The four exponentials conjecture, the six exponentials theorem and related statements.

Number Theory Seminar at NCTS October 29, 2003 The four exponentials conjecture, the six exponentials theorem and related statements. Michel Waldschmidt miw@math.jussieu.fr http://www.math.jussieu.fr/ miw/ 1

Transcendental Number Theory Liouville (1844). Hermite (1873). Transcendental numbers exist. The number e is transcendental. Lindemann (1882). The number π is transcendental. Corollary: possible. Squaring the circle using rule and compass only is not http://www.math.jussieu.fr/ miw/ 2

Theorem (Hermite-Lindemann). Let β be a non zero complex number. Set α = e β. Then one at least of the two numbers α, β is transcendental. Corollary 1. If β is a non zero algebraic number, then e β is transcendental. Example. The numbers e and e 2 are transcendental. Corollary 2. If α is a non zero algebraic number and log α a non zero logarithm of α, then log α is transcendental. Example. The numbers log 2 and π are transcendental. http://www.math.jussieu.fr/ miw/ 3

Recall: the set Q of algebraic numbers is a subfield of C - it is the algebraic closure of Q into C. The exponential map exp : C C z e z is a morphism from the additive group of C onto the multiplicative group of C. Hermite-Lindemann: Q exp(q) = {1}. Also, if we define L = exp 1 (Q ), then Q L = {0}. http://www.math.jussieu.fr/ miw/ 4

Theorem (Gel fond-schneider). (1) Let λ 1 and λ 2 be two elements in L which are linearly independent over Q. Then λ 1 and λ 2 are linearly independent over Q. (2) Let λ C, λ 0 and let β C \ Q. Then one at least of the three numbers e λ, β, e βλ is transcendental. Remark (1) (2) follows from (λ 1, λ 2 ) (λ, βλ) and (λ 1, λ 2 /λ 1 ) (λ, β). http://www.math.jussieu.fr/ miw/ 5

Remarks. (1) means: log α 1 log α 2 Q \ Q. (2) means: for α C \ {0}, β C \ Q and any log α 0, one at least of the three numbers is transcendental. α, β and α β = e β log α http://www.math.jussieu.fr/ miw/ 6

Corollaries. Transcendence of log 2 log 3, 2 2, e π. Proof. Take respectively λ 1 = log 2, λ 2 = log 3, and λ 1 = log 2, λ 2 = 2 log 2, λ 1 = iπ, λ 2 = π. http://www.math.jussieu.fr/ miw/ 7

Theorem (A. Baker). Let λ 1,..., λ n be elements in L which are linearly independent over Q. Then 1, λ 1,..., λ n are linearly independent over Q. Corollary. Transcendence of numbers like β 1 log α 1 + + β n log α n and e β 0 α β 1 1 αβ n n. http://www.math.jussieu.fr/ miw/ 8

Corollary 1. (Hermite-Lindemann) Transcendence of e β. Proof. Take n = 1 in Baker s Theorem. Corollary 2. (Gel fond-schneider) Transcendence of α β. Proof. Take n = 2 in Baker s Theorem. Baker s Theorem means: The injection of Q + L into C extends to an injection of (Q + L) Q Q into C. The image is the Q-vector space L C spanned by 1 and L: (Q + L) Q Q L. http://www.math.jussieu.fr/ miw/ 9

Algebraic independence of logarithms of algebraic numbers Conjecture Let α 1,..., α n be non zero algebraic numbers. For 1 j n let λ j C satisfy e λ j = α j. Assume λ 1,..., λ n are linearly independent over Q. Then λ 1,..., λ n are algebraically independent. Write λ j = log α j. If log α 1,..., log α n are Q-linearly independent then they are algebraically independent. http://www.math.jussieu.fr/ miw/ 10

Recall that L is the Q vector space of logarithms of algebraic numbers: L = {λ C ; e λ Q} = {log α ; α Q } = exp 1 (Q ). The conjecture on algebraic independence of logarithms of algebraic numbers can be stated: The injection of L into C extends to an injection of the symmetric algebra Sym Q (L) on L into C. http://www.math.jussieu.fr/ miw/ 11

A. Selberg (50 s). Four Exponentials Conjecture History. Th. Schneider(1957) - first problem. S Lang (60 s). K. Ramachandra (1968). Leopoldt s Conjecture on the p-adic rank of the units of an algebraic number field (non vanishing of the p-adic regulator). http://www.math.jussieu.fr/ miw/ 12

Quadratic relations between logarithms of algebraic numbers Homogeneous: Four Exponentials Conjecture For i = 1, 2 and j = 1, 2, let α ij be a non zero algebraic number and λ ij a complex number satisfying e λ ij = α ij. Assume λ 11, λ 12 are linearly independent over Q and also λ 11, λ 21 are linearly independent over Q. Then λ 11 λ 22 λ 12 λ 21. http://www.math.jussieu.fr/ miw/ 13

Quadratic relations between logarithms of algebraic numbers Example. Transcendence of 2 (log 2)/ log 3 : (log 2) 2 = (log 3)(log α)? Other open problem: Transcendence of 2 log 2. Non homogeneous quadratic relations (log α)(log β) = log γ. (log 2) 2 = log γ? http://www.math.jussieu.fr/ miw/ 14

Quadratic relations between logarithms of algebraic numbers Non homogeneous: Three exponentials Conjecture Let λ 1, λ 2, λ 3 be three elements in L satisfying λ 1 λ 2 = λ 3. Then λ 3 = 0. Example. Special case of the open question on the transcendence of α log α : transcendence of e π2? http://www.math.jussieu.fr/ miw/ 15

Four Exponentials Conjecture For i = 1, 2 and j = 1, 2, let α ij be a non zero algebraic number and λ ij a complex number satisfying e λ ij = α ij. Assume λ 11, λ 12 are linearly independent over Q and also λ 11, λ 21 are linearly independent over Q. Then λ 11 λ 22 λ 12 λ 21. Notice: λ 11 λ 22 λ 12 λ 21 = det λ 11 λ 12 λ 12 λ 22. http://www.math.jussieu.fr/ miw/ 16

Four Exponentials Conjecture (again) Let x 1, x 2 be two Q- linearly independent complex numbers and y 1, y 2 also two Q- linearly independent complex numbers. Then one at least of the four numbers is transcendental. e x 1y 1, e x 1y 2, e x 2y 1, e x 2y 2 http://www.math.jussieu.fr/ miw/ 17

Four Exponentials Conjecture (again) Let x 1, x 2 be two Q- linearly independent complex numbers and y 1, y 2 also two Q- linearly independent complex numbers. Then one at least of the four numbers e x 1y 1, e x 1y 2, e x 2y 1, e x 2y 2 is transcendental. Hint: Set x i y j = λ ij (i = 1, 2; j = 1, 2). A rank one matrix is a matrix of the form ( ) x1 y 1 x 1 y 2. x 2 y 1 x 2 y 2 http://www.math.jussieu.fr/ miw/ 18

Sharp Four Exponentials Conjecture. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, are two complex numbers which are Q-linearly independent and if β 11, β 12, β 21, β 22 are four algebraic numbers such that the four numbers e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22 are algebraic, then x i y j = β ij for i = 1, 2 and j = 1, 2. http://www.math.jussieu.fr/ miw/ 19

Sharp Four Exponentials Conjecture. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, are two complex numbers which are Q-linearly independent and if β 11, β 12, β 21, β 22 are four algebraic numbers such that the four numbers e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22 are algebraic, then x i y j = β ij for i = 1, 2 and j = 1, 2. If x i y j = λ ij + β ij then x 1 x 2 y 1 y 2 = (λ 11 + β 11 )(λ 22 + β 22 ) = (λ 12 + β 12 )(λ 21 + β 21 ). http://www.math.jussieu.fr/ miw/ 20

Recall that L is the Q-vector space spanned by 1 and L (linear combinations of logarithms of algebraic numbers with algebraic coefficients): L = { β 0 + l h=1 β h log α h ; l 0, α s in Q, β s in Q } Strong Four Exponentials Conjecture. If x 1, x 2 are Q- linearly independent and if y 1, y 2, are Q-linearly independent, then one at least of the four numbers does not belong to L. x 1 y 1, x 1 y 2, x 2 y 1, x 2 y 2 http://www.math.jussieu.fr/ miw/ 21

Six Exponentials Theorem. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, y 3 are three complex numbers which are Q-linearly independent, then one at least of the six numbers is transcendental. e x iy j (i = 1, 2, j = 1, 2, 3) http://www.math.jussieu.fr/ miw/ 22

Sharp Six Exponentials Theorem. If x 1, x 2 are two complex numbers which are Q-linearly independent, if y 1, y 2, y 3 are three complex numbers which are Q-linearly independent and if β ij are six algebraic numbers such that e x iy j β ij Q for i = 1, 2, j = 1, 2, 3, then x i y j = β ij for i = 1, 2 and j = 1, 2, 3. http://www.math.jussieu.fr/ miw/ 23

Strong Six Exponentials Theorem (D. Roy). If x 1, x 2 are Q-linearly independent and if y 1, y 2, y 3 are Q-linearly independent, then one at least of the six numbers x i y j (i = 1, 2, j = 1, 2, 3) does not belong to L. http://www.math.jussieu.fr/ miw/ 24

Five Exponentials Theorem. If x 1, x 2 are Q-linearly independent, y 1, y 2 are Q-linearly independent and γ is a non zero algebraic number, then one at least of the five numbers is transcendental. e x 1y 1, e x 1y 2, e x 2y 1, e x 2y 2, e γx 2/x 1 This is a consequence of the sharp six exponentials Theorem: set y 3 = γ/x 1 and use Baker s Theorem for checking that y 1, y 2, y 3 are linearly independent over Q. http://www.math.jussieu.fr/ miw/ 25

Sharp Five Exponentials Conjecture. If x 1, x 2 are Q- linearly independent, if y 1, y 2 are Q-linearly independent and if α, β 11, β 12, β 21, β 22, γ are six algebraic numbers with γ 0 such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij γx 2 = αx 1. for i = 1, 2, j = 1, 2 and also http://www.math.jussieu.fr/ miw/ 26

Sharp Five Exponentials Conjecture. If x 1, x 2 are Q- linearly independent, if y 1, y 2 are Q-linearly independent and if α, β 11, β 12, β 21, β 22, γ are six algebraic numbers with γ 0 such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij γx 2 = αx 1. for i = 1, 2, j = 1, 2 and also Difficult case: when y 1, y 2, γ/x 1 are Q-linearly dependent. Example: x 1 = y 1 = γ = 1. http://www.math.jussieu.fr/ miw/ 27

Sharp Five Exponentials Conjecture. If x 1, x 2 are Q- linearly independent, if y 1, y 2 are Q-linearly independent and if α, β 11, β 12, β 21, β 22, γ are six algebraic numbers with γ 0 such that e x 1y 1 β 11, e x 1y 2 β 12, e x 2y 1 β 21, e x 2y 2 β 22, e (γx 2/x 1 ) α are algebraic, then x i y j = β ij γx 2 = αx 1. for i = 1, 2, j = 1, 2 and also Consequence: Transcendence of the number e π2. Proof. Set x 1 = y 1 = 1, x 2 = y 2 = iπ, γ = 1, α = 0, β 11 = 1, β ij = 0 for (i, j) (1, 1). http://www.math.jussieu.fr/ miw/ 28

Known: One at least of the two statements is true. e π2 is transcendental. The two numbers e and π are algebraically independent. http://www.math.jussieu.fr/ miw/ 29

Other consequence of the sharp five exponentials conjecture: Transcendence of the number e λ2 = α log α for λ L, e λ = α Q. Proof. Set x 1 = y 1 = 1, x 2 = y 2 = λ, γ = 1, α = 0, β 11 = 1, β ij = 0 for (i, j) (1, 1). http://www.math.jussieu.fr/ miw/ 30

Other consequence of the sharp five exponentials conjecture: Transcendence of the number e λ2 = α log α for λ L, e λ = α Q. Proof. Set x 1 = y 1 = 1, x 2 = y 2 = λ, γ = 1, α = 0, β 11 = 1, β ij = 0 for (i, j) (1, 1). Known: One at least of the two numbers e λ2 = α log α, e λ3 (log α)2 = α is transcendental. Also a consequence of the sharp six exponentials Theorem! http://www.math.jussieu.fr/ miw/ 31

Strong Five Exponentials Conjecture. Let x 1, x 2 be Q-linearly independent and y 1, y 2 be Q-linearly independent. Then one at least of the five numbers does not belong to L. x 1 y 1, x 1 y 2, x 2 y 1, x 2 y 2, x 1 /x 2 http://www.math.jussieu.fr/ miw/ 32

9 statements Four exponentials sharp strong Five exponentials Six exponentials http://www.math.jussieu.fr/ miw/ 33

sharp strong 9 statements Four exponentials Five exponentials Six exponentials Conjecture Theorem http://www.math.jussieu.fr/ miw/ 34

sharp strong 9 statements Four exponentials Five exponentials Six exponentials Conjecture Theorem Four exponentials: three conjectures Six exponentials: Five exponentials: three theorems two conjectures (for sharp and strong) one theorem http://www.math.jussieu.fr/ miw/ 35

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. http://www.math.jussieu.fr/ miw/ 36

Consequence of the sharp 4 exponentials Conjecture Let λ ij (i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume Then λ 11 λ 12λ 21 λ 22 Q. λ 11 λ 22 = λ 12 λ 21. http://www.math.jussieu.fr/ miw/ 37

Proof. Assume λ 11 λ 12λ 21 λ 22 = β Q. Use the sharp four exponentials conjecture with (λ 11 β)λ 22 = λ 12 λ 21. http://www.math.jussieu.fr/ miw/ 38

Consequence of the strong 4 exponentials Conjecture Let λ ij (i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume Then λ 11 λ 22 λ 12 λ 21 Q. λ 11 λ 22 λ 12 λ 21 Q. http://www.math.jussieu.fr/ miw/ 39

Proof: Assume λ 11 λ 22 λ 12 λ 21 = β Q. Use the strong four exponentials conjecture with λ 11 λ 22 = βλ 12 λ 21. http://www.math.jussieu.fr/ miw/ 40

Consequence of the strong 4 exponentials Conjecture Let λ ij (i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume λ 11 λ 21 Q. λ 12 λ 22 Then either λ 11 /λ 12 Q and λ 21 /λ 22 Q or λ 12 /λ 22 Q and λ 11 λ 12 λ 21 λ 22 Q. http://www.math.jussieu.fr/ miw/ 41

Remark: λ 11 λ 12 bλ 11 aλ 12 bλ 12 = a b Proof: Assume λ 11 λ 12 λ 21 λ 22 = β Q. Use the strong four exponentials conjecture with λ 12 (βλ 22 + λ 21 ) = λ 11 λ 22. http://www.math.jussieu.fr/ miw/ 42

Question: Let λ ij (i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume λ 11 λ 22 λ 12 λ 21 Q. Deduce λ 11 λ 22 = λ 12 λ 21. http://www.math.jussieu.fr/ miw/ 43

Question: Let λ ij (i = 1, 2, j = 1, 2) be four non zero logarithms of algebraic numbers. Assume λ 11 λ 22 λ 12 λ 21 Q. Deduce λ 11 λ 22 = λ 12 λ 21. Answer: This is a consequence of the Conjecture on algebraic independence of logarithms of algebraic numbers. http://www.math.jussieu.fr/ miw/ 44

Consequences of the strong 6 exponentials Theorem Let λ ij (i = 1, 2, j = 1, 2, 3) be six non zero logarithms of algebraic numbers. Assume λ 11, λ 21 are linearly independent over Q and λ 11, λ 12, λ 13 are linearly independent over Q. http://www.math.jussieu.fr/ miw/ 45

One at least of the two numbers λ 12 λ 11λ 22 λ 21, λ 13 λ 11λ 23 λ 21 is transcendental. One at least of the two numbers λ 12 λ 21 λ 11 λ 22, λ 13 λ 21 λ 11 λ 23 is transcendental. http://www.math.jussieu.fr/ miw/ 46

One at least of the two numbers λ 12 λ 11 λ 22 λ 21, λ 13 λ 11 λ 23 λ 21 is transcendental. Also one at least of the two numbers λ 21 λ 11 λ 22 λ 12, λ 21 λ 11 λ 23 λ 13 is transcendental. http://www.math.jussieu.fr/ miw/ 47

Replacing λ 21 by 1. One at least of the two numbers is transcendental. The same holds for λ 12 λ 11 λ 22, λ 13 λ 11 λ 23 λ 12 λ 11 λ 22, λ 13 λ 11 λ 23. is transcendental. http://www.math.jussieu.fr/ miw/ 48

Finally one at least of the two numbers λ 11 λ 22 λ 12, λ 11 λ 23 λ 13 is transcendental, and also one at least of the two numbers 1 λ 11 λ 22 λ 12, 1 λ 11 λ 23 λ 13 is transcendental. http://www.math.jussieu.fr/ miw/ 49

Theorem 1. Let λ ij (i = 1, 2, j = 1, 2, 3, 4, 5) be ten non zero logarithms of algebraic numbers. Assume λ 11, λ 21 are linearly independent over Q and λ 11,..., λ 15 are linearly independent over Q. Then one at least of the four numbers is transcendental. λ 1j λ 21 λ 2j λ 11, (j = 2, 3, 4, 5) http://www.math.jussieu.fr/ miw/ 50

Elliptic Analogue Theorem 2. Let and be two non isogeneous Weierstraß elliptic functions with algebraic invariants g 2, g 3 and g2, g3 respectively. For 1 j 9 let u j (resp. u j ) be a non zero logarithm of an algebraic point of (resp. ). Assume u 1,..., u 9 are Q-linearly independent. Then one at least of the eight numbers u j u 1 u ju 1 (j = 2,..., 9) is transcendental Motivation: periods of K3 surfaces. http://www.math.jussieu.fr/ miw/ 51

Sketch of Proofs Theorem 3. Let G = G d 0 a G d 1 m G 2 be a commutative algebraic group over Q of dimension d = d 0 + d 1 + d 2, V a hyperplane of the tangent space T e (G), Y a finitely generated subgroup of V of rank l 1 such that exp G (Y ) G(Q) with l 1 > (d 1)(d 1 + 2d 2 ). Then V contains a non zero algebraic Lie subalgebra of T e (G) defined over Q. http://www.math.jussieu.fr/ miw/ 52

Sketch of Proof of Theorem 1 Assume λ 1j λ 21 λ 2j λ 11 = γ j for 1 j l 1. Take G = G a G 2 m, d 0 = 1, d 1 = 2, G 2 = {0}, V is the hyperplane z 0 = λ 21 z 1 λ 11 z 2 and Y = Zy 1 + + Zy l1 with y j = (γ j, λ 1j, λ 2j ), (1 j l 1 ). Since (d 1)(d 1 + 2d 2 ) = 4, we need l 1 5. http://www.math.jussieu.fr/ miw/ 53

Assume Sketch of Proof of Theorem 2 u j u 1 u ju 1 = γ j for 1 j l 1. Take G = G a E E, d 0 = 1, d 1 = 0, d 2 = 2, V is the hyperplane z 0 = u 1z 1 u 1 z 2 and Y = Zy 1 + + Zy l1 with y j = (γ j, u j, u j), (1 j l 1 ). Since (d 1)(d 1 + 2d 2 ) = 8, we need l 1 9. http://www.math.jussieu.fr/ miw/ 54

In both cases we need to check that V does not contain a non zero Lie subalgebra of T e (G). For Theorem 1 this follows from the assumption λ 11, λ 21 are Q-linearly independent, while for Theorem 2 this follows from the assumption E, E are not isogeneous. http://www.math.jussieu.fr/ miw/ 55

Further developments Abelian varieties in place of product of elliptic curves. Semi abelian varieties. Commutative algebraic groups. Taking periods into account. Conjectures (A. Grothendieck, Y. André, C. Bertolin.) Quadratic relations among logarithms of algebraic numbers. http://www.math.jussieu.fr/ miw/ 56