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This article is cited in 2 scientific papers (total in 2 papers) Automorphism groups of T. Bandmana, Yu. G. Zarhinb a Bar-Ilan University, Department of Mathematics, Ramat Gan, Israel b Pennsylvania State University, Department of Mathematics, University Park, PA, USA
Russian version:
Uspekhi Matematicheskikh Nauk, 2023, Volume 78, Issue 1(469), DOI: https://doi.org/10.4213/rm10093 
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    1. IntroductionIn this survey we consider the groups $\operatorname{Aut}(X)$ and $\operatorname{Bim}(X)$ of all biregular and bimeromorphic self-maps, respectively, for a compact complex connected Kähler manifold $X$. If $X$ is projective, then $\operatorname{Bim}(X)=\operatorname{Bir}(X)$ is the group of all birational transformations of $X$ (see [74]). The manifolds we are going to deal with are of special type: $X$ has to be a $\mathbb{P}^1$-bundle over a non-uniruled compact complex connected manifold $Y$. In general, the groups $\operatorname{Bim}(X)$ may be very huge and non-algebraic (for example, the Cremona group $\operatorname{Cr}_n$ of birational transformation of the $n$-dimensional projective space). Thus one is tempted to study the properties of a group via its finite and/or abelian subgroups. Namely, we are interested in the following properties of groups. Definition 1.1. (a) A group $G$ is called bounded if the orders of its finite subgroups are bounded by a universal constant that depends only on $G$ (see [50], Definition 2.9). (b) A group $G$ is called Jordan if there is a positive integer $J$ such that every finite subgroup $B$ of $G$ contains an abelian subgroup $A$ that is normal in $B$ and such that the index satisfies $[B:A]\leqslant J$. The least such $J$ is called the Jordan constant of $G$ and denoted by $J_G$ (see [76], Question 6.1, [58], Definition 2.1, and [59]). (c) A Jordan group $G$ is called strongly Jordan [62], [5] if there is a positive integer $m$ such that every finite subgroup of $G$ is generated by at most $m$ elements. (d) A group $G$ is very Jordan [7] if there exist a commutative normal subgroup $G_0$ of $G$ and a bounded group $F$ that sit in a short exact sequence In what follows, by a Jordan property we mean any one of the properties described in Definition 1.1. The study of these properties has been inspired by the following fundamental results. Theorem 1.2 (Jordan [32] (1878); also see [77], Theorem 9.9). Let $\mathbb{C}$ be the field of complex numbers. Then $\operatorname{GL}(n,\mathbb{C})$ is strongly Jordan. Theorem 1.3 (Serre (2009), see [76], Theorem 5.3). $\operatorname{Cr}_2= \operatorname{Bir}(\mathbb{P}^2)$ is Jordan, $J_{\operatorname{Cr}_2}\leqslant 2^{10}\,3^4\,5^2\,7$. It was Popov who asked in [58] a question whether for an algebraic variety $X$ the groups $\operatorname{Aut}(X)$ and $\operatorname{Bir}(X)$ are Jordan. This question originated an intensive and fruitful activity. It was proved that there are vast classes of manifolds (varieties) with Jordan groups $\operatorname{Aut}(X)$, $\operatorname{Bim}(X)$, and $\operatorname{Bir}(X)$; see § 4. In particular, the Cremona group $\operatorname{Cr}_n=\operatorname{Bir}(\mathbb{P}^n)$ appeared to be Jordan for all $n$ ([62] and [10]; this is a positive answer to a question formulated by Serre). In § 4 we take a glimpse on the richness of the known facts about the Jordan properties of $\operatorname{Aut}(X)$, $\operatorname{Bim}(X)$, or $\operatorname{Bir}(X)$ for various types of varieties $X$. We do not pretend to give a complete picture. Our aim is to demonstrate that the ‘worst’ manifolds from this point of view are uniruled but not rationally connected ones. For example, the group $\operatorname{Bim}(X)$ is not Jordan if $X$ is bimeromorphic to a product of a complex torus of positive algebraic dimension and the projective space $\mathbb{P}^N$, $N>0$ (see [85] and [87]). In this survey we concentrate on the manifolds of this kind. Namely, our main object of consideration are $\mathbb{P}^1$-bundles over non-uniruled manifolds, that is, triples $(X,p,Y)$ such that
We say that such a triple $(X,p,Y)$ has an almost section $D$ if an irreducible analytic subset $D\subset X$, $\operatorname{codim}(D)=1$, meets a general fibre of $p$ at precisely one point (see Definition 6.5). We say that such a triple $(X,p,Y)$ (or $X$, or the morphism $p$) is scarce if $X$ does not admit three distinct almost sections $A_1$, $A_2$, and $A_3$ such that $A_1\cap A_2=A_1\cap A_3=A_2\cap A_3$ (see Definition 11.5). We say that a connected compact complex manifold $Y$ is poor (Definition 13.1) if it contains neither rational curves nor analytic subsets of codimension $1$. The facts that we know about Jordan properties of $\mathbb{P}^1$-bundles $(X,p,Y)$ over non-uniruled Kähler manifolds are summarized below. Summary1) $\operatorname{Aut}(X)$ is always Jordan ([34], for surfaces see also [86]) and even strongly Jordan (see Remark 4.1). 2) If morphism $p$ is scarce, then $\operatorname{Aut}(X)$ is very Jordan (Theorem 12.1 of this paper). 3) If $Y$ is a torus and $X$ is not a projectivization of a decomposable vector bundle of rank $2$ on $Y$, then the group $\operatorname{Aut}(X)$ is strongly Jordan [78]. 4) If $X$ and $Y$ are projective, and $X$ is not birational to $Y\times\mathbb{P}^1$, then $\operatorname{Bir}(X)$ is strongly Jordan [5]. 5) If $Y$ is a poor manifold (see Definition 13.1), then $\operatorname{Bim}(X)$ coincides with $\operatorname{Aut}(X)$ and is very Jordan [7]. 6) If $Y$ is a complex torus and there exists no almost section of $p$, then $\operatorname{Bim}(X)$ is Jordan [78]. In particular, if $X$ is not the projectivization of a rank $2$ vector bundle on $Y$, then the group $\operatorname{Bim}(X)$ is strongly Jordan. 7) If $Y$ is a complex torus of positive algebraic dimension and $X$ is bimeromorphic (birational, if $Y$ is projective) to a direct product $Y\times\mathbb{P}^1$, then the group $\operatorname{Bim}(X)$ (respectively, $\operatorname{Bir}(X)$) is not Jordan [85], [87]. 8) If $Y$ is a complex torus of positive algebraic dimension, $Y_a$ is its algebraic reduction, $\mathcal{L}$ is the lift to $Y$ of a holomorphic line bundle on $Y_a$, and $X$ is the projectivization of the rank $2$ vector bundle $\mathcal{L}\oplus\boldsymbol{1}$, then $\operatorname{Bim}(X)$ is not Jordan [87]. 9) Open question. Assume that $Y$ is a complex torus of positive algebraic dimension and $X$ has no representation as in previous item. Is $\operatorname{Bim}(X)$ Jordan? Our goal is to give a review of the methods used to prove these facts. The results unpublished previously are provided with full proofs. All manifolds are compact complex, and connected unless otherwise stated. All algebraic varieties are complex, projective, irreducible, and reduced. $\mathbb{P}^n$ and $\mathbb{C}^n$ are complex projective and affine spaces, respectively; $\mathbb{P}^n_k$ and $\mathbb{C}^n_k$ are projective and affine spaces, respectively, over an algebraically closed field $k$. The structure of the survey is as follows. In § 2 we provide facts and examples concerning bounded, Jordan, and very Jordan groups. In § 3 we enumerate the assumptions and notation and recall the notions related to manifolds and their maps. In § 4 we give examples of the known facts about the Jordan properties of $\operatorname{Aut}(X)$, $\operatorname{Bim}(X)$, and $\operatorname{Bir}(X)$ for various types of manifolds $X$. Our aim is to demonstrate a special role of $\mathbb{P}^1$-bundles over a non-uniruled base in this field. In § 5 we provide some generalities on maps of $\mathbb{P}^1$-bundles. In § 6 we deal with the group $\operatorname{Bim}(X)$ of a non-trivial rational bundle (in particular, a projective conic bundle). In Chap. 3 we deal with certain $\mathbb{P}^1$-bundles over complex tori. We present a unified approach to proving results of [85] and [87]. It is based on symplectic algebra, which offers highly useful tools for studying line bundles over tori, and is inspired by Mumford’s work [46]. In Chap. 4 we consider $\mathbb{P}^1$-bundles $(X,p,Y)$ with scarce sets of sections over a non-uniruled Kähler base. That chapter presents a generalization and a modification of [7]. First, in § 11, for a $\mathbb{P}^1$-bundle $(X,p,Y)$ we consider the group $\operatorname{Aut}(X)_p$ of those automorphisms of $X$ that leave every fibre of $p$ fixed. In three subsections we describe three different types of such automorphisms. In § 12, under the assumption that $Y$ is Kähler and not uniruled and $p$ is scarce, we prove that the neutral component $\operatorname{Aut}_0(X)$ of the complex Lie group $\operatorname{Aut}(X)$ is commutative, hence $\operatorname{Aut}(X)$ is very Jordan. In § 13 we prove that if $Y$ is poor, then $p$ is scarce and $\operatorname{Aut}(X)$ is very Jordan. Acknowledgements. We are deeply grateful to Frédéric Campana, Igor Dolgachev, Lei Ni, Constantin Shramov, and Vladimir L. Popov for helpful stimulating discussions. Our special thanks go to the referee, whose numerous valuable comments helped us to improve the exposition. Chapter 1. PreliminariesIn this chapter we provide some backgrounds: the properties of Jordan groups, the notation, assumptions, and definitions. 2. Jordan properties of groupsIn this section we recall the general facts about the Jordan properties of groups. The following properties are easy consequences of Definition 1.1. 1) Every finite group is bounded, Jordan, and very Jordan. 2) Every commutative group is Jordan and very Jordan. 3) Every finitely generated commutative group is bounded. Indeed, such a group is isomorphic to a finite direct sum with every summand isomorphic to $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$, where $n$ is a positive integer. 4) A subgroup of a Jordan group is Jordan. A subgroup of a very Jordan group is very Jordan. 5) “Bounded” implies “very Jordan”, “very Jordan” implies “Jordan”. 6) “Bounded” implies “strongly Jordan”. On the other hand, “very Jordan” does not imply “strongly Jordan”. For example, a direct sum of infinitely many copies of $\mathbb{Z}/2\mathbb{Z}$ is commutative but has finite subgroups with any given minimal number of generators. Example 2.1. The group $\operatorname{GL}(n,\mathbb{Z})$ is bounded. This is a consequence of the following theorem of Minkowski ([77], § 9.1). Theorem 2.2 (Minkowski, 1887). If an element $a\in\operatorname{GL}(n,\mathbb{Z})$ is periodic, and $a=\boldsymbol{1}\operatorname{mod}m$ for $m\geqslant 3$, then $a=1$. It follows that every finite subgroup $H\subset \operatorname{GL}(n,\mathbb{Z})$ embeds into $\operatorname{GL}(n,\mathbb{Z}/{3\mathbb{Z}})$ (there are much more precise bounds: see [75], Theorem 1.1). Since every finite subgroup of $\operatorname{GL}(n,\mathbb{Q})$ is conjugate to a subgroup of $\operatorname{GL}(n,\mathbb{Z})$ (see [75], Lecture 1), the group $\operatorname{GL}(n,\mathbb{Q})$ is also bounded. Example 2.3. The multiplicative group $\mathbb{C}^*$ of $\mathbb{C}$ is commutative and very Jordan but not bounded. The same is valid for the group of translations of a complex torus of positive dimension. Example 2.4. From Theorem 1.2 it follows that the group $\operatorname{GL}(n,k)$ is strongly Jordan for every field $k$ of characteristic zero. Moreover, every linear algebraic group over $k$ is strongly Jordan. On the other hand $\operatorname{GL}(n,k)$ is obviously not very Jordan if $n\geqslant 2$. The following precise values of Jordan constants for groups $\operatorname{GL}(n,\mathbb{C})$ were found by Collins. Theorem 2.5 ([18], Theorems A and B). For the Jordan constants of groups $\operatorname{GL}(n,\mathbb{C})$ the following relations hold: (i) $J_{\operatorname{GL}(n,\mathbb{C})}=(n+1)!$ if $n\geqslant 71$ or $n=63,65,67,69$; (ii) $J_{\operatorname{GL}(n,\mathbb{C})}=60^r\cdot r!$ if $20\leqslant n\leqslant 62$ or $n=64,66,68,70$ where $r=[n/2]$. The information on the values of Jordan constants for the groups $\operatorname{GL}(n,\mathbb{C})$, $n<20$, is given in extensive tables provided in the same paper. Example 2.6. We use below some analogues of the Heisenberg groups, which were used by Mumford [46]. Let Mumford’s theta group $\mathfrak{G}_{\mathbf{K}}$ for $\mathbf{K}$ is the group of matrices of type
$$
\begin{equation*}
\begin{pmatrix} 1 & \alpha &\gamma \\ 0 & 1 & \beta \\ 0 & 0 & 1 \end{pmatrix}
\end{equation*}
\notag
$$
where $\alpha\in\widehat{\mathbf{K}}$, $\gamma\in \mathbb{C}^*$, and $\beta\in\mathbf{K}$. The product of $\alpha\in\widehat{\mathbf{K}}$ and $\beta\in\mathbf{K}$ is defined by a certain natural non-degenerate alternating bilinear form $e_{\mathbf{K}}$ on $\mathbf{H}_{\mathbf{K}}=\mathbf{K}\times\widehat{\mathbf{K}}$ with values in $\mathbb{C}^*$ ([85], p. 302). This group can be included in a short exact sequence
$$
\begin{equation*}
1 \to \mathbb{C}^* \to \mathfrak{G}_{\mathbf{K}} \to \mathbf{H}_{\mathbf{K}} \to 1
\end{equation*}
\notag
$$
where the image of $\mathbb{C}^*$ is the center of $\mathfrak{G}_{\mathbf{K}}$. The properties of $\mathfrak{G}_{\mathbf{K}}$ ([85], p. 302) imply that it is a theta group attached to the non-degenerate symplectic pair $(\mathbf{H}_{\mathbf{K}}, e_{\mathbf{K}})$ in the sense of Chapter 3 below. By Theorem 7.17 below, $\mathfrak{G}_{\mathbf{K}}$ is Jordan and In particular, if $K=\mathbb{Z}/N\mathbb{Z}$, then $J_{\mathfrak{G}_{\mathbb{Z}/N\mathbb{Z}}}=N$.Example 2.7. The example of a non-Jordan group is given by $\operatorname{SL}(2,\overline{\mathbb{F}}_p)$, where $\overline{\mathbb{F}}_p$ is the algebraic closure of a prime finite field $\mathbb{F}_p$ with $p$ elements. Indeed, if $q=p^n\geqslant 4$, then $\operatorname{SL}(2,{\mathbb{F}_q})\subset\operatorname{SL}(2,\overline{\mathbb{F}}_p)$ (here $\mathbb{F}_q$ is the $q$-element field). The group $\operatorname{SL}(2,\mathbb{F}_q)$ is non-commutative, finite, and has order $(q^2-1)q$. Every normal subgroup $C\subsetneq \operatorname{SL}(2,\mathbb{F}_q)$ consists of one or two scalars, thus the indices are unbounded as $n$ tends to infinity. Hence $\operatorname{SL}(2,\overline{\mathbb{F}}_p)$ is not Jordan.Remark 2.8. An analogue of Jordan’s theorem holds for matrix groups over fields $k$ of prime characteristic $p$ if one considers only finite subgroups whose order is prime to $p$. On the other hand there are generalizations of Jordan’s theorem (Brauer–Feit [14], Larsen–Pink [38]) that deal with arbitrary finite subgroups and take the orders of their Sylow $p$-subgroups into account. These results have led to the following definition ([30], Definition 1.6; it will be used in Remark 4.3(iv) below). A group $G$ is called $p$-Jordan if there exist positive integers $J$ and $e$ such that every finite subgroup $B$ of $G$ contains an abelian $p'$-subgroup $A$ that is normal in $B$ and such that the index satisfies $[B:A]\leqslant |B_p|^e J$. Here $|B_p|$ is the order of a Sylow $p$-subgroup of $B$. Remark 2.9. Let $G$ be a group. Assume that it can be included in the following exact sequence of groups Then the following hold: We will need the following modification of Lemma 5.3 in [7]. Lemma 2.10. Consider a short exact sequence of connected complex Lie groups: Here $i$ is a closed holomorphic embedding and $j$ is surjective and holomorphic. Assume that $D$ is a complex torus and $A$ is isomorphic as a complex Lie group either to $(\mathbb{C}^+)^n$ or to $\mathbb{C}^*$. Then $B$ is commutative. Proof. The proof of this lemma coincides verbatim with the proof of Lemma 5.3 in [7] where the case $n=1$ is treated.
Step 1. First, let us prove that $A$ is a central subgroup in $B$. Take any element $b\in B$. Define a holomorphic map $\phi_b\colon A\to A$ by $\phi_b(a)= bab^{-1}\in A$ for an element $a\in A$. Since it depends holomorphically on $b$, we have a holomorphic map $\xi\colon B\to\operatorname{Aut}(A)$, $b\to\phi_b$. Since $A$ is commutative, for every $c\in A$ we have $\phi_{bc}=\phi_b$. Thus there is a well-defined map $\psi$ fitting in the following commutative diagram: The map $\psi=\xi\mathrel{\circ} j^{-1}$ is defined at every point of $D$. It is holomorphic (see, for example, [55], § 3).Since $D$ is a complex torus, and $\operatorname{Aut}(A)$ is either $\operatorname{GL}(n,\mathbb{C})$ (if $A=(\mathbb{C}^+)^n$) or consists of two elements, $\operatorname{id}$ and $a\mapsto a^{-1}$ (if $A=\mathbb{C}^*$), we have $\psi(D)=\{\operatorname{id}\}$. It follows that $A$ is a central subgroup of $B$. Step 2. Now let us show that $B$ is commutative. Consider a holomorphic map $\operatorname{com}\colon B\times B\to A$ defined by $\operatorname{com}(x,y)=xyx^{-1}y^{-1}$. Since $A$ is a central subgroup of $B$, similarly to Step 1 we obtain a holomorphic map $D\times D\to A$. It has to be constant since $D$ is a torus and $A$ is either $(\mathbb{C}^+)^n$ or $\mathbb{C}^*$. $\Box$ 3. Complex manifoldsThis section contains preliminaries, the notation, and the assumptions that will be used further on. By a (projective) variety we mean an algebraic variety that is a Zariski closed subset of a projective space $\mathbb{P}^n$. Let $U\subset\mathbb{C}^n$ be an open subset. An analytic subset of $U$ is a closed subset $X\subset U$ such that for any $x\in X$ there exist an open neighbourhood $x \in V\subset U$ and holomorphic functions $f_1,\dots,f_k\colon V\to\mathbb{C}$ such that $X\cap V=\{f_1=0,\dots,f_k=0\}$ ([31], Definition 1.1.23). A complex space consists of a Hausdorff topological space $X$ and a sheaf of rings $\mathcal{O}_X$ such that $(X,\mathcal{O}_X)$ is locally isomorphic to an analytic subset $Z\subset U\subset\mathbb{C}^n$ endowed with the sheaf $\mathcal{O}_U/\mathcal{I}$, where $\mathcal{I}$ is a sheaf of holomorphic functions such that $Z=Z(\mathcal{I})$ (see [31], Definition 6.2.8). By Chow’s theorem any closed analytic subset of complex projective space is a projective variety ([28], Chap. V, § D, Theorem 7; [28], Proposition 13). A complex manifold is a complex space which is locally modelled on $Z=U\subset\mathbb{C}^n$ and $\mathcal{I}=\{0\}$ ([31], Example 6.2.9). In particular, it is smooth. We will use the following notation and assumptions. Notation and assumptions
Definition 3.1. Following [25], we define a covering family of rational curves for a compact complex connected manifold $X$ as a pair of morphisms $p\colon Z\to T$ and $\psi\colon Z\to X$ of compact irreducible complex spaces if the following conditions are satisfied: Manifolds $X$ admitting a covering family with this property are called uniruled ([25], Definition 2.1 and Lemma 2.2). Remark 3.2. The Kodaira dimension $\kappa(X)$ equals $-\infty$ if $X$ is a uniruled compact complex manifold ([25], the remark on p. 691; [35], Corollary IV.1.11). In low dimensions the converse is true. Theorem 3.3 ([45] for projective manifolds, [29] for non-projective ones). Let $X$ be a compact Kähler manifold of dimension at most $3$. Then $X$ is uniruled if and only if $\kappa(X)=-\infty$. Remark 3.4 (Fujiki’s theorems). It was proved by Fujiki for a compact connected complex manifold $Y$ that The next statement (see [7], Proposition 1.4) is similar to Lemma 3.1 of [34]. Proposition 3.5. Let $X$ be a connected complex compact Kähler manifold, and let $F=\operatorname{Aut}(X)/\operatorname{Aut}_0(X)$. Then the group $F$ is bounded. Remark 3.6. Lemma 3.1 of Kim [34] states the following. Let $X$ be a normal compact Kähler variety. Then there exists a positive integer $l$, depending only on $X$, such that for any finite subgroup $G$ of $\operatorname{Aut}(X)$ acting biholomorphically and meromorphically on $X$ we have $[G : G \cap \operatorname{Aut}_0(X)]\leqslant l$. We cannot use this lemma straightforwardly, since it is not clear why every finite subgroup of $\operatorname{Aut}(X)/\operatorname{Aut}_0(X)$ should be isomorphic to $G/(G\cap \operatorname{Aut}_0(X))$ for some finite subgroup $G$ of $\operatorname{Aut}(X)$. Corollary 3.7. Let $X$ be a compact connected complex Kähler manifold that is either non-uniruled or with Kodaira dimension $\kappa(X)\geqslant 0$. Then $\operatorname{Aut}(X)$ is very Jordan. Proof. In view of Proposition 3.5 it is sufficient to prove that $\operatorname{Aut}_0(X)$ is commutative. But this assertion follows from [23], Proposition 5.10, if $X$ is non-uniruled and from [23], Corollary 5.11, if $\kappa(X)\geqslant 0$ (see Remark 3.4). $\Box$ In general, let $Z$ be a compact complex connected Kähler manifold. The analogue of the Chevalley decomposition for algebraic groups is valid for complex Lie group $\operatorname{Aut}_0(Z)$:
$$
\begin{equation}
1 \to L(Z) \to \operatorname{Aut}_0(Z) \to \operatorname{Tor} (Z) \to 1
\end{equation}
\tag{2}
$$
where $L(Z)$ is bimeromorphically isomorphic to a linear group, and $ \operatorname{Tor} (Z)$ is a complex torus ([23], Theorem 5.5, [40], Theorem 3.12, [16], Theorem 3.28). Remark 3.8. If $L(Z)$ in (2) is not trivial, then $Z$ contains a rational curve. Moreover, according to [23], Proposition 5.10, $Z$ is bimeromorphic to a fibre space whose general fibre is $\mathbb{P}^1$, that is, $X$ is uniruled. Chapter 2. Rational bundlesIn § 4, we want to persuade the reader that uniruled manifolds (in particular, $\mathbb{P}^1$-bundles) are of special interest from the point of view of their Jordan properties. To this end we give a very brief and certainly non-complete overview of known facts in this field. In § 5 we provide some general properties of maps of manifolds endowed with fibration over a non-uniruled base with general fibre $\mathbb{P}^1$. In § 6 we deal with projective non-trivial conic bundles. 4. Uniruled vs non-uniruled: Jordan properties of the groups $\operatorname{Aut}(X)$, $\operatorname{Bim}(X)$, and $\operatorname{Bir}(X)$In order to demonstrate the special role of uniruled manifolds from the point of view of Jordan properties, we present samples of results on the Jordan properties of $\operatorname{Aut}(X)$ and $\operatorname{Bim}(X)$ for various types of compact complex manifolds $X$. The group $\operatorname{Aut}(X)$ is known to be Jordan if
Remark 4.1. For the group $\operatorname{Aut}(X)$ ‘Jordan’ implies ‘strongly Jordan’ because for every compact complex manifold $X$ there is a constant $C=C(X)$ such that every finite subgroup $G\subset\operatorname{Aut}(X)$ can be generated by at most $C$ elements. One can find the proof of this fact in [48], Theorem 1.3. It is based on the same property for elementary abelian $p$-groups that was proved for a much wider class of topological spaces in [42], and on group-theoretic arguments (which, according to the author, had been explained to him by Khukhro and Jaikin). Thus this fact is also valid in a much more general situation. Moreover, the connected identity component $\operatorname{Aut}_0(X)$ of $\operatorname{Aut}(X)$ is Jordan for every compact complex space $X$ ([61], Theorems 5 and 7). An example of $X=E$, where $E$ is an elliptic curve, shows that $\operatorname{Aut}(X)$ can be Jordan but not bounded. A classification of complex compact surfaces with bounded automorphisms group was done in [69]. As follows from Corollary 3.7, the group $\operatorname{Aut}(X)$ is very Jordan for any compact connected complex Kähler non-uniruled manifold $X$. For uniruled manifolds the situation changes: if $X=E\times\mathbb{P}^1$, then $\operatorname{Aut}(X)\cong\operatorname{PSL}(2,\mathbb{C})\times\operatorname{Aut}(E)$ is neither bounded nor very Jordan. The groups $\operatorname{Bir}(X)$ and $\operatorname{Bim}(X)$ of birational and bimeromorphic transformations, respectively, are more complicated. Low-dimensional cases are well understood. Consider the following list:
Here are examples of results for low-dimensional cases. (a) If $X$ is a complex compact surface with non-negative Kodaira dimension, then $\operatorname{Bir}(X)$ is bounded unless it appears in the above list ([67], Theorem 1.1). (b) If $X$ is a projective surface, then $\operatorname{Bir}(X)$ is Jordan if $X$ is not birational to a product of an elliptic curve and $\mathbb{P}^1$ [58]. (The case of $X=\mathbb{P}^2$ was done previously by Serre [76].) (c) If $X$ is birational to a product of an elliptic curve and $\mathbb{P}^1$, then $\operatorname{Bir}(X)$ is not Jordan [85]. (d) If $X$ is a projective threefold, then $\operatorname{Bir}(X)$ is not Jordan if and only if $X$ is birational to a direct product $E\times\mathbb{P}^2$ or $S\times\mathbb{P}^1$, where $S$ is a surface from the above list [65]. (e) The group $\operatorname{Bim}(X)$ is Jordan for any non-uniruled compact complex connected Kähler manifold of dimension $3$ (see [70] and [26]). (f) If $X$ is a non-algebraic uniruled compact Kähler threefold with non-Jordan group $\operatorname{Bim}(X)$, then $X$ is bimeromorphic to $\mathbb{P}(\mathcal{E})$ for a holomorphic rank $2$ vector bundle $\mathcal{E}$ on a two-dimensional complex torus $S$ with $a(S)=1$. Moreover, if $a(X)=2$, then $X\approx S\times\mathbb{P}^1$ (see [68]). The following theorem for complex projective varieties was proved by Prokhorov and Shramov (for $\dim(X)>3$, under the assumption of the so-called BAB- conjecture, named after A. Borisov, L. Borisov, and Alexeev), and Birkar (who proved this conjecture): see [61], Theorem 1.8, and [10]. Theorem 4.2. Let $X$ be a projective variety of dimension $n$. Then the following hold: Here $q(X)=\dim_{\mathbb{C}}H^1(X,\mathcal{O}_X)$ is the irregularity of $X$. In particular, the Cremona group $\operatorname{Cr}_n$ of any rank $n$ is Jordan [63]. The exact value of $J_{\operatorname{Cr}_2}$ is $7200$ (Yasinsky [84]). The Jordan constant for $\operatorname{Bir}(X)$ for a rationally connected threefold $X$ can be found in [64]. Let us sketch the proof of assertions (i) and (ii) of Theorem 4.2. First, using the MMP (Minimal Model Program) the authors reduce the problem to considering the group $ \operatorname{PAut} (X_m)$, where $X_m$ is a special (relatively minimal) model of $X$ and $ \operatorname{PAut} (Z)$ stands for the group of birational self-maps of a variety $Z$ that are isomorphisms in codimension $1$. This means that $f\in \operatorname{PAut} (X_m)$ moves a divisor to a divisor and induces an automorphism $f_*=\psi(f)$ of the finitely generated abelian group $\operatorname{NS}^W(X_m)=\operatorname{Cl}(X_m)/\operatorname{Cl}^0(X_m)$, were $\operatorname{Cl}(X_m)$ stands for the group of Weil divisors on $X_m$ modulo linear equivalence and $\operatorname{Cl}^0(X_m)$ consists of those Weil divisors that are algebraically equivalent to zero. Thus there is a short exact sequence
$$
\begin{equation}
0 \to G_i \xrightarrow{i} G \xrightarrow{\psi}\operatorname{Aut}(\operatorname{NS}^W(X)),
\end{equation}
\tag{3}
$$
where $G_i=\ker(\psi)$ acts on each equivalence class in $\operatorname{Cl}(X_m)$. Since $\operatorname{NS}^W(X_m)$ is a finitely generated abelian group, $\operatorname{Aut}(\operatorname{NS}(X))$ is bounded. Take a very ample divisor $L$ and denote by $\operatorname{Cl}_L(X_m)$ the equivalence class containing $L$. It is an abelian variety of dimension $q(X_m)=q(X)$. Let $G_L$ be the kernel of the action of $G_i$ on $\operatorname{Cl}_L(X_m)$. Then there is a short exact sequence where $G_{ab}\subset\operatorname{Aut}(\operatorname{Cl}_L(X_m))$ is a subgroup of automorphisms (as a variety, but not as a group) of the abelian variety $\operatorname{Cl}_L(X_m)$. The group $\operatorname{Aut}(\operatorname{Cl}_L(X_m))$ is strongly Jordan. Let $V$ be a linear space of sections of $L$ and $\mathbb{P}(V)$ be its projectivization. Let $F_L$ be the subgroup of those linear transformations of the projective space $\mathbb{P}(V)$ that preserve $X_m\subset\mathbb{P}(V)$. Since $F_L$ is a linear group and $X$ (and also $X_m$) is non-uniruled, $F_L$ has to be finite (see Remark 3.8). Thus $G_L\subset F_L$ is finite.Therefore,
Remark 4.3. (i) One can ask similar questions about the group $\operatorname{Diff}(M)$ of all diffeomorphisms of a smooth manifold $M$. There was the conjecture of Ghys (1997): If $M$ is a compact smooth manifold, then $\operatorname{Diff}(M)$ is Jordan. It was answered in the negative by Csikós, Pyber, and Szabó [19], whose approach was based on an algebraic geometry construction from [85] (see also Chap. 3 below). In works of Winkelmann [83] and Popov [60] it was proved that there is a connected non-compact Riemann surface $M$ such that $\operatorname{Aut}(M)$ contains an isomorphic copy of every finitely presented (in particular, every finite) group $G$. In particular, $\operatorname{Diff}(M)$ is not Jordan. Zimmerman [88] proved that if $M$ is compact and $\dim(M)\leqslant 3$, then $\operatorname{Diff}(M)$ is Jordan. The Jordan properties of $\operatorname{Diff}(M)$ were deeply studied by Mundet i Riera [47], [49]–[53]. It was proved there, in particular, that $\operatorname{Diff}(M)$ is Jordan if $M$ is one of the following:
(ii) The question on the Jordan properties of algebraic groups over various fields was considered in [61], [44], and [80] (also see [5]). (iii) The Jordan properties of $\operatorname{Aut}(X)$ and $\operatorname{Bir}(X)$ for open subsets of certain projective $\mathbb{P}^1$-bundles were considered in [4] and [6]. (iv) In the case of algebraic varieties $X$ over algebraically closed fields of prime characteristic $p$ one should not expect the Jordan properties to hold (see Example 2.7). However, there are analogues of several important results over $\mathbb{C}$ that deal instead with $p$-Jordan properties (see Remark 2.8) of $\operatorname{Aut}(X)$ and $\operatorname{Bir}(X)$ [30], [17], [36]. On the other hand, it is known that the Cremona group of rank $2$ over a finite field is Jordan [71]. For compact complex manifolds, roughly speaking, from the point of view of Jordan properties the uniruled varieties are the worst and may be divided in several categories. First, manifolds $X$ that are rationally connected (or with $q(X)=0 $). For projective varieties, thanks to Theorem 4.2, $\operatorname{Bir}(X)$ is Jordan. Second, manifolds that are fibred over a non-uniruled base $Y$ with rationally connected fibres, with $q(X)\ne 0$, and that are not bimeromorphic (birational) to a direct product $Y\times\mathbb{P}^N$. In many special cases $\operatorname{Bim}(X)$ (or $\operatorname{Bir}(X)$) is Jordan. Moreover, $\operatorname{Aut}(X)$ often appears to be very Jordan. We discuss some of these special cases in Chap. 4. Third, $X$ is isomorphic (bimeromorphic) to the direct product $Y\times \mathbb{P}^N$. If $Y$ is a torus and $a(Y)>0$, then $\operatorname{Bir}(Y)$ is not Jordan. This case is the subject of Chapter 3. 5. Rational bundlesIn this section we provide some useful facts about $\mathbb{P}^1$-bundles and their morphisms. We start with a slightly more general construction. Definition 5.1. We say that a triple $(X,p,Y)$ is a rational bundle over $Y$ if If for an open subset $U\subset Y$ and every $y\in U$ the fibre $P_y\sim \mathbb{P}^1$, then, by a theorem of Fischer and Grauert [22], $p^{-1}(U)\subset X$ is a holomorphically locally trivial fibre bundle over $U$. If $U=X$, then the triple $(X,p,Y)$ is a $\mathbb{P}^1$-bundle over $Y$. If $(X,p,Y)$ is a rational bundle over a non-uniruled Kähler manifold $Y$, then $p\colon X\to Y$ is by definition a maximal rationally connected (MRC) fibration of $X$ (for the definition and discussion, see [15], Theorem 2.3 and Remark 2.8, and [35], § IV.5). Bimeromorphic self-maps preserve the MRC fibration. This is a well-known fact, but we have not found a suitable reference for the proof of this fact in the complex analytic case. We provide it here. In the case when the Kodaira dimension satisfies $\kappa(Y)\geqslant 0$, the desired result follows from Theorem 1.1.5 in [41]. For automorphisms the detailed exposition CAN be found in [1], § 2.4. Lemma 5.2. Let $X$, $Y$, and $Z$ be three complex compact connected manifolds, and let $p\colon X\to Y$ and $q\colon X\to Z$ be surjective holomorphic maps. Assume that Then there is a meromorphic map $\tau\colon Y\dashrightarrow Z$ such that $\tau\mathrel{\circ} p=q$, that is, the following diagram commutes: Proof. Let $\Phi\colon X\to Y\times Z$ be defined by $\Phi(x)=(p(x),q(x))$. The image $T=\Phi(X)$ is an irreducible compact analytic subspace of $Y\times Z$ (see, for example, [54], Chap. VII, Theorem 2). We denote by $\operatorname{pr}_Y$ and $\operatorname{pr}_Z$ the natural projections of $T$ onto the first and second factor, respectively. Both projections are evidently surjective. The set is an analytic subset of $T\subset Y\times Z$ ([73] and [21], Theorem 3.6 on p. 137). Its projections $T_Y=\operatorname{pr}_Y(T_1)\subset Y$ and $T_Z=\operatorname{pr}_Z(T_1)\subset Z$ onto the first and second factors are analytic subsets of $Y$ and $Z$, respectively ([73], and [54], Chap. VII, Theorem 2).
If $T_Y\ne Y$, then $V:=(Y\setminus T_Y)\cap U$ is an analytical Zariski open dense subset of $Y$. For each $y\in V$ we have $p^{-1}(y)\sim\mathbb{P}^1$ and $\dim(q(p^{-1}(y)))>0$. Thus the pair $p\colon X\to Y$, $q\colon X\to Z$ would provide a covering family for $Z$, which is impossible since $Z$ is not uniruled. Thus $T_Y=Y$. Take $u\in U$. Since $T_Y=Y$, there is $z\in Z$ such that
$$
\begin{equation*}
(u,z)\in T\quad \text{and} \quad \dim\Phi^{-1}(u,z)\geqslant 1.
\end{equation*}
\notag
$$
Moreover, Since $P_u\sim\mathbb{P}^1$ and $\dim(\Phi^{-1}(u,z))\geqslant 1$, we have $P_u=\Phi^{-1}(u,z)$. Hence $q|_{P_u}=z$ for every $u\in U$ and some $z\in Z$, and there is only one $z\in Z$ such that $(u,z)\in T$. Thus, It follows that $T$ is the graph of a meromorphic map, which we denote by $\tau$. $\Box$ Remark 5.3. The fact that $q$ contracts every fibre of $p$ over an analytic Zariski open non-empty subset of $Y$ was proved in [27], Proposition 6.2. Lemma 5.4. Let $(X,p_X,Y)$ and $(W,p_W,Y)$ be two rational bundles over a non- uniruled (compact connected) manifold $Y$. Let $f\colon X\to W$ be a surjective meromorphic map. Then there exists a meromorphic map $\tau(f)\colon Y\to Y$ that can be included in the following commutative diagram: In addition, if $f$ is holomorphic, so is $\tau(f)$. Proof. Let $a\colon \widetilde X \to X$ be a modification of $X$ such that the following diagram is commutative: where $b\colon \widetilde X \to W$ is a holomorphic map (it always exists: see [56], Theorem 1.9).
Consider the holomorphic maps
$$
\begin{equation*}
\tilde p_X:=p_X\mathrel{\circ} a\colon \widetilde X\to Y \quad\text{and}\quad \tilde f:=p_W\mathrel{\circ} b\colon \widetilde X\to Y.
\end{equation*}
\notag
$$
We apply Lemma 5.2 to $\widetilde X$, $Y=Z$, $\tilde p_X\colon \widetilde X\to Y$, and $\tilde f\colon=\widetilde X\to Y$ and obtain the needed map $\tau(f)\in\operatorname{Bim}(Y)$ which can be included in the following commutative diagram: If $f$ is holomorphic, then one can take $\widetilde X=X$ and $U=Y$ (in the notation of Lemma 5.2). Then $\tau(f)$ will be defined at every point of $Y$. $\Box$ Corollary 5.5. For a rational bundle $(X,p,Y)$ over a non-uniruled (complex connected compact) manifold $Y$ there are natural group homomorphisms such that for every $f\in\operatorname{Aut}(X)$ or $f\in\operatorname{Bim}(X)$, respectively. Remark 5.6. If $Y$ is Kähler non-uniruled, then the restriction group homomorphism In what follows we use heavily the following classical theorems. Theorem 5.7 (Remmert–Stein theorem; see, for example, [5], Chap. VII, Theorem of Remmert–Stein). Let $X$ be a complex space, $Y$ be an analytic subset of $X$, and $A$ be an analytic subset of $X\setminus Y$. Suppose that there is an integer $p>0$ such that $\dim(Y)\leqslant p-1$, while $\dim_a(A)\geqslant p$ for any $a\in A$ ($\dim(Y)\leqslant -1$ means that $Y= \varnothing$). Then the closure $\bar A$ of $A$ in $X$ is an analytic set in $X$. Theorem 5.8 (Riemann’s second removable singularity theorem; [21], Chap. 2, the appendix. Assume that $X$ is a complex manifold and $A\subset X$ is an analytic subset such that Then any holomorphic function $f\colon X\setminus A\to\mathbb{C}$ has a unique holomorphic extension $\tilde f\colon X\to\mathbb{C}$. Theorem 5.9 (Levi’s continuation theorem; [39], also see [54], Chap. VII, Theorem 4, or [21], § 4.8). Let $X$ be a normal complex space and $Y$ be an analytic subset of $X$ such that for any $a\in X$ we have $\dim_a(Y)\leqslant \dim_a(X)-2$. Then any meromorphic function on $X\setminus Y$ has an extension to a meromorphic function on $X$. Remark 5.10. It follows from the Riemann’s second theorem that a holomorphic map $f\colon X\setminus\Sigma\to Z$, where $X$ is a complex manifold, $\Sigma$ is an analytic subset of codimension at least $2$, and $Z\subset\mathbb{C}^N$ is an affine complex set, can be extended to a holomorphic map $\tilde f\colon X\to Z$. Indeed, let $z_1,\dots,z_N$ be coordinates in $\mathbb{C}^N$. Then $f$ consists of $N$ holomorphic functions $z_i(x)$, $i=1,\dots,N$, defined on $X\setminus\Sigma$. By Theorem 5.8 the functions $z_i$ can be extended to holomorphic functions $\tilde z_i$ defined on $X$. Since $Z$ is a closed subset of $\mathbb{C}^N$, we have $\tilde f(x)=(\tilde z_1(y),\dots,\tilde z_N(x))\in Z$ for every $x\in X$. This fact is a particular case of the extension theorem due to Andreotti and Stoll [2]. Recall that a subset $M\subset X$ of a complex space $X$ is thin if in a neighbourhood of every point $m\in M$ it is contained in an analytic subset of codimension $1$. Theorem 5.11 (Andreotti–Stoll theorem). Let $\tau\colon A\to Y$ be a holomorphic map of the open subset $A$ of a normal complex space $X$ into a Stein space $Y$. Let $M:=X\setminus A$ be a thin set. If $M$ has topological codimension at least $3$, then $\tau$ can be extended to a holomorphic map of $X$ into $Y$. We use this fact to prove the following lemma. Lemma 5.12. Let $(X,p,Y)$ and $(Z,q,Y)$ be two $\mathbb{P}^1$-bundles over a connected complex manifold $Y$. Let $\Sigma\subset Y$ be an analytic subset of codimension at least $2$, and let $U=Y\setminus\Sigma$, $V_X=p^{-1}(U)$, and $V_Z=q^{-1}(U)$. Let $f\colon X\to Y$ be a meromorphic map such that $q\mathrel{\circ} f=p$ and the induced map $f\colon V_X\to V_Z$ is an isomorphism. Then $f\colon X\to Z$ is a biholomorphic isomorphism. Proof. By construction, for every $u\in U$ the map $f$ induces an isomorphism $f|_{P_y}\colon P_y\to Q_y$, where $P_y=p^{-1}(y)$ and $Q_y=q^{-1}(y)$. Consider a point $s\in\Sigma$ and an open neighbourhood $U_s$ of it such that there are isomorphisms $\psi_X\colon p^{-1}(U_s)\to U_s\times\mathbb{P}^1$ and $\psi_Z\colon q^{-1}(U_s)\to U_s\times\mathbb{P}^1$ compatible with the projection maps $p$ and $q$, respectively. Then for every $y\in U_s\cap U$ we have an element of $\operatorname{PSL}(2,\mathbb{C})$ representing $f|_{P_y}\colon P_y\to Q_y$, which is an automorphism of $\mathbb{P}^1$. Thus we have a holomorphic map $U_s\cap U\to \operatorname{PSL}(2,\mathbb{C})$. Since the target space is an affine set, this map extends to a holomorphic map $U_s\to\operatorname{PSL}(2,\mathbb{C})$. Hence we have an extension of $f$ to an isomorphism $\tilde f_s\colon p^{-1}(U_s)\to q^{-1}(U_s)$, which coincides with $f$ in $V_X\cap p^{-1}(U_s)$, hence everywhere. $\Box$ Lemma 5.13. Let $(X,p,Y)$ and $(Z,q,Y)$ be two $\mathbb{P}^1$-bundles over a compact connected complex manifold $Y$ with $\dim(Y)=n$. Let $\Sigma\subset Y$ be an analytic subset of codimension at least $2$, and let $U=Y\setminus\Sigma$, $V_X=p^{-1}(U)$, and $V_Z=q^{-1}(U)$. Let $f\colon V_X\to V_Z$ be a meromorphic map such that $q\mathrel{\circ} f=p$. Then there exists a meromorphic map $\tilde f\colon X\to Y$ such that $\tilde f\big|_U=f$ and $q\mathrel{\circ}\tilde f=p$. For Kähler manifold $Y$ this lemma is a consequence of the following general theorem of Siu [81]. Theorem 5.14 (Siu’s extension theorem). Let $X$ be a complex manifold, $A$ be a subvariety of codimension at least $1$ in $X$, and $G$ be an open subset of $X$ which intersects every branch of $A$ of codimension $1$. If $M$ is a compact Kähler manifold, then every meromorphic map $f$ from $(X-A)\cup G$ to $M$ can be extended to a meromorphic map from $X$ to $M$. At this stage we do not require that $Y$ (and, a fortiori, $Z$) be Kähler, but we use the fact that $X$ and $Z$ are $\mathbb{P}^1$-bundles. Proof of Lemma 5.13. Consider the fibre product
$$
\begin{equation*}
W =X\times_{Y}Z =\{(x,z)\in X\times Z\mid p(x)=q(z)\} \subset X\times Z
\end{equation*}
\notag
$$
and its subsets
$$
\begin{equation*}
\begin{aligned} \, \Gamma_f & =\{(x,z)\in V_X\times V_Z\mid p(x)=q(z),\,z\in f(x)\}\subset W \\ \text{and} \qquad \widetilde\Sigma & =\{(x,z)\in X\times Z\mid p(x)=q(z)\in\Sigma\}\subset W. \end{aligned}
\end{equation*}
\notag
$$
By construction $\dim(\widetilde\Sigma)\leqslant n$ and $\dim(\Gamma_f)=\dim(X)=n+1$. Thus, according to the Remmert–Stein theorem (Theorem 5.7) the closure $\overline{\Gamma}_f$ of $\Gamma_f$ in $W$ is an analytic subset of $W$. Let $U_1\subset U$ be an open subset such that $f$ is defined at every point of $V_1:=p^{-1}(U_1)$. We have
It follows that $\overline{\Gamma}_f$ is a graph of a meromorphic map $\tilde f\colon X\to Z$ (see [3], p. 75). We also use the following lemma. Lemma 5.15. Assume that $Y$ is a compact connected complex manifold, $\Sigma\subset Y$ is an analytic subset of codimension at least $2$, and let $U=Y\setminus\Sigma$. Let $(\mathcal{L},\pi,Y)$ be a holomorphic line bundle over $Y$ such that $\mathcal{L}|_U$ is trivial. Then $\mathcal{L}$ is trivial. Proof. Indeed, $V:=\pi^{-1}(U)\sim U\times\mathbb{C}_z$, thus $z=F(v)$ is a holomorphic function on $V$. The set $\widetilde\Sigma:=\pi^{-1}(\Sigma)$ has codimension at least $2$ in $\mathcal{L}$. By Riemann’s second removable singularity theorem (Theorem 5.8) $F$ can be extended to a holomorphic function $\overline{F}$ on $\mathcal{L}$. Thus we have a holomorphic map $\Phi\colon \mathcal{L}\to Y\times\mathbb{C}_z$, $x\in\mathcal{L}\to(p(x),\overline{F}(x))$, which is an isomorphism outside $\widetilde\Sigma$. Let $S$ be the set of all points in $\mathcal{L}$ where the differential $d\Phi$ of $\Phi$ does not have the maximum rank. The sets $S$ and $\widetilde S=p(S)$ are analytic subsets of $\mathcal{L}$ and $Y$, respectively (see, for instance, [54], Chap. VII, Theorem 2, [57], Theorem 1.22, and [73]). Moreover, $\operatorname{codim}(\widetilde S)=1$ (see [72]). But $\widetilde S\subset \Sigma$, hence $\widetilde S=\varnothing$. It follows that $\Phi$ is an isomorphism. $\Box$ 6. Non-trivial rational bundlesIn this section we consider non-trivial $\mathbb{P}^1$-bundles over a non-uniruled base. It appears that the fact that $X\not\approx Y\times\mathbb{P}^1$ imposes significant restrictions on the structure of the groups $\operatorname{Aut}(X)$ and $\operatorname{Bim}(X)$. We start with the projective case. Definition 6.1. A regular surjective map $f\colon X\to Y$ of smooth irreducible projective complex varieties is a conic bundle over $Y$ if there is a Zariski open dense subset $U\subset Y$ such that the fibre $f^{-1}(y)\sim\mathbb{P}^1$ for all $y\in U$. The generic fibre of $f$ is an irreducible smooth projective curve $\mathcal{X}_f$ over the field $K:=\mathbb{C}(Y)$ such that its field of rational functions $K(\mathcal{X}_f)$ coincides with $\mathbb{C}(X)$. (The genus of $\mathcal{X}_f$ is $0$.) Theorem 6.2 [5]. Let $X$ be a conic bundle over a non-uniruled smooth irreducible projective variety $Y$ with $\dim(Y)\geqslant 2$. If $X$ is not birational to $Y\times\mathbb{P}^1$, then $\operatorname{Bir}(X)$ is strongly Jordan. Let us sketch the proof of Theorem 6.2. Let $f\colon X\to Y$ be a conic bundle, and assume that $Y$ is non-uniruled. According to Corollary 5.5, every $\phi\in\operatorname{Bir}(X)$ is fibrewise: there is a homomorphism $\tilde\tau\colon\operatorname{Bir}(X)\to\operatorname{Bir}(Y)$ such that $\tilde\tau(\phi)\mathrel{\circ} f=f\mathrel{\circ}\phi$: It follows that there is an exact sequence of groups
$$
\begin{equation}
0 \to \operatorname{Bir}_{\mathbb{C}(Y)}(\mathcal{X}_f) \to \operatorname{Bir}(X) \to \operatorname{Bir}(Y).
\end{equation}
\tag{6}
$$
Since $Y$ is non-uniruled, the group $\operatorname{Bir}(Y)$ is strongly Jordan thanks to Theorem 4.2 (also see [5], Corollary 3.8 and its proof). Let us compute $\operatorname{Bir}_{K}(\mathcal{X}_f)$. We have: 1) $\operatorname{Bir}_K(\mathcal{X}_f)=\operatorname{Aut}(\mathcal{X}_f)$ since $\dim(\mathcal{X}_f)=1$; 2) since $X\not\approx Y\times\mathbb{P}^1$, the genus $0$ curve $\mathcal{X}_f$ has no $K$-points, and therefore there exists a ternary quadratic form over $K$ such that
3) $K$ is a field of characteristic zero that contains all roots of unity. Now we consider a quadric, that is, a hypersurface in a projective space defined by one irreducible quadratic equation over $K$. It is anisotropic if it has no point defined over $K$. The following theorem was proved in [5]. Theorem 6.3 [5]. Suppose that $K$ is a field of characteristic zero that contains all roots of unity, let $d\geqslant 3$ be an odd integer, $V$ be a $d$-dimensional $K$-vector space, and $q\colon V\to K$ be a quadratic form such that $q(v)\ne 0$ for all non-zero $v\in V$. Consider the projective quadric $X_{q}\subset\mathbb{P}(V)$ defined by the equation $q=0$, which is a smooth projective irreducible $(d-2)$-dimensional variety over $K$. Let $\operatorname{Aut}(X_{q})$ be the group of biregular automorphisms of $X_{q}$. Let $G$ be a finite subgroup of $\operatorname{Aut}(X_{q})$. Then $G$ is commutative, all of its non-identity elements have order $2$, and the order of $G$ divides $2^{d-1}$. Thus, if $G$ is a non-trivial finite subgroup of $\operatorname{Aut}(\mathcal{X}_f)$, then either $G\cong\mathbb{Z}/2\mathbb{Z}$ or $G\cong(\mathbb{Z}/2\mathbb{Z})^2$. Now applying Remark 2.9(iv), from (6) we get that $\operatorname{Bir}(X)$ is Jordan. Remark 6.4. Actually, in Theorem 6.2 the variety $X$ is considered as a pointless ($X(K)=\varnothing$) rational curve defined over a field $K$, where $K$ contains all roots of unity. ‘Pointless surfaces’ were studied by Shramov and ologodsky in [79] and [80]. For complex compact manifolds the absence of points in the generic fibre has to be reformulated in terms of sections. Let $(X,p,Y)$ be a rational bundle over a compact complex connected non-uniruled manifold $Y$ (see Definition 5.1), that is,
According to Lemma 5.4, every map $f\in\operatorname{Bim}(X)$ maps the general fibre of $p$ to a fibre of $p$. Let
$$
\begin{equation*}
\operatorname{Aut}(X)_p=\{f\in\operatorname{Aut}(X)\mid \tau(f)=\operatorname{id}\}\ \ \text{and}\ \ \operatorname{Bim}(X)_p=\{f\in\operatorname{Bim}(X)\mid \tilde \tau(f)=\operatorname{id}\}
\end{equation*}
\notag
$$
be the kernels of $\tau$ and $\tilde\tau$, respectively. Then we have the following short exact sequences:
$$
\begin{equation}
0 \to\operatorname{Aut}(X)_p \to\operatorname{Aut}(X) \xrightarrow{\tau}\operatorname{Aut}(Y),
\end{equation}
\tag{7}
$$
$$
\begin{equation}
0 \to\operatorname{Bim}(X)_p \to\operatorname{Bim}(X) \xrightarrow{\tilde{\tau}}\operatorname{Bim}(Y).
\end{equation}
\tag{8}
$$
Definition 6.5. Let $(X,p,Y)$ be an equidimensional rational bundle over a compact complex connected non-uniruled manifold $Y$. We call an irreducible analytic subspace $D$ of $X$ an almost section if the intersection number $(D,F)$ of $D$ with a fibre $F=p^{-1}(y)$, $y\in Y$, is $1$. Remark 6.6. For $f\in \operatorname{Bim}(X)_p$ let $\widetilde S_f$ be the indeterminacy locus of $f$, which is an analytic subspace of $X$ of codimension at least $2$ [73], p. 369. Let $S_f={p(\widetilde S_f)}$, which is an analytic subset of $Y$ ([73], [54], Chap. VII, Theorem 2). Since the dimension of a fibre of $p$ is one, $Y\setminus S_f$ is an analytical Zariski open dense subset $U$ of $Y$. Hence the restriction $f|_{P_y}$ of $f$ to the fibre $P_y=p^{-1}(y)$ of $p$ over a general point $y\in Y$ belongs to $\operatorname{Aut}(P_y)$. Thus $f$ induces an automorphism of $V=p^{-1}(U)$ onto itself. Let $D$ be an almost section of $X$. (i) Let $a\colon \widetilde X \to X$ be such a modification of $X$ that the following diagram is commutative: where $b\colon \widetilde X \to X$ is a holomorphic map (it always exists: [56], Theorem 1.9). Then $f(D)=ba^{-1}(D)$ is an analytic subset ([73], [21], Theorem 3.6) which is a union of finite number of irreducible components $D_1,\dots,D_n$.(ii) We may assume (maybe after shrinking $U$) that $D$ meets every fibre $P_y$, $ y\in U$, at precisely one point. Then $f(D)$ also meets $P_y$, $y\in U$, at precisely one point. (iii) It follows from (ii) that precisely one irreducible component of $f(D)$, say, $D_1$, meets a fibre $P_y$, $y\in U$. The intersection $D_1\cap P_y$, $y\in U$, consists of a unique point. Thus $D_1$ is an almost section. It follows that the image of an almost section under $f\in\operatorname{Bim}(X)_p$ contains precisely one almost section. In particular, $f$ cannot contract an almost section. Similarly, if $\Phi\colon X\to Z$ is a bimeromorphic map of a $\mathbb{P}^1$-bundle $(X,p,Y)$ to a $\mathbb{P}^1$-bundle $(Z,q,Y)$ such that $q\mathrel{\circ}\Phi=p$, then the image of an almost section contains an almost section. The following results were proved by Prokhorov and Shramov in a more general setting. We formulate below an application of these results to the case of $\mathbb{P}^1$-bundles. Theorem 6.7. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle over a compact complex connected non-uniruled manifold $Y$. Let $P_y=p^{-1}(y)$ be a fibre of $p$ over a general point $y\in Y$. Then the following hold. To this we add the following lemma. Lemma 6.8. In the notation of Theorem 6.7, assume that there exists precisely one almost section on $X$. If $\operatorname{Bim}(Y)$ is Jordan, then so is $\operatorname{Bim}(X)$. Proof. Assume that $D$ is the only almost section. Let $f\in\operatorname{Bim}(X)_p$, $f\ne\operatorname{id}$. The set $f(D)$ contains an irreducible component $D_1$ that is an almost section (see Remark 6.6). Therefore, $D=D_1$ and $D$ is contained in the set $\operatorname{Fix}(f)$ of fixed points of $f$. Let $V\subset Y$ be an analytical Zariski open dense subset such that the restriction $f_v$ of $f$ to the fibre $P_v$ is a non-identical automorphism of $P_v$ for all $v\in V$. Since $f_v$ has at most two fixed points, we have the following alternative:
Thus, every element $f\in\operatorname{Bim}(X)_p$ different from $\operatorname{id}$ has infinite order. Therefore, $G\cap\operatorname{Bim}(X)_p=\{\operatorname{id}\}$ for every finite group $G\subset \operatorname{Bim}(X)$ and $\tilde\tau\colon G\to \operatorname{Bim}(Y)$ is a group embedding. Hence the Jordan index $J_{\operatorname{Bim}(X)}\leqslant J_{\operatorname{Bim}(Y)}$. $\Box$ The opposite case, when the $\mathbb{P}^1$-bundle has many almost sections, is when $X\cong Y\times\mathbb{P}^1$. It is considered in the next chapter. Chapter 3. $\mathbb{P}^{ {\rm \scriptsize 1}}$-bundles over complex toriIn this chapter we deal with $\mathbb{P}^1$-bundles of special type, namely, $(\bar{\mathcal{L}},p,T)$, where $\mathcal{L}$ is a holomorphic line bundle over a complex torus $T$ and $\bar{\mathcal{L}}=\mathbb{P}(\mathcal{L}\oplus\boldsymbol{1}_T)$. Most examples of compact complex connected manifolds with non-Jordan group $\operatorname{Bim}(X)$ (at least for dimensions greater than $3$) are $\mathbb{P}^1$-bundles of this type. Manifolds of this type were studied by one of these authors in [85] (the projective case) and [87] (the non-algebraic case). The goal of this chapter is to present a unified approach to both situations. It is based on a construction motivated by symplectic geometry and inspired by an algebraic approach to theta functions developed in [46]. The chapter starts with symplectic constructions, then theta groups follow, and then we arrive at the description of certain subgroups of $\operatorname{Bim}(\bar{\mathcal{L}})$. 7. Symplectic group theoryThis section contains elementary but useful facts about the Jordan properties of central extensions of commutative groups by $\mathbb{C}^*$. Consistent with the tradition, some groups are written in the multiplicative and some in the additive form. We hope that no confusion will arise. Definition 7.1. A symplectic pair is a pair $(A,e)$ that consists of a commutative group $A$ and an alternating bilinear pairing Here alternating means that Bilinearity means that These properties imply that for all $a,b\in A$ that is, As usual, $e$ gives rise to the group homomorphism
$$
\begin{equation}
\Psi_e\colon A\to \operatorname{Hom}(A,\mathbb{C}^*),\quad b\mapsto \{\Psi_e(b)\colon A\to \mathbb{C}^*,\,a \mapsto e(a,b)\}.
\end{equation}
\tag{9}
$$
A subgroup $B$ of $A$ is called isotropic with respect to $e$ if We define the kernel of $e$ by this is a subgroup of $A$, which is isotropic with respect to $e$.We say that $e$ is non-degenerate if $\ker(e)=\{0\}$, that is, is an injective homomorphism. If $e$ is non-degenerate, then we call $(A,e)$ a non- degenerate symplectic pair.Example 7.2. Let $d$ be a positive integer, let $\mathbf{S}_d=(\frac{1}{d}\mathbb{Z}/\mathbb{Z})^2\cong (\mathbb{Z}/d\mathbb{Z})^2$, and let Then $(\mathbf{S}_d,\mathbf{e}_d)$ is a non-degenerate symplectic pair. Remark 7.3. Let $(A_1e_1)$ and $(A_2,e_2)$ be non-degenerate symplectic pairs. Consider the bilinear alternating form Then $(A_1\oplus A_2, e_1 e_2)$ is a non-degenerate symplectic pair. Remark 7.4. If $(A,e)$ is a symplectic pair and $B$ is a subgroup of $A$, then $(B,e|_B)$ is also a symplectic pair. Here $e|_B$ is the restriction of $e$ to $B \times B$. Remark 7.5. (i) Each symplectic pair $(A,e)$ gives rise to a non-degenerate symplectic pair $(\bar{A},\bar{e})$, where (ii) Clearly, a subgroup $B$ of $A$ is isotropic with respect to $e$ if and only if its image $\bar{B}$ in $\bar{A}$ is isotropic with respect to $\bar{e}$. In particular, $B$ is isotropic if and only if $B+\ker(e)$ is isotropic. (iii) Let $B$ be a subgroup of $A$. One can restate the property of $B$ to be isotropic with respect to $e$ as follows. The composition of $\Psi_e\colon A\to\operatorname{Hom}(A,\mathbb{C}^*)$ with the restriction map $\operatorname{Hom}(A,\mathbb{C}^*)\to\operatorname{Hom}(B,\mathbb{C}^*)$ is a group homomorphism
$$
\begin{equation}
A \xrightarrow{\Psi_e}\operatorname{Hom}(A,\mathbb{C}^*) \to \operatorname{Hom}(B,\mathbb{C}^*).
\end{equation}
\tag{11}
$$
Clearly, the kernel $B^{\bot}$ of this homomorphism (which is the orthogonal complement of $B$ in $A$ with respect to $e$) contains $B$ if and only if $B$ is isotropic. (iv) Suppose that $B$ coincides with $B^{\bot}$. This means that if $a\in A\setminus B$, then $e(B,a)\ne\{1\}$. In other words, $B$ is a maximal isotropic subgroup of $A$ with respect to $e$. Conversely, suppose that $B$ is a maximal isotropic subgroup of $A$ with respect to $e$. Since $B$ is isotropic, it follows that
$$
\begin{equation*}
B\subset B^{\bot}\subset A,\quad e(B^{\bot},B)=\{1\}.
\end{equation*}
\notag
$$
If $B^{\bot} \ne B$, then there is $a\in B^{\bot} \setminus B$ such that $e(a,B)=\{1\}$. This implies that the subgroup $B_1$ of $A$ generated by $B$ and $a$ is isotropic, which contradicts the maximality of $B$. It follows that $B=B^{\bot}$ if and only if $B$ is a maximal isotropic subgroup of $A$. Remark 7.6. Suppose that $A$ is finite. Then the finite groups $A$ and $\operatorname{Hom}(A,\mathbb{C}^*)$ are (non-canonically) isomorphic; in particular, they have the same order. It follows that in the case of finite $A$ the pairing $e$ is non-degenerate if and only if $\Psi_e$ is a group isomorphism. Lemma 7.7 (useful lemma). Let $(A,e)$ be a symplectic pair such that $A/\ker(e)$ is a finite group. If $B$ is a maximal isotropic subgroup of $A$, then the index $[A:B]$ equals $\sqrt{\#(A/\ker(e))}$. In particular, if $e$ is non-degenerate, then Proof. In light of Remark 7.5, $B$ contains $\ker(e)$, and therefore it suffices to prove the desired result for non-degenerate $(\bar{A},\bar{e})$ (instead of $(A,e)$). In other words, without loss of generality we may assume that $\ker(e)=\{0\}$, that is, $A=\bar{A}$ is finite and $e=\bar{e}$ is non-degenerate.
Since $\mathbb{C}^*$ is a divisible group, every group homomorphism $B\to\mathbb{C}^*$ extends to a group homomorphism $A\to\mathbb{C}^*$. This means that the restriction map
$$
\begin{equation*}
\operatorname{Hom}(A,\mathbb{C}^*)\to\operatorname{Hom}(B,\mathbb{C}^*)
\end{equation*}
\notag
$$
is surjective. Since $A$ is finite, the non-degeneracy of $e$ means (in light of Remark 7.6) that $\operatorname{Hom}(A,\mathbb{C}^*)=\Psi_e(A)$. On the other hand the maximality of $B$ means that the kernel of the surjective composition
$$
\begin{equation*}
A \overset{\Psi_e}{\cong} \operatorname{Hom}(A, \mathbb{C}^*) \twoheadrightarrow \operatorname{Hom}(B,\mathbb{C}^*)
\end{equation*}
\notag
$$
coincides with $B$ (see Remark 7.5), and therefore there is an injective group homomorphism which is also surjective and therefore an isomorphism. This implies that which finishes the proof once we take into account that $\#(A/B)=\#(A)/\#(B)$. $\Box$ Remark 7.8. Suppose that $\ker(e)$ is either finite or divisible. Then every finite subgroup $\bar{B}$ of $\bar{A}$ is the image of a finite subgroup $B\subset A$ under the projection $A \twoheadrightarrow \bar{A}$. Indeed, if $\ker(e)$ is finite, then one can take as $B$ the preimage of $\bar{B}$ in $A$. If $\ker(e)$ is divisible, then it is a direct summand of $A$, that is, $A$ splits into a direct sum $A=\ker(e)\oplus A'$ and the map $A\to\bar{A}$ induces an isomorphism $A'\cong\bar{A}$. Now one can take as $B$ the (isomorphic) preimage of $\bar{B}$ in $A'$. Definition 7.9. A symplectic pair $(A,e)$ is called almost isotropic if there exists a positive integer $D$ that enjoys the following property. Each finite subgroup $\mathcal{B}$ of $A$ contains an isotropic (with respect to $e$) subgroup $\mathcal{A}$ such that the index satisfies $[\mathcal{B}:\mathcal{A}]\leqslant D$. The smallest $D$ with this property is called the isotropy defect of $(A,e)$ and denoted by $D_{A,e}$. Remark 7.11. Suppose that $\ker(e)$ is either finite or divisible. (i) It follows from Remarks 7.8 and 7.5 that $(A,e)$ is almost isotropic if and only if $(\bar{A},\bar{e})$ is almost isotropic. In addition, if this is the case, then Indeed, let $\mathcal{A}$ be a finite subgroup of $A$ and $B$ be an isotropic subgroup of largest possible order in $\mathcal{A}$. In particular, $B$ is a maximal isotropic subgroup of $\mathcal{A}$. Since $B_1=B+(\mathcal{A}\cap \ker(e))$ is an isotropic subgroup of $\mathcal{A}$ that contains $B$, the maximality of $B$ implies that $B_1=B$, that is, $B\supset\mathcal{A}\cap\ker(e)$. This implies that the index $(\mathcal{A}:B)$ equals the index $[\bar{\mathcal{A}}:\bar{B}]$ where the subgroups $\bar{\mathcal{A}}$ and $\bar{B}$ are the images in $\bar{A}$ of $\mathcal{A}$ and $B$, respectively. Taking into account that $\bar{B}$ is an isotropic (with respect to $\bar{e}$) subgroup of the finite group $\bar{\mathcal{A}}\subset \bar{A}$, we conclude thatConversely, suppose that $\bar{B}$ is an isotropic (with respect to $\bar{e}$) subgroup of maximum order of a finite group $\bar{\mathcal{A}}\subset\bar{A}$. As above, this implies that $\bar{B}$ is a maximal isotropic subgroup of $\bar{\mathcal{A}}$. By Remark 7.8, $A$ contains a finite subgroup $\mathcal{A}$ whose image in $\bar{A}$ coincides with $\bar{\mathcal{A}}$. Let $B$ the preimage of $\bar{B}$ in $\mathcal{A}$. Then $B$ is isotropic with respect to $e$ and the index $[\mathcal{A}:B]$ coincides with the index $[\bar{\mathcal{A}}:\overline{B}]$. This implies that which finishes the proof.(ii) Assume additionally that $\bar{A}$ is finite. Applying Lemma 7.7 to subgroups of $\bar{A}$ and using (12) we conclude that Definition 7.12. A theta group attached to a symplectic pair $(A,e)$ is a group $G$ that sits in a short exact sequence that enjoys the following properties. The image of $\mathbb{C}^*$ is a central subgroup of $G$, and the alternating commutator pairing
$$
\begin{equation*}
A\times A\to\mathbb{C}^*,\qquad j(g_1), j(g_2)\mapsto i^{-1}(g_1 g_2 g_1^{-1}g_2^{-1})\in \mathbb{C}^*\quad \forall g_1,g_2\in G,
\end{equation*}
\notag
$$
which is attached to the exact sequence (14) coincides with $e$. Remark 7.13. Every central extension $G$ of a commutative group $A$ by $\mathbb{C}^*$ gives rise to the symplectic pair $(A,e)$, where $e(a_1,a_2)\in\mathbb{C}^*$ is the commutator of the preimages of $a_1$ and $a_2$ in $G$ (for all $a_1,a_2\in A$). This makes $G$ a theta group attached to $(A,e)$. Remark 7.14. (i) Clearly, an element $g$ of the theta group $G$ lies in the centre of $G$ if and only if Since $j(G)=A$, the element $g$ is central if and only if $j(g)\in\ker(e)$. This implies that the centre of $G$ coincides with $j^{-1}(\ker(e))$. (ii) Clearly, a subgroup $H$ of $G$ is commutative if and only if its image $j(H)\subset A$ is an isotropic subgroup of $A$ with respect to $e$. Remark 7.15. Let $G$ be a theta group that sits in the short exact sequence (14). If $B$ is a subgroup of $A$, then the preimage $j^{-1}(B)$ is obviously a theta group attached to the symplectic pair $(B,e|_B)$. Lemma 7.16. Let $B$ be a finite subgroup of $A$. Then there exists a finite subgroup $\widetilde{B}$ of the theta group $G$ such that $j(\widetilde{B})=B$. Proof. In what follows we identify $\mathbb{C}^*$ with its image in $G$ and view it as a central subgroup of $G$. Let $d$ be the exponent of $B$.
Consider the finite multiplicative subgroups $\mu_d$ and $\mu_{d^2}$ of all $d$th roots of unity and $d^2$th roots of unity, respectively, in $\mathbb{C}^*$. We have
$$
\begin{equation*}
\mu_d \subset \mu_{d^2}\subset \mathbb{C}^* \subset G;
\end{equation*}
\notag
$$
in addition,
For every $b \in B$ choose its lifting $\tilde{b}\in G$ such that
$$
\begin{equation}
\tilde{b}^d=1,\quad \tilde{b}^{-1}=\widetilde{b^{-1}}\quad \forall b\in B;
\end{equation}
\tag{16}
$$
this is possible since $\mathbb{C}^*$ is a central divisible subgroup of $\mathbb{C}^*$. Indeed, let $\tilde{b}_1\in G$ be any lifting of $b$ to $G$, that is, $j(\tilde{b}_1)=b$. Then We choose any and put $\tilde{b}=z^{-1}\tilde{b}_1\in G$. We have
$$
\begin{equation*}
j(\tilde{b}) =j(z^{-1})+j(\tilde{b}_1) =0+b =b\quad\text{and}\quad \tilde{b}^d =(z^{-1})^d\,\tilde{b}_1^d =z_1^{-1} z_1 =1.
\end{equation*}
\notag
$$
Set
$$
\begin{equation*}
\widetilde{B} :=\{\gamma \tilde{b}\mid\gamma\in\mu_{d^2},\,b\in B\} \subset G.
\end{equation*}
\notag
$$
Clearly, $\widetilde{B}$ is finite, $j(\widetilde{B})=B$, and
$$
\begin{equation*}
1 \in \mu_{d^2}\subset \widetilde{B} =\widetilde{B}^{-1} :=\{u^{-1}\mid u\in\widetilde{B}\}
\end{equation*}
\notag
$$
(the latter equality follows from the invariance of the central subgroup $\mu_{d^2}$ and the subset $\{\tilde{b}\mid b\in B\}$ under the map $u\mapsto u^{-1}$).
Thus, in order to prove that $\widetilde{B}$ is a subgroup of $G$, it suffices to check that $\widetilde{B}$ is closed under multiplication in $G$. Let $b_1,b_2\in B$ and $b_3=b_1+b_2\in B$. We need to compare $\tilde{b}_1\tilde{b}_2$ and $\tilde{b}_3$ in $G$. Clearly, there is $\gamma\in \mathbb{C}^*$ such that Notice that
$$
\begin{equation*}
\tilde{b}_1^d=\tilde{b}_2^d=\tilde{b}_3^d=1 \in \mathbb{C}^* \subset G.
\end{equation*}
\notag
$$
On the other hand, in light of (15),
$$
\begin{equation*}
\gamma_0 :=\tilde{b}_1\tilde{b}_2 \tilde{b}_1^{-1}\tilde{b}_2^{-1} =e(b_1,b_2)\in \mu_d\subset \mathbb{C}^* \subset G.
\end{equation*}
\notag
$$
It follows that the images of $\tilde{b}_1$ and $\tilde{b}_2$ in the quotient $G/\mu_d$ commute, and therefore the image of $\tilde{b}_1\tilde{b}_2$ in $G/\mu_d$ has an order which divides $d$. This means that and therefore It follows that
$$
\begin{equation*}
1 =\tilde{b}_3^{d^2} =\bigl(\gamma \cdot \tilde{b}_1\tilde{b}_2\bigr)^{d^2} =\gamma^{d^2}\bigl(\tilde{b}_1\tilde{b}_2\bigr)^{d^2} =\gamma^{d^2}\cdot 1 =\gamma^{d^2}.
\end{equation*}
\notag
$$
This implies that $\gamma^{d^2}=1$, that is, $\gamma \in \mu_{d^2}$, and therefore Theorem 7.17. Let $(A,e)$ be a symplectic pair. Assume that $\bar{A}=A/\ker(e)$ is finite. Also assume that either $\ker(e)$ is divisible or $A$ is finite. Let $G$ be a theta group attached to $(A,e)$. Then $G$ is a Jordan group and its Jordan index is $\sqrt{\#(\bar{A})}$. Proof. Assume that $G$ sits in a short exact sequence (14). We can view $\mathbb{C}^*$ as a central subgroup of $G$. Let $\widetilde{\mathcal{A}}$ be a finite subgroup of $G$ and $\widetilde{B}$ be a commutative subgroup of maximum order in $\widetilde{\mathcal{A}}$. Then $\widetilde{B}$ contains the intersection $\widetilde{\mathcal{A}}\cap \mathbb{C}^*$, and therefore the index $[\widetilde{\mathcal{A}}\,{:}\,\widetilde{B}]$ coincides with the index $[j(\widetilde{\mathcal{A}}\,)\,{:}\,j(\widetilde{B})]$. The commutativity of $\widetilde{B}$ means that $j(\widetilde{B})$ is an isotropic subgroup in $j(\widetilde{\mathcal{A}}\,)$. This implies that
Conversely, let $\mathcal{A}$ be a finite subgroup of $A$ and $B$ be an isotropic subgroup of maximal order in $\mathcal{A}$. By Lemma 7.16 there is a finite subgroup $\widetilde{\mathcal{A}}$ of $G$ such that Let $\widetilde{B}$ be the preimage of $B$ in $\widetilde{\mathcal{A}}$. Then
$$
\begin{equation*}
j(\widetilde{B})=B\quad\text{and}\quad [\mathcal{A}:B]=[\widetilde{\mathcal{A}}:\widetilde{B}].
\end{equation*}
\notag
$$
By Remark 7.14(ii), $\widetilde{B}$ is commutative because its image $B$ is isotropic. The equality of indices implies that which, combined with the previous reverse inequality, implies that $J_G=D_{A,e}$. Now the explicit formula for $J_G$ follows from Remark 7.11. 8. Symplectic linear algebraIn this section we construct theta groups that arise from (not necessarily non- degenerate) alternating bilinear forms on integral lattices. Definition 8.1. (i) An admissible triple is a triple $(V,E,\Pi)$ that consists of a non- zero real vector space $V$ of finite positive even dimension $2g$, an alternating $\mathbb{R}$- bilinear form on $V$, and a discrete lattice $\Pi$ of rank $2g$ in $V$ such that $E(\Pi,\Pi)\subset\mathbb{Z}$. Put By definition $\Pi_E^{\bot}$ is a closed real Lie subgroup of $V$ that contains $\Pi$ as a discrete subgroup. (ii) A symplectic pair attached to the admissible triple $(V,E,\Pi)$ is a pair $(K_{E,\Pi},e_{E})$ where $K_{E,\Pi}:=\Pi_E^{\bot}/\Pi$ and the bilinear pairing $e_E$ is defined as follows: Definition 8.2. Recall that a subgroup $C$ of a commutative group $D$ is called saturated if it enjoys the following equivalent properties: Our goal is to find the isotropy index of $(K_{E,\Pi},e_E)$. In order to do this, consider the kernel of $E$, that is, the subset Clearly, $\ker(E)$ is a real even-dimensional (recall that $E$ is alternating) vector subspace of $V$ containing $\Pi_E^{\bot}$. Put Clearly, $\Pi_0$ is a saturated subgroup of $\Pi$. The integrality property of $E$ implies that the natural homomorphism of real vector spaces
$$
\begin{equation*}
\Pi_0\otimes\mathbb{R}\to\ker(E),\quad l_0\otimes\lambda\mapsto\lambda\cdot l_0\quad \forall\ l_0\in \Pi_0,\ \lambda\in \mathbb{R}
\end{equation*}
\notag
$$
is an isomorphism. In particular, the following conditions are equivalent: (a) $E$ is non-degenerate, that is, $\ker(E)=\{0\}$; (b) $\Pi_0=\{0\}$. Consider several cases. Case I. If $E\equiv 0$, then
$$
\begin{equation*}
\Pi_E^{\bot}=V,\quad K_{E,\Pi}=\Pi_E^{\bot}/\Pi=V/\Pi,\quad e_E \equiv 1,
\end{equation*}
\notag
$$
$\ker(e_E)=K_{E,\Pi}$ is divisible, and $K_{E,\Pi}/\ker(e)=\{0\}$ is finite. By Remark 7.11 the isotropy defect $D_{K_{E,\Pi}, e_E}$ equals $1$. Case II. Suppose that $E$ is a non-degenerate form. Let $\{s_1,\dots,s_{2g}\}$ be any basis of the $\mathbb{Z}$-module $\Pi$. Clearly, it is also a basis of the $\mathbb{R}$-vector space $V$. Let
$$
\begin{equation*}
\widetilde{E}=\big(E(s_j,s_k)\big) \in\operatorname{Mat}_{2g}(\mathbb{Z})
\end{equation*}
\notag
$$
be the $2g\times 2g$ skew-symmetric matrix of $E$ with integer entries with respect to this basis. Let $\det(\widetilde{E})$ and $\operatorname{Pf}(\widetilde{E})$ be the determinant of $\widetilde{E}$ and the Pfaffian of $\widetilde{E}$, respectively. Then
$$
\begin{equation*}
\det(\widetilde{E})\in \mathbb{Z},\quad \operatorname{Pf}(\widetilde{E})\in \mathbb{Z},\quad\text{and}\quad 0\ne\det(\widetilde{E})=\operatorname{Pf}(\widetilde{E})^2.
\end{equation*}
\notag
$$
In particular, $\det(\widetilde{E})$ is a positive integer. Clearly, $\det(\widetilde{E})$ does not depend on the choice of a basis of $\Pi$, and therefore $|\!\operatorname{Pf}(\widetilde{E})|$ does not depend on this choice either. That is why we denote $\det(\widetilde{E})$ by $\det(E,\Pi)$ and $|\!\operatorname{Pf}(\widetilde{E})|$ by $|\!\operatorname{Pf}(E,\Pi)|$. We claim that $\Pi_E^{\bot}/\Pi$ is finite, the form
$$
\begin{equation*}
e_E\colon \Pi_E^{\bot}/\Pi \times \Pi_E^{\bot}/\Pi \to \mathbb{C}^*
\end{equation*}
\notag
$$
is non-degenerate, and its isotropy defect is $|\!\operatorname{Pf}(E,\Pi)|$. Indeed, there exists a basis $\{f_1,h_1,\dots,f_g,h_g\}$ of $\Pi$ such that
$$
\begin{equation*}
E(f_j, h_k)=-E(h_k,f_j)=0 \quad\forall\,j\ne k \ (1\leqslant j,k\leqslant g)
\end{equation*}
\notag
$$
([37], Chap. XV, Exercise 17 on p. 598 of the English ed. or Chap. XIV, Exercise 4 on p. 426 of the Russian ed.). Put
$$
\begin{equation*}
d_j=E(f_j,h_j)\in\mathbb{Z} \quad\forall j=1,\dots,g.
\end{equation*}
\notag
$$
The non-degeneracy of $E$ means that $d_j\ne 0$ for all $j$. Replacing $h_j$ by $-h_j$ if necessary, we may and will assume that $d_j>0$ for all $j$. If $\widetilde{E}$ is the matrix of $E$ with respect to this basis, then the Pfaffian $\operatorname{Pf}(\widetilde{E})$ of $\widetilde{E}$ is $\pm\prod_{j=1}^g d_j$, and therefore We claim that
$$
\begin{equation}
\Pi_E^{\bot} =\bigoplus_{j=1}^g \frac{1}{d_j}(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j).
\end{equation}
\tag{17}
$$
Indeed, a vector
$$
\begin{equation*}
v=\biggl(\,\sum_{j=1}^g \lambda_jf_j\biggr) +\biggl(\,\sum_{j=1}^g \mu_j h_j\biggr) \quad\text{with all } \lambda_j,\mu_j\in\mathbb{R}
\end{equation*}
\notag
$$
lies in $\Pi_E^{\bot}$ if and only if
$$
\begin{equation*}
\mathbb{Z}\ni E(f_j,v)=d_j \mu_j,\quad \mathbb{Z}\ni(h_j,v)=-d_j \lambda_j\quad \forall j,
\end{equation*}
\notag
$$
which is obviously equivalent to (17). It follows from (17) that
$$
\begin{equation}
\Pi_E^{\bot}/\Pi =\bigoplus_{j=1}^g \frac{1}{d_j}(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j) /(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j).
\end{equation}
\tag{18}
$$
Clearly, different summands of $\Pi_E^{\bot}/L$ are mutually orthogonal with respect to $e_E$ while the restriction of $e_E$ to each
$$
\begin{equation*}
\frac{1}{d_j}(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j) /(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j)
\end{equation*}
\notag
$$
is isomorphic to $(\mathbf{S}_{d_j},\mathbf{e}_{d_j})$. In particular, this restriction is a non-degenerate symplectic pair. This implies that the direct sum $(\Pi_E^{\bot}/\Pi,e_E)$ is also a non-degenerate symplectic pair. On the other hand it is clear that
$$
\begin{equation*}
\Pi_E^{\bot}/\Pi \cong\bigoplus_{j=1}^g \biggl(\frac{1}{d_j}\mathbb{Z}/\mathbb{Z}\biggr)^2.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\#(\Pi_E^{\bot}/\Pi)\cong \prod_{j=1}^g d_j^2\quad\text{and}\quad \sqrt{\#(\Pi_E^{\bot}/\Pi)} =\prod_{j=1}^g d_j=|\!\operatorname{Pf}(E,\Pi)|.
\end{equation*}
\notag
$$
This implies that $(K_{E,\Pi},e_E)$ is almost isotropic and its isotropy defect is $|\!\operatorname{Pf}(E,\Pi)|$. Case IIbis. We keep the notation and assumptions of Case II. Consider the form $nE$, where $n$ is a positive integer. Then
$$
\begin{equation*}
\begin{gathered} \, \Pi_{nE}^{\bot} =\frac{1}{n} \Pi_E^{\bot} =\bigoplus_{j=1}^g \frac{1}{nd_j}(\mathbb{Z}\cdot f_j\oplus \mathbb{Z}\cdot h_j), \\ \Pi_{nE}^{\bot}/\Pi\cong \bigoplus_{j=1}^g \biggl(\frac{1}{nd_j}\mathbb{Z}/\mathbb{Z}\biggr)^2, \\ \#(\Pi_{nE}^{\bot}/\Pi)=\prod_{j=1}^g (nd_j)^2\quad\text{and}\quad \sqrt{\#(\Pi_E^{\bot}/\Pi)} =n^g\prod_{j=1}^g d_j=n^g\cdot|\!\operatorname{Pf}(E,\Pi)|. \end{gathered}
\end{equation*}
\notag
$$
Hence the corresponding isotropy index is
$$
\begin{equation*}
D_{K_{nE,\Pi}, e_{nE}} =n^g\cdot|\!\operatorname{Pf}(E,\Pi)|
\end{equation*}
\notag
$$
for all positive integers $n$. Case III. Now consider the case of degenerate non-zero $E$, that is, the case when Clearly, $\Pi_0$ is a free abelian group of some positive even rank $2g_0<2g$. Since $\Pi_0$ is a saturated subgroup of $\Pi$, it is a direct summand of $\Pi$, that is, there is a (non-zero saturated) subgroup $\Pi_1$ of $\Pi$ that is a free abelian group of rank $2g-2g_0$ and such that In other words there is a basis $\{u_1,\dots,u_{2g_0}; v_1,\dots,v_{2g-2g_0}\}$ of the $\mathbb{Z}$-module $\Pi$ such that $\{u_1,\dots,u_{2g_0}\}$ is a basis of $\Pi_0$ and $\{v_1,\dots,v_{2g-2g_0}\}$ is a basis of $\Pi_1$. Consider the real vector subspaces
$$
\begin{equation*}
V_0:=\sum_{j=1}^{2g_0}\mathbb{R} u_j\subset V\quad\text{and}\quad V_1:=\sum_{k=1}^{2g_1}\mathbb{R} v_k\subset V.
\end{equation*}
\notag
$$
Clearly,
$$
\begin{equation*}
V=V_0\oplus V_1,\qquad \Pi_0=V_0\cap \Pi,\quad\text{and}\quad \Pi_1=V_1\cap \Pi;
\end{equation*}
\notag
$$
in addition, $V_0=\ker(E)$, the subspaces $V_0$ and $V_1$ are mutually orthogonal with respect to $E$ and the restriction of $E$ to $V_1$,
$$
\begin{equation*}
E_1\colon V_1\times V_1\to\mathbb{R},\qquad u,v \mapsto E(u,v),
\end{equation*}
\notag
$$
is a non-degenerate alternating bilinear form. It is also clear that
$$
\begin{equation*}
E_1(\Pi_1,\Pi_1) =E(\Pi_1,\Pi_1) \subset E(\Pi,\Pi) \subset \mathbb{Z}.
\end{equation*}
\notag
$$
On the other hand the restriction of $E$ to $V_0$, which we denote by $E_0$, is identically equal to $0$. This implies that (as a symplectic pair)
$$
\begin{equation*}
(K_{E,\Pi},e_E)=(K_{E_0,\Pi_0}, e_{E_0})\oplus (K_{E_1,\Pi_1}, e_{E_1}).
\end{equation*}
\notag
$$
By Case I as applied to $(V_0,E_0,\Pi_0)$, the group $K_{E_0,\Pi_0}=V_0/\Pi_0$ is divisible as a quotient of a complex vector space, and $e_{E_0}\equiv 1$. By Case II as applied to $(V_1,E_1,\Pi_1)$, the group $K_{E_1,\Pi_1}$ is finite of order $|\!\operatorname{Pf}(E,\Pi)|^2$ and the pairing
$$
\begin{equation*}
e_{E_1}\colon K_{E_1,\Pi_1} \times K_{E_1,\Pi_1} \to \mathbb{C}^*
\end{equation*}
\notag
$$
is non-degenerate. Hence $\ker(e_E)=K_{E_0,\Pi_0}$, and therefore $\ker(e_E)$ is divisible and is a finite group. This implies that $(K_{E,\Pi},e_E)$ is almost isotropic and its isotropy defect is
$$
\begin{equation}
D_{K_{E,\Pi},e_E} =\sqrt{\#(K_{E,\Pi}/\ker(e_E))} =\sqrt{\#(K_{E_1,\Pi_1})} =|\!\operatorname{Pf}(E_1,\Pi_1)|
\end{equation}
\tag{19}
$$
by Theorem 7.17. Case IIIbis. We keep the notation and assumptions of Case III. Let be an alternating bilinear form that enjoys the following properties: If $n$ is an integer, then we write $\mathbf{M}(n)$ for the alternating bilinear form $nE+M$ on $V$. Clearly,
$$
\begin{equation*}
\mathbf{M}(n)(\Pi,\Pi) \subset n E(\Pi,\Pi)+M(\Pi,\Pi) \subset n \mathbb{Z}+\mathbb{Z} =\mathbb{Z}.
\end{equation*}
\notag
$$
Lemma 8.3. There exists a degree $g-g_0$ polynomial $\mathcal{P}(t)\in\mathbb{Z}[t]$ with integer coefficients and leading coefficient $|\!\operatorname{Pf}(E_1,\Pi_1)|$ that enjoys the following property: For all but finitely many positive integers $n$ the symplectic pair $(K_{\mathbf{M}(n),\Pi},e_{\mathbf{M}(n)})$ is almost isotropic with isotropy defect Proof. Indeed, let $M_1\colon V_1\times V_1 \to \mathbb{R}$ be the restriction of $M$ to $V_1$. Let $\widetilde{E}_1$ and $\widetilde{M}_1$ be the matrices of $E_1$ and $M_1$ with respect to the basis $\{f_1,\dots,f_{2g-2g_0}\}$ of $\Pi_1$. The non-degeneracy of $E_1$ implies that $\det(\widetilde{E}_1)\ne0$, and therefore the determinant
$$
\begin{equation*}
\det(n \widetilde{E}_1+\widetilde{M}_1) =\det(\widetilde{E}_1) \det(n\operatorname{I}_{2g-2g_0}+\widetilde{E}_1^{-1}\widetilde{M}_1)
\end{equation*}
\notag
$$
does not vanish for all but finitely many integers $n$. (Here and in what follows $\operatorname{I}_{2g-2g_0}$ is the identity square matrix of size $2g-2g_0$.) Taking into account that $n\widetilde{E}_1+\widetilde{M}_1$ is the matrix of the restriction of $nE+M=\mathbf{M}(n)$, for all but finitely many integers $n$ we obtain In what follows we assume that $n$ is an arbitrary integer that enjoys property (21) (this assumption excludes only finitely many integers $n$). Now we can apply the results of Case III to $\mathbf{M}(n)=nE+M$ (instead of $E$) and get that $(K_{\mathbf{M}(n),\Pi},e_{\mathbf{M}(n)})$ is almost isotropic and its isotropy defect is
$$
\begin{equation*}
\begin{aligned} \, |\!\operatorname{Pf}(nE_1+M_1,\Pi_1)| & =\sqrt{\det(nE_1+M_1)} =\sqrt{\det(\widetilde{E}_1) \det\bigl(n\operatorname{I}_{2g-2g_0}+\widetilde{E}_1^{-1} \widetilde{M}_1\bigr)}\\ & =|\!\operatorname{Pf}(E_1,\Pi_1)|\, \sqrt{\det\bigl(n\operatorname{I}_{2g-2g_0}+\widetilde{E}_1^{-1}\widetilde{M}_1\bigr)}. \end{aligned}
\end{equation*}
\notag
$$
Clearly, there is a polynomial $\mathcal{Q}(t) \in \mathbb{Z}[t]$ with integer coefficients such that for all $n$ under consideration
$$
\begin{equation*}
\mathcal{Q}(n)=\operatorname{Pf}(n\widetilde{E}_1+\widetilde{M}_1).
\end{equation*}
\notag
$$
This implies that
$$
\begin{equation*}
\mathcal{Q}(n)^2 =\det(n\widetilde{E}_1+\widetilde{M}_1) =\det(\widetilde{E}_1) \det\bigl(n\operatorname{I}_{2g-2g_0}+\widetilde{E}_1^{-1}\widetilde{M}_1\bigr).
\end{equation*}
\notag
$$
It is also clear that there exists a monic polynomial $\mathcal{R}(t)\in\mathbb{Q}[t]$ of degree $2g-2g_0$ with rational coefficients such that for all $n$ under consideration
$$
\begin{equation*}
\mathcal{R}(n)=\det\bigl(n \operatorname{I}_{2g-2g_0}+\widetilde{E}_1^{-1}\widetilde{M}_1\bigr).
\end{equation*}
\notag
$$
This implies that Since $\mathcal{R}(t)$ is monic of degree $2g-2g_0$, we have and the leading coefficient of $\mathcal{Q}(t)$ is $\pm |\!\operatorname{Pf}(E_1,\Pi_1)|$.
Let $\mathcal{P}(t)$ be the polynomial with positive leading coefficient that coincides either with $\mathcal{Q}(t)$ or with $-\mathcal{Q}(t)$. Then $\mathcal{P}(t)$ is a degree $g-g_0$ polynomial with integer coefficients and leading coefficient $|\!\operatorname{Pf}(E_1,\Pi_1)|$ such that Since the leading coefficient of $\mathcal{P}(t)$ is positive, we see that $\mathcal{P}(n)$ is positive for all but finitely many positive integers $n$. Hence for all such $n$. $\Box$Theorem 8.4. Let $g$ be a positive integer, $V$ be a $2g$-dimensional real vector space, and $(V,E,\Pi)$ and $(V,M,\Pi)$ be admissible triples such that If $n$ is an integer, then we write $\mathbf{M}(n)$ for the alternating bilinear form $nE+M$ on $V$. Let $\mathcal{G}$ be a group that enjoys the following properties: there are infinitely many positive integers $n$ such that $\mathcal{G}$ contains a subgroup $G_n$ that is a theta group attached to $(K_{\mathbf{M}(n),\Pi},e_{\mathbf{M}(n)})$. Then $\mathcal{G}$ is not Jordan. Proof. It suffices to check that the Jordan index of $G_n$ tends to infinity as $n$ tends to infinity. But this assertion follows from the results of Cases II, III, and IIIbis of this section in combination with Theorem 7.17. $\Box$ 9. Line bundles over tori and theta groupsIn this section we use results from the previous two sections in order to compute the Jordan index of certain automorphism groups of holomorphic line bundles on complex tori. Let $V$ be a complex vector space of finite positive dimension $g$, $\Pi$ be a discrete lattice of rank $2g$ in $V$, and be an Hermitian form on $V$ such that its imaginary part
$$
\begin{equation*}
E\colon V\times V\to\mathbb{R},\quad (v_1,v_2)\mapsto\operatorname{Im}(H(v_1,v_2)),
\end{equation*}
\notag
$$
satisfies One can view $V$ as the $2g$-dimensional real vector space. Then $E$ becomes an alternating $\mathbb{R}$-bilinear form on $V$ such that
$$
\begin{equation*}
E(\mathbf{i}v_1,\mathbf{i}v_2)=E(v_1,v_2)\quad \forall v_1,v_2\in V.
\end{equation*}
\notag
$$
In addition,
$$
\begin{equation*}
H(v_1,v_2)=E(\mathbf{i}v_1,v_2)+\mathbf{i}E(v_1,v_2)\quad \forall v_1,v_2\in V
\end{equation*}
\notag
$$
(see [12], Lemma 2.1.7). This implies that $H$ and $E$ have the same kernels, that is,
$$
\begin{equation*}
\ker(H) :=\{w\in V\mid H(w,V)=0\} =\{w\in V\mid E(w,V)=0\} =:\ker(E).
\end{equation*}
\notag
$$
Definition 9.1 (see [11] and [33]). A pair $(H, \alpha)$ is called Appell–Humbert data (A.-H. data) on $(V,\Pi)$ if $H$, $E$, and $\Pi$ are as above and $\alpha$ is a map (a ‘semicharacter’) Notice that a classical theorem of Appell and Humbert ([33], Theorem 1.5, cite11, Theorem 21.1) classifies holomorphic line bundles on the complex torus $V/\Pi$ in terms of A.-H. data. The construction in § 8 gives us a symplectic pair $(K_{E,\Pi},e_E)$. The aim of this section is to construct a theta group $\mathfrak{G}(H,\alpha)$ attached to this pair that corresponds to any A.-H. data $(H,\alpha)$. We define $\widetilde{\mathfrak{G}}(H,V)$ to be a certain group of biholomorphic automorphisms of $\mathcal{L}(H,\alpha)$. Here $\mathcal{L}(H,\alpha)$ is the total body of the holomorphic line bundle $\mathcal{L}(H,\alpha)$ over $V/\Pi$ that corresponds to the A.-H. data $(H,\alpha)$. First we start with a certain theta group $\widetilde{\mathfrak{G}}(H,V)$ attached to the symplectic pair $(V,\tilde{e}_E)$ where
$$
\begin{equation*}
\tilde{e}_E\colon V \times V \to \mathbb{C}^*,\quad (v_1,v_2)\mapsto\exp(2\pi \mathbf{i}E(v_2,v_1)).
\end{equation*}
\notag
$$
We define $\widetilde{\mathfrak{G}}(H,V)$ as a certain group of holomorphic automorphisms of where $\mathbb{L}$ is a one-dimensional $\mathbb{C}$-vector space. Namely, $\widetilde{\mathfrak{G}}(H,V)$ consists of the automorphisms $\mathcal{B}_{H,u,\lambda}$, indexed by $u\in V$ and $\lambda\in\mathbb{C}^*$, that are defined as follows:
$$
\begin{equation*}
\mathcal{B}_{H,u,\lambda}\colon (v,c)\mapsto\bigl(v+u, \lambda \exp(\pi H(v,u)c)\bigr)\quad \forall\,v\in V,\ c\in\mathbb{L}.
\end{equation*}
\notag
$$
One can easily check (see [87], § 2.1) that
$$
\begin{equation}
\mathcal{B}_{H,u_1,\lambda_1} \mathrel{\circ}\mathcal{B}_{H,u_2,\lambda_2} =\mathcal{B}_{H,u_1+u_2,\lambda_1\lambda_2\mu} \quad\text{where}\quad \mu=\exp(\pi H(u_2,u_1))
\end{equation}
\tag{23}
$$
and the inverse is
$$
\begin{equation}
\mathcal{B}_{H,u,\lambda}^{-1}=\mathcal{B}_{H,-u,\nu/\lambda} \quad\text{where}\quad \nu=\exp(-\pi H(u,u)).
\end{equation}
\tag{24}
$$
This implies that $\widetilde{\mathfrak{G}}(H,V)$ is indeed a subgroup of the group of biholomorphic automorphisms of $V_{\mathbb{L}}$. (Our $\mathfrak{G}(H,\alpha)$ will be defined as a subquotient of $\widetilde{\mathfrak{G}}(H,V)$.) Notice that for all $\lambda \in \mathbb{C}^*$ the automorphism $\mathcal{B}_{H,0,\lambda}$ sends every $(u,c)$ to $(u,\lambda c)$. This implies that the map
$$
\begin{equation*}
\operatorname{mult}\colon\mathbb{C}^*\to\widetilde{\mathfrak{G}}(H,V), \quad\lambda\mapsto\mathcal{B}_{H,0,\lambda},
\end{equation*}
\notag
$$
is an injective group homomorphism whose image lies in the centre of $\widetilde{\mathfrak{G}}(H,V)$. This allows us to include $\widetilde{\mathfrak{G}}(H,V)$ in a short exact sequence of groups
$$
\begin{equation*}
1 \to \mathbb{C}^* \xrightarrow{\operatorname{mult}}\widetilde{\mathfrak{G}}(H,V) \xrightarrow{\,\tilde{j}\,} V \to 0
\end{equation*}
\notag
$$
where $\tilde{j}$ sends $\mathcal{B}_{H,u,\lambda}$ to $u$. It follows from (23) and (24) (also see [87], § 2.1) that
$$
\begin{equation}
\begin{aligned} \, & \mathcal{B}_{H,u_1,\lambda_1} \mathrel{\circ}\mathcal{B}_{H,u_2,\lambda_2} \mathrel{\circ}\mathcal{B}_{H,u_1,\lambda_1}^{-1} \mathrel{\circ}\mathcal{B}_{H,u_2,\lambda_2}^{-1}\nonumber\\ &\qquad =\operatorname{mult}(\exp(2\pi \mathbf{i}E(u_2,u_1))) =\operatorname{mult}(\tilde{e}_E(u_1,u_2)). \end{aligned}
\end{equation}
\tag{25}
$$
This implies that $\widetilde{\mathfrak{G}}(H,V)$ is a theta group attached to the symplectic pair $(V,\tilde{e}_E)$. Consider the following subgroups of $\widetilde{\mathfrak{G}}(H,V)$:
$$
\begin{equation}
\widetilde{\mathfrak{G}}(H,\Pi) =\tilde{j}^{-1}(\Pi) =\{\mathcal{B}_{H,u,\lambda}\mid\lambda\in\mathbb{C}^*, u\in\Pi\};
\end{equation}
\tag{26}
$$
$$
\begin{equation}
\widetilde{\mathfrak{G}}(H,\Pi_E^{\bot}) =\tilde{j}^{-1}(\Pi_E^{\bot}) =\{\mathcal{B}_{H,u,\lambda}\mid\lambda\in\mathbb{C}^*, u\in\Pi_E^{\bot}\}.
\end{equation}
\tag{27}
$$
By Remark 7.15, $\widetilde{\mathfrak{G}}(H,\Pi)$ and $\widetilde{\mathfrak{G}}(H, \Pi_E^{\bot})$ are theta groups attached to the symplectic pairs $(\Pi,\tilde{e}|_{\Pi})$ and $(\Pi_E^{\bot},\tilde{e}|_{\Pi_E^{\bot}})$, respectively. Since $\Pi\subset\Pi_E^{\bot}$, the group $\widetilde{\mathfrak{G}}(H,\Pi)$ is a subgroup of $\widetilde{\mathfrak{G}}(H,\Pi_E^{\bot})$. It follows from (25) that $\widetilde{\mathfrak{G}}(H,\Pi)$ is actually a central subgroup of $\widetilde{\mathfrak{G}}(H, \Pi_E^{\bot})$, because We define $\mathfrak{G}(H,\alpha)$ as the quotient of $\widetilde{\mathfrak{G}}(H, \Pi_E^{\bot})$ by a certain central subgroup that depends on the ‘semicharacter’ $\alpha$. In order to define this subgroup, let us consider the discrete free action of the group $\Pi$ on $V_{\mathbb{L}}$ by holomorphic automorphisms defined as follows. An element $l$ of $\Pi$ acts as
$$
\begin{equation}
\begin{aligned} \, \mathcal{A}_{H,\alpha,l}\colon V_{\mathbb{L}}\to V_{\mathbb{L}},\quad (v,c)\mapsto\biggl(v+l,c\alpha(l) \exp\biggl(\pi H(v,l)+\frac12\pi H(l,l)\biggr)\biggr) \\ \forall\ v \in V,\ c\in\mathbb{L},\notag \end{aligned}
\end{equation}
\tag{28}
$$
that is,
$$
\begin{equation}
\mathcal{A}_{H,\alpha,l} = \operatorname {mult} (\alpha(l))\mathcal{B}_{H,l,1} \in\widetilde{\mathfrak{G}}(H,\Pi).
\end{equation}
\tag{29}
$$
Direct calculations based on (22) show that
$$
\begin{equation*}
\mathcal{A}_{H,\alpha,l_1}\mathcal{A}_{H,\alpha,l_2} =\mathcal{A}_{H,\alpha,l_1+l_2}\quad \forall l_1,l_2\in\Pi,
\end{equation*}
\notag
$$
that is,
$$
\begin{equation*}
\mathbf{A}^{\Pi}\colon\Pi\to\widetilde{\mathfrak{G}}(H,\Pi),\quad l\mapsto\mathcal{A}_{H,\alpha,l},
\end{equation*}
\notag
$$
is an injective group homomorphism, whose image we denote by
$$
\begin{equation*}
\widetilde{\Pi} =\widetilde{\Pi}(H,\alpha) :=\mathbf{A}^{\Pi}(\Pi) \subset \widetilde{\mathfrak{G}}(H,\Pi) \subset \widetilde{\mathfrak{G}}(H,\Pi_E^{\bot}).
\end{equation*}
\notag
$$
Notice that $\widetilde{\Pi}$ meets $\operatorname{mult}(\mathbb{C}^*)$ precisely at the identity element of $\widetilde{\mathfrak{G}}(H,\Pi_E^{\bot})$. Notice that the quotient $V_{\mathbb{L}}/\widetilde{\Pi}(H,\alpha)$ is precisely the total body $\mathcal{L}(H,\alpha)$ of the holomorphic vector bundle $\mathcal{L}(H,\alpha)$ over $V/\Pi$ attached to the A.-H. data $(H,\alpha)$ where the structure map
$$
\begin{equation*}
p\colon \mathcal{L}(H,\alpha) =V_{\mathbb{L}}/\widetilde{\Pi}(H,\alpha) \to V/\Pi
\end{equation*}
\notag
$$
is induced by the projection map ([12], Chap. 2, § 2.2, p. 30). Put
$$
\begin{equation}
\mathfrak{G}(H,\alpha) :=\widetilde{\mathfrak{G}}(H,\Pi_E^{\bot})/\widetilde{\Pi}(H,\alpha).
\end{equation}
\tag{30}
$$
The faithful action of $\widetilde{\mathfrak{G}}(H, \Pi_E^{\bot})$ on $V_{\mathbb{L}}$ induces a faithful action of $\mathfrak{G}(H,\alpha)$ on $\mathcal{L}(H,\alpha)$. Under this action each coset
$$
\begin{equation*}
\mathcal{B}_{H,u,\lambda}\widetilde{\Pi} \in\widetilde{\mathfrak{G}}(H, \Pi_E^{\bot})/\widetilde{\Pi}(H,\alpha) =\mathfrak{G}(H,\alpha)
\end{equation*}
\notag
$$
maps the fibre of $p$ over $v+\Pi \in V/\Pi$ to the fibre over $(v+u)\Pi \in V/\Pi$ $\mathbb{C}$-linearly and isomorphically for any pair
$$
\begin{equation*}
u+\Pi \in \Pi_E^{\bot}/\Pi\subset V/\Pi, \qquad v+\Pi\in V/\Pi, \quad\text{and}\quad \lambda\in\mathbb{C}^*.
\end{equation*}
\notag
$$
In particular, $\operatorname{mult}(\lambda)\widetilde{\Pi}$ acts as the automorphism $[\lambda]$ that leaves invariant each fibre of $p\colon \mathcal{L}(H,\alpha)\to V/\Pi$ and acts on this fibre (which is a one-dimensional $\mathbb{C}$-vector space) as multiplication by $\lambda$ (for all $\lambda\in\mathbb{C}^*$). Clearly, each $[\lambda]$ lies in the centre of $\mathfrak{G}(H,\alpha)$. Lemma 9.2. The group $\mathfrak{G}(H,\alpha)$ is a theta group attached to the symplectic pair $(K_{E,\Pi}, e_E)$. Proof. Clearly,
$$
\begin{equation*}
\operatorname{[mult]}\colon \mathbb{C}^* \to \mathfrak{G}(H,\alpha),\quad \lambda \mapsto [\lambda],
\end{equation*}
\notag
$$
is an injective group homomorphism, whose image $\operatorname{[mult]}(\mathbb{C}^*)$ is a central subgroup of $\mathfrak{G}(H,\alpha)$. On the other hand $\tilde{j}$ induces the surjective group homomorphism Clearly, the kernel of $j$ consists of all $\mathcal{B}_{H,0,\lambda}\widetilde{\Pi}=\operatorname{[mult]}(\lambda)$, that is, it coincides with $\operatorname{[mult]}(\mathbb{C}^*)$. Hence $\mathfrak{G}(H,\alpha)$ sits in the short exact sequence
It follows from (25) that $\mathfrak{G}(H,\alpha)$ is a theta group attached to the symplectic pair $(K_{E,\Pi},e_E)$. Remark 9.3. It is well known ([12], Lemma 2.2.1) that if $(H_1,\alpha_1)$ and $(H_2,\alpha_2)$ are A.-H. data on $(V,\Pi)$, then the pair $(H_1+H_2,\alpha_1\alpha_2)$ also is A.-H. data on $(V,\Pi)$ and the holomorphic vector bundles $\mathcal{L}(H_1+H_2,\alpha_1\alpha_2)$ and $\mathcal{L}(H_1,\alpha_1)\otimes\mathcal{L}(H_2,\alpha_2)$ are canonically isomorphic. 10. $\mathbb{P}^1$-bundles bimeromorphic to the direct productIn this section we prove the non-Jordanness of the groups of bimeromorphic self-maps of certain $\mathbb{P}^1$-bundles over complex tori of positive algebraic dimension. Let $V$ be a complex vector space of finite positive dimension $g$, $\Pi$ be a discrete lattice of rank $2g$ in $V$, and $T=V/\Pi$ be the corresponding complex torus. Recall that $\boldsymbol{1}_T$ denotes the trivial holomorphic line bundle $T \times \mathbb{C}$ over $T$. If $x$ is point of $T$, then we write $\mathcal{L}_x$ for the fibre of the holomorphic vector bundle $\mathcal{L}$ over $T$, which is a one-dimensional complex vector space. We write $\bar{\mathcal{L}}$ for the projectivization $\mathbb{P}(\mathcal{E})$ of the two-dimensional holomorphic vector bundle $\mathcal{E}=\mathcal{L}\oplus\boldsymbol{1}_T$. The fibre $\mathcal{E}_x$ of $\mathcal{E}$ over $x$ is the set of pairs $(s_x,c)$, where $s_x\in \mathcal{L}_x$ and $c\in\mathbb{C}$, and the fibre $\bar{\mathcal{L}}_x$ of $\bar{\mathcal{L}}$ over $x$ is the set of equivalence classes of $(s_x:c)$, where either $s_x \ne 0$ or $c\ne 0$ and the equivalence class of $(s_x:c)$ is the set of all Lemma 10.1. Suppose that $\mathcal{L}=\mathcal{L}(H,\alpha)$, where $(H,\alpha)$ is A.-H. data. Then there is a natural group embedding Proof. First, let us define the group embedding
$$
\begin{equation}
\mathfrak{G}(H,\alpha) \hookrightarrow \operatorname{Aut}(\mathcal{L}(H,\alpha)\oplus\boldsymbol{1}_T)
\end{equation}
\tag{31}
$$
by the formula
$$
\begin{equation}
\begin{gathered} \, g\colon (s_x, (x,c))\mapsto(g(s_x),(x+j(g),c)) \\ \forall \ g\in\mathfrak{G}(H,\alpha),\ x \in V/\Pi= T,\ c\in\mathbb{C}, \ s_x\in\mathcal{L}_x\subset\mathcal{L}.\notag \end{gathered}
\end{equation}
\tag{32}
$$
In particular, $g$ induces an isomorphism of two-dimensional complex vector spaces between the fibres of $\mathcal{L}(H,\alpha)\oplus\boldsymbol{1}_T$ over $x$ and over $x+j(g)$. Since $\mathfrak{G}(H,\alpha)\to \operatorname{Aut}(\mathcal{L}(H,\alpha))$ is a group embedding, we conclude that if $j(g)=0$, then $g_x$ is multiplication by a scalar if and only if $g$ is the identity element of $\mathfrak{G}(H,\alpha)$. This implies that (31) and (32) induce a group embedding such that each $g \in \mathfrak{G}(H,\alpha)$ sends every $(s_x:c)\in\mathcal{L}(H,\alpha)_x$ to $(g(s_x):c) \in \mathcal{L}(H,\alpha)_{x+j(g)}$. $\Box$ Let $\mathcal{L}$ be a holomorphic line bundle over the complex torus $T=V/\Pi$. Then $\mathcal{L} \cong \mathcal{L}(H,\alpha)$ for some (actually, unique) A.-H. data $H,\alpha)$ on $(V,\Pi)$ ([33], Theorem 1.5). Let us denote the group $\mathfrak{G}(H,\alpha)$ by $\mathfrak{G}(\mathcal{L})$. By Lemma 10.1 there exists a group embedding
$$
\begin{equation}
\mathfrak{G}(\mathcal{L}) \hookrightarrow\operatorname{Aut}(\bar{\mathcal{L}}).
\end{equation}
\tag{34}
$$
Lemma 10.2. Let $\mathcal{L}$ and $\mathcal{N}$ be holomorphic line bundles over $T=V/\Pi$. Assume that $\mathcal{L}$ admits a non-zero holomorphic section. Then the compact complex manifolds $\overline{\mathcal{N}}$ and $\overline{\mathcal{L}^n\otimes \mathcal{N}}$ are bimeromorphic for all positive integers $n$. In particular, for all such $n$ there exists a group embedding Proof. Let $t$ be a non-zero section of $\mathcal{L}$. Then $t^n$ is a non-zero section of $\mathcal{L}^n$. So it suffices to prove the lemma for $n=1$, that is, to prove that $\bar{\mathcal{L}}$ and $\overline{\mathcal{L}\otimes\mathcal{N}}$ are bimeromorphic.
The holomorphic $\mathbb{C}$-linear map of rank $2$ vector bundles
$$
\begin{equation*}
\begin{aligned} \, \mathcal{N}\oplus\boldsymbol{1}_T & \to(\mathcal{L}\otimes \mathcal{N})\oplus \boldsymbol{1}_T, \\ (s_x;(x,c)) & \mapsto(s_x\otimes t(x);(x,c))\quad \forall\ x\in T,\ s_x\in\mathcal{N}_x,\ c\in\mathbb{C}, \end{aligned}
\end{equation*}
\notag
$$
induces a bimeromorphic isomorphism of their projectivizations $\overline{\mathcal{N}}$ and $\overline{\mathcal{L}\otimes\mathcal{N}}$. Hence the groups $\operatorname{Bim}(\overline{\mathcal{N}})$ and $\operatorname{Bim}(\overline{\mathcal{L}\otimes\mathcal{N}})$ are isomorphic. Now the second assertion of our lemma follows from Lemma 10.1. $\Box$ Corollary 10.3. We keep the notation and assumptions of Lemma 10.2. In particular, $\mathcal{L}$ is isomorphic to $\mathcal{L}(H,\alpha)$ and admits a non-zero holomorphic section. Suppose that $\mathcal{N}$ is isomorphic to $\mathcal{L}(H_0,\beta)$, where the kernel $\ker(H_0)$ of the Hermitian form $H_0$ contains the kernel $\ker(H)$ of the Hermitian form $H$. Then the group $\operatorname{Bim}(\overline{\mathcal{N}})$ is not Jordan. Proof. Consider the alternating $\mathbb{R}$-bilinear forms $E\!:=\operatorname{Im}(H)$ and $M\!:=\operatorname{Im}(H_0)$ on $V$. We have and therefore $\ker(E) \subset \ker(M)$. Also notice that the alternating form $\mathbf{M}(n)=nE+M$ is the imaginary part of the Hermitian form $nH+H_0$ for all positive integers $n$; in addition the holomorphic line bundles $\mathcal{L}$ and $\mathcal{N}$ obviously satisfy
In light of Lemma 10.2 there is a group embedding
$$
\begin{equation*}
\mathfrak{G}(nH+H_0,\alpha \beta^n) \hookrightarrow \operatorname{Bim}(\overline{\mathcal{N}}).
\end{equation*}
\notag
$$
On the other hand, applying Lemma 9.2 to $(nH+H_0,\alpha \beta^n)$ (instead of $(H,\alpha)$), we conclude that $\mathfrak{G}(nH+H_0,\alpha \beta^n)$ is a theta group attached to the symplectic pair $(K_{\mathbf{M}(n),\Pi}, e_{\mathbf{M}(n)})$. Now the desired result follows from Theorem 8.4. $\Box$ Definition 10.4. Let $T=V/\Gamma$ be a complex torus. We write $T_a$ for its algebraic model, which is also a complex torus (even an abelian variety) provided with a surjective holomorphic homomorphism of complex tori with connected kernel (actually, all the fibres of $\pi_a$ are connected; [11], Chap. 2, § 6). We write $\dim_a(T)$ for $\dim(T_a)$ and call it the algebraic dimension of $T$. Clearly, equality holds if and only if $T=T_a$, that is, $T$ is an abelian variety.Theorem 10.5 (Theorem 1.7 in [87]). Suppose that a complex torus $T=V/\Pi$ has a positive algebraic dimension. Then $\operatorname{Bim}(T\times \mathbb{P}^1)$ is not Jordan. Proof. Take $\mathcal{N}=\boldsymbol{1}_T$. Then $\overline{\mathcal{N}}=T\times \mathbb{P}^1$. On the other hand $\mathcal{N}=\boldsymbol{1}_T \cong \mathcal{L}(\mathbf{0},\boldsymbol{1})$, where $\mathbf{0}$ is the zero Hermitian form on $V$ and is the constant semicharacter (actually, character) of $\Pi$ that is identically equal to $1$. Clearly,
Since $\dim_a(T)>0$, the algebraic model $T_a$ is a positive-dimensional abelian variety. Then $T_a$ admits an ample holomorphic line bundle $\mathcal{L}_a$ with a non-zero section. Since $\psi\colon T \to T_a$ is surjective, the inverse image $\mathcal{L}=\psi^*\mathcal{L}_a$ is a holomorphic line bundle on $T$ that also admits a non-zero section. We have $\mathcal{L}\cong \mathcal{L}(H,\alpha)$ for some A.-H. data $(H,\alpha)$. Obviously, Therefore, we can apply Corollary 10.3 and obtain that the group $\operatorname{Bim}(\overline{\mathcal{N}})$ is not Jordan. It remains to recall that $\overline{\mathcal{N}}=T \times \mathbb{P}^1$. $\Box$The following assertion is a generalization of Theorem 10.5. Theorem 10.6 (special case of Theorem 1.8 in [87]). Let $\psi\colon T \to A$ be a surjective holomorphic group homomorphism from a complex torus $T=V/\Pi$ to a positive- dimensional complex abelian variety $A$. Let $\mathcal{M}$ be a holomorphic line bundle over $A$ and $\mathcal{F}$ be a holomorphic line bundle over $T$ that is isomorphic to the inverse image $\psi^*\mathcal{M}$. Then the group $\operatorname{Bim}(\overline{\mathcal{F}})$ is not Jordan. Proof. A positive-dimensional complex abelian variety $A$ is a complex torus $A=W/\Gamma$ (where $W$ is a complex vector space of finite positive dimension $m$ and $\Gamma$ is a discrete lattice of rank $2m$ in $W$) that admits a polarization, that is, a positive (and therefore non-degenerate) Hermitian form whose imaginary part satisfies the condition Replacing $\mathbf{H}_A$ by $2\mathbf{H}_A$ if necessary, we may and will assume that
Then it is obvious that $(\mathbf{H}_A, \boldsymbol{1}_{\Gamma})$ is A.-H. data on $(W,\Gamma)$. The positiveness of $\mathbf{H}_A$ implies that the corresponding holomorphic line bundle $\mathcal{L}(\mathbf{H}_A,\boldsymbol{1})$ over $A$ has a non-zero holomorphic section (the corresponding theta function) (see [33], Theorem 2.1). It follows from Lemma 2.3.4 on p. 33 of [11] that every surjective holomorphic homomorphism $\psi\colon T \to A$ is induced by some surjective $\mathbb{C}$-linear map $\bar{\psi}\colon V \to W$ in the following sense:
$$
\begin{equation*}
\bar{\psi}(\Pi)\subset\Gamma;\qquad \psi(v+\Pi)=\bar{\psi}(v)+\Gamma\in W/\Gamma=A\quad \forall\,v+\Pi\in V/\Pi=T.
\end{equation*}
\notag
$$
The surjectiveness of $\psi$ implies that the induced holomorphic line bundle $\mathcal{L}=\psi^*\mathcal{L}(\mathbf{H}_A,\boldsymbol{1}_{\Gamma})$ over $T$ also has a non-zero holomorphic section.
Let $(H_A,\beta)$ be A.-H. data on $(W,\Gamma)$ and $\mathcal{L}(H_A,\beta)$ be the corresponding holomorphic line bundle over $A=W/\Gamma$. Then the inverse image $\psi^*\mathcal{L}(H_A,\beta)$ is isomorphic to $\mathcal{L}(H_A\mathrel{\circ} \bar{\psi},\beta\mathrel{\circ} \bar{\psi})$ where the A.-H. data $(H_A\mathrel{\circ} \bar{\psi},\beta\mathrel{\circ} \bar{\psi})$ for $(V,\Gamma)$ are as follows (see [33], Lemma 2.3.4):
$$
\begin{equation}
\begin{gathered} \, H_A\mathrel{\circ}\bar{\psi}\colon V\times V \to\mathbb{C}, \quad (v_1,v_2) \mapsto H_A(\bar{\psi}v_1, \bar{\psi}v_2); \\ \beta\mathrel{\circ} \bar{\psi}\colon \Pi \to\mathbf{U}(1), \quad l \mapsto \beta(\bar{\psi}(l)). \end{gathered}
\end{equation}
\tag{36}
$$
In light of the non-degeneracy of $\mathbf{H}_A$, this implies that
$$
\begin{equation}
\ker(\mathbf{H}_A\mathrel{\circ} \psi) =\ker(\bar{\psi}) \subset \ker(H_A\mathrel{\circ} \bar{\psi}) \subset V.
\end{equation}
\tag{37}
$$
Now let $(H_A,\beta)$ be the A.-H. data on $(W,\Gamma)$ such that $\mathcal{M}$ is isomorphic to $\mathcal{L}(H_A,\beta)$. In light of (36), $\mathcal{F}$ is isomorphic to $\mathcal{L}(H_A\mathrel{\circ}\bar{\psi},\beta\mathrel{\circ} \bar{\psi})$. In particular, $\mathcal{L}=\psi^*\mathcal{L}(\mathbf{H}_A,\boldsymbol{1}_{\Gamma})$ is isomorphic to $\mathcal{L}(\mathbf{H}_A\mathrel{\circ} \bar{\psi},\boldsymbol{1}_{\Pi})$. Here
$$
\begin{equation*}
\boldsymbol{1}_{\Pi} =\boldsymbol{1}_{\Gamma}\mathrel{\circ}\bar{\psi}\colon \Pi \to \{1\} \subset\mathbf{U}(1)
\end{equation*}
\notag
$$
is the trivial character of $\Pi$. Since $\mathcal{L}$ admits a non-zero holomorphic section, the inclusion (37) allows us to apply Corollary 10.3 to $\mathcal{N}=\mathcal{F}$ and $H_0=H_A\mathrel{\circ} \bar{\psi}$ and conclude that $\operatorname{Bim}(\overline{\mathcal{F}})$ is not Jordan. $\Box$
Remark 10.7. Let $V$, $\Pi$, $T$, and $\mathcal{F}$ be as in Theorem 10.6. Suppose that $\mathcal{F}\cong\mathcal{L}(H,\alpha)$. Let $\alpha'\colon\Pi\to\mathbf{U}(1)$ be a map such that $(H,\alpha')$ also is A.-H. data on $(V,\Pi)$. Let $\mathcal{F}^{\prime}$ be a holomorphic line bundle on $T$ that is isomorphic to $\mathcal{L}(H,\alpha')$. Then the same arguments as in the proof of Theorem 10.6 prove that $\operatorname{Bim}(\overline{\mathcal{F}'})$ is also non-Jordan (see Theorem 1.8 of [87]). Chapter 4. Non-trivial $\mathbb{P}^{ {\rm \scriptsize 1}}$-bundles over a non-uniruled baseIn this chapter we consider the group $\operatorname{Aut}(X)$ for a non-trivial $\mathbb{P}^1$-bundle over a non-uniruled compact complex connected Kähler manifold $Y$. Recall that there is a homomorphism $\tau\colon \operatorname{Aut}(X)\to\operatorname{Aut}(Y)$ and its kernel is denoted by $\operatorname{Aut}(X)_p$. First we classify automorphisms $f\in\operatorname{Aut}(X)_p$, that is, those automorphisms that do not move fibres of $p$. We get that if $\operatorname{Aut}(X)_p\ne\{\operatorname{id}\}$, then either $X$ or its double cover is a projectivization $\mathbb{P}(\mathcal{E})$ of rank $2$ vector bundle over $Y$ or its double cover, respectively. Thus, if $Y$ is Kähler, then $X$ is too ([82], Proposition 3.5). Thus the group $\operatorname{Aut}(X)$ is Jordan by a theorem of J. H. Kim [34]. It appears that if $X$ is scarce (that is, does not have many sections: see Definition 11.5 below), then $\operatorname{Aut}_0(X)$ is commutative and $\operatorname{Aut}(X)$ is very Jordan. This is, for example, the case when $Y$ is a torus of algebraic dimension zero. 11. Automorphisms of $\mathbb{P}^1$-bundles that preserve fibresThis section contains a classification of those automorphisms of a $\mathbb{P}^1$-bundle $X$ that preserve the fibres of $p\colon X\to Y$. There are three different types; each type is described in a separate subsection. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle over a compact complex connected manifold $Y$, that is,
Let $P_y$ denote the fibre $p^{-1}(y)$. Let $U\subset Y$ be an open non-empty subset of $Y$. We call a covering $U=\bigcup U_i$ of $Y$ by open subsets $U_i$, $i\in I$, to be fine if for every $i\in I$ there exists an isomorphism $\phi_i\colon V_i=p^{-1}(U_i)\to U_i\times\mathbb{P}^1_{(x_i:y_i)}$ such that
Definition 11.1. A $k$-section $S$ of $p$ is a codimension $1$ irreducible analytic subset $D\subset X$ such that the intersection $X\cap P_y$ is finite for every $y\in Y$ and consists of $k$ distinct points for general $y\in Y$. By a bisection we mean a $2$-section that meets every fibre at two distinct points. Obviously, a holomorphic section $S$ of $p$ is a $1$-section. A section $S$ is defined by the set $\mathbf{a}=\{a_i(y)\}$ of functions $a_i\colon U_i\to \mathbb{P}^1$ such that $p(y,a_i(y))=\operatorname{id}$, $y\in U_i$. We denote this by $S=\mathbf{a}$. Lemma 11.2. Let $A_1$, $A_2$, and $A_3$ be three distinct almost sections of $p$ (see Definition 6.5). Assume that there is an analytic subspace $\Sigma\subset Y$ of codimension at least $2$ such that the $A_k$, $k=1,2,3$, are pairwise disjoint in $V=p^{-1}(U)$, where $U=Y\setminus\Sigma$. Then there exists an isomorphism $\Phi\colon X\to Y\times\mathbb{P}^1$ such that $\operatorname{\mathbf{pr}}\mathrel{\circ} \Phi=p$, where $\operatorname{\mathbf{pr}}\colon Y\times \mathbb{P}^1\to\mathbb{P}^1$ is the natural projection (see (NA.14) in § 3). Proof. Indeed, let $\{U_i\}$ be a fine covering of $Y$ and let be the equation of $A_k\cap U$, $k=1,2,3$, over $U_i$. We define a meromorphic function $F(x)$ in every $V_i$ by Then $F(x)$ is globally everywhere defined and meromorphic in $V$. Its restrictions to $A_1\cap V$, $A_2\cap V$, and $A_3\cap V$ are equal to $0$, $\infty$, and $1$, respectively.
The fibre of $p$ has dimension $1$, thus $X\setminus V=p^{-1}(\Sigma)$ has codimension $2$ in $X$. Hence the function $F$ can be extended to a meromorphic function on the whole of $X$ by Levi’s continuation theorem (Theorem 5.9). Thus, we have the bimeromorphic map $\Phi\colon X\to Y\times\mathbb{P}^1$, $\Phi(x)=(p(x),F(x))$, which induces an isomorphism of $V$ onto $U\times \mathbb{P}^1$ that is compatible with $p$. According to Lemma 5.12, $\Phi$ is an isomorphism. $\Box$ Remark 11.3. In particular, if there are three disjoint sections in $X$, then $X\sim Y\times\mathbb{P}^1$. Remark 11.4. Note that a section is an almost section. If $A$ is an almost section but not a section, then the set has codimension at least $2$ because (a) $\widetilde \Sigma:=p^{-1}(\Sigma(A))$ is a proper analytic subset of $A$ with $\dim(A)=\dim(Y)=n$; thus $\dim(\widetilde \Sigma)\leqslant n-1$; (b) every fibre of restriction of $p$ to $\widetilde\Sigma$ has dimension $1$. Definition 11.5. We say that three sections $S_1$, $S_2$, and $S_3$ in $X$ are a good configuration if $S_1\cap S_2=S_1\cap S_3=\varnothing$ and $S_2\cap S_3\ne\varnothing$. We say that three almost sections $A_1$, $A_2$, and $A_3$ in $X$ are a special configuration if $A_1\cap A_2=A_1\cap A_3=A_2\cap A_3$. We say that $X$ is scarce if $X$ admits no special configurations. Lemma 11.6. Let $S_1$, $S_2$, $S_3$, and $S_4$ be four distinct sections of $p$ such that $S_1\cap S_2=\varnothing$ and $S_3\cap S_4=\varnothing$. Then $X\sim Y\times\mathbb{P}^1$. Proof. If $S_3\cap (S_1\cup S_2)=\varnothing$, then $X\sim Y\times\mathbb{P}^1$ (Remark 11.3). Assume that $X\not\sim Y\times\mathbb{P}^1$. Let $\varnothing\ne S_3\cap S_2=D\subset S_2$. Let $\{U_i\}_{i\in I}$ be a fine covering of $Y$. In every $V_i=p^{-1}(U_i)$ we choose coordinates $(y,z_i)$ in such a way that $S_2\cap V_i=\{z_i=0\}$ and $S_1\cap V_i=\{z_i=\infty\}$. Then $z_j=\lambda_{ij}z_i$ in $V_i\cap V_j$, where the $\lambda_{ij}$ are holomorphic functions not vanishing in $U_i\cap U_j$.
Let $S_3\cap V_i=\{(y,z_i=p_i(y)),\,y\in U_i\}$, where $p_j=\lambda_{ij}p_i$, and $S_4\cap V_i=\{(y,\,z_i=q_i(y)),\,y\in U_i\}$, where $q_j=\lambda_{ij}q_i$. Then $r(y):={p_i(y)}/{q_i(y)}$ is a globally defined meromorphic function on $Y$ which omits value $1$ (since $S_3\cap S_4=\varnothing$). Thus, $r:=r(y)=\text{const}$. But then $q_i$ vanishes at $D$ and $S_3\cap S_4\supset D$. This is a contradiction. $\Box$ Remark 11.7. We have also proved the following fact: If $X$ contains two disjoint sections $S_1$ and $S_2$, then Lemma 11.8. If there exist three distinct almost sections $A_1$, $A_2$, and $A_3$ of $p$, then there exists a bimeromorphic map $\Phi\colon X\to Y\times\mathbb{P}^1$ such that $\operatorname{\mathbf{pr}}\mathrel{\circ}\Phi=p$. Proof. We keep the notation of the proof of Lemma 11.2.
Let
$$
\begin{equation*}
\Sigma(A_i)=\{y\in Y\mid p^{-1}(y)\subset A_i\},\quad i=1,2,3,\quad \text{and}\quad \Sigma=\bigcup_1^3\Sigma(A_i).
\end{equation*}
\notag
$$
Let $\widetilde \Sigma=p^{-1}(\Sigma)$.
The function $F(x)$ in (38) is defined and meromorphic at every point outside the set Since the codimension of $D$ is at least $2$, the function $F$ can be extended to a meromorphic function on $X$ by Levi’s theorem. Consider the map $\Phi\colon X\to Y\times \mathbb{P}^1$, $x\mapsto(p(x),F(x))$. It is meromorphic and induces an isomorphism on every fibre $P_u$, $u\not\in p(D)$, to $\mathbb{P}^1$. Thus $\Phi$ is bimeromorphic. $\Box$Lemma 11.9. If $X$ admits a good configuration $S_1$, $S_2$, $S_3$, then $X$ admits a special configuration. Proof. By assumption $S_1\cap S_2=S_1\cap S_3=\varnothing$ and $S_3\cap S_2\ne\varnothing$. Recall that $S_2$ is the zero section of the line bundle $\mathcal{L}(S_1,S_2)$ (see Remark 11.7). Let $\{U_i\}_{i\in I}$ be a fine covering of $Y$ and $(u,z_i)$, $u\in U_i$, $z_i\in\overline{\mathbb{C}}$, be coordinates in $V_i$ such that $S_1\cap V_i=\{z_i=\infty\}$ and $S_2\cap V_i=\{z_i=0\}$. Let the non-zero section of $\mathcal{L}$, namely, $S_3$, have the equation $z_i=h_i(u)$ in $V_i$. For any $c\in\mathbb{C}^*$ the equations $z_i=ch_i$ also define a section $S_4\ne S_3$ of $\mathcal{L}$. By construction $S_2\cap S_3=S_2\cap S_4=S_3\cap S_4=\bigcup_{i\in I}\{h_i=0\}$. Thus, $S_2$, $S_3$, $S_4$ is a special configuration. $\Box$ Now we consider the subgroup $\operatorname{Aut}(X)_p$ of those automorphisms $f$ of $X$ that do not move fibres of $p$, that is, such that $p\mathrel{\circ} f=p$. Similarly to Lemma 11.2, every $f\in\operatorname{Aut}(X)_p$ defines locally a holomorphic map $\psi_f\colon Y\to\operatorname{PSL}(2,\mathbb{C})$ and the function
$$
\begin{equation*}
\operatorname{TD}(y),\quad y\mapsto \operatorname{TD}(\psi_f(y)) =\frac{ \operatorname{tr} ^2(\psi_f(y))}{\det(\psi_f(y))}
\end{equation*}
\notag
$$
(see (NA.15) in § 3), is everywhere defined and holomorphic, hence it is a constant on $Y$ (see [7], Remark 4.9). We denote this constant by $\operatorname{TD}(f)$. Assume that $X\not\sim Y\times\mathbb{P}^1$. Let $f\in\operatorname{Aut}(X)_p$ and $f\ne\operatorname{id}$. Recall that $\operatorname{Fix}(f)$ is the set of all fixed points of $f$. Let $\{U_i\}_{i\in I}$ be a fine covering of $Y$. In Lemmas 11.10 and 11.11 below we summarize the properties of non-identity automorphisms $f\in\operatorname{Aut}(X)_p$ with $\operatorname{TD}(f)\ne 4$ (see [7]). Lemma 11.10. Assume that $(X,p,Y)$ is a $\mathbb{P}^1$-bundle and $X\not\sim Y\times\mathbb{P}^1$. Let $f\in\operatorname{Aut}(X)_p$, where $f\ne\operatorname{id}$ and $\operatorname{TD}(f)\ne 4$. Then one of following two cases holds. Case A: the set $\operatorname{Fix}(f)$ consists of exactly two disjoint sections $S_1$ and $S_2$ of $p$. We say that $f$ is of type A with data $(S_1,S_2)$ (an ordered pair). In the notation of Remark 11.7, let $\{U_i\}_{i\in I}$, $\mathcal{L}(S_1,S_2)$, and $\mathbf{a}=\{a_{ij}\}$ be the corresponding fine covering, holomorphic line bundle, and cocycle, respectively. Then
Case C: the set $\operatorname{Fix}(f)$ is a smooth unramified double cover $S$ of $Y$. (We call such an automorphism $f$ an automorphism of type C with data $S$.) Here $S$ is a bisection of $p$. Proof. The inequality $\operatorname{TD}(f)\ne 4$ implies that $f$ has exactly two distinct fixed points in every fibre $P_y=p^{-1}(y)$, $y\in Y$. Thus $\operatorname{Fix}(f)$ is either a union of two disjoint sections or a $2$-section of $p$. In case A equality (39) follows from the fact that The constant $\lambda_f=\lambda_i\ne 0$ does not depend on the choice of the fibre, hence $f$ is determined uniquely by its restriction to any fixed fibre. On the other hand, for every $\lambda\in\mathbb{C}^*$ there exists an automorphism $f_{\lambda}\in\operatorname{Aut}(X)_p$ which is defined by in every $V_i$. Therefore, $G_0\cong \mathbb{C}^*$. $\Box$ Lemma 11.11 (see [7]). Let $S$ be a bisection of the $\mathbb{P}^1$-bundle $(X,p,Y)$. Consider
$$
\begin{equation*}
\widetilde X :=\widetilde X_S :=S\times_Y X=\{(s,x)\in S\times X \subset X\times X\mid p(s)=p(x)\}.
\end{equation*}
\notag
$$
We denote the restriction of $p$ to $S$ by the same letter $p$, and let $p_X$ and $\tilde p$ be the restrictions to $\widetilde X$ of the natural projections $S\times X\to X$ and $S\times X\to S$, respectively. We write $\operatorname{inv}\colon S\to S$ for the involution (the only non-trivial deck transformation for $p|_S$). Then $(\widetilde X,\tilde p,S)$ is a $\mathbb{P}^1$-bundle with the following properties. (a) The diagram commutes.(b) $p_X\colon \widetilde X\to X$ is an unramified double cover of $X$. (c) Every fibre $\tilde p^{-1}(s)$, $s\in S$, is isomorphic to (d) The $\mathbb{P}^1$-bundle $\widetilde X$ over $S$ has two disjoint sections, namely,
$$
\begin{equation*}
\begin{aligned} \, S_+ & :=S_+(f):=\{(s,s)\in \widetilde X\mid s\in S\subset X\}\\ \textit{and} \qquad S_- & :=S_-(f):=\{(s,\operatorname{inv}(s))\in \widetilde X\mid s\in S\subset X\}. \end{aligned}
\end{equation*}
\notag
$$
They are mapped onto $S$ isomorphically by $p_X$. (e) Every $h\in\operatorname{Aut}(X)_p$ induces an automorphism $\tilde h\in\operatorname{Aut}(\widetilde X)_{\tilde p}$ defined by (f) The involution $s\mapsto\operatorname{inv}(s)$ can be extended from $S$ to a holomorphic involution of $\widetilde X$ by (g) Every section $N=\{y,\sigma(y)\}$ of $p$ in $X$ induces the section $\widetilde N:=\{(s,\sigma(p(s)))\}$ of $\tilde p$ in $\widetilde X$. As $p_X(\widetilde N)=N$ is a section of $p$, thus $\widetilde N$ cannot coincide with $S_+$ or $S_-$. 11.1. Automorphisms with $\operatorname{TD}=4$If $f\in\operatorname{Aut}(X)_p, f\ne\operatorname{id}$, and $\operatorname{TD}(f)= 4$, then there is a unique fixed point of $f$ in the fibre $P_y=p^{-1}(y)$ over the general point $y\in Y$. This means that $\operatorname{Fix}(f)$ contains precisely one almost section $D$ of $p$. In this case we say that $f$ is of type B with data $D$. Lemma 11.12. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are compact connected complex manifolds, $\dim(Y)=n$, $f\in\operatorname{Aut}(X)_p$, $f\ne\operatorname{id}$, and $\operatorname{TD}(f)=4$. Let $D$ be the unique almost section contained in $\operatorname{Fix}(f)$. Let $\Sigma =\{y\in Y\mid P_y\subset D\}$ and $U=Y\setminus\Sigma$, $V=p^{-1}(U)\subset X$. Let $\widetilde S$ be the union of all irreducible components of $\operatorname{Fix}(f)$ distinct from $D$, and let $S=p(\widetilde S)$. Then
Proof. Recall that the set $\Sigma$ has codimension at least $2$ in $Y$ (Remark 11.4). Now we prove the assertions of the lemma.
(i) follows from the fact that $D$ is a section of $p$ over $U$. (ii) follows from the fact that $D\subset\operatorname{Fix}(f)$, thus the restriction of $f$ onto a fibre $P_y$, $y\in U_i$, is an automorphism of $\mathbb{P}^1$, which is either the identity automorphism or has the unique fixed point $z_i=\infty$. (iii) follows from the fact that $z_j$ is obtained from $z_i$ by an automorphism of $\mathbb{P}^1$ with $z=\infty$ fixed. Since $X$ admits an almost section, we see that $X\sim\mathbb{P}(\mathcal{E})$ for some rank $2$ holomorphic vector bundle $\mathcal{E}$ on $Y$ with projection $\pi\colon \mathcal{E}\to Y$ ([78], Lemma 3.5, and Theorem 6.7). This means that we have a fine covering $\{U_i\}$ and a cocycle $A_{ij}\in\operatorname{GL}(2,\mathcal{O}(U_i\cap U_j))$ of $2\times 2$ ransition matrices for $\mathcal{E}$ such that Since $D\cap V$ is a section of $p$ over $U$, we can choose a basis in $\mathbb{C}^2_{x_i,y_i}$ in such a way that the preimage of $D\cap U_i$ in $U_i\times \mathbb{C}^2_{x_i,y_i}$ is $U_i\times \{(x_i,0)\}$, $x_i\in\mathbb{C}$.
Now let $z_j=x_j/y_j$ and $z_i=x_i/y_i$. Then
$$
\begin{equation}
z_j =\frac{\lambda_{i,j}x_i+b_{ij}y_i}{y_i\tilde\lambda_{i,j}} =\mu_{ij}z_i+\nu_{ij}.
\end{equation}
\tag{43}
$$
Thus $\mu_{ij}=\lambda_{i,j}^2/d_{ij}=\lambda_{i,j}/\tilde\lambda_{i,j}$ depends on the choice of $D$ and is defined by the eigenvalue of the basis vector in the invariant subspace representing $D$. It does not depend on the choice of $f$ with the given data $D$. Note that both $\{\lambda_{i,j}\}$ and $\{\tilde\lambda_{i,j}\}$ form cocycles for a covering of $U$. Part (iv) follows from the fact that $f$ is globally defined and $D$ is fixed, thus
$$
\begin{equation*}
f(u,z_j) =(u, z_j+\tau_j(u)) =(u, \mu_{ij}z_i+\nu_{ij}+\tau_j(u)) =(u, \mu_{ij}(z_i+\tau_i(u))+\nu_{ij}).
\end{equation*}
\notag
$$
Part (v) follows from the fact that the functions $\tau_i$ are holomorphic and $S\cap U_i=\{\tau_i=0\}$. Indeed, let $\widetilde S_1\subset \widetilde S$ be an irreducible component of $\widetilde S$. It cannot be an almost section, thus $S_1=p(\widetilde S_1)$ is a proper analytic subset of $Y$. Moreover, since $\widetilde \Sigma\subset D$, we have: $\widetilde S_1\not\subset\widetilde\Sigma$, $S_1\not\subset\Sigma$. Thus, $S_1\cap U$ is a dense open subset of $S_1$. Since $S\cap U_i=\{\tau_i=0\}$ has pure codimension $1$ (if $S\cap U_i\ne \varnothing$), the same is valid for every component of it that intersects $U_i$. Thus, $\dim(S_1)=n-1$. $\Box$ Proposition 11.13. In the notation of Lemma 11.12 let $S_1,\dots,S_k$ be all the irreducible components of $S$. Then (i) For every $l, 1\leqslant l\leqslant k$, a non-negative number $n_l$ is defined that is the order of the zero of $\tau_i$ along the component $S_l$ if $S_l\cap U_i\ne \varnothing$. It depends on $l$ but not on $i$. The restriction to $U$ of the holomorphic line bundle $\mathcal{L}(f)$ corresponding to the effective divisor $\Delta_f:=\sum_{l=1}^{k}n_lS_l$ is the holomorphic line bundle defined by the cocycle $\mu_{ij}$. (ii) Let $G_D$ be the subgroup of $\operatorname{Aut}(X)_p$ of all those $g\in\operatorname{Aut}(X)_p$, that have $\operatorname{TD}(g)=4$ and $D\subset \operatorname{Fix}(g)$. Then $G_D$ is isomorphic to the additive group of $H^0(Y,\mathcal{L}(f))$. Thus $G_D\cong(\mathbb{C}^+)^n$, $n>0$. Proof. Let $S_l$ be an irreducible component of $S$. For every $U_i$ such that $S_l\cap U_i\ne \varnothing$ the order $n_{li}$ of zero of $\tau_i$ along $S_l$ is well defined. In $U_i\cap U_j$ we have $\tau_j=\tau_i\mu_{ij}$. Since $\mu_{ij}$ does not vanish, $\tau_j$ has the same order of zero along $S_l\cap U_j$. Since $S_l$ is irreducible and $U\cap S_l$ is open and dense in $S_l$, the order $n_l$ is well defined (see, for example, [31], Remarks 2.3.6). By construction the divisor of $\tau_i$ in $U_i$ is $\Delta_f\cap U_i$, thus the transition function for $\mathcal{L}(f)$ in $U_i\cap U_j$ is $\tau_j/\tau_i=\mu_{ij}$.
Let $h\in \operatorname{Aut}(X)_p$, and $\operatorname{TD}(h)=4$, and $D\subset \operatorname{Fix}(g)$. Applying Lemma 11.12(iii) we get that $h(u,z_i)=(u,z_i+h_i(u))$, where $h_j=\mu_{ij}h_i$. Thus the function defined in every $U_i$ by $G_h(u)=h_i/\tau_i$ is meromorphic in $U$. By Levi’s theorem $G_h(u)$ is meromorphic on $Y$. By construction its divisor satisfies $(G_h)\geqslant -\Delta_f$, thus $G\in H^0(Y,\mathcal{L}(f))$. On the other hand let $G$ be a meromorphic function on $Y$ with divisor $(G)\geqslant -\Delta_f$ (that is, $G\in H^0(Y,\mathcal{L}(f))$). For every $i$ the function $h_i=G\tau_i$ is holomorphic in $U_i$, hence we can define a holomorphic automorphism of every $V_i=p^{-1}(U_i)$ by Since $h_j:=\mu_{ij}h_i$, the map $h$ is an automorphism of $V$. Moreover, all the points of $D\cap V=\bigcup\{z_i=\infty\}$ are fixed by $h$. By Lemma 5.13 it can be extended to a bimeromorphic map of $X$.By Lemma 5.12$, h\in\operatorname{Aut}(X)_p$. Moreover, $\operatorname{Fix}(\tilde h)$ contains the closure of $D\cap V$, that is, $D$. In the general fibre $P_y$ of $p$ it has precisely one fixed point $D\cap P_y$, thus $\operatorname{TD}(h)=4$. Hence we obtain a one-to-one map
$$
\begin{equation*}
\phi\colon G_D\to H^0(Y,\mathcal{L}(f)), \quad h\in G_D\mapsto G_h\in H^0(Y,\mathcal{L}(f)).
\end{equation*}
\notag
$$
From Lemma 11.12(iii) we get that the composition of $g,h\in\operatorname{Aut}(X)_p$ is defined by the cocycle $g_i+h_i$ made up of the corresponding cocycles, which implies that The next lemma answers the question when an almost section $D\subset\operatorname{Fix} (f)$ is a section. We used this fact in [7], in dealing with automorphisms of type B . Lemma 11.14. In the notation of Lemma 11.12 and Proposition 11.13, if $\Delta_f=0$, then $D$ is a section. Proof. First note that $\Delta_f=0$ implies that the corresponding line bundle $\mathcal{L}_f$ is trivial and $f\ne \operatorname{id}$ in the fibre $F_y=p^{-1}(y)$ if $y\not\in \Sigma$.
Since $X$ admits an almost section, $X\sim \mathbb{P}(\mathcal{E})$ for some rank $2$ holomorphic vector bundle $\mathcal{E}$ on $Y$ ([78], Lemma 3.5; Theorem 6.7). This means that we have a fine covering $\{U_i\}_{i\in I}$ of $Y$ and a cocycle $A_{ij}$ of $2\times 2$ matrices (with entries holomorphic in $U_i\cap U_j$) such that
Since $4d_i= \operatorname{tr} (F_i)^2$ is a square, we can divide $F_i$ by $ \operatorname{tr} (F_i)/2=\sqrt{d_i}$ and assume that $d_i=1$ (we use the fact that $(x_i:y_i)$ are homogeneous coordinates in $\mathbb{P}^1_{x_i:y_i}$). Assume that $D$ is not a section, that is, $\Sigma=\{y\in Y\mid p^{-1}(y)\subset D\}\ne \varnothing$. Let the fine covering of $Y$ consist of open sets $U_0,\dots,U_N$, and let $U_0,\dots,U_k$ meet $\Sigma$, while $U=Y\setminus\Sigma=\bigcup_{i=k+1}^{N}U_i$. Then for each $i>k$ we may assume the following.
Take a point $\mathbf{s}\in\Sigma$, and let $U_0$ be a neighbourhood of $\mathbf{s}$. Let $\tilde r(\mathbf{s})$ be the number of those neighbourhoods $U_i$ with $i>k$ in our fine covering for which $U_i\cap U_0\ne\varnothing$. Let $r=\tilde r(\mathbf{s})$. Let be those neighbourhoods for which $U_i\cap U_0\ne\varnothing$, $t\leqslant i\leqslant t+r$. For $t\leqslant i,j\leqslant t+r$ we have the following:(a) $ F_0=A_{i0}(u)F_iA_{i0}(u)^{-1} =I+ W_i=I+ A_{j0}(u)VA_{j0}(u)^{-1} =I+ W_j $, where $W_i= A_{i0}(u)VA_{i0}(u)^{-1}$, $t\leqslant i\leqslant t+r$; it follows that the matrix function $W_i$, defined a priori in $U_0\cap U_i$, can be extended to a matrix function with holomorphic entries to $U_0 \setminus \Sigma$, hence to the whole of $U_0$, and (b) if $U_i\cap U_j\cap U_0\ne\varnothing$, then (c) if we let
$$
\begin{equation*}
A_{i0}(u)= \begin{bmatrix} \alpha_1(u) & \beta_1(u)\\ \gamma_1(u) & \delta_1(u) \end{bmatrix}\quad\text{and} \quad A_{j0}(u)= \begin{bmatrix} \alpha_2(u) & \beta_2(u)\\ \gamma_2(u) & \delta_2(u) \end{bmatrix} ,
\end{equation*}
\notag
$$
then
$$
\begin{equation}
W_i(u)= \begin{bmatrix} -\alpha_1(u)\gamma_1(u) & \alpha_1^2(u)\\ -\gamma_1^2(u) & \alpha_1(u)\gamma_1(u) \end{bmatrix} =W_j(u)= \begin{bmatrix} -\alpha_2(u) \gamma_2(u) & \alpha_2^2(u)\\ -\gamma_2^2(u) & \alpha_2(u)\gamma_2(u) \end{bmatrix}
\end{equation}
\tag{47}
$$
$$
\begin{equation}
\text{and} \qquad A_{i0}(u)A_{j0}^{-1}(u) =\frac{1}{d_{j0}} \begin{bmatrix} \alpha_1\delta_2-\beta_1\gamma_2 & -\alpha_1\beta_2+\beta_1\alpha_2\\ \gamma_1\delta_2-\delta_1\gamma_2 &-\gamma_1\beta_2+\delta_1\alpha_2 \end{bmatrix} = \begin{bmatrix} \lambda_{ij} & \nu_{ij}\\ 0 & \lambda_{ij} \end{bmatrix}.
\end{equation}
\tag{48}
$$
Let $\widetilde U_{ij}=U_i\cap U_j\cap U_0\ne\varnothing$. From (47) we get that $\alpha_1^2=\alpha_2^2$ and $\alpha_1(u) \gamma_1(u)=\alpha_2(u)\gamma_2(u)$ in $\widetilde U_{ij}$. Note that these equations are valid in the whole of $U_0$ since $W_i$ and $W_j$ are defined there. In $\widetilde U_{ij}$ the following three cases are possible: $\alpha_1=\alpha_2$ and $\gamma_1=\gamma_2$, $\alpha_1=-\alpha_2$ and $\gamma_1=-\gamma_2$, or $\alpha_1=\alpha_2=0$. Case 1: $\alpha_1=\alpha_2$ and $\gamma_1=\gamma_2$ in $\widetilde U_{ij}$. Plugging this into (48) we obtain the following:
$$
\begin{equation*}
\begin{aligned} \, & \frac{1}{d_{j0}} \begin{bmatrix} \alpha_1\delta_2-\beta_1\gamma_2 & -\alpha_1\beta_2+\beta_1\alpha_2\\ \gamma_1\delta_2-\delta_1\gamma_2 &-\gamma_1\beta_2+\delta_1\alpha_2 \end{bmatrix} =\frac{1}{d_{j0}} \begin{bmatrix} \alpha_1\delta_2-\beta_1\gamma_1 & -\alpha_1\beta_2+\beta_1\alpha_1\\ \gamma_1\delta_2-\delta_1\gamma_1 & -\gamma_2\beta_2+\delta_1\alpha_2 \end{bmatrix}\\ & \qquad =\frac{1}{d_{j0}} \begin{bmatrix} d_{i0}+\alpha_1 (\delta_2-\delta_1) & \alpha_1(\beta_1-\beta_2)\\ \gamma_1 (\delta_2-\delta_1) & d_{j0}-\alpha_2(\delta_2-\delta_1) \end{bmatrix} = \begin{bmatrix} \lambda_{ij} & \nu_{ij}\\ 0 & \lambda_{ij} \end{bmatrix}. \end{aligned}
\end{equation*}
\notag
$$
Thus there can be two cases again. Case 1.1: $\gamma_1\equiv0$ in $\widetilde U_{ij}$, hence $\gamma_1^2=0$ in $U_0$. Then in the whole of $U_0$
$$
\begin{equation*}
F_0= \begin{bmatrix} 1 & \alpha_1^2(u)\\ 0 & 1 \end{bmatrix}
\end{equation*}
\notag
$$
and $\alpha_1^2(u)$ does not vanish in $U_0$ since $\operatorname{codim}(\Sigma)\leqslant 2$ and $\Delta_f=0$, that is, $F_0(u)\ne\mathbf{I}$ if $u\not\in\Sigma$. Thus, $D\cap V_0=\{y_0=0\}$ and $\Sigma\cap U_0=\varnothing$. This contradicts the inclusion $\mathbf{s}\in\Sigma$.
Case 1.2: $\gamma_1\not\equiv0$ and $\delta_2\equiv\delta_1$ in $\widetilde U_{ij}$. Then $1=\lambda_{ij}=d_{i0}/d_{j0}$. Moreover,
$$
\begin{equation*}
\beta_1 =\frac{\alpha_1\delta_1-d_{i0}}{\gamma_1} =\beta_2 =\frac{\alpha_2\delta_2-d_{j0}}{\gamma_2}
\end{equation*}
\notag
$$
and $\nu_{ij}=0$ in $\widetilde U_{ij}\cap\{\gamma_1\ne 0\}$. Since this set is open in $U_i\cap U_j$, we have $\nu_{ij}\equiv 0$ and
$$
\begin{equation*}
A_{ij}\equiv \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} .
\end{equation*}
\notag
$$
It follows that there is an isomorphism $V_i\cup V_j\sim(U_i\cup U_j)\times\mathbb{P}^1_{z}$ compatible with $p$, where $z=x_i/y_i=x_j/y_j$. Thus we can replace $U_i$, $U_j$ by $U_i\cup U_j$ and obtain a new fine covering of $Y$ consisting of $N-1$ open subsets and such that $\tilde r(\mathbf{s})=r-1$. Since $U_0$ is connected, we can repeat this process (recall that $\gamma_1=\gamma_2\not\equiv 0$ in $U_i\cup U_j$, so that we remain in Case 1.2) till we obtain a covering with $\tilde r(\mathbf{s})=1$. Thus, since $U_0\setminus \Sigma$ is contained in $U_t\cup\cdots\cup U_{t+r}$, we get that $p^{-1}(U_0\setminus \Sigma)\sim(U_0\setminus \Sigma)\times \mathbb{P}^1_{z}$. By Lemma 5.12 and Lemma 5.13 this extends to an isomorphism and $D$ is the preimage of $\{z=\infty\}$. Case 2: $\alpha_1=-\alpha_2$ and $\gamma_1=-\gamma_2$. Plugging this into (48) we obtain the following:
$$
\begin{equation*}
\begin{aligned} \, & \frac{1}{d_{j0}} \begin{bmatrix} \alpha_1\delta_2-\beta_1\gamma_2 & -\alpha_1\beta_2+\beta_1\alpha_2\\ \gamma_1\delta_2-\delta_1\gamma_2 & -\gamma_1\beta_2+\delta_1\alpha_2 \end{bmatrix} =\frac{1}{d_{j0}} \begin{bmatrix} \alpha_1 \delta_2+\beta_1\gamma_1 & -\alpha_1\beta_2-\beta_1\alpha_1\\ \gamma_1 \delta_2+\delta_1\gamma_1 & \gamma_2\beta_2+\delta_1\alpha_2 \end{bmatrix} \\ & \qquad =\frac{1}{d_{j0}} \begin{bmatrix} -d_{i0}+\alpha_1(\delta_2+\delta_1) & -\alpha_1(\beta_1+\beta_2)\\ \gamma_1(\delta_2+\delta_1) & -d_{j0}-\alpha_1(\delta_2+\delta_1) \end{bmatrix} = \begin{bmatrix} \lambda_{ij} & \nu_{ij}\\ 0 & \lambda_{ij} \end{bmatrix} . \end{aligned}
\end{equation*}
\notag
$$
Similarly to Case 1 we have the two following options. Case 2.1: $\gamma_1\equiv 0$. Then
$$
\begin{equation*}
F_0= \begin{bmatrix} 1 & \alpha_1^2(u)\\ 0 & 1 \end{bmatrix}
\end{equation*}
\notag
$$
and $D$ is a section of $p$ over $U_0$.
Case 2.2: $\gamma_1\not\equiv 0$ and $\delta_2\equiv-\delta_1$ in $\widetilde U_{ij}$. Then $-1=\lambda_{ij}=-d_{i0}/d_{j0}$. Then $\beta_1=(\alpha_1\delta_1-d_{i0})/\gamma_1=-\beta_2=-(\alpha_2\delta_2-d_{j0})/\gamma_2$ and $\nu_{ij}=0$. Similarly to Case 1.2 we get that $p^{-1}(U_0\setminus\Sigma)\sim(U_0\setminus\Sigma)\times \mathbb{P}^1_{z}$ and $D$ is a section of $p$ over $U_0$. Case 3: $\alpha_1=\alpha_2=0$. According to (47) and $\gamma_1^2(u)$ does not vanish in $U_0$ since $\Delta_f=0$. Thus $D\cap V_0=\{z=0\}$, which contradicts the inclusion $\mathbf{s}\in\Sigma$. $\Box$Remark 11.15. We may assume that a fine covering of $Y$ contains a finite covering of $U$ since $U_0\setminus\Sigma$ can be covered by two neighbourhoods $U_0\cap\{\alpha_i\ne 0\}$ and $U_0\cap\{\gamma_i\ne 0\}$ (see (47)). Lemma 11.16. Let $f\in\operatorname{Aut}(X)_p$, $f\ne\operatorname{id}$, be an automorphism of type B with data $D$. Assume that there exists an almost section $A$ of $p$ distinct from $D$. Then $X$ contains a special configuration. Proof. Since $A\ne D$ and $A\not\subset \operatorname{Fix}(f)$, we have $A_1:=f(A)\ne D$ and $A_1\ne A$. Similarly, $A_2:=f(A_1)\ne D$ and $A_2\ne A_1$. Let us show that $A_2\ne A$.
If $A_2=A$, then in the fibre $P_y=p^{-1}(y)$ over the general point $y\in Y$ there is a point $a=A\cap P_y$ such that $f(a)\ne a$ but $f(f(a))=a$. But along the general fibre $P_y$ the map $f$ acts as a translation $z\to z+\tau$ where $\tau\ne 0$. This map has no periodic points except $z\ne\infty$. This contradiction shows that $A_2\ne A$. Let us show that $A$, $A_1$, $A_2$ is a special configuration. For a fibre $P_y$ we have the following options:
It follows that $A\cap A_1=A\cap A_2=A_1\cap A_2$ and $A$, $A_1$, $A_2$ is a special configuration. $\Box$ Corollary 11.17. In the notation of Lemma 11.16, if $X$ is scarce and $\operatorname{Aut}(X)_p$ contains an automorphism $f$ of type B with data $D$, then it contains neither automorphisms of type B with data different from $D$, nor automorphisms of type A . Proof. Indeed, the existence of such automorphisms would imply the existence of an almost section (in particular, a section in the case of type A ) distinct from the one contained in $\operatorname{Fix}(f)$. $\Box$ 11.2. Automorphisms of type ALemma 11.18. Assume that $X\not \sim Y\times\mathbb{P}^1$. Let $S_1$ and $S_2$ be two sections of $p$ such that $S_1\cap S_2=\varnothing$. Let $f\in\operatorname{Aut}(X)_p$. Then one of the following holds: (a) $f(S_1)\subset S_1\cup S_2$; (b) $f(S_2)\subset S_1\cup S_2$; (c) $f(S_1\cup S_2)=S_1\cup S_2$. Proof. Note that a fibrewise automorphism moves a section to a section. Let $S_3=f(S_1)$ and $S_4=f(S_2)$. Since $S_1\cap S_2=\varnothing$, we have $S_3\cap S_4=\varnothing$. According to Lemma 11.6 this can occur only if the pairs $(S_3,S_4)$ and $(S_1,S_2)$ share a section. This can occur only if one of the sections of the pair $(S_3,S_4)$ coincides with either $S_1$ or $S_2$. $\Box$ Recall that the group $G_0$ of all those $f\in\operatorname{Aut}(X)_p$ that have the data $(S_1,S_2)$ is isomorphic to $\mathbb{C}^*$ (see Lemma 11.10). Assume that the holomorphic line bundle $\mathcal{L}(S_1,S_2)$ is defined by a cocycle $\{\lambda_{ij}\}$ and $\mathcal{L}(S_1,S_2)^{\otimes 2}$ has a section $T\subset X$ defined by $\mathbf{a}:=\{a_i(y)\}$ such that $a_j(y)=\lambda_{ij}^2a_i(y)$. Set
$$
\begin{equation*}
\phi_T\colon X\to X,\quad \phi_T(y, z_i)=\biggl(y,\frac{a_i(y)}{z_i}\biggr).
\end{equation*}
\notag
$$
The fixed-point set $\operatorname{Fix}(\phi_T)=\{\phi_T(y,z_i)=(y,z_i)\}$ is defined by $T\cap V_i=\{z_i^2=a_i\}$. If $\phi_T\in\operatorname{Aut}(X)_p$, then the $a_i$ do not vanish. In this case $\mathbf{a}:=\{a_i\}$ provides a section of $\mathcal{L}_p^{\otimes 2}$ that does not meet the zero section, thus $\mathcal{L}_p^{\otimes 2}$ is a trivial bundle and we can define $z_i$ in such a way that $a_i=a=\text{const}\ne0$. Then we write $T=T_a$ and $\phi_a:=\phi_T$. Proposition 11.19. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are compact connected complex manifolds and $X\not\sim Y\times\mathbb{P}^1$. Let $S_1$ and $S_2$ be two sections of $p$ such that $S_1\cap S_2=\varnothing$. Let $\mathcal{L}:=\mathcal{L}(S_1,S_2)$ be the corresponding holomorphic line bundle over $Y$. Let Then
Proof. Let ${\mathbf{\lambda}}=\{\lambda_{ij}\}$ be the cocycle corresponding to $\mathcal{L}$. Take $f\in G_1$. Since $S_1=\{z_i=\infty\}$ is $f$-invariant, we have in $V_i$, where both $a_i$ and $b_i$ are holomorphic functions in $U_i$. Since $f$ is globally defined, we have It follows that $a_i=a_j:=a$ is a constant (as a globally defined holomorphic function) and $b_j=\lambda_{ij}b_i$, hence $\mathbf{b}:=\{b_i\}$ is a section of $\mathcal{L}$. On the other hand every section $\mathbf{b}:=\{b_i\}$ of $\mathcal{L}$ defines $f\in G_1$ by formula (49). Thus, $G_1$ is isomorphic to the group of matrices where $a\in \mathbb{C}^*$ and $\mathbf{b}\in F_1$. We have also shown that if $f\in G_1$ is defined by $\mathbf{b}:=\{b_i\}\ne 0$, then $f(S_2)\ne S_2$, and $f(S_2)\cap S_1=\varnothing$. If $f(S_2)\cap S_2=\varnothing$, then $S_1$, $f(S_2)$, and $S_2$ would be three pairwise disjoint sections, which contradicts the assumption $X\not\sim Y\times\mathbb{P}^1$.
Thus $S_1$, $f(S_2)$, $S_2$ is a good configuration. In the opposite direction, consider a good configuration $S_1$, $S_2$, $S_3$ such that $S_3\cap S_1=\varnothing$ and $S_3\cap S_2\ne\varnothing$. Since $S_3$ is a section of $p$ and does not meet $S_1$, it is defined by a section $\mathbf{b}:=\{b_i\}$ as $z_i=b_i(y)$, $y\in U_i$. Thus, $F_1\ne\{0\}$. The case of $G_2$ and sections that meet $S_1$ but do not meet $S_2$ can be treated in the same way, by interchanging $S_2$ with $S_1$ and $F_1$ with $F_2$. This proves 1)–3). Let us prove 4). If for each $f\in G$ all points in $S_1\cup S_2$ are fixed, then $G=G_0\cong\mathbb{C}^*$ by Lemma 11.10. If this is not the case, take $\phi\in G\setminus G_0$. Then $\phi(S_1)=S_2$ and $\phi(S_2)=S_1$. Thus, $\phi(y,z_i)=a_i(y)/z_i$ in every $V_i$ and
$$
\begin{equation}
\lambda_{ij}\,\frac{a_i(y)}{z_i}=\frac{a_j(y)}{\lambda_{ij}z_i}
\end{equation}
\tag{50}
$$
where the $a_i(y)$ are non-vanishing holomorphic functions in $U_i$. Thus, $\{a_i(y)\}$ define a section of $\mathcal{L}^{\otimes 2}$. Since $a_i(y)$ never vanish, we get that $\mathcal{L}^{\otimes 2}$ is trivial. Therefore, we may choose $z_i$ in such a way that $a_i=a\in\mathbb{C}^*$. Then $\phi=\phi_a$.
For any other $f\in G\setminus G_0$ the composition $f\mathrel{\circ}\phi$ belongs to $ G_0$, hence $G=G_0\sqcup\phi_a\cdot G_0$. $\Box$ Corollary 11.20. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are compact connected manifolds and $X\not\sim Y\times \mathbb{P}^1$. Assume that $p$ admits no good configurations but admits two disjoint sections $S_1$ and $S_2$. Then one of the following holds: The restriction map $\operatorname{Aut}(X)_p\to\operatorname{Aut}(P_y)$, $f\mapsto f|_{P_y}$, is a group embedding. Proof. It follows from Proposition 11.19 that $F_1\!=F_2\!=\{0\}$, thus $\operatorname{Aut}(X)_p\!= G$. $\Box$ 11.3. Automorphisms of type CLet $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are complex compact connected manifolds. Assume that $X\not\sim Y\times\mathbb{P}^1$ and $f\in\operatorname{Aut}(X)_p$, $f\ne\operatorname{id}$, has type C . The analytic subset $F\subset X$ of all fixed points of $f$ contains no sections, but contains a bisection $S$ that is a smooth unramified double cover of $Y$ (see Lemma 11.10). Below we use the notation of Lemmas 11.10 and 11.11. Lemma 11.21. Assume that $\widetilde X:=\widetilde X_S\not\sim S\times \mathbb{P}^1$. Let $N\subset \widetilde X$ be a section of $\tilde p$ distinct from $S_+$ and $S_-$. Then $N_X:=p_X(N)$ is a section of $p$ and $S_+,S_-,N$ is not a good configuration. Proof. Let us show that $p_X\!\colon N\to N_X$ is an unramified double cover. Indeed, assume that this is not the case. Since $\widetilde X$ is the unramified double cover of $X$, the preimage $p_X^{-1}(x)$ contains precisely two points for every $x\in N_X$. Thus, if $p_X^{-1}(N_X)\ne N$, then the preimage $p_X^{-1}(N_X)$ consists of two irreducible components, $N$ and $N_1$. Moreover, since $p_X$ is unramified, we have $N\cap N_1=\varnothing$. It follows that there are two distinct pairs of disjoint sections of $\tilde p$, namely, $S_+,S_-$ and $N,N_1$. According to Lemma 11.6, $\widetilde X\sim S\times \mathbb{P}^1$, which gives us a contradiction. It follows that $N$ is a double cover of $N_X$. Let $s\in S$ and $y=p(s)=p(\operatorname{inv}(s))$. Then contains two points (since $N$ meets every fibre of $\tilde p$ at a single point).
Since $N$ is double cover of $N_X$, it follows that $(N_X\cap P_y)$ contains precisely one point. Therefore, $N_X$ is a section of $p$. Assume that $N$ meets $S_+$ at a point $a=(s,s)\in\widetilde X$, $s\in S$. Then it meets $S_-$ at the point $\operatorname{inv}(a)=(\operatorname{inv}(s),s)$ since $p_X(a)=p_X(\operatorname{inv}(a))$. Thus, $N$ meets both $S_+$ and $S_-$ and the configuration is not good. $\Box$ Corollary 11.22. Assume that $(X,p,Y)$ is a $\mathbb{P}^1$-bundle that admits a non- identity automorphism $f\in\operatorname{Aut}(X)_p$ of type C with data $S$. Assume that the corresponding double cover $\widetilde X_S\not\sim S\times\mathbb{P}^1$. Then Proof. Since there are no good configurations in $\widetilde X_f $ by Lemma 11.21, assertion (i) follows from Corollary 11.20 applied to $\widetilde X$.
Take $u\in S$ and $t\in Y$, $t=p(u)$. If $f|_{P_t}=\operatorname{id}$, then by construction (a) $\tilde f\big|_{P_u}=\operatorname{id}$, hence (b) $\tilde f=\operatorname{id}$ (by Corollary 11.20 applied to $\widetilde X$), hence (c) $\tilde f\big|_{P_{s}}=\operatorname{id}$ for every $s\in S$, hence (d) $f|_{P_{y}}=\operatorname{id}$ for $y=p(s)\in Y$. Consequently, $f$ is uniquely determined by its restriction to the fibre $P_t= p^{-1}(t)$. This proves (ii). On the other hand it was shown in (ii) that $\tilde h=\operatorname{id}$ implies that $f|_{P_{y}}=\operatorname{id}$ for every $y\in Y$, that is, $h=\operatorname{id}$. Therefore, $h\mapsto\tilde h$ is an embedding. This proves (iii). $\Box$ Lemma 11.23. Assume that $f\in \operatorname{Aut}(X)_p$, $f\ne\operatorname{id}$, and $f$ is of type C with data (bisection) $S$. (i) If the corresponding double cover (see case C ) $\widetilde X:=\widetilde X_S$ is not isomorphic to $S\times \mathbb{P}^1$, then the group $\operatorname{Aut}(X)_p$ has exponent $2$ and consists of two or four elements. (ii) If $\widetilde X$ is isomorphic to $S\times \mathbb{P}^1$, then there are two disjoint sections $S_1,S_2\subset X$ of $p$. Moreover, if $X\not\sim Y\times\mathbb{P}^1$, then $\operatorname{Aut}(X)_p$ is a disjoint union of its abelian complex Lie subgroup $\Gamma\cong \mathbb{C}^*$ of index $2$ and its coset $\Gamma^{\prime}$. The subgroup $\Gamma$ consists of those $f\in\operatorname{Aut}(X)_p$ that fix $S_1$ and $S_2$. The coset $\Gamma^{\prime}$ consists of those $f\in\operatorname{Aut}(X )_p$ that interchange $S_1$ and $S_2$. Moreover, the restriction homomorphism $\operatorname{Aut}(X)_p\to\operatorname{Aut}(P_y)$, $f\mapsto f|_{P_y}$, is a group embedding for every $y\in Y$. Proof. We modify the proof of Lemma 4.7 in [7].
Choose a point $a\in S$. Let $b=p(a)\in Y$. This means that $a$ sits in the two-element set $S\cap P_b$. The lift $\tilde f$ of $f$ to $\widetilde X$ has type A with data $(S_+,S_-)\subset \widetilde X$, since the points of $S$ are fixed by $f$. It is determined uniquely by its restriction to $P_a$ (see Proposition 11.19). For the corresponding holomorphic line bundle $\widetilde{\mathcal{L}}:=\widetilde{\mathcal{L}}(S_-,S_+)$, $S_+$ is the zero section. Let
The following two assertions were proved in Lemma 4.7 of [7]. A. If we define an isomorphism $\alpha\colon \overline{\mathbb{C}}_{z_i}\to\overline{\mathbb{C}}_{z_k}$ in such a way that the diagram is commutative, then for some $\nu=\nu(a)\ne 0$.B. Consider an automorphism $h\in\operatorname{Aut}(X)_p$. Let $\tilde h$ be its pullback to $\operatorname{Aut}(\widetilde X)_{\tilde p}$ defined by $\tilde h(s,x)=(s,h(x))$. Let $n_1(z_i)=\tilde h\big|_{\widetilde P_a}$, which means that $h(a,z_i)=(a,n_1(z_i))$. Let $n_2(z_k)=\tilde h\big|_{\widetilde P_{\operatorname{inv}(a)}}$, which means that $h(\operatorname{inv}(a),z_k)=(a,n_2(z_k))$. Then
$$
\begin{equation}
\frac{\nu}{n_1(z_i)} = \alpha(n_1(z_i)) =n_2(\alpha(z_i)) =n_2\biggl(\frac{\nu}{z_i}\biggr).
\end{equation}
\tag{52}
$$
Proof of (i). Assume that $\widetilde X\not\sim S\times\mathbb{P}^1$. According to Corollary 11.22, if $\tilde h \in \operatorname{Aut}(\widetilde X)_{\tilde p}$, then either $\tilde h(s,z_j)=\lambda z_j$, or $h(s, z_j)=\lambda/z_j$ in every open set $\widetilde U_j$ of our fine covering, where $\lambda\in\mathbb{C}^*$ does not depend on $s$ or $j$. Fix $a\in S$. According to assertion B, one of following two conditions holds: (a) $n_1(z_i)=\lambda z_i$, $n_2(z_k)=\lambda z_k$, $z_k=\nu(a)/z_i$, and
$$
\begin{equation*}
\frac{\nu(a)}{\lambda z_i}=\lambda\,\frac{\nu(a)}{z_i}
\end{equation*}
\notag
$$
by (52);
(b) $n_1(z_i)=\lambda/z_i$, $n_2(z_k)=\lambda/z_k$, $z_k=\nu/z_i$, and by (52).In the first case $\lambda=\pm 1$ and in the second $\lambda=\pm\nu$. Hence at most four maps are possible. Clearly, the squares of all these maps are the identity maps. Note that all calculations are done for the fibre of $\tilde p$ over the point $a$. We use the fact that the map $\tilde h$ is defined by its restriction to a fibre. A priori $\nu$ could depend on the fibre. But since $\lambda$ does not, as a byproduct we obtain that the same is valid for $\nu$. Proof of (ii). Assume that $\widetilde X\sim S\times\mathbb{P}^1$. Let $\zeta\colon S\times\mathbb{P}^1\to \mathbb{P}^1$ be the projection onto the second factor, and let $\zeta_1=\zeta|_{S_+}$ and $\zeta_2=\zeta|_{S_-}$. Since $S_+\cap S_-=\varnothing$, the function $z=(\zeta-\zeta_1)/(\zeta-\zeta_2)$ is well defined on $\widetilde X$. Since $z=0$ on $S_+=\{(s,s)\}$ and $z=\infty$ on $S_-=\{(s,\operatorname{inv}(s))\}$, we may assume that $z_j=z$ for all $j$. Recall that for every $s$
$$
\begin{equation}
\operatorname{inv}(s,z) =(\operatorname{inv}(s),\alpha(z)) =\biggl(\operatorname{inv}(s),\frac{\nu(s)}{z}\biggr).
\end{equation}
\tag{53}
$$
This implies that $\nu(s)$ is a holomorphic function on $S$, hence $\nu=\text{const}$. From (53) we get that two disjoint sections $N_1=\{(s,\,z=\sqrt{\nu})\}$ and $N_2=\{(s,\,z=-\sqrt{\nu})\}$ (for some choice of $\sqrt{\nu}$) are invariant under the involution, which means that their images are two disjoint sections $S_1$ and $S_2$, respectively, in $X$.
Thus, $X$ has two disjoint sections. Let us show that there is no good configuration in $X$. Assume that $S_3$ is a third section (of $p$) in $X$. On $\widetilde S_3=p_X^{-1}(S_3)\subset \widetilde X$ the function $z$ is either a constant or takes all values in $\overline{\mathbb{C}}$. If it is constant, then $X$ has three disjoint sections $(S_1,S_2,S_3)$, thus $X=Y\times \mathbb{P}^1$. If $z$ takes all values on $\widetilde S_3$, then $S_3$ meets both $S_1$ and $S_2$, thus $S_1$, $S_2$, $S_3$ is not a good configuration. Now (ii) follows from Corollary 11.20, which completes the proof of Lemma 11.23. We have proved (see Lemma 11.12) that if $X\not\sim Y\times\mathbb{P}^1$ and there exists $f\in\operatorname{Aut}(X)_p$, $f\ne\operatorname{id}$, of type B , then $\operatorname{Aut}(X)_p$ contains a subgroup isomorphic to $(\mathbb{C}^+)^n$ for some positive integer $n$. Corollary 11.24. Assume that $X\not\sim Y\times\mathbb{P}^1$ and $\operatorname{Aut}(X)_p$ contains an automorphism $f\ne \operatorname{id}$ of type B . Then $\operatorname{Aut}(X)_p$ contains no automorphisms of type C . Proof. Assume that $\operatorname{Aut}(X)_p$ contains an automorphism of type C . Then by Lemma 11.23, $\operatorname{Aut}(X)_p$ is either finite or consists of two cosets isomorphic to $\mathbb{C}^*$; in either case $\operatorname{Aut}(X)_p$ does not contain a Lie subgroup $\Gamma\cong (\mathbb{C}^+)^n$ with $n>0$. $\Box$ Proposition 11.25. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are complex compact connected manifolds and $Y$ is Kähler and not uniruled. Then $\operatorname{Aut}(X)$ is Jordan. Proof. Indeed, we have proved that three cases are possible.
(a) $\operatorname{Aut}(X)_p=\{\operatorname{id}\}$; then $\operatorname{Aut}(X)$ embeds into $\operatorname{Aut}(Y)$, which is Jordan according to [34]. (b) $\operatorname{Aut}(X)_p$ contains an automorphisms of type A or B . Then $X=\mathbb{P}(\mathcal{E})$ for some rank $2$ vector bundle $\mathcal{E}$ on $Y$. Thus, $X$ is Kähler ([82], Proposition 3.5). (c) $\operatorname{Aut}(X)_p$ contains an automorphisms of type C . Then the double cover $\widetilde X$ of $X$ fits into case (b). Thus, $X$ is Kähler. In cases (b) and (c), $\operatorname{Aut}(X)$ is Jordan, again according to [34]. 12. Structure of $\operatorname{Aut}_0(X)$ and $\operatorname{Aut}(X)$In this section we prove the main result of this chapter. Namely, we prove that the group $\operatorname{Aut}(X)$ is very Jordan, provided that the $\mathbb{P}^1$-bundle $(X,p,Y)$ is scarce. Theorem 12.1. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle, where $X$ and $Y$ are complex compact connected manifolds, $X$ is not biholomorphic to the direct product $Y\times\mathbb{P}^1$, and $Y$ is Kähler and not uniruled. Assume that $(X,p,Y)$ is scarce. Then:
Proof. We know that the set of almost sections is either infinite or contains at most two of them (by Lemma 11.8 and Remark 6.6).
Consider these cases. Case 1. There exist no almost sections of $p$. Then $\operatorname{Aut}(X)_p$ is finite by Lemma 11.23. Case 2. $p$ has only two almost sections, $A_1$ and $A_2$, which meet. Assume that $f\in\operatorname{Aut}(X)_p$, $f\ne \operatorname{id}$. Since $f$ takes almost sections to almost sections, $A_1\cup A_2$ is invariant under $f$. According to Proposition 11.19, the following cases are possible:
Case 3: $p$ has precisely one almost section. Then there are no automorphisms of type A , since there are no two disjoint sections. If $\operatorname{Aut}(X)_p$ contains no automorphisms of type B , then $\operatorname{Aut}(X)_p$ is finite by Lemma 11.23. If $\operatorname{Aut}(X)_p$ contains an automorphism of type B , then, thanks to Corollary 11.24, $\operatorname{Aut}(X)_p$ contains no automorphisms of type C . Since all automorphisms of type B have to share this section in their sets of fixed points, $\operatorname{Aut}(X)_p\cong(\mathbb{C}^+)^n$ by Proposition 11.13 (unless $\operatorname{Aut}(X)_p=\{\operatorname{id}\}$). Case 4: $p$ admits precisely two almost sections $S_1$ and $S_2$ and they do not meet. Then they are sections. But $X$ admits no good configuration. Thus, by Proposition 11.19 the group $\operatorname{Aut}(X)_p$ contains a subgroup isomorphic to $\mathbb{C}^*$ of index at most $2$. Case 5: $X$ is scarce and all almost sections meet pairwise (in particular, all sections meet pairwise). Then $\operatorname{Aut}(X)_p$ contains no automorphism of type A . If $\operatorname{Aut}(X)_p$ contains an automorphism of type B , then, by Lemma 11.16, the set of sections cannot be scarce (provided that there is more than one of them), which is a contradiction. Hence $\operatorname{Aut}(X)_p$ is finite by Lemma 11.23. Case 6: $X$ is scarce and admits two disjoint sections $S_1$ and $S_2$. By Lemma 11.9, $X$ admits no good configurations, and by Lemma 11.16 it has no automorphisms of type B . By Corollary 11.20, $\operatorname{Aut}(X)_p$ contains a subgroup isomorphic to $\mathbb{C}^*$ of index at most $2$. The proof now repeats the proof of Theorem 5.4 in [7] with only one modification: $\mathbb{C}^+$ should be changed to $(\mathbb{C}^+)^n$ and, accordingly, Lemma 2.10 should be applied. The group $\operatorname{Aut}(X)_p$ can be included in the short exact sequence
$$
\begin{equation}
1 \to (\operatorname{Aut}(X)_p\mathrel{\cap}\operatorname{Aut}_0(X)) \to \operatorname{Aut}_0(X) \xrightarrow{\tau} H_0 \to 1,
\end{equation}
\tag{56}
$$
where $H_0=\tau(\operatorname{Aut}_0(X))\subset \operatorname{Tor} (Y)$ is a torus (see Remark 5.6). According to Cases 1–6, one of the following holds:
Thus, by Lemma 2.10 the group $\operatorname{Aut}_0(X)$ is commutative. Now the theorem follows from the fact that $\operatorname{Aut}(X)/\operatorname{Aut}_0(X)$ is bounded (see Proposition 3.5). $\Box$ 13. Rational bundles over poor manifoldsIn this section we consider rational bundles over poor manifolds. We prove that if $Y$ is poor, then $p$ is scarce and the results of the previous section apply. Definition 13.1. We say that a compact connected complex manifold $Y$ of positive dimension is poor if it enjoys the following properties: A complex torus $T$ with $\dim(T)\geqslant 2$ and $a(T)=0$ is a poor Kähler manifold. Indeed, a complex torus $T$ is a Kähler manifold that does not contain rational curves. If $a(T)=0$, it contains no analytic subsets of codimension $1$ ([11], Chap. 2, Corollary 6.4). An explicit example of such a torus of dimension $2$ was given in [11], Example 7.4. Explicit examples of poor tori of any dimension were presented in [8]. Another example of a poor manifold is provided by a non-algebraic $K3$ surface $S$ with Néron–Severi group $\operatorname{NS}(S)={0}$ (see [9], Chap. VIII, Proposition 3.6). Below $Y$ is assumed to be a compact connected complex manifold. Proposition 13.2 ([7], Proposition 3.6). Let $(X,p,Y)$ be an equidimensional rational bundle. Assume that $Y$ contains no analytic subsets of codimension $1$. Then $(X,p,Y)$ is a $\mathbb{P}^1$-bundle. Proof. Let $\dim(Y)=n$, and let be the set of all points in $X$ where the differential $dp$ of $p$ does not have the maximum rank. Then $S$ and $\widetilde S=p(S)$ are analytic subsets of $X$ and $Y$, respectively (see, for instance, [54], Chap. VII, Theorem 2, [57], Theorem 1.22, and [73]). Moreover, $\operatorname{codim}(\widetilde S)=1$ (see [72]). Since $Y$ contains no analytic subsets of codimension $1$, we obtain $\widetilde S=\varnothing$. Thus the holomorphic map $p$ has no singular fibres. $\Box$ Lemma 13.3. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle such that $\dim(Y)=n$. For an almost section $A$ set $\Sigma(A)=\{y\in Y\mid p^{-1}(y)\subset A\}$. If $Y$ contains no analytic subsets of codimension $1$, then (i) any $k$-section has no ramification points (that is, the intersection $X\cap P_y$ consists of $k$ distinct points for every $y\in Y$); (ii) if $A_1$ and $A_2$ are two almost sections, then $p(A_1\cap A_2)\subset \Sigma(A_1)\cap\Sigma(A_2)$; (iii) any two distinct sections of $p$ in $X$ are disjoint; (iv) if there is an almost section $A\subset X$ that is not a section, then $X$ contains neither sections nor $k$-sections. Proof. (i) Let $R$ be a $k$-section of $p$, and let $A$ be the set of all points $x\in R$ where the restriction $p|_R\colon R\to Y$ of $p$ to $R$ is not locally biholomorphic. Then the image $p(A)$ is either empty or has pure codimension $1$ in $Y$ ([20], § 1, 9, [56], Theorem 1.6, [73]). Since $Y$ contains no analytic subsets of codimension $1$, we have $p(A)=\varnothing$. Hence $A=\varnothing$.
(ii) Let $B$ be an irreducible component of $A_1\cap A_2$. Since $\dim(B)= n-1$ and $\dim(p(B))\leqslant n-2$, we have $p^{-1}(p(b))\subset B$ for every point $b\in B$. Thus, $p(b)\in\Sigma(A_1)\cap\Sigma(A_2)$. (iii) In particular, if $A_1$ and $A_2$ are distinct sections, then $\Sigma(A_1)=\Sigma(A_2)=\varnothing$ and $A:=A_1\cap A_2=\varnothing$. (iv) Since $A$ is not a section, there is a point $y\in Y$ such that $P_y=p^{-1}(y)\subset A$. Thus for any $k$-section $S$ we have $S\cap A\ne\varnothing$. This contradicts assertion (ii) since $\Sigma(S)=\varnothing$. Hence such a section $S$ does not exist. Corollary 13.4. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle such that $\dim(Y)=n$. If $Y$ contains no analytic subsets of codimension $1$, then one of the following holds: (i) $X\sim Y\times \mathbb{P}^1$; (ii) $X$ admits two disjoint sections, and $\operatorname{Aut}(X)_p$ contains a subgroup $G\cong \mathbb{C}^*$ of index at most $2$; (iii) $X$ admits two meeting almost sections, and $\operatorname{Aut}(X)_p$ is finite; (iv) $X$ admits precisely one almost section $D$; then either $\operatorname{Aut}(X)_p\cong\mathbb{C}^+$ (and $D$ is a section by Lemma 11.14) or $\operatorname{Aut}(X)_p=\{\operatorname{id}\}$; (v) $X$ admits no almost sections and $\operatorname{Aut}(X)_p$ is finite. Proof. First note that, since $Y$ does not admit meromorphic functions, given a line bundle $\mathcal{L}$ on $Y$, either $H^0(\mathcal{L})=\{0\}$ or $\mathcal{L}$ is trivial and $H^0(\mathcal{L})\cong\mathbb{C}$.
Part (i): Assume that $X$ admits $m\geqslant 3$ almost sections. By Lemma 13.3 they are disjoint over an open set $U\subset Y$ that has a complement of codimension $2$. Thus, $X\sim Y\times \mathbb{P}^1$ by Lemma 11.2. Part (ii) follows from Corollary 11.20. Part (iii) is proved in Case 3 of the proof of Theorem 12.1. Part (iv) follows from Proposition 11.13: if $\operatorname{Aut}(X)_p\ne \{\operatorname{id}\}$, then $\operatorname{Aut}(X)_p$ is isomorphic to the additive group of $\mathbb{C}^m$. This means that for corresponding line bundle we have $0<m=H^0(\mathcal{L})$. Hence $m=1$. Part (v) follows from Lemma 11.23. $\Box$ Lemma 13.5. Let $(X,p,Y)$ be a $\mathbb{P}^1$-bundle such that $\dim(Y)=n$. If $Y$ is poor, then $\operatorname{Bim}(X)=\operatorname{Aut}(X)$. Proof. Since $Y$ contains no rational curves, it is not uniruled. According to Corollary 5.5, every map $f\in\operatorname{Bim}(X)$ is $p$-fibrewise, that is, there exists a group homomorphism $\tilde\tau\colon \operatorname{Bim}(X)\to \operatorname{Bim}(Y)$ (see Lemma 5.4) such that for all $f\in\operatorname{Bim}(X)$ Since $Y$ contains no rational curves, every meromophic map into $Y$ is holomorphic ([24], also see Remark 3.4). Thus $\tilde\tau(f)\in\operatorname{Aut}(Y)$.
For $f\in \operatorname{Bim}(X)$ let $\widetilde S_f$ be the indeterminacy locus of $f$, which is an analytic subspace of $X$ of codimension at least $2$ ([73], p. 369). Let $S_f={p(\widetilde S_f)}$, which is an analytic subset of $Y$ (see [73] and [54], Chap. VII, Theorem 2). Since $Y$ contains no analytic subsets of codimension 1, $\operatorname{codim} (S_f)\geqslant 2$. Moreover, $f$ is defined at all points in $X\setminus p^{-1}(S_f)$. By Lemma 5.12 both $f\in \operatorname{Bim}(X)$ and $f^{-1}\in \operatorname{Bim}(X)$ can be extended to $X$ holomorphically, hence $\operatorname{Bim}(X)=\operatorname{Aut}(X)$. $\Box$ We summarize the results in the following theorem. Theorem 13.6. Let $(X,p,Y)$ be an equidimensional rational bundle over a poor Kähler manifold $Y$. Then
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