HEIDELBERG UNIVERSITY

SEMINAR: THE ARITHMETICS OF THE

HYPERBOLIC PLANE

Continued Fractions And Cutting

Sequences

Clemens Frubose

supervised by

Prof. Dr. Beatrice Pozzetti

Dr. Daniele Alessandrini

1 Recap of some properties of the hyperbolic half space

The geodesics on the hyperbolic plane consist of vertical lines and semicircles with centre on

R. That is because we can find a transformation T ∈ PSL(2,R) which maps I to this very

semicircle. Every triangle is then formed by three intersecting geodesics, which correspond

to straight lines in euclidean geometry. The Farey diagram F forms a tessellation of the

hyperbolic plane consisting of ideal triangles (i.e. triangles whose vertices lie at infinity)

which do not overlap. Since a lot of the characteristics of F have been discussed in the

previous talk, we just review a fascinating property:

We can successively construct F starting only with ∞ (= 10) and 0

1via the operation

a

b⊕F

c

d=a+ c

b+ d(1)

The resulting point is called neighbour of ab

and cd.

We have also seen that two numbers ab

and cd

are joined by and arc of F if and only if:

ad− bc = ±1 (2)

Figure 1: finite segment of F

1

2 Invariant Mappings on H

The Farey tessellation F is invariant under the action of those Mobius transformations which

are associated1 with the group

PSL(2,Z) =

{(a b

c d

): a, b, c, d ∈ Z, ad− cd = 1

}/± Id. (3)

”Invariant” means that the Farey tesselation does not change its shape, whereas its points do

change. A Mobius transformation T acts on the hyperbolic halfspace in the following way:

T (z) =az + b

cz + d(4)

T is said to be in PSL(2,Z) if its matrix representation

(a b

c d

)∈ PSL(2,Z). We use the

matrix representation, because the composition of the mappings T and T ′ can be written in

terms of a matrix product of their mappings T (T ′(z)) = TT ′(z).2 In order to understand the

connection between continued fractions and the Farey Diagram we will have to have a quick

look at the mappings J and P ∈ PSL(2,Z):

J =

(0 −1

1 0

), P =

(1 1

0 1

)(5)

Obviously, P maps z = a+bi to P (z) = z+1 = a+1+bi, whereas J maps z to J(z) = −1z−a+bia2+b2

.

-1.0 -0.5 0.5 1.0 1.5 2.0 2.5

0.5

1.0

1.5

-1.0 -0.5 0.5 1.0 1.5 2.0 2.5

0.5

1.0

1.5

Figure 2: Geodesic line γ before and after application of transformation J. J is a Mobiustransformation, it maps circles to circles; J(γ) is also a geodesic.

1Obviously, the Mobius transformations are in general not linear, so one should remember that we onlyrepresent a Mobius transformation by a matrix

2This result can be obtained through a simple calculation.

2

3 Connection between Continued Fractions and Cut-

ting Sequences

3.1 Basic definitions

Definition 1: Continued Fraction

We call a (possibly infinite) expression of the following form a continued fraction:

a0 +1

a1 + 1a2+

1a3+...

(6)

We require ai ∈ N and ai > 0 ∀ i > 0. An elegant notation for a continued fraction is

[a0; a1, a2, ...]3. On the one hand, every finite continued fraction represents an element x ∈ Q.

On the other hand, for every x ∈ Q we can find a representation as continued fraction using

Euclid’s algorithm:4

x = bxc+ r1,→ a0 = bxc1

r1= a1 + r2, → a1 =

⌊1

r1

⌋1

r2= a2 + r3, → a2 =

⌊1

r2

⌋...

(7)

with b·c as floor-function. The resulting continued fraction is x = [a0; a1, a2, ...].

Definition 2: Cutting Sequence

Consider a geodesic γ which connects the imaginary axis5 I to x ∈ R+. This geodesic is a

semicircle6 which cuts the tiles of F on its way to x. Since every tile of F is an ideal triangle,

γ can only cut a tile of F twice. γ cuts two geodesics of a triangle which meet at ∞ or R.

This vertex can lie either to the left or to the right with respect to the direction of γ from Ito x ∈ R 7. We denote such a cutting with L and R, respectively.

3If the expression is finite we get [a0; a1, a2, ...an−1, an] = [a0; a1, a2, ..., an−1 + 1an

] by abuse of notation,

because in general 1an

/∈ N. If now an = 1, then 1an

= 1, so that in this case we find two equivalentrepresentations in terms of continued fractions.

4Accordingly, every number x /∈ Q can be represented by an infinite continued fraction, because for thoseEuclid’s algorithm is infinite; Negative numbers will be treated below

5Choose any point iR 3 p 6= 0: We are only interested in the ray which starts at the imaginary axis withγ(∞) ∈ R+. The other part of the geodesic is irrelevant for our considerations, therefore the starting pointγ(−∞) ∈ R− can be chosen arbitrarily.

6Except for the degenerate case x = 07If γ meets exactly a vertex, we can label it either by L or by R. The reason for this freedom is the

ambiguity of the representation of a number as a continued fraction (see footnote [3])

3

Figure 3: One can easily read off the cutting sequence of the geodesic ray γ this figure as:L1R1L1R1L1....γ meets R at x = 81

50

The sequence Ln0Rn1Ln2 ... is called cutting sequence of x. It is easy to see that the cutting

sequence starts with L, if x > 1 and with R, if x < 1. We can also deduce from the fact, that

γ is a geodesic, that the cutting sequence is independent of the choice of the point p ∈ I.

3.2 Central Proposition: Connection between continued fractions

and the Cutting Sequences

Proposition 1: Let γ be a geodesic which connects an arbitrary point p ∈ I to x > 1 (x <

1), x ∈ R with cutting sequence Ln0Rn1Ln2 ... (Rn0Ln1Rn2 ...) , then x can be expressed as

x = [n0;n1, n2, ...] (resp. x = [0;n1, n2, ...]).

Proof:

We approach this proof constructively.

Firstly, we draw a geodesic semicircle which cuts I at p and connects to x. We call the

points where the cutting sequence changes from L to R along γ (clockwise) z−1, z0, z1,....

By convention, we label the crossing point p ∈ I with γ as z−1.

Obviously, γ has crossed the Farey tiles n0 times with symbol L, because x = bxc + r1 =

n0 + r1. Now we perform a shift P−n0 of γ to the left. z0 is then mapped to I. γ now reaches

the real line at remainder r1 < 1, similarly to the algorithm for a continued fraction. Thus,

P−n0 corresponds to the operation bxc (see fig. [4]).

4

The cutting sequence is now Ln0.

If we perform the transformation J onto P−n0(γ), r1 is mapped to − 1r1< −1. According to

the computational algorithm, we get 1r1

= a1 + r2 and since a1 ∈ N, it follows that a1 =⌊

1r1

⌋.

Since the properties of the hyperbolic plane are symmetric with respect to the imaginary

axis, we can now see that JP−n0(γ) crosses the Farey tiles n1 times with symbol R.8

The cutting sequence is now Ln0Rn1.

Performing the transformation P n1 now shifts γ to the right and again corresponds to the

operation⌊

1r1

⌋. Now, P n1JP−n0(z1) ∈ I.

If we apply J to P n1JP−n0(γ) we see that γ now aims at 1r2> 1, therefore it crosses the

Farey tessellation n2 times with symbol L.

The cutting sequence is now Ln0Rn1Ln2.

Finally, we have reached the initial situation again; we can perform these transformations

over and over again until after some transformations γ reaches an integer, i. e. some ri = 0;

or we continue to infinity, if x /∈ Q. �

The connection between continued fraction and cutting sequences thus arises from the fact

that the two transformations J and P leave the cutting sequence of a geodesic invariant.

With aid of the operations J and P we successively move the starting point to the end of the

geodesic ray. J can be seen as the operation 1x

whereas P can be regarded as the operation

bxc. This subtle similarity connects these two subjects, which have - at a first glance - noth-

ing in common.

8Keep in mind, that we have now with J ”changed the direction” to anticlockwise, but still according tothe initial direction of γ

5

Figure 4: Procedure of the proof with aid of the example x = [1; 1, 1, 1, 1, 1, 2] = 8150. Starting

point and direction of the geodesic ray are indicated by orange signs. (to be read row-wise)1. Initial situation. The geodesic ray γ meets R at x.2. P−1(γ) We perform the first shift P−1 to the left.3. JP−1(γ) We have reached a situation analogous to the initial situation, but mirrored at I.4. P 1JP−1(γ) We perform a shift P 1 to the right.5. JP 1JP−1(γ) We have reached a situation similar to the initial situation.6. P−1JP 1JP−1(γ) The procedure continues until one vertex is met exactly, as x ∈ Q.

3.3 Cutting Sequences for negative numbers

In order to find the cutting sequence for a negative number y < 0, we can remark that y

and J(y) = −1y

=: z have the same cutting sequence, because J ∈ PSL(2,Z). Indeed, if

we perform the transformation J , a negative number y is mapped to a positive number, for

which we already know how to find the cutting sequence.9

3.4 Negative numbers again

After having shown the connection between continued fractions and cutting sequences, we

shortly show the conclusions for negative numbers explicitly:

9In the next chapter we will find out, how the cutting sequence for a negative number naturally can beread of the geodesic ray γ. We only have to take into account that γ runs anticlockwise from p ∈ I to y < 0

6

If −1 < y < 0,10 then z = − 1y> 1. Therefore the cutting sequence for z = [b0; b1, b2, ...] (and

also for y) is Lb0Rb1Lb2 .... The continued fraction which arises from this cutting sequence for

z is b0 + 1b1+

1b2+...

. So, conclude that

y = −1

z= − 1

b0 + 1b1+

1b2+...

= −[0; b0, b1, b2, ...]. (8)

If y < −1 < 0, then z < 1. The resulting cutting sequence for z (and also for y) is:

Rb0Lb1Rb2 .... Accordingly, z = [0; b0, b1, b2, ...] = 1b0+

1

b1+1

b2+...

and therefore

y = −b0 −1

b1 + 1b2+...

= −[b0; b1, b2, ...]. (9)

As we had not required a0 to be greater than zero, we find another way to represent a negative

number as a continued fraction. We can choose a0 < 0 in such a way that:

y = a0 +1

a1 + 1a2+

1a3+...

, ai > 0 ∀i > 0, because 0 <1

a1 + 1a2+

1a3+...

< 1. (10)

This representation however destroys the connection between the cutting sequence and the

continued fraction, as the representation y = [−b0; b1, b2, ...], bi > 0 ∀i > 0 does not corre-

spond to a cutting sequence of a geodesic ray any more.

4 Convergents

Definition 3: Convergents of x

We call the series of fractions pnqn

= [a0; a1, a2, ..., an] convergents of x = [a0; a1, a2, ...].

We require pn and qn to be coprime. The convergents represent x to a certain order n.

4.1 Proposition 2:

Let x = [a0; a1, a2, ...] and pnqn

= [a0; a1, a2, ..., an], especially p0 = a0, q0 = 1. Then

det

(pn pn+1

qn qn+1

)= ±1 ∀ n ≥ 0. (11)

10The cases y = −1 and y = 0 are trivial.

7

Thus, we can conclude that neighbouring terms of the sequence are Farey neighbours, i.e.

they are connected by a semicircle of the Farey graph.

Proof:

We divide the proof in two parts:

(I) shows the property for a transition from an odd to an even index, (II) vice versa. Their

combination completes the proof.11

(I)

We chose an arbitrary x; for convenience, we assume x > 0. 12

We define a matrix M as M = JP−a2nJP+a2n−1 ...JP+a1JP−a0 . Since J and P have deter-

minant 1, M also has determinant 1.

The matrix M is clearly dependent on x, since the coefficients of the continued fraction of x

appear. We inspect the wayM acts on the continued fraction of x of order 2n, [a0; a1, a2, ..., a2n]:

M([a0; a1, a2, ..., a2n]) = 13JP−a2nJP+a2n−1 ...JP+a1J([0; a1, a2, ..., a2n])

= 14JP−a2nJP+a2n−1 ...JP+a1([−a1;−a2, ...,−a2n])

= ... = JP−a2n([a2n])

(12)

In order to discover further properties of M , we extend the bracket notation [a0; a1, a2, ..., an]

to non-integer elements ai.We can then find out that x = [a0; a1, a2, ..., an, α] = [a0; a1, a2, ..., an+1α

], α ∈ N. We have obviously not changed the value of x. Now we consider the effect of M

on [a0; a1, a2, ..., a2n, α], i.e. an expansion of x up to order 2n+ 1.

M([a0; a1, a2, ..., a2n, α] = JP−a2n([a2n +1

α]) = J([0 +

1

α]) = J(

1

α) = −α. (13)

If α = 0, then

M([a0; a1, a2, ..., a2n, α]) = JP−a2n([a2n +1

0]) = J(

1

0) = −0. (14)

11One should be aware that in [1] there are a few mistakes in the original proof of theorem 2.5 (1).12If x < 0, replace x with y = −x−1 with expansion y = [b0; b1, b2, ...].12M([a0; a1, a2, ..., an]) = JP−a2nJP+a2n−1 ...JP+a1JP−a0([a0; a1, a2, ..., an]) =

JP−a2nJP+a2n−1 ...JP+a1JP−a0(a0 + 1a1+

1

a2+ 1a3+...

) = JP−a2nJP+a2n−1 ...JP+a1J(a0 + 1a1+

1

a2+ 1a3+...

−a0) =

JP−a2nJP+a2n−1 ...JP+a1J [0; a1, a2, ..., an]14J([0; a1, a2, ..., an]) = J( 1

a1+1

a2+ 1a3+...

) = −(a1 + 1a2+

1a3+...

) = [−a1;−a2, ...,−an]

8

Accordingly, if α =∞, then

M([a0; a1, a2, ..., a2n, α]) = JP−a2n([a2n +1

∞]) = J(

1

∞) = −∞. (15)

M consists of powers and compositions of J and P and both of them are invertible with

determinant 1, thus M−1 exists.

We conclude from our findings above:

M−1(−0) = [a0; a1, a2, ..., a2n−1] =p2n−1q2n−1

, M−1(−∞) = [a0; a1, a2, ..., a2n] =p2nq2n

(16)

Since a matrix M−1 with matrix representation

M−1 =

(a b

c d

)(17)

corresponds to the mapping az+bcz+d

, we can find the entries of M−1 via ac

= a·∞+bc·∞+d

and bd

= a·0+bc·0+d .

It follows 15 that

M−1 =

(−p2n p2n−1

−q2n q2n−1

)or

(p2n −p2n−1q2n −q2n−1

). (18)

Because det(M) = 1 and det(M) · det(M−1) = 1 we know that

det

(∓p2n ±p2n−1∓q2n ±q2n−1

)= 1. (19)

We remark the following fact quickly:

p2n−1q2n − p2nq2n−1 = 1

→ p2n−1q2n − p2nq2n−1 > 0

→ p2n−1q2n > p2nq2n−1

→ p2nq2n

<p2n−1q2n−1

.

(20)

(II)

In analogy with M we define N as N = JP+a2n+1JP−a2n ...JP+a1JP−a0 . We proceed similarly

to the case above. N acts on [a0; a1, a2, ..., a2n+1] (expansion of x to order 2n + 1) in the

15The argument is lengthy and can be found in the appendix.

9

following way:

N([a0; a1, a2, ..., a2n+1]) = JP+a2n+1JP−a2n ...JP+a1J([0; a1, a2, ..., a2n+1])

= JP+a2n+1JP−a2n ...JP+a1([−a1;−a2, ...,−a2n+1])

= ... = JP a2n+1(−[a2n+1])

(21)

We also remark that:

N([a0; a1, a2, ..., a2n+1, β] = JP+a2n+1(−[a2n+1 +1

β]) = J([0− 1

β]) = J(− 1

β) = β. (22)

It follows that N−1(0) = [a0; a1, a2, ..., a2n+1−1] = p2nq2n

and N−1(∞) = [a0; a1, a2, ..., a2n+1] =p2n+1

q2n+1. In matrix form we get16

N−1 =

(p2n+1 p2n

q2n+1 q2n

)or

(−p2n+1 −p2n−q2n+1 −q2n

)(23)

with det(N−1) = 1. A similar calculation to eq. (20) leads to:

p2nq2n

<p2n+1

q2n+1

. (24)

Especially we obtain from (18) and (23):

det

(pn pn+1

qn qn+1

)= ±1 ∀ n ≥ 0. (25)

�

4.2 Proposition 3:

For n ≥ 1 we have the following recursion relation:

pn+1 = an+1pn + pn−1 and qn+1 = an+1qn + qn−1 (26)

16with a analogous argument via calculation as for the matrix M−1 (see appendix)

10

Proof:

If we use p2n+1

q2n+1= [a0; a1, a2, ..., a2n, a2n+1], we obtain M−1(−a2n+1) = p2n+1

q2n+1(see eq. (13)). On

the other hand, we obtain by directly plugging in (−a2n+1) to the mapping represented by

M−1 =

(∓p2n ±p2n−1∓q2n ±q2n−1

)(27)

the following expression:

M−1(−a2n+1) =p2n+1

q2n+1

=−p2n−1 + (−a2n+1)p2n−q2n−1 + (−a2n+1)q2n

orp2n−1 + (−a2n+1)(−p2n)

q2n−1 + (−a2n+1)(−q2n). (28)

As a result, we can see:p2n+1

q2n+1

=p2n−1 + a2n+1p2nq2n−1 + a2n+1q2n

(29)

Considering numerator and denominator separately17, we get the result:

p2n+1 = p2n−1 + a2n+1p2n, q2n+1 = q2n−1 + a2n+1q2n (30)

We now proceed in the same way as above with matrix N : N−1(a2n+2) = p2n+2

q2n+2(see eq. (22))

By plugging in a2n+2 to the mapping which is represented by

N−1 = ±

(p2n+1 p2n

q2n+1 q2n

), (31)

we get:

N−1(a2n+2) =p2n+2

q2n+2

=p2n + a2n+2p2n+1

q2n + a2n+2q2n+1

→ p2n+2 = p2n + a2n+2p2n+1, q2n+2 = q2n + a2n+2q2n+1

(32)

A combination of eq. (30) and (32) leads to the final result:

pn+1 = an+1pn + pn−1 and qn+1 = an+1qn + qn−1 (33)

�17we had required the qi and pi to be coprime

11

4.3 Proposition 4:

If we define

(an)n∈N :=p2nq2n

and (bn)n∈N :=p2n+1

q2n+1

, (34)

(an)n∈N is increasing and (bn)n∈N is decreasing. Especially:

limn→∞

an = limn→∞

bn = x (35)

Explicitly, this looks like so:

p2nq2n≤ p2n+2

q2n+2

≤ x ≤ p2n+1

q2n+1

≤ p2n−1q2n−1

∀n ≥ 1. (36)

If x ∈ Q, the continued fraction terminates. Eq. (35) implies that in that case equality holds.

Proof:

For reasons of clarity and comprehensibility we choose x /∈ Q in order to avoid a termination

of the continued fraction. An adaptation to the finite case can be made easily.

From eq. (22) and (24) we know that p0q0< p1

q1and p2

q2< p1

q1. As a base case for the claim we

show p0q0< p2

q2. This is obvious, because:

p0q0

= a0 < a0 +1

a1 + 1a2

=p2q2

→ p0q0<p2q2<p1q1.

(37)

According to eq. (22) and (24) we remark that p2q2< p1

q1and p2

q2< p3

q3. Since p2

q2and p3

q3are

Farey-connected, p3q3< p1

q1(see figure [5]). This holds, because the Farey-tessellation has no

intersections, so p3q3

cannot exceed p1q1

without the connection between p2q2

and p3q3

crossing a

Farey triangle.

The same argument repeats for p2q2< p4

q4:

According to eq. (22) and (24) we remark that p2q2< p3

q3and p4

q4< p3

q3. We conclude that

p2q2< p4

q4has to hold, because otherwise the Farey connection between p3

q3and p4

q4would cross

another Farey connection.

12

Figure 5: First five convergents of x=[1;1,1,1,1,1,2]. One can easily see the alternating

property of the sequence(pnqn

)n∈N

By repeated application of that argument we finally get the result

p2nq2n≤ p2n+2

q2n+2

≤ x ≤ p2n+1

q2n+1

≤ p2n−1q2n−1

∀ n ≥ 1. (38)

which is equivalent to

(an)n∈N :=p2nq2n

is increasing, (bn)n∈N :=p2n+1

q2n+1

is decreasing. (39)

We now show that (pnqn

)n∈N is a Cauchy-sequence and converges to x:∣∣∣∣pnqn − pn+1

qn+1

∣∣∣∣ =

∣∣∣∣pnqn+1 − pn+1qnqnqn+1

∣∣∣∣ = 18

∣∣∣∣ ±1

qnqn+1

∣∣∣∣ (40)

(qn)n∈N is an increasing sequence, because ai ≥ 1 ∀i > 0. So, for every ε > 0, we can choose

N(ε) ∈ N, so that ∣∣∣∣pnqn − pn+1

qn+1

∣∣∣∣ ≤ ∣∣∣∣ ±1

qN(ε)qN(ε)+1

∣∣∣∣ ≤ ε (41)

Thus (pnqn

)n∈N is a Cauchy-sequence. This immediately leads to the fact that (pnqn

)n∈N con-

verges, since R is a complete metric space. According to the construction of the sequence, it

is obvious that it converges to x.

18see eq. (25)

13

In particular, the sequences (an)n∈N and (bn)n∈N are Cauchy-sequences, because they are

monotonically increasing (resp. decreasing) and bounded by a0 + 1 (resp. 0). Again: since

R is a complete metric space, the Cauchy-sequences converge. Because (an)n∈N and (an)n∈N

are subsequences of the sequence (pnqn

)n∈N which converges to x, it follows that:

limn→∞

an = limn→∞

bn = limn→∞

pnqn

= x (42)

�

5 Conclusion:

We have discovered an interesting connection between the Farey Diagram F and continued

fractions. From the cutting sequence of a geodesic ray γ, we can construct the continued

fraction expansion of x. Furthermore, we have clarified that the convergents of x actually

approximate x in an alternating way. In the case that x ∈ Q, the continued fraction obviously

terminates after a finite order.

References

[1] Caroline Series : Continued fractions and hyperbolic geometry

http ://home- pages.warwick.ac.uk/ masbb/HypGeomandCntdFractions-2.pdf.

6 Appendix:

With respect to the definition of M , we see: M−1 = (JP−a2nJP+a2n−1 ...JP+a1JP−a0)−1 =

P+a0J−1P−a1J−1...P−a2n−1J−1P+a2nJ−1 with

J−1 =

(0 1

−1 0

)and P−a =

(1 +a

0 1

).

We can decompose M−1 into blocks of the form P−αiJ−1P+αjJ−1 19 and one block P+a0J−1.

We now only have to consider the signs, the entries of M−1 are given by eq. (16). Via a

simple calculation we obtain the matrix representation of the block P−αiJ−1P+αjJ−1 as(−αjαi − 1 αi

αj −1

).

19where αi represent an arbitrary index ai which is by construction an integer > 0

14

Obviously, the diagonal entries are negative, whereas the off-diagonal elements are positive.

A composition of an even number of these blocks yields a matrix of the form(Ai −Aj−Ak Al

), Am > 0.

Accordingly, a composition of an odd number of these blocks yields(−Ai Aj

Ak −Al

), Am > 0.

In order to construct the full matrix M−1, we consider the remaining block

J−1P+a0 =

(−a0 1

−1 0

).

A composition of these blocks results in a matrix with either first negative column or second

negative collum. Using eq. (16) then either leads to

M−1 =

(−p2n p2n−1

−q2n q2n−1

)or

(p2n −p2n−1q2n −q2n−1

).

15