1731 Lesson 5_CRA6

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• cours - matière potentielle : discussion
• cours - matière potentielle : plan
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Saturn Educator Guide • Cassini Program website — • EG-1998-12-008-JPL L E S S O N 5 121 T H E C A S S I N I – H U Y G E N S M I S S I O N Students begin by examining their prior notions of robots and then consider the characteristics and capabilities of a robot like the Cassini–Huygens spacecraft that would be sent into space to explore another planet.

• component functions table
• human analogy
• cassini
• questions on chart paper
• robot
• construction paper
• spacecraft
• space
• function
• students

Sujets

• Discussion
• Exposé
• Cassini
• Spacecraft
• Student
• Function

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• Publié par: idwe0
• Nombre de lectures: 15
• Langue: English

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LINFOOT TALK — MODEL THEORETIC NUMBER THEORY

t Question to keep the braying crowd occupied:Supposen∈Zfor everyn∈N. What can be said aboutt?

1.Number theory

Number theory is often concerned with relationships between numbers.This often takes the form of asking how many solutions there are to some algebraic equations up to a certain size.For example: Conjecture 1(Manin’s conjecture).LetVbe the intersection ofrhypersurfaces n of degreedinP. Thenthere is a Zariski open subsetUofVand a constant c 1 depending onVand our notion of “size” such that the number of elements in n U∩P(Q)up to sizeTis n+1−rd c 2 (v) c 1 T(logT) (1+o(1)) asT→ ∞.

2.Counting problems

Manin’s conjecture is a typical counting problem. n LetX⊂RandH:Q→R >0 Letbe a meaningful function to measure size. n N(X, T) = #{~x∈X(Q) :H(x i )6T}. The counting problem is to understand this functionN. Ostensibly it depends onX,H, andT, but really the choice ofHdoesn’t make much difference as long as we’re not purposefully oafish.We’ll useH(a/ b) = max{|a|,|b|}for hcf(a, b) = 1. So, for example, ifPis the set of primes then T ∼ N(P, T) =. logT Or if we let 3n n n X n ={(x, y, z)∈R:x+y=xyzz ,6= 0}, then N(X n , T) = 0 forn>3 whileN(X 2 , T) isn’t known exactly, determining it is Gauss’s circle problem. In 1989 Enrico Bombieri and Jonathan Pila introduced some novel techniques for the counting problem whenXis an irreducible algebraic curve or the graph 1

2 LINFOOTTALK — MODEL THEORETIC NUMBER THEORY of a transcendental analytic functionf: [0,1]→Rin their paper is the. Buried following useful result. ` nk Theorem 2.Letφ: (0,1)→(0,1)be aCfunction with |α| ∂ α |∂ φ(x)|=φ(x)61   α 1 α ` ∂x∙ ∙ ∙∂x 1 ` for|α|6k, and letX= im(φ). ThenX(Q, T)is contained in the intersection of ε Xwith ε,X Talgebraic hypersurfaces of degreed(ε)withd(ε)→ ∞asε→0.

So a common technique is to reparametrise sets by functions like this.Obviously this result is no good in the algebraic case.But iff: [0,1]→Ris transcendental and analytic, for example, then these intersections are just points and B–P showed there arec(d) =c(εSo if) of them.Xis the graph of such anfthen ε N(X, T) ε,X T .

2 The next step is whenf: [0,1]→R, but now there’s a problem.The function fcan be transcendental but still contain algebraic curves which unduly boost the value ofN(X, TLet). Thesolution is to omit them. [ alg X=U, U⊆X Usemi-alg. dim(U)>1 trans algtrans andX=X\Xhigher dimensions it makes sense to estimate. InN(TX ,). This is analogous to counting points in a Zariski open subset in Manin’s conjecture. In 2003 JP published his proof of the two-dimensional case. 2 Theorem 3.Letf: [0,1]→Rbe transcendental and analytic andε >0, and let Xbe the graph off. Then transε N(X ,T) ε,X T .

ε The basic idea of the proof is the same:intersectXwithThypersurfaces. This ε3 2 givesTcurves inRwhich can be projected down toRto apply the earlier case. But these curves depend onTand without sufficient uniformity the whole thing goes to pot. JP managed this, but the general case looked pretty intractable.Which was where Alex Wilkie and model theory entered.

3.o-minimal structures

Model theory is a branch of mathematical logic.As Tony Blair might say, model theory is too large a subject to describe with a single soundbite, but it is the study of mathematical structures by studying what is true in those structures.

I like to think of model theory as an extension of the way we learn about groups at school.We don’t do that by learning everything there is to learn about the Klein-4 group, and then in the following year take an advanced course where we

LINFOOT TALK — MODEL THEORETIC NUMBER THEORY3 learn there are also other groups.But this is almost how we learn analysis, treating the reals as the only place it happens. Model theory strips away these preconceptions and looks at whole classes of structures as single objects, studying whether the truth of a statement in one structure can be carried over to the others in its class.As a subject it thrives on generalities, but this works to its advantage.Once you’ve shown that the truth of a statement carries from one structure to another, you can prove theorems where ever it is simplest, and get the same theorem for free in the other structures. Amongst the structures which contain something likeRthere is a particularly well behaved class, the o-minimal structures. IfRis a field with ordering<then an o-minimal structure onRcan be thought of as a family (S n ) n∈N such that: + n (1)S n is a boolean algebra of subsets ofR; (2) ifA∈S n thenR×AandA×Rare inS n+1 ; 2 (3){(x, y)∈R:x < y} ∈S 2 ; n (4){x~∈R:x 1 =x n } ∈S n for eachn; n+1n (5) ifπ:R→Ris the projection map on the firstncoordinates and A∈S n+1 thenπ(A)∈S n ; and (6)S 1 consists of all finite unions of singletons and (possibly unbounded) in- tervals inR. If these conditions look like someone pulled them out of their hat then it’s because they are geometric equivalents of the natural first-order logic operations, where the definition is very natural and very simple.Elements ofS n are called definable sets. The paradigm example of an o-minimal structure is the collection of all semi- algebraic sets – sets given by polynomial equations and inequalities.The fact that π(A) is semi-algebraic if A is was proved independently by logician Alfred Tarski and algebraic geometer Abraham Seidenberg. Another example is the collection of all bounded subanalytic sets.These start with bounded semi-analytic sets, those defined locally by real power series, then allow projections down.Now condition (5) is free, but showing the complement of a subanalytic set is subanalytic is not easy.This was accomplished by Andrei ε Gabrielov. Thesets that Pila got hisTbound for are definable in this structure. A final example is the structureR exp . Thisstarts with “semi-exponential” sets given by exponential polynomial equations and inequalities, and then allowing pro- jections. AlexWilkie proved this collection is closed under taking complements and Khovanski˘ı showed semi-exponential sets have finitely many connected components, thus giving us (6). In 2006 Pila and Wilkie proved the following: n Theorem 4(Pila–Wilkie).LetX⊂Rbe a definable set in some o-minimal structure andε >0. Then transε N(X ,T) ε,X T .

4 LINFOOTTALK — MODEL THEORETIC NUMBER THEORY Using this, various new proofs of problems in the Manin–Mumford circle have beengiven,includingthefirstunconditionalproofofabunchofcasesoftheAndr´e– Oort conjecture by JP.

The Pila–Wilkie theorem applies to Diophantine geometry.It can’t really be improved in general because the Bombieri–Pila result is essentially best possible. But Wilkie conjectured an improvement of a more transcendental number theoretic nature. n Conjecture 5(Wilkie’s conjecture).LetX⊂Rbe a definable set inR exp . There are constantsc 1 (X)andc 2 (X)such that for allT>e, transc 2 (X) N(TX ,)6c 1 (X)(logT).

Known cases: •dim(X) = 1 (B., Jones–Thomas). 3 •X⊂RprovidedXcan be nicely reparametrised (Jones–Thomas).In particular... 3a b c • {(x, y, z)∈R: (logx) (logy) (logz1) =}for anya, b, c∈Q(B.).

4.A likely approach

The Pila–Wilkie proof relies on showing definable sets are the image of finitely k manyCmaps with bounded derivatives up to orderkWilkie’s conjecture we. For ∞ needCmapsφwith α C|α| |∂ φ(x~)|6α!(A|α|) ` for someA, Cand for allα∈N.

That being said, that’s not how the dim(X) = 1 case is proved.

5.Sketch proof whendim(X) = 1

n LetX⊂R, dim(X) = 1.Can project down and use the maps (x, y)7→ ±1±1 (±x ,±y) and a shed load of model theory to getXto be the graph of a smooth functionφ: (0,1)→(0,1). c There is a result that saysX(Q) lies on the intersection ofXwith(logT) c0(j) hypersurfaces of degree(logTwe can ensure) if|φ|61 andφis either≡0 or c c 6= 0 for 16j6∼(logT) . So if we can split (0,1) into about (logT) subintervals where these criteria are met we’re halfway there. One last bit of model theory tells us thatφSois implicitly exp-definable. m+1 there are exponential polynomialsf 1 , . . . , f m :R→Rand smooth functions φ=φ 1 , φ 2 , . . . , φ m such that: •f i (t, φ 1 (t), . . . , φ m (t)) = 0 for 16i6mandt∈(0,1);   ∂f i •det6= 0. ∂y j 16i,j6m

LINFOOT TALK — MODEL THEORETIC NUMBER THEORY5 0 Differentiating eachf i implicitly gives usmequations in themunknownsφ, i and we can solve this system because the matrix of coefficients in nonsingular.So 0(j) we can findφ(and inductivelyφ) in terms of∂f i /∂y j .

Thanks to Khovanski˘ı, Gabrielov, and Vorobjov, we can bound the zeros of such c expressions and so split (0,1) into (logT) subintervalswhere aforementioned result holds.

A similar process lets us estimate the number of intersections ofXwith algebraic hypersurfaces of a given degree, and multiplying everything together we end up with trans 11+6m N(X ,T) X (logT).

6.Transcendental application

Suppose we could getN(X, T)logTfor number fieldsF⊂R, and then apply this to the set 2α X α ={(x, y)∈R:y=x} for irrationalα. Now suppose (x 1 , y 1 ),(x 2 , y 2 )∈X α withx i , y i ∈Qandx 1 , x 2 multiplicatively independent. LetF=Q(x i , y i ) then fora 1 , a 2 ∈Z, a 1 a 2 a 1 a 2 2 (y yx x ,)∈X α ∩F 1 21 2 and if (a 1 , a 2 )6= (b 1 , b 2 ) then the corresponding points inX α will be different. taking all pairsa 1 , a 2 with|a 1 |+|a 2 |6logTgives 2 N(X α , T)(logT), a contradiction.This implies... Conjecture 6(Four exponentials conjecture).Letx 1 , x 2 ∈RbeQ-linearly inde- pendent andy 1 , y 2 ∈RbeQ-linearly independent.Then at least one of x 1 y 1 x 1 y 2 x 2 y 1 x 2 y 2 e ,e ,e ,e is transcendental.

t t Assuming this to be true, now lett∈Rbe such that 2,3∈Z. Applythe four-exponentials conjecture to the pairs 1, tand log2,log 3.Iftis irrational this would imply that one of t t 2,3,2,3 t was transcendental, but they’re all integers.Sot∈Q. Butift∈Qand 2∈Z thent∈N.