# Projective Geometry for Perfectoid Spaces

Gabriel Dorfsman-Hopkins

###### Abstract

To understand the structure of an algebraic variety we often embed it in
various projective spaces. This develops the notion of projective geometry
which has been an invaluable tool in algebraic geometry. We develop a
perfectoid analog of projective geometry, and explore how equipping a
perfectoid space with a map to a certain analog of projective space can be a
powerful tool to understand its geometric and arithmetic structure. In
particular, we show that maps from a perfectoid space $X$ to the perfectoid
analog of projective space correspond to line bundles on $X$ together with
some extra data, reflecting the classical theory. Along the way we give a
complete classification of vector bundles on the perfectoid unit disk, and
compute the Picard group of the perfectoid analog of projective space.

##  1 Introduction

This paper is inspired by the goal of understanding vector bundles on
perfectoid spaces, and how they behave under the so called _tilting
correspondence_ of Scholze [16]. To do so, we develop a perfectoid analog of
projective geometry. We study the perfectoid analog of projective space
defined in [16], which we call _projectivoid space_ and denote by
$\mathbb{P}^{n,\mathrm{perf}}$, and show that maps from a perfectoid space $X$
to $\mathbb{P}^{n,\mathrm{perf}}$ correspond to line bundles on $X$ together
with some extra data, giving an analog to the classical theory of maps to
projective space.

To get to this point we must first understand the theory of line bundles on
projectivoid space itself, and in particular, its Picard group. In his
dissertation [4], Das worked toward computing the Picard group of the
projectivoid line, $\mathbb{P}^{1,\mathrm{perf}}$. His proof relied on having
certain local trivializations of line bundles, requiring a perfectoid analog
of the Quillen-Suslin theorem. Therefore, in order to begin developing the
theory of so called projectivoid geometry, we must prove this first.

The Quillen-Suslin theorem says that finite dimensional vector bundles on
affine $n$-space over a field are all trivial. Equivalently, all finite
projective modules on a polynomial ring $K[T]$ are free, where $T$ is an
$n$-tuple of indeterminates. In rigid analytic geometry, we replace polynomial
rings with rings of convergent power series called _Tate algebras_ , denoted
$K\langle T\rangle$, and it can be shown that over such rings the Quillen-
Suslin theorem still holds, that is, all finite projective $K\langle
T\rangle$-modules are free [14]. The analog of these rings for perfectoid
spaces is the ring $K\langle T^{1/p^{\infty}}\rangle$ of convergent power
series where the indeterminates have all their $p$th power roots. The
difficulty in extending the theorem to this _perfectoid Tate algebra_ is that
the ring is no longer noetherian, and so the result cannot be easily reduced
to the polynomial case.

In Section 2 we study the commutative ring theoretic properties of the
perfectoid Tate algebra, computing its unit group and classifying its
projective modules, as well those over its integral subring and residue ring.
In particular, we establish the necessary Quillen-Suslin style result.

######  NameIgnored (Theorem 2.18 (The Quillen-Suslin Theorem for the
Perfectoid Tate Algebra))

Finite projective modules on the perfectoid Tate algebra $K\left\langle
T_{1}^{1/p^{\infty}},\cdots,T_{n}^{1/p^{\infty}}\right\rangle$ are all free.
Equivalently, finite dimensional vector bundles on the perfectoid unit disk
are all isomorphic to the trivial vector bundle.

This completes Das’ proof, and lays the groundwork to begin studying vector
bundles on more general perfectoid spaces.

In Section 3 we develop the theory of line bundles on projectivoid space,
extending Das’ result for $n=1$.

######  NameIgnored (Theorem 3.4 (The Picard Group of Projectivoid Space))

$\operatorname{Pic}\mathbb{P}^{n,\mathrm{perf}}\cong\mathbb{Z}[1/p]$.

We also compute the cohomology of all line bundles on projectivoid space,
completing a computation of Bedi [2] for $n=2$.

In Section 4 we compute the functor of points of projectivoid space, showing
that (much like in the classical theory) it is deeply connected to the theory
of line bundles on perfectoid spaces.

######  NameIgnored (Theorem 4.6 (The Functor of Points of Projectivoid
Space))

Let $X$ be a perfectoid space over a field $K$. Morphisms
$X\to\mathbb{P}^{n,\mathrm{perf}}$ correspond to tuples
$\left(\mathscr{L}_{i},s^{(i)}_{j},\varphi_{i}\right)$, where
$\mathscr{L}_{i}\in\operatorname{Pic}X$,
$\left\\{s^{(i)}_{0},\cdots,s^{(i)}_{n}\right\\}$ are $n+1$ global sections of
$\mathscr{L}_{i}$ which generate $\mathscr{L}_{i}$, and
$\varphi_{i}:\mathscr{L}_{i+1}^{\otimes
p}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathscr{L}_{i}$ are
isomorphisms under which $\left(s^{(i+1)}_{j}\right)^{\otimes p}\mapsto
s^{(i)}_{j}$.

We also provide refinements of this theorem in characteristic $p$ and see how
it behaves under the tilting equivalence of Scholze.

In Section 5 we test out this new theory, using it to compare the Picard
groups of a perfectoid space $X$ and its tilt $X^{\flat}$. In particular,
since the tilting equivalence builds a correspondence between maps
$X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$ and maps
$X^{\flat}\to\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}$, we can chain this
together with the correspondence of line bundles and maps to projectivoid
space to compare line bundles on $X$ and $X^{\flat}$. The main result follows.

######  NameIgnored (Theorem 5.13)

Suppose $X$ is a perfectoid space over $K$. Suppose that $X$ has a weakly
ample line bundle (cf. Definition 5.12) and that
$H^{0}(X_{\overline{K}},\mathscr{O}_{X_{\overline{K}}})=\overline{K}$. Then
there is a natural injection

$\theta:\operatorname{Pic}X^{\flat}\hookrightarrow\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{\mathscr{L}\mapsto\mathscr{L}^{p}}\operatorname{Pic}X.$

In particular, if $\operatorname{Pic}X$ has no $p$ torsion, then composing
with projection onto the first coordinate gives an injection

$\theta_{0}:\operatorname{Pic}X^{\flat}\hookrightarrow\operatorname{Pic}X.$

This paper relies heavily Huber’s theory of adic spaces developed in [10] and
[9]. We do not develop the theory here and instead refer the reader to Huber’s
original papers, or [22] for an excellent summary. A summary of the theory and
with an emphasis on perfectoid spaces can be found the author’s doctoral
dissertation ([5] Sections 2-4), or in Kedlaya’s detailed notes from the 2017
Arizona Winter School on the subject [12].

###  1.1 Notational Conventions

Throughout the paper we will fix a perfectoid field $K$ with a topologically
nilpotent unit $\varpi$ called a _pseudouniformizer_. Its valuation ring will
be denoted by $K^{\circ}$, whose maximal ideal is $K^{\circ\circ}$. The
residue field will be denoted by $k$.

If $R$ is a complete topological $K$-algebra we denote its subring of power
bounded elements by $R^{\circ}$, the ideal of topologically nilpotent elements
by $R^{\circ\circ}$, and the residue ring by $\tilde{R}$. If $R$ is perfectoid
then we denote its _tilt_ by $R^{\flat}$. We denote by $\sharp:R^{\flat}\to R$
the Teichmüller map which is a map of multipicative monoids coming from
composing projection onto the first coordinate with the isomorphism (of
monoids)
$R^{\flat}\cong\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{x\mapsto
x^{p}}R$.

###  1.2 Acknowledgments

The author thanks their doctoral advisor Max Lieblich for support while
pursuing this work during graduate study, and Kiran Kedlaya for his generosity
of ideas while serving as host for a term at the University of California in
San Diego to work on this project. The author also thanks Peter Scholze for
important comments and corrections to a previous draft. Thanks also to
Bharghav Bhatt, Daniel Bragg, Jacopo Brivio, Charles Godfrey, Tuomas Tajakka,
Lucas Van Meter, Anwesh Ray, Manar Riman, Travis Scholl, Alex Voet, Peter Wear
for helpful conversations. Much of this work was completed while the author
was supported under NSF grant DMS-1600813 during a semester visit to the
University of California, San Diego, and under NSF grant DMS-1439786 while the
author was in residence at the Institute for Computational and Experimental
Research in Mathematics in Providence, RI.

##  2 The Perfectoid Tate Algebra

Algebraic geometry studies the polynomial ring and it’s various quotients and
localizations, allowing for commutative algebraic facts to be interpreted
geometrically and vice-versa. In the world of perfectoid algebras, a natural
analog is the perfectoid Tate algebra (defined below), whose commutative
algebra controls many of the geometric structures we study. We therefore begin
with a careful study of the algebraic object underlying most of this work.

Let $K$ be a perfectoid field with pseudouniformizer $\varpi$. In particular
$K$ is a nonarchimedean field, and so we can define the _Tate algebra_
$T_{n,K}=K\langle X_{1},\cdots,X_{n}\rangle$ of convergent power series over
$K$. The module theory of the Tate algebra is well understood, in no small
part due to Lütkebohmert, see [14]. Our main object of study in this section
is the following.

######  Definition 2.1 (The Perfectoid Tate Algebra).

The _perfectoid Tate algebra_ $T_{n,K}^{\mathrm{perf}}$ is the nonarchimedean
completion

$T_{n,K}^{\mathrm{perf}}=K\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{n}^{1/p^{\infty}}\right\rangle=\widehat{\bigcup_{n\geq
0}K\left\langle
X_{1}^{1/p^{n}},\cdots,X_{n}^{1/p^{n}}\right\rangle}\cong\widehat{\operatornamewithlimits{\underset{\longrightarrow}{lim}}T_{n,K}}.$

$\clubsuit$

###### Remark 2.2

The perfectoid Tate algebra consists of formal power series over $K$ which
converge on the perfectoid unit disk. Explicitely $X=(X_{1},\cdots,X_{n})$ be
an $n$-tuple, we can write down the elements of this ring.

$T_{n,K}^{\mathrm{perf}}=\left\\{\sum_{\alpha\in\left(\mathbb{Z}[1/p]_{\geq}0\right)^{n}}a_{\alpha}X^{\alpha}:\text{
for all }\lambda\in\mathbb{R}_{>0}\text{ only finitely many
}|a_{\alpha}|\geq\lambda\right\\}.$

This ring inherits the Gauss norm,
$||\sum_{\alpha}a_{\alpha}X^{\alpha}||=\sup\\{|a_{\alpha}|\\}$.

$\left(T_{n,K}^{\mathrm{perf}}\right)^{\circ}$ is the subring $\\{||f||\leq
1\\}$ of power-bounded elements of $T_{n,K}^{\mathrm{perf}}$, and consists of
power series with coefficients in $K^{\circ}$. The ideal
$\left(T_{n,K}^{\mathrm{perf}}\right)^{\circ\circ}$ of topologically nilpotent
elements consist of power series with coefficients in $K^{\circ\circ}$. The
quotient is

$\tilde{T}_{n,K}^{\mathrm{perf}}:=\left(T_{n,K}^{\mathrm{perf}}\right)^{\circ}/\left(T_{n,K}^{\mathrm{perf}}\right)^{\circ\circ}=k\left[X_{1}^{1/p^{\infty}},\cdots,X_{n}^{1/p^{\infty}}\right]$

where $k=K^{\circ}/K^{\circ\circ}$ is the residue field. Notice that every
element in the quotient ring is a polynomial, because a power series in
$\left(T_{n,K}^{\mathrm{perf}}\right)^{\circ}$ can only have finitely many
coefficients of norm 1. $\ast$

When there will be no confusion, we omit $K$ from the notation.

######  Remark 2.3 (A Note on Convergence)

Suppose $f=\sum f_{\alpha}X^{\alpha}\in T_{n}^{\mathrm{perf}}$. Morally
speaking, saying that the sum converges should mean that evaluating $f$ at any
point in the perfectoid disk should give an element of $K$. Since are sums are
not taken over $\mathbb{Z}_{\geq 0}^{n}$, but rather $(\mathbb{Z}[1/p]_{\geq
0})^{n}$, we must be more careful in defining what convergence means. Let us
begin by studying $f(1,1,\cdots,1)=\sum f_{\alpha}$. This should converge, so
let’s begin by defining partial sums

$s_{m}=\sum_{\alpha\in\left(\frac{\mathbb{Z}}{p^{m}}\right)^{n}:0\leq\alpha_{i}\leq
m}f_{\alpha}.$

If the sequence $(s_{m})$ converges, we define the infinite sum to be the
limit. Let us check that the convergence of the power series $f$ implies
convergence of $\sum f_{\alpha}$ in this sense. Fixing some $\varepsilon>0$,
there are only finitely many $f_{\alpha}$ with $|f_{\alpha}|\geq\varepsilon$.
Therefore, there is some large $N$ such that for each such $f_{\alpha}$ we
have
$\alpha=(\alpha_{1},\cdots,\alpha_{n})\in\left(\frac{\mathbb{Z}}{p^{N}}\right)^{n}$
$0<\alpha_{i}<N$. Therefore, fixing $m\geq r>N$, the differences $s_{m}-s_{r}$
have none of the coefficients $f_{\alpha}$ with absolute value larger than
$\varepsilon$, so that by the nonarchimedean property
$|s_{m}-s_{r}|<\varepsilon$. Thus the sum converges to an element $f(1)\in K$.

We remark now that if $|g_{\alpha}|\leq 1$, the same argument would show that
$\sum f_{\alpha}g_{\alpha}$ also converges. This should imply that $f(x)$
converges to a point in $K$ whenever $x$ is in the perfectoid unit disk, but
we defer further discussion until after we have the relevant definitions (see
Remark 2.10). $\ast$

We record a useful normalization trick for further use down the line.

######  Lemma 2.4 (Normalization).

Let $f\in T_{n}^{\mathrm{perf}}$ be nonzero. There is some $\lambda\in K$ such
that $||\lambda f||=1$. $\heartsuit$

###### Proof.

Since only finitely many coefficients in $f$ have absolute value above
$||f||-\varepsilon$, the supremum of that absolute values of the coefficients
is achieved by some $f_{\alpha}$. Taking $\lambda=f_{\alpha}^{-1}$ completes
the proof.

###  2.1 The Group of Units

As a first step towards understanding the perfectoid Tate algebra, we compute
its group of units.

######  Proposition 2.5.

Let $f\in T_{n}^{\mathrm{perf}}$ with $||f||=1$. The following are equivalent:

  1. (i)

$f$ is a unit in $\left(T_{n}^{\mathrm{perf}}\right)$.

  2. (ii)

$f$ is a unit in $\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$.

  3. (iii)

The image of $\overline{f}$ of $f$ in $\tilde{T}_{n}^{\mathrm{perf}}$ is a
nonzero constant $\lambda\in k^{\times}$.

  4. (iv)

$|f(0)|=1$ and $||f-f(0)||<1$.

$\heartsuit$

###### Proof.

(i)$\iff$(ii). An inverse to $f$ must have absolute value 1, and therefore
would also lie in $\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$.

(ii)$\implies$(iii). The map
$\left(T_{n}^{\mathrm{perf}}\right)^{\circ}\to\tilde{T}_{n}^{\mathrm{perf}}$
must send units to units, and the group of units of
$\tilde{T}_{n}^{\mathrm{perf}}$ is precisely the nonzero constant polynomials.
Indeed, the inverse to any element of $\tilde{T}_{n}^{\mathrm{perf}}$ would
also have to be a polynomial (in $X^{1/p^{m}}$ for some $m$), implying that
they both must be constants.

(iii)$\iff$(iv). This is immediate.

(iv)$\implies$(i). If $|f(0)|=1$ then $f(0)\in
K^{\times}\subseteq(T_{n}^{\mathrm{perf}})^{\times}$. Therefore
$1-\frac{f}{f(0)}\in T_{n}^{\mathrm{perf}}$ and

$\left|\left|1-\frac{f}{f(0)}\right|\right|=||f(0)||\cdot\left|\left|1-\frac{f}{f(0)}\right|\right|=||f(0)-f||<1.$

Therefore $1-\frac{f}{f(0)}$ is topologically nilpotent, so that by the
geometric series converges and hence $\frac{f}{f(0)}$ is a unit. Since $f(0)$
is too, we can conclude that $f$ is a unit.

######  Corollary 2.6.

$f=\sum f_{\alpha}X^{\alpha}\in\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$ is
a unit if and only if $|f_{0}|=1$ and $|f_{\alpha}|<1$ for all $\alpha\not=0$.
$\spadesuit$

######  Corollary 2.7.

$f=\sum f_{\alpha}X^{\alpha}\in T_{n}^{\mathrm{perf}}$ is a unit if and only
if $|f_{\alpha}|<|f_{0}|$ for all $\alpha\not=0$. $\spadesuit$

###### Proof.

Using our normalization trick, we know $||\lambda f||=1$ for some $\lambda\in
K^{\times}$. Then $f$ is a unit if and only if $\lambda f$ is, if and only if
$|\lambda f_{\alpha}|<1=|\lambda f_{0}|$ for all $\alpha\not=0$. Canceling
shows this holds if and only if $|f_{\alpha}|<|f_{0}|$ for all $\alpha\not=0$.

###  2.2 Vector Bundles on the Perfectoid Unit Disk

In classical algebraic geometry the polynomial ring (and it’s various
quotients) form the local building blocks of most of the objects of study.
Phrased geometrically, the prime spectrum of the polynomial ring of $n$
variables is affine space $\mathbb{A}^{n}$ which covers or contains many of
the spaces of interest. Like schemes which are locally prime spectra of rings,
perfectoid spaces are build from perfectoid algebras using the adic spectrum
functor of Huber (see [10],[9], and [11]). The perfectoid Tate algebra plays
the role of the polynomial ring, and the perfectoid unit disk (defined below)
plays the role of affine space.

######  Definition 2.8.

The _perfectoid unit disk_ is the adic space associated to the perfectoid Tate
algebra:

$\mathbb{D}^{n,\mathrm{perf}}=\operatorname{Spa}\left(T_{n}^{\mathrm{perf}},(T_{n}^{\mathrm{perf}})^{\circ}\right).$

$\clubsuit$

###### Remark 2.9

The rigid unit disk is the adic spectrum associated to the Tate algebra. Since
the perfectoid Tate algebra is the completed union of Tate algebras, using the
_tilde limit_ formalism of [18], we have

$\mathbb{D}^{n,\mathrm{perf}}\sim\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{\varphi}\mathbb{D}^{n,\mathrm{ad}},$

where $\varphi$ is the $p$th power map on coordinates. It is worth noting that
the tilde limit is not the categorical inverse limit (since these are not in
general unique in the category of adic spaces). Nevertheless, it should be
thought of affinoid locally as corresponding to the completed directed limit,
and if such a limit exists as a perfectoid space, it is unique among all
perfectoid spaces and satisfies the usual universal property (among perfectoid
spaces). See [18] Definition 2.4.1 and subsequent discussion. $\ast$

######  Remark 2.10 ($K$-points of the Perfectoid Disk)

Let us compute the $K$-points of $\mathbb{D}^{n,\mathrm{perf}}_{K}$, and in
doing so conclude the discussion of Remark 2.3. A $K$ point $x$ is a map of
adic spaces
$x:\operatorname{Spa}(K,K^{\circ})\to\mathbb{D}^{n,\mathrm{perf}}_{K}$, which
is equivalent to a map of Huber pairs
$\varepsilon_{x}:\left(T_{K}^{n,\mathrm{perf}},(T_{K}^{n,\mathrm{perf}})^{\circ}\right)\to(K,K^{\circ})$.
This is a commutative diagram of continuous $K$-algebra homomorphisms:

${K\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{n}^{1/p^{\infty}}\right\rangle}$${K}$${K^{\circ}\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{n}^{1/p^{\infty}}\right\rangle}$${K^{\circ}.}$

In particular, it is determined by the image of the $X_{i}^{1/p^{k}}$ in
$K^{\circ}$. Let $\lambda=\varepsilon_{x}(X_{i})$. Then
$\lambda_{1}=\varepsilon_{x}({X_{i}^{1/p})}$ must be a $p$th root of
$\lambda$, and $\lambda_{2}=\varepsilon_{x}(X_{i}^{1/p^{2}})$ must be a $p$th
root of $\lambda_{1}$. Continuing in this fashion, we see that choosing the
image of the $X_{i}^{1/p^{k}}$ as $k$ varies is equivalent to fixing an
element of $\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{t\mapsto
t^{p}}K^{\circ}=K^{\flat\circ}$. We have therefore computed the $K$-points of
the perfectoid disk:

$\mathbb{D}^{n,\mathrm{perf}}_{K}(K)=(K^{\flat\circ})^{n}.$

The evaluation function $\varepsilon_{x}(f)=f(x)$ then amounts to plugging the
coordinates of $x$ as an element of $(K^{\flat\circ})^{n}$ into $f$, which we
saw in Remark 2.3 converges to a point in $K$. $\ast$

There is a well known correspondence between finite projective modules over a
ring, and finite dimensional (algebraic) vector bundles over the associated
affine scheme, and more generally, between vector bundles over a locally
ringed space and locally free sheaves on that space (see, for example, [8]
exercise 2.5.18). In [19], Serre conjectured that all finite projective
modules over the polynomial ring $A=k[x_{1},\cdots,x_{n}]$ are free. This can
be interpreted geometrically as saying there are no nontrivial algebraic
vector bundles over affine space $\mathbb{A}^{n}=\operatorname{Spec}A$. In
1976, Quillen [15] and independently Suslin proved Serre’s conjecture, which
is now known as the Quillen-Suslin theorem. Lütkebohmert in [14] was shortly
able to extend the result to the Tate algebra $K\langle
X_{1},\cdots,X_{n}\rangle$ of convergent power series over a complete
nonarchimedean field.

In what follows, we prove a perfectoid analog of the Quillen-Suslin theorem.
Specifically, we prove that all finite projective modules on the perfectoid
Tate algebra are trivial. This will imply that the perfectoid unit disk has no
nontrivial finite vector bundles. Along the way we will show that both the
subring of integral elements $\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$, and
the residue ring $\tilde{T}_{n}^{\mathrm{perf}}$ also have no nontrivial
finite projective modules. Although these results are not necessary to
establish the result for the perfectoid Tate algebra, they will be important
in asserting the effectiveness of the Čech cohomology groups of certain
sheaves in Section 3.

####  2.2.1 Finite Projective Modules on the Residue Ring

Let us begin by proving that finite projective modules are free over the
residue ring

$\tilde{T}_{n}^{\mathrm{perf}}=k\left[X_{1}^{1/p^{\infty}},\cdots,X_{n}^{1/p^{\infty}}\right]=\bigcup_{m}k\left[X_{1}^{1/p^{m}},\cdots,X_{n}^{1/p^{m}}\right].$

To see this, we first briefly review a (non unique) correspondence between
projective modules and idempotent matrices.

Let $R$ be a commutative ring, and fix a finite projective $R$ module $P$.
Consider a presentation $\pi:R^{n}\to P$, as well as a section of this
projection $\sigma$. The composition $\sigma\circ\pi$ produces an idempotent
matrix $U\in M_{n}(R)$.

${R^{n}}$${R^{n}}$${R^{n}}$${P}$${P}$$\scriptstyle{\pi}$$\scriptstyle{U}$$\scriptstyle{U}$$\scriptstyle{\pi}$$\scriptstyle{U}$$\scriptstyle{\sigma}$$\scriptstyle{\sigma}$

Conversely, the image of an idempotent matrix $U$ is always projective, with
the section just given by the natural inclusion $\operatorname{im}U\subseteq
R^{n}$. In this way, we get a (non unique) correspondence between finite
projective modules and idempotent matrices over $R$.

######  Lemma 2.11.

Suppose $R=\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}R_{i}$
is a filtered colimit of commutative rings. Then every finite projective $R$
module is the base extension of a finite projective $R_{i}$-module.
$\heartsuit$

###### Proof.

To a finite projective $R$ module $M$ we may associate a projector matrix $U$.
Each entry in the matrix is defined over some $R_{i}$, and as the limit is
filtered, $U$ is defined over $R_{i}$ for some (perhaps larger) $i$. Its image
as a map from $R_{i}^{n}$ to itself is therefore a projective $R_{i}$-module
whose base extension is $M$.

######  Corollary 2.12.

Let $R=k\left[X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right].$ Every $R$-module is
free. In particular (letting $r=n$), every
$\tilde{T}_{n}^{\mathrm{perf}}$-module is free. $\spadesuit$

###### Proof.

As $R$ is the filtered colimit of (Laurent) polynomial rings, Lemma 2.11
implies that a finite projective $R$-module $M$ is the base extension of some
$N$ over a (Laurent) polynomial ring. By the Quillen-Suslin theorem, $N$ is
free, so $M$ is too.

####  2.2.2 Finite Projective Modules on the Subring of Integral Elements

We extend Corolary 2.12 to the subring of power-bounded elements of the
perfectoid Tate algebra, $\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$, using
Nakayama’s lemma. We first fix some notation.

###### Notation 2.13

For a commutative ring $R$ and an ideal $I$ contained in the Jacobson radical
of $R$ we let $R_{0}=R/I$. For an $R$ module $M$ we will denote by $M_{0}$ the
$R_{0}$ module $M/IM$, and for a homomorphism $\phi$ of $R$-modules we denote
by $\phi_{0}$ its reduction mod $I$. If $m\in M$, then we denote by
$\overline{m}$ its image in $M_{0}$. $\ast$

######  Lemma 2.14.

Let $R$ be a commutative ring, and $I$ an ideal contained in the Jacobson
radical of $R$. If $M$ and $N$ are two projective $R$-modules such that there
exists an isomorphism $\phi:M_{0}\stackrel{{\scriptstyle\sim}}{{\to}}N_{0}$,
then $\phi$ lifts to an isomorphism
$\psi:M\stackrel{{\scriptstyle\sim}}{{\to}}N$. $\heartsuit$

###### Proof.

We have the following commutative diagram.

${M}$${N}$${M_{0}}$${N_{0}}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$$\scriptstyle{\rho}$$\scriptstyle{\phi}$

Indeed, a lift $\psi$ exists because $M$ is projective. Notice that
$\psi_{0}=\phi$. Indeed,

$\displaystyle\psi_{0}(\overline{m})$ $\displaystyle=$
$\displaystyle\overline{\psi(m)}$ $\displaystyle=$ $\displaystyle\rho\psi(m)$
$\displaystyle=$ $\displaystyle\phi\pi(m)$ $\displaystyle=$
$\displaystyle\phi(\overline{m}).$

Since $\phi$ surjects, so does $\psi$ by Nakayama’s lemma. Since $N$ is
projective, $\psi$ has a section $\sigma$ which is necessarily injective. We
claim that $\sigma_{0}=\phi^{-1}$. Indeed, we can check this after applying
$\phi$.

$\displaystyle\phi\sigma_{0}(\overline{n})$ $\displaystyle=$
$\displaystyle\phi\pi\sigma(n)$ $\displaystyle=$
$\displaystyle\rho\psi\sigma(n)$ $\displaystyle=$ $\displaystyle\rho(n)$
$\displaystyle=$ $\displaystyle\overline{n}.$

Therefore $\sigma_{0}$ surjects, so that $\sigma$ surjects by Nakayama’s
lemma. Thus $\sigma$ is an isomorphism.

######  Corollary 2.15.

With the same setup as Lemma 2.14, we let $P$ be a projective $R$-module. If
$P_{0}$ is a free $R_{0}$ module, then $P$ is free. $\spadesuit$

###### Proof.

Suppose $P_{0}\cong R_{0}^{m}$. We also have $(R^{m})_{0}\cong R_{0}^{m}$ so
that by Lemma 2.14, $P\cong R^{m}$.

We now apply this to the case at hand.

######  Lemma 2.16.

Let $R$ be one of the following:

  * •

$\left(T_{n}^{\mathrm{perf}}\right)^{\circ}$ or one of its associated Laurent
series rings given by inverting a coordinate function.

  * •

Either of the previous two rings modulo a positive power of $\varpi$.

Explicitly (and up to a rearranging of coordinates),

$R=K^{\circ}\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right\rangle,$

or

$R=K^{\circ}/\varpi^{d}\left[X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right].$

The ideal of topologically nilpotent (resp. nilpotent) elements lies in the
Jacobson radical of $R$. $\heartsuit$

###### Proof.

If $f$ is (topologically) nilpotent then so is $fg$ for all $g$. Thus the
geometric series for $\frac{1}{1-fg}$ converges to an inverse of $1-fg$, so
that it is a unit. Since $g$ was arbitrary, this shows that $f$ is in the
Jacobson radical.

The desired result follows.

######  Corollary 2.17.

Let $R$ as in Lemma 2.16. Every finite projective $R$-module is free.
$\spadesuit$

###### Proof.

Notice that $R_{0}$, the reduction of $R$ modulo the ideal of (topologically)
nilpotent elements is the ring of Corollary 2.12. Let $P$ be a finite
projective $R$ module. Then $P_{0}$ is a finite projective $R_{0}$ module and
therefore is free. Therefore by Corollary 2.15 it suffices to show that the
kernel of the reduction map is contained in the Jacobson radical, but this is
Lemma 2.16.

####  2.2.3 The Quillen-Suslin Theorem for the Perfectoid Tate Algebra

We now prove the main result of this section.

######  Theorem 2.18.

Finite projective modules on the perfectoid Tate algebra are free.
Equivalently, all finite vector bundles on the perfectoid unit disk are free.
$\diamondsuit$

In fact, this will follow from something slightly more general.

######  Theorem 2.19.

Let $R$ be a perfectoid Laurent series algebra:

$R=K\left\langle X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right\rangle.$

Then every finite projective $R$ module is free. $\diamondsuit$

A result of Gabber and Romero will do most of the heavy lifting, and state the
result here:

######  Proposition 2.20 ([6] Corollary 5.4.42).

Let $R$ be a commutative ring, $t\in R$ a nonzero divisor, and $I\subset R$ an
ideal. Let $\hat{R}$ be the $tI$-adic completion of $R$, and suppose $(R,tI)$
form a Henselian pair. Then the base extension functor
$R[t^{-1}]-\textbf{Mod}\to\hat{R}[t^{-1}]-\textbf{Mod}$ induces a bijection
between isomorphism classes of finite projective $R[t^{-1}]$-modules and
finite projective $\hat{R}[t^{-1}]$-modules. $\heartsuit$

Let us describe the objects we will feed into this result. Playing the role of
$R$ is:

$R=\bigcup_{i}K^{\circ}\left\langle
X_{1}^{1/p^{i}},\cdots,X_{r}^{1/p^{i}},X_{r+1}^{\pm 1/p^{i}},\cdots,X_{n}^{\pm
1/p^{i}}\right\rangle=:\bigcup_{i}R_{i}.$ (1)

Playing the role of $t$ we have $\varpi$, and $I$ will be the unit ideal.
Therefore

$R[1/\varpi]=\bigcup_{i}K\left\langle
X_{1}^{1/p^{i}},\cdots,X_{r}^{1/p^{i}},X_{r+1}^{\pm 1/p^{i}},\cdots,X_{n}^{\pm
1/p^{i}}\right\rangle.$

Then $\hat{R}$ is the ring of integral elements (cf. Lemma 2.16), and
$\hat{R}[1/\varpi]$ is the perfectoid Tate algebra (or Laurent series
algebra), for which we are trying to prove the Quillen-Suslin result. Theorem
2.19 will follow from the following two lemmas.

######  Lemma 2.21.

Finite projective modules on $R[1/\varpi]$ are all free. $\heartsuit$

###### Proof.

Applying Lemma 2.11, a finite projective $R[1/\varpi]$-module $M$ is the base
extention of a finite projective module $N$ over a rigid analytic Tate algebra
(or Laurent series algebra). But $N$ is free due to the rigid analytic
Quillen-Suslin theorem ([14] Satz 1), so we are done.

######  Lemma 2.22.

The pair $(R,(\varpi))$ is Henselian. $\heartsuit$

###### Proof.

Lemma 2.16 shows that $\varpi$ is in the Jacobson radical of $R$. Suppose
$f\in R$ is monic, and that $\overline{f}=g_{0}h_{0}\in R/\varpi$ with
$g_{0},h_{0}$ monic. In fact, $f\in R_{i}$ for some $i$ (using the notation of
Equation 1 above). Perhaps increasing $i$, we can take the factorization of
$f$ to take place in $R_{i}/\varpi$. As $R_{i}$ is $\varpi$-complete,
$(R_{i},\varpi)$ form a Henselian pair (see, for example, [20] Tag 0ALJ).
Therefore the factorization lifts to $f=gh$ in $R_{i}\subset R$, with $g$ and
$h$ monic.

The main result now follows easily.

###### Proof of Theorem 2.19

Lemma 2.22 allows us to apply Proposition 2.20, and conclude that each finite
projective module over $\hat{R}[1/\varpi]$ is the base extension of one on
$R[1/\varpi]$, which by Lemma 2.21 must be free.

##  3 Line Bundles and Cohomology on Projectivoid Space

In classical algebraic geometry, the notion of _projective geometry_ is a very
powerful tool to study properties of varieties and schemes. Indeed, one can
learn a lot about a scheme by understanding its maps to various projective
spaces, and this theory is intimately connected to the theory of line bundles
on that space. In this and the following section we develop an analogous
theory for perfectoid spaces. Let us begin by defining a perfectoid analog of
projective space.

######  Definition 3.1 (The Projectivoid Line).

Analogously to the construction of the Riemann sphere, we can build the
perfectoid analog of the projective line by gluing two copies of the
perfectoid unit disk along the perfectoid unit circle.

Explicitly, the inclusion $K\left\langle T^{1/p^{\infty}}\right\rangle\to
K\left\langle T^{\pm 1/p^{\infty}}\right\rangle$ corresponds to the open
immersion
$\mathbb{S}^{1,\mathrm{perf}}\hookrightarrow\mathbb{D}^{1,\mathrm{perf}}$. The
map $K\left\langle T^{-1/p^{\infty}}\right\rangle\to K\left\langle T^{\pm
1/p^{\infty}}\right\rangle$ also corresponds to an open immersion of the
circle into the disk, where now the disk has coordinate $T^{-1}$. Identifying
the circles on each of these disks and gluing produces _the projectivoid line_
, denoted $\mathbb{P}^{1,\mathrm{perf}}_{K}$. $\clubsuit$

######  Definition 3.2 (Projectivoid Space).

As with the projectivoid line, we can define projectivoid $n$-space by gluing
together $n+1$ perfectoid unit $n$-polydisks along their associated perfectoid
sphere exteriors as in Example 3.1. This equips projectivoid space with a
cover by perfectoid unit disks which we will henceforth refer to as the
_standard cover_.

In [17] Section 7, Scholze showed that we could also define projectivoid space
in the following way. Let $\mathbb{P}_{K}^{n}$ be projective space over $K$,
which can be viewed as an adic space as in by first being viewed as a rigid
space using the rigid analytification functor, and then as an adic space as in
[9]. Let $\varphi:\mathbb{P}_{K}^{n}\to\mathbb{P}_{K}^{n}$ be the morphism
given in projective coordinates by
$(T_{0}:\cdots:T_{n})\mapsto(T_{0}^{p}:\cdots:T_{n}^{p})$. Then

$\mathbb{P}_{K}^{n,\mathrm{perf}}\sim\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{\varphi}\mathbb{P}_{K}^{n}.$

As with the perfectoid disk
“$\sim\operatornamewithlimits{\underset{\longleftarrow}{lim}}$” is the ‘tilde
limit’ of [18]. $\clubsuit$

###### Remark 3.3

Notice that all the finite intersections of the standard cover will be the
adic spectra associated to what we called perfectoid Laurent series algebras
in the previous section. In particular, Theorem 2.19 shows that all their
finite vector bundles are free. $\ast$

Scholze showed in [16] that the construction of projectivoid space is
compatible with the tilting functor. In this section we begin our exploration
of so called _projectivoid geometry_ by developing the theory of line bundles
on projectivoid space. In particular, we compute the Picard group of
$\mathbb{P}^{n,\mathrm{perf}}$, as well as the sheaf cohomology of all line
bundles. We continue developing the theory in the following section, where we
will show how an arbitrary perfectoid space’s maps to projectivoid space is
intimately connected to its theory of line bundles, reflecting the situation
in classical algebraic geometry, but with an extra arithmetic twist.

###  3.1 The Picard Group of Projectivoid Space

The main result of this section follows:

######  Theorem 3.4.

$\operatorname{Pic}\mathbb{P}^{n,\mathrm{perf}}\cong\mathbb{Z}[1/p]$.
$\diamondsuit$

We begin by outlining our proof. We use that for any ringed space $X$, there
is a natural isomorphism
$\operatorname{Pic}(X)\cong\mathrm{H}^{1}(X,\mathscr{O}_{X}^{*})$ (see for
example [8] Exercise III.4.5). Let $\mathscr{O}_{X}^{+}\subset\mathscr{O}_{X}$
be the subsheaf of integral functions, defined by the rule

$U\mapsto\\{f\in\mathscr{O}_{X}(U):|f(x)|\leq 1\text{ for all }x\in U\\}.$

It is a sheaf of rings, and we let $\mathscr{O}_{X}^{+*}$ be it’s unit group.
We begin in Section 3.1.1 by establishing an isomorphism

$\mathrm{H}^{1}(X,\mathscr{O}_{X}^{*})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*}).$

Let $\mathscr{O}_{X}^{++}\subset\mathscr{O}_{X}^{+}$ be the sheaf of
topologically nilpotent functions, defined by the rule

$\mathscr{O}_{X}^{++}:U\mapsto\\{f\in\mathscr{O}_{X}(U):|f(x)|<1\text{ for all
}x\in U\\}.$

It is a sheaf of ideals in $\mathscr{O}_{X}^{+}$ and the quotient we denote by
$\tilde{\mathscr{O}}_{X}$. The map
$\mathscr{O}_{X}^{+*}\to\tilde{\mathscr{O}}_{X}^{*}$ induces a map on
cohomology,
$\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*})\longrightarrow\mathrm{H}^{1}(X,\tilde{\mathscr{O}}_{X}^{*})$.
The latter is easier to understand, and we finish Section 3.1.1 by directly
computing that it is isomorphic to $\mathbb{Z}[1/p]$. Composing all this gives
us a map

$\varphi:\operatorname{Pic}X\longrightarrow\mathbb{Z}[1/p].$

In Section 3.1.2 we show that $\varphi$ is an isomorphism by deforming
invertible modules along the sheaves of algebras
$\mathscr{A}_{d}=\mathscr{O}_{X}^{+}/\varpi^{d}$, which live between
$\mathscr{O}_{X}^{+}$ and $\tilde{\mathscr{O}}_{X}$.

We make frequent use of the fact projectivoid space comes equipped with a
standard cover by perfectoid disks whose geometry we understand well due to
the results of Section 2. In particular, any line bundle on
$\mathbb{P}^{n,\mathrm{perf}}$ becomes trivial on the standard cover and its
various finite intersections due to Theorem 2.19. We therefore use Čech
cohomology with respect to this cover to study line bundles on projectivoid
space. The passage between Čech and sheaf cohomology is safe due to the
following lemma.

######  Lemma 3.5.

There are natural isomorphisms

  * •

$\mathrm{H}^{1}(X,\mathscr{O}_{X}^{*})\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{O}_{X}^{*})$.

  * •

$\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*})\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{O}_{X}^{+*})$.

  * •

$\mathrm{H}^{1}(X,\mathscr{A}_{d}^{*})\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{A}_{d}^{*})$
for all $d$.

  * •

$\mathrm{H}^{1}(X,\tilde{\mathscr{O}}_{X}^{*})\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\tilde{\mathscr{O}}_{X}^{*})$.

$\heartsuit$

###### Proof.

Let $\mathscr{R}$ be one of the following sheaves of rings:
$\mathscr{O}_{X},\mathscr{O}_{X}^{+},\mathscr{A}_{d}$, or
$\tilde{\mathscr{O}}_{X}$, and let $\mathscr{R}^{*}$ be the associated sheaf
of units. The Čech to Derived spectral sequence ([1] Exposé V Théorème 3.2)
describes the following spectral sequence:

$E_{2}^{p,q}:\check{\mathrm{H}}^{p}(\mathfrak{U},\mathscr{H}^{q}(\mathscr{R}^{*}))\implies\mathrm{H}^{p+q}(X,\mathscr{R}^{*}).$

Let $U$ be a finite intersection of elements in the standard cover. Then
either Theorem 2.19, Corollary 2.17, or Corollary 2.12 implies that
$\mathscr{R}(U)$ has no nontrivial invertible modules. In particular
$\mathscr{H}^{1}(\mathscr{R}^{*})(U)=H^{1}(U,\mathscr{R}^{*})=0$, as such a
cohomology class would construct an invertible $\mathscr{R}(U)$ module. In
particular, the sequence of low degree terms for the spectral sequence
degenerates to:

$\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{R}^{*})\cong\mathrm{H}^{1}(X,\mathscr{R}^{*}).$

####  3.1.1 Reduction to the Cohomology of Unit Groups of the Integral and
Residue Sheaves

Lemma 3.5 implies that
$\operatorname{Pic}(X)\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{O}_{X}^{*})$.
Our first step is to compare this to the cohomology on the sheaf of integral
units.

######  Lemma 3.6.

$\operatorname{Pic}(X)\cong\check{\mathrm{H}}^{1}(\mathfrak{U},\mathscr{O}_{X}^{+*})$.
$\heartsuit$

###### Proof.

We have a short exact sequence of chain complexes:

${0}$${\prod_{i}\mathscr{O}_{X}^{+*}(U_{i})}$${\prod_{i}\mathscr{O}_{X}^{*}(U_{i})}$${\prod_{i}|K^{*}|}$${0}$${0}$${\prod_{i,j}\mathscr{O}_{X}^{+*}(U_{i}\cap
U_{j})}$${\prod_{i,j}\mathscr{O}_{X}^{*}(U_{i}\cap
U_{j})}$${\prod_{i,j}|K^{*}|}$${0}$${0}$${\prod_{i,j,k}\mathscr{O}_{X}^{+*}(U_{i}\cap
U_{j}\cap U_{k})}$${\prod_{i,j,k}\mathscr{O}_{X}^{*}(U_{i}\cap U_{j}\cap
U_{k})}$${\prod_{i,j,k}|K^{*}|}$${0}$${\vdots}$${\vdots}$${\vdots}$

The left and middle complexes are the Čech complexes for
$\mathscr{O}_{X}^{+*}$ and $\mathscr{O}_{X}^{*}$ respectively, and the map on
the right is $|\cdot|$ which is plainly surjective. Also the right hand
complex has kernel $|K^{*}|$ and is otherwise exact, so that the long exact
sequence on cohomology gives

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{O}_{X}^{+*})\cong\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{O}_{X}^{*}),$

for all $i>0$. Letting $i=1$ completes the proof.

With this in hand, we can study what happens to cohomology classes reducing
modulo the sheaf of topologically nilpotent functions. Indeed, the surjection
$\mathscr{O}_{X}^{+}\to\tilde{\mathscr{O}}_{X}$ induces a map on unit groups
$\mathscr{O}_{X}^{+*}\to\tilde{\mathscr{O}}_{X}^{*}$. Taking cohomology gives

$\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*})\to\mathrm{H}^{1}(X,\tilde{\mathscr{O}}_{X}^{*}).$

The source we know is $\operatorname{Pic}X$ by Lemmas 3.5 and 3.6. We next
compute the target.

######  Lemma 3.7.

$\mathrm{H}^{1}(X,\tilde{\mathscr{O}}_{X}^{*})\cong\mathbb{Z}[1/p]$.
$\heartsuit$

###### Proof.

By Lemma 3.5 it suffices to compute
$\check{\mathrm{H}}^{1}(\mathfrak{U},\tilde{\mathscr{O}}_{X}^{*})$. We will
make use of projective coordinates $[T_{0}:\cdots:T_{n}]$ for
$\mathbb{P}^{n,\mathrm{perf}}$, and as above we denote by $k$ the residue
field of $K$. We denote the differentials of the Čech complex
$\check{\mathrm{C}}^{*}(\mathfrak{U},\tilde{\mathscr{O}}_{X})$ by $d^{i}$.
Note that for all $U_{i}$ we have

$\tilde{\mathscr{O}}_{X}(U_{i})=k\left[\left(\frac{T_{0}}{T_{i}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{i}}\right)^{1/p^{\infty}}\right],$

so that $\tilde{\mathscr{O}}_{X}^{*}(U_{i})\cong k^{*}$, since the only
invertible polynomials are the constant functions. Therefore we have
$C^{0}(\mathfrak{U},\tilde{\mathscr{O}}_{X}^{*}(U_{i}))\cong(k^{*})^{n+1}$,
and viewing the kernel as the intersection we have $\ker d^{0}\cong k^{*}$, so
that $\operatorname{im}d^{0}\cong(k^{*})^{n}$.

The rings $\tilde{\mathscr{O}}_{X}(U_{i}\cap U_{j})$ consist of Laurent
polynomials, and the only invertible Laurent polynomials are monomials in the
invertible variable. That is,

$\tilde{\mathscr{O}}_{X}^{*}(U_{i}\cap
U_{j})=\left\\{\lambda\left(\frac{T_{i}}{T_{j}}\right)^{\alpha}:\lambda\in
k^{*},\alpha\in\mathbb{Z}[1/p]\right\\}\cong k^{*}\oplus\mathbb{Z}[1/p].$

Let $(f_{ij})\in C^{1}(\mathfrak{U},\tilde{\mathscr{O}}_{X}^{*})$, and suppose
$(f_{ij})\in\ker d^{1}$. This means that for all $i<j<k$ we have
$f_{ij}f_{jk}=f_{ik}$. That is,

$\lambda_{ij}\left(\frac{T_{i}}{T_{j}}\right)^{\alpha_{ij}}\cdot\lambda_{jk}\left(\frac{T_{j}}{T_{k}}\right)^{\alpha_{jk}}=\lambda_{ik}\left(\frac{T_{i}}{T_{k}}\right)^{\alpha_{ik}}.$

In particular, $\alpha_{ij}=\alpha_{jk}=\alpha_{ik}$, and so the degree of
every factor in an element of the kernel must match. The fact that
$\lambda_{ij}\lambda_{jk}=\lambda_{ik}$ leaves $n$ degrees of freedom for the
coefficient, so that $\ker d^{1}=(k^{*})^{n}\oplus\mathbb{Z}[1/p]$, and since
$\operatorname{im}d^{0}=(k^{*})^{n}$, we conclude that
$\check{\mathrm{H}}^{1}(\mathfrak{U},\tilde{\mathscr{O}}_{X})\cong\mathbb{Z}[1/p]$.

Making the necessary identifications gives a map
$\varphi:\operatorname{Pic}X\to\mathbb{Z}[1/p]$. It remains to prove it is an
isomorphism. For now we show it is surjective.

######  Lemma 3.8.

The map $\varphi:\operatorname{Pic}X\to\mathbb{Z}[1/p]$ has a natural section.
$\heartsuit$

###### Proof.

The $\mathbb{Z}[1/p]$ on the right can be interpreted as follows. For each
$\alpha\in\mathbb{Z}[1/p]$ we can build an invertible
$\tilde{\mathscr{O}}_{X}$-module starting with $\tilde{\mathscr{O}}_{U_{i}}$
on each open set in the standard cover, and gluing along transition maps
$\left(\frac{x_{i}}{x_{j}}\right)^{\alpha}$ on $U_{i}\cap U_{j}$. Doing the
same construction, but starting with $\mathscr{O}_{U_{i}}^{+}$ gives us a
section of $\varphi$ (call it $\sigma$). In particular, $\varphi$ is
surjective and we have an embedding
$\mathbb{Z}[1/p]\hookrightarrow\operatorname{Pic}(X)$.

###### Remark 3.9

Therefore we can consider twisting sheaves
$\mathscr{O}(d)\in\operatorname{Pic}(X)$ for every $d\in\mathbb{Z}[1/p]$.
$\ast$

We also have the following immediate corollary.

######  Corollary 3.10.

Let $A=K\left\langle
T_{0}^{1/p^{\infty}},\cdots,T_{n}^{1/p^{\infty}}\right\rangle$. Then,

$\Gamma(U_{i_{1}}\cap\cdots\cap
U_{i_{r}},\mathscr{O}(d))\cong\left(\widehat{A_{T_{i_{1}}\cdots
T_{i_{r}}}}\right)_{d}$

the degree $d$ part of the completion of the localization of $A$. $\spadesuit$

####  3.1.2 Deforming Line Bundles from the Residue

Most of the following computation will be made using Čech cohomology, with
$\mathfrak{U}$ the standard cover of $X=\mathbb{P}^{n,\mathrm{perf}}$ as
above. For the crucial deformation steps we will need the following lemma,
whose proof we defer to the following section (see Remark 3.19).

######  Lemma 3.11.

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{O}_{X}^{+})=0$ for all $i>0$.
$\heartsuit$

We will also need the following lemma of commutative algebra.

######  Lemma 3.12.

Let $R\to S$ be a surjection of rings whose kernel $I$ is contained in the
Jacobson radical of $R$. Then the induced map on unit groups, $R^{*}\to
S^{*}$, remains surjective. $\heartsuit$

###### Proof.

Fix $s\in S^{*}$ and $r\in R$ mapping to $s$. If $r\in\mathfrak{m}$ for any
maximal ideal of $r$, then its image would be contained in
$\mathfrak{m}/I\cdot\mathfrak{m}$, a proper ideal of $S$. Since its image is a
unit this cannot be the case. Since $r$ is not contained in any maximal ideal
it must be a unit.

Let us begin by enumerating a few useful exact sequences.

$0\longrightarrow\mathscr{O}_{X}^{++}\longrightarrow\mathscr{O}_{X}^{+}\longrightarrow\tilde{\mathscr{O}}_{X}\longrightarrow
0.$ (2)

Because $\mathscr{O}_{X}^{++}$ consists of toplogically nilpotent functions,
it is contained in the Jacobson radical of $\mathscr{O}_{X}^{+}$, so that by
Lemma 3.12 the right hand map of the sequence remains surjective on unit
groups.

$1\longrightarrow
1+\mathscr{O}_{X}^{++}\longrightarrow\mathscr{O}_{X}^{+*}\longrightarrow\tilde{\mathscr{O}}_{X}^{*}\longrightarrow
1.$ (3)

If $1+\mathscr{O}_{X}^{++}$ were acyclic, we could reduce finding line bundles
on $X$ to finding invertible $\tilde{\mathscr{O}}_{X}$ modules and be done,
but this acyclicity has so far been rather elusive (and seems unlikely in
general). That being said, there is a filtration of $\mathscr{O}_{X}^{++}$ by
sheaves of (principal) ideals $\varpi^{d}$ for $d>0$.

$0\longrightarrow\varpi^{d}\longrightarrow\mathscr{O}_{X}^{+}\longrightarrow\mathscr{A}_{d}\longrightarrow
0.$ (4)

Notice that $\mathscr{A}_{d}$ is a sheaf of (nonreduced)
$\mathscr{O}_{X}^{+}$-algebras, and that for every $d^{\prime}>d$, we have
surjections $\mathscr{A}_{d^{\prime}}\twoheadrightarrow\mathscr{A}_{d}$ with
kernel $\varpi^{d}/\varpi^{d^{\prime}}$. As before, $\varpi^{d}$ is contained
in the Jacobson radical, so that we also have

$1\longrightarrow
1+\varpi^{d}\longrightarrow\mathscr{O}_{X}^{+*}\longrightarrow\mathscr{A}_{d}^{*}\longrightarrow
1.$ (5)

######  Lemma 3.13.

The sheaves $\mathscr{A}_{d}^{*}$ interpolate continuously between
$\tilde{\mathscr{O}}_{X}^{*}$ and $\mathscr{O}_{X}^{+*}$. More precisely:

  * •

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}\mathscr{A}_{d}^{*}\cong\tilde{\mathscr{O}}_{X}^{*}$

  * •

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathscr{A}_{d}^{*}\cong\mathscr{O}_{X}^{+*}.$

$\heartsuit$

###### Proof.

For the first statement notice that
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}(1+\varpi^{d})=\cup_{d}(1+\varpi^{d})=1+\mathscr{O}_{X}^{++}$.
This follows if $\mathscr{O}_{X}^{++}=\cup_{d}(\varpi^{d})$, so fix
$f\in\mathscr{O}_{X}^{++}$ so that have $|f|=\lambda<1$. There is some $d$
such that $1>|\varpi^{d}|>\lambda$, so that $|f/\varpi^{d}|<1$. Thus
$f^{\prime}=f/\varpi^{d}\in\mathscr{O}_{X}^{+}+$ and $f\in\varpi^{d}$. Since
colimits of abelian sheaves are exact, applying
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{d}$ to Sequence 4
proves part 1 of the lemma.

For part 2, notice that
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}(\mathscr{A}_{d})\cong\mathscr{O}_{X}^{+}$
since $\mathscr{O}_{X}^{+}$ is $\varpi$-adically complete. Since the unit
group functor commutes with inverse limits (indeed, it is left adjoint to the
group ring functor), are done.

We next assert that Sequences 2 through 5 induce long exact sequences on Čech
cohomology.

######  Lemma 3.14.

Each of the sequences 2 through 5 remain exact when evaluated on any finite
intersection of elements in the standard cover $\mathfrak{U}$. In particular,
they each induce long exact sequences on Čech cohomology. $\heartsuit$

###### Proof.

We need only check exactness on the right. For sequences 2 and 4 these
evaluate to:

$K^{\circ}\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right\rangle\longrightarrow
k\left[X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right]$

and

$K^{\circ}\left\langle
X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right\rangle\longrightarrow
K^{\circ}/\varpi^{d}\left[X_{1}^{1/p^{\infty}},\cdots,X_{r}^{1/p^{\infty}},X_{r+1}^{\pm
1/p^{\infty}},\cdots,X_{n}^{\pm 1/p^{\infty}}\right]$

which are plainly surjective. Sequences 3 and 5 come from applying the unit
group functor to surjections above. Lemma 2.16 implies that the kernel of each
is contained in the Jacobson radical, so that the maps remain surjective on
unit groups by Lemma 3.12.

######  Lemma 3.15.

For all $i>0$ and $d>0$,
$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d})=0$. $\heartsuit$

###### Proof.

This follows from the long exact sequence on Čech cohomology associated to
Sequence 4 and Lemma 3.11, noticing that $\varpi^{d}\cong\mathscr{O}_{X}^{+}$
since it is a principal ideal.

######  Lemma 3.16.

For all $d,i>0$, the natural map

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{2d}^{*})\longrightarrow\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d}^{*})$

is an isomorphism. $\heartsuit$

###### Proof.

Consider:

${0}$${1+\varpi^{2d}}$${\mathscr{O}_{X}^{+*}}$${\mathscr{A}_{2d}^{*}}$${0}$${0}$${1+\varpi^{d}}$${\mathscr{O}_{X}^{+*}}$${\mathscr{A}_{d}^{*}}$${0}$

By the snake lemma, we have

$1\longrightarrow
1+\varpi^{d}/\varpi^{2d}\longrightarrow\mathscr{A}_{2d}^{*}\longrightarrow\mathscr{A}_{d}^{*}\longrightarrow
1.$ (6)

Notice that $1+\varpi^{d}/\varpi^{2d}$ has a natural $\mathscr{A}_{d}$-module
structure making it isomorphic to $\mathscr{A}_{d}$, given locally by the map
$1\mapsto\varpi^{d}$. Indeed, the map is well defined because
$\varpi^{d}/\varpi^{2d}$ is a square zero ideal, and the kernel is precisely
$\varpi^{d}$ (which is 0 in $\mathscr{A}_{d}$), while surjectivity is clear.
In particular, by Lemma 3.15, $1+\varpi^{d}/\varpi^{2d}$ has no higher Čech
cohomology, so the conclusion follows if there is a long exact sequence on
Čech cohomology associated Sequence 6.

Let $U$ be a finite intersection of elements of the standard cover. It remains
to show that Sequence 6 remains exact when evaluated on $U$. By Lemma 3.14,

${0}$${\Gamma(U,1+\varpi^{2d})}$${\Gamma(U,\mathscr{O}_{X}^{+*})}$${\Gamma(U,\mathscr{A}_{2d}^{*})}$${0}$${0}$${\Gamma(U,1+\varpi^{d})}$${\Gamma(U,\mathscr{O}_{X}^{+*})}$${\Gamma(U,\mathscr{A}_{d}^{*})}$${0}$

has exact rows. The ring map $\mathscr{A}_{2d}(U)\to\mathscr{A}_{d}(U)$ is
surjective with nilpotent kernel, so that again applying Lemma 3.12 we see the
vertical map on the right is surjective. Thus the snake lemma exhibits
Sequence 6 evaluated at $U$ as an exact sequence, completing the proof.

######  Lemma 3.17.

For all $d^{\prime}>d>0$ and $i>0$, the natural map:

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d^{\prime}}^{*})\longrightarrow\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d}^{*}),$

is an isomorphism. $\heartsuit$

###### Proof.

By Lemma 3.16, replacing $d$ with $2^{l}d$ will not change Čech cohomology, so
we may assume $d<d^{\prime}<2d$. Then we have the following commutative
diagram.

${\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{2d}^{*})}$${\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d}^{*})}$${\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{2d^{\prime}}^{*})}$${\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{A}_{d^{\prime}}^{*})}$$\scriptstyle{\psi}$$\scriptstyle{\sim}$$\scriptstyle{\sim}$

In particular, $\psi$ is injective and surjective, so an isomorphism, which
implies the result.

We have now established the following sequence of morphisms, whose composition
is the map $\varphi$ from Lemma 3.8. We leverage Lemma 3.5 and the fact that
colimits of abelian sheaves are exact.

$\displaystyle\operatorname{Pic}(X)$ $\displaystyle\cong$
$\displaystyle\mathrm{H}^{1}(X,\mathscr{O}_{X}^{*})$ $\displaystyle\cong$
$\displaystyle\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*})$ $\displaystyle\cong$
$\displaystyle\mathrm{H}^{1}(X,\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathscr{A}_{d}^{*})$
$\displaystyle\longrightarrow$
$\displaystyle\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathrm{H}^{1}(X,\mathscr{A}_{d}^{*})$
$\displaystyle\cong$ $\displaystyle\mathrm{H}^{1}(X,\mathscr{A}_{d}^{*})$
$\displaystyle\cong$
$\displaystyle\operatornamewithlimits{\underset{\longrightarrow}{lim}}\mathrm{H}^{1}(X,\mathscr{A}_{d}^{*})$
$\displaystyle\cong$
$\displaystyle\mathrm{H}^{1}(X,\operatornamewithlimits{\underset{\longrightarrow}{lim}}\mathscr{A}_{d}^{*})$
$\displaystyle\cong$
$\displaystyle\mathrm{H}^{1}(X,\tilde{\mathscr{O}}_{X}^{*})$
$\displaystyle\cong$ $\displaystyle\mathbb{Z}[1/p].$

We make the necessary identifications to view the map $\varphi$ and its
section $\sigma$ as maps between the following groups.

${\mathrm{H}^{1}(X,\mathscr{O}_{X}^{+*})}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathrm{H}^{1}(X,\mathscr{A}_{d}^{*})}$$\scriptstyle{\varphi}$$\scriptstyle{\sigma}$

We can view the first group as isomorphism classes of invertible
$\mathscr{O}_{X}^{+}$ modules, and the second as inverse systems of
isomorphism classes of invertible $\mathscr{A}_{d}$ modules. Under these
identifications we have:

$\displaystyle\varphi:$
$\displaystyle\mathscr{L}\mapsto\\{\mathscr{L}/\varpi^{d}\mathscr{L}\\}$
$\displaystyle\sigma:$
$\displaystyle\\{\mathscr{M}_{d}\\}\mapsto\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathscr{M}_{d}.$

In particular, there is a natural map

$\mathscr{L}\longrightarrow\operatornamewithlimits{\underset{\longleftarrow}{lim}}\mathscr{L}/\varpi^{d}\mathscr{L}=\sigma\varphi\mathscr{L}.$

Locally, on an affinoid $\operatorname{Spa}(R,R^{+})$, we associate
$\mathscr{L}$ to an invertible $R$-module $M$. Then this map becomes,

$M\longrightarrow\operatornamewithlimits{\underset{\longleftarrow}{lim}}M/\varpi^{d}M\cong\hat{M},$

which is an isomorphism since $M$ is already complete. We conclude that
$\sigma$ is surjective, and therefore an isomorphism. Putting all this
together we may conclude that $\operatorname{Pic}X\cong\mathbb{Z}[1/p]$,
completing the proof of Theorem 3.4.

###  3.2 Cohomology of Line Bundles

In his Ph.D. thesis [2], Bedi computed the cohomology of some of the twisting
sheaves $\mathscr{O}(d)$ on projectivoid space. His proof was modeled on the
computation for classical projective space in Ravi Vakil’s algebraic geometry
notes ([21] Chapter 18.2), but doesn’t explicitly take into account the
completions involved, and only includes the case for $n=2$. Instead, we adapt
the proof from EGA III [7] which relates Čech cohomology to the Koszul
complex. We also fill in the result, computing the cohomology of every line
bundle in every degree. Our rings are going to be topologically
$\mathbb{Z}[1/p]$ graded, and for a graded ring $A$ we will denote by $A_{d}$
the degree $d$ part.

######  Theorem 3.18.

Let $X=\mathbb{P}^{n,\mathrm{perf}}$ be projectivoid space, and
$\mathscr{O}_{X}(d)\in\operatorname{Pic}X$ an arbitrary line bundle. Then:

  1. 1.

If $d\geq 0$,

$\mathrm{H}^{0}\left(X,\mathscr{O}_{X}(d)\right)=K\left\langle
T_{0}^{1/p^{\infty}},\cdots,T_{n}^{1/p^{\infty}}\right\rangle_{d}.$

  2. 2.

If $d<0$, then $\mathrm{H}^{n}\left(X,\mathscr{O}_{X}(d)\right)$ is the
completion of the $K$ vector space generated by monomials of degree d, where
the degree of each indeterminate is strictly negative, that is:

$\mathrm{H}^{n}\left(X,\mathscr{O}_{X}(d)\right)=\left\langle
T_{0}^{\alpha_{0}}\cdots T_{n}^{\alpha_{n}}\hskip 5.0pt\big{|}\hskip
5.0pt\alpha_{i}\in\mathbb{Z}[1/p]_{<0}\text{ and
}\sum\alpha_{i}=d\right\rangle\widehat{}$

  3. 3.

In all other cases,

$\mathrm{H}^{r}\left(X,\mathscr{O}_{X}(d)\right)=0.$

In particular:

$h^{r}\left(X,\mathscr{O}_{X}(d)\right)=\begin{cases}1&r=d=0\\\
\infty&r=0\text{ and }d>0\\\ \infty&r=n\text{ and }d<0\\\ 0&\text{ all other
cases}\end{cases}.$

$\diamondsuit$

###### Proof.

We will leverage that colimits of abelian groups are exact, so that cohomology
of finite complexes of abelian groups commute with arbitrary direct sums. We
will therefore study the Čech sequence associated to the sheaf

$\mathscr{H}=\bigoplus_{d\in\mathbb{Z}[1/p]}\mathscr{O}_{X}(d).$

Let $A=K\left\langle
T_{0}^{1/p^{\infty}},\cdots,T_{n}^{1/p^{\infty}}\right\rangle$, and
$\mathfrak{U}=\\{U_{i}\\}$ be the standard cover of $X$.

$\check{\mathrm{C}}^{r}(\mathfrak{U},\mathscr{H})=\prod_{0\leq
i_{1}<\cdots<i_{r}\leq n}\widehat{A_{T_{i_{1}}\cdots T_{i_{r}}}}.$

Since the differentials commute with degree and cohomology commutes with
direct sums, we can conclude that

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{O}_{X}(d))=\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{H})_{d}.$

Furthermore, all finite intersections of the $U_{i}$ are affinoid, and vector
bundles on affinoids are acyclic ([12] Theorem 1.4.2). Then the Čech to
derived spectral spectral sequence degenerates to an isomorphism between Čech
cohomology and sheaf cohomology, so we have reduced to computing these Čech
cohomology groups. We first consider the 0th cohomology.

$\check{\mathrm{H}}^{0}(\mathfrak{U},\mathscr{H})=\bigcap_{i=0}^{n}\widehat{A_{T_{i}}}=A.$

This proves the first statement of the theorem. For the second, we consider
the sequence

$C^{*}(A):\hskip 10.0pt0\longrightarrow\prod A_{T_{i}}\longrightarrow\prod
A_{T_{i}T_{j}}\longrightarrow\cdots\longrightarrow A_{T_{0}\cdots
T_{n}}\longrightarrow 0.$

In each case, there is a countable fundamental system of neighborhoods of
zero, given by $(\varpi^{n})$, so that proceeding by induction from the left
to the right, we see that completion on this sequence commutes with taking
cohomology (see, for example, [20] tag 0AMQ). In particular,

$\check{\mathrm{H}}^{i}(\mathfrak{U},\mathscr{H})=\widehat{\mathrm{H}^{i}\left(C^{*}(A)\right)}.$

Let’s analyze $H^{n}(C^{*}(A))$. Notice that $A_{T_{0}\cdots T_{n}}$ is the
$K$ vector space generated by monomials $T_{0}^{\alpha_{0}}\cdots
T_{n}^{\alpha_{n}}$ for $\alpha\in\mathbb{Z}[1/p]$. The image of the $(n-1)$st
differential is the $K$ vector space generated by monomials where at least one
of the $\alpha_{i}\geq 0$. Therefore $H^{n}(C^{*}(A))$, which is the cokernel
of this differential, is the $K$ vector space generated by monomials where
each $\alpha_{i}<0$. Taking completions proves the second statement of the
theorem.

For the third statement of the theorem, the cases of $r<0$ and $r>n$ are
trivial, so we assume $0<r<n$. We will show that $H^{r}(C^{*}(A))=0$ since the
completion of 0 is 0. We point out that for any $f\in A$,

$A_{f}\cong\operatornamewithlimits{\underset{\longrightarrow}{lim}}\left(A\stackrel{{\scriptstyle\cdot
f}}{{\longrightarrow}}A\stackrel{{\scriptstyle\cdot
f}}{{\longrightarrow}}\cdots\right).$

For all $s\geq 0$, let $T^{s}=(T_{0}^{s},\cdots,T_{n}^{s})$. Then $T^{s}$ is
an $A$-regular sequence, and the associated Koszul complex $K^{*}(T^{s})$ is a
free resolution of $A/(T_{0}^{s},\cdots,T_{n}^{s})$.

$K^{*}(T^{s}):\hskip
10.0pt0\longrightarrow\Lambda^{n+1}A^{n+1}\longrightarrow\cdots\longrightarrow\Lambda^{2}A^{n+1}\longrightarrow
A^{n+1}\stackrel{{\scriptstyle(T_{0}^{s},\cdots,T_{n}^{s})}}{{\longrightarrow}}A\longrightarrow
0.$

In particular, the homology groups $\mathrm{H}_{i}(K^{*}(T^{s}))=0$ for all
$i>0$. For each $s$ we can also look at the dual Koszul complex, and take the
colimit as $s$ goes to infinity, with the above identification to the
localization in mind.

${\vdots}$${\vdots}$${\vdots}$${\vdots}$${C_{s}:}$${0}$${A}$${A^{n+1}}$${\Lambda^{2}A^{n+1}}$${\cdots}$${\Lambda^{n+1}A^{n+1}}$${0}$${C_{s+1}:}$${0}$${A}$${A^{n+1}}$${\Lambda^{2}A^{n+1}}$${\cdots}$${\Lambda^{n+1}A^{n+1}}$${0}$${\vdots}$${\vdots}$${\vdots}$${\vdots}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}C_{s}:}$${0}$${A}$${\prod
A_{T_{i}}}$${\prod A_{T_{i}T_{j}}}$${\cdots}$${A_{T_{0}\cdots
T_{n}}}$${0}$$\scriptstyle{T^{s}}$$\scriptstyle{\cdot
T}$$\scriptstyle{\cdot\wedge T^{s}}$$\scriptstyle{\cdot(T\wedge
T)}$$\scriptstyle{\cdot
T^{\wedge(n+1)}}$$\scriptstyle{T^{s+1}}$$\scriptstyle{\cdot\wedge T^{s+1}}$

For $i>0$, the bottom row is $C^{*}(A)[1]$. By the self-duality of the Koszul
complex, we have that

$\mathrm{H}^{i}(C_{s})\cong\mathrm{H}_{n+1-i}(K^{*}(T^{s})),$

so that in particular, for $i<n+1$, we have $H^{i}(C_{s})=0$. Since colimits
of finite complexes commute with cohomology, we conclude that for $0<r<n$,

$\displaystyle\mathrm{H}^{r}(C^{*}(A))$ $\displaystyle\cong$
$\displaystyle\mathrm{H}^{r+1}(\operatornamewithlimits{\underset{\longrightarrow}{lim}}C_{s})$
$\displaystyle\cong$
$\displaystyle\operatornamewithlimits{\underset{\longrightarrow}{lim}}\mathrm{H}^{r+1}(C_{s})$
$\displaystyle=$
$\displaystyle\operatornamewithlimits{\underset{\longrightarrow}{lim}}0$
$\displaystyle=$ $\displaystyle 0.$

Taking completions proves the third statement of the theorem, and so we are
done.

###### Remark 3.19

An identical argument, but replacing $K$ with $K^{\circ}$, computes the Čech
cohomology of the integral line bundles $\mathscr{O}_{X}^{+}(d)$ for all
$d>0$. In particular we see that $\mathscr{O}_{X}^{+}$ is Čech acyclic with
respect to the standard cover, which completes the proof of Lemma 3.11.

Since $\mathscr{O}_{X}^{+}$ is only a priori _almost_ acyclic on affines
(rather than acyclic on the nose like $\mathscr{O}_{X}$), we cannot conclude
from this that $\mathscr{O}_{X}^{+}$ is acyclic on projectivoid space. It is
perhaps true that on the affinoids arising in the standard cover the sheaf of
integral elements is indeed acyclic, and such a result would certainly
simplify the arguments of Section 3.1, but so far a proof has eluded the
author. $\ast$

##  4 Maps to Projectivoid Space

Suppose $S$ is a scheme over $K$. Then well known correspondence between maps
from $S\to\mathbb{P}^{n}$ and globally generated line bundles on $S$ together
with a choice of $n+1$ generating global sections (see for example [8] Theorem
II.7.1). In this section we will prove an analog of this correspondence for
perfectoid spaces.

######  Definition 4.1.

To a perfectoid space $X$ over $K$, we associate a groupoid
$\mathfrak{L}_{n}(X)$ whose objects consist of tuples
$\left(\mathscr{L}_{i},s^{(i)}_{j},\alpha_{i}\right)$ for $i\geq 0$ and
$j=0,\cdots,n$, where $\mathscr{L}_{i}$ are line bundles on $X$,
$s^{(i)}_{0},\cdots,s^{(i)}_{n}\in\Gamma(X,\mathscr{L}_{i})$ are generating
global sections, and $\alpha_{i}:\mathscr{L}_{i+1}^{\otimes
p}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathscr{L}_{i}$ are
isomorphisms mapping $\left(s^{(i+1)}_{j}\right)^{\otimes p}\mapsto
s_{j}^{(i)}$. Morphisms are isomorphisms of line bundles which are compatible
with the global sections and isomorphisms $\alpha_{i}$.

If $f:X\to Y$ is a $K$-morphism, we get a pullback functor
$f^{*}:\mathfrak{L}_{n}(Y)\to\mathfrak{L}_{n}(X)$, so that $\mathfrak{L}_{n}$
is a category fibered in groupoids. $\clubsuit$

###### Remark 4.2

We will show that $\mathfrak{L}_{n}$ is actually representable by a perfectoid
space, and therefore a stack for the analytic topology. $\ast$

###### Remark 4.3

Note that if some $\alpha_{i}$ exists, it is unique. Indeed, for each $i$ the
global sections $s_{j}^{(i)}$ generate $\mathscr{L}_{i}$, so that an
isomorphism $\mathscr{L}_{i+1}^{\otimes p}\longrightarrow\mathscr{L}_{i}$
shows that the global sections $\left(\sigma_{j}^{(i+1)}\right)^{\otimes p}$
generate $\mathscr{L}_{i+1}^{\otimes p}$. In particular, the isomorphism is
completely determined by the images of these global sections. $\ast$

###### Remark 4.4

For each $i$, the data $\left(\mathscr{L}_{i},s^{(i)}_{j}\right)$ corresponds
to a map to a projective space (as a rigid analytic variety), so that objects
of the category $\mathfrak{L}_{n}(X)$ correspond to $p$th power root systems
of maps to projective space. $\ast$

The main result of this section is that the category $\mathfrak{L}_{n}(X)$
parametrizes $K$-morphisms $X\longrightarrow\mathbb{P}^{n,\mathrm{perf}}$. In
particular, viewing $\mathfrak{L}_{n}$ as a functor to sets we construct a
natural isomorphism
$\operatorname{Hom}(\cdot,\mathbb{P}^{n,\mathrm{perf}})\cong\mathfrak{L}_{n}$
of functors from perfectoid spaces over $K$ to sets. First we introduce a bit
of notation.

###### Notation 4.5

Denote by $m_{i}:\mathscr{O}(1/p^{i+1})^{\otimes
p}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathscr{O}(1/p^{i})$ the
isomorphism of line bundles on $\mathbb{P}^{n,\mathrm{perf}}$ coming from
multiplying factors together. $\ast$

We now state the main theorem of this section (compare to [8] Theorem II.7.1).

######  Theorem 4.6.

The functor $\mathfrak{L}_{n}$ is represented by projectivoid space.

Explicitly, the natural transformation
$\operatorname{Hom}(\cdot,\mathbb{P}^{n,\mathrm{perf}})\to\mathfrak{L}_{n}$,
which evaluated on $X$ takes $\phi:X\to\mathbb{P}^{n,\mathrm{perf}}$ to the
tuple
$\left(\phi^{*}\mathscr{O}(1/p^{i}),\phi^{*}T_{j}^{1/p^{i}},\phi^{*}m_{i}\right)\in\mathfrak{L}_{n}(X)$
is an isomorphism of functors. $\diamondsuit$

Since $\left\\{T_{j}^{1/p^{i}}\right\\}_{j=0}^{n}$ generates
$\mathscr{O}(1/p^{i})$, we have that
$\left\\{\phi^{*}\left(T_{j}^{1/p^{i}}\right)\right\\}_{j=0}^{n}$ generates
$\phi^{*}\left(\mathscr{O}(1/p^{i})\right)$. Furthermore, the standard
isomorphisms $\mathscr{O}(1/p^{i+1})^{\otimes
p}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\mathscr{O}(1/p^{i})$ coming
from multiplying factors together send $(T_{j}^{1/p^{i+1}})^{\otimes p}$ to
$T_{j}^{1/p^{i}}$, so pulling back these isomorphisms along $\phi$ gives us an
element of $\mathfrak{L}_{n}(X)$. We construct an inverse to this
transformation in Section 4.2, but first we will need a bit of setup.

###  4.1 $\mathscr{L}$-Distinguished Open Sets

For this section we let $X$ be an adic space, $\mathscr{L}$ a line bundle on
$X$, and $s_{1},\cdots,s_{n}$ global sections of $\mathscr{L}$ which generate
it at every point. Let $D(s_{i})=\\{x\in X:s_{i}|_{x}$ generates
$\mathscr{L}_{x}\\}$ be the _doesn’t vanish_ set of the section $s_{i}$. Then
the map $\mathscr{O}_{X}\to\mathscr{L}$ determined by $s_{i}$ is an
isomoprhism on the stalks of every point of $D(s_{i})$, and therefore
restricts to an isomorphism on $D(s_{i})$. We suggestively denote the inverse
by $s\mapsto s/s_{i}$. Let’s validate this notation with the following lemma.

######  Lemma 4.7.

On $D(s_{i})\cap D(s_{j})$, we have the following relation.

$\frac{s_{i}}{s_{j}}\cdot\frac{s_{j}}{s_{i}}=1.$

$\heartsuit$

###### Proof.

We have two isomorphisms,

${\Gamma(D(s_{i})\cap D(s_{j}),\mathscr{O}_{X})}$${\Gamma(D(s_{i})\cap
D(s_{j}),\mathscr{L}).}$$\scriptstyle{s_{i}}$$\scriptstyle{s_{j}}$

Then we have

$\frac{s_{i}}{s_{j}}=s_{j}^{-1}\circ s_{i}(1)$
$\frac{s_{j}}{s_{i}}=s_{i}^{-1}\circ s_{j}(1).$

Since the maps $s_{i}^{-1}\circ s_{j}$ and $s_{j}^{-1}\circ s_{i}$ are
inverses to each other, we win.

For every $x\in D(s_{i})$, we can use the isomorphism $s_{i}^{-1}$ to get a
valuation on $\Gamma(X,\mathscr{L})$.

${\Gamma(X,\mathscr{L})}$${\Gamma(D(s_{i}),\mathscr{L})}$${\Gamma(D(s_{i}),\mathscr{O}_{X})}$${\Gamma_{x}\cup\\{0\\}}$${s}$${{}|(s/s_{i})(x)|}$$\scriptstyle{res}$$\scriptstyle{s_{i}^{-1}}$$\scriptstyle{x}$

With this in hand, we can define the following open subsets of $D(s_{i})$ for
each $i$.

######  Definition 4.8.

Let $X$ be a perfectoid space, $\mathscr{L}$ a line bundle on $X$ and
$s_{1},\cdots,s_{n}$ generating global sections of $\mathscr{L}$. An open set
of $X$ is called an _$\mathscr{L}$ -distinguished open set_ if it is of the
form

$X\left(\frac{s_{1},\cdots,s_{n}}{s_{i}}\right)=\\{x\in
D(s_{i}):|(s_{j}/s_{i})(x)|\leq 1\text{ for all }j\\}.$

$\clubsuit$

For the case of classical projective space, we can build a map to projective
space along the _doesn’t vanish_ sets of the given sections, and glue them
together. In the analytic topology these are not affinoid, so we must use
these smaller $\mathscr{L}$-distinguished open sets. Let’s prove these smaller
open sets cover $X$. Indeed, our notation suggests that one of
$|(s_{j}/s_{i})(x)|$ or $|(s_{i}/s_{j})(x)|$ should be less than 1, let’s
check the details.

######  Lemma 4.9.

The $\mathscr{L}$-distinguished open sets
$X_{i}=X\left(\frac{s_{1},\cdots,s_{n}}{s_{i}}\right)$ for $i=1,\cdots,n$ are
open and cover $X$. $\heartsuit$

###### Proof.

The openness of $X_{i}$ follows because it is in fact a _rational_ open in the
adic space $D(s_{i})$, which is open in $X$. To show these cover $X$, fix some
$x\in X$. We already know the $D(s_{i})$ cover $X$, because the $s_{i}$
generate $\mathscr{L}$. Therefore $I=\left\\{i\in\\{1,\cdots,n\\}:x\in
D(s_{i})\right\\}$ is nonempty and finite. Order the elements of $I$ via
$i\leq j$ if $|(s_{i}/s_{j})(x)|\leq 1$. Notice that for any $i,j\in I$, we
have $i\leq j$ or $j\leq i$. Indeed, applying Lemma 4.7 together with the
multiplicativity of the valuation given by $x$, we have either
$|(s_{i}/s_{j})(x)|\leq 1$ or $|(s_{j}/s_{i})(x)|\leq 1$. Also if $i\leq j$
and $j\leq k$ then

$|(s_{i}/s_{k})(x)|=|(s_{i}/s_{j})(x)|\cdot|(s_{j}/s_{k})(x)|\leq 1,$

so that $i\leq k$. Finally, notice that if $i\leq j$ and $j\leq i$ then we
have $|(s_{i}/s_{j})(x)|=|(s_{j}/s_{i})(x)|=1$. Therefore we can choose (not
necessarily uniquely) some $r$ which is maximal with respect to this ordering.
Then $|(s_{i}/s_{r})(x)|\leq 1$ for all $i\in I$. For all other $i$, we have
$x\notin D(s_{i})$ so that $|(s_{i}/s_{r})(x)|=0\leq 1$. Therefore $x\in
X_{r}$, completing the proof.

###### Example 4.10

The standard cover of $\mathbb{P}^{n,\mathrm{perf}}$ by perfectoid unit disks
consists of the $\mathscr{O}(1)$-distinguished open sets
$\mathbb{P}^{n,\mathrm{perf}}\left(\frac{T_{0},\cdots,T_{n}}{T_{i}}\right)$.
$\ast$

The following lemma which implies that if $\mathscr{L}^{\otimes
p}\cong\mathscr{M}$ and $s$ and $t$ are global sections of $\mathscr{L}$ and
$\mathscr{M}$ respectively, with $s^{\otimes p}=t$, then $D(s)=D(t)$. In
particular, using the notation of Definition 4.1, this implies that if the
$s^{(i)}_{j}$ generate $\mathscr{L}_{i}$, then the $s^{(i+1)}_{j}$ generate
$\mathscr{L}_{i+1}$.

######  Lemma 4.11.

Let $(R,\mathfrak{m})$ be a local ring, and $M,N$ invertible $R$-modules such
that $M^{\otimes r}\cong N$ for a positive integer $r$. Let $f\in M$, and
$g\in N$ such that under this identification $f^{\otimes r}=g$. Then if $g$
generates $N$, $f$ generates $M$. $\heartsuit$

###### Proof.

We show the contrapositive. If $f$ does not generate $M$, then by Nakayama’s
lemma, $f\in\mathfrak{m}M$. Thus $f=a\cdot s$ for some $a\in\mathfrak{m}$ and
$s\in M$. But then under the appropriate identification,

$g=f^{\otimes r}=(a\cdot s)^{\otimes r}=a^{r}\cdot s^{\otimes
r}\in\mathfrak{m}^{r}N\subseteq\mathfrak{m}N.$

Therefore $g$ cannot generate $N$.

###  4.2 Construction of the Projectivoid Morphism

We can now finish the proof of Theorem 4.6 by constructing an inverse to the
natural tranformation from the theorem. The result follows from the following
proposition.

######  Proposition 4.12.

Let $X$ be a perfectoid space over $K$ and
$\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$.
There is a unique $K$-morphism
$\phi:X\longrightarrow\mathbb{P}^{n,\mathrm{perf}}$ such that

$\left(\phi^{*}\mathscr{O}(1/p^{i}),\phi^{*}T_{j}^{1/p^{i}},\phi^{*}m_{i}\right)\cong\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right).$

$\heartsuit$

###### Proof.

Let $X_{j}=X\left(\frac{s_{0}^{(0)},\cdots,s_{n}^{(0)}}{s^{(0)}_{j}}\right)$
be the cover of $X$ by $\mathscr{L}_{0}$-distinguished opens. Let
$U_{j}=\mathbb{P}^{n,\mathrm{perf}}\left(\frac{T_{0},\cdots,T_{n}}{T_{j}}\right)\subseteq\mathbb{P}^{n,perf}$
be the standard cover by affinoids. The $U_{i}$ are isomorphic to the
perfectoid unit polydisk, and are naturally identified with

$\operatorname{Spa}\left(K\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle,K^{\circ}\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle\right).$

We build $\phi$ locally from maps $\phi_{j}:X_{j}\to U_{j}$. Since $U_{j}$ is
affinoid, it is equivalent to build a map of Huber pairs,

$\left(K\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle,K^{\circ}\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle\right)\stackrel{{\scriptstyle\gamma_{j}}}{{\longrightarrow}}\left(\mathscr{O}_{X}(X_{j}),\mathscr{O}_{X}^{+}(X_{j})\right).$

That is, a ring map

$K\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle\stackrel{{\scriptstyle\gamma_{j}}}{{\longrightarrow}}\Gamma(X_{j},\mathscr{O}_{X}),$

satisfying

$\gamma_{j}\left(K^{\circ}\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle\right)\subseteq\Gamma\left(X_{j},\mathscr{O}_{X}^{+}\right).$

We define $\gamma_{j}$ on generators by the rule

$\gamma_{j}\left(\left(\frac{T_{r}}{T_{j}}\right)^{1/p^{i}}\right)=\frac{s^{(i)}_{r}}{s_{j}^{(i)}}.$

To make sure this is a ring homomorphism, we must check that

$\left(\frac{s^{(i+1)}_{r}}{s^{(i+1)}_{j}}\right)^{p}=\frac{s^{(i)}_{r}}{s^{(i)}_{j}}.$

First notice that, under the identification
$\alpha_{i}:\mathscr{L}_{i+1}^{\otimes p}\cong\mathscr{L}_{i}$, the following
diagram commutes (keeping in mind that the horizontal maps are not
homomorphisms).

${\mathscr{O}_{X}}$${\mathscr{O}_{X}}$${\mathscr{L}_{i+1}}$${\mathscr{L}_{i}.}$$\scriptstyle{x\mapsto
x^{p}}$$\scriptstyle{s_{j}^{(i+1)}}$$\scriptstyle{s^{(i)}_{j}}$$\scriptstyle{s\mapsto
s^{\otimes p}}$

Indeed, the commutativity of this diagram follows directly from the
multilinearity of tensor product together with the identification
$\left(s_{j}^{(i+1)}\right)^{\otimes p}=s_{j}^{(i)}$. Chasing this diagram, we
see that

$\displaystyle\left(\frac{s_{r}^{(i+1)}}{s_{j}^{(i+1)}}\right)^{p}$
$\displaystyle=$
$\displaystyle\left(\left(s_{j}^{(i+1)}\right)^{-1}\left(s_{r}^{(i+1)}\right)\right)^{p}$
$\displaystyle=$
$\displaystyle\left(s_{j}^{(i)}\right)^{-1}\left(s_{r}^{(i)}\right)$
$\displaystyle=$ $\displaystyle\frac{s_{r}^{(i)}}{s_{j}^{(i)}},$

as desired. Therefore $\gamma_{j}$ is a homomorphism. Finally, the definition
of $X_{j}$ implies that for all $x\in X_{j}$,

$\left|\gamma_{j}\left(\frac{T_{i}}{T_{j}}\right)(x)\right|=\left|\frac{s_{i}^{(0)}}{s_{k}^{(0)}}(x)\right|\leq
1,$

so that

$\gamma_{j}\left(\frac{T_{i}}{T_{j}}\right)\in\Gamma\left(X_{j},\mathscr{O}_{X}^{+}\right).$

The multiplicativity of the valuation associated to $x$ shows the same holds
for all $p$th power roots so that

$\gamma_{j}\left(K^{\circ}\left\langle\left(\frac{T_{0}}{T_{j}}\right)^{1/p^{\infty}},\cdots,\left(\frac{T_{n}}{T_{j}}\right)^{1/p^{\infty}}\right\rangle\right)\subseteq\Gamma\left(X_{j},\mathscr{O}_{X}^{+}\right).$

Therefore we get a morphism $\phi_{j}:X_{j}\to
U_{j}\subseteq\mathbb{P}^{n,\mathrm{perf}}$, for each $j$. Notice also that
$s_{r}^{(i)}/s_{j}^{(i)}$ is a $p^{i}$-th root of $s_{r}^{(0)}/s_{j}^{(0)}$ as
desired.

Finally, we check that these morphisms glue to a map
$\phi:X\to\mathbb{P}^{n,\mathrm{perf}}$. This amounts to showing that the
restrictions of $\gamma_{j}$ and $\gamma_{k}$ are equal as maps from
$\Gamma(U_{j}\cap
U_{k},\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}})\longrightarrow\Gamma(X_{j}\cap
X_{k},\mathscr{O}_{X})$. That is, that

$\gamma_{j}\left(\left(\frac{T_{k}}{T_{j}}\right)^{1/p^{i}}\right)=\gamma_{k}\left(\left(\frac{T_{j}}{T_{k}}\right)^{1/p^{i}}\right)^{-1}.$

With our notation, this boils down to

$\frac{s_{k}^{(i)}}{s_{j}^{(i)}}\cdot\frac{s_{j}^{(i)}}{s_{k}^{(i)}}=1.$

But this is just Lemma 4.7.

The rest is immediate from the construction. Since
$\mathscr{O}_{\mathbb{P}^{n\mathrm{perf}}}(d)$ is generated by the monomials
of degree $d$, the construction shows that
$\phi^{*}\mathscr{O}(1/p^{i})=\mathscr{L}_{i}$, and that
$\phi^{*}\left(T_{j}^{1/p^{i}}\right)=s_{j}^{(i)}$. Furthermore, any map
$\psi:X\to\mathbb{P}^{n,\mathrm{perf}}$ with these properties is by definition
given affinoid locally on the standard cover of the target by the ring map
$(T_{j}/T_{k})^{1/p^{i}}\mapsto s^{(i)}_{j}/s^{(i)}_{k}$. That is,
$\psi|_{X_{i}}=\phi_{i}$ so that $\psi=\phi$.

###  4.3 The Positive Characteristic Case

If $X$ is a perfectoid space of characteristic $p$, then the Frobenius
morphism $Frob:\mathbb{G}_{m}\to\mathbb{G}_{m},x\mapsto x^{p}$ is an
isomorphism. Therefore the $p$th power map on $\operatorname{Pic}X$ is an
isomorphism as well, since it is $H^{1}(X,Frob)$. This means that given
$\left(\mathscr{L}_{i},s^{(i)}_{j},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$,
the $\mathscr{L}_{i}$ for $i>0$ are uniquely determined by $\mathscr{L}_{0}$.
Similarly, since $X$ is perfect, the map $\gamma_{i}$ constructed in the proof
of Proposition 4.2 is completely determined by where $T_{r}/T_{i}$ goes for
each $r\not=j$, because the $p$th roots of the image are unique. We summarize
this in the following corollary.

######  Corollary 4.13.

If $X$ is a perfectoid space over $K$ of characteristic $p$, a map
$X\to\mathbb{P}^{n,\mathrm{perf}}$ is equivalent to a line bundle
$\mathscr{L}$ on $X$ and global sections $s_{0},\cdots,s_{n}$ that generate
$\mathscr{L}$, or equivalently, to a map to classical projective space
$\mathbb{P}^{n}$ (as a rigid analytic variety). $\spadesuit$

We can now leverage the tilting equivalence to say that maps to
$X\to\mathbb{P}^{n,\mathrm{perf}}$ in any characteristic are governed by a
single line bundle on $X^{\flat}$. Indeed, by the tilting equivalence we have
that
$\operatorname{Hom}(X,\mathbb{P}^{n,\mathrm{perf}}_{K})=\operatorname{Hom}(X^{\flat},\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}})$.
This implies the following corollary to Theorem 4.6.

######  Corollary 4.14.

If $X$ is a perfectoid space over $K$ of any characteristic, a map
$X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$ is equivalent to a single line bundle
$\mathscr{L}$ on $X^{\flat}$ together $n+1$ global sections generating
$\mathscr{L}$. $\spadesuit$

Using this corollary as an intermediary, we get a natural and geometric
correspondence between certain inverse systems of line bundles on $X$ and
single line bundles on $X^{\flat}$.

######  Corollary 4.15.

An element of $\mathfrak{L}_{n}(X)$ is equivalent to the a line bundle
$\mathscr{L}\in\operatorname{Pic}X^{\flat}$ together with $n+1$ generating
global sections. $\spadesuit$

This will be a useful tool in understanding the relationship between
$\operatorname{Pic}X$ and $\operatorname{Pic}X^{\flat}$.

##  5 Untilting Line Bundles

Recall that one of our motivations was to understand the behavior of vector
bundles under the tilting equivalence. In this section, we use the tools of
projectivoid geometry developed in Section 4 to compare the Picard groups of a
perfectoid space $X$ and its tilt $X^{\flat}$. Indeed, the theory of maps to
projectivoid space allows us to pass between line bundles on $X$ and
$X^{\flat}$ by choosing (compatible) generating sections, constructing the
associated map to projectivoid space, and then using the tilting equivalence
to pass across characteristics. We remark that the theory of _pro-étale
cohomology_ on perfectoid spaces allows us to make this comparison
cohomologically, but the geometric theory we developed in the previous section
gives us a firm geometric grasp.

###  5.1 Cohomological Untilting

In [3], Bhatt and Scholze introduce the _pro-étale_ site for schemes and
perfectoid spaces. We review the definition here.

######  Definition 5.1.

A map $f:Y=\operatorname{Spa}(S,S^{+})\to X=\operatorname{Spa}(R,R^{+})$ of
affinoid perfectoid spaces is called _affinoid pro-étale_ if it can be written
as a cofiltered limit of étale maps
$Y_{i}=\operatorname{Spa}(S_{i},S_{i}^{+})\to X$ of affinoid perfectoid
spaces. More generally, a map $f:Y\to X$ of perfectoid spaces is _pro-étale_
if is locally on the source and target affinoid pro-étale.

The (small) pro-étale site of $X$ is the Grothendieck topology on the category
of perfectoid spaces $f:Y\to X$ pro-étale over $X$ on which a collection
$\\{f_{i}:Y_{i}\to X\\}_{i\in I}$ is a covering if for each quasicompact open
$U\subseteq X$ there exists a finite subset $J\subseteq I$ and quasicompact
open subsets $V_{i}\subseteq Y_{i}$ for $i\in J$ such that $U=\cup_{i\in
J}f_{i}(V_{i})$.

If $\mathscr{F}$ is a pro-étale sheaf on $X$ (that is a sheaf on the pro-étale
site of $X$), the pro-étale cohomology groups
$\mathrm{H}^{i}(X_{\text{pro-\'{e}t}},\mathscr{F})$ are the derived functor
sheaf cohomology groups on the pro-étale site. $\clubsuit$

Let $X$ be a perfectoid space over $K$. The pro-étale sheaf $\mathbb{G}_{m,X}$
maps $U\mapsto\Gamma(U,\mathscr{O}_{U})^{*}$. We have the following theorem.

######  Proposition 5.2.

$\mathrm{H}^{1}(X_{\text{pro-\'{e}t}},\mathbb{G}_{m})\cong\operatorname{Pic}X$.
$\heartsuit$

###### Proof.

For any site $S$, the cohomology group $\mathrm{H}^{1}(X_{S},\mathbb{G}_{m})$
parametrizes isomorphism classes of line bundles on $X$ with respect to the
topology of $S$. Due to [13] Theorem 3.5.8, vector bundles (of any rank) on a
perfectoid space with respect to the pro-étale, étale, and analytic topologies
coincide.

We use the equivalence of the pro-étale topologies of $X$ and $X^{\flat}$ to
construct the tilt of $\mathbb{G}_{m}$ as a pro-étale sheaf on $X$:

$\mathbb{G}_{m,X}^{\flat}:U\mapsto(\Gamma(U,\mathscr{O}_{U})^{\flat})^{*}=\Gamma(U^{\flat},\mathscr{O}_{U^{\flat}})^{*}=\Gamma(U^{\flat},\mathbb{G}_{m,X^{\flat}}).$

The equivalence of the étale topologies on $X$ and $X^{\flat}$ show that
$\mathbb{G}_{m,X}^{\flat}$ is indeed a sheaf. Better yet, the effectiveness of
Čech cohomology on the pro-étale site shows that

$\mathrm{H}^{i}(X_{\text{pro-\'{e}t}},\mathbb{G}_{m,X}^{\flat})\cong\mathrm{H}^{i}(X^{\flat}_{\text{pro-\'{e}t}},\mathbb{G}_{m,X^{\flat}}).$

In particular,
$\mathrm{H}^{1}(X_{\text{pro-\'{e}t}},\mathbb{G}_{m,X}^{\flat})\cong\operatorname{Pic}X^{\flat}$.
Now consider the Kummer sequence for various powers of $p$.

$0\longrightarrow\mu\\!\\!\mu_{p^{n}}\longrightarrow\mathbb{G}_{m,X}\longrightarrow\mathbb{G}_{m,X}\longrightarrow
0.$

This is an exact sequence of sheaves on the pro-étale site of $X$. Indeed,
this can be checked on the stalks, which on the pro-étale site are strictly
Henselian local rings. Therefore we can form an inverse system of exact
sequences:

${0}$${\mu\\!\\!\mu_{p}}$${\mathbb{G}_{m,X}}$${\mathbb{G}_{m,X}}$${0}$${\vdots}$${\vdots}$${\vdots}$${0}$${\mu\\!\\!\mu_{p^{n}}}$${\mathbb{G}_{m,X}}$${\mathbb{G}_{m,X}}$${0}$${0}$${\mu\\!\\!\mu_{p^{n+1}}}$${\mathbb{G}_{m,X}}$${\mathbb{G}_{m,X}}$${0}$${\vdots}$${\vdots}$${\vdots}$

The vertical maps on the the left and middle sides are $x\mapsto x^{p}$.
Taking this limit gives the following sequence.

$0\longrightarrow\mathbb{Z}_{p}(1)\longrightarrow\mathbb{G}_{m,X}^{\flat}\stackrel{{\scriptstyle\sharp}}{{\longrightarrow}}\mathbb{G}_{m,X}.$

The middle term is $\mathbb{G}_{m,X}^{\flat}$ essentially by definition.
Indeed, the construction the tilt of a perfectoid algebra $R$ ([16] Lemma 3.4)
induces an map of multiplicative monoids:

$R^{\flat}\cong\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{x\mapsto
x^{p}}R,$

which restricts to the desired isomorphism on unit groups. Finally, exactness
on the right can be checked explicitly in the pro-étale topology. Indeed,
adjoining a $p$th power root is étale cover so that adjoining all the missing
$p$-power roots gives a pro-étale cover on which $\sharp$ is surjective.
Therefore we have a short exact sequence of pro-étale sheaves:

$0\longrightarrow\mathbb{Z}_{p}(1)\longrightarrow\mathbb{G}_{m,X}^{\flat}\stackrel{{\scriptstyle\sharp}}{{\longrightarrow}}\mathbb{G}_{m,X}\longrightarrow
0.$

###### Remark 5.3

If $R$ is a perfectoid algebra we always get a map of monoids
$\sharp:R^{\flat}\to R$ given by projection onto the first coordinate.
Although it is not a ring homomorphism unless $R$ already had characteristic
$p$, its restriction to unit groups $(R^{\flat})^{*}\to R^{*}$ is a group
homomorphism. This construction is another way of building the map
$\sharp:\mathbb{G}_{m,X}^{\flat}\to\mathbb{G}_{m,X}$. The advantage of the
above construction is that it explicitly exhibits the Tate module
$\mathbb{Z}_{p}(1)$ as the kernel. $\ast$

Taking long exact sequences in cohomology gives us the following diagram,
where the rows are exact.

${\vdots}$${\vdots}$${\vdots}$${\vdots}$${\cdots}$${\mathrm{H}^{1}(X_{\text{pro-\'{e}t}},\mu\\!\\!\mu_{p^{n}})}$${\operatorname{Pic}X}$${\operatorname{Pic}X}$${\mathrm{H}^{2}(X_{\text{pro-\'{e}t}},\mu\\!\\!\mu_{p^{n}})}$${\cdots}$${\cdots}$${\mathrm{H}^{1}(X_{\text{pro-\'{e}t}},\mu\\!\\!\mu_{p^{n+1}})}$${\operatorname{Pic}X}$${\operatorname{Pic}X}$${\mathrm{H}^{2}(X_{\text{pro-\'{e}t}},\mu\\!\\!\mu_{p^{n+1}})}$${\cdots}$${\vdots}$${\vdots}$${\vdots}$${\vdots}$${\cdots}$${\mathrm{H}^{1}(X_{\text{pro-\'{e}t}},\mathbb{Z}_{p}(1))}$${\operatorname{Pic}X^{\flat}}$${\operatorname{Pic}X}$${\mathrm{H}^{2}(X_{\text{pro-\'{e}t}},\mathbb{Z}_{p}(1))}$${\cdots}$$\scriptstyle{\mathscr{L}\mapsto\mathscr{L}^{\otimes
p}}$$\scriptstyle{\theta_{0}}$$\scriptstyle{\theta_{n+1}}$$\scriptstyle{\theta_{n}}$

Taking the inverse limit of the $\theta_{n}$ gives a homomorphism of groups,

$\theta:\operatorname{Pic}X^{\flat}\longrightarrow\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{\mathscr{L}\mapsto\mathscr{L}^{\otimes
p}}\operatorname{Pic}X,$ (7)

and $\theta_{0}$ is this map composed with the projection onto the first
coordinate.

###### Remark 5.4

In Corollary 4.15 we established that inverse systems of $p$th roots of line
bundles (with generating sections) on $X$ correspond to individual line
bundles (with generating sections) on $X^{\flat}$. This seems to suggest that
$\theta$ could be an isomorphism in cases where we have nice maps to
projective space. $\ast$

###  5.2 Untilting Via Maps to Projectivoid Space

We now give a geometric understanding of $\theta$ and $\theta_{0}$ in terms of
maps to projectivoid space, and in doing so study whether whether the
following correspondence holds for a perfectoid space $X$.

$\left\\{\begin{tabular}[]{p{3cm}}Globally generated
$\mathscr{L}\in\operatorname{Pic}X^{\flat}$\end{tabular}\right\\}\leftrightarrow\left\\{\begin{tabular}[]{p{5cm}}Systems
of globally generated line bundles $(\mathscr{L}_{0},\mathscr{L}_{1},\cdots)$
on $X$ such that $\mathscr{L}_{i+1}^{\otimes
p}\cong\mathscr{L}_{i}$.\end{tabular}\right\\}$

We begin by constructing a map in the righthand direction. Given a globally
generated invertible sheaf $\mathscr{L}\in\operatorname{Pic}X^{\flat}$, choose
$n$ sections which generate $\mathscr{L}$. Associated to this data there is a
unique morphism
$\phi^{\flat}:X^{\flat}\longrightarrow\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}$,
which is the tilt of a unique morphism
$\phi:X\longrightarrow\mathbb{P}^{n,\mathrm{perf}}_{K}$. Let
$\mathscr{L}_{i}=\phi^{*}(\mathscr{O}(1/p^{i}))$. This gives a system of
$(\mathscr{L}_{0},\mathscr{L}_{1},\cdots)$ on the right hand side. As a first
step we show that the sheaves $\mathscr{L}_{i}$ do not depend on the choices
of global sections of $\mathscr{L}$.

######  Proposition 5.5.

The construction in the previous paragraph is well defined, and
$(\mathscr{L}_{0},\mathscr{L}_{1},\cdots)=\theta(\mathscr{L})$ where $\theta$
is the cohomological map defined in Equation 7 above. $\heartsuit$

###### Proof.

$\phi^{*}$ can be constructed cohomologically by applying cohomology to the
unit of the adjunction,
$u:\mathbb{G}_{m,\mathbb{P}^{n,\mathrm{perf}}_{K}}\to\phi_{*}\mathbb{G}_{m,X}$
and composing with the natural map
$\mathrm{H}^{1}(\mathbb{P}^{n,\mathrm{perf}},\phi_{*}\mathbb{G}_{m,X})\to\mathrm{H}^{1}(X,\mathbb{G}_{m,X})$.
Pulling $u$ back along the $p$th power map gives $\phi^{\flat*}$ the same way.
Since the $p$th power map commutes with pullback, we get the following
commutative diagram.

${\operatorname{Pic}\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}$${\operatorname{Pic}X^{\flat}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}\operatorname{Pic}\mathbb{P}^{n,\mathrm{perf}}_{K}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}\operatorname{Pic}X.}$$\scriptstyle{\phi^{\flat*}}$$\scriptstyle{\theta_{\mathbb{P}^{n,\mathrm{perf}}}}$$\scriptstyle{\theta_{X}}$$\scriptstyle{\phi^{*}}$

Since
$\mathscr{L}=\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)$
and
$\mathscr{L}_{i}=\phi^{*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K}}(1/p^{i})$,
we have reduced to proving the proposition for
$\mathbb{P}^{n,\mathrm{perf}}_{K}$. Explicitly, we must show

$\theta_{\mathbb{P}^{n,\mathrm{perf}}}(\mathscr{O}(1))=(\mathscr{O}(1),\mathscr{O}(1/p),\mathscr{O}(1/p^{2}),\cdots).$

Since $\operatorname{Pic}\mathbb{P}^{n,\mathrm{perf}}=\mathbb{Z}[1/p]$, and is
therefore uniquely $p$-divisible, it is enough to show that

$\theta_{0,\mathbb{P}^{n,\mathrm{perf}}}\mathscr{O}(1)=\mathscr{O}(1).$

Now $\theta_{0}$ is the cohomological map associated to the Teichmuller map
$\sharp:\mathbb{G}_{m}^{\flat}\longrightarrow\mathbb{G}_{m}$. Scholze showed
in [16] Proposition 5.20 that the Teichmuller map on the perfectoid Tate
algebra maps $T_{i}\mapsto T_{i}$. View $\theta_{0}$ as a map on Čech
cohomology with respect to the standard affine covers, and view
$\mathrm{H}^{1}(\mathbb{P}^{n,\mathrm{perf}},\mathbb{G}_{m})$ as descent data
for building a line bundle (and similarly for the tilt). Then we see that
$\sharp$ sends descent data for $\mathscr{O}(1)$ (which is monomials of degree
one), to monomials of degree one, which build $\mathscr{O}(1)$ on
$\mathbb{P}^{n,\mathrm{perf}}_{K}$.

Therefore untilting line bundles via maps to projectivoid is a well defined
process, as it agrees with the cohomological method which does not depend on
the choice of sections.

We split the study of whether this bijects into the two cases of injectivity
and surjectivity. We first analyze each case and see where the difficulties
may lie.

  * •

Injectivity: Suppose $\mathscr{L},\mathscr{M}\in\operatorname{Pic}(X^{\flat})$
are globally generated. Suppose choosing sections and untilting the associated
maps to projectivoid space gives us maps
$\phi:X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$ and
$\psi:X\to\mathbb{P}^{r,\mathrm{perf}}_{K}$. If
$\phi^{*}(\mathscr{O}(1/p^{i}))\cong\psi^{*}(\mathscr{O}(1/p^{i})=:\mathscr{L}_{i}$
for all $i$, can we conclude that $\mathscr{L}\cong\mathscr{M}$? We can attack
this using the methods of Section 4 by considering the tuples
$(\mathscr{L}_{i},s^{(i)}_{j},\alpha_{i})\in\mathfrak{L}_{n}(X)$ and
$(\mathscr{L}_{i},t^{(i)}_{j},\beta_{i})\in\mathfrak{L}_{r}(X)$ associated to
$\phi$ and $\psi$ respectively. If the $\alpha_{i}$ and $\beta_{i}$ agree, we
can consider
$\left(\mathscr{L}_{i},\left\\{s^{(i)}_{j},t^{(i)}_{k}\right\\},\alpha_{i}\right)\in\mathscr{L}_{n+r+1}(X)$
and consider how the associated map $X\to\mathbb{P}^{n+r+1,\mathrm{perf}}_{K}$
tilts. In this section we settle the case where $\alpha_{i}=\beta_{i}$, and
solve the general case in Section 5.3.

  * •

Surjectivity: Suppose $(\mathscr{L}_{0},\mathscr{L}_{1},\cdots)$ are globally
generated with $\mathscr{L}_{i+1}^{\otimes p}\cong\mathscr{L}_{i}$, and there
are global sections $s_{j}^{(i)}$ generating $\mathscr{L}_{i}$ such that
$\left(s_{j}^{(i)}\right)^{\otimes p}=s_{j}^{(i)}$. Then passing through the
maps to projective space we get $\mathscr{L}\in\operatorname{Pic}X^{\flat}$
which maps to $(\mathscr{L}_{0},\mathscr{L}_{1},\cdots)$ under $\theta$. But,
can we always find sections $s_{j}^{(i)}$ and isomorphisms such that
$\left(s_{j}^{(i+1)}\right)^{\otimes p}=s_{j}^{(i)}$? Restated, are there
generating global sections of $\mathscr{L}_{0}$ all of whose $p$th power roots
exist? If so, our correspondence surjects.

In the rest of this section we settle injectivity in the case where the
isomorphisms $\mathscr{L}_{i+1}^{\otimes p}\cong\mathscr{L}_{i}$ agree for the
two sets of sections.

######  Proposition 5.6.

Let $X$ be a perfectoid space over $K$. Suppose
$\left(\mathscr{L}_{i},s^{(i)}_{j},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$
and
$\left(\mathscr{L}_{i},t^{(i)}_{j},\alpha_{i}\right)\in\mathfrak{L}_{r}(X)$
correspond to maps $\phi:X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$ and
$\psi:X\to\mathbb{P}^{r,\mathrm{perf}}_{K}$ respectively. Then

$\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)\cong\psi^{\flat*}\mathscr{O}_{\mathbb{P}^{r,\mathrm{perf}}_{K^{\flat}}}(1).$

$\heartsuit$

Fix $\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)$ corresponding to a
map $\phi:X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$. As a first step, we show that
we can add one global section to each $\mathscr{L}_{i}$ that are compatible
with the $\alpha_{i}$ without changing the line bundle we get over
$X^{\flat}$. Suppose $t_{i}\in\Gamma(X,\mathscr{L}_{i})$ is a global section
such that $\alpha_{i}\left(t_{i+1}^{\otimes p}\right)=t_{i}$. For every
$\lambda=(\lambda_{0},\lambda_{1},\cdots)\in\operatornamewithlimits{\underset{\longleftarrow}{lim}}K^{*}=K^{\flat*}$,
we let $\psi_{\lambda}:X\longrightarrow\mathbb{P}^{n+1,\mathrm{perf}}_{K}$ be
the projectivoid map corresponding to adding $t_{i}$, that is, corresponding
to $\left(\mathscr{L}_{i},\\{s_{j}^{(i)},t_{i}\\},\alpha_{i}\right)$. We hope
to fit $\phi$ and $\psi_{\lambda}$ in a commutative diagram. To do so we must
develop an analog of rational maps in this analytic context.

If we want to define a map
$\mathbb{P}^{n+1,\mathrm{perf}}\longrightarrow\mathbb{P}^{n,\mathrm{perf}}$
given by
$\left(\mathscr{O}(1/p^{i}),\left\\{T_{0}^{1/p^{i}},\cdots,T_{n}^{1/p^{i}}\right\\},m_{i}\right)$
we would notice that this isn’t defined wherever $|T_{i}/T_{n+1}|>1$. In
particular, it is only defined on the open set:

$U=\bigcup_{j\not=n+1}\mathbb{P}^{n+1,\mathrm{perf}}_{K}\left(\frac{T_{0},\cdots,T_{n+1}}{T_{j}}\right).$

This is the projectivoid analog of projecting away from the hyperplane where
$T_{n+1}$ vanishes, (here we are projecting away from a perfectoid disk at the
‘north pole’). Unfortunately, the image of $\psi_{\lambda}$ does not lie in
$U$, because there may be points $x$ where
$\left|(s_{j}^{(0)}/t_{0})(x)\right|>1$ for all $i$, so that
$|(T_{i}/T_{n+1})(\psi_{\lambda}(x))|>1$. But, restricted to the open set

$V_{\lambda}=\bigcup_{j}X\left(\frac{s_{0}^{(0)},\cdots,s_{n}^{(0)},t_{0}}{s_{j}^{(0)}}\right),$

the image of $\psi_{\lambda}$ does lie in $U$. Thus we have the following
commutative diagram for every $\lambda$.

${\mathbb{P}^{n,\mathrm{perf}}_{K}}$${X}$${V_{\lambda}}$${U}$${\mathbb{P}^{n+1,\mathrm{perf}}_{K}}$$\scriptstyle{\phi}$$\scriptstyle{\psi_{\lambda}}$$\scriptstyle{\psi_{\lambda}}$$\scriptstyle{\pi}$

######  Lemma 5.7.

The sets $V_{(\varpi^{\flat})^{r}}$ form an open cover of $X$. As a
consequence the sets $V_{(\varpi^{\flat})^{r}}^{\flat}$ cover $X^{\flat}$.
$\heartsuit$

###### Proof.

Notice $(\varpi^{\flat})^{r}=(\varpi^{r},\varpi^{r/p},\cdots)$. Fix $x\in X$.
There is some $j$ such that $x\in
X\left(\frac{s_{0}^{(0)},\cdots,s_{n}^{(0)}}{s_{j}^{(0)}}\right)$.
Furthermore, since $\varpi$ is topologically nilpotent, there is some $r$ such
that

$\left|(\varpi^{r}t_{0}/s_{j}^{(0)})(x)\right|=|\varpi^{r}|\cdot\left|(t_{0}/s_{j}^{(0)})(x)\right|<1$

proving the first statement. The second is an immediate consequence of the
tilting equivalence.

######  Lemma 5.8.

For any $\lambda\in K^{\flat*}$,

$\left(\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)\right)|_{V^{\flat}_{\lambda}}\cong\psi_{\lambda}^{\flat*}\left(\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1)|_{U^{\flat}}\right)\cong\left(\psi_{\lambda}^{\flat*}\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1)\right)|_{V_{\lambda}^{\flat}}$

$\heartsuit$

###### Proof.

This follows from the commutativity of the tilt of the diagram above,
reproduced below, together with the fact that $\pi^{\flat}$ is given by the
line bundle $\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1)|_{U}$
together with the sections $T_{0},\cdots,T_{n}$.

${\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}$${X^{\flat}}$${V^{\flat}_{\lambda}}$${U^{\flat}}$${\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}$$\scriptstyle{\phi^{\flat}}$$\scriptstyle{\psi^{\flat}_{\lambda}}$$\scriptstyle{\psi^{\flat}_{\lambda}}$$\scriptstyle{\pi^{\flat}}$

######  Lemma 5.9.

Fix any
$\lambda,\xi\in\operatornamewithlimits{\underset{\longleftarrow}{lim}}K^{*}=K^{\flat*}$.
Then

$\psi^{\flat*}_{\lambda}\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1)\cong\psi^{\flat*}_{\xi}\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1).$

$\heartsuit$

###### Proof.

Let
$\tau:\mathbb{P}^{n+1,\mathrm{perf}}_{K}\to\mathbb{P}^{n+1,\mathrm{perf}}_{K}$
be the map determined by
$\left(\mathscr{O}(1/p^{i}),\left\\{T_{0}^{1/p^{i}},\cdots,T_{n}^{1/p^{i}},\frac{\gamma_{i}}{\xi_{i}}T_{n+1}^{1/p^{i}}\right\\},m_{i}\right)$.
Then $\tau$ is an isomorphism, and $\tau^{\flat}$ is the map determined by
$\mathscr{O}(1)$ and $T_{0},\cdots,T_{n},\frac{\gamma}{\xi}T_{n+1}$. We have
the following two commutative diagrams, the right hand diagram being the tilt
of the left.

${\mathbb{P}^{n+1,\mathrm{perf}}_{K}}$${\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}$${X}$${X^{\flat}}$${\mathbb{P}^{n+1,\mathrm{perf}}_{K}}$${\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}$$\scriptstyle{\tau}$$\scriptstyle{\tau^{\flat}}$$\scriptstyle{\psi_{\lambda}}$$\scriptstyle{\psi_{\xi}}$$\scriptstyle{\psi_{\lambda}^{\flat}}$$\scriptstyle{\psi_{\xi}^{\flat}}$

Since $\tau^{\flat*}\mathscr{O}(1)=\mathscr{O}(1)$, we are done.

Putting these three lemmas together, we conclude that

$\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)\cong\psi_{1}^{\flat*}\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1).$

Indeed, the pullback of $\mathscr{O}(1)$ along $\psi_{1}^{\flat}$ agrees with
the pullback along $\psi_{(\varpi^{\flat})^{r}}$, for any $r$, but this agrees
with the restriction of
$\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)$ to
$V_{(\varpi^{\flat})^{r}}^{\flat}$ for any $r$. Since these sets cover
$X^{\flat}$, we are done.

In summary, we have proved the following proposition.

######  Proposition 5.10.

Let
$\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$,
correspond to a map $\phi:X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$. Suppose
$t_{i}\in\Gamma(X,\mathscr{L}_{i})$ is a global section such that
$\alpha_{i}\left(t_{i+1}^{\otimes p}\right)=t_{i}$, and let
$\psi:X\to\mathbb{P}^{n+1}_{K}$ be the map associated to
$\left(\mathscr{L}_{i},\\{s_{j}^{(i)},t_{i}\\},\alpha_{i}\right)\in\mathfrak{L}_{n+1}(X)$.
Then

$\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)\cong\psi^{\flat*}\mathscr{O}_{\mathbb{P}^{n+1,\mathrm{perf}}_{K^{\flat}}}(1).$

$\heartsuit$

Adding sections one at a time by induction completes the proof of Proposition
5.6.

###  5.3 Injectivity of $\theta$

With these tools in hand, we can prove the injectivity of $\theta$ for certain
perfectoid spaces $X$. We will first need one more lemma.

######  Lemma 5.11.

Let
$\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$
correspond to a map $\phi:X\to\mathbb{P}^{n,\mathrm{perf}}_{K}$. Fix
$\lambda=(\lambda_{0},\lambda_{1},\cdots)\in\Gamma(X,\mathscr{O}_{X}^{\flat*})$,
that is, $\lambda_{i+1}^{p}=\lambda_{i}$, so that
$\left(\mathscr{L}_{i},\lambda_{i}s_{0}^{(i)},\lambda_{i}\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$
corresponds to a map $\psi:X\to\mathbb{P}^{n}_{K}$. Then $\phi=\psi$.
$\heartsuit$

###### Proof.

Multiplication by $\lambda_{i}$ for each $i$ gives us an isomorphism
$\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}\left(\mathscr{L}_{i},\lambda_{i}s_{0}^{(i)},\lambda_{i}\alpha_{i}\right)$
in $\mathfrak{L}_{n}(X)$. Then we are done by Theorem 4.6

Before we state the main theorem we make the following definition.

######  Definition 5.12.

A line bundle $\mathscr{L}$ on a perfectoid space $X$ is said to be _weakly
ample_ if for any other line bundle $\mathscr{M}$, there is some $N>>0$ such
that for all $r>N$ we have $\mathscr{M}\otimes\mathscr{L}^{r}$ globally
generated. $\clubsuit$

######  Theorem 5.13.

Suppose $X$ is a perfectoid space over $K$. Suppose that $X$ has a weakly
ample line bundle and that
$H^{0}(X_{\overline{K}},\mathscr{O}_{X_{\overline{K}}})=\overline{K}$, where
$\overline{K}$ is a fixed algebraic closure of $K$. Then

$\theta:\operatorname{Pic}X^{\flat}\hookrightarrow\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{\mathscr{L}\mapsto\mathscr{L}^{p}}\operatorname{Pic}X.$

In particular, if $\operatorname{Pic}X$ has no $p$ torsion, then

$\theta_{0}:\operatorname{Pic}X^{\flat}\hookrightarrow\operatorname{Pic}X.$

$\diamondsuit$

###### Proof.

Fix $\mathscr{L},\mathscr{M}\in\operatorname{Pic}X^{\flat}$ with
$\theta(\mathscr{L})=\theta(\mathscr{M})$. We first reduce to the case that
$\mathscr{L},\mathscr{M}$ are globally generated. Indeed, letting
$\mathscr{F}$ be a weakly ample line bundle, we have
$\theta(\mathscr{L}\otimes\mathscr{F}^{N})=\theta(\mathscr{M}\otimes\mathscr{F}^{N})$.
If the result holds for globally generated line bundles, for large enough $N$
we conclude that
$\mathscr{L}\otimes\mathscr{F}^{N}\cong\mathscr{M}\otimes\mathscr{F}^{N}$ so
that $\mathscr{L}\cong\mathscr{M}$.

Next we prove it for the case where $K$ contains all $p$th power roots for all
its elements. Choose generating sections $s_{0},\cdots,s_{n}$ for
$\mathscr{L}$ and $t_{0},\cdots,t_{r}$ of $\mathscr{M}$, which give us maps

$\phi^{\flat}:X^{\flat}\longrightarrow\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}},$

and

$\psi^{\flat}:X^{\flat}\longrightarrow\mathbb{P}^{r,\mathrm{perf}}_{K^{\flat}},$

respectively. These untilt to

$\phi:X\longrightarrow\mathbb{P}^{n,\mathrm{perf}}_{K},$

and

$\psi:X\longrightarrow\mathbb{P}^{r,\mathrm{perf}}_{K},$

which in turn correspond to tuples
$\left(\mathscr{L}_{i},s_{j}^{(i)},\alpha_{i}\right)\in\mathfrak{L}_{n}(X)$
and
$\left(\mathscr{L}_{i},t_{j}^{(i)},\beta_{i}\right)\in\mathfrak{L}_{r}(X)$.
Notice that $\alpha_{i}$ and $\beta_{i}$ differ by an element

$\lambda_{i}\in\operatorname{Isom}(\mathscr{L}_{i},\mathscr{L}_{i})=\Gamma(X,\mathscr{O}_{X}^{*})=K^{*}.$

That is, $\alpha_{i}=\lambda_{i}\beta_{i}$. Choose $p$th power roots
$\lambda_{i}^{1/p^{j}}$ for each $i,j$ (these exist by assumption), and for
all $j$ define:

$\displaystyle\tilde{t}_{j}^{(0)}$ $\displaystyle=$ $\displaystyle
t_{j}^{(0)}$ $\displaystyle\tilde{t}_{j}^{(1)}$ $\displaystyle=$
$\displaystyle\lambda_{0}^{-1/p}t_{j}^{(1)}$
$\displaystyle\tilde{t}_{j}^{(2)}$ $\displaystyle=$
$\displaystyle\lambda_{1}^{-1/p}\lambda_{0}^{-1/p^{2}}t_{j}^{(2)}$
$\displaystyle\vdots$ $\displaystyle\tilde{t}_{j}^{(i+1)}$ $\displaystyle=$
$\displaystyle\lambda_{i}^{-1/p}\lambda_{i-1}^{-1/p^{2}}\cdots\lambda_{0}^{-1/p^{i+1}}t_{j}^{(i+1)}$
$\displaystyle\vdots$

Then

$\displaystyle\alpha_{i}\left(\left(\tilde{t}_{j}^{(i+1)}\right)^{\otimes
p}\right)$ $\displaystyle=$
$\displaystyle\lambda_{i}\beta_{i}\left(\left(\lambda_{i}^{-1/p}\lambda_{i-1}^{-1/p^{2}}\cdots\lambda_{0}^{-1/p^{i+1}}t_{j}^{(i+1)}\right)^{\otimes
p}\right)$ $\displaystyle=$
$\displaystyle\lambda_{i}\lambda_{i}^{-1}\lambda_{i-1}^{1/p}\cdots\lambda_{0}^{1/p^{i}}\beta\left(\left(t_{j}^{(i+1)}\right)^{\otimes
p}\right)$ $\displaystyle=$
$\displaystyle\lambda_{i-1}^{1/p}\cdots\lambda_{0}^{1/p^{i}}t_{j}^{(i)}$
$\displaystyle=$ $\displaystyle\tilde{t}_{j}^{(i)}.$

Therefore the tuple
$(\mathscr{L}_{i},\tilde{t}_{j}^{(i)},\alpha_{i})\in\mathfrak{L}_{n}(X)$, and
it also corresponds to $\psi$ by Lemma 5.11. Furthermore, the isomorphisms
corresponding to this data are now $\alpha_{i}$ in both cases, so that by
Proposition 5.6

$\mathscr{L}=\phi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)\cong\psi^{\flat*}\mathscr{O}_{\mathbb{P}^{n,\mathrm{perf}}_{K^{\flat}}}(1)=\mathscr{M}.$

For the general case, we let $L/K$ be the extension given by adjoining all
$p$th power roots of all elements of $K$. We have the following diagram.

${\operatorname{Pic}X_{L}^{\flat}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}\operatorname{Pic}X_{L}}$${\operatorname{Pic}X^{\flat}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}\operatorname{Pic}X}$$\scriptstyle{\theta_{L}}$$\scriptstyle{\theta}$

$\theta_{L}$ injects by the argument we just made. Furthermore, since
$X^{\flat}_{L}\to X^{\flat}$ is a pro-étale cover of $p$th power degree, the
kernel of $\operatorname{Pic}X^{\flat}\to\operatorname{Pic}X_{L}^{\flat}$ is
$p$th power torsion. Since $X^{\flat}$ is perfect,
$\operatorname{Pic}X^{\flat}$ has no $p$th power torsion, so the map injects.
Therefore $\theta$ injects.

###### Open Problem 5.14

In which contexts does $\theta$ surject? $\ast$

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