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

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 = . R ! x −n Then [[M]] = ((x −1)), and for the family we have R R n! n∈N x −n X ∈ [[x Z]]. n! R n∈N −1 If we are to get a logarithmic exponential extention of R((x )) we would have for r ∈ R

E(log(rx)) = x r .

Hence it is natural to start with M = x R ordered in the natural way.

26/50 Pre-exponential ﬁeld

n X f In [[x R]] it is natural to define E(f ) = but only when R n! n∈N Lm(f ) < 1 = x 0. The same construction does not work for f = x. Definition (K, A, B, E) is called a pre-exponential field if 1. K is an ordered field, 2. A is an additive subgroup, 3. B is a convex subgroup, 4. K = A ⊕ B, 5. E : B → K >0 is an order-preserving group homomorphism.

27/50 Example

2 3/2 4/3 −π −4 K = R[[x R]] e.g. x + x + x + ··· + x + 3 + 2x + x + ...

B = {f ∈ K : Lm(f ) ≤ 1} ∪ {0} e.g. 3 + 2x −π + x −4 + ...

A = {f ∈ K : supp(f ) > 1} ∪ {0} e.g. x 2 + x 3/2 + x 4/3 + ··· + x

>0 P n E : B → K with E(r + ) = exp(r) n! . Note that B in this case is a local ring.

28/50 We need to extend E to all of K = A ⊕ B, so we need to ﬁgure out what E is on A.

Consider E(A) = {E(a): a ∈ A} where E(a)E(b) = E(a + b) and E(a) < E(b) if a < b. A multiplicative copy of A. in a way that

EA : A → E(A), where EA(a) = E(a), is an order preserving group homomorphism from A to E(A).

Note that E(A) is a monomial multiplicative ordered group.

29/50 The ﬁrst pre-exponential extension

By setting K 0 = K[[E(A)]],

B0 = {f ∈ K 0 : Lm(f ) ≤ 1 = E(0)} ∪ {0},

A0 = {f ∈ K 0 : supp(f ) > 1 = E(0)} ∪ {0}, and

E 0 : B0 −→ K 0>0 deﬁned by n E 0(a + b + ) = E (a)E(b) X A n! n∈N

we get that (K 0, A0, B0, E 0) is a pre-exponential ﬁeld extension of (K, A, B, E) where K ⊆ Dom(E 0).

30/50 31/50 Unfortunately E 0 is not deﬁned everywhere. For example what would E 0(E 0(x)) be? We shall go to inﬁnity! (a particular construction actually goes beyond but this lies out of the current scope). With K0 = K = R[[x R]] we construct

K1 = K0[[E(A)]]

K2 = K1[[E(A1)]] . .

Note that Kn+1 = Kn[[E(An)]] = Kn−1[[E(An−1)]][[E(An)]] can be identiﬁed with Kn−1[[E(An−1)E(An)]]

32/50 Exponential Transseries

E [ T = Kn n E E Clearly for every element f ∈ T we get E(f ) ∈ T . E But T is not a logarithmic-exponential ﬁeld! Lemma For f ∈ Kn, f > 0 E f ∈ E(T ) if and only if Lm(f ) ∈ E(A0) ··· E(An−1)

33/50 Characterizing E(TE )

Lemma For f ∈ Kn, f > 0 E f ∈ E(T ) if and only if Lm(f ) ∈ E(A0) ··· E(An−1) Proof. E (⇒) Let g ∈ Kp ⊆ T . So g = α + s + δ for α ∈ A0 ⊕ · · · ⊕ Ap−1, P δn s ∈ R, δ ∈ mp. E(g) = E(α) exp(s) n! . (⇐) Assume Lm(f ) = E(α) and Lc(f ) = c > 0. Then f = E(α)c(1 − ) for some ∈ mn. P n So f = E(α + log(c) − n ).

34/50 The substitution map Φ

E E Consider Φ : T → T deﬁned as follows. X f E(x)r for f = X f x r , f ∈  r r r R r∈ Φ(f ) = XR X  Φ(fa)E(Φ(a)) for f = faE(a), fa ∈ Kn−1  a∈An−1

Basically it behaves by precomposing with E. E Note that Φ embeds T into itself. ,

35/50 The ﬁeld of Transseries

We construct a chain of ﬁelds E E T = L0 ⊂ L1 ⊂ ... , and isomorphisms ηn : Ln → T such that for z ∈ Ln we have ηn+1(z) = Φ(ηn(z)), and η0 = idTE . S We set T = Ln.

In this way by using the previous lemma we get for f ∈ Ln, f > 0 a g ∈ Ln+1 such that E(ηn+1(g)) = ηn+1(f ). Thus the ﬁeld of Transseries, T, is a logarithmic-exponential ﬁeld. For further use we set `n to be the element in Ln that maps to x under ηn. (Think of an iterated log function).

36/50 A derivation on TE

Now we are interested in adding a derivation to make T into a differential field. E We start by defining a derivation on T .

P r For f = r∈R fr x ∈ K0, all fr ∈ R, we set 0 P r−1 D(f ) = f = r∈R rfr x .

P For f = a∈An−1 faE(a) ∈ Kn, we set 0 P 0 0 D(f ) = f = a∈An−1 (fa + faa )E(a).

37/50 The derivation on T

E Observe that for f ∈ T we have 1.Φ( f )0 = Φ(f 0)E(x) m 0 m 0 Qm i 2.Φ (f ) = Φ (f ) i=1 E (x) 1 −1 Define D : L → L by z 7→ n η ◦ D ◦ η (z). n n n Q ` n n i=1 i Clearly D0 = D and one can check that for f ∈ Ln+1, we get S Dn(f ) = Dn+1(f ) so Dn+1 extends Dn. By taking ∂ = Dn, we get that (T, ∂) is a differential field.

38/50 Integration

E 0 E For f ∈ T the equation y = f has a solution in T provided that x −1 ∈/ supp(f ). P r P fr r+1 Indeed, for f = r∈R fr x we have that g = r∈ \{−1} r+1 x is P R P a solution. For f = a∈An faE(a), we have that g = a∈An gaE(a) 0 0 is a solution provided that ga + a ga = fa for each a.

39/50 Integration

0 0 0 Hence there is a solution for y = f , if x + a x = fa has a solution. Note that the solutions to this equation are the same as those for

y 0 y − − = f , a0 a which has the solution

X n y = P (fa) n∈N where P is a small operator.

40/50 1/2 ex+x +··· x+x 1/2 3 2 −1 −3/2 f|x −1 = e + e + ··· + x + x + 3+x + x + ··· So we get an initial segment. Formally X f |m = fnn n>m

Truncation

So what’s this about?

41/50 |x −1 So we get an initial segment. Formally X f |m = fnn n>m

Truncation

So what’s this about? 1/2 f = eex+x +··· + ex+x 1/2 + ··· + x 3 + x 2 + 3+x −1 + x −3/2 + ···

41/50 +x −1 + x −3/2 + ··· So we get an initial segment. Formally X f |m = fnn n>m

Truncation

So what’s this about? 1/2 ex+x +··· x+x 1/2 3 2 f|x −1 = e + e + ··· + x + x + 3

41/50 +x −1 + x −3/2 + ···

Formally X f |m = fnn n>m

Truncation

So what’s this about? 1/2 ex+x +··· x+x 1/2 3 2 f|x −1 = e + e + ··· + x + x + 3 So we get an initial segment.

41/50 +x −1 + x −3/2 + ···

Truncation

So what’s this about? 1/2 ex+x +··· x+x 1/2 3 2 f|x −1 = e + e + ··· + x + x + 3 So we get an initial segment. Formally X f |m = fnn n>m

41/50 Truncation closed subsets

Let H be a Hahn ﬁeld. We call A ⊂ H truncation closed if for any f ∈ A and any monomial m we have f |m ∈ A. We are interested in truncation closed subsets since they appear naturally as images of germs of ﬁelds of functions under maps into generalized power series.

42/50 Known results for Hahn ﬁelds Algebraic extensions

Let H be a Hahn ﬁeld and A ⊆ H truncation closed, then

I The ring generated by A and the ﬁeld generated by A are truncation closed.

I If A is a ﬁeld then the henselization of A in H is truncation closed.

I If A is a henselian ﬁeld, any algebraic extention of A inside H is truncation closed.

43/50 Known results for Hahn ﬁelds Transcendental extensions

Theorem Let F = (Fn)n be family, where Fn ⊆ K[[Y1,..., Yn]] is closed under taking partial derivatives. Let K be a truncation closed subfield of H containing K. Then the F-extension of K is truncation closed. Here the F-extension of K is the field generated by K together with the elements of the form F (f1,..., fn) for F ∈ Fn and f1,..., fn ∈ K. Corollary The F-closure of a truncation closed subfield of H is truncation closed.

44/50 tL-closure E a notion of truncation in T

Deﬁnition E Let f ∈ T the E-support of f

r suppE (f ) = {E(a): ∃r ∈ R x E(a) ∈ supp(f )}

Deﬁnition E Let S ⊆ T . We say that S is L-closed if for each E(a) ∈ suppE (S) we have that E(a), a ∈ S. Deﬁnition E We call S ⊆ T tL-closed if it is both truncation closed and L-closed.

45/50 Examples

If 2 2 f = 7eex +x+···+x 2+4x + 5x −1 + ··· + 2e−x + e−ex , then 2 2 ex +x+···+x 2+4x −x −ex suppE (f ) = {e , 1, e , e }

S = {5eex , eex , ex , x, 3x 2e−x 4 , e−x 4 , −x 4} is L-closed.

46/50 Some immediate results

Let M(S) correspond to the multiplicative monoid generated by the support of S. If S is tL-closed then: I RS ∪ M(S) is tL-closed; I R[S] is tL-closed; I R(S) is tL-closed; h I R(S) is tL-closed; rc I R(S) is tL-closed.

47/50 More interesting results

Theorem E If S is a truncation closed subset of T , then the differential ring generated by S over R, R{S} is tL-closed. From this it follows easily that RhSi is tL-closed. Theorem E If K is a truncation closed subfield of T then the smallest field containing K and all elements of the form E(f ) for f ∈ K is tL-closed.

48/50 A few other open questions

I Are there any Hardy ﬁelds with inﬁnite functions that grow slower than any iteration of the logarithm?

I Are there asymptotic couples that do not come from an asymptotic ﬁeld?

49/50 Bibliography

Antongiulio Fornasiero, Embedding henselian ﬁelds into power series, J. Algebra 304 (2006), no. 1, 112–156. MR 2255823 (2007f:12009) Lou van den Dries, Truncation in hahn ﬁelds. Lou van den Dries, Angus Macintyre, and David Marker, Logarithmic-exponential power series, Annals of Pure and Applied Logic 11 (2001), 1–11356.

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