triangular billiards and generalized continued fractions · geometric interpretations of continued...

Geometric interpretations of continued fractionsNatural extensions

Conjugacy of the Rosen and Veech algorithms

Triangular billiardsand generalized continued fractions

Pierre Arnoux

Orleans, 26 mars 2008

Joint work with Thomas A. Schmidt

Pierre Arnoux Triangular billiards and generalized continued fractions

Classical CF and their variantsRosen CFVeech CF

Geometric Interpretations of continued fractions:Classical continued fractions

The classical expansion of real numbers in continued fractionadmits several interpretations:

I Arithmetic: approximation by rational numbers pn

I Algebraic: matrices in GL(2,Z)

I Geometric: approximation of a line in the plane by elements ofthe lattice Z2

I Dynamic: geodesic flow on the modular surface SL(2,Z)\H,or equivalently Teichmuller flow for geometric structure forthe torus.

The Klein model

2 4 6 8 10 12

The classical domain for SL(2, Z)

Variants of the classical continued fraction

I Additive Continued Fraction

I Nearest Integer Continued Fraction

I Optimal Continued Fraction

I α-Continued Fraction (Japanese continued fractions)

Occur naturally in various contexts.All linked to the group SL(2,Z), lead essentially to the sameresults (infinite number of common approximants, for example).They can be analyzed as first return maps of the modular flow onvarious sections.

It is natural to look for such interpretations for generalizedcontinued fractions.

Occur naturally in various contexts.All linked to the group SL(2,Z), lead essentially to the sameresults (infinite number of common approximants, for example).They can be analyzed as first return maps of the modular flow onvarious sections.It is natural to look for such interpretations for generalizedcontinued fractions.

An arithmetic generalization: Rosen Continued fraction

In the nearest integer continued fraction, take for the partialquotients multiples of a fixed number λ instead of integers; i.e.write real numbers as:

x = n0λ+ε1

n1λ+ ε2

n2λ+ε3

n3λ+...

with ni ∈ Z, εi = ±1

Specially interesting if λ = λq = 2 cos πqAssociated to the map:x 7→ 1

|x | − kxλ [−λ2 ,

λ2 [→ [−λ

2 ,λ2 [

where kx is the unique integer k ∈ Z such that 1|x | − kλ ∈ [−λ

2 ,λ2 [

x = n0λ+ε1

n1λ+ ε2

n2λ+ε3

n3λ+...

with ni ∈ Z, εi = ±1Specially interesting if λ = λq = 2 cos πq

Associated to the map:x 7→ 1

|x | − kxλ [−λ2 ,

λ2 [→ [−λ

2 ,λ2 [

x = n0λ+ε1

n1λ+ ε2

n2λ+ε3

n3λ+...

with ni ∈ Z, εi = ±1Specially interesting if λ = λq = 2 cos πqAssociated to the map:x 7→ 1

|x | − kxλ [−λ2 ,

λ2 [→ [−λ

2 ,λ2 [

If λ = λq, SL(2,Z) is replaced by Hecke group Hq, generated by:

(0 −11 0

(1 λq

)small problem with orientation, solved by symmetrization:

x 7→ −1

x− kxλ

written as x 7→ (S−kT ).xProblem: can we find a geometric interpretation of this generalizedcontinued fraction, for example in terms of deformation ofgeometric structure (flow on moduli spaces)?

(0 −11 0

(1 λq

x 7→ −1

x− kxλ

(0 −11 0

(1 λq

x 7→ −1

x− kxλ

(0 −11 0

(1 λq

x 7→ −1

x− kxλ

(0 −11 0

(1 λq

x 7→ −1

x− kxλ

The domain for Hq

Rosen symmetrized continued fraction

Rational triangular billiard

Billiard in a right triangle with one angle πn

The group generated by reflexions on the sides has order 2nBy identifying sides, we obtain one or two regular polygon,depending on parityOpposite sides are identified

The billard with angle π8

A geometric generalization: Veech Continued fraction

Regular polygon with q = 2n sides. By identification of oppositesides: Riemann surface with a canonical quadratic differential givenby the natural flat structure out of the vertices of the polygon.This flat structure admits a non-trivial group of affineautomorphisms, its Veech group Vq, which preserves the affinestructure.Fix µq = 2 cot πqBoth maps

(cos 2π

q − sin 2πq

sin 2πq − cos 2π

(1 µq

)preserve the affine structure; they generate the Veech group Vq,subgroup of index 2 of a group Gq conjugate to Hq.

(cos 2π

q − sin 2πq

(1 µq

(cos 2π

q − sin 2πq

(1 µq

(cos 2π

q − sin 2πq

(1 µq

Renormalization of the octogon

The domain of Veech group

The Veech Continued fraction

The Veech group acts on the lines through the origin; one definesthe Veech continued fraction, on the slope of these lines, in thefollowing way:Line with slope in [−µq

2 ,µq

2 ].Act by Rq till the slope gets out of the interval. Then act by theparabolic element Pq till the slope comes back to the interval.One obtains in this way a generalized continued fraction, theadditive Veech AlgorithmWe will prove that The Veech and Rosen continued fraction areessentially the same.More precisely: if we take a suitable conjugate the Rosenalgorithm, then it has , for a.e. initial point, an infinite number ofcommon iterates with Veech algorithm.

2 ,µq

Veech additive continued fraction

Sketch of the proof

I Find a model for the natural extension in terms of sections forthe geodesic flow in H\PSL(2,R)

I Characterize first-return maps

I Prove that, after taking the square of the Rosen map andconjugating, these two maps are first return maps of thegeodesic Teichmuller flow on the Teichmuller disc of theregular polygon

I Consider their respective induction on the intersection of thedefinition domain: they are equal.

Sketch of the proof

Heuristics for natural extensionsParametrizations for SL(2,R)Numerical experiments

Heuristics for models of natural extensions of piecewisehomographies

Remark: everything is really done in PSL(2,R) = T1H2; forsimplicity, we work in SL(2,R), allowing to replace M by −M.The Haar measure for SL(2,R) has a simple expression.

In SL(2,R) the set of matrices

(0 bc d

)has codimension 1, and

Haar measure 0. Out of this set, one can take coordinates (a, b, c)

for the matrix

(a bc d

Lemma: In these coordinates, the Haar measure m, defined up toa constant, has expression:

dm =da db dc

Proof is a trivial computationPierre Arnoux Triangular billiards and generalized continued fractions

(0 bc d

for the matrix

(a bc d

dm =da db dc

(0 bc d

for the matrix

(a bc d

dm =da db dc

(0 bc d

for the matrix

(a bc d

dm =da db dc

The diagonal parametrization

Using the diagonal flow gt =

(et/2 0

0 e−t/2

), one can write almost

all element of PSL(2,R) as:(Xet/2 (XY − 1)e−t/2

et/2 Ye−t/2

)In these coordinates, the Haar measure admits the simpleexpression:

dm = dX dY dt

The Hyperbolic plane parametrization

One can write almost any element as(Xet/2√

X−YYe−t/2√

X−Yet/2√

X−Ye−t/2√

)In these coordinates, the Haar measure admits the expression:

dm =dX dY dt

(X − Y )2

This formula admits a nice interpretation as Liouville measure onthe unit tangent of the hyperbolic plane.

The arithmetic parametrization

Taking Z = − 1Y , one obtains an expression more common in

arithmetics:

dm =dX dZ dt

(1 + XZ )2

The search for a natural extension: numerical experiments.

T : x 7→ Mx .x piecewise homography T on an interval.M is a piecewise constant map to PSL(2,R), the Mx generate adiscrete subgroup Γ of PSL(2,R). We look for an explicit model ofthe natural extension T .We look for this natural extension as a first-return map of the

geodesic flow on Γ\H to the section

(x (xy − 1)1 y

), using the

diagonal parametrization and taking x as the first coordinate.A computation shows that the natural extension must be of theform:

(x , y) 7→(

ax + b

cx + d, (cx + d)2y − c(cx + d)

(x (xy − 1)1 y

), using the

(x , y) 7→(

ax + b

cx + d, (cx + d)2y − c(cx + d)

(x (xy − 1)1 y

), using the

(x , y) 7→(

ax + b

cx + d, (cx + d)2y − c(cx + d)

(x (xy − 1)1 y

), using the

(x , y) 7→(

ax + b

cx + d, (cx + d)2y − c(cx + d)

Problem: one does not know the domain of the natural extension.Solution: since we know the explicit form of the map, and since itis ergodic for Lebesgue measure, take a generic point and computeits orbit. It is then usually easy to find the domain of the naturalextension.The simplest example is the classical continued fraction, where onerecovers the Gauss measure.Some problems are encountered if there are indifferent fixed points:infinite measure. In this case, one needs to accelerate thealgorithm near the fixed point.

Natural extension of Veech continued fraction.

This algorithm has two indifferent fixed points at the boundary ofthe interval; one accelerates the algoritm on the correspondingcontinuity interval.define β = µ/2 · (3µ2 − 4)/(5µ2 + 4) andγ = µ/2 · (5µ2 + 4)/(3µ2 − 4).The domain ΩV of the model for the natural extension is bounded:

Above by

y = 1/(x + γ ) on [−µ/2,−β[ ;

y = 1/(x + µ/2) ) on [−β, µ/2[ .

Under by

y = 1/(x − µ/2 ) on [−µ/2, β[ ;

y = 1/(x − γ ) on [β, µ/2[ ;

The area of Ωv is cv = 2 log(

8 cos2 πq

Above by

y = 1/(x + γ ) on [−µ/2,−β[ ;

y = 1/(x + µ/2) ) on [−β, µ/2[ .

Under by

y = 1/(x − µ/2 ) on [−µ/2, β[ ;

y = 1/(x − γ ) on [β, µ/2[ ;

8 cos2 πq

Above by

y = 1/(x + γ ) on [−µ/2,−β[ ;

y = 1/(x + µ/2) ) on [−β, µ/2[ .

Under by

y = 1/(x − µ/2 ) on [−µ/2, β[ ;

y = 1/(x − γ ) on [β, µ/2[ ;

8 cos2 πq

Veech multiplicative continued fraction

Natural extension for Veech continued fraction

Natural extension of Rosen continued fractionUsing numerical experiments, one finds the domain of the naturalextension; denoting m = M.1 its domain ΩR is bounded by:

y = 1x−m under [−2/µ, µ/2[ ;

y = 1x+m above [−µ/2, 2/µ[ ;

x−Rj2 .m

under [− tan (j+1)πq ,− tan jπ

x+Rj2 .m

above [tan jπq , tan (j+1)π

The conjugate algorithm is of first return type.

It uses R1/2q ; one checks that its square only uses Rq and Pq, hence

its associated natural extension is also of first return type.Pierre Arnoux Triangular billiards and generalized continued fractions

Natural extension for Rosen continued fraction

Functions of first-return typeThe conjugacyremarks

Functions of first return type

Suppose that the map T is a piecewise isometry whose associatedmatrices generate a discret group Γ.Suppose that the natural extension T is defined on a domain Ω,which can be considered as a surface in SL(2,R).We want to know whether T is the first return map F to Ω of thegeodesic flow on Γ.It is clear that one can write T (P) = F nP (P).Denote τ(x) = −1

2 log(cx + d).One has: ∫

Ωτ(x) dm ≥ πVol(Γ\PSL(2,R))

Functions of first return type

Theorem: T is the first return map to the section if and only if∫Ωτ(x) dm = πVol(Γ\PSL(2,R))

One shows that the Veech algorithm is of first return type (notvery surprising: it is defined in this way!)Theorem: if T is of first return type, T k is of first return type iff itgenerates a subgroup of index k of Γ.

Rosen and Veech algorithm correspond to different groups, and aredefined on different intervals.BUT groups Hq and Gq are conjugate by

M =1√

sinπ/q

(1 cosπ/q0 sinπ/q

)M sends interval [−λq/2, λq/2] to [0, µq]. Explicit conjugacybetween Hq and Gq

One obtains in this way a conjugate Rosen algorithm on [0, µq];acting by Pq, one recovers an algorithm on [−µq/2, µq/2].

One has the identity:

MTSqM−1 = R1/2q

.Using this identity, one computes the conjugate algorithm:

k : [−µ/2, µ/2[→ [−µ/2, µ/2[

Which sends x to

R−12 .x si 0 ≤ x < R

12 .µ2 ;

P−kR−12 .x si R

12 Pk .− µ

2 ≤ x < R12 Pk .µ2 ;

R12 .x si 0 > x ≥ R−

12 .− µ

P lR12 .x si R−

12 P−l .− µ

2 ≤ x < R−12 P−l .µ2 .

Comparison of Rosen and Veech algorithms

The intersection Ω = ΩR ∩ ΩV is defined by:

y = 1x+M.1 over x ∈ [−µ/2, 2/µ[ ;

y = 1x+µ/2 over x ∈ [ 2/µ, µ/2[ ;

y = 1x−µ/2 under x ∈ [−µ/2,−2/µ[ ;

y = 1x−M.1 under x ∈ [−2/µ, µ/2[ .

From the preceding arguments, it is clear that the induced maps,on Ω, of Veech algorithm and the square of the conjugate Rosenalgorithm are equal. By ergodicity, for almost all starting point, thetwo orbits have infinite intersection.Q.E.D.

Additional remarks

Which are the points with no common approximants? (essentially,closed trajectories that do not intersect the common section)It must be possible to give combinatorial rules: one of theviewpoints is that the two algorithms consist in taking a differentset of generators for the group Vq.There is another algorithm, using interval exchanges on 4 intervals(for octogons, q = 8). This is similar to the link betweencontinued fractions and rotations. The relation of this algorithm tothe previous two is unclear.One should be in this setting able to use symbolic dynamics (likesturmian sequences for rotations). This should give some matricesin SL(4,Z), corresponding to the subgroup V8 of SL(2,Q(

√2)).

Additional remarks

√2)).

Additional remarks

√2)).

Additional remarks

√2)).

triangular billiards and generalized continued fractions · geometric interpretations of continued...

Documents