Notes on Number Theory and Dynamical Systems - Unicam

Notes on Number Theory and Dynamical Systems Stefano Isola

∗

Contents 1 Fast and slow convergents 1.1 Connection to rotations of the circle . . . . . . . . . . . . . . . 1.2 Growth of denominators . . . . . . . . . . . . . . . . . . . . . . 2 A walk on the Farey tree 2.1 The {L, R} coding . . 2.2 The Stern-Brocot tree 2.3 The {A, B} coding . 2.4 The Farey shift and its

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3 The Minkowski question mark

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1 6 8 10 11 15 18 19 23

4 Transfer operators and partition functions 26 4.1 The partition function for negative integer temperatures . . . . . 32

1

Fast and slow convergents

Let 1

x=

1

a1 + a2 +

≡ [a1 , a2 , a3 , . . .]

(1.1)

1 a3 + · · ·

∗

Dipartimento di Matematica e Informatica, Universit`a degli Studi di Camerino (Italy), e-mail: [email protected], url: https://unicam.it/ stefano.isola/index.html

1

be the continued fraction expansion of the number x ∈ [0, 1]. By applying Euclid’s algorithm one sees that the above expansion terminates if and only if x is a rational number. For x irrational one can construct recursively a sequence pn /qn of rational approximants of x as p1 1 = q1 a1

0 p0 = , q0 1

and

pn+1 an+1 pn + pn−1 = , qn+1 an+1 qn + qn−1

We can write this recursion in matrix form as follows: letting  

1 1 1 0 and B := A := 1 0 1 1 and noting that BA we have


and




k−1

p1 q1

pn+1 qn+1




=

p0 q0

k 1

1 0

n≥1

(1.2)

(1.3)

 (1.4)

 = Aa 1

pn qn

(1.5)

 = Aa1 BAa2 −1 · · · BAan+1 −1 ,

n≥1

(1.6)

A short manipulation of (1.2) gives qn+1 pn − qn pn+1 = −(qn pn−1 − qn−1 pn ). Since q1 p0 − q0 p1 = −1 one obtains inductively the Lagrange formula qn pn−1 − qn−1 pn = (−1)n ,

n ≥ 1.

(1.7)

Another useful formula which can be easily obtained from (1.2) is the following: for all r ≥ 1 and n ≥ 1, [a1 , a2 , a3 , . . . , an +

r pn + pn−1 1 ]= · r r qn + qn−1

(1.8)

Letting r → ∞ we get in particular [a1 , a2 , a3 , . . . , an ] =

pn · qn

(1.9)

2

Note that

qn−1 = qn

1

1

=

1 an + qn−2 qn−1

an +

1 1 an−1 + qn−3 qn−2

and so forth. Therefore qn−1 = [an , an−1 , an−2 , . . . , a1 ]. qn

(1.10)

The numbers pqnn are called continued fraction convergents (CFC) of x and it turns out that the n-th CFC pqnn is the best rational approximation to x whose denominator does not exceed qn [HW]. One sees that p2n−1 p2n 0. (1.11) q2n q2n−1 Putting r = an+1 in (1.8) we get [a1 , a2 , a3 , . . . , an +

1 an+1

] ≡ [a1 , a2 , a3 , . . . , an , an+1 ] =

pn+1 · qn+1

(1.12)

But what happens if r in (1.8) takes on an intermediate value 1, 2, . . . , an+1 ?   n−1 Definition 1.1 For n ≥ 1 the sets rr pqnn +p for 1 ≤ r ≤ an+1 are the n’th +qn−1 Farey convergents (FC) for the real number x ∈ [0, 1). Example. Let x = e − 2 = [1, 2, 1, 1, 4, 1, 1, 6, . . .]. The first five CFC are p1 n=1 =1 q1 1 2 p2 = n=2 1 = q2 3 1+ 2 1 3 p3 = = n=3 1 q3 4 1 + 2+ 1 1

n=4

p4 1 = q4 1 + 2+ 1 1

=

5 7

1+ 1 1

n=5

p5 1 = q5 1 + 2+

=

1

1+

1 1 1+ 1 4

3

23 32

On the other hand, within the same accuracy, there are 2 + 1 + 1 + 4 = 8 FC’s. They are p1 + p0 1 t1,1 = = n = 1, r = 1, s1,1 q1 + q0 2 t1,2 2p1 + p0 2 n = 1, r = 2, = = s1,2 2q1 + q0 3 t2,1 p2 + p1 3 n = 2, r = 1, = = s2,1 q2 + q1 4 t3,1 p3 + p2 5 n = 3, r = 1, = = s3,1 q3 + q2 7 t4,1 p4 + p3 8 n = 4, r = 1, = = s4,1 q4 + q3 11 t4,2 2p4 + p3 13 n = 4, r = 2, = = s4,2 2q4 + q3 18 t4,3 3p4 + p3 18 n = 4, r = 3, = = s4,3 3q4 + q3 25 t4,4 4p4 + p3 23 n = 4, r = 4, = = s4,4 4q4 + q3 32 We now need some notions. Definition 1.2 The Farey sum over two rationals eration given by

a b

and

a
b


is the mediant op-

a a
a + a
a

=

· ⊕
:= b b b + b
b

(1.13)











It is easy to see that ab

falls in the interval ( ab , ab
). We say that ab and ab
are Farey neighbours if ab
− a
b = ±1. Two Farey neighbours define a Farey interval and each Farey interval can be labeled uniquely according to the mediant (child)
a

= a+a of the neighbours. b

b+b
Observe that given a pair of consecutive FC’s, say r pn + pn−1 tn,r = sn,r r qn + qn−1

and

tn,r+1 (r + 1) pn + pn−1 = sn,r+1 (r + 1) qn + qn−1

for some n ≥ 1 and 1 ≤ r < an+1 , we have tn,r+1 tn,r pn = ⊕ · sn,r+1 sn,r qn 4

(1.14)

Moreover qn tn,r − pn sn,r = qn pn−1 − pn qn−1 = (−1)n

(1.15)

for r = by Lagrange’s formula. Therefore, for every n ≥ 1, each FC stn,r n,r pn 1, . . . , an+1 is a Farey neighbour of qn , the corresponding Farey interval getting smaller and smaller as r increases. More precisely, using again Lagrange’s formula, one easily obtains    pn rpn + pn−1  1  − = · (1.16)  qn  rqn + qn−1 qn (rqn + qn−1 ) We therefore see that the FC stn,r is the best one-sided rational approximation to n,r x whose denominator does not exceed sn,r (although, if r < an+1 , there might be a CFC with denominator less than sn,r and closer to x on the other side of x). Increasing r, once we arrive at r = an+1 we hit a new CFC on the current side of x, closer than the previous CFC. Finally, using matrix notation, the FC’s can be expressed in terms of intermediate products in (1.5) for n ≥ 1 as
 tn,r pn (1.17) = Aa1 BAa2 −1 · · · BAan −1 BAr−1 , 1 ≤ r ≤ an+1 . sn,r qn The algorithm which produces the sequence of F C’s of a given real number is called slow continued fraction algorithm (see, e.g., [AO], [La]). Remark 1.3 The set F˜
of Farey fractions of order
is the set of irreducible fractions in [0, 1] with denominator ≤
, listed in order of magnitude (see [Ap]). Thus, F˜1 = ( 01 , 11 ),


   0 0 0 1 1 1 1 2 1 1 1 1 2 3 1 , , , F˜3 = , , , , , F˜4 = , , , , , , F˜2 = 1 2 1 1 3 2 3 1 1 4 3 2 3 4 1  2 and so on. In particular |F˜
| − 2 =
k=1 ϕ(k) ∼ 3π
2 with Euler totient function for ϕ(k) = |{0 < i ≤ k : gcd (i, k) = 1}|. Then we see that each stn,r n,r pn r = 1, . . . , an+1 is consecutive to in F˜
for sn,r <
≤ sn,r+1 . qn

5

1.1

Connection to rotations of the circle

One can interpret the above construction in terms of a kind of renormalization procedure for rotations of the circle [0, 1) through an angle x. With no loss  1  we take the initial point to be the origin 0 and set J0 = [0, x]. Since a1 = x we have a1 x ≤ 1 < (a1 + 1)x and thus 1 = a1 |J0 | + |J1 | with |J1 | = 1 − a1 x = p1 − xq1 . Moreover we have

(1.18)

|J1 | 1 = − a1 = [a2 , a3 , . . .] |J0 | x

and therefore a2 |J1 | ≤ |J0 | < (a2 + 1)|J1 | or, which is the same, |J0 | = a2 |J1 | + |J2 | with |J2 | = x − a2 (1 − a1 x) = q2 x − p2 .

(1.19)

Iterating this procedure, we construct a family of nested intervals Jn , n ≥ 0, such that |Jn | = [an+1 , an+2 , . . .] |Jn−1 |

(1.20)

and |Jn−1 | = an+1 |Jn | + |Jn+1 |,

n ≥ 0,

(1.21)

where we have set J−1 = [0, 1]. Using (1.18), (1.19) and (1.21) one gets inductively the formula fn := |Jn | = |qn x − pn | = (−1)n (qn x − pn ).

(1.22)

Note that 1 = qn fn−1 + qn−1 fn ,

∀n ≥ 1. 6

(1.23)

J0

J2

J0

J1

J2

J1

J0

J1

J2

J2

J3

Figure 1:

Now, if we denote by d the euclidean metric on [0, 1) then d(0, T r 0) = min |rx − p| =: rx.

(1.24)

p∈Z

Therefore d(0, T qn 0) = qn x = |qn x − pn | = fn

(1.25)

That is, the sequence of arc-lengths fn is but the sequence of successive closest distances to the initial point. This can be seen in the following way: starting from 0 and iterating a1 times one ends up at the point a1 x = q1 x which lies on the left of 0 and is the point closest to 0 up to now, being distant |J1 | from it. Iterating a1 more times one ends up at the point 2a1 x = 2q1 x which lies on the left of 0 at distance 2|J1 |, ... iterating a2 a1 times one ends up at the point a2 a1 x = a2 q1 x = (q2 − 1)x which still lies on the left of 0, at distance a2 |J1 |. One more iterate yields the point q2 x which now lies on the right of 0 at distance |J2 | and is the point closest to 0 up to now, and so on and so forth. The above implies that the first return map in the interval Jn (which is [0, fn ] or [1 − fn , 1] according whether n is even or odd) is the rotation through the angle (−1)n+1 fn+1 = qn+1 x − pn+1 . Finally, one has the equivalence fn ≈ 0

⇐⇒

x ≈ [a1 , . . . , an ] = 7

pn · qn

(1.26)

In addition, for each r = 1, . . . , an+1 , it holds fn ≈

fn−1 r

⇐⇒

tn,r 1 x ≈ [a1 , . . . , an + ] = · r sn,r

(1.27)

The three distance theorem. The points {kα} with 0 ≤ k ≤
partition the unit circle into
+ 1 intervals. A classical result (see e.g. [AB]), which can be easily obtained by induction using the above construction, is that the possible lengths of these intervals are organized according to the Farey convergents in the following way: • If 0 <
≤ q1 then there are two distinct lengths: f0 and 1 −
f0 (which become f0 and f1 when
= q1 ). • If rqn + qn−1 ≤
< (r + 1)qn + qn−1 for some n ≥ 1 and 1 ≤ r ≤ an+1 then there are at most three lengths: fn , fn−1 − rfn and fn−1 − (r − 1)fn , the last of which disappears when
= (r + 1)qn + qn−1 − 1. We point out that in the second case above there are two intervals, chosen from among those having the smallest lengths: fn = |qn x − pn | and fn−1 − rfn = |(rqn + qn−1 )x − (rpn + pn−1 )| which have 0 as their common endpoint. We then see that the approximations (1.26) and (1.27) are the same as shrinking one of these intervals to zero. Moreover, the fractions pn and rpn +pn−1 are the two successive elements of F˜n having qn

rqn +qn−1

x between them (see also Remark 1.3).

1.2

Growth of denominators

The Gauss map G : [0, 1] → [0, 1] is defined as G(x) = {1/x} for x > 0 and G(0) = 0.

(1.28)

It is well known that G has an invariant ergodic probability measure µ given by µ(dx) =

1 dx · . log 2 (1 + x)

(1.29)

A short reflection shows that x = [a1 , a2 , . . . , an + Gn (x)] or else x=

(Gn (x))−1 pn + pn−1 · (Gn (x))−1 qn + qn−1

(1.30) 8

From this we obtain at once qn x − p n fn =− , Gn (x) = − qn−1 x − pn−1 fn−1 where the numbers fn have been introduced in (1.22). Therefore n Gk (x) fn =

(1.31)

(1.32)

k=0

and, by the ergodic theorem, we have for µ-almost all x ∈ [0, 1] and then almost everywhere,

1 1 π2 lim log fn = log x µ(dx) = − · (1.33) n→∞ n 12 log 2 0 Since [(Gn (x))−1 ] = an+1 and thus an+1 < (Gn (x))−1 < an+1 + 1 another consequence of (1.30) is that 1 qn 1 qn < qn fn < < (1.34) < an+1 + 2 qn + qn+1 qn+1 an+1 and therefore using (1.31) 1 (1.35) < qn fn−1 < 1. 2 Putting together (1.33) and (1.35) we get the classical theorem of L´evy log qn π2 → almost everywhere (1.36) n 12 log 2 On the other hand we may expect the growth of FC’s denominator to be subextn,r tm ponential. Indeed, let sm ≡ sn,r with m = nk=2 ak + r be the m-th FC. Its denominator satisfies qn < sm ≤ qn+1 . It is a result of Khinchin and L´evy (see [Kh]) that n 1  1 in measure (1.37) ak → n log n k=1 log 2 Combining the above we get the following result Lemma 1.4

log sm π2 ∼ m 12 log m

in measure

√ Of course there are special behaviours: take x = ( 5 − 1)/2 = [1, 1, 1, . . .], then sn = qn and both are equal to the n-th Fibonacci number. Hence n−1 log qn converge to x−1 . 9

2

A walk on the Farey tree

Having fixed
≥ 1, let F
be the ascending sequence of irreducible fractions between 0 and 1 constructed inductively in the following way: set first F1 = ( 01 , 11 ), then F
is obtained from F
−1 by inserting among each pair of neighbours


a and ab
in F
−1 their child ab

as in (1.13). Thus b


   0 1 1 0 1 1 2 1 0 1 1 2 1 3 2 3 1 , , , F3 = , , , , , F4 = , , , , , , , , F2 = 1 2 1 1 3 2 3 1 1 4 3 5 2 5 3 4 1 and so on. The elements of F
are called again Farey fractions. Evidently F
⊇ F˜
. Remark 2.1 It has been shown in ([CG], Thm 2.6) that the set {{log b} | ab ∈ F
} becomes  equidistributed as
→ ∞. More specifically, the probability mea1 sure 2
+1 a ∈F
+1 δ{log b} converges to the Lebesgue measure on [0, 1]. b

Definition 2.2 For
≥ 1 we say that a Farey fraction x has rank
if x ∈ F
+1 \ F
. We also define the rank( 01 ) = rank( 11 ) = 0. For
≥ 1 there are exactly 2
−1 Farey fractions of rank
and their sum is equal to 2
−2 . Recall that every rational number x ∈ (0, 1) has a unique finite continued fraction expansion x = [a1 , . . . , an ] with an > 1 [HW]. The validity of the following relation will arise straightforwardly in the sequel:1 Lemma 2.3

x = [a1 , . . . , an ] −→ rank(x) =

n 

ai − 1

i=1 1 It is also easy to realize that all Farey fractions which fall in the interval (
+1 , 1
) have rank greater than or equal to
+ 1, whereas their continued fraction expansion starts with a1 =
.

An interesting object is the Farey tree F whose vertex-set is Q ∩ (0, 1) and which is constructed as follows: 1 We observe that, according to (2.1), the number 2
−1 of entries of F of rank
can be interpreted as the number of choices integers a1 , . . . , an , with 1 ≤ n ≤
and so that ai ≥ 1 of n for i = 1, . . . , n − 1, an > 1 and i=1 ai =
+ 1. Indeed, for each fixed n the number of
−1 , then sum over n = 1, . . . ,
. such choices is n−1

10

(2.1)

• every column in F contains one entry (vertex or node); • for
≥ 1 the
-th row is F
+1 \ F
;





• the node a+a , representing the interval [ ab , ab
], is connected by edges to b+b


its left child 2a+a and right child a+2a in the underlying row. 2b+b
b+2b
Note that the fractions 01 and 11 play the role of ancestors when using the Farey sum to obtain one row from the previous one. Besides the Farey sum, an alternative way to construct recursively the entries of F is as follows. Definition 2.4 Given ab ∈ F its descendants are the symmetrical entries of F a b given by a+b and a+b , respectively. Lemma 2.5 The collection of all descendants of the entries of a given row in F is precisely the underlying row. a b Proof. If ab = [a1 , . . . , an ] then a+b = [a1 + 1, . . . , an ] and a+b = [1, a1 , . . . , an ]. a b a Therefore rank( a+b ) = rank( a+b ) = rank( b ) + 1 and the claim easily follows. ♦

Remark 2.6 If ab = [a1 , a2 . . . , an ] and
rank( ab
) and b = b
.

2.1

a
b


= [an , an−1 . . . , a1 ] then rank( ab ) =

The {L, R} coding

Every rational number in (0, 1) appears exactly once in the above construction and corresponds to a unique finite path on F starting at the root node 12 and whose number of vertices equals the rank of the rational number. We can code this path in the following way: first, any rl ∈ F can be uniquely decomposed as2 l m n = ⊕ r s t

with ns − mt = 1

(2.2)

and the unimodular relations sl − rm = rn − lt = ns − mt = 1 2

All fractions are supposed in lowest terms.

11

(2.3)

1 2

2 3

1 3

2 5

1 4

1 5

2 7

3 4

3 5

3 7

3 8

4 7

5 8

5 7

4 5

Figure 2: The first four levels of the Farey tree

plainly hold. The neighbours may accordingly identify 
l n m ∈U ⇐⇒ t s r with


U=

a b c d



m s

and

n t

are thus the ‘parents’ of

l r

in F and we

(2.4)

 | 0 < a ≤ c, 0 ≤ b < d, ad − bc = 1 ·

Note that the left column bears on the right parent and viceversa. Thus 
1 1 0 ⇐⇒ 1 1 2 On the other hand, any given by m m+n ⊕ s s+t

l r

and

(2.5)

(2.6)

as above has a unique pair of (left and right) children, m+n n ⊕ s+t t

(2.7)

respectively. In order to generate them we set  

1 1 1 0 and R := L := 0 1 1 1 12

(2.8)

Note that for k ∈ Z 
1 0 k L = k 1


k

and R =

1 k 0 1

and also




k

R S =SL

k

with S = S

−1

=

 (2.9)

0 1 1 0

 .

Moreover, we have  


m m+n m+n m 1 0 n m ⇐⇒ = ⊕ s+t s 1 1 t s s s+t and




n m t s




1 1 0 1




=

n m+n t s+t

 ⇐⇒

m+n n ⊕ s+t t

(2.10)

(2.11)

(2.12)

In other words, the matrices L and R, when acting from the right, move to the left and right child in F, respectively. Moreover, it is plain that given Y ∈ V we have Y R ∈ V and Y L ∈ V. We have thus proved the following Proposition 2.7 To each entry x ∈ F there corresponds a unique element X ∈ U which, in turn, can be uniquely presented as X=L Mi (2.13) i

 where the number of terms in the product L i Mi is equal to rank(x) and Mi = L or Mi = R according whether the i-th turn, along the descending path in F which starts from the root node 12 and reaches x, goes left or right. Remark 2.8 By the way, the matrices L and R induce the so called Farey tesselation of the upper half plane H = {z : Im z > 0} (see [Kn1]). 3 Example. 10 is the right child of 27 , which is the right child of 14 , which is the left child of 13 , which is the left child of 12 . Thus 
3 1 2 = LLLRR ⇐⇒ 3 7 10

13

For

3 , 11

which is the left child of 27 , we find 
3 2 1 = LLLRL ⇐⇒ 7 4 11

3 3 Note that rank( 10 ) = rank( 11 ) = 5. To any given irrational number x ∈ [0, 1] we may associate a unique infinite path on F, and thus a unique semi-infinite word in {L, R}N . Bearing in mind the continued fraction expansion (1.1) of x, let

1 t1,1 = s1,1 a1 + 1 the first FC of x. In order to reach it from the top of F we need the block La1 −1 . Whence we code x through the map φ : [0, 1] → {L, R}N defined by φ(x) = La1 M1 M2 . . .

(2.14)

where Mi = L or Mi = R according whether the i-th turn along the infinite path in F which starts from st1,1 and approaches x along the sequence of successive 1,1 FC’s goes left or right. This coding is faithful to the binary structure of F but apparently not so much to the continued fraction expansion of x. To make the latter more transparent we may note that, according to the characterization of the FC’s given above (see (1.15) and (1.16)), the symbols L and R in (2.14) come in blocks whose lengths are given by nothing but the partial quotients ai of x. More precisely, a short reflection shows that the following rule is in force: the first block is such that Mi = R if 1 ≤ i ≤ a2 . Moreover, for k ≥ 2 let bk =

k 

ai ,

i=2

then we have  L, if Mi = R, if

b2k−2 < i ≤ b2k−1 , b2k−1 < i ≤ b2k .

(2.15)

In other words, we have the coding x = [a1 , a2 , a3 , . . .]

←→

φ(x) = La1 Ra2 La3 . . .

(2.16)

Furthermore we set φ(0) = L∞ and φ(1) = LR∞ . More generally, we note that each rational x has two infinite paths which agree down to node x: they are those 14

starting with the finite sequence coding the path to reach x from the root node and terminating with either RL∞ or LR∞ . We shall agree that φ(x) terminates with RL∞ or LR∞ according whether the number of its (finite) partial quotients is even or odd. On the other hand, for notational simplicity’ sake we shall assume this agreement only implicitly. We summarize the above in the following Theorem 2.9 To every x ∈ [0, 1] with continued fraction expansion x = [a1 , a2 , a3 , . . .] there corresponds a unique sequence φ(x) ∈ {L, R}N given by φ(x) = La1 Ra2 La3 . . . which represents an infinite path on F whose sequence of vertices starting from the a1 -th is precisely the sequence ( stm )m≥1 of FC’s of x. Moreover, if  denotes m N the lexicographic order on {L, R} then x > y =⇒ φ(x)  φ(y). An simple consequence of the above construction is the following result. Proposition 2.10 Let x = [a1 , . . . , an ] with an > 1 and n even. Then its left and right children in F are given by x
= [a1 , . . . , an − 1, 2] and x

= [a1 , . . . , an + 1], respectively. If instead n is odd the expansions for x
and x

have to be interchanged. Proof. Since n is even we can write x = [a1 , . . . , an ] ←→ φ(x) = La1 Ra2 · · · Ran −1 .

(2.17)

Therefore x
= φ−1 (La1 Ra2 · · · Ran −1 L) and x

= φ−1 (La1 Ra2 · · · Ran ) which yield the claim. A similar reasoning applies for n odd. ♦

2.2

The Stern-Brocot tree

The Farey tree extends to the Stern-Brocot tree, which contains every positive rational number exactly once. The latter is constructed as the Farey tree but instead of using as ancestors the fractions 01 and 11 , one uses 01 and 10 (that is ∞ ‘in lowest terms’). The root node is thus 11 and its left sub-tree starting at the second row is just the Farey tree. Each row is thus obtained by reflecting the corresponding row of the Farey tree w.r.t the middle column and interchanging numerator and denominator of each fraction of the right hand side (see also [GKP]). 15

1 1

2 1

1 2

2 3

1 3

1 4

2 5

3 1

3 2

3 4

3 5

4 3

5 3

5 2

4 1

Figure 3: The first four levels of the Stern-Brocot tree

One can easily extend the {L, R} coding to the Stern-Brocot tree. In the spirit of (2.4) let us start by identifying the root node as 
1 0 1 1 0 , = ⊕ ⇐⇒ 0 1 1 1 0 and more generally l m n = ⊕ ⇐⇒ r s t




n m t s

 ∈ SL (2, Z).

Note that given M ∈ SL(2, Z) which represents the fraction rl , the matrix S M S represents the symmetric fraction rl . Now, to a number x ∈ R+ with continued fraction expansion 1

x = a0 +

≡ [a0 ; a1 , a2 , a3 , . . .]

1

a1 + a2 +

1 a3 + · · ·

there corresponds a unique sequence in {L, R}N given by Ra0 La1 Ra2 La3 . . . representing the infinite path on the Stern-Brocot tree which converges to x. There is an interesting way to construct this structure dynamically. To show 16

this we first introduce the notion of depth of a number x ∈ Q+ : first define depth( 01 ) = depth( 10 ) = 0 and n 

x = [a0 ; a1 , . . . , an ] =⇒ depth(x) =

ai

(2.18)

i=0

Thus depth( 11 ) = 1, depth( 12 ) = depth( 21 ) = 2 and so on. Its relation with the rank is depth(x) = [x] + rank({x}) + 1

(2.19)

Consider the map T : R+ ∪ {∞} → R+ ∪ {∞} given by T : x →

1 1 − {x} + [x]

(2.20)

The first 33 iterates T i ( 10 ), i = 0, 1, . . . , 32, are 1 0 3 4

0 1 4 1

→ →

1 1 1 5

→ 5 4

2 1

1 2 4 7

7 3

→ 3 8

3 2

1 3 8 5

5 7

2 3 7 2

→

3 1 2 7

7 5

5 8

1 4 8 3

4 3 3 7

3 5 7 4

5 2 4 5

2 5 5 1

5 3

each block corresponding to the elements of the tree with fixed depth. An interesting item is the ordering of the iterates of T within each block of given depth. To clarify this point let kX ∈ SL (2, Z) represent a number x ∈ Q+ as above and write it as X = i=1 Mi where Mi ∈ {L, R} and k = depth(x). ˆ Let 1 us denote by xˆ the positive rational number represented by the matrix X = x) but x = xˆ if and only if the sequence i=k Mi . Clearly depth(x) = depth(ˆ M1 . . . Mk is a palindrome. Proposition 2.11 The map T has the following properties: 1. T is an automorphism of R+ ; 2. for any x ∈ R+ , T (x) ∈ Q+ if and only if x ∈ Q+ ; 3. the forward T -orbit of 10 is Q+ ranged with non-decreasing depth. More precisely, for 2k−1 < i ≤ 2k , let xi be the increasing sequence of elements of the Stern-Brocot tree with depth k ≥ 1. Then the corresponding elements of the T -orbit of 10 are T i ( 10 ) = xˆi .

17

Proof. One easily checks that T is one-to-one and onto, with inverse T −1 (x) = 2n + 1 −

1 x

,

1 1 ≤x< n+1 n

,

n ≥ 0.

(2.21)

This proves 1. Statement 2 is immediate. Now, if x ∈ N then T (x) = 1/(x + 1) so that depth(T (x)) = depth(x) + 1. If instead x = [a0 ; a1 , . . . , an ] with n ≥ 1 then T (x) = 1/(a0 + 1 − {x}) so that depth(T (x)) = depth(x) since depth(1/x) = depth(x) and rank({x}) = rank(1 − {x}). This yields the first part of statement 3. ♦ Remark 2.12 Although the T -orbit of 10 is dense in R+ , it ‘diffuses’ only logarithmically. Indeed we have T i ( 10 ) = n for i = 2n and therefore 1 sup T i ( ) = O(log n) 0 i≤n Remark 2.13 According to the above discussion one may argue that all orbits {T i (x) : i ≥ 0}, x ∈ R+ , are dense in R+ .

The {A, B} coding

2.3

Using (2.10) we can write La1 Ra2 La3 . . . = La1 SLa2 SLa3 . . . On the other hand we have L ≡ A and (see (1.4)) 
k 1 k = BAk−1 SL = 1 0

(2.22)

(2.23)

This defines a recoding ψ : [0, 1] → {A, B}N so that σ(x) = Aa1 BAa2 −1 BAa3 −1 . . . (2.24)  of x, which has rank
= ni=1 ai + r, will then be expressed as The FC stn,r+1 n,r+1
  a a tn,r+1 L 1 R 2 · · · Lan Rr , n odd, φ (2.25) = La1 Ra2 · · · Ran Lr , n even, sn,r+1 x = [a1 , a2 , a3 , . . .]

or else ψ




tn,r+1 sn,r+1

←→

 = Aa1 BAa2 −1 · · · BAan −1 BAr−1 . 18

(2.26)

Note that both expansions have exactly
terms and the latter agrees with (1.17) once we interpret the l.h.s. of (1.17) as the FC stn,r+1 of x, that is taking the n,r+1 Farey sum of the columns in the same spirit as (2.4). Example. The example with x = e − 2 = [1, 2, 1, 1, 4, 1, 1, 6, . . .] discussed above, which yields φ(e−2) = LRRLRLLLLRL . . .

or else ψ(e−2) = ABABBBAAABB . . .

can be used to check step by step what we are claiming here. For example its FC st4,2 = 13 = [1, 2, 1, 1, 2], which has rank 6, can be expressed as 18 4,2
φ

2.4

13 18




= LRRLRL or else ψ

13 18

 = ABABBB

The Farey shift and its relatives

So far, a sequence in {L, R}N starting with the symbol R has no image in [0, 1] with φ−1 . Let us make the identification Rn1 Ln2 Rn3 . . . = SLn1 Rn2 Ln3 . . . ≡ Ln1 Rn2 Ln3 . . .

(2.27)

and denote by Σ the half-space of {L, R}N so obtained. We can write Σ = {L, R}N /S

(2.28)

We see that the map φ is a bijection between [0, 1] and Σ. Let Φ : Σ → Σ be the Farey shift map defined by Φ(La1 Ra2 La3 Ra4 . . .) = La1 −1 Ra2 La3 Ra4 . . .

(2.29)

Note that, besides L∞ the only fixed point of Φ is given by the sequence LRLRLR . . . which is the image with φ of x = [1, 1, . . .], the golden mean. This map acts

on points in F by reducing their rank of one unit. For example, since φ 13 = LRRLRL, with the identifications made above we have 18


 5 13 = RRLRL ≡ LLRLR = φ Φ φ 18 13


 5 5 Φ φ = LRLR = φ 13 8 19




 3 5 = RLR ≡ LRL = φ Φ φ 8 5


 3 2 Φ φ = RL ≡ LR = φ 5 3


 2 1 Φ φ = R≡L=φ 3 2 Let us define the Farey map F : [0, 1] → [0, 1] given by  x  , if 0 ≤ x ≤ 12 , 1 − x F (x) = 1 − x  , if 12 < x ≤ 1 . x

(2.30)

Its name can be related to the easily verified observation that the set of preimages ∪
k=0 F −k {0} coincides with F
for all
≥ 1. Note also that the
-th row of the Farey tree is precisely F −(
−1) ( 12 ). In particular, this implies that −k ∪∞ {0} = Q ∩ [0, 1]. k=0 F Proposition 2.14 Let φ : [0, 1] → Σ be the coding described above. Then Φ ◦ φ = φ ◦ F. Proof. If 1/2 < x ≤ 1 then a1 = 1 and F (x) = then a1 > 1 and F (x) = 1/( x1 − 1). Therefore,

1 x

− a1 . If instead 0 < x ≤ 1/2

if x = [a1 , a2 , a3 , . . .] then F (x) = [a1 − 1, a2 , a3 , . . .] ,

(2.31)

with [0, a2 , a3 , . . .] ≡ [a2 , a3 , . . .]. The claim now follows from (2.29) and (2.27). ♦

20

The Gauss and Fibonacci maps The map F has (at least) two induced versions: the first one is the Gauss map G : [0, 1] → [0, 1] already introduced in (1.28), which for x > 0 can be written as G(x) = F [1/x] = {1/x}.

(2.32)

Recall that if x = [a1 , a2 , a3 , . . .] then G(x) = [a2 , a3 , . . .] . Noting that




G(x) = F (x) if x ∈ An = n

1 1 , n+1 n

(2.33)

 (2.34)

we see that G is obtained by iterating F once plus the number of times necessary to reach the interval [1/2, 1]. The second one is the Fibonacci map H and is defined by iterating F once plus the number of times necessary to reach the interval [0, 1/2]. Let F0 = 0, F1 = 1 and Fn+1 = Fn + Fn−1 for n ≥ 1 be the Fibonacci numbers. Then, for n ≥ 0,  F 2n+1 x − F2n  , if x ∈ B2n ,  F −F x (2.35) H(x) = F2n+1 − F2n+2 x 2n+2   2n+1 , if x ∈ B2n+1 , F2n+5 x − F2n+4 with



B2n

F2n F2n+2 = , F2n+1 F2n+3




 ,

B2n+1 =

 F2n+3 F2n+1 , . F2n+4 F2n+2

(2.36)

In this case it is easy to check that if x = [a1 , a2 , a3 , . . .] then H(x) = [ar − 1, ar+1 , . . .],
Given M =

a b c d

where r = min { i : ai > 1}.

(2.37)

 we may define the M¨obius transformation

ˆ (x) := ax + b x→M cx + d

(2.38)

By the above, given x ∈ [0, 1/2] the point φ−1 ◦ L−1 ◦ φ(x) is but F (x) and ˆ −1 (x) (recall that for x ∈ [0, 1/2] we have F (x) = φ−1 ◦ L−1 ◦ φ(x) = L 21

Figure 4: The Farey map and its induced Fibonacci (upper) and Gauss (lower) maps.  1 0 ). But what happens if x ∈ (1/2, 1] so that a1 = 1? To see L = −1 1 this we put  

1 0 0 1 and I1 = SR = LS = . (2.39) I0 ≡ L = 1 1 1 1


−1

We have I1 Rn1 Ln2 Rn3 . . . = I1 Ln1 Rn2 Ln3 . . . (2.40) 
−1 1 , for x ∈ (1/2, 1] we have F (x) = Therefore, noting that I1−1 = 1 0 φ−1 ◦ I1−1 ◦ φ(x) = Iˆ1−1 (x). To summarize we can represent the action of F as  −1 Iˆ (x), if x ∈ [0, 1/2] , F (x) = ˆ0−1 (2.41) I1 (x), if x ∈ (1/2, 1] , that of G as −(n−1)

G(x) = Iˆ1−1 ◦ Iˆ0

(x),

x ∈ An .

(2.42)

−(n−1) (x), H(x) = Iˆ0−1 ◦ Iˆ1

x ∈ Bn .

(2.43)

and that of H as

22

The modified Farey map Finally we introduce the modified Farey map F˜ : [0, 1] → [0, 1] given by  x  , if 0 ≤ x ≤ 12 , 1 − x ˜ F (x) = (2.44)  2 − 1 , if 1 < x ≤ 1 . 2 x This map preserves orientation and has two indifferent fixed points, at 0 and 1. The advantage of using F˜ instead of F is that one can retrace the path from a leaf x ∈ F back  to the root 1/2. More precisely, for x ∈ F let (cf. Proposition 2.7) X = L i Mi be the element which uniquely represents x in U. Then one easily sees that the following rule is in force: if F˜ (i−1) (x) < 1/2 then Mi = L, F˜ (i−1) (x) > 1/2 then Mi = R, for i = 1, . . . , k with k = rank(x) so that F˜ k (x) = 1/2.

3

The Minkowski question mark

Given a number x ∈ (0, 1) with continued fraction expansion x = [a1 , a2 , a3 , . . .], one may ask what is the number obtained by interpreting the sequence φ(x) (see (2.16)) as the binary expansion of a real number in (0, 1). The number so obtained is denoted ?(x) and writes ?(x) = 0.00 . . . 0 11 . . . 1 00 . . . 0 · · ·       a1 −1

a2

(3.1)

a3

or, which is the same,  (−1)k−1 2−(a1 +···+ak −1) . ?(x) =

(3.2)

k≥1

For instance ?(1/n) = 1/2n−1 , for all n ≥ 1. Setting ?(0) = 0 and ?(1) = 1 one has the following properties for the function ? : [0, 1] → [0, 1] (see [Sa], [Ki]): • ?(x) is strictly increasing from 0 to 1 and H¨older continuous of order 2 β = √log ; 5+1 • x is rational iff ?(x) is of the form k/2s , with k and s integers; • x is a quadratic irrational iff ?(x) is a (non-dyadic) rational; 23

?x 1 0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1

x

Figure 5: The ? function

• ?(x) is a singular function: its derivative vanishes Lebesgue-almost everywhere.

The following additional property easily follows from the definition. Lemma 3.1 ?(x) satisfies the functional equation ?(1 − x) =?(1)−?(x)

(3.3)

Proof. Assuming that x ∈ [1/2, 1] we write x = 1/(y + 1) with y ∈ [0, 1] and 1 − x = y/(y + 1) ∈ [0, 1/2]. Setting moreover y = [a1 , a2 , . . .] we have x = [1, a1 , a2 , . . .] and 1 − x = [1 + a1 , a2 , . . . , ]. The assertion now follows by direct application of (3.2). ♦ Let us now see how ? acts on Farey fractions. We have already seen that

 1 0+1 1 1 ? =? = [?(0)+?(1)] = 2 1+1 2 2 More generally, for any pair a/b and a
/b
of consecutive Farey fractions the function ? equates their child to the arithmetic average:
 

 a + a
1 a a ? = (3.4) ? +? b + b
2 b b


24

One sees that the function ? maps the Farey tree F to the dyadic tree D defined as follows: having fixed
≥ 1, let D
be the ascending sequence of fractions of the form k/2
−1 , k = 0, 1, . . . , 2
−1 . We have


   0 1 0 1 1 0 1 1 3 1 D1 = , , D2 = , , , D3 = , , , , 1 1 1 2 1 1 4 2 4 1 and so on. Then D is the same graph as F with the
-th row replaced by D
+1 \ D
. An immediate consequence of the fact that ?(F) = D is that ?(x) is the asymptotic distribution function of the sequence of Farey fractions3 : Theorem 3.2 Since #{ ab ∈ D
+1 : x = lim
→∞ 2


a b

≤ x}

(3.5)

then #{ ab ∈ F
+1 : ?(x) = lim
→∞ 2


a b

≤ x}

·

(3.6)

As a further immediate consequence we get that the Fourier-Stieltjes coefficients of ?(x) are as in the following Corollary 3.3 Let

1 cn = e2 π i n x d?(x)

(3.7)

0

then 1  2πin a b. e
→∞ 2
a

cn = lim

b

(3.8)

∈F
+1

Finally, a short reflection using the definition (3.1) shows that ? conjugates the Farey map F and the modified Farey map F˜ to the tent map  2x, if 0 ≤ x < 1/2, (3.9) T (x) = 2(1 − x) if 1/2 ≤ x ≤ 1 3

This result can be also deduced as a consequence of a more general result obtained in [VPB] using a suitable enumeration of the rationals in (0, 1). As for the convergence of the atomic measure concentrated on Fn to d? see [CG] and [Go].

25

and the doubling map D(x) = 2x mod 1, respectively. Indeed, for any ω = ω1 ω2 . . . with ωi ∈ {0, 1} we have . . . 0 1 ω) = 0.00 . . . 0 1 ω T i (0.00     k

for i ≤ k

(3.10)

k−i

and (3.11)

T (0. 1 ω) = 0.ω

where ω = ω 1 ω 2 . . . and ω i = 1 − ωi . A similar reasoning applies for D. Putting together the above, (2.31) and (3.1) we then get the following commutative diagrams Theorem 3.4 F˜

F

[0,1] −→ [0,1]   ? ? T [0, 1] −→ [0, 1]

,

[0,1] −→ [0,1]   ? ? D [0, 1] −→ [0, 1]

This implies that the measure d?(x) is invariant under both maps F and F˜ , and its entropy is equal to log 2. This makes d?(x) the measure of maximal entropy for F and F˜ . Being zero at every rational point d? is of course singular w.r.t. Lebesgue. More specifically, d? is concentrated on a subset X ⊂ [0, 1] having Hausdorff dimension α ≈ 0.875 (see [Ki]). It is however easy to realize that F and F˜ have (not normalizable) absolutely continuous invariant measures, with densities 1/x and 1/x(1 − x), respectively.

4

Transfer operators and partition functions

 a b ∈ U and complex parameter q one can To a given matrix M = c d associate the positive operator πq (M ) acting on the right as [LeZa]
 ax + b −2q ˆ (x)). ˆ
(x)|q · f (M (4.1) = |M πq (M ) f (x) := (cx + d) f cx + d


For example we have 1 πq (S)f (x) = 2q f x


 1 . x

(4.2) 26

The operator Pq associated in this way to the map ([0, 1], F ) turns out to be the transfer operator acting as Pq f (x) := [ πq (I0 ) + πq (I1 ) ] f (x) 
 
1 1 x = +f . f (x + 1)2q x+1 x+1

(4.3)

Of special significance is the (Perron-Frobenius) operator P1 which satisfies

1

1 g ◦ F (x) f (x)dx = g(x) P1 f (x)dx (4.4) 0

0

and has norm at most one in the Banach space L1 ([0, 1], dx). A function h is the density of an absolutely continuous invariant measure for F if and only if P1 h = h. In this case we find h(x) = 1/x (which however does not lie in L1 ([0, 1], dx)!). Let f be an eigenfunction of Pq analytic in the half-plane Re x > 0. It satisfies
2q 

 1 x 1 f +f (4.5) λf (x) = Pq f (x) = x+1 x+1 x+1 and also πq (S)f (x) = f (x)

(4.6)

Therefore the eigenvalue equation is equivalent to the three-term equation 
1 1 (4.7) λf (x) = f (x + 1) + 2q f 1 + x x which is a generalisation of the Lewis functional equation (with λ = 1) studied in number theory (see [Le], [LeZa]). Remark 4.1 In the context of the thermodynamic formalism, once a one-sided shift Φ : Σ → Σ and a potential function ϕ ∈ C(Σ) are given one defines a transfer operator Lϕ on C(Σ) by  eϕ(η) g(η) (4.8) Lϕ g(ξ) = η ∈ Φ−1 ξ

which plays a key role in the study of equilibrium states for (Σ, Φ, ϕ) and their properties [B], [Ru]. In particular, one defines vark ϕ = sup{|ϕ(ξ) − ϕ(ξ
)| : ξi = ξi
, 0 < i ≤ k} 27

(4.9)

and it turns out that if vark ϕ decays exponentially then there is a unique mixing equilibrium state. Relying on the above discussion it is now easy to see that φ∗ Pq = Lϕ with ϕ(ξ) = −2q log(1 + φ−1 (ξ)).

(4.10)

In order to compute vark ϕ we have to consider points sharing the same path up to the k-th row of F. Take for instance ξ = L∞ and ξ
= Lk R . . .. Then a short reflection yields, for q = 0, |ϕ(ξ) − ϕ(ξ
)| ≥

C k

(4.11)

We therefore see that although vark ϕ → 0 (so that ϕ is uniformly continuous) ϕ it is not even of summable variation. This entails that (Σ, Φ, ϕ) has indeed two equilibrium states, thus exhibiting a phase transition [Is]. Next, we express the n-th iterate of Pq as  ! n−1  −2q     1 + Iˆωi ···ω1 (x) (1 + x) f Iˆωn ···ω1 (x) (4.12) Pqn f (x) = ω∈{0,1}n

i=1

where Iωi ···ω1 := Iωi · · · Iω1 . We have Iˆωi ···ω1 (x) = Iˆωi ◦ Iˆωi−1 ···ω1 (x) so that, in particular, putting x = 0 we get Iˆ0 (0) = 0,

Iˆ1 (0) = 1

(4.13)

and a if Iˆωi−1 ···ω1 (0) = b

  

a , if ωi = 0, a + b ˆ then Iωi ···ω1 (0) = b   , if ωi = 1 . a+b

(4.14)

Lemma 4.2 Let (hn )n≥1 be the sequence of functions defined by h1 ≡ 1 and hn (ω) :=

n−1 

 ˆ 1 + Iωi ···ω1 (0) ,

ω ∈ {0, 1}n−1 ,

n > 1.

i=1

For each fixed n ≥ 1 we have that hn determines a bijection between {0, 1}n−1 and the set of denominators of the elements of Fn \{ 01 } (considered as an ordered set). 28

Proof. The proof is just a straightforward verification. Suppose for instance that ω = 0k−1 1u with u ∈ {0, 1}
−1 , so that k +
= n. Then by (4.13) and (4.14) hn (ω) is given by a product with
factors of the type (1 + 1)(1 + 12 ) · · · (1 + a r )(1 + a+b ) = a + b + r where r = a if u
−1 = 0, r = b otherwise. The result b now readily follows by lemma 2.5. ♦ Remark 4.3 The rank of the elements of Fn \ { 01 } with denominator hn (ω) is given by
= n − min{1 ≤ i < n, ωi = 1} with the convention min ∅ = n. The smallest of the above denominators is 1, it has rank 0 and is obtained as hn (0n−1 ). The √two largest ones are equal to the (n + 1)-st Fibonacci number √ 1+ 5 1− 5 1 n+1 n+1 − ( 2 ) ]. They are symmetrical w.r.t 12 , have rank fn+1 = √5 [( 2 ) n−1 and are obtained as hn (101n−3 ) and hn (1n−1 ), respectively. More generally, it is not difficult to see that the following equivalence is in force: suppose that the element ab ∈ Fn has rank
> 1 so that b = hn (0n−
−1 1xu) for some x ∈ {0, 1} and u ∈ {0, 1}
−2 , then the same denominator b, but corresponding to the symmetrical fraction 1 − ab , is obtained as b = hn (0n−
−1 1xu) with x = 1 − x. A straightforward consequence of the above lemma is the following  b−2q Theorem 4.4 Pqn 1(0) = 2

(4.15)

a ∈Fn \{ 01 } b

Remarkably, the above sum is equal to the partition function Zn−1 (2q) at (inverse) temperature 2q of the number-theoretical spin chain introduced by Andreas Knauf in [Kn2]. For Re q > 1 we have (see [Kn3]) lim Pqn 1(0) = 2

n→∞

∞  ϕ(k) ζ(2q − 1) =2 . 2q ζ(2q) k k=1

(4.16)

Note that for q = 1 the above limit diverges. This reflects the fact that the invariant density for the Farey map F , that is the fixed point of the operator P1 , is the function 1/x.

29

Let us define the pressure function p(q) as4 1 (4.17) log Pqn 1(0) n Since the sum in (4.15) has 2n−1 terms we see that p(0) = log 2 (this is the topological entropy of the map F ). More generally let {di,n , i = 1, . . . , 2n−1 } denote the sequence of denominators of the elements of Fn \{ 01 } when the latters are arranged in increasing order in [0, 1], so that p(q) := lim

n→∞

2 1 n Zn−1 (2q) = Pq 1(0) = d−2q i,n . 2 i=1 n−1

(4.18)

The ratio Zn−1 (2q)/2n−1 can be interpreted as the moment of order −2q of the size of the denominators in Fn \ { 01 }. Zn (2q) is plainly non-increasing and for q ≤ 0 satisfies Zn (2q) ≥ 2n . Moreover we have Zn (2q) = Zn−1 (2q) + =

n−1 2

i=1

 d−2q i,n

n−1 2

(di−1,n + di,n )−2q

i=1




di−1 1+ 1+ di,n

−2q !

d

} = n we get for q > 0 with d0,n ≡ 1 for all n. Noting that maxi { i−1,n di,n 
2q ! 1 Zn (2q) ≥ 1 + Zn−1 (2q) 1+n Since Z0 (2q) = 1 this yields 
2q !n 1 , Zn (2q) ≥ 1 + 1+n

n ≥ 1.

(4.19)

It holds p(q) = −2q f (2q) where f (β) is the free energy of the Knauf model. In the context of thermodynamic formalism the pressure p(q) is a central object. In particular it is used as a generator of averages: its first derivative p (q), wherever it exists, yields the mean of the function ϕ(x)/q = −2 log(1 + x) w.r.t. the equilibrium measure µq , which can be defined as the weak ∗-limit point of atomic measures supported on periodic points of F weighted with the function eϕ [B]. Note that p (q) = 0 for q > 1 and p (q)  0 as q  1. On the other hand we have already seen that p(0) = log 2 and µ0 is called measure of maximal entropy. Higher derivatives of p(q) are connected to (sums of) higher correlation functions, see [Ru], [KS]. 4

30

Thus, for all q ∈ R, 1 log 2Zn−1 (2q) ≥ 0. n→∞ n

(4.20)

p(q) = lim

In addition, since p(q) is non-increasing and p(1) = 0 (because the spectral radius of P1 is 1, see above) we have p(q) = 0 for q ≥ 1. Note that the same conclusion follows at once from the fact that Zn (2q) is finite for Re q > 1 (see (4.16)). Let us study the asymptotic behaviour of Zn (2q) for q = 1. To this end, we notice that if, instead of x = 0, we evaluate the iterate Pqn 1 at x = 1, all sequences ω ∈ {0, 1}n in (4.12) yield paths which end up at the same row of the Farey tree. The same argument leading to Theorem 4.4 now yields the following  Corollary 4.5 Pqn 1(1) = 2 b−2q , n ≥ 1.

(4.21)

rank( ab )=n

By (4.15) and (4.21) we obtain Pqn

1(0) = 1 +

n−1 

Pq
1(1)

(4.22)


=0

so that we can directly apply the results obtained by Thaler in [Tha] to get5 Lemma 4.6

P1n

1(0) = 1 +

n−1 

P1
1(1) = 2 Zn−1 (2) ∼


=0

n · log n

(4.23)

We finally obtain a general expression for Pq
1(x) with x ∈ R+ . Denoting by Rn the n-th row of the Stern-Brocot tree we have the following (see [DEIK] for further generalisations) Theorem 4.7 For all x ∈ R+ and n ≥ 1 we have  (a x + b)−2q . Pqn 1(x) = 2

(4.24)

a ∈Rn b

We say that an and bn are asymptotically equivalent, denoted as an ∼ bn , if the quotient an /bn tends to unity as n approaches ∞. 5

31

Proof. The proof is a straightforward generalization of the arguments used above: first note that   Pq 1(x) = 2(1 + x)−2q and Pq2 1(x) = 2 (1 + 2x)−2q + (2 + x)−2q . One may then use Iˆωi ···ω1 (x) = Iˆωi ◦ Iˆωi−1 ···ω1 (x) along with lemma 2.5 to proceed inductively with f ≡ 1 in (4.12) and obtains  [(a + b x)−2q + (a x + b)−2q ] Pqn 1(x) = 2 rank( ab )=n−1

which is the same as (4.24). ♦

4.1

The partition function for negative integer temperatures

Lemma 4.8 We have, for all n ≥ 1, Zn ( 0 ) = 2n , Zn (−1) = 3n ,

$

√

5 + 17 1 Zn (−2) = √  2 17

$

%n+1 −

5−

√ 2

%n+1  17 

Proof. The first identity is trivial. The second one follows immediately from Z0 (−1) = 1 along with (4.19), which gives the recursion Zn (−1) = 3Zn−1 (−1). As the third one,  wencan reason as follows: let us denote An ≡ Zn (−2) = 2for n 2 2 d and B = n i=1 i,n i=1 di−1,n di,n . Then (4.19) yields An = 3An−1 + 2Bn−1 . Moreover, we have 2  n

Bn =

di−1,n+1 di,n+1

i=1

=

n−1 2

di−1,n (di−1,n + di,n ) +

i=1

=

n−1 2

n−1 2

(di−1,n + di,n )di,n

i=1

(di−1,n + di,n )2 = An − An−1

i=1

32

This yields the recursion An+1 = 5An − 2An−1 with A0 = 1 and A1 = 5 and the claim easily follows ♦. The above result indicates a general argument to work out Zn (−k) for any k ∈ N: n (k) (k,r) setting An = Zn (−k) and Bn = 2i=1 dri−1,n dk−r i,n one has k−1
  k (k) (k,r) (k) (4.25) Bn−1 An = 3An−1 + r r=1 and 2  n

Bn(k,r)

=

dri−1,n+1 dk−r i,n+1

i=1

=

n−1 2

 r  di−1,n (di−1,n + di,n )k−r + (di−1,n + di,n )r dk−r i,n

i=1 r


  k−r
 r k−r (k,s) (k,r+s) = (4.26) Bn−1 + Bn−1 s s s=0 s=0  r−1
 k−1
  r k−r (k) (k,s) (k,r) (k,s) Bn−1 + 2Bn−1 + Bn−1 = 2An−1 + s s − r s=1 s=r+1 This yields a k-dimensional recursion  (k)   (k)  An−1 An (k,1)  (k,1)   Bn   Bn−1    = V   .. k ..     . . (k,k−1) (k,k−1) Bn Bn−1 with k × k matrix k k  ··· ··· 3 1 2 k−1 2 2 ··· ··· 1   .. .. .. .. . . . . . .. .. ..  .. . . . . .. .. .. .. Vk :=  . . .  r r r 2 · · ·  1 2 r−1 . .. .. ..  .. . . .  k−2  2 k−2 · · · · · · 1 2 k−1 k−1 2 ··· ··· 1 2 33

(4.27)

··· ··· .. . .. . .. . 2 .. . ··· ···

··· ··· .. . .. . .. .

k−r 1

.. . ··· ···

··· ···

··· 2 k−1 k−2

k  k−1

k−1  k−2   ..  .  ..  .  .. (4.28) . k−r   k−r−1   ..  .  2  2

and initial condition     (k) A0 1  B (k,1)   1  0     = 1I :=  .  ...  ..   (k,k−1) 1 B0

(4.29)

By Perron-Frobenius theorem the matrix Vk has a simple real positive maximal eigenvalue λ1 (k) whose eigenvector v1 has strictly positive components. This immediately yields p(−k/2) = log λ1 (k).

(4.30)

More specifically, by the above the exact behaviour of Zn (−k) can be obtained by standard linear algebra. If for instance Vk can be diagonalized with spectrum sp(Vk )  = (λi )1≤i≤k and corresponding eigenvectors (vi )1≤i≤k , then we can expand 1I = i ai vi so that (4.27) and (4.29) yield Zn (−k) =

k 

(1)

n ≥ 0,

ai vi λni ,

(4.31)

i=1 (1)

where vi denotes the first component of vi . On the other hand, as we shall see in the forthcoming example, Vk is not always diagonalizable. Examples. For 2q = −2 we find
V2 =

3 2 2 2

.

 ,

sp(V2 ) =

5+

√ 2

17 5 − ,

√

17

/

2

 √  so that by (4.30) p(−1) = log 5+2 17 and using (4.31) one easily recover the result of Lemma 4.8 for Zn (−2). For 2q = −3 we get   3 3 3 sp(V3 ) = {7, 0, 0} . V3 =  2 2 2  , 2 2 2 In this case (4.31) does not hold but one easily finds Zn (−3) = 9 · 7n−1 , 34

n > 1,

and p(−3/2) = log 7. The case 2q = −4 is still different, yielding   3 4 6 4 / . √ √ 2 2 3 3 11 − 11 + 113 113  V4 =  , , −1, −1 .  2 2 2 2  , sp(V4 ) = 2 2 2 3 3 2 Results closely related to the above are discussed in [BGI].

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37