#  Residue of special functions of Anderson $A$-modules at the characteristic
graph

Quentin Gazda,
and Andreas Maurischat Max Planck Institut für Mathematik, Vivatsgasse 7,
D-53111 BonnFH Aachen University of Applied Sciences & RWTH Aachen University,
D-52062 Aachen

###### Contents

  1. 1 Motivations and results
  2. 2 Preliminaries
    1. 2.1 Notations
    2. 2.2 The characteristic graph ideal $\mathfrak{j}$
  3. 3 Objects from rigid analytic geometry
    1. 3.1 The rigid analytic space $\mathfrak{X}$
    2. 3.2 Sheaves and divisors on $\mathfrak{X}$
    3. 3.3 Costable subsets and costable sheaves
  4. 4 Anderson modules over rigid analytic spaces
    1. 4.1 $E(\mathbb{C}_{\infty})$-valued meromorphic functions
    2. 4.2 Action of $A\otimes A$
    3. 4.3 Special functions as $E(\mathbb{C}_{\infty})$-valued functions
  5. 5 Continuation and residues of special functions
    1. 5.1 Residue at $\mathfrak{j}$
      1. 5.1.1 Residue map for meromorphic functions on $\mathfrak{X}$
      2. 5.1.2 Residue of $E(\mathbb{C}_{\infty})$-valued functions at $\mathfrak{j}$
    2. 5.2 Proofs of the main results
    3. 5.3 Filtration on $\mathfrak{sf}(E)$
      1. Summary

##  1 Motivations and results

Let $(C,\mathcal{O}_{C})$ be a geometrically irreducible smooth projective
curve over a finite field $\mathbb{F}$ and let $\infty$ be a closed point on
it. Let $A=\mathcal{O}_{C}(C\setminus\\{\infty\\})$ be the
$\mathbb{F}$-algebra of rational functions on $C$ that are regular away from
$\infty$. Let $K$ be the function field of $C$.

In [GM21], the authors constructed an $A$-module $\mathfrak{sf}(E)$ of special
functions attached to an Anderson $A$-module $E$ of generic characteristic,
assembling prior existing construction in the literature. The purpose of
elements of $\mathfrak{sf}(E)$ is to play the same role as that of the
Anderson-Thakur function $\omega$ for the Carlitz module. Some fascinating
features of $\mathfrak{sf}(E)$ are its relation to the Betti realization of
$E$ (Thm. 3.3 in loc. cit.), to function fields Gauss sums (known as Gauss-
Thakur sums), as well as to the period lattice $\Lambda_{E}$. For the later
fact, given $u\in A$ an element such that $K/\mathbb{F}(u)$ is a finite
separable field extension (we say that $u$ is separating), the authors
exhibited an isomorphism of $A$-modules:

$\delta_{u}:\mathfrak{d}_{A/\mathbb{F}[u]}\otimes_{A}\Lambda_{E}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathfrak{sf}(E)$
(1.1)

where $\mathfrak{d}_{A/\mathbb{F}[u]}$ is the different ideal relative to the
extension $K/\mathbb{F}(u)$ (see Thm. 3.11 in loc. cit. for more details).
Although (1.1) is functorial in $E$, it is non canonical as it depends on the
choice of $u$.

Since the class of $\mathfrak{d}_{A/\mathbb{F}[u]}$ is the inverse of that of
$\Omega^{1}_{A/\mathbb{F}}$ in $\operatorname{Pic}(A)$, it is natural to ask
whether (1.1) admits a ”canonical inverse”, taking the form

$\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathfrak{sf}(E)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\Lambda_{E}.$
(1.2)

That the above does not depend on $u$ was already noticed by the authors in
the $A=\mathbb{F}[t]$-case (Prop. 3.13 in loc. cit.).

The goal of this note is to provide a positive answer to the above question,
and to interpret (1.2) as a residue map. In this respect, we extend the result
of the second author in [Mau18a, Prop. 3.8] to the general coefficient ring
$A$ setting.

Once one agrees with [GM21], the main difficulty of this note lies in the
signification of what we mean by a residue map. Let us fix some notations in
order to be more precise: let $(E,\varphi)$ be an Anderson $A$-module over
$\mathbb{C}_{\infty}$. The module of special functions of $E$ is defined as
the submodule of

$A\hat{\otimes}E(\mathbb{C}_{\infty})$ (1.3)

consisting of elements $\omega$ satisfying $(a\otimes
1)\omega=(1\otimes\varphi_{a})\omega$ for all $a\in A$. The module (1.3) is
defined as follows: once an isomorphism $\kappa:E\cong\mathbb{G}_{a}^{d}$ of
$\mathbb{F}$-vector spaces $\mathbb{C}_{\infty}$-schemes is fixed, $d$ being
$\dim E$, we obtain a non-canonical identification among $A\otimes
E(\mathbb{C}_{\infty})$ and $(A\otimes\mathbb{C}_{\infty})^{d}$. The Gauss
norm on the latter induces a metric on the former, and one shows that this
metric is independent of the choice of $\kappa$. Taking completion, we obtain
(1.3) thusly, interpreting it as a module of ”$E(\mathbb{C}_{\infty})$-valued
functions on an affinoid domain”.

If one wishes to extend this definition to more general admissible opens
$\mathfrak{U}$ of the rigid analytic variety
$\mathfrak{X}:=(\operatorname{Spec}A\otimes\mathbb{C}_{\infty})^{\text{rig}}$,
one encounters the following issue: surely, one wants
”$E(\mathbb{C}_{\infty})$-valued functions on $\mathfrak{U}$” to be isomorphic
to $\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})^{d}$ after the choice of
$\kappa$. Yet, assume $d>1$ and suppose we have chosen another isomorphism
$\kappa^{\prime}:E\cong\mathbb{G}_{a}^{d}$ which differs from $\kappa$ by

$\begin{pmatrix}1&&&\tau\\\ &1&&\\\ &&\ddots&\\\
&&&1\end{pmatrix}\in\operatorname{GL}_{d}(\mathbb{C}_{\infty}\\{\tau\\})\cong\operatorname{Aut}(\mathbb{G}_{a}^{d}).$

Then, the module of ”$E(\mathbb{C}_{\infty})$-valued functions on
$\mathfrak{U}$” will also be identified with
$\mathcal{O}_{\mathfrak{X}}(\mathfrak{U}\cap\operatorname{Frob}_{\mathbb{F}}(\mathfrak{U}))^{d}$,
a fact we surely not want to happen in general.

To make things independent on the choice of coordinates $\kappa$, we thus
require to work with admissible opens $\mathfrak{U}$ that are $\tau$-costable,
meaning that
$\mathfrak{U}\subseteq\operatorname{Frob}_{\mathbb{F}}(\mathfrak{U})$. If
$\underline{\mathfrak{X}}=(\operatorname{Cat}\mathfrak{X},\operatorname{Cov}\mathfrak{X})$
denote the site on $\mathfrak{X}$ corresponding to the strong Grothendieck
topology, we denote by
$\,{}^{\tau}\underline{\mathfrak{X}}=(\operatorname{Cat}^{\tau}\mathfrak{X},\operatorname{Cov}^{\tau}\mathfrak{X})$
the induced site compound of $\tau$-costable admissible opens and coverings of
$\underline{\mathfrak{X}}$. This happens to be enough for our purpose (see
Section 4.2):

######  Proposition 1.1.

Up to canonical isomorphism, there exists a unique pair $(\mathcal{E},\iota)$
consisting of a sheaf $\mathcal{E}(\mathcal{O}_{\mathfrak{X}})$ of $A\otimes
A$-modules on $\,{}^{\tau}\underline{\mathfrak{X}}$ and a family
$\iota=(\iota(\mathfrak{U}))_{\mathfrak{U}}$ of $A\otimes A$-linear morphisms
indexed by objects $\mathfrak{U}$ of $\operatorname{Cat}^{\tau}\mathfrak{X}$
such that, for any
$\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ and any such
$\mathfrak{U}$, there exists an $A$-linear isomorphism
$m_{\kappa}(\mathfrak{U}):\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{U})\stackrel{{\scriptstyle\sim}}{{\to}}\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})^{d}$
inserting in the pushout square of $A$-modules:

${(A\otimes\mathbb{C}_{\infty})^{d}}$${A\otimes
E(\mathbb{C}_{\infty})}$${\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})^{d}}$${\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{U})}$$\scriptstyle{\kappa^{-1}}$$\scriptstyle{\sim}$$\scriptstyle{\iota(\mathfrak{U})}$$\scriptstyle{m_{\kappa}(\mathfrak{U})^{-1}}$
(1.4)

We call $\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{U})$ the $A\otimes
A$-module of $E(\mathbb{C}_{\infty})$-valued holomorphic functions on
$\mathfrak{U}$. More generally, if $\mathcal{M}_{\mathfrak{X}}$ denotes the
sheaf of meromorphic functions on $\underline{\mathfrak{X}}$ and
$\mathcal{L}\subset\mathcal{M}_{\mathfrak{X}}$ is a $\tau$-costable
sub-$\mathcal{O}_{\mathfrak{X}}$-modules sheaf (Definition 3.10), there exists
a sheaf $\mathcal{E}(\mathcal{L})$ of $A\otimes A$-modules on
$\,{}^{\tau}\underline{\mathfrak{X}}$ verifying Proposition 1.1 with
$\mathcal{L}$ in place of $\mathcal{O}_{\mathfrak{X}}$ in diagram (1.4).

We subsequently generalize the definition of the module of special functions
(Def. 3.1 in loc. cit.) as follows (Definition 4.10).

######  Definition 1.2.

Let $\mathfrak{U}$ be an object of $\operatorname{Cat}^{\tau}\mathfrak{X}$.
Given $\mathcal{L}$ a $\tau$-costable sub-$\mathcal{O}_{\mathfrak{X}}$-module
sheaf of $\mathcal{M}_{\mathfrak{X}}$, we let
$\mathfrak{sf}_{E}(\mathcal{L})(\mathfrak{U})$ be the sub-$A$-module of
$\mathcal{E}(\mathcal{L})(\mathfrak{U})$ consisting in elements $\omega$
satisfying $(a\otimes 1)\omega=(1\otimes a)\omega$ for all $a\in A$.

One verifies that $\mathfrak{sf}_{E}(\mathcal{L})$ is a sheaf of $A$-modules
on $\,{}^{\tau}\underline{\mathfrak{X}}$ (Proposition 4.11), and that the
inclusion
$\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})\subset\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$
\- where $\mathfrak{D}:=\operatorname{Sp}(A\hat{\otimes}\mathbb{C}_{\infty})$
is the unit disc - is canonically isomorphic to $\mathfrak{sf}(E)\subset
A\hat{\otimes}E(\mathbb{C}_{\infty})$ (Lemma 4.9).

It has been observed in several situations that special functions not only
live in the Tate algebra, but can also be analytically continued to
meromorphic functions on the full $\mathfrak{X}$ with only poles at
$\mathfrak{j}$ and its iterates by $\tau$ (denoted $\tau\mathfrak{j}$,
$\tau^{2}\mathfrak{j}$, etc.) and of order at most $d=\dim E$. In our setting,
the statement takes the following form (Proposition 5.5):

######  Theorem 1.3 (Meromorphic continuation of special functions).

Let $J$ be the rigid analytic divisor
$\mathfrak{j}+\tau\mathfrak{j}+\tau^{2}\mathfrak{j}+...$ on $\mathfrak{X}$.
Then, the map

$\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(d\cdot
J))(\mathfrak{X})\to\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(d\cdot
J))(\mathfrak{D})=\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})=\mathfrak{sf}(E),$

where the first map is given by restriction, is an $A$-linear isomorphism.

To state the main theorem of this note, we require the notion of residue map
at $\mathfrak{j}$ for $E(\mathbb{C}_{\infty})$-valued meromorphic functions.
Denote by $\mathcal{J}$ the sheaf of meromorphic functions on
$\underline{\mathfrak{X}}$ that are holomorphic away from the support of $J$.
Let $\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ be a
choice of coordinates. The residue map at $\mathfrak{j}$ is defined thusly: we
let $\operatorname{res}_{\mathfrak{j}}$ be the unique map of $A$-modules
making the following diagram commute:

${\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathcal{E}(\mathcal{J})(\mathfrak{X})}$${\operatorname{Lie}_{E}\mathbb{C}_{\infty}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathcal{J}(\mathfrak{X})^{d}}$${\mathbb{C}_{\infty}^{d}}$$\scriptstyle{\kappa}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{res}_{\mathfrak{j}}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{Lie}\kappa}$$\scriptstyle{\operatorname{residue}_{\mathfrak{j}}}$
(1.5)

Our main result is as follows:

######  Theorem 1.4.

The morphism
$\operatorname{res}_{\mathfrak{j}}:\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathcal{E}(\mathcal{J})(\mathfrak{X})\to\operatorname{Lie}_{E}\mathbb{C}_{\infty}$
is independent on the choice of $\kappa$. Further, through
$\mathfrak{sf}(E)=\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(d\cdot
J))(\mathfrak{X})\subseteq\mathcal{E}(\mathcal{J})(\mathfrak{X})$, the map
$\operatorname{res}_{\mathfrak{j}}$ induces an isomorphism
$\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathfrak{sf}(E)\stackrel{{\scriptstyle\sim}}{{\to}}\Lambda_{E}$.

######  Remark 1.5.

Surprisingly, we shall never use in this note the assumption that $E$ is
uniformizable nor abelian. Theorem 1.4 hence provides a general method to
reconstruct the period lattice of $E$ without any reference to its exponential
function.

An immediate consequence of Theorem 1.4 is that there are no non-zero special
functions that are holomorphic on the whole $\mathfrak{X}$:

######  Corollary 1.6.

The module $\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{X})$ is
trivial.

Compiling the results of this note, we notice that the $A$-module
$\mathfrak{sf}(E)$ carries a natural increasing finite filtration indexed by
the order of poles along the divisor $J$:

$(0)\stackrel{{\scriptstyle\text{Cor~{}}\ref{cor:no-pole-no-
function}}}{{=}}\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{X})\hookrightarrow\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(J))(\mathfrak{X})\hookrightarrow\cdots\hookrightarrow\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(d\cdot
J))(\mathfrak{X})\stackrel{{\scriptstyle\text{Thm~{}}\ref{thm:intro-
continuation}}}{{=}}\mathfrak{sf}(E).$ (1.6)

The above has by design at most $d$ jumps. We computed some examples of this
filtration in Subsection 5.3. We leave the following question open for future
research:

######  Question.

To what filtration on $\Lambda_{E}$ does (1.6) correspond?

##### Context and acknowledgments:

The main ideas of this paper were already in the mind of the authors at the
time of [GM21]. The preprint form of this note is due to the interest of
Giacomo Hermes Ferraro and his will to read a detailed version of the main
theorem. In preparing this manuscript, the authors noticed the fundamental
need of a costable site \- an addition to the theory compared to loc. cit. \-
in order to state the results. Apart from the costable theory, the fact that
periods should appear as the residue at $t=\theta$ of special functions was
already in vogue: for instance, this was already noticed by Pellarin [Pel08,
§4.2] for Drinfeld modules in the $\mathbb{F}[t]$-case; for $A$ being the
coordinate ring of an elliptic curve over $\mathbb{F}$ and $E$ a Drinfeld-
Hayes module, an explicit formula was found by Green-Papanikolas [GP18, Thm.
4.6]. Parts of this picture were also present at various stages of [HJ20],
though in a different language assuming particular coefficient rings,
abelianess and uniformizability.

The first author is grateful to Max Planck Institute for Mathematics in Bonn
for its hospitality and financial support.

##  2 Preliminaries

###  2.1 Notations

Let $\mathbb{F}$ be a finite field of characteristic $p$. Let
$(C,\mathcal{O}_{C})$ be a geometrically irreducible smooth projective curve
over $\mathbb{F}$ and let $\infty$ be a closed point on it. Let

$A=H^{0}(C\setminus\\{\infty\\},\mathcal{O}_{C})$

be the $\mathbb{F}$-algebra of rational functions on $C$ that are regular away
from $\infty$. We let $K=\mathbb{F}(C)$ be the function field of
$(C,\mathcal{O}_{C})$ or, equivalently, the fraction field of $A$. We let
$K_{\infty}$ be the completion of $K$ with respect to the valuation given by
the order of vanishing at the closed point $\infty$, $K_{\infty}^{s}$ be a
fixed separable closure of $K_{\infty}$, and $\mathbb{C}_{\infty}$ be the
completion of $K_{\infty}^{s}$.111It equals the completion of an algebraic
closure of $K_{\infty}$. Let $\ell:A\to\mathbb{C}_{\infty}$ denote the
inclusion and call it the characteristic morphism.

Let $q=|\mathbb{F}|$. By $\mathbb{C}_{\infty}\\{\tau\\}$, we mean the non-
commutative ring of polynomials over $\mathbb{C}_{\infty}$ in the variable
$\tau$ subject to the commutation rule $\tau\cdot c=c^{q}\tau$ for all
$c\in\mathbb{C}_{\infty}$. Let $\mathbb{G}_{a}$ be the $\mathbb{F}$-vector
space scheme over $\mathbb{C}_{\infty}$ assigning to a
$\mathbb{C}_{\infty}$-algebra $R$ its underlying $\mathbb{F}$-vector space
structure (as a group scheme, it is canonically isomorphic to the additive
group scheme over $\mathbb{C}_{\infty}$). There is a canonical isomorphism of
rings:

$\mathbb{C}_{\infty}\\{\tau\\}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}),\quad\sum_{i}{c_{i}\tau^{i}}\longmapsto\sum_{i}{c_{i}\cdot\operatorname{Frob}_{\mathbb{G}_{a}}^{q^{i}}}.$

By an Anderson $A$-module of dimension $d$ over $\mathbb{C}_{\infty}$ we mean
an $A$-module scheme $E$ over $\mathbb{C}_{\infty}$ which satisfies:

  1. $(i)$

As an $\mathbb{F}$-vector space scheme over $\mathbb{C}_{\infty}$, $E$ is
isomorphic to $\mathbb{G}_{a}^{d}$;

  2. $(ii)$

Let
$\varphi:A\to\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(E)$
be the ring homomorphism induced by the $A$-module structure on $E$ seen as an
$\mathbb{F}$-vector space scheme, and let
$\partial\varphi:A\to\operatorname{End}_{\mathbb{C}_{\infty}}(\operatorname{Lie}_{E}(\mathbb{C}_{\infty}))$
be the morphism obtained by functoriality on the Lie algebra. Then, for all
$a\in A$, $\partial\varphi(a)-\ell(a)$ is nilpotent on
$\operatorname{Lie}_{E}(\mathbb{C}_{\infty})$.

We will refer to an isomorphism
$\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ of
$\mathbb{F}$-vector spaces schemes over $\mathbb{C}_{\infty}$ as a choice of
coordinates for $E$.

###  2.2 The characteristic graph ideal $\mathfrak{j}$

We again denote by $\tau:A\otimes\mathbb{C}_{\infty}\to
A\otimes\mathbb{C}_{\infty}$ the ring endomorphism which acts $A$-linearly but
as the $q$-Frobenius on $\mathbb{C}_{\infty}$. We let $\mathfrak{j}$ be the
characteristic graph ideal, i.e. the kernel of the multiplication map:

$\mathfrak{j}:=\ker(A\otimes\mathbb{C}_{\infty}\stackrel{{\scriptstyle\ell\otimes\operatorname{id}}}{{\longrightarrow}}\mathbb{C}_{\infty}).$

It is a maximal ideal of $A\otimes\mathbb{C}_{\infty}$ whose support
$V(\mathfrak{j})$ is the graph of $\ell$. We denote by $\tau^{i}\mathfrak{j}$
the $i$th iterate of $\mathfrak{j}$ by $\tau$; namely $\tau^{i}\mathfrak{j}$
is the kernel of the map $A\otimes\mathbb{C}_{\infty}\to\mathbb{C}_{\infty}$,
$a\otimes b\mapsto\ell(a)b^{q^{i}}$. In this section we recall some classical
facts about $\mathfrak{j}$.

We call separating any element $u$ in $A$ such that the extension
$K/\mathbb{F}(u)$ is finite and separable.

######  Lemma 2.1.

Any non separating element in $A$ is the $p$-th power of an element in $A$. In
particular, for any $a\in A\setminus\mathbb{F}$, there exists a separating
element $u\in A$ and a non-negative integer $k$ such that $a=u^{p^{k}}$.

###### Proof.

Let $a\in A$ be non separating. If $a$ is constant, it clearly is a $p$-th
power, hence we assume that the extension $K/\mathbb{F}(a)$ is finite. There
is an intermediate extension $L$ such that $K/L$ is purely inseparable of
degree $p^{k}$ for some $k\geq 1$, and $L/\mathbb{F}(a)$ is separable. As
$K^{p^{k}}\subseteq L$, it is in fact an equality by comparing degrees. Hence,
$\mathbb{F}(a)\subset K^{p}$ and there exists $b\in K$ such that $b^{p}=a$. As
the divisor of $a$ is the the $p$-th power of the divisor of $b$, we obtain
that $b\in A$. ∎

######  Lemma 2.2.

The ideal $\mathfrak{j}$ is generated by the differences $u\otimes
1-1\otimes\ell(u)$ where $u$ runs over separating elements in $A$.

###### Proof.

We first claim that $\mathfrak{j}$ is generated by the set $\\{a\otimes
1-1\otimes\ell(a)|a\in A\\}$. It is clear that $a\otimes
1-1\otimes\ell(a)\in\mathfrak{j}$ for all $a\in A$. Conversely, let
$\mathfrak{g}$ be the ideal of $A\otimes\mathbb{C}_{\infty}$ generated by
$\\{a\otimes 1-1\otimes\ell(a)|a\in A\\}$. Every element $x\in
A\otimes\mathbb{C}_{\infty}$ can be written as a sum of elementary tensors
$a_{i}\otimes c_{i}$ for some $a_{i}\in A$, $c_{i}\in\mathbb{C}_{\infty}$. We
have

$x=\sum_{i}{a_{i}\otimes
c_{i}}\equiv\sum_{i}{1\otimes\ell(a_{i})c_{i}}\pmod{\mathfrak{g}}.$

If $x\in\mathfrak{j}$, the right-hand sum is zero, hence $x\in\mathfrak{g}$.

Now, by Lemma 2.1, for any $a\in A\setminus\mathbb{F}$, there exists a
separating element $u$ and a positive integer $k$ such that $a\otimes
1-1\otimes\ell(a)=(u\otimes 1-1\otimes\ell(u))^{p^{k}}$. Thus
$\mathfrak{j}\subset\langle\\{u\otimes
1-1\otimes\ell(u)~{}|~{}u~{}\text{separating}\\}\rangle$. ∎

######  Lemma 2.3.

If $u$ is a separating element, then $u\otimes 1-1\otimes\ell(u)$ is a
uniformizing parameter for $\mathfrak{j}$.

###### Proof.

Recall by definition of the module of Kähler differentials that we have an
isomorphism
$\Omega^{1}_{A/\mathbb{F}}\otimes\mathbb{C}_{\infty}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathfrak{j}/\mathfrak{j}^{2}$,
$da\mapsto a\otimes 1-1\otimes\ell(a)$. Thus, to prove the lemma, it suffices
to show that $du\neq 0$ for $u$ separating. Yet, applying the relative
cotangent sequence to $\mathbb{F}\to\mathbb{F}(u)\to K$, we obtain an exact
sequence

$Kdu\longrightarrow\Omega^{1}_{K/\mathbb{F}}\longrightarrow\Omega^{1}_{K/\mathbb{F}(u)}$

whose right-hand side term is zero as $K/\mathbb{F}(u)$ is finite separable.
Hence, the first map is surjective and the image of $du$ is nonzero. ∎

######  Lemma 2.4.

Let $\mathfrak{m}$ be a maximal ideal of $A\otimes\mathbb{C}_{\infty}$
different from $\mathfrak{j}$, $\tau\mathfrak{j}$, $\tau^{2}\mathfrak{j}$, …
There exists a separating element $u$ such that, for all non-negative integer
$i$, $u\otimes 1-1\otimes\ell(u)^{q^{i}}$ does not belong to $\mathfrak{m}$.

###### Proof.

Let $t$ be a separating element. We first compute the prime ideal
decomposition of $(t\otimes 1-1\otimes\ell(t))$ in the Dedekind domain
$A\otimes\mathbb{C}_{\infty}$. The inclusion of Dedekind rings
$\mathbb{F}[t]\otimes\mathbb{C}_{\infty}\subset A\otimes\mathbb{C}_{\infty}$
makes $A\otimes\mathbb{C}_{\infty}$ a free
$\mathbb{F}[t]\otimes\mathbb{C}_{\infty}$-module of rank $[K:\mathbb{F}(t)]$.
In particular, there are at most $[K:\mathbb{F}(t)]$ prime divisors of
$(t\otimes 1-1\otimes\ell(t))$. For $\sigma:\ell(K)\to\mathbb{C}_{\infty}$ an
$\mathbb{F}(t)$-algebra morphism, the ideal $\mathfrak{j}^{\sigma}$ of
$A\otimes\mathbb{C}_{\infty}$ generated by the set $\\{a\otimes
1-1\otimes\sigma(\ell(a))|a\in A\\}$ is maximal and divides the principal
ideal $(t\otimes 1-1\otimes\ell(t))$. There are
$\\#\operatorname{Hom}_{\mathbb{F}(t)}(\ell(K),\mathbb{C}_{\infty})=[K:\mathbb{F}(t)]$
such ideals, hence

$(t\otimes 1-1\otimes\ell(t))=\prod_{\sigma}\mathfrak{j}^{\sigma}$

where the product runs over
$\sigma\in\operatorname{Hom}_{\mathbb{F}(t)}(\ell(K),\mathbb{C}_{\infty})$.

We turn to the proof of the lemma. Assume the converse, that is, for all
separating elements $v$ there exists $j\geq 0$ such that $v\otimes
1-1\otimes\ell(v)^{q^{j}}\in\mathfrak{m}$. This means that there exists a non-
negative integer $i$ for which $\mathfrak{m}\supset(t\otimes
1-1\otimes\ell(t)^{q^{i}})=\tau^{i}\left(\prod_{\sigma}{\mathfrak{j}^{\sigma}}\right)$.
By uniqueness of the prime ideal decomposition, there exists
$\sigma\in\operatorname{Hom}_{\ell(\mathbb{F}(t))}(\ell(K),\mathbb{C}_{\infty})$
such that $\mathfrak{m}=\tau^{i}(\mathfrak{j}^{\sigma})$. Because
$\mathfrak{m}$ is distinct from $\mathfrak{j}$, $\tau\mathfrak{j}$,
$\tau^{2}\mathfrak{j}$, …, the morphism $\sigma$ is not the inclusion
$\ell(K)\subset\mathbb{C}_{\infty}$. Because $K$ is generated by separating
elements over $\mathbb{F}$, there exists a separating element $u$ such that
$\sigma(\ell(u))\neq\ell(u)$. From our converse assumption, there exists a
non-negative integer $j$ such that $u\otimes
1-1\otimes\ell(u)^{q^{j}}\in\mathfrak{m}=\tau^{i}(\mathfrak{j}^{\sigma})$.
Hence, both $u\otimes 1-1\otimes\ell(u)^{q^{j}}$ and $u\otimes
1-1\otimes\sigma(\ell(u))^{q^{i}}$ are in $\mathfrak{m}$. Since
$\mathfrak{m}\neq A\otimes\mathbb{C}_{\infty}$, this implies
$\sigma(\ell(u))^{q^{i}}=\ell(u)^{q^{j}}$.

This is a contradiction. Indeed, $\ell(u)^{q^{i}}$ and
$\sigma(\ell(u))^{q^{i}}=\ell(u)^{q^{j}}$ have the same minimal polynomial
over $\ell(\mathbb{F}(t))$ so that either the latter polynomial has
coefficients in $\mathbb{F}$ or $i=j$. The first option is impossible as it
would imply $\ell(u)\in\bar{\mathbb{F}}\cap\ell(A)=\mathbb{F}$. The second
option is also impossible as we chose $u$ such that
$\sigma(\ell(u))\neq\ell(u)$. ∎

##  3 Objects from rigid analytic geometry

Let $X:=\operatorname{Spec}A\otimes\mathbb{C}_{\infty}$. In this section, we
develop in more details the definition of the rigid analytic variety
$\mathfrak{X}:=X^{\text{rig}}$, obtained by applying the rigid analytic GAGA
on $X$ (see [Bos14, §5.4]).

###  3.1 The rigid analytic space $\mathfrak{X}$

To begin with, we define Gauss norms on $A\otimes\mathbb{C}_{\infty}$. Let
$c\in\mathbb{C}_{\infty}^{\times}$ and let $\rho:=|c|>0$. For $f\in
A\otimes\mathbb{C}_{\infty}$, we set:

$\|f\|_{\rho}:=\inf\left(\max_{i}\\{|c_{i}|\rho^{\deg a_{i}}\\}\right)$

where the infinimum is taken over all representations of $f$ as finite sums
$\sum_{i}{a_{i}\otimes c_{i}}$ in $A\otimes\mathbb{C}_{\infty}$.

######  Remark 3.1.

For $\rho=1$, $\|\cdot\|_{1}$ was denoted by $\|\cdot\|$ in [GM21] and the
completed ring denoted by $\mathbb{T}$. We will rather use the notation
$\mathbb{C}_{\infty}\langle A\rangle$ instead of $\mathbb{T}$ below.

We prove next that $\|\cdot\|_{\rho}$ is indeed a Gauss norm. For this, we
construct a presentation of $A$ as follows. First observe that $A$ has a
canonical filtration by finite dimensional sub-$\mathbb{F}$-vectors spaces
$D_{n}$ of elements of degree $\leq n$. Clearly $D_{0}=\mathbb{F}$ and the
union of all $D_{n}$ is $A$. Let $n_{1}$ be the least positive integer such
that the inclusion $D_{0}\subset D_{n_{1}}$ is strict, and let
$(a_{1},...,a_{s_{1}})$ be a basis of $D_{n_{1}}$. We let $A_{n_{1}}$ be the
sub-$\mathbb{F}$-algebra of $A$ generated by $(a_{1},...,a_{s_{1}})$. Proceed
by induction to define $n_{i}$ as the least integer $>n_{i-1}$ such that
$D_{n_{i}}$ is not contained in $A_{n_{i-1}}$, let
$(a_{s_{1}+1},...,a_{s_{i}})$ be a basis of a supplementary subspace of
$(A_{n_{i-1}}\cap D_{n_{i}})$ in $D_{n_{i}}$, and defined $A_{n_{i}}$ as the
sub-$\mathbb{F}$-algebra of $A$ generated by $(a_{1},...,a_{s_{i}})$.

This procedure eventually terminates by the Riemann-Roch Theorem, and we
obtain a finite generating family $(a_{1},...,a_{s})$ of $A$. We obtain a
morphism

$\mathbb{F}[t_{1},...,t_{s}]\longrightarrow A,\quad t_{i}\longmapsto a_{i}$
(3.1)

and a new degree $d$ given as $d(t_{i}):=\deg(a_{i})$ on monomials of
$\mathbb{F}[t_{1},...,t_{s}]$. By design, (3.1) bears the property that $\deg
P(a_{1},...,a_{s})=d(P(t_{1},...,t_{s}))$ for any
$P(t_{1},...,t_{s})\in\mathbb{F}[t_{1},...,t_{s}]$. We denote by
$\mathfrak{a}$ the kernel of (3.1), and we let
$(f_{1}(t_{1},...,t_{s}),...,f_{t}(t_{1},...,t_{s}))$ be a family of
generators of $\mathfrak{a}$. We denote by
$\mathfrak{a}_{c}\subset\mathbb{C}_{\infty}[t_{1},...,t_{s}]$ the rescaled
ideal

$\left(f_{1}\left(c^{\deg t_{1}}t_{1},\cdots,c^{\deg
t_{s}}t_{s}\right),...,f_{g}\left(c^{\deg t_{1}}t_{1},\cdots,c^{\deg
t_{1}}t_{s}\right)\right)\subset\mathbb{C}_{\infty}[t_{1},...,t_{s}].$

######  Proposition 3.2.

Through the above presentation of $A$, the ring homomorphism

$\iota:\mathbb{C}_{\infty}[t_{1},...,t_{s}]/\mathfrak{a}_{c}\to
A\otimes\mathbb{C}_{\infty},\quad t_{i}\mapsto a_{i}\otimes c^{-\deg a_{i}}$

is an isomorphism which makes the Gauss norm on the left-hand side coincide
with $\|\cdot\|_{\rho}$ on the right.

######  Remark 3.3.

Recall that the Gauss norm on the quotient
$\mathbb{C}_{\infty}[t_{1},...,t_{s}]/\mathfrak{a}_{c}$ is defined as

$f\in\mathbb{C}_{\infty}[t_{1},...,t_{s}],\quad\|f\\!\pmod{\mathfrak{a}_{c}}\|:=\inf_{a\in\mathfrak{a}_{c}}\|f-a\|$

where $\|\cdot\|$ is the classical Gauss norm on
$\mathbb{C}_{\infty}[t_{1},...,t_{s}]$; i.e. given by the maximum norm of
coefficients.

###### Proof of Proposition 3.2.

That $\iota$ is bijective is clear, so we prove the statement on the norms.
Let $x\in\mathbb{C}_{\infty}[t_{1},...,t_{s}]/\mathfrak{a}_{c}$. For every
$\varepsilon>0$, there exists
$\tilde{x}=\sum{\alpha_{i_{1},...,i_{s}}t_{1}^{i_{1}}\cdots
t^{i_{s}}_{s}}\in\mathbb{C}_{\infty}[t_{1},...,t_{s}]$ lifting $x$ such that
$\|x\|_{c}\geq\|\tilde{x}\|-\varepsilon$. Hence,

$\displaystyle\|\iota(x)\|_{\rho}$
$\displaystyle=\left\|\sum_{i_{1},...,i_{s}}{a_{1}^{i_{1}}\cdots
a_{s}^{i_{s}}\otimes\alpha_{i_{1},...,i_{s}}c^{-\deg(a_{1}^{i_{1}}\cdots
a_{s}^{i_{s}})}}\right\|_{\rho}\leq\max_{i_{1},...,i_{s}}|\alpha_{i_{1},...,i_{s}}|=\|\tilde{x}\|$
$\displaystyle\leq\|x\|_{c}+\varepsilon.$

Being true for all $\varepsilon>0$, we obtain
$\|\iota(x)\|_{\rho}\leq\|x\|_{c}$. Conversely, let $\iota(x)\in
A\otimes\mathbb{C}_{\infty}$. Similarly, for all $\varepsilon>0$, there exists
a representation of $\iota(x)$ as $\sum_{i}{b_{i}\otimes c_{i}}\in
A\otimes\mathbb{C}_{\infty}$ for which

$\|\iota(x)\|_{\rho}\geq\max_{i}\\{|c_{i}|c^{\deg b_{i}}\\}-\varepsilon.$

Let us write $b_{i}$ as $P_{i}(a_{1},...,a_{s})$ for a polynomial $P_{i}$ with
coefficients in $\mathbb{F}$. Then,

$x=\sum_{i}{c_{i}P(c^{\deg a_{1}}t_{1},...,c^{\deg
a_{s}}t_{s})}\in\mathbb{C}_{\infty}[t_{1},...,t_{s}]/\mathfrak{a}_{c}$

and $\|x\|_{c}$ is therefore bounded by $\max_{i}\\{|c_{i}|\|P_{i}(c^{\deg
a_{1}}t_{1},...,c^{\deg a_{s}}t_{s})\|\\}$. Now, by design of the chosen
presentation, one has

$\|P_{i}(c^{\deg a_{1}}t_{1},...,c^{\deg a_{s}}t_{s})\|=|c|^{\deg
P_{i}(a_{1},...,a_{s})},$

and we conclude that $\|x\|_{c}\leq\|\iota(x)\|_{\rho}+\varepsilon$. ∎

In virtue of Proposition 3.2, we deduce that $\|\cdot\|_{\rho}$ is a norm, and
we let $\mathbb{C}_{\infty}\langle A\rangle_{\rho}$ denote the completion of
$A\otimes\mathbb{C}_{\infty}$ with respect to it. Also from Proposition 3.2,
$\mathbb{C}_{\infty}\langle A\rangle_{\rho}$ is an affinoid
$\mathbb{C}_{\infty}$-algebra, hence
$\mathfrak{X}_{\rho}:=\operatorname{Sp}\mathbb{C}_{\infty}\langle
A\rangle_{\rho}$ is a $\mathbb{C}_{\infty}$-affinoid variety. If
$\rho<\rho^{\prime}$, there is a canonical ring homomorphism
$\mathbb{C}_{\infty}\langle
A\rangle_{\rho^{\prime}}\to\mathbb{C}_{\infty}\langle A\rangle_{\rho}$, and it
induces a morphism of affinoid varieties denoted
$t_{\rho,\rho^{\prime}}:\mathfrak{X}_{\rho}\to\mathfrak{X}_{\rho^{\prime}}$.

######  Definition 3.4.

Let $\mathfrak{X}$ be the rigid analytic variety over $\mathbb{C}_{\infty}$
resulting as the glueing of $(\mathfrak{X}_{\rho})_{\rho}$ along the morphisms
$(t_{\rho,\rho^{\prime}})_{\rho,\rho^{\prime}}$ (for $\rho$, $\rho^{\prime}$
running over $|\mathbb{C}_{\infty}^{\times}|$).

The next lemma is immediate (e.g. [Bos14, §5.3 Prop. 5]).

######  Lemma 3.5.

The family $(\mathfrak{X}_{\rho}\to\mathfrak{X})_{\rho}$ is an admissible
covering of $\mathfrak{X}$.

By definition, we have $\mathfrak{X}:=\varinjlim_{\rho}\mathfrak{X}_{\rho}$ as
ringed spaces. In particular, the ring of global sections of $\mathfrak{X}$ is

$\mathbb{C}_{\infty}\langle\\!\langle
A\rangle\\!\rangle:=\varprojlim_{\rho}\mathbb{C}_{\infty}\langle
A\rangle_{\rho}.$

For $f\in A\otimes\mathbb{C}_{\infty}$, observe that we have
$\|f^{(1)}\|_{\rho}=\|f\|_{\rho^{1/q}}^{q}$, so that twisting induces an
isomorphism of rings and spaces:

$\tau:\mathbb{C}_{\infty}\langle
A\rangle_{\rho}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathbb{C}_{\infty}\langle
A\rangle_{\rho^{1/q}},\quad\tau:\mathfrak{X}_{\rho^{1/q}}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathfrak{X}_{\rho}.$
(3.2)

Compiling these for all $\rho>0$, we obtain isomorphisms

$\tau:\mathbb{C}_{\infty}\langle\\!\langle
A\rangle\\!\rangle\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathbb{C}_{\infty}\langle\\!\langle
A\rangle\\!\rangle,\quad\tau:\mathfrak{X}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathfrak{X}.$
(3.3)

######  Remark 3.6.

We refer to [Bos14, §5.4] for the fact that $\mathfrak{X}$ is isomorphic to
$(\operatorname{Spec}A\otimes\mathbb{C}_{\infty})^{\operatorname{rig}}$ as
rigid analytic spaces.

######  Example 3.7.

Let $C=\mathbb{P}^{1}_{\mathbb{F}}$ and let $\infty$ be the point $[0:1]$. We
identify $A$ with $\mathbb{F}[t]$ and $A\otimes\mathbb{C}_{\infty}$ with the
polynomial ring $\mathbb{C}_{\infty}[t]$. Denoting $\ell(t)$ by $\theta$, we
have $\mathfrak{j}=(t-\theta)$ and, more generally,
$\tau^{i}\mathfrak{j}=(t-\theta^{q^{i}})$. We have

$\displaystyle\mathbb{C}_{\infty}\langle A\rangle_{\rho}$
$\displaystyle\cong\left\\{\sum_{n=0}^{\infty}{a_{n}t^{n}}\in\mathbb{C}_{\infty}[\\![t]\\!]~{}\bigg{|}~{}\lim_{n\to\infty}|a_{n}|\rho^{n}\to
0\right\\},$ $\displaystyle\mathbb{C}_{\infty}\langle\\!\langle
A\rangle\\!\rangle$
$\displaystyle\cong\left\\{\sum_{n=0}^{\infty}{a_{n}t^{n}}\in\mathbb{C}_{\infty}[\\![t]\\!]~{}\bigg{|}~{}\forall\rho>0:~{}\lim_{n\to\infty}|a_{n}|\rho^{n}\to
0\right\\},$

where $\tau$ acts as
$\sum_{n=0}^{\infty}{a_{n}t^{n}}\mapsto\sum_{n=0}^{\infty}{a_{n}^{q}t^{n}}$.
The one-dimensional Tate algebra over $\mathbb{C}_{\infty}$ is recovered by
$\mathbb{C}_{\infty}\langle A\rangle_{1}$.

###  3.2 Sheaves and divisors on $\mathfrak{X}$

Given a commutative ring $R$ with unit, we denote by $\operatorname{Q}(R)$ its
total ring of fractions; i.e. the localization of $R$ by the set of non-zero
divisors. Given an admissible open $\mathfrak{U}$ in $\mathfrak{X}$, we denote
by $\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})$ the total ring of fractions
$\operatorname{Q}(\mathcal{O}_{\mathfrak{X}}(\mathfrak{U}))$. According to
Bosch [Bos82, §2], the assignment

$\mathfrak{U}\longmapsto\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})$

is a sheaf on $\underline{\mathfrak{X}}$; i.e with respect to the strong
Grothendieck topology. We call $\mathcal{M}_{\mathfrak{X}}$ the sheaf of
meromorphic functions. A fractional ideal sheaf on $\underline{\mathfrak{X}}$
is an invertible sub-$\mathcal{O}_{\mathfrak{X}}$-module sheaf of
$\mathcal{M}_{\mathfrak{X}}$. Lacking references, we recall briefly the
dictionary among fractional ideal sheaves and rigid analytic divisors.

For any $x\in|\mathfrak{X}|=|X|$, we have
$\mathcal{M}_{\mathfrak{X},x}=\operatorname{Q}(\mathcal{O}_{\mathfrak{X},x})$
where $\mathcal{O}_{\mathfrak{X},x}$ is a discrete valuation ring with residue
field $\mathbb{C}_{\infty}$ which coincides with $\mathcal{O}_{X,x}$. Given
$f\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})$, a non zero meromorphic
function on an admissible open $\mathfrak{U}$, we define its divisor as
follows. For $x\in\mathfrak{U}$, we let $\operatorname{ord}_{x}f$ be the order
of $f$ in the discrete valued field $\mathcal{M}_{\mathfrak{X},x}$. The
divisor of $f$ on $\mathfrak{U}$ is the formal (possibly infinite) sum:

$\operatorname{div}f:=\sum_{x\in\mathfrak{U}}{(\operatorname{ord}_{x}f)\cdot
x}.$

Let $D=\sum_{x\in\mathfrak{U}}{n_{x}\cdot x}$ be a formal $\mathbb{Z}$-sum of
points of $\mathfrak{X}$. We say that $D$ is a rigid analytic divisor if there
exists an admissible open cover
$(\mathfrak{U}_{i}\stackrel{{\scriptstyle\iota_{i}}}{{\to}}\mathfrak{U})_{i\in
I}$ such that, for each $i\in I$, there exists
$f_{i}\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{U}_{i})$ for which

$D_{|\mathfrak{U}_{i}}:=\sum_{x\in\mathfrak{U}_{i}}{n_{x}\cdot
x}=\operatorname{div}f_{i}.$

More generally, we call divisor on $\mathfrak{X}$ any formal $\mathbb{Z}$-sum
of points of $\mathfrak{X}$. To any divisor $D$ on $\mathfrak{X}$, one
associates a presheaf $\mathcal{O}_{\mathfrak{X}}(D)$ given on an admissible
open $\mathfrak{U}$ as

$\mathcal{O}_{\mathfrak{X}}(D)(\mathfrak{U}):=\\{g\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})~{}|~{}\operatorname{div}g+D\geq
0\\}.$

Whenever $D$ is rigid analytic, $\mathcal{O}_{\mathfrak{X}}(D)$ is further a
fractional ideal sheaf on $\underline{\mathfrak{X}}$. Conversely, to any
fractional ideal sheaf $\mathcal{I}$, there is a rigid analytic divisor $D$
such that $\mathcal{I}=\mathcal{O}_{\mathfrak{X}}(D)$. Observe in addition
that given $f\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{X})$ non zero, we have

$\mathcal{O}_{\mathfrak{X}}(\operatorname{div}f)(\mathfrak{U})=\\{g\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})~{}|~{}g\cdot
f_{|\mathfrak{U}}\in\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})\\}.$

Lemma 2.4 has the following counterpart:

######  Proposition 3.8.

For $u\in A$ a non constant element, let $\mathfrak{j}_{u}$ denote the
principal ideal $(u\otimes 1-1\otimes\ell(u))$ of
$A\otimes\mathbb{C}_{\infty}$. Also, consider the divisor
$J:=\mathfrak{j}+\tau\mathfrak{j}+\tau^{2}\mathfrak{j}+...$ and
$J_{u}:=\mathfrak{j}_{u}+\tau\mathfrak{j}_{u}+\tau^{2}\mathfrak{j}_{u}+...$ on
$\mathfrak{X}$. Then, both $J$ and $J_{u}$ are rigid analytic and, for any
$n\in\mathbb{Z}$ and $\mathfrak{U}$ an affinoid subdomain of $\mathfrak{X}$,
we have the equality:

$\mathcal{O}_{\mathfrak{X}}(n\cdot
J)(\mathfrak{U})=\bigcap_{\begin{subarray}{c}u\in A\\\
\text{separating}\end{subarray}}\mathcal{O}_{\mathfrak{X}}(n\cdot
J_{u})(\mathfrak{U})$

where the intersection is taken over separating elements $u\in A$. In
particular, $\bigcap_{u}\mathcal{O}_{\mathfrak{X}}(n\cdot J_{u})$ is a sheaf.

######  Lemma 3.9.

Let $\rho\in|\mathbb{C}_{\infty}^{\times}|$ and let
$t=\lfloor\log_{q}\rho\rfloor$. Let $a\in A$ be non constant. Then $a\otimes
1-1\otimes\ell(a)^{q^{j}}$ is invertible in $\mathbb{C}_{\infty}\langle
A\rangle_{\rho}$ for all $j>t$. In particular, $\tau^{j}\mathfrak{j}$ is
invertible in $\mathbb{C}_{\infty}\langle A\rangle_{\rho}$ for all $j>t$.

###### Proof.

It suffices to note that, for $j>t$, the series

$\frac{-1}{1\otimes\ell(a)^{q^{j}}}\sum_{n=0}^{\infty}{\left(\frac{a\otimes
1}{1\otimes\ell(a)^{q^{j}}}\right)^{n}}$

converges in $\mathbb{C}_{\infty}\langle A\rangle_{\rho}$ to an inverse of
$a\otimes 1-1\otimes\ell(a)^{q^{j}}$. ∎

###### Proof of Proposition 3.8.

Let $\mathfrak{U}$ be an affinoid subdomain of $\mathfrak{X}$. By Lemma 3.5,
there exists $\rho\in|\mathbb{C}_{\infty}^{\times}|$ such that
$\mathfrak{U}\to\mathfrak{X}$ factors through
$\mathfrak{X}_{\rho}\to\mathfrak{X}$. Let $t=\lfloor\log_{q}\rho\rfloor$.

On one hand, we have by Lemma 3.9,

$\mathcal{O}_{\mathfrak{X}}(n\cdot
J)(\mathfrak{U})=\\{f\in\operatorname{Q}(\mathcal{O}_{\mathfrak{X}}(\mathfrak{U}))~{}|~{}\forall
h\in(\mathfrak{j}\cdot\tau\mathfrak{j}\cdots\tau^{t}\mathfrak{j})^{n}:~{}hf\in\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})\\}.$
(3.4)

To prove that $J$ is rigid analytic, observe that the ideal
$\mathfrak{j}\cdot\tau\mathfrak{j}\cdots\tau^{t}\mathfrak{j}$ of
$A\otimes\mathbb{C}_{\infty}$ is locally principal (as
$A\otimes\mathbb{C}_{\infty}$ is a Dedekind domain). There thus exists a
Zariski cover $(U_{i}\to X)_{i\in I}$ such that
$\mathfrak{j}\cdot\tau\mathfrak{j}\cdots\tau^{t}\mathfrak{j}$ is principal on
each $U_{i}$. Rigidifying, we obtain a covering
$(U_{i}^{\operatorname{rig}}\times_{\mathfrak{X}}\mathfrak{U}\to\mathfrak{U})_{i\in
I}$ such that $\mathfrak{j}\cdot\tau\mathfrak{j}\cdots\tau^{t}\mathfrak{j}$ is
principal on each admissible open. Hence, $J$ is rigid analytic.

On the other hand,

$\bigcap_{u}\mathcal{O}_{\mathfrak{X}}(n\cdot
J_{u})(\mathfrak{U})=\left\\{f\in\operatorname{Q}(\mathcal{O}_{\mathfrak{X}}(\mathfrak{U}))~{}\bigg{|}~{}\forall
u:\left(\prod_{j=0}^{t}{(u\otimes
1-1\otimes\ell(u)^{q^{j}})}\right)^{\\!n}\\!f\in\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})\right\\}.$
(3.5)

Hence $J_{u}$ is also a rigid analytic divisor.

We now prove equality among (3.4) and (3.5). Clearly, the former is included
in the latter. Conversely, let $f=ab^{-1}$ be an element of (3.5), where
$a,b\in\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})$ with $b$ a non-zero divisor.
Let $\mathfrak{m}$ be a maximal ideal in
$\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})$ for which $b\in\mathfrak{m}$ but
$a\notin\mathfrak{m}$. As the rigid analytic GAGA induces a bijection on
closed points, $\mathfrak{m}$ arises from a maximal ideal of
$A\otimes\mathbb{C}_{\infty}$ (which we still denote by $\mathfrak{m}$). By
assumption, for all $u\in A$ separating,

$\left(\prod_{j=0}^{t}{(u\otimes
1-1\otimes\ell(u)^{q^{j}})}\right)^{\\!n}a\subset\mathfrak{m}.$

Since $\mathfrak{m}$ is prime and $a\notin\mathfrak{m}$, there exists
$j\in\\{0,...,t\\}$ for which $u\otimes
1-1\otimes\ell(u)^{q^{j}}\in\mathfrak{m}$. By Lemma 2.4, this implies
$\mathfrak{m}\in\\{\mathfrak{j},\tau\mathfrak{j},\tau^{2}\mathfrak{j},...\\}$.
Hence $f$ belongs to (3.4). ∎

###  3.3 Costable subsets and costable sheaves

We will be faced with maps on $(A\otimes\mathbb{C}_{\infty})^{d}$ and its
extensions $\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})^{d}$ which come from
elements in
$\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}^{d})\cong\mathsf{Mat}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$.
However, if $F\in\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})^{d}$ and
$\varphi_{\kappa}(a)=A_{0}+A_{1}\tau+...+A_{s}\tau^{s}$ for some
$A_{i}\in\mathsf{Mat}_{d}(\mathbb{C}_{\infty})$, then $\varphi_{\kappa}(a)(F)$
only belongs to
$\mathcal{M}_{\mathfrak{X}}(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})^{d}$.

The notion of costability defined below gives us the correct setting where
this issue disappears. A similar issue occurs when we consider more general
fractional ideal sheaves $\mathcal{L}$ instead of
$\mathcal{M}_{\mathfrak{X}}$, leading us to the notion of costability for such
sheaves.

Denote by $\tau:\underline{\mathfrak{X}}\to\underline{\mathfrak{X}}$ the
morphism of sites given by the $A$-linear Frobenius, and denote by
$\tau^{-1}\mathcal{M}_{\mathfrak{X}}$ the inverse image sheaf.

######  Definition 3.10.

Let $\mathfrak{U}\subset\mathfrak{X}$ be admissible open,
$(\mathfrak{U}_{i}\to\mathfrak{U})_{i\in I}$ an admissible covering.

  1. 1.

We say that $\mathfrak{U}$ is costable if
$\mathfrak{U}\subseteq\tau\mathfrak{U}$.

  2. 2.

We say that $(\mathfrak{U}_{i}\to\mathfrak{U})_{i\in I}$ is costable if
$\mathfrak{U}$ and $\mathfrak{U}_{i}$ are costable for all $i\in I$.

  3. 3.

Given any costable admissible open $\mathfrak{U}$, we denote by $f\mapsto
f^{(1)}$ the composition

${\bullet^{(1)}:\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})}$${\tau^{-1}\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})=\mathcal{M}_{\mathfrak{X}}(\tau\mathfrak{U})}$${\mathcal{M}_{\mathfrak{X}}(\mathfrak{U}).}$$\scriptstyle{f\mapsto
f^{(1)}}$$\scriptstyle{\text{restr~{}}\tau\mathfrak{U}\to\mathfrak{U}}$ (3.6)

We say that $\mathcal{L}$ is costable if it verifies
$\mathcal{L}(\mathfrak{U})^{(1)}\subset\mathcal{L}(\mathfrak{U})$ for any
costable admissible open $\mathfrak{U}$.

######  Example 3.11.

  1. $(a)$

The admissible open $\mathfrak{D}:=\operatorname{Sp}\mathbb{C}_{\infty}\langle
A\rangle=\operatorname{Sp}\mathbb{C}_{\infty}\langle A\rangle_{1}$ is
costable. More generally,
$\mathfrak{X}_{\rho}=\operatorname{Sp}\mathbb{C}_{\infty}\langle
A\rangle_{\rho}$ is costable if and only if $\rho\geq 1$.

  2. $(b)$

Both $\mathcal{O}_{\mathfrak{X}}$ and $\mathcal{M}_{\mathfrak{X}}$ are
costable. The sheaf $\mathcal{O}_{\mathfrak{X}}(D)$ is costable whenever $D$
is an effective rigid analytic divisor that verifies $D\geq\tau D$.

Let $\operatorname{Cat}^{\tau}\mathfrak{X}$ be the full subcategory of
$\operatorname{Cat}\mathfrak{X}$ consisting of costable objects, and let
$\operatorname{Cov}^{\tau}\mathfrak{X}$ be the subset of
$\operatorname{Cov}\mathfrak{X}$ consisting of costable coverings. It is
formal to verify that the pair
$(\operatorname{Cat}^{\tau}\mathfrak{X},\operatorname{Cov}^{\tau}\mathfrak{X})$
forms a site, which we denote by $\,{}^{\tau}\underline{\mathfrak{X}}$ and
refer to as the costable site of $\mathfrak{X}$.

##  4 Anderson modules over rigid analytic spaces

###  4.1 $E(\mathbb{C}_{\infty})$-valued meromorphic functions

Let $E$ be an Anderson module of dimension $d$ over $\mathbb{C}_{\infty}$.
This whole section is devoted to the construction of the sheaf of
$E(\mathbb{C}_{\infty})$-valued functions: it will consist in a sheaf of
$A\otimes A$-modules on a certain subsite of $\underline{\mathfrak{X}}$ which
would produce $A\hat{\otimes}E(\mathbb{C}_{\infty})$ on the rigid analytic
unit disk $\mathfrak{D}:=\operatorname{Sp}\mathbb{C}_{\infty}\langle
A\rangle$. To anticipate generalizations to polar divisor and residues in the
next sections, we not only focus on the holomorphic version, but also divisor
versions. To materialize this, we fix an effective fractional ideal sheaf
$\mathcal{L}$ on $\mathfrak{X}$, that is, a subsheaf of
$\mathcal{O}_{\mathfrak{X}}$-modules $\mathcal{L}$ of
$\mathcal{M}_{\mathfrak{X}}$ on $\underline{\mathfrak{X}}$ that contains
$\mathcal{O}_{\mathfrak{X}}$. The latter assumption ensures that, for any
admissible open $\mathfrak{U}$, there is a map

$A\otimes\mathbb{C}_{\infty}\hookrightarrow\mathcal{O}_{\mathfrak{X}}(\mathfrak{U})\hookrightarrow\mathcal{L}(\mathfrak{U}).$
(4.1)

In practice, $\mathcal{L}$ will either be $\mathcal{O}_{\mathfrak{X}}$,
$\mathcal{O}_{\mathfrak{X}}(D)$ for a certain effective rigid analytic divisor
$D$, or $\mathcal{M}_{\mathfrak{X}}$ itself.

######  Definition 4.1.

Let $\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ be an
isomorphism of $\mathbb{F}$-vector space schemes over $\mathbb{C}_{\infty}$.
We define $\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$ and
$\iota_{\kappa}(\mathcal{L})(\mathfrak{U})$ as resulting from the pushout
square

${(A\otimes\mathbb{C}_{\infty})^{d}}$${A\otimes
E(\mathbb{C}_{\infty})}$${\mathcal{L}(\mathfrak{U})^{d}}$${\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})}$$\scriptstyle{\kappa^{-1}}$$\scriptstyle{\sim}$$\scriptstyle{\eqref{eq:canonical-
inclusion-sheaf}}$$\scriptstyle{\iota_{\kappa}(\mathcal{L})(\mathfrak{U})}$
(4.2)

in the category of $A$-modules.

As the top horizontal map of diagram (4.2) is an isomorphism, we get by the
universal property of the pushout square that the lower horizontal map is an
isomorphism as well. As its inverse will appear below, we give it a name:

######  Definition 4.2.

We denote by
$m_{\kappa}(\mathfrak{U}):\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})\to\mathcal{L}(\mathfrak{U})^{d}$
the inverse of the lower horizontal map in diagram (4.2).

Let $\mathfrak{U}\to\mathfrak{V}$ be a morphism in
$\operatorname{Cat}\mathfrak{X}$ (i.e. an inclusion of admissible opens). The
restriction map
$\mathcal{M}_{\mathfrak{X}}(\mathfrak{V})\to\mathcal{M}_{\mathfrak{X}}(\mathfrak{U})$
together with the universality of diagram (4.2) produces a morphism

$\rho_{\kappa}(\mathfrak{U}\to\mathfrak{V}):\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{V})\longrightarrow\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$
(4.3)

of $A$-modules. The data of $\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$
for all $\mathfrak{U}$ and $\rho_{\kappa}$ forms a contravariant functor from
$\operatorname{Cat}\mathfrak{X}$ to $\mathbf{Mod}_{A}$, and hence a presheaf
on the site $\underline{\mathfrak{X}}$.

######  Lemma 4.3.

$\mathcal{E}_{\kappa}(\mathcal{L})$ is a sheaf of $A$-modules on
$\underline{\mathfrak{X}}$.

###### Proof.

The data of $m_{\kappa}=(m_{\kappa}(\mathfrak{U}))_{\mathfrak{U}}$ indexed
over objects of $\operatorname{Cat}\mathfrak{X}$ defines an equivalence of
functors among $\mathcal{E}_{\kappa}(\mathcal{L})$ and $\mathcal{L}^{d}$. As
$\mathcal{L}^{d}$ is a sheaf, so is $\mathcal{E}_{\kappa}(\mathcal{L})$. ∎

###  4.2 Action of $A\otimes A$

In this subsection, we show that under some costability assumptions diagram
(4.2) can be upgraded to a diagram of $A\otimes A$-modules preserving the
$A\otimes A$-module structure on $A\otimes E(\mathbb{C}_{\infty})$. Throughout
this subsection, $\mathcal{L}$ is a costable effective fractional ideal sheaf
on $\mathfrak{X}$.

Let $\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ be a
morphism of $\mathbb{F}$-vector space schemes over $L$. Through the choice of
$\kappa$, we obtain a ring morphism
$\varphi_{\kappa}:A\to\mathsf{Mat}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$ as
resulting from the composition

$A\longrightarrow\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(E)\stackrel{{\scriptstyle\text{via~{}}\kappa}}{{\cong}}\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}^{d})\cong\mathsf{Mat}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$

where $\mathbb{C}_{\infty}\\{\tau\\}$ is the non-commutative ring of
$\tau$-polynomials. We thus obtain on $(A\otimes\mathbb{C}_{\infty})^{d}$ an
$A\otimes A$-module structure where, for $a\in A$, $a\otimes 1$ acts by
multiplication and where $1\otimes a$ acts by
$\operatorname{id}\otimes\varphi_{\kappa}(a)$. This makes the top arrow of
diagram (4.2) a morphism of $A\otimes A$-modules.

If $\mathfrak{U}$ is a costable admissible open, then
$\mathcal{L}(\mathfrak{U})^{d}$ admits an action of $\varphi_{\kappa}$
described as follows: An element $a\in A$ acts on
$F\in\mathcal{L}(\mathfrak{U})^{d}$ as

$\varphi_{\kappa}(a)(F)=A_{0}F+A_{1}F^{(1)}+...+A_{s}F^{(s)}\in\mathcal{L}(\mathfrak{U})^{d}$
(4.4)

according to the notations of Definition 3.10.

######  Remark 4.4.

If $\mathfrak{U}$ was not assumed to be costable, then identity (4.4) would
only make sense in
$\mathcal{L}(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})^{d}$.

This provides to $\mathcal{L}(\mathfrak{U})^{d}$ an $A\otimes A$-module
structure. The next facts are immediate:

######  Lemma 4.5.

Let $\mathfrak{U}$ be a costable admissible open. Then, there exists a unique
$A\otimes A$-module structure on
$\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$ making (4.2) a diagram of
$A\otimes A$-modules with respects to the structure defined above. If
$\mathfrak{U}\to\mathfrak{V}$ is a morphism in
$\operatorname{Cat}^{\tau}\mathfrak{X}$, the morphism (4.3) is a morphism of
$A\otimes A$-modules. Further, $\mathcal{E}_{\kappa}(\mathcal{L})$ is a sheaf
of $A\otimes A$-modules on $\,{}^{\tau}\underline{\mathfrak{X}}$.

We now wish to make the previous notion $\kappa$-free, and construct
$\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$ in a way independent of
$\kappa$. Let $\kappa^{\prime}$ be another choice of coordinates, and let

$M=M_{0}+M_{1}\tau+...+M_{s}\tau^{s}\in\operatorname{GL}_{d}(\mathbb{C}_{\infty}\\{\tau\\}),\quad
M_{i}\in\operatorname{GL}_{d}(\mathbb{C}_{\infty}),$

be the matrix obtained from $\kappa^{\prime}\circ\kappa^{-1}$ in
$\operatorname{End}_{\mathbb{F}\text{-vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}^{d})\cong\operatorname{GL}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$.
Let $\mathfrak{U}$ be an admissible open. By costability of $\mathcal{L}$, the
morphism obtained from $M$ maps $\mathcal{L}(\mathfrak{U})^{d}$ to
$\mathcal{L}(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})^{d}$.
In particular, the universal property of the pushout produces a morphism

$\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})\longrightarrow\mathcal{E}_{\kappa^{\prime}}(\mathcal{L})(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})$
(4.5)

via the commutative diagram of $A\otimes A$-modules:

${(A\otimes\mathbb{C}_{\infty})^{d}}$${A\otimes
E(\mathbb{C}_{\infty})}$${(A\otimes\mathbb{C}_{\infty})^{d}}$${A\otimes
E(\mathbb{C}_{\infty})}$${\mathcal{L}(\mathfrak{U})^{d}}$${\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})}$${\mathcal{L}(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})^{d}}$${\mathcal{E}_{\kappa^{\prime}}(\mathcal{L})(\mathfrak{U}\cap\tau\mathfrak{U}\cap\cdots\cap\tau^{s}\mathfrak{U})}$$\scriptstyle{\kappa^{-1}~{}~{}}$$\scriptstyle{\sim}$$\scriptstyle{M}$$\scriptstyle{(\kappa^{\prime})^{-1}}$$\scriptstyle{\sim}$$\scriptstyle{M}$

If $\mathfrak{U}$ is costable, (4.5) becomes an isomorphism

$t_{\kappa,\kappa^{\prime}}:\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathcal{E}_{\kappa^{\prime}}(\mathcal{L})(\mathfrak{U}).$
(4.6)

######  Definition 4.6.

Let $\mathfrak{U}\in\operatorname{Cat}^{\tau}\mathfrak{X}$,
$(\mathfrak{U}_{i}\to\mathfrak{U})_{i\in
I}\in\operatorname{Cov}^{\tau}\mathfrak{X}$ an admissible covering.

  1. 1.

We define $\mathcal{E}(\mathcal{L})(\mathfrak{U})$ as the $A\otimes A$-module
given by the inverse limit

$\mathcal{E}(\mathcal{L})(\mathfrak{U})=\varprojlim_{\kappa}\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$

taken over the projective system compound of isomorphisms of
$\mathbb{F}$-vector space schemes
$\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ over
$\mathbb{C}_{\infty}$, the transition maps being $t_{\kappa,\kappa^{\prime}}$
in (4.6).

  2. 2.

We denote by
$\varepsilon_{\kappa}(\mathcal{L})(\mathfrak{U}):\mathcal{E}(\mathcal{L})(\mathfrak{U})\to\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})$
the isomorphism given by universality of the inverse limit.

  3. 3.

We let $\iota(\mathfrak{U}):A\otimes
E(\mathbb{C}_{\infty})\to\mathcal{E}(\mathcal{L})(\mathfrak{U})$ be the
composition

$A\otimes
E(\mathbb{C}_{\infty})\stackrel{{\scriptstyle\iota_{\kappa}(\mathfrak{U})}}{{\longrightarrow}}\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})\stackrel{{\scriptstyle\varepsilon_{\kappa}(\mathfrak{U})^{-1}}}{{\longrightarrow}}\mathcal{E}(\mathcal{L})(\mathfrak{U}).$

######  Remark 4.7.

Note that $\iota$ is independent of the choice of $\kappa$, as for another
choice of $\kappa^{\prime}$, the diagram of $A\otimes A$-modules

${A\otimes
E(\mathbb{C}_{\infty})}$${\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})}$${\mathcal{E}(\mathcal{L})(\mathfrak{U})}$${\mathcal{E}_{\kappa^{\prime}}(\mathcal{L})(\mathfrak{U})}$$\scriptstyle{\iota_{\kappa}(\mathfrak{U})}$$\scriptstyle{\iota_{\kappa^{\prime}}(\mathfrak{U})}$$\scriptstyle{t_{\kappa,\kappa^{\prime}}}$$\scriptstyle{\varepsilon_{\kappa}(\mathfrak{U})}$$\scriptstyle{\varepsilon_{\kappa^{\prime}}(\mathfrak{U})}$

commutes.

Proposition 1.1 of the Introduction is then an easy consequence of the
definitions and the following fact:

######  Proposition 4.8.

$\mathcal{E}(\mathcal{L})$ is a sheaf of $A\otimes A$-modules on
$\,{}^{\tau}\underline{\mathfrak{X}}$.

###### Proof.

Let $(\mathfrak{U}_{i}\to\mathfrak{U})_{i\in
I}\in\operatorname{Cov}^{\tau}\mathfrak{X}$. We need to verify that the
sequence

$\mathcal{E}(\mathcal{L})(\mathfrak{U})\to\prod_{i\in
I}{\mathcal{E}(\mathcal{L})(\mathfrak{U}_{i})}\rightrightarrows\prod_{i,j\in
I}{\mathcal{E}(\mathcal{L})(\mathfrak{U}_{i}\times_{\mathfrak{U}}\mathfrak{U}_{j})}$

is exact. But, given $\kappa$, the above is canonically isomorphic to

$\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U})\to\prod_{i\in
I}{\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U}_{i})}\rightrightarrows\prod_{i,j\in
I}{\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{U}_{i}\times_{\mathfrak{U}}\mathfrak{U}_{j})}$

which is exact by Lemma 4.3. ∎

As one verifies, this definition is consistent with the definition of
$A\hat{\otimes}E(\mathbb{C}_{\infty})$ present in [GM21]. Recall that
$\mathfrak{D}=\operatorname{Sp}\mathbb{C}_{\infty}\langle A\rangle$. The
relevant statement is the following:

######  Lemma 4.9.

The admissible open $\mathfrak{D}\subset\mathfrak{X}$ is costable. Further,
the morphism

$\iota(\mathfrak{D}):A\otimes
E(\mathbb{C}_{\infty})\to\mathcal{E}(\mathcal{L})(\mathfrak{D})$

factors uniquely through the completion map $A\otimes
E(\mathbb{C}_{\infty})\to A\hat{\otimes}E(\mathbb{C}_{\infty})$. The image of
the latter is $\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$.

###### Proof.

That $\operatorname{Sp}\mathbb{C}_{\infty}\langle A\rangle$ is costable only
means that $\tau\mathfrak{m}$ is again maximal in $\mathbb{C}_{\infty}\langle
A\rangle$ for any maximal ideal $\mathfrak{m}\subset\mathbb{C}_{\infty}\langle
A\rangle$. Now, recall that the choice of $\kappa$ gives rise to a metric on
$A\otimes E(\mathbb{C}_{\infty})$ which is independent of $\kappa$, and that
$A\hat{\otimes}E(\mathbb{C}_{\infty})$ is the completion associated to the
latter metric. We obtain by definition a pushout square of $A\otimes
A$-modules:

${(A\otimes\mathbb{C}_{\infty})^{d}}$${A\otimes
E(\mathbb{C}_{\infty})}$${\mathbb{C}_{\infty}\langle
A\rangle^{d}}$${A\hat{\otimes}E(\mathbb{C}_{\infty})}$$\scriptstyle{\kappa^{-1}}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$
(4.7)

As $\mathcal{O}_{\mathfrak{X}}(\mathfrak{D})=\mathbb{C}_{\infty}\langle
A\rangle$, we obtain a unique map
$c_{\kappa}:A\hat{\otimes}E(\mathbb{C}_{\infty})\to\mathcal{E}_{\kappa}(\mathcal{L})(\mathfrak{D})$,
and this map is an isomorphism. Now, one verifies that

$A\hat{\otimes}E(\mathbb{C}_{\infty})\stackrel{{\scriptstyle(c_{\kappa})_{\kappa}}}{{\longrightarrow}}\varprojlim_{\kappa}\mathcal{E}_{\kappa}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})=\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$

is a well-defined isomorphism of $A\otimes A$-modules. ∎

###  4.3 Special functions as $E(\mathbb{C}_{\infty})$-valued functions

We are now in position to introduce the object of study.

In [GM21, Def. 3.1], we defined the $A$-module of special functions of $E$ as

$\mathfrak{sf}(E)=\left\\{\omega\in
A\hat{\otimes}E(\mathbb{C}_{\infty})\,\,\middle|\,\,\forall a\in A:(a\otimes
1)\omega=(1\otimes a)\omega\right\\}.$

According to Lemma 4.9, the $A$-module $\mathfrak{sf}(E)$ admits the following
generalization:

######  Definition 4.10.

Let $\mathcal{L}$ be a costable effective fractional ideal sheaf on
$\mathfrak{X}$, and let $\mathfrak{U}$ be an object of
$\operatorname{Cat}^{\tau}\mathfrak{X}$. The module of special functions of
$\mathcal{E}(\mathcal{L})$ on $\mathfrak{U}$ is the $A$-module

$\mathfrak{sf}_{E}(\mathcal{L})(\mathfrak{U})=\left\\{\omega\in\mathcal{E}(\mathcal{L})(\mathfrak{U})\,\,\middle|\,\,\forall
a\in A:~{}(a\otimes 1)\omega=(1\otimes a)\omega\right\\}.$

In particular as
$\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})=A\hat{\otimes}E(\mathbb{C}_{\infty})$,
we have
$\mathfrak{sf}(E)=\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$
as $A$-modules.
We have the following:

######  Proposition 4.11.

$\mathfrak{sf}_{E}(\mathcal{L})$ is a sheaf of $A$-modules on
$\,{}^{\tau}\underline{\mathfrak{X}}$.

###### Proof.

This follows immediately from Definition 4.10 and the fact that $\mathcal{E}$
is a sheaf on $\,{}^{\tau}\underline{\mathfrak{X}}$. ∎

##  5 Continuation and residues of special functions

###  5.1 Residue at $\mathfrak{j}$

####  5.1.1 Residue map for meromorphic functions on $\mathfrak{X}$

We begin with a brief recall of residues for meromorphic functions on
$\mathfrak{X}$. Recall that we have a morphism of locally ringed spaces
$f:\mathfrak{X}\to X$ given by the GAGA functor, and that any sheaf
$\mathcal{F}$ on $X$ induces a sheaf
$\mathcal{F}^{\operatorname{rig}}=f^{*}\mathcal{F}$ on $\mathfrak{X}$. The
sheaf of holomorphic differentials, denoted
$\Omega^{\operatorname{hol}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$, is defined
as

$\Omega^{\operatorname{hol}}_{\mathfrak{X}/\mathbb{C}_{\infty}}:=\left(\Omega_{X/\mathbb{C}_{\infty}}^{1}\right)^{\operatorname{rig}}=\mathcal{O}_{\mathfrak{X}}\otimes_{\mathcal{O}_{X}}\Omega^{1}_{\mathcal{O}_{X}/\mathbb{C}_{\infty}}.$

######  Remark 5.1.

In [FvdP04, §3.6],
$\Omega^{\operatorname{hol}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$ is rather
called the universal finite differential module and denoted by
$\Omega^{\operatorname{f}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$.

Similarly, one defines the sheaf of meromorphic differentials, denoted
$\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$, as

$\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}:=\mathcal{M}_{\mathfrak{X}}\otimes_{\mathcal{O}_{\mathfrak{X}}}\Omega^{\operatorname{hol}}_{\mathfrak{X}/\mathbb{C}_{\infty}}=\mathcal{M}_{\mathfrak{X}}\otimes_{\mathcal{O}_{X}}\Omega^{1}_{\mathcal{O}_{X}/\mathbb{C}_{\infty}}.$

Identifying $\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$
with $\mathcal{M}_{\mathfrak{X}}\otimes_{A}\Omega_{A/\mathbb{F}}^{1}$, we
recognize that there is a map

$\tau:\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}\longrightarrow\tau^{-1}\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}$

given as $\tau\otimes_{A}\operatorname{id}_{\Omega_{A}}$ on
$\mathcal{M}_{\mathfrak{X}}\otimes_{A}\Omega_{A/\mathbb{F}}^{1}$. Therefore,
if $\mathfrak{U}$ is a costable admissible open of $\mathfrak{X}$, we obtain a
morphism
$\tau:\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}(\mathfrak{U})\to\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}}(\mathfrak{U})$.

Given $x\in|\mathfrak{X}|=|X|$, we hence have a residue map

$\operatorname{residue}_{x}:(\Omega^{\operatorname{mer}}_{\mathfrak{X}/\mathbb{C}_{\infty}})_{x}\cong(\Omega_{X/\mathbb{C}_{\infty}}^{1})_{x}\longrightarrow\mathbb{C}_{\infty},\quad\left(\sum_{i\geq-h}{a_{i}\pi_{x}^{i}}\right)d\pi_{x}\longmapsto
a_{-1}$

obtained from the classical algebraic residue map, and independent on the
choice of a uniformizer $\pi_{x}\in\mathcal{O}_{\mathfrak{X},x}$.

####  5.1.2 Residue of $E(\mathbb{C}_{\infty})$-valued functions at
$\mathfrak{j}$

Now that we have constructed a sheaf of $E(\mathbb{C}_{\infty})$-valued
meromorphic functions, let us discuss in which sense one might take the
residue at $\mathfrak{j}\in\mathfrak{X}$. We denote by $\mathcal{J}$ the
subsheaf of $\mathcal{M}_{\mathfrak{X}}$ of functions that are meromorphic on
$\mathfrak{X}$, yet holomorphic away from the support of $J$; namely,

$\mathcal{J}:=\operatorname{colim}_{n}\mathcal{O}_{\mathfrak{X}}(n\cdot J).$

As all sheaves $\mathcal{O}_{\mathfrak{X}}(n\cdot J)$ are costable, so is
$\mathcal{J}$.

######  Remark 5.2.

Note that $\mathcal{J}(\mathfrak{U})$ does generally not correspond to
$\bigcup_{n}\mathcal{O}_{\mathfrak{X}}(n\cdot J)(\mathfrak{U})$ for an
arbitrary admissible open $\mathfrak{U}$: the meromorphic functions with only
poles the points $\tau^{n}\mathfrak{j}$ with order $n$, for all positive $n$,
will typically belong to $\mathcal{J}(\mathfrak{X})$ but not to
$\bigcup_{n}\mathcal{O}_{\mathfrak{X}}(n\cdot J)(\mathfrak{X})$.

######  Definition 5.3.

Let $\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$ be a
choice of coordinates for $E$. We define $\operatorname{res}_{\mathfrak{j}}$
as the unique map of $A$-modules making the following diagram commute:

${\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathcal{E}(\mathcal{J})(\mathfrak{X})}$${\operatorname{Lie}_{E}\mathbb{C}_{\infty}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathcal{J}(\mathfrak{X})^{d}}$${\mathbb{C}_{\infty}^{d}}$$\scriptstyle{m_{\kappa}\circ\varepsilon_{\kappa}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{res}_{\mathfrak{j}}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{Lie}\kappa}$$\scriptstyle{\operatorname{residue}_{\mathfrak{j}}}$
(5.1)

The next proposition justifies the absence of ”$\kappa$” in the notation for
$\operatorname{res}_{\mathfrak{j}}$.

######  Proposition 5.4.

The map $\operatorname{res}_{\mathfrak{j}}$ is independent of $\kappa$.

###### Proof.

First observe that if $f\in\mathcal{J}(\mathfrak{X})$ then $f^{(1)}$ belongs
to $\mathcal{J}(\mathfrak{X})^{(1)}$ which consists in functions that are
meromorphic on $\mathfrak{X}$ and holomorphic on the admissible open
$\mathfrak{X}\setminus\\{\tau\mathfrak{j},\tau^{2}\mathfrak{j},...\\}$, and
$f^{(1)}$ is thus regular at $\mathfrak{j}$. In particular, given
$F\in(\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathcal{J}(\mathfrak{X}))^{d}$ and
a matrix
$M=M_{0}+M_{1}\tau+...\in\operatorname{GL}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$
where $M_{i}\in\mathsf{Mat}_{d}(\mathbb{C}_{\infty})$ and
$M_{0}\in\operatorname{GL}_{d}(\mathbb{C}_{\infty})$, we have

$\displaystyle\operatorname{residue}_{\mathfrak{j}}\left(M\cdot F\right)$
$\displaystyle=\operatorname{residue}_{\mathfrak{j}}\left(M_{0}\cdot
F+\underbrace{M_{1}\cdot F^{(1)}+M_{2}\cdot
F^{(2)}+...}_{\text{regular~{}at~{}}\mathfrak{j}}\right)=\operatorname{residue}_{\mathfrak{j}}(M_{0}\cdot
F)$ $\displaystyle=M_{0}\cdot\operatorname{residue}_{\mathfrak{j}}(F).$

Now, for another choice $\kappa^{\prime}$ of coordinates for $E$,
$\kappa^{\prime}$ differs from $\kappa$ by an element in
$\operatorname{Aut}_{\mathbb{F}_{q}-\text{vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}^{d})$,
and hence is represented by a matrix
$M\in\operatorname{GL}_{d}(\mathbb{C}_{\infty}\\{\tau\\})$. We hence have a
diagram:

${\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathcal{E}(\mathcal{J})(\mathfrak{X})}$${\operatorname{Lie}_{E}\mathbb{C}_{\infty}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathcal{J}(\mathfrak{X})^{d}}$${\mathbb{C}_{\infty}^{d}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathcal{J}(\mathfrak{X})^{d}}$${\mathbb{C}_{\infty}^{d}}$$\scriptstyle{m_{\kappa}\circ\varepsilon_{\kappa}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{res}_{\mathfrak{j}}}$$\scriptstyle{m_{\kappa^{\prime}}\circ\varepsilon_{\kappa^{\prime}}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{Lie}\kappa}$$\scriptstyle{\operatorname{Lie}\kappa^{\prime}}$$\scriptstyle{\operatorname{residue}_{\mathfrak{j}}}$$\scriptstyle{M}$$\scriptstyle{M_{0}}$$\scriptstyle{\operatorname{residue}_{\mathfrak{j}}}$

where all inner squares commute. In particular, the outer square commutes,
which proves the claim. ∎

###  5.2 Proofs of the main results

The next result, already announced in the introduction, translates the fact
that special functions admit a unique meromorphic continuation to
$\mathfrak{X}$, with poles at $\mathfrak{j}$ and its iterates of order at most
$d=\dim E$. For a rigid analytic divisor $D$, we rather use the notation
$\mathfrak{sf}_{E}(D)$ for $\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}}(D))$.

######  Theorem 5.5 (Meromorphic continuation of special functions).

The canonical $A$-module map $\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{X})\to\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{D})=\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})=\mathfrak{sf}(E)$
is an isomorphism.

We begin with some preliminary results. Recall that there exists a unique
$A$-linear morphism

$\exp_{E}:\operatorname{Lie}_{E}(\mathbb{C}_{\infty})\longrightarrow
E(\mathbb{C}_{\infty})$

which is analytic in the following sense: for any choice of
$\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$, the map
$\exp_{E}^{\kappa}:\mathbb{C}_{\infty}^{d}\to\mathbb{C}_{\infty}^{d}$ obtained
as $\kappa\circ\exp_{E}\circ(\operatorname{Lie}_{E}\kappa)^{-1}$ can be
written in the form

$\exp^{\kappa}_{E}(x)=e_{0}^{\kappa}x+e_{1}^{\kappa}\tau(x)+e_{2}^{\kappa}\tau^{2}(x)+...\quad\text{where}~{}e_{i}^{\kappa}\in\mathsf{Mat}_{d}(\mathbb{C}_{\infty}),$

the series converging for any $x\in\mathbb{C}_{\infty}^{d}$. By continuity,
$\exp_{E}$ extends $A$-linearly to

$\widehat{\exp}_{E}:A\hat{\otimes}\operatorname{Lie}_{E}(\mathbb{C}_{\infty})\longrightarrow
A\hat{\otimes}E(\mathbb{C}_{\infty})$

(see [GM21, §2]).

Let $u\in A$ be a separating element. We recall from [GM21, Thm. 3.11] that we
have an isomorphism of $A$-modules

$\delta_{u}:\mathfrak{d}_{A/\mathbb{F}[u]}\otimes_{A}\Lambda_{E}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathfrak{sf}(E)$

mapping an element
$\lambda\in\mathfrak{d}_{A/\mathbb{F}[u]}\otimes_{A}\Lambda_{E}$ to
$\omega_{\lambda}:=\widehat{\exp}_{E}((u\otimes
1-1\otimes\partial\varphi_{u})^{-1}(\lambda))$ seen in
$A\hat{\otimes}E(\mathbb{C}_{\infty})$.

######  Remark 5.6.

The map $\delta_{u}$ is not exactly the one that appears in [GM21], but minus
that one.

######  Lemma 5.7.

Let $n_{u}$ be the $\mathbb{C}_{\infty}$-linear endomorphism of
$\operatorname{Lie}_{E}(\mathbb{C}_{\infty})$ given by
$\partial\varphi_{u}-\ell(u)$. Then, the composition

$\mathfrak{sf}(E)\hookrightarrow\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathcal{O}_{\mathfrak{X}}(\mathfrak{D})^{d}=\mathbb{C}_{\infty}\langle
A\rangle^{d},$

where the second arrow is
$(m_{\kappa}\circ\varepsilon_{\kappa})(\mathfrak{D})$, maps $\omega_{\lambda}$
to the converging sum

$\sum_{j=0}^{d-1}{\sum_{i=0}^{\infty}{\frac{(1\otimes
e_{i}^{\kappa})[(1\otimes\operatorname{Lie}_{E}\kappa)(1\otimes
n_{u}^{j})(\lambda)]^{(i)}}{(u\otimes 1-1\otimes\ell(u)^{q^{i}})^{j+1}}}}.$

###### Proof.

Let us first contemplate the diagram

${A\hat{\otimes}\operatorname{Lie}_{E}(\mathbb{C}_{\infty})}$${A\hat{\otimes}E(\mathbb{C}_{\infty})}$${\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})}$${\mathbb{C}_{\infty}\langle
A\rangle^{d}}$${\mathbb{C}_{\infty}\langle
A\rangle^{d}}$${\mathcal{O}_{\mathfrak{X}}(\mathfrak{D})^{d}}$$\scriptstyle{\widehat{\exp}_{E}}$$\scriptstyle{1\otimes\operatorname{Lie}_{E}\kappa}$$\scriptstyle{\sim}$$\scriptstyle{1\otimes\kappa}$$\scriptstyle{(m_{\kappa}\circ\varepsilon_{\kappa})(\mathfrak{D})}$$\scriptstyle{\widehat{\exp}_{E}^{\kappa}}$$\scriptstyle{=}$
(5.2)

where we defined $\widehat{\exp}^{\kappa}_{E}$ as the unique map making the
left-hand square commute. That the right-hand square commute is a matter of
definition.

Now, observe that $n_{u}$ is nilpotent of order at most $d$ by definition of
an Anderson $A$-module. Hence, we have

$(u\otimes 1-1\otimes\partial\varphi_{u})^{-1}=\sum_{j=0}^{d-1}{\frac{1\otimes
n_{u}^{j}}{(u\otimes 1-1\otimes\ell(u))^{j+1}}}.$

and then, seeing $\omega_{\lambda}\in A\hat{\otimes}E(\mathbb{C}_{\infty})$,

$\displaystyle(1\otimes\kappa)(\omega_{\lambda})$
$\displaystyle=((1\otimes\kappa)\circ\widehat{\exp}_{E})\left((u\otimes
1-1\otimes\partial\varphi_{u})^{-1}(\lambda)\right)$
$\displaystyle=[\widehat{\exp}_{E}^{\kappa}\circ(1\otimes\operatorname{Lie}_{E}\kappa)]\left(\sum_{j=0}^{d-1}{\frac{1\otimes
n_{u}^{j}}{(u\otimes 1-1\otimes\ell(u))^{j+1}}}\right)(\lambda)$
$\displaystyle=\sum_{j=0}^{d-1}{\sum_{i=0}^{\infty}{\frac{(1\otimes
e_{i}^{\kappa})[(1\otimes\operatorname{Lie}_{E}\kappa)(1\otimes
n_{u}^{j})(\lambda)]^{(i)}}{(u\otimes 1-1\otimes\ell(u)^{q^{i}})^{j+1}}}}.$
(5.3)

That the upper series converge is immediate from the convergence of
$\exp^{\kappa}_{E}$. ∎

###### Proof of Theorem 5.5.

Let $u\in A$ be a separating element. We have a diagram of $A$-modules

${\mathfrak{sf}_{E}(d\cdot J_{u})(\mathfrak{X})}$${\mathcal{E}(d\cdot
J_{u})(\mathfrak{X})}$${\mathcal{O}_{\mathfrak{X}}(d\cdot
J_{u})(\mathfrak{X})^{d}}$${\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})}$${\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})}$${\mathcal{O}_{\mathfrak{X}}(\mathfrak{D})^{d}=\mathbb{C}_{\infty}\langle
A\rangle^{d}}$$\scriptstyle{(m_{\kappa}\circ\varepsilon_{\kappa})(\mathfrak{X})}$$\scriptstyle{\sim}$$\scriptstyle{(m_{\kappa}\circ\varepsilon_{\kappa})(\mathfrak{D})}$$\scriptstyle{\sim}$
(5.4)

where, by the previous Lemma 5.7, the composition of the lower horizontal maps
factors through the right-hand vertical map (represented as the dashed arrow).
The map $(m_{\kappa}\circ\varepsilon_{\kappa})(\mathfrak{X})$ being an
isomorphism, the latter factors through $\mathcal{E}(d\cdot
J_{u})(\mathfrak{X})$ (represented as the dashdotted arrow). As by
construction, $\mathfrak{sf}_{E}(d\cdot J_{u})(\mathfrak{X})$ is the
intersection of $\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$
and $\mathcal{E}(d\cdot J_{u})(\mathfrak{X})$ in
$\mathcal{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$, we thus obtain that
the inclusion

$\mathfrak{sf}_{E}(d\cdot
J_{u})(\mathfrak{X})\hookrightarrow\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$

is an equality. As this holds for all $u\in A$ separating, Proposition 3.8
implies

$\mathfrak{sf}_{E}(d\cdot J)=\bigcap_{\begin{subarray}{c}u\in A\\\
u~{}\text{separating}\end{subarray}}\mathfrak{sf}_{E}(d\cdot J_{u})$

hence that $\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{X})=\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{D})$
as desired. ∎

Let $u$ be a separating element in $A$. The relative different ideal
$\mathfrak{d}_{A/\mathbb{F}[u]}$ of the separable extension $K/\mathbb{F}(u)$
is generated by the elements of the form $f_{s}^{\prime}(s)$ where $s\in A$ is
such that $K=\mathbb{F}(s,u)$ and $f_{s}$ is the minimal polynomial of $s$
over $\mathbb{F}(u)$. It is also described as

$\mathfrak{d}_{A/\mathbb{F}[u]}=\\{x\in A~{}|~{}\forall y\in A:~{}xdy\in
Adu\\}.$

As such, we have an $A$-linear isomorphism
$\mu:\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{d}_{A/\mathbb{F}[u]}\stackrel{{\scriptstyle\sim}}{{\to}}Adu$
given by multiplication $\omega\otimes x\mapsto x\omega$. The main result of
this note is the following:

######  Theorem 5.8.

We have
$\operatorname{res}_{\mathfrak{j}}(\Omega^{1}_{A/\mathbb{F}}\otimes_{A}\mathfrak{sf}_{E}(d\cdot
J))=\Lambda_{E}$ as submodules of
$\operatorname{Lie}_{E}(\mathbb{C}_{\infty})$. In addition, the diagram

${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{d}_{A/\mathbb{F}[u]}\otimes_{A}\Lambda_{E}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{sf}(E)}$${\Lambda_{E}}$${\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{X})}$$\scriptstyle{\mu\otimes\operatorname{id}_{\Lambda}}$$\scriptstyle{\wr}$$\scriptstyle{\operatorname{id}_{\Omega}\otimes_{A}\delta_{u}}$$\scriptstyle{\sim}$$\scriptstyle{\operatorname{res}_{\mathfrak{j}}}$$\scriptstyle{\operatorname{Thm}.~{}\ref{thm:continuation-
sf}}$$\scriptstyle{\wr}$ (5.5)

commutes.

###### Proof.

The differential $du$ has a unique preimage
$\mu^{-1}(du)\in\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{d}_{A/\mathbb{F}[u]}$
which we write as

$Du:=\sum_{i}{dt_{i}\otimes_{A}f_{s_{i}}^{\prime}(s_{i})}.$

The inverse
$\Lambda_{E}\stackrel{{\scriptstyle\sim}}{{\to}}\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{d}_{A/\mathbb{F}[u]}\otimes_{A}\Lambda_{E}$
of the left-vertical map in diagram (5.5) maps $\lambda$ to
$Du\otimes_{A}\lambda=\sum_{i}{dt_{i}\otimes_{A}f_{s_{i}}^{\prime}(s_{i})\otimes_{A}\lambda}$.
Once composed with $\operatorname{id}_{\Omega}\otimes_{A}\delta_{u}$, it is
mapped to

$\sum_{i}{dt_{i}\otimes_{A}\widehat{\exp}_{E}\left((u\otimes
1-1\otimes\partial\varphi_{u})^{-1}(f^{\prime}_{s_{i}}(s_{i})\otimes
1)(\lambda)\right)}\in\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{X}).$

To conclude, we need to compute $\operatorname{res}_{\mathfrak{j}}$ of the
above. Through a choice of coordinates
$\kappa:E\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{G}_{a}^{d}$, it boils
down thanks to Lemma 5.7 to the computation of

$\sum_{i}{\operatorname{residue}_{\mathfrak{j}}\left(\sum_{j=0}^{d-1}{\sum_{n=0}^{\infty}{dt_{i}\otimes_{A}\frac{f^{\prime}_{s_{i}}(s_{i})\otimes\operatorname{Lie}_{E}\kappa^{-1}\left(e_{n}^{\kappa}(\operatorname{Lie}_{E}\kappa(n_{u}^{j}(\lambda)))^{(n)}\right)}{(u\otimes
1-1\otimes\ell(u)^{q^{n}})^{j+1}}}}\right)}.$

Only the terms in $(j,n)=(0,0)$ are relevant, and we are left with:

$\sum_{i}{\operatorname{residue}_{\mathfrak{j}}\left(dt_{i}\otimes_{A}\frac{f^{\prime}_{s_{i}}(s_{i})\otimes\lambda}{u\otimes
1-1\otimes\ell(u)}\right)}=\operatorname{residue}_{\mathfrak{j}}\left(\frac{du\otimes\lambda}{u\otimes
1-1\otimes\ell(u)}\right)=\lambda,$ (5.6)

since $u\otimes 1-1\otimes\ell(u)$ is a uniformizer at $\mathfrak{j}$. ∎

######  Corollary 5.9.

We have $\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{X})=(0)$.

###### Proof.

Since $\Omega_{A/\mathbb{F}}^{1}$ is invertible as an $A$-module, it suffices
to show that
$\Omega_{A/\mathbb{F}}^{1}\otimes_{A}\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{X})$
is trivial. Yet, any element $\eta$ in the latter module satisfies
$\operatorname{res}_{\mathfrak{j}}(\eta)=0$. This implies $\eta=0$ by Theorem
5.8. ∎

###  5.3 Filtration on $\mathfrak{sf}(E)$

As already mentioned in the introduction, the $A$-module $\mathfrak{sf}(E)$
carries a natural increasing finite filtration indexed by the order of poles
along the divisor $J$:

$(0)\stackrel{{\scriptstyle\text{Cor~{}}\ref{cor:no-pole-no-
sf}}}{{=}}\mathfrak{sf}_{E}(\mathcal{O}_{\mathfrak{X}})(\mathfrak{X})\hookrightarrow\mathfrak{sf}_{E}(J)(\mathfrak{X})\hookrightarrow\cdots\hookrightarrow\mathfrak{sf}_{E}(d\cdot
J)(\mathfrak{X})\stackrel{{\scriptstyle\text{Thm~{}}\ref{thm:continuation-
sf}}}{{=}}\mathfrak{sf}(E).$ (5.7)

We conclude this text by giving a few examples of the above filtration built
out from the Carlitz $t$-module. These examples show that the filtration steps
can be quite arbitrary.

#### Carlitz tensor powers

Let $C=\mathbb{P}^{1}_{\mathbb{F}}$ and let $\infty$ be the point $[0:1]$. We
identify $A$ with $\mathbb{F}[t]$ and $A\otimes\mathbb{C}_{\infty}$ with the
polynomial ring $\mathbb{C}_{\infty}[t]$. We also denote $\ell(t)$ by
$\theta$. Let $n>0$. The $n$th tensor power of the Carlitz $t$-module over
$\mathbb{C}_{\infty}$ is the unique $A$-module $\mathsf{C}^{\otimes n}$ which
equals $\mathbb{G}_{a}^{n}$ as an $\mathbb{F}$-vector space scheme, and for
which

$\varphi(t)=\begin{pmatrix}\theta&1&&\\\ &\theta&\ddots&\\\ &&\ddots&1\\\
\tau&&&\theta\end{pmatrix}\in\mathsf{Mat}_{n}(\mathbb{C}_{\infty}\\{\tau\\})\cong\operatorname{End}_{\mathbb{F}-\text{vs}/\mathbb{C}_{\infty}}(\mathbb{G}_{a}^{n}).$

As sub-$\mathbb{F}[t]$-modules of $A\hat{\otimes}\mathsf{C}^{\otimes
n}(\mathbb{C}_{\infty})=\mathbb{C}_{\infty}\langle t\rangle^{n}$, we have

$\mathfrak{sf}(\mathsf{C}^{\otimes n})=\mathbb{F}[t]\cdot\begin{pmatrix}1\\\
(t-\theta)\\\ \vdots\\\ (t-\theta)^{n-1}\end{pmatrix}\omega(t)^{n},$

where $\omega$ is the Anderson-Thakur function:

$\omega(t)=(-\theta)^{\frac{1}{q-1}}\prod_{i=0}^{\infty}{\left(1-\frac{t}{\theta^{q^{i}}}\right)^{-1}}\in\mathbb{C}_{\infty}\langle
t\rangle.$

Since $\omega(t)^{n}$ has a pole of order $n$ at $\mathfrak{j}=(t-\theta)$,
the filtration (5.7) has only one jump at the $n$th degree:

$(0)=\mathfrak{sf}_{\mathsf{C}^{\otimes n}}((n-1)\cdot
J)(\mathfrak{X})\subsetneq\mathfrak{sf}_{\mathsf{C}^{\otimes n}}(n\cdot
J)(\mathfrak{X})=\mathfrak{sf}(\mathsf{C}^{\otimes n}).$

#### Sum of Carlitz tensor powers

Let $E$ be the product of $m$ copies of $\mathsf{C}^{\otimes n}$. Then, $E$
has dimension $mn$, and $\mathfrak{sf}(E)=\mathfrak{sf}(\mathsf{C}^{\otimes
n})^{\oplus m}$. In particular, the filtration (5.7) has only one jump at the
$n$th degree:

$(0)=\mathfrak{sf}_{E}((n-1)\cdot
J)(\mathfrak{X})\subsetneq\mathfrak{sf}_{E}(n\cdot
J)(\mathfrak{X})=\mathfrak{sf}_{E}(mn\cdot J)(\mathfrak{X})=\mathfrak{sf}(E).$

#### Prolongations of the Carlitz module

For $k\geq 0$, let $\rho_{k}\mathsf{C}$ be the $k$th prolongation of
$\mathsf{C}$ as defined in [Mau18b]. It is an Anderson $A$-module which equals
$\mathbb{G}_{a}^{k+1}$ as an $\mathbb{F}$-module scheme. The module
$\mathfrak{sf}(\rho_{k}\mathsf{C})$ is free rank $k+1$ over $\mathbb{F}[t]$,
having the elements

$\omega_{1}:=\begin{pmatrix}\omega(t)\\\ 0\\\ \vdots\\\
0\end{pmatrix},\quad\omega_{2}:=\begin{pmatrix}\partial_{t}^{(1)}\omega(t)\\\
\omega(t)\\\ \vdots\\\
0\end{pmatrix},~{}\cdots~{},\omega_{k+1}:=\begin{pmatrix}\partial_{t}^{(k)}\omega(t)\\\
\partial_{t}^{(k-1)}\omega(t)\\\ \vdots\\\ \omega(t)\end{pmatrix}.$

for basis (here $\partial_{t}^{(j)}$ refers to the $j$th higher derivative).
Therefore, the filtration (5.7) has exactly $(k+1)$-jumps, and

$\mathfrak{sf}_{\rho_{k}\mathsf{C}}(d\cdot
J)(\mathfrak{X})=\bigoplus_{i=1}^{d}\mathbb{F}[t]\cdot\omega_{i}.$

#### Summary

For appropriate choices of $n$, $m$, and $k$ in these three cases, we obtain
$t$-modules of dimension $d$ with different filtration steps.

  * •

For the $d$th Carlitz tensor power $\mathsf{C}^{\otimes d}$, there is one step
and this one is at the $d$th degree.

  * •

For the product of $d$ copies of $\mathsf{C}$, i.e. $\mathsf{C}^{\oplus d}$,
there is also one step, but this time at degree $1$.

  * •

For the $(d-1)$st prolongation, there is a step at each degree.

Of course, these examples can easily be extended, e.g. to products of
different tensor powers of the Carlitz module, to obtain even more variation
in the steps. But this is not in the scope of this manuscript.

## References

  * [Bos82] Siegfried Bosch.  Meromorphic functions on proper rigid analytic varieties.  Séminaire de Théorie des Nombres de Bordeaux, pages 1–22, 1982\.
  * [Bos14] Siegfried Bosch.  Lectures on formal and rigid geometry, volume 2105 of Lecture Notes in Mathematics.  Springer, Cham, 2014.
  * [FvdP04] Jean Fresnel and Marius van der Put.  Rigid analytic geometry and its applications, volume 218 of Progress in Mathematics.  Birkhäuser Boston, Inc., Boston, MA, 2004.
  * [GM21] Quentin Gazda and Andreas Maurischat.  Special functions and gauss–thakur sums in higher rank and dimension.  Journal für die reine und angewandte Mathematik (Crelles Journal), 2021(773):231–261, 2021.
  * [GP18] Nathan Green and Matthew A. Papanikolas.  Special $L$-values and shtuka functions for Drinfeld modules on elliptic curves.  Res. Math. Sci., 5(1):Paper No. 4, 47, 2018.
  * [HJ20] Urs Hartl and Ann-Kristin Juschka.  Pink’s theory of Hodge structures and the Hodge conjecture over function fields.  In $t$-motives: Hodge structures, transcendence and other motivic aspects, EMS Ser. Congr. Rep., pages 31–182. EMS Publ. House, Berlin, [2020] ©2020.
  * [Mau18a] Andreas Maurischat.  Periods of t-modules as special values.  Journal of Number Theory, 2018.  DOI:10.1016/j.jnt.2018.09.024.
  * [Mau18b] Andreas Maurischat.  Prolongations of $t$-motives and algebraic independence of periods.  Doc. Math., 23:815–838, 2018.
  * [Pel08] Federico Pellarin.  Aspects de l’indépendance algébrique en caractéristique non nulle (d’après Anderson, Brownawell, Denis, Papanikolas, Thakur, Yu, et al.).  Astérisque, (317):Exp. No. 973, viii, 205–242, 2008.  Séminaire Bourbaki. Vol. 2006/2007.


Since $r>0$ we can drop it up to a change in the standard deviation of $x$. We
want to show that its derivative is non-positive:

$\displaystyle\sigma(x)\mathbb{E}[\sigma^{\prime}(x+r^{\prime}z)|x]\leq\sigma^{\prime}(x)\mathbb{E}[\sigma(x+r^{\prime}z)|x]\
\Leftrightarrow\
\frac{\sigma(x)}{\sigma^{\prime}(x)}\leq\frac{\mathbb{E}[\sigma(x+r^{\prime}z)|x]}{\mathbb{E}[\sigma^{\prime}(x+r^{\prime}z)|x]}.$
(153)

From the property (iii) of softplus (Lemma G.8), we have that
$R(x)=\sigma(x)/\sigma^{\prime}(x)$ is convex. Applying Jensen, we get

$\displaystyle\frac{\sigma(x)}{\sigma^{\prime}(x)}\leq\mathbb{E}\left[\frac{\sigma(x+r^{\prime}z)}{\sigma^{\prime}(x+r^{\prime}z)}\Big{|}x\right].$

What remains to show is that

$\displaystyle\mathbb{E}\left[\frac{\sigma(x+r^{\prime}z)}{\sigma^{\prime}(x+r^{\prime}z)}\Big{|}x\right]\mathbb{E}\left[\sigma^{\prime}(x+r^{\prime}z)|x\right]\leq\mathbb{E}\left[\sigma(x+r^{\prime}z)|x\right].$
(154)

Note that $\mathbb{E}[\sigma^{\prime}(x+r^{\prime}z)|x]$ is increasing in $x$
since $\sigma^{\prime}$ is increasing. Moreover, the function

$\mathbb{E}\left[\frac{\sigma(x+r^{\prime}z)}{\sigma^{\prime}(x+r^{\prime}z)}\Big{|}x\right]$

is increasing in $x$ since its integrand $R$ is increasing from the property
(ii) of softplus (Lemma G.8). Then we conclude by the FKG inequality that Eq.
154 holds. Therefore, for a solution $(r,u)$ of the fixed point Eq. 117, we
have $r\\!\leq\\!u=\frac{1}{k}$.

####  G.5.3 Bounding the Outgoing Weight

To get a bound on $a$, let us analyze the ratio of interactions in Eq. 117

$\displaystyle\frac{g_{\sigma}(r,1,u)}{g_{\sigma}(r,r,1)}=\frac{a}{k}.$ (155)

Using the convexity of softplus (property (i) of Lemma G.8), we get

$\displaystyle\frac{\mathbb{E}[\sigma(rx)\sigma(ux+u^{\prime}z)]}{\mathbb{E}[\sigma(rx)^{2}]}$
$\displaystyle\geq\frac{\mathbb{E}[\sigma(rx)\sigma(rx)]+\mathbb{E}[\sigma(rx)((u-r)x+u^{\prime}z)\sigma^{\prime}(rx)]}{\mathbb{E}[\sigma(rx)^{2}]}$
$\displaystyle=1+(u-r)\frac{\mathbb{E}[\sigma^{\prime}(rx)\sigma(rx)x]}{\mathbb{E}[\sigma(rx)^{2}]}.$
(156)

We can transform the numerator using Stein’s lemma with
$v(x)=\sigma(rx)\sigma^{\prime}(rx)$

$\displaystyle\mathbb{E}[\sigma(rx)\sigma^{\prime}(rx)x]=r\mathbb{E}[\sigma^{\prime}(rx)^{2}+\sigma(rx)\sigma^{\prime\prime}(rx)]$
(157)

which is positive since softplus is positive, increasing, and convex.
Combining it with $u\geq r$, we get that the ratio is bounded below by $1$
which yields $a\geq k$.

####  G.5.4 Helper Lemmas

In this subsection, we provide helper lemmas used in the proof of Theorem 5.3.
We present Lemma G.6 which shows that $\tilde{G}$ is increasing in correlation
and Lemma G.7 used in the proof of the former. Finally, we present several
properties of the softplus family in Lemma G.8 that are used throughout the
proof.

######  Lemma G.6.

The following function is increasing in $u\in[0,1]$

$\displaystyle\tilde{G}(u,r)=\frac{\mathbb{E}[\sigma^{\prime}(rx)\sigma(y)x]}{\mathbb{E}[\sigma(rx)\sigma(y)]}$
(158)

for any $r\geq 0$, where $x$ and $y$ are standard Gaussians with correlation
$\mathbb{E}[xy]=u$. Moreover, $\tilde{G}(u,r)<\tilde{G}(0,r)$ for $u<0$.

###### Proof.

Let us assume $0\leq u_{1}<u_{2}\leq 1$. For the first part of the statement,
we want to show

$\displaystyle\frac{\mathbb{E}[\sigma^{\prime}(rx)\sigma(y_{1})x]}{\mathbb{E}[\sigma(rx)\sigma(y_{1})]}<\frac{\mathbb{E}[\sigma^{\prime}(rx)\sigma(y_{2})x]}{\mathbb{E}[\sigma(rx)\sigma(y_{2})]}$
(159)

where $\mathbb{E}[xy_{1}]=u_{1}$ and $\mathbb{E}[xy_{2}]=u_{2}$. Changing the
measure from the standard Gaussian $p(x)$ to

$\displaystyle\tilde{p}(x)=p(x)\frac{\sigma(rx)\mathbb{E}[\sigma(y_{2})|x]}{\mathbb{E}[\sigma(rx)\sigma(y_{2})]},$

we get the following equivalent inequality

$\displaystyle\mathbb{E}\left[\frac{\sigma^{\prime}(rx)x}{\sigma(rx)}\frac{\mathbb{E}[\sigma(y_{1})|x]}{\mathbb{E}[\sigma(y_{2})|x]}\right]<\mathbb{E}\left[\frac{\sigma^{\prime}(rx)x}{\sigma(rx)}\right]\mathbb{E}\left[\frac{\mathbb{E}[\sigma(y_{1})|x]}{\mathbb{E}[\sigma(y_{2})|x]}\right].$
(160)

Thanks to the property (iv) of softplus (Lemma G.8), we have that
$\sigma^{\prime}(rx)x/\sigma(rx)$ is increasing in $x$ after a substitution
$x\leftarrow rx$ for $r>0$. For $r=0$, the function reduces to $\gamma x$ with
some $\gamma>0$, hence increasing. We will next show that (all integrations
are w.r.t $z$ hereafter, hence we drop the conditioning on $x$)

$\displaystyle\frac{\mathbb{E}[\sigma(y_{1})]}{\mathbb{E}[\sigma(y_{2})]}$
(161)

is decreasing in $x$. Computing the derivative w.r.t $x$, we want to show that
it is negative

$\displaystyle\frac{\mathbb{E}[\sigma^{\prime}(y_{1})]u_{1}}{\mathbb{E}[\sigma(y_{2})]}-\frac{\mathbb{E}[\sigma(y_{1})]\mathbb{E}[\sigma^{\prime}(y_{2})]u_{2}}{\mathbb{E}[\sigma(y_{2})]^{2}}<0\
\ \Leftrightarrow\ \
\frac{\mathbb{E}[\sigma^{\prime}(y_{1})]u_{1}}{\mathbb{E}[\sigma(y_{1})]}<\frac{\mathbb{E}[\sigma^{\prime}(y_{2})]u_{2}}{\mathbb{E}[\sigma(y_{2})]}.$
(162)

Note that this is equivalent to showing

$\displaystyle\frac{d}{du}\frac{\mathbb{E}[\sigma^{\prime}(y)]u}{\mathbb{E}[\sigma(y)]}=\frac{d^{2}}{dudx}\log(\mathbb{E}[\sigma(y)])>0$

for all $u\in[0,1)$ and $x\in\mathbb{R}$. Changing the order of derivatives,
it is sufficient to show

$\displaystyle\frac{d}{dx}\left(\frac{\mathbb{E}[\sigma^{\prime}(y)]x}{\mathbb{E}[\sigma(y)]}-\bigl{(}\frac{u}{1-u^{2}}\bigr{)}\frac{\mathbb{E}[u^{\prime}z\sigma^{\prime}(y)]}{\mathbb{E}[\sigma(y)]}\right)>0.$
(163)

The first function

$\displaystyle
s_{1}(x)=\frac{\mathbb{E}[\sigma^{\prime}(y)]x}{\mathbb{E}[\sigma(y)]}$ (164)

is shown to be increasing in $x$ in Lemma G.7 where we need to substitute
$x\to xu_{1}$ for $u_{1}>0$, and for $u_{1}=0$, we have $s_{1}(x)=\gamma x$
for some $\gamma>0$ hence it is increasing. The remaining part is to show that
the second function

$\displaystyle
s_{2}(x)=\frac{\mathbb{E}[u^{\prime}z\sigma^{\prime}(y)]}{\mathbb{E}[\sigma(y)]}$
(165)

is decreasing. We will consider $z\leftarrow u^{\prime}z$ and $x\leftarrow ux$
in what follows. We have

$\displaystyle\frac{d}{dx}\frac{\mathbb{E}[z\sigma^{\prime}(x+z)]}{\mathbb{E}[\sigma(x+z)]}<0\quad\Leftrightarrow\quad\frac{d}{dx}\frac{\mathbb{E}[\sigma(x+z)]}{\mathbb{E}[\sigma^{\prime\prime}(x+z)]}>0$

due to first applying Stein’s Lemma to the numerator and then inverting the
ratio. Using the chain rule, it is sufficient to show that

$\displaystyle
f_{1}(x)=\frac{\mathbb{E}[\sigma(x+z)]}{\mathbb{E}[\sigma^{\prime}(x+z)]},\quad\text{and}\quad
f_{2}(x)=\frac{\mathbb{E}[\sigma^{\prime}(x+z)]}{\mathbb{E}[\sigma^{\prime\prime}(x+z)]}$
(166)

are increasing, since both functions are positive.

Interestingly, $f_{1}$ is increasing in $x$ if $\sigma$ is a log-concave
function. Because its derivative is positive

$\displaystyle\frac{d}{dx}f_{1}(x)=1-\frac{\mathbb{E}[\sigma(x+z)]\mathbb{E}[\sigma^{\prime\prime}(x+z)]}{\mathbb{E}[\sigma^{\prime}(x+z)]^{2}}>0$

if $\mathbb{E}[\sigma(x+z)]$ is log-concave. This is the case since a centered
Gaussian distribution $p(z)$ is log-concave, therefore $\sigma(x+z)p(z)$ is
jointly log-concave, and marginalization preserves log-concavity.

Similarly, $f_{2}$ is increasing since $\sigma^{\prime}$ is also log-concave
due to property (v) of softplus (Lemma G.8). Hence we showed that

$\displaystyle
r(u)=\frac{\mathbb{E}[\sigma^{\prime}(y)]x}{\mathbb{E}[\sigma(y)]}$ (167)

is increasing for $u\in[0,1)$. The derivative of $r$ explodes at $1$, however,
we can conclude by contradiction that $r(1)>r(u)$ for $u<1$: if $r(u)\geq
r(1)$ for some $0\leq u<1$, then there exists $u_{0}\in(u,1)$ where the
function is decreasing. Therefore, $r$ is increasing for $u\in[0,1]$. We can
conclude the first part of the proof by the FKG inequality
$\nicefrac{{\sigma^{\prime}(rx)x}}{{\sigma(rx)}}$ and
$\nicefrac{{\mathbb{E}[\sigma(y_{1})]}}{{\mathbb{E}[\sigma(y_{2})]}}$ are
negatively correlated.

For the second part of the statement, we need to show

$\displaystyle\mathbb{E}[\sigma^{\prime}(rx)\sigma(ux+u^{\prime}z)x]\mathbb{E}[\sigma(rx)]<\mathbb{E}[\sigma^{\prime}(rx)x]\mathbb{E}[\sigma(rx)\sigma(ux+u^{\prime}z)]$
(168)

for $u<0$. Changing the measure from standard Gaussian $p(x)$ to

$\displaystyle\tilde{p}(x)=p(x)\frac{\sigma(rx)}{\mathbb{E}[\sigma(rx)]},$
(169)

the above inequality is equivalent to

$\displaystyle\mathbb{E}_{x\sim\tilde{p}}\left[\frac{\sigma^{\prime}(rx)x}{\sigma(rx)}\sigma(ux+u^{\prime}z)\right]<\mathbb{E}_{x\sim\tilde{p}}\left[\frac{\sigma^{\prime}(rx)x}{\sigma(rx)}\right]\mathbb{E}_{x\sim\tilde{p}}[\sigma(ux+u^{\prime}z)].$
(170)

This holds since $\sigma^{\prime}(rx)x/\sigma(rx)$ is increasing in $x$,
however, $\sigma(ux+u^{\prime}z)$ is decreasing in $x$ since $u$ is negative
which implies a negative correlation due to the FKG inequality. ∎

######  Lemma G.7.

The following function

$\frac{\mathbb{E}[\sigma^{\prime}(x+z)|x]x}{\mathbb{E}[\sigma(x+z)|x]}$

is increasing in $x$ where the integrations are w.r.t a centered Gaussian $z$.

###### Proof.

Since all integrals are w.r.t $z$, we drop the conditioning with respect to
$x$ in the proof. Taking the derivative w.r.t $x$, and arranging the terms, it
suffices to show

$\displaystyle\left(\frac{\mathbb{E}[\sigma^{\prime\prime}(x+z)]x}{\mathbb{E}[\sigma^{\prime}(x+z)]}+1\right)\mathbb{E}[\sigma(x+z)]>\mathbb{E}[\sigma^{\prime}(x+z)]x$
(171)

which is equivalent to the following due to the property
$\sigma^{\prime\prime}(z)=\beta\sigma^{\prime}(z)(1-\sigma^{\prime}(z))$

$\displaystyle\left(\beta
x\left(1-\frac{\mathbb{E}[\sigma^{\prime}(x+z)^{2}]}{\mathbb{E}[\sigma^{\prime}(x+z)]}\right)+1\right)\mathbb{E}[\sigma(x+z)]>\mathbb{E}[\sigma^{\prime}(x+z)]x.$
(172)

In the case $x\geq 0$, the LHS is bigger than $\mathbb{E}[\sigma(x+z)]$ since
$\sigma^{\prime}(\cdot)$ is upper bounded by $1$. Moreover, since
$\sigma(x)>x$ and from the convexity of softplus, we get
$\mathbb{E}[\sigma(x+z)]>x$. This yields the above inequality by again noting
that $\mathbb{E}[\sigma^{\prime}(x+z)]$ is upper bounded by $1$.

In the case $x<0$, we need another strategy. We have thanks to Cauchy-Schwartz

$\displaystyle\frac{\mathbb{E}[\sigma^{\prime}(x+z)^{2}]}{\mathbb{E}[\sigma^{\prime}(x+z)]}\geq\mathbb{E}[\sigma^{\prime}(x+z)],$
(173)

thus it suffices to show

$\displaystyle\left(\beta x-\beta
x\mathbb{E}[\sigma^{\prime}(x+z)]+1\right)\mathbb{E}[\sigma(x+z)]>\mathbb{E}[\sigma^{\prime}(x+z)]x.$
(174)

We will now show the following

$\displaystyle\mathbb{E}[\sigma^{\prime}(x+z)]\geq\sigma^{\prime}(x)$ (175)

for which it suffices to show that
$v(z):=\sigma^{\prime}(x+z)+\sigma^{\prime}(x-z)\geq 2\sigma^{\prime}(x)$ for
all $z$ since the centered Gaussian measure $p(z)$ is even and the integration
can be done over the integrand $v(z)$. We have

$\displaystyle
v^{\prime}(z)=\sigma^{\prime\prime}(x+z)-\sigma^{\prime\prime}(x-z)$ (176)

that is zero iff either $x+z=x-z$ or $x+z=-x+z$ where the latter is not
possible since $x<0$. Hence we get that a critical point of $v(z)$ at $z=0$
which is a minimizer since
$v^{\prime\prime}(0)=2\sigma^{\prime\prime\prime}(x)>0$ for $x<0$. Hence
$v(z)\geq v(0)=2\sigma^{\prime}(x)$ for all $z$ which completes the argument.

Finally, it remains to show

$\displaystyle\left(\frac{\beta x}{e^{\beta
x}+1}+1\right)\mathbb{E}[\sigma(x+z)]>\mathbb{E}[\sigma^{\prime}(x+z)]x.$
(177)

From the proof of Lemma G.8, we have that $\beta\sigma(x)>\sigma^{\prime}(x)$,
which in combination with the following trivial observation for all $x<0$
(note that $+1$ is not needed for the following to hold)

$\displaystyle\frac{\beta x}{e^{\beta x}+1}+1>\beta x$ (178)

shows that Eq. 177 holds, hence the proof is complete. ∎

######  Lemma G.8.

The softplus family has the following properties

  1. i.

$\sigma(x)$ is increasing and convex,

  2. ii.

$\sigma(x)$ is log-concave (equivalently, $\sigma(x)/\sigma^{\prime}(x)$ is
increasing),

  3. iii.

$\sigma(x)/\sigma^{\prime}(x)$ is convex,

  4. iv.

$\sigma^{\prime}(x)x/\sigma(x)$ is increasing,

  5. v.

$\sigma^{\prime}(x)$ is log-concave.

###### Proof.

For the property (i), see the formulas of $\sigma^{\prime}$ and
$\sigma^{\prime\prime}$ in the proof of Lemma F.2. We next prove each one of
the properties one after the other. Let us start with property (ii). First
note that $\sigma(x)$ is log-concave if and only if
$\sigma(x)/\sigma^{\prime}(x)$ is increasing since

$\displaystyle\frac{d}{dx}\frac{\sigma(x)}{\sigma^{\prime}(x)}=1-\frac{\sigma(x)\sigma^{\prime\prime}(x)}{\sigma^{\prime}(x)^{2}}>0\
\Leftrightarrow\ \sigma^{\prime}(x)^{2}>\sigma(x)\sigma^{\prime\prime}(x)$
(179)

where the second inequality is a characterization of log-concavity. We will
prove that $R(x):=\sigma(x)/\sigma^{\prime}(x)$ is increasing.

Let us write out the ratio explicitly

$\displaystyle R(x)=\frac{1}{\beta}\left(\log(e^{\beta
x}+1)+\frac{\log(e^{\beta x}+1)}{e^{\beta x}}\right).$ (180)

The first derivative of $R$ is given by

$\displaystyle R^{\prime}(x)$
$\displaystyle=\sigma^{\prime}(x)+\frac{\sigma^{\prime}(x)-\beta\sigma(x)}{e^{\beta
x}}=\frac{e^{\beta x}-\beta\sigma(x)}{e^{\beta x}}.$ (181)

Since $\log$ is concave, expanding it around $1$ we get $\log(y+1)<y$ for all
$y>0$. Substituting $y=e^{\beta x}$, we get that the numerator of $R^{\prime}$
is positive, thus $R$ is increasing. This completes the proof of property
(ii). Computing the second derivative of $R$, we get

$\displaystyle R^{\prime\prime}(x)$
$\displaystyle=\frac{\sigma^{\prime\prime}(x)(e^{\beta
x}+1)-2\beta\sigma^{\prime}(x)+\beta^{2}\sigma(x)}{e^{\beta
x}}=\beta\left(\frac{-\sigma^{\prime}(x)+\beta\sigma(x)}{e^{\beta x}}\right).$
(182)

What remains to show is that $\beta\sigma(x)>\sigma^{\prime}(x)$. Using the
fundamental theorem of calculus, we get

$\displaystyle\frac{\log(y+1)}{y}=\frac{1}{y}\int_{0}^{y}\frac{1}{t+1}dt>\frac{1}{y+1}$
(183)

since $1/(y+1)$ is a lower bound of the integrand which completes the proof of
the property (iii). Let us prove the property (iv) by taking the derivative of
the function of interest

$\displaystyle\frac{d}{dx}\frac{\sigma^{\prime}(x)x}{\sigma(x)}=\frac{(\sigma^{\prime\prime}(x)x+\sigma^{\prime}(x))\sigma(x)-\sigma^{\prime}(x)^{2}x}{\sigma(x)^{2}}$
(184)

Using $\sigma^{\prime\prime}(x)=\beta\sigma^{\prime}(x)(1-\sigma^{\prime}(x))$
and dropping the positive term $\sigma^{\prime}(x)$, the numerator of the
derivative is

$\displaystyle((1-\sigma^{\prime}(x))\beta x+1)\sigma(x)-\sigma^{\prime}(x)x$
$\displaystyle=\left(\frac{\beta x}{e^{\beta
x}+1}+1\right)\sigma(x)-\frac{e^{\beta x}}{e^{\beta x}+1}x$ (185)
$\displaystyle=\frac{e^{\beta x}}{e^{\beta x}+1}\left(\frac{1}{e^{\beta
x}}(\beta x+e^{\beta x}+1)\sigma(x)-x\right)$ (186)

For the case $x\geq 0$, we have $\sigma(x)>x$ and $(\beta x+1)/e^{\beta x}>0$,
hence the derivative is positive. For the case $x<0$, we want to show

$\displaystyle\left(e^{\beta x}+\beta x+1\right)\frac{\log(e^{\beta
x}+1)}{e^{\beta x}}>\beta x.$ (187)

If $e^{\beta x}+\beta x+1>0$, it is done since the LHS is positive. If
$e^{\beta x}+\beta x+1\leq 0$, we have

$\displaystyle\left(e^{\beta x}+\beta x+1\right)\frac{\log(e^{\beta
x}+1)}{e^{\beta x}}\geq\left(e^{\beta x}+\beta
x+1\right)\sup\frac{\log(e^{\beta x}+1)}{e^{\beta x}}$ (188)

since $\log(e^{\beta x}+1)/e^{\beta x}$ is positive. We will next show that
$\log(e^{\beta x}+1)/e^{\beta x}$ is a decreasing function therefore its
supremum is achieved at $x\\!\to\\!-\infty$. From the integral expression in
Eq. 183, we deduce that $\log(y+1)/y$ is a decreasing function since adding
smaller terms in the average decreases it. Thus the following limit gives us
the supremum using L’Hôpital’s rule

$\displaystyle\lim_{y\to 0}\frac{\log(y+1)}{y}=\lim_{y\to 0}\frac{1}{y+1}=1.$
(189)

Combining it with the Eq. 188 after the substitution $y=e^{\beta x}$, we get
the desired Ineq. 187 which implies that the derivative is positive in this
case too. This completes the proof of property (iv).

For the property (v), we first give a formula for the third derivative of
softplus

$\displaystyle\sigma^{\prime\prime\prime}(x)=\beta\sigma^{\prime\prime}(x)(1-2\sigma^{\prime}(x)).$
(190)

$\sigma^{\prime}$ is log-concave if and only if we have

$\displaystyle\sigma^{\prime\prime\prime}(x)\sigma(x)<\sigma^{\prime\prime}(x)\sigma^{\prime}(x)\Leftrightarrow$
$\displaystyle\beta\sigma^{\prime\prime}(x)(1-2\sigma^{\prime}(x))\sigma(x)<\sigma^{\prime\prime}(x)\sigma^{\prime}(x)$
(191)

which is equivalent to

$\displaystyle(1-e^{\beta x})\log(e^{\beta x}+1)<e^{\beta x}.$ (192)

This is equivalent to $(1-y)\log(y+1)<\log(y+1)<y$ where $y=e^{\beta x}>0$;
the second inequality holds due to $y+1<e^{y}$.

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