
This article is cited in 3 scientific papers (total in 3 papers)
Finite groups of bimeromorphic selfmaps of nonuniruled Kähler threefolds
Yu. G. Prokhorov^{ab}, С. A. Shramov^{ab} ^{a} Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
^{b} Laboratory of Algebraic Geometry and Its Applications, National Research University Higher School of Economics, Moscow, Russia
Abstract:
We prove the Jordan property for groups of bimeromorphic selfmaps of threedimensional compact Kähler varieties of nonnegative Kodaira dimension and positive irregularity.
Bibliography: 32 titles.
Keywords:
group of bimeromorphic selfmaps, Kähler variety.
Received: 15.03.2022 and 16.09.2022
§ 1. Introduction In the theory of automorphism groups of complex varieties the following notion plays an important role. Definition 1.1 (see [1], Definition 2.1). A group $\Gamma$ is called Jordan (alternatively, $\Gamma$ is said to have the Jordan property) if there exists a constant $J=J(\Gamma)$ such that any finite subgroup $G\subset\Gamma$ contains a normal abelian subgroup $A\subset G$ of index at most $J$. It turns out that the Jordan property holds for many groups of geometric origin, including the automorphism groups and groups of bimeromorphic selfmaps of many compact complex varieties. For instance, if $X$ is a compact complex surface, then the automorphism group of $X$ is always Jordan (see [2], Theorem 1.6), while its group of bimeromorphic selfmaps is Jordan if and only if $X$ is not bimeromorphic to a product of $\mathbb{P}^1$ and an elliptic curve (see [2], Theorem 1.7). As for higher dimensions, much less is understood. For instance, it is unclear whether or not the automorphism group of an arbitrary compact complex variety is Jordan; a positive answer is known only for the neutral component of such a group (see [3], Theorem 5). However, there are several important results in the case of Kähler varieties. To start with, the following theorem was proved in [4]. Theorem 1.2. Let $X$ be a compact Kähler variety. Then the automorphism group of $X$ is Jordan. Theorem 1.2 was recently generalized to the case of the automorphism groups of compact complex spaces of Fujiki’s class $\mathcal{C}$ [5] (cf. [6]). In [7] groups of bimeromorphic selfmaps of uniruled compact Kähler varieties of dimension $3$ were considered from the point of view of the Jordan property. In higher dimensions, partial results for $\mathbb{P}^1$bundles over appropriate bases were obtained in [8] and [9] (cf. [10]). In particular, [11] provides plenty of examples of uniruled compact Kähler varieties with nonJordan groups of bimeromorphic selfmaps. Denote the Kodaira dimension of a compact complex variety $X$ by $\varkappa(X)$, and its irregularity by $\operatorname{q}(X)$; we refer the reader to § 2 for details. Recall that a threedimensional compact Kähler variety is uniruled if and only if its Kodaira dimension is negative (see [12], Corollary 1.4): this is a very deep fact, which follows from the results in [13], [14] and [12]. The purpose of this paper is to prove the following theorem concerning the groups of bimeromorphic selfmaps of nonuniruled threedimensional compact Kähler varieties; we refer the reader to § 2 for the notation and conventions. Theorem 1.3. Let $X$ be a threedimensional compact Kähler variety such that $\varkappa(X)\geqslant 0$ and $\operatorname{q}(X)>0$. Then the group of bimeromorphic selfmaps of $X$ is Jordan. For a nonuniruled algebraic variety $X$ of arbitrary dimension a much stronger result than Theorem 1.3 is valid: the group of birational selfmaps of $X$ is Jordan without any restriction on irregularity; see [15], Theorem 1.8, (ii). The proof of this result in [15] uses the equivariant Minimal Model Program (see [16] for a brief introduction). The proofs in this paper also use the MMP for threedimensional compact Kähler varieties, which exists in view of [12]. We expect that an analogue of Theorem 1.3 is valid in an arbitrary dimension and can be proved independently of the MMP. Furthermore, our proof of the Jordan property for pseudoautomorphism groups is based on Fujiki’s result in [17] (see Theorem 4.4 below), which is known only for smooth Kähler varieties. It is desirable to generalize it to Kähler varieties with terminal singularities. This would allow one to simplify largely our arguments and to remove the assumption $\operatorname{q}(X)>0$ in Theorem 1.3.^{1}^{[x]}^{1}After this paper was written, Golota [18] generalized Fujiki’s result to the case of singular Kähler varieties and proved a stronger version of Theorem 1.3. The plan of the paper is as follows. In § 2 we collect the basic definitions and auxiliary facts concerning compact complex varieties and manifolds. In § 3 we study groups of bimeromorphic selfmaps that preserve fibrations on compact complex varieties. In § 4 we consider the groups of pseudoautomorphisms of compact Kähler manifolds. In particular, we show in Proposition 4.5 that the group of pseudoautomorphisms of a compact Kähler manifold is always Jordan; a projective analogue of this result is well known to experts. In § 5 we collect some information about the Albanese maps of compact complex varieties. In § 6 we discuss some technical facts on the indeterminacy loci of pseudoautomorphisms. In § 7 we make a couple of additional observations on compact complex surfaces and their automorphism groups. Finally, in § 8 we complete the proof of Theorem 1.3. Acknowledgements We are grateful to Ch. Hacon, who pointed out a gap in the first draft of the paper. We also thank an anonymous referee for helpful suggestions.
§ 2. Preliminaries In this section we collect some auxiliary facts about compact complex varieties and manifolds. We refer the reader to [19] for the most basic facts and definitions. In particular, by a complex variety we mean an irreducible reduced complex space. A morphism of complex varieties is a holomorphic map between them. A smooth complex variety is called a complex manifold. A complex surface is a complex manifold of dimension $2$. Given a compact complex variety $X$, we denote the algebraic dimension of $X$, that is, the transcendence degree of the field of meromorphic functions on $X$, by $\operatorname{a}(X)$. We let $\operatorname{q}(X)$ denote the irregularity of $X$, that is, the dimension of $H^1(X',\mathscr{O}_{X'})$, where $X'$ is an arbitrary compact complex manifold bimeromorphic to $X$. Similarly, the Kodaira dimension $\varkappa(X)$ is defined as the Kodaira dimension of its smooth compact model $X'$ (see [19], Definition 6.5). We denote the group of biholomorphic selfmaps of $X$ by $\operatorname{Bim}(X)$. A Zariski open subset of $X$ is a subset of the form $X\setminus\Sigma$, where $\Sigma$ is a closed (analytic) subset of $X$. A typical point of $X$ is a point in some nonempty Zariski open subset of $X$; a typical fibre of a (meromorphic) map $\phi\colon X\dashrightarrow Y$ is a fibre over a typical point of $Y$. Given a meromorphic map $\chi\colon X\dashrightarrow Y$, we denote by $\operatorname{Ind}(\chi)$ the indeterminacy locus of $\chi$, that is, the minimal closed analytic subset $V\subset X$ such that the restriction of $\chi$ to $X\setminus V$ is holomorphic. Recall that the canonical class of a normal complex variety $X$ is the reflexive rank $1$ sheaf
$$
\begin{equation*}
\boldsymbol{\omega}_X=j_* \Omega_{X_0}^{\dim(X)},
\end{equation*}
\notag
$$
where $X_0\subset X$ is the smooth locus and $j\colon X_0 \hookrightarrow X$ is the embedding (see [20], § 1 A). Note that in contrast to the projective case the canonical class $\boldsymbol{\omega}_X$ need not be represented by a Weil divisor. However, $\boldsymbol{\omega}_X$ is always represented by a Weil divisor locally, in a small analytic neighbourhood of any point. Thus, we will occasionally abuse the notation and write $K_X$ instead of $\boldsymbol{\omega}_X$. Note also that throughout this paper we consider varieties of nonnegative Kodaira dimension; for such a variety $X$ the reflexive sheaf
$$
\begin{equation*}
\boldsymbol{\omega}_X^{[n]}=(\boldsymbol{\omega}_X^{\otimes n})^{\vee\vee}
\end{equation*}
\notag
$$
is represented by a Weil divisor $nK_X$ for some positive integer $n$. Indeed, for such a divisor one can take $nK_X=f_* D$, where $f\colon\widetilde X\to X$ is a resolution of singularities, and $D$ is an effective divisor in the nonempty linear system $nK_{\widetilde X}$. For the basic terminology, definitions and facts concerning the Kähler Minimal Model Program we refer to [12]. We emphasize only the following distinctions. Since the canonical class $\boldsymbol{\omega}_X$ need not be represented by a Weil divisor, in the case of nonalgebraic complex varieties the definition of a normal $\mathbb{Q}$factorial singularity includes the assumption that the sheaf $\boldsymbol{\omega}_X^{[n]}$ is invertible for some $n$. We also recall that a terminal singularity is a normal $\mathbb{Q}$Gorenstein singularity of a complex variety such that all of its discrepancies are positive. In particular, if a variety $X$ has only terminal (or just normal $\mathbb{Q}$Gorenstein) singularities, then the intersection numbers of $\boldsymbol{\omega}_X$ with all curves on $X$ are well defined. A canonical singularity is a normal $\mathbb{Q}$Gorenstein singularity of a complex variety such that all of its discrepancies are nonnegative. All canonical (and, in particular, terminal) singularities are rational; see, for example, [21], Theorem 5.22. One says that a group $\Gamma$ has bounded finite subgroups if there exists a constant $B=B(\Gamma)$ such that every finite subgroup of $\Gamma$ has order at most $B$. Lemma 2.1. Let
$$
\begin{equation*}
1\to \Gamma'\to \Gamma\to \Gamma''
\end{equation*}
\notag
$$
be an exact sequence of groups. Assume that the group $\Gamma''$ has bounded finite subgroups. Then the group $\Gamma$ is Jordan if and only if $\Gamma'$ is Jordan. The proof is obvious. A group $\Gamma$ is called strongly Jordan if it is Jordan and there exists a constant $r=r(\Gamma)$ such that every finite subgroup of $\Gamma$ is generated by at most $r$ elements. Lemma 2.2 (see [15], Lemma 2.8). Let
$$
\begin{equation*}
1\to \Gamma'\to \Gamma\to \Gamma''
\end{equation*}
\notag
$$
be an exact sequence of groups. Assume that the group $\Gamma''$ is strongly Jordan and $\Gamma'$ has bounded finite subgroups. Then the group $\Gamma$ is Jordan. The following classical theorem was proved by Minkowski. Theorem 2.3 (see, for instance, [22], Theorem 1). For every positive integer $n$ the group $\mathrm{GL}_n(\mathbb{Z})$ has bounded finite subgroups. Corollary 2.4. Let $\Lambda$ be a finitely generated abelian group. Then the group $\operatorname{Aut}(\Lambda)$ has bounded finite subgroups. The following assertion is well known; see, for instance, [23], Proposition 1.2.1, or [2], Theorem 8.4. Theorem 2.5. Let $T$ be a complex torus of dimension $n$. Then
$$
\begin{equation*}
\operatorname{Aut}(T)\cong T\rtimes\Gamma,
\end{equation*}
\notag
$$
where $\Gamma$ is a subgroup of $\mathrm{GL}_{2n}(\mathbb{Z})$. The next result follows from Lemma 9.11 in [19] and Lemma 3.5 in [24]. Proposition 2.6. Let $X$ be a compact complex variety with rational singularities, and let $T$ be a complex torus. Let $\zeta\colon X\dashrightarrow T$ be a meromorphic map. Then $\zeta$ is holomorphic. Corollary 2.7. Let $T$ be a complex torus of dimension $n$. Then the group $\operatorname{Bim}(T)$ is Jordan. Moreover, there exists a positive integer $r=r(n)$ which depends only on $n$ but not on $T$ such that every finite subgroup of $\operatorname{Bim}(T)$ is generated by at most $r$ elements. In particular, $\operatorname{Bim}(T)$ is strongly Jordan. Proof. One has $\operatorname{Bim}(T)=\operatorname{Aut}(T)$ by Proposition 2.6. Thus, the first assertion follows from Theorems 2.5 and 2.3 in combination with Lemma 2.1 (alternatively, it can be obtained from Theorem 1.2). The second assertion follows directly from Theorems 2.5 and 2.3. The corollary is proved. Theorem 2.8 (see [19], Corollary 14.3). Let $X$ be a compact complex variety with $\dim(X)=\varkappa(X)$. Then the group $\operatorname{Bim}(X)$ is finite. Lemma 2.9. Let $X$ and $Y$ be compact complex varieties, and let $\phi\colon X\dashrightarrow Y$ be a dominant meromorphic map. Assume that $\varkappa(X)\geqslant 0$. Let $F$ be a typical fibre of $\phi$, and let $F'$ be an irreducible component of $F$. Then $\varkappa(F')\geqslant 0$. Proof. Applying the resolution of singularities and indeterminacies, we may assume that $X$ and $Y$ are smooth and the map $\phi$ is holomorphic. In particular, this means that $F$ is a compact complex manifold, and $F'$ is a connected component of $F$. By adjunction
$$
\begin{equation*}
\boldsymbol{\omega}_{F'}\cong\boldsymbol{\omega}_X_{F'}.
\end{equation*}
\notag
$$
Therefore, since $\boldsymbol{\omega}_X^{\otimes n}$ is represented by an effective divisor for some positive integer $n$, the same holds for $\boldsymbol{\omega}_{F'}$. Thus, we have $\varkappa(F')\geqslant 0$.
The lemma is proved. For most compact complex surfaces their groups of bimeromorphic selfmaps are Jordan. More precisely, the following is known. Theorem 2.10 (see [2], Theorem 1.7). Let $X$ be a compact complex surface with $\varkappa(X)\geqslant 0$. Then the group $\operatorname{Bim}(X)$ is strongly Jordan. Remark 2.11. Let $X$ be a (smooth connected) compact complex curve. It is easy to see that the group $\operatorname{Bim}(X)=\operatorname{Aut}(X)$ is strongly Jordan. If a group acts with a fixed point on a compact Kähler manifold of nonnegative Kodaira dimension, then it has bounded finite subgroups. Theorem 2.12 (see [25], Theorem 1.5). Let $X$ be a compact Kähler manifold of nonnegative Kodaira dimension, and let $P$ be a point on $X$. Then the stabilizer of $P$ in $\operatorname{Aut}(X)$ has bounded finite subgroups.
§ 3. Equivariant fibrations In this section we make several observations about groups of bimeromorphic selfmaps that preserve fibrations on complex varieties. Given a dominant meromorphic map $\alpha\colon X\dashrightarrow Y$ of compact complex varieties, we denote by $\operatorname{Bim}(X; \alpha)$ the subgroup in $\operatorname{Bim}(X)$ that consists of all bimeromorphic selfmaps of $X$ that map fibres of $\alpha$ to fibres of $\alpha$ again. In other words, if $\chi\in\operatorname{Bim}(X;\alpha)$, then for any two points $P,Q\in X\setminus\operatorname{Ind}(\chi)$ such that $\alpha(P)=\alpha(Q)$ we have $\alpha(\chi(P))=\alpha(\chi(Q))$. There is a natural homomorphism
$$
\begin{equation*}
h_\alpha\colon \operatorname{Bim}(X;\alpha)\to \operatorname{Bim}(Y),
\end{equation*}
\notag
$$
and the map $\alpha$ is equivariant with respect to $\operatorname{Bim}(X;\alpha)$. Denote the kernel of the homomorphism $h_\alpha$ by $\operatorname{Bim}(X)_\alpha$. The proof of the following lemma is similar to the proof of Lemma 4.1 in [7]. Lemma 3.1. Let $X$ and $Y$ be compact complex varieties, and let $\alpha\colon X\dashrightarrow Y$ be a dominant meromorphic map. Then there is a constant $I=I(\alpha)$ with the following property. Let $G_i$, $i\in\mathbb{N}$, be a countable family of finite subgroups of $\operatorname{Bim}(X)_{\alpha}$. Then there exists a reduced fibre $F$ of the map $\alpha$ and its irreducible component $F'$ of dimension $\dim(X)\dim(Y)$ such that in every group $G_i$ there is a subgroup of index at most $I$ that is isomorphic to a subgroup of $\operatorname{Bim}(F')$. Moreover, if $\dim(Y)>0$, then, given a countable union $\Xi$ of proper closed analytic subsets of $Y$, the fibre $F$ can be chosen so that the point $\alpha(F)$ does not lie in $\Xi$. Proof. Let $\Delta\subset Y$ be the minimal closed subset of $Y$ such that every point $P$ of $Y\setminus\Delta$ is smooth, the fibre $\alpha^{1}(P)$ is reduced, and every irreducible component of $\alpha^{1}(P)$ has dimension $\dim(X)\dim(Y)$. Then $Y\setminus \Delta$ is a dense open subset of $Y$. Enlarging $\Delta$ if necessary we may assume that the fibres over all the points of $Y\setminus\Delta$ have the same number $N$ of irreducible components. Set
$$
\begin{equation*}
\mathcal{G}=\bigcup_i G_i;
\end{equation*}
\notag
$$
thus, $\mathcal{G}$ is a countable set of elements of $\operatorname{Bim}(X)_\alpha$.
Let $\gamma$ be an element of the group $\operatorname{Bim}(X)_{\alpha}$. Consider the set $\nabla_\gamma\subset Y$ of all points $P$ for which $\operatorname{Ind}(\gamma)$ contains an irreducible component of the fibre $\alpha^{1}(P)$. Thus, the map $\gamma$ is defined at a typical point of every irreducible component of the fibre $\alpha^{1}(P)$ over every point $P\in Y\setminus \nabla_\gamma$. Moreover, for any point
$$
\begin{equation*}
P\in Y\setminus(\Delta\cup\nabla_\gamma\cup\nabla_{\gamma^{1}})
\end{equation*}
\notag
$$
the restriction $\gamma_{F'}$ of the map $\gamma$ to every irreducible component $F'$ of $\alpha^{1}(P)$ is a bimeromorphic map of $F'$ onto its image $\gamma(F')$, and $\gamma(F')$ does not coincide with the image of any other irreducible component of $\alpha^{1}(P)$.
Consider the subset $D_\gamma\subset Y\setminus (\Delta\cup\nabla_\gamma)$ of all points $P$ such that for a typical point $Q$ of some irreducible component of the fibre $\alpha^{1}(P)$ one has $\gamma(Q)=Q$. Let $\overline{D_\gamma}$ be the closure of $D_\gamma$ in $Y$. Then for any point
$$
\begin{equation*}
P\in Y\setminus(\Delta\cup\nabla_\gamma\cup\nabla_{\gamma^{1}} \cup\overline{D_\gamma})
\end{equation*}
\notag
$$
the restriction $\gamma_{F'}$ of the map $\gamma$ to every irreducible component $F'$ of the fibre $\alpha^{1}(P)$ is distinct from the identity map of $F'$, provided that $\gamma$ itself is not the identity map of $X$.
The sets $\Delta$, $\nabla_\gamma$ and $\overline{D_\gamma}$ are proper closed analytic subsets of $Y$. If $\dim(Y)>0$, then also fix a subset $\Xi$ that is a countable union of proper closed analytic subsets of $Y$. Since the field $\mathbb{C}$ is uncountable, $Y$ cannot be represented as a countable union of proper closed subsets. Hence the complement
$$
\begin{equation*}
U=Y\setminus\biggl(\Xi\cup\Delta\cup\bigcup_{\gamma\in\mathcal{G} \setminus\{\mathrm{id}\}}(\nabla_\gamma\cup\overline{D_\gamma})\biggr)
\end{equation*}
\notag
$$
is nonempty.
Let $P$ be a point in $U$, and let $F$ be the fibre of $\alpha$ over $P$. Every element $\gamma\in \mathcal{G}$ defines a permutation of the set of $N$ irreducible components of $F$. Thus, for every $i$ we have a homomorphism $G_i\to\mathfrak{S}_N$ to the symmetric group of degree $N$. Denote the kernel of this homomorphism by $K_i\subset G_i$. Then the index of $K_i$ in $G_i$ is at most $N!=\mathfrak{S}_N$. Moreover, every element of $K_i$ maps every irreducible component $F'$ of $F$ to itself, and every nontrivial element of $K_i$ restricts to a nontrivial bimeromorphic selfmap of $F'$. Therefore, all groups $K_i$ are embedded into the group $\operatorname{Bim}(F')$. Lemma 3.2. Let $X$ and $Y$ be compact complex varieties, and let $\alpha\colon X\dashrightarrow Y$ be a dominant meromorphic map. Let $F$ be a typical fibre of $\alpha$. Assume that for any irreducible component $F'$ of $F$ the group $\operatorname{Bim}(F')$ is Jordan. Also assume that the image $h_\alpha(\operatorname{Bim}(X;\alpha))\subset\operatorname{Bim}(Y)$ has bounded finite subgroups. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. Proof. Suppose that the group $\operatorname{Bim}(X;\alpha)$ is not Jordan. By Lemma 2.1 this means that the group $\operatorname{Bim}(X)_\alpha$ is not Jordan either. Hence $\operatorname{Bim}(X)_\alpha$ contains a countable family of subgroups $G_i$, $i\in\mathbb{N}$, such that the minimum indices $J_i$ of normal abelian subgroups of $G_i$ form an unbounded sequence. On the other hand, by Lemma 3.1 there exists a constant $I$, a typical fibre $F$ of the map $\alpha$, and an irreducible component $F'$ of $F$ such that every $G_i$ contains a subgroup $K_i$ of index at most $I$ that can be embedded into $\operatorname{Bim}(F')$. Since $\operatorname{Bim}(F')$ is a Jordan group, we conclude that the minimum index of an abelian subgroup of $K_i$ is bounded by a constant $J$ independent of $i$. Therefore, the minimum index of an abelian subgroup of $G_i$ is bounded by $IJ$, and so the minimum index of a normal abelian subgroup of $G_i$ is also bounded. This contradicts the unboundedness of the indices $J_i$.
The lemma is proved. Corollary 3.3. Let $X$ be a threedimensional compact complex variety, let $Z$ be a compact complex surface, and let $\alpha\colon X\dashrightarrow Z$ be a dominant meromorphic map. Assume that the image $h_\alpha(\operatorname{Bim}(X;\alpha))\!\subset\!\operatorname{Aut}(Z)$ has bounded finite subgroups. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. Proof. Let $F$ be a typical fibre of $\alpha$, and let $F'$ be an irreducible component of $F$. Then $F'$ is a smooth projective curve. Hence $\operatorname{Bim}(F')=\operatorname{Aut}(F')$ is Jordan (see Remark 2.11). Therefore, the required assertion follows from Lemma 3.2. The corollary is proved. Corollary 3.4. Let $X$ be a threedimensional compact complex variety such that $\varkappa(X)\geqslant 0$, let $B$ be a curve, and let $\alpha\colon X\dashrightarrow B$ be a dominant meromorphic map. Assume that the image $h_\alpha(\operatorname{Bim}(X;\alpha))\!\subset\!\operatorname{Aut}(B)$ has bounded finite subgroups. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. Proof. Let $F$ be a typical fibre of $\alpha$, and let $F'$ be an irreducible component of $F$. Then $\varkappa(F')\geqslant 0$ by Lemma 2.9. Hence $\operatorname{Bim}(F')$ is Jordan by Theorem 2.10. Therefore, the required assertion follows from Lemma 3.2. The corollary is proved. Corollary 3.5. Let $X$ be a threedimensional compact complex variety such that $\varkappa(X)\geqslant 0$, let $A$ be a smooth projective curve, and let $\alpha\colon X\dashrightarrow A$ be a dominant meromorphic map. Assume that either the genus of $A$ is at least $2$, or $A$ is elliptic and the image $h_\alpha(\operatorname{Bim}(X;\alpha))\subset\operatorname{Aut}(A)$ preserves a nonempty finite subset of $A$. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. We see from the assumptions that the group $h_\alpha(\operatorname{Bim}(X;\alpha))$ is finite. Therefore, this assertion follows from Corollary 3.4. Corollary 3.6. Let $X$ be a threedimensional compact Kähler variety such that $\varkappa(X)\geqslant 0$, and let $\alpha\colon X \dashrightarrow T$ be a dominant meromorphic map to a twodimensional complex torus $T$. Assume that there exists a subvariety $V\subsetneq T$ which is invariant with respect to $h_\alpha(\operatorname{Bim}(X;\alpha))$. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. Proof. Recall that the action of $h_\alpha(\operatorname{Bim}(X;\alpha))$ on $T$ is regular by Proposition 2.6. If the group $h_\alpha(\operatorname{Bim}(X;\alpha))$ has bounded finite subgroups, then $\operatorname{Bim}(X;\alpha)$ is Jordan by Corollary 3.3.
Let $V_1$ be an irreducible component of $V$, and let $\Gamma_1\subset h_\alpha(\operatorname{Bim}(X;\alpha))$ be the stabilizer of $V_1$. Then $\Gamma_1$ is a finiteindex subgroup of $h_\alpha(\operatorname{Bim}(X;\alpha))$. If $\Gamma_1$ acts on $V_1$ with a fixed point, then it has bounded finite subgroups by Theorem 2.12 (or by Theorems 2.5 and 2.3). Hence $h_\alpha(\operatorname{Bim}(X;\alpha))$ has bounded finite subgroups too. In particular, this applies to the case when $V_1$ is a point itself.
Assume that $V_1$ is a curve; then $V_1$ is not rational. If $V_1$ is singular, then a subgroup of finite index of $\Gamma_1$ acts on $T$ with a fixed point, and so $\Gamma_1$ and $h_\alpha(\operatorname{Bim}(X;\alpha))$ have bounded finite subgroups. Thus, we assume that $V_1$ is smooth. If $\operatorname{g}(V_1)>1$, then the group $\operatorname{Aut}(V_1)$ is finite. This means that a subgroup of finite index of $\Gamma_1$ acts trivially on $V_1$ and, in particular, has a fixed point. As before, we conclude that $\Gamma_1$ and $h_\alpha(\operatorname{Bim}(X;\alpha))$ have bounded finite subgroups in this case.
Therefore, we may assume that $V_1$ is an elliptic curve. Consider the quotient torus $T_1=T/V_1$, where the group structure on $T$ is chosen in such a way that $V_1$ is a subgroup of $T$ (that is, $V_1$ contains the neutral element of $T$). Note that $T_1$ and the quotient map $T\to T_1$ do not depend on this choice, and therefore the map $T\to T_1$ is $\Gamma_1$equivariant. Thus the composition $X \to T\to T_1$ is equivariant with respect to the preimage $\widetilde{\Gamma}_1$ of $\Gamma_1$ in $\operatorname{Bim}(X;\alpha)$. Moreover, the image of $\widetilde{\Gamma}_1$ (or $\Gamma_1$) in $\operatorname{Aut}(T_1)$ fixes a point on the elliptic curve $T_1$. Therefore, the group $\widetilde{\Gamma}_1$ is Jordan by Corollary 3.5. Since $\widetilde{\Gamma}_1$ has finite index in $\operatorname{Bim}(X;\alpha)$, the latter group is Jordan too.
The corollary is proved. Lemma 3.7. Let $X$ and $Y$ be compact complex varieties, and let $\alpha\colon X\dashrightarrow Y$ be a dominant meromorphic map. Let $F$ be a typical fibre of $\alpha$. Assume that for any irreducible component $F'$ of $F$ the group $\operatorname{Bim}(F')$ has bounded finite subgroups. Also assume that the image $h_\alpha(\operatorname{Bim}(X;\alpha))\subset\operatorname{Bim}(Y)$ is strongly Jordan. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan. Proof. As in the proof of Lemma 3.2, suppose that the group $\operatorname{Bim}(X;\alpha)$ is not Jordan. By Lemma 2.2 this means that the group $\operatorname{Bim}(X)_\alpha$ has unbounded finite subgroups. In other words, $\operatorname{Bim}(X)_\alpha$ contains a countable family of subgroups $G_i$, $i\in\mathbb{N}$, such that the orders of $G_i$ are unbounded. On the other hand, by Lemma 3.1 there exists a constant $I$, a typical fibre $F$ of the map $\alpha$, and an irreducible component $F'$ of $F$ such that every $G_i$ contains a subgroup $K_i$ of index at most $I$ that can be embedded into $\operatorname{Bim}(F')$. Since $\operatorname{Bim}(F')$ has bounded finite subgroups, we see that the orders of the $K_i$ are bounded. This contradicts the unboundedness of the orders of the groups $G_i$.
The lemma is proved. Lemma 3.7 immediately implies the following. Corollary 3.8. Let $X$ and $Y$ be compact complex varieties, and let $\alpha\colon X\dashrightarrow Y$ be a dominant meromorphic map such that a typical fibre of $\alpha$ is finite. Assume that the group $\operatorname{Bim}(Y)$ is strongly Jordan. Then the group $\operatorname{Bim}(X;\alpha)$ is Jordan. In particular, if under these assumptions the map $\alpha$ is equivariant with respect to the whole group $\operatorname{Bim}(X)$, then $\operatorname{Bim}(X)$ is Jordan.
§ 4. Pseudoautomorphisms In this section we consider groups of pseudoautomorphisms of compact Kähler manifolds and establish the Jordan property for the groups of bimeromorphic selfmaps of threedimensional compact Kähler varieties of Kodaira dimension zero. Recall that a biholomorphic selfmap $f\colon X\dashrightarrow X$ of a compact complex variety $X$ is called a pseudoautomorphism if there exist nonempty Zariski open subsets $U_1,U_2\subset X$ such that $\operatorname{codim}_X(X\setminus U_i)\geqslant 2$, and $f$ restricts to an isomorphism
$$
\begin{equation*}
f_{U_1}\colon U_1\xrightarrow{\sim} U_2.
\end{equation*}
\notag
$$
The pseudoautomorphisms of $X$ form a subgroup of $\operatorname{Bim}(X)$, which we denote by $\operatorname{PAut}(X)$. Lemma 4.1 (negativity lemma; see [26], § 1.1, [27], Lemma 1.3). Let $f\colon \widetilde V\to V$ be a proper bimeromorphic morphism between normal complex varieties. Let $D$ be a Cartier divisor on $\widetilde V$ such that $D$ is $f$nef. Then $D$ is effective if and only if $f_*D$ is effective. The following assertion is well known to experts (see, for instance, [28], Lemma 4.3). We provide its proof for the convenience of the reader. Lemma 4.2. Let $\chi\colon X \dashrightarrow X'$ be a bimeromorphic map of compact complex varieties with terminal singularities such that $\boldsymbol{\omega}_{X'}$ is nef. Then $\chi^{1}$ does not contract any divisors. Proof. By the definition of terminal singularities, for a sufficiently divisible positive integer $m$ the sheaves $\boldsymbol{\omega}_X^{[m]}$ and $\boldsymbol{\omega}_{X'}^{[m]}$ are invertible. Let be a common resolution, that is, a commutative diagram where $Y$ is a compact complex manifold, and $p$ and $q$ are proper bimeromorphic morphisms. Making further blowups if necessary we may assume that the exceptional sets $\operatorname{Exc}(p)$ and $\operatorname{Exc}(q)$ are of pure codimension $1$. Set $\Upsilon_p=p(\operatorname{Exc}(p))$ and $\Upsilon_q=q(\operatorname{Exc}(q))$. Then $p$ induces an isomorphism of open subsets $Y\setminus\operatorname{Exc}(p)$ and $X\setminus \Upsilon_p$; similarly, $q$ induces an isomorphism of $Y\setminus\operatorname{Exc}(q)$ and $X'\setminus \Upsilon_q$. This implies that we can write
$$
\begin{equation*}
\begin{aligned} \, \boldsymbol{\omega}_Y^{\otimes m} &\cong p^*(\boldsymbol{\omega}_X^{[m]})\otimes \mathscr{O}_Y\Bigl(m\Bigl(\sum a_i E_i +\sum b_j F_j\Bigr)\Bigr) \\ &\cong q^*(\boldsymbol{\omega}_{X'}^{[m]}) \otimes\mathscr{O}_Y\Bigl(m\Bigl(\sum c_iE_i +\sum d_k G_k\Bigr)\Bigr), \end{aligned}
\end{equation*}
\notag
$$
where $E_i$ (respectively, $F_j$; respectively, $G_k$) are exceptional divisors with respect to both $p$ and $q$ (respectively, $p$exceptional but not $q$exceptional; respectively, $q$exceptional but not $p$exceptional). Thus, none of the three divisors $\sum E_i$, $\sum F_j$ and $\sum G_k$ has a common irreducible component with any of the other two. Since $X$ and $X'$ have only terminal singularities, the rational numbers $a_i$, $b_j$, $c_i$ and $d_k$ are strictly positive.
Consider the Cartier divisor
$$
\begin{equation*}
D= m\Bigl(\sum (c_ia_i) E_i+\sum d_k G_k\sum b_j F_j\Bigr).
\end{equation*}
\notag
$$
Clearly, its pushforward $p_*D$ is effective. Since
$$
\begin{equation*}
\mathscr{O}_Y(D)\cong q^* \boldsymbol{\omega}_{X'}^{[m]}\otimes p^*\boldsymbol{\omega}_X^{[m]},
\end{equation*}
\notag
$$
we see that the divisor $D$ is $p$nef. Thus, $D$ is effective by Lemma 4.1. Hence $\sum F_j=0$, that is, the map $\chi^{1}$ does not contract any divisors.
The lemma is proved. Corollary 4.3. Let $X$ be a compact complex variety with terminal singularities and $\boldsymbol{\omega}_X$ nef. Then $\operatorname{Bim}(X)$ acts on $X$ by pseudoautomorphisms. The following result is due to Fujiki. Theorem 4.4 ([17], Corollary 3.3). Let $X$ be a compact Kähler manifold, and let $g$ be a pseudoautomorphism of it. Assume that for a Kähler class $\alpha$ on $X$ its pushforward $g_*\alpha$ is a Kähler class again. Then $g^{1}$ is a morphism. Theorem 4.4 allows one to investigate the groups of pseudoautomorphisms of compact Kähler manifolds. Proposition 4.5. Let $X$ be a compact Kähler manifold. Then the group of its pseudoautomorphisms is Jordan. Proof. The group $\operatorname{PAut}(X)$ acts naturally on $H^2(X,\mathbb{Z})$. Let $\operatorname{PAut}'(X)$ be the kernel of this action. Then $\operatorname{PAut}'(X)$ preserves a Kähler class. By Theorem 4.4 the group $\operatorname{PAut}'(X)$ consists of biholomorphic automorphisms. Hence the group $\operatorname{PAut}'(X)$ is Jordan by Theorem 1.2. On the other hand the quotient $\operatorname{PAut}(X)/\operatorname{PAut}'(X)$ acts faithfully on the finitely generated abelian group $H^2(X,\mathbb{Z})$, and thus it has bounded finite subgroups by Corollary 2.4. Therefore, the group $\operatorname{PAut}(X)$ is Jordan by Lemma 2.1.
The proposition is proved. Remark 4.6. In [17], Theorem 4.4 was proved for manifolds (smooth varieties). We do not know if this result, and thus Proposition 4.5, can be generalized to the case of singular Kähler varieties. Using Proposition 4.5 in combination with Corollary 4.3 we obtain the following. Corollary 4.7. Let $X$ be a compact Kähler manifold such that $\boldsymbol{\omega}_X$ is nef. Then the group $\operatorname{Bim}(X)$ is Jordan.
§ 5. Albanese map In this section we collect some information about the Albanese maps of compact complex manifolds. Let $X$ be a compact complex manifold. We denote by
$$
\begin{equation*}
\boldsymbol{\alpha}\colon X\to \operatorname{Alb}(X)
\end{equation*}
\notag
$$
the Albanese morphism of $X$ (see [19], Definition 9.6). The map $\boldsymbol{\alpha}$ has the following universal property. If $\zeta\colon X\to T$ is a morphism to an arbitrary complex torus $T$, then there exists a unique homomorphism of complex tori $\xi\colon \operatorname{Alb}(X)\to T$ that fits in the commutative diagram Remark 5.1. More generally, the Albanese morphism is well defined (and has the above universal property) for compact complex varieties with rational singularities (see [24], Theorem 3.3 and Remark 3.4). The next assertion is well known to experts, but we provide its proof for the reader’s convenience. We will use it several times without further reference in what follows. Proposition 5.2. Let $X$ be a compact complex manifold. Then there is a natural biregular action of the group $\operatorname{Bim}(X)$ on $\operatorname{Alb}(X)$ such that the morphism $\boldsymbol{\alpha}$ is equivariant. Proof. Let $\varphi\colon X\dashrightarrow X$ be an arbitrary bimeromorphic map. The composition $\boldsymbol{\alpha}\circ\varphi$ is holomorphic by Proposition 2.6. By the universal property of $\boldsymbol{\alpha}$ there exist a unique translation
$$
\begin{equation*}
\mathrm{t}_a\colon \operatorname{Alb}(X) \to \operatorname{Alb}(X)
\end{equation*}
\notag
$$
by an element $a\in \operatorname{Alb}(X)$ and a unique homomorphism of complex tori
$$
\begin{equation*}
\psi\colon \operatorname{Alb}(X) \to\operatorname{Alb}(X)
\end{equation*}
\notag
$$
such that
$$
\begin{equation*}
\boldsymbol{\alpha}\mathbin{\circ} \varphi= \mathrm{t}_a\mathbin{\circ} \psi \mathbin{\circ}\boldsymbol{\alpha}.
\end{equation*}
\notag
$$
In other words, there exists a unique morphism of abstract varieties
$$
\begin{equation*}
\theta_{\varphi}= \mathrm{t}_a\mathbin{\circ} \psi\colon \operatorname{Alb}(X) \to\operatorname{Alb}(X),
\end{equation*}
\notag
$$
that fits in the following commutative diagram It is easy to see that this correspondence is functorial: for any two bimeromorphic maps $\varphi'\colon X\dashrightarrow X$ and $\varphi''\colon X\dashrightarrow X$ one has
$$
\begin{equation*}
\theta_{\varphi'}\mathbin{\circ} \theta_{\varphi''}=\theta_{\varphi'\mathbin{\circ} \varphi''}.
\end{equation*}
\notag
$$
In particular, the correspondence $\varphi \longmapsto \theta_{\varphi}$ defines a group homomorphism
$$
\begin{equation*}
\operatorname{Bim}(X) \to \operatorname{Aut}(\operatorname{Alb}(X)).
\end{equation*}
\notag
$$
Remark 5.3. If $X$ is a normal compact complex variety with rational singularities and $\operatorname{q}(X)=1$, then fibres of the Albanese map $\boldsymbol{\alpha}\colon X\to A=\operatorname{Alb}(X)$ are connected. Indeed, otherwise $\boldsymbol{\alpha}$ admits a nontrivial Stein factorisation
$$
\begin{equation*}
\boldsymbol{\alpha}\colon X\xrightarrow{\boldsymbol{\alpha}'} A'\to A.
\end{equation*}
\notag
$$
Let $\beta$ be the composition of $\boldsymbol{\alpha}'$ with the embedding of $A'$ into its Jacobian $J(A')$. Then the map $\beta\colon X\to J(A')$ must factor through $\boldsymbol{\alpha}$ by the universal property of the Albanese map, which is clearly impossible. Recall that a compact complex subvariety $Y$ of a complex torus $T$ is said to generate $T$ if for some positive integer $n$ every point of $T$ can be represented as a sum of $n$ points of $Y$; see [19], Definition 9.13. For instance, a proper subtorus of $T$ (containing the neutral element of the group $T$) does not generate $T$. Proposition 5.4. Let $T$ be a complex torus, and let $Y\subset T$ be a compact complex subvariety. Then there exists a canonically defined fibration $Y\to Z$ whose typical fibre is a subtorus $T_1\subset T$ and the base $Z$ is a (possibly singular) compact complex variety with $\dim(Z)=\varkappa(Z)$. Moreover, $Z$ is a point if and only if $Y$ is a translation of a subtorus $T_1\subset T$. Take the largest subtorus of $T_1\subset T$ such that $Y$ is invariant under the translations by all elements of $T_1$, and then set $Z=Y/T_1$. For details, see [19], Theorem 10.9 and its proof. Lemma 5.5. Let $X$ be a threedimensional compact complex manifold such that $\varkappa(X)\geqslant 0$. Assume that $\boldsymbol{\alpha}$ is not surjective. Then the group $\operatorname{Bim}(X)$ is Jordan. Proof. Let $Y=\boldsymbol{\alpha}(X)\subsetneq\operatorname{Alb}(X)$. By Proposition 5.4 there exists a canonically defined fibration
$$
\begin{equation*}
\gamma\colon Y \to Z,
\end{equation*}
\notag
$$
where $\dim(Z)=\varkappa(Z)$. The group $\operatorname{Bim}(Z)$ is finite by Theorem 2.8. Recall that $Y$ generates $\operatorname{Alb}(X)$ (see [19], Lemma 9.14). Since $Y$ contains the neutral element of the group $\operatorname{Alb}(X)$, this implies that it is not contained in a proper subtorus of $\operatorname{Alb}(X)$. Hence $Z$ is not a point by Proposition 5.4.
Now we have the morphism
$$
\begin{equation*}
\lambda=\gamma\mathbin{\circ}\boldsymbol{\alpha}\colon X \to Z,
\end{equation*}
\notag
$$
which is equivariant with respect to the group $\operatorname{Bim}(X)$. Let $F$ be a typical fibre of $\lambda$, and let $F'$ be its connected component. Then $F'$ is a compact complex manifold of dimension at most $2$, and $\varkappa(F')\geqslant 0$ by Lemma 2.9. Thus the group $\operatorname{Bim}(F')$ is Jordan by Theorem 2.10 and Remark 2.11. Therefore, the group $\operatorname{Bim}(X)$ is Jordan by Lemma 3.2.
The lemma is proved. Corollary 5.6. Let $X$ be a threedimensional compact complex manifold such that $\varkappa(X)\geqslant0$. Assume that $\operatorname{Bim}(X)$ is not Jordan. Then $\dim(\operatorname{Alb}(X))<\dim(X)$. In particular, if $X$ is Kähler, then $\operatorname{q}(X)<\dim(X)$. Proof. We know from Lemma 5.5 that $\boldsymbol{\alpha}$ is surjective. In particular, this implies that $\dim(\operatorname{Alb}(X))\leqslant \dim(X)$. Suppose that $\dim(\operatorname{Alb}(X))=\dim(X)$. Then a typical fibre of $\boldsymbol{\alpha}$ is finite. Applying Corollaries 3.8 and 2.7 we see that the group $\operatorname{Bim}(X)$ is Jordan, which is not the case by assumption.
If $X$ is a compact Kähler manifold, then $\dim(\operatorname{Alb}(X))=\operatorname{q}(X)$.
The corollary is proved.
§ 6. Indeterminacy loci In this section we make some observations concerning the indeterminacy loci of bimeromorphic maps of threedimensional compact complex varieties. Lemma 6.1. Let $X$ be a normal threedimensional compact complex variety, and let $\chi\colon X \dashrightarrow X$ be a pseudoautomorphism. Let $\beta\colon X\to Z$ be a morphism to a curve $Z$ such that $\chi$ maps fibres of $\beta$ to fibres of $\beta$ again. Then the indeterminacy locus $\operatorname{Ind}(\chi)$ is contained in a finite set of fibres of $\beta$. Proof. Assume the contrary, that is, suppose there is an irreducible curve $C\subset \operatorname{Ind}(\chi)$ that dominates $Z$. Let $\widetilde Y$ be the normalization of the graph of $\chi$. Thus, we have a commutative diagram Since $\chi$ is not defined at a typical point of $C$, there exists a twodimensional component $E$ of the $p$exceptional set that dominates $C$. Since the map $\widetilde Y\to X\times X$ is finite onto its image, no curves on $\widetilde Y$ are contracted by both $p$ and $q$. Hence $C'=q(E)$ is a curve, and a typical fibre $\Gamma$ of $q_E\colon E\to C'$ dominates $C$. Thus $\Gamma$ meets the proper transform $\widetilde F$ of any fibre $F=\beta^{1}(z)$, and so $q(\widetilde F)$ contains $C'$. In other words, $C'$ is contained in every fibre of $\beta$, which gives a contradiction.
The lemma is proved. The next proposition is a relative analogue of the wellknown assertion about the decomposition of some maps into flops (see, for instance, [28], Theorem 4.9). In our case the proof follows the same scheme. We outline it for the convenience of the reader. Proposition 6.2. Let $f\colon X\to S$ and $f'\colon X'\to S$ be proper morphisms with onedimensional fibres, where $X$ and $X'$ are threedimensional Kähler varieties with terminal $\mathbb{Q}$factorial singularities and $S$ is a smooth surface. Assume that both $K_X$ and $K_{X'}$ are nef over $S$. Let $\chi\colon X\dashrightarrow X'$ be a bimeromorphic map such that it is an isomorphism in codimension one, and the indeterminacy loci $\operatorname{Ind}(\chi)$ and $\operatorname{Ind}(\chi^{1})$ are contained in (a finite set of) fibres of $f$ and $f'$, respectively. Assume that there exists a commutative diagram where $\zeta$ is a bimeromorphic map of $S$. Then $\zeta$ is an isomorphism, and $\chi$ is a composition of flops in curves contained in (a finite set of) fibres over $S$. Outline of the proof. By assumption $\chi$ is an automorphism outside a finite union of fibres of $f$. This means that $\zeta$ induces an automorphism of a smooth open subset $S^o\subset S$ that is the complement of a finite subset of $S$. Therefore, by Hartogs’s extension theorem $\zeta$ is an isomorphism on the whole of $S$. To prove that $\chi$ is a composition of flops in curves contracted by $f$, we may assume that $S$ is a small analytic neighbourhood of some point $s\in S$. Thus, we may assume that $f'$ is a projective morphism, because it has onedimensional fibres. In other words, there exists a divisor $D'$ on $X'$ that is very ample over $S$: one can construct $D'$ as a union of discs meeting the fibre $f'^{1}(s)$ transversely. In a similar way we see that the morphism $f$ is projective. Let $D\subset X$ be the proper transform of $D'$. For small positive $\varepsilon$, run the $(K_X+\varepsilon D)$MMP over $S$. This is possible because $f$ is projective (see [29], § 4). Every step of this MMP is a flop, and we end up with $X'$. Corollary 6.3. In the notation of Proposition 6.2 set $\Theta=f(\operatorname{Sing}(X))$. Then $\zeta(\Theta)=\Theta$. Proof. A flop preserves the analytic types of singularities of the ambient variety (see [28], Theorem 2.4). Thus, the required assertion follows from Proposition 6.2. The following fact is well known. Lemma 6.4 (see, for instance, [30], Lemma 1.8). Let $X$ be a threedimensional compact complex variety with at worst terminal singularities, and let $\chi\colon X \dashrightarrow X$ be a pseudoautomorphism. Then any irreducible component of the indeterminacy locus $\operatorname{Ind}(\chi)$ is a rational curve.
§ 7. More on compact complex surfaces In this section we make some additional observations on compact complex surfaces and their automorphism groups. Lemma 7.1. Let $S$ be a smooth compact Kähler surface such that $0\leqslant \varkappa(S)\leqslant 1$ and $\operatorname{q}(S)>0$. Then $\chi_{\mathrm{top}}(S)=0$ if and only if $S$ contains no rational curves. Proof. We may assume that $S$ is minimal. Indeed, according to the EnriquesKodaira classification (see, for example, [31], Ch. VI, § 1), every minimal surface of nonnegative Kodaira dimension has a nonnegative topological Euler characteristic. Thus, any nonminimal compact complex surface of nonnegative Kodaira dimension has a positive topological Euler characteristic (and contains rational curves).
Assume that $\varkappa(S)=0$. Since $\operatorname{q}(S)>0$, we conclude that $S$ is either a complex torus, or a bielliptic surface. In both cases $\chi_{\mathrm{top}}(S)=0$ and $S$ contains no rational curves.
Assume that $\varkappa(S)=1$. Let $\boldsymbol{\pi}\colon S\to B$ be the pluricanonical map; thus, $\boldsymbol{\pi}$ is an elliptic fibration over a curve $B$. If the genus $\operatorname{g}(B)$ is positive, then all rational curves on $S$ are contained in a finite set of fibres of $\boldsymbol{\pi}$, and $\chi_{\mathrm{top}}(S)=\sum\chi_{\mathrm{top}}(F_i)$, where the $F_i$ are degenerate fibres. Clearly, $\chi_{\mathrm{top}}(S)>0$ if and only if $\boldsymbol{\pi}$ has a fibre that is not an elliptic curve. Such a fibre is a union of rational curves.
Assume that $B\cong \mathbb{P}^1$. We are going to show that in this case $\chi_{\mathrm{top}}(S)=0$ and $S$ does not contain rational curves. Indeed, the Albanese map $\boldsymbol{\alpha}\colon S\to A$ cannot be constant on the fibres of $\boldsymbol{\pi}$, because otherwise it would factor through the Albanese map of $\mathbb{P}^1$, which is impossible since $\operatorname{q}(S)>0$. Therefore, each fibre $S_b$, $b\in B$, of the map $\boldsymbol{\pi}$ contains no rational curves; in other words, it is of type ${}_m\mathrm{I}_0$ in the Kodaira classification of degenerate fibres of elliptic fibrations (see, for instance, [31], Ch. V, § 7). Hence
$$
\begin{equation*}
\chi_{\mathrm{top}}(S)=\sum_{b\in B}\chi_{\mathrm{top}}(S_b)=0.
\end{equation*}
\notag
$$
We claim that $\operatorname{q}(S)=1$. Indeed, given a typical fibre $S_o$ of $\boldsymbol{\pi}$, the image $\boldsymbol{\alpha}(S_o)$ is an elliptic curve on $A$. Consider the quotient map $A\to A'=A/\boldsymbol{\alpha}(S_o)$. The image of $\boldsymbol{\alpha}(S_o)$ is a point in $A'$, and so for any fibre $S_b$ of $\boldsymbol{\pi}$ the image of $\boldsymbol{\alpha}(S_b)$ in $A'$ is also a point. Thus, the image of $\boldsymbol{\alpha}(S)$ in $A'$ is a curve dominated by the (rational) base of the fibration $\boldsymbol{\pi}$. Since $A'$ contains no rational curves, we conclude that the image of $\boldsymbol{\alpha}(S)$ in $A'$ is a point. Hence $\boldsymbol{\alpha}(S)=\boldsymbol{\alpha}(S_o)=A$.
Therefore, $A$ is an elliptic curve and $\boldsymbol{\alpha}$ has connected fibres (see Remark 5.3). Since $S$ is a Kähler surface, we know that $\operatorname{b}_1(S)=\operatorname{b}_3(S)=2\operatorname{q}(S)$ and
$$
\begin{equation*}
\operatorname{b}_2(S)=\chi_{\mathrm{top}}(S)2+4\operatorname{q}(S)=2.
\end{equation*}
\notag
$$
Hence all the fibres of $\boldsymbol{\alpha}$ are irreducible.
Let $F_1,\dots, F_r$ be all the singular fibres of $\boldsymbol{\alpha}$, let $m_1,\dots, m_r$ be their multiplicities, and let $F$ be a typical fibre of $\boldsymbol{\alpha}$. Thus, $F$ is numerically equivalent to $m_iF_i$, and $\chi_{\mathrm{top}}(F)=22\operatorname{g}(F)=K_S\cdot F$. In a similar way
$$
\begin{equation*}
\chi_{\mathrm{top}}(F_i)\geqslant 22\operatorname{p}_{\mathrm{a}}(F_i)=K_S\cdot F_i=\frac{1}{m_i}K_S\cdot F.
\end{equation*}
\notag
$$
Thus we have
$$
\begin{equation*}
0=\chi_{\mathrm{top}}(S)=\sum_{i}\bigl(\chi_{\mathrm{top}}(F_i)\chi_{\mathrm{top}}(F)\bigr)\geqslant K_S\cdot F \cdot \sum_{i} \biggl(1\frac 1{m_i}\biggr).
\end{equation*}
\notag
$$
Since $S$ is not covered by rational curves, we know that $K_S\cdot F>0$. This shows that $m_i=1$ and $\chi_{\mathrm{top}}(F_i)= 22\operatorname{p}_{\mathrm{a}}(F_i)$ for all $i$. Therefore, $\boldsymbol{\alpha}$ is a smooth morphism. Now, if $S$ contains a rational curve $C$, then $C$ must be a (smooth) fibre of $\boldsymbol{\alpha}$. But then
$$
\begin{equation*}
2=2\operatorname{g}(C)2=K_S\cdot C=K_S\cdot F\geqslant 0,
\end{equation*}
\notag
$$
which is a contradiction.
Lemma 7.1 is proved. The next result was proved in [25]. Proposition 7.2 (see [25], Corollary 4.1). Let $S$ be a compact Kähler surface such that $\varkappa(S)\geqslant 0$. Assume that the group $\operatorname{Bim}(S)$ has unbounded finite subgroups. Then either $\varkappa(S)=1$, or $S$ is bimeromorphic to a complex torus or a bielliptic surface. The following fact is a version of Proposition 7.2 for automorphism groups. Proposition 7.3. Let $S$ be a minimal compact Kähler surface with $\varkappa(S)\geqslant 0$. Assume that the group $\operatorname{Aut}(S)$ has unbounded finite subgroups. Then $\operatorname{q}(S)>0$ and $\chi_{\mathrm{top}}(S)=0$. Moreover, $S$ is either a complex torus, or a bielliptic surface, or a surface with $\varkappa(S)=1$. Proof. We have $\varkappa(S)<2$ by Theorem 2.8. If $\varkappa(S)= 0$, then by Proposition 7.2 the surface $S$ is either a complex torus, or a bielliptic surface. In both cases the required assertions clearly hold.
Assume that $\varkappa(S)=1$. Let $\boldsymbol{\pi}\colon S\to B$ be the pluricanonical fibration. If $\boldsymbol{\pi}$ has a fibre $S_b=\boldsymbol{\pi}^{1}(b)$ over some point $b\in B$ which is not of type ${}_m\mathrm{I}_0$, then a finite index subgroup $\Gamma\subset \operatorname{Aut}(S)$ fixes a singular point of $F_{\mathrm{red}}$. Since $S$ is Kähler, this implies that $\operatorname{Aut}(S)$ has bounded finite subgroups by Theorem 2.12. Thus we may assume that all the fibres of $\boldsymbol{\pi}$ are of type ${}_m\mathrm{I}_0$. Hence $\chi_{\mathrm{top}}(S)=0$. By Noether’s formula this gives $\chi(\mathscr{O}_S)=0$, so that $\operatorname{q}(S)=1+\operatorname{p}_{\mathrm{g}}(S)>0$.
The proposition is proved.
§ 8. Proof of the main theorem In this section we complete the proof of Theorem 1.3. Lemma 8.1. Let $X$ be a threedimensional compact Kähler variety such that ${\varkappa(X)\geqslant0}$. Assume that there exists a dominant $\operatorname{Bim}(X)$equivariant meromorphic map $f\colon X \dashrightarrow Z$ to a smooth projective curve $Z$ of positive genus. Then $\operatorname{Bim}(X)$ is Jordan. Proof. Running the MMP on $X$ we may assume that $X$ has at worst terminal $\mathbb{Q}$factorial singularities and $\boldsymbol{\omega}_X$ is nef. Since $Z$ is not a rational curve, $f$ is holomorphic: otherwise its composition with the embedding of $Z$ into its Jacobian is a nonholomorphic map to a complex torus, which is impossible by Proposition 2.6. Applying Stein factorization we may assume that the fibres of $f$ are connected. Thus, any smooth fibre $F$ of $f$ is a minimal surface of nonnegative Kodaira dimension; in particular, one has $\operatorname{Bim}(F)=\operatorname{Aut}(F)$. Note also that $\operatorname{Bim}(X)$ acts on $Z$ by automorphisms.
Suppose that $\operatorname{Bim}(X)$ is not Jordan. Then it follows from Lemma 3.7 and Remark 2.11 that, given a typical fibre $F$ of $f$, the group $\operatorname{Aut}(F)$ has unbounded finite subgroups. Therefore, by Proposition 7.3 we have $0\leqslant \varkappa(F)\leqslant 1$, $\chi_{\mathrm{top}}(F)=0$ and $\operatorname{q}(F)>0$. Recall that the topological Euler characteristic is constant on smooth families of compact manifolds. Hence for any smooth fibre $F$ of $f$ one has $\chi_{\mathrm{top}}(F)=0$. Furthermore, since the irregularity of a compact complex surface is uniquely determined by its first Betti number (see, for example, [31], Ch. IV, Theorem 2.7), it is also constant on smooth families, so that $\operatorname{q}(F)>0$. Also, we see that $\varkappa(F)\geqslant 0$ by the adjunction formula, and $K_F^2=0$, so that $\varkappa(F)\leqslant 1$. This means that any smooth fibre of $f$ contains no rational curves by Lemma 7.1.
Let $F$ be a smooth fibre of $f$, and let $\gamma\in \operatorname{Bim}(X)$ be an arbitrary element. By Lemma 6.4 any irreducible component $C$ of the indeterminacy locus $\operatorname{Ind}(\gamma)$ is a rational curve. Since $\operatorname{g}(Z)>0$, the curve $C$ cannot dominate $Z$. On the other hand we know already that $C\not\subset F$. Hence $C$ is disjoint from $F$, that is, $\gamma$ is holomorphic near $F$. Then $\gamma(F)$ does not contain rational curves either, and so $\gamma^{1}$ is holomorphic near $\gamma(F)$. This implies that $\gamma$ is an isomorphism near $F$. Hence $\gamma(F)$ is a smooth fibre, that is, any element $\gamma\in \operatorname{Bim}(X)$ maps smooth fibres to smooth fibres (and vice versa, it maps singular fibres to singular fibres).
Now let $F_1,\dots,F_r$, $r\geqslant 0$, be all the singular fibres of $f$. Then $\operatorname{Bim}(X)$ acts biholomorphically on the complement $X\setminus \bigcup F_i$. If $r=0$, then $\operatorname{Bim}(X)$ acts on the whole of $X$ by automorphisms. In this case $\operatorname{Bim}(X)=\operatorname{Aut}(X)$ is Jordan by Theorem 1.2. If $r>0$, then $\operatorname{Bim}(X)$ permutes the points $f(F_1),\dots,f(F_r)$. Therefore, $\operatorname{Bim}(X)$ is Jordan by Corollary 3.5.
The lemma is proved. Lemma 8.2. Let $X$ be a threedimensional compact Kähler variety such that $\varkappa(X)\geqslant0$. Assume that there exists a dominant $\operatorname{Bim}(X)$equivariant meromorphic map $f\colon X \dashrightarrow Z$ to a compact complex surface $Z$ such that $\varkappa(Z)\geqslant 0$. Then $\operatorname{Bim}(X)$ is Jordan. Proof. Running the MMP, we may assume that singularities of $X$ are at worst terminal $\mathbb{Q}$factorial and $\boldsymbol{\omega}_X$ is nef. Then $\operatorname{Bim}(X)$ acts on $X$ by pseudoautomorphisms: see Corollary 4.3. Furthermore, we may assume that $Z$ is a minimal surface. So the induced action of $\operatorname{Bim}(X)$ on $Z$ is biholomorphic. We may also assume that the group $\operatorname{Aut}(Z)$ has unbounded finite subgroups, because otherwise $\operatorname{Bim}(X)$ is Jordan by Corollary 3.3. Note that the surface $Z$ is Kähler by Theorem 5 in [32]. Thus, we know from Proposition 7.3 that $Z$ is ether a complex torus, or a bielliptic surface, or a surface with $\varkappa(Z)=1$.
Assume that $\varkappa(Z)=1$. Let $\psi\colon X\dashrightarrow B$ be the composition of $f$ with the pluricanonical fibration $Z\to B$. The image of $\operatorname{Bim}(X)$ in $\operatorname{Aut}(B)$ is finite by Proposition 1.2 in [25]. Hence $\operatorname{Bim}(X)$ is Jordan by Corollary 3.4. The same argument works if $Z$ is a bielliptic surface: in this case there is an $\operatorname{Aut}(Z)$equivariant elliptic fibration $Z\to \mathbb{P}^1$ (see [31], Ch. V, § 5), and by Kodaira’s canonical bundle formula (see [31], Ch. V, Theorem 12.1) it has $3$ or $4$ multiple fibres. Hence the image of $\operatorname{Bim}(X)$ in $\operatorname{Aut}(\mathbb{P}^1)$ is finite and $\operatorname{Bim}(X)$ is again Jordan.
Finally, assume that $Z$ is a complex torus and $\operatorname{Bim}(X)$ is not Jordan. The map $f$ is holomorphic in this case: see Proposition 2.6.
Assume that $f$ has a twodimensional fibre. Denote by $\Sigma$ the image under $f$ of the union of all twodimensional fibres of $f$. Then $\Sigma$ is a finite nonempty subset of $Z$. Since a pseudoautomorphism cannot contract a twodimensional fibre, we conclude that $\Sigma$ is invariant under the action of the image of $\operatorname{Bim}(X)$ on $Z$. Thus, the group $\operatorname{Bim}(X)$ is Jordan by Corollary 3.6.
Now assume that all fibres of $f$ are onedimensional. By Lemma 6.1 the indeterminacy locus of every bimeromorphic selfmap of $X$ is contained in a finite set of fibres of $f$. Therefore, it follows from Corollary 6.3 that the group $\operatorname{Bim}(X)$ preserves the image of the singular locus of $X$ on $Z$. Thus, $\operatorname{Bim}(X)$ preserves the image of $\operatorname{Sing}(X)$ on $Z$, which is a finite nonempty subset of $Z$ by assumption. By Corollary 3.6 again, this implies that the group $\operatorname{Bim}(X)$ is Jordan.
The lemma is proved. Now we are ready to prove Theorem 1.3. Proof of Theorem 1.3. Let $\boldsymbol{\alpha}\colon X\to A$ be the Albanese map. By Corollary 5.6 we may assume that $\operatorname{q}(X)\leqslant 2$, and by Lemma 5.5 we may assume that $\boldsymbol{\alpha}$ is surjective. Now the assertion of the theorem is given by Lemma 8.1 if $\operatorname{q}(X)=1$ and by Lemma 8.2 if $\operatorname{q}(X)=2$.
The theorem is proved.



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Citation:
Yu. G. Prokhorov, С. A. Shramov, “Finite groups of bimeromorphic selfmaps of nonuniruled Kähler threefolds”, Mat. Sb., 213:12 (2022), 86–108; Sb. Math., 213:12 (2022), 1695–1714
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