Number theory and mathematical logic

Some number theory

Some model theory

Tales from the Dark Side
When mathematical logic meets number theory

Lee A. Butler
Department of Mathematics
University of Bristol
lee.butler@bris.ac.uk

November 10, 2010

Counting them points

Some number theory

Some model theory

Counting them points


It’s always a pleasure to introduce ideas from model
theory to people who do real mathematics.
Boris Zilber

Some number theory

Some model theory

Number theory is counting
• Number theory is all about counting things.

Counting them points

Some number theory

Some model theory

Number theory is counting
• Number theory is all about counting things.

• For example:

Counting them points

Some number theory

Some model theory

Counting them points

Number theory is counting
• Number theory is all about counting things.
• For example:
• The number of primes less than x, π(x),
• The number of rational solutions to x n + y n = z n with
xyz 6= 0 for n ∈ N+ ,
• The number of solutions to ζ(s) = 0 with s 6= 21 + it,
• The number of algebraic numbers α such that eα is an
algebraic number.


Some number theory

Some model theory

Counting them points

Number theory is counting
• Number theory is all about counting things.
• For example:
• The number of primes less than x, π(x),
• The number of rational solutions to x n + y n = z n with
xyz 6= 0 for n ∈ N+ ,
• The number of solutions to ζ(s) = 0 with s 6= 21 + it,
• The number of algebraic numbers α such that eα is an
algebraic number.
• Answer may be:
• “there are infinitely many”,
• “there are none”,
• “there are only finitely many”.


Some number theory

Some model theory

Counting them points

Number theory is counting
• Number theory is all about counting things.
• For example:
• The number of primes less than x, π(x),
• The number of rational solutions to x n + y n = z n with
xyz 6= 0 for n ∈ N+ ,
• The number of solutions to ζ(s) = 0 with s 6= 21 + it,
• The number of algebraic numbers α such that eα is an
algebraic number.
• Answer may be:
• “there are infinitely many”,
• “there are none”,
• “there are only finitely many”.
• “There are infinitely many” isn’t usually a very good answer.


Some number theory

Some model theory

How dense can you get?
• Density results improve “there are infinitely many”.

Counting them points

Some number theory

Some model theory

How dense can you get?
• Density results improve “there are infinitely many”.

• Let

X = {x : x satisfies our property}


Counting them points

Some number theory

Some model theory

Counting them points

How dense can you get?
• Density results improve “there are infinitely many”.

• Let

X = {x : x satisfies our property}

• Count elements of X up to “size” T , then let T → ∞.

Some number theory


Some model theory

Counting them points

How dense can you get?
• Density results improve “there are infinitely many”.

• Let

X = {x : x satisfies our property}

• Count elements of X up to “size” T , then let T → ∞.
• For example,

X = {p ∈ N+ : p is prime}.
Count p ∈ X such that p 6 T . This is π(T ).

• Prime number theorem: π(T ) ∼ T / log T .

Some number theory


Some model theory

Another example
• Density results can imply finiteness results.

Counting them points

Some number theory

Some model theory

Another example
• Density results can imply finiteness results.
• Let E(T ) = {n ∈ N : en ∈ Q, n 6 T }.

Counting them points

Some number theory


Some model theory

Counting them points

Another example
• Density results can imply finiteness results.
• Let E(T ) = {n ∈ N : en ∈ Q, n 6 T }.
• Suppose we can show that #E(T ) 6

large T .



T for all sufficiently

Some number theory

Some model theory

Counting them points


Another example
• Density results can imply finiteness results.
• Let E(T ) = {n ∈ N : en ∈ Q, n 6 T }.
• Suppose we can show that #E(T ) 6



T for all sufficiently

large T .

• Now suppose a ∈ N, a 6= 0, and ea ∈ Q. Then

 
T
0, a, 2a, 3a, . . . ,
a ∈ E(T )
a


so for large enough T , #E(T ) > ⌊T /a⌋ > T .

Some number theory

Some model theory

Counting them points

Another example
• Density results can imply finiteness results.
• Let E(T ) = {n ∈ N : en ∈ Q, n 6 T }.
• Suppose we can show that #E(T ) 6



T for all sufficiently

large T .

• Now suppose a ∈ N, a 6= 0, and ea ∈ Q. Then

 
T
0, a, 2a, 3a, . . . ,
a ∈ E(T )
a

so for large enough T , #E(T ) > ⌊T /a⌋ > T .

• Density results can give insights into transcendental

number theory.

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

• If no such P exists, α is transcendental.

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

• If no such P exists, α is transcendental.

• Cantor: The algebraic numbers are countable.

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

• If no such P exists, α is transcendental.

• Cantor: The algebraic numbers are countable.

• Experience: It’s outrageously difficult to prove that a given

number α ∈ C is transcendental.

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

• If no such P exists, α is transcendental.

• Cantor: The algebraic numbers are countable.

• Experience: It’s outrageously difficult to prove that a given

number α ∈ C is transcendental.

• If α ∈ R can restrict to just proving there’s no linear

polynomial P with P(α) = 0 (i.e. α is irrational).

Some number theory

Some model theory

Counting them points

Transcendental number theory

• α ∈ C is algebraic if there is P ∈ Q[x], P 6= 0, such that

P(α) = 0.

• If no such P exists, α is transcendental.

• Cantor: The algebraic numbers are countable.

• Experience: It’s outrageously difficult to prove that a given

number α ∈ C is transcendental.

• If α ∈ R can restrict to just proving there’s no linear

polynomial P with P(α) = 0 (i.e. α is irrational).

• This is really hard too!

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:

Counting them points

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5

Counting them points

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5
• ζ(7), ζ(9), ζ(11), . . .

Counting them points

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5
• ζ(7), ζ(9), ζ(11), . . .
• γ = limn→∞

Pn

1
m=1 m

− log n



Counting them points

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5
• ζ(7), ζ(9), ζ(11), . . .
• γ = limn→∞

• π + e, πe, π e

Pn

1
m=1 m

− log n



Counting them points

Some number theory

Some model theory

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5
• ζ(7), ζ(9), ζ(11), . . .
• γ = limn→∞

• π + e, πe, π e

Pn

e
• 2e , ee , ee , . . .

1
m=1 m

− log n



Counting them points

Some number theory

Some model theory

Counting them points

Irrational woes

We don’t know if the following are rational or irrational:
P
1
• ζ(5) = ∞
n=1 n5
• ζ(7), ζ(9), ζ(11), . . .
• γ = limn→∞

• π + e, πe, π e

Pn

e
• 2e , ee , ee , . . .

• G=

1
m=1 m

(−1)n
n=0 (2n+1)2

P∞

=

− log n

1
12



1
32



+

1
52



1
72

+ ...

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:
• 1

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:
• 1, 11

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:
• 1, 11, 21

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:
• 1, 11, 21, 1211

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:
• 1, 11, 21, 1211, 111221

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

• 1, 11, 21, 1211, 111221, 312211,. . .

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

• 1, 11, 21, 1211, 111221, 312211,. . .

• Let ℓn be the number of digits in the nth term.

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

• 1, 11, 21, 1211, 111221, 312211,. . .

• Let ℓn be the number of digits in the nth term.

• λ = limn→∞ n+1
≈ 1.303577269034296 . . .

n

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

• 1, 11, 21, 1211, 111221, 312211,. . .

• Let ℓn be the number of digits in the nth term.

• λ = limn→∞ n+1
≈ 1.303577269034296 . . .

n

• Conway: λ is algebraic.

Some number theory

Some model theory

Counting them points

Look-and-say
All the numbers on the previous slide are conjectured to be
transcendental.
Tempting to think that “interesting” numbers are either rational
or transcendental.
• Conway’s look-and-say constant:

• 1, 11, 21, 1211, 111221, 312211,. . .

• Let ℓn be the number of digits in the nth term.

• λ = limn→∞ n+1
≈ 1.303577269034296 . . .

n

• Conway: λ is algebraic.

• Its minimal polynomial is...

Some number theory

Some model theory

Counting them points

λ

x 71 − x 69 − 2x 68 − x 67 + 2x 66 + 2x 65 + x 64 − x 63 − x 62 − x 61

− x 60 − x 59 + 2x 58 + 5x 57 + 3x 56 − 2x 55 − 10x 54 − 3x 53 − 2x 52
+ 6x 51 + 6x 50 + x 49 + 9x 48 − 3x 47 − 7x 46 − 8x 45 − 8x 44

+ 10x 43 + 6x 42 + 8x 41 − 5x 40 − 12x 39 + 7x 38 − 7x 37 + 7x 36
+ x 35 − 3x 34 + 10x 33 + x 32 − 6x 31 − 2x 30 − 10x 29 − 3x 28

+ 2x 27 + 9x 26 − 3x 25 + 14x 24 − 8x 23 − 7x 21 + 9x 20 + 3x 19

− 4x 18 − 10x 17 − 7x 16 + 12x 15 + 7x 14 + 2x 13 − 12x 12 − 4x 11
− 2x 10 + 5x 9 + x 7 − 7x 6 + 7x 5 − 4x 4 + 12x 3 − 6x 2 + 3x − 6.

Some number theory

Some model theory

As if that wasn’t bad enough

Counting them points

Some number theory

Some model theory

Counting them points

As if that wasn’t bad enough
• Proving that given numbers are transcendental is the easy

part.

Some number theory

Some model theory

Counting them points

As if that wasn’t bad enough
• Proving that given numbers are transcendental is the easy

part.

• Transcendental number theory is more concerned with

algebraic relations, or their absence.

Some number theory

Some model theory

Counting them points

As if that wasn’t bad enough
• Proving that given numbers are transcendental is the easy

part.

• Transcendental number theory is more concerned with

algebraic relations, or their absence.

• Say α1 , . . . , αn in a commutative ring A ⊃ K are

algebraically dependent over K if there is
P ∈ K [x1 , . . . , xn ] \ {0} such that P(α1 , . . . , αn ) = 0.

Some number theory

Some model theory

Counting them points

As if that wasn’t bad enough
• Proving that given numbers are transcendental is the easy

part.

• Transcendental number theory is more concerned with

algebraic relations, or their absence.

• Say α1 , . . . , αn in a commutative ring A ⊃ K are

algebraically dependent over K if there is
P ∈ K [x1 , . . . , xn ] \ {0} such that P(α1 , . . . , αn ) = 0.

• The transcendence degree of A over K is the biggest

n ∈ N such that there are algebraically independent
α1 , . . . , αn ∈ A.

Some number theory

Some model theory

Counting them points

As if that wasn’t bad enough
• Proving that given numbers are transcendental is the easy

part.

• Transcendental number theory is more concerned with

algebraic relations, or their absence.

• Say α1 , . . . , αn in a commutative ring A ⊃ K are

algebraically dependent over K if there is
P ∈ K [x1 , . . . , xn ] \ {0} such that P(α1 , . . . , αn ) = 0.

• The transcendence degree of A over K is the biggest

n ∈ N such that there are algebraically independent
α1 , . . . , αn ∈ A.
α ∈ C transcendental



trdegQ (Q(α)) = 1.

Some number theory

Some model theory

Counting them points

Indiana Jones and the holy grail (of number theory)
Proving that “natural” collections of numbers are algebraically
independent is oh-so difficult.

Some number theory

Some model theory

Counting them points

Indiana Jones and the holy grail (of number theory)
Proving that “natural” collections of numbers are algebraically
independent is oh-so difficult.
Theorem (Nesterenko, 1996)
The following collections are algebraically independent over Q:
• {π, eπ },

• {π, eπ , Γ(1/4)},


• {π, eπ 3 , Γ(1/3)}.

Some number theory

Some model theory

Counting them points

Indiana Jones and the holy grail (of number theory)
Proving that “natural” collections of numbers are algebraically
independent is oh-so difficult.
Theorem (Nesterenko, 1996)
The following collections are algebraically independent over Q:
• {π, eπ },

• {π, eπ , Γ(1/4)},


• {π, eπ 3 , Γ(1/3)}.

Theorem (Lindemann–Weierstrass, 1885)
If α1 , . . . , αn are algebraic and linearly independent over Q, then
trdegQ (Q(eα1 , . . . , eαn )) = n.

Some number theory

Some model theory

Counting them points

The Lindemann–Weierstrass theorem is a very special case
of...
Schanuel’s conjecture (1960s)
If α1 , . . . , αn ∈ C are linearly independent over Q, then
trdegQ (Q(α1 , . . . , αn , eα1 , . . . , eαn )) > n.

Some number theory

Some model theory

Counting them points

The Lindemann–Weierstrass theorem is a very special case
of...
Schanuel’s conjecture (1960s)
If α1 , . . . , αn ∈ C are linearly independent over Q, then
trdegQ (Q(α1 , . . . , αn , eα1 , . . . , eαn )) > n.
Known cases:

Some number theory

Some model theory

Counting them points

The Lindemann–Weierstrass theorem is a very special case
of...
Schanuel’s conjecture (1960s)
If α1 , . . . , αn ∈ C are linearly independent over Q, then
trdegQ (Q(α1 , . . . , αn , eα1 , . . . , eαn )) > n.
Known cases:
• α1 , . . . , αn all algebraic (Lindemann–Weierstrass),
• n = 1 (Hermite–Lindemann).

Some number theory

Some model theory

Schanuel schmanuel
Schanuel implies a lot:

Counting them points

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function,

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function, in particular any
relation between e and π is explained by eπi = −1.

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function, in particular any
relation between e and π is explained by eπi = −1.
⇒ Hermite–Lindemann, Lindemann–Weierstrass,
Gelfond–Schneider, Baker’s theorem, and the four
exponentials conjecture.

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function, in particular any
relation between e and π is explained by eπi = −1.
⇒ Hermite–Lindemann, Lindemann–Weierstrass,
Gelfond–Schneider, Baker’s theorem, and the four
exponentials conjecture.

⇒ real exponentiation is decidable, so given any statement
about real numbers that involves addition, multiplication,
and exponentiation, one can decide whether it’s true.

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function, in particular any
relation between e and π is explained by eπi = −1.
⇒ Hermite–Lindemann, Lindemann–Weierstrass,
Gelfond–Schneider, Baker’s theorem, and the four
exponentials conjecture.

⇒ real exponentiation is decidable, so given any statement
about real numbers that involves addition, multiplication,
and exponentiation, one can decide whether it’s true. (Cf.
Gödel’s theorem: N with addition, and multiplication is
undecidable.)

Some number theory

Some model theory

Counting them points

Schanuel schmanuel
Schanuel implies a lot:
⇒ no “unexpected” algebraic relations between complex
numbers and the exponential function, in particular any
relation between e and π is explained by eπi = −1.
⇒ Hermite–Lindemann, Lindemann–Weierstrass,
Gelfond–Schneider, Baker’s theorem, and the four
exponentials conjecture.

⇒ real exponentiation is decidable, so given any statement
about real numbers that involves addition, multiplication,
and exponentiation, one can decide whether it’s true. (Cf.
Gödel’s theorem: N with addition, and multiplication is
undecidable.)
⇒ the quantum torus is a “stable structure”, and thus easily
studied using a plethora of model theoretic tools.

Some number theory

Some model theory

Counting them points

Woe is we

Let L = {λ ∈ C : eλ ∈ Q}.
Special case of Schanuel
If β1 , . . . , βn ∈ L are linearly independent over Q, then they’re
algebraically independent over Q.

Some number theory

Some model theory

Counting them points

Woe is we

Let L = {λ ∈ C : eλ ∈ Q}.
Special case of Schanuel
If β1 , . . . , βn ∈ L are linearly independent over Q, then they’re
algebraically independent over Q.
Current progress: haven’t proved the algebraic independence
of any two linearly independent elements of L.

Some number theory

Some model theory

Counting them points

A different approach

Schanuel tells us sets defined using the exponential function
don’t have many algebraic points in them unless there are
trivial reasons for their presence.

Some number theory

Some model theory

Defining sets
Work in the real ordered exponential field
Rexp = (R, +, −, ·, exp, Tj
for some sequence Tj → ∞.

Some number theory

Some model theory

Counting them points

Must try harder
The Pila–Wilkie theorem can’t be much improved in general.
Let ε(T ) → 0 as slowly as you want. Then there is a definable
set X in some o-minimal structure such that
ε(Tj )

N(X , Tj ) > Tj
for some sequence Tj → ∞.

In particular, can’t get N(X , T ) ≪ (log T )c for definable sets in
general.

Some number theory

Some model theory

Counting them points

Must try harder
The Pila–Wilkie theorem can’t be much improved in general.
Let ε(T ) → 0 as slowly as you want. Then there is a definable
set X in some o-minimal structure such that
ε(Tj )

N(X , Tj ) > Tj
for some sequence Tj → ∞.

In particular, can’t get N(X , T ) ≪ (log T )c for definable sets in
general.
What about definable sets in Rexp ?

Some number theory

Some model theory

Counting them points

Wilkie’s conjecture

Conjecture (Wilkie, 2006)
Let X ⊆ Rn be definable in Rexp . Then there is a constant c(X ) >
0 such that for any T > e,
N(X \ X alg , T ) ≪X (log T )c(X ) .

Some number theory

Some model theory

Why Rexp ?
What’s so special about Rexp ?

Counting them points

Some number theory

Some model theory

Counting them points

Why Rexp ?
What’s so special about Rexp ?
• It’s model complete. (Definable sets are projections of

exponential varieties.)

Some number theory

Some model theory

Counting them points

Why Rexp ?
What’s so special about Rexp ?
• It’s model complete. (Definable sets are projections of

exponential varieties.)

• It has smooth and analytic cell decomposition. (Definable

functions are piecewise smooth and analytic, definable
sets have smooth boundaries.)

Some number theory

Some model theory

Counting them points

Why Rexp ?
What’s so special about Rexp ?
• It’s model complete. (Definable sets are projections of

exponential varieties.)

• It has smooth and analytic cell decomposition. (Definable

functions are piecewise smooth and analytic, definable
sets have smooth boundaries.)

• The exponential function is Pfaffian. (Its derivative is a

polynomial in itself.)

Some number theory

Some model theory

Counting them points

Why Rexp ?
What’s so special about Rexp ?
• It’s model complete. (Definable sets are projections of

exponential varieties.)

• It has smooth and analytic cell decomposition. (Definable

functions are piecewise smooth and analytic, definable
sets have smooth boundaries.)

• The exponential function is Pfaffian. (Its derivative is a

polynomial in itself.)

• And, just maybe, all its definable sets can be very nicely

reparametrised.

Some number theory

Some model theory

Counting them points

Mildness

A smooth function f : (0, 1)n → (0, 1) is called (A, C)-mild if for
every ~z ∈ (0, 1)n , and every µ
~ ∈ Nn ,




|~
µ| f



~
(
z
)
~ !(A|~
µ|C )|~µ| .

µ1

∂x1 · ∂xnµn

Some number theory

Some model theory

Counting them points

Mildness

A smooth function f : (0, 1)n → (0, 1) is called (A, C)-mild if for
every ~z ∈ (0, 1)n , and every µ
~ ∈ Nn ,




|~
µ| f



~
(
z
)
~ !(A|~
µ|C )|~µ| .

µ1

∂x1 · ∂xnµn
Pila: If a set X ⊂ (0, 1)n can be reparametrised by mild
functions, then its rational points of height 6 T lie on
≪ (log T )c(X ) hypersurfaces.

Some number theory

Some model theory

Counting them points

Where we’re at

Take for granted that there is a height function on algebraic
numbers in some fixed number field F . Let N(X , F , T ) be the
number of elements in X ∩ F n of height at most T .

Some number theory

Some model theory

Counting them points

Where we’re at

Take for granted that there is a height function on algebraic
numbers in some fixed number field F . Let N(X , F , T ) be the
number of elements in X ∩ F n of height at most T .
Theorem (B., 2009)
Let X ⊆ R2 be definable in Rexp and f ∈ N+ . There are constants c1 (X , f ) and c2 (X ) such that for any number field F ⊂ R
of degree f and any T > e,
N(X \ X alg , F , T ) 6 c1 (X , f )(log T )c2 (X ) .

Some number theory

Some model theory

Counting them points

Proof

• Use some model theory (analytic cell decomposition,

extension of model completeness) to lift to a simpler but
higher dimensional set.

Some number theory

Some model theory

Counting them points

Proof

• Use some model theory (analytic cell decomposition,

extension of model completeness) to lift to a simpler but
higher dimensional set.

• Infer information about our original set, and descend back

to R2 .

Some number theory

Some model theory

Counting them points

Proof

• Use some model theory (analytic cell decomposition,

extension of model completeness) to lift to a simpler but
higher dimensional set.

• Infer information about our original set, and descend back

to R2 .

• Use this new information and tools from Bombieri–Pila for

counting points on planar graphs to get the bound.

Some number theory

Some model theory

Counting them points

Where we’re at (reprise)
Theorem (B., 2009)
Let a, b, c ∈ R and f ∈ N+ . Let
X = {(x, y , z) ∈ (0, ∞)3 : (log x)a (log y )b (log z)c = 1}.
Then there is a constant c3 = c3 (a, b, c, f ) such that for any
number field F ⊂ R of degree f and any T > e,
N(X \ X alg , F , T ) 6 c3 (log T )107 .

Some number theory

Some model theory

Counting them points

Where we’re at (reprise)
Theorem (B., 2009)
Let a, b, c ∈ R and f ∈ N+ . Let
X = {(x, y , z) ∈ (0, ∞)3 : (log x)a (log y )b (log z)c = 1}.
Then there is a constant c3 = c3 (a, b, c, f ) such that for any
number field F ⊂ R of degree f and any T > e,
N(X \ X alg , F , T ) 6 c3 (log T )107 .
• Show X has a mild reparametrisation.

Some number theory

Some model theory

Counting them points

Where we’re at (reprise)
Theorem (B., 2009)
Let a, b, c ∈ R and f ∈ N+ . Let
X = {(x, y , z) ∈ (0, ∞)3 : (log x)a (log y )b (log z)c = 1}.
Then there is a constant c3 = c3 (a, b, c, f ) such that for any
number field F ⊂ R of degree f and any T > e,
N(X \ X alg , F , T ) 6 c3 (log T )107 .
• Show X has a mild reparametrisation.

• Use an explicit, Pfaffian reparametrisation then intersect

with hypersurfaces.

Some number theory

Some model theory

Counting them points

Where we’re at (reprise)
Theorem (B., 2009)
Let a, b, c ∈ R and f ∈ N+ . Let
X = {(x, y , z) ∈ (0, ∞)3 : (log x)a (log y )b (log z)c = 1}.
Then there is a constant c3 = c3 (a, b, c, f ) such that for any
number field F ⊂ R of degree f and any T > e,
N(X \ X alg , F , T ) 6 c3 (log T )107 .
• Show X has a mild reparametrisation.

• Use an explicit, Pfaffian reparametrisation then intersect

with hypersurfaces.

• Project these intersections onto R2 .

Some number theory

Some model theory

Counting them points

Where we’re at (reprise)
Theorem (B., 2009)
Let a, b, c ∈ R and f ∈ N+ . Let
X = {(x, y , z) ∈ (0, ∞)3 : (log x)a (log y )b (log z)c = 1}.
Then there is a constant c3 = c3 (a, b, c, f ) such that for any
number field F ⊂ R of degree f and any T > e,
N(X \ X alg , F , T ) 6 c3 (log T )107 .
• Show X has a mild reparametrisation.

• Use an explicit, Pfaffian reparametrisation then intersect

with hypersurfaces.

• Project these intersections onto R2 .

• We have a logarithmic bound for “Pfaff curves”.

Some number theory

Some model theory

Counting them points

Everything has consequences

Pila–Wilkie gives us number theoretic results in Diophantine
geometry:

Some number theory

Some model theory

Counting them points

Everything has consequences

Pila–Wilkie gives us number theoretic results in Diophantine
geometry:
• New proof of Manin–Mumford conjecture,

Some number theory

Some model theory

Counting them points

Everything has consequences

Pila–Wilkie gives us number theoretic results in Diophantine
geometry:
• New proof of Manin–Mumford conjecture,

• Unconditional proof of some cases of the André–Oort

conjecture,

Some number theory

Some model theory

Counting them points

Everything has consequences

Pila–Wilkie gives us number theoretic results in Diophantine
geometry:
• New proof of Manin–Mumford conjecture,

• Unconditional proof of some cases of the André–Oort

conjecture,

• Proofs of some cases of Pink’s conjecture.

Some number theory

Some model theory

Counting them points

Everything has consequences

Pila–Wilkie gives us number theoretic results in Diophantine
geometry:
• New proof of Manin–Mumford conjecture,

• Unconditional proof of some cases of the André–Oort

conjecture,

• Proofs of some cases of Pink’s conjecture.

Wilkie’s conjecture is more related to transcendental number
theory.

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

• Suppose (xi , yi ) ∈ Xα , i = 1, 2, . . . , 18, are algebraic points

with the xi multiplicatively independent.

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

• Suppose (xi , yi ) ∈ Xα , i = 1, 2, . . . , 18, are algebraic points

with the xi multiplicatively independent.

• Let F contain all these xi and yi .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

• Suppose (xi , yi ) ∈ Xα , i = 1, 2, . . . , 18, are algebraic points

with the xi multiplicatively independent.

• Let F contain all these xi and yi .

Q
xiai , yiai ) ∈ Xα ∩ F 2 , and they’re distinct for
distinct 18-tuples of integers ~a.
Q

• Then (

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

• Suppose (xi , yi ) ∈ Xα , i = 1, 2, . . . , 18, are algebraic points

with the xi multiplicatively independent.

• Let F contain all these xi and yi .

Q
xiai , yiai ) ∈ Xα ∩ F 2 , and they’re distinct for
distinct 18-tuples of integers ~a.
Q

• Then (

• So N(Xα , F , T ) ≫ (log T )18 .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem

• Let α ∈ R \ Q. The set Xα = {(x, y ) : y = x α } is definable

in Rexp .

• Suppose (xi , yi ) ∈ Xα , i = 1, 2, . . . , 18, are algebraic points

with the xi multiplicatively independent.

• Let F contain all these xi and yi .

Q
xiai , yiai ) ∈ Xα ∩ F 2 , and they’re distinct for
distinct 18-tuples of integers ~a.
Q

• Then (

• So N(Xα , F , T ) ≫ (log T )18 .

• But N(Xα , F , T ) ≪ (log T )17 . Contradiction!

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .
• Six exponential theorem: this works with just three

numbers w1 , w2 , w3 .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .
• Six exponential theorem: this works with just three

numbers w1 , w2 , w3 .

• Four exponentials conjecture: it works with just two

numbers w1 , w2 .

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .
• Six exponential theorem: this works with just three

numbers w1 , w2 , w3 .

• Four exponentials conjecture: it works with just two

numbers w1 , w2 .

• Better bounds in the exponent mean better results.

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .
• Six exponential theorem: this works with just three

numbers w1 , w2 , w3 .

• Four exponentials conjecture: it works with just two

numbers w1 , w2 .

• Better bounds in the exponent mean better results.

• Need N(Xα , F , T ) ≪ log T for four exponential conjecture.

Some number theory

Some model theory

Counting them points

The 36-exponentials theorem
Theorem
Let w1 , . . . , w18 ∈ R be linearly independent over Q and
α ∈ R \ Q, then one of the following 36 numbers is transcendental:
ew1 , . . . , ew18 , eαw1 , . . . , eαw18 .
• Six exponential theorem: this works with just three

numbers w1 , w2 , w3 .

• Four exponentials conjecture: it works with just two

numbers w1 , w2 .

• Better bounds in the exponent mean better results.

• Need N(Xα , F , T ) ≪ log T for four exponential conjecture.

Refinements of current proof can’t hope to get exponent
below 11.

Some number theory

Some model theory

Onwards

Counting them points

Some number theory

Some model theory

Onwards

• Don’t reparametrise the whole set.

Counting them points

Some number theory

Some model theory

Counting them points

Onwards

• Don’t reparametrise the whole set.

• Find other structures that might satisfy Wilkie’s conjecture.

Some number theory

Some model theory

Counting them points

Onwards

• Don’t reparametrise the whole set.

• Find other structures that might satisfy Wilkie’s conjecture.
• Finish thesis.

Some number theory

Some model theory

Counting them points

Onwards

• Don’t reparametrise the whole set.

• Find other structures that might satisfy Wilkie’s conjecture.
• Finish thesis.

Thank you.