Hardy Fields Santiago Camacho University of Illinois at Urbana-Champaign June 26, 2016 1/50 Germs at +∞ Let us consider the germs of real valued functions over R at ∞. That is,the equivalence classes of the equivalence relation E¯ given by f Eg¯ :⇔ ∃a ∈ R ∀r > a f (r) = g(r) We can give this set the operations of a ring, and we denote such a structure by G. We say that a germ g ∈ G is differentiable if it is the germ of some differentiable function. 2/50 Hardy Fields Definition A Hardy field is a subrfield K of G such that for any g ∈ K We have that g 0 ∈ K. Hardy fields come equipped with a natural ordering. Example I Any subring of R. I R(x) The field of rational functions. 3/50 Diferential fields Definition We say that K = (K; 0, 1, +, −, ×,−1 , ∂) is a differential field if (K; 0, 1, +, −, ×,−1 ) is a field and ∂ : K → K is an additive function satisfying the Liebniz rule. That is, f , g ∈ K 1. ∂(a + b) = ∂(a) + ∂(b) 2. ∂(ab) = ∂(a)b + a∂(b) As usual if understood by context we write f 0 := ∂(f ) and f † := f 0/f 4/50 Hardy fields as differential fields By definition Hardy fields are differential fields when we consider ∂ to be the previously mentioned derivation.Note that for f in a Hardy field we have, f † = log(f )0. Thus f † is known as the logarithmic derivative. 5/50 Simple Hardy field Extensions Theorem If K is a Hardy field F ∈ K[X] and g ∈ G is such that F (g) = 0, then K[g] is a Hardy field. Theorem Let K be a Hardy field, F (X), G(X) ∈ K[X] and g ∈ G differentiable such thatG(g) is eventually different from 0 and F (g) g 0 = . Then K(g) is a Hardy field. G(g) Thus; R(x R), R(x, exp), R(x, log), R(x, exp, log), R(x R, exp, log) are Hardy fields. 6/50 f (x + 1) = ef (x) Two results on Extending Hardy fields Theorem (Boshernitzan) There exist Hardy fields with transexponential functions. Moreover there are Hardy fields that are not bounded by the germ of any continuous function. Theorem (Rolin, Sanz y Schaefke) There exist Hardy fields that are not the Hardy field of an o-minimal expansion of R. 7/50 Two results on Extending Hardy fields Theorem (Boshernitzan) There exist Hardy fields with transexponential functions. Moreover there are Hardy fields that are not bounded by the germ of any continuous function. f (x + 1) = ef (x) Theorem (Rolin, Sanz y Schaefke) There exist Hardy fields that are not the Hardy field of an o-minimal expansion of R. 7/50 Dominance Relations Less common but more intuitive for our cases of study are dominance relations. Definition A dominance relation over a field K is a binary relation 4 over K such that for all f , g, h ∈ K; DR11 46 0, DR2 f 4 f , DR3 f 4 g, g 4 h ⇒ f 4 h, DR4 f 4 g or g 4 f , DR5 f 4 g ⇒ hf 4 hg, DR6 f 4 h, g 4 h ⇒ f − g 4 h We write f ≺ g : ⇐⇒ g 46 f 8/50 Equivalence classes on fields with dominance relations For f , g ∈ K × we write f g : ⇐⇒ (f 4 g & g 4 f ), and f ∼ g : ⇐⇒ f − g ≺ f Note that and ∼ are equivalence relations and ∼ is a refinement of . Moreover K ×/ has the structure of an abelian group. 9/50 The dominance relation on Hardy fields We can define a natural dominance relation on Hardy fields, 4, that comes from the natural ordering ¡. We define f 4 g :⇔ ∃n such that |f | ≤ n|g|. From the definition we obtain f f g ⇐⇒ = r + ε for r ∈ × g R f f ∼ g ⇐⇒ = 1 + ε g 10/50 Examples 2 2 2 2 I x 4x + x but x 6∼ 4x + x x 5 3 1 −x I e x + x log(x) 1 x e x x 2 x I 2xe 2xe + x + log(x) ∼ 2xe + 5 11/50 Valued Fields Definition We say that (K, v) is a valued field if there is an ordered group Γ where v : K × → Γ is a surjective function such that for f , g ∈ K; (V1) if f + g 6= 0, then v(f + g) ≥ min{v(f ), v(g)}, (V2) v(fg) = v(f ) + v(g) Three structures rise up naturally from this valuation. I O = {f ∈ K : v(f ) ≥ 0} the valuation ring of K, I O = {f ∈ K : v(f ) > 0} the maximal ideal of O, I k = O/O the residue field. 12/50 The relation between dominance relations and valuations Given a dominance relation in a field K we can define a valuation ring of K given by O = {f ∈ K|f 4 1}. We can take Γ := K ×/ With the reverse order with respect to the one suggested by 4. That is f < g : ⇐⇒ g 4 f Valued field theory has been studied at large. 13/50 The dominance relation and ultrametric Given a field K with a dominance relation 4 we can think of K, 4 as an ultrametric space. Here the group K/ plays the role of R, where the distance between two elements of K can be interpreted as the asymptotic class of their difference. Ultrametric spaces can be thought of as spaces of isosceles triangles with small bases. 14/50 Asymptotic couples Definition An asymptotic couple is a pair (Γ, ψ) where Γ is an ordered abelian group and ψ : γ6= → γ such that for all α, β ∈ Γ6= (AC1) α + β 6= 0 =⇒ ψ(α + β) ≥ min{ψ(α), ψ(β)}; 6= (AC2) For all k ∈ Z we have that ψ(kα) = ψ(α); (AC3) α > 0 =⇒ ψ(β) < α + ψα If moreover (Γ, ψ) satisfies (HC)0 < α < β =⇒ ψ(β) ≤ ψ(α) We say that (Γ, ψ) es of H type. 15/50 Asymptotic fields Definition We say that a valued differential field K is asymptotic if for all f , g ∈ K ≺1 we have that 0 0 f 4 g ⇐⇒ f 4 g We say that K is H-asymptotic, or of H type if moreover † † f ≺ g ≺ 1 ⇒ g 4 f 16/50 A characterization of Asymptotic Fields Theorem Let K be a valued differential field. The following are equivalent. 1. K is an asymptotic field; 2. There is an asymptotic couple (Γ, ψ) and a valuation v : K × → Γ such that ψ(v(f )) = v(f )†. 17/50 Asymptotic couples Definition An asymptotic couple is a pair (Γ, ψ) where Γ is an ordered abelian group and ψ : γ6= → γ such that for all α, β ∈ Γ6= (AC1) α + β 6= 0 =⇒ ψ(α + β) ≥ min{ψ(α), ψ(β)}; 6= (AC2) For all k ∈ Z we have that ψ(kα) = ψ(α); (AC3) α > 0 =⇒ ψ(β) < α + ψα.(Ψ < (Γ>)0) If moreover (Γ, ψ) satisfies (HC)0 < α < β =⇒ ψ(β) ≤ ψ(α) We say that (Γ, ψ) es of H type.( ψ is a convex valuation on Γ) 18/50 Hardy Fields are Asymptotic Fields Theorem Let H be a Hardy field f , g 6 1. Then 0 0 f 4 g ⇐⇒ f 4 g . L’Hospital’s Rule! 19/50 Hardy fields are of H type Theorem × Let H be a Hardy field, and f , g ∈ H such that 1 ≺ g 4 f . Then † † g 4 f . Proof. Assume that f , g are positive. g < nf for some n ∈ N log(g) < log(f ) + log(n) log(g) 4 log(f ) 0 0 log(g) 4 log(f ) 20/50 Asymptotic couples Trichotomy Theorem Given an asymptotic couple of H type (Γ, ψ), one and only one of the following holds; 1.Ψ has a maximum element; 2. There exists (a unique) β ∈ Γ such that Ψ < β < (Γ>)0. 3.(Γ , ψ) Has asymptotic integration. 21/50 Logarithmic Exponential Field Definition A exponential field is a tuple (K, E), where K is an ordered field and E is a strictly increasing group homomorphism from E :(K, +) → (K >0, ×) In principle E need not be surjective. but if E(K) = K >0 we call (K, E) a logarithmic-exponential field. A familiar example is (R, exp) 22/50 Hahn Field Let M be an ordered monomial multiplicative group and K an ordered field. Consider the set {f ∈ KM : supp(f ) is noetherian }. We will write its elements as X X f (m)m = fmm, m∈M m∈M We will denote by Lm(f ) = max{supp(f )} the leading monomial of f and by Lc(f ) = fLm(f ) its leading coefficient. 23/50 We endow the above mentioned set with pointwise addition X X X fmm + gmm = (fm + gm)m, m∈M m∈M m∈M and product given by ! X X X X fmm · gmm = (fm1 gm2 ) m m∈M m∈M m∈M m1m2=m With this operations we have constructed a ring, K[[M]]. One checks that in general it is a field. Together with the ordering f > 0 ⇔ Lc(f ) >K 0 we have constructed an ordered field. 24/50 Given a family (fi )i∈I with fi ∈ K[[M]], it makes sense to talk about X fi i∈I as long as for each m ∈ M the set {i ∈ I : m ∈ supp(fi )} is finite, S and supp(fi ) is noetherian. 25/50 Example k k k Let M = x Z = {x : k ∈ Z} with the order x 1 > x 2 ⇔ k1 > k2 K = .