MichelWaldschmidt x + y + z =3 xyz OntheMarko ﬀ Equation (ICART2008) May29,2008 InternationalConferenceonAlgebraandRelatedTopics Abstract Abstract(continued) ThesequenceofMarko ﬀ numbers

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In ternational Conference on Algebra and Related T opics (ICAR T 2008) May 29, 2008

http://www.math.sc.chula.ac.th/∼icart2008/

On the Mark o ff Equation x 2 + y 2 + z 2 =3 xy z

Michel Waldschmidt

http://www.math.jussieu.fr/∼miw/

1/79

Abstract

It is easy to chec k that the equation x

2

+ y

2

+ z

2

=3 xy z , where the three unkno wns x , y , z are p ositiv e in tegers, has infinitely man y solutions. There is a simple algorithm whic h pro duces all of them. Ho w ev er, this do es not answ er to all questions on this equation : in particular F rob enius ask ed whether it is true that for eac h in teger z> 0 , there is at most one pair ( x, y ) suc h that x < y < z and ( x, y ,z ) is a solution. This question is an activ e researc h topic no w ada ys.

2/79

Abstract (con tin ued)

Mark o ff ’s equation o ccurred initially in the study of minima of quadratic forms at the end of the XIX–th cen tury and the b eginning of the XX–th cen tury . It w as in vestigated b y man y a mathematician, including Lagrange , Hermite , Korkine , Zolotarev , Mark o ff , F rob enius , Hurwitz , Cassels . The solutions are related with the L agr ange-Marko ff sp ectrum , whic h consists of those quadratic n um b ers whic h are badly appro ximable b y rational n um b ers. It o ccurs also in other parts of mathematics, in particular free groups, F uc hsian groups and h yp erb olic Riemann surfaces ( F ord , Lehner , Cohn , Rankin , Con w ay , Co xeter , Hirzebruc h and Zagier .. .). W e discuss some asp ects of this topic without trying to co ver all of them.

3

The sequence of Mar ko ff num b ers

A Marko ff numb er is a p ositiv e in teger z suc h that there exist tw o p ositiv e in tegers x and y satisfying

x

2

+ y

2

+ z

2

=3 xy z.

F or instance 1 is a Mark o ff n um b er, since ( x, y ,z ) = (1 , 1 , 1) is a solution. Andrei Andrey evic h Mark o ff (1856–1922)

Photos:http://www-history.mcs.st-andrews.ac.uk/history/

4

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The On-Line Encyclop edia of In teger Sequences

1,2,5,13,29,34,89,169,194,233,433,610,985,1325,1597,2897,4181,5741,6466,7561,9077,10946,14701,28657,33461,37666,43261,51641,62210,75025,96557,135137,195025,196418,294685,...

The sequence of Mark o ff n um b ers is av ailable on the w eb The On-Line Encyclop edia of In teger Sequences Neil J. A. Sloane

http ://www.research.att.com/ ∼ njas/sequences/A002559

5/79

In teger p oin ts on a surface

Giv en a Mark o ff n um b er z , there exist infinitely man y pairs of p ositiv e in tegers x and y satisfying

x

2

+ y

2

+ z

2

=3 xy z.

This is a cubic equation in the 3 variables ( x, y ,z ) , of whic h w e kno w a solution (1 , 1 , 1) .

There is an algorithm pro ducing all in teger solutions.

6/79

Mark o ff ’s cubic variet y

The surface defined b y Mark o ff ’s equation

x

2

+ y

2

+ z

2

=3 xy z.

is an algebraic variet y with man y automorphisms : p erm utations of the variables, changes of signs and

( x, y ,z ) "→ (3 yz − x, y ,z ) . A.A. Mark o ff (1856–1922)

Algorithm pro ducing all solutions

Let ( m ,m

1

,m

2

) b e a solution of Mark o ff ’s equation :

m

2

+ m

21

+ m

22

=3 mm

1

m

2

. Fix tw o co ordinates of this solution, sa y m

1

and m

2

. W e get a quadratic equation in the third co ordinate m , of whic h w e kno w a solution, hence the equation

x

2

+ m

21

+ m

22

=3 xm

1

m

2

. has tw o solutions, x = m and, sa y, x = m

!

, with m + m

!

=3 m

1

m

2

and mm

!

= m

21

+ m

22

. This is the cor d and tangente pr oc ess.

Hence another solution is ( m

!

,m

1

,m

2

) with m

!

=3 m

1

m

2

− m .

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Three solutions deriv ed from one

Starting with one solution ( m ,m

1

,m

2

) , w e deriv e three new solutions :

( m

!

,m

1

,m

2

) , ( m ,m

!1

,m

2

) , ( m ,m

1

,m

!2

) .

If the solution w e start with is (1 , 1 , 1) , w e pro duce only one new solution, (2 , 1 , 1) (up to p erm utation).

If w e start from (2 , 1 , 1) , w e pro duce only tw o new solutions, (1 , 1 , 1) and (5 , 2 , 1) (up to p erm utation).

A new solution means distinct fr om the one we start with.

9/79

New solutions

W e shall see that an y solution di ff eren t from (1 , 1 , 1) and from (2 , 1 , 1) yields three new di ff eren t solutions – and w e shall see also that in eac h other solution the three n um b ers m , m

1

and m

2

are pairwise distinct.

Tw o solutions are neighb ors if they share tw o comp onen ts.

10/79

Mark o ff ’s tree

Assume w e start with ( m ,m

1

,m

2

) satisfying m > m

1

>m

2

. W e shall chec k m

!2

>m

!1

>m>m

!

.

W e order the solution according to the largest co ordinate. Then tw o of the neigh b ors of ( m ,m

1

,m

2

) are larger than the initial solution, the third one is smaller.

Hence if w e start from (1 , 1 , 1) , w e pro duce infinitely man y solutions, whic h w e organize in a tree : this is Marko ff ’s tr ee .

11

This algorithm yields all the solutions

Con vers ely , starting from an y solution other than (1 , 1 , 1) , the algorithm pro duces a smal ler solution.

Hence b y induction w e get a sequence of smaller and smaller solutions, un til w e reac h (1 , 1 , 1) .

Therefore the solution w e started from w as in Mark o ff ’s tree.

12

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First branc hes of Mar ko ff ’s tree

13/79

Mark o ff ’s tree starting from (2 , 5 , 29)

DON ZAGIER

FIGURE 2 Markoff triples ( p, q,

r

) with max( p, q) 100000

Conversely, given a Markoff triple (p, q, r) with r > 1, one checks easily that 3pq

-

r < r; and from ths it follows by induction that all Markoff triples occur, and occur only once, on this tree (for a fuller discussion of ths and other properties of the Markoff tree, see [2]). To prove the theorem we must analyze the asymptotic behavior of the Markoff tree. From the Markoff equation (1) we find that 3r2

2

3pqr or r

2

pq; if p is large (which will happen for all but a small portion of the tree, contributing O(log

x)

to M(x)), then this implies that r is much larger than q and hence (1) gives r2 < 3pqr < r2 + o(r2) or r - 3pq. Multiplying both sides of this equation by 3 and taking logarithms gives

log(3p) + log(3q)

=

log(3r) + o(1) ( p large)

14/79

Mark o ff ’s tree up to 100 000

DON ZAGIER

FIGURE2 Markoff triples (p, q, r ) with max( p, q) 100000

Conversely, given a Markoff triple (p, q, r) with r > 1, one checks easily that 3pq -r <r; and from ths it follows by induction that all Markoff triples occur, and occur only once, on this tree (for a fuller discussion of ths and other properties of the Markoff tree, see [2]). To prove the theorem we must analyze the asymptotic behavior of the Markoff tree. From the Markoff equation (1) we find that 3r2 2 3pqr or r 2pq; if p is large (which will happen for all but a small portion of the tree, contributing O(log x) to M(x)), then this implies that r is much larger than q and hence (1) gives r2 < 3pqr < r2 + o(r2) or r -3pq. Multiplying both sides of this equation by 3 and taking logarithms gives

log(3p) + log(3q) = log(3r) + o(1) (p large)

Don Zagier , On the numb er of Marko ff numb ers below a given bound. Mathematics of Computation, 39 160 (1982), 709–723.

15

Mark o ff ’s tree

16

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a 2 + b 2 + c 2 =3 abc

X

2

− 3 abX + a

2

+ b

2

= ( X − c )( X − 3 ab + c )

17/79

The Fib onacci sequence and the Mark o ff equation

The smallest Mark o ff n um b er is 1 . When w e imp ose z =1 in the Mark o ff equation x

2

+ y

2

+ z

2

=3 xy z , w e obtain the equation x

2

+ y

2

+ 1 = 3 xy .

Going along the Mark o ff ’s tree starting from (1 , 1 , 1) , w e obtain the subsequence of Mark o ff n um b ers

1 , 2 , 5 , 13 , 34 , 89 , 233 , 610 , 1597 , 4181 , 10946 , 28657 , .. .

whic h is the sequence of Fib onacci n um b ers with o dd indices

F

1

=1 ,F

3

=2 ,F

5

=5 ,F

7

= 13 ,F

9

= 34 ,F

11

= 89 , .. .

18/79

Leonardo Pisano (Fib onacci)

The Fib onacci sequence ( F

n

)

n≥0

:

0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 ,

34 , 55 , 89 , 144 , 233 .. .

is defined b y

F

0

=0 ,F

1

=1 ,

F

n

= F

n−1

+ F

n−2

( n ≥ 2) . Leonardo Pisano (Fib onacci) (1170–1250)

19/79

Encyclop edia of in teger sequences (again)

0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,..

The Fib onacci sequence is av ailable online The On-Line Encyclop edia of In teger Sequences Neil J. A. Sloane

http ://www.research.att.com/ ∼ njas/sequences/A000045

20/79

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Fib onacci num b ers with odd indices

Fib onacci n um b ers with o dd indices are Mark o ff ’s n um b ers :

F

m+3

F

m−1

− F

2m+1

=( − 1)

m

for m ≥ 1 and F

m+3

+ F

m−1

=3 F

m+1

for m ≥ 1 .

Set y = F

m+1

, x = F

m−1

, x

!

= F

m+3

, so that, for ev en m ,

x + x

!

=3 y , xx

!

= y

2

+1 and X

2

− 3 yX + y

2

+ 1 = ( X − x )( X − x

!

) .

21/79

Order of the new solutions

Let ( m ,m

1

,m

2

) b e a solution of Mark o ff ’s equation

m

2

+ m

21

+ m

22

=3 mm

1

m

2

. Denote b y m

!

the other ro ot of the quadratic p olynomial

X

2

− 3 m

1

m

2

X + m

21

+ m

22

.

Hence

X

2

− 3 m

1

m

2

X + m

21

+ m

22

=( X − m )( X − m

!

) and m + m

!

=3 m

1

m

2

, m m

!

= m

21

+ m

22

.

22/79

m 1 & = m 2

Let us chec k that if m

1

= m

2

, then m

1

= m

2

=1 : this holds only for the tw o exceptional solutions (1 , 1 , 1) , (2 , 1 , 1) .

Assume m

1

= m

2

. W e ha ve m

2

+2 m

21

=3 mm

21

hence m

2

= (3 m − 2) m

21

. Therefore m

1

divides m . Let m = km

1

. W e ha ve k

2

=3 km

1

− 2 , hence k divise 2 . F or k =1 w e get m = m

1

=1 . F or k =2 w e get m

1

=1 , m =2 .

Consider no w a solution distinct from (1 , 1 , 1) or (2 , 1 , 1) : hence m

1

& = m

2

.

23

Tw o larger, one smaller

Assume m

1

>m

2

. Question : Do we have m

!

>m

1

or else m

!

<m

1

? Consider the n um b er a =( m

1

− m )( m

1

− m

!

) . Since m + m

!

=3 m

1

m

2

, and mm

!

= m

21

+ m

22

, w e ha ve

a = m

21

− m

1

( m + m

!

)+ mm

!

=2 m

21

+ m

22

− 3 m

21

m

2

= (2 m

21

− 2 m

21

m

2

)+( m

22

− m

21

m

2

) . Ho w ev er 2 m

21

< 2 m

21

m

2

and m

22

<m

21

m

2

, hence a< 0 . This means that m

1

is in the in terv al defined b y m and m

!

.

24

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Order of the solutions

If m > m

1

, w e ha ve m

1

>m

!

and the new solution ( m

!

,m

1

,m

2

) is smaller than the initial solution ( m ,m

1

,m

2

) . If m < m

1

, w e ha ve m

1

<m

!

and the new solution ( m

!

,m

1

,m

2

) is larger than the initial solution ( m ,m

1

,m

2

) .

25/79

Prime factors

R emark. Let m b e a Mark o ff n um b er with

m

2

+ m

21

+ m

22

=3 mm

1

m

2

. The same pro of sho ws that the gcd of m , m

1

and m

2

is 1 : indeed, if p divides m

1

, m

2

and m , then p divides the new solutions whic h are pro duced b y the preceding pro cess – going do wn in the tree sho ws that p w ould divide 1 .

The o dd prime factors of m are all congruen t to 1 mo dulo 4 (since they divide a sum of tw o relativ ely prime squares).

If m is ev en, then the n um b ers

m 2 , 3 m − 2 4 , 3 m +2 8 ,

are o dd in tegers.

26/79

Mark o ff ’s Conjecture

The previous algorithm pro duces the sequence of Mark o ff n um b ers. Eac h Mark o ff n um b er o ccurs infinitely often in the tree as one of the comp onen ts of the solution.

According to the definition, for a Mark o ff n um b er m> 2 there exist a pair ( m

1

,m

2

) of p ositiv e in tegers with m > m

1

>m

2

suc h that m

2

+ m

21

+ m

22

=3 mm

1

m

2

.

Question : Given m , is such a pair ( m

1

,m

2

) unique ? The answ er is yes, as long as m ≤ 10

105

.

27

F rob enius’s w ork

Marko ff ’s Conje ctur e do es not o ccur in Mark o ff ’s 1879 and 1880 pap ers but in F rob enius ’s one in 1913. F erdinand Georg F rob enius (1849–1917)

28

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Sp ecial cases

The Conjecture has b een pro ved for certain classes of Mark o ff n um b ers m lik e

p

n

, p

n

± 2 3 for p prime. A. Baragar (1996), P . Sc hm utz (1996), J.O. Button (1998), M.L. Lang, S.P . T an (2005), Ying Zhang (2007). Arth ur Baragar

http://www.nevada.edu/baragar/

29/79

P ow ers of a prime num b er

Anitha Sriniv asan , 2007 A re al ly simple pr oof of the Marko ff conje ctur e for prime powers

Num b er Theory W eb Created and main tained b y Keith Matthews , Brisbane, Australia www.numbertheory.org/pdfs/simpleproof.pdf

30/79

The state of the art

10/09/2007, 04/12/2007 : Norb ert Riedel h ttp ://fr.arxiv.org/abs/0709.1499v2 http ://fr.arxiv.org/abs/0709.1499

A triple ( a, b, c ) of positive inte gers is cal le d a Marko ff triple i ff it satisfies the diophantine equation a

2

+ b

2

+ c

2

= abc . R ec asting the Marko ff tr ee, whose vertic es ar e Marko ff triples, in the fr amework of inte gr al upp er triangular 3 × 3 matric es, it wil l be shown that the lar gest memb er of such a triple determines the other two uniquely. This answers a question which has be en op en for almost 100 ye ars.

Fla w in the pro of disco vered b y Serge P errine .

31

Wh y the co e ffi cien t 3 ?

Let n b e a p ositiv e in teger. If the equation x

2

+ y

2

+ z

2

= n xy z has a solution in positive inte gers, then either n =3 and x , y , z ar e relatively prime, or n =1 and the gcd of the numb ers x , y , z is 3 .

F riedric h Hirzebruc h & Don Zagier , The A tiyah–Singer The or em and elementary numb er the ory, Publish or P erish (1974)

32

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Mark o ff typ e equations

Bije ction betwe en the solutions for n =1 and those for n =3 :

• if x

2

+ y

2

+ z

2

=3 xy z , then (3 x, 3 y, 3 z ) is solution of X

2

+ Y

2

+ Z

2

= X Y Z , since (3 x )

2

+ (3 y )

2

+ (3 z )

2

= (3 x )(3 y )(3 z ) .

• if X

2

+ Y

2

+ Z

2

= X Y Z , then X , Y , Z are m ultiples of 3 and ( X/ 3)

2

+( Y/ 3)

2

+( Z/ 3)

2

= 3( X/ 3)( Y/ 3)( Z/ 3) .

The squares mo dulo 3 are 0 and 1 . If X , Y and Z are not m ultiples of 3 , then X

2

+ Y

2

+ Z

2

is a m ultiple of 3 .

If one or tw o (not three) in tegers among X , Y , Z are m ultiples of 3 , then X

2

+ Y

2

+ Z

2

is not a m ultiple of 3 .

33/79

Equations x 2 + ay 2 + bz 2 = (1 + a + b ) xy z If w e insist that (1 , 1 , 1) is a solution, then up to p erm utations there are only tw o more Diophan tine equations of the typ e

x

2

+ ay

2

+ bz

2

= (1 + a + b ) xy z

ha ving infinitely man y in teger solutions, namely those with ( a, b ) = (1 , 2) and (2 , 3) :

x

2

+ y

2

+2 z

2

=4 xy z and x

2

+2 y

2

+3 z

2

=6 xy z

• x

2

+ y

2

+ z

2

: tessalation of the plane b y equilateral triangles • x

2

+ y

2

+2 z

2

=4 xy z : tessalation of the plane b y iso celes rectangle triangles • x

2

+2 y

2

+3 z

2

=6 xy z : tessalation ?

34/79

Lauren t’s phenomenon Connection with Lauren t p olynomials. James Propp, The combinatorics of frieze patterns and Marko ff numb ers,

http://fr.arxiv.org/abs/math/0511633

If f , g , h are Lauren t p olynomials in tw o variables x and y , i.e., p olynomials in x , x

−1

, y , y

−1

, in general

h ! f ( x, y ) ,g ( x, y ) "

is not a Lauren t p olynomial :

f ( x )= x

2

+1 x = x + 1 x ,

f ! f ( x ) " = # x + 1 x $

2

+1

x + 1 x = x

4

+3 x

2

+1 x ( x

2

+ 1) ·

35

Hurwitz’s equation (1907)

F or eac h n ≥ 2 the set K

n

of p os itiv e in tegers k for whic h the equation

x

21

+ x

22

+ ·· · + x

2n

= kx

1

·· · x

n

has a solution in p ositiv e in tegers is finite. The largest value of k in K

n

is n — with the solution

(1 , 1 ,. .. , 1) .

Examples :

K

3

= { 1 , 3 } , K

4

= { 1 , 4 } , K

7

= { 1 , 2 , 3 , 5 , 7 } .

36

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Hurwitz’s equation x 2 1 + x 2 2 + ·· · + x 2 n = kx 1 ·· · x n

When there is a solution in p ositiv e in tegers, there are infinitely man y solutions, whic h can b e organized in finitely man y trees.

A. Baragar pro ved that ther e exists such equations which re quir e an arbitr arily lar ge numb er of tr ees J. Num b er Theory (1994), 49 No 1, 27-44.

The analog for the rank of elliptic curv es ov er the rational n um b er field is yet a conjecture.

37/79

Gro wth of Mark o ff ’s sequence

1978 : order of magnitude of m , m

1

and m

2

for m

2

+ m

21

+ m

22

=3 mm

1

m

2

with m

1

<m

2

<m , log (3 m

1

) + log (3 m

2

) = log (3 m )+ o (1) .

T o iden tify primitiv e w ords in a free group with tw o generators, H. Cohn used Marko ff forms. Harv ey Cohn

x "→ log (3 x ) : ( m

1

,m

2

,m ) "→ ( a, b, c ) with a + b ∼ c .

38/79

Euclidean tree Start with (0 , 1 , 1) . F rom a triple ( a, b, c ) satisfying a + b = c and a ≤ b ≤ c , one pro duces tw o larger suc h triples ( a, c, a + c ) and ( b, c, b + c ) and a smaller one ( a, b − a, b ) or ( b − a, a, b ) .

⁽⁰^,₁

,1)|(1,1,2)|(1,2,3)|||(1,3,4)(2,3,5)

||||||

(1,4,5)(3,4,7)(2,5,7)(3,5,8)

|... ... |... ... |... ... |... ...39

Mark o ff and Euclidean trees

T om Cusik & Mary Flahiv e , The Marko ff and L agr ange sp ectr a, Math. Surv eys and Monographs 30 , AMS (1989).

40

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Gro wth of Mark o ff ’s sequence

Don Zagier (1982) : estimating the n um b er of the Mark o ff triples b ounded by x :

c (log x )

2

+ O ! log x (log log x )

2

" ,

c =0 , 18071704711507 .. .

Conjecture : the n -th Mark o ff n um b er m

n

is

m

n

∼ A

√n

with A = 10 , 5101504 ·· ·

41/79

Historical origin : rational appro ximation

Hurwitz ’s Theor em (1891) : F or any re al irr ational numb er x , ther e exist infinitely many rational numb ers p/q such that % % % % x − p q % % % % ≤ 1 √ 5 q

2

· Golden ratio Φ = (1 + √ 5) / 2= 1 , 6180339887498948482 .. . Hurwitz ’s result is optimal. Adolf Hurwitz (1859–1919)

42/79

The Fib onacci sequence and the Golden ratio

F orm ula of A. De Moivre (1730), L. Euler (1765), J.P .M. Binet (1843) :

F

n

= 1 √ 5 & 1+ √ 5 2 '

n

− 1 √ 5 & 1 − √ 5 2 '

n

.

43

F orm ul a of De Moivre–Euler–Binet

Abraham de Moivre (1667–1754) Leonhard Euler (1707–1783) Jacques Philipp e Marie Binet (1786–1856)

F

n

is the nearest in teger to 1 √ 5 Φ

n

.

44

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Quadratic relation

One chec ks b y induction

F

2n+1

− F

n+1

F

n

− F

2n

=( − 1)

n

for all n ≥ 0 . The left hand side is the value at ( F

n+1

,F

n

) of the quadratic form

X

2

− XY − Y

2

=( X − Φ Y )( X + Φ

−1

Y ) . The sequence u

n

= F

n+1

/F

n

, n ≥ 1 con verges to the Golden ratio Φ and

F

2n+1

− F

n+1

F

n

− F

2n

= F

2n

( u

n

− Φ )( u

n

+ Φ

−1

) .

45/79

Quotien ts of consecut iv e Fib onacci num b ers

One deduces

F

2n

| Φ − u

n

| = 1 Φ

−1

+ u

n

→ 1 Φ

−1

+ Φ = 1 √ 5 ·

Hence

lim

n→∞

F

2n

% % % % Φ − F

n+1

F

n

% % % % = 1 √ 5 ·

46/79

Con tin ued fractions

The sequence u

n

= F

n+1

/F

n

is also defined b y

u

1

=1 ,u

n

= 1 + 1 u

n−1

, ( n ≥ 2) .

Hence

u

n

= 1 + 1

1+ 1 u

n−2

= 1 + 1

1+ 1

1+ 1 u

n−3

= ·· ·

= [1 , 1 ,. .. , 1] n times

=[ 1]

47

Hurwitz’s result is optimal Hurwitz ’s result

lim inf

q→∞

! q min

p∈Z

| qx − p | " ≤ 1 √ 5 for all x ∈ R \ Q

is optimal : there is equalit y for x = Φ . F or | q Φ − p | ≤ 1 , w e ha ve

1 ≤ | q

2

+ pq − p

2

| = | q Φ − p |· ( q Φ

−1

+ p )

with

q Φ

−1

+ p = q ( Φ + Φ

−1

)+ p − q Φ ≤ q √ 5+1 , hence 1 ≤ | q Φ − p |· ( q √ 5 + 1) .

Notice that P ( X )= X

2

− X − 1 has discriminan t 5 and P

!

( Φ )= √ ∆ = √ 5 .

48

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Liouville’s inequalit y

Liouville ’s inequalit y . L et α be an algebr aic numb er of de gr ee d ≥ 2 , P ∈ Z [ X ] its minimal polynomial, c = | P

!

( α ) | and " > 0 . Ther e exists q

0

such that, for any p/q ∈ Q with q ≥ q

0

, % % % % α − p q % % % % ≥ 1 ( c + " ) q

d

· Joseph Liouville , 1844

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Mark o ff ’s constan t

F or x ∈ R \ Q denote b y λ ( x ) ∈ [ √ 5 , + ∞ ] the least upp er b ound of the n um b ers γ > 0 suc h that there exist infinitely man y p/q ∈ Q satisfying % % % % x − p q % % % % ≤ 1 γ q

2

· This means

1 λ ( x ) = lim inf

q→∞

! q min

p∈Z

| qx − p | " .

Hurwitz : λ ( x ) ≥ √ 5 for an y x and λ ( Φ )= √ 5 .

50/79

Mark o ff ’s constan t

An irrational real n um b er x is bad ly appr oximable by rational n um b ers if its Mark o ff ’s constan t is finite. This means that there exists γ > 0 suc h that, for an y p/q ∈ Q , % % % % x − p q % % % % ≥ 1 γ q

2

· F or instance Liouville ’s n um b ers ha ve an infinite Mark o ff ’s constan t. A real n um b er is badly appro ximable if and only if the sequence ( a

n

)

n≥0

of partial quotien ts in its con tin ued fraction expansion

x =[ a

0

,a

1

,a

2

, .. ., a

n

,. .. ]

is b ounded.

51

Badly appro ximable num b ers

An y quadratic irrational real n um b er has a finite Mark o ff ’s constan t (= is badly appro ximable).

It is not kno wn whether there exist real algebraic n um b ers of degree ≥ 3 whic h are badly appro ximable.

It is not kno wn whether there exist real algebraic n um b ers of degree ≥ 3 whic h are not badly appro ximable .. .

One conjectures that any irr ational re al numb er which is not quadr atic and which is bad ly appr oximable is tr ansc endental.

52

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Leb esgue measure

The set of badly appro ximable real n um b ers has zero measure for Leb esgue ’s measure. Henri L ´eon Leb esgue (1875–1941)

53/79

Prop erties of the Mark o ff ’s constan t

W e ha ve

λ ( x + 1) = λ ( x ): % % % % x +1 − p q % % % % = % % % % x − p + q q % % % %

and

λ ( − x )= λ ( x ): % % % % − x − p q % % % % = % % % % x + p q % % % % ,

Also λ (1 /x )= λ ( x ) :

p

2

% % % % 1 x − q p % % % % = q

2

% % % % p qx % % % % · % % % % x − p q % % % % .

54/79

The mo dular group

The m ultiplicativ e group generated b y the three matrices # 11 01 $ , # 10 0 − 1 $ , # 01 10 $ is the group GL

2

( Z ) of 2 × 2 matrices # ab cd $ with co e ffi cien ts in

Z and determinan t ± 1 .

J-P. Serre – Cours d’arithm ´etique , Coll. SUP , Presses Univ ersitaires de F rance, P aris, 1970.

55

# ab cd $ x = ax + b cx + d

# 11 01 $ x = x +1 # 10 0 − 1 $ x = − x # 01 10 $ x = 1 x

λ ( x + 1) = λ ( x ) λ ( − x )= λ ( x ) λ (1 /x )= λ ( x )

Consequence : L et x ∈ R \ Q and let a , b , c , d be rational inte gers satisfying ad − bc = ± 1 . Set

y = ax + b cx + d ·

Then λ ( x )= λ ( y ) .

56

(15)

Hurwitz’s w ork (con tin ued)

The inequalit y λ ( x ) ≥ √ 5 for al l re al irr ational x is optimal for the Golden ratio and for all the noble irrational n um b ers whose con tin ued fraction expansion ends with an infinite sequence of 1 ’s – these n um b ers are the ro ots of the quadratic p olynomials ha ving discriminan t 5 :

Φ = [1 , 1 , 1 ,. .. ]=[ 1] . Adolf Hurwitz , 1891

57/79

The first elemen ts of the sp ectrum

Hurwitz ’s inequalit y λ ( x ) ≥ √ 5 is optimal for the Golden ratio Φ and all the n um b ers related to Φ b y a homograph y of determinan t ± 1 :

a Φ + b c Φ + d with # ab cd $ ∈ GL

2

( Z ) . F or all the other real n um b ers w e ha ve λ ( x ) ≥ 2 √ 2 . This is optimal for √ 2=1 , 414213562373095048801688724209698078 .. .

whose con tin ued fraction expansion is

[1; 2 , 2 , .. ., 2 , .. . ] = [1; 2] .

58/79

Minima of quadratic forms

Let f ( X ,Y )= aX

2

+ bX Y + cY

2

b e a quadratic form with real co e ffi cien ts. Denote b y ∆ ( f ) its discriminan t b

2

− 4 ac . Consider the minim um m ( f ) of | f ( x, y ) | on Z

2

\{ (0 , 0) } . Assume ∆ ( f ) & =0 and set

C ( f )= m ( f ) / ( | ∆ ( f ) | . Let α and α

!

b e the ro ots of f ( X, 1) :

f ( X ,Y )= a ( X − α Y )( X − α

!

Y ) , { α , α

!

} = ) 1 2 a ! − b ± ( ∆ ( f )

" *

.

59

Example with ∆ < 0 The form f ( X ,Y )= X

2

+ XY + Y

2

has discriminan t ∆ ( f )= − 3 and minim um m ( f ) = 1 , hence

C ( f )= m ( f ) ( | ∆ ( f ) | = 1 √ 3 ·

F or ∆ < 0 , the form

f ( X ,Y )= + | ∆ | 3 ( X

2

+ XY + Y

2

) has discriminan t ∆ and minim um ( | ∆ | / 3 . Again

C ( f )= 1 √ 3 ·

60

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Definite quadratic forms ( ∆ < 0)

If the discriminan t is negativ e, J.L. Lagrange and Ch. Hermite (letter to Jacobi , August 6, 1845) pro ved C ( f ) ≤ 1 / √ 3 with equalit y for f ( X ,Y )= X

2

+ XY + Y

2

. F or eac h % ∈ (0 , 1 / √ 3] , there exists suc h a form f with C ( f )= % .

Joseph-Louis Lagrange (1736–1813) Charles Hermite (1822–1901) Carl Gusta v Jacob Jacobi (1804–1851)

61/79

Example with ∆ > 0 The form f ( X ,Y )= X

2

− XY − Y

2

has discriminan t ∆ ( f )=5 and minim um m ( f )=1 , hence

C ( f )= m ( f ) ( ∆ ( f ) = 1 √ 5 ·

F or ∆ > 0 , the form

f ( X ,Y )= + ∆ 5 ( X

2

− XY − Y

2

) has discriminan t ∆ and minim um ( ∆ / 5 . Again

C ( f )= 1 √ 5 ·

62/79

Indefinite quadratic forms ( ∆ > 0 )

Assume ∆ > 0

A. Korkine and E.I.. Zolotarev pro ved in 1873 C ( f ) ≤ 1 / √ 5 with equalit y for f

0

( X ,Y )= X

2

− XY − Y

2

. F or all forms whic h are not equiv alen t to f

0

under GL(2 , Z ) , they pro ve C ( f ) ≤ 1 / √ 8 . 1 / √ 5 = 0 , 447 213 595 .. . 1 / √ 8 = 0 , 353 553 391 .. .

Gap ! Egor Iv ano vic h Zolotarev (1847–1878)

63

Indefinite quadratic forms ( ∆ > 0 ).

The w orks b y Korkine and Zolotarev inspired M ark o ff who pursued the study of this question. He pro duced infinitely man y values C ( f

i

) , i =0 , 1 ,. .. , b et w een 1 / √ 5 and 1 / 3 , with the same prop ert y as f

0

. These values form a sequence whic h con verges to 1 / 3 . He constructed them b y means of the tree of solutions of the Mark o ff equation. A. Mark o ff , 1879 and 1880.

64

(17)

Indefinite quadratic forms ( ∆ > 0 )

Assume f (( X ,Y )= aX

2

+ bX Y + cY

2

∈ R [ X ,Y ] with a> 0 has discriminan t ∆ > 0 .

If | f ( x, y ) | is small with y & =0 , then x/y is close to a ro ot of f ( X, 1) , sa y α .

Then | x − α

!

y | ∼ | y |·| α − α

!

| and α − α

!

= √ ∆ /a .

Hence

| f ( x, y ) | = | a ( x − α y )( x − α

!

y ) | ∼ y

2

√ ∆ % % % % α − x y % % % % .

65/79

Lagrange sp ectrum and Mark o ff sp ectrum Mark o ff sp ectrum = set of values tak en b y

1 C ( f ) = ( ∆ ( f ) /m ( f ) when f runs ov er the set of quadratic forms ax

2

+ bxy + cy

2

with real co e ffi cien ts of discriminan t ∆ ( f )= b

2

− 4 ac > 0 and m ( f ) = inf

(x,y)∈Z2\{0}

| f ( x, y ) | . Lagrange sp ectrum = set of values tak en b y Mark o ff ’s constan t ( !)

λ ( x ) = 1 / lim inf

q→∞

q (min

p∈Z

| qx − p | ) when x runs ov er the set of real n um b ers. The Mark o ff sp ectrum con tains the Lagrange sp ectrum. The in tersection with the in terv all [ √ 5 , 3[ is the same for b oth of them, and is a discrete sequence.

66/79

F uc hsian groups and hyp erb olic R iemann surfaces

Mark o ff ’s tree can b e seen as the dual of the triangulation of the h yp erb olic upp er half plane b y the images of the fundamen tal domain of the mo dular in varian t under the action of the mo dular group . Lazarus Imman uel F uc hs (1833–1902)

67

T riangulation of p olygons, metric prop erties of p olytop es

Harold Scott MacDonald Co xeter (1907–2003) Rob ert Alexander Rankin (1915–2001) John Horton Con w ay

68

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F ord ci rcles

The F ord circle asso ciated to the irreducible fraction p/q is tangen t to the real axis at the p oin t p/q and has radius 1 / 2 q

2

.

F ord circles asso ciated to tw o consecutiv e elemen ts in a F arey sequence are tangen t. Lester Randolph F ord (1886–1967)

Amer. Math. Mon thly (1938).

69/79

F arey sequence of order 5

0 1 , 1 5 , 1 4 , 1 3 , 2 5 , 1 2 , 3 5 , 2 3 , 3 4 , 4 5 , 1 1

70/79

Complex con tin ued fracti on

The third generation of Asm us Sc hmidt ’s complex con tin ued fraction metho d.

http://www.maa.org/editorial/mathgames/mathgames−03−15−04.html

71

Con tin ued fractions and hyp erb olic geometry

72

(19)

The Geometry of Mark o ff Num b ers

Caroline Series , The Ge ometry of Marko ff Numb ers, The Mathematical In telligencer 7 N.3 (1985), 20–29.

73/79

F ric ke groups

The subgroup Γ of SL

2

( Z ) generated b y the tw o matrices # 11 12 $ and # 21 11 $

is the free group with tw o generators.

The Riemann surface quotien t of the P oincar ´e upp er half plane b y Γ is a punctur ed torus . The minimal lengths of the closed geo desics are related to the C ( f ) , for f indefinite quadratic form.

74/79

F ree groups. ^F ^ric _ke

pro ved that if A and B are tw o generators of Γ , then their traces satisfy

(tr A )

2

+ (tr B )

2

+ (tr AB )

2

= (tr A )(tr B )(tr AB ) Harv ey Cohn sho w ed that quadratic forms with a Mark o ff constan t C ( f ) ∈ ]1 / 3 , 1 / √ 5] are equiv alen t to

cx

2

+( d − a ) xy − by

2

where # ab cd $

is a generator of Γ .

75

F undamen tal domain of a punctured disc

76

(20)

A simple curv e on a punctured disc

77/79

Mark o ff and Diophan tine appro ximati on

J.W.S. Cassels, A n intr oduction to Diophantine appr oximation, Cam bridge Univ. Press (1957) John William Scott Cassels

78/79

In ternational Conference on Algebra and Related T opics (ICAR T 2008) May 29, 2008

http://www.math.sc.chula.ac.th/∼icart2008/

On the Mark o ff Equation x 2 + y 2 + z 2 =3 xy z

Michel Waldschmidt

http://www.math.jussieu.fr/∼miw/

79