March 12 – 18, 2021

Online conference on Analytic and Combinatorial number theory
in honour of Professor R. Balasubramanian’s 70th Birthday

Schanuel’s Conjecture:
algebraic independenceof transcendental numbers

Michel Waldschmidt
Institut de Mathématiques de Jussieu — Sorbonne Université
http://www.math.jussieu.fr/∼miw/

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Abstract
Schanuel’s conjecture asserts that for Q–linearly independent
complex numbers x1 , ..., xn , there are at least n algebraicallyindependent numbers among the 2n numbers
x1 , . . . , xn , exp(x1 ), . . . , exp(xn ).
This simple statement has many remarkable consequences ; wewill explain some of them. We will also present the state of theart on this topic.
Note : We write exp(z) for ez .

2 / 106

Linear independence over Q
Given complex numbers, we may ask whether they are linearly
independent over Q.
For instance given a number x, the linear independence of 1, x
over Q is equivalent to the irrationality of x.
As an example, the numberslog 2, log 3, log 5, . . . , log p, . . .
are linearly independent over Q : for bi ∈ Z,
b1 log p1 + · · · + bn log pn = 0 =⇒ b1 = · · · = bn = 0.
pb11 · · · pbnn = 1 =⇒ b1 = · · · = bn = 0.
3 / 106

Linear independence over Q
The set of algebraic numbers is a subfield of C (sums andproducts of algebraic numbers are algebraic).
Given complex numbers, we may ask whether they are linearly
independent over the field Q of algebraic numbers.
For instance, given a number x, the linear independence of
1, x over Q is equivalent to the transcendence of x.
It has been proved by A. Baker in 1968 that the numbers
1, log 2, log 3, log 5, . . . , log p, . . .
are linearly independent over Q : for βi ∈ Q,
β0 + β1 log p1 + · · · + βn log pn = 0 =⇒ β0 = · · · = βn = 0.
4 / 106

Algebraic independence
Given complex numbers x1 , . . . , xn , we may ask whether they
are algebraically independent over Q : this means that there isno nonzero polynomial P ∈ Q[X 1 , . . . , X n ] such that
P (x1 , . . . , xn ) = 0.
This is equivalent to saying that x1 , . . . , xn are algebraicallyindependent over Q : if a nonzero polynomial
Q ∈ Q[X 1 , . . . , X n ] satisfies Q(x1 , . . . , xn ) = 0, then bytaking for P the product of the conjugates of Q over Q one
gets a nonzero polynomial P ∈ Q[X 1 , . . . , X n ] such that
P (x1 , . . . , xn ) = 0.
For n = 1, x1 is algebraically independent over Q if and only if
x1 is transcendental over Q.
If x1 , . . . , xn are algebraically independent, each of thesenumbers xi is transcendental.
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Transcendence degree
The transcendence degree of a field extension K 1 ⊂ K 2 ,denoted either trdeg (K 2 /K 1 ) or trdeg K 1 (K 2 ), is the
maximal number of elements in K 2 which are algebraicallyindependent over K 1 . The transcendence degree trdeg K of afield K of characteristic zero is the transcendence degree of K
over Q.
Given complex numbers x1 , . . . , xm , the maximal number ofalgebraically independent elements in the set {x1 , . . . , xm } isthe same as the transcendence degree of the field
Q(x1 , . . . , xm ) (over Q).
The transcendence degree of the field Q(x1 , . . . , xm ) is m if
and only if x1 , . . . , xm are algebraically independent.
For m = 1, the transcendence degree of the field Q(x) is 0 if
x is algebraic, 1 if x is transcendental.
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Addtivity of the transcendence degree
For K 1 ⊂ K 2 ⊂ K 3 , we havetrdeg (K 3 /K 1 ) = trdeg (K 3 /K 2 ) + trdeg (K 2 /K 1 ).

Also K 2 is an algebraic extension of K 1 if and only if
trdeg (K 2 /K 1 ) = 0.

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Lindemann–Weierstraß Theorem (1885)
Let β1 , . . . , βn be algebraic numbers which are linearly
independent over Q. Then the numbers eβ1 , . . . , eβn are
algebraically independent (over Q or over Q).
Ferdinand von Lindemann(1852 – 1939)

Karl Weierstraß(1815 - 1897)

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Equivalent forms of the Lindemann–Weierstraß
TheoremLet β1 , . . . , βn be algebraic. Then the numbers eβ1 , . . . , eβn are
algebraically independent (over Q or over Q) if and only if
β1 , . . . , βn are linearly independent over Q.
Let γ1 , . . . , γn be algebraic numbers. Then the numbers
eγ1 , . . . , eγn are linearly independent (over Q or over Q) if andonly if γ1 , . . . , γn are pairwise distinct.
(eβ1 )b1 · · · (eβn )bn = eb1 β1 +···+bn βn .

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A.O. Gel’fond
CRAS 1934

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Statement by Gel’fond (1934)
Let P (X 1 , . . . , X n , Y 1 , . . . , Y m ) be a polynomial with rational
integer coefficients and α1 , α2 , . . . , αn , β1 , β2 , . . . , βm algebraic
numbers, βi 6= 0, 1.
The equality
P (eα1 , eα2 , . . . , eαn , ln β1 , ln β2 , . . . , ln βm ) = 0
is impossible ; the numbers α1 , α2 , . . . , αn , as well as thenumbers ln β1 , ln β2 , . . . , ln βm are linearly independent in thefield of rational numbers.

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Statement by Gel’fond (1934)

This theorem includes as special cases, the theorems ofHermite and Lindemann, the complete solution of Hilbert’s 7thωproblem, the transcendence of numbers eω1 e 2 (where ω1 andω2 are algebraic numbers), the theorem on the relativetranscendence of the numbers e and π.

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Second statement by A.O. Gel’fond
The numbers
ω eeω1 e 2.

ωn.
. ωn−1 eαα2 3and α1

.

.
. αm

,

where ω1 6= 0, ω2 , . . . , ωn and α1 6= 0, 1, α2 6= 0, 1, α3 6= 0,. . . , αm are algebraic numbers, are transcendental numbers,
and among numbers of this form there is no nontrivial
algebraic relations with rational integer coefficients.
The proof of this result and a few other results ontranscendental numbers will be given in another journal.
Remark by Mathilde Herblot : the condition on α2 should be that it isirrational.
13 / 106Schanuel’s Conjecture

If x1 , . . . , xn are Q–linearly independent complex numbers,
then n at least of the 2n numbers x1 , . . . , xn , ex1 , . . . , exn are
algebraically independent.
Equivalently :If x1 , . . . , xn are Q-linearly independent complex numbers, then
trdeg Q x1 , . . . , xn , ex1 , . . . , exn ≥ n.
14 / 106

Origin of Schanuel’s Conjecture

Course given by Serge Lang
(1927–2005) at Columbia inthe 60’s

S. Lang – Introduction to transcendental numbers,
Addison-Wesley 1966.

15 / 106Power series
J. Ax’s Theorem (1968) :Version of Schanuel’s
Conjecture for power series
over C.

James Ax
1937 – 2006

R. Coleman (1980) for power
series over Q.

Samit Dasgupta, Introduction to TranscendenceTheory.
Math 790, 3/2/2021. Lecture 3: Schanuel’s Conjecture and Ax’s Theorem
https://services.math.duke.edu/~dasgupta/Transcendence/index.html

16 / 106Formal analogs

Joint work by W.D.
Brownawell and K. Kubota(1975) on the elliptic analog
of Ax’s Theorem.

W.D. Brownawell17 / 106

Dale Brownawell and Stephen Schanuel

18 / 106

Methods from logic
Ehud Hrushovski

Boris Zilber

Jonathan Kirby

pre-dimension function (E. Hrushovski)
B. Zilber : pseudo-exponentiation
Also : A. Macintyre, D.E. Marker, G. Terzo, A.J. Wilkie,
D. Bertrand. . .
19 / 106

Kirby, Zilber, Wilkie
Kirby, Jonathan ; Zilber, Boris
The uniform Schanuel conjecture over the real numbers.
Bull. London Math. Soc. 38 (2006), no. 4, 568–570.
Wilkie, A. J.
Schanuel’s conjecture and the decidability of the real exponential
field.
Algebraic model theory (Toronto, ON, 1996), 223–230,
NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 496, Kluwer Acad.
Publ., Dordrecht, 1997.

20 / 106Daniel Bertrand

Daniel Bertrand,
Schanuel’s conjecture for
non-isoconstant elliptic curves
over function fields.

Model theory with applications to algebra and analysis. Vol. 1,
41–62, London Math. Soc. Lecture Note Ser., 349, Cambridge
Univ. Press, Cambridge, 2008.

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Ax-Schanuel• for Shimura varieties,• for variations of Hodge structures.,• for the modular invariant,• for linear differential equations,• and o-minimality...

22 / 106Lindemann–Weierstraß Theorem (1885)
According to the Lindemann–Weierstraß Theorem, Schanuel’s
Conjecture is true for algebraic x1 , . . . , xn : in this case the transcendence degree of the field Q x1 , . . . , xn , ex1 , . . . , exn
is n.
Ferdinand von Lindemann(1852 – 1939)

Karl Weierstraß(1815 - 1897)

23 / 106Transcendence degree ≤ n

If we select ex1 , . . . , exs to be algebraic (this means that thexi ’s are logarithms of algebraic numbers) and xs+1 , . . . , xn
also to be algebraic, then the transcendence degree of the field
Q(x1 , . . . , xn , ex1 , . . . , exn )is the same as the transcendence degree of the field
Q(x1 , . . . , xs , exs+1 , . . . , exn ),hence is ≤ n.
The conjecture (A.O. Gel’fond) is that it is n : the numbers
log α1 , . . . , log αs , eβs+1 , . . . eβn
are expected to be algebraically independent.
24 / 106Baire and Lebesgue
René Baire
1874 – 1932

Henri Léon Lebesgue
1875 – 1941

The set of tuples (x1 , . . . , xn ) in Cn such that the 2n numbers
x1 , . . . , xn , ex1 , . . . , exn are algebraically independent• is a Gδ set (countable intersection of dense open sets) in
Baire’s classification (a generic set for dynamical systems)• and has full Lebesgue measure.
True for any transcendental function in place of the exponentialfunction.

25 / 106Mathematical genealogy
René Baire (1899)|
Arnaud Denjoy (1909)|
Charles Pisot (1938)|
Yvette Amice (1965)|
Jean Fresnel (1967)|
Michel Waldschmidt (1972)
http://genealogy.math.ndsu.nodak.edu
26 / 106

Joint work with Senthil Kumar and Thangadurai

Given two integers m and n with 1 ≤ m ≤ n, there exist
uncountably many tuples (x1 , . . . , xn ) in Rn such that
x1 , . . . , xn and ex1 , . . . , exn are all Liouville numbers and thetranscendence degree of the field
Q(x1 , . . . , xn , ex1 , . . . , exn )is n + m.
27 / 106m = 0?
1≤m≤n:
trdeg Q(x1 , . . . , xn , ex1 , . . . , exn ) = n + m.

We do not know whether there are Liouville numbers x suchthat ex is also a Liouville number and the two numbers x andex are algebraically dependent.

28 / 106Schanuel’s Conjecture for n = 1
For n = 1, Schanuel’s Conjecture is the Hermite–Lindemann
Theorem :If x is a non–zero complex numbers, then one atleast of the two numbers x, ex is transcendental.
Equivalently, if x is a non–zero algebraic number, then ex is atranscendental number.
Another equivalent statement is that if α is a non–zeroalgebraic number and log α any non–zero logarithm of α, thenlog α is a transcendental number.
Consequence : transcendence of numbers like√e, π, log 2, e 2 .
Proof: take for x respectively
1, iπ, log 2,√

2.
29 / 106Schanuel’s Conjecture for n = 2
For n = 2, Schanuel’s Conjecture is not yet known :? If x1 , x2 are Q–linearly independent complex
numbers, then among the 4 numbers x1 , x2 , ex1 , ex2 ,at least two are algebraically independent.
A few consequences (open problems) :With x1 = 1, x2 = iπ : algebraic independence of e and π.
With x1 = 1, x2 = e : algebraic independence of e and ee .
With x1 = log 2, x2 = (log 2)2 : algebraic independence oflog 2 and 2log 2 .
With x1 = log 2, x2 = log 3 : algebraic independence of log 2
and log 3.

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Easy consequence of Schanuel’s Conjecture
A consequence of Schanuel’s Conjecture is that the following
numbers are algebraically independent :
2

2

e

2

π

e + π, eπ, π e , eπ , ee , ee , . . . , ee , . . . , π π , π π , . . . π π . . .
log π, log(log 2), π log 2, (log 2)(log 3), 2log 2 , (log 2)log 3 . . .

31 / 106Lang’s exercise
Define E 0 = Q. Inductively,
for n ≥ 1, define E n as thealgebraic closure of the field
generated over E n−1 by the
numbers exp(x) = ex , wherex ranges over E n−1 . Let E be
the union of E n , n ≥ 0.
Then Schanuel’s Conjecture implies that the number π does
not belong to E.
More precisely : Schanuel’s Conjecture implies that the
numbers π, log π, log log π, log log log π, . . . are algebraicallyindependent over E.
32 / 106

A variant of Lang’s exercise
Define L0 = Q. Inductively, for n ≥ 1, define Ln as thealgebraic closure of the field generated over Ln−1 by the
numbers y, where y ranges over the set of complex numbers
such that ey ∈ Ln−1 . Let L be the union of Ln , n ≥ 0.
Then Schanuel’s Conjecture implies that the number e does
not belong to L.

More precisely : Schanuel’s
Conjecture implies that theee ee eenumbers e, e , e , e . . . are algebraically independent over L.

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Arizona Winter School AWS2008, Tucson
Theorem [Chuangxun Cheng, Brian Dietel, Mathilde
Herblot, Jingjing Huang, Holly Krieger, Diego Marques,
Jonathan Mason, Martin Mereb and Robert Wilson.]
Schanuel’s Conjecture implies that the fields E and L arelinearly disjoint over Q.
Definition Given a field extension F /K and two
subextensions F 1 , F 2 ⊆ F , we say F 1 , F 2 are linearly disjoint
over K when the following holds : any set {x1 , . . . , xn } ⊆ F 1of K– linearly independent elements is linearly independent
over F 2 .
Some consequences of Schanuel’s conjecture. J. Number
Theory 129 (2009), no. 6, 1464–1467.
arXiv.0804.3550
Contribution of Georges Racinet.
34 / 106Abelian analog
Philippon, Patrice ; Saha, Biswajyoti ; Saha, Ekata.
An abelian analogue of Schanuel’s conjecture and applications.
Ramanujan J. 52 (2020), no. 2, 381–392.

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Semi-abelian analog
Cristiana Bertolin, Patrice Philippon, Biswajyoti Saha, Ekata
Saha
Semi-abelian analogues of Schanuel Conjecture andapplications, 2020.
https://arxiv.org/abs/2010.15170

36 / 106Algebraic independence
Towards Schanuel’s Conjecture :Ch. Hermite, F. Lindemann, C.L. Siegel, A.O. Gel’fond,Th. Schneider, A. Baker, S. Lang, W.D. Brownawell,
D.W. Masser, D. Bertrand, G.V. Chudnovsky, P. Philippon,
G. Wüstholz, Yu.V. Nesterenko, D. Roy. . .

N.I. Feldman and Yu. V. Nesterenko Yu. V. Nesterenko and P. Philippon
37 / 106Algebraic independence : A.O. Gel’fond 1948√
3The
two numbers 2 2 and√3
2 4 are algebraicallyindependent.

More generally, if α is an algebraic number, α 6= 0, α 6= 1 andif β is an algebraic number of degree d ≥ 3, then two at leastof the numbers2d−1
αβ , αβ , . . . , αβare algebraically independent.
38 / 106Problem of Gel’fond and Schneider
Raised by A.O. Gel’fond in 1948 and Th. Schneider in 1952.
Conjecture : if α is an algebraic number, α 6= 0, α 6= 1 and if
β is an irrational algebraic number of degree d, then the d − 1
numbers2d−1
αβ , αβ , . . . , αβare algebraically independent.
Special case of Schanuel’s Conjecture : takexi = β i−1 log α. The conclusion is, for β algebraic number ofdegree d,2d−1 trdeg Q log α, αβ , αβ , . . . , αβ= d.

39 / 106

Large transcendence degree
Let α be a nonzero complex algebraic number, log α a nonzero
logarithm of α and β an algebraic number of degree d ≥ 2. Forz ∈ C, define αz := exp(z log α). Then


d+1
β
β2
β d−1
.
trdeg Q Q(α , α , . . . , α)≥2This result is due to the work of A.O. Gel’fond, G.V. Chudnovsky,
P. Philippon, Yu. V. Nesterenko, G. Diaz (1989).
References :N.I. Fel’dman and Yu.V. Nesterenko. Transcendental numbers. Number
theory, IV, Encyclopaedia Math. Sci., 44, Springer, Berlin, 1998.
Yu.V. Nesterenko and P. Philippon (ed). Introduction to algebraic
independence theory. Lecture Notes in Mathematics, 1752.
Springer-Verlag, Berlin, 2001.
Yu.V. Nesterenko. Algebraic independence. Published for the Tata
Institute of Fundamental Research, Bombay ; by Narosa Publishing
House, New Delhi, 2009.
40 / 106p–adic transcendental numbers
Two open problems : the radius of convergence of expp is finite :n! → ∞ but |n!|p → 0.
Problem 1 : p–adic analogue of the Lindemann – Weierstraß
Theorem.
Problem 2 : p–adic analogue of Gel’fond’s Theorem on the2
algebraic independence of αβ and αβ for β cubic irrational.
W.W. Adams (1966) :For [Q(β) : Q] = d ≥ 4, two of the numbers
2αβ , αβ , . . . αβd−1

are algebraically independent.
2
For [Q(β) : Q] = 3, two of the numbers log α, αβ and αβ are
algebraically independent.
41 / 106Transcendence in finite characteristic

Federico Pellarin
On a variant of Schanuelconjecture for the Carlitz
exponential.
J. Théor. Nombres Bordeaux
29 (2017), no. 3, 845–873.

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Criteria for transcendence
A.O. Gel’fond, W.D. Brownawell, P. Philippon,
G.V. Chudnovsky, Yu.V. Nesterenko, M. Laurent, D. Roy. . .

Patrice Philippon

Michel Laurent

Damien Roy

43 / 1062eπ , e and π (1972)

W.D. Brownawell

One at least of the two
following statements is true :2• the number eπ istranscendental• the two numbers e and π
are algebraically independent.

Schanuel’s Conjecture implies that both statements are true !

44 / 106Algebraic independence
G.V. Chudnovsky (1978)

The numbers π andΓ(1/4) = 3.625 609 908 2 . . .
are algebraically independent.

Also π and Γ(1/3) = 2.678 938 534 7 . . . are algebraicallyindependent.

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On the number eπ
Yu.V.Nesterenko (1996)
Algebraic independence ofΓ(1/4), π and eπ .
Also : Algebraic
independence of √Γ(1/3), π and eπ 3 .

Corollary : The numbers π = 3.141 592 653 5 . . . andeπ = 23.140 692 632 7 . . . are algebraically independent.
The proof uses modular functions.

46 / 106

On the number eπ
Nesterenko : algebraic independence of π and eπ .
Open problem : eπ is not a Liouville number :
eπ −

p
1
> κ·
q
q

G.V. Chudnovsky : algebraic independence of π and Γ(1/4).
Yu.V. Nesterenko : Algebraic independence of π, Γ(1/4) andeπ .
Open problem : algebraic independence of π and e.
Expected : e, π and eπ are algebraic independent.
47 / 106Conjecture of algebraic independence of logarithms
of algebraic numbers
The most important special case of Schanuel’s Conjecture is :Conjecture. Let λ1 , . . . , λn be Q-linearly independent
complex numbers. Assume that the numbers eλ1 , . . . , eλn are
algebraic. Then the numbers λ1 , . . . , λn are algebraicallyindependent over Q.
Not yet known that the transcendence degree is ≥ 2.

48 / 106Baker’s linear independence Theorem

Let λ1 , . . . , λn be Q–linearly independent logarithms ofalgebraic numbers. Then the numbers 1, λ1 , . . . , λn are linearly
independent over the field Q of algebraic numbers.

Schanuel’s Conjecture deals with algebraic independence (overQ or Q), Baker’s Theorem deals with linear independence.
Baker’s Theorem is a special case of Schanuel’s Conjecture.

49 / 106

Serre’s reformulation of Baker’s Theorem
Denote by L the set of complex numbers λ for which eλ isalgebraic (set of logarithms of algebraic numbers). Hence L isa Q-vector subspace of C.

J-P. Serre(Bourbaki seminar) :the injection of L into C
extends to a Q–linear map
ι : Q + L ⊗Q Q → C, andBaker’s Theorem means that
ι is an injective map.

50 / 106

Reformulation by D. Roy
Instead of taking logarithms of algebraic numbers and looking
for the algebraic independence relations, D. Roy fixes apolynomial and looks at the points, with coordinates
logarithms of algebraic numbers, on the corresponding
hypersurface.
Recall that L denotes the Q–vector subspace of C of all
logarithms of all nonzero algebraic numbers.
The Conjecture on (homogeneous) algebraic independence oflogarithms of algebraic numbers is equivalent to :Conjecture (Roy). For any algebraic subvariety V of Cn
defined over the field Q of algebraic numbers, the set V ∩ Ln
is the union of the sets E ∩ Ln , where E ranges over the setof vector subspaces of Cn which are contained in V .
51 / 106Points with coordinates logarithms of algebraic
numbers
Damien Roy : Grassmanian varieties.

Stéphane Fischler : orbit of an affine algebraic group G over Q
related to a linear representation of G on a vector space with aQ–structure.

52 / 106

Quadratic relations among logarithms of algebraic
numbers
One does not know yet how to prove that there is nonontrivial quadratic relations among logarithms of algebraic
numbers, like(log α1 )(log α2 ) = log β.
2

Example: Assume eπ = β is algebraic. Then(−iπ)(iπ) = log β.

2• Open problem : is the number eπ transcendental ?
53 / 106

Homogeneous quadratic relations among
logarithms of algebraic numbers
Any homogeneous quadratic relation among logarithms ofalgebraic numbers(log α1 )(log α4 ) = (log α2 )(log α3 )should be trivial.
Example of a trivial relation : (log 2)(log 9) = (log 4)(log 3).
The Four Exponentials Conjecture can be stated as : any
quadratic relation (log α1 )(log α4 ) = (log α2 )(log α3 ) among
logarithms of algebraic numbers is trivial : either log α1 / log α2
is rational, or log α1 / log α3 is rational.
54 / 106

S. Ramanujan,
C.L. Siegel, S. Lang, K. Ramachandra

Ramanujan : Highly composite numbers.
Alaoglu and Erdős (1944), Siegel.

55 / 106Four exponentials conjecture (special case)
Let t be a positive real number. Assume 2t and 3t are both
integers. Prove that t is an integer.
Set n = 2t . Then t = (log n)/(log 2) and
3t = et log 3 = e(log n)(log 3)/(log 2) = n(log 3)/(log 2) .
Equivalently :If n is a positive integer such thatn(log 3)/(log 2)
is an integer, then n is a power of 2 :2k(log 3)/(log 2) = 3k .
Siegel : if 2t , 3t and 5t are integers, then t is an integer.
56 / 106The four exponentials Conjecture
Let x1 , x2 be two linearly independent complex numbers and also
y 1 , y 2 be two linearly independent complex numbers. Then one atleast of the four numbers exi yj = αij , i = 1, 2, j = 1, 2 istranscendental.
A 2 × 2 matrix has rank 1 if and only if it can be written
x1 y 1 x1 y 2
x2 y 1 x2 y 2
The four exponentials conjecture is equivalent to : Let
log α11 log α12
log α21 log α22
be a 2 × 2 matrix with Q–linearly independent rows and Q–linearly
independent columns. If the entries are logarithms of algebraic
numbers, then the rank of this matrix is 2.
57 / 106Four exponentials and six exponentials
Let t ∈ C ; the matrix

log 2 log 3t log 2 t log 3

has rank one. Hence if t is irrational, the Four exponentials
Conjecture implies that one at least of 2t , 3t is transcendental.
Six exponentials Theorem : exi yj = αij , i = 1, 2, j = 1, 2, 3.
Matrix log αij i=1,2; j=1,2,3 .
Example : let t ∈ C be irrational ; the matrix
log 2 log 3 log 5t log 2 t log 3 t log 5
has rank one. Hence one at least of 2t , 3t , 5t is transcendental.
58 / 106Structural rank of a matrix with entries in L
Let M be a matrix with entries in L (logarithms of algebraic
numbers). Take a basis of the Q–space spanned by the entries– let m be the dimension. Write the entries of M as linear
combinations with rational coefficients of the elements of thebasis. Replace the basis by variables X 1 , . . . , X m ; one gets amatrix with entries in Q(X 1 , . . . , X m ). The rank of thatmatrix is the structural rank of M . It does not depend on the
choice of the basis.
The conjecture of algebraic independence of logarithms (thehomogeneous form suffices) implies that both matrices havethe same rank.

59 / 106Structural rank of a matrix
Let K be a field, k a subfield and M a matrix with entries inK. Following D. Roy, we define structural rank of M withrespect to k
Consider the k-vector subspace E of K spanned by the entriesof M. Choose an injective morphism ϕ of E into a k-vector
space kX 1 + · · · + kX n . The image ϕ(M) of M is a matrix withentries in the field k(X 1 , . . . , X n ) of rational fractions. Its
rank does not depend on the choice of ϕ.
This is the structural rank of M with respect to k.

60 / 106

Homogeneous algebraic independence of logarithms
According to D. Roy, the homogeneous case of the conjectureon algebraic independence of logarithms of algebraic numbers
is equivalent to :Conjecture. Let M be a matrix with logarithms of algebraic
numbers entries. Then the rank of M is equal to its structuralrank with respect to Q.
Half of this conjecture is known :The rank of a matrix with logarithms of algebraic numbers
entries is at least half its structural rank with respect to Q
Samit Dasgupta, Introduction to TranscendenceTheory.
Math 790, 3/4/2021. Lecture 4: Ranks of matrices of logarithms
https://services.math.duke.edu/~dasgupta/Transcendence/index.html

61 / 106Algebraic independence of logarithms
According to D. Roy, the conjecture on algebraic independenceof logarithms of algebraic numbers is equivalent to :Conjecture. Any matrix
bij + λij 1≤i≤d 1 j≤`≤with bij ∈ Q and λij ∈ L has a rank equal to its structuralrank.

62 / 106

Any Polynomial is the Determinant of a Matrix

Roy’s proof of the equivalence between the two conjecturesuses the nice auxiliary result :For any P ∈ k[X 1 , . . . , X n ] there exists a square matrix withentries in the k-vector space k + kX 1 + · · · + kX n , thedeterminant of which is P .

63 / 106The Strong Four Exponentials Conjecture
e the Q-vector space spanned by 1 and L : hence
Denote by L
e is the set of linear combinations with algebraic coefficientsLof logarithms of algebraic numbers :e = {β0 + β1 λ1 + · · · + βn λn ; n ≥ 0, βi ∈ Q, λi ∈ L}.
L
e such that
Conjecture. If Λ1 , Λ2 , Λ3 , Λ4 are elements in L
Λ1 /Λ2 and Λ1 /Λ3 are transcendental, thenΛ1 Λ4 6= Λ2 Λ3 .
Means :
det

Λ1 Λ2
6= 0.
Λ3 Λ4
64 / 106Consequence of the Strong Four Exponentials
Conjecture
Assume the strong Four Exponentials Conjecture.
ee \ Q then the quotient 1/Λ is not in L.
• If Λ is in L
e \ Q, then the product Λ1 Λ2 is not in• If Λ1 and Λ2 are in LeL.
e with Λ1 and Λ2 /Λ1 transcendental,• If Λ1 and Λ2 are in Lethen this quotient Λ2 /Λ1 is not in L.

65 / 106

Partial result
D. Roy proved results related to the Strong four exponentials
Conjecture including the Strong six exponentials Theorem aswell as higher dimensional results like :The rank of a matrix, with entries which are linear
combinations of logarithms of algebraic numbers withalgebraic coefficients, is at least half its structural rank withrespect to the field Q of algebraic numbers.

66 / 106

Quadratic independence of logarithms of algebraic
numbers
One does not know yet how to prove that there is nonontrivial quadratic relations among logarithms of algebraic
numbers, like(log α1 )(log α2 ) = γ.
Diaz’ Conjecture. Let u ∈ C× . Assume |u| is algebraic. Theneu is transcendental.
If |u|2 = γ and eu = α are algebraic numbers, then(log α)(log α) = γ.

ix

For instance with |u| = 1, the curve x 7→ ee should not meet
any algebraic point.
ix
Similarly, the curve x 7→ e2iπe should not meet any algebraic
point.
67 / 106Conjectures of G. Diaz
• For λ ∈ L \ {0}, is it true that λλ 6∈ L ?
2Previous example : λ = iπ (transcendence of eπ ).
• More generally : for λ1 and λ2 in L \ {0}, is it true that
λ1 λ2 6∈ L ?
e?
• For λ1 and λ2 in L \ {0}, is it true that λ1 λ2 6∈ L

G. Diaz – Utilisation de la conjugaison complexe dans l’étude de latranscendance de valeurs de la fonction exponentielle, J. Théor.
Nombres Bordeaux 16 (2004), p. 535–553.
G. Diaz – Produits et quotients de combinaisons linéaires delogarithmes de nombres algébriques : conjectures et résultats
partiels , J. Théor. Nombres Bordeaux 19 (2007), no. 2, 373–391.

68 / 106

Damien Roy

Strategy suggested by D. Roy
in 1999, Journées
Arithmétiques, Roma :Conjecture equivalent to
Schanuel’s Conjecture.

69 / 106

Roy’s approach to Schanuel’s Conjecture (1999)
Let D denote the derivation
D=

∂
∂
+ X1
∂X 0
∂X 1

over the ring C[X 0 , X 1 ]. The height of a polynomial
P ∈ C[X 0 , X 1 ] is defined as the maximum of the absolutevalues of its coefficients.
Let k be a positive integer, y 1 , . . . , y k complex numbers which
are linearly independent over Q, α1 , . . . , αk non-zero complex
numbers and s0 , s1 , t0 , t1 , u positive real numbers satisfyingmax{1, t0 , 2t1 } < min{s0 , 2s1 }and1max{s0 , s1 + t1 } < u < (1 + t0 + t1 ).
2
70 / 106

Roy’s Conjecture
Assume that, for any sufficiently large positive integer N ,there exists a non-zero polynomial P N ∈ Z[X 0 , X 1 ] withpartial degree ≤ N t0 in X 0 , partial degree ≤ N t1 in X 1 andheight ≤ eN which satisfies
k

D PN

k
 X
j=1

mj y j ,

k
Y
j=1

αj mj



≤ exp(−N u )for any non-negative integers k, m1 , . . . , mk with k ≤ N s0 andmax{m1 , . . . , mk } ≤ N s1 . Thentrdeg Q(y 1 , . . . , y k , α1 , . . . , αk ) ≥ k.

71 / 106Equivalence between Schanuel and Roy
Let (y, α) ∈ C × C× , and let s0 , s1 , t0 , t1 , u be positive real
numbers satisfying the inequalities of Roy’s Conjecture. Thenthe following conditions are equivalent :(a) The number αe−y is a root of unity.
(b) For any sufficiently large positive integer N , there exists a
nonzero polynomial QN ∈ Z[X 0 , X 1 ] with partial degree≤ N t0 in X 0 , partial degree ≤ N t1 in X 1 and heightH(QN ) ≤ eN such that(∂ k QN )(my, αm ) ≤ exp(−N u )for any k, m ∈ N with k ≤ N s0 and m ≤ N s1 .

72 / 106Further progress by D. Roy
Ga ,

Gm ,

Ga × Gm .

Small value estimates for the additive group. Int. J. Number
Theory 6 (2010), 919–956.
Small value estimates for the multiplicative group. Acta Arith.
135 (2008), 357–393.
A small value estimate for Ga × Gm . Mathematika 59 (2013),
333–363.

73 / 106Further developments
Roy’s Conjecture deals with polynomials vanishing on somesubsets of C × C× with multiplicity along the space associatedwith the derivation ∂/∂X + Y ∂/∂Y .
Roy’s conjecture depends on parameters s0 , s1 , t0 , t1 , u in acertain range. D. Roy proved that if his conjecture is true forone choice of values of these parameters in the given range,
then Schanuel’s Conjecture is true, and that conversely, if
Schanuel’s Conjecture is true, then his conjecture is true for all
choices of parameters in the same range.
Nguyen Ngoc Ai Van
extended the range of theseparameters.
A refined criterion for Schanuel’s
conjecture. Chamchuri J. Math. 1(2009), no. 2, 25–29.
74 / 106

Ubiquity of Schanuel’s Conjecture
Other contexts : p–adic numbers, Leopoldt’s Conjecture on thep–adic rank of the units of an algebraic number field.
Non-vanishing of Regulators.
Non–degenerescence of heights.
Conjecture of B. Mazur on rational points.
Diophantine approximation on tori.
Dipendra Prasad
Gopal Prasad

Dipendra Prasad. An analogue of a conjecture of Mazur : a question in Diophantine
approximation on tori. Contributions to automorphic forms, geometry, and number
theory, 699– 709, Johns Hopkins Univ. Press, Baltimore, MD, 2004.
Gopal Prasad, Andrei S. Rapinchuk. Zariski-dense subgroups and transcendental
number theory. Math. Res. Lett. 12 (2005), no. 2-3, 239–249.
75 / 106Peter Bundschuh (1979).
For p/q ∈ Q with
0 < |p/q| < 1, the sum of theseries∞X

ζ(n)(p/q)nn=2

is a transcendental number.
For p/q ∈ Q \ Z,

Γ0
Γ

 
p
+γ
q

is transcendental (where γ is Euler’s constant).
76 / 106Golomb arithmetic function (1973)
γ(n) : number of distinct
representations of n in theform ab with positive integers
a and b.

Solomon W. Golomb
1932–2016

Peter Bundschuh (1979). A consequence of Schanuel’s
Conjecture is the transcendence of∞X

γ(n)n−sn=2

for s ≥ 4.

77 / 106Peter Bundschuh (1979)(P. Bundschuh) : As a consequence of Nesterenko’s Theorem,
the number∞X
1 π eπ + e−π1=+ ·= 2.076 674 047 4 . . .
n2 + 12 2 eπ − e−πn=0
is transcendental, while∞X
n=2(telescoping series).
Hence the number

1n2 − 1

=

∞
X

1

n=2

ns − 13
4

is transcendental over Q for s = 4. The transcendence of thisnumber for even integers s ≥ 4 would follow as a consequenceof Schanuel’s Conjecture.
78 / 106Sum of values of a rational function
Arithmetic nature ofX P (n)
n≥1

Q(n)

where
P /Q ∈ Q(X).
In case Q has distinct zeroes, by decomposing P /Q in simple
fractions one gets linear combinations of logarithms ofalgebraic numbers (Baker’s method).
The example P (X)/Q(X) = 1/X 3 shows that the generalcase is hard :X 1
ζ(3) =·
3nn≥1
79 / 106

S. Gun, R. Murty, P. Rath
Multiple Gamma function of Barnes, defined by Γ0 (z) = 1/z,
Γ1 (z) = Γ(z),Γn+1 (z),Γn+1 (z + 1) =Γn (z)
with Γn (1) = 1.
Assuming Schanuel’s Conjecture, one at least of the two
following statements is true :(i) The two numbers π and ζ(3) are algebraically independent.
(ii) The number Γ2 (1/2)/Γ3 (1/2) is transcendental.

80 / 106Sum of values of a rational function (2001)
Let P and Q be non-zero polynomials having rationalcoefficients and deg Q ≥ 2 + deg P . Consider
X P (n)·Q(n)n≥0
Q(n)6=0

S.D. Adhikari N. Saradha

T.N. Shorey

R. Tijdeman

Transcendental infinite sums. Indag. Math. (N.S.) 12 (2001), no. 1, 1–14.
81 / 106Sum of values of a rational functionMore recent references :• N. Saradha and R. Tijdeman,
On the transcendence of infinite sums of values of rational
functions.
J. London Math. Soc. (2) 67 (2003), no. 3, 580–592.
• R. Tijdeman,
On irrationally and transcendency of infinite sums of rational
numbers.
in Diophantine equations, 279–296, Tata Inst. Fund. Res.
Stud. Math., 20, Tata Inst. Fund. Res., Mumbai, 2008.
• Ram Murty ; Chester J. Weatherby.
On the transcendence of certain infinite series.
Int. J. Number Theory 7 (2011), no. 2, 323–339.
82 / 106Telescoping series
Examples
∞
X

1
= 1,
n(n
+
1)
n=1

∞ 
X
n=0

∞
X

1

n=2

n2 − 1

3
= ,
4

3
1
1
1
−
+
+
4n + 1 4n + 2 4n + 3 4n + 4

∞ 
X
n=0

1
3
1
−
+
5n + 2 5n + 7 5n − 3


=


=0

5
6

83 / 106Transcendental values∞X

1= log 2,
(2n
+
1)(2n
+
2)
n=0

∞
X
1

π2
,
=
2
n
6
n=1

∞
X
1
,
ζ(2m) =
2m
n
n=1

∞
X

1
π
=
(n + 1)(2n + 1)(4n + 1)
3
n=0

are transcendental.

84 / 106Transcendental values∞
X1(6n + 1)(6n + 2)(6n + 3)(6n + 4)(6n + 5)(6n + 6)
n=0
=

√
1
(192 log 2 − 81 log 3 − 7π 3)
4320

∞
X

1
1 π eπ + e−π=+ ·= 2.076 674 047 4 . . .
n2 + 12 2 eπ − e−π
n=0
∞
X
(−1)n
n=0

n2 + 1=2πeπ − e−π= 0.272 029 054 982 . . .
85 / 106Catalan’s ConstantCatalan’s constant is
X (−1)nG=(2n + 1)2n≥1
= 0.915 965 594 177 219 015 0 . . .
Is it an irrational number ?
Eugène Catalan(1814 - 1894)

86 / 106Catalan’s constant, Dirichlet and Kronecker
Catalan’s constant is the value at s = 2 of the Dirichlet
L–function L(s, χ−4 ) associated with the Kronecker character
if n is even,
0
n 
= 1
χ−4 (n) =if n ≡ 1 (mod 4) ,4

−1 if n ≡ −1 (mod 4) .

Johann Peter Gustav Lejeune Dirichlet
1805 – 1859

Leopold Kronecker
1823 – 1891
87 / 106Catalan’s constant, Dedekind and Riemann
The Dirichlet L–function L(s, χ−4 ) associated with the
Kronecker character χ−4 is the quotient of the Dedekind zetafunction of Q(i) and the Riemann zeta function :ζQ(i) (s) = L(s, χ−4 )ζ(s)Julius Wilhelm Richard
Dedekind1831 – 1916

G = L(2, χ−4 ).

Georg Friedrich Bernhard
Riemann
1826 – 1866

88 / 106

S. Gun, R. Murty, P. Rath

Assuming Schanuel’s Conjecture, one at least of the two next
statements is true :(i) The two numbers π and G are algebraically independent.
(ii) The number Γ2 (1/4)/Γ2 (3/4) is transcendental.
89 / 106Further applications of Schanuel’s Conjecture
Ram Murty ; Purusottam Rath.
Transcendental numbers. Springer, New York, 2014.

• Transcendental sums related to the zeros of zeta functions.
• Linear and algebraic independence of generalized
Euler-Briggs constants.
• Special values of L-functions.
• Transcendence of the log gamma function and some discrete
periods.
90 / 106Conjectures by S. Schanuel and A. Grothendieck• Schanuel : if x1 , . . . , xn are Q–linearly independent complex
numbers, then n at least of the 2n numbers x1 , . . . , xn ,ex1 , . . . , exn are algebraically independent.
• Periods conjecture by Grothendieck : Dimension of the
Mumford–Tate group of a smooth projective variety.

91 / 106

Motives
Y. André : generalization ofGrothendieck’s conjecture to
motives.
Implies Schanuel’s Conjecture.

Case of 1–motives :Elliptico-Toric Conjecture of
C. Bertolin.
(+ Philippon & Saha2 )92 / 106Francis Brown
Irrationality proofs for zeta values, moduli spaces and dinner parties
arXiv:1412.6508 http://www.ihes.fr/~brown/IrratModuliMotivesv8.pdf
Slides : http://www.ihes.fr/~brown/IrrationalitySlidesPrintable.pdf

A simple geometric construction on the moduli spaces M0,n ofcurves of genus 0 with n ordered marked points is describedwhich gives a common framework for many irrationality proofs
for zeta values. This construction yields Apéry’s
approximations to ζ(2) and ζ(3), and for larger n, an infinite
family of small linear forms in multiple zeta values with aninteresting algebraic structure. It also contains a generalisationof the linear forms used by Ball and Rivoal to prove that
infinitely many odd zeta values are irrational.
93 / 106Francis Brown
For k, s1 , . . . , sk positive integers with s1 ≥ 2, we set
s = (s1 , . . . , sk ) andX
1ζ(s) =·
s1 · · · n sk
n1kn >n >···>n ≥1
1

2

k

The Q–vector space Z spanned by the numbers ζ(s) is also a
Q–algebra. For n ≥ 2, denote by Zn the Q-subspace of Z
spanned by the real numbers ζ(s) where s has weights1 + · · · + sk = n.
The numbers ζ(s1 , . . . , sk ),s1 + · · · + sk = n, where eachsi is 2 or 3, span Zn over Q.

94 / 106Chester Weatherby
Let q 1 , . . . , q k be natural
numbers. For a complex
valued function f on thegroup
G = Z/q 1 Z × · · · × Z/q k Z,
consider the multiple Dirichlet
seriesF (s1 , . . . , sk ) =∞X

f (n1 , . . . , nk ).
n1 s1 · · · nk sk
n ,...,n =1
1

kAssuming the conjecture on algebraic independence oflogarithms of algebraic numbers, Chester Weatherby provesthat F (1, . . . , 1) is either zero or transcendental.
Ram Murty ; Chester J. Weatherby. On the transcendence of certain
infinite series. Int. J. Number Theory 7 (2011), no. 2, 323–339.

95 / 106Periods : Maxime Kontsevich and Don Zagier
Periods,
Mathematics
unlimited—2001
and beyond,Springer 2001,
771–808.

A period is a complex number whose real and imaginary parts
are values of absolutely convergent integrals of rational
functions with rational coefficients, over domains in Rn given
by polynomial inequalities with rational coefficients.
96 / 106The number π
Basic example of a period :ez+2iπ = ez

Z
2iπ =

Z ∞

Z Z
π =

dz
|z|=1 zdx
1 − x2−∞Z 1√=2
1 − x2 dx.

dxdy =

Z 1 x2 +y2 ≤1dx√=
1 − x2−1−1

97 / 106Further examples of periods
√

Z
2=

dx
2x2 ≤1and all algebraic numbers.
Z
log 2 =

dx
1<x<2 xand all logarithms of algebraic numbers.
Z
π=
dxdy,
x2 +y 2 ≤1

A product of periods is a period (subalgebra of C), but 1/π is
expected not to be a period.
98 / 106

Numbers which are not periods
Problem (Kontsevich–Zagier) : To produce an explicit example
of a number which is not a period.

Several levels :1 analog of Cantor : the set of periods is countable.
Hence there are real and complex numbers which are notperiods (most of them).

99 / 106

Numbers which are not periods
2 analog of Liouville
Find a property which should be satisfied by all periods, andconstruct a number which does not satisfies that property.
Masahiko Yoshinaga, Periods and elementary real numbers
arXiv:0805.0349
Compares the periods with hierarchy of real numbers induced
from computational complexities.
In particular, he proves that periods can be effectively
approximated by elementary rational Cauchy sequences.
As an application, he exhibits a computable real number which
is not a period.
100 / 106

Numbers which are not periods
3 analog of Hermite
Prove that given numbers are not periods
Candidates : 1/π, e, Euler constant.
M. Kontsevich : exponential periods
The last chapter, which is at a more advanced level and also more
speculative than the rest of the text, is by the first author only.
M. Kontsevich and D. Zagier. Periods, Mathematics unlimited—2001 and beyond,Springer 2001, 771–808.

101 / 106Relations among periods
1Additivity(in the integrand and in the domain of integration)Z b

f (x) + g(x) dx =Z bf (x)dx +

a

a

Z b

Z c
f (x)dx =

a

2

Z b
g(x)dx,aZbf (x)dx +af (x)dx.
c

Change of variables :if y = f (x) is an invertible change of variables, thenZ f (b)Z b
F (y)dy =f (a)F f (x) f 0 (x)dx.

a102 / 106Relations among periods (continued)

3

Newton–Leibniz–Stokes FormulaZ b
f 0 (x)dx = f (b) − f (a).
a103 / 106Conjecture of Kontsevich and Zagier
A widely-held belief, based on a
judicious combinationof experience, analogy,
and wishful thinking,is the following
Conjecture (Kontsevich–Zagier). If a period has two integral
representations, then one can pass from one formula toanother by using only rules 1 , 2 , 3 in which all functions
and domains of integration are algebraic with algebraic
coefficients.

104 / 106

Period Conjecture of Kontsevich and Zagier
We do not expect any miraculous coı̈ncidence of two
integrals of algebraic functions which will not be
possible to prove using three simple rules.
This conjecture, which is similar in spirit to the
Hodge conjecture, is one of the central conjectures
about algebraic independence and transcendental
numbers, and is related to many of the results andideas of modern arithmetic algebraic geometry andthe theory of motives.

Advice : if you wish to prove a number is transcendental, firstprove it is a period.

105 / 106iniya piranthanaal nalvaazhthukkal

இ"ய $ற&த நா*ந+வா-./0க*This file is available on the internet at the URL
http://www.math.jussieu.fr/⇠miw/
106 / 106