L.8

# Surreal fields stable under exponential, logarithmic, derivative and anti-
derivative functions

Olivier<EMAIL_ADDRESS>École Polytechnique, LIX,
91128 Palaiseau Cedex, France
This work was partially supported by ANR Project $\partial$IFFERENCE. Quentin
<EMAIL_ADDRESS>École Polytechnique, LIX, 91128
Palaiseau Cedex, France
This work was partially supported by ANR Project $\partial$IFFERENCE.

###### Abstract

The class of surreal numbers, denoted by No, initially proposed by Conway, is
a universal ordered field in the sense that any ordered field can be embedded
in it. They include in particular the real numbers and the ordinal numbers.
They have strong relations with other fields such as field of transseries.
Following Gonshor, surreal numbers can be seen as signs sequences of ordinal
length, with some exponential and logarithmic functions that extend the usual
functions over the reals. No can actually be seen as an elegant (generalized)
power series field with real coefficients, namely Hahn series with exponents
in No itself.

Some years ago, Berarducci and Mantova considered derivation over the surreal
numbers, seeing them as germs of functions, in correspondence to transseries.
In this article, following our previous work, we exhibit a sufficient
condition on the structure of a surreal field to be stable under all
operations among exponential, logarithm, derivation and anti-derivation.
Motivated, in the long term, by computability considerations, we also provide
a non-trivial application of this theorem: the existence of a pretty
reasonable field that only requires ordinals up to $\varepsilon_{\omega}$,
which is far smaller than $\omega_{1}^{\textrm{CK}}$ (resp. $\omega_{1}$), the
first non-computable (resp. uncountable) ordinal.

##  1 Introduction

Conway introduced in [Con00] the class of surreal numbers. They were later on
popularized by Knuth [Knu74], and then formalized later on by Gonshor [Gon86],
and by many other authors. The general initial idea is to define a class of
numbers, based on a concept of “simplicity”. This permits to obtain a real
closed field that both contains the real numbers and the ordinals, as this
provides an unification of Dedekind’s construction of real numbers in terms of
cuts of the rational numbers, and of von Neumann’s construction of ordinal
numbers by transfinite induction in terms of set membership.

Following the alternative presentation from Gonshor in [Gon86], a surreal
number can also be seen as an ordinal-length sequence over $\\{+,-\\}$, that
we call a signs sequence. Basically, the idea is that such sequences are
ordered lexicographically, and have a tree-like structure. Namely, a $+$
(respectively $-$) added to a sequence $x$ denotes the simplest number greater
(resp. smaller) than $x$ but smaller (resp. greater) than all the prefixes of
$x$ which are greater (resp. smaller) than $x$. With this definition of
surreal numbers, it is possible to define operations such as addition,
substraction, multiplication, division, obtaining a real closed field.
Following Gonshor [Gon86], based on ideas from Kruskal, it is also possible to
define consistently classical functions such as the exponential function and
the logarithmic function over No, and to do analysis of this fields of
numbers.

It can be considered as “the” field that includes “all numbers great and
small” [Ehr12]. In particular, any divisible ordered Abelian group is
isomorphic to an initial subgroup of No, and any real closed field is
isomorphic to an initial subfield of No [Ehr01, Theorems 9 and 19], [Con00,
Theorems 28 and 29]. This leads to the fact that it can be considered as “the”
field that includes “all numbers great and small” [Ehr12]. No can also be
equipped with a derivation, so that it can be considered as a fields of
transseries [BM18a]. See example [MM17] for a survey of fascinating recent
results in all these directions.

More concretely, No can also be seen as a field of (generalized) power series
with real coefficients, namely as Hahn series where exponents are surreal
numbers themselves. More precisely, write
$\mathbb{K}\left(\left(G\right)\right)$ for the set of Hahn series with
coefficients in $\mathbb{K}$ and terms corresponding to elements of $G$, where
$\mathbb{K}$ is a field, and $G$ is some divisible ordered Abelian group: This
means that $\mathbb{K}\left(\left(G\right)\right)$ corresponds to formal power
series of the form $s=\sum_{g\in S}a_{g}t^{g}$, where $S$ is a well-ordered
subset of $G$ and $a_{g}\in\mathbb{K}.$ The support of $s$ is
$\operatorname{supp}(s)=\left\\{\left.g\in S\ \vphantom{a_{g}\neq 0}\right|\
a_{g}\neq 0\right\\}$ and the length of the serie of $s$ is the order type of
$\operatorname{supp}(s)$. The field operations on
$\mathbb{K}\left(\left(G\right)\right)$ are defined as expected, considering
elements of $\mathbb{K}\left(\left(G\right)\right)$ as formal power series. We
have $\textnormal{No}=\mathbb{R}\left(\left(\textnormal{No}\right)\right)$.

Our previous work [BG22] has shown that some acceptable subfields of No are
stable by both exponential and logarithm. This fields are built with some
restriction on ordinals allowed in the ordinal sum. In the current article, we
pursue the work by exhibiting some acceptable subfields of No are stable by
both exponential and logarithm, derivation and anti-derivation. We actually
provide some sufficient condition on the structure of a surreal field to be
stable under all these operations among exponential, logarithm, derivation and
anti-derivation, and use these to derive such subfields.

##### More precise statements

Given some ordinal $\gamma$ (or more generally a class of ordinals), we write
$\mathbb{K}\left(\left(G\right)\right)_{\gamma}$ for the restriction of
$K\left(\left(G\right)\right)$ to formal power series whose support has an
order type in $\gamma$ (that is to say, corresponds to some ordinal less than
$\gamma$). We have of course
$\textnormal{No}=\mathbb{R}\left(\left(\textnormal{No}\right)\right)_{\textnormal{Ord}}$.
In this point of view, $\varepsilon$-numbers, i.e ordinals $\lambda$, such
that $\omega^{\lambda}=\lambda$, play a major role as they are such limit on
ordinal we can accept in our fields.

As we will often play with exponents of formal power series considered in the
Hahn series, we propose to introduce the following notation: We denote

$\mathbb{R}_{\lambda}^{\Gamma}=\mathbb{R}\left(\left(\Gamma\right)\right)_{\lambda}$

when $\lambda$ is an $\varepsilon$-number and $\Gamma$ a divisible Abelian
group.

As a consequence of MacLane’s theorem (Theorem 3.1 below from [Mac39], see
also [All87, section 6.23]), we know that
$\mathbb{R}_{\lambda}^{\textnormal{No}_{\mu}}$ is a real-closed field when
$\mu$ is a multiplicative ordinal (i.e. $\mu=\omega^{\omega^{\alpha}}$ for
some ordinal $\alpha$) and $\lambda$ an $\varepsilon$-number.

Furthermore:

######  Theorem 1.1 ([P]roposition 4.7).

DriesEhrlich01] Let $\lambda$ be an $\varepsilon$-number. Then

  1. 1.

The field $\textnormal{No}_{\lambda}$ can be expressed as

(1)
$\textnormal{No}_{\lambda}=\bigcup_{\mu}\mathbb{R}_{\lambda}^{\textnormal{No}_{\mu}},$

where $\mu$ ranges over the additive ordinals less than $\lambda$
(equivalently, $\mu$ ranges over the multiplicative ordinals less than
$\lambda$ ).

  2. 2.

$\textnormal{No}_{\lambda}$ is a real closed subfield of No, and is closed
under the restricted analytic functions of No.

  3. 3.

$\textnormal{No}_{\lambda}=\mathbb{R}_{\lambda}^{\textnormal{No}_{\lambda}}$
if and only if $\lambda$ is a regular cardinal.

Actually, even if we always can write $\textnormal{No}_{\lambda}$ as an
increasing union of fields by Equation (1), and even if
$\textnormal{No}_{\lambda}$ is stable under exponential and logarithmic
functions (Theorem 5.4) none of the fields in this union has stability
property beyond the fact that they are fields. Indeed:

######  Proposition 1.1 ([BG22, Proposition 1.5]).

$\mathbb{R}_{\lambda}^{\textnormal{No}_{\mu}}$ is never closed under
exponential function for $\mu<\lambda$ a multiplicative ordinal.

In our previous work [BG22], we studied stability of subfields of No by
exponential and logarithm. For the sake of effectiveness and representations
for ordinal Turing machines, we kept ordinals as small as possible to identify
natural subfields stable by both these functions. This paper will keep the
same spirit and we will give an example construction that only involves
ordinals up to $\varepsilon_{\omega}$, which is much smaller than
$\omega_{1}$, the first uncountable ordinal, and even
$\omega_{1}^{\textrm{CK}}$, the first non-computable ordinal.

To achieve that purpose, we will have to handle carefully the
$\varepsilon$-numbers that are involved. Recall that there is some enumeration
$(\varepsilon_{\alpha})_{\alpha\in\textnormal{Ord}}$of $\varepsilon$-numbers:
Any $\varepsilon$-number ordinal $\lambda$ is $\varepsilon_{\alpha}$ for some
ordinal $\alpha$.

######  Definition 1.1 (Canonical sequence defining an $\varepsilon$-number).

Let $\lambda$ be an $\varepsilon$-number. Ordinal $\lambda$ can always be
written as $\lambda=\sup\left(e_{\beta}\right)_{\beta<\gamma_{\lambda}}$ for
some canonical sequence, where $\gamma_{\lambda}$ is the length of this
sequence, and this sequence is defined as follows:

  * •

If $\lambda=\varepsilon_{0}$ then we can write
$\varepsilon_{0}=\sup\\{\omega,\omega^{\omega},\omega^{\omega^{\omega}},\dots\\}$
and we take$\omega,\omega^{\omega},\omega^{\omega^{\omega}},\dots$ as
canonical sequence for $\varepsilon_{0}$. Its length is $\omega$, and for
$\beta<\lambda$, $e_{\beta}$ is $\omega^{\iddots^{\omega}}$ where there are
$\beta$ occurrences of $\omega$ in the exponent.

  * •

If $\lambda=\varepsilon_{\alpha}$, where $\alpha$ is a non-zero limit ordinal,
then we can write $\lambda=\underset{\beta<\alpha}{\sup}\varepsilon_{\beta}$
and we take $\left(\varepsilon_{\beta}\right)_{\beta<\alpha}$ as the canonical
sequence of $\lambda$. Its length is $\alpha$ and for $\beta<\alpha$,
$e_{\beta}=\varepsilon_{\beta}$.

  * •

If $\lambda=\varepsilon_{\alpha}$, where $\alpha$ is a successor ordinal, then
we can write

$\lambda=\sup\\{\varepsilon_{\alpha-1},{\varepsilon_{\alpha-1}}^{\varepsilon_{\alpha-1}},{\varepsilon_{\alpha-1}}^{{\varepsilon_{\alpha-1}}^{\varepsilon_{\alpha-1}}},\dots\\}$

and we take
$\varepsilon_{\alpha-1},{\varepsilon_{\alpha-1}}^{\varepsilon_{\alpha-1}},{\varepsilon_{\alpha-1}}^{{\varepsilon_{\alpha-1}}^{\varepsilon_{\alpha-1}}},\dots$
as the canonical sequence of $\lambda$. Its length is $\omega$, and for
$\beta<\omega$,
$e_{\beta}={\varepsilon_{\alpha-1}}^{\iddots^{\varepsilon_{\alpha-1}}}$ where
there are $\beta$ occurrences of $\varepsilon_{\alpha-1}$ in the exponent.

For example, the canonical sequence defining $\varepsilon_{1}$ is
$\varepsilon_{0},{\varepsilon_{0}}^{\varepsilon_{0}},{\varepsilon_{0}}^{{\varepsilon_{0}}^{\varepsilon_{0}}},\dots$,
the canonical sequence defining $\varepsilon_{\omega}$ is
$\varepsilon_{0},\varepsilon_{1},\varepsilon_{2},\dots$, the canonical
sequence of $\varepsilon_{\omega 2}$ is
$\varepsilon_{0},\varepsilon_{1},\varepsilon_{2},\dots,\varepsilon_{\omega},\varepsilon_{\omega+1},\dots$
and the canonical sequence of $\varepsilon_{\omega 2+1}$ is
${\varepsilon_{\omega 2}},{\varepsilon_{\omega 2}}^{\varepsilon_{\omega
2}},{\varepsilon_{\omega 2}}^{{\varepsilon_{\omega 2}}^{\varepsilon_{\omega
2}}},\dots$

######  Definition 1.1.

Let $\Gamma$ be an Abelian subgroup of No and $\lambda$ be an
$\varepsilon$-number whose canonical sequence is
$\left(e_{\beta}\right)_{\beta<\gamma_{\lambda}}$. We denote
$\Gamma^{\uparrow\lambda}$ for the family of group
$\left(\Gamma_{\beta}\right)_{\beta<\gamma_{\lambda}}$ defined as follows:

  * •

$\Gamma_{0}=\Gamma$;

  * •

$\Gamma_{\beta+1}$ is the group generated by the groups $\Gamma_{\beta}$,
$\mathbb{R}_{e_{\beta}}^{g\left((\Gamma_{\beta})^{*}_{+}\right)}$ and the set
$\left\\{\left.h(a_{i})\
\vphantom{\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\in\Gamma_{\beta}}\right|\
\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\in\Gamma_{\beta}\right\\}$
where $g$ and $h$ are Gonshor’s functions associated to exponential and
logarithm (see Section 5 below for some details);

  * •

For a limit ordinal number $\beta$,
$\Gamma_{\beta}=\underset{\gamma<\beta}{\overset{}{\bigcup}}\Gamma_{\gamma}$.

When considering a family of set $(S_{i})_{i\in I}$, we denote

$\mathbb{R}_{\lambda}^{(S_{i})_{i\in I}}=\underset{i\in
I}{\overset{}{\bigcup}}\mathbb{R}_{\lambda}^{S_{i}}$

In particular,
$\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}=\underset{i<\gamma_{\lambda}}{\overset{}{\bigcup}}\mathbb{R}_{\lambda}^{\Gamma_{i}}$

######  Remark 1.1.

By construction, if $\Gamma\subseteq\Gamma^{\prime}$ then
$\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}\subseteq\mathbb{R}_{\lambda}^{{\Gamma^{\prime}}^{\uparrow\lambda}}$.

The idea behind the definition of $\Gamma^{\uparrow}$ is that at step $i+1$ we
add new elements to close $\mathbb{R}_{\lambda}^{\Gamma_{i}}$ under
exponential and logarithm. The reason why we add
$\mathbb{R}_{e_{\beta}}^{g\left((\Gamma_{\beta})^{*}_{+}\right)}$ to
$\Gamma_{\beta}$ rather than
$\mathbb{R}_{\lambda}^{g\left((\Gamma_{\beta})^{*}_{+}\right)}$ is that we
want to keep control on what we add in the new group. In our previous work
[BG22] we came up with the following three statements:

######  Lemma 1.1 ([BG22, Lemma 5.7]).

Write
$\Gamma^{\uparrow\lambda}=\left(\Gamma_{\beta}\right)_{\beta<\gamma_{\lambda}}$,
and let

$L=\left\\{\left.\exp_{n}x,\ln_{n}x\
\vphantom{\begin{array}[]{c}x\in\mathbb{L},\ n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}}\right|\
\begin{array}[]{c}x\in\mathbb{L},\ n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}\right\\}$

we have for all $i<\gamma_{\lambda}$,

$L=\left\\{\left.\exp_{n}x,\ln_{n}x\
\vphantom{\begin{array}[]{c}x\in\mathbb{L},n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma_{i}}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}}\right|\
\begin{array}[]{c}x\in\mathbb{L},n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma_{i}}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}\right\\}$

######  Corollary 1.1 ([BG22, Corollary 5.8]).

Let $\Gamma$ be an Abelian additive subgroup of No and

$L=\left\\{\left.\exp_{n}x,\ln_{n}x\
\vphantom{\begin{array}[]{c}x\in\mathbb{L},n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}}\right|\
\begin{array}[]{c}x\in\mathbb{L},n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma}\ \exists P\in\mathcal{P}(y)\ \exists
k\in\mathbb{N}\quad P(k)=x\end{array}\right\\}$

Then,

$L=\left\\{\left.\exp_{n}x,\ln_{n}x\
\vphantom{\begin{array}[]{c}x\in\mathbb{L},\quad n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}\ \exists
P\in\mathcal{P}(y)\ \exists k\in\mathbb{N}\quad P(k)=x\end{array}}\right|\
\begin{array}[]{c}x\in\mathbb{L},\quad n\in\mathbb{N},\\\ \exists
y\in\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}\ \exists
P\in\mathcal{P}(y)\ \exists k\in\mathbb{N}\quad P(k)=x\end{array}\right\\}$

######  Theorem 1.2 ([T]heorem 1.10).

bournez2022surreal] Let $\Gamma$ be an Abelian subgroup of No and $\lambda$ be
an $\varepsilon$-number, then
$\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}$ is stable under exponential
and logarithmic functions.

With such a notion, we managed to make a link between the two types of field
involved in Theorems 5.4 and 1.2. More precisely, the fields
$\mathbb{R}_{\lambda}^{\Gamma^{\uparrow\lambda}}$ are part of the fields
$\textnormal{No}_{\lambda}$.

######  Theorem 1.3 ([T]heorem 1.11).

bournez2022surreal]
$\textnormal{No}_{\lambda}=\bigcup_{\mu}\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu}}^{\uparrow\lambda}}$,
where $\mu$ ranges over the additive ordinals less than $\lambda$
(equivalently, $\mu$ ranges over the multiplicative ordinals less $\lambda$),

Notice that now, $\textnormal{No}_{\lambda}$ is expressed as a increasing
union of fields, each of them closed by $\exp$ and $\ln$. Indeed, by
definition, if $\mu<\mu^{\prime}$ then
$\textnormal{No}_{\mu}\subseteq\textnormal{No}_{\mu^{\prime}}$ and Remark 1
gives
$\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu}}^{\uparrow\lambda}}\subseteq\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu^{\prime}}}^{\uparrow\lambda}}$.

Finally, we proved that each field
$\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu}}^{\uparrow\lambda}}$ is
interesting for itself since none of them is $\textnormal{No}_{\lambda}$. More
precisely:

######  Theorem 1.4 ([T]heorem 1.12).

bournez2022surreal] For all $\varepsilon$-number $\lambda$, the hierarchy in
previous theorem is strict:

$\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu}}^{\uparrow\lambda}}\subsetneq\mathbb{R}_{\lambda}^{{\textnormal{No}_{\mu^{\prime}}}^{\uparrow\lambda}}$

for all multiplicative ordinals $\mu$ and $\mu^{\prime}$ such that
$\omega<\mu<\mu^{\prime}<\lambda$.

In this article, we will go further and investigate the case of stability
under derivative and anti-derivative.

######  Main Theorem 1.4.

Let $\alpha$ be a limit ordinal and
$\left(\Gamma_{\beta}\right)_{\beta<\alpha}$ be a sequence of Abelian
subgroups of No such that

  * •

$\forall\beta<\alpha\quad\forall\gamma<\beta\qquad\Gamma_{\gamma}\subseteq\Gamma_{\beta}$

  * •

$\forall\beta<\alpha\qquad\omega^{(\Gamma_{\beta})^{*}_{+}}\succ^{K}\kappa_{-\varepsilon_{\beta}}$

  * •

$\forall\beta<\alpha\quad\forall\gamma<\varepsilon_{\beta}\qquad\kappa_{-\gamma}\in\omega^{\Gamma_{\beta}}$

  * •

$\forall\beta<\alpha\quad\exists\eta_{\beta}<\varepsilon_{\beta}\quad\forall
x\in\omega^{\Gamma_{\beta}}\qquad\operatorname{NR}(x)<\eta_{\beta}$

Then
$\underset{\beta<\alpha}{\overset{}{\bigcup}}\mathbb{R}_{\varepsilon_{\beta}}^{\Gamma_{\beta}^{\uparrow\varepsilon_{\beta}}}$
is stable under $\exp$, $\ln$, $\partial$ and anti-derivation (see section 6).

Actually, we mainly focus on properties of the derivation suggested by
Beraducci and Mantova and its anti-derivation. In particular, we establish
various bounds that are useful to find fields stable under derivation and
anti-derivation. We also prove the following:

######  Proposition 1.4.

For any $x\in\textnormal{No}$, the set $\mathcal{P}_{\mathbb{L}}(x)$ is well-
ordered with order type
$\beta<\omega^{\omega^{\omega(\operatorname{NR}(x)+1)}}$. In particular,

$\nu(\partial x)<\omega^{\omega^{\omega(\operatorname{NR}(x)+1)}}$

In the above proposition and the above theorem, $\operatorname{NR}$ is the
nested truncation rank which is defined in Definition 6.2.1 and $\nu(x)$ is
the length of the series of the normal form the surreal number $x$ (see
Definition 3.3.3). The previous proposition is essential to control
derivatives of surreal numbers and then get field stable under derivation. To
handle anti-derivation, we came up with the following proposition:

######  Proposition 1.4.

Let $x$ be a surreal number. Let $\gamma$ be the smallest ordinal such that
$\kappa_{-\gamma}\prec^{K}P(k_{P})$ for all path
$P\in\mathcal{P}_{\mathbb{L}}(x)$. Let $\lambda$ be the least
$\varepsilon$-number greater than $\operatorname{NR}(x)$ and $\gamma$. Then
$\underset{i\in\mathbb{N}}{\overset{}{\bigcup}}\operatorname{supp}\Phi^{i}(x)$
(see Definition 6.5) is reverse well-ordered with order type less than
$\omega^{\omega^{\lambda+2}}$.

Thanks to Propositions 1 and 1, we will be able to prove our main theorem:

##### Organization of the paper

This article is organized as follows. Section 2 is a quick reminder of some
lemmas about order types that will be useful at the end of this article.
Section 3 recalls basics of the concepts and definitions of the theory of
surreal numbers, and fixes the notations used in the rest of the paper.
Section 4 recalls what is known about the stability properties of various
subfields of No according to their signs sequence representation or Hahn
series representation. In Section 5 we recall the definitions and properties
of exponential and logarithm. In Section 6, we recall some existing literature
about log-atomic numbers and derivation and established some result about the
nested truncation rank, a notion of rank related to the structure of the
surreal numbers and to log-atomic numbers. Finally, in Section 7, we build
surreal fields that are stable under exponentiation, logarithm, derivation and
anti-derivation. We also show how this construction can lead too an example
which only uses “small” ordinals, which is good from a Computability Theory
point of view.

##  2 Order type toolbox

In this section, we quickly take a look at some useful lemma about order type
of well ordered sets. In all the following, circled operators
($\oplus,\otimes$) stand for usual operations over ordinal numbers. The usual
symbols ($+,\times$) stand for natural operations, which are commutative.

Our first proposition is about the union of well ordered sets. This result is
already knows but we still provide a proof since it is hard to find it in the
literature.

######  Lemma 2.0 (Folklore).

Let $\Gamma$ be a totally ordered set, $A\subseteq\Gamma$ be a well-ordered
subset with order type $\alpha$. Let $g\in\Gamma$. Then the set $A\cup\\{g\\}$
is well ordered with order type at most $\alpha+1$.

###### Proof.

We prove it by induction on $\alpha$.

  * •

If $\alpha=0$ then $A\cup\\{g\\}$ has only one element, and then has order
type $1=\alpha+1$.

  * •

If $\alpha=\gamma+1$ is a successor ordinal. Let $u$ the largest element in
$A$. If $u\leq g$ then $A\cup\\{g\\}$ has indeed order type at most
$\alpha+1$. If not, then, by induction hypothesis,
$\left(A\setminus\\{u\\}\right)\cup\\{g\\}$ has order type at most
$\gamma+1=\alpha$. Then
$A\cup\\{g\\}=\left(\left(A\setminus\\{u\\}\right)\cup\\{g\\}\right)\cup\\{u\\}$
has order type at most $\alpha+1$.

  * •

If $\alpha$ is a limit ordinal. If $g$ is larger than any element of $A$, then
$A\cup\\{g\\}$ has order type $\alpha+1$. If not, let $a_{0}\in A$ such that
$a_{0}\geq g$. For $a\in A$ such that $a>a_{0}$ set

$B_{a}=\\{g\\}\cup\left\\{\left.a^{\prime}\in A\
\vphantom{a^{\prime}<a}\right|\ a^{\prime}<a\right\\}$

Since $\alpha$ is limit, we have
$A\cup\\{g\\}=\underset{a>a_{0}}{\overset{}{\bigcup}}B_{a}$

and each of the element in the union is an initial segment of $A\cup\\{g\\}$.
We also denote $\alpha_{a}$ the order type of the set
$\left\\{\left.a^{\prime}\in A\ \vphantom{a^{\prime}<a}\right|\
a^{\prime}<a\right\\}$. In particular, $\alpha_{a}<\alpha$. Using induction
hypothesis, $B_{a}$ has order type at most $\alpha_{a}+1$. Then, since we have
an increasing union of initial segments, the order type of $A\cup\\{g\\}$ is
at most

$\sup\left\\{\left.\alpha_{a}+1\ \vphantom{a>a_{0}}\right|\
a>a_{0}\right\\}=\sup\left\\{\left.\alpha^{\prime}+1\
\vphantom{\alpha^{\prime}<\alpha}\right|\
\alpha^{\prime}<\alpha\right\\}=\alpha$

since $\alpha$ is a limit ordinal.

We conclude thanks to the induction principle. ∎

######  Proposition 2.0 (Union of well-ordered sets, folklore).

Let $\Gamma$ be a totally ordered set $A,B\subseteq\Gamma$ be non-empty well-
ordered subsets with respective order types $\alpha$ and $\beta$. Then the
subset $A\cup B$ is well ordered with order type at most $\alpha+\beta$.

###### Proof.

$A\cup B$ is well-ordered. Indeed, if we have an infinite decreasing sequence
of $A\cup B$, then we can extract either an infinite one for either $A$ or $B$
which is not possible. It remains to show the bound on its order type. We do
it by induction over $\alpha$ and $\beta$.

  * •

If $\alpha=\beta=1$, then $A\cup B$ has at most two elements. Then, its order
type is at most $2=\alpha+\beta$.

  * •

If $\alpha$ or $\beta$ is a successor ordinal. Since both cases are symmetric,
we assume without loss of generality that $\beta=\gamma+1$. Let $u$ be the
largest element of $B$ and $C=B\setminus\\{u\\}$. Then, by induction
hypothesis, $A\cup C$ has order type at most $\alpha+\gamma$. Using Lemma 2,
we get that the order type of $A\cup B$ is at most
$\alpha+\gamma+1=\alpha+\beta$.

  * •

If $\alpha$ and $\beta$ are limit ordinal. $A$ or $B$ must be cofinal with
$A\cup B$. For instance say it is $A$. For $a\in A$, let

$A_{a}=\left\\{\left.a^{\prime}\in A\ \vphantom{a^{\prime}<a}\right|\
a^{\prime}<a\right\\}\qquad\text{and}\qquad B_{a}=\left\\{\left.b\in B\
\vphantom{b<a}\right|\ b<a\right\\}$

We have $A\cup B=\underset{a\in A}{\overset{}{\bigcup}}A_{a}\cup B_{a}$

Since $A$ is cofinal with $A\cup B$, it is an increasing union of initial
segments. Let $\alpha_{a}$ be the order type of $A_{a}$ and $\beta_{a}$ the
one of $B_{a}$. We have $\alpha_{a}<\alpha$ and $\beta_{a}\leq\beta$. By
induction hypothesis, $A_{a}\cup B_{a}$ has order type at most
$\alpha_{a}+\beta_{a}$. Then $A\cup B$ has order type at most

$\sup\left\\{\left.\alpha_{a}+\beta_{a}\ \vphantom{a\in A}\right|\ a\in
A\right\\}\leq\alpha+\beta$

We conclude the proof using the induction principle. ∎

We know move to addition of well ordered subset of a group. Again this result
in know but its proof is not easily findable in the literature.

######  Proposition 2.0 (Folklore).

Let $\Gamma$ be an ordered Abelian additive monoid and $A,B\subseteq\Gamma$ be
non-empty well-ordered subsets with respective order types $\alpha$ and
$\beta$. Then the subset $A+B=\left\\{\left.a+b\ \vphantom{a\in A\quad B\in
B}\right|\ a\in A\quad B\in B\right\\}$ is well ordered with order type at
most $\alpha\beta$.

###### Proof.

We do it by induction over $\alpha$ and $\beta$.

  * •

If $\alpha=\beta=1$, then $A+B$ has only one element, then has order type
$1=\alpha\beta$.

  * •

If $\alpha$ or $\beta$ is not an additive ordinal. Let say
$\beta=\gamma+\delta$ with $\gamma,\delta<\beta$. We choose $\gamma,\delta$
such that $\gamma+\delta=\gamma\oplus\delta$. Let $B_{1}$ the initial segment
of length $\gamma$ of $B$. Let $B_{2}=B\setminus B_{1}$. $B_{2}$ has order
type $\delta$. Then, by induction hypothesis, $A+B_{1}$ has order type at most
$\alpha\gamma$ and $A+B_{2}$ has order type at most $\alpha\delta$. Then,
using Proposition 2, $A+B$ has order type at most
$\alpha\gamma+\alpha\delta=\alpha\beta$.

  * •

If both $\alpha$ and $\beta$ are additive ordinals. Assume $A+B$ has order
type more than $\alpha\beta$. Let $a+b\in A+B$ such that the set $C$ defined
by

$C:=\left\\{\left.c\in A+B\ \vphantom{c<a+b}\right|\ c<a+b\right\\}$

has order type $\alpha\beta$. Let

$A_{0}=\left\\{\left.a^{\prime}\in A\ \vphantom{a^{\prime}<a}\right|\
a^{\prime}<a\right\\}$ and $B_{0}=\left\\{\left.b^{\prime}\in B\
\vphantom{b^{\prime}<b}\right|\ b^{\prime}<b\right\\}$

and $\alpha_{0}$ and $\beta_{0}$ their respective order types. We have

$C\subseteq\left(A_{0}+B\right)\cup\left(A+B_{0}\right)$

Using induction hypothesis and Proposition 2, $C$ has order type at most
$\alpha_{0}\beta+\alpha\beta_{0}$. Since $\alpha_{0}<\alpha$ and
$\beta_{0}<\beta$, we have $\alpha_{0}\beta<\alpha\beta$ and
$\alpha\beta_{0}<\alpha\beta$. $\alpha$ and $\beta$ being additive ordinal,
$\alpha\beta$ is itself an additive ordinal and then $C$ has order type less
than $\alpha\beta$, what is a contradiction. Then $A+B$ has order type at most
$\alpha\beta$.

We conclude thanks to the induction principle. ∎

In the same idea, we can take a look at a well ordered non-negative subset of
an ordered group. The proof is less easy so we refer to [Wei09] for the
details.

######  Proposition 2.0 ([Wei09, Corollary 1]).

Let $\Gamma$ be an ordered Abelian group and $S\subseteq\Gamma_{+}$ be a well-
ordered subset with order type $\alpha$. Then, $\left\langle S\right\rangle$,
the monoid generated by $S$ in $\Gamma$ is itself well-ordered with order type
at most $\omega^{\widehat{\alpha}}$ where, if the Cantor normal form of
$\alpha$ is

$\alpha=\underset{i=1}{\overset{n}{\sum}}\omega^{\alpha_{i}}n_{i}$

then
$\widehat{\alpha}=\underset{i=1}{\overset{n}{\sum}}\omega^{\alpha_{i}^{\prime}}n_{i}$

and $\beta^{\prime}=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\beta+1&\text{if
$\beta$ is an $\varepsilon$-number}\\\
\beta&\text{otherwise}\end{array}\right.$

In particular, $\left\langle S\right\rangle$ has order type at most
$\omega^{\omega\alpha}$ (commutative multiplication).

Finally, we consider finite sequences over a well ordered set.

######  Theorem 2.1 ([dP77, Theorem 3.11] and [Sch20, Theorem 2.9]).

Let $(X,\leq)$ be a well ordered set with order type $\alpha$. Let $X^{*}$ be
the set of finite sequences over $X$. Let $\beta$ the order type of $X^{*}$.
We have

$\beta\leq\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\omega^{\omega^{\alpha-1}}&\text{if
}\alpha\text{ is finite}\\\ \omega^{\omega^{\alpha+1}}&\textit{if
}\varepsilon\leq\alpha<\varepsilon+\omega\text{ for some $\varepsilon$-number
}\varepsilon\\\ \omega^{\omega^{\alpha}}&\text{ otherwise}\end{array}\right.$

##  3 Surreal numbers

We assume some familiarity with the ordered field of surreal numbers (refer to
[Con00, Gon86] for presentations) which we denote by No. In this section we
give a brief presentation of the basic definitions and results, and we fix the
notations that will be used in the rest of the paper.

###  3.1 Order and simplicity

The class No of surreal numbers can be defined either by transfinite
recursion, as in [Con00] or by transfinite length sequences of $+$ and $-$ as
done in [Gon86]. We will mostly follow [Gon86], as well as [BM18a] for their
presentation.

We introduce the class $\textnormal{No}=2^{<\textnormal{On}}$ of all binary
sequences of some ordinal length $\alpha\in\textnormal{On}$, where On denotes
the class of the ordinals. In other words, No corresponds to functions of the
form $x:\alpha\to\\{-,+\\}$. The length (sometimes also called birthday in
literature) of a surreal number $x$ is the ordinal number
$\alpha=\operatorname{dom}(x)$. We will also write
$\alpha=\left|x\right|_{+-}$ (the point of this notation is to “count” the
number of pluses and minuses). Note that No is not a set but a proper class,
and all the relations and functions we shall define on No are going to be
class-relations and class-functions, usually constructed by transfinite
induction.

We say that $x\in\textnormal{No}$ is simpler than $y\in\textnormal{No}$,
denoted $x\sqsubset y$, i.e., if $x$ is a strict initial segment (also called
prefix) of $y$ as a binary sequence. We say that $x$ is simpler than or equal
to $y$, written $x\sqsubseteq y$, if $x\sqsubset y$ or $x=y$ i.e., $x$ is an
initial segment of $y$. The simplicity relation is a binary tree-like partial
order on No, with the immediate successors of a node $x\in\textnormal{No}$
being the sequences $x_{-}$ and $x_{+}$ obtained by appending $-$ or $+$ at
the end of the signs sequence of $x$. Observe in particular that the
simplicity relation $\sqsubset$ is well-founded, and the empty sequence, which
will play the role of the number zero, is simpler than any other surreal
number.

We can introduce a total order $<$ on No which is basically the lexicographic
order over the corresponding sequences: More precisely, we consider the order
$-<\square<+$ where $\square$ is the blank symbol. Now to compare two signs
sequences, append blank symbols to the shortest so that they have the same
length. Then, just compare them with the corresponding lexicographic order to
get the total order $<$.

Given two sets $A\subseteq\textnormal{No}$ and $B\subseteq\textnormal{No}$
with $A<B$ (meaning that $a<b$ for all $a\in A$ and $b\in B$), it is quite
easy to understand why there is a simplest surreal number, denoted
$\left[\left.A\ \vphantom{B}\right|\ B\right]$ such that $A<\left[\left.A\
\vphantom{B}\right|\ B\right]<B$. However, a formal proof is long. See [Gon86,
Theorem 2.1] for details. If $x=\left[\left.A\ \vphantom{B}\right|\ B\right]$,
we say that Such a pair $\left[\left.A\ \vphantom{B}\right|\ B\right]$ is
representation of $x$.

Every surreal number $x$ has several different representations
$x=\left[\left.A\ \vphantom{B}\right|\ B\right]=\left[\left.A^{\prime}\
\vphantom{B^{\prime}}\right|\ B^{\prime}\right]$, for instance, if $A$ is
cofinal with $A^{\prime}$ and $B$ is coinitial with $B^{\prime}$. In this
situation, we shall say that $\left[\left.A\ \vphantom{B}\right|\
B\right]=\left[\left.A^{\prime}\ \vphantom{B^{\prime}}\right|\
B^{\prime}\right]$ by cofinality. On the other hand, as discussed in [BM18a],
it may well happen that $\left[\left.A\ \vphantom{B}\right|\
B\right]=\left[\left.A^{\prime}\ \vphantom{B^{\prime}}\right|\
B^{\prime}\right]$ even if $A$ is not cofinal with $A^{\prime}$ or $B$ is not
coinitial with $B^{\prime}$. The canonical representation $x=\left[\left.A\
\vphantom{B}\right|\ B\right]$ is the unique one such that $A\cup B$ is
exactly the set of all surreal numbers strictly simpler than $x$. Indeed it
turns out that is $A=\left\\{\left.y\sqsubset x\ \vphantom{y<x}\right|\
y<x\right\\}$ and $B=\left\\{\left.y\sqsubset x\ \vphantom{y>x}\right|\
y>x\right\\}$, then $x=\left[\left.A\ \vphantom{B}\right|\ B\right]$.

######  Remark 3.0.

By definition, if $x=\left[\left.A\ \vphantom{B}\right|\ B\right]$ and
$A<y<B$, then $x\sqsubseteq y$.

To make the reading easier we may forget $\\{\\}$ when writing explicitly $A$
and $B$. For instance $\left[\left.x\ \vphantom{y}\right|\ y\right]$ will
often stand for $\left[\left.\\{x\\}\ \vphantom{\\{y\\}}\right|\
\\{y\\}\right]$ when $x,y\in\textnormal{No}$.

###  3.2 Field operations

Ring operations $+$, $·$ on No are defined by transfinite induction on
simplicity as follows:

$x+y:=\left[\left.x^{\prime}+y,x+y^{\prime}\
\vphantom{x^{\prime\prime}+y,x+y^{\prime\prime}}\right|\
x^{\prime\prime}+y,x+y^{\prime\prime}\right]$
$xy:=\left[\left.\begin{array}[]{c}x^{\prime}y+xy^{\prime}-x^{\prime}y^{\prime}\\\
x^{\prime\prime}y+xy^{\prime\prime}-x^{\prime\prime}y^{\prime\prime}\end{array}\
\vphantom{\begin{array}[]{c}x^{\prime}y+xy^{\prime\prime}-x^{\prime}y^{\prime\prime}\\\
x^{\prime\prime}y+xy^{\prime}-x^{\prime\prime}y^{\prime}\end{array}}\right|\
\begin{array}[]{c}x^{\prime}y+xy^{\prime\prime}-x^{\prime}y^{\prime\prime}\\\
x^{\prime\prime}y+xy^{\prime}-x^{\prime\prime}y^{\prime}\end{array}\right]$

where $x^{\prime}$ (resp. $y^{\prime}$) ranges over the numbers simpler than
$x$ (resp. $y$) such that $x^{\prime}<x$ (resp. $y^{\prime}<y$) and
$x^{\prime\prime}$ (resp. $y^{\prime\prime}$) ranges over the numbers simpler
than $x$ (resp. $y$) such that $x<x^{\prime\prime}$ (resp.
$y<y^{\prime\prime}$); in other words, when $x=\left[\left.x^{\prime}\
\vphantom{x^{\prime\prime}}\right|\ x^{\prime\prime}\right]$ and
$y=\left[\left.y^{\prime}\ \vphantom{y^{\prime\prime}}\right|\
y^{\prime\prime}\right]$ are the canonical representations of $x$ and $y$
respectively. The expression for the product may seem not intuitive, but
actually, it is basically inspired by the fact that we expect
$(x-x^{\prime})(y-y^{\prime})>0$,
$(x-x^{\prime\prime})(y-y^{\prime\prime})>0$,
$(x-x^{\prime})(y-y^{\prime\prime})<0$ and
$(x-x^{\prime\prime})(y-y^{\prime})<0$.

######  Remark 3.0.

The definitions of sum and product are uniform in the sense of [Gon86, page
15]. Namely the equations that define $x+y$ and $xy$ does not require the
canonical representations of $x$ and $y$ but any representation. In
particular, if $x=\left[\left.A\ \vphantom{B}\right|\ B\right]$ and
$y=\left[\left.C\ \vphantom{D}\right|\ D\right]$, the variables
$x^{\prime},x^{\prime\prime},y^{\prime},y^{\prime\prime}$ may range over $A$,
$B$, $C$, $D$ respectively.

It is an early result that these operations, together with the order, give No
a structure of ordered field, and even a structure of real closed field (see
[Gon86, Theorem 5.10]). Consequently, there is a unique embedding of the
rational numbers in No so we can identify $\mathbb{Q}$ with a subfield of No.
Actually, the subgroup of the dyadic rationals $m/2^{n}\in\mathbb{Q}$, with
$m\in\mathbb{Z}$ and $n\in\mathbb{N}$, correspond exactly to the surreal
numbers $s:k\to\\{-,+\\}$ of finite length $k\in\mathbb{N}.$

The field $\mathbb{R}$ can be isomorphically identified with a subfield of No
by sending $x\in\mathbb{R}$ to the number $\left[\left.A\ \vphantom{B}\right|\
B\right]$ where $A\subseteq\textnormal{No}$ is the set of rationals
(equivalently: dyadics) lower than $x$ and $B\subseteq\textnormal{No}$ is the
set of (equivalently: dyadics) greater than $x$. This embedding is consistent
with the one of $\mathbb{Q}$ into No. We may thus write
$\mathbb{Q}\subseteq\mathbb{R}\subseteq\textnormal{No}$. By [Gon86, page 33],
the length of a real number is at most $\omega$ (the least infinite ordinal).
There are however surreal numbers of length $\omega$ which are not real
numbers, such as $\omega$ itself or its inverse that is a positive
infinitesimal.

The ordinal numbers can be identified with a subclass of No by sending the
ordinal $\alpha$ to the sequence $s:\alpha\rightarrow\\{+,-\\}$ with constant
value $+$. Under this identification, the ring operations of No, when
restricted to the ordinals $\textnormal{Ord}\subseteq\textnormal{No}$,
coincide with the Hessenberg sum and product (also called natural operations)
of ordinal numbers. Similarly, the sequence $s:\alpha\rightarrow\\{+,-\\}$
with constant value $-$ corresponds to the opposite (inverse for the additive
law) of the ordinal $\alpha$, namely $-\alpha$. We remark that
$x\in\textnormal{Ord}$ if and only if $x$ admits a representation of the form
$x=\left[\left.A\ \vphantom{B}\right|\ B\right]$ with $B=\varnothing$, and
similarly $x\in-\textnormal{Ord}$ if and only if we can write
$x=\left[\left.A\ \vphantom{B}\right|\ B\right]$ with $A=\varnothing$.

Under the above identification of $\mathbb{Q}$ as a subfield of No, the
natural numbers $\mathbb{N}\subseteq\mathbb{Q}$ are exactly the finite
ordinals.

###  3.3 Hahn series

####  3.3.1 Generalities

Let $\mathbb{K}$ be a field, and let $G$ be a divisible ordered Abelian group.

######  Definition 3.0 (Hahn series [Hah95]).

The Hahn series (obtained from $\mathbb{K}$ and $G$) are formal power series
of the form $s=\sum_{g\in S}a_{g}t^{g}$, where $S$ is a well-ordered subset of
$G$ and $a_{g}\in\mathbb{K}.$ The support of s is
$\operatorname{supp}(s)=\left\\{\left.g\in S\ \vphantom{a_{g}\neq 0}\right|\
a_{g}\neq 0\right\\}$ and the length of $s$ is the order type of
$\operatorname{supp}(s)$.

We write $\mathbb{K}\left(\left(G\right)\right)$ for the set of Hahn series
with coefficients in $\mathbb{K}$ and terms corresponding to elements of $G$.

######  Definition 3.0 (Operations on
$\mathbb{K}\left(\left(G\right)\right)$).

The operations on $K\left(\left(G\right)\right)$ are defined in the natural
way: Let $s=\sum_{g\in S}a_{g}t^{g},s^{\prime}=\sum_{g\in
S^{\prime}}a_{g}^{\prime}t^{g}$, where $S,S^{\prime}$ are well ordered.

  * •

$s+s^{\prime}=\sum_{g\in S\cup
S^{\prime}}\left(a_{g}+a_{g}^{\prime}\right)t^{g}$, where $a_{g}=0$ if
$g\notin S$, and $a_{g}^{\prime}=0$ if $g\notin S^{\prime}$.

  * •

$s\cdot s^{\prime}=\sum_{g\in T}b_{g}t^{g}$, where
$T=\left\\{\left.g_{1}+g_{2}\ \vphantom{g_{1}\in S\wedge g_{2}\in
S^{\prime}}\right|\ g_{1}\in S\wedge g_{2}\in S^{\prime}\right\\}$, and for
each $g\in T$, we set $b_{g}=\underset{g_{1}\in S,g_{2}\in
S^{\prime}|g_{1}+g_{2}=g}{\overset{}{\sum}}b_{g_{1}}\cdot b_{g_{2}}$

Hahn fields inherits a lot of from the structure of the coefficient field. In
particular if $\mathbb{K}$ is algebraically closed, and if $G$ is some
divisible (i.e. for any $n\in\mathbb{N}$ and $g\in G$ there is some
$g^{\prime}\in G$ such that $ng^{\prime}=g$) ordered Abelian group, then the
corresponding Hahn field is also algebraically closed. More precisely:

######  Theorem 3.1 (Generalized Newton-Puiseux Theorem, Maclane [Mac39]).

Let $G$ be a divisible ordered Abelian group, and let $\mathbb{K}$ be a field
that is algebraically closed of characteristic $0$. Then
$\mathbb{K}\left(\left(G\right)\right)$ is also algebraically closed.

As noticed in [All87], we can deduce the following:

######  Corollary 3.1.

Let $G$ be a divisible ordered Abelian group, and let $\mathbb{K}$ be a field
that is real closed of characteristic $0$. Then
$\mathbb{K}\left(\left(G\right)\right)$ is also real closed.

###### Proof.

$\mathbb{K}$ is real closed. That is to say that $-1$ is not a square in
$\mathbb{K}$ and that $\mathbb{K}[i]$ is algebraically closed. Notice that
$\mathbb{K}[i]((G))=\left(\mathbb{K}((G))\right)[i]$. Therefore, Theorem 3.1
ensures that $\left(\mathbb{K}((G))\right)[i]$ is algebraically closed. Also,
$-1$ is not a square in $\mathbb{K}((G))$. Therefore, $\mathbb{K}((G))$ is
real closed. ∎

####  3.3.2 Restricting length of ordinals

In this article, will often restrict the class of ordinals allowed in the
ordinal sum, namely by restricting to ordinals up to some ordinal $\lambda$.
We then give the following notation:

######  Definition 3.1 ($\mathbb{K}\left(\left(G\right)\right)_{\gamma}$).

Let $\lambda$ be some ordinal. We define
$\mathbb{K}\left(\left(G\right)\right)_{\gamma}$ for the restriction of
$\mathbb{K}\left(\left(G\right)\right)$ to formal power series whose support
has an order type in $\gamma$ (that is to say, corresponds to some ordinal
less than $\gamma$).

######  Theorem 3.2.

Assume $\gamma$ is some $\varepsilon$-number. Then
$\mathbb{K}\left(\left(G\right)\right)_{\gamma}$ is a field.

###### Proof.

This basically relies on the observation that the length of the inverse of
some Hahn series in this field remains in the field: This is basically a
consequence of Proposition 2. ∎

We also get:

######  Proposition 3.2 ([vE01, Lemma 4.6]).

Assume $\mathbb{K}$ is some real closed field, and $G$ is some abelian
divisible group. Then $\mathbb{K}\left(\left(G\right)\right)_{\gamma}$ is real
closed.

Actually, this was stated in [vE01, Lemma 4.6] for the case
$\mathbb{K}=\mathbb{R}$, but the proof ony uses the fact that $\mathbb{R}$ is
real-closed.

####  3.3.3 Normal form theorem for surreal numbers

######  Definition 3.2.

For $a$ and $b$ two surreal numbers, we define the following relations:

  * •

$a\prec b$ if for all $n\in\mathbb{N}$, $n|a|<|b|$.

  * •

$a\preceq b$ if there is some natural number $n\in\mathbb{N}$ such that
$|a|<n|b|$.

  * •

$a\asymp b$ if $a\preceq b$ and $b\preceq a$.

With this definition, $\preceq$ is a preorder and $\prec$ is the corresponding
strict preorder. The associated equivalence relation is $\asymp$ and the
equivalence classes are the Archimedean classes.

######  Theorem 3.3 ([Gon86, Theorem 5.1]).

For all surreal number $a$ there is a unique positive surreal $x$ of minimal
length such that $a\asymp x$.

The unique element of minimal length in its Archimedean class has many
properties similar to those of exponentiation:

######  Definition 3.3.

For all surreal number $a$ written in canonical representation
$a=\left[\left.a^{\prime}\ \vphantom{a^{\prime\prime}}\right|\
a^{\prime\prime}\right]$, we define

$\omega^{a}=\left[\left.0,\left\\{\left.n\omega^{a^{\prime}}\
\vphantom{n\in\mathbb{N}}\right|\ n\in\mathbb{N}\right\\}\
\vphantom{\left\\{\left.\frac{1}{2^{n}}\omega^{a^{\prime\prime}}\
\vphantom{n\in\mathbb{N}}\right|\ n\in\mathbb{N}\right\\}}\right|\
\left\\{\left.\frac{1}{2^{n}}\omega^{a^{\prime\prime}}\
\vphantom{n\in\mathbb{N}}\right|\ n\in\mathbb{N}\right\\}\right]$

we call such surreal numbers monomials.

Actually this definition is uniform ([Gon86, Corollary 5]) and therefore, we
can use any representation of $a$ in this definition. Another point is that we
can easily check that this notation is consistent with the ordinal
exponentiation. More precisely, if $a$ is an ordinal, $\omega^{a}$ is indeed
the ordinal corresponding to the ordinal exponentiation (see [Gon86, Theorem
5.4]). Finally, as announced, this definition gives the simplest elements
among the Archimedean classes.

######  Theorem 3.4 ([Gon86, Theorem 5.3]).

A surreal number is of the form $\omega^{a}$ if and only if it is simplest
positive element in its Archimedean class. More precisely,

$\forall a\in\textnormal{No}\qquad(\exists c\in\textnormal{No}\quad
a=\omega^{c})\implies(\forall b\in\textnormal{No}\quad b\asymp a\implies
a\sqsubseteq|b|)$

Elements of the form $\omega^{a}$ are by definition positive and have the
following property:

######  Proposition 3.4 ([Gon86, Theorem 5.4]).

We have

  * •

$\omega^{0}=1$

  * •

$\forall a,b\in\textnormal{No}\qquad\omega^{a}\omega^{b}=\omega^{a+b}$

Thanks to this definition of the $\omega$-exponentiation, we are now ready to
expose a normal form for surreal numbers which is analogous to the Cantor
normal form for ordinal normal.

######  Definition 3.4 ([Gon86, Section 5C, page 59]).

For $\nu$ an ordinal number, $\left(r_{i}\right)_{i<\nu}$ a sequence of non-
zero real numbers and $\left(a_{i}\right)_{i<\nu}$ a decreasing sequence of
surreal numbers, we define
$\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ inductively as
follows:

  * •

If $\nu=0$, then $\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}=0$

  * •

If $\nu=\nu^{\prime}+1$ then
$\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}=\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+r_{\nu^{\prime}}\omega^{a_{\nu^{\prime}}}$

  * •

If $\nu$ is a limit ordinal,
$\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ is defined as the
following bracket:

$\left[\left.\left\\{\left.\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+s\omega^{a_{\nu^{\prime}}}\
\vphantom{\begin{array}[]{c}\nu^{\prime}<\nu\\\
s<r_{\nu^{\prime}}\end{array}}\right|\ \begin{array}[]{c}\nu^{\prime}<\nu\\\
s<r_{\nu^{\prime}}\end{array}\right\\}\
\vphantom{\left\\{\left.\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+s\omega^{a_{\nu^{\prime}}}\
\vphantom{\begin{array}[]{c}\nu^{\prime}<\nu\\\
s>r_{\nu^{\prime}}\end{array}}\right|\ \begin{array}[]{c}\nu^{\prime}<\nu\\\
s>r_{\nu^{\prime}}\end{array}\right\\}}\right|\
\left\\{\left.\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+s\omega^{a_{\nu^{\prime}}}\
\vphantom{\begin{array}[]{c}\nu^{\prime}<\nu\\\
s>r_{\nu^{\prime}}\end{array}}\right|\ \begin{array}[]{c}\nu^{\prime}<\nu\\\
s>r_{\nu^{\prime}}\end{array}\right\\}\right]$

Note that if $0$ is seen as a limit ordinal, then both definition are
consistent.

######  Theorem 3.5 ([Gon86, Theorem 5.6]).

Every surreal number can has a unique writing of the form
$\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$. This expression will
be called its normal form.

Note that if $a$ is an ordinal number, then its normal form coincides with its
Cantor normal form. In such a sum, elements $r_{i}\omega^{a_{i}}$ will be
called the terms of the series.

######  Definition 3.5.

The length of the series in the normal form of a surreal number $x$ is denoted
$\nu(x)$.

######  Definition 3.5.

A surreal number $a$ in normal form
$a=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ is

  * •

purely infinite if for all $i<\nu$, $a_{i}>0$. $\textnormal{No}_{\infty}$ will
stand for the class of purely infinite numbers.

  * •

infinitesimal if for all $i<\nu$, $a_{i}<0$ (or equivalently if $a\prec 1$).

  * •

appreciable if for all $i<\nu$, $a_{i}\leq 0$ (or equivalently if $a\preceq
1$).

If $\nu^{\prime}\leq\nu$ is the first ordinal such that $a_{i}\leq 0$, then
$\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ is called the
purely infinite part of $a$. Similarly, if $\nu^{\prime}\leq\nu$ is the first
ordinal such that $a_{i}<0$, $\underset{\nu^{\prime}\leq
i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ is called the infinitesimal part
of $a$.

######  Theorem 3.6 ([Gon86, Theorems 5.7 and 5.8]).

Operation over surreal numbers coincides with formal addition and formal
multiplication over the normal forms. More precisely,

$\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+\underset{i<\nu^{\prime}}{\overset{}{\sum}}s_{i}\omega^{b_{i}}=\underset{x\in\textnormal{No}}{\overset{}{\sum}}t_{x}\omega^{x}$

where

  * •

$t_{x}=r_{i}$ if $i$ is such that $a_{i}=x$ and there is no $i$ such that
$b_{i}=x$.

  * •

$t_{x}=s_{i}$ if $i$ is such that $b_{i}=x$ and there is no $i$ such that
$a_{i}=x$.

  * •

$t_{x}=r_{i}+s_{j}$ if $i$ is such that $a_{i}=x$ and $j$ is such that
$b_{j}=x$

and

$\left(\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\right)\left(\underset{i<\nu^{\prime}}{\overset{}{\sum}}s_{i}\omega^{b_{i}}\right)=\underset{x\in\textnormal{No}}{\overset{}{\sum}}\left(\underset{\tiny\left\\{\left.\begin{array}[]{c}i<\nu\\\
j<\nu^{\prime}\end{array}\ \vphantom{a_{i}+b_{j}=x}\right|\
a_{i}+b_{j}=x\right\\}}{\overset{}{\sum}}r_{i}s_{j}\right)\omega^{x}$

We stated that every surreal number has a normal form. However, in the other
direction, it is possible to get back the sign expansion from a normal form.

######  Definition 3.6 (Reduced sign expansion, Gonshor, [Gon86]).

Let $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ be a surreal
number. The reduced sign expansion of $a_{i}$, denoted $a_{i}^{\circ}$ is
inductively defined as follows:

  * •

$a_{0}^{\circ}=a_{0}$

  * •

For $i>0$, if $a_{i}(\delta)=-$ and if there is there is $j<i$ such that for
$\gamma\leq\delta$, $a_{j}(\gamma)=a_{i}(\gamma)$, then we discard the minus
in position $\delta$ in the sign expansion of $a_{i}$.

  * •

If $i>0$ is a non-limit ordinal and $(a_{i-1})_{-}$ (as a sign expansion) is a
prefix of $a_{i}$, then we discard this minus after $a_{i-1}$ if $r_{i-1}$ is
not a dyadic rational number.

More informally, $a_{i}^{\circ}$ is the sign expansion obtained when copying
$a_{i}$ omitting the minuses that have already been treated before, in an
other exponent of the serie. We just keep the new one brought by $a_{i}$.
However, the later case give a condition where even a new minus can be
omitted.

######  Theorem 3.7 ([Gon86], Theorems 5.11 and 5.12).

For $\alpha$ an ordinal and a surreal $a$, we write $|a[:\alpha]|_{+}$ for the
(ordinal) number of pluses in $\alpha[:\alpha]$ the prefix of length of
$\alpha$ of $x$. Then,

  * •

The sign expansion of $\omega^{a}$ is as follows: we start with a plus and the
for any ordinal $\alpha<|a|$ we add $\omega^{|a[:\alpha]|_{+}+1}$ occurrences
of $a(\alpha)$ (the sign in position $\alpha$ in the signs sequence of $a$).

  * •

The sign expansion of $\omega^{a}n$ is the signs sequence of $\omega^{a}$
followed by $\omega^{|a|_{+}}(n-1)$ pluses.

  * •

The sign expansion of
$\omega^{a}\mathchoice{\dfrac{1}{2^{n}}}{\dfrac{1}{2^{n}}}{\frac{1}{2^{n}}}{\frac{1}{2^{n}}}$
is the sign expansion of $\omega^{a}$ followed by $\omega^{|a|_{+}}n$ minuses.

  * •

The sign expansion of $\omega^{a}r$ for $r$ a positive real is the sign
expansion of $\omega^{a}$ to which we add each sign of $r$ $\omega^{|a|_{+}}$
times excepted the first plus which is omitted.

  * •

The sign expansion of $\omega^{a}r$ for $r$ a negative real is the sign
expansion of $\omega^{a}(-r)$ in which we change every plus in a minus and
conversely.

  * •

The sign expansion of $\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$
is the juxtaposition of the sign expansions of the
$\omega^{a_{i}^{\circ}}r_{i}$

As a final note of this subsection, we give some bounds on the length of
monomials and terms.

######  Lemma 3.7 ([vE01, Lemma 4.1]).

For all surreal number $a\in\textnormal{No}$,

$\left|a\right|_{+-}\leq\left|\omega^{a}\right|_{+-}\leq\omega^{\left|a\right|_{+-}}$

######  Lemma 3.7 ([Gon86, Lemma 6.3]).

Let $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ a surreal
number. We have for all $i<\nu$,
$\left|r_{i}\omega^{a_{i}}\right|_{+-}\leq\left|x\right|_{+-}$.

####  3.3.4 Hahn series and surreal numbers

As a consequence of Theorems 3.5 and 3.6, the field No in in fact a Hahn serie
field. More precisely,

######  Corollary 3.7.

The fields No and $\mathbb{R}((t^{\textnormal{No}}))$ are isomorphic.

###### Proof.

Sending $t^{a}$ to $\omega^{-a}$ for all surreal number $a$, we notice that
all the definitions match to each other. ∎

Notice that we have of course
$\textnormal{No}=\mathbb{R}\left(\left(\textnormal{No}\right)\right)_{\textnormal{Ord}}$.

##  4 Surreal subfields

###  4.1 Subfields defined by Gonshor’s representation

Let $\textnormal{No}_{\lambda}$ denote the set surreal number whose signs
sequences have length less than $\lambda$ where $\lambda$ is some ordinal. We
have of course
$\textnormal{No}=\bigcup_{\lambda\in\textnormal{On}}\textnormal{No}_{\lambda}$.

Van den Dries and Ehrlich have proved the following:

######  Theorem 4.1 ([vE01, vdDE01]).

The ordinals $\lambda$ such that $\textnormal{No}_{\lambda}$ is closed under
the various fields operations of No can be characterised as follows:

  * •

$\textnormal{No}_{\lambda}$ is an additive subgroup of No iff
$\lambda=\omega^{\alpha}$ for some ordinal $\alpha$.

  * •

$\textnormal{No}_{\lambda}$ is a subring of No iff
$\lambda=\omega^{\omega^{\alpha}}$ for some ordinal $\alpha$.

  * •

$\textnormal{No}_{\lambda}$ is a subfield of No iff
$\omega^{\lambda}=\lambda$.

The ordinals $\lambda$ satisfying first (respectively: second) item are often
said to be additively (resp. multiplicatively) indecomposable but for the sake
of brevity we shall just call them additive (resp. multiplicative).
Multiplicative ordinals are exactly the ordinals $\lambda>1$ such that
$\mu\nu<\lambda$ whenever $\mu,\nu<\lambda$. The ordinal satisfying third item
are called $\varepsilon$-numbers. The smallest $\varepsilon$-number is usually
denoted by $\varepsilon_{0}$ and is given by

$\varepsilon_{0}:=\sup\\{\omega,\omega^{\omega},\omega^{\omega^{\omega}},\dots\\}.$

######  Remark 4.1.

Since rational numbers have length at most $\omega$, we have that if $\lambda$
is multiplicative, then $\textnormal{No}_{\lambda}$ is a divisible group.

If $\lambda$ is an $\varepsilon$-number, $\textnormal{No}_{\lambda}$ is
actually more than only a field:

######  Theorem 4.2 ([vE01, vdDE01]).

Let $\lambda$ be any $\varepsilon$-number. Then $\textnormal{No}_{\lambda}$ is
a real closed field.

###  4.2 Subfields defined from Hahn’s series representation

As we will often play with exponents of formal power series consided in the
Hahn series, we propose to introduce the following notation:

######  Definition 4.2.

If $\lambda$ is an $\varepsilon$-number and $\Gamma$ a divisible Abelian
group, we denote

$\mathbb{R}_{\lambda}^{\Gamma}=\mathbb{R}\left(\left(\Gamma\right)\right)_{\lambda}$

As a consequence of Proposition 3.3.2 we have

######  Corollary 4.2.

$\mathbb{R}_{\lambda}^{\textnormal{No}_{\mu}}$ is a real-closed field when
$\mu$ is a multiplicative ordinal and $\lambda$ an $\varepsilon$-number.

This fields are somehow the atoms constituting the fields
$\textnormal{No}_{\lambda}$.

See 1.1

######  Remark 4.2.

The fact that if $\lambda$ is not a regular cardinal, then
$\textnormal{No}_{\lambda}\neq\mathbb{R}_{\lambda}^{\textnormal{No}_{\lambda}}$
can be seen as follows: Suppose that $\lambda$ is not a regular cardinal. This
means that we can take some strictly increasing sequence
$(\mu_{\alpha})_{\alpha<\beta}$ that is cofinal in $\lambda$ with
$\beta<\lambda$. Then $\sum_{\alpha<\beta}\omega^{-\mu_{\alpha}}$ is in
$\mathbb{R}_{\lambda}^{\textnormal{No}_{\lambda}}$ by definition, but is not
in $\textnormal{No}_{\lambda}$.

##  5 Exponentiation and logarithm

###  5.1 Gonshor’s exponentiation

The field surreal numbers No admits an exponential function $\exp$ defined as
follows.

######  Definition 5.0 (Function $\exp$, [Gon86, page 145]).

Let $x=\left[\left.x^{\prime}\ \vphantom{x^{\prime\prime}}\right|\
x^{\prime\prime}\right]$ be the canonical representation of $x$. We define
inductively

$\exp x=\left[\left.\begin{array}[]{c}0,\exp(x^{\prime})[x-x^{\prime}]_{n},\\\
\exp(x^{\prime\prime})[x-x^{\prime\prime}]_{2n+1}\end{array}\
\vphantom{\mathchoice{\dfrac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\dfrac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}},\mathchoice{\dfrac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\dfrac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}}\right|\
\mathchoice{\dfrac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\dfrac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime})}{[x^{\prime}-x]_{2n+1}}},\mathchoice{\dfrac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\dfrac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}{\frac{\exp(x^{\prime\prime})}{[x^{\prime\prime}-x]_{2n+1}}}\right]$

where $n$ ranges in $\mathbb{N}$ and

$[x]_{n}=1+\frac{x}{1!}+\dots+\frac{x}{n!},$

with the further convention that the expressions containing terms of the form
$[y]_{2n+1}$ are to be considered only when $[y]_{2n+1}>0$.

It can be shown that the function $\exp$ is a surjective homomorphism from
$(\textnormal{No},+)$ to $(\textnormal{No}^{>0},·)$ which extends $\exp$ on
$\mathbb{R}$ and makes $(\textnormal{No},+,·,\exp)$ into an elementary
extension of $(\mathbb{R},+,·,\exp)$ (see [vdDMM94, Corollaries 2.11 and 4.6],
[vE01] and [Res93]). As $\exp$ is surjective, and from its properties, it can
be shown that it has some inverse $\ln:\textnormal{No}^{>0}\to\textnormal{No}$
(called logarithm).

######  Definition 5.0 (Functions $\log$, $\log_{n}$, $\exp_{n}$).

Let $\ln:\textnormal{No}^{>0}\to\textnormal{No}$ (called logarithm) be the
inverse of $\exp$. We let $\exp_{n}$ and $\ln_{n}$ be the $n$-fold iterated
compositions of $\exp$ and $\ln$ with themselves.

We recall some other basic properties of the exponential functions:

######  Theorem 5.1 ([Gon86, Theorems 10.2, 10.3 and 10.4]).

For all $r\in\mathbb{R}$ and $\varepsilon$ infinitesimal, we have

$\exp
r=\underset{k=0}{\overset{\infty}{\sum}}{\mathchoice{\dfrac{r^{k}}{k!}}{\dfrac{r^{k}}{k!}}{\frac{r^{k}}{k!}}{\frac{r^{k}}{k!}}}\quad\text{and}\quad\exp\varepsilon=\underset{k=0}{\overset{\infty}{\sum}}\mathchoice{\dfrac{\varepsilon^{k}}{k!}}{\dfrac{\varepsilon^{k}}{k!}}{\frac{\varepsilon^{k}}{k!}}{\frac{\varepsilon^{k}}{k!}}$

and
$\exp(r+\varepsilon)=\exp(r)\exp(\varepsilon)=\underset{k=0}{\overset{\infty}{\sum}}\mathchoice{\dfrac{(r+\varepsilon)^{k}}{k!}}{\dfrac{(r+\varepsilon)^{k}}{k!}}{\frac{(r+\varepsilon)^{k}}{k!}}{\frac{(r+\varepsilon)^{k}}{k!}}$

Moreover for all purely infinite number $x$,

$\exp(x+r+\varepsilon)=\exp(x)\exp(r+\varepsilon)$

######  Proposition 5.1 ([Gon86, Theorem 10.7]).

If $x$ is purely infinite, then $\exp x=\omega^{a}$ for some surreal number
$a$.

More precisely:

######  Proposition 5.1 (Function $g$, [Gon86, Theorem 10.13]).

If $x$ is purely infinite, i.e.
$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ with $a_{i}>0$ for
all $i$, then

$\exp x=\omega^{\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{g(a_{i})}},$

for some function $g:\textnormal{No}^{>0}\to\textnormal{No}$. Function $g$
satisfies for all $x$,

$g(x)=\left[\left.c(x),g(x^{\prime})\ \vphantom{g(x^{\prime\prime})}\right|\
g(x^{\prime\prime})\right]$

where $c(x)$ is the unique number such that $\omega^{c(x)}$ and $x$ are in the
same Archimedean class [Gon86, Thm. 10.11] (i.e. such that
$x\asymp\omega^{c(x)}$), where $x^{\prime}$ ranges over the lower non-zero
prefixes of $x$ and $x^{\prime\prime}$ over the upper prefixes of $x$.

###  5.2 About some properties of function $g$

######  Proposition 5.1 ([Gon86, Theorem 10.14]).

If $a$ is an ordinal number then

$g(a)=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}a+1&\text{if }\lambda\leq
a<\lambda+\omega\text{ for some }\varepsilon\text{-number }\lambda\\\
a&\text{otherwise}\end{array}\right.$

Note that in the previous proposition, $a\neq 0$ since $g$ is defined only for
positive elements.

######  Proposition 5.1 ([Gon86, Theorem 10.15]).

Let $n$ be a natural number and $b$ be an ordinal. We have
$g(2^{-n}\omega^{-b})=-b+2^{-n}.$

######  Proposition 5.1 ([Gon86, Theorems 10.17, 10.19 and 10.20]).

If $b$ is a surreal number such that for some $\varepsilon$-number
$\varepsilon_{i}$, some ordinal $\alpha$ and for all natural number $n$,
$\varepsilon_{i}+n<b<\alpha<\varepsilon_{i+1}$, then $g(b)=b$. This is also
true if there is some ordinal $\alpha<\varepsilon_{0}$ such that for all
natural number $b$, $n\omega^{-1}<b<\alpha<\varepsilon_{0}$.

######  Proposition 5.1 ([Gon86, Theorem 10.18]).

If $\varepsilon\leq b\leq\varepsilon+n$ for some $\varepsilon$-number
$\varepsilon$ and some integer $n$. In particular, the sign expansion of $b$
is the sign expansion of $\varepsilon$ followed by some sign expansion $S$.
Then, the sign expansion of $g(b)$ is the sign expansion of $\varepsilon$
followed by a $+$ and then $S$. In particular, $g(b)=b+1$.

It is possible to bound the length of $g(a)$ depending on the length of $a$.

######  Lemma 5.1 ([vE01, Lemma 5.1]).

For all $a\in\textnormal{No}$,
$\left|g(a)\right|_{+-}\leq\left|a\right|_{+-}+1$.

The function $g$ has a inverse function, $h$ defined as follows

$h(b)=\left[\left.0,h(b^{\prime})\
\vphantom{h(b^{\prime\prime}),\mathchoice{\dfrac{\omega^{b}}{n}}{\dfrac{\omega^{b}}{n}}{\frac{\omega^{b}}{n}}{\frac{\omega^{b}}{n}}}\right|\
h(b^{\prime\prime}),\mathchoice{\dfrac{\omega^{b}}{n}}{\dfrac{\omega^{b}}{n}}{\frac{\omega^{b}}{n}}{\frac{\omega^{b}}{n}}\right]$

This expression is uniform (see [Gon86]) and then does not depend of the
expression of $b$ as $\left[\left.b^{\prime}\
\vphantom{b^{\prime\prime}}\right|\ b^{\prime\prime}\right]$.

######  Corollary 5.1.

If $a$ is an ordinal number then $h(-a)=\omega^{-a-1}$.

###### Proof.

It is a direct consequence of Proposition 5.2 and the fact that $h=g^{-1}$. ∎

As for $g$, we can bound the length of $h(a)$ in function of the length of
$a$.

######  Lemma 5.1 ([AVDDVDH19, Proposition 3.1]).

For all $a\in\textnormal{No}$ we have,

$\left|h(a)\right|_{+-}\leq\omega^{\left|a\right|_{+-}+1}$

We will also prove another lemma, Lemma 5.2, that looks like the previous
lemma but that is better in many cases but not always. To do so we first prove
another technical lemma.

######  Lemma 5.1.

For all $c$, denote $c_{+}$ the surreal number whose signs sequence is the one
of $c$ followed by a plus. Assume $g(a)<c$ for all $a\sqsubset\omega^{c}$ such
that $0<a<\omega^{c}$. Then $g(\omega^{c})$ is $c_{+}$ if $c$ does not have a
longest prefix greater than itself, otherwise,
$g(\omega^{c})=c^{\prime\prime}$ where $c^{\prime\prime}$ is the longest
prefix of $c$ such that $c^{\prime\prime}>c$.

###### Proof.

By induction on $c$:

  * •

For $c=0$, $g(\omega^{0})=g(1)=1$ whose signs sequence is indeed the one of
$0$ followed by a plus.

  * •

Assume the property for $b\sqsubset c$. Assume $g(a^{\prime})<c$ for all
$a^{\prime}\sqsubset\omega^{c}$ such that $0<a^{\prime}<\omega^{c}$. Then,

$g(\omega^{c})=\left[\left.c\ \vphantom{g(a^{\prime\prime})}\right|\
g(a^{\prime\prime})\right]$

where $a^{\prime\prime}$ ranges over the elements such that
$a^{\prime\prime}\sqsubset\omega^{c}$ and $a^{\prime\prime}>\omega^{c}$.

    * ➢

First case: $c$ has a longest prefix $c_{0}$ such that $c_{0}>c$. Then, for
all $a^{\prime\prime}$ such that $a^{\prime\prime}\sqsubset\omega^{c}$ and
$a^{\prime\prime}>\omega^{c}$, $a^{\prime\prime}\succeq\omega^{c_{0}}$, hence
$g(a^{\prime\prime})>c_{0}$. Since $c<c_{0}<g(a^{\prime\prime})$, the
simplicity property ensures $g(\omega^{c})\sqsubseteq c_{0}\sqsubset c$. Then
$g(\omega^{c})$ is some prefix $c^{\prime\prime}$ of $c$, greater than $c$. We
look at $\omega^{c^{\prime\prime}}$. Notice that for all $b\sqsubset
c^{\prime\prime}$ is such that $0<b<c^{\prime\prime}$, $b\sqsubset c$ and
$b<c$, hence $g(b)<c<c^{\prime\prime}$. Therefore we can apply the induction
hypothesis to $c^{\prime\prime}$ and $g(\omega^{c^{\prime\prime}})$ is
$c^{\prime\prime}_{+}$ if the signs sequence of $c$ does not end with only
minuses, otherwise, $g(\omega^{c^{\prime\prime}})$ is the last (strict) prefix
of $c^{\prime\prime}$ greater than $c^{\prime\prime}$.

      * $\because$

First subcase: $g(\omega^{c^{\prime\prime}})=c^{\prime\prime}_{+}$. If there
is some $b$ such that $c^{\prime\prime}\sqsubset b\sqsubset c$ and $b>c$, then
$g(\omega^{b})$ is a prefix of $g(\omega^{c})=c^{\prime\prime}$. But,
$c^{\prime\prime}=g(\omega^{c})<g(\omega^{b})<g(\omega^{c^{\prime\prime}})=c^{\prime\prime}_{+}$.
Then $c^{\prime\prime}$ must be a strict prefix of $g(\omega^{b})$ which is a
contradiction. Then $c^{\prime\prime}$ is indeed the last strict prefix of $c$
greater than $c$.

      * $\because$

Second subcase: $g(\omega^{c^{\prime\prime}})$ is the last (strict) prefix of
$c^{\prime\prime}$ greater than $c^{\prime\prime}$. If there is some $b$ such
that $c^{\prime\prime}\sqsubset b\sqsubset c$ and $b>c$, then $g(\omega^{b})$
is a prefix of $g(\omega^{c})=c^{\prime\prime}$. Since
$g(b)<g(c^{\prime\prime})$, $g(b)$ is prefix of $c^{\prime\prime}$ smaller
than $c^{\prime\prime}$. But this contradicts the fact that
$g(\omega^{b})>g(\omega^{c})=c^{\prime\prime}$. Therefore, $c^{\prime\prime}$
is the last prefix of $c$ greater than $c$.

    * ➢

Second case: $c$ does not have a longest prefix greater than $c$. Then,

$g(\omega^{c})=\left[\left.c\ \vphantom{g(\omega^{c^{\prime\prime}})}\right|\
g(\omega^{c^{\prime\prime}})\right]$

where $c^{\prime\prime}$ ranges over the prefixes of $c$ greater than $c$. Let
$d\sqsubset c$ such that $d>c$. Then there is $d_{1}$ or minimal length such
that $d\sqsubset d_{1}\sqsubset c$ and $d_{1}>c$. By minimality of $d_{1}$,
$d$ is the longest prefix of $d_{1}$ greater than $d_{1}$. As in the first
case, we can apply the induction hypothesis on $d_{1}$ and get
$g(\omega^{d_{1}})=d$. Therefore, again by induction hypothesis,

$g(\omega^{c})=\left[\left.c\
\vphantom{c^{\prime\prime},c^{\prime\prime}_{+}}\right|\
c^{\prime\prime},c^{\prime\prime}_{+}\right]=\left[\left.c\
\vphantom{c^{\prime\prime}}\right|\ c^{\prime\prime}\right]$

where $c^{\prime\prime}$ ranges over the prefixes of $c$ greater than $c$. We
finally conclude that $g(\omega^{c})=c_{+}$.

∎

In the following we denote $\oplus$ the usual addition over the ordinal
numbers and $\otimes$ the usual product over ordinal numbers.

######  Lemma 5.1.

For all $a>0$,
$\left|a\right|_{+-}\leq\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)$.

###### Proof.

We proceed by induction on $\left|a\right|_{+-}$.

  * •

For $a=1$, $g(a)=1$ and we indeed have $1\leq\omega^{2}$.

  * •

Assume the property for all $b\sqsubset a$. Let $c$ such that
$\omega^{c}\asymp a$. Then

$g(a)=\left[\left.c,g(a^{\prime})\ \vphantom{g(a^{\prime\prime})}\right|\
g(a^{\prime\prime})\right]$

We split into two cases:

    * ➢

If there is some $a_{0}\sqsubset a$ such that $a_{0}<a$ and $g(a_{0})\geq c$
then

$g(a)=\left[\left.g(a^{\prime})\ \vphantom{g(a^{\prime\prime})}\right|\
g(a^{\prime\prime})\right]$

and if $S$ stand for the signs sequence such that $a$ is the signs sequence of
$a_{0}$ followed by $S$, $g(a)$ is the signs sequence of $g(a_{0})$ followed
$S$. Let $\alpha$ the length of $S$. Therefore using Theorem 3.7,

$\left|\omega^{g(a)}\right|_{+-}\geq\left|\omega^{g(a_{0})}\right|_{+-}\oplus(\omega\otimes\alpha)$

and then,

$\displaystyle\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)$
$\displaystyle\geq\left|\omega^{g(a_{0})}\right|_{+-}\otimes\omega\oplus\left|\omega^{g(a_{0})}\right|_{+-}\oplus\alpha$
$\displaystyle\geq\left|\omega^{g(a_{0})}\right|_{+-}\otimes(\omega+1)\oplus\alpha$

and by induction hypothesis on $a_{0}$,

$\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)\geq\left|a_{0}\right|_{+-}\oplus\alpha=\left|a\right|_{+-}$

    * ➢

Otherwise, for any $a_{0}\sqsubset a$ such that $a_{0}<a$, $g(a_{0})<c$.
Therefore,

$g(a)=\left[\left.c\ \vphantom{g(a^{\prime\prime})}\right|\
g(a^{\prime\prime})\right]$

Also, since $a>0$, we can write the signs sequence of $a$ as the one of
$\omega^{c}$ followed by some signs sequence $S$. If $S$ contains a plus, then
there is a prefix of $a$, $a_{0}$ such that $a_{0}<a$ and still
$a_{0}\asymp\omega^{c}$ and then $g(a_{0})>c$ what is not the case by
assumption. Then, $S$ is a sequence of minuses. If $S$ is not the empty
sequence, let $\alpha$ be the length of $S$. Then the signs sequence of $g(a)$
is the one of $g(\omega^{c})$ followed by $S$. Hence,

$\left|\omega^{g(a)}\right|_{+-}\geq\left|\omega^{g(\omega^{c})}\right|_{+-}\oplus(\omega\otimes\alpha)$

As in the previous case, but using the induction hypothesis on $\omega^{c}$,

$\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)\geq\left|\omega^{c}\right|_{+-}\oplus\alpha=\left|a\right|_{+-}$

Now if $S$ is the empty sequence, $a=\omega^{c}$. Applying Lemma 5.2 to $c$ we
get that either $g(a)=c_{+}$ or $g(a)$ is the last prefix of $c$ greater than
$c$. If the first case occurs then $a$ is a prefix of $\omega^{g(a)}$ and then
$\left|\omega^{g(a)}\right|_{+-}\geq\left|a\right|_{+-}$. Now assume that the
second case occurs. Then for any $b$ such that $g(a)\sqsubset b\sqsubset c$,
$b<c$. If for all $b^{\prime}\sqsubset b$ such that $b^{\prime}<b$,
$g(b^{\prime})<b$, then Lemma 5.2 applies. Since $b$ has a last prefix greater
than itself, $g(a)$, $g(\omega^{b})=g(a)$ and we reach a contradiction since
$b<c$ and therefore $\omega^{b}<\omega^{c}=a$. Then for all $b$ such that
$g(a)\sqsubset b\sqsubset c$, there is some $b^{\prime}\sqsubset b$,
$b^{\prime}<b$ such that $g(\omega^{b^{\prime}})>b$. Since the signs sequence
of $b$ consists in the one of $g(a)$ a minus and then a bunch of pluses, and
since $g(\omega^{b^{\prime}})$ must also a a prefix of $c$,
$g(\omega^{b^{\prime}})\sqsubseteq g(a)\sqsubset b$. Therefore to ensure
$g(b^{\prime})>b$, we must have $g(\omega^{b^{\prime}})\geq g(a)$. Since
$\omega^{b^{\prime}}$ is a prefix of $a$ lower than $a$, it is a
contradiction. Therefore, there is no $b$ such that $g(a)\sqsubset b\sqsubset
c$ and $b<c$, and finally, the signs sequence of $c$ is the one $g(a)$
followed by a minus. In particular, $g(a)$ and $c$ have the same amount of
pluses, say $\alpha$. Then, using Theorem 3.7,

$\displaystyle\left|a\right|_{+-}$
$\displaystyle=\left|\omega^{g(a)}\right|_{+-}\oplus\omega^{\alpha+1}$
$\displaystyle\leq\left|\omega^{g(a)}\right|_{+-}\oplus\left|\omega^{g(a)}\right|_{+-}\otimes\omega=\left|\omega^{g(a)}\right|_{+-}\otimes\omega$
$\displaystyle\leq\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)$

The induction principle concludes.

∎

######  Corollary 5.1.

For all $a>0$ and for all multiplicative ordinal greater than $\omega$, if
$\left|a\right|_{+-}\geq\mu$, then $\left|\omega^{g(a)}\right|_{+-}\geq\mu$.

###### Proof.

Assume the that $\left|\omega^{g(a)}\right|_{+-}<\mu$. Then using Lemma 5.2,
$\mu\leq\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)$. Since $\mu$ is a
multiplicative ordinal greater than $\omega$, we have $\omega+1<\mu$. $\mu$ is
a multiplicative ordinal, hence
$\left|\omega^{g(a)}\right|_{+-}\otimes(\omega+1)<\mu$ and we reach a
contradiction. ∎

###  5.3 Gonshor’s logarithm

We already know that a logarithm exist over positive surreal numbers.
Nevertheless we were very elliptical and we now get deeper into it.

######  Definition 5.1.

For a surreal number $a$ in canonical representation
$a=\left[\left.a^{\prime}\ \vphantom{a^{\prime\prime}}\right|\
a^{\prime\prime}\right]$, we define

$\ln\omega^{a}=\left[\left.\begin{array}[]{c}\left\\{\left.\ln\omega^{a^{\prime}}+n\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime}\sqsubset a\\\
a^{\prime}<a\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime}\sqsubset a\\\ a^{\prime}<a\end{array}\right\\}\\\
\left\\{\left.\ln\omega^{a^{\prime\prime}}-\omega^{\frac{a^{\prime\prime}-a}{n}}\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime\prime}\sqsubset a\\\
a<a^{\prime\prime}\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime\prime}\sqsubset a\\\
a<a^{\prime\prime}\end{array}\right\\}\end{array}\\!\\!\
\vphantom{\\!\\!\begin{array}[]{c}\left\\{\left.\ln\omega^{a^{\prime\prime}}-n\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime\prime}\sqsubset a\\\
a<a^{\prime\prime}\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime\prime}\sqsubset a\\\ a<a^{\prime\prime}\end{array}\right\\}\\\
\left\\{\left.\ln\omega^{a^{\prime}}+\omega^{\frac{a-a^{\prime}}{n}}\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime}\sqsubset a\\\
a^{\prime}<a\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime}\sqsubset a\\\ a^{\prime}<a\end{array}\right\\}\end{array}}\right|\
\\!\\!\begin{array}[]{c}\left\\{\left.\ln\omega^{a^{\prime\prime}}-n\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime\prime}\sqsubset a\\\
a<a^{\prime\prime}\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime\prime}\sqsubset a\\\ a<a^{\prime\prime}\end{array}\right\\}\\\
\left\\{\left.\ln\omega^{a^{\prime}}+\omega^{\frac{a-a^{\prime}}{n}}\
\vphantom{\begin{array}[]{c}n\in\mathbb{N}\\\ a^{\prime}\sqsubset a\\\
a^{\prime}<a\end{array}}\right|\ \begin{array}[]{c}n\in\mathbb{N}\\\
a^{\prime}\sqsubset a\\\ a^{\prime}<a\end{array}\right\\}\end{array}\right]$

As often with this kind of definitions, the uniformity property holds.

######  Lemma 5.1 ([Gon86, Lemma 10.1]).

The definition of $\ln\omega^{a}$ does not require $a$ in canonical
representation.

######  Proposition 5.1 ([Gon86, Theorem 10.8]).

For all surreal number $a$, $\ln\omega^{a}$ is purely infinite.

Purely infinite numbers are a special case in the definition of the
exponential function. We can state the previous definition of $\ln$ is
consistent with the one of $\exp$.

######  Theorem 5.2 ([Gon86, Theorem 10.9]).

For all surreal number $a$,

$\exp\ln\omega^{a}=\omega^{a}$

######  Theorem 5.3 ([Gon86, Theorem 10.12]).

For all surreal number $a$,

$\ln\omega^{\omega^{a}}=\omega^{h(a)}$

The above theorem is not actually stated like this in [Gon86] but this
statement follows from the proof there.

As a consequence of Theorems 5.2 and 5.3 and Propositions 5.3 and 5.1, we have

######  Corollary 5.3.

For all surreal number
$a=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$, we have

$\ln\omega^{a}=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{h(a_{i})}$

Finally, since for appreciable numbers $\exp$ is defined by its usual serie,
$\ln(1+x)$ is also defined by its usual serie when $x$ in infinitesimal. More
precisely,

######  Definition 5.3.

For $x$ an infinitesimal,

$\ln(1+x)=\underset{i=1}{\overset{\infty}{\sum}}\frac{(-1)^{i-1}x^{i}}{i}$

And thanks to Theorem 5.1,

######  Corollary 5.3.

Let $a=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ a positive
surreal number. Then

$\ln a=\ln\omega^{a_{0}}+\ln r_{0}+\ln\left(1+\underset{1\leq
i<\nu}{\overset{}{\sum}}\frac{r_{i}}{r_{0}}\omega^{a_{i}-a_{0}}\right)$

where the last term is defined in Definition 5.3.

###  5.4 Stability of $\textnormal{No}_{\lambda}$ by exponential and logarithm

We first recall some result by van den Dries and Ehrlich.

######  Lemma 5.3 ([vE01, Lemmas 5.2, 5.3 and 5.4]).

For all surreal number $a\in\textnormal{No}$,

  * •

$\left|\exp a\right|_{+-}\leq\omega^{\omega^{2\left|a\right|_{+-}\oplus 3}}$

  * •

$\left|\ln\omega^{a}\right|_{+-}\leq\omega^{4\omega\left|a\right|_{+-}\left|a\right|_{+-}}$

  * •

$\left|\ln a\right|_{+-}\leq\omega^{\omega^{3\left|a\right|_{+-}\oplus 3}}$

######  Corollary 5.3 ([vE01, Corollary 5.5]).

For $\lambda$ an $\varepsilon$-number, $\textnormal{No}_{\lambda}$ is stable
under $\exp$ and $\ln$.

We have

######  Theorem 5.4 ([BG22, Theorem 1.3]).

The following are equivalent:

  * •

$\textnormal{No}_{\lambda}$ is a subfield of No stable by $\exp$, and $\ln$

  * •

$\textnormal{No}_{\lambda}$ is a subfield of No

  * •

$\lambda$ is some $\varepsilon$-number.

###  5.5 A hierarchy of subfields of No stable by exponential and logarithm

In this subsection we recall our previous work on a hierarchy of surreal
subfields stable under exponential and logarithm.

We start by Theorem 1.2 repeated here for readability:

See 1.2

This result is actually a consequence of a more general proposition which is
the following.

######  Proposition 5.4 ([BG22, Proposition 5.1]).

Let $\lambda$ be an $\varepsilon$-number and $\left(\Gamma_{i}\right)_{i\in
I}$ be a family of Abelian subgroups of No. Then
$\mathbb{R}_{\lambda}^{\left(\Gamma_{i}\right)_{i\in I}}$ is stable under
$\exp$ and $\ln$ if and only if

$\underset{i\in I}{\overset{}{\bigcup}}\Gamma_{i}=\underset{i\in
I}{\overset{}{\bigcup}}\mathbb{R}_{\lambda}^{g\left(\left(\Gamma_{i}\right)^{*}_{+}\right)}$

Note that a consequence of Proposition 5.5 is also the following:

######  Corollary 5.4 ([BG22, Corollary 5.2]).

Let $\lambda$ be an $\varepsilon$-number and $\Gamma$ be an abelian subgroup
of No. Then $\mathbb{R}_{\lambda}^{\Gamma}$ is stable under $\exp$ and $\ln$
if and only if $\Gamma=\mathbb{R}_{\lambda}^{g\left(\Gamma^{*}_{+}\right)}$.

This result is quite similar to Theorem 1.2 but in the particular very
particular case where
$\underset{G\in\Gamma^{\uparrow\lambda}}{\overset{}{\bigcup}}G=\Gamma$. This
apply for instance when $\Gamma=\\{0\\}$. In this case, we get
$\mathbb{R}_{\lambda}^{\Gamma}=\mathbb{R}$. If $\lambda$ is a regular cardinal
we get an other example considering
$\mathbb{R}_{\lambda}^{\Gamma}=\Gamma=\textnormal{No}_{\lambda}$.

Theorem 1.2 enables us to consider a lot of fields stable under exponential
and logarithm and enabled us to prove that we can express
$\textnormal{No}_{\lambda}$ as a strictly increasing hierarchy of fields
stable under $\exp$ and $\ln$.

See 1.3

See 1.4

##  6 The class of log-atomic numbers, derivation and anti-derivation

###  6.1 Log-atomic numbers

We now introduce the concept of log-atomic numbers. Log-atomic numbers were
first introduced by Schmeling in [Sch01, page 30] about transseries. Such
number are basically number whose series of iterated logarithm have all length
$1$.

######  Definition 6.0 (Log-atomic).

A positive surreal number $x\in\textnormal{No}^{*}_{+}$ is said log-atomic iff
for all $n\in\mathbb{N}$, there is a surreal number $a_{n}$ such that
$\ln_{n}x=\omega^{a_{n}}$. We denote $\mathbb{L}$ the class of log-atomic
numbers.

For instance, $\omega$ is a log-atomic number and we can check that for all
$n\in\mathbb{N}$,
$\ln_{n}\omega=\omega^{\mathchoice{\dfrac{1}{\omega^{n}}}{\dfrac{1}{\omega^{n}}}{\frac{1}{\omega^{n}}}{\frac{1}{\omega^{n}}}}$.
Log-atomic number are the number we cannot divide into simpler numbers when
considering exponential and logarithm and are the fundamental blocs we end up
with when writing $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$
and then each $\omega^{a_{i}}$ as $\omega^{a_{i}}=\exp x_{i}$ with $x_{i}$ a
purely infinite number and then doing the same thing with each of the
$x_{i}$s. The use of the word “simpler” is not innocent. Indeed, log-atomic
numbers are also the simplest elements for some equivalence relation
introduced by Beraducci and Mantova [BM18b].

######  Definition 6.0 ([BM18b, Definition 5.2]).

Let $x,y$ be two positive infinite surreal numbers. We write

  * •

$x\asymp^{L}y$ iff there are some natural numbers $n,k$ such that

$\exp_{n}\left(\mathchoice{\dfrac{1}{k}}{\dfrac{1}{k}}{\frac{1}{k}}{\frac{1}{k}}\ln_{n}y\right)\leq
x\leq\exp_{n}\left(k\ln_{n}y\right)$

Equivalently, we ask that the is a natural number $n$ such that
$\ln_{n}x\asymp\ln_{n}y$. For such $n$ we notice that
$\ln_{n+1}x\sim\ln_{n+1}y$.

  * •

$x\prec^{L}y$ iff for all natural numbers $n$ and $k$,

$x<\exp_{n}\left(\mathchoice{\dfrac{1}{k}}{\dfrac{1}{k}}{\frac{1}{k}}{\frac{1}{k}}\ln_{n}y\right)$

Equivalently, we ask that for all $n\in\mathbb{N}$, $\ln_{n}x\prec\ln_{n}y$.

  * •

$a\preceq^{L}b$ iff there are some natural numbers $n$ and $k$,

$x\leq\exp_{n}\left(\mathchoice{\dfrac{1}{k}}{\dfrac{1}{k}}{\frac{1}{k}}{\frac{1}{k}}\ln_{n}y\right)$

Equivalently, we ask that for some $n\in\mathbb{N}$,
$\ln_{n}x\preceq\ln_{n}y$.

Log-atomic number are closely related to this equivalence relation since they
representatives of each equivalence classes.

######  Proposition 6.0 ([BM18b, Propositions 5.6 and 5.8]).

For all positive infinite $x$ there is unique log-atomic number
$y\in\mathbb{L}$ such that $y\sqsubseteq x$ and such that $y\asymp^{L}x$. In
particular, if $x,y\in\mathbb{L}$ with $x<y$ then $x\prec^{L}y$.

This proposition shows in particular that not even log-atomic are
representative of the equivalence classes of $\asymp^{L}$, they also are the
simplest element (i.e the shortest in terms of length) in their respective
equivalence classes. This make them a canonic class of representatives.

As we can parametrize additive ordinal, multiplicative ordinal or even
$\varepsilon$-numbers (for which a generalization for surreal numbers exists
in Gonshor’s book [Gon86]), we can parametrized epsilon numbers by a an
increasing function $\lambda_{\cdot}$. A first conjecture was to consider
$\kappa$-numbers which are defined by Kuhlmann and Matusinski as follows:

######  Definition 6.0 ([KM14, Definition 3.1]).

Let $x$ be a surreal number and write it in canonical representation as
$x=\left[\left.x^{\prime}\ \vphantom{x^{\prime\prime}}\right|\
x^{\prime\prime}\right]$. Then we define

$\kappa_{x}=\left[\left.\mathbb{R},\exp_{n}\kappa_{x^{\prime}}\
\vphantom{\ln_{n}\kappa_{x^{\prime\prime}}}\right|\
\ln_{n}\kappa_{x^{\prime\prime}}\right]$

Intuitively, $x<y$ iff every iterated exponential of $\kappa_{x}$ is less than
$\kappa_{y}$ and we try to build them as simple as possible. As an example, it
is quite easy to see that $\kappa_{0}=\omega$,
$\kappa_{-1}=\omega^{\omega^{-\omega}}$ and $\kappa_{1}=\varepsilon_{0}$. It
was conjectured that $\mathbb{L}$ consists in $\kappa$-number and there
iterated exponentials and logarithms. As shown by Berarducci and Mantova, it
turns out that it is not true. They then suggest a more general map which is
the following:

######  Definition 6.0 ([BM18b, Definition 5.12]).

Let $x$ be a surreal number and write it in canonical representation
$x=\left[\left.x^{\prime}\ \vphantom{x^{\prime\prime}}\right|\
x^{\prime\prime}\right]$. Then we define

$\lambda_{x}=\left[\left.\mathbb{R},\exp_{n}\left(k\ln_{n}\lambda_{x^{\prime}}\right)\
\vphantom{\exp_{n}\left(\mathchoice{\dfrac{1}{k}}{\dfrac{1}{k}}{\frac{1}{k}}{\frac{1}{k}}\ln_{n}\lambda_{x^{\prime\prime}}\right)}\right|\
\exp_{n}\left(\mathchoice{\dfrac{1}{k}}{\dfrac{1}{k}}{\frac{1}{k}}{\frac{1}{k}}\ln_{n}\lambda_{x^{\prime\prime}}\right)\right]$

where $n,k\in\mathbb{N}^{*}$.

######  Proposition 6.0 ([BM18b, Proposition 5.13 and Corollary 5.15]).

The function $x\mapsto\lambda_{x}$ is well defined, increasing, satisfies the
uniformity property and if $x<y$ then $\lambda_{x}\prec^{L}\lambda_{y}$.

######  Proposition 6.0 ([BM18b, Proposition 5.16]).

For every $x\in\textnormal{No}$ with $x>\mathbb{R}$ there is a unique
$y\in\textnormal{No}$ such that $x\asymp^{L}\lambda_{y}$ and
$\lambda_{y}\sqsubseteq x$. In particular, $\lambda_{y}$ is the simplest
number in its equivalence class for $\asymp^{L}$. As a consequence,
$\lambda_{\textnormal{No}}=\mathbb{L}$.

Moreover, the $\lambda$ map behaves very nicely with exponential and
logarithm.

######  Proposition 6.0 ([AVDDVDH19, Proposition 2.5]).

For all surreal number $x$,

$\exp\lambda_{x}=\lambda_{x+1}\qquad\text{and}\qquad\ln\lambda_{x}=\lambda_{x-1}$

######  Lemma 6.0 ([AVDDVDH19, Lemma 2.6]).

For all ordinal $\alpha$, $\lambda_{-\alpha}=\omega^{\omega^{-\alpha}}$.

######  Lemma 6.0 ([AVDDVDH19, Aschenbrenner, van den Dries and van der
Hoeven, Corollary 2.9]).

For all ordinal number $\alpha$,

$\kappa_{-\alpha}=\lambda_{-\omega\otimes\alpha}=\omega^{\omega^{-\omega\otimes\alpha}}$

###  6.2 Nested truncation rank

####  6.2.1 Definition

Log-atomic number are the base case (up to minor changes) of a notion of rank
over surreal numbers, the nested truncation rank. As expected, it is based on
some well partial order. This one has been defined by Berarducci and Mantova
as follows:

######  Definition 6.0 ([BM18b, Definition 4.3]).

For all natural number $n\in\mathbb{N}$, we define the relation
$\trianglelefteq_{n}$ as follows:

  * •

Writing $y\trianglelefteq_{0}x$ if any only if
$y=\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ and
$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ with
$\nu^{\prime}\leq\nu$. We say that $y$ is a truncation of $x$.

  * •

Let $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ Since
$\omega^{\textnormal{No}}=\exp(\textnormal{No}_{\infty})$, we can write

$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\exp(x_{i})$

where $\exp(x_{i})=\omega^{a_{i}}$. For a surreal number $y$, we say
$y\trianglelefteq_{n+1}x$ if there is $\nu^{\prime}<\nu$ and
$y^{\prime}\trianglelefteq_{n}x_{\nu^{\prime}}$ such that

$y=\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\exp(x_{i})+\operatorname{sign}(r_{\nu^{\prime}})\exp
y^{\prime}$

We say that $y$ is a nested truncation of $x$.

We also write $y\trianglelefteq x$ is there is some natural number $n$ such
that $y\trianglelefteq_{n}x$. We also introduce the corresponding strict
relations $\triangleleft_{n}$ and $\triangleleft$.

######  Definition 6.0 (Nested truncation rank [BM18b, Definition 4.27]).

The nested truncation rank of $x\in\textnormal{No}^{*}$ is defined by

$\operatorname{NR}(x)=\sup\left\\{\left.\operatorname{NR}y+1\
\vphantom{y\triangleleft x}\right|\ y\triangleleft x\right\\}$

By convention, we also set $\operatorname{NR}(0)=0$.

####  6.2.2 Properties

We know investigate some properties of the nested truncation rank. More
precisely, we provide compatibility properties with the operations over
surreal numbers and bounds on some particular nested truncation ranks. First
of all, the nested truncation rank is unaffected by the exponential.

######  Proposition 6.0 ([BM18b, Proposition 4.28]).

If $\gamma\in\textnormal{No}_{\infty}$, then

$\operatorname{NR}(\pm\exp\gamma)=\operatorname{NR}(\gamma)$

######  Corollary 6.0.

For all $a\in\textnormal{No}^{*}$,
$\operatorname{NR}(a)=\operatorname{NR}\left(-a\right)$

###### Proof.

Without loss of generality, we assume that $a>0$. Then

| $\operatorname{NR}\left(a\right)$ | $\ =\ $ | $\operatorname{NR}\left(\ln a\right)$ | (Proposition 6.2.2)
---|---|---|---|---
|  | $\ =\ $ | $\operatorname{NR}(-\exp\ln a)$ | (Proposition 6.2.2)
|  | $\ =\ $ | $\operatorname{NR}\left(-a\right)$ |

∎

######  Corollary 6.0.

For all $a\in\textnormal{No}^{*}$,
$\operatorname{NR}(a)=\operatorname{NR}\left(\mathchoice{\dfrac{1}{a}}{\dfrac{1}{a}}{\frac{1}{a}}{\frac{1}{a}}\right)$

###### Proof.

| $\operatorname{NR}\left(\mathchoice{\dfrac{1}{a}}{\dfrac{1}{a}}{\frac{1}{a}}{\frac{1}{a}}\right)$ | $\ =\ $ | $\operatorname{NR}\left(\ln\mathchoice{\dfrac{1}{a}}{\dfrac{1}{a}}{\frac{1}{a}}{\frac{1}{a}}\right)$ | (Proposition 6.2.2)
---|---|---|---|---
|  | $\ =\ $ | $\operatorname{NR}(-\ln a)$ |
|  | $\ =\ $ | $\operatorname{NR}\left(\ln a\right)$ | (Corollary 6.2.2)
|  |  |  |
|  | $\ =\ $ | $\operatorname{NR}(a)$ | (Proposition 6.2.2)

∎

######  Lemma 6.0.

For all $x\in\textnormal{No}$, $\operatorname{NR}(x)=0$ iff either
$x\in\mathbb{R}$ or $x=\pm\lambda^{\pm 1}$ for some log-atomic number
$\lambda$.

###### Proof.

  * $\overset{\text{SC}}{\Leftarrow}$

Note that if $x\in\mathbb{R}$ then there is no $y\in\textnormal{No}$ such that
$y\triangleleft x$. Therefore $\operatorname{NR}(x)=0$. Now assume that there
is some $x=\pm\lambda^{\pm 1}$ with $\lambda\in\mathbb{L}$ such that
$\operatorname{NR}(x)\neq 0$. Therefore there is some $y\in\textnormal{No}$
such that $y\triangleleft x$. Let $n\in\mathbb{N}$ minimal such that there is
$y\in\textnormal{No}$ and $\lambda\in\mathbb{L}$ such that
$y\triangleleft_{n}\pm\lambda^{\pm 1}$. Note that since $\pm\lambda^{\pm 1}$
is a term, $n>0$. Then $y=\pm\exp(\pm y^{\prime})$ with
$y^{\prime}\triangleleft_{n-1}\ln\lambda\in\mathbb{L}$. But this contradicts
the minimality of $n$. hence, for all $\lambda\in\mathbb{L}$,
$\operatorname{NR}\left(\pm\lambda^{\pm 1}\right)=0$.

  * $\overset{\text{NC}}{\Rightarrow}$

Assume $\operatorname{NR}(x)=0$ and $x$ is not a real number. If $x$ is not a
term, then there is $y\triangleleft_{0}x$ and in particular
$\operatorname{NR}(x)\geq 1$, what is impossible. Therefore there is some
$r\in\mathbb{R}^{*}$ and some $x_{1}\in\mathbb{J}$ such that $x=r\exp(x_{1})$.
If $r\neq\pm 1$ then $\operatorname{sign}(x)\exp(x^{\prime})\triangleleft x$
what is again impossible. Hence, $x=\pm\exp(x_{1})$. Proposition 6.2.2 ensures
that $\operatorname{NR}(x_{1})=0$. We then can apply the same work to $x_{1}$
so that there is some $x_{2}\in\mathbb{J}$ such that $x_{1}=\pm\exp(x_{2})$.
By induction, we can always define $x_{n}=\pm\exp(x_{n+1})$ with
$x_{n+1}\in\mathbb{J}$. For $n\geq 1$ we have $x_{n}\in\mathbb{J}$, therefore
$x_{n+1}>0$. In particular

$\forall n\geq 2\qquad x_{n}=\exp(x_{n+1})$

So, for all $n\in\mathbb{N}$, $\ln_{n}x_{2}$ is a monomial, this means that
$x_{2}\in\mathbb{L}$. We also have

$x=\pm\exp\left(\pm\exp x_{2}\right)=\pm\left(\exp_{2}(x_{2})\right)^{\pm 1}$

Since $\exp_{2}x_{2}\in\mathbb{L}$, we have the expected result.

∎

######  Lemma 6.0.

Let $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ and
$r\in\mathbb{R}^{*},a\in\textnormal{No}$ such that for all $i<\nu$,
$r\omega^{a}\prec\omega^{a_{i}}$. Then

$\operatorname{NR}(x+r\omega^{a})=\operatorname{NR}(x)\oplus
1\oplus\operatorname{NR}(\omega^{a})\oplus\mathds{1}_{r\neq\pm 1}$

where the $\oplus$ is the usual sum on ordinal numbers.

###### Proof.

Let $y\triangleleft x+r\omega^{a}$. Then $y\trianglelefteq x$ or
$y=x+\operatorname{sign}(r)\exp(\delta)$ with
$\delta\triangleleft\ln\omega^{a}$ or, if $r\neq\pm 1$,
$y=x+\operatorname{sign}(r)\omega^{a}$. Let

$A=\left\\{\left.y\ \vphantom{y\trianglelefteq x}\right|\ y\trianglelefteq
x\right\\}\qquad\text{and}\qquad
B=\left\\{\left.x+\operatorname{sign}(r)\exp(\delta)\
\vphantom{\delta\triangleleft\ln\omega^{a}}\right|\
\delta\triangleleft\ln\omega^{a}\right\\}$

and $C=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\varnothing&r=\pm 1\\\
x+\operatorname{sign}(r)\omega^{a}&r\neq\pm 1\end{array}\right.$

One can easily see that

$\forall y\in A\quad\forall y^{\prime}\in B\quad\forall y^{\prime\prime}\in
C\qquad y\triangleleft y^{\prime}\wedge y\triangleleft y^{\prime\prime}\wedge
y^{\prime}\triangleleft y^{\prime\prime}$

We now proceed by induction on $\operatorname{NR}(\omega^{a})$.

  * •

If $\operatorname{NR}(\omega^{a})=0$, using Lemma 6.2.2, either
$\omega^{a}=\pm\lambda^{\pm 1}$ for some log-atomic number $\lambda$ or $a=0$.
In both cases, there is no $\delta\triangleleft\ln\omega^{a}$.

| $\operatorname{NR}(x+r\omega^{a})$ | $\ =\ $ | $\sup\left\\{\left.\operatorname{NR}(y)+1\ \vphantom{y\in A\cup C}\right|\ y\in A\cup C\right\\}$
---|---|---|---
|  | $\ =\ $ | $\sup\left(\left\\{\left.\underbrace{\operatorname{NR}(y)+1}_{\leq\operatorname{NR}(x)}\ \vphantom{y\triangleleft x}\right|\ y\triangleleft x\right\\}\right.$
|  |  | $\left.\qquad\vphantom{\underbrace{\operatorname{NR}(y)+1}_{\leq\operatorname{NR}(x)}}\cup\\{\operatorname{NR}(x)+1\\}\cup\left\\{\left.\operatorname{NR}(y)+1\ \vphantom{y\in C}\right|\ y\in C\right\\}\right)$
|  | $\ =\ $ | $\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\operatorname{NR}(x)+1&r=\pm 1\\\ \operatorname{NR}(x+\operatorname{sign}(r)\omega^{a})&r\neq\pm 1\end{array}\right.$
|  | $\ =\ $ | $\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\operatorname{NR}(x)+1&r=\pm 1\\\ \operatorname{NR}(x)+2&r\neq\pm 1\end{array}\right.$
| $\operatorname{NR}(x+r\omega^{a})$ | $\ =\ $ | $\operatorname{NR}(x)+1+\operatorname{NR}(\omega^{a})+\mathds{1}_{r\neq\pm 1}$

  * •

For heredity now. Let $\delta\triangleleft\ln\omega^{a}$. Since
$\ln\omega^{a}$ is a purely infinite number, so is $\delta$. Then $\exp\delta$
is of the form $\omega^{b}$ for some surreal $b\in\textnormal{No}$. Moreover

$\operatorname{NR}(\omega^{b})\underset{\text{Proposition
}\ref{prop:NRStableExp}}{=}\operatorname{NR}(\delta)<\operatorname{NR}(\ln\omega^{a})\underset{\text{Proposition
}\ref{prop:NRStableExp}}{=}\operatorname{NR}(\omega^{a})$

From the induction hypothesis, we have that for any
$\delta\triangleleft\ln\omega^{a}$

$\operatorname{NR}(x+\operatorname{sign}(r)\exp(\delta))=\operatorname{NR}(x)\oplus
1\oplus\operatorname{NR}(\exp\delta)$

| $\operatorname{NR}(x+r\omega^{a})$ | $\ =\ $ | $\sup\left\\{\left.\operatorname{NR}(y)+1\ \vphantom{y\in B\cup C}\right|\ y\in B\cup C\right\\}$
---|---|---|---
|  | $\ =\ $ | $\sup\left(\left\\{\left.\underbrace{\operatorname{NR}(x+\operatorname{sign}(r)\exp\delta)+1}_{\leq\operatorname{NR}(x+\operatorname{sign}(r)\omega^{a})}\ \vphantom{\delta\triangleleft\ln\omega^{a}}\right|\ \delta\triangleleft\ln\omega^{a}\right\\}\right.$
|  |  | $\left.\qquad\cup\left\\{\left.\operatorname{NR}(y)+1\ \vphantom{y\in C}\right|\ y\in C\right\\}\vphantom{\underbrace{\operatorname{NR}(x+\operatorname{sign}(r)\exp\delta)+1}_{\leq\operatorname{NR}(x+\operatorname{sign}(r)\omega^{a})}}\right)$
|  | $\ =\ $ | $\sup\left\\{\left.\operatorname{NR}(x)\oplus 1\oplus\operatorname{NR}(\exp\delta)\oplus 1\ \vphantom{\delta\triangleleft\ln\omega^{a}}\right|\ \delta\triangleleft\ln\omega^{a}\right\\}+\mathds{1}_{r\neq\pm 1}$
|  | $\ =\ $ | $\operatorname{NR}(x)\oplus 1\oplus\sup\left\\{\left.\operatorname{NR}(\exp\delta)+1\ \vphantom{\delta\triangleleft\ln\omega^{a}}\right|\ \delta\triangleleft\ln\omega^{a}\right\\}\oplus\mathds{1}_{r\neq\pm 1}$
| $\operatorname{NR}(x+r\omega^{a})$ | $\ =\ $ | $\operatorname{NR}(x)\oplus 1\oplus\operatorname{NR}(\omega^{a})\oplus\mathds{1}_{r\neq\pm 1}$

∎

######  Lemma 6.0.

Let $x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ such that for
all $i<\nu$, $r_{i}=\pm 1$ and $\omega^{a_{i}}=\lambda_{i}^{\pm 1}$ for some
$\lambda\in\mathbb{L}$. Then

$\operatorname{NR}(x)=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}\nu+1&\nu<\omega\\\
\nu&\nu\geq\omega\end{array}\right.$

###### Proof.

If $\nu<\omega$, we just proceed by induction using Lemma 6.2.2. Now we prove
by induction the remaining.

  * •

If $\nu=\omega$. Then

$\operatorname{NR}(x)=\sup\left\\{\left.\operatorname{NR}\left(\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\right)+1\
\vphantom{\nu^{\prime}<\nu}\right|\
\nu^{\prime}<\nu\right\\}=\sup\left\\{\left.\nu^{\prime}+2\
\vphantom{\nu^{\prime}<\omega}\right|\ \nu^{\prime}<\omega\right\\}=\omega$

  * •

Assume for $\omega\leq\nu^{\prime}<\nu$,
$\operatorname{NR}\left(\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\right)=\nu^{\prime}$.
If $\nu$ is a non-limit ordinal, then Lemma 6.2.2 concludes. Otherwise

$\operatorname{NR}(x)=\sup\left\\{\left.\operatorname{NR}\left(\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\right)+1\
\vphantom{\nu^{\prime}<\nu}\right|\
\nu^{\prime}<\nu\right\\}=\sup\left\\{\left.\nu^{\prime}+^{1}\
\vphantom{\omega\leq\nu^{\prime}<\nu}\right|\
\omega\leq\nu^{\prime}<\nu\right\\}=\nu$

∎

######  Lemma 6.0.

Let
$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}(x)\omega^{a_{i}(x)}\in\textnormal{No}$.
Then $\nu\leq\operatorname{NR}(x)+1$. The equality stands iff $x$ is a finite
sum of numbers of the form $\pm y^{\pm 1}$ with $y\in\mathbb{L}$ and possibly
one non-zero real number.

###### Proof.

Using induction on $\nu$ it is trivial. For $0$, $\nu=0=\operatorname{NR}(0)$.
Now assume $\nu\neq 0$. Then, by definition

$\operatorname{NR}(x)+1\geq\sup\left\\{\left.\operatorname{NR}(y)+1\
\vphantom{y\triangleleft_{0}x\quad y\neq 0}\right|\ y\triangleleft_{0}x\quad
y\neq 0\right\\}+1\underset{\text{induction
hypothesis}}{\geq}\sup\left\\{\left.\nu(y)\ \vphantom{y\triangleleft_{0}x\quad
y\neq 0}\right|\ y\triangleleft_{0}x\quad y\neq 0\right\\}+1\geq\nu(x)$

Now assume $\nu(x)=\operatorname{NR}(x)+1$ and write
$x=\underset{i<\nu(x)}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$. We use induction
on $\textnormal{No}^{*}$ with the well partial order $\triangleleft_{0}$.

  * •

If $x$ is a monomial, $\nu(x)=1$ and $\operatorname{NR}(x)=0$. That is $x=\pm
y^{\pm 1}$ for some $y\in\mathbb{L}$ or $x\in\mathbb{R}$ (using Lemma 6.2.2).

  * •

If $x$ is not a monomial. Assume $r_{i}\omega^{a_{i}}\notin\pm\mathbb{L}^{\pm
1}\cup\mathbb{R}^{*}$ with $i$ minimal for that property. Let
$x^{\prime}=\underset{j<i}{\overset{}{\sum}}r_{j}\omega^{a_{j}}$.

    * ➢

If $i=0$ then $\operatorname{NR}(r_{0}\omega^{a_{0}})\geq 1$. A simple
induction shows that
$\operatorname{NR}\left(\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\right)\geq\nu^{\prime}$
for all $\nu^{\prime}\leq\nu$. What is a contradiction.

    * ➢

Otherwise $x^{\prime}\neq 0$ and $x^{\prime}\triangleleft_{0}x$. If
$\operatorname{NR}(x^{\prime})+1\neq i$ then
$\operatorname{NR}(x^{\prime})\geq i$ and

$\operatorname{NR}(x)\geq\operatorname{NR}(x^{\prime})\oplus(\nu\ominus
i)\geq\nu$

where $\nu\ominus i$ is the ordinal such that $i\oplus(\nu\ominus i)=\nu$.
what is a contraction. Then by induction hypothesis,
$i=\operatorname{NR}(x^{\prime})+1$ is finite. Now consider $y\triangleleft
x^{\prime}+r_{i}\omega^{a_{i}}$. Then $y\trianglelefteq_{0}x^{\prime}$
($y\triangleleft_{n}x^{\prime}$ with $n\geq 1$ is impossible since
$x^{\prime}$ has only terms in $\pm\mathbb{L}^{\pm 1}\cup\mathbb{R}$) or
$y=x^{\prime}+\operatorname{sign}(r_{i})\exp(\delta)$ with
$\delta\trianglelefteq\ln(\omega^{a_{i}})$. Since
$r_{i}\omega^{a_{i}}\notin\pm\mathbb{L}^{\pm 1}\cup\mathbb{R}$, there is such
a $y$ of the later form such that $y\neq x^{\prime}+r_{i}\omega^{a_{i}}$. From
Lemma 6.2.2, we have
$\operatorname{NR}(y)\geq\operatorname{NR}(x^{\prime})+1$. Then
$\operatorname{NR}(x^{\prime}+r_{i}\omega^{a_{i}})\geq\operatorname{NR}(y)+1\geq\operatorname{NR}(x^{\prime})+2$.
By induction we then can show that

$\operatorname{NR}(x)\geq\operatorname{NR}(x^{\prime}+r_{i}\omega^{a_{i}})\oplus(\nu-(i+1))\geq\operatorname{NR}(x^{\prime})\oplus
2\oplus(\nu\ominus(i\oplus 1))=i\oplus 1+(\nu\ominus(i\oplus 1))=\nu$

and we get a contradiction.

Then, every term of $x$ is in $\pm\mathbb{L}^{\pm 1}\cup\mathbb{R}$ and by
definition only one can be a non-zero real number. It remains to show that
there are finitely many terms, what follows from Lemma 6.2.2.

∎

######  Remark 6.0.

For all $x\in\textnormal{No}$, $\operatorname{NR}(x)\leq\left|x\right|_{+-}$

###### Proof.

Assume the converse and take $x$ with minimal length that contradicts the
property then there is $y\triangleleft x$ such that
$\operatorname{NR}(y)\geq\left|x\right|_{+-}$. Since
$\left|x\right|_{+-}>\left|y\right|_{+-}$, then $y$ reaches contradiction with
the minimality of $x$. ∎

######  Proposition 6.0 ([BM18b, Berarducci and Mantova, Proposition 4.29]).

For all $a\in\textnormal{No}^{*}$, for all $r\in\mathbb{R}\setminus\\{\pm
1\\}$, we have
$\operatorname{NR}(r\omega^{a})=\operatorname{NR}(\omega^{a})+1$.

######  Proposition 6.0 ([BM18b, Berarducci and Mantova, Proposition 4.30]).

Let
$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}\in\textnormal{No}^{*}$.
Then

  * •

$\forall
i<\nu\qquad\operatorname{NR}(r_{i}\omega^{a_{i}})\leq\operatorname{NR}(x)$

  * •

$\forall i<\nu\qquad
i+1<\nu\Rightarrow\operatorname{NR}(r_{i}\omega^{a_{i}})<\operatorname{NR}(x)$

We can also say something about the nested truncation rank of a sum of surreal
number.

######  Lemma 6.0.

For $a,b\in\textnormal{No}$,
$\operatorname{NR}(a+b)\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$
(natural sum of ordinal, which correspond to the surreal sum).

###### Proof.

We prove it by induction on the couple
$(\operatorname{NR}(a),\operatorname{NR}(b))$.

  * •

If $\operatorname{NR}(a)=\operatorname{NR}(b)=0$ then, by Lemma 6.2.2 both
$a,b$ are in $\pm\mathbb{L}^{\pm 1}\cup\mathbb{R}$. If $a\in\mathbb{R}$ or
$b\in\mathbb{R}$ then $\operatorname{NR}(a+b)\leq 1$ by Lemmas 6.2.2 and
6.2.2. Otherwise, either $a=\pm b$ and then $\operatorname{NR}(a+b)=0$ or
$a\neq\pm b$ and Lemma 6.2.2 ensure that $\operatorname{NR}(a+b)=1$.

  * •

Assume the property for all $x,y$ such that

$(\operatorname{NR}(x),\operatorname{NR}(y))<_{lex}(\operatorname{NR}(a),\operatorname{NR}(b))$

Then, consider $y\triangleleft a+b$. Write
$a+b=\underset{i<\nu}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$.

    * ➢

If $y=\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}$ with
$\nu^{\prime}<\nu$. Let $z_{a}$ be the series constituted of the terms of $a$
which asolute value is infintely larger than $\omega^{a_{\nu}^{\prime}}$. We
define the same way $z_{b}$. Then $y=z_{a}+z_{b}$. We have
$(\operatorname{NR}(z_{a}),\operatorname{NR}(z_{b}))<_{lex}(\operatorname{NR}(a),\operatorname{NR}(b))$
since there is term with order of magnitude $\omega^{a_{\nu^{\prime}}}$ in
either $a$ or $b$. Then, applying induction hypothesis,

$\operatorname{NR}(y)\leq\operatorname{NR}(z_{a})+\operatorname{NR}(z_{b})+1$

Since we have at least one of the following inequalities
$z_{a}\triangleleft_{0}a$ or $z_{b}\triangleleft_{0}b$, then
$\operatorname{NR}(z_{a})+1\leq\operatorname{NR}(a)$ or
$\operatorname{NR}(z_{b})+1\leq\operatorname{NR}(b)$. In all cases

$\operatorname{NR}(y)+1\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$

    * ➢

If
$y=\underset{i<\nu^{\prime}}{\overset{}{\sum}}r_{i}\omega^{a_{i}}+\operatorname{sign}(r_{\nu^{\prime}})\exp(y^{\prime})$
with $\nu^{\prime}<\nu$ and
$y^{\prime}\trianglelefteq\ln\omega^{a_{\nu^{\prime}}}$ (and
$y\triangleleft\ln\omega^{a_{\nu^{\prime}}}$ if $r_{\nu^{\prime}}=\pm 1$). Let
$z_{a}$ be the series constituted of the terms of $a$ which absolute value is
infinitely larger than $\omega^{a_{\nu}^{\prime}}$. We define the same way
$z_{b}$. Then
$y=z_{a}+z_{b}+\operatorname{sign}(r_{\nu^{\prime}})\omega^{a_{\nu^{\prime}}}$.
Since there is term with order of magnitude $\omega^{a_{\nu^{\prime}}}$ with
the same sign as $r_{\nu^{\prime}}$ in either $a$ or $b$. Without loss of
generality, assume it is $a$. Then
$z_{a}+\operatorname{sign}(r_{\nu^{\prime}})\exp y^{\prime}\trianglelefteq a$.
We have

$(\operatorname{NR}(z_{a}+\operatorname{sign}(r_{\nu^{\prime}})\exp
y^{\prime}),\operatorname{NR}(z_{b}))<_{lex}(\operatorname{NR}(a),\operatorname{NR}(b))$

otherwise $y=a+b$ what is not the case. Then, applying induction hypothesis,

$\operatorname{NR}(y)\leq\operatorname{NR}(z_{a}+\operatorname{sign}(r_{\nu^{\prime}})\exp
y^{\prime})+\operatorname{NR}(z_{b})+1$

Since we have at least one of the following inequalities
$z_{a}+\operatorname{sign}(r_{\nu^{\prime}})\exp y^{\prime}\triangleleft a$ or
$z_{b}\triangleleft_{0}b$, then we have either

$\operatorname{NR}(z_{a}+\operatorname{sign}(r_{\nu^{\prime}})\exp
y^{\prime})+1\leq\operatorname{NR}(a)$

or $\operatorname{NR}(z_{b})+1\leq\operatorname{NR}(b)$

In all cases

$\operatorname{NR}(y)+1\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$

Then, for any $y\triangleleft a+b$,
$\operatorname{NR}(y)+1\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$. This
proves that

$\operatorname{NR}(a+b)\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$

∎

######  Corollary 6.0.

For all $a,b\in\textnormal{No}$,
$\operatorname{NR}(ab)\leq\operatorname{NR}(a)+\operatorname{NR}(b)+1$.

###### Proof.

We have | $\operatorname{NR}(ab)$ | $\ =\ $ | $\operatorname{NR}(\ln\left(ab\right))$ | (Proposition 6.2.2)
---|---|---|---|---
|  | $\ =\ $ | $\operatorname{NR}(\ln a+\ln b)$ |
|  | $\ \leq\ $ | $\operatorname{NR}(\ln a)+\operatorname{NR}(\ln b)+1$ | (Lemma 6.2.2)
|  | $\ \leq\ $ | $\operatorname{NR}(a)+\operatorname{NR}(b)+1$ | (Proposition 6.2.2)

∎

####  6.2.3 Paths

Surreal numbers can be seen as trees. More precisely, it is possible to
associate to each surreal number a tree (with an ordinal numbers of node at
each layers) whose leaves are labeled by log-atomic numbers or $0$. This gives
us some information about the structure of the surreal number. With this
notion of tree we can look at the paths from the root (labeled by the surreal
number itself) to the leaves that are not labeled by $0$ (actually there is at
most one such a leaf). More precisely, the tree associated to a surreal number
$x$ is built as follows:

  * •

Base case: if $x\in\mathbb{L}$ or $x=0$ just create a node labeled by $x$ and
stop the construction.

  * •

Otherwise:

    1. 1.

Put a node at the root and label is t $x$. Write $x$ under the form
$x=\underset{i<\nu}{\overset{}{\sum}}r_{i}\exp(x_{i})$ where
$r_{i}\in\mathbb{R}^{*}$, $\nu$ is an ordinal and
$x_{i}\in\textnormal{No}_{\infty}$ form a decreasing sequence.

    2. 2.

For all $i<\nu$ create built the tree for $x_{i}$ and link its root to $x$ by
an edge labeled by $r_{i}$.

With the a notion, it is possible to have a geometric interpretation of the
well partial order $\triangleleft$.

The dotted arrows from “$\operatorname{sign}$” are to be understood by the
fact that we can apply the sign function or not to this arrow. The plain one
means that we must apply it. Thanks to this figure we can understand
$y\triangleleft x$ by the fact that the tree representation of $y$ is a left-
part of the tree representation of $x$.

######  Remark 6.0.

The reason why we stop the construction on log-atomic numbers is because if we
proceed the construction, we would get an infinite path where each node as
exactly one child and where every edge is labeled by $1$.

This notion of tree comes with a notion of path inside the tree.

######  Definition 6.0.

Let $x$ be a surreal number. A path $P$ of $x$ is sequence
$P:\mathbb{N}\to\textnormal{No}$ such that

  * •

$P(0)$ is a term of $x$

  * •

For all $i\in\mathbb{N}$, $P(i+1)$ is an infinite term of $\ln|P(i)|$

We denote $\mathcal{P}(x)$ the set of all paths of $x$.

We also denote $\ell(x)$ to be the purely infinite part of $\ln|x|$. Then
$P(i+1)$ is an infinite term of $\ell(P(i))$.

######  Definition 6.0.

The dominant path of $x$ is the path such that

  * •

$P(0)$ is the leading term of $x$

  * •

$P(i+1)$ is the leading term of $\ln|P(i)|$.

In a more graphical point of view, the dominant path of $x$ is the left most
path in the tree of $x$ that does not end on the lead $0$. This reduce to the
left most path if $x\not\asymp 1$.

######  Proposition 6.0.

Let $x\in\textnormal{No}$ and $P\in\mathcal{P}(x)$. Then for any
$n\in\mathbb{N}$, the length of the serie of $\ell(P(n))$, $\nu(\ell(P(n)))$
satisfies

$\nu(\ell(P(n)))\leq\operatorname{NR}(x)+1$

###### Proof.

For any $x\in\textnormal{No}$ we write
$x=\underset{i<\nu(x)}{\overset{}{\sum}}r_{i}(x)\omega^{a_{i}(x)}$ in
Gonshor’s normal form. Now fix $x\in\textnormal{No}$. Let
$P\in\mathcal{P}(x)$. We set $x_{0}=x$, and $\alpha_{0}<\nu(x)$ such
$P(0)=r_{\alpha_{0}}(x)\omega^{a_{\alpha_{0}}(x_{0})}$ and for any natural
number $n$,

$x_{n+1}=\ln\omega^{a_{\alpha_{n}}(x_{n})}=\ell(P(n))$

and $P(n+1)=r_{\alpha_{n+1}}\omega^{a_{\alpha_{n+1}}(x_{n+1})}$

Using Proposition 6.2.2, we get

$\operatorname{NR}(x_{n+1})=\operatorname{NR}\left(\omega^{a_{\alpha_{n}}(x_{n})}\right)$

By definition $x_{n+1}$ is purely infinite. Then $a_{\alpha_{n+1}}(x_{n+1})>0$
for all natural number $n$. Since $P$ is path, $P(0)\notin\mathbb{R}$
(otherwise $P(1)$ is not defined) and then $a_{\alpha_{0}}(x_{0})\neq 0$. We
then can apply Proposition 6.2.2 and get for all natural number $n$

$\operatorname{NR}(x_{n+1})\leq\operatorname{NR}\left(r_{\alpha_{n}}(x_{n})\omega^{a_{\alpha_{n}}(x_{n})}\right)$

Now using Proposition 6.2.2,
$\operatorname{NR}(x_{n+1})\leq\operatorname{NR}(x_{n})$

Then for any natural number $n$ we have
$\operatorname{NR}(x_{n})\leq\operatorname{NR}(x_{0})=\operatorname{NR}(x)$.
Applying Lemma 6.2.2, we get

$\forall
n\in\mathbb{N}\qquad\nu(x_{n})\leq\operatorname{NR}(x_{n})+1\leq\operatorname{NR}(x)+1$

∎

######  Remark 6.0.

Actually, we often have $\nu(\ell(P(n)))\leq\operatorname{NR}(x)$. Indeed,
using the notations of the proof and assuming that
$\nu(x_{n+1})=\operatorname{NR}(x)+1$, we have

$\operatorname{NR}(x)+1=\nu(x_{n+1})\underset{\text{Proposition
\ref{prop:cheminMajorationNu}}}{\leq}\operatorname{NR}(x_{n+1})+1\leq\cdots\leq\operatorname{NR}(x)+1$

Then, all the inequalities are equalities and from Proposition 6.2.3 we get
that $x_{n+1}$ is a finite sum of terms of the form $\pm\mathbb{L}^{\pm 1}$,
in particular $\nu(x_{n+1})<\omega$ and $\operatorname{NR}(x)$ is finite.

###  6.3 Derivative of a surreal number

######  Definition 6.0 (Summable family).

Let $\left\\{x_{i}\right\\}_{i\in I}$ be a family of surreal numbers. For
$i\in I$ write

$x_{i}=\underset{a\in\textnormal{No}}{\overset{}{\sum}}r_{i,a}\omega^{a}$

The family $\left\\{x_{i}\right\\}_{i\in I}$ is summable iff

  1. (i)

$\underset{i\in I}{\overset{}{\bigcup}}\operatorname{supp}x_{i}$ is a reverse
well ordered set.

  2. (ii)

For all $a\in\underset{i\in I}{\overset{}{\bigcup}}\operatorname{supp}x_{i}$,
$\left\\{\left.i\in I\ \vphantom{a\in\operatorname{supp}x_{i}}\right|\
a\in\operatorname{supp}x_{i}\right\\}$ is a finite set.

In this case, its sum is defined as $\underset{i\in
I}{\overset{}{\sum}}x_{i}=\underset{a\in\textnormal{No}}{\overset{}{\sum}}s_{a}\omega^{a}$
where for all $a\in\textnormal{No}$,

$s_{a}=\underset{i\in I\ |\
a\in\operatorname{supp}x_{i}}{\overset{}{\sum}}r_{i,a}$

which is a finite sum.

######  Definition 6.0 ([BM18b, Berarducci and Mantova, Definition 6.1]).

A derivation $D$ over a totally ordered exponential (class)-field
$\mathbb{K}\supseteq\mathbb{R}$ is a function $D:\mathbb{K}\to\mathbb{K}$ such
that

  1. D1.

It satisfies $\forall x,y\in\mathbb{K}\qquad D(xy)=xD(y)+D(x)y$ (Liebniz Rule)

  2. D2.

If $\left\\{x_{i}\right\\}_{i\in I}$ is summable, $D\left(\underset{i\in
I}{\overset{}{\sum}}x_{i}\right)=\underset{i\in I}{\overset{}{\sum}}D(x_{i})$
(Strong additivity)

  3. D3.

$\forall x\in\mathbb{K}\qquad D(\exp x)=D(x)\exp x$

  4. D4.

$\ker D=\mathbb{R}$

  5. D5.

$\forall x>\mathbb{N}\qquad D(x)>0$

######  Remark 6.0.

We can replace Axiom D2. by

  1. D2’.

If $\left\\{x_{i}\right\\}_{i\in I}$ is summable and
$\left\\{r_{i}\right\\}_{i\in I}$ is a family of real numbers,

$D\left(\underset{i\in I}{\overset{}{\sum}}r_{i}x_{i}\right)=\underset{i\in
I}{\overset{}{\sum}}r_{i}D(x_{i})$ (Strong lineraity)

Indeed, we have

$\ref{ax:D2p}\implies\ref{ax:D2}\qquad\text{and}\qquad\ref{ax:D1}\wedge\ref{ax:D2}\wedge\ref{ax:D4}\implies\ref{ax:D2p}$

Berarducci and Mantova [BM18b] provided a general way to define derivation
over the class-field No. We recall quickly some of their results.

######  Proposition 6.0 ([BM18b, Berarducci and Mantova, Proposition 6.4]).

We have the following properties for a derivation $D$:

  * •

$\forall x,y\in\mathbb{K}\qquad 1\not\asymp x\succ y\Rightarrow D(x)\succ
D(y)$

  * •

$\forall x,y\in\mathbb{K}\qquad 1\not\asymp x\sim y\Rightarrow D(x)\sim D(y)$

  * •

$\forall x,y\in\mathbb{K}\qquad 1\not\asymp x\asymp y\Rightarrow D(x)\asymp
D(y)$

If $\mathbb{K}\subseteq\textnormal{No}$ is stable under $\exp$ and $\ln$, we
can get a nice property satisfied by a general derivation.

######  Proposition 6.0 ([BM18b, Berarducci and Mantova, Proposition 6.5]).

Let $\mathbb{K}\subseteq\textnormal{No}$ be a field of surreal number stable
by $\exp$ and $\ln$. Let $D$ be a derivation over $\mathbb{K}$. For all
$x,y>\mathbb{N}$ such that $x-y>\mathbb{N}$,

$\ln D(x)-\ln D(y)\prec x-y\preceq\max(x,y)$

To define the derivation, Berarducci and Mantova started by defining it on
log-atomic numbers and then extending it on all surreal numbers. More
precisely, a derivation on log-atomic number must satisfy the following:

######  Definition 6.0 ([BM18b, Berarducci and Mantova, Definition 9.1]
Prederivation).

Let $\mathbb{K}$ be a field of surreal numbers stale under $\exp$ and $\ln$
and such that for all $x\in\mathbb{K}$, for all path $P\in\mathcal{P}(x)$, for
all $k\in\mathbb{N}$, if $P(k)\in\mathbb{L}$, then $P(k)\in\mathbb{K}$. A
prederivation over $\mathbb{K}$ is a function
$D_{\mathbb{L}}:\mathbb{L}\cap\mathbb{K}\to\mathbb{K}$ such that

  * D3.

$\forall\lambda\in\mathbb{L}\cap\mathbb{K}\qquad
D_{\mathbb{L}}\exp\lambda=(D_{\mathbb{L}}\lambda)\exp\lambda$

  * PD1.

For all $\lambda\in\mathbb{L}\cap\mathbb{K}$, $D_{\mathbb{L}}\lambda$ is a
positive term.

  * PD2.

$\forall\lambda,\mu\in\mathbb{L}\cap\mathbb{K}\qquad\ln
D_{\mathbb{L}}\lambda-\ln D_{\mathbb{L}}\ln\mu\prec\max(\lambda,\mu)$

They key notion to define the derivation from the the prederivation is the
notion of path derivative. This notion look at all the paths of the surreal
number to say how it contributes to the derivative of the surreal number.

######  Definition 6.0 ([BM18b, Berarducci and Mantova, Definition 6.13] Path
derivative).

Let $P$ be a path. We define the path derivative $\partial
P\in\mathbb{R}\omega^{\textnormal{No}}$ by

$\partial P=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}P(0)\cdots
P(k-1)D_{\mathbb{L}}P(k)&P(k)\in\mathbb{L}\\\ 0&\forall k\in\mathbb{N}\quad
P(k)\notin\mathbb{L}\end{array}\right.$

We denote $\mathcal{P}_{\mathbb{L}}(x)=\left\\{\left.P\in\mathcal{P}(x)\
\vphantom{\partial P\neq 0}\right|\ \partial P\neq 0\right\\}$, which is the
set of paths that indeed reach log-atomic numbers at some point.

One can notice that for any $P\in\mathcal{P}_{\mathbb{L}}(x)$, $\partial
P=r\omega^{a}$ for some $r\in\mathbb{R}^{*}$ and $a\in\textnormal{No}$.
Indeed, every $P(k)$ is a term and $D_{\mathbb{L}}P(k)$, when
$P(k)\in\mathbb{L}$ is an exponential of a purely infinite number, hence, it
is a monomial. For $P\in\mathcal{P}_{\mathbb{L}}(x)$ there is a minimum
$k_{P}\in\mathbb{N}$ such that $P(k_{P})\in\mathbb{L}$. Then $P$ is entirely
determined by $P(0),\dots,P(k_{P})$. We then define
$\alpha_{0}(P),\dots,\alpha_{k_{P}}(P)$ as follows :

  * •

Writing $x=\underset{i<\nu(x)}{\overset{}{\sum}}r_{i}(x)\omega^{a_{i}(x)}$,
then define $\alpha_{0}(P)<\nu(x)$ such that
$P(0)=r_{\alpha_{0}(P)}(x)\omega^{a_{\alpha_{0}(P)}}$.

  * •

For $0\leq i<k$, write $P(i)=r\omega^{a}$. Then $P(i+1)$ is a term of
$\ln\omega^{a}$. Write
$\ln\omega^{a}=\underset{i<\nu(a)}{\overset{}{\sum}}r_{i}(a)\omega^{h(a_{i}(a))}$.
Then set $\alpha_{i+1}(P)$ such that
$P(i+1)=r_{\alpha_{i+1}(P)}(a)\omega^{h(a_{\alpha_{i+1}(P)}(a))}$

Using Proposition 6.2.3, we get that
$\left(\alpha_{i}(P)\right)_{i\in\left\llbracket\,0\ ;\
k_{P}\,\right\rrbracket}$ is a finite sequence over ordinal less than
$\operatorname{NR}(x)+1$. In particular, we can give
$\mathcal{P}_{\mathbb{L}}(x)$ a lexicographic order inherited from the one
over finite sequences.

######  Definition 6.0.

We define the order $<_{lex}$ on paths by

$P<_{lex}Q\Longleftrightarrow(\alpha_{0}(P),\dots,\alpha_{k_{P}}(P))<_{lex}(\alpha_{0}(Q),\dots,\alpha_{k_{Q}}(Q))$

This order will be useful later when we will try to understand better what is
going on to get some bounds about the derivatives. For now, the path-
derivative being defined, we can recall a theorem by Berarducci and Mantova
which explains how to build a general derivation from a prederivation.

######  Lemma 6.0 ([BM18b, Berarducci and Mantova, Corollary 6.17]).

Let $P,Q\in\mathcal{P}(x)$ such that $\partial P,\partial Q\neq 0$. If there
is $i\in\mathbb{N}$ such that

  1. 1.

$\forall j\leq i\qquad P(i)\preceq Q(i)$

  2. 2.

$P(i+1)$ is not a term of $\ell(Q(i))$,

then $\partial P\prec\partial Q$

######  Lemma 6.0 ([BM18b, Berarducci and Mantova, Lemma 6.18]).

Given $P\in\mathcal{P}(x)$ a path of $x$ we have for all $i$
$\operatorname{NR}(P(i+1))\leq\operatorname{NR}(P(i))$ with equality if and
only if $P(i)$ is the last term of $\ell(P(i))$. We also have
$\operatorname{NR}(P(0)\leq\operatorname{NR}(x)$ with equality if and only if
$P(0)$ is the last term of $x$.)

######  Theorem 6.1 ([BM18b, Berarducci and Mantova, Proposition 6.20, Theorem
6.32]).

Let $D_{\mathbb{L}}$ be a prederivation over a surreal field $\mathbb{K}$
stable under $\exp$ and $\ln$. Then $D_{\mathbb{L}}$ extends to a derivation
$\partial:\mathbb{K}\to\textnormal{No}$ such that

$\forall x\in\mathbb{K}\qquad\partial
x=\underset{P\in\mathcal{P}(x)}{\overset{}{\sum}}\partial P$

In particular, $\left\\{\partial P\right\\}_{P\in\mathcal{P}(x)}$ is summable
(see Definition 6.3).

The study would not be complete without an example. Berarducci and Mantova
provided such a derivation and even more: it is the simplest in some sense.

######  Definition 6.1 ([BM18b, Berarducci and Mantova, Definition 6.7]).

We define $\partial_{\mathbb{L}}:\mathbb{L}\to\textnormal{No}$ by

$\forall\lambda\in\mathbb{L}\qquad\partial_{\mathbb{L}}\lambda=\exp\left(-\underset{\alpha\in\textnormal{Ord}|\kappa_{-\alpha}\succeq^{K}\lambda}{\overset{}{\sum}}\
\underset{n=1}{\overset{{\mbox{\raisebox{0.56905pt}{\tiny{$+$}}}\infty}}{\sum}}\ln_{n}\kappa_{\alpha}+\underset{n=1}{\overset{{\mbox{\raisebox{0.56905pt}{\tiny{$+$}}}\infty}}{\sum}}\ln_{n}\lambda\right)$

For example, we have:

$\displaystyle\partial_{\mathbb{L}}\omega$ $\displaystyle=1$
$\displaystyle\partial_{\mathbb{L}}\exp\omega$ $\displaystyle=\exp\omega$
$\displaystyle\partial_{\mathbb{L}}\ln\omega$
$\displaystyle=\exp(-\ln\omega)=\mathchoice{\dfrac{1}{\omega}}{\dfrac{1}{\omega}}{\frac{1}{\omega}}{\frac{1}{\omega}}$
$\displaystyle\partial_{\mathbb{L}}\ln_{n}\omega$
$\displaystyle=\mathchoice{\dfrac{1}{\underset{k=0}{\overset{n-1}{\prod}}\ln_{k}\omega}}{\dfrac{1}{\underset{k=0}{\overset{n-1}{\prod}}\ln_{k}\omega}}{\frac{1}{\underset{k=0}{\overset{n-1}{\prod}}\ln_{k}\omega}}{\frac{1}{\underset{k=0}{\overset{n-1}{\prod}}\ln_{k}\omega}}$
$\displaystyle\partial_{\mathbb{L}}\kappa_{1}=\partial_{\mathbb{L}}\varepsilon_{0}$
$\displaystyle=\exp\left(\underset{n=1}{\overset{{\mbox{\raisebox{0.56905pt}{\tiny{$+$}}}\infty}}{\sum}}\ln_{n}\kappa_{1}\right)$
$\displaystyle\partial_{\mathbb{L}}\kappa_{-1}$
$\displaystyle=\exp\left(-\underset{n=1}{\overset{{\mbox{\raisebox{0.56905pt}{\tiny{$+$}}}\infty}}{\sum}}\ln_{n}\omega\right)$

In fact, $\kappa_{1}$ is intuitively $\exp_{\omega}\omega$. Therefore it is
also quite intuitive that
$\partial_{\mathbb{L}}\kappa_{1}=\kappa_{1}\ln(\kappa_{1})\ln\ln(\kappa_{1})\cdots$.
The same happens for $\kappa_{-1}$ which is intuitively $\ln_{\omega}\omega$.
We indeed have
$\partial_{\mathbb{L}}\kappa_{-1}=\mathchoice{\dfrac{1}{\omega\ln(\omega)\ln\ln(\omega)\cdots}}{\dfrac{1}{\omega\ln(\omega)\ln\ln(\omega)\cdots}}{\frac{1}{\omega\ln(\omega)\ln\ln(\omega)\cdots}}{\frac{1}{\omega\ln(\omega)\ln\ln(\omega)\cdots}}$.

######  Proposition 6.1 ([BM18b, Berarducci and Mantova, Propositions 6.9 and
6.10]).

$\partial_{\mathbb{L}}$ is a prederivation.

The previous proposition ensures that the associated function $\partial$
defined by Theorem 6.1 is indeed a derivation over surreal numbers. It turns
out that it the simplest for the order $\sqsubseteq$.

We now explain what is meant when saying that $\partial$ is the simplest
derivation. In fact, we mean that $\partial_{\mathbb{L}}$ is the simplest
prederivation with respect to the order $\sqsubseteq$.

######  Theorem 6.2 (Berarducci and Mantova[BM18b, Theorem 9.6]).

Let $D_{\lambda}$ be a prederivation. Let $\lambda\in\mathbb{L}$, minimal (in
$\mathbb{L}$) for $\sqsubseteq$ such that
$D_{\mathbb{L}}\lambda\neq\partial_{\mathbb{L}}\lambda$. Then
$\partial_{\mathbb{L}}\lambda\sqsubset D_{\mathbb{L}}\lambda$.

###  6.4 A first bound about the derivative

We give here some bound on the length of the series of a derivative.

See 1

###### Proof.

We know that $\left\\{\partial P\right\\}_{P\in\mathcal{P}(x)}$ is summable
(see Definition 6.3). In particular $\left\\{\partial
P\right\\}_{P\in\mathcal{P}_{\mathbb{L}}(x)}$ is summable. By definition of
summability (in this context) for any $P\in\mathcal{P}_{\mathbb{L}}(x)$, there
are finitely many $Q\in\mathcal{P}_{\mathbb{L}}(x)$ such that $\partial
P\asymp\partial Q$.

By definition of summability, $<_{\mathcal{P}}$ is a well total order over
$\mathcal{P}_{\mathbb{L}}(x)$ and if $\beta$ is its order type, then
$\omega\otimes\nu(\partial x)<\beta$ (usual ordinal product). Then, to
complete the proof, we just need to show that
$\beta<\omega^{\omega^{\omega(\operatorname{NR}(x)+1)}}$. We proceed by
induction on $\operatorname{NR}(x)$.

  * •

$\operatorname{NR}(x)$=0 : then $x=0$ or $x=\pm y^{\pm 1}$ for some
$y\in\mathbb{L}$ and $\nu\left(\partial x\right)\leq
1<\omega^{\omega^{\omega}}$ and we conclude the proof.

  * •

Assume that for any $y$ such that $\operatorname{NR}(y)<\operatorname{NR}(x)$,
$\mathcal{P}_{\mathbb{L}}(y)$ has order type less than
$\omega^{\omega^{\omega(\operatorname{NR}(y)+1)}}$. Assume for contradiction
that $\beta\geq\omega^{\omega^{\omega(\operatorname{NR}(x)+1)}}$. Then for any
multiplicative ordinal $\mu<\omega^{\omega^{\omega(\operatorname{NR}(x)+1)}}$,
there is some $P_{\mu}\in\mathcal{P}_{\mathbb{L}}(x)$, minimum with respect to
$<_{lex}$, such that the set

$\mathcal{E}_{\mu}(x)=\left\\{\left.Q\in\mathcal{P}_{\mathbb{L}}(x)\
\vphantom{Q<_{\mathcal{P}}P_{\mu}}\right|\ Q<_{\mathcal{P}}P_{\mu}\right\\}$

has order type $\beta_{\mu}\geq\mu$. Let us select any $\mu$ such that
$\mu\geq\omega^{\omega^{\omega\operatorname{NR}(x)+1}}$. Now define

| $\mathcal{E}_{\mu}^{(1)}(x)$ | $\ =\ $ | $\left\\{\left.Q\in\mathcal{P}_{\mathbb{L}}(x)\ \vphantom{Q<_{\mathcal{P}}P_{\mu}\quad Q<_{lex}P_{\mu}}\right|\ Q<_{\mathcal{P}}P_{\mu}\quad Q<_{lex}P_{\mu}\right\\}$
---|---|---|---
| $\mathcal{E}_{\mu}^{(2)}$ | $\ =\ $ | $\left\\{\left.Q\in\mathcal{P}_{\mathbb{L}}(x)\ \vphantom{Q>_{lex}P_{\mu}}\right|\ Q>_{lex}P_{\mu}\right\\}$

Theses sets are disjoints and
$\mathcal{E}_{\mu}=\mathcal{E}_{\mu}^{(1)}\cup\mathcal{E}_{\mu}^{(2)}$

Let $\beta_{\mu}^{(i)}$ be the order type of $\mathcal{E}_{\mu}^{(i)}$. We
then have

$\mu\leq\beta_{\mu}\leq\beta_{\mu}^{(1)}+\beta_{\mu}^{(2)}$

where the addition is the surreal addition of ordinal numbers. Since $\mu$ is
multiplicative ordinal, hence, an additive one, at least one of the
$\beta_{\mu}^{(i)}\geq\mu$.

    * ➢

First case : $\beta_{\mu}^{(2)}\geq\mu$. Since $\mu$ is additive, there is an
$i\in\left\llbracket\,0\ ;\ k_{P}\,\right\rrbracket$ such that the well
ordered set

$\mathcal{E}_{\mu}^{(2,i)}=\left\\{\left.Q\in\mathcal{E}_{\mu}^{(2)}\
\vphantom{\forall j<i\ Q(j)=P_{\mu}(j)\quad Q(i)\prec P_{\mu}(i)}\right|\
\forall j<i\ Q(j)=P_{\mu}(j)\quad Q(i)\prec P_{\mu}(i)\right\\}$

has order type at least $\mu$. We take such an $i$. For
$Q\in\mathcal{E}_{\mu}^{(2,i)}$, we consider the path
$Q^{\prime}(n)=Q(n+i+1)$. Since $\partial Q\succeq\partial P_{\mu}$, Lemma 6.3
gives us that $Q(i+1)$ is a term of $\ell(P_{\mu}(i))$. We then have
$Q^{\prime}\in\mathcal{P}\left(\ell(P_{\mu}(i))\right)$ and

$\partial Q^{\prime}=\mathchoice{\dfrac{\partial Q}{Q(0)\cdots
Q(i)}}{\dfrac{\partial Q}{Q(0)\cdots Q(i)}}{\frac{\partial Q}{Q(0)\cdots
Q(i)}}{\frac{\partial Q}{Q(0)\cdots Q(i)}}=\mathchoice{\dfrac{\partial
Q}{P_{\mu}(0)\cdots P_{\mu}(i-1)Q(i)}}{\dfrac{\partial Q}{P_{\mu}(0)\cdots
P_{\mu}(i-1)Q(i)}}{\frac{\partial Q}{P_{\mu}(0)\cdots
P_{\mu}(i-1)Q(i)}}{\frac{\partial Q}{P_{\mu}(0)\cdots P_{\mu}(i-1)Q(i)}}$

In particular
$Q^{\prime}\in\mathcal{P}_{\mathbb{L}}\left(\ell(P_{\mu}(i))\right)$. Since
$Q(i)\prec P_{\mu}(i)$, $P_{\mu}(i)$ is not the last term of
$\ell(P_{\mu}(i-1))$ (or $x$ if $i=0$). Then Proposition 6.2.2 ensures that

$\operatorname{NR}(\ell(P_{\mu}(i)))\leq\operatorname{NR}(P_{\mu}(i))<\operatorname{NR}(x)$

Applying the induction hypothesis on $\ell(P_{\mu}(i))$, the order type of
$\mathcal{P}_{\mathbb{L}}(\left(\ell(P_{\mu}(i)))\right)$ has order type
$\gamma$ such that

$\gamma<\omega^{\omega^{\omega\left(\operatorname{NR}(\ell(P(i)))+1\right)}}\leq\omega^{\omega^{\omega\operatorname{NR}(x)}}<\omega^{\omega^{\omega\operatorname{NR}(x)+1}}\leq\mu$

For $Q,R\in\mathcal{E}_{\mu}^{(2,i)}$, $Q<_{\mathcal{P}}R$ iff

| $(Q(i)\partial Q^{\prime}\succ R(i)\partial R^{\prime})$ | $\ \vee\ $ | $(Q(i)\partial Q^{\prime}\asymp R(i)\partial R^{\prime}\wedge Q(i)\partial Q^{\prime}>R(i)\partial R^{\prime})$
---|---|---|---
|  |  | $\vee(Q(i)\partial Q^{\prime}=R(i)\partial R\wedge Q<_{lex}R)$

what we can also write

| $Q<_{\mathcal{P}}R$ | $\ \Leftrightarrow\ $ | $\big{(}\ell(Q(i))+\ell(\partial Q^{\prime})>\ell(R(i))+\ell(\partial R^{\prime})\big{)}$
---|---|---|---
|  |  | $\vee\big{(}\ell(Q(i))+\ell(\partial Q^{\prime})=\ell(R(i))+\ell(\partial R^{\prime})\wedge Q(i)\partial Q^{\prime}>R(i)\partial R^{\prime}\big{)}$
|  |  | $\vee(Q(i)\partial Q^{\prime}=R(i)\partial R\wedge Q<_{lex}R)$

where the two later cases occur finitely may times for $Q$ or $R$ fixed. Let
$\delta$ denote the order type of the possible values for $Q(i)$ and
$\beta_{\mu}^{(2,i)}$ the order type of $\mathcal{E}_{\mu}^{(2,i)}$. Since
$\ell$ is non-decreasing, the set $\left\\{\left.\ell(\partial Q^{\prime})\
\vphantom{Q\in\mathcal{E}_{\mu}^{(2,i)}}\right|\
Q\in\mathcal{E}_{\mu}^{(2,i)}\right\\}$ has order type at most $\gamma$ and
the set $\left\\{\left.\ell(Q(i))\
\vphantom{Q\in\mathcal{E}_{\mu}^{(2,i)}}\right|\
Q\in\mathcal{E}_{\mu}^{(2,i)}\right\\}$ has order type at most
$\operatorname{NR}(x)$. Using Proposition 2,

$\beta_{\mu}^{(2,i)}\leq(\gamma\operatorname{NR}(x))\otimes\omega<\mu$

Finally $\mu\leq\beta_{\mu}^{(2,i)}<\mu$

and we reach the contradiction.

    * ➢

Second case : $\beta_{\mu}^{(2)}<\mu$. Then $\beta_{\mu}^{(1)}\geq\mu$. Let us
define for $i\in\left\llbracket\,0\ ;\ k_{P}\,\right\rrbracket$

$\mathcal{E}_{\mu}^{(1,i)}=\left\\{\left.Q\in\mathcal{E}_{\mu}^{(1)}\
\vphantom{\forall j<i\ P_{\mu}(j)=Q(j)\quad P_{\mu}(i)\prec Q(i)}\right|\
\forall j<i\ P_{\mu}(j)=Q(j)\quad P_{\mu}(i)\prec Q(i)\right\\}$

Since there are finitely many of them, that they form a partition of
$\mathcal{E}_{\mu}^{(1)}$ and $\mu$ is multiplicative, hence additive, there
is at least one of them which has order type at least $\mu$. We consider such
an $i\in\left\llbracket\,0\ ;\ k_{P}\,\right\rrbracket$. Now define

$x_{j}=\left\\{\begin{array}[]{@{}r@{\quad}l@{}}x&i=j\\\
\ell(P(i-j-1))&j<i\end{array}\right.$

Writing
$x_{0}=\underset{n<\nu(x_{0})}{\overset{}{\sum}}r_{n}(x_{0})\omega^{a_{n}(x_{0})}$
and $P_{\mu}(i)=r_{\alpha_{0}}(x_{0})\omega^{a_{\alpha_{0}}(x_{0})}$ we set

$y_{0}=\underset{n<\alpha_{0}}{\overset{}{\sum}}r_{n}(x_{0})\omega^{a_{n}(x_{0})}$

Now for $0\leq j<i$, we define $y_{j+1}$ has follows. $P_{\mu}(i-j-1)$ is a
term of $x_{j+1}$. Write
$P_{\mu}(i-j-1)=r_{\alpha_{j+1}}(x_{j+1})\omega^{a_{\alpha_{j+1}}(x_{j+1})}$
for some ${\alpha_{j+1}}<\nu(x_{j+1})$. Then set

$y_{j+1}=\underset{n<\alpha_{j+1}}{\overset{}{\sum}}r_{n}(x_{j+1})\omega^{a_{n}(x_{j+1})}+\operatorname{sign}(r_{\alpha_{j+1}}(x_{j+1}))\exp(y_{j})$

Denote $y=y_{i}$. For $Q\in\mathcal{E}_{\mu}^{(1,i)}$. For any
$Q\in\mathcal{E}_{\mu}^{(1,i)}$ we will build
$Q^{\prime}\in\mathcal{P}_{\mathbb{L}}(y)$. We expect to use the induction
hypothesis on $y$. First we prove that
$\operatorname{NR}(y)<\operatorname{NR}(x)$. In fact, by trivial induction, we
have $y_{j}\triangleleft_{j}x_{j}$. So $y\triangleleft_{i}x$ and by definition
of $\operatorname{NR}$ we have $\operatorname{NR}(y)<\operatorname{NR}(x)$.
Now consider the path $Q^{\prime}$ defined as follows :

      * $\because$

$\forall j<i\qquad
Q^{\prime}(j)=\operatorname{sign}(r_{\alpha_{j}}(x_{i-j}))\exp(y_{i-j-1})$

      * $\because$

$\forall j\geq i\qquad Q^{\prime}(j)=Q(j)$

We then have $Q^{\prime}\in\mathcal{P}(y)$. We can even say
$Q^{\prime}\in\mathcal{P}_{\mathbb{L}}(y)$. Moreover, since we change only the
common terms of the path, and the changes do not depend on $Q$, we have

$\forall Q,R\in\mathcal{E}_{\mu}^{(1,i)}\qquad
Q<_{\mathcal{P}}R\Leftrightarrow Q^{\prime}<_{\mathcal{P}}R^{\prime}$

We then have an increasing function

$\Phi:\left\\{\begin{array}[]{c c
c}\mathcal{E}_{\mu}^{(1,i)}&\rightarrow&\mathcal{P}_{\mathbb{L}}(y)\\\