Number theory and mathematical logic

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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

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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.

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Number theory is counting

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Number theory is counting

• Number theory is all about counting things.

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Number theory is counting

• For example:

• The number of primes less thanx,π(x),

• The number of rational solutions toxn+yn=znwith xyz ₆=0 forn_∈N+,

• The number of solutions toζ(s) =0 withs6= 1₂+it, • The number of algebraic numbersαsuch thateαis an

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Number theory is counting

• For example:

algebraic number.

• Answer may be:

• “there are infinitely many”, • “there are none”,

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Number theory is counting

• For example:

algebraic number.

• Answer may be:

• “there are infinitely many”, • “there are none”,

• “there are only finitely many”.

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How dense can you get?

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How dense can you get?

• Density results improve “there are infinitely many”.

• Let

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How dense can you get?

• Let

X =_{x : x satisfies our property}

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How dense can you get?

• Let

X =_{x : x satisfies our property}

• Count elements ofX up to “size”T, then letT _{→ ∞}.

• For example,

X =_{p_∈N+ : pis prime_}.

Countp _∈X such thatp 6T. This isπ(T).

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Another example

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Another example

• Density results can imply finiteness results.

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Another example

• LetE(T) =_{n_∈N : en_∈_Q_,_n₆_T_}_.

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Another example

• LetE(T) =_{n_∈N : en_∈_Q_,_n₆_T_}_.

• Suppose we can show that#E(T)6√T for all sufficiently largeT.

• Now supposea_∈N,a₆=0, andea_∈Q. Then

0,a,2a,3a, . . . ,

T a

a_∈E(T)

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Another example

• LetE(T) =_{n_∈N : en_∈_Q_,_n₆_T_}_.

• Suppose we can show that#E(T)6√T for all sufficiently largeT.

• Now supposea_∈N,a₆=0, andea_∈Q. Then

0,a,2a,3a, . . . ,

T a

a_∈E(T)

so for large enoughT,#E(T)>_⌊T/a_⌋>√T.

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Transcendental number theory

• α_∈Cisalgebraicif there isP _∈Q[x],P ₆=0, such that

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Transcendental number theory

P(α) =0.

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P(α) =0.

• If no suchPexists,αistranscendental.

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P(α) =0.

• Cantor: The algebraic numbers are countable.

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Transcendental number theory

P(α) =0.

• Experience: It’s outrageously difficult to prove that a given numberα_∈Cis transcendental.

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P(α) =0.

• Experience: It’s outrageously difficult to prove that a given numberα_∈Cis transcendental.

• Ifα_∈Rcan restrict to just proving there’s no linear polynomialP withP(α) =0 (i.e.αis irrational).

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Irrational woes

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Irrational woes

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

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Irrational woes

• ζ(5) =P∞ n=1n15

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Irrational woes

• ζ(5) =P∞ n=1n15

• ζ(7), ζ(9), ζ(11), . . .

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Irrational woes

• ζ(5) =P∞ n=1n15

• ζ(7), ζ(9), ζ(11), . . .

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Irrational woes

• ζ(5) =P∞ n=1n15

• ζ(7), ζ(9), ζ(11), . . .

• γ =limn_→∞ P_mn₌₁_m1 −logn • π+e,πe,πe

• 2e_,_ee_,_eee

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Irrational woes

• ζ(5) =P∞ n=1n15

• ζ(7), ζ(9), ζ(11), . . .

• γ =limn_→∞ P_mn₌₁_m1 −logn • π+e,πe,πe

• 2e_,_ee_,_eee

,. . . • G=P∞

n=0

(−1)n

(2n+1)2 =

1 12 −

1 32 +

1 52 −

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Look-and-say

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Look-and-say

All the numbers on the previous slide are conjectured to be transcendental.

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Look-and-say

Tempting to think that “interesting” numbers are either rational or transcendental.

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Look-and-say

• Conway’s look-and-say constant:

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Look-and-say

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Look-and-say

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Look-and-say

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Look-and-say

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Look-and-say

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Look-and-say

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

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Look-and-say

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

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Look-and-say

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

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Look-and-say

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

• Letℓn be the number of digits in thenth term. • λ=limn_→∞ ℓn_ℓ+_n1 ≈1.303577269034296. . . • Conway: λis algebraic.

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λ

x71₋x69₋2x68₋x67+2x66+2x65+x64₋x63₋x62₋x61

−x60₋x59+2x58+5x57+3x56₋2x55₋10x54₋3x53₋2x52 +6x51+6x50+x49+9x48₋3x47₋7x46₋8x45₋8x44

+10x43+6x42+8x41₋5x40₋12x39+7x38₋7x37+7x36

+x35₋3x34+10x33+x32₋6x31₋2x30₋10x29₋3x28

+2x27+9x26₋3x25+14x24₋8x23₋7x21+9x20+3x19

−4x18₋10x17₋7x16+12x15+7x14+2x13₋12x12₋4x11

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As if that wasn’t bad enough

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As if that wasn’t bad enough

• Proving that given numbers are transcendental is the easy part.

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As if that wasn’t bad enough

• Transcendental number theory is more concerned with algebraic relations, or their absence.

• Sayα1, . . . , αnin a commutative ringA⊃K are algebraically dependent over K if there is

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As if that wasn’t bad enough

P _∈K[x1, . . . ,xn]_{\ {}0_}such thatP(α1, . . . , αn) =0. • Thetranscendence degreeofAoverK is the biggest

n_∈Nsuch that there are algebraically independent

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As if that wasn’t bad enough

P _∈K[x1, . . . ,xn]_{\ {}0_}such thatP(α1, . . . , αn) =0. • Thetranscendence degreeofAoverK is the biggest

n_∈Nsuch that there are algebraically independent

α1, . . . , αn∈A.

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Indiana Jones and the holy grail (of number theory)

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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 overQ:

• {π,eπ_}_,

• {π,eπ_,_Γ(₁_/₄₎_}_,

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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 overQ:

• {π,eπ_}_,

• {π,eπ_,_Γ(₁_/₄₎_}_,

• {π,eπ√3,Γ(1/3)_}.

Theorem (Lindemann–Weierstrass, 1885)

Ifα1, . . . , αnare algebraic and linearly independent overQ, then

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The Lindemann–Weierstrass theorem is a very special case of...

Schanuel’s conjecture (1960s)

Ifα1, . . . , αn∈Care linearly independent overQ, then

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trdeg_Q(Q(α1, . . . , αn,eα1, . . . ,eαn))>n.

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trdeg_Q(Q(α1, . . . , αn,eα1, . . . ,eαn))>n.

Known cases:

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Schanuel schmanuel

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Schanuel schmanuel

Schanuel implies a lot:

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Schanuel schmanuel

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Schanuel schmanuel

⇒ no “unexpected” algebraic relations between complex numbers and the exponential function, in particular any relation betweeneandπis explained byeπi =₋1.

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Schanuel schmanuel

⇒ Hermite–Lindemann, Lindemann–Weierstrass, Gelfond–Schneider, Baker’s theorem, and the four exponentials conjecture.

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Schanuel schmanuel

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Schanuel schmanuel

⇒ 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: Nwith addition, and multiplication is undecidable.)

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Woe is we

LetL=_{λ_∈C : eλ _∈_Q_}_.

Special case of Schanuel

If β1, . . . , βn ∈ L are linearly independent over Q, then they’re

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Woe is we

LetL=_{λ_∈C : eλ _∈_Q_}_.

Special case of Schanuel

If β1, . . . , βn ∈ L are linearly independent over Q, then they’re

algebraically independent overQ.

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A different approach

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Defining sets

Work in the real ordered exponential field

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Defining sets

Rexp= (R,+,−,·,exp, <,0,1).

Define sets using:

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Defining sets

Rexp= (R,+,−,·,exp, <,0,1).

Define sets using:

+,₋,_·,exp, <, elements ofR, variables,_∀,_∃,_∨,_∧,_¬,_⇒,_⇔,(,).

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Some definable sets

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Some definable sets

• {(x,y)_∈R2 : y =ex_}

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Some definable sets

• {(x,y)_∈R2 : y =ex_}

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Some definable sets

• {(x,y)_∈R2 : y =ex_}

• {(x,y)_∈R2 : _∃u_∃v (x =eu _∧ _y ₌_ev _∧ _u_·_v ₌₁₎_} • {(x,y)_∈R2 : logxlogy =1_}

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Some definable sets

• {(x,y)_∈R2 : y =ex_}

• {(x,y)_∈R2 : _∃u_∃v (x =eu _∧ _y ₌_ev _∧ _u_·_v ₌₁₎_} • {(x,y)_∈R2 : logxlogy =1_}

• {(x,y)_∈R2 : _∃u_∃v ((x₋eu)2+(y₋ev)2+(u_·v₋1)2=0)_}

Wilkie’s theorem

Every definable set inRexp can be defined by an expression of

the form

∃u1. . ._∃uℓ,P(x1, . . . ,xn,u1, . . . ,uℓ,ex1, . . . ,exn,eu1, . . . ,euℓ) =0

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A dictionary

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A dictionary

LetX =_{~x _∈Rn : φ(~x)_},Y =_{~y _∈Rn : ψ(~y)_}. Then:

Set theory Logic

X_∪Y _{~x _∈Rn : φ(~x) _∨ ψ(~x)_} X_∩Y _{~x _∈Rn : φ(~x) _∧ ψ(~x)_} Rn_\X _{~x _∈Rn :_¬φ(~x)_}

π(X) _{~x _∈Rn−1 : _∃t, φ(x1, . . . ,xn₋1,t)}

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A dictionary

LetX =_{~x _∈Rn : φ(~x)_},Y =_{~y _∈Rn : ψ(~y)_}. Then:

Set theory Logic

X_∪Y _{~x _∈Rn : φ(~x) _∨ ψ(~x)_} X_∩Y _{~x _∈Rn : φ(~x) _∧ ψ(~x)_} Rn_\X _{~x _∈Rn :_¬φ(~x)_}

π(X) _{~x _∈Rn−1 : _∃t, φ(x1, . . . ,xn₋1,t)}

whereπ:Rn_→Rn−1,π(x1, . . . ,xn) = (x1, . . . ,xn₋1).

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Khovanski˘ı’s theorem

Theorem (Khovanski˘ı)

An exponential variety, i.e. a setU ⊆Rndefined by an exponen-tial polynomial

P(x1, . . . ,xn,ex1, . . . ,exn) =0,

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Khovanski˘ı’s theorem

Theorem (Khovanski˘ı)

An exponential variety, i.e. a setU ⊆Rndefined by an exponen-tial polynomial

P(x1, . . . ,xn,ex1, . . . ,exn) =0,

has only finitely many connected components.

Corollary

The subsets of R that are definable in Rexp are precisely the

finite unions of points and intervals:

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O-minimal? Oh yeah!

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O-minimal? Oh yeah!

An expansion of the real ordered field(R,+,₋,_·, <,0,1, . . .)is calledo-minimalif the only sets that are definable inRare finite unions of points and intervals.

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O-minimal? Oh yeah!

{a1_{} ∪}. . ._{∪ {}an} ∪(b1,c1)∪. . .∪(bm,cm).

n _

i=1

(x =ai) _∨ m _

j=1

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O-minimal? Oh yeah!

{a1_{} ∪}. . ._{∪ {}an} ∪(b1,c1)∪. . .∪(bm,cm).

n _

i=1

(x =ai) _∨ m _

j=1

(bj <x ∧ x <cj).

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O-minimal? Oh yeah!

{a1_{} ∪}. . ._{∪ {}an} ∪(b1,c1)∪. . .∪(bm,cm).

n _

i=1

(x =ai) _∨ m _

j=1

(bj <x ∧ x <cj).

O-minimal structures realise Grothendieck’s “tame topology”.

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Count, Dracula

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Count, Dracula

Some notation for counting.

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Count, Dracula

• ht(a/b) =max(_|a_|,_|b_|).

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Count, Dracula

• ht(a/b) =max(_|a_|,_|b_|).

• If(α1, . . . , αn)_∈Qnthen ht(α~) =maxiht(αi). • LetX _⊆RnandT >0. Then

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Begone, foul semi-algebraic subset

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Begone, foul semi-algebraic subset

We want estimates onN(X,T)for sets definable inRexp.

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{(x,y)_∈R2 : y =ex _∨ y =0_}

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Begone, foul semi-algebraic subset

{(x,y)_∈R2 : y =ex _∨ y =0_}

• Interested in the party =ex,

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{(x,y)_∈R2 : y =ex _∨ y =0_}

• Interested in the party =ex,

• AsT _{→ ∞}, we only see the contribution ofy =0.

Let

Xalg= [ U_⊆X Usemi-algebraic

dim(U)>1 U.

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Counting points in three easy steps

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Counting points in three easy steps

Step 1 Reparametrise the set by nice functions.

Step 2 Show these nice functions only hit rational points at

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Counting points in three easy steps

Step 1 Reparametrise the set by nice functions.

Step 2 Show these nice functions only hit rational points at

intersections with hypersurfaces.

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Step 1

(0,1)

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The Pila–Wilkie theorem

LetX _⊆Rn be definable in some o-minimal structure and

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The Pila–Wilkie theorem

T, ε >0. Then:

• There arec1(ε,X)mapsφ: (0,1)dim(X)_→_X _{that together}

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The Pila–Wilkie theorem

T, ε >0. Then:

coverX, all of them with bounded derivatives up to some prescribed order.

• The rational points ofφ((0,1)dim(X)₎_{of height at most}_T _lie

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The Pila–Wilkie theorem

T, ε >0. Then:

in at mostc2(ε,X)Tε_{hypersurfaces of degree}₆_d₍_ε₎_.

• Hypersurfaces of degree6d(ε)can intersectX in at most

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The Pila–Wilkie theorem

T, ε >0. Then:

in at mostc2(ε,X)Tε_{hypersurfaces of degree}₆_d₍_ε₎_.

• Hypersurfaces of degree6d(ε)can intersectX in at most

c3(ε,X)points.

Theorem (Pila–Wilkie, 2006)

IfX _⊆ Rn is definable in some o-minimal structure, and ε > 0 then for anyT >0,

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Must try harder

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Must try harder

The Pila–Wilkie theorem can’t be much improved in general.

Letε(T)_→0 as slowly as you want. Then there is a definable setX in some o-minimal structure such that

N(X,Tj)>Tε(Tj) j

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Must try harder

N(X,Tj)>Tε(Tj) j

for some sequenceTj → ∞.

In particular, can’t getN(X,T)_≪(logT)c _{for definable sets in}

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Must try harder

N(X,Tj)>Tε(Tj) j

for some sequenceTj → ∞.

In particular, can’t getN(X,T)_≪(logT)c _{for definable sets in}

general.

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Wilkie’s conjecture

Conjecture (Wilkie, 2006)

LetX _⊆Rnbe definable inRexp. Then there is a constantc(X)>

0 such that for anyT >e,

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Why

R

_exp

?

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Why

R

_exp

?

What’s so special aboutRexp?

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Why

R

_exp

?

• It’s model complete. (Definable sets are projections of exponential varieties.)

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Why

R

_exp

?

• It has smooth and analytic cell decomposition. (Definable functions are piecewise smooth and analytic, definable sets have smooth boundaries.)

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Why

R

_exp

?

• 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.)

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Mildness

A smooth functionf : (0,1)n_→₍₀_,₁₎_{is called}₍_A,_C₎_{-mild if for}

every~z _∈(0,1)n, and every~µ_∈Nn,

∂|µ~_|_f

∂xµ1

1 ·∂x

µn n

(~z)

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Mildness

A smooth functionf : (0,1)n_→₍₀_,₁₎_{is called}₍_A,_C₎_{-mild if for}

every~z _∈(0,1)n, and every~µ_∈Nn,

∂|µ~_|_f

∂xµ1

1 ·∂x

µn n

(~z)

6~µ!(A|~µ_|C)|~µ|.

Pila: If a setX ⊂(0,1)n_{can be reparametrised by mild}

functions, then its rational points of height6T lie on

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Where we’re at

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Where we’re at

Take for granted that there is a height function on algebraic numbers in some fixed number fieldF. LetN(X,F,T)be the number of elements inX∩Fnof height at mostT.

Theorem (B., 2009)

LetX _⊆ R2 be definable in Rexp and f ∈ N+. There are

con-stantsc1(X,f) andc2(X) such that for any number fieldF ⊂R

of degreef and anyT >e,

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Proof

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Proof

• Use some model theory (analytic cell decomposition, extension of model completeness) to lift to a simpler but higher dimensional set.

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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 toR2.

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Where we’re at (reprise)

Theorem (B., 2009)

Leta,b,c ∈Randf ∈N+. Let

X =_{(x,y,z)_∈(0,_∞)3 : (logx)a(logy)b(logz)c =1_}.

Then there is a constant c3 = c3(a,b,c,f) such that for any number fieldF _⊂Rof degreef and anyT >e,

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Where we’re at (reprise)

Theorem (B., 2009)

N(X _\Xalg,F,T)6c3(logT)107.

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Where we’re at (reprise)

Theorem (B., 2009)

• ShowX has a mild reparametrisation.

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Where we’re at (reprise)

Theorem (B., 2009)

• Use an explicit, Pfaffian reparametrisation then intersect with hypersurfaces.

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Where we’re at (reprise)

Theorem (B., 2009)

• Use an explicit, Pfaffian reparametrisation then intersect with hypersurfaces.

• Project these intersections ontoR2.

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Everything has consequences

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Everything has consequences

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

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• New proof of Manin–Mumford conjecture,

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Everything has consequences

• Unconditional proof of some cases of the André–Oort conjecture,

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• Unconditional proof of some cases of the André–Oort conjecture,

• Proofs of some cases of Pink’s conjecture.

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The 36-exponentials theorem

• Letα_∈R_\Q. The setXα ={(x,y) : y =xα}is definable

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The 36-exponentials theorem

inRexp.

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

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inRexp.

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The 36-exponentials theorem

inRexp.

with thexi multiplicatively independent. • LetF contain all thesexi andyi. • Then(Q

xai i ,

Q yai

i )∈Xα∩F

2_{, and they’re distinct for}

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The 36-exponentials theorem

inRexp.

xai i ,

Q yai

i )∈Xα∩F

distinct 18-tuples of integers~a.

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inRexp.

xai i ,

Q yai

i )∈Xα∩F

distinct 18-tuples of integers~a.

• SoN(Xα,F,T)≫(logT)18.

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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 transcen-dental:

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Theorem