Complex Exponential Field

Paola D’Aquino Universit`adegli Studi della Campania ”L. Vanvitelli”

Logic Colloquium 2018

Udine, 25-28 July 2018 Exponential Function

The model theoretic analysis of the exponential function over a field started with a problem left open by Tarski in the 30’s, about the decidability of the reals with exponentiation. Only in the mid 90’s Macintyre and Wilkie gave a positive answer to this question assuming Schanuel’s Conjecture. Complex exponentiation involves much deeper issues, and it is much harder to approach, as it inherits the Gödelincompleteness and undecidability phenomena via the definition of the set of periods. Despite this negative results there are still many interesting and natural model-theoretic aspects to analyze. Exponential rings

Deﬁnition: An exponential ring, or E-ring, is a pair (R, E) where R is a ring (commutative with 1) and

E :(R, +) → (U(R), ·)

a morphism of the additive group of R into the multiplicative group of units of R satisfying • E(x + y) = E(x) · E(y) for all x, y ∈ R • E(0) = 1.

1 (R, exp); (C, exp); 2 (K, E) where K is any ring and E(x) = 1 for all x ∈ K. Model-theoretic analysis of (C, +, ·, 0, 1, )

The model theory of the ﬁeld of complex numbers is very tame

C is a canonical model of ACF (0), i.e. it is the unique algebraically closed ﬁeld of characteristic 0 and cardinality 2ℵ0

C is strongly minimal, i.e. the definable subsets of C are either finite or cofinite.

The deﬁnable sets have a geometrical interpretation in terms of algebraic varieties. Comparing (R, exp) and (C, exp)

Th(R, exp) decidable Th(C, exp) undecidable modulo (SC) Z = {x : ∀y(E(y) = 1 → E(xy) = 1)} (Macintyre-Wilkie ’96)

Th(R, exp) Th(C, exp) model-complete not model-complete (Wilkie ’96) (Macintyre, Marker)

Th(R, exp) o-minimal • Is Th(C, exp) quasi-minimal? good description of • What are the automorphisms deﬁnable sets of (C, exp)? (Wilkie ’96) • Is R deﬁnable in (C, exp)? (C, exp)

Macintyre 1996 x ∈ Q iﬀ ∃t, u, v((v − u)t = 1 ∧ E(v) = E(u) = 1 ∧ (vx = u))

Laczkovich 2002 For any x ∈ Q

x ∈ Z iﬀ ∃z(E(z) = 2 ∧ E(zx) ∈ Q)

Th∃(C, exp) is undecidable. Model-theoretic analysis of (C, exp)

What can we say about (C, exp)?

Zilber Conjecture: (C, exp) is quasi-minimal, i.e. every subset of C deﬁnable in (C, exp) is either countable or co-countable.

In 2004 Zilber introduces a new class of E-fields, the pseudo-exponential fields (or Zilber fields), and finds an

axiomatization of this class in Lω1ω(Q) - Q is the quantiﬁer exist uncountably many

- Lω1ω allows countable ∨ and ∧ Axiomatization of Zilber

(K, E) is a pseudo-exponential ﬁelds (or Zilber ﬁelds) if:

K is an algebraically closed ﬁeld of characteristic 0;

E :(K, +, 0) −→ (K ×, ·, 1) is a surjective homomorphism and there is ω ∈ K transcendental over Q such that ker E = Zω;

Schanuel’s Conjecture (SC) Let λ1, . . . , λn ∈ K be linearly independent over Q. Then Q(λ1, . . . , λn, E(λ1),..., E(λn)) has transcendence degree (t.d.) at least n over Q; Axiomatization of Zilber

(Strong Exponential Closure) For all ﬁnite A ⊆ K if n ∗ n V ⊆ Gn(K) = K × (K ) is irreducible, free and normal with dim V = n there is (z, E(z)) ∈ V generic over A.

(Countable Closure Property) For all ﬁnite A ⊆ K , the exponential algebraic closure eclK (A) of A in K is countable

V ⊆ Gn(K) is normal if dim[M]V ≥ k for any k × n integer matrix M of rank k.

V ⊆ Gn(K) is free if there are no m1,..., mn ∈ Z and a, b ∈ K where b 6= 0 such that V is contained in m1 mn {(x, y): m1x1 + ... + mnxn = a} or {(x, y): y1 · ... · yn = b}. Pseudo-exponential ﬁelds - Categoricity

THEOREM (Zilber 2000/2005 - Bays and Kirby 2013/2018) Up to isomorphism, there is exactly one model of the axioms in each uncountable cardinality.

THEOREM (Zilber, Bays and Kirby) Each pseudo exponential field is quasiminimal, i.e. every definable set is countable or co-countable. If (K, E) is pseudo-exponential field, |K| = k then there are 2k many automorphisms of (K, E). Zilber’s Conjecture

Zilber’s Conjecture: The unique pseudo-exponential ﬁeld of ℵ cardinality 2 0 is (C, exp).

A positive answer would imply

• Is (C, exp) quasi-minimal? YES

• Are there automorphisms of (C, exp) diﬀerent from identity and conjugation? YES

• Is R deﬁnable in (C, exp)NO (C, exp) vs (C, E)

Does (C, exp) satisfy properties which follow directly from Zilber’s axioms?

Does a pseudo-exponential ﬁeld (K, E) satisfy properties which are known for (C, exp)?

Analytic method and results cannot be applied over pseudo-exponential ﬁelds, no topology except an obvious exponential Zariski topology Schanuel’s conjecture

Generalization of Lindemann-Weierstrass Theorem: Let α1, . . . , αn be algebraic numbers which are linearly independent α α over Q.Then e 1 ,..., e n are algebraic independent over Q.

1 α = 1 transcendence of e (Hermite 1873)

2 α = 2πi transcendence of π (Lindemann 1882)

iπ π 3 α = (π, iπ) then tr.d.(π, iπ, e, e ) = 2, i.e. π, e are algebraically independent over Q (Nesterenko 1996)

4 (SC) is true for power series C[[t]] (Ax 1971) (C, exp) vs (C, E)

THEOREM (Zilber 2005) (C, exp) satisﬁes the countable closure property.

REMARK Assuming Schanuel’s Conjecture the axiom of Strong Exponential Closure for (C, exp) is the only impediment to prove Zilber’s Conjecture.

THEOREM (Mantova 2011) If (K, E) is a pseudo-exponential field of cardinality up to the continuum then there exists an involution σ on K, i.e. there is a field automorphism σ : K −→ K of order 2 such that σ ◦ E = E ◦ σ Exponential polynomials over C with finitely many roots

THEOREM (Henson and Rubel 1984) E Let f (X ) ∈ C[X ] .

g(X ) f (X ) has no solution in C iﬀ f (X ) = e

E where g(X ) ∈ C[X ] .

THEOREM (Katzberg 1983) E A non constant exponential polynomial f (z) ∈ C[z] has always inﬁnitely many zeros unless it is of the form

n1 nk g(z) f (z) = (z − α1) · ... · (z − αk ) e ,

E where α1, . . . , αk ∈ C, n1,..., nk ∈ N, and g(z) ∈ C[z] .

Both proofs use Nevanlinna Theory (C, exp) vs (K, E)

THEOREM (D’A., Macintyre and Terzo, 2010) Let (K, E) be a Zilber ﬁeld and f (X ) ∈ K[X ]E .

f (X ) has no solution in K iﬀ f (X ) = eg(X )

where g(X ) ∈ K[X ]E .

THEOREM (D’A., Macintyre and Terzo, 2010) A non constant exponential polynomial f (z) ∈ K[z]E has always inﬁnitely many zeros unless it is of the form

n1 nk g(z) f (z) = (z − α1) · ... · (z − αk ) e ,

E where α1, . . . , αk ∈ K, n1,..., nk ∈ N, and g(z) ∈ K[z] . Algebraic methods and Zilber’s axioms (C, exp) vs (K, E)

COROLLARY (Picard’s Little Theorem) Let f (x) ∈ K[x]E . If f (x) is non constant then f (x) cannot omit two values. Generic solutions

e (z) Notation. Let e0(z) = z, and for every k ∈ N, ek+1(z) = e k . Fix 1 ≤ k ∈ N, let x = (x0,..., xk ) and p(x) ∈ C[x]. Let f (z) = p(z, e1(z),..., ek (z)) over C.

DEFINITION A solution a of f (z) = 0 is generic over L (for L a ﬁnitely generated extension of Q containing the coeﬃcients of p) if

t.d.L(a, e1(a),..., ek (a)) = k,

where k is the number of iterations of exponentiation which appear in the polynomial p. Strong exponential closure - simplest case

THEOREM (Marker 2006) 1) If p(x, y) ∈ C[x, y] is irreducible and depends on x and y then f (z) = p(z, ez ) has inﬁnitely many zeros.

alg 2) (SC) If p(x, y) ∈ Q [x, y] then there are inﬁnitely many algebraically independent solutions over Q.

Proof 1) Existence of inﬁnitely many zeros follows from Hadamard Factorization theorem and Henson and Rubel’s result.

2) Schanuel’s Conjecture is crucial. Strong exponential closure - simplest case

THEOREM (Mantova 2016) (SC) If p(x, y) ∈ C[x, y] is irreducible and depends on x and y z z then there is z ∈ C such that p(z, e ) = 0 and tr.d.(z, e /K) = 1 for any ﬁnitely generated K ⊂ C. Proof uses some number theory results due to Zannier in order to show that there are only ﬁnitely many solutions of p in K alg .

Ideas due also to Gunaydin and Martin-Pizarro. Iterated exponentials

alg Question: Let p(x, y1,..., yn) ∈ Q [x, y1,..., yn] a nonzero irreducible polynomial depending on x and the last variable yn. Does ez z e... p(z, ez , ee ,..., ee )) = 0 have a generic solution?

REMARK Strong Exponential Closure in C would imply a positive answer. Iterated exponentials

THEOREM (D’A., Fornasiero and Terzo, 2017) alg (SC) Let p(x, y1, y2, y3) ∈ Q [x, y1, y2, y3] be a nonzero irreducible polynomial depending on x and y3. Then, there exists a generic solution of

z ez p(z, ez , ee , ee ) = 0.

Due to Katzberg’s result the polynomial has infinitely many solutions, unless it is of a certain form. No restrictions on the coefficients and no (SC). In the proof of the theorem we use a refinement of a result due to Masser on the existence of zeros of systems of exponential equations Masser’s result

THEOREM

Let P1(x),..., Pn(x) ∈ C[x], where x = (x1,..., xn), and Pi (x) are non zero polynomials in C[x]. Then there exist z1,..., zn ∈ C such that  z e 1 = P1(z1,..., zn)  z  e 2 = P2(z1,..., zn) (1) .  .  zn  e = Pn(z1,..., zn)

n n We have to show that the function F : C → C x xn F (x1,..., xn) = (e 1 − P1(x1,..., xn),..., e − Pn(x1,..., xn)) n has a zero in C . LEMMA (Kantorovich) n n Let F : C → C be an entire function, and p0 be such that J(p0), −1 the Jacobian of F at p0, is non singular. Let η = |J(p0) F (p0)| and U the closed ball of center p0 and radius 2η. Let M > 0 be such that |H(F )|2 ≤ M2 (where H(F ) denotes the Hessian of F ). −1 If 2Mη|J(p0) | < 1 then there is a zero of F in U.

LEMMA

Let P1(x),..., Pn(x) ∈ C[x], where x = (x1,..., xn) and d1,..., dn be the total degrees of P1(x),..., Pn(x), respectively. n There exists a constant c > 0 and an inﬁnite set S ⊆ Z such that n X dj |Pj (2πik1,..., 2πikn)| ≥ c(1 + |kl |) l=1 for all k = (k1,..., kn) ∈ S, j = 1,..., n. First example

ez Let f (z) = p(z, e ) where 0 6= p(x, y) ∈ C[x, y], and both x and y appear in p. We want to show that it has a generic solution in C. For this purpose we consider the corresponding system in four variables (z1, z2, w1, w2):

p(z , w ) = 0 V = 1 2 (2) w1 = z2 2 ∗ 2 thought of as an algebraic set V in G2(C) = C × (C ) . First example

THEOREM

alg (SC) If p(x, y) ∈ Q [x, y] then the variety V intersects the graph of exponentation in a generic point (w, ew , ew , eew ), (i.e. w w ew t.d.Q(w, e , e , e ) = 2 = dimV ). Proof: Wlog let w be a non-zero solution. Case 1: w and ew are linearly independent. Schanuel’s Conjecture implies: w w ew t.d.Q(w, e , e , e ) ≥ 2. Indeed, the transcendence degree is exactly 2 since w and eew are algebraically dependent. Hence, (w, ew , ew , eew ) ∈ V and w w ew t.d.Q(w, e , e , e ) = 2, which is the dimension of V , and so the point (w, ew , ew , eew ) is generic for V . w Case 2. w, e are linearly dependent over Q, i.e.

new = mw (3) for some m, n ∈ Z and (m, n) = 1, and w 6= 0 implies n 6= 0. Moreover, w is transcendental over Q, otherwise we have a contradiction with Lindemann Weierstrass Theorem. Applying exponentation to relation (3) it follows

w ene = emw , i.e. w m m (ee )n = (ew )m = ( w)m = ( )mw m. n n Subcase 2.1: If n, m > 0 then (w, eew ) is a root of m n q(x, y) = sx − y where s ∈ Q

Subcase 2.2: If n > 0 and m < 0 then (w, eew ) is a root of −m n q(x, y) = x y − r, where r ∈ Q

In both cases q(x, y) is irreducible, and it can be showed that p divides q. Since p and q are irreducible polynomials they diﬀer by a non-zero constant. So, wlog p(x, y) = sxm − y n (or p(x, y) = x−my n − r).

For any solution (w, eew ) of p(x, y) = 0 the linear dependence between w and ew is uniquely determined by the degrees of x and y in p, hence s is uniquely determined. Using a system a’ la Masser it is always possible to ﬁnd a point (w, ew , ew , eew ) of V with w, ew linearly independent. Indeed, we ez z want p(z, e ) = 0 and z 6= se , s ∈ Q.

It is enough to solve the following system a la Masser

 z  e = A(z, t, u) eu = B(z, t, u) (4)  et = C(z, t, u)

t t zn where A(z, t, u) = m , B(z, t, u) = m − sz, and C(z, t, u) = s . By Masser’s result there exists a solution of system (4) which gives a generic solution of the original polynomial, generic thanks to the second equation in (4) which guarantees that there is no linear dependence between a solution z and its exponential ez . Generalization of Masser’s result

n A cone is an open subset U ⊆ C s. t. for every 1 ≤ t ∈ R, if x ∈ U then tx ∈ U.

DEFINITION An algebraic function is an analytic function f : U → C s. t. there exists a nonzero polynomial q(x, u) ∈ C[x, u] with q(x, f (x)) = 0 on all x ∈ U. If, moreover, the polynomial q is monic in u, we say that f is integral algebraic.

THEOREM

Let f1,..., fn : U → C be nonzero algebraic functions, deﬁned on ∗ n n some cone U. Assume that U ∩ (2πiZ ) is Zariski dense in C . Then  z1 e = f1(z),  ... (5)

 zn  e = fn(z) has a solution a ∈ U. Second example

z ez Let f (z) = p(z, e , e ) ∈ C[x, y], and consider the corresponding system in four variables (z1, z2, w1, w2):

p(z , z , w ) = 0 V = 1 2 2 (6) w1 = z2

thought of as an algebraic set V in G2(C).

THEOREM

alg (SC) If p(x, y, z) ∈ Q [x, y, z], then the variety V deﬁned in (6) intersects the graph of exponentation in a generic point.

Proof: Let a 6= 0 be a root of f . If a and ea are linearly a a ea independent then as before by (SC) t.d.Q(a, e , e , e ) = 2, so that (a, ea, ea, eea ) is a generic point of V . Second example

a If e = ra for some r ∈ Q then a is necessarily transcendental by Lindemann-Weierstrass. a Call r ∈ Q “bad” if there exists a ∈ C solution of (6), s. t. e = ra. Using an argument similar to that of the previous case we can prove that there exist only ﬁnitely many “bad” r ∈ Q, {r1,..., rk }. We construct a generalization of the system a la Masser whose solution guarantees the existence of a point in V whose t.d. is 2. Second example

 z  e = f1(z, t, u1,..., uk )  t  e = f2(z, t, u1,..., uk )  u1 e = f3(z, t, u1,..., uk )   ...   uk  e = fk+2(z, t, u1,..., uk )

where f1 = t, f3 = t − r1z,... fk+2 = t − rk z, and f2 is the algebraic function which solves z in the original polynomial p(x, y, z) = 0. By Masser’s generalization the above system has a solution (b, eb, eb, eeb ) which turns out to be a generic solution for (6), since the last k equations guarantee that there is no linear dependence between b and eb. Three iterations

In an analogous way we can prove

z Let f (z) = p(z, ez , eez , eee ). The corresponding system in six variables (z1, z2, z3, w1, w2, w3) is:   p(z1, z2, z3, w3) = 0 V = w1 = z2 (7)  w2 = z3.

thought of as an algebraic set V in G2(C).

THEOREM (DFT ’17) alg (SC) If p(x, y, z, w) ∈ Q [x, y, z, w] then the variety V has a generic point. Strong exponential closure

We now consider the more general case of a variety and not a single polynomial.

THEOREM (DFT ’18) n ∗ n alg (SC) Let V ⊆ C × (C ) an irreducible variety over Q with dim V = n. If π1(V ) and π2(V ) are dominant, then there exists n a a a ∈ C such that (a, e ) ∈ V and (a, e ) is generic in V , i.e. a t.d.Q(a, e ) = dim V = n.

REMARK The result implies many cases of Zilber’s Conjecture Polynomials with arbitrary iterations of exponentiations

REMARK alg Let p(x1, y1,..., yn) ∈ Q [x1, y1,..., yn] a nonzero irreducible polynomial depending on x and the last variable yn. Let z ...e p(z, ez , eez ,..., eee ) = 0, the corrisponding system in 2n variables is: p(x , y ,..., y ) = 0 V = 1 1 n (8) xi+1 = yi for all i = 1,..., n − 1. V is a variety of dim V = n.

1 There exists a solution is unconditionally and no restriction on the coeﬃcients, just avoid polynomials of the form k1 kn p(x, y1,..., yn) = g(x)y1 · ... · yn , where g(x) ∈ C[x]. 2 There is a generic solution modulo (SC) Proof of Theorem DFT

Existence of solutions:

THEOREM (Masser Brownawell-DFT ’17) n ∗ n Let V ⊆ C × (C ) be an irreducible algebraic variety such that the projection onto the ﬁrst coordinates π1(V ) is Zariski dense in n n a n C . Then the set {a ∈ C :(a, e ) ∈ V } is Zariski dense in C .

REMARK This is unconditionally, no use of Schanuel’s conjecture, no restriction on the parameters deﬁning V . Proof of Theorem DFT

Existence of generic solutions:

1 Schanuel’s conjecture

2 Masser’s system Proof of genericity

a a Let (a, e ) ∈ V and suppose that t.d.Q(a, e ) = m < n, i.e. it is not generic. Without loss of generality we can assume

ea |Va| < ∞ and |V | < ∞

By Schanuel’s Conjecture

a l.d.(a) ≤ t.d.Q(a, e ) = m < n.

n n−m So, there exists M ∈ Z × Z such that M · a = 0. Applying exponentiation we have the relation eaM = 1. The above relations deﬁne:

M LM = {x : M · x = 0} and TM = {y : y = 1}. It turns out that LM is tangent to TM in 1 and so

m = dim LM = dim TM .

ea Moreover, |Va| and |V | < ∞ imply

a a e is algebraic over Q(a), and a is algebraic over Q(e ), i.e.

a a m = t.d.Q(a) = t.d.Q(a, e ) = t.d.Q(e ). Proof of genericity

a In other words a is generic in LM and e is generic in TM . We consider

y WN = {(x, y) ∈ V : x ∈ LN ∧ | Vx |< ∞∧ | V |< ∞},

n n−m a where N ∈ C × C . If N = M then (a, e ) ∈ WM . | Va |< ∞ implies

dim WM = dim π1(WM ) ≤ dim LM .

a Moreover (a, e ) ∈ WM so, dim WM ≥ dim LM . We have

dim WM = dim LM . 0 Let WM be the irreducible components of the Zariski closure of a WM containing the point (a, e ).

a 0 a 0 Since (a, e ) ∈ WM is generic and e ∈ π2(WM ) then 0 π2(WM ) ⊆ TM , so its Zariski closure is contained in TM . a a 0 Moreover, e is generic in TM and e ∈ π2(WM ), then we have

0 Zar TM = π2(WM ) . Proof of genericity

0 Zar 0 Zar We consider U = {π2(WM ) : π2(WM ) = TM } ∗ n Since U is a countable definable family in (C ) , and C is ω1-saturated then U is either finite or co-countable, and since it is countable then U is necessarily finite, i.e. U = {H1,..., Hl }. So, TM = Hi , for some i = 1,..., l. We can avoid such tori by adding finitely many inequalities in the Masser’s system which guarantees that the solution is a generic. Next goals:

alg 1) Eliminate the hypothesis that V is deﬁned over Q (work going on with Fornasiero, Gunaydin and Terzo)

2) Weaken the hypothesis that π1(V ) and π2(V ) are dominant.