derivatives. (These conditions are satisfied if $K$ is an algebraically closed
differential field.) In the next subsection we show that up to isomorphism
over $K$ there is a unique exponential extension $R$ of $K$ satisfying
$C_{R}=C$ and ${(R^{\times})^{\dagger}=K}$. By Lemma II.2.3 we must then have
a group embedding $e\colon K\to R^{\times}$ such that $e(b)^{\dagger}=b$ for
all $b\in K$; this motivates the construction below.

#### The universal exponential extension

We first describe a certain exponential extension of $K$. For this, take a
complement $\Lambda$ of $K^{\dagger}$, that is, a ${\mathbb{Q}}$-linear
subspace of $K$ such that $K=K^{\dagger}\oplus\Lambda$ (internal direct sum of
${\mathbb{Q}}$-linear subspaces of $K$). Below $\lambda$ ranges over
$\Lambda$. Let $\operatorname{e}(\Lambda)$ be a multiplicatively written
abelian group, isomorphic to the additive subgroup $\Lambda$ of $K$, with
isomorphism
$\lambda\mapsto\operatorname{e}(\lambda)\colon\Lambda\to\operatorname{e}(\Lambda)$.
Put

${\operatorname{U}}\ :=\ K\big{[}\\!\operatorname{e}(\Lambda)\big{]},$

the group ring of $\operatorname{e}(\Lambda)$ over $K$, an integral domain. As
$K$-linear space,

${\operatorname{U}}\ =\
\bigoplus_{\lambda}K\operatorname{e}(\lambda)\qquad(\text{an internal direct
sum of $K$-linear subspaces}).$

For every $f\in{\operatorname{U}}$ we have a unique family $(f_{\lambda})$ in
$K$ such that

$f\ =\ \sum_{\lambda}f_{\lambda}\operatorname{e}(\lambda),$

with $f_{\lambda}=0$ for all but finitely many $\lambda$; we call
$(f_{\lambda})$ the spectral decomposition of $f$ (with respect to $\Lambda$).
We turn ${\operatorname{U}}$ into a differential ring extension of $K$ by

$\operatorname{e}(\lambda)^{\prime}\ =\
\lambda\operatorname{e}(\lambda)\qquad\text{for all $\lambda$.}$

(Think of $\operatorname{e}(\lambda)$ as $\exp(\int\lambda)$.) Thus for
$f\in{\operatorname{U}}$ with spectral decomposition $(f_{\lambda})$,

$f^{\prime}\ =\ \sum_{\lambda}\big{(}f_{\lambda}^{\prime}+\lambda
f_{\lambda}\big{)}\operatorname{e}(\lambda),$

so $f^{\prime}$ has spectral decomposition $(f_{\lambda}^{\prime}+\lambda
f_{\lambda})$. Note that ${\operatorname{U}}$ is exponential over $K$ by Lemma
II.1.1: ${\operatorname{U}}^{\times}=K^{\times}\operatorname{e}(\Lambda)$, so
$({\operatorname{U}}^{\times})^{\dagger}=K^{\dagger}+\Lambda=K$.

######  Example II.2.4.

Let $K=C(\\!(t^{{\mathbb{Q}}})\\!)$ be as in Example I.2.12, so
$K^{\dagger}=({\mathbb{Q}}\oplus\smallo)t$. Take a ${\mathbb{Q}}$-linear
subspace $\Lambda_{\operatorname{c}}$ of $C$ with
$C={\mathbb{Q}}\oplus\Lambda_{\operatorname{c}}$ (internal direct sum of
${\mathbb{Q}}$-linear subspaces of $C$), and let

$K_{\succ}\ :=\ \big{\\{}f\in K:\ \operatorname{supp}(f)\succ 1\big{\\}},$

a $C$-linear subspace of $K$. Then
$\Lambda:=(K_{\succ}\oplus\Lambda_{\operatorname{c}})t$ is a complement to
$K^{\dagger}$, and hence
$t^{-1}\Lambda=K_{\succ}\oplus\Lambda_{\operatorname{c}}$ is a complement to
$(K^{t})^{\dagger}$ in $K^{t}$. Moreover, if $L:=\operatorname{P}(C)\subseteq
K$ is the differential field of Puiseux series over $C$ and
$L_{\succ}:=K_{\succ}\cap L$, then $L_{\succ}\oplus\Lambda_{\operatorname{c}}$
is a complement to $(L^{t})^{\dagger}$.

A subgroup $\Lambda_{0}$ of $\Lambda$ yields a differential subring
$K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}$ of ${\operatorname{U}}$ that
is exponential over $K$ as well. These differential subrings have a useful
property. Recall from [ADH, 4.6] that a differential ring is said to be simple
if $\\{0\\}$ is its only proper differential ideal.

######  Lemma II.2.5.

Let $\Lambda_{0}$ be a subgroup of $\Lambda$. Then the differential subring
$K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}$ of ${\operatorname{U}}$ is
simple. In particular, the differential ring ${\operatorname{U}}$ is simple.

###### Proof.

Let $I\neq R$ be a differential ideal of
$R:=K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}$. Let
$f_{1},\dots,f_{n}\in K^{\times}$ and let
$\lambda_{1},\dots,\lambda_{n}\in\Lambda_{0}$ be distinct such that
$f=\sum_{i=1}^{n}f_{i}\operatorname{e}(\lambda_{i})\in I$. If $n\geqslant 1$,
then Lemma II.2.1 yields $i\neq j$ with
$\operatorname{e}(\lambda_{i})/\operatorname{e}(\lambda_{j})=cg$ for some
constant $c$ in the differential fraction field of ${\operatorname{U}}$ and
some $g\in K^{\times}$, so by taking logarithmic derivatives,
$\lambda_{i}-\lambda_{j}\in K^{\dagger}$ and thus $\lambda_{i}=\lambda_{j}$, a
contradiction. Thus $f=0$. ∎

######  Corollary II.2.6.

Any morphism $K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}\to R$ of
differential $K$-algebras, with $\Lambda_{0}$ a subgroup of $\Lambda$ and $R$
a differential ring extension of $K$, is injective.

The differential ring ${\operatorname{U}}$ is the directed union of its
differential subrings of the form
${\operatorname{U}}_{0}=K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}$ where
$\Lambda_{0}$ is a finitely generated subgroup of $\Lambda$. These
${\operatorname{U}}_{0}$ are simple by Lemma II.2.5 and finitely generated as
a $K$-algebra, hence their differential fraction fields have constant field
$C$ by [ADH, 4.6.12]. Thus the differential fraction field of
${\operatorname{U}}$ has constant field $C$.

######  Lemma II.2.7.

Suppose $R$ is an exponential extension of $K$ and $R_{0}$ is a differential
subring of $R$ with $C_{R}^{\times}\subseteq C_{R_{0}}$ and
$K\subseteq(R_{0}^{\times})^{\dagger}$. Then $R_{0}=R$.

###### Proof.

Let $E$ be the group of units of $R$ that are exponential over $K$; so
$R=K[E]$. Given $u\in E$ we have $u^{\dagger}\in
K\subseteq(R_{0}^{\times})^{\dagger}$, hence we have $u_{0}\in R_{0}^{\times}$
with $u^{\dagger}=u_{0}^{\dagger}$, so $u=cu_{0}$ with $c\in
C_{R}^{\times}\subseteq C_{R_{0}}$. Thus $E\subseteq R_{0}$ and so $R_{0}=R$.
∎

######  Corollary II.2.8.

Every endomorphism of the differential $K$-algebra ${\operatorname{U}}$ is an
automorphism.

###### Proof.

Injectivity holds by Corollary II.2.6, and surjectivity by Lemma II.2.7. ∎

Every exponential extension of $K$ with constant field $C$ embeds into
${\operatorname{U}}$, and hence is an integral domain. More precisely:

######  Lemma II.2.9.

Let $R$ be an exponential extension of $K$ such that $C_{R}^{\times}$ is
divisible, and set $\Lambda_{0}:=\Lambda\cap(R^{\times})^{\dagger}$, a
subgroup of $\Lambda$. Then there exists a morphism
$K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}\to R$ of differential
$K$-algebras. Any such morphism is injective, and if $C_{R}=C$, then any such
morphism is an isomorphism.

###### Proof.

Let $E$ be as in the proof of Lemma II.2.7, and let $e_{E}\colon
E^{\dagger}\to E$ be the map $e$ from Lemma II.2.3. Since
$E^{\dagger}=K^{\dagger}+\Lambda_{0}$ we have

(II.2.1) $E\ =\ C_{R}^{\times}\,e_{E}(E^{\dagger})\ =\
C_{R}^{\times}\,e_{E}(K^{\dagger})\,e_{E}(\Lambda_{0})\ =\
C_{R}^{\times}\,K^{\times}\,e_{E}(\Lambda_{0}).$

The group morphism $\operatorname{e}(\lambda_{0})\mapsto
e_{E}(\lambda_{0})\colon\operatorname{e}(\Lambda_{0})\to E$
($\lambda_{0}\in\Lambda_{0}$) extends uniquely to a $K$-algebra morphism
$\iota\colon K\big{[}\\!\operatorname{e}(\Lambda_{0})\big{]}\to R=K[E]$. One
verifies easily that $\iota$ is a differential ring morphism. The injectivity
claim follows from Corollary II.2.6. If $C_{R}=C$, then
$E=K^{\times}e_{E}(\Lambda_{0})$ by (II.2.1), whence surjectivity. ∎

Recall that ${\operatorname{U}}$ is an exponential extension of $K$ with
$C_{{\operatorname{U}}}=C$ and $({\operatorname{U}}^{\times})^{\dagger}=K$. By
Lemma II.2.9, this property characterizes ${\operatorname{U}}$ up to
isomorphism:

######  Corollary II.2.10.

If $U$ is an exponential extension of $K$ such that $C_{U}=C$ and
$K\subseteq(U^{\times})^{\dagger}$, then $U$ is isomorphic to
${\operatorname{U}}$ as a differential $K$-algebra.

Now ${\operatorname{U}}$ is also an exponential extension of $K$ with
$C_{{\operatorname{U}}}=C$ and with the property that every exponential
extension $R$ of $K$ with $C_{R}=C$ embeds into ${\operatorname{U}}$ as a
differential $K$-algebra. This property determines ${\operatorname{U}}$ up to
isomorphism as well:

######  Corollary II.2.11.

Suppose $U$ is an exponential extension of $K$ with $C_{U}=C$ such that every
exponential extension $R$ of $K$ with $C_{R}=C$ embeds into $U$ as a
differential $K$-algebra. Then $U$ is isomorphic to ${\operatorname{U}}$ as a
differential $K$-algebra.

###### Proof.

Any embedding ${\operatorname{U}}\to U$ of differential $K$-algebras gives
$K\subseteq(U^{\times})^{\dagger}$. ∎

The results above show to what extent ${\operatorname{U}}$ is independent of
the choice of $\Lambda$. We call ${\operatorname{U}}$ the universal
exponential extension of $K$. If we need to indicate the dependence of
${\operatorname{U}}$ on $K$ we denote it by ${\operatorname{U}}_{K}$. By [ADH,
5.1.40] every ${y\in{\operatorname{U}}=K\\{\operatorname{e}(\Lambda)\\}}$
satisfies a linear differential equation $A(y)=0$ where $A\in K[\der]^{\neq}$;
in the next section we isolate conditions on $K$ which ensure that every $A\in
K[\der]^{\neq}$ has a zero
$y\in{\operatorname{U}}^{\times}=K^{\times}\operatorname{e}(\Lambda)$.

Corollary II.2.10 gives for $\phi\in K^{\times}$ an isomorphism
${\operatorname{U}}_{K^{\phi}}\cong({\operatorname{U}}_{K})^{\phi}$ of
differential $K^{\phi}$-algebras. Next we investigate how
${\operatorname{U}}_{K}$ behaves when passing from $K$ to a differential field
extension. Therefore, in the rest of this subsection $L$ is a differential
field extension of $K$ with algebraically closed constant field $C_{L}$, and
$L^{\dagger}$ is divisible. The next lemma relates the universal exponential
extension ${\operatorname{U}}_{L}$ of $L$ to ${\operatorname{U}}_{K}$:

######  Lemma II.2.12.

The inclusion $K\to L$ extends to an embedding
$\iota\colon{\operatorname{U}}_{K}\to{\operatorname{U}}_{L}$ of differential
rings. The image of any such embedding $\iota$ is contained in $K[E]$ where
$E:=\\{u\in{\operatorname{U}}_{L}^{\times}:u^{\dagger}\in K\\}$, and if
$C_{L}=C$, then $\iota({\operatorname{U}}_{K})=K[E]$.

###### Proof.

The differential subring $R:=K[E]$ of ${\operatorname{U}}_{L}$ is exponential
over $K$ with $(R^{\times})^{\dagger}=K$, hence Lemma II.2.9 gives an
embedding ${\operatorname{U}}_{K}\to R$ of differential $K$-algebras. Let
$\iota\colon{\operatorname{U}}_{K}\to{\operatorname{U}}_{L}$ be any embedding
of differential $K$-algebras. Then
$\iota\big{(}\\!\operatorname{e}(\Lambda)\big{)}\subseteq E$, so
$\iota({\operatorname{U}}_{K})\subseteq R$; if $C_{L}=C$, then
$\iota({\operatorname{U}}_{K})=R$ by Lemma II.2.7. ∎

######  Corollary II.2.13.

If $L^{\dagger}\cap K=K^{\dagger}$ and
$\iota\colon{\operatorname{U}}_{K}\to{\operatorname{U}}_{L}$ is an embedding
of differential $K$-algebras, then
$L^{\times}\cap\iota({\operatorname{U}}_{K}^{\times})=K^{\times}$.

###### Proof.

Assume $L^{\dagger}\cap K=K^{\dagger}$ and identify ${\operatorname{U}}_{K}$
with a differential $K$-subalgebra of ${\operatorname{U}}_{L}$ via an
embedding ${\operatorname{U}}_{K}\to{\operatorname{U}}_{L}$ of differential
$K$-algebras. Let $a\in L^{\times}\cap{\operatorname{U}}_{K}^{\times}$; then
$a^{\dagger}\in L^{\dagger}\cap K=K^{\dagger}$, so $a=bc$ where $c\in
C_{L}^{\times}$, $b\in K^{\times}$. Now $c=a/b\in
C_{L}^{\times}\cap{\operatorname{U}}_{K}^{\times}=C^{\times}$, since
${\operatorname{U}}_{K}$ has ring of constants $C$. So $a\in K^{\times}$ as
required. ∎

Suppose $L^{\dagger}\cap K=K^{\dagger}$. Then the subspace $L^{\dagger}$ of
the ${\mathbb{Q}}$-linear space $L$ has a complement
$\Lambda_{L}\supseteq\Lambda$. We fix such $\Lambda_{L}$ and extend
$\operatorname{e}\colon\Lambda\to\operatorname{e}(\Lambda)$ to a group
isomorphism $\Lambda_{L}\to\operatorname{e}(\Lambda_{L})$, also denoted by
$\operatorname{e}$, with $\operatorname{e}(\Lambda_{L})$ a multiplicatively
written commutative group extending $\operatorname{e}(\Lambda)$. Let
${\operatorname{U}}_{L}:=L\big{[}\\!\operatorname{e}(\Lambda_{L})\big{]}$ be
the corresponding universal exponential extension of $L$. Then the natural
inclusion ${\operatorname{U}}_{K}\to{\operatorname{U}}_{L}$ is an embedding of
differential $K$-algebras.

#### Automorphisms of ${\operatorname{U}}$

These are easy to describe: the beginning of Section II.1 gives a group
embedding

$\chi\mapsto\sigma_{\chi}\colon\operatorname{Hom}(\Lambda,K^{\times})\to\operatorname{Aut}\\!\big{(}K[\operatorname{e}(\Lambda)]|K\big{)}$

into the group of $K$-algebra automorphisms of
$K\big{[}\\!\operatorname{e}(\Lambda)\big{]}$, given by

$\sigma_{\chi}(f)\ :=f_{\chi}\ =\
\sum_{\lambda}f_{\lambda}\chi(\lambda)\operatorname{e}(\lambda)\qquad(\chi\in\operatorname{Hom}(\Lambda,K^{\times}),\
f\in K[\operatorname{e}(\Lambda)]).$

It is easy to check that if
$\chi\in\operatorname{Hom}(\Lambda,C^{\times})\subseteq\operatorname{Hom}(\Lambda,K^{\times})$,
then $\sigma_{\chi}\in\operatorname{Aut}({\operatorname{U}}|K)$, that is,
$\sigma_{\chi}$ is a differential $K$-algebra automorphism of
${\operatorname{U}}$. Moreover:

######  Lemma II.2.14.

The map
$\chi\mapsto\sigma_{\chi}\colon\operatorname{Hom}(\Lambda,C^{\times})\to\operatorname{Aut}({\operatorname{U}}|K)$
is a group isomorphism. Its inverse assigns to any
$\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$ the function
$\chi\colon\Lambda\to C^{\times}$ given by
$\chi(\lambda):=\sigma\big{(}\\!\operatorname{e}(\lambda)\big{)}\operatorname{e}(-\lambda)$.
In particular, $\operatorname{Aut}({\operatorname{U}}|K)$ is commutative.

###### Proof.

Let $\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$ and let
$\chi\colon\Lambda\to{\operatorname{U}}^{\times}$ be given by
$\chi(\lambda):=\sigma\big{(}\\!\operatorname{e}(\lambda)\big{)}\operatorname{e}(-\lambda)$.
Then $\chi(\lambda)^{\dagger}=0$ for all $\lambda$. It follows easily that
$\chi\in\operatorname{Hom}(\Lambda,C^{\times})$ and $\sigma_{\chi}=\sigma$. ∎

The proof of the next result uses that the additive group ${\mathbb{Q}}$
embeds into $C^{\times}$.

######  Corollary II.2.15.

If $f\in{\operatorname{U}}$ and $\sigma(f)=f$ for all
$\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$, then $f\in K$.

###### Proof.

Suppose $f\in U$ and $\sigma(f)=f$ for all
$\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$. For
$\chi\in\operatorname{Hom}(\Lambda,C^{\times})$ we have $f_{\chi}=f$, that is,
$f_{\lambda}\chi(\lambda)=f_{\lambda}$ for all $\lambda$, so $\chi(\lambda)=1$
whenever $f_{\lambda}\neq 0$. Now use that for $\lambda\neq 0$ there exists
$\chi\in\operatorname{Hom}(\Lambda,C^{\times})$ such that $\chi(\lambda)\neq
1$, so $f_{\lambda}=0$. ∎

######  Corollary II.2.16.

Every automorphism of the differential field $K$ extends to an automorphism of
the differential ring ${\operatorname{U}}$.

###### Proof.

Lemma II.2.3 yields a group morphism $\mu\colon
K\to{\operatorname{U}}^{\times}$ such that $\mu(a)^{\dagger}=a$ for all $a\in
K$. Let $\sigma\in\operatorname{Aut}(K)$. Then $\sigma$ extends to an
endomorphism, denoted also by $\sigma$, of the ring ${\operatorname{U}}$, such
that
$\sigma\big{(}\\!\operatorname{e}(\lambda)\big{)}=\mu\big{(}\sigma(\lambda)\big{)}$
for all $\lambda$. Then

$\sigma\big{(}\\!\operatorname{e}(\lambda)^{\prime}\big{)}\ =\
\sigma\big{(}\lambda\operatorname{e}(\lambda)\big{)}\ =\
\sigma(\lambda)\mu\big{(}\sigma(\lambda)\big{)}\ =\
\mu\big{(}\sigma(\lambda)\big{)}^{\prime}\ =\
\sigma\big{(}\\!\operatorname{e}(\lambda)\big{)}^{\prime},$

hence $\sigma$ is an endomorphism of the differential ring
${\operatorname{U}}$. By Lemma II.2.5, $\sigma$ is injective, and by Lemma
II.2.7, $\sigma$ is surjective. ∎

#### The real case

In this subsection $K=H[\imag]$ where $H$ is a real closed differential
subfield of $K$ and $\imag^{2}=-1$. Set $S_{C}:=\big{\\{}c\in
C:\,|c|=1\big{\\}}$, a subgroup of $C^{\times}$. Then by Lemmas II.1.7 and
II.2.14:

######  Corollary II.2.17.

For $\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$ we have the
equivalence

$\sigma(f^{*})=\sigma(f)^{*}\text{ for all
$f\in{\operatorname{U}}$}\quad\Longleftrightarrow\quad\sigma=\sigma_{\chi}\text{
for some $\chi\in\operatorname{Hom}(\Lambda,S_{C})$.}$

Corollaries II.2.17 and II.1.8 together give:

######  Corollary II.2.18.

Let $\sigma\in\operatorname{Aut}({\operatorname{U}}|K)$ satisfy
$\sigma(f^{*})=\sigma(f)^{*}$ for all $f\in{\operatorname{U}}$. Then
$\big{\langle}\sigma(f),\sigma(g)\big{\rangle}=\langle f,g\rangle$ for all
$f,g\in{\operatorname{U}}$, hence $\lVert\sigma(f)\rVert=\lVert f\rVert$ for
all $f\in{\operatorname{U}}$.

Next we consider the subgroup

$S\ :=\ \\{a+b\imag:\ a,b\in H,\ a^{2}+b^{2}=1\\}$

of $K^{\times}$, which is divisible, hence so is the subgroup $S^{\dagger}$ of
$K^{\dagger}$. Lemma I.2.4 yields $K^{\dagger}=H^{\dagger}\oplus S^{\dagger}$
(internal direct sum of ${\mathbb{Q}}$-linear subspaces of $K$) and
$S^{\dagger}\subseteq H\imag$. Thus we can (and do) take the complement
$\Lambda$ of $K^{\dagger}$ in $K$ so that
$\Lambda=\Lambda_{\operatorname{r}}+\Lambda_{\operatorname{i}}\imag$ where
$\Lambda_{\operatorname{r}},\Lambda_{\operatorname{i}}$ are subspaces of the
${\mathbb{Q}}$-linear space $H$ with $\Lambda_{\operatorname{r}}$ a complement
of $H^{\dagger}$ in $H$ and $\Lambda_{\operatorname{i}}\imag$ a complement of
$S^{\dagger}$ in $H\imag$. The automorphism
$a+b\imag\mapsto\overline{a+b\imag}:=a-b\imag$ (${a,b\in H}$) of the
differential field $K$ now satisfies in
${\operatorname{U}}=K[\operatorname{e}(\Lambda)]$ the identity

$\operatorname{e}(\overline{\lambda+\mu})\ =\
\operatorname{e}(\overline{\lambda})\operatorname{e}(\overline{\mu})\qquad(\lambda,\mu\in\Lambda),$

so it extends to an automorphism $f\mapsto\overline{f}$ of the ring
${\operatorname{U}}$ as follows: for $f\in{\operatorname{U}}$ with spectral
decomposition $(f_{\lambda})$, set

$\overline{f}\ :=\
\sum_{\lambda}\overline{f_{\lambda}}\operatorname{e}(\overline{\lambda})\ =\
\sum_{\lambda}\overline{f_{\overline{\lambda}}}\operatorname{e}(\lambda),$

so
$\overline{\operatorname{e}(\lambda)}=\operatorname{e}(\overline{\lambda})$,
and $\overline{f}$ has spectral decomposition
$(\overline{f_{\overline{\lambda}}})$. We have $\overline{\overline{f}}=f$ for
$f\in{\operatorname{U}}$, and $f\mapsto\overline{f}$ lies in
$\operatorname{Aut}({\operatorname{U}}|H)$. If $H^{\dagger}=H$, then
$\Lambda_{r}=\\{0\\}$ and hence $\overline{f}=f^{*}$ for
$f\in{\operatorname{U}}$, where $f^{*}$ is as defined in Section II.1. For
$f\in{\operatorname{U}}$ we set

$\operatorname{Re}f\ :=\
\textstyle\frac{1}{2}(f+\overline{f}),\qquad\operatorname{Im}f\ :=\
\textstyle\frac{1}{2\imag}(f-\overline{f}).$

(For $f\in K$ these agree with the usual real and imaginary parts of $f$ as an
element of $H[\imag]$.) Consider the differential $H$-subalgebra

${\operatorname{U}}_{\operatorname{r}}\ :=\
\big{\\{}f\in{\operatorname{U}}:\overline{f}=f\big{\\}}$

of ${\operatorname{U}}$. For $f\in{\operatorname{U}}$ with spectral
decomposition $(f_{\lambda})$ we have
$f\in{\operatorname{U}}_{\operatorname{r}}$ iff
$f_{\overline{\lambda}}=\overline{f_{\lambda}}$ for all $\lambda$; in
particular ${\operatorname{U}}_{\operatorname{r}}\cap K=H$. For
$f\in{\operatorname{U}}$ we have
$f=(\operatorname{Re}f)+(\operatorname{Im}f)\imag$ with
$\operatorname{Re}f,\operatorname{Im}f\in{\operatorname{U}}_{\operatorname{r}}$,
hence

${\operatorname{U}}\ =\
{\operatorname{U}}_{\operatorname{r}}\oplus{\operatorname{U}}_{\operatorname{r}}\imag\quad\text{
(internal direct sum of $H$-linear subspaces).}$

Let $D$ be a subfield of $H$ (not necessarily the constant field of $H$), so
$D[\imag]$ is a subfield of $K$. Let $V$ be a $D[\imag]$-linear subspace of
${\operatorname{U}}$; then
$V_{\operatorname{r}}:=V\cap{\operatorname{U}}_{\operatorname{r}}$ is a
$D$-linear subspace of $V$. If $\overline{V}=V$ (that is, $V$ is closed under
$f\mapsto\overline{f}$), then $\operatorname{Re}f,\operatorname{Im}f\in
V_{\operatorname{r}}$ for all $f\in V$, hence $V=V_{\operatorname{r}}\oplus
V_{\operatorname{r}}\imag$ (internal direct sum of $D$-linear subspaces of
$V$), so any basis of the $D$-linear space $V_{\operatorname{r}}$ is a basis
of the $D[\imag]$-linear space $V$.

Suppose now that $V=\bigoplus_{\lambda}V_{\lambda}$ (internal direct sum of
subspaces of $V$) where $V_{\lambda}$ is for each $\lambda$ a
$D[\imag]$-linear subspace of $K\operatorname{e}(\lambda)$. Then
$\overline{V}=V$ iff $V_{\overline{\lambda}}=\overline{V_{\lambda}}$ for all
$\lambda$. Moreover:

######  Lemma II.2.19.

Assume $H=H^{\dagger}$, $V_{0}=\\{0\\}$, and $\overline{V}=V$. Let
$\mathcal{V}\subseteq{\operatorname{U}}^{\times}$ be a basis of the subspace
$\sum_{\operatorname{Im}\lambda>0}V_{\lambda}$ of $V$. Then the maps
$v\mapsto\operatorname{Re}v,\ v\mapsto\operatorname{Im}v\colon\mathcal{V}\to
V_{\operatorname{r}}$ are injective, $\operatorname{Re}\mathcal{V}$ and
$\operatorname{Im}\mathcal{V}$ are disjoint, and
$\operatorname{Re}\mathcal{V}\cup\operatorname{Im}\mathcal{V}$ is a basis of
$V_{\operatorname{r}}$.

###### Proof.

Note that $\Lambda=\Lambda_{\operatorname{i}}\imag$. Let $\mu$ range over
$\Lambda_{\operatorname{i}}^{>}$ and set $\cal{V}_{\mu}=\cal{V}\cap
K^{\times}\operatorname{e}(\mu\imag)$, a basis of the $D[\imag]$-linear space
$V_{\mu\imag}$. Then $\cal{V}=\bigcup_{\mu}\cal{V}_{\mu}$, a disjoint union.
For $v\in\cal{V}_{\mu}$ we have $v=a\operatorname{e}(\mu\imag)$ with
$a=a_{v}\in K^{\times}$, so

$\operatorname{Re}v\ =\
\textstyle\frac{a}{2}\operatorname{e}(\mu\imag)+\textstyle\frac{\overline{a}}{2}\operatorname{e}(-\mu\imag),\qquad\operatorname{Im}v\
=\
\textstyle\frac{a}{2i}\operatorname{e}(\mu\imag)-\textstyle\frac{\overline{a}}{2\imag}\operatorname{e}(-\mu\imag),$

from which it is clear that the two maps $\cal{V}\to V_{\operatorname{r}}$ in
the statement of the lemma are injective. It is also easy to check that
$\operatorname{Re}\mathcal{V}$ and $\operatorname{Im}\mathcal{V}$ are
disjoint.

As $\mathcal{V}$ is a basis of the $D[\imag]$-linear space
$\sum_{\mu}V_{\mu\imag}=\sum_{\operatorname{Im}\lambda>0}V_{\lambda}$, its set
of conjugates $\overline{\mathcal{V}}$ is a basis of the $D[\imag]$-linear
space
$\sum_{\mu}\overline{V_{\mu\imag}}=\sum_{\mu}V_{-\mu\imag}=\sum_{\operatorname{Im}\lambda<0}V_{\lambda}$,
and so $\mathcal{V}\cup\overline{\mathcal{V}}$ (a disjoint union) is a basis
of $V$. Thus $\operatorname{Re}\mathcal{V}\cup\operatorname{Im}\mathcal{V}$ is
a basis of $V$ as well. As
$\operatorname{Re}\mathcal{V}\cup\operatorname{Im}\mathcal{V}$ is contained in
$V_{\operatorname{r}}$, it is a basis of the $D$-linear space
$V_{\operatorname{r}}$. ∎

If $H=H^{\dagger}$, then $V:=\sum_{\lambda\neq 0}K\operatorname{e}(\lambda)$
gives $\overline{V}=V$, so Lemma II.2.19 gives then for $D:=H$ the basis of
the $H$-linear space $V_{\operatorname{r}}$ consisting of the elements

$\operatorname{Re}\\!\big{(}\\!\operatorname{e}(\lambda)\big{)}\ =\
\textstyle\frac{1}{2}\big{(}\\!\operatorname{e}(\lambda)+\operatorname{e}(\overline{\lambda})\big{)},\qquad\operatorname{Im}\\!\big{(}\\!\operatorname{e}(\lambda)\big{)}\
=\
\textstyle\frac{1}{2\imag}\big{(}\\!\operatorname{e}(\lambda)-\operatorname{e}(\overline{\lambda})\big{)}\qquad(\operatorname{Im}\lambda>0).$

######  Corollary II.2.20.

Suppose $H=H^{\dagger}$. Set
$\operatorname{c}(\lambda):=\operatorname{Re}\\!\big{(}\\!\operatorname{e}(\lambda)\big{)}$
and
$\operatorname{s}(\lambda):=\operatorname{Im}\\!\big{(}\\!\operatorname{e}(\lambda)\big{)}$,
for $\operatorname{Im}\lambda>0$. Then for $V:=\sum_{\lambda\neq
0}K\operatorname{e}(\lambda)$ we have
${\operatorname{U}}_{\operatorname{r}}=H+V_{\operatorname{r}}$, so

${\operatorname{U}}_{\operatorname{r}}\ =\
H\oplus\bigoplus_{\operatorname{Im}\lambda>0}\big{(}H\operatorname{c}(\lambda)\oplus
H\operatorname{s}(\lambda)\big{)}\quad\text{{(}internal direct sum of
$H$-linear subspaces{)},}$

and thus
${\operatorname{U}}_{\operatorname{r}}=H\big{[}\operatorname{c}(\Lambda_{\operatorname{i}}^{>}\imag)\cup\operatorname{s}(\Lambda_{\operatorname{i}}^{>}\imag)\big{]}$.

###  II.3. The Spectrum of a Differential Operator

In this section $K$ is a differential field, $a$, $b$ range over $K$, and $A$,
$B$ over $K[\der]$. This and the next two sections are mainly differential-
algebraic in nature, and deal with splittings of linear differential
operators. In the present section we introduce the concept of eigenvalue of
$A$ and the spectrum of $A$ (the collection of its eigenvalues). In Section
II.4 we show how the eigenvalues of $A$ relate to the behavior of $A$ over the
universal exponential extension of $K$.

#### Twisting

Let $L$ be a differential field extension of $K$ with $L^{\dagger}\supseteq
K$. Let $u\in L^{\times}$ be such that $u^{\dagger}=a\in K$. Then the twist
$A_{\ltimes u}=u^{-1}Au$ of $A$ by $u$ has the same order as $A$ and
coefficients in $K$ [ADH, 5.8.8], and only depends on $a$, not on $u$ or $L$;
in fact, $\operatorname{Ri}(A_{\ltimes u})=\operatorname{Ri}(A)_{+a}$ [ADH,
5.8.5]. Hence for each $a$ we may define

$A_{a}\ :=\ A_{\ltimes u}\ =\ u^{-1}Au\in K[\der]$

where $u\in L^{\times}$ is arbitrary with $u^{\dagger}=a$. The map $A\mapsto
A_{\ltimes u}$ is an automorphism of the ring $K[\der]$ that is the identity
on $K$ (with inverse $B\mapsto B_{\ltimes u^{-1}}$); so $A\mapsto A_{a}$ is an
automorphism of the ring $K[\der]$ that is the identity on $K$ (with inverse
$B\mapsto B_{-a}$). Note that ${}_{a}=\der+a$, and that

$(a,A)\mapsto A_{a}\ :\ K\times K[\der]\to K[\der]$

is an action of the additive group of $K$ on the set $K[\der]$, in particular,
$A_{a}=A$ for $a=0$. For $b\neq 0$ we have $(A_{a})_{\ltimes
b}=A_{a+b^{\dagger}}$.

#### Eigenvalues

In the rest of this section $A\neq 0$ and $r:=\operatorname{order}(A)$. We
call

$\operatorname{mult}_{a}(A)\ :=\ \dim_{C}\ker_{K}A_{a}\in\\{0,\dots,r\\}$

the multiplicity of $A$ at $a$. If $B\neq 0$, then
$\operatorname{mult}_{a}(B)\leqslant\operatorname{mult}_{a}(AB)$, as well as

(II.3.1) $\operatorname{mult}_{a}(AB)\ \leqslant\
\operatorname{mult}_{a}(A)+\operatorname{mult}_{a}(B),$

with equality if and only if $B_{a}(K)\supseteq\ker_{K}A_{a}$; see [ADH,
remarks before 5.1.12]. For $u\in K^{\times}$ we have an isomorphism

$y\mapsto yu\ \colon\ \ker_{K}A_{\ltimes u}\to\ker_{K}A$

of $C$-linear spaces, hence

$\operatorname{mult}_{a}(A)\ =\ \operatorname{mult}_{b}(A)\qquad\text{whenever
$a-b\in K^{\dagger}$.}$

Thus we may define the multiplicity of $A$ at the element $[a]:=a+K^{\dagger}$
of $K/K^{\dagger}$ as
$\operatorname{mult}_{[a]}(A):=\operatorname{mult}_{a}(A)$.

In the rest of this section $\alpha$ ranges over $K/K^{\dagger}$. We say that
$\alpha$ is an eigenvalue of $A$ if $\operatorname{mult}_{\alpha}(A)\geqslant
1$. Thus for $B\neq 0$: if $\alpha$ is an eigenvalue of $B$ of multiplicity
$\mu$, then $\alpha$ is an eigenvalue of $AB$ of multiplicity $\geqslant\mu$;
if $\alpha$ is an eigenvalue of $AB$, then it is an eigenvalue of $A$ or of
$B$; and if $B_{a}(K)\supseteq\ker_{K}(A_{a})$, then $\alpha=[a]$ is an
eigenvalue of $AB$ if and only if it is an eigenvalue of $A$ or of $B$.

######  Example II.3.1.

Suppose $A=\der-a$. Then for each element $u\neq 0$ in a differential field
extension of $K$ with $b:=u^{\dagger}\in K$ we have $A_{b}=A_{\ltimes
u}=\der-(a-b)$, so $\operatorname{mult}_{b}(A)\geqslant 1$ iff $a-b\in
K^{\dagger}$. Hence the only eigenvalue of $A$ is $[a]$.

The spectrum of $A$ is the set $\Sigma(A)=\Sigma_{K}(A)$ of its eigenvalues.
Thus $\Sigma(A)=\emptyset$ if $r=0$, and for $b\neq 0$ we have
$\operatorname{mult}_{a}(A)=\operatorname{mult}_{a}(bA)=\operatorname{mult}_{a}(A_{\ltimes
b})$, so $A$, $bA$, and $Ab=bA_{\ltimes b}$ all have the same spectrum. By
[ADH, 5.1.21] we have

(II.3.2) $\Sigma(A)=\big{\\{}\alpha:A\in K[\der](\der-a)\text{ for some $a$
with $[a]=\alpha$}\big{\\}}.$

Hence for irreducible $A$: $\ \Sigma(A)\neq\emptyset\ \Leftrightarrow\ r=1$.
From (II.3.1) we obtain:

######  Lemma II.3.2.

Suppose $B\neq 0$ and set $s:=\operatorname{order}B$. Then

$\operatorname{mult}_{\alpha}(B)\ \leqslant\ \operatorname{mult}_{\alpha}(AB)\
\leqslant\ \operatorname{mult}_{\alpha}(A)+\operatorname{mult}_{\alpha}(B),$

where the second inequality is an equality if $K$ is $s$-linearly surjective.
Hence

$\Sigma(B)\ \subseteq\ \Sigma(AB)\ \subseteq\ \Sigma(A)\cup\Sigma(B).$

If $K$ is $s$-linearly surjective, then $\Sigma(AB)=\Sigma(A)\cup\Sigma(B)$.

######  Example.

For $n\geqslant 1$ we have
$\Sigma\big{(}(\der-a)^{n}\big{)}=\big{\\{}[a]\big{\\}}$. (By induction on
$n$, using Example II.3.1 and Lemma II.3.2.)

It follows from Lemma II.3.2 that $A$ has at most $r$ eigenvalues. More
precisely:

######  Lemma II.3.3.

We have $\sum_{\alpha}\operatorname{mult}_{\alpha}(A)\leqslant r$. If
$\sum_{\alpha}\operatorname{mult}_{\alpha}(A)=r$, then $A$ splits over $K$;
the converse holds if $r=1$ or $K$ is $1$-linearly surjective.

###### Proof.

By induction on $r$. The case $r=0$ is obvious, so suppose $r>0$. We may also
assume $\Sigma(A)\neq\emptyset$: otherwise
$\sum_{\alpha}\operatorname{mult}_{\alpha}(A)=0$ and $A$ does not split over
$K$. Now (II.3.2) gives $a$, $B$ with $A=B(\der-a)$. By Example II.3.1 we have
$\Sigma(\der-a)=\big{\\{}[a]\big{\\}}$ and
$\operatorname{mult}_{a}(\der-a)=1$. By the inductive hypothesis applied to
$B$ and the second inequality in Lemma II.3.2 we thus get
$\sum_{\alpha}\operatorname{mult}_{\alpha}(A)\leqslant r$.

Suppose that $\sum_{\alpha}\operatorname{mult}_{\alpha}(A)=r$. Then
$\sum_{\alpha}\operatorname{mult}_{\alpha}(B)=r-1$ by Lemma II.3.2 and the
inductive hypothesis applied to $B$. Therefore $B$ splits over $K$, again by
the inductive hypothesis, and so does $A$. Finally, if $K$ is $1$-linearly
surjective and $A$ splits over $K$, then we arrange that $B$ splits over $K$,
so $\sum_{\alpha}\operatorname{mult}_{\alpha}(B)=r-1$ by the inductive
hypothesis, hence $\sum_{\alpha}\operatorname{mult}_{\alpha}(A)=r$ by Lemma
II.3.2. ∎

Section II.4 gives a more explicit proof of Lemma II.3.3, under additional
hypotheses on $K$. Next, let $L$ be a differential field extension of $K$.
Then $\operatorname{mult}_{a}(A)$ does not strictly decrease in passing from
$K$ to $L$ [ADH, 4.1.13]. Hence the group morphism

$a+K^{\dagger}\mapsto a+L^{\dagger}\colon K/K^{\dagger}\to L/L^{\dagger}$

restricts to a map $\Sigma_{K}(A)\to\Sigma_{L}(A)$; in particular, if
$\Sigma_{K}(A)\neq\emptyset$, then $\Sigma_{L}(A)\neq\emptyset$. If
$L^{\dagger}\cap K=K^{\dagger}$, then
$\lvert\Sigma_{K}(A)\rvert\leqslant\lvert\Sigma_{L}(A)\rvert$, and
$\sum_{\alpha}\operatorname{mult}_{\alpha}(A)$ also does not strictly decrease
if $K$ is replaced by $L$.

######  Lemma II.3.4.

Let $a_{1},\dots,a_{r}\in K$ and

$A\ =\ (\der-a_{r})\cdots(\der-
a_{1}),\quad\sum_{\alpha}\operatorname{mult}_{\alpha}(A)\ =\ r.$

Then the spectrum of $A$ is $\big{\\{}[a_{1}],\dots,[a_{r}]\big{\\}}$, and for
all $\alpha$,

$\operatorname{mult}_{\alpha}(A)\ =\ \big{|}\big{\\{}i\in\\{1,\dots,r\\}:\
\alpha=[a_{i}]\big{\\}}\big{|}.$

###### Proof.

Let $i$ range over $\\{1,\dots,r\\}$. By Lemma II.3.2 and Example II.3.1,

$\operatorname{mult}_{\alpha}(A)\ \leqslant\
\sum_{i}\operatorname{mult}_{\alpha}(\der-a_{i})\ =\
\big{|}\big{\\{}i:\alpha=[a_{i}]\big{\\}}\big{|}$

and hence

$r\ =\ \sum_{\alpha}\operatorname{mult}_{\alpha}(A)\ \leqslant\
\sum_{\alpha}\big{|}\big{\\{}i:\alpha=[a_{i}]\big{\\}}\big{|}\ =\ r.$

Thus for each $\alpha$ we have
$\operatorname{mult}_{\alpha}(A)=\big{|}\big{\\{}i:\alpha=[a_{i}]\big{\\}}\big{|}$
as required. ∎

Recall from [ADH, 5.1.8] that $D^{*}\in K[\der]$ denotes the adjoint of $D\in
K[\der]$, and that the map $D\mapsto D^{*}$ is an involution of the ring
$K[\der]$ with $a^{*}=a$ for all $a$ and ${}^{*}=-\der$. If $A$ splits over
$K$, then so does $A^{*}$. Furthermore, $(A_{a})^{*}=(A^{*})_{-a}$ for all
$a$. By Lemmas II.3.3 and II.3.4:

######  Corollary II.3.5.

Suppose $K$ is $1$-linearly surjective and
$\sum_{\alpha}\operatorname{mult}_{\alpha}(A)=r$. Then
$\operatorname{mult}_{\alpha}(A)=\operatorname{mult}_{-\alpha}(A^{*})$ for all
$\alpha$. In particular, the map $\alpha\mapsto-\alpha$ restricts to a
bijection $\Sigma(A)\to\Sigma(A^{*})$.

Let $\phi\in K^{\times}$. Then $(A^{\phi})_{a}=(A_{\phi a})^{\phi}$ and hence

$\operatorname{mult}_{a}(A^{\phi})\ =\ \operatorname{mult}_{\phi a}(A),$

so the group isomorphism

(II.3.3) $[a]\mapsto[\phi a]\ \colon\ K^{\phi}/\phi^{-1}K^{\dagger}\to
K/K^{\dagger}$

maps $\Sigma(A^{\phi})$ onto $\Sigma(A)$.

Note that $K[\der]/K\der]A$ as a $K$-linear space has dimension
$r=\operatorname{order}A$. Recall from [ADH, 5.1] that $A$ and $B\neq 0$ are
said to have the same type if the (left) $K[\der]$-modules $K[\der]/K[\der]A$
and $K[\der]/K[\der]B$ are isomorphic (and so $\operatorname{order}B=r$). By
[ADH, 5.1.19]:

######  Lemma II.3.6.

The operators $A$ and $B\neq 0$ have the same type iff
$\operatorname{order}B=r$ and there is $R\in K[\der]$ of order $<r$ with $1\in
K[\der]R+K[\der]A$ and $BR\in K[\der]A$.

Hence if $A$, $B$ have the same type, then they also have the same type as
elements of $L[\der]$, for any differential field extension $L$ of $K$. Since
$B\mapsto B_{a}$ is an automorphism of the ring $K[\der]$, Lemma II.3.6 and
[ADH, 5.1.20] yield:

######  Lemma II.3.7.

If $A$ and $B\neq 0$ have the same type, then so do $A_{a}$, $B_{a}$, for all
$a$, and thus $A$, $B$ have the same eigenvalues, with same multiplicity.

By this lemma the spectrum of $A$ depends only on the type of $A$, that is, on
the isomorphism type of the $K[\der]$-module $K[\der]/K[\der]A$, suggesting
one might try to associate a spectrum to each differential module over $K$.
(Recall from [ADH, 5.5] that a differential module over $K$ is a
$K[\der]$-module of finite dimension as $K$-linear space.) We do not develop
this point of view further in the present monograph, where our focus is on
linear differential operators. (There will be more on this in [7].) But we
remark here that it motivates the terminology of “eigenvalues” originating in
the case of the differential field of Puiseux series over ${\mathbb{C}}$
treated in [20].

###  II.4. Eigenvalues and Splittings

In this section $K$ is a differential field such that $C$ is algebraically
closed and $K^{\dagger}$ is divisible. We let $A$, $B$ range over $K[\der]$,
and we assume $A\neq 0$ and set $r:=\operatorname{order}A$.

#### Spectral decomposition of differential operators

Fix a complement $\Lambda$ of the subspace $K^{\dagger}$ of the
${\mathbb{Q}}$-linear space $K$, let
${\operatorname{U}}:=K\big{[}\\!\operatorname{e}(\Lambda)\big{]}$ be the
universal exponential extension of $K$, let $\Omega$ be the differential
fraction field of the differential $K$-algebra ${\operatorname{U}}$, and let
$\lambda$ range over $\Lambda$. Then

$A_{\lambda}\ =\ A_{\ltimes\\!\operatorname{e}(\lambda)}\ =\
\operatorname{e}({-\lambda})A\operatorname{e}(\lambda)\in K[\der].$

Moreover, for every $a\in K$ there is a unique $\lambda$ with $a-\lambda\in
K^{\dagger}$, so
$\operatorname{mult}_{[a]}(A)=\operatorname{mult}_{\lambda}(A)$. Call
$\lambda$ an eigenvalue of $A$ with respect to our complement $\Lambda$ of
$K^{\dagger}$ in $K$ if $[\lambda]$ is an eigenvalue of $A$; thus the group
isomorphism $\lambda\mapsto[\lambda]\colon\Lambda\to K/K^{\dagger}$ maps the
set of eigenvalues of $A$ with respect to $\Lambda$ onto the spectrum of $A$.
For $f\in{\operatorname{U}}$ with spectral decomposition $(f_{\lambda})$ we
have

$A(f)\ =\ \sum_{\lambda}A_{\lambda}(f_{\lambda})\,\operatorname{e}(\lambda),$

so
$A({\operatorname{U}}^{\times})\subseteq{\operatorname{U}}^{\times}\cup\\{0\\}$.
We call the family $(A_{\lambda})$ the spectral decomposition of $A$ (with
respect to $\Lambda$). Given a $C$-linear subspace $V$ of
${\operatorname{U}}$, we set $V_{\lambda}:=V\cap K\operatorname{e}(\lambda)$,
a $C$-linear subspace of $V$; the sum $\sum_{\lambda}V_{\lambda}$ is direct.
For $V:={\operatorname{U}}$ we have
${\operatorname{U}}_{\lambda}=K\operatorname{e}(\lambda)$, and
${\operatorname{U}}=\bigoplus_{\lambda}{\operatorname{U}}_{\lambda}$ with
$A({\operatorname{U}}_{\lambda})\subseteq{\operatorname{U}}_{\lambda}$ for all
$\lambda$. Taking $V:=\ker_{{\operatorname{U}}}A$, we obtain
$V_{\lambda}=(\ker_{K}A_{\lambda})\operatorname{e}(\lambda)$ and hence
$\dim_{C}V_{\lambda}=\operatorname{mult}_{\lambda}(A)$, and
$V=\bigoplus_{\lambda}V_{\lambda}$. Thus

(II.4.1) $\lvert\Sigma(A)\rvert\ \leqslant\
\sum_{\lambda}\operatorname{mult}_{\lambda}(A)\ =\
\dim_{C}\ker_{{\operatorname{U}}}A\ \leqslant\ r.$

Moreover:

######  Lemma II.4.1.

The $C$-linear space $\ker_{{\operatorname{U}}}A$ has a basis contained in
${\operatorname{U}}^{\times}=K^{\times}\operatorname{e}(\Lambda)$.

######  Example.

We have a $C$-algebra isomorphism $P(Y)\mapsto P(\der)\colon C[Y]\to C[\der]$.
Suppose $A\in C[\der]\subseteq K[\der]$, let $P(Y)\in C[Y]$, $P(\der)=A$, and
let $c_{1},\dots,c_{n}\in C$ be the distinct zeros of $P$, of respective
multiplicities $m_{1},\dots,m_{n}\in{\mathbb{N}}^{\geqslant 1}$ (so $r=\deg
P=m_{1}+\cdots+m_{n}$). Suppose also $C\subseteq\Lambda$, and $x\in K$
satisfies $x^{\prime}=1$. (This holds in Example II.2.4.) Then the
$x^{i}\operatorname{e}(c_{j})\in{\operatorname{U}}$ ($1\leqslant j\leqslant
n$, $0\leqslant i<m_{j}$) form a basis of the $C$-linear space
$\ker_{{\operatorname{U}}}A$ by [ADH, 5.1.18]. So the eigenvalues of $A$ with
respect to $\Lambda$ are $c_{1},\dots,c_{n}$, with respective multiplicities
$m_{1},\dots,m_{n}$.

######  Corollary II.4.2.

Suppose $\dim_{C}\ker_{{\operatorname{U}}}A=r\geqslant 1$ and
$A={}^{r}+a_{r-1}{}^{r-1}+\cdots+a_{0}$ where $a_{0},\dots,a_{r-1}\in K$. Then

$\sum_{\lambda}\operatorname{mult}_{\lambda}(A)\lambda\ \equiv\ -a_{r-1}\mod
K^{\dagger}.$

In particular, $\sum_{\lambda}\operatorname{mult}_{\lambda}(A)\lambda=0$ iff
$a_{r-1}\in K^{\dagger}$.

###### Proof.

Take a basis $y_{1},\dots,y_{r}$ of $\ker_{{\operatorname{U}}}A$ with
$y_{j}=f_{j}\operatorname{e}(\lambda_{j})$, $f_{j}\in K^{\times}$,
$\lambda_{j}\in\Lambda$. The Wronskian matrix
$\operatorname{Wr}(y_{1},\dots,y_{r})$ of $(y_{1},\dots,y_{r})$ [ADH, p. 206]
equals

$\operatorname{Wr}(y_{1},\dots,y_{r})\ =\
M\begin{pmatrix}\operatorname{e}(\lambda_{1})&&\\\ &\ddots&\\\
&&\operatorname{e}(\lambda_{r})\end{pmatrix}\qquad\text{where
$M\in\operatorname{GL}_{n}(K)$.}$

Then
$w:=\operatorname{wr}(y_{1},\dots,y_{r})=\det\operatorname{Wr}(y_{1},\dots,y_{r})\neq
0$ by [ADH, 4.1.13] and

$-a_{r-1}\ =\ w^{\dagger}\ =\ (\det
M)^{\dagger}+\lambda_{1}+\cdots+\lambda_{r}$

where we used [ADH, 4.1.17] for the first equality. ∎

If $A$ splits over $K$, then so does $A_{\lambda}$. Moreover, if
$A_{\lambda}(K)=K$, then
$A({\operatorname{U}}_{\lambda})={\operatorname{U}}_{\lambda}$: for $f,g\in K$
with $A_{\lambda}(f)=g$ we have
$A\big{(}f\operatorname{e}(\lambda)\big{)}=g\operatorname{e}(\lambda)$. Thus:

######  Lemma II.4.3.

Suppose $K$ is $r$-linearly surjective, or $K$ is $1$-linearly surjective and
$A$ splits over $K$. Then
$A({\operatorname{U}}_{\lambda})={\operatorname{U}}_{\lambda}$ for all
$\lambda$ and hence $A({\operatorname{U}})={\operatorname{U}}$.

In the next subsection we study the connection between splittings of $A$ and
bases of the $C$-linear space $\ker_{{\operatorname{U}}}A$ in more detail.

#### Constructing splittings and bases

Recall that $\operatorname{order}A=r\in{\mathbb{N}}$. Set
${\operatorname{U}}={\operatorname{U}}_{K}$, so
${\operatorname{U}}^{\times}=K^{\times}\operatorname{e}(\Lambda)$. Let
$y_{1},\dots,y_{r}\in{\operatorname{U}}^{\times}$. We construct a sequence
$A_{0},\dots,A_{n}$ of monic operators in $K[\der]$ with $n\leqslant r$ as
follows. First, set $A_{0}:=1$. Next, given $A_{0},\dots,A_{i-1}$ in
$K[\der]^{\neq}$ ($1\leqslant i\leqslant r$), set $f_{i}:=A_{i-1}(y_{i})$; if
$f_{i}\neq 0$, then $f_{i}\in{\operatorname{U}}^{\times}$, so
$f_{i}^{\dagger}\in K$, and the next term in the sequence is

$A_{i}\ :=\ (\der-a_{i})A_{i-1},\qquad a_{i}\ :=\ f_{i}^{\dagger},$

whereas if $f_{i}=0$, then $n:=i-1$ and the construction is finished.

######  Lemma II.4.4.

$\ker_{{\operatorname{U}}}A_{i}=Cy_{1}\oplus\cdots\oplus Cy_{i}$ (internal
direct sum) for $i=0,\dots,n$.

###### Proof.

By induction on $i\leqslant n$. The case $i=0$ being trivial, suppose
$1\leqslant i\leqslant n$ and the claim holds for $i-1$ in place of $i$. Then
$A_{i-1}(y_{i})=f_{i}\neq 0$, hence
$y_{i}\notin\ker_{{\operatorname{U}}}A_{i-1}=Cy_{1}\oplus\cdots\oplus
Cy_{i-1}$, and $A_{i}=(\der-f_{i}^{\dagger})A_{i-1}$, so by [ADH, 5.1.14(i)]
we have $\ker_{{\operatorname{U}}}A_{i}=\ker_{{\operatorname{U}}}A_{i-1}\oplus
Cy_{i}=Cy_{1}\oplus\cdots\oplus Cy_{i}$. ∎

We denote the tuple $(a_{1},\dots,a_{n})\in K^{n}$ just constructed by
$\operatorname{split}(y_{1},\dots,y_{r})$, so $A_{n}=(\der-a_{n})\cdots(\der-
a_{1})$. Suppose $r\geqslant 1$. Then $n\geqslant 1$, $a_{1}=y_{1}^{\dagger}$,
$A_{1}=\der-a_{1}$,
$A_{1}(y_{2}),\dots,A_{1}(y_{n})\in{\operatorname{U}}^{\times}$, and we have

$(a_{2},\dots,a_{n})\ =\
\operatorname{split}\\!\big{(}A_{1}(y_{2}),\dots,A_{1}(y_{n})\big{)}.$

By Lemma II.4.4, $n\leqslant r$ is maximal such that $y_{1},\dots,y_{n}$ are
$C$-linearly independent. In particular, $y_{1},\dots,y_{r}$ are $C$-linearly
independent iff $n=r$.

######  Corollary II.4.5.

If $A(y_{i})=0$ for $i=1,\dots,n$, then $A\in K[\der]A_{n}$. Thus if $n=r$ and
$A(y_{i})=0$ for $i=1,\dots,r$, then $A=a(\der-a_{r})\cdots(\der-a_{1})$ where
$a\in K^{\times}$.

This follows from [ADH, 5.1.15(i)] and Lemma II.4.4.

Suppose that $H$ is a differential subfield of $K$ and
$y_{1}^{\dagger},\dots,y_{r}^{\dagger}\in H$. Then we have
$\operatorname{split}(y_{1},\dots,y_{r})\in~{}H^{n}$: use that $y^{\prime}\in
Hy$ with $y\in U$ gives $y^{(m)}\in Hy$ for all $m$, so $B(y)\in Hy$ for all
$B\in H[\der]$, hence for such $B$, if $f:=B(y)\neq 0$, then $f^{\dagger}\in
H$.

######  Corollary II.4.6.

Suppose $\dim_{C}\ker_{{\operatorname{U}}}A=r$. Then
$\ker_{{\operatorname{U}}}A=\ker_{\Omega}A$ and $A$ splits over $K$. If
$A=(\der-a_{r})\cdots(\der-a_{1}),\ a_{1},\dots,a_{r}\in K$, then the spectrum
of $A$ is $\big{\\{}[a_{1}],\dots,[a_{r}]\big{\\}}$, and for all $\alpha\in
K/K^{\dagger}$,

$\operatorname{mult}_{\alpha}(A)\ =\ \big{|}\big{\\{}i\in\\{1,\dots,r\\}:\
\alpha=[a_{i}]\big{\\}}\big{|}.$

###### Proof.

$A$ splits over $K$ by Lemma II.4.1 and Corollary II.4.5. The rest follows
from Lemma II.3.4 in view of
$\sum_{\lambda}\operatorname{mult}_{\lambda}(A)=\dim_{C}\ker_{{\operatorname{U}}}A$.
∎

Conversely, we can associate to a given splitting of $A$ over $K$ a basis of
$\ker_{{\operatorname{U}}}A$ consisting of $r$ elements of
${\operatorname{U}}^{\times}$, provided $K$ is $1$-linearly surjective when
$r\geqslant 2$:

######  Lemma II.4.7.

Assume $K$ is $1$-linearly surjective in case $r\geqslant 2$. Let

$A\ =\ (\der-a_{r})\cdots(\der-a_{1})\qquad\text{where
$a_{i}=b_{i}^{\dagger}+\lambda_{i}$, $b_{i}\in K^{\times}$,
$\lambda_{i}\in\Lambda$ {(}$i=1,\dots,r${)}.}$

Then there are $C$-linearly independent
$y_{1},\dots,y_{r}\in\ker_{{\operatorname{U}}}A$ with $y_{i}\in
K^{\times}\operatorname{e}(\lambda_{i})$ for $i=1,\dots,r$ and
$\operatorname{split}(y_{1},\dots,y_{r})=(a_{1},\dots,a_{r})$.

###### Proof.

By induction on $r$. The case $r=0$ is trivial, and for $r=1$ we can take
$y_{1}=b_{1}\operatorname{e}(\lambda_{1})$. Let $r\geqslant 2$ and suppose
inductively that for

$B\ :=\ (\der-a_{r})\cdots(\der-a_{2})$

we have $C$-linearly independent
$z_{2},\dots,z_{r}\in\ker_{{\operatorname{U}}}B$ with $z_{i}\in
K^{\times}\operatorname{e}(\lambda_{i})$ for ${i=2,\dots,r}$ and
$\operatorname{split}(z_{2},\dots,z_{r})=(a_{2},\dots,a_{r})$. For
$i=2,\dots,r$, Lemma II.4.3 gives ${y_{i}\in
K^{\times}\operatorname{e}(\lambda_{i})}$ with $(\der-a_{1})(y_{i})=z_{i}$.
Set $y_{1}:=b_{1}\operatorname{e}(\lambda_{1})$, so
$\ker_{{\operatorname{U}}}(\der-a_{1})=~{}Cy_{1}$. Then
$y_{1},\dots,y_{r}\in\ker_{{\operatorname{U}}}A$ are $C$-linearly independent
such that $y_{i}\in K^{\times}\operatorname{e}(\lambda_{i})$ for
${i=1,\dots,r}$, and one verifies easily that
$\operatorname{split}(y_{1},\dots,y_{r})=(a_{1},\dots,a_{r})$. ∎

######  Corollary II.4.8.

Assume $K$ is $1$-linearly surjective when $r\geqslant 2$. Then

$\text{$A$ splits over~{}$K$}\ \Longleftrightarrow\
\dim_{C}\ker_{{\operatorname{U}}}A\ =\ r.$

######  Remark.

If $\dim_{C}\ker_{{\operatorname{U}}}A=r$ and $\lambda_{1},\dots,\lambda_{d}$
are the eigenvalues of $A$ with respect to $\Lambda$, then the differential
subring
$K\big{[}\\!\operatorname{e}(\lambda_{1}),\operatorname{e}(-\lambda_{1}),\dots,\operatorname{e}(\lambda_{d}),\operatorname{e}(-\lambda_{d})\big{]}$
of ${\operatorname{U}}$ is the Picard-Vessiot ring for $A$ over $K$; see [20,
Section 1.3]. If $K$ is linearly closed and linearly surjective, then
${\operatorname{U}}$ is by Corollary II.4.8 the universal Picard-Vessiot ring
of the differential field $K$ as defined in [20, Chapter 10]. Our construction
of ${\operatorname{U}}$ above is modeled on the description of the universal
Picard-Vessiot ring of the algebraic closure of $C(\\!(t)\\!)$ given in [20,
Chapter 3].

Recalling our convention that $r=\operatorname{order}A$, here is a complement
to Lemma II.4.1:

######  Corollary II.4.9.

Let $V$ be a $C$-linear subspace of ${\operatorname{U}}$ with $r=\dim_{C}V$.
Then there is at most one monic $A$ with $V=\ker_{{\operatorname{U}}}A$.
Moreover, the following are equivalent:

  1. (i)

$V=\ker_{{\operatorname{U}}}A$ for some monic $A$ that splits over $K$;

  2. (ii)

$V=\ker_{{\operatorname{U}}}B$ for some $B\neq 0$;

  3. (iii)

$V=\sum_{\lambda}V_{\lambda}$;

  4. (iv)

$V$ has a basis contained in ${\operatorname{U}}^{\times}$.

###### Proof.

The first claim follows from [ADH, 5.1.15] applied to the differential
fraction field of ${\operatorname{U}}$ in place of $K$. The implication (i)
$\Rightarrow$ (ii) is clear, (ii) $\Rightarrow$ (iii) was noted before Lemma
II.4.1, and (iii) $\Rightarrow$ (iv) is obvious. For (iv) $\Rightarrow$ (i),
let $y_{1},\dots,y_{r}\in{\operatorname{U}}^{\times}$ be a basis of $V$. Then
$\operatorname{split}(y_{1},\dots,y_{r})=(a_{1},\dots,a_{r})\in K^{r}$, so
$V=\ker_{{\operatorname{U}}}A$ for $A=(\der-a_{r})\cdots(\der-a_{1})$ by Lemma
II.4.4, so (i) holds. ∎

Let $y_{1},\dots,y_{r}\in{\operatorname{U}}^{\times}$ and
$(a_{1},\dots,a_{n}):=\operatorname{split}(y_{1},\dots,y_{r})$. We finish this
subsection with some remarks about $(a_{1},\dots,a_{n})$ for use in [7]. Let
$A_{0},\dots,A_{n}\in K[\der]$ be as above and recall that $n\leqslant r$ is
maximal such that $y_{1},\dots,y_{n}$ are $C$-linearly independent.

######  Lemma II.4.10.

Assume $n=r$. Let $z_{1},\dots,z_{r}\in{\operatorname{U}}^{\times}$. The
following are equivalent:

  1. (i)

$z_{1},\dots,z_{r}$ are $C$-linearly independent and
$(a_{1},\dots,a_{r})=\operatorname{split}(z_{1},\dots,z_{r})$;

  2. (ii)

for $i=1,\dots,r$ there are $c_{ii},c_{i,i-1},\dots,c_{i1}\in C$ such that

$z_{i}\ =\ c_{ii}y_{i}+c_{i,i-1}y_{i-1}+\cdots+c_{i1}y_{1}\ \text{and}\
c_{ii}\neq 0.$

###### Proof.

The case $r=0$ is trivial. Let $r=1$. If (i) holds, then
$y_{1}^{\dagger}=a_{1}=z_{1}^{\dagger}$, hence $z_{1}\in C^{\times}\,y_{1}$,
so (ii) holds. The converse is obvious. Let $r\geqslant 2$, and assume (i)
holds. Put $\widetilde{y}_{i}:=A_{1}(y_{i})$ and
$\widetilde{z}_{i}:=A_{1}(z_{i})$ for $i=2,\dots,r$. Then

$\operatorname{split}(\widetilde{y}_{2},\dots,\widetilde{y}_{r})\ =\
(a_{2},\dots,a_{r})\ =\
\operatorname{split}(\widetilde{z}_{2},\dots,\widetilde{z}_{r}),$

so we can assume inductively to have $c_{ij}\in C$ ($2\leqslant j\leqslant
i\leqslant r$) with

$\widetilde{z}_{i}\ =\
c_{ii}\widetilde{y}_{i}+c_{i,i-1}\widetilde{y}_{i-1}+\cdots+c_{i2}\widetilde{y}_{2}\quad\text{and}\quad
c_{ii}\neq 0\qquad(2\leqslant i\leqslant r).$

Hence for $2\leqslant i\leqslant r$,

$z_{i}\in
c_{ii}y_{i}+c_{i,i-1}y_{i-1}+\cdots+c_{i2}y_{2}+\ker_{{\operatorname{U}}}A_{1}.$

Now use $\ker_{{\operatorname{U}}}A_{1}=Cy_{1}$ to conclude (ii). For the
converse, let $c_{ij}\in C$ be as in (ii). Then clearly $z_{1},\dots,z_{r}$
are $C$-linearly independent. Let
$(b_{1},\dots,b_{r}):=\operatorname{split}(z_{1},\dots,z_{r})$ and
$B_{r-1}:=(\der-b_{r-1})\cdots(\der-b_{1})$. Then $a_{r}=f_{r}^{\dagger}$
where $f_{r}=A_{r-1}(y_{r})\neq 0$, and $b_{r}=g_{r}^{\dagger}$ where
$g_{r}:=B_{r-1}(z_{r})\neq 0$. Now inductively we have $a_{j}=b_{j}$ for
$j=1,\dots,r-1$, so $A_{r-1}=B_{r-1}$, and $A_{r-1}(y_{i})=0$ for
$i=1,\dots,r-1$ by Lemma II.4.4. Hence $g_{r}=c_{rr}f_{r}$, and thus
$a_{r}=b_{r}$. ∎

######  Lemma II.4.11.

Let $z\in{\operatorname{U}}^{\times}$. Then
$\operatorname{split}(y_{1}z,\dots,y_{r}z)=(a_{1}+z^{\dagger},\dots,a_{n}+z^{\dagger})$.

###### Proof.

Since for $m\leqslant r$, the units $y_{1}z,\dots,y_{m}z$ of
${\operatorname{U}}$ are $C$-linearly independent iff $y_{1},\dots,y_{m}$ are
$C$-linearly independent, we see that the tuples
$\operatorname{split}(y_{1}z,\dots,y_{r}z)$ and
$\operatorname{split}(y_{1},\dots,y_{r})$ have the same length $n$. Let
$(b_{1},\dots,b_{n}):=\operatorname{split}(y_{1}z,\dots,y_{r}z)$; we show
$(b_{1},\dots,b_{n})=(a_{1}+z^{\dagger},\dots,a_{n}+z^{\dagger})$ by induction
on $n$. The case $n=0$ is obvious, so suppose $n\geqslant 1$. Then
$a_{1}=y_{1}^{\dagger}$ and $b_{1}=(y_{1}z)^{\dagger}=a_{1}+z^{\dagger}$ as
required. By remarks following the proof of Lemma II.4.4 we have

$(a_{2},\dots,a_{n})\ =\
\operatorname{split}\\!\big{(}A_{1}(y_{2}),\dots,A_{1}(y_{n})\big{)}\qquad\text{where
$A_{1}:=\der-a_{1}$.}$

Now $B_{1}:=\der-b_{1}=(A_{1})_{\ltimes z^{-1}}$, so likewise

$(b_{2},\dots,b_{n})\ =\
\operatorname{split}\\!\big{(}B_{1}(y_{2}z),\dots,B_{1}(y_{n}z)\big{)}\ =\
\operatorname{split}\\!\big{(}A_{1}(y_{2})z,\dots,A_{1}(y_{n})z\big{)}.$

Hence $b_{2}=a_{2}+z^{\dagger},\dots,b_{n}=a_{n}+z^{\dagger}$ by our inductive
hypothesis. ∎

For $f\in\der K$ we let $\int f$ denote an element of $K$ such that $(\int
f)^{\prime}=f$.

######  Lemma II.4.12.

Let $g_{1},\dots,g_{r}\in K^{\times}$ and

$A\ =\ g_{1}\cdots g_{r}(\der g_{r}^{-1})(\der g_{r-1}^{-1})\cdots(\der
g_{1}^{-1}),$

and suppose the integrals below can be chosen such that

$y_{1}\ =\ g_{1},\quad y_{2}\ =\ g_{1}\textstyle\int g_{2},\quad\dots,\quad
y_{r}\ =\ g_{1}\int(g_{2}\int g_{3}(\cdots(g_{r-1}\int g_{r})\cdots)),$

Then $y_{1},\dots,y_{r}\in K^{\times}$, $n=r$, and $a_{i}=(g_{1}\cdots
g_{i})^{\dagger}$ for $i=1,\dots,r$.

###### Proof.

Let $b_{i}:=(g_{1}\cdots g_{i})^{\dagger}$ for $i=1,\dots,r$. By induction on
$i=0,\dots,r$ we show $n\geqslant i$ and
$(a_{1},\dots,a_{i})=(b_{1},\dots,b_{i})$. This is clear for $i=0$, so suppose
${i\in\\{1,\dots,r\\}}$, $n\geqslant i-1$, and
$(a_{1},\dots,a_{i-1})=(b_{1},\dots,b_{i-1})$. Then

$A_{i-1}=(\der-a_{i-1})\cdots(\der-a_{1})=(\der-b_{i-1})\cdots(\der-
b_{1})=g_{1}\cdots g_{i-1}(\der g_{i-1}^{-1})\cdots(\der g_{1}^{-1}),$

using Lemma I.1.3 for the last equality. So $A_{i-1}(y_{i})=g_{1}\cdots
g_{i}\neq 0$, and thus $n\geqslant i$ and
$a_{i}=A_{i-1}(y_{i})^{\dagger}=b_{i}$. ∎

#### The case of real operators

We now continue the subsection The real case of Section II.2. Thus
$K=H[\imag]$ where $H$ is a real closed differential subfield of $K$ and
${\imag^{2}=-1}$, and
$\Lambda=\Lambda_{\operatorname{r}}+\Lambda_{\operatorname{i}}\imag$ where
$\Lambda_{\operatorname{r}}$, $\Lambda_{\operatorname{i}}$ are subspaces of
the ${\mathbb{Q}}$-linear space $H$. The complex conjugation automorphism
$z\mapsto\overline{z}$ of the differential field $K$ extends uniquely to an
automorphism $B\mapsto\overline{B}$ of the ring $K[\der]$ with
$\overline{\der}=\der$. We have $\overline{A(f)}=\overline{A}(\overline{f})$
for $f\in{\operatorname{U}}$, from which it follows that
$\dim_{C}\ker_{K}A=\dim_{C}\ker_{K}\overline{A}$,
$(\overline{A})_{\lambda}=\overline{(A_{\overline{\lambda}})}$,
$\operatorname{mult}_{\lambda}\overline{A}=\operatorname{mult}_{\overline{\lambda}}A$,
and $f\mapsto\overline{f}\colon{\operatorname{U}}\to{\operatorname{U}}$
restricts to a $C_{H}$-linear bijection
$\ker_{{\operatorname{U}}}A\to\ker_{{\operatorname{U}}}\overline{A}$.

In the rest of this subsection we assume $H=H^{\dagger}$ $($so
$\Lambda=\Lambda_{\operatorname{i}}\imag)$ and $A\in H[\der]$ $($and by
earlier conventions, $A\neq 0$ and $r:=\operatorname{order}A)$. Then
$A=\overline{A}$, hence for all $\lambda$ we have
$A_{\overline{\lambda}}=\overline{A_{\lambda}}$ and
$\operatorname{mult}_{\lambda}A=\operatorname{mult}_{\overline{\lambda}}A$.
Thus with $\mu$ ranging over $\Lambda_{\operatorname{i}}^{>}$:

$\sum_{\lambda}\operatorname{mult}_{\lambda}(A)\ =\
\operatorname{mult}_{0}(A)+2\sum_{\mu}\operatorname{mult}_{\mu\imag}(A).$

Note that $0$ is an eigenvalue of $A$ iff $\ker_{H}A\neq\\{0\\}$.

Let $V:=\ker_{{\operatorname{U}}}A$, a subspace of the $C$-linear space
${\operatorname{U}}$ with $\overline{V}=V$ and $\dim_{C}V\leqslant r$. Recall
that we have the differential $H$-subalgebra
${\operatorname{U}}_{\operatorname{r}}=\\{f\in{\operatorname{U}}:\overline{f}=f\\}$
of ${\operatorname{U}}$ and the $C_{H}$-linear subspace
$V_{\operatorname{r}}=\ker_{{\operatorname{U}}_{\operatorname{r}}}A$ of
${\operatorname{U}}_{\operatorname{r}}$. Now $V=V_{\operatorname{r}}\oplus
V_{\operatorname{r}}\imag$ (internal direct sum of $C_{H}$-linear subspaces),
so $\dim_{C}V=\dim_{C_{H}}V_{\operatorname{r}}$. Combining Lemma II.4.1 and
the remarks preceding it with Lemma II.2.19 and its proof yields:

######  Corollary II.4.13.

The $C$-linear space $V$ has a basis

$a_{1}\operatorname{e}(\mu_{1}\imag),\,\overline{a_{1}}\operatorname{e}(-\mu_{1}\imag),\
\dots,\
a_{m}\operatorname{e}(\mu_{m}\imag),\,\overline{a_{m}}\operatorname{e}(-\mu_{m}\imag),\
h_{1},\ \dots,\ h_{n}\qquad(2m+n\leqslant r),$

where $a_{1},\dots,a_{m}\in K^{\times}$,
$\mu_{1},\dots,\mu_{m}\in\Lambda_{\operatorname{i}}^{>}$,
$h_{1},\dots,h_{n}\in H^{\times}$. For such a basis,

$\operatorname{Re}\\!\big{(}a_{1}\operatorname{e}(\mu_{1}\imag)\big{)},\,\operatorname{Im}\\!\big{(}a_{1}\operatorname{e}(\mu_{1}\imag)\big{)},\
\dots,\
\operatorname{Re}\\!\big{(}a_{m}\operatorname{e}(\mu_{m}\imag)\big{)},\,\operatorname{Im}\\!\big{(}a_{m}\operatorname{e}(\mu_{m}\imag)\big{)},\
h_{1},\ \dots,\ h_{n}$

is a basis of the $C_{H}$-linear space $V_{\operatorname{r}}$, and
$h_{1},\dots,h_{n}$ is a basis of the $C_{H}$-linear subspace $\ker_{H}A=V\cap
H$ of $H$.

Using $H=H^{\dagger}$, arguments as in the proof of Lemma II.4.7 show:

######  Lemma II.4.14.

Assume $H$ is $1$-linearly surjective when $r\geqslant 2$. Let
$a_{1},\dots,a_{r}\in H$ be such that $A=(\der-a_{r})\cdots(\der-a_{1})$. Then
the $C_{H}$-linear space $\ker_{H}A$ has a basis $y_{1},\dots,y_{r}$ such that
$\operatorname{split}(y_{1},\dots,y_{r})=(a_{1},\dots,a_{r})$.

Recall from Lemma II.3.3 that if $r=1$ or $K$ is $1$-linearly surjective, then

$\text{$A$ splits over
$K$}\quad\Longleftrightarrow\quad\sum_{\lambda}\operatorname{mult}_{\lambda}(A)=r.$

Now
$\operatorname{mult}_{\lambda}(A)=\operatorname{mult}_{\overline{\lambda}}(A)$
for all $\lambda$, so if $\operatorname{mult}_{\lambda}(A)=r\geqslant 1$, then
$\lambda=0$. Also, for $W:=V\cap K=\ker_{K}A$ and
$W_{\operatorname{r}}:=W\cap{\operatorname{U}}_{\operatorname{r}}$ we have
$W_{\operatorname{r}}=\ker_{H}A$ and

$W\ =\ W_{\operatorname{r}}\oplus W_{\operatorname{r}}\imag\quad\text{
(internal direct sum of $C_{H}$-linear subspaces)},$

so $\operatorname{mult}_{0}(A)=\dim_{C}\ker_{K}A=\dim_{C_{H}}\ker_{H}A$. If
$y_{1},\dots,y_{r}$ is a basis of the $C_{H}$-linear space $\ker_{H}A$, then
$\operatorname{split}(y_{1},\dots,y_{r})\in H^{r}$ in reversed order is a
splitting of $A$ over $H$ by Corollary II.4.5. These remarks and Lemma II.4.14
now yield:

######  Corollary II.4.15.

If $\operatorname{mult}_{0}(A)=r$, then $A$ splits over $H$. The converse
holds if $H$ is $1$-linearly surjective or $r=1$.

######  Corollary II.4.16.

Suppose $r\geqslant 1$, and $K$ is $1$-linearly surjective if $r\geqslant 2$.
Then

$\text{$A$ splits
over~{}$H$}\quad\Longleftrightarrow\quad\operatorname{mult}_{0}(A)=r\quad\Longleftrightarrow\quad\lvert\Sigma(A)\rvert=1.$

We now focus on the order $2$ case:

######  Lemma II.4.17.

Suppose $r=2$ and $A$ splits over $K$ but not over $H$. Then

$\dim_{C}\ker_{{\operatorname{U}}}A\ =\ 2.$

If $H$ is $1$-linearly surjective, then $A$ has two distinct eigenvalues.

###### Proof.

We can assume $A$ is monic, so $A=(\der-f)(\der-g)$ with $f,g\in K$ and
$g=a+b\imag$, $a,b\in H$, $b\neq 0$. Then $g=d^{\dagger}+\mu\imag$ with $d\in
K^{\times}$ and $\mu\in\Lambda_{\operatorname{i}}$, and so
$d\operatorname{e}(\mu\imag)\in\ker_{\operatorname{U}}A$. From
$A=\overline{A}$ we obtain
$\overline{d}\operatorname{e}(-\mu\imag)\in\ker_{\operatorname{U}}A$. These
two elements of $\ker_{{\operatorname{U}}}A$ are $C$-linearly independent,
since

$d\operatorname{e}(\mu\imag)/\overline{d}\operatorname{e}(-\mu\imag)\ =\
(d/\overline{d})\operatorname{e}(2\mu\imag)\notin C:$

this is clear if $\mu\neq 0$, and if $\mu=0$, then $d^{\dagger}=g$, so
${(d/\overline{d})^{\dagger}=g-\overline{g}=2b\imag\neq 0}$, and hence
$d/\overline{d}\notin C$. Thus $\dim_{C}\ker_{\operatorname{U}}A\ =\ 2$, and
$\mu\imag$, $-\mu\imag$ are eigenvalues of $A$ with respect to $\Lambda$. Now
assume $H$ is $1$-linearly surjective. Then we claim that ${\mu\neq 0}$. To
see this note that [ADH, 5.1.21, 5.2.10] and the assumption that $A$ does not
split over $H$ yield $\dim_{C_{H}}\ker_{H}A=\dim_{C}\ker_{K}A=0$, hence
$g\notin K^{\dagger}$ and thus ${\mu\imag=g-d^{\dagger}\neq 0}$. ∎

Combining Lemmas II.4.14 and II.4.17 yields:

######  Corollary II.4.18.

If $H$ is $1$-linearly surjective, $A$ has order $2$, and $A$ splits over $K$,
then $\dim_{C}\ker_{\operatorname{U}}A\ =\ 2$.

In the rest of this subsection $H$ is $1$-linearly surjective and
$A=4{}^{2}+f$, $f\in H$. Let $\omega\colon H\to H$ and $\sigma\colon
H^{\times}\to H$ be as in (0.5) and (0.6). Then by (0.7) and (0.8):

$A$ splits over $H$ $\displaystyle\quad\Longleftrightarrow\quad
f\in\omega(H),$ $A$ splits over $K$
$\displaystyle\quad\Longleftrightarrow\quad
f\in\sigma(H^{\times})\cup\omega(H).$

If $A$ splits over $H$, then $\Sigma(A)=\\{0\\}$ and
$\operatorname{mult}_{0}(A)=2$, by Corollary II.4.16. Suppose $A$ splits over
$K$ but not over $H$, and let $y\in H^{\times}$ satisfy
$\sigma(y)=f\notin\omega(H)$. Then by [ADH, p. 262] we have
$A=4(\der+g)(\der-g)$ where $g=\frac{1}{2}(-y^{\dagger}+y\imag)$. Hence the
two distinct eigenvalues of $A$ are $(y/2)\imag+K^{\dagger}$ and
$-(y/2)\imag+K^{\dagger}$.

#### The case of oscillating transseries

We now apply the results above to the algebraically closed differential field
$K={\mathbb{T}}[\imag]$. Note that ${\mathbb{T}}[\imag]$ has constant field
${\mathbb{C}}$ and extends the (real closed) differential field ${\mathbb{T}}$
of transseries. After (0.12) in the introduction, we already remarked:

######  Lemma II.4.19.

${\mathbb{T}}[\imag]$ is linearly closed and linearly surjective.

Now applying Corollary II.4.8 and Lemma II.4.1 to ${\mathbb{T}}[\imag]$ gives:

######  Corollary II.4.20.

For $K={\mathbb{T}}[\imag]$, there are $\mathbb{C}$-linearly independent units
$y_{1},\dots,y_{r}$ of ${\operatorname{U}}_{{\mathbb{T}}[\imag]}$ with
$A(y_{1})=\cdots=A(y_{r})=0$.

Next we describe another incarnation of
${\operatorname{U}}_{{\mathbb{T}}[\imag]}$, namely as a ring $\mathbb{O}$ of
“oscillating” transseries. Towards this goal we first note that by [ADH,
11.5.1, 11.8.2] we have

$\displaystyle\operatorname{I}({\mathbb{T}})\ $ $\displaystyle=\
\big{\\{}y\in{\mathbb{T}}:\ \text{$y\preccurlyeq f^{\prime}$ for some $f\prec
1$ in ${\mathbb{T}}$}\big{\\}}$ $\displaystyle=\ \big{\\{}y\in{\mathbb{T}}:\
\text{$y\prec 1/(\ell_{0}\cdots\ell_{n})$ for all $n$}\big{\\}},$

so a complement $\Lambda_{{\mathbb{T}}}$ of $\operatorname{I}({\mathbb{T}})$
in ${\mathbb{T}}$ is given by

$\Lambda_{{\mathbb{T}}}\ :=\ \big{\\{}y\in{\mathbb{T}}:\
\text{$\operatorname{supp}(y)\succ 1/(\ell_{0}\cdots\ell_{n-1}\ell_{n}^{2})$
for all $n$}\big{\\}}.$

Since ${\mathbb{T}}^{\dagger}={\mathbb{T}}$ and
$\operatorname{I}\\!\big{(}{\mathbb{T}}[\imag]\big{)}\subseteq{\mathbb{T}}[\imag]^{\dagger}$
we have
${\mathbb{T}}[\imag]^{\dagger}={\mathbb{T}}\oplus\operatorname{I}({\mathbb{T}})\imag$
by Lemmas I.2.4 and I.2.13. We now take $\Lambda=\Lambda_{{\mathbb{T}}}\imag$
as our complement $\Lambda$ of ${\mathbb{T}}[\imag]^{\dagger}$ in
${\mathbb{T}}[\imag]$ and explain how the universal exponential extension
${\operatorname{U}}$ of ${\mathbb{T}}[\imag]$ for this $\Lambda$ was
introduced in [15, Section 7.7] in a different way. Let

${\mathbb{T}}_{\succ}\ :=\ \\{f\in{\mathbb{T}}:\operatorname{supp}f\succ
1\\},$

and similarly with $\prec$ in place of $\succ$; then
${\mathbb{T}}_{\prec}=\smallo_{{\mathbb{T}}}$ and ${\mathbb{T}}_{\succ}$ are
${\mathbb{R}}$-linear subspaces of ${\mathbb{T}}$, and ${\mathbb{T}}$
decomposes as an internal direct sum

(II.4.2) ${\mathbb{T}}\ =\
{\mathbb{T}}_{\succ}\oplus{\mathbb{R}}\oplus{\mathbb{T}}_{\prec}$

of ${\mathbb{R}}$-linear subspaces of ${\mathbb{T}}$. Let
$\operatorname{e}^{\imag{\mathbb{T}}_{\succ}}=\\{\operatorname{e}^{\imag
f}:f\in{\mathbb{T}}_{\succ}\\}$ be a multiplicative copy of the additive group
${\mathbb{T}}_{\succ}$, with isomorphism $f\mapsto\operatorname{e}^{\imag f}$.
Then we have the group ring

$\mathbb{O}\ :=\ K\big{[}\operatorname{e}^{\imag{\mathbb{T}}_{\succ}}\big{]}$

of $\operatorname{e}^{\imag{\mathbb{T}}_{\succ}}$ over
$K={\mathbb{T}}[\imag]$. We make $\mathbb{O}$ into a differential ring
extension of $K$ by

$(\operatorname{e}^{\imag f})^{\prime}\ =\ \imag
f^{\prime}\operatorname{e}^{\imag f}\qquad(f\in{\mathbb{T}}_{\succ}).$

Hence $\mathbb{O}$ is an exponential extension of $K$. The elements of
$\mathbb{O}$ are called oscillating transseries. For each $f\in{\mathbb{T}}$
there is a unique $g\in{\mathbb{T}}$, to be denoted by $\int f$, such that
$g^{\prime}=f$ and $g$ has constant term $g_{1}=0$. The injective map
$\int\colon{\mathbb{T}}\to{\mathbb{T}}$ is ${\mathbb{R}}$-linear; we use this
map to show that ${\operatorname{U}}$ and $\mathbb{O}$ are disguised versions
of each other:

######  Proposition II.4.21.

There is a unique isomorphism
${\operatorname{U}}=K\big{[}\\!\operatorname{e}(\Lambda)\big{]}\to\mathbb{O}$
of differential $K$-algebras sending $\operatorname{e}(h\imag)$ to
$\operatorname{e}^{\imag\int h}$ for all $h\in\Lambda_{{\mathbb{T}}}$.

This requires the next lemma. We assume familiarity with [ADH, Appendix A],
especially with the ordered group $G^{\operatorname{LE}}$ (a subgroup of
${\mathbb{T}}^{\times}$) of logarithmic-exponential monomials and its subgroup
$G^{\operatorname{E}}=\bigcup_{n}G_{n}$ of exponential monomials.

######  Lemma II.4.22.

If ${\mathfrak{m}}\in G^{\operatorname{LE}}$ and ${\mathfrak{m}}\succ 1$, then
$\operatorname{supp}{\mathfrak{m}}^{\prime}\ \subseteq\
\Lambda_{{\mathbb{T}}}$.

###### Proof.

We first prove by induction on $n$ a fact about elements of
$G^{\operatorname{E}}$:

$\text{ if ${\mathfrak{m}}\in G_{n}$, ${\mathfrak{m}}\succ 1$, then
$\operatorname{supp}{\mathfrak{m}}^{\prime}\succ 1/x$}.$

For $r\in{\mathbb{R}}^{>}$ we have $(x^{r})^{\prime}=rx^{r-1}\succ 1/x$, so
the claim holds for $n=0$. Suppose the claim holds for a certain $n$. Now
$G_{n+1}=G_{n}\exp(A_{n})$, $G_{n}$ is a convex subgroup of $G_{n+1}$, and

$A_{n}\ =\ \big{\\{}f\in{\mathbb{R}}[[G_{n}]]:\ \operatorname{supp}f\succ
G_{n-1}\big{\\}}\qquad\text{ (where $G_{-1}:=\\{1\\}$).}$

Let ${\mathfrak{m}}={\mathfrak{n}}\exp(a)\in G_{n+1}$ where ${\mathfrak{n}}\in
G_{n}$, $a\in A_{n}$; then

${\mathfrak{m}}\succ 1\quad\Longleftrightarrow\quad\text{$a>0$, or $a=0$,
${\mathfrak{n}}\succ 1$.}$

Suppose ${\mathfrak{m}}\succ 1$. If $a=0$, then
${\mathfrak{m}}={\mathfrak{n}}$, and we are done by inductive hypothesis, so
assume $a>0$. Then
${\mathfrak{m}}^{\prime}=({\mathfrak{n}}^{\prime}+{\mathfrak{n}}a^{\prime})\exp(a)$
and
$({\mathfrak{n}}^{\prime}+{\mathfrak{n}}a^{\prime})\in{\mathbb{R}}[[G_{n}]]$,
a differential subfield of ${\mathbb{T}}$, and
$\exp(a)>{\mathbb{R}}[[G_{n}]]$, hence
$\operatorname{supp}{\mathfrak{m}}^{\prime}\succ 1\succ 1/x$ as required.

Next, suppose ${\mathfrak{m}}\in G^{\operatorname{LE}}$ and
${\mathfrak{m}}\succ 1$. Take $n\geqslant 1$ such that
${\mathfrak{m}}{\uparrow}^{n}\in G^{\operatorname{E}}$. We have
$({\mathfrak{m}}{\uparrow}^{n})^{\prime}=({\mathfrak{m}}^{\prime}\cdot\ell_{0}\ell_{1}\cdots\ell_{n-1}){\uparrow}^{n}$.
For ${\mathfrak{n}}\in\operatorname{supp}{\mathfrak{m}}^{\prime}$ and using
${\mathfrak{m}}{\uparrow}^{n}\succ 1$ this gives

$({\mathfrak{n}}\cdot\ell_{0}\ell_{1}\cdots\ell_{n-1}){\uparrow}^{n}\ \succ\
1/x$

by what we proved for monomials in $G^{\operatorname{E}}$. Applying
${\downarrow}_{n}$ this yields ${\mathfrak{n}}\succ
1/(\ell_{0}\ell_{1}\cdots\ell_{n})$, hence
${\mathfrak{n}}\in\Lambda_{{\mathbb{T}}}$ as claimed. ∎

###### Proof of Proposition II.4.21.

Applying to the decomposition (II.4.2) gives

${\mathbb{T}}\ =\ \der({\mathbb{T}}_{\succ})\oplus\der({\mathbb{T}}_{\prec}).$

Now $\der({\mathbb{T}}_{\succ})\subseteq\Lambda_{{\mathbb{T}}}$ by Lemma
II.4.22, and
$\der({\mathbb{T}}_{\prec})\subseteq\operatorname{I}({\mathbb{T}})$, and so
these two inclusions are equalities. Thus
$\int\Lambda_{{\mathbb{T}}}={\mathbb{T}}_{\succ}$, from which the proposition
follows. ∎

######  Proposition II.4.23.

There is a unique group morphism $\exp\colon
K={\mathbb{T}}[\imag]\to\mathbb{O}^{\times}$ that extends the given
exponential maps $\exp\colon{\mathbb{T}}\to{\mathbb{T}}^{\times}$ and
$\exp\colon\mathbb{C}\to\mathbb{C}^{\times}$, and such that $\exp(\imag
f)=\operatorname{e}^{\imag f}$ for all $f\in{\mathbb{T}}_{\succ}$ and
$\exp(\varepsilon)=\sum_{n}\frac{\varepsilon^{n}}{n!}$ for all
$\varepsilon\in\smallo$. It is surjective, has kernel
$2\pi\imag{\mathbb{Z}}\subseteq\mathbb{C}$, and satisfies
$\exp(f)^{\prime}=f^{\prime}\exp(f)$ for all $f\in K$.

###### Proof.

The first statement follows easily from the decompositions

$K\ =\ {\mathbb{T}}\oplus\imag{\mathbb{T}}\ =\
{\mathbb{T}}\oplus\imag{\mathbb{T}}_{\succ}\oplus\imag{\mathbb{R}}\oplus\imag\smallo_{{\mathbb{T}}},\qquad\mathbb{C}\
=\ {\mathbb{R}}\oplus\imag{\mathbb{R}},\qquad\smallo\ =\
\smallo_{{\mathbb{T}}}\oplus\imag\smallo_{{\mathbb{T}}}$

of $K$, $\mathbb{C}$, and $\smallo=\smallo_{K}$ as internal direct sums of
${\mathbb{R}}$-linear subspaces. Next,

$\mathbb{O}^{\times}\ =\
K^{\times}\operatorname{e}^{\imag{\mathbb{T}}_{\succ}}=\ {\mathbb{T}}^{>}\cdot
S_{\mathbb{C}}\cdot(1+\smallo)\cdot\operatorname{e}^{\imag{\mathbb{T}}_{\succ}},\qquad
S_{\mathbb{C}}\ :=\ \big{\\{}z\in\mathbb{C}:\ |z|=1\big{\\}},$

by Lemmas II.1.1 and I.2.4, and Corollary I.2.7. Now
${\mathbb{T}}^{>}=\exp({\mathbb{T}})$ and
$S_{\mathbb{C}}=\exp(\imag{\mathbb{R}})$, so surjectivity follows from
$\exp(\smallo)=1+\smallo$, a consequence of the well-known bijectivity of the
map $\varepsilon\mapsto\sum_{n}\frac{\varepsilon^{n}}{n!}\colon\smallo\to
1+\smallo$, whose inverse is given by

$1+\delta\mapsto\log(1+\delta):=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\delta^{n}\qquad(\delta\in\smallo).$

That the kernel is $2\pi\imag{\mathbb{Z}}$ follows from the initial
decomposition of the additive group of $K$ as
${\mathbb{T}}\oplus\imag{\mathbb{T}}_{\succ}\oplus\imag{\mathbb{R}}\oplus\imag\smallo_{{\mathbb{T}}}$.
The identity $\exp(f)^{\prime}=f^{\prime}\exp(f)$ for $f\in K$ follows from it
being satisfied for $f\in{\mathbb{T}}$, $f\in\imag{\mathbb{T}}_{\succ}$,
$f\in\mathbb{C}$, and $f\in\smallo$. ∎

To integrate oscillating transseries, note first that the
${\mathbb{R}}$-linear operator $\int\colon{\mathbb{T}}\to{\mathbb{T}}$ extends
uniquely to a ${\mathbb{C}}$-linear operator
$\int\colon{\mathbb{T}}[\imag]\to{\mathbb{T}}[\imag]$. This in turn extends
uniquely to a ${\mathbb{C}}$-linear operator
$\int\colon\mathbb{O}\to\mathbb{O}$ such that $(\int\Phi)^{\prime}=\Phi$ for
all $\Phi\in\mathbb{O}$ and
$\int{\mathbb{T}}[\imag]\operatorname{e}^{\phi\imag}\subseteq{\mathbb{T}}[\imag]\operatorname{e}^{\phi\imag}$
for all $\phi\in{\mathbb{T}}_{\succ}$: given
$\phi\in{\mathbb{T}}_{\succ}^{\neq}$ and ${g\in{\mathbb{T}}[\imag]}$, there is
a unique $f\in{\mathbb{T}}[\imag[$ such that
$(f\operatorname{e}^{\phi\imag})^{\prime}=g\operatorname{e}^{\phi\imag}$:
existence holds because ${y^{\prime}+y\phi^{\prime}\imag=g}$ has a solution in
${\mathbb{T}}[\imag]$, the latter being linearly surjective, and uniqueness
holds by Lemma I.2.3 applied to $K=L={\mathbb{T}}[\imag]$, because
$\phi^{\prime}\imag\notin{\mathbb{T}}[\imag]^{\dagger}$ in view of remarks
preceding Lemma I.2.13.

The operator $\int$ is a right-inverse of the linear differential operator on
$\mathbb{O}$. To extend this to other linear differential operators, make the
subgroup
$G^{\mathbb{O}}:=G^{\operatorname{LE}}\operatorname{e}^{\imag{\mathbb{T}}_{\succ}}$
of $\mathbb{O}^{\times}$ into an ordered group so that the ordered subgroup
$G^{\operatorname{LE}}$ of ${\mathbb{T}}^{>}$ is a convex ordered subgroup of
$G^{\mathbb{O}}$ and $\operatorname{e}^{\imag\phi}\succ G^{\operatorname{LE}}$
for $\phi>0$ in ${\mathbb{T}}_{\succ}$. (Possible in only one way.) Next,
extend the natural inclusion
${\mathbb{T}}[\imag]\to{\mathbb{C}}[[G^{\operatorname{LE}}]]$ to a
${\mathbb{C}}$-algebra embedding
${\mathbb{O}\to{\mathbb{C}}[[G^{\mathbb{O}}]]}$ by sending
$\operatorname{e}^{\imag\phi}\in\mathbb{O}$ to
$\operatorname{e}^{\imag\phi}\in
G^{\mathbb{O}}\subseteq{\mathbb{C}}[[G^{\mathbb{O}}]]$. Identify $\mathbb{O}$
with a subalgebra of ${\mathbb{C}}[[G^{\mathbb{O}}]]$ via this embedding, so
$\operatorname{supp}f\subseteq G^{\mathbb{O}}$ for $f\in\mathbb{O}$. It makes
the Hahn space ${\mathbb{C}}[[G^{\mathbb{O}}]]$ over ${\mathbb{C}}$ an
immediate extension of its valued subspace $\mathbb{O}$. The latter is in
particular also a Hahn space over ${\mathbb{C}}$.

Let $A\in{\mathbb{T}}[\imag][\der]^{\neq}$. Then $A(\mathbb{O})=\mathbb{O}$ by
Lemmas II.4.3, II.4.19, and Proposition II.4.21. The proof of [ADH, 2.3.22]
now gives for each $g\in\mathbb{O}$ a unique element
${f=:A^{-1}(g)\in\mathbb{O}}$ with $A(f)=g$ and
$\operatorname{supp}(f)\cap{\mathfrak{d}}\big{(}\\!\ker_{\mathbb{O}}^{\neq}A\big{)}=\emptyset$.
This requirement on $\operatorname{supp}A^{-1}(g)$ yields a
${\mathbb{C}}$-linear operator $A^{-1}$ on $\mathbb{O}$ with $A\circ
A^{-1}=\operatorname{id}_{\mathbb{O}}$; we call it the distinguished right-
inverse of the operator $A$ on $\mathbb{O}$. With this definition -1 is the
operator $\int$ on $\mathbb{O}$ specified earlier.

In the next section we explore various valuations on universal exponential
extensions (such as $\mathbb{O}$) with additional properties.

###  II.5. Valuations on the Universal Exponential Extension

In this section $K$ is a valued differential field with algebraically closed
constant field ${C\subseteq\mathcal{O}}$ and divisible group $K^{\dagger}$ of
logarithmic derivatives. Then $\Gamma=v(K^{\times})$ is also divisible, since
we have a group isomorphism

$va\mapsto a^{\dagger}+(\mathcal{O}^{\times})^{\dagger}\ :\ \Gamma\to
K^{\dagger}/(\mathcal{O}^{\times})^{\dagger}\qquad(a\in K^{\times}).$

Let $\Lambda$ be a complement of the ${\mathbb{Q}}$-linear subspace
$K^{\dagger}$ of $K$, let $\lambda$ range over $\Lambda$, let
${{\operatorname{U}}=K\big{[}\\!\operatorname{e}(\Lambda)\big{]}}$ be the
universal exponential extension of $K$ constructed in Section II.2 and set
$\Omega:=\operatorname{Frac}({\operatorname{U}})$. Thus $\Omega$ is a
differential field with constant field $C$.

#### The gaussian extension

We equip ${\operatorname{U}}$ with the gaussian extension
$v_{\operatorname{g}}$ of the valuation of $K$ as defined in Section II.1; so
for $f\in{\operatorname{U}}$ with spectral decomposition $(f_{\lambda})$:

$v_{\operatorname{g}}(f)\ =\ \min_{\lambda}v(f_{\lambda}),$

and hence

$v_{\operatorname{g}}(f^{\prime})\ =\
\min_{\lambda}v(f_{\lambda}^{\prime}+\lambda f_{\lambda}).$

The field $\Omega$ with the valuation extending $v_{\operatorname{g}}$ is a
valued differential field extension of $K$, but it can happen that $K$ has
small derivation, whereas $\Omega$ does not:

######  Example.

Let $K=C(\\!(t^{{\mathbb{Q}}})\\!)$ and $\Lambda$ be as in Example II.2.4, so
$t\prec 1\prec x=t^{-1}$ and $t^{\prime}=-t^{2}$. Then $K$ is
$\operatorname{d}$-valued of $H$-type with small derivation, but in $\Omega$
with the above valuation,

$t\operatorname{e}(x)\ \prec\
1,\qquad\big{(}t\operatorname{e}(x)\big{)}{}^{\prime}\ =\
-t^{2}\operatorname{e}(x)+\operatorname{e}(x)\ \sim\ \operatorname{e}(x)\
\asymp\ 1.$

To obtain an example where $K=H[\imag]$ for a Liouville closed $H$-field $H$
and $\imag^{2}=-1$, take $K:={\mathbb{T}}[\imag]$ and
$\Lambda:=\Lambda_{{\mathbb{T}}}\imag$ as at the end of Section II.4. Now
$x\in\Lambda_{{\mathbb{T}}}$ and in $\Omega$ equipped with the above valuation
we have for $t:=x^{-1}$:

$t\operatorname{e}(x\imag)\ \prec\
1,\qquad\big{(}t\operatorname{e}(x\imag)\big{)}{}^{\prime}\ =\
-t^{2}\operatorname{e}(x\imag)+\imag\operatorname{e}(x\imag)\ \sim\
\imag\operatorname{e}(x\imag)\ \asymp\ 1,$

so $\big{(}t\operatorname{e}(x\imag)\big{)}{}^{\prime}\not\prec t^{\dagger}$,
hence $\Omega$ is neither asymptotic nor has small derivation.

However, we show next that under certain assumptions on $K$ with small
derivation, $\Omega$ has also a valuation which does make $\Omega$ a valued
differential field extension of $K$ with small derivation. For this we rely on
results from [ADH, 10.4]. Although such a valuation is less canonical than
$v_{\operatorname{g}}$, it is useful for harnessing the finiteness statements
about the set ${\mathscr{E}}^{\operatorname{e}}(A)$ of eventual exceptional
values of $A\in K[\der]^{\neq}$ from Section I.4 to obtain similar facts about
the set of ultimate exceptional values of $A$ introduced later in this
section.

#### Spectral extensions

In this subsection $K$ is $\operatorname{d}$-valued of $H$-type with
$\Gamma\neq\\{0\\}$ and with small derivation.

######  Lemma II.5.1.

The valuation of $K$ extends to a valuation on the field $\Omega$ that makes
$\Omega$ a $\operatorname{d}$-valued extension of $K$ of $H$-type with small
derivation.

###### Proof.

Applying [ADH, 10.4.7] to an algebraic closure of $K$ gives a
$\operatorname{d}$-valued algebraically closed extension $L$ of $K$ of
$H$-type with small derivation and $C_{L}=C$ such that $L^{\dagger}\supseteq
K$. Let $E:=\\{y\in L^{\times}:\,y^{\dagger}\in K\\}$, so $E$ is a subgroup of
$L^{\times}$, $E^{\dagger}=K$, and $K[E]$ is an exponential extension of $K$
with $C_{K[E]}=C$. Then Corollary II.2.10 gives an embedding
${{\operatorname{U}}\to L}$ of differential $K$-algebras with image $K[E]$,
which extends to an embedding $\Omega\to L$ of differential fields. Using this
embedding to transfer the valuation of $L$ to $\Omega$ gives a valuation as
required. ∎

A spectral extension of the valuation of $K$ to $\Omega$ is a valuation on the
field $\Omega$ with the properties stated in Lemma II.5.1. If $K$ is
$\upomega$-free, then so is $\Omega$ equipped with any spectral extension of
the valuation of $K$, by [ADH, 13.6] (and then $\Omega$ has rational
asymptotic integration by [ADH, 11.7]). We do not know whether this goes
through with “$\uplambda$-free” instead of “$\upomega$-free”. Here is
something weaker:

######  Lemma II.5.2.

Suppose $K$ is algebraically closed and $\uplambda$-free. Then some spectral
extension of the valuation of $K$ to $\Omega$ makes $\Omega$ a
$\operatorname{d}$-valued field with divisible value group and asymptotic
integration.

###### Proof.

Take $L$, $E$ and an embedding $\Omega\to L$ as in the proof of Lemma II.5.1.
Use this embedding to identify $\Omega$ with a differential subfield of $L$,
so $U=K[E]$ and $\Omega=K(E)$, and equip $\Omega$ with the spectral extension
of the valuation of $K$ obtained by restricting the valuation of $L$ to
$\Omega$. Since $L$ is algebraically closed, $E$ is divisible, and
$\Gamma_{L}=\Gamma+v(E)$ by [ADH, 10.4.7(iv)]. So $\Gamma_{\Omega}=\Gamma_{L}$
is divisible. Let $a\in K^{\times}$, $y\in E$. Then $K(y)$ has asymptotic
integration by Proposition I.3.12, hence
$v(ay)\in(\Gamma_{K(y)}^{\neq})^{\prime}\subseteq(\Gamma_{\Omega}^{\neq})^{\prime}$.
Thus $\Omega$ has asymptotic integration. ∎

In the rest of this subsection $\Omega$ is equipped with a spectral extension
$v$ (with value group $\Gamma_{\Omega}$) of the valuation of $K$. The proof of
Lemma II.5.1 and [ADH, 10.4.7] show that we can choose $v$ so that
$\Psi_{\Omega}\subseteq\Gamma$; but under suitable hypotheses on $K$, this is
automatic:

######  Lemma II.5.3.

Suppose $K$ has asymptotic integration and $\operatorname{I}(K)\subseteq
K^{\dagger}$. Then ${\Psi_{\Omega}\subseteq\Gamma}$, the group morphism

(II.5.1) $\lambda\mapsto v\big{(}\\!\operatorname{e}(\lambda)\big{)}\ \colon\
\Lambda\to\Gamma_{\Omega}$

is injective, and $\Gamma_{\Omega}$ is divisible with
$\Gamma_{\Omega}=\Gamma\oplus v\big{(}\\!\operatorname{e}(\Lambda)\big{)}$
(internal direct sum of ${\mathbb{Q}}$-linear subspaces of $\Gamma_{\Omega}$).
Moreover, $\Psi_{\Omega}=\Psi^{\downarrow}$ in $\Gamma$.

###### Proof.

For $a\in K^{\times}$ we have
$\big{(}a\operatorname{e}(\lambda)\big{)}{}^{\dagger}=a^{\dagger}+\lambda\in
K$, and if $a\operatorname{e}(\lambda)\asymp 1$, then

$a^{\dagger}+\lambda\ =\
\big{(}a\operatorname{e}(\lambda)\big{)}{}^{\dagger}\in(\mathcal{O}^{\times}_{\Omega})^{\dagger}\cap
K\ \subseteq\ \operatorname{I}(\Omega)\cap K\ =\ \operatorname{I}(K),$

so
$\lambda\in\Lambda\cap\big{(}\operatorname{I}(K)+K^{\dagger}\big{)}=\Lambda\cap
K^{\dagger}=\\{0\\}$ and $a\asymp 1$. Thus for $a_{1},a_{2}\in K^{\times}$ and
distinct $\lambda_{1},\lambda_{2}\in\Lambda$ we have
$a_{1}\operatorname{e}(\lambda_{1})\not\asymp
a_{2}\operatorname{e}(\lambda_{2})$, and so for $f\in{\operatorname{U}}$ with
spectral decomposition $(f_{\lambda})$ we have
$vf=\min_{\lambda}v\big{(}f_{\lambda}\operatorname{e}(\lambda)\big{)}$. Hence

$\Psi_{\Omega}\ \subseteq\big{\\{}v(a^{\dagger}+\lambda):\ a\in K^{\times},\
\lambda\in\Lambda\big{\\}}\ =\ v(K)\ =\ \Gamma_{\infty},$

the map (II.5.1) is injective and $\Gamma\cap
v\big{(}\\!\operatorname{e}(\Lambda)\big{)}=\\{0\\}$, and so
$\Gamma_{\Omega}=\Gamma\oplus v\big{(}\\!\operatorname{e}(\Lambda)\big{)}$
(internal direct sum of subgroups of $\Gamma_{\Omega}$). Since $\Gamma$ and
$\Lambda$ are divisible, so is $\Gamma_{\Omega}$. Now
$\Psi_{\Omega}=\Psi^{\downarrow}$ follows from
$K=({\operatorname{U}}^{\times})^{\dagger}\subseteq\Omega^{\dagger}$ and $K$
having asymptotic integration. ∎

We can now improve on Lemma II.4.1:

######  Corollary II.5.4.

Suppose $K$ has asymptotic integration and $\operatorname{I}(K)\subseteq
K^{\dagger}$, and let $A\in K[\der]^{\neq}$. Then the $C$-linear space
$\ker_{\operatorname{U}}A$ has a basis
$\mathcal{B}\subseteq\operatorname{U}^{\times}$ such that $v$ is injective on
$\mathcal{B}$ and $v(\mathcal{B})=v(\ker^{\neq}_{\operatorname{U}}A)$, and
thus $\lvert
v(\ker^{\neq}_{\operatorname{U}}A)\rvert=\dim_{C}\ker_{\operatorname{U}}A$.

###### Proof.

By [ADH, 5.6.6] we have a basis $\mathcal{B}_{\lambda}$ of the $C$-linear
space $\ker_{K}A_{\lambda}$ such that $v$ is injective on
$\mathcal{B}_{\lambda}$ and
$v(\mathcal{B}_{\lambda})=v(\ker^{\neq}_{K}A_{\lambda})$. Then
$\mathcal{B}:=\bigcup_{\lambda}\mathcal{B}_{\lambda}\operatorname{e}(\lambda)$
is a basis of $\ker_{\operatorname{U}}A$. It has the desired properties by
Lemma II.5.3. ∎

######  Corollary II.5.5.

Suppose $K$ is $\uplambda$-free and $\operatorname{I}(K)\subseteq
K^{\dagger}$. Then $\Omega$ has asymptotic integration, and so its
$H$-asymptotic couple is closed by Lemma II.5.3.

###### Proof.

By Lemma II.5.3,
$\Gamma_{\Omega}=\Gamma+v\big{(}\\!\operatorname{e}(\Lambda)\big{)}$. Using
Proposition I.3.12 as in the proof of Lemma II.5.2, with
$\operatorname{e}(\Lambda)$ in place of $E$, shows $\Omega$ has asymptotic
integration. ∎

#### Ultimate exceptional values

In this subsection $K$ is $H$-asymptotic with small derivation and asymptotic
integration. Also $A\in K[\der]^{\neq}$ and $r:=\operatorname{order}(A)$, and
$\gamma$ ranges over $\Gamma=v(K^{\times})$. We have
$v(\ker^{\neq}A_{\lambda})\subseteq{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})$,
so if $\lambda$ is an eigenvalue of $A$ with respect to $\lambda$, then
${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\neq\emptyset$. We call the
elements of the set

${\mathscr{E}}^{\operatorname{u}}(A)\ =\
{\mathscr{E}}^{\operatorname{u}}_{K}(A)\ :=\
\bigcup_{\lambda}\,{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\ =\
\big{\\{}\gamma:\,\text{$\operatorname{nwt}_{A_{\lambda}}(\gamma)\geqslant 1$
for some $\lambda$}\big{\\}}$

the ultimate exceptional values of $A$ with respect to $\Lambda$. The
definition of ${\mathscr{E}}^{\operatorname{u}}_{K}(A)$ involves our choice of
$\Lambda$, but we are leaving this implicit to avoid complicated notation. In
Section IV.4 we shall restrict $K$ and $\Lambda$ so that
${\mathscr{E}}^{\operatorname{u}}(A)$ does not depend any longer on the choice
of $\Lambda$. There we shall use the following observation:

######  Lemma II.5.6.

Let $a,b\in K$ be such that $a-b\in(\mathcal{O}^{\times})^{\dagger}$. Then for
all $\gamma$ we have
$\operatorname{nwt}_{A_{a}}(\gamma)=\operatorname{nwt}_{A_{b}}(\gamma)$; in
particular,
${\mathscr{E}}^{\operatorname{e}}(A_{a})={\mathscr{E}}^{\operatorname{e}}(A_{b})$.

###### Proof.

Use that if $u\in\mathcal{O}^{\times}$ and $a-b=u^{\dagger}$, then
$A_{a}=(A_{b})_{\ltimes u}$. ∎

######  Corollary II.5.7.

Let $\Lambda^{*}$ be a complement of the ${\mathbb{Q}}$-linear subspace
$K^{\dagger}$ of $K$ and let
$\lambda\mapsto\lambda^{*}\colon\Lambda\to\Lambda^{*}$ be the group
isomorphism with $\lambda-\lambda^{*}\in K^{\dagger}$ for all $\lambda$. If
$\lambda-\lambda^{*}\in(\mathcal{O}^{\times})^{\dagger}$ for all $\lambda$,
then
$\operatorname{nwt}_{A_{\lambda}}(\gamma)=\operatorname{nwt}_{A_{\lambda^{*}}}(\gamma)$
for all $\gamma$, so
${\mathscr{E}}^{\operatorname{u}}(A)=\bigcup_{\lambda}\,{\mathscr{E}}^{\operatorname{e}}(A_{\lambda^{*}})$.

######  Remark II.5.8.

For $a\in K^{\times}$ we have
${\mathscr{E}}^{\operatorname{u}}(aA)={\mathscr{E}}^{\operatorname{u}}(A)$ and
${\mathscr{E}}^{\operatorname{u}}(Aa)={\mathscr{E}}^{\operatorname{u}}(A)-va$.
Note also that
${\mathscr{E}}^{\operatorname{e}}(A)={\mathscr{E}}^{\operatorname{e}}(A_{0})\subseteq{\mathscr{E}}^{\operatorname{u}}(A)$.
Let $\phi\in K^{\times}$ be active in $K$, and set
$\lambda^{\phi}:=\phi^{-1}\lambda$. Then $\Lambda^{\phi}:=\phi^{-1}\Lambda$ is
a complement of the ${\mathbb{Q}}$-linear subspace
$(K^{\phi})^{\dagger}=\phi^{-1}K^{\dagger}$ of $K^{\phi}$, and
$(A^{\phi})_{\lambda^{\phi}}=(A_{\lambda})^{\phi}$. Hence
${\mathscr{E}}^{\operatorname{u}}_{K}(A)$ agrees with the set
${\mathscr{E}}^{\operatorname{u}}_{K^{\phi}}(A^{\phi})$ of ultimate
exceptional values of $A^{\phi}$ with respect to $\Lambda^{\phi}$.

######  Remark II.5.9.

Suppose $L$ is an $H$-asymptotic extension of $K$ with asymptotic integration
and algebraically closed constant field $C_{L}$ such that $L^{\dagger}$ is
divisible, and $\Psi$ is cofinal in $\Psi_{L}$ or $K$ is $\uplambda$-free.
Then
${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})={\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})\cap\Gamma$,
by Lemma I.4.1 and Corollary I.7.10. Hence if $\Lambda_{L}\supseteq\Lambda$ is
a complement of the subspace $L^{\dagger}$ of the ${\mathbb{Q}}$-linear space
$L$, and ${\mathscr{E}}^{\operatorname{u}}_{L}(A)$ is the set of ultimate
exceptional values of $A$ (viewed as an element of $L[\der]$) with respect to
$\Lambda_{L}$, then
${\mathscr{E}}^{\operatorname{u}}(A)\subseteq{\mathscr{E}}^{\operatorname{u}}_{L}(A)$.
(Note that such a complement $\Lambda_{L}$ exists iff $L^{\dagger}\cap
K=K^{\dagger}$.)

In the rest of this subsection we equip ${\operatorname{U}}$ with the gaussian
extension $v_{\operatorname{g}}$ of the valuation of $K$. Recall that we have
a decomposition $\ker_{{\operatorname{U}}}A=\bigoplus_{\lambda}(\ker
A_{\lambda})\operatorname{e}(\lambda)$ of the $C$-linear space
$\ker_{{\operatorname{U}}}A$ as an internal direct sum of subspaces, and hence

(II.5.2) $v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)\ =\
\bigcup_{\lambda}\,v(\ker^{\neq}A_{\lambda})\ \subseteq\
\bigcup_{\lambda}\,{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\ =\
{\mathscr{E}}^{\operatorname{u}}(A).$

Here are some consequences:

######  Lemma II.5.10.

Suppose $K$ is $r$-linearly newtonian. Then
$v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{u}}(A)$.

###### Proof.

By Proposition I.4.2 we have
$v(\ker^{\neq}A_{\lambda})={\mathscr{E}}^{\operatorname{e}}(A_{\lambda})$ for
each $\lambda$. Therefore
$v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{u}}(A)$
by (II.5.2). ∎

######  Lemma II.5.11.

Suppose $K$ is $\operatorname{d}$-valued. Then $\lvert
v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)\rvert\leqslant\dim_{C}\ker_{{\operatorname{U}}}A\leqslant
r$.

###### Proof.

By [ADH, 5.6.6(i)] applied to $A_{\lambda}$ in place of $A$ we have

$\lvert v(\ker^{\neq}A_{\lambda})\rvert\ =\ \dim_{C}\ker A_{\lambda}\ =\
\operatorname{mult}_{\lambda}(A)\qquad\text{ for all~{}$\lambda$}$

and thus by (II.5.2),

$\lvert v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)\rvert\
\leqslant\ \sum_{\lambda}\,\lvert v(\ker^{\neq}A_{\lambda})\rvert\ =\
\sum_{\lambda}\,\operatorname{mult}_{\lambda}(A)\ =\
\dim_{C}\ker_{{\operatorname{U}}}A\ \leqslant\ r$

as claimed. ∎

######  Lemma II.5.12.

Suppose $\operatorname{I}(K)\subseteq K^{\dagger}$ and $r=1$. Then

$v_{\operatorname{g}}(\ker^{\neq}_{{\operatorname{U}}}A)\ =\
{\mathscr{E}}^{\operatorname{u}}(A),\qquad\lvert{\mathscr{E}}^{\operatorname{u}}(A)\rvert\
=\ 1.$

###### Proof.

Arrange $A=\der-g$, $g\in K$, and take $f\in K^{\times}$ and $\lambda$ such
that $g=f^{\dagger}+\lambda$. Then
$u:=f\operatorname{e}(\lambda)\in{\operatorname{U}}^{\times}$ satisfies
$A(u)=0$, hence $\ker^{\neq}_{{\operatorname{U}}}A=Cu$ and thus
$v_{\text{g}}(\ker^{\neq}_{{\operatorname{U}}}A)=\\{vf\\}$. By Lemma I.4.9 we
have $v(\ker^{\neq}A_{\lambda})={\mathscr{E}}^{\operatorname{e}}(A_{\lambda})$
for all $\lambda$ and hence
$v_{\operatorname{g}}(\ker^{\neq}_{{\operatorname{U}}}A)={\mathscr{E}}^{\operatorname{u}}(A)$
by (II.5.2). ∎

######  Corollary II.5.13.

If $\operatorname{I}(K)\subseteq K^{\dagger}$ and $a\in K^{\times}$, then
${\mathscr{E}}^{\operatorname{e}}(\der-a^{\dagger})={\mathscr{E}}^{\operatorname{u}}(\der-a^{\dagger})=\\{va\\}$.

Proposition II.5.15 below partly extends Lemma II.5.12.

#### Spectral extensions and ultimate exceptional values

In this subsection $K$ is $\operatorname{d}$-valued of $H$-type with small
derivation, asymptotic integration, and $\operatorname{I}(K)\subseteq
K^{\dagger}$. Also $A\in K[\der]^{\neq}$ has order $r$ and $\gamma$ ranges
over $\Gamma$.

Suppose $\Omega$ is equipped with a spectral extension $v$ of the valuation of
$K$. Let ${g\in K^{\times}}$ with $vg=\gamma$. The Newton weight of
$A_{\lambda}g\in K[\der]$ does not change in passing from $K$ to $\Omega$,
since $\Psi$ is cofinal in $\Psi_{\Omega}$ by Lemma II.5.3; see [ADH, 11.1].
Thus

$\operatorname{nwt}_{A_{\lambda}}(\gamma)\ =\
\operatorname{nwt}(A_{\lambda}g)\ =\
\operatorname{nwt}\\!\big{(}Ag\operatorname{e}(\lambda)\big{)}\ =\
\operatorname{nwt}_{A}\\!\big{(}v(g\operatorname{e}(\lambda)\big{)}\ =\
\operatorname{nwt}_{A}\\!\big{(}\gamma+v(\operatorname{e}(\lambda)\big{)}\big{)}.$

In particular, using $\Gamma_{\Omega}=\Gamma\oplus
v\big{(}\\!\operatorname{e}(\Lambda)\big{)}$,

(II.5.3) ${\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)\ =\
\bigcup_{\lambda}\,{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})+v\big{(}\\!\operatorname{e}(\lambda)\big{)}\qquad\text{(a
disjoint union)}.$

Thus
${\mathscr{E}}^{\operatorname{u}}(A)=\pi\big{(}{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)\big{)}$
where $\pi\colon\Gamma_{\Omega}\to\Gamma$ is given by
$\pi\big{(}{\gamma+v\big{(}\\!\operatorname{e}(\lambda)}\big{)}\big{)}=\gamma$.

######  Lemma II.5.14.

We have
$\dim_{C}\ker_{{\operatorname{U}}}A\leqslant\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert$,
and

$\dim_{C}\ker_{{\operatorname{U}}}A\ =\
\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\
\Longleftrightarrow\ v(\ker^{\neq}A_{\lambda})\ =\
{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\text{ for all $\lambda$.}$

Moreover, if
$\dim_{C}\ker_{{\operatorname{U}}}A=\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert$,
then
$v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{u}}(A)$.

###### Proof.

Clearly, $\dim_{C}\ker_{{\operatorname{U}}}A\leqslant\dim_{C}\ker_{\Omega}A$.
Equip $\Omega$ with a spectral extension of the valuation of $K$. Then
$\dim_{C}\ker_{\Omega}A=\lvert v(\ker_{\Omega}^{\neq}A)\rvert$ and
$v(\ker_{\Omega}^{\neq}A)\subseteq{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)$
by [ADH, 5.6.6(i)] and [ADH, p. 481], respectively, applied to $\Omega$ in the
role of $K$. Also
$|{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)|=\sum_{\lambda}|{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})|$
(a sum of cardinals) by the remarks preceding the lemma. This yields the first
claim of the lemma.

Next, note that
$v(\ker^{\neq}A_{\lambda})\subseteq{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})$
for all $\lambda$. Hence from (II.5.3) and

$v(\ker_{{\operatorname{U}}}^{\neq}A)\ =\
\bigcup_{\lambda}v(\ker^{\neq}A_{\lambda})+v\big{(}\\!\operatorname{e}(\lambda)\big{)}\qquad\text{(a
disjoint union)}$

we obtain:

$v(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)\quad\Longleftrightarrow\quad
v(\ker^{\neq}A_{\lambda})\ =\
{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\text{ for all $\lambda$.}$

Also $\lvert
v(\ker_{{\operatorname{U}}}^{\neq}A)\rvert=\dim_{C}\ker_{{\operatorname{U}}}A$
by [ADH, 2.3.13], and

$v(\ker_{{\operatorname{U}}}^{\neq}A)\ \subseteq\ v(\ker_{\Omega}^{\neq}A)\
\subseteq\
{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A),\qquad|{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)|\
=\ \sum_{\lambda}|{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})|.$

This yields the displayed equivalence.

Suppose
$\dim_{C}\ker_{{\operatorname{U}}}A=\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert$;
we need to show
$v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{u}}(A)$.
We have
$\pi\big{(}{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)\big{)}={\mathscr{E}}^{\operatorname{u}}(A)$
for the above projection map $\pi\colon\Gamma_{\Omega}\to\Gamma$, so it is
enough to show
$\pi\big{(}v(\ker_{{\operatorname{U}}}^{\neq}A)\big{)}=v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)$.
For that, note that for $\mathcal{B}\subseteq
K^{\times}\operatorname{e}(\Lambda)$ in Corollary II.5.4 we have

$\pi\big{(}v(\ker_{{\operatorname{U}}}^{\neq}A)\big{)}\ =\ \pi(v\mathcal{B})\
=\ v_{\operatorname{g}}(\mathcal{B})\ =\
v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A),$

using for the last equality the details in the proof of Corollary II.5.4. ∎

######  Proposition II.5.15.

Suppose $K$ is $\upomega$-free. Then
$\operatorname{nwt}_{A_{\lambda}}(\gamma)=0$ for all but finitely many pairs
$(\gamma,\lambda)$ and

$\lvert{\mathscr{E}}^{\operatorname{u}}(A)\rvert\ \leqslant\
\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\ =\
\sum_{\gamma,\lambda}\operatorname{nwt}_{A_{\lambda}}(\gamma)\ \leqslant\ r.$

If $\dim_{C}\ker_{{\operatorname{U}}}A=r$, then
$\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert=r$
and
$v_{\operatorname{g}}(\ker_{{\operatorname{U}}}^{\neq}A)={\mathscr{E}}^{\operatorname{u}}(A)$.

###### Proof.

Equip $\Omega$ with a spectral extension $v$ of the valuation of $K$. Then
$\Omega$ is $\upomega$-free, so
$\sum_{\lambda}|{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})|=|{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)|\leqslant
r$ by the remarks preceding Lemma II.5.14 and Corollary I.4.5 applied to
$\Omega$ in place of $K$. These remarks also give
$\operatorname{nwt}_{A_{\lambda}}(\gamma)=0$ for all but finitely many pairs
$(\gamma,\lambda)$, and so

$\sum_{\gamma,\lambda}\operatorname{nwt}_{A_{\lambda}}(\gamma)\ =\
\sum_{\gamma,\lambda}\operatorname{nwt}_{A}\\!\big{(}\gamma+v(\operatorname{e}(\lambda)\big{)}\
=\ \lvert{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)\rvert\ \leqslant\ r.$

Corollary I.4.5 applied to $A_{\lambda}$ in place of $A$ yields
$\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert=\sum_{\gamma}\operatorname{nwt}_{A_{\lambda}}(\gamma)$
and so
$\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert=\sum_{\gamma,\lambda}\operatorname{nwt}_{A_{\lambda}}(\gamma)$.
This proves the first part (including the display). The rest follows from this
and Lemma II.5.14. ∎

In the next lemma (to be used in the proof of Proposition III.1.27), as well
as in Corollary II.5.21, $L$ is a $\operatorname{d}$-valued $H$-asymptotic
extension of $K$ with algebraically closed constant field and asymptotic
integration (so $L$ has small derivation), such that $L^{\dagger}$ is
divisible, $L^{\dagger}\cap K=K^{\dagger}$, and $\operatorname{I}(L)\subseteq
L^{\dagger}$ . We also fix there a complement $\Lambda_{L}$ of the
${\mathbb{Q}}$-linear subspace $L^{\dagger}$ of $L$ with
$\Lambda\subseteq\Lambda_{L}$. Let
${\operatorname{U}}_{L}=L\big{[}\\!\operatorname{e}(\Lambda_{L})\big{]}$ be
the corresponding universal exponential extension of $L$ containing
${\operatorname{U}}=K\big{[}\\!\operatorname{e}(\Lambda)\big{]}$ as a
differential subring, as described in the remarks following Corollary II.2.13,
with differential fraction field $\Omega_{L}$.

######  Lemma II.5.16.

Assume $C_{L}=C$. Let $\Omega_{L}$ be equipped with a spectral extension of
the valuation of $L$, and take $\Omega$ as a valued subfield of $\Omega_{L}$;
so the valuation of $\Omega$ is a spectral extension of the valuation of $K$.
Suppose $\Psi$ is cofinal in $\Psi_{L}$ or $K$ is $\uplambda$-free. Then
${\mathscr{E}}^{\operatorname{e}}_{\Omega_{L}}(A)\cap\Gamma_{\Omega}\ =\
{\mathscr{E}}^{\operatorname{e}}_{\Omega}(A)$.

###### Proof.

Let $\mu$ range over $\Lambda_{L}$. We have

$\Gamma_{\Omega_{L}}\ =\ \Gamma_{L}\oplus
v\big{(}\\!\operatorname{e}(\Lambda_{L})\big{)},\qquad\Gamma_{\Omega}\ =\
\Gamma\oplus v\big{(}\\!\operatorname{e}(\Lambda)\big{)}$

by Lemma II.5.3 and

${\mathscr{E}}^{\operatorname{e}}_{\Omega_{L}}\ =\
\bigcup_{\mu}{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\mu})+v\big{(}\\!\operatorname{e}(\mu)\big{)},\qquad{\mathscr{E}}^{\operatorname{e}}_{\Omega}\
=\
\bigcup_{\lambda}{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})+v\big{(}\\!\operatorname{e}(\lambda)\big{)}$

by (II.5.3). Hence

${\mathscr{E}}^{\operatorname{e}}_{\Omega_{L}}(A)\cap\Gamma_{\Omega}\ =\
\bigcup_{\lambda}\big{(}{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})\cap\Gamma\big{)}+v\big{(}\\!\operatorname{e}(\lambda)\big{)}\
=\
\bigcup_{\lambda}{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})+v\big{(}\\!\operatorname{e}(\lambda)\big{)}\
=\ {\mathscr{E}}^{\operatorname{e}}_{\Omega}(A),$

where we used the injectivity of $\mu\mapsto
v\big{(}\\!\operatorname{e}(\mu)\big{)}$ for the first equality and Remark
II.5.9 for the second. ∎

The material in the rest of this section is not used later in this monograph.

#### Terminal operators

In this subsection we continue to work in the setting of the previous
subsection. Call $A$ terminal with respect to $\Lambda$ if
$\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert=r$.
We omit “with respect to $\Lambda$” if it is clear from the context what
$\Lambda$ is. In Section IV.4 we shall restrict $K$, $\Lambda$ anyway so that
this dependence on $\Lambda$ disappears. Recall also that for a given spectral
extension of the valuation of $K$ to $\Omega$ we have
$\lvert{\mathscr{E}}_{\Omega}(A)\rvert=\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert$.
If $A$ is terminal and $\phi\in K^{\times}$ is active in $K$, then
$A^{\phi}\in K^{\phi}[\derdelta]$ is terminal with respect to $\Lambda^{\phi}$
(cf. remarks after Corollary II.5.7). If $A$ is terminal and $a\in
K^{\times}$, then $aA$ is terminal. If $r=0$, then $A$ is terminal.

######  Lemma II.5.17.

If $r=1$, then $A$ is terminal.

###### Proof.

Assume $r=1$. Then $\dim_{C}\ker_{{\operatorname{U}}}A=1$, so
$\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\geqslant
1$ by Lemma II.5.14. Equip $\Omega$ with a spectral extension of the valuation
of $K$. Then $\Omega$ is ungrounded by Lemma II.5.3, and $r=1$ gives
$\lvert{\mathscr{E}}_{\Omega}(A)\rvert\leqslant 1$. Now use
$\lvert{\mathscr{E}}_{\Omega}(A)\rvert=\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert$.
∎

######  Lemma II.5.18.

Suppose $A$ and $B\in K[\der]^{\neq}$ are terminal, and each operator
$B_{\lambda}$ is asymptotically surjective. Then $AB$ is terminal.

###### Proof.

Use that $(AB)_{\lambda}=A_{\lambda}B_{\lambda}$, and that
$\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda}B_{\lambda})\rvert=\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert+\lvert{\mathscr{E}}^{\operatorname{e}}(B_{\lambda})\rvert$
by Corollary I.4.19. ∎

Thus if $A$ is terminal and $a\in K^{\times}$, then $aA,Aa$, and $A_{\ltimes
a}$ are terminal. From Lemmas II.5.17, II.5.18, and Corollary I.4.25 we
conclude:

######  Corollary II.5.19.

If $K$ is $\uplambda$-free and $A$ splits over $K$, then $A$ is terminal.

######  Corollary II.5.20.

Suppose $K$ is $\upomega$-free and $B\in K[\der]^{\neq}$. Then $A$ and $B$ are
terminal iff $AB$ is terminal.

###### Proof.

The “only if” part follows from Lemma II.5.18 and Corollary I.4.26. For the
“if” part, use Corollary I.4.19 and Proposition II.5.15. ∎

######  Corollary II.5.21.

Suppose $A$ is terminal, $\Psi$ is cofinal in $\Psi_{L}$ or $K$ is
$\uplambda$-free, and $L$ is $\upomega$-free. Then, with respect to the
complement $\Lambda_{L}$ of $L^{\dagger}$ in $L$, we have:

  1. (i)

as an element of $L[\der]$, $A$ is terminal;

  2. (ii)

${\mathscr{E}}^{\operatorname{e}}(A_{\mu})=\emptyset$ for all
$\mu\in\Lambda_{L}\setminus\Lambda$;

  3. (iii)

${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})={\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})$
for all $\lambda$; and

  4. (iv)

${\mathscr{E}}^{\operatorname{u}}(A)={\mathscr{E}}^{\operatorname{u}}_{L}(A)$.

###### Proof.

By the remarks after Corollary II.5.7 we have
${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\subseteq{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})$
for each $\lambda$, and so with $\mu$ ranging over $\Lambda_{L}$, by
Proposition II.5.15 applied to $L$ in place of $K$, we have
$r=\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\leqslant\sum_{\mu}\lvert{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\mu})\rvert\leqslant
r$. This yields (i)–(iv). ∎

In [8] we shall study other situations where $A$ is terminal.

#### The real case

In this subsection $H$ is a real closed $H$-field with small derivation,
asymptotic integration, and $H^{\dagger}=H$; also $K=H[\imag]$,
$\imag^{2}=-1$, for our valued differential field $K$. We also assume
$\operatorname{I}(H)\imag\subseteq K^{\dagger}$. Then $K$ is
$\operatorname{d}$-valued of $H$-type with small derivation, asymptotic
integration, $K^{\dagger}=H+\operatorname{I}(H)\imag$, and
$\operatorname{I}(K)\subseteq K^{\dagger}$. Note that $H$ and thus $K$ is
$\uplambda$-free by [ADH, remark after 11.6.2, and 11.6.8]. Let $A$ in
$K[\der]^{\neq}$ have order $r$ and let $\gamma$ range over $\Gamma$.

######  Lemma II.5.22.

If the real closed $H$-asymptotic extension $F$ of $H$ has asymptotic
integration and convex valuation ring, then $L^{\dagger}\cap K=K^{\dagger}$
for the algebraically closed $H$-asymptotic field extension $L:=F[\imag]$ of
$K$.

###### Proof.

Use Corollary I.2.15 and earlier remarks in the same subsection. ∎

######  Corollary II.5.23.

The $H$-field $H$ has an $H$-closed extension $F$ with $C_{F}=C_{H}$, and for
any such $F$, the algebraically closed $\operatorname{d}$-valued field
extension $L:=F[\imag]$ of $H$-type of $K$ is $\upomega$-free with $C_{L}=C$,
$\operatorname{I}(L)\subseteq L^{\dagger}$, and $L^{\dagger}\cap
K=K^{\dagger}$.

###### Proof.

Use [ADH, 16.4.1, 9.1.2] to extend $H$ to an $\upomega$-free $H$-field with
the same constant field as $H$, next use [ADH, 11.7.23] to pass to its real
closure, and then use [ADH, 14.5.9] to extend further to an $H$-closed $F$,
still with the same constant field as $H$. For any such $F$, the
$\operatorname{d}$-valued field $L:=F[\imag]$ of $H$-type is $\upomega$-free
by [ADH, 11.7.23] and newtonian by (0.12). Hence $\operatorname{I}(L)\subseteq
L^{\dagger}$ by Lemma I.2.9, and $L^{\dagger}\cap K=K^{\dagger}$ by Lemma
II.5.22. ∎

This leads to a variant of Proposition II.5.15:

######  Proposition II.5.24.

The conclusion of Proposition II.5.15 holds. In particular:

$\dim_{C}\ker_{{\operatorname{U}}}A=r\ \Longrightarrow\ A\text{ is terminal}.$

###### Proof.

Corollary II.5.23 gives an $H$-closed extension $F$ of $H$ with $C_{F}=C_{H}$,
so $L:=F[\imag]$ is $\upomega$-free, $C_{L}=C$, $\operatorname{I}(L)\subseteq
L^{\dagger}$, and $L^{\dagger}\cap K=K^{\dagger}$. Take a complement
$\Lambda_{L}\supseteq\Lambda$ of the subspace $L^{\dagger}$ of the
${\mathbb{Q}}$-linear space $L$. By Remark II.5.9 we have
${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})={\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})\cap\Gamma$.
Hence Proposition II.5.15 applied to $K$, $\Lambda$ replaced by $L$,
$\Lambda_{L}$, respectively, and $A$ viewed as element of $L[\der]$, yields
$\sum_{\lambda}\,\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\leqslant
r$. Corollary I.7.10 applied to $A_{\lambda}$ in place of $A$ gives
$\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert=\sum_{\gamma}\operatorname{nwt}_{A_{\lambda}}(\gamma)$.
This yields the conclusion of Proposition II.5.15 as in the proof of that
proposition. ∎

Let now $F$ be a Liouville closed $H$-field extension of $H$ and suppose
$\operatorname{I}(L)\subseteq L^{\dagger}$ where $L:=F[\imag]$. Lemma II.5.22
yields $L^{\dagger}\cap K=K^{\dagger}$, so $L$ is as described just before
Lemma II.5.16, and we have a complement $\Lambda_{L}\supseteq\Lambda$ of the
subspace $L^{\dagger}$ of the ${\mathbb{Q}}$-linear space $L$. Note that if
$A$ splits over $K$, then $A$ is terminal by Corollary II.5.19.

######  Corollary II.5.25.

Suppose $A$ is terminal. Then, with respect to the complement $\Lambda_{L}$ of
$L^{\dagger}$ in $L$, the conclusions (i)–(iv) of Corollary II.5.21 hold.

###### Proof.

By the remarks after Corollary II.5.7 we have
${\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\subseteq{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\lambda})$
for all $\lambda$, and so with $\mu$ ranging over $\Lambda_{L}$, Proposition
II.5.24 applied to $L$ in place of $K$, we have
$r=\sum_{\lambda}\lvert{\mathscr{E}}^{\operatorname{e}}(A_{\lambda})\rvert\leqslant\sum_{\mu}\lvert{\mathscr{E}}^{\operatorname{e}}_{L}(A_{\mu})\rvert\leqslant
r$. This yields (i)–(iv). ∎

##  Part III Normalizing Holes and Slots

In this introduction $K$ is an $H$-asymptotic field with small derivation and
rational asymptotic integration. In Section III.2 we introduce holes in $K$: A
hole in $K$ is a triple $(P,{\mathfrak{m}},\widehat{a})$ with $P\in
K\\{Y\\}\setminus K$, ${\mathfrak{m}}\in K^{\times}$, and
$\widehat{a}\in\widehat{K}\setminus K$ for some immediate asymptotic extension
$\widehat{K}$ of $K$, such that $\widehat{a}\prec{\mathfrak{m}}$ and
$P(\widehat{a})=0$. The main goal of Part III is a normalization theorem,
namely Theorem III.3.33, that allows us to transform under reasonable
conditions a hole $(P,{\mathfrak{m}},\widehat{a})$ in $K$ into a “normal”
hole; this helps to pin down the location of $\widehat{a}$ relative to $K$.
The notion of $(P,{\mathfrak{m}},\widehat{a})$ being normal involves the
linear part of the differential polynomial $P_{\times{\mathfrak{m}}}$, in
particular the span of this linear part. We introduce the span in the
preliminary Section III.1. In Section III.4 we study isolated holes
$(P,{\mathfrak{m}},\widehat{a})$ in $K$, which under reasonable conditions
ensure the uniqueness of the isomorphism type of $K\langle\widehat{a}\rangle$
as a valued differential field over $K$; see Proposition III.4.9. In Section
III.5 we focus on holes $(P,{\mathfrak{m}},\widehat{a})$ in $K$ where
$\operatorname{order}P=\deg P=1$. For technical reasons we actually work in
Part III also with slots in $K$, which are a bit more general than holes in
$K$.

First some notational conventions. Let $\Gamma$ be an ordered abelian group.
For $\gamma,\delta\in\Gamma$ with $\gamma\neq 0$ the expression
“$\delta=o(\gamma)$” means “$n|\delta|<|\gamma|$ for all $n\geqslant 1$”
according to [ADH, 2.4], but here we find it convenient to extend this to
$\gamma=0$, in which case “$\delta=o(\gamma)$” means “$\delta=0$”. Suppose
$\Gamma=v(E^{\times})$ is the value group of a valued field $E$ and
${\mathfrak{m}}\in E^{\times}$. Then we denote the archimedean class
$[v{\mathfrak{m}}]\subseteq\Gamma$ of $v{\mathfrak{m}}\in\Gamma$ by just
$[{\mathfrak{m}}]$. Suppose ${\mathfrak{m}}\not\asymp 1$. Then we have a
proper convex subgroup

$\Delta({\mathfrak{m}})\ :=\
\big{\\{}\gamma\in\Gamma:\,\gamma=o(v{\mathfrak{m}})\big{\\}}\ =\
\big{\\{}\gamma\in\Gamma:\,[\gamma]<[{\mathfrak{m}}]\big{\\}},$

of $\Gamma$. If
${\mathfrak{m}}\asymp_{\Delta({\mathfrak{m}})}{\mathfrak{n}}\in E$, then
$0\neq{\mathfrak{n}}\not\asymp 1$ and
$\Delta({\mathfrak{m}})=\Delta({\mathfrak{n}})$. In particular, if
${\mathfrak{m}}\asymp{\mathfrak{n}}\in E$, then $0\neq{\mathfrak{n}}\not\asymp
1$ and $\Delta({\mathfrak{m}})=\Delta({\mathfrak{n}})$. Note that for $f,g\in
E$ the meaning of “$f\preccurlyeq_{\Delta({\mathfrak{m}})}g$” does not change
in passing to a valued field extension of $E$, although
$\Delta({\mathfrak{m}})$ can increase as a subgroup of the value group of the
extension.

###  III.1. The Span of a Linear Differential Operator

In this section $K$ is a valued differential field with small derivation and
$\Gamma:=v(K^{\times})$. We let $a$, $b$, sometimes subscripted, range over
$K$, and ${\mathfrak{m}}$, ${\mathfrak{n}}$ over $K^{\times}$. Consider a
linear differential operator

$A\ =\ a_{0}+a_{1}\der+\cdots+a_{r}{}^{r}\in K[\der],\qquad a_{r}\neq 0.$

We shall use below the quantities $\operatorname{dwm}(A)$ and
$\operatorname{dwt}(A)$ defined in [ADH, 5.6]. We also introduce a measure
${\mathfrak{v}}(A)$ for the “lopsidedness” of $A$ as follows:

${\mathfrak{v}}(A)\ :=\ a_{r}/a_{m}\in\ K^{\times}\qquad\text{where
$m:=\operatorname{dwt}(A)$.}$

So $a_{r}\asymp{\mathfrak{v}}(A)A$ and ${\mathfrak{v}}(A)\preccurlyeq 1$, with

${\mathfrak{v}}(A)\asymp
1\quad\Longleftrightarrow\quad\operatorname{dwt}(A)=r\quad\Longleftrightarrow\quad{\mathfrak{v}}(A)=1.$

Also note that ${\mathfrak{v}}(aA)={\mathfrak{v}}(A)$ for $a\neq 0$. Moreover,

${\mathfrak{v}}(A_{\ltimes{\mathfrak{n}}})A_{\ltimes{\mathfrak{n}}}\ \asymp\
a_{r}\ \asymp\ {\mathfrak{v}}(A)A$

since $A_{\ltimes{\mathfrak{n}}}=a_{r}{}^{r}+\text{lower order terms in
$\der$}$.

######  Example.

${\mathfrak{v}}(a+\der)=1$ if $a\preccurlyeq 1$, and
${\mathfrak{v}}(a+\der)=1/a$ if $a\succ 1$.

We call ${\mathfrak{v}}(A)$ the span of $A$. We are mainly interested in the
valuation of ${\mathfrak{v}}(A)$. This is related to the gaussian valuation
$v(A)$ of $A$: if $A$ is monic, then $v\big{(}{\mathfrak{v}}(A)\big{)}=-v(A)$.
An important property of the span of $A$ is that its valuation is not affected
by small additive perturbations of $A$:

######  Lemma III.1.1.

Suppose $B\in K[\der]$, $\operatorname{order}(B)\leqslant r$ and
$B\prec{\mathfrak{v}}(A)A$. Then:

  1. (i)

$A+B\sim A$, $\operatorname{dwm}(A+B)=\operatorname{dwm}(A)$, and
$\operatorname{dwt}(A+B)=\operatorname{dwt}(A)$;

  2. (ii)

$\operatorname{order}(A+B)=r$ and ${\mathfrak{v}}(A+B)\sim{\mathfrak{v}}(A)$.

###### Proof.

From $B\prec{\mathfrak{v}}(A)A$ and ${\mathfrak{v}}(A)\preccurlyeq 1$ we
obtain $B\prec A$, and thus (i). Set $m:=\operatorname{dwt}(A)$, let $i$ range
over $\\{0,\dots,r\\}$, and let $B=b_{0}+b_{1}\der+\cdots+b_{r}{}^{r}$. Then
$a_{i}\preccurlyeq a_{m}$ and $b_{i}\prec{\mathfrak{v}}(A)A\asymp
a_{r}\preccurlyeq a_{m}$. Therefore, if $a_{i}\asymp a_{m}$, then
$a_{i}+b_{i}\sim a_{i}$, and if $a_{i}\prec a_{m}$, then $a_{i}+b_{i}\prec
a_{m}$. Hence ${\mathfrak{v}}(A+B)=(a_{r}+b_{r})/(a_{m}+b_{m})\sim
a_{r}/a_{m}={\mathfrak{v}}(A)$. ∎

For $b\neq 0$, the valuation of ${\mathfrak{v}}(Ab)$ only depends on $vb$; it
is enough to check this for $b\asymp 1$. More generally:

######  Lemma III.1.2.

Let $B\in K[\der]^{\neq}$ and $b\asymp B$. Then
${\mathfrak{v}}(AB)\asymp{\mathfrak{v}}(Ab){\mathfrak{v}}(B)$.

###### Proof.

Let $B=b_{0}+b_{1}\der+\cdots+b_{s}{}^{s}$, $b_{s}\neq 0$. Then

$AB\ =\ a_{r}b_{s}{}^{r+s}+\text{lower order terms in}\ \der,$

so by [ADH, 5.6.1(ii)] for $\gamma=0$:

$\displaystyle v\big{(}{\mathfrak{v}}(AB)\big{)}\ $ $\displaystyle=\
v(a_{r}b_{s})-v(AB)\ =\ v(a_{r}b_{s})-v(Ab)$ $\displaystyle=\
v(a_{r}b)-v(Ab)+v(b_{s})-v(B)$ $\displaystyle=\
v\big{(}{\mathfrak{v}}(Ab){\mathfrak{v}}(B)\big{)}.\qed$

######  Corollary III.1.3.

Let $B\in K[\der]^{\neq}$. If ${\mathfrak{v}}(AB)=1$, then
${\mathfrak{v}}(A)={\mathfrak{v}}(B)=1$. The converse holds if $B$ is monic.

This is clear from from Lemma III.1.2, and in turn gives:

######  Corollary III.1.4.

Suppose $A=a(\der-b_{1})\cdots(\der-b_{r})$. Then

${\mathfrak{v}}(A)=1\quad\Longleftrightarrow\quad
b_{1},\dots,b_{r}\preccurlyeq 1.$

######  Remark.

Suppose $K=C(\\!(t)\\!)$ with the $t$-adic valuation and derivation
$\der=t\frac{d}{dt}$. In the literature, $A$ is called regular singular if
${\mathfrak{v}}(A)=1$, and irregular singular if ${\mathfrak{v}}(A)\prec 1$;
see [20, Definition 3.14].

######  Lemma III.1.5.

Let $B\in K[\der]^{\neq}$. Then
${\mathfrak{v}}(AB)\preccurlyeq{\mathfrak{v}}(B)$, and if $B$ is monic, then
${\mathfrak{v}}(AB)\preccurlyeq{\mathfrak{v}}(A)$.

###### Proof.

Lemma III.1.2 and ${\mathfrak{v}}(Ab)\preccurlyeq 1$ for $b\neq 0$ yields
${\mathfrak{v}}(AB)\preccurlyeq{\mathfrak{v}}(B)$. Suppose $B$ is monic, so
$v(B)\leqslant 0$. To show ${\mathfrak{v}}(AB)\preccurlyeq{\mathfrak{v}}(A)$
we arrange that $A$ is also monic. Then $AB$ is monic, and
${\mathfrak{v}}(AB)\preccurlyeq{\mathfrak{v}}(A)$ is equivalent to
$v(AB)\leqslant v(A)$. Now

$v(AB)\ =\ v_{AB}(0)\ =\ v_{A}\big{(}v_{B}(0)\big{)}\ =\
v_{A}\big{(}v(B)\big{)}\ \leqslant\ v_{A}(0)\ =\ v(A)$

by [ADH, 4.5.1(iii), 5.6.1(ii)]. ∎

######  Corollary III.1.6.

If $A=a(\der-b_{1})\cdots(\der-b_{r})$, then $b_{1},\dots,b_{r}\ \preccurlyeq\
{\mathfrak{v}}(A)^{-1}.$

Let $\Delta$ be a convex subgroup of $\Gamma$, let $\dot{\mathcal{O}}$ be the
valuation ring of the coarsening $v_{\Delta}$ of the valuation $v$ of $K$ by
$\Delta$, with maximal ideal $\dot{\smallo}$, and
$\dot{K}=\dot{\mathcal{O}}/\dot{\smallo}$ be the valued differential residue
field of $v_{\Delta}$. The residue morphism $\dot{\mathcal{O}}\to\dot{K}$
extends to the ring morphism $\dot{\mathcal{O}}[\der]\to\dot{K}[\der]$ with
$\der\mapsto\der$. If $A\in\dot{\mathcal{O}}[\der]$ and $\dot{A}\neq 0$, then
$\operatorname{dwm}(\dot{A})=\operatorname{dwm}(A)$ and
$\operatorname{dwt}(\dot{A})=\operatorname{dwt}(A)$. We set
${\mathfrak{v}}:={\mathfrak{v}}(A)$.

######  Lemma III.1.7.

If $A\in\dot{\mathcal{O}}[\der]$ and $\operatorname{order}(\dot{A})=r$, then
${\mathfrak{v}}(\dot{A})=\dot{\mathfrak{v}}$.

#### Behavior of the span under twisting

Recall that $o(\gamma):=0\in\Gamma$ for $\gamma=0\in\Gamma$. With this
convention, here is a consequence of [ADH, 6.1.3]:

######  Lemma III.1.8.

Let $B\in K[\der]^{\neq}$. Then $v(AB)=v(A)+v(B)+o\big{(}v(B)\big{)}$.

###### Proof.

Take $b$ with $b\asymp B$. Then

$v(AB)=v_{AB}(0)=v_{A}\big{(}v_{B}(0)\big{)}=v_{A}(vb)=v(Ab)$

by [ADH, 5.6.1(ii)]. Moreover, $v(Ab)=v(A)+vb+o(vb)$, by [ADH, 6.1.3]. ∎

We have
${\mathfrak{v}}(A_{\ltimes{\mathfrak{n}}})={\mathfrak{v}}(A{\mathfrak{n}})$,
so $v(A_{\ltimes{\mathfrak{n}}})=v(A)+o(v{\mathfrak{n}})$ by Lemma III.1.8.
Moreover:

######  Lemma III.1.9.

$v\big{(}{\mathfrak{v}}(A{\mathfrak{n}})\big{)}=v\big{(}{\mathfrak{v}}(A)\big{)}+o(v{\mathfrak{n}})$.

###### Proof.

Replacing $A$ by $a_{r}^{-1}A$ we arrange $A$ is monic, so
$A_{\ltimes{\mathfrak{n}}}$ is monic, and thus

$v\big{(}{\mathfrak{v}}(A{\mathfrak{n}})\big{)}\ =\
v\big{(}{\mathfrak{v}}(A_{\ltimes{\mathfrak{n}}})\big{)}=-v(A_{\ltimes{\mathfrak{n}}})=-v(A)+o(v{\mathfrak{n}})=v\big{(}{\mathfrak{v}}(A)\big{)}+o(v{\mathfrak{n}})$

by remarks preceding the lemma. ∎

Recall: we denote the archimedean class $[v{\mathfrak{n}}]\subseteq\Gamma$ by
$[{\mathfrak{n}}]$. Lemma III.1.9 yields:

######  Corollary III.1.10.

$\big{[}{\mathfrak{v}}(A)\big{]}<[{\mathfrak{n}}]\ \Longleftrightarrow\
\big{[}{\mathfrak{v}}(A{\mathfrak{n}})\big{]}<[{\mathfrak{n}}]$.

Under suitable conditions on $K$ we can say more about the valuation of
${\mathfrak{v}}(A_{\ltimes{\mathfrak{n}}})$: Lemma III.1.12 below.

######  Lemma III.1.11.

Let ${\mathfrak{n}}^{\dagger}\succcurlyeq 1$ and
${\mathfrak{m}}_{0},\dots,{\mathfrak{m}}_{r}\in K^{\times}$ be such that

$v({\mathfrak{m}}_{i})+v(A)\ =\ \min_{i\leqslant j\leqslant
r}v(a_{j})+(j-i)v({\mathfrak{n}}^{\dagger}).$

Then with $m:=\operatorname{dwt}(A)$ we have

${\mathfrak{m}}_{0}\ \succcurlyeq\ \cdots\ \succcurlyeq\
{\mathfrak{m}}_{r}\quad\text{and}\quad({\mathfrak{n}}^{\dagger})^{m}\
\preccurlyeq\ {\mathfrak{m}}_{0}\ \preccurlyeq\
({\mathfrak{n}}^{\dagger})^{r}.$

(In particular, $[{\mathfrak{m}}_{0}]\leqslant[{\mathfrak{n}}^{\dagger}]$,
with equality if $m>0$.)

###### Proof.

From $v({\mathfrak{n}}^{\dagger})\leqslant 0$ we obtain
$v({\mathfrak{m}}_{0})\leqslant\cdots\leqslant v({\mathfrak{m}}_{r})$. We have
$0\leqslant v(a_{j}/a_{m})$ for $j=0,\dots,r$ and so

$rv({\mathfrak{n}}^{\dagger})\ \leqslant\ \min_{0\leqslant j\leqslant
r}v(a_{j}/a_{m})+jv({\mathfrak{n}}^{\dagger})\ =\ v({\mathfrak{m}}_{0})\
\leqslant\ mv({\mathfrak{n}}^{\dagger})$

as required. ∎

######  Lemma III.1.12.

Suppose $\der\mathcal{O}\subseteq\smallo$. Then

${\mathfrak{n}}^{\dagger}\preccurlyeq 1\ \Longrightarrow\
v(A_{\ltimes{\mathfrak{n}}})=v(A),\qquad{\mathfrak{n}}^{\dagger}\succ 1\
\Longrightarrow\ \lvert v(A_{\ltimes{\mathfrak{n}}})-v(A)\rvert\leqslant-
rv({\mathfrak{n}}^{\dagger}).$

###### Proof.

Let $R:=\operatorname{Ri}A$. Then
$v(A_{\ltimes{\mathfrak{n}}})=v(R_{+{\mathfrak{n}}^{\dagger}})$ by [ADH,
5.8.11]. If ${\mathfrak{n}}^{\dagger}\preccurlyeq 1$, then
$v(R_{+{\mathfrak{n}}^{\dagger}})=v(R)$ by [ADH, 4.5.1(i)], hence
$v(A_{\ltimes{\mathfrak{n}}})=v(R)=v(A)$ by [ADH, 5.8.10]. Now suppose
${\mathfrak{n}}^{\dagger}\succ 1$. Claim:
$v(A_{\ltimes{\mathfrak{n}}})-v(A)\geqslant rv({\mathfrak{n}}^{\dagger})$. To
prove this claim we replace $A$ by $a^{-1}A$, where $a\asymp A$, to arrange
$A\asymp 1$. Let $i$, $j$ range over $\\{0,\dots,r\\}$. We have
$R_{+{\mathfrak{n}}^{\dagger}}=\sum_{i}b_{i}R_{i}$ where

$b_{i}\ =\ \sum_{j\geqslant i}{j\choose
i}a_{j}R_{j-i}({\mathfrak{n}}^{\dagger}).$

Take ${\mathfrak{m}}_{i}\in K^{\times}$ as in Lemma III.1.11. By Lemma I.1.17
we have $R_{n}({\mathfrak{n}}^{\dagger})\sim({\mathfrak{n}}^{\dagger})^{n}$
for all $n$; hence $v(b_{i})\geqslant v({\mathfrak{m}}_{i})$ for all $i$. Thus

$v(A_{\ltimes{\mathfrak{n}}})-v(A)\ =\ v(A_{\ltimes{\mathfrak{n}}})\ =\
v(R_{+{\mathfrak{n}}^{\dagger}})\ \geqslant\ \min_{i}v(b_{i})\ \geqslant\
v({\mathfrak{m}}_{0})\ \geqslant\ rv({\mathfrak{n}}^{\dagger})$

by Lemma III.1.11, proving our claim. Applying this claim with
$A_{\ltimes{\mathfrak{n}}}$, ${\mathfrak{n}}^{-1}$ in place of $A$,
${\mathfrak{n}}$ also yields $v(A_{\ltimes{\mathfrak{n}}})-v(A)\leqslant-
rv({\mathfrak{n}}^{\dagger})$, thus $\lvert
v(A_{\ltimes{\mathfrak{n}}})-v(A)\rvert\leqslant-
rv({\mathfrak{n}}^{\dagger})$. ∎

######  Remark.

Suppose that $\der\mathcal{O}\subseteq\smallo$ and
${\mathfrak{n}}^{\dagger}\succ 1$. Then Lemma III.1.12 improves on Lemma
III.1.9, since $v({\mathfrak{n}}^{\dagger})=o(v{\mathfrak{n}})$ by [ADH,
6.4.1(iii)].

######  Lemma III.1.13.

Suppose $\der\mathcal{O}\subseteq\smallo$ and
${\mathfrak{n}}^{\dagger}\preccurlyeq{\mathfrak{v}}(A)^{-1}$. Let $B\in
K[\der]$ and $s\in{\mathbb{N}}$ be such that $\operatorname{order}(B)\leqslant
s$ and $B\prec{\mathfrak{v}}(A)^{s+1}A$. Then
$B_{\ltimes{\mathfrak{n}}}\prec{\mathfrak{v}}(A_{\ltimes{\mathfrak{n}}})A_{\ltimes{\mathfrak{n}}}$.

###### Proof.

We may assume $B\neq 0$ and $s=\operatorname{order}(B)$. It suffices to show
$B_{\ltimes{\mathfrak{n}}}\prec{\mathfrak{v}}(A)A$. If
${\mathfrak{n}}^{\dagger}\preccurlyeq 1$, then Lemma III.1.12 applied to $B$
in place of $A$ yields $B_{\ltimes{\mathfrak{n}}}\asymp
B\prec{\mathfrak{v}}(A)A$. Suppose ${\mathfrak{n}}^{\dagger}\succ 1$. Then
Lemma III.1.12 gives $\lvert v(B_{\ltimes{\mathfrak{n}}})-v(B)\rvert\leqslant-
sv({\mathfrak{n}}^{\dagger})\leqslant sv({\mathfrak{v}}(A))$ and hence
$B_{\ltimes{\mathfrak{n}}}\preccurlyeq{\mathfrak{v}}(A)^{-s}B\prec{\mathfrak{v}}(A)A$.
∎

If $\der\mathcal{O}\subseteq\smallo$, then we have functions
$\operatorname{dwm}_{A},\operatorname{dwt}_{A}\colon\Gamma\to{\mathbb{N}}$ as
defined in [ADH, 5.6]. Combining Lemmas III.1.1 and III.1.13 yields a variant
of [ADH, 6.1.7]:

######  Corollary III.1.14.

Suppose $\der\mathcal{O}\subseteq\smallo$ and
${\mathfrak{n}}^{\dagger}\preccurlyeq{\mathfrak{v}}(A)^{-1}$. Let $B\in
K[\der]$ be such that $\operatorname{order}(B)\leqslant r$ and
$B\prec{\mathfrak{v}}(A)^{r+1}A$. Then
$\operatorname{dwm}_{A+B}(v{\mathfrak{n}})=\operatorname{dwm}_{A}(v{\mathfrak{n}})$
and
$\operatorname{dwt}_{A+B}(v{\mathfrak{n}})=\operatorname{dwt}_{A}(v{\mathfrak{n}})$.
In particular,

$v{\mathfrak{n}}\in{\mathscr{E}}(A+B)\ \Longleftrightarrow\
v{\mathfrak{n}}\in{\mathscr{E}}(A).$

#### About $A({\mathfrak{n}}^{q})$ and $A{\mathfrak{n}}^{q}$

Suppose ${\mathfrak{m}}^{l}=\pm{\mathfrak{n}}^{k}$ where $k,l\in{\mathbb{Z}}$,
$l\neq 0$. Then ${\mathfrak{m}}^{\dagger}=q{\mathfrak{n}}^{\dagger}$ with
$q=k/l\in{\mathbb{Q}}$. In particular, if $K$ is real closed or algebraically
closed, then for any ${\mathfrak{n}}$ and $q\in{\mathbb{Q}}$ we have
${\mathfrak{m}}^{\dagger}=q{\mathfrak{n}}^{\dagger}$ for some
${\mathfrak{m}}$.

Below in this subsection $K$ is $\operatorname{d}$-valued and ${\mathfrak{n}}$
is such that for all $q\in{\mathbb{Q}}^{>}$ we are given an element of
$K^{\times}$, denoted by ${\mathfrak{n}}^{q}$ for suggestiveness, with
$({\mathfrak{n}}^{q})^{\dagger}=q{\mathfrak{n}}^{\dagger}$.

Let $q\in{\mathbb{Q}}^{>}$; then $v({\mathfrak{n}}^{q})=qv({\mathfrak{n}})$:
to see this we may arrange that $K$ is algebraically closed by [ADH, 10.1.23],
and hence contains an ${\mathfrak{m}}$ such that
$v{\mathfrak{m}}=q\,v{\mathfrak{n}}$ and
${\mathfrak{m}}^{\dagger}=q{\mathfrak{n}}^{\dagger}=({\mathfrak{n}}^{q})^{\dagger}$,
and thus $v({\mathfrak{n}}^{q})=v{\mathfrak{m}}=q\,v{\mathfrak{n}}$.

######  Lemma III.1.15.

Suppose ${\mathfrak{n}}^{\dagger}\succcurlyeq 1$. Then for all but finitely
many $q\in{\mathbb{Q}}^{>}$,

$v\big{(}A({\mathfrak{n}}^{q})\big{)}\ =\
v({\mathfrak{n}}^{q})+\min_{j}v(a_{j})+jv({\mathfrak{n}}^{\dagger}).$

###### Proof.

Let $q\in{\mathbb{Q}}^{>}$ and take $b_{0},\dots,b_{r}\in K$ with
$A{\mathfrak{n}}^{q}=b_{0}+b_{1}\der+\cdots+b_{r}{}^{r}$. Then

$b_{0}\ =\ A({\mathfrak{n}}^{q})\ =\
{\mathfrak{n}}^{q}\big{(}a_{0}R_{0}(q{\mathfrak{n}}^{\dagger})+a_{1}R_{1}(q{\mathfrak{n}}^{\dagger})+\cdots+a_{r}R_{r}(q{\mathfrak{n}}^{\dagger})\big{)}.$

Let $i$, $j$ range over $\\{0,\dots,r\\}$. By Lemma I.1.17,
$R_{i}(q{\mathfrak{n}}^{\dagger})\sim q^{i}({\mathfrak{n}}^{\dagger})^{i}$ for
all $i$. Take ${\mathfrak{m}}$ (independent of $q$) such that
$v({\mathfrak{m}})=\min_{j}v(a_{j})+jv({\mathfrak{n}}^{\dagger})$, and let $I$
be the nonempty set of $i$ with ${\mathfrak{m}}\asymp
a_{i}({\mathfrak{n}}^{\dagger})^{i}$. For $i\in I$ we take $c_{i}\in
C^{\times}$ such that $a_{i}({\mathfrak{n}}^{\dagger})^{i}\sim
c_{i}{\mathfrak{m}}$, and set $R:=\sum_{i\in I}c_{i}Y^{i}\in C[Y]^{\neq}$.
Therefore, if $R(q)\neq 0$, then

$\sum_{i\in I}a_{i}R_{i}(q{\mathfrak{n}}^{\dagger})\ \sim\
{\mathfrak{m}}R(q).$

Assume $R(q)\neq 0$ in what follows. Then

$\sum_{i=0}^{r}a_{i}R_{i}(q{\mathfrak{n}}^{\dagger})\ \sim\ \sum_{i\in
I}a_{i}R_{i}(q{\mathfrak{n}}^{\dagger})\ \sim\ {\mathfrak{m}}R(q)\ \asymp\
{\mathfrak{m}},$

hence $b_{0}\asymp{\mathfrak{m}}{\mathfrak{n}}^{q}$, in particular, $b_{0}\neq
0$. ∎

######  Lemma III.1.16.

Assume ${\mathfrak{n}}^{\dagger}\succcurlyeq 1$ and
$[{\mathfrak{v}}]<[{\mathfrak{n}}]$ for ${\mathfrak{v}}:={\mathfrak{v}}(A)$.
Then $\big{[}{\mathfrak{v}}(A{\mathfrak{n}}^{q})\big{]}<[{\mathfrak{n}}]$ for
all $q\in{\mathbb{Q}}^{>}$, and for all but finitely many
$q\in{\mathbb{Q}}^{>}$ we have
${\mathfrak{v}}(A{\mathfrak{n}}^{q})\preccurlyeq{\mathfrak{v}}$, and thus
$[{\mathfrak{v}}]\leqslant\big{[}{\mathfrak{v}}(A{\mathfrak{n}}^{q})\big{]}$.

###### Proof.

Let $q\in{\mathbb{Q}}^{>}$. Then
$[{\mathfrak{v}}]<[{\mathfrak{n}}]=[{\mathfrak{n}}^{q}]$, so
$[{\mathfrak{v}}(A{\mathfrak{n}}^{q})]<[{\mathfrak{n}}^{q}]=[{\mathfrak{n}}]$
by Corollary III.1.10. To show the second part, let $m=\operatorname{dwt}(A)$.
Replacing $A$ by $a_{m}^{-1}A$ we arrange $a_{m}=1$, so
$a_{r}={\mathfrak{v}}$, $A\asymp 1$. Take $b_{0},\dots,b_{r}$ with
$A{\mathfrak{n}}^{q}=b_{0}+b_{1}\der+\cdots+b_{r}{}^{r}$. As in the proof of
Lemma III.1.15 we obtain an ${\mathfrak{m}}$ and a polynomial $R(Y)\in
C[Y]^{\neq}$ (both independent of $q$) such that
$v({\mathfrak{m}})=\min_{j}v(a_{j})+jv({\mathfrak{n}}^{\dagger})$, and
$b_{0}\asymp{\mathfrak{m}}{\mathfrak{n}}^{q}$ if $R(q)\neq 0$. Assume
$R(q)\neq 0$ in what follows; we show that then
${\mathfrak{v}}(A{\mathfrak{n}}^{q})\preccurlyeq{\mathfrak{v}}$. For
$n:=\operatorname{dwt}(A{\mathfrak{n}}^{q})$,

$b_{0}{\mathfrak{v}}(A{\mathfrak{n}}^{q})\ \preccurlyeq\
b_{n}{\mathfrak{v}}(A{\mathfrak{n}}^{q})\ =\ b_{r}\ =\
{\mathfrak{n}}^{q}{\mathfrak{v}},$

hence
${\mathfrak{v}}(A{\mathfrak{n}}^{q})\preccurlyeq{\mathfrak{v}}/{\mathfrak{m}}$.
It remains to note that ${\mathfrak{m}}\succcurlyeq
a_{m}({\mathfrak{n}}^{\dagger})^{m}=({\mathfrak{n}}^{\dagger})^{m}\succcurlyeq
1$. ∎

######  Lemma III.1.17.

Assume ${\mathfrak{n}}^{\dagger}\succcurlyeq 1$ and ${\mathfrak{m}}$ satisfies

$v{\mathfrak{m}}+v(A)\ =\ \min_{0\leqslant j\leqslant
r}v(a_{j})+jv({\mathfrak{n}}^{\dagger}).$

Then $[{\mathfrak{m}}]\leqslant[{\mathfrak{n}}^{\dagger}]$, with equality if
$\operatorname{dwt}(A)>0$, and for all but finitely many
$q\in{\mathbb{Q}}^{>}$,

$A{\mathfrak{n}}^{q}\ \asymp\
{\mathfrak{m}}\,{\mathfrak{n}}^{q}\,A,\qquad{\mathfrak{v}}(A)/{\mathfrak{v}}(A{\mathfrak{n}}^{q})\
\asymp\ {\mathfrak{m}}.$

###### Proof.

Replacing $A$ by $a_{m}^{-1}A$ where $m=\operatorname{dwt}(A)$ we arrange
$a_{m}=1$, so $a_{r}={\mathfrak{v}}:={\mathfrak{v}}(A)$ and $A\asymp 1$. Let
$i$, $j$ range over $\\{0,\dots,r\\}$. Let $q\in{\mathbb{Q}}^{>}$, and take
$b_{i}\in K$ such that $A{\mathfrak{n}}^{q}=\sum_{i}b_{i}{}^{i}$. By [ADH,
(5.1.3)] we have

$b_{i}\ =\ \frac{1}{i!}A^{(i)}({\mathfrak{n}}^{q})\ =\
{\mathfrak{n}}^{q}\frac{1}{i!}\operatorname{Ri}(A^{(i)})(q{\mathfrak{n}}^{\dagger})\
=\ {\mathfrak{n}}^{q}\sum_{j\geqslant i}{j\choose
i}a_{j}R_{j-i}(q{\mathfrak{n}}^{\dagger}).$

Take ${\mathfrak{m}}_{i}\in K^{\times}$ as in Lemma III.1.11. Then
${\mathfrak{m}}_{0}\asymp{\mathfrak{m}}$ (so
$[{\mathfrak{m}}]\leqslant[{\mathfrak{n}}^{\dagger}]$, with equality if
$m>0$), and ${\mathfrak{m}}_{r}\asymp{\mathfrak{v}}$. Lemma III.1.15 applied
to $A^{(i)}/i!$ instead of $A$ gives that for all but finitely
$q\in{\mathbb{Q}}^{>}$ we have
$b_{i}\asymp{\mathfrak{m}}_{i}{\mathfrak{n}}^{q}$ for all $i$. Assume that
$q\in{\mathbb{Q}}^{>}$ has this property. From
$v({\mathfrak{m}})=v({\mathfrak{m}}_{0})\leqslant\cdots\leqslant
v({\mathfrak{m}}_{r})=v({\mathfrak{v}})$ we obtain

$v({\mathfrak{m}})+qv({\mathfrak{n}})\ =\ v(b_{0})\ \leqslant\ \cdots\
\leqslant\ v(b_{r})\ =\ v({\mathfrak{v}})+q\,v({\mathfrak{n}}).$

With $n=\operatorname{dwt}(A{\mathfrak{n}}^{q})$ this gives
$v(b_{0})=\cdots=v(b_{n})=v(A{\mathfrak{n}}^{q})$. Thus

${\mathfrak{v}}(A{\mathfrak{n}}^{q})\ =\ b_{r}/b_{n}\ \asymp\ b_{r}/b_{0}\
\asymp\ ({\mathfrak{n}}^{q}{\mathfrak{v}})/({\mathfrak{n}}^{q}{\mathfrak{m}})\
=\ {\mathfrak{v}}/{\mathfrak{m}}$

as claimed. ∎

The next lemma (not used later) is a more precise version of Lemma III.1.17,
but with an additional hypothesis on ${\mathfrak{n}}^{\dagger}$:

######  Lemma III.1.18.

Assume ${\mathfrak{n}}^{\dagger}\succ{\mathfrak{v}}(A)^{-1}$. Then

$A({\mathfrak{n}})\sim A{\mathfrak{n}}\sim
a_{r}{\mathfrak{n}}({\mathfrak{n}}^{\dagger})^{r}\sim
a_{r}{\mathfrak{n}}^{(r)},\qquad{\mathfrak{v}}(A{\mathfrak{n}})\sim({\mathfrak{n}}^{\dagger})^{-r}.$

###### Proof.

Let $i$, $j$ range over $\\{0,\dots,r\\}$ and take $b_{i}\in K$ such that
$A{\mathfrak{n}}=\sum_{i}b_{i}{}^{i}$, so
$b_{i}={\mathfrak{n}}\sum_{j\geqslant i}{j\choose
i}a_{j}R_{j-i}({\mathfrak{n}}^{\dagger})$. By Lemma I.1.16 we have
$R_{j-i}({\mathfrak{n}}^{\dagger})\sim({\mathfrak{n}}^{\dagger})^{j-i}$ for
${i\leqslant j}$. From
${\mathfrak{n}}^{\dagger}\succ{\mathfrak{v}}^{-1}\succcurlyeq 1$ we get for
$i\leqslant j<r$:

$a_{r}({\mathfrak{n}}^{\dagger})^{r-i}\ \succcurlyeq\
a_{r}{\mathfrak{n}}^{\dagger}({\mathfrak{n}}^{\dagger})^{j-i}\ \succ\
a_{r}{\mathfrak{v}}^{-1}({\mathfrak{n}}^{\dagger})^{j-i}\ \succcurlyeq\
a_{j}({\mathfrak{n}}^{\dagger})^{j-i}.$

Therefore $b_{i}\sim{\mathfrak{n}}{r\choose
i}a_{r}({\mathfrak{n}}^{\dagger})^{r-i}$, from which the first displayed
equivalences follow. Now $\operatorname{dwt}(A{\mathfrak{n}})=0$ and so
${\mathfrak{v}}(A{\mathfrak{n}})=b_{r}/b_{0}=({\mathfrak{n}}a_{r})/A({\mathfrak{n}})\sim({\mathfrak{n}}^{\dagger})^{-r}$
as claimed. ∎

Let ${\mathfrak{v}}\in K^{\times}$ with ${\mathfrak{v}}\not\asymp 1$; so we
have the proper convex subgroup of $\Gamma$ given by

$\Delta({\mathfrak{v}})\ =\
\big{\\{}\gamma\in\Gamma:\,\gamma=o(v{\mathfrak{v}})\big{\\}}\ =\
\big{\\{}\gamma\in\Gamma:\,[\gamma]<[{\mathfrak{v}}]\big{\\}}.$

If $K$ is $H$-asymptotic, then we also have the convex subgroup

$\Delta\ =\
\big{\\{}\gamma\in\Gamma:\,\gamma^{\dagger}>v({\mathfrak{v}}^{\dagger})\big{\\}}$

of $\Gamma$ with $\Delta\subseteq\Delta({\mathfrak{v}})$. If $K$ is
$H$-asymptotic of Hardy type (Section I.2), then we have
${\Delta=\Delta({\mathfrak{v}})}$, and hence the relations
$\preccurlyeq_{\Delta({\mathfrak{v}})}$, $\prec_{\Delta({\mathfrak{v}})}$,
$\asymp_{\Delta({\mathfrak{v}})}$ agree with $\preccurlyeq_{\mathfrak{v}}$,
$\prec_{{\mathfrak{v}}}$, $\asymp_{{\mathfrak{v}}}$, respectively, from [ADH,
p. 407].

######  Corollary III.1.19.

Suppose ${\mathfrak{n}}^{\dagger}\succcurlyeq 1$ and
$[{\mathfrak{n}}^{\dagger}]<[{\mathfrak{v}}]$ where
${\mathfrak{v}}:={\mathfrak{v}}(A)$ (so $0\neq{\mathfrak{v}}\prec 1)$. Let
$B\in K[\der]$ and $w\geqslant r$ be such that
$B\prec_{\Delta({\mathfrak{v}})}{\mathfrak{v}}^{w}A$. Then for all but
finitely many $q\in{\mathbb{Q}}^{>}$ we have
${\mathfrak{w}}:={\mathfrak{v}}(A{\mathfrak{n}}^{q})\asymp_{\Delta({\mathfrak{v}})}{\mathfrak{v}}$
and
$B{\mathfrak{n}}^{q}\prec_{\Delta({\mathfrak{w}})}{\mathfrak{w}}^{w}A{\mathfrak{n}}^{q}$.

###### Proof.

The case $B=0$ is trivial, so assume $B\neq 0$. Take ${\mathfrak{m}}$ as in
Lemma III.1.17, and take ${\mathfrak{m}}_{B}$ likewise with $B$ in place of
$A$. By this lemma,
$[{\mathfrak{m}}],[{\mathfrak{m}}_{B}]\leqslant[{\mathfrak{n}}^{\dagger}]<[{\mathfrak{v}}]$,
hence ${\mathfrak{m}},{\mathfrak{m}}_{B}\asymp_{\Delta({\mathfrak{v}})}1$.