Cutting sequences,
regular polygons
and Veech groups
Corinna Ulcigrai
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
D
B
C
C
D
B
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
D
B
C
C
D
B
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
D
B
C
C
D
B
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
D
B
C
C
D
B
1
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
D
B
2
C
C
B
D
2
1
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
D
8
6
A
1
4
3
B
2
5
5
C
C
7
7
3
B
D
2
6
8
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
D
8
6
A
1
4
3
B
2
5
5
C
C
7
7
3
B
D
2
6
8
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
D
8
6
A
1
4
3
B
2
5
5
C
C
7
7
3
B
D
2
6
8
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is:
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
D
B
C
C
D
B
1
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
D
B
2
C
C
B
D
2
1
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
D
3
B
2
C
C
3
B
D
2
1
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B B
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
4
D
3
B
2
C
C
3
B
D
2
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B B A
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
A
1
4
D
3
B
2
5
5
C
C
3
B
D
2
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B B A C
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
D
6
A
1
4
3
B
2
5
5
C
C
7
3
B
D
2
6
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B B A C D
Linear trajectories and cutting sequences
Consider a regular polygon, for simplicity
with 2n sides.
As an example, in the talk we will consider
a regular octagon.
Glue opposite sides.
Label pairs of sides by {A, B, C , D}.
Let ϕθt be the linear flow in direction θ:
trajectories which do not hit singularities
are straight lines in direction θ.
D
6
A
1
4
3
B
2
5
5
C
C
7
7
3
B
D
2
6
1
4
A
Definition (Cutting sequence)
The cutting sequence in {A, B, C , D}Z that codes a bi-infinite linear
trajectory of ϕθt consists of the sequence of labels of the sides hit by the
trajectory.
Example
The cutting sequence of the trajectory in the example is: A B B A C D
A classical case: Sturmian sequences
Consider the special case in which the polygon is a square.
A
B
B
A
Minima complessita’
Sturmian sequences are characterized by having the smallest possible
complexity among non-periodic sequences.
(Let Pw (n) the number of words of lenght n which appear in the
sequence w : Pw (n) = n iff w is periodic. Sturmian sequences satisfy
Pw (n) = n + 1.)
A classical case: Sturmian sequences
Consider the special case in which the polygon is a square.
A
B
B
B
B
A
B
B
A
A
A
In this case the cutting sequence correspond to the sequence of
horizontal (letter A) and vertical (letter B) sides crossed by a line in
direction θ in a square grid.
A classical case: Sturmian sequences
Consider the special case in which the polygon is a square.
A
B
B
B
B
A
B
B
A
A
A
Square cutting sequences are Sturmian sequences. They were studied
since Hedlund e Morse.
Minimal complexity
Sturmian sequences are characterized by having the smallest possible
complexity among non-periodic sequences.
(Let Pw (n) the number of words of lenght n which appear in the
sequence w : Pw (n) = n iff w is periodic. Sturmian sequences satisfy
Pw (n) = n + 1.)
A classical case: Sturmian sequences
Consider the special case in which the polygon is a square.
A
B
B
B
B
A
B
B
A
A
A
Square cutting sequences are Sturmian sequences. They were studied
since Hedlund e Morse.
Minimal complexity
Sturmian sequences are characterized by having the smallest possible
complexity among non-periodic sequences.
(Let Pw (n) the number of words of lenght n which appear in the
sequence w : Pw (n) = n iff w is periodic. Sturmian sequences satisfy
Pw (n) = n + 1.)
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
Translation surfaces
I
A
D
B
C
C
D
B
A
Glueing opposite sides one gets a surface of genus 2, with a flat
metric with a singularity (it’s a translation surface); ϕθt is the
geodesic flow with respect to the flat metric;
I
Poligonal Billiards
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
For example: A billiard with angles π2 , π8 and 3π
8 (motion of a particle
with elastic reflections at sides) by a procedure called unfolding is
equivalent to the flow ϕθt in the octagon.
I
Interval exchange transformations
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
B
C
A
D
D
A
C
B
The Poincaré first return map on a section is an interval exchange
transformation (IET). As θ changes, one has a one-parameter family
which is not generic.
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
B
C
A
A
B
C
D
D
T
D
A
D
C
B A
C
B
The Poincaré first return map on a section is an interval exchange
transformation (IET). As θ changes, one has a one-parameter family
which is not generic.
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
B
C
A
A
B
C
D
D
T
D
A
D
C
B A
C
B
The Poincaré first return map on a section is an interval exchange
transformation (IET). As θ changes, one has a one-parameter family
which is not generic.
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
Remark
All these systems have entropy zero: cutting sequences have linear
complexity.
Remark
Translation surfaces and IETs which come from regular polygons are not
generic: techniques that are used for the generic setting do not apply
here.
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
Remark
All these systems have entropy zero: cutting sequences have linear
complexity.
Remark
Translation surfaces and IETs which come from regular polygons are not
generic: techniques that are used for the generic setting do not apply
here.
Motivation
Cutting sequences give a symbolic coding of the following systems:
I
Translation surfaces
I
Poligonal Billiards
I
Interval exchange transformations
Remark
All these systems have entropy zero: cutting sequences have linear
complexity.
Remark
Translation surfaces and IETs which come from regular polygons are not
generic: techniques that are used for the generic setting do not apply
here.
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Characterization of cutting sequences
Problem:
Describe explicitely the symbolic sequences which are cutting sequences
of trajectories.
In particular, answer the following questions:
Q1) Which sequences in {A, B, C , D}Z are cutting sequences?
Q2) Given a cutting sequence, can one recover the direction of the
trajectory?
Given a finite piece of a cutting sequence, can one recover a sector
of possible directions?
Q3) Given a direction or more in general a sector of directions, can one
produce all cutting sequences of trajectories in that direction or
sector?
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
Seminar outline:
I
The classical case: the square (Sturmian sequences)
I
I
I
I
Caracterization of Sturmian sequences;
(revisiting Caroline Series work)
Connection with Continued Fractions;
Sketch of proof for the square;
Regular polygons with 2n sides:
joint work with John Smillie (Cornell University)
I
I
Formulation of results in the case of the octagon;
Continued Fractions for regular polygons;
[Further works: D. Davis-S. Tabachnikov-D. Fuchs (pentagon),
D. Davis (double n-gons with n-odd and Bouw-Möller surfaces)]
I
Renormalization, Fuchsian groups and Teichmüller flow;
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
0
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: isometries and sectors
A
Let D4 be the group of isometries of the
square.
The letters {A, B} are invariant under
vertical symmetry and horizontal symmetry
and are exachanged if we reflect diagonally
B
B
A
WLOG we can assume
that θ ∈ 0, π2 and, up to permuting {A, B}, let
θ ∈ Σ0 := 0, π4 .
1
0
Let Σ1 :=
π
π
4, 2
be the other sector.
The square: possible transitions
I
If θ ∈ Σ0 := 0, π4 , AA does not appear:
A
B
B
A
I
If θ ∈ Σ1 :=
π
π
4, 2
, BB does not appear:
The square: possible transitions
I
If θ ∈ Σ0 := 0, π4 , AA does not appear:
A
B
A
B
A
I
If θ ∈ Σ1 :=
π
π
4, 2
, BB does not appear:
B
The square: possible transitions
I
If θ ∈ Σ0 := 0, π4 , AA does not appear:
A
B
B
A
B
A
B
A
I
If θ ∈ Σ1 :=
π
π
4, 2
, BB does not appear:
A
B
B
A
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissibile if it gives a infinite path on one of
these two diagrams:
A
B
A
B
D0
In this case, we say that w is admissible in D0 or D1 respectively.
Lemma
A square cutting sequence is admissible.
If θ ∈ Σ0 = 0, π4 , w is admissible in D0 , if θ ∈ Σ1 = π4 , π2 , w is
admissibile in D1 .
D1
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissibile if it gives a infinite path on one of
these two diagrams:
A
B
A
B
D0
In this case, we say that w is admissible in D0 or D1 respectively.
Lemma
A square cutting sequence is admissible.
If θ ∈ Σ0 = 0, π4 , w is admissible in D0 , if θ ∈ Σ1 = π4 , π2 , w is
admissibile in D1 .
D1
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissibile if it gives a infinite path on one of
these two diagrams:
A
B
A
B
D0
In this case, we say that w is admissible in D0 or D1 respectively.
Lemma
A square cutting sequence is admissible.
If θ ∈ Σ0 = 0, π4 , w is admissible in D0 , if θ ∈ Σ1 = π4 , π2 , w is
admissibile in D1 .
D1
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissibile if it gives a infinite path on one of
these two diagrams:
A
B
A
B
D0
In this case, we say that w is admissible in D0 or D1 respectively.
Lemma
A square cutting sequence is admissible.
If θ ∈ Σ0 = 0, π4 , w is admissible in D0 , if θ ∈ Σ1 = π4 , π2 , w is
admissibile in D1 .
D1
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Derivable sequences
Definition (Derived sequence)
Let w be the cuttings sequence of a trajectory with θ ∈ Σ0 . The derived
sequence w 0 is obtained erasing one B from each block of Bs. If
θ ∈ Σ1 , w 0 is obtained erasing one A from each block of As.
Example
w = . . . ABBBBABBBABBBBABBBABBB . . . ,
w 0 = . . . ABBB ABB ABBB ABB ABB . . .
Definition (Derivable Sequences)
A sequence w ∈ {A, B}Z is derivable if it is admissible and the derived
sequence is admissible.
Definition
A sequence w ∈ {A, B}Z is infinitely derivable if it is admissible and all
its derived sequences are still admissible.
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Characterization of Sturmian sequences
Theorem
A square cutting sequence is infinitely derivable.
Corollary
If w is a square cutting sequence with θ ∈ Σ0 , the blocks of Bs have
length n o n + 1.
Example
The sequence w = . . . ABBBBABBABBBBA . . . is NOT a square
cutting sequence. Indeed: w 0 = . . . ABBBABABBBA . . . and
w 00 = . . . ABBAABBA . . . which is not admissible.
Theorem
Let w be infinitely derivable. Then w belongs to the closure of square
cutting sequences.
Example
(infinitely derivable sequence which is not a square cutting sequence)
. . . AAAAAABAAAAA . . .
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
w” = . . . BAAAABAAABAAA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
w” = . . . BAAAABAAABAAA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
w” = . . . BAAAABAAABAAA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
w” = . . . BAAAABAAABAAA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Direction recognition and continued fractions
Let w a square cutting sequence, for example:
w = . . . BBBABBABBABBABBBABBAABBBABBABBABBA. . .
w’ = . . . BBABABABABBABAABBABABA. . .
w” = . . . BAAAABAAABAAA. . .
Let us define {an }n ∈ NN as follows:
let a0 such that the blocks of Bs in w have length a0 or a0 + 1 (in
the example a0 = 2: BB o BBB)
let a1 such that the blocks of As in w (a0 ) (in the example w”) have
length a1 or a1 + 1 (in the example a1 = 3: AAA o AAAA)
...
let an such that the blocks of As (n odd) or Bs (n even) in
w (a0 +···+an−1 ) have length an or an + 1;
Theorem (Direction Recognition)
The direction θ of the trajectory with cutting sequence w is given by:
θ=
1
a0 +
1
a1 +
···+
1
1
an +...
Proof for Sturmian sequences
The key step is given by the following Lemma:
Lemma
If w is a square cutting sequence, also the derived sequence w 0 is a
square cutting sequence.
Recalling that a square cutting sequence is clearly admissible, we have:
Corollary
Square cutting sequences are infinitely derivable.
Lemma
If w is a cutting sequence of a trajectory in
direction θ, the derived sequence w 0 is a
cutting sequence of a trajectory in direction
θ0 , where θ0 = F (θ) and F is the Farey map
in Figure.
Proof for Sturmian sequences
The key step is given by the following Lemma:
Lemma
If w is a square cutting sequence, also the derived sequence w 0 is a
square cutting sequence.
Recalling that a square cutting sequence is clearly admissible, we have:
Corollary
Square cutting sequences are infinitely derivable.
Lemma
If w is a cutting sequence of a trajectory in
direction θ, the derived sequence w 0 is a
cutting sequence of a trajectory in direction
θ0 , where θ0 = F (θ) and F is the Farey map
in Figure.
Proof for Sturmian sequences
The key step is given by the following Lemma:
Lemma
If w is a square cutting sequence, also the derived sequence w 0 is a
square cutting sequence.
Recalling that a square cutting sequence is clearly admissible, we have:
Corollary
Square cutting sequences are infinitely derivable.
Lemma
If w is a cutting sequence of a trajectory in
direction θ, the derived sequence w 0 is a
cutting sequence of a trajectory in direction
θ0 , where θ0 = F (θ) and F is the Farey map
in Figure.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
A
in the example:
w = ... A B B
ABBB
ABB
ABB
ABBB
...
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
A
in the example:
w = ... A B B
ABBB
ABB
ABB
ABBB
...
Let us add the diagonal C.
Let w̃ be the extended sequence:
Each BB becomes BCB; AB stays AB
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
A
in the example:
w = ... A B B
ABBB
ABB
ABB
ABBB
...
Let us add the diagonal C.
Let w̃ be the extended sequence:
Each BB becomes BCB; AB stays AB
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
A
in the example:
w = ... A B B
ABBB
ABB
ABB
ABBB
...
Let us add the diagonal C.
Let w̃ be the extended sequence:
Each BB becomes BCB; AB stays AB
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
A
in the example:
w = ... A B B A B B B
ABB ABB ABBB
...
w̃ = . . . A B C B A B C B C B A B C B A B C B A B C B C B . . .
Let us add the diagonal C.
Let w̃ be the extended sequence:
Each BB becomes BCB; AB stays AB
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
A
B
c
B
c
A
A
in the example:
w = ... A B B A B B B
ABB ABB ABBB
...
w̃ = . . . A B C B A B C B C B A B C B A B C B A B C B C B . . .
Let us cut and paste the rectangle.
Consider the cutting sequence u with respect to the parallelogram Π.
To obtain u from w̃ it is enough to drop the Bs.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
A
B
c
B
c
A
A
in the example:
w = ... A B B A B B B
ABB ABB ABBB
...
w̃ = . . . A B C B A B C B C B A B C B A B C B A B C B C B . . .
Let us cut and paste the rectangle.
Consider the cutting sequence u with respect to the parallelogram Π.
To obtain u from w̃ it is enough to drop the Bs.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
A
B
c
B
c
A
in
w
w̃
u
A
the example:
= ... A B B A B B B
ABB ABB ABBB
...
= ... A B C B A B C B C B A B C B A B C B A B C B C B ...
= ... A C A C C A C A C A C C ...
Let us cut and paste the rectangle.
Consider the cutting sequence u with respect to the parallelogram Π.
To obtain u from w̃ it is enough to drop the Bs.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in
w
w̃
u
A
A
A
c
c
c
0
A
the example:
= ... A B B A B B B
ABB ABB ABBB
...
= ... A B C B A B C B C B A B C B A B C B A B C B C B ...
= ... A C A C C A C A C A C C ...
Let us renormalize:
we can transform Π in a square by the shear
1 −1
. Let us transform back the Cs into Bs.
0 1
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in
w
w̃
u
A
A
A
c
c
c
0
A
the example:
= ... A B B A B B B
ABB ABB ABBB
...
= ... A B C B A B C B C B A B C B A B C B A B C B C B ...
= ... A C A C C A C A C A C C ...
= ... A B A B B A B A B A B B ...
Let us renormalize:
we can transform Π in a square by the shear
1 −1
. Let us transform back the Cs into Bs.
0 1
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in the example:
w = ... A B B
w̃ = . . . A B C B
u = ... A C
w0 = . . . A B
A
AB
AB
A
A
A
A
c
c
c
0
A
BB
ABB ABB ABBB
...
C B C B A B C B A B C B A B C B C B ...
C C A C A C A C C ...
B B A B A B A B B ...
Let
we can transform Π in a square by the shear
us renormalize:
1 −1
. Let us transform back the Cs into Bs.
0 1
The sequence thus obtained is the derived sequence.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in the example:
w = ... A B B
w̃ = . . . A B C B
u = ... A C
w0 = . . . A B
A
AB
AB
A
A
A
A
c
c
c
0
A
BB
ABB ABB ABBB
...
C B C B A B C B A B C B A B C B C B ...
C C A C A C A C C ...
B B A B A B A B B ...
To check it:
A BBB A → A BCBCB A → A CC A → A BB A
act as derivation.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in the example:
w = ... A B B
w̃ = . . . A B C B
u = ... A C
w0 = . . . A B
A
AB
AB
A
A
A
A
c
c
c
0
A
BB
ABB ABB ABBB
...
C B C B A B C B A B C B A B C B C B ...
C C A C A C A C C ...
B B A B A B A B B ...
To check it:
A BBB A → A BCBCB A → A CC A → A BB A
act as derivation.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in the example:
w = ... A B B
w̃ = . . . A B C B
u = ... A C
w0 = . . . A B
A
AB
AB
A
A
A
A
c
c
c
0
A
BB
ABB ABB ABBB
...
C B C B A B C B A B C B A B C B C B ...
C C A C A C A C C ...
B B A B A B A B B ...
To check it:
A BBB A → A BCBCB A → A CC A → A BB A
act as derivation.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
in the example:
w = ... A B B
w̃ = . . . A B C B
u = ... A C
w0 = . . . A B
A
AB
AB
A
A
A
A
c
c
c
0
A
BB
ABB ABB ABBB
...
C B C B A B C B A B C B A B C B C B ...
C C A C A C A C C ...
B B A B A B A B B ...
To check it:
A BBB A → A BCBCB A → A CC A → A BB A
act as derivation.
The new direction θ0 is obtained applying to θ a shear. One can verify
that θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
A
A
A
c
c
c
0
A
Summarizing:
We showed that the sequence w 0 is the cutting sequence of a new
trajectory in the square (thus still a square cutting sequence). The new
direction θ0 is obtained applying to θ a shear. One can verify that
θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
A
A
A
c
c
c
0
A
Summarizing:
We showed that the sequence w 0 is the cutting sequence of a new
trajectory in the square (thus still a square cutting sequence). The new
direction θ0 is obtained applying to θ a shear. One can verify that
θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
Renormalization and derivation
Let w be the cutting sequence of a trajectory in direction θ ∈ Σ0 ;
A
c
B
B
c
B
A
A
A
A
c
c
c
0
A
Summarizing:
We showed that the sequence w 0 is the cutting sequence of a new
trajectory in the square (thus still a square cutting sequence). The new
direction θ0 is obtained applying to θ a shear. One can verify that
θ0 = F (θ) where F is the Farey map. The Farey map is the additive
version of the continued fraction algorithm.
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
180
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
180
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
0
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
The octagon: isometries and sectors
Let D8 be the isometries group of the
octagon.
The letters {A, B, C , D} are invariant with
respect to a central symmetry. The other
elements induce permutations of
{A, B, C , D} for example:
A 7→ C , C 7→ A, B 7→ B, D 7→ D .
A
D
B
C
C
D
B
A
Let us assume that the direction of the trajectory
is θ ∈ 0, π8 .
A fundamental domain for D8 is Σ0 := 0, π8 . Thus, acting by an
element of D8 , up to a permutation
of the letters {A, B, C , D},
we can consider θ ∈ Σ0 := 0, π8 .
5 4 3 2
6
0
7
The other sectors of angle π/8 are in order Σ1 , Σ2 , . . . , Σ7 .
1
0
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
A
D
B
C
C
D
B
A
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
O1
O2
O8
O3
O7
O4
AD
O6 A
O5
A
D
D
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
O2
O1 A
O8
D
DA
B
DB
O3
O4
O7
O6
O5
A
D
B
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
O1
O2
O8
O3
C
BC
O7
B
O6
BD
A
D
B
C
O4
O5
D
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
O1
O2
CB
O8
O3
CC
C
B
C
A
D
B
O4
O7
O6
O5
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
C
The octagon: allowed transitions in Σ0
Let θ ∈ Σ0 := 0, π8 .
0
The transitions (pairs of consecutive letters) which can appear are:
A
D
B
C
C
A
D
B
D
B
A
Each octagon cutting sequence of a trajectory in direction θ ∈ Σ0
determines a path in the diagram in Figure.
C
The octagon: possible transitions
Permuting the letters we obtain the diagrams corresponding to the other
sectors:
A
5 4 3 2
6
D
B
C
5 4 3 2
6
1
7
0
5 4 3 2
1
7
0
D
A
C
B
6
5 4 3 2
0
D
C
A
B
7
0
C
D
B
A
7
D
A
B
C
A
D
1
B
A
C
D
A
B
D
C
0
5 4 3 2
6
B
0
6
1
7
7
1
7
5 4 3 2
6
C
0
5 4 3 2
6
1
7
5 4 3 2
6
1
1
0
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissible if it gives an infinite path on one
of the following diagrams:
D0
D1
D2
D3
A
D
B
C
D4
D
A
C
B
D5
D
C
A
B
D6
C
D
B
A
D7
Lemma
An octagon cutting sequence is admissible.
C
B
D
A
B
C
A
D
B
A
C
D
A
B
D
C
Admissible sequences
Definition
A sequence w ∈ {A, B}Z is admissible if it gives an infinite path on one
of the following diagrams:
D0
D1
D2
D3
A
D
B
C
D4
D
A
C
B
D5
D
C
A
B
D6
C
D
B
A
D7
Lemma
An octagon cutting sequence is admissible.
C
B
D
A
B
C
A
D
B
A
C
D
A
B
D
C
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
B
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
B
A
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
B
A
C DBD C
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Derived sequences
Definition
A letter in {A, B, C , D} is sandwiched if it is preceeded and followed by
the same letter.
Example
If w = . . . D A D B C C B C C B D A D B C B D B D B C B D . . . ,
w’ = . . .
A
B
A
C DBD C
Definition (Derivable sequences)
A sequence w ∈ {A, B, C , D}Z is derivable if it is admissible and its
derivative is still admissible. The sequence w is infinitely derivable if each
of its derivatives is derivable.
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
A
D
B
C
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
A
D
B
C
Necessary condition and sequence of sectors
Theorem
An octagon cutting sequence is infinitly derivable.
The converse is not true, but we can describe exactly the condition which
one needs to add.
Definition
Let w be infinitly derivable and let w (n) be the nth derived sequence. The
sequence {sk }k∈N ∈ {0, 1, . . . , 7}N is a sequence of sectors for w if for
each k, w (k) gives a path on the diagram Dsk .
Example
w= C C C B C C B D B
s0 = 0
w’= C B D B D
s1 = 4
A
D
B
C
C
B
D
A
Octagon Continued Fractions
Let F : [0, π] → [π/8, π] the following map, that we call Octagon Farey
map:
Definition
The octagon continued fraction expansion of θ is
θ = [s0 , s1 , s2 , . . . , sk , . . . ]O
iff
{θ} = ∩k Fs−1
Fs−1
. . . Fs−1
[0, π].
0
1
k
In this case we have F k (θ) ∈ Σsk for all k.
Octagon Continued Fractions
Let F : [0, π] → [π/8, π] the following map, that we call Octagon Farey
map:
Definition
The octagon continued fraction expansion of θ is
θ = [s0 , s1 , s2 , . . . , sk , . . . ]O
iff
{θ} = ∩k Fs−1
Fs−1
. . . Fs−1
[0, π].
0
1
k
In this case we have F k (θ) ∈ Σsk for all k.
Direction Recognitionn
Let w be an octagon cutting sequence.
Lemma
If w is not a periodic sequence, the sequence of sectors {sk }k∈N is
uniquely determined. In particular, each derivative w (k) is admissible in
an unique Dsk .
Theorem
If w is not periodic, there is a unique sequence of sectors {sk }k∈N for w
and the direction of the trajectories with cutting sequence w is given by
θ = [s0 , s1 , s2 , ....]O .
Direction Recognitionn
Let w be an octagon cutting sequence.
Lemma
If w is not a periodic sequence, the sequence of sectors {sk }k∈N is
uniquely determined. In particular, each derivative w (k) is admissible in
an unique Dsk .
Theorem
If w is not periodic, there is a unique sequence of sectors {sk }k∈N for w
and the direction of the trajectories with cutting sequence w is given by
θ = [s0 , s1 , s2 , ....]O .
Ideas from proofs
As in the case of the square, the theorems follow if we prove that:
Theorem
If w is an octagon cutting sequence, also the derived sequence w 0 is an
octagon cutting sequence.
Furthermore, if w is the cutting sequence of a trajectory in direction θ,
the derived sequence w 0 is a cutting sequence of a trajectory in direction
θ0 = F (θ), where F is the octagon Farey map.
To prove it, one uses an argument in the same spirit of:
i.e. one uses renormalization in the space of affine deformations of the
octagon.
Regular polygons are rich of affine automorphisms.
Ideas from proofs
As in the case of the square, the theorems follow if we prove that:
Theorem
If w is an octagon cutting sequence, also the derived sequence w 0 is an
octagon cutting sequence.
Furthermore, if w is the cutting sequence of a trajectory in direction θ,
the derived sequence w 0 is a cutting sequence of a trajectory in direction
θ0 = F (θ), where F is the octagon Farey map.
To prove it, one uses an argument in the same spirit of:
i.e. one uses renormalization in the space of affine deformations of the
octagon.
Regular polygons are rich of affine automorphisms.
Ideas from proofs
As in the case of the square, the theorems follow if we prove that:
Theorem
If w is an octagon cutting sequence, also the derived sequence w 0 is an
octagon cutting sequence.
Furthermore, if w is the cutting sequence of a trajectory in direction θ,
the derived sequence w 0 is a cutting sequence of a trajectory in direction
θ0 = F (θ), where F is the octagon Farey map.
To prove it, one uses an argument in the same spirit of:
A
c
B
A
A
c
c
c
0
A
i.e. one uses renormalization in the space of affine deformations of the
octagon.
Regular polygons are rich of affine automorphisms.
Ideas from proofs
As in the case of the square, the theorems follow if we prove that:
Theorem
If w is an octagon cutting sequence, also the derived sequence w 0 is an
octagon cutting sequence.
Furthermore, if w is the cutting sequence of a trajectory in direction θ,
the derived sequence w 0 is a cutting sequence of a trajectory in direction
θ0 = F (θ), where F is the octagon Farey map.
To prove it, one uses an argument in the same spirit of:
A
c
B
A
A
c
c
c
0
A
i.e. one uses renormalization in the space of affine deformations of the
octagon.
Regular polygons are rich of affine automorphisms.
Affine automorphism and Veech group
Let SO be the surface obtained glueing opposite sides of the octagon by
translations: SO is an example of a translation surface, i.e. it has an atlas
whose changes of coordinates are of the form z 7→ z + c.
Definition
An automorphism Ψ : S 7→ S of a translation surface S is an affine
automorphism if it is affine in each chart and DΨ(z) is independent on
z ∈ S. Let
Aff (S) = {Ψ, Ψ affine automorphism}.
Affine automorphism and Veech group
Let SO be the surface obtained glueing opposite sides of the octagon by
translations: SO is an example of a translation surface, i.e. it has an atlas
whose changes of coordinates are of the form z 7→ z + c.
Definition
An automorphism Ψ : S 7→ S of a translation surface S is an affine
automorphism if it is affine in each chart and DΨ(z) is independent on
z ∈ S. Let
Aff (S) = {Ψ, Ψ affine automorphism}.
Affine automorphism and Veech group
Let SO be the surface obtained glueing opposite sides of the octagon by
translations: SO is an example of a translation surface, i.e. it has an atlas
whose changes of coordinates are of the form z 7→ z + c.
Definition
An automorphism Ψ : S 7→ S of a translation surface S is an affine
automorphism if it is affine in each chart and DΨ(z) is independent on
z ∈ S. Let
Aff (S) = {Ψ, Ψ affine automorphism}.
Example (1)
A
D
Se Ψ ∈ D8
is an isometry of O,
clearly one has
Ψ ∈ Aff (O).
A
D
B
B
90
C
C
C
C
90
D
B
A
D
B
A
Affine automorphism and Veech group
Let SO be the surface obtained glueing opposite sides of the octagon by
translations: SO is an example of a translation surface, i.e. it has an atlas
whose changes of coordinates are of the form z 7→ z + c.
Definition
An automorphism Ψ : S 7→ S of a translation surface S is an affine
automorphism if it is affine in each chart and DΨ(z) is independent on
z ∈ S. Let
Aff (S) = {Ψ, Ψ affine automorphism}.
Example (2)
The matrix
1
0
2(1 +
1
√ 2)
∈ V (O).
Affine automorphism and Veech group
Let SO be the surface obtained glueing opposite sides of the octagon by
translations: SO is an example of a translation surface, i.e. it has an atlas
whose changes of coordinates are of the form z 7→ z + c.
Definition
An automorphism Ψ : S 7→ S of a translation surface S is an affine
automorphism if it is affine in each chart and DΨ(z) is independent on
z ∈ S. Let
Aff (S) = {Ψ, Ψ affine automorphism}.
Example (2)
The matrix
1
0
2(1 +
1
√ 2)
∈ V (O).
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
<
0
1
,
1
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
0
1
−1
>
0
= SL(2, Z).
Ex 2
V (O) =
1
<
0
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
<
0
1
,
1
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
0
1
−1
>
0
= SL(2, Z).
Ex 2
V (O) =
1
<
0
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
<
0
1
,
1
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
0
1
−1
>
0
= SL(2, Z).
Ex 2
V (O) =
1
<
0
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
<
0
1
,
1
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
0
1
−1
>
0
= SL(2, Z).
Ex 2
V (O) =
1
<
0
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
,
1
1
<
0
Ex 2
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
−1
>
0
0
1
A
D
A
D
B
C
C
D
B
V (O) =
1
<
0
B
C
C
D
B
A
= SL(2, Z).
A
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
,
1
1
<
0
Ex 2
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
−1
>
0
0
1
A
D
A
D
B
C
C
D
B
V (O) =
1
<
0
B
C
C
D
B
A
= SL(2, Z).
A
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
The Veech group of the octagon
The Veech group V (S) is the group of linear parts of Aff (S):
V (S) = {DΨ,
Ex 1
2 +
V (T ) =
1
,
1
1
<
0
Ex 2
Ψ ∈ Aff (S)} ⊂ SL(2, R)± .
−1
>
0
0
1
A
D
A
D
B
C
C
D
B
V (O) =
1
<
0
B
C
C
D
B
A
= SL(2, Z).
A
0
,
−1
√1
2
− √12
√1
2
√1
2
!
,
1
0
2(1 +
1
√ 2)
If S is a translation surface glued from a regular polygon, V (S) is a
lattice in SL(2, R)± (Veech)
The surfaces S for which V (S) is a lattice are actively researched in
Teichmüller dynamics.
>
Derivation and renormalization in the octagon
Let w be the cutting
of a trajectory in direction θ ∈ Σ0 . Let
√ sequence
−1
2(1
+
2)
O0 =
O.
0
1
Lemma
The derived sequence w 0 coincides with the cutting sequence of the same
trajectory in direction θ with respect to the sides of O 0 .
Let us renormalize:
O 0 7→ O
θ 7→ θ0
Lemma
The derived sequence w 0 is an octagon
cutting sequence in direction θ0 = FO (θ).
Derivation and renormalization in the octagon
Let w be the cutting
of a trajectory in direction θ ∈ Σ0 . Let
√ sequence
−1
2(1
+
2)
O0 =
O.
0
1
Lemma
The derived sequence w 0 coincides with the cutting sequence of the same
trajectory in direction θ with respect to the sides of O 0 .
Let us renormalize:
O 0 7→ O
θ 7→ θ0
[C ℄ = =2
0
11111111111
00000000000
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
=8
00000000000000000000000000000D
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
000000000000000 C
111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
000000000000000 111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
000000000000000 B
111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
A = 0
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
[D ℄ = =4
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
4
[B ℄ = 3=11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
00000000000
11111111111
0
1
0
0
Lemma
The derived sequence w 0 is an octagon
cutting sequence in direction θ0 = FO (θ).
0000000000
1111111111
00000000000
11111111111
11111111111
00000000000
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
4
511111111111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
00000
11111
0000000000000000
1111111111111111
2
3
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
1
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
6 1111111111
7
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
[
0
0
[ 11111111111111111111111111111
00000000
11111111
[ 11111111
00000000000000000000000000000
00000000
00000000000000000000000000000
11111111111111111111111111111
0
[A ℄ = 0
00000000
11111111
11111111111111111111111111111
00000000000000000000000000000
00000000
00000000000000000000000000000
11111111111111111111111111111
00000000 [=8℄ = =8
11111111
11111111
00000000000000000000000000000
11111111111111111111111111111
00000000
11111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
Derivation and renormalization in the octagon
Let w be the cutting
of a trajectory in direction θ ∈ Σ0 . Let
√ sequence
2)
−1
2(1
+
O.
O0 =
0
1
Lemma
The derived sequence w 0 coincides with the cutting sequence of the same
trajectory in direction θ with respect to the sides of O 0 .
[C ℄ = =2
Let us renormalize:
O 0 7→ O
θ 7→ θ0
0
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
11111111111
00000000000
=8
11111111111111111111111111111D
00000000000000000000000000000
00000000000000000000000000000
11111111111111111111111111111
000000000000000 111111111111111
000000000000000
111111111111111
C
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000 B
111111111111111
0000000000000000
1111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
000000000000000
111111111111111
0000000000000000
1111111111111111
00000000000000000000000000000
11111111111111111111111111111
A = 0
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
[D ℄ = =4
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
4
[B ℄ = 3=11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000000000
00000000000
11111111111
0
1
0
0000000000
1111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
4
511111111111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
00000
11111
0000000000000000
1111111111111111
2
3
0000000000
1111111111
00000000000
11111111111
00000
11111
1111111111111111
0000000000000000
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
1
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
6
7
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
0000000000000000
1111111111111111
0000000000
1111111111
00000000000
11111111111
00000
11111
00000000
11111111
11111111111111111111111111111
00000000000000000000000000000
00000000 [=8℄ = =8
11111111
00000000000000000000000000000
11111111111111111111111111111
00000000
11111111
00000000000000000000000000000
11111111111111111111111111111
00000000
11111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
00000000000000000000000000000
11111111111111111111111111111
[
0
0
0
[ 11111111111111111111111111111
00000000
11111111
[ 11111111
00000000000000000000000000000
00000000
00000000000000000000000000000
11111111111111111111111111111
0
[A ℄ = 0
Lemma
P
0
The derived sequence w is an octagon
cutting sequence in direction θ0 = FO (θ).
P/8
0
P/8
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
0
1
1
1
1
0
1
S1 =
S2 =
1
1
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
1
1
Q square
1 1
0 1
1
1
S1 =
S2 =
0
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
0
1
1
1
1
0
1
S1 =
S2 =
1
1
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
0
1
1
1
1
0
1
S1 =
S2 =
1
1
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
0
1
1
1
1
0
1
S1 =
S2 =
1
1
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization, modular surface and continued fractions
The space of lattices is H/SL(2, Z) (moduli space of tori with a flat
metric).
1 1
1 0
Farey Tessellation: images of T (0, 1, ∞) via V (Q) = <
,
>
0
1
0
1
1
1
1
0
1
S1 =
S2 =
1
1
1
Given θ, gtθ geodesics
θ=
1
a0 +
1
a1 +...
gtθ is approximated by
S1 a0 S2 a1 S1 a2 . . .
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Renormalization and dynamics on the Teichmüller disk
SL(2, R) · O = {A · O, A ∈ SL(2, R)} affine deformations of the octagon
affine deformations
SL(2, R)
=
affine automorphisms
V (O)
(Teichmueller disk)
V (O) · O centers of ideal
octagons
tree of renormalization moves
Given θ, gtθ geodesics
is approximated by a
sequence of renormalization
moves
the Farey map of the octagon
is given by the action on ∂D.
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
Example
w
g02 w
= ... D C A B B B A C ...
= ... D B C ...
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C A B B B A C ...
= ... D B C ...
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C A B B B A C ...
= ... D B C ...
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C A B B B A C ...
= ... D B C ...
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C
A B B B A C ...
= . . . D B C BD A . . .
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C
A
B B B A C ...
= . . . D B C BD A DBCC B . . .
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C
A
B
B
B
A
C ...
= . . . D B C BD A DBCC B CC B CC B CCBD A DB C . . .
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
Generation of cutting sequences
Let us define operators gij that invert derivation.
The operator gij interpolates a sequence w admissible in Dj producing a
sequence ammissible in Di and such that (gij (w ))0 = w . For example:
Definition (Generation operator g02 )
Let w be ammissible in D0 . The sequence g02 w is obtained interpolating
the letters corresponding to vertices in Figure with the words on the
arrows:
BD
B
D
C
B
DBCC
A
DB
B
CC
CCBD
Example
w
g02 w
= ... D C
A
B
B
B
A
C ...
= . . . D B C BD A DBCC B CC B CC B CCBD A DB C . . .
The only sandwiched letters are the coloured ones, thus (g02 w )0 = w .
The interpolation operators
DB
D
A
C
g10
B
g30
BCC
D
B
C
B
g50
B
D
CC
g20
BD
C
C
C
DBCCBD
g40
CCB
DBCC
CBD
DBCCB
D
g60
DBC DCCBD
D
A
g70
DCCBD
D
C
BCC
DBC
A
A
DBCCBD
DBCCB
CB
C
CBD
CC
CCBD
CCB
D
D
B
D
BCCB
BC
BC
B
D
A
B
B
BCCB
DBCC
DB
C
A
A
C
B
CCBD
B
C
BD
B
D
C
CB
The other operators are obtained from these ones by permuting the
letters.
Characterization of the closure of cutting sequences
Lemma
If w is a cutting sequence and {sn }n∈N a sequence of sectors, we have
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
This condition, together with infinite derivability, is necessary and
sufficient to characterize the closure of octagon cutting sequences in
{A, B, C , D}Z :
Theorem (Smillie-U)
A sequence w ∈ {A, B, C , D}Z belongs to the closure of cutting
sequences in the octagon iff there exists a squence {sn }n∈N ∈ {0, . . . , 7}N
such that
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
An analogous theorem holds for every regular polygon.
Characterization of the closure of cutting sequences
Lemma
If w is a cutting sequence and {sn }n∈N a sequence of sectors, we have
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
This condition, together with infinite derivability, is necessary and
sufficient to characterize the closure of octagon cutting sequences in
{A, B, C , D}Z :
Theorem (Smillie-U)
A sequence w ∈ {A, B, C , D}Z belongs to the closure of cutting
sequences in the octagon iff there exists a squence {sn }n∈N ∈ {0, . . . , 7}N
such that
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
An analogous theorem holds for every regular polygon.
Characterization of the closure of cutting sequences
Lemma
If w is a cutting sequence and {sn }n∈N a sequence of sectors, we have
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
This condition, together with infinite derivability, is necessary and
sufficient to characterize the closure of octagon cutting sequences in
{A, B, C , D}Z :
Theorem (Smillie-U)
A sequence w ∈ {A, B, C , D}Z belongs to the closure of cutting
sequences in the octagon iff there exists a squence {sn }n∈N ∈ {0, . . . , 7}N
such that
w ∈ ∩n gss10 gss21 . . . gssnn−1 {A, B, C , D}Z .
An analogous theorem holds for every regular polygon.