. That is, there are realnumbers a,b, c,d such that on local charts

f (x + iy) = (ax + by) + i(cx + dy).

Definition #2

We call an affine homeomorphism from (X , ω) to itself anaffine automorphism. Let Aff (X , ω) be the collection of allaffine automorphisms of (X , ω). This forms a group under

the composition.





Definition #1

Let (X , ω) and (X0, ω0) be two translation surfaces. Ahomeomorphism f : X → X0 is called affine if it preserves

underlying affine structures. That is, there are realnumbers a,b, c,d such that on local charts

f (x + iy) = (ax + by) + i(cx + dy).

Definition #2

We call an affine homeomorphism from (X , ω) to itself anaffine automorphism. Let Aff (X , ω) be the collection of allaffine automorphisms of (X , ω). This forms a group under

the composition.





Definition #3

A derivative map D : Aff (X , ω)→ GL(2,R) is given on

local charts by f 7−→(

a bc d

).

Definition #4

Let GL(X , ω) be the image D(Aff (X , ω)) ⊂ GL(2,R). AVeech group is the orientation preserving part of

GL(X , ω), namelySL(X , ω) = D+(Aff (X , ω)) = GL(X , ω) ∩ SL(2,R). Let

PGL(X , ω) and PSL(X , ω) be the projectivizations of thesegroups.





Definition #3

A derivative map D : Aff (X , ω)→ GL(2,R) is given on

local charts by f 7−→(

a bc d

).

Definition #4

Let GL(X , ω) be the image D(Aff (X , ω)) ⊂ GL(2,R). AVeech group is the orientation preserving part of

GL(X , ω), namelySL(X , ω) = D+(Aff (X , ω)) = GL(X , ω) ∩ SL(2,R). Let

PGL(X , ω) and PSL(X , ω) be the projectivizations of thesegroups.




More Definitions

Definition #5

A translation surface (X , ω) is said to have the latticeproperty if SL(X , ω) has finite co-volume.

Definition #6

Two subgroups Γ1, Γ2 ⊂ SL(2,R) are commensurable ifthere are finite index subgroups G1 ⊂ Γ1 and G2 ⊂ Γ2

which are conjugate in SL(2,R). A subgroup of SL(2,R)is arithmetic if it is commensurable to SL(2,Z).




More Definitions

Definition #5

A translation surface (X , ω) is said to have the latticeproperty if SL(X , ω) has finite co-volume.

Definition #6

Two subgroups Γ1, Γ2 ⊂ SL(2,R) are commensurable ifthere are finite index subgroups G1 ⊂ Γ1 and G2 ⊂ Γ2

which are conjugate in SL(2,R). A subgroup of SL(2,R)is arithmetic if it is commensurable to SL(2,Z).




From bipartite ribbon graphsFrom grid graphs

Outline

1 Preliminaries



4 Veech groups





Bipartite Ribbon Graph

Definition #7A bipartite ribbon graph is a finite simple connected graph Gwith vertex set V and edge set E , equipped with twopermutations n, e : E → E that satisfy the following conditions.

The vertex set V is a disjoint union of two sets A and B.There are functions α : E → A and β : E → B such thatevery edge e ∈ E joins vertex α(e) ∈ A the vertexβ(e) ∈ B.For all e ∈ E , the orbit Oe(e) = {ek (e) : k ∈ N} satisfiesα(Oe(e)) = α(e). Similarly, the orbitOn(e) = {nk (e) : k ∈ N} satisfies β(On(e)) = β(e).n, e act transitively on E .





How it looks like?





Constructing a surface

Say we have a function w : V → R>0 with the aboveinformation. For e ∈ E , define a rectangleRe = [0,w ◦ β(e)]× [0,w ◦ α(e)]. We isometrically identify theright side of Re to the left side of Re(e), and the top side of Re tothe bottom side of Rn(e).





Example





Example





Constructing a surface

The translation surface we constructed from a bipartiteribbon graph G and a positive real-valued function w on Vis denoted by (XG,w , ωG,w )Note that (XG,w , ωG,w ) has both vertical and horizontalcylinder decomposition.A horizontal cylinder comes from an orbit Oe(e) of an edgee which shares a vertex in A. Similar observation could bemade for vertical cylinders, so we have the followingbijective correspondences:

(horizontal cylinders)↔ (orbits of e)↔ (elements in A)(vertical cylinders)↔ (orbits of n)↔ (elements in B)

Given a ∈ A and b ∈ B, the respective cylinders are∪e∈α−1(a)Re and ∪e∈β−1(a)Re, respectively.





Eigenfunction

Definition #8A function w : V → R>0 is an eigenfunction of G correspondingto the eigenvalue λ if for all x ∈ V,

∑xy∈E w(y) = λw(x).

For a given bipartite (ribbon) graph, there is a unique positiveeigenfunction up to scalar multiplication. It could be proved bythe following theorem.





Perron-Frobenius theorem

Theorem #1 [Perron-Frobenius]Let A be an essentially positive square matrix, ie., there existsk ∈ N such that all entries of Ak are positive numbers. Thenthere exists a real positive eigenvalue λ1 which is the simpleroot of the characteristic equation of A such that any othereigenvalue λj with j > 1 satisfies λ1 > |λj |. The eigenvector v1

for the eigenvalue λ1 can be chosen with all entries strictlypositive. In fact, all other eigenvectors have components of bothsigns.

By the theorem, λ is just the larger positive real eigenvalue ofthe incidence matrix of the graph, and w is given by thecorresponding strictly positive eigenvector.





Parabolic elements corresponding to multi-twists

If w is a positive eigenfunction with eigenvalue λ, then theVeech group of (XG,w , ωG,w ) contains two parabolic elements

P0 =

(1 λ0 1

)and Q0 =

(1 0−λ 1

)which corresponds to two

multi-twists along the boundaries of horizontal and verticalcylinder decomposition. Very same conclusion was alreadyshown in Chenxi’s part.





Dehn Twist





Dehn Twist

Assume that we have |Oe(e)| = 4 for some edge e. Under theDehn twist for the horizontal cylinder for α(e): (1,0) 7−→ (1,0)and (0,w(α(e))) 7−→ (λw(α(e)),w(α(e)), so (0,1) 7−→ (λ,1).

Note that this happens for any horizontal cylinder in the exactlysame way. So, the horizonal Dehn twist acts on the surface as

an affine homeomorphism P0 =

(1 λ0 1

).





Outline

1 Preliminaries



4 Veech groups





Grid Graph

Definition #9 Grid graph

For given m ≥ 2,n ≥ 2 with mn ≥ 6, a (m,n) grid graph Gm,n isa a graph with V = {vi,j : 1 ≤ i < m,1 ≤ j < n} andE = {vi,jvk ,l : (i − k)2 + (j − l)2 = 1} with the followings:

A = {vi,j ∈ V : i + j is even} andB = {vi,j ∈ V : i + j is odd}.The permutations e, n are determined by cyclic ordering forthe edges around each vertex vi,j . Embedding Gm,n into R2

in a trivial way, we choose the clockwise ordering aroundvi,j when i is even, and the counter-clockwise orderingwhen i is odd.





Grid Graph Example

It is not hard to show that the eigenfunction is given by theequation w(vi,j) = sin( iπ

m ) sin( jπn ). Using the construction of a

translation surface from a bipartite ribbon graph given inprevious part, we get a translation surface which would bedenoted by (Xm,n, ωm,n).




Semi-regular polygonsSemi-regular Decomposition

Outline

1 Preliminaries



4 Veech groups





Semi-regular polygon

Definition #10 Semi-regular polygon

The (a,b)-semiregular 2n-gon is the 2n-gon whose edgevectors (oriented counter-clockwise) are given by

vi =

{a(cos iπ

n , sin iπn ) if i is even

b(cos iπn , sin iπ

n ) if i is oddfor i = 0, . . . ,2n − 1.

Denote this 2n-gon by Pn(a,b). We are restricting the caseswhere a ≥ 0,b ≥ 0 but a 6= 0 or b 6= 0. In the case where oneof a or b is zero, Pn(a,b) degenerates to a regular n-gon. In thecase where a = 0 or b = 0 and n = 2, it degenerated to anedge.





Semi-regular polygons

Fix m and n, and define the polygons P(k) for k = 0, . . . ,m − 1by

P(k) =

Pn(sin (k+1)π

m , sin kπn ) if n is odd

Pn(sin kπm , sin (k+1)π

n ) if n is even and k is evenPn(sin (k+1)π

m , sin kπn ) if n is even and k is odd

We form a translation surface by identifying the edges of thepolygons P(k) in pairs. For k odd, we identify the even sides ofP(k) with the opposite side of P(k + 1), and identify the oddsides of P(k) with the opposite side of P(k − 1). The cases inthe definition of P(k) are chosen so that this gluing makessense. Let (Ym,n, ηm,n) denote the resulting translation surface.





Example for m = 4,n = 5

k = 0P(0) = P5(sin π

4 ,0)vi =

1√2(cos iπ

5 , sin iπ5 ) for i = 0,2,4,6,8.

v0 = ( 1√2,0), so P(0) is a regular pentagon where the

bottom edge is parallel to x-axis and each edge has length1√2.

k = 1P(1) = P5(sin 2π

4 , sin π4 ) = P5(1, 1√

2).

vi =

{(cos iπ

5 , sin iπ5 ) if i = 0,2,4,6,8

1√2(cos iπ

5 , sin iπ5 ) if i = 1,3,5,7,9






In this way, we can get P(0),P(1),P(2),P(3) and they look likethe following:






And the next one shows the gluing information:





Singularities

Theorem #2 [Singularities of (Ym,n, ηm,n)]

Let γ = gcd(m,n). There are γ equivalence classes of verticesof the decomposition into polygons. Each of these points hascone angle 2π(mn −m − n)/γ.

We know the semi-regular polygonal decomposition(Ym,n, ηm,n) = ∪m−1

i=0 P(i) has m faces and (m − 1)n edges,since n edges of P(i) and n edges of P(i + 1) are identified.Plus, as a corollary of the above theorem, we know know thishas γ vertices. Hence the Euler characteristic of (Ym,n, ηm,n) ism − (m − 1)n + γ = m + n + γ −mn.





Singularities

proof. Let v0 be a vector based at a vertex v of P(0), which ispointing along the boundary of P(0) in a counterclockwisedirection. We will rotate this vector counterclockwise around thepoint v of (Ym,n, ηm,n). Since P(0) is a regular n-gon, we reachP(1) after we rotate by π − 2π

n . Inside P(1), we need to rotateby π − 2π

2n to reach P(2). In this way, we reach to P(m − 1) afterwe go through P(2), . . . ,P(m − 2). Hence We have rotated by(π − 2π

n ) + (m − 2)π − 2πn . Now P(m − 1) is again a regular

n-gon and now we go through the polygons backward. Soagain we need to rotate by (π − 2π

n ) + (m − 2)π − 2πn to reach

P(0). So far, we have rotated by 2(π − 2πn ) + 2(m − 2)π − 2π

nuntil we go back to P(0).





Singularities

proof continued. We keep rotating until the total angle ofrotation becomes a multiple of 2π. (We have a translationsurface, and the cone angle should a multiple of 2π.) Thus thecone angle at v is x(2(π− 2π

n ) + 2(m− 2)π− 2πn ) = 2xπmn−m−n

nwhere x is the smallest positive integer for which this number isa multiple of 2π. But x mn−m−n

n = x(m − 1− mn ) so it is an

integer if and only if x mn is an integer. Now the smallest such x

is obviously nγ . Now we conclude that the cone angle at v is

2π(mn−m− n)/γ. Since any singularity is of this form, and thesum of all the angles of polygons P(0), . . . ,P(m − 1) is2π(nm −m − n). Hence there are γ total singularities. �.





Outline

1 Preliminaries



4 Veech groups





Decomposition into semi-regular polygons

Theorem #3 [Main Theorem]There are affine homeomorphismsµ : (Xm,n, ωm,n)→ (Ym,n, ηm,n) andν : (Xm,n, ωm,n)→ (Yn,m, ηn,m) with derivatives

D(µ) =

(csc π

n cot πn0 −1

)and D(ν) =

(− csc π

m cot πm0 −1

)





Step1. Xm,n → Xn,m

LemmaThere is an affine homeomorphism

ρ : (Xm,n, ωm,n)→ (Xn,m, ωn,m) with derivative E =

(−1 00 1

).

First consider the graph homomorphism ρ : Gm,n → Gn,m withvi,j 7−→ vj,i . It satisfies

ρ(A) = A and ρ(B) = B.ρ ◦ em,n = e−1

n,m ◦ ρ.ρ ◦ nm,n = nn,m ◦ ρwm,n ◦ ρ = wn,m.





Step1. Xm,n → Xn,m

Then from the picture, we can see that the map ρ extends to anaffine automorphism ρ : (Xm,n, ωm,n)→ (Xn,m, ωn,m).





Step2. Augmented Grid graph

Recall that the nodes of Gm,n are in bijection with thecoordinates (i , j) ∈ Z2 with 0 < i < m,0 < j < n. Thenodes of G′m,n will be in bijection with the coordinates(i , j) ∈ Z2 with 0 ≤ i ≤ m,0 ≤ j ≤ n. Edges are defined asbefore.E ′ and V ′ denote the set of all edges and the set of alledges of G′m,n, respectively.e′, n′ : E ′ → E ′ are defined as before.The added nodes are called degenerate nodes and thenew edges are degenerate edges.





Step3. Polygonal decomposition of Xm,n

For e ∈ E ′, let d(e) denote the vector witch points along thepositive diagonal of Re, oriented rightward and upward.L(e) denotes the lower triangle below d(e), and U(e)denotes the upper triangle above d(e).Hk == {vk ,ivk+1,i ∈ E ′ : 0 < i < n} for k = 0, . . . ,m − 1.For each k , let Q(k) =∪e∈Hk Re ∪ L(n′(e)) ∪ L(e′−1(e)) ∪ U(n′−1(e)) ∪ U(e′(e)).






Let k be a fixed odd integer. Let ei = vk ,ivk+1,i for 0 ≤ i ≤ n. If0 < i < n, then ei ∈ Hk . Since k is odd,n′(e′(ei)) = e′(n′(ei)) = ei+1.So, e′(ei) = n′−1(ei+1) and n′(ei) = e′−1(ei+1). ThusU(e′(ei)) = U(n′−1(ei+1)) and L(n′(ei)) = L(e′−1(ei+1)).Hence many triangles in the definition of Q(k) are mentionedtwice. And then, for i = 0, . . . ,n − 2, the top right coordinatevertex of R(ei) is nonsingular. So the rectangles form a chainwhich goes up and right.











In our example, Q(1) should look like the following:






When k = m− 1, then the half of the edges degenerate. To seethis more explicitly, let’s see Q(3) for our example.






In case of k even, similar analysis showsn′−1(e′−1(ei)) = e′−1(n′−1(ei)) = ei+1. So, e′−1(ei) = n′(ei+1)and n′−1(ei) = e′(ei+1). Thus L(e′−1(ei)) = L(n′(ei+1)) andU(n′−1(ei)) = U(e′(ei+1)). At the bottom left coordinate vertexof R(ei) should be like the following:

So Q(2) is a 10-gon which is similar with Q(1), and Q(0) isagain a pentagon as Q(3), since half of the edges collapse.





Step4. Q(k) vs. P(k) when k is odd

Let M =

(csc π

n cot πn0 −1

)and fix k odd.

Let ui for i = 0, . . . ,2n − 1 be the edge vectors of Q(k)oriented counterclockwise around Q(k).u0 is the lower horizontal edge of the rectangle R(e1)

most of ui are actually positive diagonals of somerectanglesIn fact all of them are. u0 = d(n′−1(e1)) = d(αk+1,0βk+1,1)






We have the following:

ui =

d(αk+1,iβk+1,i+1) if i < n and i evend(αk ,iβk ,i+1) if i < n and i odd−d(βk+1,2n−1−iαk+1,2n−i) if i ≥ n and i even−d(βk ,2n−1−iαk ,2n−i) if i ≥ n and i odd

We can

actually get the vectors explicitly, using the eigenfunctionw(vi,j) = sin( iπ

m ) sin( jπn )






I will do it for d(αk+1,iβk+1,i+1) as an example.Here we know thatR(αk+1,iβk+1,i+1) = [0,w(βk+1,i+1)]× [0,w(αk+1,i)]. Alsow(αk+1,i) = sin (k+1)π

m sin iπn and

w(βk+1,i+1) = sin (k+1)πm sin (i+1)π

n . Thusd(αk+1,iβk+1,i+1) = sin (k+1)π

m (sin (i+1)πn , sin iπ

n ).If we do this job for all other cases, we get the following:

ui =

{sin (k+1)π


n ) if i evensin kπ


n ) if i odd






And by east computation,

Mui =

{sin (k+1)π

m (cos iπn , sin iπ

n ) if i evensin kπ

m (cos iπn , sin iπ

n ) if i odd

These are exactly same with the edge vectors of P(k). So wegot M(Qk ) = Pk for k odd. The case of k even is more or lessthe same.





Step5. Gluing information

The only thing left is to check whether Q(k) are glued togetherin the same how P(k) are glued together. We will see whichedge is identified with ui of Q(k) for some chosen i , k . Thereare four cases: i is even and i < n, i is even and i ≥ n, i is oddand i < n, i is odd and i ≥ n. All cases are similar so I will doonly for the case of i is even and i < n. Up to sign, ui is thepositive diagonal of the rectangle R(αk+1,iβk+1,i+1). Theopposite side of Q(k + 1) is the edge vector wi+n, and this isthe positive diagonal of the rectangle R(αk+1,iβk+1,i+1) with theopposite direction. Hence they are identified, and this is samewith the gluing rule for P(k).





Step6. Lemma revisited.

Let µ be an affine homeomorphism from (Xm,n, ωm,n) to(Ym,n, ηm,n) whose derivative is M. We just showed theexistence of such a map. By the Lemma in Step 1, there is anaffine homeomorphism ρ : (Xm,n, ωm,n)→ (Xn,m, ωn,m) with

derivative E =

(−1 00 1

). The previous part guarantees that

there exits an affine homeomorphism µ′ : Xn,m → Yn,m. Nowdefine ν as µ′ ◦ ρ. Then we have D(ν) = D(µ′) · D(ρ) =(

csc πm cot πm

0 −1

)(−1 00 1

)=

(− csc π

n cot πn0 −1

). Now we finally

get our main theorem.




Triangle Groups

Definition #11The triangle group 4(m,n,∞) is the subgroup ofPGL(2,R) = Isom(H2) generated by reflections in the sides ofa hyperbolic triangle with one ideal vertex and two angles ofπ/m, π/n. It has the presentation< a,b, c : a2 = b2 = c2 = (ac)m = (bc)n = e >.

A =

(−1 −2 cos( πm )0 1

), B =

(−1 2 cos(πn )0 1

)C =

(0 −1−1 0

), E =

(−1 00 1

).




Triangle Groups

A,B,C,E are 2× 2 matrices of determinant −1 witheigenvalues ±1.it is easy to check thatA2 = B2 = C2 = I, (AC)m = (BC)n = −I, while AB isparabolic.the projection of the subgroup < A,B,C >⊂ GL(2,R) toPGL(2,R) is conjugate to the triangle group 4(m,n,∞).




Orthogonal Groups

Definition #12

Let Yn =

(cos π

n − sin πn

− sin πn − cos π

n

).

Let O(X , ω) = GL(X , ω) ∩O(2,R)..

Proposition

Suppose (Ym,n, ηm,n) is not a torus. If m and n are not botheven, then O(Ym,n, ηm,n) =< E ,Yn >. If both m and n are even,then O(Ym,n, ηm,n) =< E ,YnEYn >.




Orthogonal Groups

Definition #12

Let Yn =

(cos π

n − sin πn

− sin πn − cos π

n

).

Let O(X , ω) = GL(X , ω) ∩O(2,R)..

Proposition

Suppose (Ym,n, ηm,n) is not a torus. If m and n are not botheven, then O(Ym,n, ηm,n) =< E ,Yn >. If both m and n are even,then O(Ym,n, ηm,n) =< E ,YnEYn >.




Orthogonal Groups Continued

Proof of the Proposition: Recall that Ym,n is a union of thesemiregular 2n-gons P(0), . . . ,P(m − 1). Both E and YnEYnare symmetries of every semiregular 2n-gon.

YnEYn =

(sin2(πn )− cos2(πn ) 2 sin(πn ) cos(πn )

2 sin(πn ) cos(πn ) − sin2(πn ) + cos2(πn )

).

By elementary trigonometry, we get

YnEYn =

(cos(2π

n ) sin(2πn )

sin(2πn ) − cos(2π

n )

).





Pick any P(k) of Ym,n and let vi be its edge vector. Thenvi = α(cos( iπ

n ), sin( iπn )) where α = sin kπ

n when i is odd, andα = sin (k+1)π

n when i is even. Then,

YnEYn · vi = α

(cos(2π

n ) sin(2πn )

sin(2πn ) − cos(2π

n )

)·(

cos( iπn )

sin( iπn )

)=

α(cos( (2−i)πn ), sin( (2−i)π

n )). Noting that vi = vj where i = j(mod 2n), we know that they are affine automorphisms of Ym,nwhich preserves each of the polygons P(k). In addition, whenm or n is odd, then Yn(P(i)) = P(m − 1− i), up to translation.This action extends to an affine automorphism of Xm,n withderivative Yn.





Conversely, suppose M ∈ O(Ym,n). Then associatedautomorphism must permute the shortest saddle connections(a geodesic segment which intersects the zeros of η preciselyat its endpoints), which are just boundaries of P(0) andP(m − 1). (Since we assumed that Ym,n is not a torus, allvertices are singularities.) In particular, M must preserve theset of directions in which these shortest saddle connectionspoint. When m,n are not both even, the group of matrices withthis property is < E ,Yn >, and when m,n are even, it is< E ,YnEYn >. So the converse holds.




Veech Groups

Theorem #4 [Veech Groups]Let m ≥ 2,n ≥ 2 be integers with mn ≥ 6.

if m 6= n, then- When m and n are not both even,GL(Xm,n, ωm,n) =< A,B,C >.- When m and n are both even,GL(Xm,n, ωm,n) =< A,B,CAC,CBC >. (This is an indextwo subgroup of < A,B,C >.)if m = n, then- When m is odd, GL(Xm,n, ωm,n) =< A,C,E >.- When m is even, GL(Xm,n, ωm,n) =< A,E ,CAC >. (Thisis a reflection group in an (m

2 ,∞,∞) triangle.)




Veech Groups

First note that the matrices satisfy the relations:A = D(ν)−1 ◦ E ◦ D(ν),B = D(µ)−1 ◦ E ◦ D(µ) andC = −I ◦ D(µ)−1 ◦ Yn ◦ D(µ) = D(ν)−1 ◦ Yn ◦ D(ν). ThenCAC = D(ν)−1YnEYnD(ν) and CBC = −D(µ)−1YnEYnD(µ).First of all, by the main theorem, we know that A,B are inGL(Xm,n, ωm,n). (Just by pulling back the automorphism.)Similarly, when m and n are not both even, by the proposition,Yn ∈ O(Ym,n, ηm,n) so that C ∈ GL(Xm,n, ωm,n). And when mand n are both even, YnEYn ∈ O(Ym,n, ηm,n) so thatCAC,CBC ∈ GL(Xm,n, ωm,n). This shows the Veech groupcontains the groups appearing in the above theorem. Actuallythese are the whole Veech group, but the proof is not givenhere.




References

Grid graphs and lattice surfaces, W. Patrick HooperOn groups generated by two positive multi-twists:Teichm̈uller curves and Lehmer’s number, Christopher JLeiningerOn the geometry and dynamics on diffeomorphims ofsurfaces, William P. Thurston


Documents

Grid graphs and Lattice surfaces - Cornell Universitypi.math.cornell.edu/~hrbaik/math711_nov2010.pdf · Veech groups Grid graphs and Lattice surfaces Based on Patrick Hooper’s paper