non-geometric veering triangulations...we found examples of non-geometric veering triangulations. i...
TRANSCRIPT
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Non-geometric veering triangulations
Ahmad Issa
University of Texas at Austin
Joint work with Craig Hodgson and Henry Segerman
1
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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
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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
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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
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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
![Page 7: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/7.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
![Page 18: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/18.jpg)
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
![Page 19: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/19.jpg)
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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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}
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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}
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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
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I What happens to the dual ideal triangulation when a traintrack splits?
I A diagonal exchange occurs!
splits
τ
τ’
τ’
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I What happens to the dual ideal triangulation when a traintrack splits?
I A diagonal exchange occurs!
splits
τ
τ’
τ’
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I What happens to the dual ideal triangulation when a traintrack splits?
I A diagonal exchange occurs!
11
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I What happens to the dual ideal triangulation when a traintrack splits?
I A diagonal exchange occurs!
T
T’
T’
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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?
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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
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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
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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
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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
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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
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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
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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?
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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.)
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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
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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
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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...
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Here is the the triangulation of the cusp (the shaded tetrahedron isnegatively oriented for the complete hyperbolic structure.)
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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.)
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![Page 70: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/70.jpg)
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.)
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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
![Page 72: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/72.jpg)
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
![Page 73: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/73.jpg)
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
![Page 74: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/74.jpg)
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
![Page 75: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/75.jpg)
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
![Page 76: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/76.jpg)
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.
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![Page 77: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/77.jpg)
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
![Page 78: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/78.jpg)
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
![Page 79: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/79.jpg)
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
![Page 80: Non-geometric veering triangulations...We found examples of non-geometric veering triangulations. I Conjecture: Every cusped hyperbolic 3-manifold admits a geometric ideal triangulation](https://reader033.vdocument.in/reader033/viewer/2022060416/5f13e3143ea0c47c74545a4e/html5/thumbnails/80.jpg)
Thanks!
Veering webpage: http://www.ms.unimelb.edu.au/~veering/ 26