Non-geometric veering triangulations

Ahmad Issa

University of Texas at Austin

Joint work with Craig Hodgson and Henry Segerman

1

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

I Today we consider ideal triangulations. An ideal triangulationof a 3-manifold M is a decomposition of M into tetrahedrawith their vertices removed.

I In 2010 Ian Agol gave an elegant construction of idealtriangulations of a subclass of hyperbolic 3-manifolds,characterised by a combinatorial condition which he calledveering.

I He posed the question: is every veering triangulationgeometric, i.e. realised by positive volume ideal hyperbolictetrahedra in the complete hyperbolic structure?

I Previously, this was verified for many examples.

I Answer: No! We found examples of non-geometric veeringtriangulations.

I Conjecture: Every cusped hyperbolic 3-manifold admits ageometric ideal triangulation.

2

Mapping torus of a homeomorphism

I The veering triangulations we are interested in aretriangulations of a subclass of mapping tori.

I The mapping torus of a homeomorphism ϕ : S → S of asurface S is the 3-manifold

Mϕ = S × [0, 1]/{(x , 0) ∼ (ϕ(x), 1)}.

φ

S x {0}

S x {1}

Hyperbolization theorem for mapping tori

The mapping torus Mϕ admits a hyperbolic structure if and only ifϕ is (isotopic to) a pseudo-Anosov homeomorphism.

3

Mapping torus of a homeomorphismI The veering triangulations we are interested in are

triangulations of a subclass of mapping tori.

I The mapping torus of a homeomorphism ϕ : S → S of asurface S is the 3-manifold

Mϕ = S × [0, 1]/{(x , 0) ∼ (ϕ(x), 1)}.

φ

S x {0}

S x {1}

Hyperbolization theorem for mapping tori

The mapping torus Mϕ admits a hyperbolic structure if and only ifϕ is (isotopic to) a pseudo-Anosov homeomorphism.

3

Mapping torus of a homeomorphismI The veering triangulations we are interested in are

triangulations of a subclass of mapping tori.I The mapping torus of a homeomorphism ϕ : S → S of a

surface S is the 3-manifold

Mϕ = S × [0, 1]/{(x , 0) ∼ (ϕ(x), 1)}.

φ

S x {0}

S x {1}

Hyperbolization theorem for mapping tori

The mapping torus Mϕ admits a hyperbolic structure if and only ifϕ is (isotopic to) a pseudo-Anosov homeomorphism.

3

Mapping torus of a homeomorphismI The veering triangulations we are interested in are

triangulations of a subclass of mapping tori.I The mapping torus of a homeomorphism ϕ : S → S of a

surface S is the 3-manifold

Mϕ = S × [0, 1]/{(x , 0) ∼ (ϕ(x), 1)}.

φ

S x {0}

S x {1}

Hyperbolization theorem for mapping tori

The mapping torus Mϕ admits a hyperbolic structure if and only ifϕ is (isotopic to) a pseudo-Anosov homeomorphism.

3

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

A

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

A

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.

I Starting with v instead of u gives another measured foliation(F s , µs).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I Consider A = ( 2 11 1 ) : R2 → R2.

I Descends to a homeomorphism ϕ : S → S whereS = (R2 − Z2)/Z2 is the once-punctured torus.

I A has eigenvectors u and v with eigenvalues λ > 1 and 1λ < 1.

I Lines parallel to u foliate (partition) R2 − Z2. Projectingthese lines onto S gives a foliation Fu of S into leaves(curves).

I Measure µ on F assigns a positive real number to an arctransverse to F on S .

I ϕ(Fu) = Fu and ϕ(µu) = λµu.I Starting with v instead of u gives another measured foliation

(F s , µs).

-4 -2 0 2 4

-4

-2

0

2

4

-4 -2 0 2 4

-4

-2

0

2

4

A

stretch

contract

u

v

4

I A singular foliation F of a surface S is a foliation of S awayfrom a finite set P of singular points. Points of P are locallymodelled on a k-prong with k ≥ 3. Punctures of S are locallymodelled on a k-prong with k ≥ 1.

Figure : k-prong for k = 1 and k = 3, singular point (or puncture) shownin red.

5

A homeomorphism ϕ : S → S (with (S) < 0), is pseudo-Anosovif there exist measured foliations (Fu, µu), (F s , µs) of S withtransverse leaves and a common set of singular points withϕ(Fu, µu) = (Fu, λµu) and ϕ(F s , µs) = (F s , 1λµs), where λ > 1is called the dilatation of ϕ.

I (Fu, µu) is called the unstable foliation of ϕ and (F s , µs) iscalled the stable foliation.

I A finite segment of a leaf of Fu can be thought of asstretched by a factor of λ.

6

Train tracksI A (measured) train track on a surface S is a finite 1-complexτ ⊂ S with C 1 embedded edges and a positive real numberassigned to each edge such that each vertex is modelled on aswitch (see Figure).

a

b

c

a, b, c > 0

Switch condition:

a = b + c

I We require that each component of S\τ is a disk with at least3 cusps, or a punctured disk with at least 1 cusp.

12.618..

1.618..

7

Train tracksI A (measured) train track on a surface S is a finite 1-complexτ ⊂ S with C 1 embedded edges and a positive real numberassigned to each edge such that each vertex is modelled on aswitch (see Figure).

a

b

c

a, b, c > 0

Switch condition:

a = b + c

I We require that each component of S\τ is a disk with at least3 cusps, or a punctured disk with at least 1 cusp.

12.618..

1.618..

7

Train track to foliationI Step 1: Replace each edge of τ by a foliated Euclidean

rectangle.

1.618...1.618...

anything

I Step 2: Perform identifications of the boundary of foliatedregions to get a foliation (F , µ).

I We say that τ carries the foliation (F , µ).

1.618...2.618...

1

8

Train track to foliationI Step 1: Replace each edge of τ by a foliated Euclidean

rectangle.

1.618...1.618...

anything

I Step 2: Perform identifications of the boundary of foliatedregions to get a foliation (F , µ).

I We say that τ carries the foliation (F , µ).

1.618...2.618...

1

8

Train track to foliationI Step 1: Replace each edge of τ by a foliated Euclidean

rectangle.

1.618...1.618...

anything

I Step 2: Perform identifications of the boundary of foliatedregions to get a foliation (F , µ).

I We say that τ carries the foliation (F , µ).

1.618...2.618...

1

8

Train track to foliationI Step 1: Replace each edge of τ by a foliated Euclidean

rectangle.

1.618...1.618...

anything

I Step 2: Perform identifications of the boundary of foliatedregions to get a foliation (F , µ).

I We say that τ carries the foliation (F , µ).

1.618...2.618...

1

8

Splitting of train tracksI A split of a train track τ on a surface S is a move which

produces a train track τ ′ by ‘splitting’ a large edge of τ in oneof two ways.

I A maximal split is a move which splits every edge of maximalmeasure (weight) simultaneously. We write τ ⇀ τ ′ to denotea maximal split.

e

a

b

c

d

If max(a, d) > max(b, c)

If max(b, c) > max(a, d)

a c

ca

b d

e’ = c - a = b - d

b d

e’ = a - c = d - b

e = a + b = c + d

split edge

9

Splitting of train tracksI A split of a train track τ on a surface S is a move which

produces a train track τ ′ by ‘splitting’ a large edge of τ in oneof two ways.

I A maximal split is a move which splits every edge of maximalmeasure (weight) simultaneously. We write τ ⇀ τ ′ to denotea maximal split.

e

a

b

c

d

If max(a, d) > max(b, c)

If max(b, c) > max(a, d)

a c

ca

b d

e’ = c - a = b - d

b d

e’ = a - c = d - b

e = a + b = c + d

split edge

9

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

S = punctured torus

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

S = punctured torus

τ

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

S = punctured torus

τ

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

S = punctured torus

τ

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

S = (R2 u ∞) \ {4 points}

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

τ

S = (R2 u ∞) \ {4 points}

10

Dual triangulation to a train trackI Let τ be a train track on a surface S .I Assume that every component of S\τ is homeomorphic to a

punctured disk. Then there is an ideal triangulation of S dualto τ .

τ

S = (R2 u ∞) \ {4 points}

10

I What happens to the dual ideal triangulation when a traintrack splits?

I A diagonal exchange occurs!

splits

τ

τ’

τ’

11

I What happens to the dual ideal triangulation when a traintrack splits?

I A diagonal exchange occurs!

splits

τ

τ’

τ’

11

I What happens to the dual ideal triangulation when a traintrack splits?

I A diagonal exchange occurs!

11

I What happens to the dual ideal triangulation when a traintrack splits?

I A diagonal exchange occurs!

T

T’

T’

11

Triangulating a mapping torusA layered triangulation of a mapping torus Mϕ, where ϕ : S → Sis a homeomorphism, is one obtained as follows:

1. Start with an ideal triangulation T0 of the surface S .

2. Pick an edge and replace it by the ‘other’ diagonal edge to geta triangulation T1 of S .

3. Insert a tetrahedron which interpolates between T0 and T1.4. Repeat this to get a sequence of triangulations T0,T1, . . . ,Tn

such that ϕ(T0) = Tn.5. Glue T0 to Tn via ϕ.

T0

12

Triangulating a mapping torusA layered triangulation of a mapping torus Mϕ, where ϕ : S → Sis a homeomorphism, is one obtained as follows:

1. Start with an ideal triangulation T0 of the surface S .2. Pick an edge and replace it by the ‘other’ diagonal edge to get

a triangulation T1 of S .

3. Insert a tetrahedron which interpolates between T0 and T1.4. Repeat this to get a sequence of triangulations T0,T1, . . . ,Tn

such that ϕ(T0) = Tn.5. Glue T0 to Tn via ϕ.

T0

T1

diagonal exchange

12

Triangulating a mapping torusA layered triangulation of a mapping torus Mϕ, where ϕ : S → Sis a homeomorphism, is one obtained as follows:

1. Start with an ideal triangulation T0 of the surface S .2. Pick an edge and replace it by the ‘other’ diagonal edge to get

a triangulation T1 of S .3. Insert a tetrahedron which interpolates between T0 and T1.

4. Repeat this to get a sequence of triangulations T0,T1, . . . ,Tn

such that ϕ(T0) = Tn.5. Glue T0 to Tn via ϕ.

T0

T1ideal tetrahedron

12

Triangulating a mapping torusA layered triangulation of a mapping torus Mϕ, where ϕ : S → Sis a homeomorphism, is one obtained as follows:

1. Start with an ideal triangulation T0 of the surface S .2. Pick an edge and replace it by the ‘other’ diagonal edge to get

a triangulation T1 of S .3. Insert a tetrahedron which interpolates between T0 and T1.4. Repeat this to get a sequence of triangulations T0,T1, . . . ,Tn

such that ϕ(T0) = Tn.

5. Glue T0 to Tn via ϕ.

T0

T1

Tn

φ

12

Triangulating a mapping torusA layered triangulation of a mapping torus Mϕ, where ϕ : S → Sis a homeomorphism, is one obtained as follows:

1. Start with an ideal triangulation T0 of the surface S .2. Pick an edge and replace it by the ‘other’ diagonal edge to get

a triangulation T1 of S .3. Insert a tetrahedron which interpolates between T0 and T1.4. Repeat this to get a sequence of triangulations T0,T1, . . . ,Tn

such that ϕ(T0) = Tn.5. Glue T0 to Tn via ϕ.

T0

T1

Tn

φ

12

Periodic splitting sequence

Theorem (Agol)

Let τ be a train track carrying the unstable foliation Fu of apseudo-Anosov homeomorphism ϕ : S → S . There exist n,m > 0such that

τ ⇀ τ1 ⇀ τ2 ⇀ · · ·⇀ τn ⇀ · · ·⇀ τn+m,

such that ϕ( 1λτn) = τn+m, where λ > 1 is the dilatation of ϕ. We

call τn ⇀ · · ·⇀ τn+m = ϕ( 1λτn) a periodic splitting sequence.

Idea of proof: Train tracks which carry the same foliation have acommon maximal split. We have that ϕ( 1

λτ) carriesϕ(Fu, 1λµu) = (Fu, µu). So τ and ϕ( 1

λτ) have a common maximalsplit, namely ϕ( 1

λτn) = τn+m for some n,m > 0.

13

Periodic splitting sequence

Theorem (Agol)

Let τ be a train track carrying the unstable foliation Fu of apseudo-Anosov homeomorphism ϕ : S → S . There exist n,m > 0such that

τ ⇀ τ1 ⇀ τ2 ⇀ · · ·⇀ τn ⇀ · · ·⇀ τn+m,

such that ϕ( 1λτn) = τn+m, where λ > 1 is the dilatation of ϕ. We

call τn ⇀ · · ·⇀ τn+m = ϕ( 1λτn) a periodic splitting sequence.

Idea of proof: Train tracks which carry the same foliation have acommon maximal split. We have that ϕ( 1

λτ) carriesϕ(Fu, 1λµu) = (Fu, µu). So τ and ϕ( 1

λτ) have a common maximalsplit, namely ϕ( 1

λτn) = τn+m for some n,m > 0.

13

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

I Assume that the unstable foliation Fu of ϕ : S → S has nosingular points. We construct a layered triangulation of Mϕ.

I Note: if Fu has singular points on S , we can remove thosesingular points to get a surface S◦. Then ϕ restricts toϕ◦ : S◦ → S◦, and we can triangulate Mϕ◦ instead.

I Then to each train track τi in the periodic splitting sequencethere is a dual ideal triangulation Ti of the surface.

τn ⇀ τn+1 ⇀ · · ·⇀ τn+m = ϕ(1

λτn)

Tn → Tn+1 → · · · → Tn+m = ϕ(Tn).

I Each maximal split τi ⇀ τi+1 corresponds to a sequence ofdiagonal exchanges, for which we attach tetrahedrainterpolating between Ti and Ti+1.

I Since τn+m = ϕ( 1λτn) we have Tn+m = ϕ(Tn). So we can

glue Tn to Tn+m.

I We have constructed a layered triangulation of Mϕ.

14

A taut tetrahedron is an ideal tetrahedron with a coorientationassigned to each face, such that precisely two faces are coorientedinto the tetrahedron, and precisely two are cooriented outwards.Each edge is assigned an angle of either π if the coorientations onthe adjacent faces agree, or 0 if they disagree.

A taut triangulation of M is an ideal triangulation of M with acoorientation assigned to each ideal triangle so that eachtetrahedron is taut and the sum of the angles around each edge is2π.

15

We colour the zero angle edges blue and red as shown. Then thered edges are called right veering and the blue edges are calledleft veering.

As viewed from a red edge the triangles at the red edge move tothe right going from bottom to top. As viewed from a blue edgethe triangles at the edge move to the left going from bottom totop.A veering triangulation of M is a taut triangulation T with anassignment of red or blue to the edges of T so that the zero anglesof each tetrahedron are coloured as above.

16

I The fact that this construction produces veeringtriangulations follows from the combinatorics of train tracks.

I Each tetrahedron is naturally taut.

I When an edge is born do we colour it red or blue?

17

I The fact that this construction produces veeringtriangulations follows from the combinatorics of train tracks.

I Each tetrahedron is naturally taut.

I When an edge is born do we colour it red or blue?

17

I The fact that this construction produces veeringtriangulations follows from the combinatorics of train tracks.

I Each tetrahedron is naturally taut.

I When an edge is born do we colour it red or blue?

red, right veering

blue, left veering

17

I The fact that this construction produces veeringtriangulations follows from the combinatorics of train tracks.

I Each tetrahedron is naturally taut.

I When an edge is born do we colour it red or blue?

red, right veering

blue, left veering

17

I We wrote a computer program called Veering whichautomates this construction of veering triangulations.

I Relies on the program Trains by Toby Hall, an implementationof the Bestvina-Handel algorithm for constructing a traintrack carrying the unstable foliation of a pseudo-Anosovhomeomorphism.

I Homeomorphisms are specified as a composition of Dehntwists in the curves shown below, and permutations of thepunctures.

b1

ap

a1

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e1

e2

b2b

3

d1

d2

d3

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c3

cg c

2c1

bg e

g-1

1

2

p

18

I We wrote a computer program called Veering whichautomates this construction of veering triangulations.

I Relies on the program Trains by Toby Hall, an implementationof the Bestvina-Handel algorithm for constructing a traintrack carrying the unstable foliation of a pseudo-Anosovhomeomorphism.

I Homeomorphisms are specified as a composition of Dehntwists in the curves shown below, and permutations of thepunctures.

b1

ap

a1

a2

e1

e2

b2b

3

d1

d2

d3

d4

c3

cg c

2c1

bg e

g-1

1

2

p

18

I We wrote a computer program called Veering whichautomates this construction of veering triangulations.

I Relies on the program Trains by Toby Hall, an implementationof the Bestvina-Handel algorithm for constructing a traintrack carrying the unstable foliation of a pseudo-Anosovhomeomorphism.

I Homeomorphisms are specified as a composition of Dehntwists in the curves shown below, and permutations of thepunctures.

b1

ap

a1

a2

e1

e2

b2b

3

d1

d2

d3

d4

c3

cg c

2c1

bg e

g-1

1

2

p

18

I The shape of an oriented hyperbolic ideal tetrahedron in H3 isdetermined by a complex number z ∈ C with Im(z) > 0.

I Given an oriented ideal triangulation of a hyperbolic3-manifold M, we can realise each tetrahedron Ti as ahyperbolic tetrahedron determined by zi with Im(z) > 0.

I The hyperbolic tetrahedra glue together coherently to give thecomplete hyperbolic structure on M if and only if they satisfya system of equations in the zi ’s called Thurston’s gluingand completeness equations.

I If there exists such a solution, the ideal triangulation is calledgeometric.

Q: Are veering triangulations always geometric?

19

A necessary (but insufficient!) condition for an (oriented) idealtriangulation to be geometric is that it admits a strict anglestructure, that is, we can find positive volume ideal hyperbolictetrahedron shapes on the tetrahedra so that the angle aroundeach edge is 2π.

Hodgson, Rubinstein, Segerman and Tillmann introduced a largerclass of ‘veering triangulations’ and showed:

Theorem (HRST)

Veering triangulations admit strict angle structures.

(Gueritaud and Futer have given another proof of this.)

20

I We found examples of veering triangulations which are notgeometric.

I The smallest example we’ve found is a 13 tetrahedrontriangulation of the bundle Mϕ, where ϕ is the compositionTc1 ◦ Tb2 ◦ Ta1 ◦ Ta1 ◦ Ta1 ◦ Tb1 ◦ Ta1 of Dehn twists of theonce punctured genus 2 surface (Tγ is a Dehn twist in γ.)

I Mϕ is the manifold s479 in the SnapPea census. Thedilatation of ϕ is approx 2.89005.., and Mϕ has hyperbolicvolume 4.85117...

c2

b2

c1

a1

b1

21

I We found examples of veering triangulations which are notgeometric.

I The smallest example we’ve found is a 13 tetrahedrontriangulation of the bundle Mϕ, where ϕ is the compositionTc1 ◦ Tb2 ◦ Ta1 ◦ Ta1 ◦ Ta1 ◦ Tb1 ◦ Ta1 of Dehn twists of theonce punctured genus 2 surface (Tγ is a Dehn twist in γ.)

I Mϕ is the manifold s479 in the SnapPea census. Thedilatation of ϕ is approx 2.89005.., and Mϕ has hyperbolicvolume 4.85117...

c2

b2

c1

a1

b1

21

I We found examples of veering triangulations which are notgeometric.

I The smallest example we’ve found is a 13 tetrahedrontriangulation of the bundle Mϕ, where ϕ is the compositionTc1 ◦ Tb2 ◦ Ta1 ◦ Ta1 ◦ Ta1 ◦ Tb1 ◦ Ta1 of Dehn twists of theonce punctured genus 2 surface (Tγ is a Dehn twist in γ.)

I Mϕ is the manifold s479 in the SnapPea census. Thedilatation of ϕ is approx 2.89005.., and Mϕ has hyperbolicvolume 4.85117...

c2

b2

c1

a1

b1

21

Here is the the triangulation of the cusp (the shaded tetrahedron isnegatively oriented for the complete hyperbolic structure.)

22

I Although there are non-geometric veering triangulations, wemight still hope for weaker conclusions, e.g.Question: Given a veering triangulation, is there a (possiblyincomplete) hyperbolic structure obtained by gluing togetherpositive volume ideal tetrahedra? (This would give a point inThurston’s hyperbolic Dehn surgery space.)

I Example: Here is the Dehn surgery space for the aboveexample, with green indicating all tetrahedra positivelyoriented, blue some negatively oriented solutions (as found bySnapPy.)

23

I Although there are non-geometric veering triangulations, wemight still hope for weaker conclusions, e.g.Question: Given a veering triangulation, is there a (possiblyincomplete) hyperbolic structure obtained by gluing togetherpositive volume ideal tetrahedra? (This would give a point inThurston’s hyperbolic Dehn surgery space.)

I Example: Here is the Dehn surgery space for the aboveexample, with green indicating all tetrahedra positivelyoriented, blue some negatively oriented solutions (as found bySnapPy.)

23

I We constructed all veering triangulations (up to conjugationand inversion) of a once-punctured genus 2 surface, given by aproduct of at most 7 Dehn twists in the curves shownpreviously.

I Out of about 600 triangulations produced, 48 are numericallyreported non-geometric by the computer program SnapPy.

I In the 13 tetrahedron example we verified rigorously that it isnon-geometric, expect all 48 to be non-geometric.

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Figure : Vertical axis: no. of tetrahedra in the veering triangulation.Horizontal axis: no. of tetrahedra after the veering triangulation issimplified by SnapPy.

24

I We constructed all veering triangulations (up to conjugationand inversion) of a once-punctured genus 2 surface, given by aproduct of at most 7 Dehn twists in the curves shownpreviously.

I Out of about 600 triangulations produced, 48 are numericallyreported non-geometric by the computer program SnapPy.

I In the 13 tetrahedron example we verified rigorously that it isnon-geometric, expect all 48 to be non-geometric.

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Figure : Vertical axis: no. of tetrahedra in the veering triangulation.Horizontal axis: no. of tetrahedra after the veering triangulation issimplified by SnapPy.

24

I We constructed all veering triangulations (up to conjugationand inversion) of a once-punctured genus 2 surface, given by aproduct of at most 7 Dehn twists in the curves shownpreviously.

I Out of about 600 triangulations produced, 48 are numericallyreported non-geometric by the computer program SnapPy.

I In the 13 tetrahedron example we verified rigorously that it isnon-geometric, expect all 48 to be non-geometric.

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Figure : Vertical axis: no. of tetrahedra in the veering triangulation.Horizontal axis: no. of tetrahedra after the veering triangulation issimplified by SnapPy.

24

I We constructed all veering triangulations (up to conjugationand inversion) of a once-punctured genus 2 surface, given by aproduct of at most 7 Dehn twists in the curves shownpreviously.

I Out of about 600 triangulations produced, 48 are numericallyreported non-geometric by the computer program SnapPy.

I In the 13 tetrahedron example we verified rigorously that it isnon-geometric, expect all 48 to be non-geometric.

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Figure : Vertical axis: no. of tetrahedra in the veering triangulation.Horizontal axis: no. of tetrahedra after the veering triangulation issimplified by SnapPy. 24

Conjugacy in mapping class group

I Let ϕ and ψ be pseudo-Anosov homeomorphisms of a surfaceS .

I Denote by MCG(S) the mapping class group of S .

Question: are ϕ and ψ conjugate in MCG(S), that is, does thereexist h ∈ MCG(S) such that

ϕ = h ◦ ψ ◦ h−1?

I ϕ and ψ are conjugate if and only if they have’combinatorially isomorphic’ periodic splitting sequences.

I Veering implements this algorithmically.

25

Conjugacy in mapping class group

I Let ϕ and ψ be pseudo-Anosov homeomorphisms of a surfaceS .

I Denote by MCG(S) the mapping class group of S .

Question: are ϕ and ψ conjugate in MCG(S), that is, does thereexist h ∈ MCG(S) such that

ϕ = h ◦ ψ ◦ h−1?

I ϕ and ψ are conjugate if and only if they have’combinatorially isomorphic’ periodic splitting sequences.

I Veering implements this algorithmically.

25

Conjugacy in mapping class group

I Let ϕ and ψ be pseudo-Anosov homeomorphisms of a surfaceS .

I Denote by MCG(S) the mapping class group of S .

Question: are ϕ and ψ conjugate in MCG(S), that is, does thereexist h ∈ MCG(S) such that

ϕ = h ◦ ψ ◦ h−1?

I ϕ and ψ are conjugate if and only if they have’combinatorially isomorphic’ periodic splitting sequences.

I Veering implements this algorithmically.

25

Conjugacy in mapping class group

I Let ϕ and ψ be pseudo-Anosov homeomorphisms of a surfaceS .

I Denote by MCG(S) the mapping class group of S .

Question: are ϕ and ψ conjugate in MCG(S), that is, does thereexist h ∈ MCG(S) such that

ϕ = h ◦ ψ ◦ h−1?

I ϕ and ψ are conjugate if and only if they have’combinatorially isomorphic’ periodic splitting sequences.

I Veering implements this algorithmically.

25

Conjugacy in mapping class group

I Let ϕ and ψ be pseudo-Anosov homeomorphisms of a surfaceS .

I Denote by MCG(S) the mapping class group of S .

Question: are ϕ and ψ conjugate in MCG(S), that is, does thereexist h ∈ MCG(S) such that

ϕ = h ◦ ψ ◦ h−1?

I ϕ and ψ are conjugate if and only if they have’combinatorially isomorphic’ periodic splitting sequences.

I Veering implements this algorithmically.

25

Thanks!

Veering webpage: http://www.ms.unimelb.edu.au/~veering/ 26