# Twistor fibers in $(1,d)$-surfaces of the flag threefold

Amedeo Altavilla  Dipartimento di Matematica, Università degli Studi di Bari
‘Aldo Moro’, via Edoardo Orabona, 4, 70125, Bari, Italia
<EMAIL_ADDRESS>,  Edoardo Ballico  Dipartimento Di Matematica,
Università di Trento, Via Sommarive 14, 38123, Povo, Trento, Italia
<EMAIL_ADDRESS>and  Maria Chiara Brambilla  Università Politecnica
delle Marche, via Brecce Bianche, I-60131 Ancona, Italia
<EMAIL_ADDRESS>
###### Abstract.

We study surfaces of bidegree $(1,d)$ contained in the flag threefold under
the action of the twistor projection. First, we prove that there is no
integral surfaces of bidegree $(1,d)$ containing $d+2$ twistor fibers such
that three of them are not collinear. Then, fixed any union of $0\leq n\leq
d+1$ non-three-by-three collinear twistor fibers, we show that there is an
integral $(1,d)$-surface containing them and no other twistor fibers. The
result is also true for $d+2$ twistor fibers with additional suitable
hypotheses. Later, we focus on surfaces of low bidegrees and prove that, for
any set of $0\leq n\leq 3$ (resp. $n=4$) twistor fibers, there is a smooth
(resp. integral) surface of bidegree $(1,2)$ containing them and no other
twistor fiber. Finally, we prove that there is no integral $(1,d)$-surface,
for $d=2,3$, containing $d+3$ twistor fibers.

###### Key words and phrases:

flag threefold, twistor projection, twistor fiber, surfaces, bidegree

###### 2010 Mathematics Subject Classification:

Primary: 32L25, 14M15; Secondary: 14D21, 14J26

All the authors are partially supported by GNSAGA. The first named author is
partially supported by the INdAM project ‘Teoria delle funzioni ipercomplesse
e applicazioni’.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries and first results
    1. 2.1 Curves in $\mathbb{F}$ and smooth conics
    2. 2.2 Surfaces of bidegree $(1,0)$ and $(0,1)$
    3. 2.3 Surfaces of bidegree $(0,d)$ and $(1,d)$
    4. 2.4 Non-collinear smooth conics
  3. 3 Surfaces of bidegree $(1,d)$
  4. 4 Surfaces of bidegree $(1,2)$ and $(1,3)$
    1. 4.1 Surfaces of bidegree $(1,2)$ containing $0\leq n\leq 4$ twistor fibers
    2. 4.2 Non existence results for surfaces of bidegree $(1,2)$ and $(1,3)$

##  1\. Introduction

The flag threefold $\mathbb{F}$ can be seen as the twistor space of the
complex projective plane $\mathbb{P}^{2}$ endowed with all its standard
structures except for its orientation which, in this case, is the opposite of
the usual one [6, 11],

$\pi:\mathbb{F}\to\mathbb{P}^{2}.$

The study of the twistor geometry of the flag is motivated by the search of
Riemannian $4$-manifolds admitting several integrable complex structures
compatible with the prescribed metric (see e.g. [8, 12, 13]). In this context,
a recent trend is that of study particular cases, in order to find explicit
examples [1, 2, 5, 7, 9].

In [4] we have started a detailed analysis of the geometry of the algebraic
curves and surfaces contained in $\mathbb{F}$, in relation with the twistor
projection. In particular, twistor fibers are smooth integral curves of
bidegree (1,1) (called smooth conics). In [3] we gave a first bound on the
maximum number of smooth conics contained in a smooth surface
$S\subset\mathbb{F}$.

Here, by focusing our attention to a specific family of surfaces, we obtain
more precise results. In fact we analyze the case of bidegree $(1,d)$ surfaces
in $\mathbb{F}$ and study the number and the arrangements of twistor fibers
contained in them.

In order to explain in details our results, we need to clarify some
definitions. The flag threefold can be defined as

$\mathbb{F}:=\\{(p,\ell)\in\mathbb{P}^{2}\times\mathbb{P}^{2}\,|\,p\ell=0\\},$

where
$p=[p_{0}:p_{1},p_{2}],\ell=[\ell_{0}:\ell_{1}:\ell_{2}]\in\mathbb{P}^{2}$ and
$p\ell=p_{0}\ell_{0}+p_{1}\ell_{1}+p_{2}\ell_{2}$. The notation $(p,\ell)$,
would recall a couple (point,line), and the condition $p\ell=0$ translates as
$p$ belongs to $\ell$. In order to simplify the notation, we identify the
second factor $\mathbb{P}^{2\vee}$ with $\mathbb{P}^{2}$.

We have three projection maps:
$\pi_{1},\pi_{2},\pi:\mathbb{F}\to\mathbb{P}^{2}$, defined as

$\pi_{1}(p,\ell)=p,\quad\pi_{2}(p,\ell)=\ell,\quad\pi(p,\ell)=\overline{p}\times\ell=[p_{1}\ell_{2}-p_{2}\ell_{1},p_{2}\ell_{0}-p_{0}\ell_{2},p_{0}\ell_{1}+p_{1}\ell_{0}],$

where the third one is the twistor map. The fibers of such three maps are the
object of our investigation.

Being embedded in $\mathbb{P}^{2}\times\mathbb{P}^{2}$, it is possible to
define a natural notion of bidegree for surfaces and a (bit less natural) one
for curves in $\mathbb{F}$ (see Definition 2.1). In particular, for any
$q\in\mathbb{P}^{2}$, the three fibers $\pi_{1}^{-1}(q),\pi_{2}^{-1}(q)$ and
$\pi^{-1}(q)$ are curves of bidegree $(0,1),(1,0)$ and $(1,1)$, respectively.
While the fibers of $\pi_{1}$ and $\pi_{2}$ exhaust the family of bidegrees
$(1,0)$ and $(0,1)$ curves, twistor fibers are only a (non-open Zariski dense)
subset of those of bidegree $(1,1)$. Since the curves of bidegree $(1,1)$ are
rational, we call them conics. There are only two types of bidegree $(1,1)$
curves: the reducible ones (union of a bidegree $(1,0)$ and of a bidegree
$(0,1)$ curves intersecting at a point), or smooth conics. All of them can be
described as

$L_{q,m}:=\\{(p,\ell)\in\mathbb{F}\,|\,pm=0,\,q\ell=0\\},$

where $q,m\in\mathbb{P}^{2}$ and, the reducible and smooth cases are obtained
for $qm=0$ or $qm\neq 0$, respectively.

In twistor theory there is an antiholomorphic involution without fixed points
that plays important roles in identifying twistor fibers [6, 11]. In our case,
this map can be defined as $j:\mathbb{F}^{2}\to\mathbb{F}^{2}$, where

$j(p,\ell)=(\overline{\ell},\overline{p}).$

A smooth conic $L_{q,m}$ is a twistor fiber if and only if
$j(L_{q,m})=L_{q,m}$ if and only if $m=\overline{q}$.

As explained in [6], the geometry of $\mathbb{F}$ as twistor space does not
change if we consider a conformal copy of its base space $\mathbb{P}^{2}$.
Therefore, it is natural to classify objects in $\mathbb{F}$ up to conformal
transformations, which, in this case, means up to projective transformations
of $\mathbb{F}$ coming from the lift via $\pi$ of a conformal transformation
of $\mathbb{P}^{2}$. It is possible to see that such transformations of
$\mathbb{F}$ are exactly those which commute with $j$ (see [4] for the case of
the flag manifold). Hence, in particular, the number of twistor fibers
contained in a given surface and their arrangement are conformal invariants.

In order to state our main results we need some more notation. We denote by
$\cal{C}(1)=\cal{C}$ the set of smooth conics in $\mathbb{F}$ and by
$\cal{C}(n)$, $n\geq 2$, the set of $n$ pairwise disjoint smooth conics. In an
analogous way we define $\cal{T}(1)=\cal{T}\subset\cal{C}$ as the set of
twistor fibers and $\cal{T}(n)\subset\cal{C}(n)$, $n\geq 2$, as the set of $n$
pairwise disjoint twistor fibers.

We will see in Remark 2.5 that for any couple of different smooth conics,
there is a unique bidegree $(1,0)$ curve $L=\pi_{2}^{-1}(q_{2})$ and a unique
bidegree $(0,1)$ curve $R=\pi_{1}^{-1}(q_{1})$ such that $L$ and $R$ intersect
both smooth conics. In the case of a couple of twistor fibers we also have
$R=j(L)$. We say that three or more smooth conics are collinear if there is a
$(1,0)$ curve $L$ which intersects all of them. To be collinear, for three or
more smooth conics, is a Zariski closed condition.

To be more precise, in Definition 2.7, we define the set $\cal{C}^{*}(n)$
which parametrizes all $A\in\cal{C}(n)$ such that $\\#(L\cap A)\leq 2$ for all
curves $L$ of bidegree $(1,0)$. Clearly $\cal{C}^{*}(1)=\cal{C}(1)$, and
$\cal{C}^{*}(2)=\cal{C}(2)$, while for $n\geq 3$ the open set $\cal{C}^{*}(n)$
is given by the set of disjoint smooth conics such that no three of them are
collinear. Moreover, we set $\cal{T}^{*}(n):=\cal{T}(n)\cap\cal{C}^{*}(n)$. In
Theorem 2.19 we characterize the elements $A\in\cal{C}^{*}(d+1)$ to be those
which do not obstruct the linear system $|\cal{I}_{A}(1,d)|$.

We now summarize the main results of the paper. In Section 3, we study
surfaces of bidegree $(1,d)$ containing a certain number of smooth conics or
twistor fibers, and we prove the following two theorems.

######  Theorem 1.1.

For any $d\in\mathbb{N}$ and $A\in\cal{T}^{*}(d+2)$, there is no integral
surfaces of bidegree $(1,d)$ containing $A$.

######  Theorem 1.2.

Fix integer $d\geq 1$ and $0\leq n\leq{d+2}$. There is an integral
$S\in|\cal{O}_{\mathbb{F}}(1,d)|$ containing exactly $n$ twistor fibers.

We also show, in Theorem 3.4, that the first result is sharp. Indeed, for any
$A\in\cal{T}^{*}(n)$, with $0\leq n\leq d+1$, we are able to find an integral
surface of bidegree $(1,d)$ containing $A$ and no other twistor fibers. This
last issue requires some effort and the proof is divided into several
particular case.

Theorem 1.2 is a consequence of Theorem 3.4 and Theorem 3.8. More precisely,
in Theorem 3.4, for $0\leq n\leq d+1$, we prove that fixed any union $A$ of
$0\leq n\leq d+1$ non-three-by-three collinear twistor fibers, there is an
integral $(1,d)$-surface containing $A$ and no other twistor fibers. The
extremal case $n=d+2$ is considered in Theorem 3.8, where we prove that given
$d+2$ general collinear twistor fibers there is an integral surface of
bidegree $(1,d)$ containing them.

In Section 4, we focus on surfaces of bidegree $(1,2)$ and $(1,3)$. The main
results are summarized by the following statements:

######  Theorem 1.3.

Fix $0\leq n\leq 3$. There is a smooth $S\in|\cal{O}_{\mathbb{F}}(1,2)|$
containing exactly $n$ twistor fibers. Moreover, there exists a bidegree
$(1,2)$ integral surface containing exactly $4$ twistor fibers.

######  Theorem 1.4.

There is no integral $S\in|\cal{O}_{\mathbb{F}}(1,2)|$ containing at least
${5}$ twistor fibers.

######  Theorem 1.5.

There is no integral $S\in|\cal{O}_{\mathbb{F}}(1,3)|$ containing at least $6$
twistor fibers.

The first existence result (Theorem 1.3) follows from Theorem 4.6, for $0\leq
n\leq 3$, and Theorem 4.10, for the case $n=4$. In the extremal case $n=4$, we
will also show that the surfaces are singular along a line.

The two non-existence results (Theorems 1.4 and 1.5) are proved in the last
Section 4.2. An essential tool is Lemma 4.11 which states that if a surface of
bidegree $(1,d)$ contains $d+3$ or more collinear twistor fibers, then this
surface is reducible and one of its components is a surface of bidegree
$(1,1)$ containing $4$ of the prescribed twistor fibers.

We conclude here with a comparison with the other (smooth) algebraic twistor
space of a Riemannian $4$-manifold, which is the complex projective space.
This is the twistor space of the standard $4$-sphere [11]. In this case,
surfaces of degree $2$ and $3$ were studied in some details. In particular,
analogously to our case of surfaces of bidegree $(1,1)$, surfaces of degree
$2$ in $\mathbb{P}^{3}$ might contain $0,1$ or $2$ twistor fibers. If a
surface of degree $2$ contains more than $2$ twistor fibers, then it contains
infinitely many of them [13]. For degree $3$ surfaces, such a maximum is
realized for $5$ twistor fibers [5] which is more than our maximum of $4$ for
surfaces of bidegree $(1,2)$. This difference could be explained by observing
a certain unbalancedness of the case of “total degree” $3$ in the flag
threefold. On the other hand, this particular unbalancedness allows us to
compute all the Betti numbers in the next section as well as the use of the
geometry of the Hirzebruch surfaces.

##  2\. Preliminaries and first results

In this section, we collect some known results about algebraic curves and
surfaces in the flag. Then, we give first results on the space of bidegree
$(0,d)$ and $(1,d)$ surfaces containing a certain amounts of twistor fibers.
In particular, we introduce the concept of collinear smooth conics and give a
topological characterization in terms of cohomology of certain ideal sheaves.

For most of the known material about $\mathbb{F}$, we refer mainly to [4] and
to [3, Section 2]. However, we recall here some basic notion and results in
order to be as more self-contained as possible.

Let us consider the multi projective space
$\mathbb{P}^{2}\times\mathbb{P}^{2}$; an element
$(p,\ell)\in\mathbb{P}^{2}\times\mathbb{P}^{2}$ will be a couple written in
the following form $p=[p_{0}:p_{1}:p_{2}]$,
$\ell=[\ell_{0}:\ell_{1}:\ell_{2}]^{\top}$, so that
$p\ell=p_{0}\ell_{0}+p_{1}\ell_{1}+p_{2}\ell_{2}$. Even if it is classically
embedded in $\mathbb{P}^{2}\times\mathbb{P}^{2\vee}$, we might see
$\mathbb{F}:=\\{(p,\ell)\in\mathbb{P}^{2}\times\mathbb{P}^{2}\,|\,p\ell=0\\}$
as a hypersurface of bidegree $(1,1)$ of $\mathbb{P}^{2}\times\mathbb{P}^{2}$.
We denote by $\Pi_{1}$ and $\Pi_{2}$ the two standard projections of
$\mathbb{P}^{2}\times\mathbb{P}^{2}$ and we will use small letters for their
restrictions, i.e. $\pi_{i}=\Pi_{i|\mathbb{F}}$, $i=1,2$. Thus, the two
natural projections define a natural notion of bidegree for algebraic surfaces
in $\mathbb{F}$. Moreover, for all $(a,b)\in\mathbb{Z}^{2}$ we have the
following natural exact sequence

(1)
$0\to\cal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(a-1,b-1)\to\cal{O}_{\mathbb{P}^{2}\times\mathbb{P}^{2}}(a,b)\to\cal{O}_{\mathbb{F}}(a,b)\to
0,$

and, for any $(a,b)\in\mathbb{N}^{2}$, we get (see e.g. [4, Lemma 2.3])

(2) $h^{0}(\cal{O}_{\mathbb{F}}(a,b))=\frac{(a+1)(b+1)(a+b+2)}{2}\qquad\mbox{
and }\qquad h^{1}(\cal{O}_{\mathbb{F}}(a,b))=0.$

It will be useful to recall from [4, Proposition 3.11] the multiplication
rules in the Chow ring:

(3)
$\begin{split}\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(1,0)=0,&\qquad\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(1,0)=1,\\\
\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(0,1)=1,&\qquad\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(0,1)=0.\end{split}$

###  2.1. Curves in $\mathbb{F}$ and smooth conics

Let us recall a notion of bidegree for the family of algebraic curves in
$\mathbb{F}$ already given in [4, 3].

######  Definition 2.1.

Let $C\subset\mathbb{F}$ be an integral algebraic curve. We define the
bidegree of $C$ as the couple of positive integer numbers $(d_{1},d_{2})$,
where $d_{i}=0$ if $\pi_{i}(C)=\\{x\\}$, otherwise
$d_{i}=\deg(\pi_{i}(C))\deg({\pi_{i}}_{|C})$.

If a curve $D$ has irreducible components $C_{1},\dots,C_{s}$ then the
bidegree of $D$ is taken to be the sum of the bidegrees of
$C_{1},\dots,C_{s}$.

Recall from [3, Remark 2.4] that if a curve $C$ is such that
$C\cdot\cal{O}_{\mathbb{F}}(1,0)=d_{1}$ and
$C\cdot\cal{O}_{\mathbb{F}}(0,1)=d_{2}$, then it has bidegree $(d_{1},d_{2})$.

From the previous table of multiplication, we can easily derive the following
formula.

######  Lemma 2.2.

For any choice of non-negative integers $a,b,c,d$, the one-dimensional cycle
$\cal{O}_{\mathbb{F}}(a,b)\cdot\cal{O}_{\mathbb{F}}(c,d)$ has bidegree

$(ad+b(c+d),a(c+d)+bc).$

###### Proof.

We have
$\cal{O}_{\mathbb{F}}(a,b)\cdot\cal{O}_{\mathbb{F}}(c,d)=ac\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(1,0)+(ad+bc)\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(0,1)+bd\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(0,1)$.
Hence the thesis is easily obtained by recalling that
$\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(1,0)$ (resp.
$\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(0,1)$, resp.
$\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(0,1)$) is a one-
dimensional cycle of bidegree $(0,1)$ (resp. bidegree $(1,1)$, resp. bidegree
$(1,0)$). ∎

######  Remark 2.3.

Notice that the fibers of $\pi_{1}$ are algebraic curves of bidegree $(0,1)$,
while those of $\pi_{2}$ have bidegree $(1,0)$ (see, e.g. [4, Section 3]).
Moreover, all bidegree $(0,1)$ curves can be seen as complete intersections
between two different $(1,0)$ surfaces (and analogously for bidegree $(1,0)$
curves).

Among all algebraic curves in $\mathbb{F}$ we focus our attention on the
family of bidegree $(1,1)$ curves. These are geometrically described in [4,
Section 3.1] and are parameterized by
$(q,m)\in\mathbb{P}^{2}\times\mathbb{P}^{2}$. In fact, as anticipated in the
introduction, any of these curves can be written as

$L_{q,m}:=\\{(p,\ell)\in\mathbb{F}\,|\,p\in m,\,\ell\ni
q\\}=\\{(p,\ell)\in\mathbb{F}\,|\,q\ell=0,\,pm=0\\}.$

There are two types of these curves: the smooth and irreducible ones (when
$qm\neq 0$) and the union of a $(1,0)$ and of a $(0,1)$ intersecting at a
point (when $qm=0$, i.e. $(q,m)\in\mathbb{F}$). In any case, each bidegree
$(1,1)$ curve can be seen as the complete intersection of a surface of
bidegree $(1,0)$ with one of bidegree $(0,1)$. As already mentioned in the
introduction, the $4$-dimensional family of smooth irreducible $(1,1)$ curves
will be denoted by $\cal{C}$. The elements of $\cal{C}$ will be called smooth
conics.

######  Remark 2.4.

From the very definition of smooth integral conics, it is clear that, for any
$C\in\cal{C}$ we have that $\pi_{i}(C)$ is a line in $\mathbb{P}^{2}$.

######  Remark 2.5.

Notice that for any two different elements
$L_{q,m},L_{q^{\prime},m^{\prime}}\in\cal{C}$ there exist a unique curve $L$
of bidegree $(1,0)$ and a unique $R$ of bidegree $(0,1)$ such that $L$ and $R$
meets both $L_{q,m}$ and $L_{q^{\prime},m^{\prime}}$ at a point (see Figure
1). From the analysis made in [4, Section 3.1] it is easy to see that
$L=\pi_{2}^{-1}(q\times q^{\prime})$ and $R=\pi_{1}^{-1}(m\times m^{\prime})$,
where $\times$ stands for the standard (formal) cross product. Equivalently,
$L=\pi_{2}^{-1}(\mathrm{Sing}(\pi_{2}(A))$ and
$R=\pi_{1}^{-1}(\mathrm{Sing}(\pi_{1}(A))$.

Figure 1. Any two smooth conics are connected by a curve of bidegree $(1,0)$
and by a curve of $(0,1)$.

We say that three disjoint smooth conics are collinear if they intersect the
same $(1,0)$ curve $L$.

The fibers of the twistor projection $\pi:\mathbb{F}\to\mathbb{P}^{2}$ (see
[4, Section 5]) form a subset $\cal{T}$ of the family of conics $\cal{C}$. The
twistor fibers are also characterized to be the irreducible elements in
$\cal{C}$ that are fixed by the anti-holomorphic involution
$j:\mathbb{F}\to\mathbb{F}$ defined as

$j(p,\ell)=(\overline{\ell},\overline{p}).$

Being the set of fixed point of $j$, a curve $L_{q,m}$ belongs to $\cal{T}$ if
and only if $m=\overline{q}$. Moreover, the set $\cal{T}$ is a Zariski dense
in $\cal{C}$ (see, e.g. [3, Section 4]).

######  Remark 2.6.

If $L$ is the curve of bidegree $(1,0)$ connecting two different twistor
fibers (see Remark 2.5), then the curve of bidegree $(0,1)$ connecting them is
exactly $R=j(L)$. Hence if three twistor fibers are collinear, then they
intersect the same $(1,0)$ curve and the same $(0,1)$ curve.

Recall from the introduction that, for any positive integer $n$, $\cal{C}(n)$
denotes the $4n$-dimensional set of $n$ pairwise disjoint elements of
$\cal{C}$ and $\cal{T}(n)$ the set of $n$ pairwise disjoint elements of
$\cal{T}$. As before, $\cal{T}(n)$ is a Zariski dense in $\cal{C}(n)$ (see
again [3, Section 4]). We now introduce the following crucial definition.

######  Definition 2.7.

For any $n\geq 1$ let $\cal{C}^{*}(n)$ be the set of all $A\in\cal{C}(n)$ such
that for any curve $L$ of bidegree $(1,0)$, it holds $\\#(L\cap D)\leq 2$. Set
$\cal{T}^{*}(n):=\cal{T}(n)\cap\cal{C}^{*}(n)$.

Clearly we have $\cal{C}^{*}(n)=\cal{C}(n)$ and $\cal{T}^{*}(n)=\cal{T}(n)$,
for $n=1,2$. For $n\geq 3$ the set $\cal{C}(n)\setminus\cal{C}^{*}(n)$
parametrizes unions of $n$ disjoint smooth conics such that at least three of
them are collinear, hence $\cal{C}^{*}(n)$ is an open Zariski dense in
$\cal{C}(n)$, as well as $\cal{T}^{*}(n)$ in $\cal{T}(n)$. Therefore, for any
$n\geq 1$, all the following inclusions are Zariski dense:
$\cal{T}^{*}(n)\subset\cal{T}(n)\subset\cal{C}(n)$,
$\cal{T}^{*}(n)\subset\cal{C}^{*}(n)\subset\cal{C}(n)$.

###  2.2. Surfaces of bidegree $(1,0)$ and $(0,1)$

We now turn our attention back to surfaces. We recall from [4, Section 3.2]
and [3, Section 2] that $(1,0)$ and $(0,1)$ surfaces are Hirzebruch surfaces
of first type. In particular, a surface $X$ of bidegree $(1,0)$ can be seen as
the lift, via $\pi_{1}$ of a line (and analogously for a bidegree $(0,1)$
surface $Y$). Using this description, it is easy to see that any of these
surfaces represent the blow up of $\mathbb{P}^{2}$ at a point. Let $F_{1}$ be
a Hirzebruch surface of type $1$; we now describe the relation between the
generators of the Picard group of $F_{1}$ and the family of curves in
$\mathbb{F}$ previously described. We recall that
$\mathrm{Pic}(F_{1})=\mathbb{Z}h\oplus\mathbb{Z}f$, where

$h^{2}=-1,\qquad f^{2}=0,\qquad hf=1.$

######  Notation 2.8.

For the following analysis and the rest of the paper $X$ will denote a surface
of bidegree $(1,0)$, while $Y$ one of bidegree $(0,1)$.

Identifying a surface $X$ with $F_{1}$ we obtain that
$\cal{O}_{X}(1,0)\simeq\cal{O}_{F_{1}}(f)$ which in turn corresponds to the
set of curves in $\mathbb{F}$ of bidegree (0,1) contained in $X$. On the other
hand we have that $\cal{O}_{X}(0,1)\simeq\cal{O}_{F_{1}}(h+f)$ which
corresponds to elements of $\cal{C}$.

Hence, for any $a,b\in\mathbb{Z}$ and for any $\alpha,\beta\in\mathbb{Z}$, we
obtain the following two relations

(4) $\cal{O}_{X}(a,b)\cong\cal{O}_{F_{1}}(bh+(a+b)f),\quad\mbox{ and
}\quad\cal{O}_{F_{1}}(\alpha h+\beta f)\cong\cal{O}_{X}(\beta-\alpha,\alpha).$

For a surface $Y$ of bidegree $(0,1)$ we can derive similar formulæ:

(5) $\cal{O}_{Y}(a,b)\cong\cal{O}_{F_{1}}(ah+(a+b)f),\quad\mbox{ and
}\quad\cal{O}_{F_{1}}(\alpha h+\beta f)\cong\cal{O}_{Y}(\alpha,\beta-\alpha),$

for any $a,b\in\mathbb{Z}$ and for any $\alpha,\beta\in\mathbb{Z}$.

######  Remark 2.9.

Let $X$ be a surface of bidegree $(1,0)$. Then $X$ does not contain any
element of $\cal{C}(2)$. In fact, any element of $\cal{C}$ in $X$ corresponds
to an element of $\cal{O}_{F_{1}}(h+f)$ and any two elements of type $h+f$
meets. The same holds for surfaces $Y$ of bidegree $(0,1)$. In particular, for
each bidegree $(1,0)$ or $(0,1)$ surface there is exactly one twistor fiber
contained in it.

We now recall from [3, Lemma 2.5] that, for any $a,b\geq 0$, using the exact
sequence

$0\to\cal{O}_{\mathbb{F}}(a-1,b)\to\cal{O}_{\mathbb{F}}(a,b)\to\cal{O}_{X}(a,b)\to
0,$

and its analogous for $Y$, we have that

$h^{0}(\cal{O}_{X}(a,b))=a(b+1)+\binom{b+2}{2},\quad
h^{0}(\cal{O}_{Y}(a,b))=b(a+1)+\binom{a+2}{2},$

while

$h^{1}(\cal{O}_{X}(a,b))=h^{1}(\cal{O}_{Y}(a,b))=0.$

Moreover, if $a>0,b>0$, the line bundles $\cal{O}_{X}(a,b)$ and
$\cal{O}_{Y}(a,b)$ are very ample.

###  2.3. Surfaces of bidegree $(0,d)$ and $(1,d)$

We now pass to study higher bidegree surfaces. We start with some
consideration about bidegree $(0,d)$ surfaces.

######  Remark 2.10.

As described in [4, Section 3.3], any integral surface $S$ of bidegree $(0,d)$
is equal to $\pi_{1}^{-1}(C)$ for some degree $d$ integral curve $C$.
Therefore, for $d\geq 2$, no integral $S\in|\cal{O}_{\mathbb{F}}(0,d)|$
contains a smooth conic. Otherwise, thanks to Remark 2.4, $\pi_{i}(S)$ would
contain a line, but $\pi_{1}(S)$ is an integral curve of degree $d$.

For $n\geq 2$ we now compute how many (non integral) bidegree $(0,d)$ surfaces
contain a fixed element of $\cal{C}(n)$. First we set the following notation.

######  Notation 2.11.

We will denote by $\cal{I}_{U,V}$ the ideal sheaf of a scheme $U$ contained in
a projective variety $V$; whenever $V=\mathbb{F}$ we will omit it. So, in
particular, if $A\in\cal{C}(n)$ we will write
$\cal{I}_{A}:=\cal{I}_{A,\mathbb{F}}$.

######  Lemma 2.12.

Fix $d\geq 0$, $n\geq 1$, and $A\in\cal{C}(n)$. We have

$h^{0}(\cal{I}_{A}(0,d))=\frac{(d-n+2)(d-n+1)}{2}$

and

$h^{1}(\cal{I}_{A}(0,d))=\begin{cases}\frac{n(n-1)}{2}&\mbox{ if }n\leq d+1\\\
n(d+1)-\frac{(d+2)(d+1)}{2}&\mbox{ if }n\geq d+1.\end{cases}$

###### Proof.

Recall that
$\cal{O}_{\mathbb{F}}(0,d)=\pi_{1}^{\ast}(\cal{O}_{\mathbb{P}^{2}}(d))$, and
$\cal{I}_{A}(0,d)=\pi_{1}^{\ast}(\cal{I}_{T,\mathbb{P}^{2}}(d))$ where
$T=\pi_{1}(A)$ is a union of $n$ distinct lines in $\mathbb{P}^{2}$. In
general, we have that:

$h^{0}(\cal{O}_{\mathbb{F}}(0,d))=\binom{d+2}{2},\quad
h^{0}(\cal{I}_{A}(0,d))=\binom{d+2-n}{2},\quad
h^{0}(\cal{O}_{A}(0,d))=n(d+1),$

hence, using the exact sequence

$0\to\cal{I}_{A}(0,d)\to\cal{O}_{\mathbb{F}}(0,d)\to\cal{O}_{A}(0,d)\to 0,$

since $h^{1}(\cal{O}_{\mathbb{F}}(0,d))=0$ and $\binom{d+2-n}{2}=0$ if $n\geq
d+1$, we get the result. ∎

######  Remark 2.13.

As a direct consequence of the previous lemma, we can state that, for any
$C\in\cal{C}$, there is only one surface $Y$ in $|\cal{I}_{C}(0,1)|$ and,
analogously, only one surface $X$ in $|\cal{I}_{C}(1,0)|$.

######  Remark 2.14.

If $A\in\cal{C}(n)$, we consider the following exact sequence:

$0\to\cal{I}_{A}(a,b)\to\cal{O}_{\mathbb{F}}(a,b)\to\cal{O}_{A}(a,b)\to 0.$

Since the conics in $A$ are all disjoint, we have that for any
$(a,b)\in\mathbb{N}^{2}$

(6) $h^{0}(\cal{O}_{A}(a,b))=n(a+b+1),$

(see, e.g. [3, Sequence (7) and proof of Theorem 4.4]). Recall from Formula
(2) that $h^{0}(\cal{O}_{\mathbb{F}}(a,b))=\frac{(a+1)(b+1)(a+b+2)}{2}$.
Therefore, as $h^{1}(\cal{O}_{A}(a,b))=h^{1}(\cal{O}_{\mathbb{F}}(a,b))=0$,
for any $A\in\cal{C}(n)$ and $d\geq 0$, we have

(7)
$\chi(\cal{I}_{A}(1,d))=h^{0}(\cal{I}_{A}(1,d))-h^{1}(\cal{I}_{A}(1,d))=(d+1)(d+3)-n(d+2).$

Hence, if $n=d+2$, we have $h^{0}(\cal{O}_{A}(1,d))=(d+2)^{2}$ and so

(8)
$h^{0}(\cal{O}_{\mathbb{F}}(1,d))=(d+1)(d+3)=(d+2)^{2}-1=h^{0}(\cal{O}_{A}(1,d))-1$

and, if $n=d+1$:

$h^{0}(\cal{O}_{\mathbb{F}}(1,d))=h^{0}(\cal{O}_{A}(1,d))+d+1.$

In the following we will implicitly make use of the following simple
observation.

######  Remark 2.15.

Since the elements of $\cal{T}(1)$ are the fibers of a map, the twistor map,
each $p\in\mathbb{F}$ is contained in a unique element, $C$, of $\cal{T}$.
Since $j(C)=C$, $j(p)\in C$. Analogously, note that $p$ is contained in a
unique curve of bidegree $(1,0)$ and a unique curve of bidegree $(0,1)$,
$\pi_{2}^{-1}(\pi_{2}(p))$ and $\pi_{1}^{-1}(\pi_{1}(p))$.

The following remark will be used several times in the next pages.

######  Remark 2.16.

Fix a positive integer $d$ and an integral $S\in|\cal{O}_{\mathbb{F}}(1,d)|$.
Since $\pi_{1|X}$ is birational onto its image, $S$ is rational. By Bézout
Theorem, for any $p\in\mathbb{P}^{2}$, the bidegree $(1,0)$ curve
$\pi_{2}^{-1}(p)$ either is contained in $S$, or intersects $S$ in a single
point (scheme-theoretically).

We close this subsection with a technical result that will be used in the next
pages. In [3, Proposition 4.1] we proved that there exists at most one surface
of bidegree $(a,b)$ containing a number equal or greater than $a^{2}+ab+b^{2}$
of smooth conics. We will now generalize this result to a more general
context.

######  Proposition 2.17.

Fix $(a,b,c,d)\in\mathbb{N}^{4}$ such that $(a,b)\neq(0,0)$, $c>0$, $d>0$.
Take $A\in\cal{C}(n)$. Assume the existence of an integral
$S\in|\cal{O}_{\mathbb{F}}(a,b)|$ containing $A$ and assume one of the
following conditions:

  1. (1)

$ad+b(c+d)<n$;

  2. (2)

$a(c+d)+bc<n$;

  3. (3)

$ad+b(c+d)=a(c+d)+bc=n$.

Then each element of $|\cal{I}_{A}(c,d)|$ contains $S$ and in particular
$c\geq a$ and $d\geq b$.

###### Proof.

Assume by contradiction the existence of $S^{\prime}\in|\cal{I}_{A}(c,d)|$
such that $S^{\prime}\nsupseteq S$. We have

$\displaystyle\cal{O}_{\mathbb{F}}(a,b)\cdot\cal{O}_{\mathbb{F}}(c,d)=$
$\displaystyle
ac\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(1,0)+(ad+bc)\cal{O}_{\mathbb{F}}(1,0)\cdot\cal{O}_{\mathbb{F}}(0,1)$
$\displaystyle+bd\cal{O}_{\mathbb{F}}(0,1)\cdot\cal{O}_{\mathbb{F}}(0,1).$

Since $S^{\prime}\nsupseteq S$, the intersection $S\cap S^{\prime}$ is a curve
of bidegree $(ad+b(c+d),a(c+d)+bc)$ (see Lemma 2.2). Since $c>0$ and $d>0$,
then $\cal{O}_{\mathbb{F}}(c,d)$ is ample. Moreover, since $S$ is irreducible,
$A$ has bidegree $(n,n)$ and $A$ is not connected, then the intersection
contains some more component in addition to $A$. Hence, either $ad+bc+bd>n$
and $ac+ad+bc\geq n$ or $ad+bc+bd\geq n$ and $ac+ad+bc>n$. ∎

###  2.4. Non-collinear smooth conics

We now want to characterize the conics in $\cal{C}^{*}(n)$ in terms of
cohomology. We start showing that the vanishing of certain cohomology groups
implies that an element $A\in\cal{C}(n)$ lies in $\cal{C}^{*}(n)$.

######  Lemma 2.18.

Fix $d\geq 0$, $3\leq n\leq d+1$ and $A\in\cal{C}(n)$. If
$h^{1}(\cal{I}_{A}(1,d))=0$ then $A\in\cal{C}^{*}(n)$.

###### Proof.

Assume, by contradiction, that there exists a curve $L$ of bidegree $(1,0)$
such that $\\#(L\cap A)\geq 3$ and consider the exact sequence defining
$\cal{I}_{A}$:

(9) $0\to\cal{I}_{A}(1,d)\to\cal{O}_{\mathbb{F}}(1,d)\to\cal{O}_{A}(1,d)\to
0.$

We will prove that the restriction map $H^{0}(\cal{O}_{\mathbb{F}}(1,d))\to
H^{0}(\cal{O}_{A}(1,d))$ is not surjective, which implies that
$h^{1}(\cal{I}_{A}(1,d))>0$. In fact, assume that it is surjective. Then,
consider the following diagram

$\begindc{\commdiag}\obj(50,25)[SA]{\cal{O}_{\mathbb{F}}(1,d)}\obj(50,10)[CP]{\cal{O}_{L}(1,d)}\obj(75,25)[H]{\cal{O}_{A}(1,d)}\obj(75,10)[T]{\cal{O}_{L\cap
A}(1,d)}\mor{SA}{CP}{}[\atright,\solidarrow]\mor{SA}{H}{}[\atleft,\solidarrow]\mor{CP}{T}{}[\atright,\solidarrow]\mor{H}{T}{}[\atleft,\solidarrow]\enddc\,.$

As the vertical maps are surjective, then the map
$H^{0}(\cal{O}_{\mathbb{F}}(1,d))\to H^{0}(\cal{O}_{A\cap L}(1,d))$, induced
by the composition, is surjective too and hence of rank at least 3 (since
$L\cap A$ has cardinality at least $3$ and the irreducible components of $A$
are pairwise disjoint). On the other hand the restriction map
$H^{0}(\cal{O}_{\mathbb{F}}(1,d))\to H^{0}(\cal{O}_{L}(1,d))$ has rank $2$ and
this gives a contradiction. ∎

Before completing the characterization of the conics in $\cal{C}^{*}(d+1)$, we
expose a general construction that will be used in several following
discussions. Let $A\in\cal{C}(n)$ and let $C$ be any connected component of
$A$. Set $B:=A\setminus C$. Then, for any $a,b\geq 0$, if
$Y\in|\cal{O}_{C}(0,1)|$, we have the following residual exact sequence

$0\to\cal{I}_{Res_{Y}(A)}(a,b-1)\to\cal{I}_{A}(a,b)\to\cal{I}_{A\cap
Y,Y}(a,b)\to 0,$

but as $Res_{Y}(A)=B$ and $A\cap Y=(B\cap Y)\cup C$, we have

(10) $0\to\cal{I}_{B}(a,b-1)\to\cal{I}_{A}(a,b)\to\cal{I}_{(B\cap Y)\cup
C,Y}(a,b)\to 0.$

Clearly an analogous sequence can be written for $X\in|\cal{O}_{C}(1,0)|$.

######  Theorem 2.19.

Fix $d\geq 0$ and $A\in\cal{C}(d+1)$. We have $A\in\cal{C}^{*}(d+1)$ if and
only if $h^{1}(\cal{I}_{A}(1,d))=0$.

###### Proof.

By the previous lemma we need only to prove that $A\in\cal{C}^{*}(d+1)$
satisfies $h^{1}(\cal{I}_{A}(1,d))=0$. We use induction on $d\geq 0$. The case
$d=0$ is true by Lemma 2.12. Therefore, we may assume $d>0$ and use induction
on $d$. Let $C$ be a connected component of $A$, set $B:=A\setminus C$ and cal
$Y$ the only element of $|\cal{I}_{C}(0,1)|$. Consider the residual exact
sequence (10), with $a=1$ and $b=d$. Since $A\in\cal{C}(d+1)$, $C\cap
B=\emptyset$ and $B\cap Y$ is formed by $d$ different points, up to the
identification of $D$ with $F_{1}$ we have

$\cal{I}_{(B\cap Y)\cup C,Y}(1,d)\cong\cal{I}_{(B\cap Y)\cup
C,F_{1}}(1,d)(h+(d+1)f)\cong\cal{I}_{B\cap Y,F_{1}}(df)\,.$

Using (10) and induction we are left to prove that $h^{1}(\cal{I}_{B\cap
Y,F_{1}}(df))=0$ if and only if $A\in\cal{C}^{*}(d+1)$.

Consider now the following exact sequence

(11) $0\to\cal{I}_{B\cap Y,F_{1}}(df)\to\cal{O}_{F_{1}}(df)\to\cal{O}_{B\cap
Y}(df)\to 0.$

Since $h^{0}(\cal{O}_{F_{1}}(df))=d+1$ and $h^{0}(\cal{O}_{B\cap Y}(df))=d$,
we have $h^{1}(\cal{I}_{B\cap Y,F_{1}}(df))>0$ if and only if
$h^{0}(\cal{I}_{B\cap Y,F_{1}}(df))\geq 2$. The last inequality means that
there exist at least two different sets of $d$ fibers containing the set of
$d$ points $B\cap Y$. This is equivalent to the fact that there exists a fiber
$L\in|f|$ such that $\\#(B\cap L)\geq 2$. Since $L$ is a curve of bidegree
$(1,0)$ in $\mathbb{F}$, then $L\cdot Y=0$ in the intersection ring of
$\mathbb{F}$, therefore $L\subset Y$ and hence we get $L\cap C\neq\emptyset$.
Thus $\\#(L\cap A)\geq 3$, which means that $A\not\in\cal{C}^{*}(d+1)$. ∎

######  Corollary 2.20.

Fix $d\geq 0$, $0\leq n\leq d+1$ and $A\in\cal{C}^{*}(n)$. Then
$h^{1}(\cal{I}_{A}(1,d))=0$. In particular, for $n=0,1,2$ and for any
$A\in\cal{C}(n)$, we have

$h^{1}(\cal{I}_{A}(1,1))=0\quad\text{ and }\quad
h^{0}(\cal{I}_{A}(1,1))=8-3n.$

###### Proof.

By using [3, Remark 4.3], we easily get the first part of the statement. The
second one, follows from Formula (7). ∎

We point out that since $\cal{T}(n)$ is a Zariski dense of $\cal{C}(n)$, then
the characterization given by Theorem 2.19 also holds for the set
$\cal{T}^{*}(n)$.

######  Lemma 2.21.

Fix an integer $d\geq 0$ and $A\in\cal{C}^{*}(d+2)$. Then
$h^{1}(\cal{I}_{A}(1,d))\leq 1$.

###### Proof.

The lemma is true for $d=0$, because $h^{0}(\cal{I}_{A}(1,0))=0$ (see Remark
2.9). We assume $d>0$ and use induction on $d$. Let $C$ be a connected
component of $A$, set $B:=A\setminus C$ and call $Y$ the only element of
$|\cal{I}_{C}(0,1)|$. Consider the residual exact sequence (10), with $a=1$
and $b=d$. Since $A\in\cal{C}(d+2)$, $C\cap B=\emptyset$ and $B\cap Y$ is
formed by $d+1$ different points, up to the identification of $D$ with $F_{1}$
we have

$\cal{I}_{(B\cap Y)\cup C,Y}(1,d)\cong\cal{I}_{(B\cap Y)\cup
C,F_{1}}(1,d)(h+(d+1)f)\cong\cal{I}_{B\cap Y,F_{1}}(df)\,.$

Using (10) and induction we are left to prove that $h^{1}(\cal{I}_{B\cap
Y,F_{1}}(df))=0$ if $A\in\cal{C}^{*}(d+2)$.

Consider now the following exact sequence

(12) $0\to\cal{I}_{B\cap Y,F_{1}}(df)\to\cal{O}_{F_{1}}(df)\to\cal{O}_{B\cap
Y}(df)\to 0\,.$

Since $h^{0}(\cal{O}_{F_{1}}(df))=d+1$ and $h^{0}(\cal{O}_{B\cap Y}(df))=d+1$,
we have $h^{1}(\cal{I}_{B\cap Y,F_{1}})>0$ if and only if
$h^{0}(\cal{I}_{B\cap Y,F_{1}}(df))>0$. This is equivalent to the fact that
there exists a fiber $L\in|f|$ such that $\\#(B\cap L)\geq 2$. Since $L$ is a
curve of bidegree $(1,0)$ in $\mathbb{F}$, then $L\cdot Y=0$ in the
intersection ring of $\mathbb{F}$, therefore $L\subset Y$ and hence we get
$L\cap C\neq\emptyset$. Thus $\\#(L\cap A)\geq 3$, which means that
$A\not\in\cal{C}^{*}(d+2)$. ∎

As said in Remark 2.5, for any element $A\in\cal{C}(2)$, there is a unique
curve $L$ of bidegree $(1,0)$ and a unique $R$ of bidegree $(0,1)$ such that
both intersect the elements of $A$ at a point. As described in the following
result, it turns out that $A\cup L\cup R$ is the base locus of
$|\cal{I}_{A}(1,1)|$.

######  Proposition 2.22.

For any $A\in\cal{C}(2)$, we have that

  1. (1)

the general element in $|\cal{I}_{A}(1,1)|$ is integral;

  2. (2)

the base locus $\cal{B}$ of $|\cal{I}_{A}(1,1)|$ is $A\cup L\cup R$, where $L$
is and $R$ are the curves described in Remark 2.5.

###### Proof.

Since $A\in\cal{C}(2)$, by Corollary 2.20 we have $h^{1}(\cal{I}_{A}(1,1))=0$
and $h^{0}(\cal{I}_{A}(1,1))=2$. Call $C_{1}$ and $C_{2}$ the connected
components of $A$. Denote by $X_{i}$ the only element of
$|\cal{O}_{\mathbb{F}}(1,0)|$ containing $C_{i}$ and by $Y_{i}$ the only
element of $|\cal{O}_{\mathbb{F}}(0,1)|$ containing $C_{i}$. The surfaces
$X_{1}\cup Y_{2}$ and $X_{2}\cup Y_{1}$ are the only reducible elements of
$|\cal{I}_{A}(1,1)|$ and hence, the general element in $|\cal{I}_{A}(1,1)|$ is
irreducible and (1) is proved.

To prove (2) we analyze the base locus $\cal{B}$ of $|\cal{I}_{A}(1,1)|$. If
$S,S^{\prime}\in|\cal{I}_{A}(1,1)|$ are irreducible and $S\neq S^{\prime}$,
then the one-dimensional cycle $S\cap S^{\prime}$ has bidegree $(3,3)$ and it
contains $A$, which has bidegree $(2,2)$. Take $R:=X_{1}\cap X_{2}$ and
$L:=Y_{1}\cap Y_{2}$, where $X_{i}$ and $Y_{i}$ are the surfaces defined in
the first part of this proof. The curves $L$ and $R$ are exactly the ones
stated in Remark 2.5. In particular, $\\#(L\cap A)=\\#(R\cap A)=2$ and hence,
by Bézout and Remark 2.16, $L\cup R\subset\cal{B}$. Moreover, recall that the
reducible surfaces, $X_{1}\cup Y_{2}$ and $X_{2}\cup Y_{1}$ belong to
$|\cal{I}_{A}(1,1)|$ and their intersection $(X_{1}\cup Y_{2})\cap(X_{2}\cup
Y_{1})$ is $A\cup L\cup R$. Hence the base locus of $|\cal{I}_{A}(1,1)|$ is
exactly $\cal{B}=A\cup L\cup R$. ∎

######  Remark 2.23.

By generalizing the proof of Proposition 2.22 we can say something about the
base locus of $\cal{I}_{A}(1,d)$, for $A\in\cal{C}(n)$. Fix integers $d>0$ and
$n\geq 2$ and take any $A\in\cal{C}(n)$. Then, the base locus $\cal{B}$ of
$\cal{I}_{A}(1,d)$ contains all curves $L$ of bidegree $(1,0)$ such that
$\\#(L\cap A)\geq 2$. If $A\in\cal{C}^{*}(n)$, then there are exactly
$\binom{n}{2}$ such curves $L$ (the number of lines joining two points in a
set of $n$ general points). If $A\in\cal{T}(n)$, then $\\#(j(L)\cap A)\geq 2$
for all $L$ such that $\\#(L\cap A)\geq 2$, since $j(A)=A$.

##  3\. Surfaces of bidegree $(1,d)$

In this section we prove Theorems 1.1 and 1.2. In particular, we give several
results for the case of a surface of bidegree $(1,d)$. Later, in the following
section, we will specialize to the cases $d=2,3$. The case $d=1$ was studied
in many details in [4] and here we add a simple lemma useful for what follows.

######  Lemma 3.1.

For any $A\in\cal{T}^{*}(3)$, there is no integral surfaces of bidegree
$(1,1)$ containing $A$.

###### Proof.

Assume the existence of an integral surface $M$ of bidegree $(1,1)$ containing
$A\in\cal{T}^{*}(3)$. First of all, thanks to [4, Corollary 8.4], we have that
$M=j(M)$. Then, as explained at the beginning of Section 7.1 in [4], either
$M$ is smooth or reducible. But if $M$ is a smooth $j$-invariant surface of
bidegree $(1,1)$ containing $3$ twistor fibers, then it contains infinitely
many of them and these are parametrized by a circle (see [4, Theorem 7.2]).
However, smooth surfaces of bidegree $(1,1)$ can be seen as the blow-up of
$\mathbb{P}^{2}$ at three points both via $\pi_{1}$ and $\pi_{2}$. In
particular, up to unitary transformation it is possible to write $M$ as the
set
$\\{([p_{0}:p_{1}:p_{2}],[\ell_{0}:\ell_{1}:\ell_{2}])\in\mathbb{F}\,|\,p_{1}\ell_{1}+\lambda
p_{2}\ell_{2}\\}$, with $\lambda\in\mathbb{R}\setminus\\{0,1\\}$. In these
coordinates, $M$ contains
$\pi_{\mu}^{-1}([1:0:0]),\pi_{\mu}^{-1}([0:1:0]),\pi_{\mu}^{-1}([0:0:1])$, for
$\mu=1,2$ and, the family of twistor fibers $\pi^{-1}([q_{0}:q_{1}:q_{2}])$
defined by:

$\begin{cases}q_{0}=0\mbox{ and }|q_{1}|^{2}\lambda+|q_{2}|^{2}=0&\mbox{ if
}\lambda<0\,,\\\ q_{1}=0\mbox{ and
}|q_{2}|^{2}-|q_{0}|^{2}(\lambda-1)=0&\mbox{ if }\lambda>1\,,\\\ q_{2}=0\mbox{
and }|q_{1}|^{2}\lambda+|q_{0}|^{2}(\lambda-1)=0&\mbox{ if
}0<\lambda<1\,.\end{cases}$

Take for instance $\lambda<0$, then any twistor fiber in $M$ intersects the
line $L=\pi_{2}^{-1}([1:0:0])$ of bidegree $(1,0)$. An analogous consideration
holds if $0<\lambda<1$ or $\lambda>1$. Hence we get a contradiction. ∎

In the previous lemma we show that an integral $(1,1)$ surface cannot contain
three general twistor fibers. On the other hand if $M$ is a $(1,1)$ surface
containing a given $A\in\cal{T}(3)\setminus\cal{T}^{*}(3)$, then by [4,
Corollary 8.3] we have that $M$ is $j$-invariant, hence either it is smooth or
reducible. Moreover, if $M$ contains infinitely many twistor fibers, then all
of them intersect a bidegree $(1,0)$ curve $L$ and its associated $(0,1)$
curve $R=j(L)$.

######  Remark 3.2.

In [4, Section 8.1] we gave examples of bidegree $(1,1)$ smooth surfaces
containing exactly $0$, $1$ or $2$ twistor fibers.

######  Remark 3.3.

As a smooth surface of bidegree $(1,1)$ is a Del Pezzo surface of degree $6$,
then this is characterized either by the three bidegree $(1,0)$ curves that
contains or by the three bidegree $(0,1)$ curves that contains. In fact,
recall that these surfaces represent the blow-up of $\mathbb{P}^{2}$ at three
points with respect to either $\pi_{1}$ or $\pi_{2}$. We notice that if a
smooth surface of bidegree $(1,1)$ is $j$-invariant, then, this is uniquely
determined by three twistor fibers contained in it and not by the curves $L$
and $R=j(L)$ (of bidegree $(1,0)$ and $(0,1)$, respectively), which intersect
all the twistor fibers. In fact, in [4], we proved that these kind of surfaces
are uniquely determined by the “circle” defined by the infinite family of
twistor fibers contained in it and, such a “circle” is properly contained in
in the $(1,0)$ line $L$ which intersects all fibers.

Thanks to these first considerations about bidegree $(1,1)$ surfaces, we are
ready to give the proof of our first main theorem.

###### Proof of Theorem 1.1:.

Thanks to Remark 2.9 and Lemma 3.1, the result is true for $d=0$ and $d=1$.

Assume now that $d\geq 2$ and, by contradiction, that $S$ is an integral
$(1,d)$ surface containing $A\in\cal{T}^{*}(d+2)$. Call $C$ a connected
component of $A$ and set $B:=A\setminus C$. Call $Y$ the only (by Lemma 3.2)
element of $|\cal{I}_{C}(0,1)|$ and consider the following exact sequence
(which is a particular case of the one in Formula (10)):

(13) $0\to\cal{I}_{B}(1,d-1)\to\cal{I}_{A}(1,d)\to\cal{I}_{(B\cap Y)\cup
C,Y}(1,d)\to 0.$

From Formulæ (6) and (8), we have
$h^{0}(\cal{O}_{A}(1,d))=(d+2)^{2}=h^{0}(\cal{O}_{\mathbb{F}}(1,d))+1$.
Clearly, we have that $B\cap Y$ is formed by $d+1$ points, and, up to the
identification of $Y$ with $F_{1}$ given in Formula (5), as the curve $C$
corresponds to an element of type $h+f$ in $F_{1}$, we can write
$\cal{I}_{(B\cap Y)\cup C,Y}(1,d)\cong{\cal{I}_{(B\cap Y)\cup
C,F_{1}}(h+(d+1)f)\cong}\cal{I}_{B\cap Y,F_{1}}(df)$. Since
$A\in\cal{T}^{*}(d+2)$ and every element of $|f|$ meets $C$ (indeed
$(h+f)f=1$), the restriction to $B\cap Y$ of the ruling morphism
$D\to\mathbb{P}^{1}$ associated to $|f|$ is injective. Thus,
$h^{1}(Y,\cal{I}_{A\cap Y,Y}(1,d))=0$ and the exact sequence (13) gives
$h^{1}(\cal{I}_{B}(1,d-1))\geq h^{1}(\cal{I}_{A}(1,d))\geq 2$, where the last
is greater or equal than $2$ because $\chi(\cal{I}_{A}(1,d))=-1$ (see Formula
(7)), and we are assuming that $h^{0}(\cal{I}_{A}(1,d))\geq 1$. Thus, we also
have $h^{0}(\cal{I}_{B}(1,d-1))>0$.

Recall that $B\in\cal{T}^{*}(d+1)$ and hence, by the inductive assumption, $B$
is not contained in any integral $E\in|\cal{O}_{\mathbb{F}}(1,d-1)|$.
Therefore, thanks to Remarks 2.9 and 2.10, there must be an integral
$M\in|\cal{O}_{\mathbb{F}}(1,1)|$ containing at least $3$ connected components
of $B$, say $B^{\prime}\subset M$ with $B^{\prime}$ of bidegree $(3,3)$.
Hence, by Lemma 3.1, there is a curve $L$ of bidegree $(1,0)$ such that
$\\#(L\cap B^{\prime})=3$. Thus $A\notin\cal{T}^{*}(d+2)$, a contradiction. ∎

Having proved that an integral bidegree $(1,d)$ surface cannot contains $d+2$
(or more) non collinear twistor fibers, we now pass to prove that all the
other cases can actually arise. In particular, we prove in the following
result a stronger version of Theorem 1.2 in the case $n\leq d+1$.

######  Theorem 3.4.

Fix integers $d\geq 1$ and $0\leq n\leq{d+1}$. Then, for any
$A\in\cal{T}^{*}(n)$ there is an integral $S\in|\cal{O}_{\mathbb{F}}(1,d)|$
containing $A$. Moreover, the general $S\in|\cal{I}_{A}(1,d)|$ contains no
other twistor fibers.

###### Proof.

We use induction on the integer $d$. If $d=1$ the statement is true by Remark
3.2.

Assume now $d\geq 2$ and take an element $A\in\cal{T}^{*}(n)$. Since
$\cal{T}^{*}(n)$ is Zariski dense in $\cal{C}(n)$, $A$ has the bigraded
Hilbert function of a general element of $\cal{C}(n)$. Thus, thanks to
Corollary 2.20, we have that $h^{1}(\cal{I}_{A}(1,d))=0$ and, by Formula (7)

$h^{0}(\cal{I}_{A}(1,d))=(d+1)(d+3)-n(d+2)=:N_{n}+1.$

Fix a connected component $C$ of $A$ and set $B:=A\setminus C$. If $Y$ denotes
be the only $(0,1)$ surface containing $C$, Corollary 2.20 entails that
$h^{1}(\cal{I}_{B}(1,d-1))=0$ and, again by Formula 7

$h^{0}(\cal{I}_{B}(1,d-1))=d(d+2)-(n-1)(d+1).$

By the inductive assumption, we know that $|\cal{I}_{B}(1,d-1)|\neq\emptyset$
and a general $W\in|\cal{I}_{B}(1,d-1)|$ is irreducible. Thus $Y\cup
W\in|\cal{I}_{A}(1,d)|$ and $Y\cup W$ has $2$ irreducible components, one of
them having bidegree $(1,d-1)$.

Let us denote by $C_{1},\ldots,C_{n}$ the connected components of $A$, by
$B_{i}:=A\setminus C_{i}$ and by $Y_{i}$ the unique element in
$|\cal{I}_{C_{i}}(0,1)|$. The set of all the reducible surfaces $W\cup
Y_{i}\in|\cal{I}_{A}(1,d)|$, where $W\in|\cal{I}_{B_{i}}(1,d-1)|$, is the
union of $e$ projective spaces (one for each choice of $C_{i}$), each of them
of codimension $h^{0}(\cal{I}_{A}(1,d))-h^{0}(\cal{I}_{B}(1,d-1))=d+2-n>0$ in
$|\cal{I}_{A}(1,d)|{=\mathbb{P}^{N_{n}}}$ (in particular of codimension $1$ if
$n=d+1$). Therefore, they do not cover all $|\cal{I}_{A}(1,d)|$.

We now want to exclude other possible splittings. In particular, we consider
reducible surfaces of the form $W_{1}\cup D_{1}$ with $W_{1}$ integral,
$D_{1}$ possible reducible of bidegree $(0,x)$ for some $x\geq 2$ and hence
$W_{1}$ of bidegree $(1,d-x)$. Remark 2.10 shows that only irreducible
components of $D_{1}$ of bidegree $(0,1)$ may contain some component of $A$.
We obtain that the surface is of the form $W_{1}\cup D_{2}\cup D_{3}$ with
$W_{1}\cup D_{2}$ of bidegree $(1,d-1)$, but, as showed before, these kind of
surfaces do not cover all $|\cal{I}_{A}(1,d)|$, hence we have the thesis.

Now we prove that a general $S\in|\cal{I}_{A}(1,d)|$ contains no other twistor
fibers. We start by analyzing the case $d=2$ and discussing the cases
$n=1,2,3$ separately.

Assume $n=1$. Fix $C\in\cal{T}(1)$. Let $\cal{C}_{C}(1)$ denote the set of all
$B\in\cal{C}(1)$ such that $B\cap C=\emptyset$. Note that
$\cal{T}(1)\setminus\\{C\\}=\cal{C}_{C}(1)\cap\cal{T}(1)$. For any
$B\in\cal{C}_{C}(1)$ we have $h^{1}(\cal{I}_{C\cup B}(1,2))=0$ and hence
$h^{0}(\cal{I}_{C\cup B}(1,2))=h^{0}(\cal{I}_{C}(1,2))-4$. Let $\cal{X}_{C}$
be the set of all smooth and integral surfaces of bidegree $(1,2)$ containing
$C$. It is a non-empty Zariski open subset of $|\cal{I}_{C}(1,2)|$. For any
$B\in\cal{C}_{C}(1)$ the set of all $S\in\cal{X}_{C}$ containing $B$ has
complex codimension $4$ and hence real codimension $8$ as real manifolds.
Since $\cal{T}(1)$ has real dimension $4$, a general $S\in\cal{X}_{C}$
contains no other twistor fiber.

Let now $n=2$. Fix $A\in\cal{T}(2)$ and let $\cal{C}_{A}(1)$ denote the set of
all $B\in\cal{C}(1)$ such that $B\cap A=\emptyset$ and $B\cup
A\in\cal{C}^{*}(3)$. Note that $h^{1}(\cal{I}_{A\cup B}(1,2))=0$. So as in the
previous step, we get that a sufficiently general $S\in|\cal{I}_{A}(1,2)|$
contains no element of $\cal{T}_{A}(1)$. Assume that $S$ contains
$B\in\cal{T}(1)$ such that there is a curve of bidegree $(1,0)$ intersecting
each connected component of $A\cup B$. Note that $L$ is uniquely determined by
$A$. Let $\cal{C}(A,L)$ denote the set of all $B\in\cal{C}(1)$ such that
$B\cap A=\emptyset$ and $L$ meets $B$ and set
$\cal{T}(A,L):=\cal{C}(A,L)\cap\cal{T}(1)$. For any $o\in L\setminus(L\cap C)$
the set of all $B\in\cal{C}(A,L)$ containing $o$ is a non-empty family of
complex dimension $1$, while there is a unique twistor fiber containing $o$.
The set $\cal{C}(A,L)$ is a complex manifold of dimension $2$, while
$h^{1}(\cal{I}_{A\cup B}(1,2))=1$ and hence $h^{0}(\cal{I}_{A\cup
B}(1,2))=h^{0}(\cal{I}_{A}(1,2))-3$. Since $\dim\cal{C}(A,L)=2$, a general
$S\in|\cal{I}_{A}(1,2)|$ contains no element of $\cal{C}(A,L)$. Hence, there
is no twistor fiber $B$ such that $A\cup B\in\cal{T}^{*}(3)$.

Assume now that $n=3$. Start by considering a general $A\in\cal{T}^{*}(3)$.
Then we have $h^{1}(\cal{I}_{A}(1,2))=0$ and $h^{0}(\cal{I}_{A}(1,2))=3$. For
any $x\in\\{0,1,2,3\\}$ let $\cal{C}(A,x)$ denote the set of all
$C\in\cal{C}(1)$ such that $A\cap C=\emptyset$ and $h^{0}(\cal{I}_{A\cup
C}(1,2))=x$ and set $\cal{T}(A,x):=\cal{C}(A,x)\cap\cal{T}(1)$. For a general
$D\in\cal{C}(4)$ we have $h^{0}(\cal{I}_{D}(1,1))=0$ (but
$h^{1}(\cal{I}_{D}(1,1))=1$). Fix $C\in\cal{C}(1)$ such that $C\cap
A=\emptyset$, call $Y$ the only element of $|\cal{I}_{C}(0,1)|$. Since $C\cap
A=\emptyset$, no connected component of $A$ is contained in $Y$ and $Y\cap A$
is formed by $3$ points, all of them in $Y\setminus C$. We easily deal with
the case $x=0$ as any curve $C\in\cal{C}(A,0)$ is not contained in any element
of $|\cal{I}_{A}(1,2)|$. We have
$\cal{O}_{Y}(1,2)(-C)\cong\cal{O}_{F_{1}}(2f)$ and hence $C\in\cal{C}(A,0)$ if
no curve of bidegree $(1,0)$ $L\in|f|$ intersects $2$ of the components of $A$
(since $h^{1}(\cal{I}_{A}(1,1))=1$ the last statement is only an “if” and not
an “if and only if”). A necessary condition for being $C\in\cal{C}(A,2)$ is
that $L$ intersects all connected components of $A$, but this is excluded
because $A\in\cal{T}^{*}(3)$. Since $A\in\cal{T}^{*}(3)$ there are exactly $3$
curves $L_{1},L_{2},L_{3}$ of bidegree $(1,0)$ intersecting $2$ of the
connected components of $A$. We claim that a general $S\in|\cal{I}_{A}(1,2)|$
contains no $C\in\cal{C}(1)$ such that $C\cap A=\emptyset$. The family of
smooth conics which intersect $L_{i}$ has complex dimension $2$, while the
family of twistor fibers intersecting $L_{i}$ has real dimension $2$. As the
general $S\in|\cal{I}_{A}(1,2)|$ has only finitely many conics, it only has a
finite number of elements in $\cal{C}(A,1)$ and, for the general, none of them
is a twistor fiber.

We now pass to analyze the case $d\geq 3$. Assume the general surface of
$|\cal{I}_{A}(1,d)|$ contains the twistor fiber $C\nsubseteq A$. Thus $A\cap
C=\emptyset$. Set $A^{\prime}:=A\cup C$. Take $Y\in|\cal{O}_{Y}(0,1)|$
containing $C$ and consider the residual exact sequence

(14) $0\to\cal{I}_{A}(1,d-1)\to\cal{I}_{A^{\prime}}(1,d)\to\cal{I}_{(Y\cap
A)\cup C,Y}(1,d)\to 0.$

We have $\cal{I}_{C,Y}(1,d)\cong\cal{O}_{F_{1}}(df)$.

Assume first $n\leq d$. If $A^{\prime}\in\cal{T}^{*}(n+1)$, then
$h^{1}(\cal{I}_{A}^{\prime}(1,d))=0$ and hence
$h^{0}(\cal{I}_{A}^{\prime}(1,d))=h^{0}(\cal{I}_{A}(1,d))-d-2$. Since
$\dim\cal{C}(1)=4$, for $d\geq 3$ the general $S\in|\cal{I}_{A}(1,d)|$
contains no $C$ such that $A\cup C\in\cal{T}^{*}(n+1)$.

Now assume $A^{\prime}\notin\cal{T}^{*}(n+1)$. Thus there are connected
components $C^{\prime}$ and $C^{\prime\prime}$ of $A$ such that $C\cup
C^{\prime}\cup C^{\prime\prime}\notin\cal{T}^{*}(3)$, i.e. $C$ intersects the
unique line $L$ meeting $C^{\prime}$ and $C^{\prime\prime}$. Since $\dim L=1$,
to exclude this case it is sufficient to prove that
$h^{0}(\cal{I}_{B}(1,d))\leq h^{0}(\cal{I}_{A}(1,d))-2$, i.e.
$h^{0}(F_{1},\cal{I}_{A\cap Y}(df))\leq d$. But since
$h^{0}(\cal{O}_{F_{1}}(df))=d+1$, then $\cal{O}_{F_{1}}(df)$ is globally
generated and $A\cap Y\neq\emptyset$, therefore $h^{0}(F_{1},\cal{I}_{A\cap
Y}(df))\leq d$.

Assume now $n=d+1$ and that $A^{\prime}\in\cal{C}^{*}(d+2)$. By Lemma 2.21 we
have $h^{1}(\cal{I}_{A^{\prime}}(1,d))\leq 1$ and hence
$h^{0}(\cal{I}_{A^{\prime}}(1,d))\leq h^{0}(\cal{I}_{A}(1,d))-d-1$. Now assume
$A^{\prime}\notin\cal{C}^{*}(d+2)$. We need $h^{0}(F_{1},\cal{I}_{A\cap
Y}(df))\leq d-1$. Let $\pi:F_{1}\to\mathbb{P}^{1}$ denote the ruling of
$F_{1}$. Since $\cal{O}_{\mathbb{P}^{1}}(d)$ is very ample,
$h^{0}(F_{1},\cal{I}_{A\cap Y}(df))\leq d-1$ if and only if $\\#\pi(A\cap
Y)\geq 2$, which is true because $\\#(A\cap Y)=d+1\geq 3$ and (since
$A\in\cal{C}^{*}(d+1)$ no fiber $F$ of $\pi$ contains at least $3$ points of
$A$). The only remaining case is now $d=3$ and $n=4$, which can be dealt just
by adapting the previous argument for $d=2$ and $n=3$. ∎

Having proved Theorem 1.2 in the case $n\leq d+1$ for non collinear twistor
fibers, we focus now to the case $n=d+2$. In this setting we need some
preliminary results.

######  Lemma 3.5.

Fix $d>0$, $n\leq d+2$ and consider a general $A\in\cal{T}(n)$. Then
$h^{1}(\cal{I}_{A}(1,d))=0$.

###### Proof.

Since $\cal{T}(n)$ is Zariski dense in $\cal{C}(n)$, it is sufficient to prove
the statement for a general $A\in\cal{C}(n)$. Clearly, it is sufficient to
prove the case $n=d+2$. Take a connected component $C$ of $A$ and set
$B:=A\setminus C$ and $Y\in|\cal{I}_{C}(0,1)|$. Consider the residual exact
sequence of $Y$, as in Formula (13). Recall the correspondence $Y\cong F_{1}$
given in Formula (5) and let $\rho:Y\to\mathbb{P}^{1}$ denote its ruling.
Since $A$ is general $\\#\rho(A\cap Y)=d+1$ and hence
$h^{i}(F_{1},\cal{I}_{B\cap Y}(df))=0$, for $i=0,1$. Therefore
$h^{1}(\cal{I}_{A}(1,d))=0$. ∎

By Theorem 1.1, we already know that an integral $(1,d)$ surface may contain a
union of $d+2$ twistor fibers only if it belongs to
$\cal{T}(d+2)\setminus\cal{T}^{*}(d+2)$. Therefore, we introduce the following
notation for sets of disjoint smooth conics which are collinear. Given a curve
$L$ of bidegree $(1,0)$ and an integer $n>0$, let $\cal{C}(n,L)$ denote the
set of all $A\in\cal{C}(n)$ such that each connected component of $A$ meets
$L$. The set $\cal{C}(n,L)$ is isomorphic (as real algebraic variety) to the
set $\cal{S}(L,n)$ of all subsets of $L$ with cardinality $n$ and hence it is
irreducible. An analogous definition and observation can be done for a curve
$R$ of bidegree $(0,1)$ and, of course, for the family $\cal{T}$ instead of
$\cal{C}$.

######  Lemma 3.6.

Fix integers $n\geq 3$ and $d\geq 1$ and take any $A\in\cal{T}(n,L)$. Then,

$h^{1}(\cal{I}_{A}(1,d))\geq n-2+\max\\{0,n-(d+1)\\}.$

###### Proof.

Call $C_{1},\dots,C_{n}$ the connected component of $A$. Let $S_{i}\subset
C_{i}$ be any union of $d+2$ distinct points on each conic and
$S:=S_{1}\cup\cdots\cup S_{n}$. Since $C_{i}$ is a smooth rational curve, the
restriction map $H^{0}(\cal{O}_{C_{i}}(1,d))\to H^{0}(\cal{O}_{S_{i}}(1,d))$
is bijective. Thus, the restriction map $H^{0}(\cal{O}_{A}(1,d))\to
H^{0}(\cal{O}_{S}(1,d))$ is bijective and
$\chi(\cal{O}_{A}(1,d))=\chi(\cal{O}_{S}(1,d))$. Thus we get
$\chi(\cal{I}_{A}(1,d))=\chi(\cal{I}_{S}(1,d))$.

Since $A\in\cal{T}(n,L)$, we know that there exists $L$ of bidegree $(1,0)$
which intersects each conic in $A$ (and the $(0,1)$ curve $j(L)$ does the
same). Hence we can choose $S$ such that $n$ points are on $L$ and $n$ points
are on $j(L)$. In other words we assume $(A\cap L)\cup(A\cap j(L))\subseteq
S$. By using Bézout and the fact that the bidegree is $(1,d)$, we get that
$(n-2)+\max\\{0,n-(d+1)\\}$ of these points may be omitted without changing
the set $|\cal{I}_{S}(1,d)|$, and hence $H^{0}(\cal{I}_{S}(1,d))$. It follows
that $h^{1}(\cal{I}_{A}(1,d))\geq n-2+\max\\{0,n-(d+1)\\}$. ∎

######  Remark 3.7.

Thanks to the previous lemma, if $A\in\cal{T}(3,L)$, for some bidegree $(1,0)$
curve $L$, then $h^{1}(\cal{I}_{A}(1,1))\geq 2$, and hence
$h^{0}(\cal{I}_{A}(1,1))\geq 1$. However, since no surface of bidegree $(1,0)$
or $(0,1)$ contains an element of $\cal{T}(2)$, then every
$S\in|\cal{I}_{A}(1,1)|$ is irreducible. Therefore, Proposition 2.17 gives
that $|\cal{I}_{A}(1,1)|=\\{S\\}$ and so, thanks to Formula (7),
$h^{1}(\cal{I}_{A}(1,1))=2$.

Finally, the following result completes the proof of Theorem 1.2, in the case
of $d+2$ twistor fibers.

######  Theorem 3.8.

Fix an integer $d\geq 2$ and take a general $A\in\cal{T}(d+2,L)$. Then
$h^{0}(\cal{I}_{A}(1,d))\geq d$ and the general $S\in|\cal{I}_{A}(1,d)|$ is
integral.

###### Proof.

By applying Lemma 3.6 with $n=d+2$, we get $h^{1}(\cal{I}_{A}(1,d))\geq d+1$.
Since $\chi(\cal{I}_{A}(1,d))=(d+1)(d+3)-(d+2)^{2}=-1$, we get
$h^{0}(\cal{I}_{A}(1,d))\geq d$ and hence, $|\cal{I}_{A}(1,d)|\neq\emptyset$.

We now prove that a general element in $|\cal{I}_{A}(1,d)|$ is integral. Take
$S\in|\cal{I}_{A}(1,d)|$. Every surface of bidegree $(1,0)$ or $(0,1)$
contains at most one connected component of $A$. Therefore, $S$ cannot be the
union of a surface of bidegree $(1,0)$ and $d$ of bidegree $(0,1)$. No
integral surface of bidegree $(0,x)$, for $x\geq 2$, contains a twistor fiber.
If $d=2$, then $h^{0}(\cal{I}_{A}(1,2))\geq 2$. However, thanks to Remark 3.7,
for every choice of $C\in A$, there is only one $M\in|\cal{I}_{(A\setminus
C)}(1,1)|$. Since we have considered all the possible reducible elements of
$|\cal{I}_{A}(1,2)|$, we get the thesis.

Now assume $d>2$. Since $A$ has finitely many components, it is sufficient to
prove that for any $x\in\\{3,\dots,d-1\\}$, any union $E$ of $x$ connected
components of $A$ and any connected component $C$ of $A\setminus E$ we have

(15) $h^{0}(\cal{I}_{E}(1,x-2))<h^{0}(\cal{I}_{E\cup C}(1,x-1)),$

and then proceed as in the proof of Theorem 3.4.

Let $Y$ be the only element of $|\cal{O}_{\mathbb{F}}(0,1)|$ containing $C$.
The exact sequence in Formula (10) gives $h^{0}(\cal{I}_{E}(1,x-2))\leq
h^{0}(\cal{I}_{E\cup C}(1,x-1))$ and equality holds if and only if $Y$ is in
the base locus $\cal{B}$ of $|\cal{I}_{E\cup C}(1,x-1)|$. Call
$C_{1},\dots,C_{x}$ the components of $E$ and $Y_{i}$ the only surface of
bidegree $(0,1)$ containing $C_{i}$. By Remark 2.13 the irreducible surfaces
$Y,Y_{1},\dots,Y_{x}$ are all different one eachother. For a general $A$ we
get that each integer $h^{0}(\cal{I}_{E}(1,x-2))$ is the same for all union of
$x$ connected components of $A$. Thus if the inequality is false, then
$\cal{B}$ contains the surface $Y\cup Y_{1}\cup\cdots\cup Y_{x}$ of bidegree
$(0,x+1)$, which is a contradiction. ∎

##  4\. Surfaces of bidegree $(1,2)$ and $(1,3)$

In this section we specialize our study to the case of surfaces of bidegree
$(1,2)$ and $(1,3)$. In particular, we will prove Theorems 1.3, 1.4 and 1.5.

Recall, from Formula (7), that for any $A\in\cal{C}(n)$, we have
$\chi(\cal{I}_{A}(1,2))=15-4n,$ and hence, if $n\leq 3$, we get
$h^{0}(\cal{I}_{A}(1,2))>0$. We also recall that a general
$S\in|\cal{O}_{\mathbb{F}}(1,2)|$ contains finitely many smooth conics and,
thanks to Theorem 2.19, for every $B\in\cal{C}(2)$ we have
$h^{1}(\cal{I}_{B}(1,1))=0$.

###  4.1. Surfaces of bidegree $(1,2)$ containing $0\leq n\leq 4$ twistor
fibers

In this section, we show the existence of a smooth surface of bidegree $(1,2)$
containing exactly $0,1,2,3$ or $4$ twistor fibers. In order to analyze the
space $|\cal{I}_{A}(1,2)|$ when $A$ is in $\cal{C}(n)$ (or in $\cal{T}(n)$),
for $0\leq n\leq 4$, we will need some preliminary results. Note that the
extremal case, when $n=4$, will be treated in a different way.

We start by considering $(1,2)$-surfaces containing three disjoint smooth
conics.

######  Proposition 4.1.

Take $A\in\cal{C}(3)$ such that $h^{0}(\cal{I}_{A}(1,1))>0$. Then

  1. (1)

there exists a curve $L$ of bidegree $(1,0)$ and a curve $R$ of bidegree
$(0,1)$ such that $A\in\cal{C}(3,L)$ and $A\in\cal{C}(3,R)$

  2. (2)

there is an integral element in $|\cal{I}_{A}(1,2)|$;

  3. (3)

$h^{1}(\cal{I}_{A}(1,2))=1$;

  4. (4)

the base locus $\cal{B}$ of $|\cal{I}_{A}(1,2)|$ is $A\cup L\cup R$, where $L$
is and $R$ are the curves defined in (1).

###### Proof.

We start arguing as in Remark 3.7. Thanks to Remark 2.9, each surface of
bidegree $(1,0)$ or $(0,1)$ does not contain any element of $\cal{C}(2)$,
hence, as $A\in\cal{C}(3)$, any element in $|\cal{I}_{A}(1,1)|$ is
irreducible. Proposition 2.17 (with $a=b=c=d=1$) gives that
$h^{0}(\cal{I}_{A}(1,1))=1$ and hence we set $|\cal{I}_{A}(1,1)|=\\{M\\}$ and
by Formula (7) we compute $h^{1}(\cal{I}_{A}(1,1))=2$.

We prove now the first statement. Let $C$ be a connected component of $A$, set
$B:=A\setminus C$ and denote by $X$ the only element of $|\cal{I}_{C}(1,0)|$.
Performing the same construction that leads to Formula (10), we have

$0\to\cal{I}_{B}(0,1)\to\cal{I}_{A}(1,1)\to\cal{I}_{A\cap X,X}(1,1)\to 0.$

Since $h^{0}(\cal{I}_{B}(0,1))=0$ and $h^{0}(\cal{I}_{A}(1,1))=1$, the
previous residual exact sequence gives

(16) $h^{0}(\cal{I}_{A\cap X,X}(1,1))\geq 1.$

Thanks to Formula (4), we have that
$\cal{O}_{X}(1,1)\simeq\cal{O}_{F_{1}}(h+2f)$; moreover, recall from Remark
2.9 that $C$ is identified with an element of $|\cal{O}_{F_{1}}(h+f)|$. Since
$A\cap X$ is the union of $C$ and the two points $B\cap X$, there is a fiber
$L\in|f|$ of the ruling of $F_{1}$ containing $B\cap X$. Since $f(h+f)=1$, we
have that $L$ meets $C$. Thus $L$ meets each connected component of $A$.
Taking instead of $X$ the only element of $|\cal{I}_{C}(0,1)|$ we get the
existence of $R$.

We now prove (2). We will show that there is an integral element in
$|\cal{I}_{A}(1,2)|$ by showing that the possible reducible cases do not cover
the whole family. Remark 2.10 shows that $A$ is not contained in a surface of
bidegree $(1,2)$ with an irreducible component of bidegree $(0,2)$. By Remark
2.9, a bidegree $(0,1)$ or $(0,1)$ surface does not contain any element of
$\cal{C}(n)$, with $n\geq 2$. Hence, there are only finitely many elements of
$|\cal{O}_{\mathbb{F}}(1,2)|$ with at least $3$ irreducible components.

Since $h^{0}(\cal{O}_{\mathbb{F}}(0,1))=3$, the set of all reducible elements
of $|\cal{O}_{\mathbb{F}}(1,2)|$ with an irreducible component of bidegree
$(1,1)$ containing $A$ is isomorphic to $\mathbb{P}^{2}$. Thus, in order to
prove the existence of an integral element in $|\cal{I}_{A}(1,2)|$, it is
sufficient to prove that $h^{0}(\cal{I}_{A}(1,2))\geq 4$. But, using the exact
sequence (9), this is equivalent to prove that $h^{1}(\cal{I}_{A}(1,{2}))\geq
1$, and the last inequality is true, by Theorem 2.19 because $\\#(L\cap A)=3$.

To prove (3), i.e. $h^{1}(\cal{I}_{A}(1,2))=1$, it is sufficient to prove that
$h^{1}(\cal{I}_{A}(1,2))\leq 1$. As before, take a connected component $C$ of
$A$ and set $B=A\setminus C$. Let $Y$ be the only element of
$|\cal{I}_{C}(0,1)|$. In the identification (5) of $Y$ with $F_{1}$ we have

$\cal{I}_{(B\cap Y)\cup C,Y}(1,2)\cong\cal{I}_{(B\cap Y)\cup
C,F_{1}}(h+3f)\cong\cal{O}_{(B\cap Y),F_{1}}(2f).$

Since $\\#(B\cap Y)=2$ and $\cal{O}_{F_{1}}(2f)$ is globally generated,
$h^{1}(F_{1},\cal{I}_{B\cap Y,F_{1}}(2f))\leq 1$. We have
$\operatorname{Res}_{Y}(A)=B$. Thanks to Theorem 2.19, we have
$h^{1}(\cal{I}_{B}(1,1))=0$, and the residual exact sequence of $Y$

$0\to\cal{I}_{B}(1,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{(B\cap Y)\cup C,Y}(1,2)\to
0,$

gives $h^{1}(\cal{I}_{A}(1,2))\leq 1$.

Finally, we discuss the base locus of $|\cal{I}_{A}(1,2)|$ in order to prove
(4). First of all, for any surface $S\in|\cal{I}_{A}(1,2)|$, we clearly have
$A\subset S$. Moreover, as $\\#(L\cap A)=3$ and $\\#(R\cap A)=3$, then by
Bézout, both curves are contained in $S$: in fact, thanks to Remark 2.3 and
Formula (3), the general intersection between a curve of bidegree $(1,0)$ and
$S$ consists of one point while the intersection of a curve of bidegree
$(0,1)$ and $S$ consists of two points (see also Remark 2.16). Therefore,
$L\cup R\subset S$ and $A\cup L\cup R\subset\cal{B}$.

We now prove that $\cal{B}\subset A\cup L\cup R$. Fix
$p\in\cal{B}\setminus(A\cup L\cup R)$. Take a connected component $C_{i}$,
$i=1,2,3$, of $A$ and set $B_{i}:=A\setminus C_{i}$. Let $Y_{i}$ be the only
element of $|\cal{O}_{\mathbb{F}}(0,1)|$ containing $C_{i}$. By Proposition
2.22 $B_{i}\cup L\cup R$ is the base locus of $|\cal{I}_{B_{i}}(1,1)|$. Thus
there is $S_{i}\in|\cal{I}_{B_{i}}(1,1)|$ such that $p\notin S_{i}$. If
$p\notin Y_{i}$, then $p\notin\cal{B}$. Since $S_{1}\cap S_{2}\cap S_{3}=L\cup
R$, we may take $i\in\\{1,2,3\\}$ such that $p\notin Y_{i}$. Thus
$\cal{B}=A\cup L\cup R$. ∎

The following remark shows that if $A\in\cal{C}(3)$ satisfies the condition
(1) of Theorem 4.1, then the existence of a $(1,1)$-surface containing $A$ is
granted. In particular, there exists a $(1,1)$-surface containing any triplets
of collinear twistor fibers.

######  Remark 4.2.

Take $A\in\cal{C}(3)$ and assume the existence of curves $L$ of bidegree
$(1,0)$ and $R$ of bidegree $(0,1)$ intersecting each connected component of
$A$. By adapting the proof of Lemma 3.6, since $\\#(L\cap A)=3$ and $\\#(R\cap
A)=3$, we have that $h^{1}(\cal{I}_{A}(1,1))\geq 2$. Thus
$h^{0}(\cal{I}_{A}(1,1))\geq 1$ and $A$ satisfies the assumptions of
Proposition 4.1.

We can even be more specific and say that if $A\in\cal{C}(3)$ (with no
assumption on $L$ or $R$), then $h^{0}(\cal{I}_{A}(1,1))\leq 1$ and if
$|\cal{I}_{A}(1,1)|\neq\emptyset$, then the only element of
$|\cal{I}_{A}(1,1)|$ is integral. This is true because, thanks to Remark 2.9,
any reducible element of $|\cal{O}_{\mathbb{F}}(1,1)|$ contains at most $2$
disjoint smooth conics.

Note that if $A\in\cal{T}(3)$ and $L$ exists, then we may take $R:=j(L)$. Thus
if $A\in\cal{T}(3)$ to get $h^{0}(\cal{I}_{A}(1,1))>0$ it is sufficient to
assume $A\notin\cal{T}^{*}(3)$.

The following lemma is a sort of vice versa of the previous remark.

######  Lemma 4.3.

Take $A\in\cal{C}^{*}(3)$. Then $h^{0}(\cal{I}_{A}(1,1))=0$ and
$h^{1}(\cal{I}_{A}(1,1))=1$.

###### Proof.

If $A\in\cal{C}(3)$, thanks to Formula (7), $\chi(\cal{I}_{A}(1,1))=-1$. Hence
$h^{0}(\cal{I}_{A}(1,1))=0$ if and only if $h^{1}(\cal{I}_{A}(1,1))=1$. We
assume $h^{0}(\cal{I}_{A}(1,1))\neq 0$ and will prove that
$A\notin\cal{C}^{*}(3)$. Let $B\subset A$ be the union of $2$ connected
components of $A$ and set $C:=A\setminus B$. Let $L$ and $R$ be the curves
defined in Remark 2.5 for $B\in\cal{C}(2)$. Take any element
$D\in|\cal{I}_{B}(1,1)|$. Since $\\#(B\cap(L\cup R))=2$, $B\subset D$ and $D$
has bidegree $(1,1)$, then Bézout theorem implies $L\cup R\subset D$.

By Theorem 2.19 and Proposition 2.22, we have $h^{1}(\cal{I}_{B}(1,1))=0$,
$h^{0}(\cal{I}_{B}(1,1))=2$, and the general element $M$ in
$|\cal{I}_{B}(1,1)|$ is integral. Since $h^{0}(\cal{I}_{B}(1,1))=2$ and $M$ is
general, $C\nsubseteq M$. Consider the following residual exact sequence:

(17) $0\to\cal{I}_{C}\to\cal{I}_{A}(1,1)\to\cal{I}_{B\cup(M\cap C),M}(1,1)\to
0$

Since $M\in|\cal{I}_{B}(1,1)|$ and $h^{1}(\cal{O}_{\mathbb{F}})=0$, the exact
sequence

$0\to\cal{O}_{\mathbb{F}}\to\cal{I}_{B}(1,1)\to\cal{I}_{B,M}(1,1)\to 0$

gives $h^{0}(M,\cal{I}_{B,M}(1,1))=1$. Moreover, $h^{0}(\cal{I}_{C})=0$, and
so the sequence (17) and the assumption $h^{0}(\cal{I}_{A}(1,1))\geq 1$ imply
$h^{0}(M,\cal{I}_{B\cup(M\cap C),M}(1,1))\geq 1$. By Proposition 2.22, the
curve $A\cup L\cup R$ is the base locus of $|\cal{I}_{B}(1,1)|$ and hence the
base locus of $H^{0}(M,\cal{I}_{B,M}(1,1))$ is the curve $B\cup L\cup R$.
Since $B\cap C=\emptyset$, the degree $2$ scheme $C\cap M$ is contained in
$L\cup R$. To get $A\notin\cal{C}^{*}(3)$ we need to prove that $C\cap
L\neq\emptyset$. It is sufficient to observe that $\deg(C\cap T)\leq 1$ for
any curve $T$ of bidegree $(0,1)$. Indeed, this is true by Remark 2.3 and the
fact that $C$ is the intersection of a surface of bidegree $(1,0)$ and a
surface of bidegree $(0,1)$. ∎

We now discuss the case of $A\in\cal{C}(2)$ contained in a smooth bidegree
$(1,1)$ surface. In this case we will also prove smoothness for the general
element in $|\cal{I}_{A}(1,2)|$.

######  Proposition 4.4.

Take any $A\in\cal{C}(2)$ contained in a smooth element of
$|\cal{O}_{\mathbb{F}}(1,1)|$. Then we have:

  1. (1)

$h^{1}(\cal{I}_{A}(1,2))=0$ and $h^{0}(\cal{I}_{A}(1,2))=7$;

  2. (2)

the set $A\cup L$ is contained in the base locus $\cal{B}$ of
$|\cal{I}_{A}(1,2)|$, where $L$ is the bidegree $(1,0)$ curve described in
Remark 2.5;

  3. (3)

a general $S\in|\cal{I}_{A}(1,2)|$ is smooth.

###### Proof.

To prove (1) it is sufficient to apply Corollary 2.20 giving
$h^{1}(\cal{I}_{A}(1,2))=0$ and Formula (7), which entails
$h^{0}(\cal{I}_{A}(1,2))=7$.

We now pass to point (2). Take $L$ and $R$ as in Remark 2.5. Since
$\cal{O}_{\mathbb{F}}(0,1)$ is globally generated, $\cal{B}\subseteq A\cup
L\cup R$; moreover, thanks to Remark 2.16 we also have $A\cup
L\subseteq\cal{B}$.

We are left to prove (3). By Bertini’s theorem $\mathrm{Sing}(S)\subseteq
A\cup L\cup R$ for a general $S\in|\cal{I}_{A}(1,2)|$. Fix a smooth
$M\in|\cal{I}_{A}(1,1)|$. Take a general
$Y^{\prime}\in|\cal{O}_{\mathbb{F}}(0,1)|$. Since $Y^{\prime}$ is general
$L\cap Y^{\prime}=\emptyset$ (and hence it is not singular at any $p\in L$).
Thus, up to small deformation, we can say that $S$ (which is general) is
smooth in a neighborhood of $L$. We are left to exclude the case
$\mathrm{Sing}(S)\subseteq A\cup R$. Fix $p\in A\cup R$ and let $2p$ be the
$0$-dimensional scheme of $\mathbb{F}$ defined by the ideal
$\cal{I}^{2}_{p,\mathbb{F}}$. $S$ is singular at $p$ if and only if
$S\in|\cal{I}_{2p\cup A}(1,2)|$. To conclude our proof we need to prove that

$h^{0}(\cal{I}_{2p\cup A}(1,2))=h^{0}(\cal{I}_{A}(1,2))-2,$

for all $p\in(A\cup R)\setminus A\cap R$ and that, for $p\in A\cap R$,
$h^{0}(\cal{I}_{2p\cup A}(1,2))<h^{0}(\cal{I}_{A}(1,2))$. These two statements
give the thesis because $(A\cup R)\setminus A\cap R$ and $A\cap R$ are
$1$-dimensional and $0$-dimensional, respectively, and we are saying that the
set of bidegree $(1,2)$ surfaces containing $A$ and a singular points has
codimension $2$ in the first case and positive codimension in the second one.

Let us start by taking $p\in(A\cup R)\setminus A\cap R$. Since $p$ is a smooth
point of $A\cup R$, $\deg(2p\cap(A\cup R))=2$. Consider the exact sequence

(18) $0\to\cal{I}_{(A\cup R)\cup 2p}(1,2)\to\cal{I}_{A\cup
R}(1,2)\to\cal{I}_{A\cup R}\otimes\cal{O}_{2p}(1,2)\to 0.$

Since $\deg(2p)=4$ and $A$ is smooth, we have $h^{0}(\cal{I}_{A\cup
R}\otimes\cal{O}_{2p}(1,2))=2$ if $p\in A\cup R$ and
$h^{0}(\cal{I}_{A}\otimes\cal{O}_{2p}(1,2))=4$ if $p\in R$. Hence it is
sufficient to prove that

$h^{1}(\cal{I}_{(A\cup R)\cup 2p}(1,2))=0.$

First of all, assume that $p\in A\setminus R$. Let $C$ be the connected
component of $A$ containing $p$. Set the following notation $E:=A\setminus C$.
As $R$ is in the base locus of $\cal{I}_{A}(1,1)$ we have that
$h^{0}(\cal{I}_{A}(1,1))=h^{0}(\cal{I}_{A\cup R}(1,1))$ (see [3, proof of
Theorem 1.1]). Moreover, thanks to part (1) and to [3, Remark 4.3], we have
$h^{0}(\cal{I}_{A\cup
R}(1,1))=h^{0}(\cal{I}_{A}(1,1))=h^{0}(\cal{I}_{E}(1,1))-3$. Thus $p$ is not
in the base locus of $|\cal{I}_{E}(1,1)|$. Fix $M\in|\cal{I}_{E}(1,1)|$ such
that $p\notin S$. Let $Y$ be the surface of $|\cal{O}_{\mathbb{F}}(0,1)|$
containing $C$ and consider the residual exact sequence with respect to $Y$:

(19) $0\to\cal{I}_{E\cup p}(1,1)\to\cal{I}_{A\cup 2p}(1,2)\to\cal{I}_{{(E\cap
Y)}\cup C\cup(2p\cap Y),Y}(1,2)\to 0.$

Now we prove that

(20) $h^{1}(\cal{I}_{E\cup p}(1,1))=0.$

Recall that $A=E\cup C$ and $p\in C$, hence we have the exact sequence

$0\to\cal{I}_{A}(1,1)\to\cal{I}_{E\cup p}(1,1)\to\cal{I}_{p,C}(2)\to 0.$

Thanks to Theorem 2.19 we have that $h^{1}(\cal{I}_{A}(1,1))=0$; on the other
hand, since $C$ is a smooth rational curve, we have
$h^{1}(\cal{I}_{p,C}(2))=h^{1}(\cal{O}_{C}(1))=0$ and this proves (20).

In order to conclude it is sufficient to prove now that

(21) $h^{1}(\cal{I}_{(E\cap Y)\cup C\cup(2p\cap Y),Y}(1,2))=0.$

Note that $\cal{I}_{C,Y}(1,2)\cong\cal{O}_{F_{1}}(2f)\cong\cal{O}_{Y}(0,1),$
hence, by [3, Remark 2.11], we know that $\cal{I}_{C,Y}(1,2)$ is very ample.
Therefore we get $h^{1}(\cal{I}_{C\cup(2p\cap Y),Y}(1,2))=0$. Since $E\cap Y$
consists of a point we conclude that (21) holds.

Thus the exact sequence (19) gives $h^{1}(\cal{I}_{A\cup
2p,\mathbb{F}}(1,2))=0$, concluding the proof in the case $p\in A\setminus R$.

Fix $p\in R\setminus(A\cap R)$ and ecall that we need to prove that
$h^{0}(\cal{I}_{A\cup 2p}(1,2))=5$. Fix a general
$Y^{\prime}\in|\cal{I}_{p}(0,1)|$. Since $Y^{\prime}$ is general, $R\nsubseteq
Y^{\prime}$ (and also $L\nsubseteq Y^{\prime}$). Since $Y^{\prime}$ is smooth,
$Y^{\prime}\cap 2p=(2p,Y^{\prime})$ is a degree $3$ scheme and
$\operatorname{Res}_{Y^{\prime}}(2p)=\\{p\\}$. As $p\in R$, we have that
$h^{0}(\cal{I}_{A\cup\\{p\\}}(1,1))=h^{0}(\cal{I}_{A}(1,1))=2$. Thus, by the
residual exact sequence of $Y^{\prime}$ it is sufficient to prove that

$h^{0}(Y^{\prime},\cal{I}_{(A\cap Y^{\prime})\cup(2p,Y^{\prime})}(1,2))\leq
3.$

Since $\cal{O}_{Y^{\prime}}(1,2)$ is very ample, we have
$h^{0}(Y^{\prime},\cal{I}_{(2p,Y^{\prime})}(1,2))=h^{0}(Y^{\prime},\cal{O}_{Y^{\prime}}(1,2))-3=4$.
Thus it is sufficient to prove that $A\cap Y^{\prime}$ is not contained in the
base locus, $\cal{B}^{\prime}$, of $|\cal{O}_{(2p,Y^{\prime})}(1,2)|$. In the
identification between $Y^{\prime}$ and $F_{1}$ we have
$\cal{O}_{Y^{\prime}}(1,2)\cong\cal{O}_{F_{1}}(h+3f)$. Let $N$ be the only
element of $|f|$ containing $p$. We have $N\cong\mathbb{P}^{1}$ and
$h^{1}(N,\cal{I}_{2p\cap N}(1,2))=0$, but $N\subseteq\cal{B}^{\prime}$. Since
$\cal{O}_{F_{1}}(h+2f)$ is very ample, $\cal{I}_{p}(h+2f)$ has only $p$ in its
base locus. Thus $\cal{B}^{\prime}=N$ and so, since $R\nsubseteq Y^{\prime}$
(and also $L\nsubseteq Y^{\prime}$), $\cal{B}^{\prime}$ cannot contain both
points of $A\cap Y^{\prime}$.

The last case is $p\in A\cap R$. To prove our claim, i.e. that
$h^{0}(\cal{I}_{2p\cup A}(1,2))<h^{0}(\cal{I}_{A}(1,2))$, it is sufficient to
use (18). Hence, a general $S$ is smooth. ∎

The following result is analogous to Proposition 4.1, when we choose the
conics to be twistor fibers.

######  Proposition 4.5.

Take $A\in\cal{T}(3)$ such that $h^{0}(\cal{I}_{A}(1,1))=0$. Then we have the
following:

  1. (1)

$h^{1}(\cal{I}_{A}(1,2))=0$ and hence $h^{0}(\cal{I}_{A}(1,2))=3$;

  2. (2)

there is an integral $S\in|\cal{I}_{A}(1,2)|$;

  3. (3)

the base locus of $|\cal{I}_{A}(1,2)|$ is contained in the union of $A$ and
$3$ distinct curves of bidegree $(1,0)$;

  4. (4)

for a sufficiently general $A$ (contained in a dense euclidean open subset of
$\cal{T}(3)$), we may take a smooth $S\in|\cal{I}_{A}(1,2)|$.

Thanks to Lemma 3.1, the hypothesis $A\in\cal{T}(3)$ such that
$h^{0}(\cal{I}_{A}(1,1))=0$ in the previous statement, implies that the conics
in $A$ do not belong to any infinite family of twistor fibers contained in a
smooth $j$-invariant surface of bidegree $(1,1)$.

###### Proof.

We start by proving (1). Fix a connected component $C$ of $A$ and call $D$ the
only element of $|\cal{I}_{C}(0,1)|$. Set $B:=A\setminus C$. To get
$h^{1}(\cal{I}_{A}(1,2))=0$ mimicking the proof of Proposition 4.1 it is
sufficient to prove that $h^{1}(F_{1},\cal{I}_{B\cap D}(2f))=0$. Assume
$h^{1}(F_{1},\cal{I}_{B\cap D}(2f))>0$, i.e. assume the existence of
$T\in|\cal{O}_{F_{1}}(f)|$ containing the $2$ points $B\cap D$. Since
$C\in|\cal{O}_{F_{1}}(h+f)|$, $C\cap T\neq\emptyset$. Thus $T$ meets each
connected component of $A$. Remark 4.2 gives $h^{0}(\cal{I}_{A}(1,1))>0$, a
contradiction.

To prove (2) it is sufficient to show that the reducible cases do not cover
the whole $|\cal{I}_{A}(1,2)|$. In fact, reasoning as in the proof of
Proposition 4.1, the only possible splitting are of the form
$(1,0)+(0,1)+(0,1)$, which are in a finite number, or $(1,1)+(0,1)$, where the
bidegree $(1,1)$ component contains $2$ connected components of $A$ and the
remainder bidegree $(0,1)$ part is uniquely determined. Now,
$h^{0}(\cal{I}_{B}(1,1))=2$, so, the set of all reducible elements of
$|\cal{I}_{A}(1,2)|$ with an irreducible component of bidegree $(1,1)$ does
not cover $|\cal{I}_{A}(1,2)|$.

We now prove (3) and (4). Since $h^{0}(\cal{I}_{A}(1,1))=0$ and $A$ is
$j$-invariant, neither $\pi_{1}(A)$ nor $\pi_{2}(A)$ has a triple points (both
have $3$ double points). Set $L_{1}\cup L_{2}\cup
L_{3}:=\pi_{2}^{-1}(\mathrm{Sing}(\pi_{2}(A)))$ and $R_{1}\cup R_{2}\cup
R_{3}:=\pi_{1}^{-1}(\mathrm{Sing}(\pi_{1}(A)))$. Since $\\#(L_{i}\cap
A)=\\#(R_{i}\cap A)={2}$, $L_{1}\cup L_{2}\cup L_{3}$ are in the base locus of
$|\cal{I}_{A}(1,2)|$ and each $L_{i}$ and each $R_{i}$ meets exactly $2$
connected components of $A$.

To prove the existence of a smooth element, it is sufficient to reason as in
the proof of Proposition 4.4 case (3). ∎

We are now ready to prove the first part of Theorem 1.3.

######  Theorem 4.6.

Fix $n\in\\{0,1,2,3\\}$. There is a smooth $S\in|\cal{O}_{\mathbb{F}}(1,2)|$
containing exactly $n$ twistor fibers.

###### Proof.

A general $S\in|\cal{O}_{\mathbb{F}}(1,2)|$ contains only finitely many smooth
conics. Since the set of all twistor fibers has real codimension $4$ in the
space of all smooth conics, a general $S\in|\cal{O}_{\mathbb{F}}(1,2)|$
contains no twistor fiber.

Now we prove the case $n=1$. Fix a twistor fiber $C$ and take a general
$S\in|\cal{I}_{C}(1,2)|$. Assume that $S$ contains another twistor fiber, $E$.
We have $h^{1}(\cal{I}_{C}(1,2))=h^{1}(\cal{I}_{C\cup E}(1,2))=0$ (Theorem
2.19 and Remark 2.14). Thus $|\cal{I}_{C\cup E}(1,2)|$ is a $4$-codimensional
complex projective subspace of $|\cal{I}_{C}(1,2)|$ (this is explained by the
equality $h^{0}(\cal{I}_{C\cup E}(1,2))=h^{0}(\cal{I}_{C}(1,2))-4$ contained
in [3, proof of Theorem 1.1]). However $\cal{T}(1)$ is a real $4$-dimensional
space. So a general $S\in|\cal{I}_{C}(1,2)|$ contains no other twistor fiber.

Note that $C$ is the base locus of $|\cal{I}_{C}(1,2)|$. By Bertini theorem a
general $S\in|\cal{I}_{C}(1,2)|$ is smooth outside $C$. Fix $p\in C$ and let
$2p$ the closed subscheme of $\mathbb{F}$ with $(\cal{I}_{p})^{2}$ as its
ideal sheaf. Recall that $2p\subset S$ if and only if $p\in\mathrm{Sing}(S)$.
Since $\dim C=1$ to get that $S$ is smooth it is sufficient to prove that
$h^{0}(\cal{I}_{2p\cup C}(1,2))\leq h^{0}(\cal{I}_{C}(1,2))-2=9$. This follows
from the proof of Proposition 4.4 case (3).

The case $n=2$ is true by Proposition 4.4 with $\cal{T}(2)$ instead of
$\cal{C}(2)$.

The case $n=3$ is true by Proposition 4.5. ∎

In the remainder of the section, we will construct a smooth $(1,2)$-surface
containing $4$ twistor fibers. The following lemma, in the case $d=2$, says
that if an integral $(1,2)$-surface contains $4$ disjoint smooth conics, then
these conics are not general, because three of them must be collinear.

######  Lemma 4.7.

Let $d\geq 2$ and $A\in\cal{C}(d+2)$. If there is an integral
$S\in|\cal{O}_{\mathbb{F}}(1,d)|$, then $A\notin\cal{C}^{*}(d+2)$

###### Proof.

We prove the lemma by induction on $d$. We start with the case $d=2$. Assume
that $A\in\cal{C}^{*}(4)$, i.e. there is no union $B$ of $3$ of the connected
components of $A$ such that $\\#(L\cap B)=3$ for some curve $L$ of bidegree
$(1,0)$. Fix a connected component $C$ of $A$ and set $B:=A\setminus C$. Call
$Y$ the only element of $|\cal{I}_{C}(0,1)|$. Remark 2.9 gives
$\operatorname{Res}_{Y}(A)=B$. By assumption and Lemma 4.3,
$h^{0}(\cal{I}_{B}(1,1))=0$. Since $h^{0}(\cal{I}_{A}(1,2))\neq 0$, the
residual exact sequence

$0\to\cal{I}_{B}(1,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{A\cap Y,Y}(1,2)\to 0\,,$

gives $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,2))>0$ (otherwise
$|\cal{I}_{A}(1,2)|=\emptyset$). The scheme $A\cap Y$ is the union of $C$ and
the $3$ points $B\cap Y$. In the identification of $Y$ with $F_{1}$ the line
bundle $\cal{O}_{Y}(1,2)$ goes to the line bundle $\cal{O}_{F_{1}}(h+3f)$ and
$C$ goes to an element of $|h+f|$. Thus $h^{0}(F_{1},\cal{I}_{B\cap
Y,F_{1}}(2f))>0$. Hence at least $2$ of the $3$ points $B\cap Y$ are in the
same fiber $\hat{L}$ of the ruling $|f|$ of $F_{1}$. Since $\hat{L}\cap
C\neq\emptyset$, $\hat{L}$ is a curve of bidegree $(1,0)$ meeting at least $3$
connected components of $A$. Call $B^{\prime}$ a union of $3$ components of
$A$ intersecting $\hat{L}$. The curves $B^{\prime}$ and $\hat{L}$ give a
contradiction.

Assume now that the result is true for $d+1$. Notice that, as a byproduct of
the previous part, if $B\in\cal{C}^{*}(d+1)$, then
$h^{0}(\cal{I}_{B}(1,d-1))=0$.

Assume $A\in\cal{C}^{*}(d+2)$ and that there is an integral
$S\in|\cal{O}_{\mathbb{F}}(1,d)|$. Fix a connected component $C$ of $A$ and
set $B:=A\setminus C$. Take a surface $Y$ of bidegree $(0,1)$ containing $C$.
By means of the sequence in Formula (10), we either have
$h^{0}(\cal{I}_{B}(1,d-1))>0$ or $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,d))>0$. Since
$A\in\cal{C}^{*}(d+2)$, $B\in\cal{C}^{*}(d+1)$ and hence, thanks to the
inductive assumption, we have $h^{0}(\cal{I}_{B}(1,d-1))=0$. The scheme $A\cap
Y$ is the union of $C$ and the scheme $B\cap Y$ with $A\cap B\cap
Y=\emptyset$. Up to the identification of $Y$ and $F_{1}$ we have
$\cal{O}_{Y}(1,d)(-C)\cong\cal{O}_{F_{1}}(df)$. Since $Y$ has bidegree $(0,1)$
each connected component of $B$ is either contained in $Y$ or it intersects
transversely $Y$ at a unique point. By Remark 2.9, the set $B\cap Y$ is formed
by $d+1$ points. Thus $\cal{I}_{A\cap Y,Y}(1,d)\cong\cal{I}_{B\cap Y}(df)$. We
saw $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,d))>0$ and this is true if and only if
there are $u_{1},\dots u_{d+1}\in B\cap Y$ and $F\in|f|$ such that that
$u_{i}\neq u_{j}$, for $i\neq j$ and $\\{u_{1},\dots,u_{d+1}\\}\subset F$. The
set $F\cap C$ is a unique point, $o$, and $o\notin\\{u_{1},\dots,u_{d+1}\\}$,
because $B\cap C=\emptyset$. The curve $F$ has bidegree $(0,1)$ and hence
$A\notin\cal{C}^{*}(d+2)$, a contradiction. ∎

Thanks to the previous result, if an integral $(1,2)$-surface contains $4$
disjoint smooth conics, then these are in special position. We now show that
if these $4$ conics are twistor fibers, then their position is very special.
We begin by introducing the following notation. For $n\geq 4$, we denote by
$\cal{C}(n)^{-}$ the set of elements $A\in\cal{C}(n)$ for which there exists a
bidegree $(1,0)$ curve $L$ such that $A\in\cal{C}(n,L)$. The set
$\cal{C}(n)^{-}$ parametrizes the families of $n$ collinear disjoint smooth
conics. For $n\geq 4$ we also write
$\cal{T}(n)^{-}:=\cal{T}(n)\cap\cal{C}(n)^{-}$. The families $\cal{C}(n)^{-}$
and $\cal{T}(n)^{-}$ are Zariski closed in $\cal{C}(n)$ and $\cal{T}(n)$,
respectively.

The following lemma shows that if an integral $(1,2)$-surface contains $4$
twistor fibers, then they are all collinear.

######  Lemma 4.8.

Take an integral $S\in|\cal{O}_{\mathbb{F}}(1,2)|$ containing
$A\in\cal{T}(4)$. Then $A\in\cal{T}(4)^{-}$

###### Proof.

Assume the existence of an integral $S\in|\cal{O}_{\mathbb{F}}(1,2)|$
containing $A\in\cal{T}(4)$. By Lemma 4.7 there is a union $B$ of $3$ of the
connected components of $A$ such that $B\in\cal{T}(3)\setminus\cal{T}^{*}(3)$,
i.e., there exists a bidegree $(1,0)$ curve $L$, such that $B\in\cal{T}(3,L)$
and hence, thanks to Remark 3.7 $h^{0}(\cal{I}_{B}(1,1))>0$. However, the same
remark tells us that $h^{0}(\cal{I}_{B}(1,1))=1$, $h^{1}(\cal{I}_{B}(1,1))=2$
and that the only element $M$ of $|\cal{I}_{B}(1,1)|$ is integral.

As usual, set $C:=A\setminus B$. As in Remark 4.2 since $L$ of bidegree
$(1,0)$ meets each connected components of $B$, then $R:=j(L)$, of bidegree
$(0,1)$, do the same.

Thanks to Remark 2.16 we get $B\cup L\cup R\subset M$. Since $S$ and $M$ are
integral, thanks to Lemma 2.2, the one-dimensional scheme $S\cap M$ has
bidegree $(5,4)$. Since $B\cup L\cup R$ has bidegree $(4,4)$, then
$C\nsubseteq M$. Let $Y$ be only element of $|\cal{I}_{C}(0,1)|$. Since
$B\subset M\cup Y$, then $M\cup Y\in|\cal{I}_{A}(1,2)|$. Moreover, as $S$ is
irreducible, then $S\neq M\cup Y$, and hence $h^{0}(\cal{I}_{A}(1,2))\geq 2$,
i.e. $h^{1}(\cal{I}_{A}(1,2))\geq 3$. Since $h^{1}(\cal{I}_{B}(1,1))=2$, the
residual exact sequence

$0\to\cal{I}_{B}(1,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{A\cap Y,Y}(1,2)\to 0,$

gives $h^{1}(Y,\cal{I}_{A\cap Y,Y}(1,2))>0$. As in the proof of Lemma 4.7 we
obtain the following inequality $h^{1}(F_{1},\cal{I}_{B\cap Y}(2f))>0$, i.e.
there is a curve $\hat{L}\in|f|$ of bidegree $(1,0)$ intersecting at least $2$
of the connected components of $B$.

Call $B^{\prime}$ the union of $2$ of the connected components of $B$
intersecting $\hat{L}$. Since $\hat{L}\cap C\neq\emptyset$ and $B^{\prime}\cup
C$ is $j$-invariant, each connected component of $B^{\prime}\cup C$ meets
$j(\hat{L})$. Remark 4.2, Proposition 4.1 and Bézout imply the existence of an
integral surface $M^{\prime}$ of bidegree $(1,1)$ containing $B^{\prime}\cup
C\cup\hat{L}\cup j(\hat{L})$. Since $B^{\prime}\subset M^{\prime}$, $\hat{L}$
and $j(\hat{L})$ contain at least $2$ points of $M^{\prime}$, then
$B^{\prime}\cup\hat{L}\cup j(\hat{L})\subset M$. But by Remark 2.5 there is a
unique curve of bidegree $(1,0)$ intersecting two different smooth conics,
hence $\hat{L}=L$ and both $L$ and $j(L)$ intersect each connected component
of $B$. Thus $L$ intersects each connected component of $A$, i.e.
$A\in\cal{C}(4)^{-}$.∎

As a byproduct of the proof of the previous result, we get the following
lemma. It essentially says that there are infinitely many integral
$(1,2)$-surfaces containing $4$ collinear twistor fibers.

######  Lemma 4.9.

Take $A\in\cal{T}(4)^{-}$ and assume that $A$ is not contained in a surface of
bidegree $(1,1)$. Then $\dim|\cal{I}_{A}(1,2)|=1$ and $|\cal{I}_{A}(1,2)|$
contains exactly $4$ reducible elements of $|\cal{O}_{\mathbb{F}}(1,2)|$.

###### Proof.

Call $L$ the curve of bidegree $(1,0)$ intersecting each connected component
of $A$. Since each connected component of $A$ is $j$-invariant, $j(L)$
intersects each connected component of $A$. In the proof of Lemma 4.8 we
showed that $h^{0}(\cal{I}_{A}(1,2))\geq 2$. From the lines of that proof, it
is possible to derive that only $4$ elements in $|\cal{I}_{A}(1,2)|$ are
reducible and they are all obtained fixing a connected component $C$ of $A$
and taking the union of the unique surface $M_{C}$ of bidegree $(1,1)$
containing $A\setminus C$ and the unique surface $Y_{C}$ of bidegree $(0,1)$
containing $C$. To conclude the proof it is sufficient to prove that
$h^{0}(\cal{I}_{A}(1,2))\leq 2$. Take a connected component $C$ of $A$ and
consider the residual exact sequence

(22) $0\to\cal{I}_{C}(0,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{M_{C}\cap
A,M_{C}}(1,2)\to 0.$

We have $h^{0}(\cal{I}_{C}(0,1))=1$, because the intersection of $2$ different
elements of $|\cal{O}_{Y}(0,1)|$ is a curve of bidegree $(1,0)$. Thus by (22)
to conclude the proof it is sufficient to prove that the image $\cal{V}$ of
$H^{0}(\cal{I}_{A}(1,2))$ in $H^{0}(M_{C},\cal{I}_{M_{C}\cap A,M_{C}}(1,2))$
has dimension at most $1$. Bezóut gives that $A\cup L\cup j(L)$ is contained
in the base locus of $|\cal{I}_{B,M_{C}}(1,2)|$. Every $D\in|\cal{V}|$ has
bidegree $(5,4)$ as a curve of $\mathbb{F}$ and hence a general
$D\in|\cal{V}|$ is the union (counting multiplicities as divisors of the
smooth surface $M_{C}$) of $A\cup L\cup j(L)$ and a curve $E$ of bidegree
$(1,0))$ as a curve of $\mathbb{F}$. Recall that $M_{C}$ is the blow up of
$\mathbb{P}^{2}$ at $3$ non collinear points and that these $3$ exceptional
divisors are the only curve of $M_{C}$ with bidegree $(1,0)$. Since $M_{C}$
has only finitely many curves of bidegree $(1,0)$, $D$ is the same for all
non-zero elements of $\cal{V}$ and hence $\dim\cal{V}=1$. ∎

The following result completes the proof of Theorem 1.3.

######  Theorem 4.10.

There are integral $S\in|\cal{O}_{\mathbb{F}}(1,2)|$ containing exactly $4$
twistor fibers and for any such $S$ and $A\in\cal{T}(4)$ with $A\subset S$,
there is a curve $L$ of bidegree $(1,0)$ intersecting each connected component
of $A$. Moreover, $h^{0}(\cal{I}_{A}(1,2))=2$ and each
$S\in|\cal{I}_{A}(1,2)|$ is singular along $L$.

###### Proof.

By Theorem 1.4 no integral surface of bidegree $(1,2)$ contains at least $5$
twistor fiber. The curve $L$ exists by Lemma 4.8. Now we reverse the
construction. We start with $A\in\cal{T}(4,L)^{-}$. Let $2L$ denote the closed
subscheme of the “double line”. To prove that each $S\in|\cal{I}_{A}(1,2)|$ is
singular at each point of $L$ it is sufficient to prove that
$h^{0}(\cal{I}_{A}(1,2))=h^{0}(\cal{I}_{A\cup 2L}(1,2))$. Lemma 4.9 gives
$h^{0}(\cal{I}_{A}(1,2))=2$. Hence, it is sufficient to prove that
$h^{0}(\cal{I}_{A\cup 2L}(1,2))>1$. For any connected component $C$ of $A$ let
$M_{C}$ the only surface of bidegree $(1,1)$ containing $A\setminus C$ and let
$Y_{C}$ the only surface of bidegree $(0,1)$. Since $C\cap L\neq\emptyset$,
$L\cap Y_{C}\neq\emptyset$. Since $Y_{C}$ has bidegree $(0,1)$ and $L$
bidegree $(1,0)$, we get $L\subset Y_{C}$. Thus $L\subseteq M_{C}\cap Y_{C}$
and hence $|\cal{I}_{A\cup 2L}(1,2)|$ contains at least the $4$ reducible
elements of $|\cal{I}_{A}(1,2)|$. Hence $h^{0}(\cal{I}_{A\cup 2L}(1,2))>1$.∎

###  4.2. Non existence results for surfaces of bidegree $(1,2)$ and $(1,3)$

In this last part, we prove our two last main results, i.e. Theorems 1.4 and
1.5.

For any $A\in\cal{T}(n)^{-}$, $n\geq 4$, let us call $L$ and $R:=j(L)$ the
curves of bidegree $(1,0)$ e $(0,1)$, respectively, intersecting all the
connected components of $A$.

In view of our goal, we need to discuss the reducibility of some surfaces
containing a certain amount of twistor fibers. First of all, fix an integer
$n\geq 2$, take $B\in\cal{T}(4)$ such that $h^{0}(\cal{I}_{B}(1,1))>0$ and
call $M$ the unique (see e.g. Remark 4.2) surface of bidegree $(1,1)$
containing $B$. Since each element of $\cal{C}(1)$ is contained in an element
of $|\cal{O}_{\mathbb{F}}(0,1)|$ for each $E\in\cal{T}(n-1)$ there is a
reducible element $W\in|\cal{O}_{\mathbb{F}}(1,k)|$, union of $M$ and $n-1$
surfaces of bidegree $(0,1)$ such that $B\cup E\subset W$. The following lemma
is a sort of viceversa of this remark. Moreover, it will be a key tool in the
last two proofs.

######  Lemma 4.11.

If $d\geq 2$ and $A\in\cal{T}(d+3)^{-}$ is such that
$h^{0}(\cal{I}_{A}(1,d))>0$, then each element of $|\cal{I}_{A}(1,d)|$ has an
irreducible component $M$ of bidegree $(1,1)$ containing at least $4$
connected components of $A$. In particular, for any $n\geq d+3$, there is no
integral $S\in|\cal{O}_{\mathbb{F}}(1,d)|$ containing $A\in\cal{T}(n)^{-}$.

###### Proof.

In order to prove the last statement, it is sufficient to do the case $n=d+3$
and thus it is sufficient to prove the first assertion.

We use induction on $d\geq 2$. Let us assume first $d=2$. Take
$A\in\cal{T}(5)^{-}$ and let $L$ and $j(L)$ be the curves of bidegree $(1,0)$
and $(0,1)$ intersecting all the connected components of $A$. Fix a connected
component $C$ of $A$ and set $B:=A\setminus C$. Since $C\cap L\neq\emptyset$,
the curve $C\cup L$ is a connected and nodal curve of bidegree $(2,1)$ with
arithmetic genus $0$. Hence $h^{0}(\cal{O}_{C\cup L}(0,1))=2$. Thus there is
$Y\in|\cal{I}_{C\cup L}(0,1)|$ and such a $Y$ is unique. Since any two smooth
conics of $Y$ meet, no component of $B$ is contained in $Y$. Hence $B\cap Y$
is formed by $4$ points of $L\setminus(L\cap C)$. Recall that
$\cal{O}_{Y}(1,2)\cong\cal{O}_{F_{1}}(h+3f)$ and that
$C\in|\cal{O}_{F_{1}}(h+f)|$ and thus $\cal{I}_{A\cap Y,Y}\cong\cal{I}_{B\cap
L,Y}(2f)$. Since each element of $|f|$ contains a unique point of $L$ we have
that $h^{0}(D,\cal{I}_{A\cap Y,Y}(1,2))=0$. The residual exact sequence of $Y$

$0\to\cal{I}_{B}(1,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{A\cap Y,Y}(1,2)\to 0,$

gives an isomorphism $\varphi:H^{0}(\cal{I}_{B}(1,1))\to
H^{0}(\cal{I}_{A}(1,2))$. If $h^{0}(\cal{I}_{B}(1,1))=0$, then
$h^{0}(\cal{I}_{A}(1,2))=0$. Now assume $h^{0}(\cal{I}_{B}(1,1))\neq 0$. The
isomorphism $\varphi$ says that each $W\in|\cal{I}_{A}(1,2)|$ has $Y$ as an
irreducible component, say $W=Y\cup W_{1}$ with $W_{1}\in|\cal{I}_{B}(1,1)|$,
and hence we have the thesis.

Assume now $d\geq 3$ and use induction on $d$. By reasoning as in the base
case, take $A\in\cal{T}(d+3)^{-}$ and use the exact sequence

$0\to\cal{I}_{B}(1,d-1)\to\cal{I}_{A}(1,d)\to\cal{I}_{A\cap Y,Y}(1,d)\to 0,$

to prove that $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,d))=0$ and hence that there is an
isomorphism $\varphi:H^{0}(\cal{I}_{B}(1,d-1))\to H^{0}(\cal{I}_{A}(1,d))$.
Now, again, if $h^{0}(\cal{I}_{B}(1,d-1))=0$, then
$h^{0}(\cal{I}_{A}(1,d))=0$. Hence, assume $h^{0}(\cal{I}_{B}(1,d-1))\neq 0$.
The isomorphism $\varphi$ says that each $S\in|\cal{I}_{A}(1,d)|$ has $Y$ as
an irreducible component, i.e. $S=D\cup S_{1}$ with
$S_{1}\in|\cal{I}_{B}(1,d-1)|$. The inductive assumption says that $S_{1}$ has
an irreducible component $M$ of bidegree $(1,1)$ containing at least $4$
components of $B$. ∎

We now have all the ingredients to prove Theorems 1.4 and 1.5. First we prove
that no integral surface of bidegree $(1,2)$ contains 5 twistor fibers.

###### Proof of Theorem 1.4.

Assume the existence of an integral $S\in|\cal{O}_{\mathbb{F}}(1,2)|$
containing $A\in\cal{T}(5)$. Lemma 4.8 shows that for any union
$A^{\prime}\subset A$ of $4$ components of $A$ there is a union
$A^{\prime\prime}\subset A^{\prime}$ of $3$ connected components intersecting
some $L$ of bidegree $(1,0)$. Let $L$ be a curve of bidegree $(1,0)$
intersecting the maximal number, $z$, of components of $A$. Clearly $z\geq 3$.
By Lemma 4.11 to get a contradiction it is sufficient to prove that $z\geq 5$.

Assume $z\in\\{3,4\\}$. Take any ordering $C_{1},\dots,C_{5}$ of the connected
components of $A$ and set $B_{i}:=\pi_{1}(C_{i})$, $1\leq i\leq 5$. Each
$B_{i}$ is a line of $\mathbb{P}^{2}$. Since any two conics contained in an
element of $|\cal{O}_{\mathbb{F}}(1,0)|$ meet, $B_{1},\dots,B_{5}$ are $5$
different lines of $\mathbb{P}^{2}$. For any $i<j<h$, there is a curve $T$ of
bidegree $(1,0)$ intersecting $C_{i}$, $C_{j}$ and $C_{h}$ if and only if
$B_{h}$ contains the point $B_{i}\cap B_{j}$ and in this case
$L=\pi_{1}^{-1}(C_{i}\cap C_{j})$. With no loss of generality we may assume
that $L$ meets $C_{1},\dots,C_{z}$.

(a) Assume $z=3$ and hence $B_{1}\cap B_{2}\in B_{3}$. Applying Lemma 4.8 to
$C_{1}\cup C_{2}\cup C_{4}\cup C_{5}$ we have one of the following mutually
esclusive relations:

$\begin{matrix}B_{1}\cap B_{2}\cap B_{4}\neq\emptyset,&&B_{1}\cap B_{2}\cap
B_{5}\neq\emptyset,\\\ B_{1}\cap B_{4}\cap B_{5}\neq\emptyset,&&B_{2}\cap
B_{4}\cap B_{5}\neq\emptyset.\end{matrix}$

Since $B_{1}\cap B_{2}\in B_{3}$ and $z=3$, we can exclude the first two
cases, i.e. we have

$B_{1}\cap B_{2}\cap B_{4}=B_{1}\cap B_{2}\cap B_{5}=\emptyset.$

Thus, either $B_{1}\cap B_{4}\cap B_{5}\neq\emptyset$ or $B_{2}\cap B_{4}\cap
B_{5}\neq\emptyset$. Exchanging if necessary $C_{1}$ and $C_{2}$ we may assume
$B_{1}\cap B_{4}\cap B_{5}\neq\emptyset$, i.e. $B_{4}\cap B_{5}\in B_{1}$, and
hence $B_{2}\cap B_{4}\cap B_{5}=\emptyset$. Since $B_{2}\cap B_{4}\cap
B_{5}=\emptyset$, applying Lemma 4.8 to $C_{2}\cup C_{3}\cup C_{4}\cup C_{5}$
we have one of the following mutually esclusive relations

$\begin{matrix}B_{2}\cap B_{3}\cap B_{4}\neq\emptyset,&&B_{2}\cap B_{3}\cap
B_{5}\neq\emptyset,\\\ B_{3}\cap B_{4}\cap B_{5}\neq\emptyset.&&\end{matrix}$

Since $B_{4}\cap B_{5}\in B_{1}$ and $z=3$, $B_{3}\cap B_{4}\cap
B_{5}=\emptyset$. Since $B_{2}\cap B_{3}\in B_{1}$ and $z=3$, $B_{2}\cap
B_{3}\cap B_{5}=\ B_{2}\cap B_{3}\cap B_{4}=\emptyset$, a contradiction.

(b) Assume $z=4$. Since $C_{1}\cap L\neq\emptyset$, the curve $C_{1}\cup L$ is
a connected and nodal curve of arithmetic genus $0$ and bidegree $(2,1)$. Thus
$h^{0}(\cal{O}_{C_{1}\cup L}(0,1))=2$. Thus there is $Y\in|\cal{I}_{C_{1}\cup
L}(0,1)|$. Since $S$ is irreducible, $Y$ is not an irreducible component of
$S$. Thus the residual exact sequence of $Y$

$0\to\cal{I}_{B}(1,1)\to\cal{I}_{A}(1,2)\to\cal{I}_{A\cap Y,Y}(1,2)\to 0,$

gives $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,2))\neq 0$. Up to the isomorphism of $Y$
and $F_{1}$ we have $L=h$, $C_{1}\in|\cal{O}_{F_{1}}(h+f)|$ and
$\cal{O}_{Y}(1,2)\cong\cal{O}_{F_{1}}(h+3f)$. Since $(A\setminus C_{1})\cap
C_{1}=\emptyset$, $\cal{I}_{A\cap Y,Y}(1,3)\cong\cal{I}_{(A\setminus
C_{1})\cap D,D}(2f)$. Since $L$ meets $C_{1},\dots,C_{4}$, $(A\setminus
C_{1})\cap D$ contains a set $F\subset L$ such that $\\#F=3$. Since each
element of $|f|$ contains a unique point of $L$, $h^{0}(Y,\cal{I}_{A\setminus
C)\cap Y,Y}(2f))=0$, a contradiction. ∎

We now conclude our paper with the proof of Theorem 1.5 which concerns
surfaces of bidegree $(1,3)$.

###### Proof of Theorem 1.5:.

Assume the existence of $A\in\cal{T}(6)$ and of an integral
$S\in|\cal{O}_{\mathbb{F}}(1,3)|$ containing $A$. By Lemma 4.11 to get a
contradiction it is sufficient to prove the existence of a curve $L$ of
bidegree $(1,0)$ such that all the components of $A$ intersects $L$. By Lemma
4.7 for any union $A^{\prime}\subset A$ of $5$ components of $A$ there is a
union $A^{\prime\prime}\subset A^{\prime}$ of $3$ connected components
intersecting some $L$ of bidegree $(1,0)$. Let $L$ be a curve of bidegree
$(1,0)$ intersecting the maximal number, $z$, of components of $A$. We have
that $z\geq 3$. Hence, by Lemma 4.11 it is sufficient to prove that $z\geq 6$.
Assume then that $z\leq 5$. We will now exclude all the cases $z=3,4,5$.

For any connected component $C$ of $A$, Lemma 4.7 tells us that there is a
curve $L$ of bidegree $(1,0)$ intersecting at least $3$ connected components
of $A\setminus C$. In particular there is an integral
$M\in|\cal{O}_{\mathbb{F}}(1,1)|$ containing at least $3$ components of $A$
(see Remark 4.2). Note that $j(M)=M$. We may take $M$ with the additional
condition that it contains the maximal number $e$ of components of $A$. Let
$E$ be the union of the components of $A$ contained in $M$. Thus $3\leq e\leq
z\leq 5$.

Since each twistor fiber is $j$-invariant, $j(L)$ meets each connected
component of $E$. Bézout gives $L\cup j(L)\subset M$ and $L\subset S$. If
$e\geq 4$ Bézout gives $j(L)\subset S$. However, the one-dimensional cycle
$M\cap S$ has bidegree $(7,5)$ and thus $e\leq 4$. Set $\Sigma:=S\cap M$ (as a
scheme-theoretic intersection). Since the one-dimensional scheme $\Sigma$ is
the complete intersection of $\mathbb{F}$ with $2$ very ample divisors,
$h^{0}(\cal{O}_{\Sigma})=1$. Set $F:=A\setminus E$.

(a) Assume $e=4$. Hence $E\cup L\cup j(L)\subset\Sigma$. Since $E\cup L\cup
j(L)$ has bidegree $(5,5)$ and $h^{0}(\cal{O}_{\Sigma})=1$, $\Sigma$ is the
union of $E\cup L\cup j(L)$ and a multiple structure on $L$. Note that
$\Sigma\in|\cal{O}_{\mathbb{F}}(1,3)|$ and that $\Sigma$ contains $E\cup j(L)$
with multiplicity $1$ and $L$ with multiplicity $3$ (as divisors of the smooth
surface $M$). Since $\Sigma$ has multidegree $(7,5)$, $\Sigma=3L\cup j(L)\cup
E$. Note that $\Sigma$ contains the degree $4$ zero-dimensional scheme $F\cap
M$. Since $F\cap E=\emptyset$, $F\cap(j(L)\cup L)\neq\emptyset$. Thus at least
one irreducible component, $T$, of $F$ meets $L\cup j(L)$. Since $j(T)=T$,
$T\cap L\neq\emptyset$. Thus $z=5$. Let $C$ be a component of $E$. Since
$C\cap L\neq\emptyset$, $C\cup L$ is a connected and nodal curve of bidegree
$(2,1)$ with arithmetic genus $0$. Thus $h^{0}(\cal{O}_{C\cup L}(0,1))=2$.
Thus there is $Y\in|\cal{I}_{C\cup L}(0,1)|\neq 0$. Since $S$ is irreducible,
$Y$ is not an irreducible component of $S$. Thus the residual exact sequence
of $Y$ gives $h^{0}(Y,\cal{I}_{A\cap Y,Y}(1,3))\neq 0$. Up to the isomorphism
of $Y$ and $F_{1}$ we have $L=h$, $C\in|\cal{O}_{F_{1}}(h+f)|$ and
$\cal{O}_{Y}(1,3)\cong\cal{O}_{F_{1}}(h+4f)$. Since $(A\setminus C)\cap
C=\emptyset$, $\cal{I}_{A\cap Y,Y}(1,3)\cong\cal{I}_{(A\setminus C)\cap
Y,Y}(3f)$. Since $z=5$, $(A\setminus C)\cap Y$ contains a set $H\subset L$
such that $\\#H=4$. Since each element of $|f|$ contains a unique point of
$L$, $h^{0}(Y,\cal{I}_{A\setminus C)\cap Y,Y}(3f))=0$, a contradiction.

(b) Assume $e=3$. Fix a connected component $C$ of $E$ and set $B:=A\setminus
C$. Set $\\{Y\\}:=|\cal{I}_{C}(0,1)|$. As in step (a) we have that $L\subset
Y$. The following exact sequence

(23) $0\to\cal{I}_{B}(1,2)\to\cal{I}_{A}(1,3)\to\cal{I}_{C\cup(B\cap
Y),Y}(1,3)\to 0$

is the residual exact sequence of $Y$. Since $Y$ is not an irreducible
component of $S$, we have $h^{0}(Y,\cal{I}_{C\cup(B\cap Y),Y}(1,3))>0$. As in
step (a) we have $\cal{I}_{C\cup(B\cap Y),Y}(1,3)\cong\cal{I}_{B\cap
Y,F_{1}}(3f)$. We now have two possibilities: either
$h^{0}(\cal{I}_{B}(1,2))=0$ or $h^{0}(\cal{I}_{B}(1,2))>0$.

(b1) Assume for the moment $h^{0}(\cal{I}_{B}(1,2))=0$. Thus
$h^{0}(Y,\cal{I}_{C\cup(B\cap Y),Y}(1,3))\geq 2$ and
$h^{1}(Y,\cal{I}_{C\cup(B\cap Y),Y}(1,3))\geq 3$. Up to the identification of
$Y$ and $F_{1}$ we have $\cal{I}_{C,Y}(1,3)\cong\cal{O}_{F_{1}}(3f)$. Hence
the $5$ points $B\cap Y$ give at most one condition to the linear system
$|\cal{O}_{F_{1}}(3f)|$. Thus there is $J\in|\cal{O}_{F_{1}}(f)|$ such that
$B\cap Y\subset J$. Note that $J$ is a curve of bidegree $(1,0)$. The
maximality of the integer $e$ gives a contradiction.

(b2) Assume that $h^{0}(\cal{I}_{B}(1,2))>0$. By Theorem 1.4 any surface
containing B is reducible, say $M_{1}\cup Y$ with $M_{1}$ irreducible of
bidegree $(1,1)$ containing at least 4 components of B. Thus $e\geq 4$, a
contradiction. ∎

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