WIEN,ESIOctober7,2003SYMPOSIUMONDIOPHANTINEPROBLEMSINHONOUROFWOLFGANGSCHMIDT MichelWaldschmidt Dependenceoflogarithmsoncommutativealgebraicgroups

WIEN, ESI October 7, 2003 SYMPOSIUM ON DIOPHANTINE PROBLEMS

IN HONOUR OF WOLFGANG SCHMIDT

Dependence of logarithms on commutative algebraic groups

Michel Waldschmidt

miw@math.jussieu.fr http://www.math.jussieu.fr/∼miw/

www.esi.ac.at 1

Algebraic independence of logarithms of algebraic numbers Conjecture Let α

, . . . , α

be non zero algebraic numbers. For 1 ≤ j ≤ n let λ

∈ C satisfy e

= α

. Assume λ

, . . . , λ

are linearly independent over Q. Then λ

, . . . , λ

are algebraically independent.

Write λ

= log α

.

If log α

, . . . , log α

are Q-linearly independent then they are

algebraically independent.

Denote by L the Q vector space of logarithms of algebraic numbers:

L = {λ ∈ C ; e

∈ Q} = {log α ; α ∈ Q

} = exp

(Q

).

The conjecture on algebraic independence of logarithms of algebraic numbers can be phrased:

The injection of L into C extends into an injection of the symmetric algebra Sym

(L) on L into C.

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Four Exponentials Conjecture For i = 1, 2 and j = 1, 2, let α

be a non zero algebraic number and λ

a complex number satisfying e

= α

. Assume λ

, λ

are linearly independent over Q and also λ

, λ

are linearly independent over Q. Then

λ

λ

6= λ

λ

.

Four Exponentials Conjecture For i = 1, 2 and j = 1, 2, let α

be a non zero algebraic number and λ

a complex number satisfying e

= α

. Assume λ

, λ

are linearly independent over Q and also λ

, λ

are linearly independent over Q. Then

λ

λ

6= λ

λ

.

Notice:

λ

λ

− λ

λ

= det

λ

λ

λ

λ

.

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Four Exponentials Conjecture (again) Let x

, x

be two Q- linearly independent complex numbers and y

, y

also two Q- linearly independent complex numbers. Then one at least of the four numbers

e

, e

, e

, e

is transcendental.

Four Exponentials Conjecture (again) Let x

, x

be two Q- linearly independent complex numbers and y

, y

also two Q- linearly independent complex numbers. Then one at least of the four numbers

e

, e

, e

, e

is transcendental.

Hint: Set

x

y

= λ

(i = 1, 2; j = 1, 2).

A rank one matrix is a matrix of the form x

y

x

y

x

y

x

y

.

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Sharp Four Exponentials Conjecture. If x

, x

are two complex numbers which are Q-linearly independent, if y

, y

, are two complex numbers which are Q-linearly independent and if β

, β

, β

, β

are four algebraic numbers such that the four numbers

e

, e

, e

, e

are algebraic, then x

y

= β

for i = 1, 2 and j = 1, 2.

Sharp Four Exponentials Conjecture. If x

, x

are two complex numbers which are Q-linearly independent, if y

, y

, are two complex numbers which are Q-linearly independent and if β

, β

, β

, β

are four algebraic numbers such that the four numbers

e

, e

, e

, e

are algebraic, then x

y

= β

for i = 1, 2 and j = 1, 2.

If x

y

= λ

+ β

then

x

x

y

y

= (λ

+ β

)(λ

+ β

) = (λ

+ β

)(λ

+ β

).

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Denote by L ˜ the Q-vector space spanned by 1 and L (linear combinations of logarithms of algebraic numbers with algebraic coefficients) :

L ˜ = (

β

+

`

X

h=1

β

log α

; ` ≥ 0, α’s in Q

, β ’s in Q )

Strong Four Exponentials Conjecture. If x

, x

are Q- linearly independent and if y

, y

, are Q-linearly independent, then one at least of the four numbers

x

y

, x

y

, x

y

, x

y

does not belong to L. ˜

Six Exponentials Theorem. If x

, x

are two complex numbers which are Q-linearly independent, if y

, y

, y

are three complex numbers which are Q-linearly independent, then one at least of the six numbers

e

(i = 1, 2, j = 1, 2, 3) is transcendental.

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Sharp Six Exponentials Theorem. If x

, x

are two complex numbers which are Q-linearly independent, if y

, y

, y

are three complex numbers which are Q-linearly independent and if β

are six algebraic numbers such that

e

∈ Q for i = 1, 2, j = 1, 2, 3,

then x

y

= β

for i = 1, 2 and j = 1, 2, 3.

Strong Six Exponentials Theorem (D. Roy). If x

, x

are Q-linearly independent and if y

, y

, y

are Q-linearly independent, then one at least of the six numbers

x

y

(i = 1, 2, j = 1, 2, 3) does not belong to L. ˜

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Five Exponentials Theorem. If x

, x

are Q-linearly independent, y

, y

are Q-linearly independent and γ is a non zero algebraic number, then one at least of the five numbers

e

, e

, e

, e

, e

is transcendental.

This is a consequence of the sharp 6 exponentials Theorem: set

y

= γ/x

and use Baker’s Theorem for checking that y

, y

, y

are linearly independent over Q.

Sharp Five Exponentials Conjecture. If x

, x

are Q- linearly independent, if y

, y

are Q-linearly independent and if α, β

, β

, β

, β

, γ are six algebraic numbers such that

e

, e

, e

, e

, e

are algebraic, then x

y

= β

for i = 1, 2, j = 1, 2 and also γx

= αx

.

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Sharp Five Exponentials Conjecture. If x

, x

are Q- linearly independent, if y

, y

are Q-linearly independent and if α, β

, β

, β

, β

, γ are six algebraic numbers such that

e

, e

, e

, e

, e

are algebraic, then x

y

= β

for i = 1, 2, j = 1, 2 and also γx

= αx

.

Difficult case: when y

, y

, γ/x

are Q-linearly dependent.

Example: x

= y

= γ = 1.

Sharp Five Exponentials Conjecture. If x

, x

are Q- linearly independent, if y

, y

are Q-linearly independent and if α, β

, β

, β

, β

, γ are six algebraic numbers such that

e

, e

, e

, e

, e

are algebraic, then x

y

= β

for i = 1, 2, j = 1, 2 and also γx

= αx

.

Consequence: Transcendence of the number e

.

Proof. Set x

= y

= 1, x

= y

= iπ, γ = 1, α = 0, β

= 1, β

= 0 for (i, j ) 6= (1, 1).

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Known: One at least of the two statements is true.

• e

is transcendental.

• The two numbers e and π are algebraically independent.

Other consequence of the sharp five exponentials conjecture: Transcendence of the number e

= α

for λ ∈ L, e

= α ∈ Q

.

Proof. Set x

= y

= 1, x

= y

= λ, γ = 1, α = 0, β

= 1, β

= 0 for (i, j ) 6= (1, 1).

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Other consequence of the sharp five exponentials conjecture: Transcendence of the number e

= α

for λ ∈ L, e

= α ∈ Q

.

Proof. Set x

= y

= 1, x

= y

= λ, γ = 1, α = 0, β

= 1, β

= 0 for (i, j ) 6= (1, 1).

Known: One at least of the two numbers e

= α

, e

= α

is transcendental.

Also a consequence of the sharp six exponentials Theorem!

Strong Five Exponentials Conjecture. Let x

, x

be Q-linearly independent and y

, y

be Q-linearly independent.

Then one at least of the five numbers

x

y

, x

y

, x

y

, x

y

, x

/x

does not belong to L. ˜

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9 statements

Four exponentials

sharp Five exponentials

strong Six exponentials

9 statements Four exponentials

Conjecture sharp Five exponentials

Theorem strong Six exponentials

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9 statements Four exponentials

Conjecture sharp Five exponentials

Theorem strong Six exponentials

Four exponentials: three conjectures Six exponentials: three theorems

Five exponentials: two conjectures (for sharp and strong)

one theorem

Alg. indep. C

⇓

Strong 4 exp C ⇒ Strong 5 exp C ⇒ Strong 6 exp T

⇓ ⇓ ⇓

Sharp 4 exp C ⇒ Sharp 5 exp C ⇒ Sharp 6 exp T

⇓ ⇓ ⇓

4 exp C ⇒ 5 exp T ⇒ 6 exp T

Remark. The sharp 6 exponentials Theorem implies the 5 exponentials Theorem.

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Consequence of the sharp 4 exponentials Conjecture

Let λ

(i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers.

Assume

λ

− λ

λ

λ

∈ Q.

Then

λ

λ

= λ

λ

.

Proof. Assume

λ

− λ

λ

λ

= β ∈ Q.

Use the sharp four exponentials conjecture with (λ

− β )λ

= λ

λ

.

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Consequence of the strong 4 exponentials Conjecture

Let λ

(i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers.

Assume

λ

λ

λ

λ

∈ Q.

Then

λ

λ

λ

λ

∈ Q.

Proof: Assume

λ

λ

λ

λ

= β ∈ Q.

Use the strong four exponentials conjecture with λ

λ

= βλ

λ

.

http://www.math.jussieu.fr/∼miw/articles/pdf/Wien-ESI.pdf 29

Consequence of the strong 4 exponentials Conjecture

Let λ

(i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers.

Assume

λ

λ

− λ

λ

∈ Q.

Then

• either λ

/λ

∈ Q and λ

/λ

∈ Q

• or λ

/λ

∈ Q and

λ

λ

− λ

λ

∈ Q.

Remark:

λ

λ

− bλ

− aλ

bλ

= a b ·

Proof: Assume

λ

λ

− λ

λ

= β ∈ Q.

Use the strong four exponentials conjecture with λ

(βλ

+ λ

) = λ

λ

.

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Question: Let λ

(i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume

λ

λ

− λ

λ

∈ Q.

Deduce

λ

λ

= λ

λ

.

Question: Let λ

(i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume

λ

λ

− λ

λ

∈ Q.

Deduce

λ

λ

= λ

λ

.

Answer: This is a consequence of the Conjecture on algebraic independence of logarithms of algebraic numbers.

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Consequences of the strong 6 exponentials Theorem

Let λ

(i = 1, 2, j = 1, 2, 3) be six non zero logarithms of algebraic numbers. Assume

• λ

, λ

are linearly independent over Q and

• λ

, λ

, λ

are linearly independent over Q.

• One at least of the two numbers

λ

− λ

λ

λ

, λ

− λ

λ

λ

is transcendental.

• The same holds for

λ

λ

λ

λ

, λ

λ

λ

λ

• and for

λ

λ

− λ

λ

, λ

λ

− λ

λ

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• Also one at least of the two numbers

λ

λ

− λ

λ

, λ

λ

− λ

λ

is transcendental,

• similarly for

λ

− λ

λ

, λ

− λ

λ

• and for

λ

λ

− λ

, λ

λ

− λ

.

Theorem 1. Let λ

(i = 1, 2, j = 1, 2, 3, 4, 5) be ten non zero logarithms of algebraic numbers. Assume

• λ

, λ

are linearly independent over Q and

• λ

, . . . , λ

are linearly independent over Q.

Then one at least of the four numbers

λ

λ

− λ

λ

, (j = 2, 3, 4, 5) is transcendental.

http://www.math.jussieu.fr/∼miw/articles/pdf/Wien-ESI.pdf 37

Elliptic Analogue

Theorem 2. Let ℘ and ℘

be two non isogeneous Weierstraß elliptic functions with algebraic invariants g

, g

and g

, g

respectively. For 1 ≤ j ≤ 9 let u

(resp. u

) be a non zero logarithm of an algebraic point of ℘ (resp. ℘

). Assume u

, . . . , u

are Q-linearly independent. Then one at least of the eight numbers

u

u

− u

u

(j = 2, . . . , 9) is transcendental

Motivation: periods of K 3 surfaces.

Sketch of Proofs

Theorem 3. Let G = G

× G

× G

be a commutative algebraic group over Q of dimension d = d

+ d

+ d

, V a hyperplane of the tangent space T

(G), Y a finitely generated subgroup of V of rank `

such that exp

(Y ) ⊂ G(Q) with

`

> (d − 1)(d

+ 2d

).

Then V contains a non zero algebraic Lie subalgebra of T

(G) defined over Q.

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Sketch of Proof of Theorem 1

Assume

λ

λ

− λ

λ

= γ

for 1 ≤ j ≤ `

.

Take G = G

× G

, d

= 1, d

= 2, G

= {0}, V is the hyperplane

z

= λ

z

− λ

z

and Y = Zy

+ · · · + Zy

with

y

= (γ

, λ

, λ

), (1 ≤ j ≤ `

).

Since (d − 1)(d

+ 2d

) = 4, we need `

≥ 5.

Sketch of Proof of Theorem 2 Assume

u

u

− u

u

= γ

for 1 ≤ j ≤ `

.

Take G = G

× E × E

, d

= 1, d

= 0, d

= 2, V is the hyperplane

z

= u

z

− u

z

and Y = Zy

+ · · · + Zy

with

y

= (γ

, u

, u

), (1 ≤ j ≤ `

).

Since (d − 1)(d

+ 2d

) = 8, we need `

≥ 9.

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In both cases we need to check that V does not contain a non zero Lie subalgebra of T

(G). For Theorem 1 this follows from the assumption

λ

, λ

are Q-linearly independent, while for Theorem 2 this follows from the assumption

E , E

are not isogeneous.

WIEN, ESI, October 7, 2003

Erwin Schr¨ odinger Institute – Wolfgang Schmidt Conference

Happy Birthday Wolfgang!

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