# Jordan property for groups of bimeromorphic self-maps of complex manifolds
with large Kodaira dimension

Konstantin Loginov  Steklov Mathematical Institute of Russian Academy of
Sciences, Moscow, Russia; Laboratory of Algebraic Geometry, HSE University,
Russian Federation; Laboratory of AGHA, Moscow Institute of Physics and
Technology<EMAIL_ADDRESS>
###### Abstract.

We prove that the image of the pluricanonical representation of a group of
bimeromorphic automorphisms of a complex manifold has bounded finite
subgroups. As a consequence, we show that the group of bimeromorphic
automorphisms of an $n$-dimensional complex manifold whose Kodaira dimension
is at least $n-2$, satisfies the Jordan property.

##  1\. Introduction

In this note, we study the groups of bimeromorphic automorphisms of complex
manifolds. More precisely, we concentrate on the finite subgroups of such
groups. One way to study their structure is to establish the so called _Jordan
property_ , that is, to show that any finite subgroup of a given group admits
a normal abelian subgroup of bounded index.

In what follows, we denote by $\mathrm{Aut}(X)$ the group of automorphisms of
a complex manifold $X$, by $\mathrm{Bim}(X)$ the group of bimeromorphic self-
maps of $X$, and by $\kappa(X)$ the Kodaira dimension of $X$.

We list several results on the Jordan property, see Definition 2.3. If $X$ is
a compact complex surface, then $\mathrm{Aut}(X)$ is Jordan, while
$\mathrm{Bim}(X)$ is Jordan if and only if $X$ is not bimeromorphic to a
product of $\mathbb{P}^{1}$ and an elliptic curve, see [PSh21, Theorems 1.6
and 1.7]. If $X$ is a projective variety over a field of characteristic zero,
then the group $\mathrm{Aut}(X)$ is Jordan [MZh18].

For Kähler varieties, there are the following results. If $X$ is a compact
Kähler variety, then the group $\mathrm{Aut}(X)$ is Jordan [Kim18]. If $X$ be
a compact $3$-dimensional Kähler variety with $\kappa(X)\geq 0$ and
$q(X)=\mathrm{H}^{1}(X,\mathcal{O}_{X})>0$, then the group $\mathrm{Bim}(X)$
is Jordan [PSh21b]. Recently, A. Golota showed that it is possible to get rid
of the assumption on the irregularity of $X$:

######  Theorem 1.1 ([Go21]).

Let $X$ be a $3$-dimensional compact Kähler variety with $\kappa(X)\geq~{}0$.
Then the group $\mathrm{Bim}(X)$ is Jordan.

Our goal is to derive an analogous result for complex varieties without the
assumption that $X$ is Kähler. We prove the following

######  Theorem 1.2.

Let $X$ be an $n$-dimensional compact complex variety with $\kappa(X)\geq
n-2$. Then the group $\mathrm{Bim}(X)$ is Jordan.

######  Corollary 1.3.

Let $X$ be a $3$-dimensional compact complex variety with $\kappa(X)>0$, then
the group $\mathrm{Bim}(X)$ is Jordan.

Our proof is based on the study of bounded finite subgroup property for the
pluricanonical representation of the group of bimeromorphic self-maps. More
precisely, let us denote by $\rho$ the _pluricanonical representation_ of the
group $\mathrm{Bim}(X)$, that is, the action of $\mathrm{Bim}(X)$ on the space
of holomorphic $m$-forms. Also, by $\overline{\rho}(\mathrm{Bim}(X))$ we
denote the projectivization of the pluricanonical representation which acts on
image of the pluricanonical map, see section 2.1 for details. For convenience,
we call it the _projective pluricanonical representation_. We say that a group
has _bounded finite subgroups_ , if the cardinality of any finite subgroup of
such group is bounded by a constant which is independent of the subgroup. Then
we prove the following

######  Theorem 1.4.

Let $X$ be a compact complex manifold. Then both the image of the
pluricanonical representation $\rho(\mathrm{Bim}(X))$ and the image of the
projective pluricanonical representation $\overline{\rho}(\mathrm{Bim}(X))$
have bounded finite subgroups.

Combining Theorem 1.1 and Theorem 1.4, we deduce

######  Corollary 1.5.

Let $X$ be an $n$-dimensional compact Kähler variety with $\kappa(X)\geq n-3$.
Then the group $\mathrm{Bim}(X)$ is Jordan.

In the proof of Theorem 1.4, we follow the ideas used by Ueno in [Ue75, §14]
where the following result, whose proof is attributed to Deligne, was
established:

######  Theorem 1.6 ([Ue75, Corollary 14.10]).

Let $X$ be a compact complex manifold. Assume that $X$ is Moishezon. Then the
group $\rho(\mathrm{Bim}(X))$ is finite.

In contrast, in Theorem 1.4 we do not need the assumption that $X$ is
Moishezon, but instead of finiteness, we establish a weaker property of
$\rho(\mathrm{Bim}(X))$. However, this is the best we can get as there exist
compact complex manifolds with infinite $\rho(\mathrm{Bim}(X))$, see Example
4.8. We can also ask about finiteness of the group
$\overline{\rho}(\mathrm{Bim}(X))$. If $X$ is a compact complex surface, then
$\overline{\rho}(\mathrm{Bim}(X))$ is finite. Indeed, finiteness of
$\overline{\rho}(\mathrm{Bim}(X))$ is trivial for $\kappa(X)=0$, for
$\kappa(X)=1$ its finiteness is proven in [PSh21, Lemma 8.2], and for
$\kappa(X)=2$ the group $\mathrm{Bim}(X)$ is itself finite by [Ue75, Corollary
14.3]. In higher dimensions, we do not know about finiteness results for
$\overline{\rho}(\mathrm{Bim}(X))$ when the Kodaira dimension is not maximal
possible. Hence we formulate the following

######  Question 1.7.

Does there exist a compact complex manifold $X$ such that the image of the
projective pluricanonical representation $\overline{\rho}(\mathrm{Bim}(X))$ is
infinite?

We refer to [HX16] and references therein for finiteness results in the case
of projective pairs.

Acknowledgements. The work was performed at the Steklov International
Mathematical Center and supported by the Ministry of Science and Higher
Education of the Russian Federation (agreement no. 075-15-2022-265), by the
HSE University Basic Research Program, Russian Academic Excellence Project
“5-100”, and the Simons Foundation. He is Young Russian Mathematics award
winner and would like to thank its sponsors and jury. The author is grateful
to Constantin Shramov for suggesting the problem, careful reading of the draft
of the paper, and useful discussions. The author thanks Andrey Ryabichev,
Andrey Soldatenkov, Aleksei Golota and Alexander Kuznetsov for helpful
conversations.

##  2\. Preliminaries

We work over the field of complex numbers. We refer the reader to [Ue75] for
the basic facts and definitions.

We recall some of them. By a _complex variety_ we mean an irreducible reduced
complex space. A smooth complex variety is called a _complex manifold_. A
_complex surface_ is a complex manifold of dimension $2$. By a (Zariski) open
subset of complex manifold $X$ we mean a subset of the form $X\setminus Z$,
where $Z$ is a closed analytic subset in $X$. By a _typical point_ of $X$ we
mean a point in some non-empty Zariski open subset of $X$.

A proper surjective holomorphic map $f\colon X\to Y$ of reduced complex spaces
is called a _modification_ if there exist nowhere dense closed analytic
subsets $V\subset X$ and $W\subset Y$ such that $f$ restricts to a
biholomorphic map $X\setminus V\to Y\setminus W$. A _meromorphic map_ $f\colon
X\dashrightarrow Y$ of reduced complex analytic spaces $X$ and $Y$ is a
holomorphic map defined outside a nowhere dense subset such that the closure
of its graph $\overline{\Gamma_{f}}\subset X\times Y$ is a closed analytic
subset of $X\times Y$, and the natural projection $\overline{\Gamma_{f}}\to X$
is a modification. A meromorphic map $f$ is called _bimeromorphic_ if the
natural projection $\overline{\Gamma_{f}}\to Y$ is a modification as well.

Given a reduced complex space $X$, by $\mathrm{Bim}(X)$ we denote its group of
bimeromorphic self-maps, and by $\mathrm{Aut}(X)$ we denote its group of
biholomorphic self-maps.

######  Example 2.1.

A bimeromorphic self-map of an open subset $U$ in a complex variety $X$ may
not induce a bimeromorphic self-map of $X$. Indeed, by [VT87, Section 2] there
exist complex manifolds $X_{1}$ and $X_{2}$ such that they are two non-
bimeromorphic compactifications of $(\mathbb{C}^{\times})^{2}$. Put
$(\mathbb{C}^{\times})^{2}=U_{1}\subset X_{1}$,
$(\mathbb{C}^{\times})^{2}=U_{2}\subset X_{2}$, and

$X=X_{1}\times X_{2}\supset U=U_{1}\times U_{2}.$

Define a biholomorphic map $f$ on $U=U_{1}\times U_{2}$ by the formula
$f(x_{1},x_{2})=(x_{2},x_{1})$ when $x_{i}\in U_{i}$ for $i=1,2$, and $U_{1}$
is identified with $U_{2}$. Assume that $f$ induces a bimeromomorphic map on
$X$. Consider the closure of its graph

$\overline{\Gamma_{f}}\subset X\times X=X_{1}\times X_{2}\times X_{1}\times
X_{2}.$

Then both projections

$p_{i}=\mathrm{pr}_{i}|_{\overline{\Gamma_{f}}}\colon\overline{\Gamma_{f}}\to
X=X_{1}\times X_{2}$

are modifications. Consider the subset $U_{1}\times\\{q\\}\subset
X_{1}\times\\{q\\}$ where $q\in U_{2}$. We have
$f(U_{1}\times\\{q\\})=\\{q\\}\times U_{2}$. Hence

$p_{1}^{-1}(U_{1}\times\\{q\\})=U_{1}\times\\{q\\}\times\\{q\\}\times
U_{1}\subset\overline{\Gamma_{f}},$

Put

$\Gamma_{g}={X_{1}\times\\{q\\}\times\\{q\\}\times
X_{2}}\cap\overline{\Gamma_{f}}.$

Consider $\Gamma_{g}$ as a closed analytic subset that defines a graph of a
bimeromorphic map $g\colon X_{1}\dashrightarrow X_{2}$. Indeed, this follows
from the fact that the proper maps $\Gamma_{g}\to X_{i}$ are modifications for
$i=1,2$. This contradiction shows that $f$ does not induce a bimeromorphic map
on $X$.

###  2.1. Pluricanonical representation

Let $X$ be a complex manifold. Assume that for the Kodaira dimension of $X$ we
have $\kappa(X)\geq 0$. For any $m\geq 1$, consider the map

$\sigma_{m}\colon
X\dasharrow\mathbb{P}(\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))^{\vee}).$

For a sufficiently big and divisible $m$, we have
$\dim\sigma_{m}(X)=\kappa(X)$ and the fibers of $\sigma_{m}$ are connected.
For such $m$, we put $\sigma=\sigma_{m}$. Passing to a modification of $X$, we
may assume that $\sigma$ is a morphism. Obviously, this does not change the
group $\mathrm{Bim}(X)$. Then a typical fiber of $\sigma$ is smooth, and hence
irreducible (cf. [Ue75, Theorem 5.10]). We have the following

######  Proposition 2.2 (see e.g. [Ue75, Lemma 6.3]).

The group $\mathrm{Bim}(X)$ acts on $\sigma(X)$ biholomorphically.

Put $V=\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$ and consider the induced
homomorphism

(2.1) $\rho=\rho_{m}\colon\mathrm{Bim}(X)\to\mathrm{GL}(V),$

which is called a _pluricanonical representation_ of $\mathrm{Bim}(X)$.
Consider the natural exact sequence of groups

$1\to\mathbb{C}^{\times}\to\mathrm{GL}(V)\xrightarrow{p}\mathrm{PGL}(V)\to 1.$

Put $\overline{\rho}=p\circ\rho$. Then by the _projective pluricanonical
representation_ of $\mathrm{Bim}(X)$ we mean the image
$\overline{\rho}(\mathrm{Bim}(X))$. Observe that the action of
$\overline{\rho}(\mathrm{Bim}(X))$ on $\sigma(X)$ is faithful.

###  2.2. Group theory

We start with the following definition.

######  Definition 2.3.

We say that a group $\Gamma$ has _bounded finite subgroups_ , if there is a
constant $C=~{}C(\Gamma)$ such that for any finite subgroup $G\subset\Gamma$
we have $|G|\leq C$.

Following [Po09, Definition 2.1], we say that a group $\Gamma$ is _Jordan_ (or
has _Jordan property_) if there is a constant $J=J(\Gamma)$ such that every
finite subgroup $G\subset\Gamma$ contains a normal abelian subgroup $A\subset
G$ of index at most $J$.

Note that the Jordan property (as well as the property to have bounded finite
subgroups) behaves badly with respect to passing to the quotient group.
Indeed, any group $G$ which does not enjoy this property can be realised as a
quotient of a free group on some number of generators, and the free group
clearly satisfies Jordan property (and has bounded finite subgroups) since the
only finite subgroup in it is the trivial group. See also Example 2.10 in the
case when $G$ is a subgroup of the general linear group. Now, let us make the
following straightforward observation.

######  Remark 2.4.

Suppose that the following sequence of groups is exact

$1\to G_{1}\to G\to G_{2}\to 1.$

Assume that $G_{2}$ has bounded finite subgroups. Then $G$ is Jordan if and
only if $G_{1}$ is Jordan.

######  Example 2.5.

A classical result of C. Jordan says that for every positive integer $n$ the
group $\mathrm{GL}_{n}(\mathbb{C})$ is Jordan, see e.g. [CR62, Theorem 36.13].
As a consequence, any linear algebraic groups over an arbitrary field of
characteristic zero is Jordan.

The group $\mathrm{GL}_{n}(\mathbb{\mathbb{K}})$ has bounded finite subgroups
for any field $\mathbb{K}$ which is finitely generated over $\mathbb{Q}$. This
theorem was proven by H. Minkowski, see e.g. [Ser07, Theorem 5] and [Ser07,
§4.3]. It follows that $\mathrm{GL}_{n}(\mathbb{Z})$ has bounded finite
subgroups.

The next result can be seen as an effective version of Burnside theorem for
subgroups of the general linear group.

######  Theorem 2.6 ([HP76, Theorem 1]).

Let $n$ and $d$ be positive integers. Let
$\Gamma\subset\mathrm{GL}_{n}(\mathbb{C})$ be a subgroup. Suppose that for
every $g\in\Gamma$ one has $g^{d}=1$. Then $\Gamma$ is finite and $|\Gamma|$
divides $d^{n}$.

######  Corollary 2.7.

Let $\Gamma\subset\mathrm{PGL}_{n}(\mathbb{C})$ be a subgroup. Suppose that
for every $g\in\Gamma$ one has $g^{d}=1$. Then $\Gamma$ is finite and
$|\Gamma|$ divides $(nd)^{n}$.

###### Proof.

Write the following exact sequence

$0\to\mathbb{Z}/n\mathbb{Z}\to\mathrm{SL}(V)\xrightarrow{p^{\prime}}\mathrm{PGL}(V)\to
1.$

Consider the group $\widetilde{\Gamma}=p^{\prime-1}(\Gamma)$. Pick any element
$\widetilde{g}\in\widetilde{\Gamma}$ and put $g=p^{\prime}(\widetilde{g})$.
Note that $\operatorname{ord}(\widetilde{g})$ divides $n\operatorname{ord}(g)$
which divides $nd$ by assumption. By Theorem 2.6, we have that
$\widetilde{\Gamma}$ is finite and $|\widetilde{\Gamma}|$ divides $(nd)^{n}$.
Hence $\Gamma$ is finite and $|\Gamma|$ divides $(nd)^{n}$ as well. ∎

Now we establish some boundedness properties of subgroups of the general
linear group that we will use later.

######  Lemma 2.8.

Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$. Let
$g\in\mathrm{GL}(V)$ be an element of finite order $d$. Assume that there
exists a constant $C$ such that for any eigenvalue $\alpha$ of $g$ the degree
of the field extension $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is bounded by $C$.
Then

$d\leq 2^{n}C^{2n}.$

###### Proof.

Let $\alpha_{1},\ldots,\alpha_{n}$ be the eigenvalues of $g$. Then each
$\alpha_{i}$ is a root of unity whose degree we denote by $d_{i}$, so that we
have $\alpha_{i}^{d_{i}}=1$. Then $d=\operatorname{lcm}(d_{i})_{1\leq i\leq
n}$. It is well known that

$[\mathbb{Q}(\alpha_{i}):\mathbb{Q}]=\varphi(d_{i})$

where $\varphi(d_{i})$ is the Euler’s totient function. One easily checks that
$d_{i}\leq 2\varphi(d_{i})^{2}$, so we estimate

$\sqrt{d_{i}/2}\leq\varphi(d_{i})=[\mathbb{Q}(\alpha_{i}):\mathbb{Q}]\leq C.$

Consequently,

$d=\operatorname{lcm}(d_{i})_{1\leq i\leq n}\leq d_{1}\cdot\ldots\cdot
d_{n}\leq 2^{n}C^{2n},$

which completes the proof. ∎

######  Lemma 2.9.

Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$, and let
$\Gamma\subset\mathrm{GL}(V)$ be a subgroup. Consider an exact sequence of
groups

$1\to\mathbb{C}^{\times}\to\mathrm{GL}(V)\xrightarrow{p}\mathrm{PGL}(V)\to 1.$

Assume that there exists a constant $C$ such that for any eigenvalue $\alpha$
of $g$ the degree of the field extension $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is
bounded by $C$. Then both $\Gamma$ and $p(\Gamma)$ have bounded finite
subgroups.

###### Proof.

First we show that $\Gamma$ has bounded finite subgroups. Consider a finite
subgroup $H\subset\Gamma$. By Lemma 2.8, given any element $g\in H$, for its
order $d=\operatorname{ord}(g)$ we have

$d\leq 2^{n}C^{2n}.$

Using Theorem 2.6, we conclude that $|H|$ is bounded by $2^{n^{2}}C^{2n^{2}}$.
Hence $\Gamma$ has bounded finite subgroups.

Now we prove that $p(\Gamma)$ has bounded finite subgroups. Consider a finite
subgroup $G\subset p(\Gamma)$. By Corollary 2.7, it is enough to bound the
order $d$ of an arbitrary element $g\in G$ by a constant which is independent
of $g$. Pick any element $\widetilde{g}\in\mathrm{GL}(V)$ in
$p^{-1}(g)\cap\Gamma$. Note that $\widetilde{g}$ may have infinite order.
Denote by $\alpha_{1},\ldots,\alpha_{n}$ the eigenvalues of $\widetilde{g}$.
Since $g^{d}=\mathrm{id}$, we have $\widetilde{g}^{d}=\lambda\cdot\mathrm{id}$
for some $\lambda\in\mathbb{C}^{\times}$. In particular,
$\alpha_{i}^{d}=\lambda$ for any $1\leq i\leq n$. Thus, the element
$\alpha_{1}^{-1}\widetilde{g}$ satisfies

$\operatorname{ord}(\alpha_{1}^{-1}\widetilde{g})=d.$

Note that its eigenvalues are $1$ and $\alpha_{i}/\alpha_{1}$ for $2\leq i\leq
n$. Each $\alpha_{i}/\alpha_{1}$ belongs to the field
$\mathbb{Q}(\alpha_{1},\alpha_{i})$, and the degree of the field extension
$[\mathbb{Q}(\alpha_{1},\alpha_{i}):\mathbb{Q}]$ is bounded by $C^{2}$.
Applying Lemma 2.8 to the element $\alpha_{0}^{-1}\widetilde{g}$, we see that

$d\leq 2^{n}C^{4n}.$

Using Corollary 2.7, we conclude that $|G|$ is bounded by
$2^{n^{2}}C^{4n^{2}}n^{n}$. Hence $p(\Gamma)$ has bounded finite subgroups. ∎

The following example due to C. Shramov shows that in general if
$\Gamma\subset\mathrm{GL}(V)$ has bounded finite subgroups, it might be not
true that $\Gamma^{\prime}=p(\Gamma)\subset\mathrm{PGL}(V)$ enjoys the same
property.

######  Example 2.10.

Let $V$ be a $2$-dimensional vector space over $\mathbb{C}$. By $U$ we denote
the set of all complex roots of unity. Start with a group

$\Gamma^{\prime}=\left\\{\begin{pmatrix}\xi&0\\\ 0&1\end{pmatrix}\Big{|}\
\,\xi\in U\,\right\\}\subset\mathrm{PGL}(V).$

Observe that $\Gamma^{\prime}\simeq\mathbb{Q}/\mathbb{Z}$. Now, we define a
subgroup $\Gamma\subset\mathrm{GL}(V)$ with $p(\Gamma)=\Gamma^{\prime}$.
Choose a countable subset in $\mathbb{C}$ consisting of elements
$\\{\zeta(\xi)\\}_{\xi\in U}$, indexed by all roots of unity $\xi\in U$, that
are algebraically independent over $\mathbb{Q}$. Then we define

$\Gamma=\left\\{\,\zeta(\xi)\cdot\begin{pmatrix}\xi&0\\\
0&1\end{pmatrix}\Big{|}\ \,\xi\in U\,\right\\}\subset\mathrm{GL}(V).$

It is clear that $\Gamma$ is a free Abelian group, and hence $\Gamma$ has
bounded finite subgroups, while $\Gamma^{\prime}$ has finite subgroups of
arbitrarily large order.

##  3\. Resolution of singularities

In this section, we construct some special “fiberwise” resolution of
singularities which we will use in the proof of Theorem 1.4. Similar
construction for Moishezon varieties was proposed by Ueno in the proof of
[Ue75, Theorem 14.10]. However, since we work in the analytic category,
instead of his argument involving scheme-theoretic generic point, we propose
another purely analytic construction.

######  Definition 3.1.

By an _embedded smooth model_ of a possibly reducible and non-reduced complex
subspace $Y^{\prime}$ of a complex manifold $M^{\prime}$ we mean a possibly
reducible smooth complex subspace $Y$ of a complex manifold $M$ together with
a commutative digram

${M}$${M^{\prime}}$${Y}$${Y^{\prime}}$$\scriptstyle{h}$$\scriptstyle{g}$

where $h$ and $g$ are proper surjective holomorphic maps, the map $h$ is
bimeromorphic, and $Y$ is the strict transform of $Y^{\prime}$ via the map
$g$.

The last condition in the above definition means that for each irreducible
component $F$ of $Y$ there exists an irreducible component $F^{\prime}$ of
$Y^{\prime}$ considered with the reduced structure such that the map
$g|_{F}\colon F\to F^{\prime}$ is bimeromorphic, and the map $g$ induces a
bijection between the irreducible components of $Y$ and the irreducible
components of $Y^{\prime}$. In particular, each component of $Y$ is a strict
transform of the corresponding irreducible component of $Y$ via the map $g$.

Note that an embedded smooth model always exists by the usual embedded
resolution of singularities [Wl08, Theorem 2.0.2].

In what follows, we call a map $\pi\colon Y\to W$ _equidimensional over_
$U\subset W$, if all the components of all its fibers over $U$ have the same
dimension. We say that $\pi$ is _equidimensional_ , if it is equidimensional
over $W$.

######  Lemma 3.2.

Consider the following commutative diagram

(3.1) ${W\times
M^{\prime}}$${Y^{\prime}}$${W}$$\scriptstyle{\mathrm{pr}_{1}}$$\scriptstyle{\pi}$

where

  * •

$Y^{\prime}$ is a reduced, possibly reducible compact analytic subspace of
$W\times M^{\prime}$,

  * •

$M^{\prime}$ is a smooth (possibly non-compact) complex manifold,

  * •

$W$ is an irreducible reduced (possibly non-compact) complex space,

  * •

$\pi$ is a proper surjective equidimensional map.

Then there exists an open subset $U\subset W$ and a commutative diagram

(3.2) ${\widetilde{M_{U}}}$${U\times
M^{\prime}}$${\widetilde{\pi^{-1}(U)}}$${\pi^{-1}(U)}$${U\subset
W}$$\scriptstyle{h_{U}}$$\scriptstyle{\mathrm{pr}_{1}}$$\scriptstyle{g_{U}}$$\scriptstyle{\pi}$

such that

  * •

$\widetilde{\pi^{-1}(U)}\subset\widetilde{M_{U}}$ is an embedded smooth model
of $\pi^{-1}(U)\subset U\times M^{\prime}$,

  * •

the map $\pi\circ g_{U}$ is smooth,

  * •

for any point $\phi\in U$ we have that $Y_{\phi}\subset M_{\phi}$ is an
embedded smooth model of $Y^{\prime}_{\phi}\subset M^{\prime}$:

(3.3) ${M_{\phi}}$${M^{\prime}}$${Y_{\phi}}$${Y^{\prime}_{\phi}}$${\phi\in
U}$$\scriptstyle{h_{\phi}}$$\scriptstyle{g_{\phi}}$$\scriptstyle{\pi}$

where $Y_{\phi}=(\pi\circ g_{U})^{-1}(\phi)$,
$Y^{\prime}_{\phi}=\pi^{-1}(\phi)$ and $M_{\phi}=(\mathrm{pr}_{1}\circ
h_{U})^{-1}(\phi)$.

###### Proof.

Passing to an open subset of $W$, if needed, we may assume that $W$ is smooth.
By [Wl08, Theorem 2.0.2], there exists a resolution of singularities of
$Y_{0}=Y^{\prime}$, which is realized as a sequence of blow-ups
$g_{1},\ldots,g_{t}$ in smooth centers $Z_{i}\subset M_{i}$ starting from
$M_{0}=W\times M^{\prime}$ as shown in the next diagram

(3.4)
${Z_{t-1}}$${\ldots}$${Z_{1}}$${Z_{0}}$${M=M_{t}}$${M_{t-1}}$${\ldots}$${M_{1}}$${M_{0}=W\times
M^{\prime}}$${Y=Y_{t}}$${Y_{t-1}}$${\ldots}$${Y_{1}}$${Y_{0}=Y^{\prime}}$${W}$$\scriptstyle{h_{t}}$$\scriptstyle{h_{t-1}}$$\scriptstyle{h_{2}}$$\scriptstyle{h_{1}}$$\scriptstyle{g_{t}}$$\scriptstyle{g_{t-1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{1}}$$\scriptstyle{\pi}$

Here $h_{i}$ for $1\leq i\leq t$ is the blow-up of $M_{i}$ in a smooth center
$Z_{i-1}\subset Y_{i-1}$, each $Y_{i}$ is the strict transform of $Y_{0}$, and
$g_{i}=h_{i}|_{Y_{i}}$; here by the strict transform we mean the union of
strict transforms of all the irreducible components of $Y_{0}$. Moreover, by
[Wl08, Theorem 2.0.2] we may assume that each center $Z_{i}$ is disjoint from
the set $\mathrm{Reg}(Y_{i})$ of smooth points of $Y_{i}$. Put

$h=h_{1}\circ\ldots\circ h_{t},\quad\quad\quad\quad g=g_{1}\circ\ldots\circ
g_{t},\quad\quad\quad\quad\pi_{i}=\pi\circ g_{1}\circ\ldots\circ
g_{i}\quad\text{for}\ 0\leq i\leq t.$

To start with, we define an open subset $U\subset W$ by the condition that the
map $\pi_{t}=\pi\circ g$ is smooth over $U$. Then $W\setminus U$ is a proper
closed analytic subset in $W$, cf. [Ue75, Corollary 1.7].

For an arbitrary point $\phi\in U$, we consider the diagram

(3.5) ${Z_{t-1}\cap Y_{t-1,\phi}}$${\ldots}$${Z_{1}\cap
Y_{1,\phi}}$${Z_{0}\cap
Y_{\phi}}$${M_{\phi}=M_{t,\phi}}$${M_{t-1,\phi}}$${\ldots}$${M_{1,\phi}}$${M_{0,\phi}=M^{\prime}}$${Y_{\phi}=Y_{t,\phi}}$${Y_{t-1,\phi}}$${\ldots}$${Y_{1,\phi}}$${Y_{0,\phi}=Y^{\prime}_{\phi}}$${\phi\in
U}$$\scriptstyle{h_{t,\phi}}$$\scriptstyle{h_{t-1,\phi}}$$\scriptstyle{h_{2,\phi}}$$\scriptstyle{h_{1,\phi}}$$\scriptstyle{g_{t,\phi}}$$\scriptstyle{g_{t-1,\phi}}$$\scriptstyle{g_{2,\phi}}$$\scriptstyle{g_{1,\phi}}$$\scriptstyle{\pi}$

where $Y_{i,\phi}=\pi_{i}^{-1}(\phi)$ and $g_{i,\phi}=g_{i}|_{Y_{i,\phi}}$ for
$1\leq i\leq t$. Put $g_{\phi}=g|_{Y_{\phi}}$.

We claim that, possibly after shrinking $U$ at each of $t$ steps in diagram
(3.4), each map $\pi_{i}$ is equidimensional over $U$. Indeed, $\pi_{0}=\pi$
is equidimensional by assumption. By induction, for a fixed $i$ with $1\leq
i\leq t$ we may assume that $\pi_{j}$ is equidimensional for all $j\leq i-1$.
We show that, possibly after shrinking $U$, the map $\pi_{i}$ is
equidimensional. Consider the exceptional set $E_{i}=\mathrm{Exc}(g_{i})$
which maps to $Z_{i-1}$. By [BS76, Theorem V.4.10], the subset of points in
$U$ over which the maps $\pi_{i}$ and $\pi_{i}|_{E_{i}}$ are not flat, is a
proper closed analytic subset of $U$. Hence, possibly after shrinking $U$, we
see that these maps are flat over $U$. By [BS76, Proposition V.2.4] the maps
$\pi_{i}$ and $\pi_{i}|_{E_{i}}$ are equidimensional over $U$. In particular,
$Z_{i}$ either maps to $U$ surjectively, or $\pi_{i}(Z_{i})\cap U=\emptyset$.
(In the latter case, $h_{i}$ induces an isomorphism of $\pi_{i}^{-1}(U)$ with
$\pi_{i-1}^{-1}(U)$, hence we may skip this step in our construction).

Consequently, for any $i$ with $1\leq i\leq t$ and any $\phi\in U$, we can
write

$\dim Y^{\prime}-1\geq\dim E_{i}=\dim U+\dim(E_{i}\cap Y_{i,\phi}).$

In particular, if $E_{i}$ is a divisor on $Y_{i}$, then the above inequality
becomes an equality. On the other hand,

$\dim Y^{\prime}=\dim U+\dim Y^{\prime}_{\phi}=\dim U+\dim Y_{i,\phi}.$

We conclude that

(3.6) $\dim(E_{i}\cap Y_{i,\phi})\leq\dim Y_{i,\phi}-1.$

Hence, for any $1\leq i\leq t$ the intersection $E_{i}\cap Y_{i,\phi}$ is a
proper closed subset of $Y_{i,\phi}$ which has codimension at least $1$ in any
irreducible component of $Y_{i,\phi}$.

As a consequence, for each $1\leq i\leq t$ we have that $Y_{i,\phi}$ is the
strict transform of $Y_{i-1,\phi}$ via the map $g_{i,\phi}$. Hence $Y_{\phi}$
is the strict transform of $Y^{\prime}_{\phi}$ via the map $g_{\phi}$. Observe
that $Y_{\phi}$ is smooth (possibly, reducible), since by construction the map
$\pi_{t}|_{U}$ is smooth. Hence $Y_{\phi}\subset M_{\phi}$ is an embedded
smooth model of $Y^{\prime}_{\phi}\subset M^{\prime}$ for any $\phi\in U$. Put

$\widetilde{M_{U}}=(h\circ\mathrm{pr_{1}})^{-1}(U),\quad\quad\quad\widetilde{\pi^{-1}(U)}=\pi_{t}^{-1}(U),\quad\quad\quad
h_{U}=h|_{\widetilde{M_{U}}},\quad\quad\quad g_{U}=g|_{\pi_{t}^{-1}(U)}.$

This completes the proof. ∎

######  Lemma 3.3.

In the assumptions and notation of Lemma 3.2, if the complex space $W$ is
compact then for any point $\phi\in W$ we can find an embedded smooth model
$Y_{\phi}\subset M_{\phi}$ of $Y^{\prime}_{\phi}=\pi^{-1}(\phi)\subset
M^{\prime}_{\phi}$ as in the next diagram

(3.7) ${M_{\phi}}$${M^{\prime}}$${Y_{\phi}}$${Y^{\prime}_{\phi}}$${\phi\in
U}$$\scriptstyle{h_{\phi}}$$\scriptstyle{g_{\phi}}$$\scriptstyle{\pi}$

such that the Betti numbers of $Y_{\phi}$ are bounded uniformly in $\phi\in
W$.

###### Proof.

By Lemma 3.2, there exists an open subset $U\subset W$ and a resolution of
singularities of

$g_{U}\colon\widetilde{\pi^{-1}(U)}\to\pi^{-1}(U)$

such that it gives an embedded smooth model $g_{\phi}\colon Y_{\phi}\to
Y^{\prime}_{\phi}$ for any $\phi\in U$, where $Y_{\phi}$ is embedded into
$M_{\phi}$ as in diagram 3.7 (see also diagram 3.5). Put $U_{0}=U$. Consider
the following decomposition of the complement of $U_{0}$ in $W$ into the union
of irreducible components:

$W\setminus U_{0}=\bigcup_{j=1}^{k_{1}}W_{j},\quad\quad\quad\quad\quad
Y^{\prime}\setminus\pi^{-1}(U_{0})=\bigcup_{j=1}^{k_{1}}Y^{\prime}_{j},\quad\quad\quad\quad\quad\pi_{j}=\pi|_{Y^{\prime}_{j}}\colon
Y^{\prime}_{j}\to W_{j}.$

Applying Lemma 3.2 to each $Y^{\prime}_{j}$ and $\pi_{j}\colon
Y^{\prime}_{j}\to W_{j}$ and iterating this process, we obtain a decomposition
of $W$ into finitely many locally closed analytic subsets $U_{j}$

$W=\bigcup_{j=0}^{k}U_{j},\quad\quad\quad\quad\quad\quad\quad\quad\quad
Y^{\prime}=\bigcup_{j=0}^{k}Y^{\prime}_{j},\quad\quad\quad\quad\quad\quad\quad\quad\quad
Y^{\prime}_{j}=\pi^{-1}(U_{j})$

and resolutions of singularities $g_{j}\colon Y_{j}\to Y^{\prime}_{j}$ such
that each fiber $Y_{\phi}$ of the induced map

$\pi\circ g_{j}\colon Y_{j}\to U_{j}$

over $\phi\in U_{j}$ is a smooth model of $Y^{\prime}_{\phi}$, and $Y_{\phi}$
is embedded into $M_{\phi}$ as in diagram 3.7. By construction, each map
$\pi\circ g_{j}$ is smooth, so by Ehresmann’s theorem it is differentially
locally trivial. Hence, we see that for each $n$ the Betti numbers
$b_{n}(Y_{\phi})$ are constant for any $\phi\in U_{j}$. Since there are only
finitely many $U_{j}$, we conclude that the Betti numbers of $Y_{\phi}$ are
uniformly bounded for each $\phi\in W$. This completes the proof. ∎

##  4\. Proof of main results

We use the notation of section 2.1. First we prove Theorem 1.4 in the simple
case when the linear system $|K_{X}|$ defines the pluricanonical map.

Recall that for a compact smooth (possibly, reducible) equidimensional complex
space $X$ one can define the canonical class $K_{X}$ and the vector spaces of
the holomorphic forms $\mathrm{H}^{0}(X,\mathcal{O}_{X}(K_{X}))$ in the usual
way. Also, the group of bimeromorphic self-maps $\mathrm{Bim}(X)$ is also
well-defined in this situation.

######  Proposition 4.1.

Let $X$ be a compact smooth (possibly, reducible) equidimensional complex
space. Let $g\in\mathrm{Bim}(X)$, and let $\beta$ be an eigenvalue of $g$
considered as an operator acting on

$\mathrm{H}^{0}(X,\mathcal{O}_{X}(K_{X})).$

Then the degree of the field extension $[\mathbb{Q}(\beta):\mathbb{Q}]$ is
bounded by the Betti number

$b_{n}(X)=\mathrm{rk}(\mathrm{H}^{n}(X,\mathbb{Z})/\mathrm{tors}).$

where $n$ is the dimension of $X$.

###### Proof.

Consider the map

(4.1)
$\mathrm{H}^{0}(X,\mathcal{O}_{X}(K_{X}))\xhookrightarrow{}\mathrm{H}^{n}(X,\mathbb{C})=(\mathrm{H}^{n}(X,\mathbb{Z})/\mathrm{tors})\otimes\mathbb{C}$

which is injective by de Rham theorem. It is well-known that any element
$g\in\mathrm{Bim}(X)$ acts as a linear operator on
$\mathrm{H}^{n}(X,\mathbb{Z})/\text{tors}$, see e.g. [Ue75, 14.3]. Since the
natural action of $g\in\mathrm{Bim}(X)$ on the space of holomorphic forms is
compatible with its action on $\mathrm{H}^{n}(X,\mathbb{Z})/\mathrm{tors}$, we
conclude that $\beta$ is an algebraic integer, and
$[\mathbb{Q}(\beta):\mathbb{Q}]$ is bounded by the Betti number $b_{n}(X)$. ∎

######  Corollary 4.2.

Let $X$ be a compact complex manifold of dimension $n$. If the linear system
$|K_{X}|$ defines the pluricanonical map, then both $\rho(\mathrm{Bim}(X))$
and $\overline{\rho}(\mathrm{Bim}(X))$ have bounded finite subgroups.

###### Proof.

Let $g\in\mathrm{Bim}(X)$, and let $\beta$ be an eigenvalue of $g$ considered
as a linear operator acting on $\mathrm{H}^{0}(X,\mathcal{O}_{Y}(K_{X}))$. By
Proposition 4.1, the degree of the field extension
$[\mathbb{Q}(\beta):\mathbb{Q}]$ is bounded by the Betti number $b_{n}(X)$.
Thus from Lemma 2.9 it follows that both $\rho(\mathrm{Bim}(X))$ and
$\overline{\rho}(\mathrm{Bim}(X))$ have bounded finite subgroups. ∎

Now we are going to prove Theorem 1.4 in the general case.

######  Construction 4.3.

We follow [Ue75, 14.4]. Let $X$ be a complex manifold, and let $M^{\prime}$ be
the total space of the line bundle $\mathcal{O}_{X}(K_{X})$ considered as a
non-compact complex manifold. Let $\tau\colon M^{\prime}\to X$ be the natural
projection. Then the _$m$ -canonical cover_ of $X$ associated with a
holomorphic $m$-form

$\phi\in\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$

is a compact complex subspace $Y^{\prime}_{\phi}$ of $M^{\prime}$ constructed
as follows.

Let $\\{U_{i}\\}_{i\in I}$ be an open covering of $X$ with local coordinates
$z_{i}^{1},\ldots,z_{i}^{n}$ on $U_{i}$. Locally on $U_{i}$ we can write

$\phi=\phi_{i}(z_{i}^{1},\ldots,z_{i}^{n})(dz_{i}^{1}\wedge\ldots\wedge
dz_{i}^{n})^{\otimes m},$

where $\phi_{i}\in\mathcal{O}_{X}(U_{i})$. Note that $M^{\prime}$ admits an
open covering $\\{V_{i}\\}_{i\in I}$ where $V_{i}=U_{i}\times\mathbb{C}$ with
local coordinates $z_{i}^{1},\ldots,z_{i}^{n},w_{i}$. For each $i\in I$,
define a closed analytic subset $Y^{\prime}_{\phi}\subset M^{\prime}$ locally
on $V_{i}$ by the equations

(4.2) $w_{i}^{m}=\phi_{i}(z_{i}^{1},\ldots,z_{i}^{n}).$

######  Remark 4.4.

Note that the complex space $Y^{\prime}_{\phi}$ constructed above may be not
connected, reducible and even non-reduced. However, by construction it is
equidimensional.

For the reader’s convenience, we reproduce the proof of the following lemma.

######  Lemma 4.5 ([Ue75, 14.4]).

Let $X$ be a compact complex manifold of dimension $n$, and $m$ be a number
such that the linear system $|mK_{X}|$ defines the pluricanonical map. Let

$\phi\in\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$

be an eigenvector of an element $g\in\mathrm{Bim}(X)$ acting on
$\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$ with an eigenvalue $\alpha$. Then
the degree of the field extension $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is bounded
by $C=C(X,\phi)$ depending only on $X$ and $\phi$, but not on $g$.

More precisely, we can take $C=b_{n}(Y_{\phi},\mathbb{Z})$ where
$Y_{\phi}\subset M_{\phi}$ is any embedded smooth model of the $m$-canonical
cover $Y^{\prime}_{\phi}\subset M^{\prime}$.

###### Proof.

Let $f\colon Y^{\prime}_{\phi}\to X$ be the $m$-canonical cover of $X$
associated with an $m$-form $\phi$ as in Construction 4.3. Note that
$Y^{\prime}_{\phi}$ is embedded into the total space $M^{\prime}$ of the line
bundle $\mathcal{O}_{X}(K_{X})$. Define a holomorphic $n$-form $\psi$ on
$M^{\prime}$ in the following way. Let $\\{U_{i}\\}_{i\in I}$ be a
sufficiently fine open covering of $X$ with local coordinates
$z_{i}^{1},\ldots,z_{i}^{n}$ on $U_{i}$. Then $M^{\prime}$ admits an open
covering $\\{V_{i}\\}_{i\in I}$ where $V_{i}=U_{i}\times\mathbb{C}$ with local
coordinates $z_{i}^{1},\ldots,z_{i}^{n},w_{i}$. For each $i\in I$, in the
chart $V_{i}$ put

$\psi=w_{i}dz_{i}^{1}\wedge\ldots\wedge dz_{i}^{n}.$

The holomorphic forms in the right-hand side agree on the intersections
$V_{i}\cap V_{j}$, hence they define a global homomorphic $n$-form $\psi$ on
$M^{\prime}$.

Note that any bimeromorphic map $g\in\mathrm{Bim}(X)$ induces a bimeromorphic
map $g_{M^{\prime}}\in\mathrm{Bim}(M^{\prime})$. More precisely, if
$g(U_{i})\cap U_{j}\neq\emptyset$, then the map

$g_{M^{\prime}}|_{V^{\prime}_{j}}\colon V^{\prime}_{j}=g^{-1}(g(U_{i})\cap
U_{j})\times\mathbb{C}\to U_{j}\times\mathbb{C}$

is given by

$(z_{i}^{1},\ldots,z_{i}^{n},w_{i})\mapsto\left(g^{1}(z_{i}),\ldots,g^{n}(z_{i}),\det\left(\frac{\partial(g^{1}(z_{i}),\ldots,g^{n}(z_{i}))}{\partial(z_{i}^{1},\ldots,z_{i}^{n})}\right)^{-1}w_{i}\right).$

Define a holomorphic automorphism $\mu_{\beta}\in\mathrm{Aut}(M^{\prime})$
locally given by

$\mu_{\beta}\colon(z_{i}^{1},\ldots,z_{i}^{n},w_{i})\mapsto(z_{i}^{1},\ldots,z_{i}^{n},\beta
w_{i}),$

where $\beta\in\mathbb{C}$ satisfies $\beta^{m}=\alpha$. Since by assumption
we have $g^{*}\phi=\alpha\phi$, it follows that

$\phi_{i}(g^{1}(z_{i}),\ldots,g^{n}(z_{i}))=\alpha\phi_{i}(z_{i}^{1},\ldots,z_{i}^{n})\det\left(\frac{\partial(g^{1}(z_{i}),\ldots,g^{n}(z_{i}))}{\partial(z_{i}^{1},\ldots,z_{i}^{n})}\right)^{-m}.$

Thus, from (4.2) in Construction 4.3 it follows that the map
$h^{\prime}=\mu_{\beta}\circ g_{M^{\prime}}$ preserves $Y^{\prime}_{\phi}$,
and hence $h^{\prime}$ induces a bimeromorphic automorphism of
$Y^{\prime}_{\phi}$.

Consider any embedded smooth model $Y_{\phi}$ of $Y^{\prime}_{\phi}$ (we know
that it exists by [Wl08, Theorem 2.0.2]), where $Y^{\prime}_{\phi}$ is
embedded into $M^{\prime}$, and $Y_{\phi}$ is embedded into a smooth manifold
$M$. We denote the corresponding map as $f_{2}\colon Y_{\phi}\to
Y^{\prime}_{\phi}$. Let $f_{1}\colon Y^{\prime}_{\phi}\to X$ be the
restriction of the natural projection $\tau\colon M^{\prime}\to X$ to
$Y^{\prime}_{\phi}\subset M^{\prime}$. Finally, put

$f=f_{1}\circ f_{2}\colon Y_{\phi}\to X.$

Denote by $\omega$ a holomorphic form on $Y_{\phi}$ which is obtained by
pulling back $\psi$ to $M$ and then restricting to $Y_{\phi}$. We have

$\omega^{m}=f^{*}\phi.$

Recall that $g^{*}\phi=\alpha\phi$. Note that the bimeromorphic map
$h^{\prime}=\mu_{\beta}\circ g_{M^{\prime}}$ induces a bimeromorphic map $h$
on $M$. Since by the above $h^{\prime}$ restricts to a bimeromorphic
automorphism on $Y^{\prime}_{\phi}$, it follows that $h$ restricts to a
bimeromorphic automorphism on $Y_{\phi}$. Then

$(\mu_{\beta}\circ g_{M^{\prime}})^{*}\psi=\beta\psi,\quad\quad
h^{*}(\omega)=\beta\omega.$

Hence $\beta$ is an eigenvalue of an operator $h$ acting on
$\mathrm{H}^{0}(Y_{\phi},\mathcal{O}_{Y_{\phi}}(K_{Y_{\phi}})).$ Then by
Proposition 4.1, the degree of the field extensions
$[\mathbb{Q}(\beta):\mathbb{Q}]$ is bounded by the Betti number
$b_{n}(Y_{\phi})$, and by construction $Y_{\phi}$ depends only on $X$ and
$\phi$. Since $\beta^{m}=\alpha$, the degree of the field extension
$[\mathbb{Q}(\alpha):\mathbb{Q}]$ is also bounded by $b_{n}(Y_{\phi})$. ∎

######  Construction 4.6.

Let $X$ be a compact complex manifold. Following [Ue75, 14.10], we construct
its _global $m$-canonical cover_ $Y^{\prime}$ together with a map $f\colon
Y^{\prime}\to X$. It can be seen as a global version of the $m$-canonical
cover 4.3. The construction is as follows.

Let $\phi_{0},\ldots,\phi_{N}$ be a basis of the vector space
$\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$, and let

$\mathbb{P}^{N}=\mathbb{P}(\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))).$

be the corresponding projective space. Consider a product
$\mathbb{P}^{N}\times M^{\prime}$ where $M^{\prime}$ is the total space of
$\mathcal{O}_{X}(K_{X})$ considered as a non-compact complex manifold. Let
$\\{U_{i}\\}_{i\in I}$ be a sufficiently fine open covering of $X$ with local
coordinates $z_{i}^{1},\ldots,z_{i}^{n}$ on $U_{i}$. Let

$u_{0},\ldots,u_{k-1},u_{k+1},\ldots,u_{N}$

be the local coordinates on some affine chart $V_{k}$ of $\mathbb{P}^{N}$ for
$0\leq k\leq N$. Locally on $U_{i}$ for any $0\leq j\leq N$ we can write

$\phi_{j}=\phi_{j,i}(z^{i}_{1},\ldots,z^{i}_{n})dz^{i}_{1}\wedge\ldots\wedge
dz^{i}_{n}.$

We define $Y^{\prime}$ as a possibly singular analytic subspace of
$\mathbb{P}^{N}\times M^{\prime}$ locally in $V_{k}\times
U_{i}\times\mathbb{C}$ defined by the equations

$w_{i}^{m}=u_{1}\phi_{0,i}+\ldots+u_{k-1}\phi_{k-1,i}+\phi_{k,i}+u_{k+1}\phi_{k+1,i}+\ldots+u_{n}\phi_{N,i},$

where $w_{i}$ is the coordinate along the factor $\mathbb{C}$. These equations
agree on the intersections of charts and hence they define an analytic
subspace $Y^{\prime}$ of $M^{\prime}$.

######  Remark 4.7.

Consider the natural projection

$\pi\colon Y^{\prime}\to\mathbb{P}^{N},$

induced by the projection map $\mathrm{pr}_{1}\colon\mathbb{P}^{N}\times
M^{\prime}\to\mathbb{P}^{N}$. Note that for any
$\phi\in\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$, if we consider the
corresponding point $\phi\in\mathbb{P}^{N}$, we have $\pi^{-1}(\phi)\simeq
Y^{\prime}_{\phi}$ where $Y^{\prime}_{\phi}$ is the $m$-canonical cover
associated with an $m$-form $\phi$ as in Construction 4.3. Conversely, any
$m$-canonical cover $Y^{\prime}_{\phi}$ associated with an $m$-form $\phi$ is
isomorphic to $\pi^{-1}(\phi)$ for $\phi\in\mathbb{P}^{N}$. Note that the map
$\pi$ is proper, and by construction its fibers are equidimensional.

We are ready to prove Theorem 1.4. Note that unlike in Theorem 1.6, we do not
need the assumption that $X$ is Moishezon.

###### Proof of Theorem 1.4.

Let $X$ be a compact complex variety of dimension $n$. Passing to a resolution
of singularities if needed, we may assume that $X$ is smooth. Let $m$ be a
number such that the linear system $|mK_{X}|$ defines the pluricanonical map.

Put $V=\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X}))$. We are going to prove that
the groups $\Gamma=\rho(\mathrm{Bim}(X))\subset\mathrm{GL}(V)$ and
$\Gamma^{\prime}=\overline{\rho}(\mathrm{Bim}(X))\subset\mathrm{PGL}(V)$ have
bounded finite subgroups. By Lemma 2.9, it is enough to show that for any
$g\in\Gamma$ and for any eigenvalue $\alpha$ of $g$ the degree of the field
extension $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is bounded by some constant which
is independent of $g$ and $\alpha$.

By Lemma 4.5, we know that $[\mathbb{Q}(\alpha):\mathbb{Q}]$ is bounded by the
Betti number $b_{n}({Y_{\phi}})$ where $Y_{\phi}$ is any embedded smooth model
of the $m$-canonical cover $Y^{\prime}_{\phi}$ associated to $\phi$ as in
Construction 4.3. Hence it is enough to find embedded smooth models of
$Y^{\prime}_{\phi}$ such that the Betti numbers $b_{n}({Y_{\phi}})$ are
uniformly bounded for any

$\phi\in\mathrm{H}^{0}(X,\mathcal{O}_{X}(mK_{X})).$

To this end, we apply Lemma 3.3 to the global $m$-canonical cover $f\colon
Y^{\prime}\to X$ as in Construction 4.6. More precisely, in the notation of
Lemma 3.3, put $\mathbb{P}^{N}=W$ and $\pi\colon Y^{\prime}\to W$ to conclude
that the Betti numbers of some embedded smooth models $Y_{\phi}\subset
M_{\phi}$ of $Y^{\prime}_{\phi}\subset M^{\prime}$ are bounded for any $\phi$.
This finishes the proof. ∎

The following example shows that we may not have finite
$\rho(\mathrm{Bim}(X))$ when $X$ is a compact complex compact manifold.

######  Example 4.8.

By [MS08], there exists a non-algebraic K3 surface $X$ that admits a non-
symplectic automorphism of infinite order acting on a holomorphic two-form via
multiplication by a complex number of infinite order. Consequently,
$\rho(\mathrm{Bim}(X))$ is not a finite group. Note also that by [Ue75, Remark
14.6], there exists a $3$-dimensional complex torus such that the group
$\rho(\mathrm{Bim}(X))$ is infinite.

To prove Theorem 1.2, we will use the following technical results.

######  Lemma 4.9 ([PSh21b, Lemma 2.9]).

Let $X$ and $Y$ be compact complex varieties and let $\sigma\colon
X\dasharrow~{}Y$ be a dominant meromorphic map. Suppose that $\kappa(X)\geq
0$. Let $F$ be a typical fiber of $\varphi$, and let $F^{\prime}$ be an
irreducible component of $F$. Then $\kappa(F^{\prime})\geq 0$.

######  Lemma 4.10 ([PSh21b, Lemma 3.2]).

Let $X$ and $Y$ be compact complex varieties and let $\sigma\colon X\dasharrow
Y$ be a dominant meromorphic map. Let $F$ be a typical fiber of $\varphi$.
Suppose that for any irreducible component $F^{\prime}$ of $F$ the group
$\mathrm{Bim}(F^{\prime})$ is Jordan. Assume that the map $\varphi$ is
equivariant with respect to the group $\mathrm{Bim}(X)$, and that the image of
$\mathrm{Bim}(X)$ in $\mathrm{Bim}(Y)$ has bounded finite subgroups. Then the
group $\mathrm{Bim}(X)$ is Jordan.

Now, we are ready to prove Theorem 1.2.

###### Proof of Theorem 1.2.

Let $X$ be a compact complex variety of dimension $n$ with $\kappa(X)\geq
n-2$. Passing to a resolution of singularities if needed, we may assume that
$X$ is smooth. Let $F$ be a typical fiber of the pluricanonical map. Then by
section 2.1 we may assume that $F$ is irreducible and smooth. By Lemma 4.9, we
have $\kappa(F)\geq 0$. By assumption, we have $\dim F\leq 2$. Consider two
cases. The first possibility is that $\dim F\leq 1$, and hence
$\mathrm{Bim}(F)=\mathrm{Aut}(F)$ is obviously Jordan. The second possibility
is that $\dim F=2$, and since $\kappa(F)\geq 0$, it follows that $F$ is not
bimeromorphic to a product of $\mathbb{P}^{1}$ and some other curve. Hence by
[PSh21, Theorem 1.7] the group $\mathrm{Bim}(F)$ is Jordan. Since by Theorem
1.4 the group $\rho(\mathrm{Bim}(X))$ has bounded finite subgroups, applying
Lemma 4.10 we conclude that $\mathrm{Bim}(X)$ is Jordan. ∎

Then, Corollary 1.3 follows immediately.

###### Proof of Corollary 1.5.

Let $X$ be a compact complex variety of dimension $n$ with $\kappa(X)\geq
n-3$. Passing to a resolution of singularities if needed, we may assume that
$X$ is smooth. By Theorem 1.2, it is enough to consider the case
$\kappa(X)=n-3$. By Lemma 4.9 a typical fiber $F$ is a $3$-dimensional Kähler
manifold, and by Theorem 1.1 we have that $\mathrm{Bim}(F)$ is Jordan. By
Theorem 1.4 the image of projective pluricanonical representation
$\overline{\rho}(\mathrm{Bim}(X))$ has bounded finite subgroups. Applying
Lemma 4.10 to the pluricanonical map, we conclude that $\mathrm{Bim}(X)$ is
Jordan. ∎

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