# The structure of exceptional sequences on toric varieties of Picard rank two

Klaus Altmann  Institut für Mathematik, FU Berlin, Königin-Luise-Str. 24-26,
D-14195 Berlin<EMAIL_ADDRESS>and  Frederik Witt  Fachbereich
Mathematik, U Stuttgart, Pfaffenwaldring 57, D-70569 Stuttgart
<EMAIL_ADDRESS>
###### Abstract.

For a smooth projective toric variety of Picard rank two we classify all
exceptional sequences of invertible sheaves which have maximal length. In
particular, we prove that unlike non-maximal sequences, they (a) remain
exceptional under lexicographical reordering (b) satisfy strong spatial
constraints in the Picard lattice (c) are full, that is, they generate the
derived category of the variety.

MSC 2020: 14F08, 14M25, 52C05; Key words: toric variety, derived category,
exceptional sequence

##  1\. Introduction

###  1.1. Fullness of exceptional sequences

Let $X$ be a smooth projective variety over an algebraically closed field $K$
and ${\mathcal{D}}(X)$ the bounded category of coherent sheaves on $X$. In his
ICM talk [Kuz14], Kuznetsov posed the following

Fullness Conjecture. If ${\mathcal{D}}(X)$ is generated by an exceptional
sequence, then any exceptional sequence of the same length is full.

Though a counterexample to Kuznetsov’s conjecture was recently given on a
rational surface, cf. the preprint [Kra23], this question is still of
interest. Namely, if the fullness property holds for $X$, then it implies the
absence of so-called phantom categories appearing in [GKMS15] and [BGKS15].

In this paper we shall address the question of fullness in the case of
exceptional sequences of line bundles on a toric variety of Picard rank $2$.
Let us first explain the general setting before we comment on our assumptions.
A sequence of elements ${\mathcal{E}}_{1},\ldots,{\mathcal{E}}_{N}$ in
${\mathcal{D}}(X)$ is called exceptional if if the derived homomorphisms
satisfy

$\operatorname{\mathbb{R}Hom}({\mathcal{E}}_{i},{\mathcal{E}}_{i})=K\hskip
8.00003pt\mbox{and}\hskip
8.00003pt\operatorname{\mathbb{R}Hom}({\mathcal{E}}_{j},{\mathcal{E}}_{i})=0\;\mbox{for
all }j>i.$

For instance, if the ${\mathcal{E}}_{i}$ are given by invertible sheaves and
$X$ satisfies $\mathbb{R}\Gamma(X,{\mathcal{O}}_{X})=K$, the first condition
holds automatically. Further, the second condition becomes
$\operatorname{Ext}^{\scriptscriptstyle\bullet}({\mathcal{E}}_{j},{\mathcal{E}}_{i})=0$.

An exceptional sequence is full if it generates the derived category. This
means that $\langle{\mathcal{E}}_{1},\ldots,{\mathcal{E}}_{N}\rangle$, the
smallest triangulated full subcategory of ${\mathcal{D}}(X)$ containing the
${\mathcal{E}}_{i}$, is ${\mathcal{D}}(X)$ itself. The length $N$ equals the
rank of the $K$-group $K_{0}(X)$. Any other exceptional sequence has at most
$N$ elements; it is called maximal if it attains this bound. In particular,
full sequences are maximal.

The simplest example is Beilinson’s full exceptional sequence on
$\mathbb{P}^{n}$ [Bei78], namely

${\mathcal{D}}(\mathbb{P}^{n})=\langle{\mathcal{O}}_{\mathbb{P}^{n}}(0),\ldots,{\mathcal{O}}_{\mathbb{P}^{n}}(n)\rangle.$

More generally, Kawamata proved existence of full exceptional sequences
consisting of complexes of coherent sheaves on any smooth projective toric
variety $X$ (or even a smooth toric DM-stack for that matter) [Kaw06, Kaw13,
Kaw16]. Note that the rank of $K_{0}(X)$ equals the number of $d$-dimensional
cones $\\#\Sigma(d)$ of the underlying fan $\Sigma$. Equivalently, this
coincides with the number of vertices of any polytope associated with an ample
divisor on $X$.

In view of Beilinson’s example one could even hope of finding full exceptional
sequences consisting of invertible sheaves ${\mathcal{L}}_{i}$ instead of
general complexes ${\mathcal{E}}_{i}$, but this fails even for toric Fano
varieties [Efi14]. However, existence of such sequences was established in
[CM04] if $\Sigma$ is a splitting fan. Geometrically, this means that the
toric variety $X=\mathbb{T}\mathbb{V}(\Sigma)$ arises as the total space of a
sequence of successive fibrations via
$X_{0}=\mathbb{P}^{n},X_{1},\ldots,X_{r}=X$ with $X_{i}=\mathbb{P}(E_{i-1})$
for a sum of invertible sheaves $E_{i-1}$ on $X_{i-1}$.

###  1.2. Exceptional sequences of line bundles on toric varieties of Picard
rank two

From now on we solely consider exceptional sequences of line bundles on smooth
projective toric varieties of Picard rank $2$, the easiest examples apart from
$\mathbb{P}^{n}$ for toric varieties $X(\Sigma)$ with splitting fan $\Sigma$.
The basic invariant of $X$ is

the pair $(\ell_{1},\ell_{2})$ of integers $\ell_{1}$, $\ell_{2}\geq 2$

which indicates that $X$ fibres over $\mathbb{P}^{\ell_{1}-1}$ with fibre
$\mathbb{P}^{\ell_{2}-1}$. Moreover, $d=\dim X=\ell_{1}+\ell_{2}-2$, and the
defining fan $\Sigma$ contains exactly two rays more than $d$; see Subsection
(3.1) for further details. In particular, $\\#\Sigma(d)=\ell_{1}\ell_{2}$. We
refer to the trivial fibration
$\mathbb{P}^{\ell_{1}-1}\times\mathbb{P}^{\ell_{2}-1}$ as the product case and
to a nontrivial fibration as the twisted case. For the latter, the order of
the two numbers $\ell_{1},\ell_{2}\geq 2$ really matters. In dimension two
where $\ell_{1}=\ell_{2}=2$, we just find the family of Hirzebruch surfaces
but the complexity quickly increases with dimension. The fibration structure
also implies that we have a canonical identification

$\operatorname{Pic}(X)\cong\mathbb{Z}^{2}$

given by the primitive generators of the nef cone, see Subsection (3.1.3).
Geometrically, these generators are given by the pullback of
${\mathcal{O}}_{\mathbb{P}^{\ell_{1}-1}}(1)$ and a relative hyperplane section
of the fibration.

Since
$\operatorname{Ext}^{\scriptscriptstyle\bullet}({\mathcal{L}}_{j},{\mathcal{L}}_{i})=\mathrm{H}^{\scriptscriptstyle\bullet}(X,{\mathcal{L}}_{j}^{-1}\otimes{\mathcal{L}}_{i})$
a sequence of line bundles ${\mathcal{L}}_{0},\ldots,{\mathcal{L}}_{N}$ is
exceptional if and only if ${\mathcal{L}}_{j}^{-1}\otimes{\mathcal{L}}_{i}$,
$i<j$, lies in the locus of cohomologically trivial line bundles inside the
Picard group. This locus is explicitly known for toric varieties given by a
splitting fan [ABKW20]. Second, we also understand the extensions provided by
nontrivial cohomology, cf. [AP20] and [AFH23].

###  1.3. Maximal exceptional sequences are lexicographic

Since properties of exceptional sequences such as fullness only depend on
their underlying set, it is natural to look for a canonical order. We call an
exceptional sequence vertically respectively horizontally orderable if it
remains exceptional for the lexicographic order on
$\operatorname{Pic}(X)\cong\mathbb{Z}^{2}$ where priority is given either to
the “vertical” or “horizontal” direction. In general, lexicographic reordering
destroys exceptionality of the sequence, but remarkably, this does not happen
for maximal exceptional sequences.

Theorem A [see Theorems 6.4 and 8.2]. Let $s\subseteq\mathbb{Z}^{2}$ be a
maximal exceptional sequence of invertible sheaves on a smooth projective
toric variety $X$ of Picard rank two. In the product case, $s$ is either
vertically or the horizontally orderable. In the twisted case, $s$ is
vertically orderable.

In contrast, it was shown in [AA22, Example 3.5] that there are maximal
exceptional sequences on
$\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ that are not
orderable with respect to any of the six possible lexicographic orders on
$\operatorname{Pic}(\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1})\cong\mathbb{Z}^{3}$.

###  1.4. Maximal exceptional sequences are densely packed

It is well-known that the shape of exceptional sequences also impacts on the
derived category, see for instance [KS20], [KS21] or [Mir21]. Our next
structure result concerns the spatial size of maximal exceptional sequences.

Theorem B [see Theorems 6.1 and 9.8]. Let
$s\subseteq\mathbb{Z}^{2}=\operatorname{Pic}(X)$ be a maximal exceptional
sequence of invertible sheaves on a projective toric variety $X$ of Picard
rank two. In the twisted case, the height, which is the minimal number of rows
of a horizontal strip containing $s$, is bounded by $2\ell_{2}$. In the
product case, either the height or the width (the minimal number of columns of
a vertical strip containing $s$) is bounded by $2\ell_{2}-1$ or $2\ell_{1}-1$,
respectively.

Again, it is easy to construct counterexamples for non maximal sequences. What
is striking about this result is that it is false for higher Picard rank. In
[AA22, Example 3.4] it was shown that maximal exceptional sequences on
$\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$ can spread
arbitrarily far and simultaneously in all three directions.

###  1.5. The classification of maximal exceptional sequences

Let

${\mathcal{R}}_{\ell_{1},\ell_{2}}:=\\{(a,b)\in\mathbb{Z}^{2}\mid 0\leq
a<\ell_{1},\,0\leq b<\ell_{2}\\}$

be the standard rectangle associated with the pair $(\ell_{1},\ell_{2})$. The
sequence given by

$s_{a+b\ell_{1}}:=(a,b)\in{\mathcal{R}}$

is maximal exceptional with respect to the vertical lexicographical order, and
so is any sequence obtained by a global shift, or by shifting the rows of
points $(a,b)$ by $(a_{b},0)$. We refer to these maximal exceptional sequences
as vertically trivial, see also Subsection (5.4). Similarly, there are also
horizontally trivial sequences in the product case. Composing the well-known
helix operator (e.g. [Rud90]) with lexicographical reordering yields the heLex
operator $\hslash$, see Subsection (7.1).

Theorem C [see Theorems 7.2 and 10.1]. Any vertically orderable maximal
exceptional sequence can be transformed into a vertically trivial maximal
exceptional sequence by applying $\hslash_{\mbox{\tiny lex}}$ at most
$\ell_{1}\ell_{2}$ times. Mutatis mutandis, the statement holds for
horizontally maximal exceptional sequences in the product case.

Theorem C can be recast into a more constructive version:

Theorem D [see Theorems 7.7 and 10.2]. Any maximal exceptional sequence is
determined by a so-called admissible set
$X\subseteq(-\beta,\ell_{2})+{\mathcal{R}}_{\ell_{1},\ell_{2}}$ which consists
either of horizontal or vertical lines of consecutive points.

The precise definition of admissible set is given in Definition 7.3 and 10.2.
In this way we can classify the totality of maximal exceptional sequences.

###  1.6. Fullness of maximal exceptional sequences

Finally, we show that maximality is sufficient for fullness. Viewing
exceptionality as an orthogonality condition in the derived category, this is
comparable to the fact that in a finitely generated vector space any linearly
independent set of maximal cardinality generates the space. This follows
either directly from Theorem A (admitting that the helix (!) operator
preserves fullness from the general theory), or from a combinatorial argument
building on Theorems A, B and D.

Theorem E [see Theorems 7.9 and 11.2. An exceptional sequence of invertible
sheaves on a smooth projective toric variety of Picard rank two is full if and
only if it is maximal.

For toric Fano varieties of Picard rank two and dimension less than five this
was shown in [Lee23] by direct calculations in a case-by-case analysis
(special cases were already considered in [LYY19]). On the other hand, [AA22]
required aid of a computer to prove the same result for
$\mathbb{P}^{1}\times\mathbb{P}^{1}\times\mathbb{P}^{1}$, the easiest example
of Picard rank three. On the other hand, our structural Theorems A-E yield a
completely general and conceptional treatment for Picard rank two. As a final
comment we note that our arguments are completely combinatorial (at the
expense of possible shortcuts, cf. for instance Remark 4.6). A more
geometrical approach is given in the sequel [AHW23].

###  1.7. The example $\mathbb{P}^{1}\times\mathbb{P}^{1}$

As an illustration of our theorems we consider
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ and show how we generate the whole
lattice $\operatorname{Pic}(X)=\mathbb{Z}^{2}$ out of the maximal exceptional
sequence

$s=(s_{0},\,s_{1},\,s_{2},\,s_{3},)=\big{(}0,\,(-3,1),\,(-2,1),\,(1,2)\big{)},$

see also (a) of Figure 1. It is vertically ordered and of height $3$ in
accordance with Theorems A and B (for horizontally ordered examples of
unbounded height see 4.2).

The main tool is the Beilinson sequence

$0\to{\mathcal{O}}_{\mathbb{P}^{1}}\to{\mathcal{O}}_{\mathbb{P}^{1}}(1)\oplus{\mathcal{O}}_{\mathbb{P}^{1}}(1)\to{\mathcal{O}}_{\mathbb{P}^{1}}(2)\to
0$

on $\mathbb{P}^{1}$. As we explain in Example 4.5 it allows us to generate or
“fill” the whole (horizontal or vertical) line whenever it carries two
consecutive points of $\langle s\rangle$.

(a)

(b)

(c)

(d)

(e)

Figure 1. Filling
$\operatorname{Pic}(\mathbb{P}^{1}\times\mathbb{P}^{1})=\mathbb{Z}^{2}$ from
$s$. The green point marks the origin.

Here and in the sequel, let $[x=a]$ and $[y=b]$ denote the lines $\\{(a,j)\mid
j\in\mathbb{Z}\\}$ and $\\{(i,b)\mid i\in\mathbb{Z}\\}$ in $\mathbb{Z}^{2}$.
We have a consecutive pair of elements in $s$ on $[y=1]$ so that we can
generate or “fill” the entire line $[y=1]$, cf. (b). Hence
$[y=1]\subseteq\langle s\rangle$.

Next, we fill the vertical line $[x=1]$ by using the consecutive pair $(1,1)$,
$(1,2)$ in (c) – we showed in (b) that $(1,1)\in\langle s\rangle$. Similarly,
we can fill the line $[x=0]$ since $(0,1)\in\langle s\rangle$, cf. (d).

It follows that we can fill any horizontal line $[y=b]$ for $(0,b)$,
$(1,b)\in\langle s\rangle$, $b\in\mathbb{Z}$, see (e). Hence
$\operatorname{Pic}(\mathbb{P}^{1}\times\mathbb{P}^{1})\subseteq\langle
s\rangle$. As $\operatorname{Pic}(\mathbb{P}^{1}\times\mathbb{P}^{1})$
generates ${\mathcal{D}}(\mathbb{P}^{1}\times\mathbb{P}^{1})$, $s$ is full.

The proof of Theorem E for
$\mathbb{P}^{\ell_{1}-1}\times\mathbb{P}^{\ell_{2}-1}$ is a direct
generalisation of this example. The key step consists in proving existence of
sufficiently many “horizontal” and “vertical” consecutive points inside the
sequence, see Section 6. The twisted case works differently. In particular, it
requires a suitable “vertical” Beilinson sequence which reflects the fine
structure of the twist, see Subsection (3.2).

###  1.8. Plan of the paper

The first Sections 2 to 5 provide the necessary background and establish the
needed technical tools. Though the product and the twisted case work for large
parts similarly and are equally technical we found it more perspicuous to
treat them separately with the product case serving as guideline.
Consequently, Sections 6 and 7 prove Theorems A-E for the product case, while
the remaining Sections 8 to 11 are devoted to proving the twisted versions.

##  2\. Some background on toric geometry

We briefly review some features of toric geometry which we shall use in the
paper. For a short introduction to toric geometry, see [Ful93].

###  2.1. Torus invariant divisors

Let $X=\mathbb{T}\mathbb{V}(\Sigma)$ be a smooth toric variety with underlying
fan $\Sigma$. The $r$-dimensional cones in $\Sigma$ form the subset
$\Sigma(r)$. These live in the real vector spaces
$N_{\mathbb{R}}=N\otimes_{\mathbb{Z}}\mathbb{R}$, where $N$ is the lattice of
one-parameter groups of rank equal to $d=\dim X$. It is dual to the character
lattice $M$ of $X$.

These lattices link to the group of torus invariant Weil divisors
$\mathrm{Div}_{T}(X)$ and the class group $\operatorname{Cl}(X)$ of $X$ as
follows. Any ray, that is, an element $\rho\in\Sigma(1)$, corresponds to a
unique torus orbit $\operatorname{orb}(\rho)$ of codimension one. We let
$D_{\rho}=\overline{\operatorname{orb}}(\rho)$ be its closure. For $m\in M$ we
define $\rho^{\ast}(m)=\operatorname{div}(\chi^{m})=\sum_{\rho}\langle
m,\rho\rangle D_{\rho}$, where $\rho$ denotes both the ray and its primitive
generator in $N$.

Moreover, let $\nabla$ be a lattice polytope in $M$ which is compatible with
$\Sigma$, i.e., $\Sigma$ is a refinement of the normal fan
$\mathcal{N}(\nabla)$ of $\nabla$. This comes with an associated Weil divisor

$D_{\nabla}=-\sum_{\rho}\min\langle\nabla,\rho\rangle\cdot D_{\rho}.$

The induced line bundle ${\mathcal{O}}(\nabla):={\mathcal{O}}(D_{\nabla})$ is
globally generated by the monomials $\chi^{m}$ with $m\in\nabla\cap M$.
Further, for any $\sigma\in\Sigma$ we have an associated vertex
$v(\sigma)\in\nabla\cap M$ which is characterised by

$\langle v(\sigma),\rho\rangle=\min\langle\nabla,\rho\rangle\hskip
6.99997pt\mbox{for}\hskip 6.99997pt\rho\in\sigma(1).$

It gives rise to a local generator of ${\mathcal{O}}(\nabla)$ on
$\mathbb{T}\mathbb{V}(\sigma)$, namely

${\mathcal{O}}(\nabla)|_{\mathbb{T}\mathbb{V}(\sigma)}=\chi^{v(\sigma)}\cdot{\mathcal{O}}_{\mathbb{T}\mathbb{V}(\sigma)}.$

Finally, if the primitive generators span $N_{\mathbb{R}}$ (for instance if
$X$ is complete), then we have the exact sequence

(1)
$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{\ast}}$$\textstyle{\Big{[}\mathrm{Div}_{T}(X)\cong\bigoplus\limits_{\rho\in\Sigma(1)}\mathbb{Z}D_{\rho}\Big{]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{\operatorname{Cl}(X)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

In particular, $\,\operatorname{rk}\operatorname{Cl}(X)=\\#\Sigma(1)-\dim X$.

###  2.2. Exact sequences reflecting polyhedral covers

We recall a method from [AFH23] to transform polyhedral inclusion / exclusion
sequences into exact sequences of sheaves on toric varieties
$X=\mathbb{T}\mathbb{V}(\Sigma)$.

We start with a so-called $\Sigma$-family of lattice polytopes
$S=\\{\nabla_{\\!i}\mid i=1,\ldots,n\\}$ in $M_{\mathbb{R}}$. This means that
for all $I\subseteq[n]:=\\{1,\ldots,n\\}$ the intersections

$\textstyle\nabla_{\\!I}:=\bigcap_{i\in I}\nabla_{\\!i}\mbox{ for
}I\neq\varnothing$

as well as

$\textstyle\nabla_{\varnothing}:=\bigcup_{i\in[n]}\nabla_{\\!i}$

are compatible with $\Sigma$, i.e., $\Sigma\leq\mathcal{N}(\nabla_{\\!I})$.

The second ingredient is the standard Koszul complex
$(\Lambda^{\scriptscriptstyle\bullet}K^{n},d)$ where

$d:\Lambda^{p+1}K^{n}\to\Lambda^{p}K^{n},\quad e_{I}\mapsto\sum_{i\in
I}(-1)^{i}e_{I\setminus\\{i\\}}$

for any $I\subseteq[n]$ with $\\#I=p+1$, and $e_{I}:=\wedge_{i\in I}e_{i}$ for
the standard basis vectors $e_{1},\ldots,e_{n}\in K^{n}$. Tensoring with
$K[M]$ yields the exact complex

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K[M]\cdot
e_{[n]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\oplus_{\begin{array}[]{c}\scriptstyle
I\subseteq[n]\\\ \scriptstyle\\#I=n-1\end{array}}K[M]\cdot
e_{I}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ldots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\oplus_{i\in[n]}K[M]\cdot
e_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K[M]\cdot
e_{\varnothing}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

For each $I\subseteq[n]$ the vector space $K[M]\cdot e_{I}$ appears as a
direct summand inside this complex and contains the finite-dimensional
subspace

$S(I):=K[\nabla_{\\!I}\cap M]\cdot e_{I}:=\oplus_{m\in\nabla_{\\!I}\cap
M}\,K\cdot\chi^{m}\cdot e_{I}.$

It follows from [AFH23, Section 3] that these subvector spaces define an exact
subcomplex $S^{\scriptscriptstyle\bullet}\subseteq
K[M]\otimes\Lambda^{\scriptscriptstyle\bullet}K^{n}$. Moreover, we have the

######  Theorem 2.1.

[AFH23, Theorem 12] $S^{\scriptscriptstyle\bullet}$ is the complex of global
sections of the equivariant, exact complex
${\mathcal{S}}^{\scriptscriptstyle\bullet}$ given by the globally generated
sheaves

${\mathcal{S}}^{k}=\bigoplus_{\scriptstyle
I\subseteq[n],\,\\#I=k}{\mathcal{O}}_{X}(\nabla_{\\!I})$

on $X=\mathbb{T}\mathbb{V}(\Sigma)$, namely

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{O}}_{X}(\nabla_{\\![n]})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\oplus_{i=1}^{n}{\mathcal{O}}_{X}(\nabla_{\\![n]\setminus\\{i\\}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ldots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\oplus_{i=1}^{n}{\mathcal{O}}_{X}(\nabla_{\\!i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{\mathcal{O}}_{X}(\nabla_{\\!\varnothing})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

######  Example 2.2.

Let $X={\mathcal{H}}_{1}$ be the first Hirzebruch surface, see also the
picture on the left hand side of Figure 8. We consider the $\Sigma$-family
$S=\\{\nabla_{\\!1},\,\nabla_{\\!2}\\}$ provided by the triangle and the
quadrangle in the middle box of the polyhedral exact sequence

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Here, the red dots indicate the position of the origin in each of the
polyhedra. Using the notation from Subsection (3.1.2), the leftmost polyhedron
$\nabla_{\\!1}\cap\nabla_{\\!2}$ equals $U$, and the triangle $\nabla_{\\!1}$
itself is just $V$. The sequence may be therefore translated into

$0\to{\mathcal{O}}_{{\mathcal{H}}_{1}}(U)\to{\mathcal{O}}_{{\mathcal{H}}_{1}}(V)\oplus{\mathcal{O}}_{{\mathcal{H}}_{1}}(U+V)\to{\mathcal{O}}_{{\mathcal{H}}_{1}}(2V)\to
0.$

###  2.3. Dealing with primitive collections

Next we apply the formalism of Subsection (2.2) and fix an arbitrary subset
$S\subseteq\Sigma(1)$. Let $n=\\#S$. To make contact with the previous
Subsection (2.2), we choose an order on $S$. Hence we may think of $S$ as a
sequence $\rho_{1},\ldots,\rho_{n}$ and identify $\rho_{i}\in S$ with
$i\in[n]$.

For subsets $I\subseteq S$ we define integral tuples
$k_{I}\in\mathbb{Z}^{\Sigma(1)}$ by

$(k_{I})_{\rho}:=\left\\{\begin{array}[]{ll}1&\mbox{if }\rho\in S\setminus
I\\\ 0&\mbox{if otherwise, i.e., }\rho\in I\cup(\Sigma(1)\setminus
S).\end{array}\right.$

Interpreting $k_{I}\in\mathbb{Z}^{\Sigma(1)}$ as a $T$-invariant, effective
and reduced Weil divisor on $X=\mathbb{T}\mathbb{V}(\Sigma)$ we denote by
${\mathcal{O}}_{X}(k_{I})$ the associated sheaf. Since $X$ is smooth, these
divisors are automatically Cartier. In particular, the sheaves
${\mathcal{O}}_{X}(k_{I})$ are invertible albeit not nef in general.

######  Proposition 2.3.

Assume that $S\subseteq\Sigma(1)$ is a non-face, that is, $S$ does not define
a cone in $\Sigma$. Then, with $I$ running through the subsets of $S$, the
following complex ${\mathcal{C}}_{S}^{\scriptscriptstyle\bullet}$ of
invertible sheaves with the usual Koszul-like differentials is exact:

$0\to{\mathcal{O}}_{X}(k_{S})\to\oplus_{\\#I=k-1}{\mathcal{O}}_{X}(k_{I})\to\ldots\to\oplus_{\\#I=1}{\mathcal{O}}_{X}(k_{I})\to{\mathcal{O}}_{X}(k_{\varnothing})\to
0.$

Here, ${\mathcal{O}}_{X}(k_{S})={\mathcal{O}}_{X}$ and
${\mathcal{O}}_{X}(k_{\varnothing})={\mathcal{O}}_{X}(\underline{1}_{S})$
which is the sheaf associated with the effective and reduced divisor
$\sum_{\rho\in S}D_{\rho}$.

###### Proof.

We choose a sufficiently ample polytope $\Delta$ such that all polytopes

$\nabla_{\\!I}:=\Delta(k_{I}):=\\{m\in M_{\mathbb{R}}\,|\;\langle
m,\rho\rangle\geq\min\langle\Delta,\rho\rangle-k_{I}\\}$

are at least nef. It follows immediately that

$\nabla_{\\!I}\cap\nabla_{\\!J}=\Delta(k_{I})\cap\Delta(k_{J})=\Delta(k_{I\cup
J})=\nabla_{\\!I\cup J}$

for all subsets $I$, $J\subseteq[n]$. In particular,
$\nabla_{\\!I}=\bigcap_{i\in I}\nabla_{\\!i}$ if $I\not=\varnothing$. We
assert that $\bigcup_{i\in[n]}\nabla_{\\!i}=\nabla_{\\!\varnothing}$ whence
$\\{\nabla_{\\!i}\\}$ is a $\Sigma$-family. This immediately implies the claim
of the proposition by tensoring the sequence in Theorem 2.1 with
${\mathcal{O}}(\Delta)^{-1}$.

We claim that for sufficiently ample $\Delta$,
$\bigcup_{i\in[n]}\Delta(k_{i})=\Delta(k_{\varnothing})$ if and only if $S$ is
a non-face.

Let $\bigcup_{i\in[n]}\Delta(k_{i})=\Delta(k_{\varnothing})$. Further, assume
that $S\subseteq\sigma(1)$ for some (smooth) and full-dimensional cone
$\sigma\in\Sigma$ with set of rays $\sigma(1)$. The vertex
$v_{\varnothing}(\sigma)$ of $\nabla_{\\!\varnothing}=\Delta(k_{\varnothing})$
associated with $\sigma$ satisfies

$\langle
v_{\varnothing}(\sigma),\rho\rangle=\left\\{\begin{array}[]{ll}\min\langle\Delta,\rho\rangle-1&\mbox{if
}\rho\in S\subseteq\sigma(1)\\\ \min\langle\Delta,\rho\rangle&\mbox{if
}\rho\in\sigma(1)\setminus S.\end{array}\right.$

However, this contradicts the inequality $\langle
v_{\varnothing}(\sigma),\rho_{i}\rangle\geq\min\langle\Delta,\rho_{i}\rangle$
of $\Delta(k_{i})$ for $\rho_{i}\in S$.

Conversely, assume that $S\notin\Sigma$. For each $\rho\in\Sigma(1)$ we
consider the associated facet

$\operatorname{face}(\Delta,\rho):=\\{r\in\Delta\,|\;\langle
r,\rho\rangle=\min\langle\Delta,\rho\rangle\\}$

and define the “thickened $\rho$-facet” by

$F(\Delta,\rho):=\\{r\in\Delta\,|\;\langle
r,\rho\rangle\leq\min\langle\Delta,\rho\rangle+1\\}.$

More generally, these definitions work for all cones
$\sigma\in\Sigma\setminus\\{0\\}$: The face associated to $\sigma$ is

$\operatorname{face}(\Delta,\sigma):=\bigcap_{\rho\in\sigma(1)}\operatorname{face}(\Delta,\rho)$

and the corresponding “thickened $\sigma$-face” is

$F(\Delta,\sigma):=\bigcap_{\rho\in\sigma(1)}F(\Delta,\rho).$

The usual one-to-one correspondence between faces of $\Delta$ and cones of
$\Sigma=\mathcal{N}(\Delta)$ implies that
$\sigma,\sigma^{\prime}\in\Sigma\setminus\\{0\\}$ are not contained in some
common cone $\widetilde{\sigma}\in\Sigma$ if and only if the
$\operatorname{face}(\Delta,\sigma)$ and
$\operatorname{face}(\Delta,\sigma^{\prime})$ are disjoint. Therefore, the
thickened faces $F(\Delta,\sigma)$ and $F(\Delta,\sigma^{\prime})$ are also
disjoint for sufficiently ample $\Delta$.

Applying this to the polytope
$\nabla_{\\!\varnothing}=\Delta(k_{\varnothing})$ shows that for a non-face
$S\subseteq\Sigma(1)$ the corresponding thickened facets are disjoint, that
is, $\bigcap_{\rho\in
S}F\big{(}\nabla_{\\!\varnothing},\rho\big{)}=\varnothing$. On the other hand,

$\nabla_{\\!\varnothing}\;\setminus\;F\big{(}\nabla_{\\!\varnothing},\rho\big{)}\;\subseteq\;\nabla_{\\!\rho},$

implies

$\textstyle\bigcup_{\rho\in S}\nabla_{\\!\rho}\;\supseteq\;\bigcup_{\rho\in
S}\big{[}\nabla_{\\!\varnothing}\setminus
F(\nabla_{\\!\varnothing},\rho)\big{]}=\nabla_{\\!\varnothing}\setminus\bigcap_{\rho\in
S}F(\nabla_{\\!\varnothing},\rho)=\nabla_{\\!\varnothing}.\vspace{-4ex}$

∎

######  Remark 2.4.

(i) Though we we will not use this observation in our later arguments, we note
in passing that for a primitive collection $S$ the exact sequence of
Proposition 2.3 represents the unique extension arising from
$\operatorname{Ext}^{n-1}\big{(}{\mathcal{O}}_{X}(k_{\varnothing}),{\mathcal{O}}_{X}\big{)}=K$.

(ii) Proposition 2.3 is the homological counterpart to the multiplicative
Stanley-Reisner presentation of the equivariant K-theory ring of a smooth
toric variety, see [VV03].

##  3\. Toric varieties of Picard rank two

###  3.1. Kleinschmidt’s classification

Let $X=\mathbb{T}\mathbb{V}(\Sigma)$ be a complete and smooth toric variety of
dimension $d$ and Picard rank two. These varieties are characterised by the
following data [Kle88]:

  1. (i)

Natural numbers $\ell_{1},\ell_{2}\geq 2$ with $\ell_{1}+\ell_{2}=d+2$.

  2. (ii)

An integer vector $c\in\mathbb{Z}^{\ell_{2}}$ with nonpositive components

$0=c^{1}\geq c^{2}\geq\ldots\geq{c^{\ell_{2}}}.$

We write the corresponding variety $X=(\ell_{1},\ell_{2};c)$. Here are some
key properties.

####  3.1.1. The class map

The varieties $(\ell_{1},\ell_{2};c)$ arise as fibre bundles over
$\mathbb{P}^{\ell_{1}-1}$ with typical fibre $\mathbb{P}^{\ell_{2}-1}$. The
fibration is trivial (product case) if and only if $c=0$. Identifying
$\operatorname{Div}_{T}(\ell_{1},\ell_{2};c)$ with
$\mathbb{Z}^{\ell_{1}+\ell_{2}}$ we can rearrange this data in terms of the
$2\times(\ell_{1}+\ell_{2})$-matrix

$\pi_{c}:=\left(\begin{array}[]{@{}*{3}{r}|*{4}{r}@{}}1&\ldots&1&0&c^{2}&\ldots&{c^{\ell_{2}}}\\\
0&\ldots&0&1&1&\ldots&1\end{array}\right):\mathbb{Z}^{\ell_{1}+\ell_{2}}\rightarrow\rightarrow\mathbb{Z}^{2}.$

For $X=(\ell_{1},\ell_{2};c)$ this provides the map
$\pi:\mathrm{Div}_{T}(X)\to\operatorname{Cl}(X)$ in the exact sequence (1) on
page 1. In particular, $M\cong\ker\pi_{c}$ and
$N\cong\operatorname{coker}\pi^{\ast}_{c}$ where
$\pi^{\ast}_{c}:\mathbb{Z}^{2}\hookrightarrow\mathbb{Z}^{\ell_{1}+\ell_{2}}$
is the transpose.

####  3.1.2. The fan

Let $\\{e_{1},\ldots,e_{\ell_{1}},f_{1},\ldots,f_{\ell_{2}}\\}$ and
$\\{e^{1},\ldots,e^{\ell_{1}},f^{1},\ldots,f^{\ell_{2}}\\}$ be the mutually
dual bases of $\mathrm{Div}_{T}(\ell_{1},\ell_{2};c)$ and
$\mathrm{Div}_{T}(\ell_{1},\ell_{2};c)^{\ast}$, respectively. Under $\rho_{c}$
from the exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{Cl}(X)^{\ast}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\ast}_{c}}$$\textstyle{\Big{[}\mathrm{Div}_{T}(X)^{\ast}\cong\mathbb{Z}^{\Sigma(1)}\Big{]}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{c}}$$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

dual to (1) the latter vectors are mapped to
$\\{u^{1},\ldots,u^{\ell_{1}},v^{1},\ldots,v^{\ell_{2}}\\}$. These images
define the rays of the fan $\Sigma=\Sigma(\ell_{1},\ell_{2};c)$ generated by
the $d$-dimensional smooth cones

$\sigma_{ij}:=\langle\Sigma(1)\setminus\\{u^{i},v^{j}\\}\rangle,\quad
i=1,\ldots,\ell_{1}\mbox{ and }j=1,\ldots,\ell_{2}.$

In particular, $\\#\Sigma(d)=\ell_{1}\ell_{2}$ as stated before.

####  3.1.3. The nef divisors

Regarded as a map $\operatorname{Div}_{T}(X)\to\operatorname{Cl}(X)$,
$\pi_{c}$ sends the equivariant prime divisors
$\overline{\operatorname{orb}}(u^{i})$ and
$\overline{\operatorname{orb}}(v^{j})$ to their classes. The effective cone
$\operatorname{Eff}\subseteq\operatorname{Cl}(X)$ is generated by
$[\overline{\operatorname{orb}}(u^{1})]$ and
$[\overline{\operatorname{orb}}(v^{\ell_{2}})]$. The nef cone
$\operatorname{Nef}\subseteq\operatorname{Eff}$ is generated by
$[\overline{\operatorname{orb}}(u^{1})]$ and
$[\overline{\operatorname{orb}}(v^{1})]$. This provides a natural
identification ${\mathcal{L}}(X)\cong\mathbb{Z}^{2}$ by sending
$[\overline{\operatorname{orb}}(u^{1})]$ to $(1,0)$ and
$[\overline{\operatorname{orb}}(v^{1})]$ to $(0,1)$. In particular,
$[\overline{\operatorname{orb}}(v^{\ell_{2}})]$ is sent to
$({c^{\ell_{2}}},1)$. These classes are also represented by the lattice
polytopes

$U:=\Delta(u^{1})=\operatorname{conv}\\{e_{i}-e_{1}\mid
i=1,\ldots,\ell_{1}\\}$

and

$V:=\Delta(v^{1})=\operatorname{conv}\\{f_{j}-f_{1}-c^{j}e_{i}\mid
i=1,\ldots,\ell_{1};\,j=1,\ldots,\ell_{2}\\}$

in $M_{\mathbb{R}}\subseteq\mathbb{R}^{\ell_{1}+\ell_{2}}$. Note that $U$
equals the $(\ell_{1}-1)$-dimensional standard simplex $\Delta^{\ell_{1}-1}$
while $V$ can be understood as the Caley product

$V=(-c^{1}\cdot\Delta^{\ell_{1}-1})*\ldots*(-{c^{\ell_{2}}}\cdot\Delta^{\ell_{1}-1}).$

####  3.1.4. The anti-canonical divisor

The divisor class of $-K_{X}$ is

$[-K_{X}]=\pi_{c}(1,\ldots,1)=(\ell_{1}-\beta,\,\ell_{2})$

with

$\beta:=-\sum_{j}c^{j}\geq 0.$

Hence, $(\ell_{1},\ell_{2};c)$ is Fano if and only if $\,\beta<\ell_{1}$.

######  Remark 3.1.

Rather than the full vector $c$ the non-negative parameters

$\textstyle\alpha:=-{c^{\ell_{2}}}\hskip 10.00002pt\mbox{and}\hskip
10.00002pt\beta=-\sum_{j=1}^{\ell_{2}}c^{j}$

are the most important ones for our purposes. Almost by definition we get the
inequalities

(2) $0\leq\alpha\leq\beta\leq\alpha(\ell_{2}-1).$

The latter will be referred to as the basic inequality. For further
simplification we also introduce

${\gamma}:=\alpha\ell_{2}-\beta$

which satisfies $\gamma\geq\alpha\geq 0$.

###  3.2. Application of Subsection (2.3) to the situation of Picard rank two

Next we apply Proposition 2.3 to the toric varieties
$X=(\ell_{1},\ell_{2};c)$. From the description of their fan in Subsection
(3.1.2) we derive that there are exactly two primitive collections, namely

$p_{u}=\\{u^{1},\ldots,u^{\ell_{1}}\\}\hskip 6.99997pt\mbox{and}\hskip
6.99997ptp_{v}=\\{v^{1},\ldots,v^{\ell_{2}}\\}.$

######  Corollary 3.2.

The primitive collections $p_{u}$ and $p_{v}$ give rise to the “$U$-sequence”

$0\to{\mathcal{O}}_{X}\to\ldots\to{\mathcal{O}}_{X}(U)^{\otimes\ell_{1}}\to 0$

and the “$V$-sequence”

$0\to{\mathcal{O}}_{X}\to{\mathcal{F}}_{1}\to\ldots\to{\mathcal{F}}_{\ell_{2}-1}\to{\mathcal{O}}_{X}(U)^{\otimes(-\beta)}\otimes{\mathcal{O}}_{X}(V)^{\otimes\ell_{2}}\to
0,$

where the sheaves ${\mathcal{F}}_{k}$ are direct sums indexed by subsets
$I\subseteq\\{1,\ldots,\ell_{2}\\}$, namely

$\textstyle{\mathcal{F}}_{k}=\oplus_{\\#I=k\,}{\mathcal{O}}_{X}(c^{I}\,U+k\,V)\hskip
10.00002pt\mbox{with}\hskip 5.0ptc^{I}:=\sum_{i\in I}c^{i}\leq 0.$

######  Remark 3.3.

Setting ${\mathcal{F}}_{0}:={\mathcal{O}}_{X}$ and
${\mathcal{F}}_{\ell_{2}}:={\mathcal{O}}_{X}(-\beta U+\ell_{2}V)$ we can
extend this notation to $k\in\\{0,\ldots,\ell_{2}\\}$. Moreover, using the
identification $\operatorname{Pic}X\cong\mathbb{Z}^{2}$ with $U=(1,0)$ and
$V=(0,1)$ from Subsection (3.1.3) we usually write
${\mathcal{F}}_{k}=\oplus_{\\#I=k\,}{\mathcal{O}}_{X}(c^{I},k)$.

###### Proof.

We deal with the primitive collection $S=\\{v^{1},\ldots,v^{\ell_{2}}\\}$
giving rise to the $V$-sequence; the case of the $U$-sequence works similarly.

Proposition 2.3 implies that we obtain an exact sequence for
${\mathcal{F}}_{k}=\oplus_{\\#J=\ell_{2}-k\,}{\mathcal{O}}_{X}(k_{J})$ with
$k_{J}=\underline{1}_{S\setminus J}$. Renaming $I:=S\setminus J$ this becomes
${\mathcal{F}}_{k}=\oplus_{\\#I=k\,}{\mathcal{O}}_{X}(\underline{1}_{I})$. On
the other hand, assigning the Weil divisor $\underline{1}_{I}=\sum_{\rho\in
I}D_{\rho}$ to its class means applying $\pi_{c}$ from Subsection (3.1.1).
Therefore, $\underline{1}_{I}$ becomes $(c^{I},k)$ since
$\pi_{c}(f_{i})=(c^{i},1)$. ∎

######  Example 3.4.

One of the very first examples one comes across when computing the cohomology
of invertible sheaves on toric varieties is $\operatorname{Ext}^{1}(2V,U)=K$
on the Hirzebruch surface $X={\mathcal{H}}_{1}=(2,2;(0,1))$. This is
represented by the exact sequence from Example 2.2. After twisting with
${\mathcal{O}}(-U)$ this becomes the $V$-sequence of Corollary 3.2.

###  3.3. The immaculate locus

From now on we will work with toric varieties of the form
$X=(\ell_{1},\ell_{2};c)$ and identify $\operatorname{Pic}(X)$ with
$\mathbb{Z}^{2}$ via the map $\pi_{c}$ from Subsection (3.1.1).

As pointed out in the introduction the locus of invertible sheaves with
vanishing cohomology plays a crucial rôle in this paper. We call

$\operatorname{{\mathcal{I}}}(X):=\\{{\mathcal{L}}\in\operatorname{Pic}(X)\mid\operatorname{H}^{j}(X,{\mathcal{L}}^{-1})=0,\,j\in\mathbb{Z}_{\geq
0}\\}.$

the immaculate locus of $\operatorname{Pic}(X)$. This distinguished subset was
referred to as the negative immaculate locus in [ABKW20] and [AA22]. We
believe, however, that the present notation increases the readability and that
no confusion should arise. In passing we remark that here and in the sequel we
shall not distinguish between invertible sheaves and their isomorphism classes
in the Picard group.

Rewriting [ABKW20, Theorem VI.2] in terms of our version of the (negative)
immaculate locus, we can describe
$\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)=\mathcal{H}\cup\mathcal{P}$
as the union of the horizontal strip

$\mathcal{H}:=\\{(a,b)\in\mathbb{Z}^{2}\mid 0<b<\ell_{2}\\}$

and the parallelogram

$\displaystyle\mathcal{P}$
$\displaystyle\phantom{:}=\\{(a,b)\in\mathbb{Z}^{2}\mid-\beta<a<\ell_{1}\hskip
5.0pt\mbox{and}\hskip
5.0pt0<\langle(a,b),\,(1,\alpha)\rangle<\ell_{1}+{\gamma}\\}.$

The immaculate locus is point symmetric with respect to
$(\ell_{1}-\beta,\ell_{2})/2$. In particular, the origin is point symmetric to
the anti-canonical class $[-K]=(\ell_{1}-\beta,\ell_{2})$. Figure 2
illustrates the typical shape of the immaculate locus for $\ell_{1}=7$ and
$\ell_{2}=4$. The parallelogram is indicated by the shaded area; lattice
points in $\mathcal{H}$ are in grey while lattice points in $\mathcal{P}$ but
not in $\mathcal{H}$ are blue.

$\scriptstyle 0$$\scriptstyle-[K]=(\ell_{1},\,\ell_{2})$$\scriptstyle
1$$\scriptstyle\ell_{2}-1$$\scriptstyle\vdots$$\scriptstyle
0$$\scriptstyle\ell_{2}$$\scriptstyle
0$$\scriptstyle\ell_{1}$$\scriptstyle\ell_{1}-1$$\scriptstyle\ldots$$\scriptstyle
1$$\scriptstyle-K/2$

$\scriptstyle 0$$\scriptstyle-K/2$$\scriptstyle
1$$\scriptstyle\ell_{2}-1$$\scriptstyle\vdots$$\scriptstyle
0$$\scriptstyle\ell_{2}$$\scriptstyle-[K]=(\ell_{1}-\beta,\,\ell_{2})$$\scriptstyle
0$$\scriptstyle\ell_{1}$$\scriptstyle\ldots$$\scriptstyle 1$

Figure 2. The immaculate locus of $(7,4;0)$ (left hand side) and
$\big{(}7,4;(0,0,-1,-2)\big{)}$ (right hand side).

######  Remark 3.5.

  1. (i)

The sequel of this article entirely depends on the combinatorics of the
immaculate locus, not the underlying fan. The former depends solely on
$\ell_{1}$, $\ell_{2}$, $\alpha=-{c^{\ell_{2}}}$ and $\beta=-\sum_{j}c^{j}$.
The immaculate locus will be therefore written as
$\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$ (or simply
$\operatorname{{\mathcal{I}}}$ depending on the context).

  2. (ii)

The immaculate locus is horizontally integral convex, that is, if $s$,
$t\in\operatorname{{\mathcal{I}}}\cap\ell$ for any horizontal line
$\ell\subseteq\mathbb{Z}^{2}$, then the segment $[s,t]\subseteq\ell$ is
contained in $\operatorname{{\mathcal{I}}}$, too.

  3. (iii)

Since $[-K]=(\ell_{1}-\beta,\ell_{2})$ the anti-canonical class sits always to
the left of the line $[x=\ell_{1}+1]$. By Subsection (3.1.4) it sits to the
right of $[x=0]$ if and only if $(\ell_{1},\ell_{2};c)$ is Fano.

###  3.4. The associated lattice

The boundary points $(\ell_{1},0)$ and $(-\beta,\ell_{2})$ of the immaculate
locus $\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$, define
the associated lattice

$L:=\mathbb{Z}(\ell_{1},0)\oplus\mathbb{Z}(-\beta,\ell_{2}).$

We denote by ${\mathcal{T}}=\mathbb{Z}^{2}/L$ the associated quotient and by
$\Phi:\mathbb{Z}^{2}\to{\mathcal{T}}$ the projection map which sends $(a,b)$
to its class $[a,b]$ in ${\mathcal{T}}$, cf. also [AA22, Section 4].

######  Lemma 3.6.

We have
$L\cap\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)=\varnothing$.

###### Proof.

This is obvious for $c=0$ so assume that $c\not=0$.

Suppose we could pick $(a,b)\in L\cap\operatorname{{\mathcal{I}}}$ so that
$(a,b)=(n\ell_{1}-m\beta,m\ell_{2})\in\operatorname{{\mathcal{I}}}$ for some
$n$, $m\in\mathbb{Z}$. In particular, we necessarily have
$(a,b)\in\mathcal{P}$. If $m\leq 0$, this implies

$0\leq-m\alpha\ell_{2}<n\ell_{1}-m\beta<\ell_{1}$

and therefore

${-m{\gamma}=}m(\beta-\alpha\ell_{2})<n\ell_{1}<\ell_{1}.$

By the basic inequality (2) on page 2 we have $0\leq-m{\gamma}$ whence a
contradiction for both cases $n\geq 1$ and $n\leq 0$. Similarly, $m\geq 1$
implies

$0\leq(m-1)\beta<n\ell_{1}<\ell_{1}-(m-1){\gamma}\leq\ell_{1}$

leading again to a contradiction. ∎

##  4\. Exceptional sequences of invertible sheaves

###  4.1. The exceptionality condition

Recall from the introduction that a sequence
${\mathcal{L}}_{0},\ldots,{\mathcal{L}}_{N}$ of invertible sheaves on a
variety $X$ is said to be exceptional if all backward $\mathrm{Ext}$-groups
vanish. Equivalently,

$\operatorname{H}^{k}(X,{\mathcal{L}}_{i}\otimes{\mathcal{L}}_{j}^{-1})=0\mbox{
if }i<j.$

Consequently, we can rephrase the exceptionality condition as follows.
Denoting the isomorphism classes in $\operatorname{Pic}(X)$ by
$s_{i}:={\mathcal{L}}_{i}$, the sequence
$s_{0},s_{1},\ldots,s_{N}\in\operatorname{Pic}(X)$ is exceptional on $X$ if
and only if

$\overrightarrow{s_{i}\,s_{j}}=s_{j}-s_{i}\in\operatorname{{\mathcal{I}}}(X)$

for all $i<j$, or equivalently,

$s_{j}\in\bigcap_{i<j}\big{(}s_{i}+\operatorname{{\mathcal{I}}}(X)\big{)}\quad\mbox{for
all }j\geq 1.$

This condition persists under a simultaneous shift so that we may replace the
original sequence by
${\mathcal{L}}_{i}^{\prime}:={\mathcal{L}}_{i}\otimes{\mathcal{L}}_{0}^{-1}$.
Therefore, we may assume that ${\mathcal{L}}_{0}={\mathcal{O}}_{X}$ is trivial
whenever this is convenient. In particular, $s_{0}=0$ which implies
$s_{i}\in\operatorname{{\mathcal{I}}}(X)$ for all $i\geq 1$.

######  Example 4.1.

For $X=\mathbb{P}^{\ell-1}$, Serre duality and the well-known vanishing
theorems for invertible sheaves on projective space yield

$\operatorname{{\mathcal{I}}}(\mathbb{P}^{\ell-1})=\\{{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(1),\ldots,{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(\ell-1)\\}\mathop{\widehat{=}}\,\\{1,2,\ldots,\ell-1\\}\subseteq\mathbb{Z}.$

######  Example 4.2.

The previous example behaves well under products. Figure 3 visualises the case
of $X=\mathbb{P}^{1}\times\mathbb{P}^{1}$. Let us determine on
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ all possible types of exceptional
sequences of maximal length which is $4$. The shape of the immaculate locus
immediately implies that we have at most two elements on the lines $[x=1]$ and
$[y=1]$, and any two elements on the same line must be consecutive. Computing
the sets
$\operatorname{{\mathcal{I}}}\cap\big{(}(a,1)+\operatorname{{\mathcal{I}}}\big{)}$
and
$\operatorname{{\mathcal{I}}}\cap\big{(}(1,b)+\operatorname{{\mathcal{I}}}\big{)}$
for $a$, $b\in\mathbb{Z}$ shows that we have four families
$s^{1},\ldots,s^{4}$ of maximal exceptional sequence given by

$\begin{array}[]{llll}s^{1}&=\big{(}(0,0),\,(1,0),\,(a,1),\,(a+1,1)\big{)}&s^{2}&=\big{(}(0,0),\,(0,1),\,(1,b),\,(1,b+1)\big{)}\\\\[5.0pt]
s^{3}&=\big{(}(0,0),\,(a,1),\,(a+1,1),\,(1,2)\big{)}&s^{4}&=\big{(}(0,0),\,(1,b),\,(1,b+1),\,(2,1)\big{)}.\end{array}$

Examples are displayed in Figure 4.

$0$$\ldots$$\ldots$$\vdots$$\vdots$ Figure 3. The immaculate locus of
$\mathbb{P}^{1}\times\mathbb{P}^{1}$ is given by the grey points. Note that
the origin does not belong to
$\operatorname{{\mathcal{I}}}(\mathbb{P}^{1}\times\mathbb{P}^{1})$.

Figure 4. $s^{1}$, $s^{2}$, $s^{3}$, $s^{4}$ for $a=b=-3$.

###  4.2. The maximality condition

Using the identification
$\operatorname{Pic}(\mathbb{T}\mathbb{V}(\Sigma))\cong\mathbb{Z}^{2}$ we think
of exceptional sequences as subsets of $\mathbb{Z}^{2}$ together with a
certain ordering.

######  Definition 4.3.

Consider the smooth toric variety of Picard rank two $(\ell_{1},\ell_{2};c)$
which is of dimension $d:=\dim X=\ell_{1}+\ell_{2}-2$, cf. Subsection (3.1).
The subset inside $\mathbb{Z}^{2}$ underlying an exceptional sequence
$s=(s_{0},s_{1},\ldots,s_{N})$ is called

  1. (i)

non-extendable if it is not strictly contained in some other subset of
$\mathbb{Z}^{2}$ underlying an exceptional sequence.

  2. (ii)

maximal if $N+1$ equals $\\#\Sigma(d)=\ell_{1}\ell_{2}$.

By abuse of language, we refer to the sequence $s$ itself as non-extendable or
maximal if the underlying set is non-extendable or maximal.

Note that “maximal” implies “non-extendable”, the converse being false, see
Example 8.1. A crucial property of maximal exceptional sequences is this.

######  Lemma 4.4.

If $s$ is a maximal exceptional sequence, then the restriction of
$\Phi:\mathbb{Z}^{2}\to{\mathcal{T}}$ from Subsection (3.4) defines a
bijection.

###### Proof.

If $\Phi(s_{i})=\Phi(s_{j})$ for some pair $s_{i}<s_{j}$ in $s$, then
$s_{j}-s_{i}\in\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)\cap
L$ which is empty by Lemma 3.6. Hence $\Phi|_{s}$ is injective.

Since $s$ and ${\mathcal{T}}$ have the same cardinality, $\Phi|_{s}$ is also
surjective. ∎

###  4.3. The fullness condition

Recall from the introduction that an exceptional sequence $s$ is said to be
full if the underlying set $\\{s_{0},\ldots,s_{N}\\}$ generates
${\mathcal{D}}(X)$. Since $X$ is smooth, ${\mathcal{D}}(X)$ is generated by
$\operatorname{Pic}(X)$. Regarding $s$ as a subset of $\operatorname{Pic}(X)$
it is therefore sufficient to show that we can generate any invertible sheaf.

######  Example 4.5.

The celebrated Beilinson exact sequence [Bei78] yields

(3)
$0\to\Lambda^{\ell}K^{\ell}\otimes{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(0)\to\Lambda^{\ell-1}K^{\ell}\otimes{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(1)\to\ldots\to\Lambda^{0}K^{\ell}\otimes{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(\ell)\to
0.$

In our language, this is just the exact sequence from Theorem 2.1 for the
standard $(\ell-1)$-simplex in $\mathbb{R}^{\ell}$.

At any rate, given the sequence
${\mathcal{O}}_{\mathbb{P}^{\ell-1}}(1),\ldots,{\mathcal{O}}_{\mathbb{P}^{\ell-1}}(\ell-1)$
the bundle ${\mathcal{O}}_{\mathbb{P}^{\ell-1}}(0)$ generates
${\mathcal{O}}_{\mathbb{P}^{\ell-1}}(\ell)$ in
${\mathcal{D}}(\mathbb{P}^{\ell-1})$ and vice versa. Thus, any sequence of
$\ell$ consecutive classes of invertible sheaves generates
$\operatorname{Pic}(\mathbb{P}^{\ell-1})\cong\mathbb{Z}$ and thus the derived
category ${\mathcal{D}}(\mathbb{P}^{\ell-1})$. Note that any such sequence
actually defines a maximal exceptional sequence.

###  4.4. The helix operator

Let $s=(s_{0},s_{1},\ldots,s_{N})$ be a maximal exceptional sequence on
$(\ell_{1},\ell_{2};c)$. We define

$\hslash(s):=(s_{1},s_{2},\ldots,s_{N},-[K]+s_{0})=(s_{1},s_{2},\ldots,s_{N},\,(\ell_{1}-\beta,\ell_{2})+s_{0})$

and call $\hslash$ the helix operator. It preserves exceptionality as follows
directly from the point symmetry of $\operatorname{{\mathcal{I}}}$ with
respect to $-[K]/2$, cf. Subsection (3.3).

######  Remark 4.6.

Helixing is a standard operation in the theory of exceptional sequences, cf.
for instance [Rud90], and it is well-known that the helix operator also
preserves fullness. From our combinatorial point of view, this follows as a
corollary to Theorem E; assuming the general theory, Theorem E is a simple
corollary of Theorem C.

##  5\. Lexicographical order and chains

###  5.1. Lexicographically orderable exceptional sequences

Let $s=(s_{0},\ldots,s_{N})$ be an exceptional sequence on
$X=(\ell_{1},\ell_{2};c)$, that is,
$s_{j}-s_{i}\in\operatorname{{\mathcal{I}}}=\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$
for all $0\leq i<j\leq N$. The underlying set $\\{s_{0},\ldots,s_{N}\\}$ can
give rise to exceptional sequences for various orders. Still, we necessarily
have the

######  Lemma 5.1.

Let $s=(s_{0},\ldots,s_{N})$ be an exceptional sequence on
$(\ell_{1},\ell_{2};c)$. Then for any $s_{i}=(a_{i},b)$ and $s_{j}=(a_{j},b)$
in $s\cap[y=b]$ we have $a_{i}<a_{j}$ if and only if $i<j$. Moreover,
$|a_{j}-a_{i}|\leq\ell_{1}-1$.

###### Proof.

By definition of exceptionality, $i<j$ implies

$s_{j}-s_{i}=(a_{j}-a_{i},\,0)\in\operatorname{{\mathcal{I}}}$

which holds if and only if $0<a_{j}-a_{i}<\ell_{1}$. ∎

######  Definition 5.2.

The subset underlying an exceptional sequence $s$ is called vertically
orderable if it also defines an exceptional sequence with respect to the
vertical (lexicographical) order

$(a_{1},b_{1})<(a_{2},b_{2})$ if and only if $b_{1}<b_{2}$ or
$\big{(}b_{1}=b_{2}$ and $a_{1}<a_{2}\big{)}$.

Similarly, it is called horizontally orderable if it defines an exceptional
sequence with respect to the horizontal (lexicographical) order

$(a_{1},b_{1})<(a_{2},b_{2})$ if and only if $a_{1}<a_{2}$ or
$\big{(}a_{1}=a_{2}$ and $b_{1}<b_{2}\big{)}$.

In general, exceptionality cannot be expected to be preserved under
lexicographic reordering. For instance, the sequence
$s=(s_{0},\,s_{1},\,s_{2})$ on $(\ell_{1},\ell_{2};c)$ with $c\not=0$ and
$\alpha\ell_{2}+1<\ell_{1}$ which is given by

$s_{0}=(0,0),\quad s_{1}=(\alpha\ell_{2}+1,-\ell_{2}),\quad
s_{2}=(\alpha\ell_{2}+2-\beta,0){\;=({\gamma}+2,0)}$

is certainly exceptional, but neither the horizontally ordered sequence
$(s_{0},s_{2},s_{1})$ nor the vertically ordered sequence
$(s_{1},s_{0},s_{2})$ are as $\beta<\alpha\ell_{2}$ in virtue of the basic
inequality (2). However, these “downward jumps” are the only obstruction
against vertical order.

######  Proposition 5.3.

An exceptional sequence $s=(s_{0},\ldots,s_{N})$ on $(\ell_{1},\ell_{2};c)$
with $s_{i}=(a_{i},b_{i})$ is vertically orderable if and only if

(4) $b_{j}-b_{i}>-\ell_{2}$

for all $0\leq i<j\leq N$.

###### Proof.

Assume that $s$ is vertically orderable and that we have a pair $s_{i}<s_{j}$
with

$-\ell_{2}\geq b:=b_{j}-b_{i}.$

In particular,

$-\alpha b<a:=a_{j}-a_{i}.$

On the other hand, we also have $s_{i}-s_{j}\in\operatorname{{\mathcal{I}}}$.
Since $-b\geq\ell_{2}$, the difference $s_{i}-s_{j}=(-a,-b)$ must be in the
parallelogram $\mathcal{P}$ and thus $-\beta<-a$. By the basic inequality (2),
$a>-\alpha b>\beta$ whence $-a<-\beta$, contradiction.

Conversely, assume that the bound (4) holds. We show that there exists a
permutation $\sigma:\\{0,\ldots,N\\}\to\\{0,\ldots,N\\}$ such that
$s_{\sigma(0)},\ldots,s_{\sigma(N)}$ is an exceptional sequence with respect
to lexicographical order.

From Lemma 5.1, $i<j$ implies $a_{i}<a_{j}$ if $s_{i}=(a_{i},b)$ and
$s_{j}=(a_{j},b)$. In addition, we show that we can interchange the order of
two adjacent sequence elements $s_{i}=(a_{i},b_{i})$ and
$s_{i+1}=(a_{i+1},b_{i+1})$ with $b_{i}>b_{i+1}$ while keeping the sequence
exceptional.

Indeed, the difference
$\overrightarrow{s_{i}\,s_{i+1}}=(a_{i+1}-a_{i},\;b_{i+1}-b_{i})$ lies in
$\operatorname{{\mathcal{I}}}$. The assumptions imply

$0<b_{i}-b_{i+1}\leq\ell_{2}-1.$

Hence $\overrightarrow{s_{i+1}\,s_{i}}=-\overrightarrow{s_{i}\,s_{i+1}}$
belongs to the horizontal strip and therefore lies in
$\operatorname{{\mathcal{I}}}$, too.

Replacing in the sequence the pair $s_{i},s_{i+1}$ by $s_{i+1},s_{i}$ does not
affect the remaining difference vectors of $\\{s_{i},s_{i+1}\\}$ with elements
from $\\{s_{0},\ldots,s_{i-1}\\}$ and $\\{s_{i+2},\ldots,s_{N}\\}$ so that
$s_{0},\ldots,s_{i-1},s_{i+1},s_{i},s_{i+2},\ldots,s_{N}$ is still
exceptional. Furthermore, it still satisfies (4) since $b_{i+1}-b_{i}\leq 0$.
After a finite number of pairwise permutations which preserve (4) and
exceptionality, we obtain the desired permutation. ∎

######  Corollary 5.4 (“no horizontal gaps”).

Let $s=(s_{0},s_{1},\ldots,s_{N})$ define an exceptional sequence which is not
extendable. If $s$ is vertically orderable, then the restriction $s\cap[y=b]$
to any horizontal line has no gaps. Namely, if $(a,b)$ and $(a^{\prime},b)\in
s$ with $a<a^{\prime}$, then $(a^{\prime\prime},b)\in s\cap[y=b]$ for all
$a\leq a^{\prime\prime}\leq a^{\prime}$.

###### Proof.

Reordering if necessary we may assume that $s$ is vertically ordered. Suppose
we have $s_{i}=(a_{i},b)$, $s_{i+1}=(a_{i+1},b)$ with $a_{i}+2\leq a_{i+1}$.
Then we can choose a point $(a,b)\in\mathbb{Z}^{2}$ and $a_{i}<a<a_{j}$. We
define a new sequence $s^{\prime}$ by

$s^{\prime}_{k}:=\left\\{\begin{array}[]{ll}s_{k}&\mbox{if }k\leq i\\\
(a,b)&\mbox{if }k=i+1\\\ s_{k-1}&\mbox{if }k\geq i+2\end{array}\right..$

We show that $s^{\prime}$ is exceptional, contradicting the nonextendability
of $s$. Consider the vectors
$\overrightarrow{s^{\prime}_{k}s^{\prime}_{k+r}}=s^{\prime}_{k+r}-s^{\prime}_{k}$
for $k\geq 1$. If none of the indices $k$ or $k+r$ equals $i+1$, then
$s^{\prime}_{k}$ and $s^{\prime}_{k+r}$ belong to the original sequence $s$,
whence
$\overrightarrow{s^{\prime}_{k}s^{\prime}_{k+r}}\in\operatorname{{\mathcal{I}}}$.

If $k\leq i$, consider $s^{\prime}_{i+1}-s^{\prime}_{k}=(a,b)-s_{k}$ which
sits between $s_{i}-s_{k}$ and $s_{i+1}-s_{k}$ on the same horizontal line.
Hence $(a,b)-s_{k}\in\operatorname{{\mathcal{I}}}$ by horizontal integral
convexity, cf. Subsection (3.3).

Finally, if $k\geq i+1$, we can reason as before and conclude that the
difference $s^{\prime}_{k}-s^{\prime}_{i+1}$ belongs to
$\operatorname{{\mathcal{I}}}$. ∎

The following lemma is useful when dealing with sequences which are not
vertically orderable.

######  Lemma 5.5.

Let $s=(s_{0},\ldots,s_{N})$ be an exceptional sequence on
$(\ell_{1},\ell_{2};c)$ which is not vertically orderable. Then we can reorder
$s$ in such a way that the new sequence is still exceptional and there is a
consecutive pair $s_{i}<s_{i+1}$ with $b_{i}-b_{i+1}\geq\ell_{2}$.

###### Proof.

We write $s_{i}=(a_{i},b_{i})$. Let $s_{i}<s_{j}$ be a pair with
$b_{j}-b_{i}\leq-\ell_{2}$.

Decreasing $j$ if necessary yields a new pair still denoted $s_{i}<s_{j}$ such
that $b_{j}-b_{i}\leq-\ell_{2}$ and $b_{k}-b_{i}\geq-\ell_{2}+1$ for all
$s_{i}<s_{k}<s_{j}$. In particular, we can permute the elements
$s_{i},\ldots,s_{j-1}$ to obtain an exceptional sequence where $s_{j}$ is an
immediate successor of $s_{i}$, cf. also the proof of Proposition 5.3. ∎

######  Remark 5.6.

For $c=0$ the $\mathbb{Z}^{2}$-involution $(a,b)\mapsto(b,a)$ maps any
$(\ell_{1},\ell_{2};0)$-exceptional sequence to an
$(\ell_{2},\ell_{1};0)$-exceptional sequence. In particular, it follows that
$s$ is horizontally orderable if and only if

(5) $a_{j}-a_{i}>-\ell_{1}$

for all $0\leq i<j\leq N$. Similarly, Lemma 5.1, 5.5 and Corollary 5.4 hold
mutatis mutandis.

###  5.2. Orderable varieties

For $c\not=0$ we define the integral depth of $\operatorname{{\mathcal{I}}}$
as the smallest integer $d_{\mbox{\rm\tiny int}}$ such that the line
$[y=d_{\mbox{\rm\tiny int}}]$ meets $\operatorname{{\mathcal{I}}}$, namely

(6) $d_{\mbox{\rm\tiny
int}}=-\Big{\lfloor}{\frac{\ell_{1}-2}{\alpha}}\Big{\rfloor}.$

Proposition 5.3 immediately implies the

######  Corollary 5.7.

If $d_{\mbox{\rm\tiny int}}\geq 1-\ell_{2}$, then any exceptional sequence on
$(\ell_{1},\ell_{2};c)$ with $c\not=0$ can be vertically ordered.

We call the toric variety $(\ell_{1},\ell_{2};c)$ itself vertically orderable
if every exceptional sequence is vertically orderable. Subsection (3.1.4) and
the basic inequality (2) immediately imply the

######  Proposition 5.8.

A toric variety $(\ell_{1},\ell_{2};c)$ with $c\not=0$ and
$\beta\geq\ell_{1}-2$ is vertically orderable.

This holds, for instance, for varieties of dimension less than or equal to
three as well as for all non-Fano varieties.

###  5.3. Existence

To generalise the generation strategy from Subsection (1.7) we make the
following

######  Definition 5.9.

A horizontal chain in $\mathbb{Z}^{2}$ is a subset of the form

$(a,b)+\\{(0,0),\ldots,(\ell_{1}-1,0)\\}.$

Similarly, a vertical chain is a subset of the form
$(a,b)+\\{(0,0),\ldots,(0,\ell_{2}-1)\\}$.

######  Remark 5.10.

Thinking of $\mathbb{Z}^{2}$ as the Picard group
$\operatorname{Pic}(\ell_{1},\ell_{2};c)$ we can fill, that is, generate in
the derived category any line $[y=b]$ which contains a horizontal chain via
the Beilinson sequence (3) in Subsection (4.3). The product case also requires
vertical chains.

Given an exceptional sequence $s$ we quantify its “spatial size” as follows.
We let

$\underline{a}:=\underline{a}(s):=\min\\{a\mid(a,b)\in
s\\},\quad\underline{b}:=\underline{b}(s):=\min\\{b\mid(a,b)\in s\\}$

and

$\overline{a}:=\overline{a}(s):=\max\\{a\mid(a,b)\in
s\\},\quad\overline{b}:=\overline{b}(s):=\max\\{b\mid(a,b)\in s\\},$

The height and the width of $s$ are then defined by

$H(s):=\overline{b}-\underline{b}+1\quad\mbox{and}\quad
W(s):=\overline{a}-\underline{a}+1.$

######  Proposition 5.11.

Let $s$ be a maximal exceptional sequence on $(\ell_{1},\ell_{2};c)$ and
$H(s)\leq 2\ell_{2}$. For all integers $b$ with $0\leq
b-\underline{b}<\ell_{2}$, the sets

$Y_{b+\ell_{2}}:=(-\beta,\ell_{2})+([y=b]\cap s)\hskip
8.00003pt\mbox{and}\hskip 8.00003ptX_{b+\ell_{2}}:=[y=b+\ell_{2}]\cap s$

define the horizontal chain

$S_{b+\ell_{2}}:=Y_{b+\ell_{2}}\cup X_{b+\ell_{2}}\subseteq\mathbb{Z}^{2}.$

In particular, $Y_{b+\ell_{2}}\cap X_{b+\ell_{2}}=\varnothing$ and
$\\#S_{b+\ell_{2}}=\ell_{1}$.

###### Proof.

Consider the lattice $L$ with associated map
$\Phi:\mathbb{Z}^{2}\to{\mathcal{T}}$ from Subsection (3.4). By Lemma 4.4 its
restriction to $s$ is bijective. We denote by $[a,b]=\Phi(a,b)$ the
equivalence class of $(a,b)\in s$ in ${\mathcal{T}}$.

To ease notation we assume that $\underline{b}=0$. Since $H(s)\leq 2\ell_{2}$,
$0\leq b\leq 2\ell_{2}-1$ whenever $(a,b)\in s$. Hence, the $\ell_{1}$ classes
$[a_{1},b],\ldots,[a_{\ell_{1}},b]$ come either from $[y=b]\cap s$ or
$[y=b+\ell_{2}]\cap s$. In particular, the union $Y_{b+\ell_{2}}\cup
X_{b+\ell_{2}}$ is disjoint and has precisely $\ell_{1}$ elements.

Let $Y_{b}^{\prime}:=\\{y\in\mathbb{Z}\mid(y,b+\ell_{2})\in Y_{b+\ell_{2}}\\}$
and $X_{b}^{\prime}:=\\{x\in\mathbb{Z}\mid(x,b+\ell_{2})\in
X_{b+\ell_{2}}\\}$. Since both sets are disjoint, we can define

$s(z):=\left\\{\begin{array}[]{ll}(y+\beta,b)&\mbox{if }z=y\in
Y_{b}^{\prime}\\\ (x,b+\ell_{2})&\mbox{if }z=x\in
X_{b}^{\prime}\end{array}\right.$

which is just the element of $s$ giving rise to $z$. By Lemma 5.1, $x$,
$x^{\prime}\in X_{b}^{\prime}$ satisfy $s(x)<s(x^{\prime})$ if and only if
$x<x^{\prime}$, and similarly for $Y_{b}^{\prime}$. A more involved
characterisation holds for mixed pairs $(x,y)\in X_{b}^{\prime}\times
Y_{b}^{\prime}$, namely

$s(y)<s(x)\,\Longleftrightarrow\,0<x-y<\ell_{1}\hskip
8.00003pt\mbox{and}\hskip
8.00003pts(x)<s(y)\,\Longleftrightarrow\,{\gamma}<y-x<\ell_{1}-\beta.$

Indeed, $s(y)<s(x)$ if and only if
$(x-y-\beta,\ell_{2})\in\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$.
Further, $s(x)<s(y)$ if and only if
$(y+\beta-x,-\ell_{2})\in\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$.
The claim follows from the inequalities defining the immaculate locus as well
as ${\gamma}\geq 0$, cf. Remark 3.1. In particular, $Y_{b}^{\prime}\cup
X_{b}^{\prime}$ forms a sequence of $\ell_{1}$ consecutive integers. ∎

######  Remark 5.12.

If in addition $c\not=0$, the proof of Proposition 5.11 also yields a
horizontal no-gap lemma for $Y_{b+\ell_{2}}\cup X_{b+\ell_{2}}$ without
assuming that the maximal exceptional sequence $s$ is orderable, compare
Corollary 5.4. Moreover, it implies for this case that $Y_{b+\ell_{2}}$ lies
to the left of $X_{b+\ell_{2}}$. Indeed, if there exists $x\in X_{b}^{\prime}$
with $y=x+1\in Y_{b}^{\prime}$, then $1=y-x>{\gamma}\geq 1$, contradiction.

For the sequel we say that an element $s_{i}$ of $s$ is at level $h$, if
$s_{i}=(a,\underline{b}+h)$.

######  Corollary 5.13.

Let $s$ be a maximal exceptional sequence with ${l}:=2\ell_{2}-H(s)>0$. Then
there exists horizontal chains in $s$ at level
$\ell_{2}-{l},\ldots,\ell_{2}-1$.

###### Proof.

The assertion is invariant under shifts so we may assume that
$\underline{b}=\underline{b}(s)=0$. So if $(a,b)\in s$, then
$\underline{b}=0\leq b\leq\overline{b}=2\ell_{2}-{l}-1$. Therefore,
$[y=2\ell_{2}-{l}]\cap s,\ldots,[y=2\ell_{2}-1]\cap s=\varnothing$. By
Proposition 5.11 we must have horizontal chains in $[y=\ell_{2}-{l}]\cap
s,\ldots,[y=\ell_{2}-1]\cap s$. ∎

The converse statement also holds. For this, let

$\Delta^{\mbox{\tiny
up}}:=\\{(a,b)\in\mathcal{P}\,|\;b\geq\ell_{2}\\}=\\{(a,b)\in\mathbb{Z}^{2}\mid
b\geq\ell_{2},\;-\beta<a\mbox{ and }a+b\alpha<\ell_{1}+{\gamma}\\}.$

Note that the inequalities $b\geq\ell_{2}$ and $-\beta<a$ alone already imply
that $a+b\alpha>{\gamma}$. The subsequent lemma states, roughly speaking, that
a point of $s$ at sufficiently high level prevents horizontal chains in $s$.

######  Lemma 5.14.

Let $s$ be an exceptional sequence on $(\ell_{1},\ell_{2};c)$ which is
vertically orderable. If $(a^{\prime\prime},b^{\prime\prime})\in s$, then for
all pairs $(a,b)<(a^{\prime},b)$ with $b^{\prime\prime}-b\geq\ell_{2}$ we have
$a^{\prime}-a<\ell_{1}-1$ (instead of $\leq\ell_{1}-1$ as asserted in Lemma
5.1).

###### Proof.

From the definition of the immaculate locus it follows that

$\Delta^{\mbox{\tiny
up}}\subseteq\\{(a,b)\in\mathbb{Z}^{2}\mid-\beta<a<\ell_{1}-\beta\\}.$

Reordering vertically if necessary we have
$(a,b)<(a^{\prime\prime},b^{\prime\prime})$ for any $(a,b)\in s$ hence
$[y=b]\cap s$ is contained in

$(a^{\prime\prime},b^{\prime\prime})-\Delta^{\mbox{\tiny
up}}\subseteq\\{(a,b)\in\mathcal{P}\mid\beta+a^{\prime\prime}-\ell_{1}<a<\beta+a^{\prime\prime}\\}.$

Consequently, $a^{\prime}-a<\ell_{1}-1$. ∎

If an exceptional sequence $s$ starts at $s_{0}=0$, any horizontal chain in
$s$ must be necessarily located at level $h$ with $0\leq h\leq\ell_{2}-1$. We
therefore immediately deduce the

######  Corollary 5.15.

Let $s$ be an exceptional sequence on $(\ell_{1},\ell_{2};c)$ which is
vertically orderable. If there exists a horizontal chain in $s$ at level $h$,
then $H(s)\leq\ell_{2}+h\leq 2\ell_{2}-1$.

######  Corollary 5.16.

Let $s$ be a maximal exceptional sequence on $(\ell_{1},\ell_{2};c)$ which is
vertically orderable. If $H(s)\leq 2\ell_{2}$, then $[y=b]\cap
s\not=\varnothing$ for all $\underline{b}\leq b\leq\overline{b}$. In
particular, $H(s)=\\#$ of rows occupied by $s$.

###### Proof.

Since the assertion concerns only the underlying set we may suppose that $s$
is vertically ordered and $s_{0}=0$. Assume that $[y=b]\cap s=\varnothing$. If
$0\leq b<\ell_{2}$, then $[y=b+\ell_{2}]\cap s$ must have $\ell_{1}$ elements
by Proposition 5.11. But $\ell_{2}\leq b+\ell_{2}$ whence $([y=b+\ell_{2}]\cap
s)\subseteq\Delta^{\mbox{\tiny up}}$ which is impossible.

On the other hand, $b\geq\ell_{2}$ implies that $[y=b-\ell_{2}]\cap s$ has
$\ell_{1}$ elements. By Corollary 5.15, $H(s)\leq b$ which contradicts $b\leq
H(s)-1$. ∎

######  Remark 5.17.

For $c=0$ Proposition 5.11 and Corollaries 5.4 5.13, 5.15 and 5.16 hold
mutatis mutandis for the horizontal case.

###  5.4. The trivial maximal exceptional sequences

In Subsection (1.5) we introduced the standard rectangle

${\mathcal{R}}_{\ell_{1},\ell_{2}}=\\{(a,b)\in\mathbb{Z}^{2}\mid 0\leq
a<\ell_{1},\,0\leq b<\ell_{2}\\}.$

With respect to the vertical lexicographical order this induces the maximal
exceptional sequence given by

$s_{a+b\ell_{1}}:=(a,b)\in{\mathcal{R}}.$

Indeed, $\\#{\mathcal{R}}=\ell_{1}\ell_{2}$, and the difference of $s_{i}$ and
$s_{i+k}$ sitting in a common row is $(k,0)\in\operatorname{{\mathcal{I}}}$.
In all other cases, the difference is of the form $(\cdot,b)$ with
$b\in\\{1,2,\ldots,\ell_{2}-1\\}$. It is thus contained in
$\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};\alpha,\beta)$, too.

Furthermore, we obtain maximal exceptional sequences by

  1. (i)

applying an overall shift $(a,b)$ to the entire sequence.

  2. (ii)

shifting any of the rows at level $b=1,\,2,\ldots,\ell_{1}-1$ by
$(a_{b},0)\in\mathbb{Z}^{2}$ depending on the level $b$.

The resulting maximal exceptional sequences will be referred to as vertically
trivial. They are always vertically orderable and have the maximal number
$z=\ell_{2}$ of horizontal chains. In particular, a maximal exceptional
sequence $s$ is vertically trivial if and only if $H(s)=\ell_{2}$.

Mutatis mutandis we also have horizontally trivial maximal exceptional
sequences in the product case $c=0$.

##  6\. The dichotomy of the product case

Choose two integers $\ell_{1}$, $\ell_{2}\geq 2$. We set out to tackle the
Theorems A-E from the introduction for the product case
$(\ell_{1},\ell_{2};0)=\mathbb{P}^{\ell_{1}-1}\times\mathbb{P}^{\ell_{2}-1}$.
The reader which is solely interested in the twisted case can continue with
Section 8. For the remainder of this section
$\operatorname{{\mathcal{I}}}:=\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2};0,0)$.

###  6.1. Exceptional sequences are semi-bounded

######  Theorem 6.1 (Theorem B, product version).

An exceptional sequence $s$ on $(\ell_{1},\ell_{2};0)$ is semi-bounded, that
is, we have either $H(s)\leq 2\ell_{2}-1$ or $W(s)\leq 2\ell_{1}-1$.

###### Proof.

Suppose to the contrary that both $H(s)\geq 2\ell_{2}$ and $W(s)\geq
2\ell_{1}$.

We consider (a priori not necessarily distinct points)
$A=(\underline{a},b_{A})$, $B=(a_{B},\underline{b})$,
$C=(\overline{a},b_{C})$, $D=(a_{D},\overline{b})\in s$, see Figure 5. We
decompose

$R:=\\{(a,b)\in\mathbb{Z}^{2}\mid\underline{a}\leq
a\leq\overline{a},\,\underline{b}\leq b\leq\overline{b}\\}$

into pairs of horizontal and vertical strips, namely

$H^{\mbox{\rm\tiny down}}:=\\{(a,b)\in R\mid b<\underline{b}+\ell_{2}\\},\quad
H^{\mbox{\rm\tiny up}}:=\\{(a,b)\in R\mid b\geq\underline{b}+\ell_{2}\\}$

and

$V^{\mbox{\rm\tiny left}}:=\\{(a,b)\in R\mid a<\underline{a}+\ell_{1}\\},\quad
V^{\mbox{\rm\tiny right}}:=\\{(a,b)\in R\mid a\geq\underline{a}+\ell_{1}\\}.$

see Figure 5. In particular, $A\in V^{\mbox{\rm\tiny left}}$ and $B\in
H^{\mbox{\rm\tiny down}}$. We distinguish two cases.

$A$$B$$C$$D$$H^{\mbox{\rm\tiny up}}$$H^{\mbox{\rm\tiny down}}$

$A$$B$$C$$D$$V^{\mbox{\rm\tiny right}}$$V^{\mbox{\rm\tiny left}}$

Figure 5. The horizontal and the vertical subdivision of $R$.

Case 1: $A\in H^{\mbox{\rm\tiny up}}$. Then $A\not=B$ and the exceptionality
condition imply that either $A-B$ or $B-A$ lies in
$\operatorname{{\mathcal{I}}}$. However,
$A-B=(\underline{a}-a_{B},b_{A}-\underline{b})\in\operatorname{{\mathcal{I}}}$
is impossible for $b_{A}-\underline{b}\geq\ell_{2}$ and
$\underline{a}-a_{B}\leq 0$. Hence $A<B$ and in particular $B\in
V^{\mbox{\rm\tiny left}}$ as $\underline{b}-b_{A}<0$, so
$a_{B}-\underline{a}<\ell_{1}$. We conclude in a similar way that necessarily
$B<C$ and $C\in H^{\mbox{\rm\tiny down}}$. It follows that $A<C$, but this is
impossible since $\overline{a}-\underline{a}\geq W(s)-1\geq\ell_{1}$ while
$b_{C}-b_{A}<0$.

Case 2: $A\in H^{\mbox{\rm\tiny down}}$. Since $\underline{a}-a_{D}\leq 0$ and
$b_{A}-\overline{b}<0$ we necessarily have $A<D$ and $D\in V^{\mbox{\rm\tiny
left}}$. We conclude in a similar way that $C<D$ and $C\in H^{\mbox{\rm\tiny
up}}$, thus $B<C$ and $B\in V^{\mbox{\rm\tiny right}}$. But then $B<D$ which
is impossible for $a_{D}-a_{B}<0$ and $\overline{b}-\underline{b}\geq
H(s)-1\geq\ell_{2}$. ∎

###  6.2. An inductive argument

Our mainstream development for proving fullness will pursue a rather
algorithmic approach based on the lexicographical order from Subsection (6.3)
and the classification of maximal exceptional sequences in Subsection (7.2).
As an aside, we briefly sketch an inductive approach to Theorem E which is
based on a collapsing procedure.

######  Lemma 6.2.

For $\ell_{2}\geq 3$ let $s$ be an exceptional sequence on
$(\ell_{1},\ell_{2};0)$ with $s_{0}=0$ and $H(s)\leq 2\ell_{2}-1$. Then we
obtain a sequence $s^{\prime}$ on

$(\ell_{1}^{\prime},\ell_{2}^{\prime};0)=(\ell_{1},\ell_{2}-1;0)$

via the following procedure:

  1. (i)

Remove the horizontal line at level $\ell_{2}-1$ from $s$.

  2. (ii)

For every $s_{i}=(a_{i},b_{i})$ with $b_{i}\geq\ell_{2}$ put
$s^{\prime}_{i}:=s_{i}-(0,1)$.

  3. (iii)

For all remaining $s_{i}$ put $s_{i}^{\prime}:=s_{i}$.

If we endow $s^{\prime}$ with the order induced by $s$, then $s^{\prime}$
defines an exceptional sequence, too.

###### Proof.

Consider $s_{i}=(a_{i},b_{i})<s_{j}=(a_{j},b_{j})$ and assume that $b_{i}$,
$b_{j}\neq\ell_{2}-1$. We denote the immaculate locus of
$(\ell_{1}^{\prime},\ell_{2}^{\prime};0)$ by
$\operatorname{{\mathcal{I}}}^{\prime}$.

Case 1. If $b_{i}\geq b_{j}$ or $b_{i}+2\leq b_{j}\leq b_{i}+\ell_{2}-2$, then
$s_{1}^{\prime}<s_{2}^{\prime}$ is immediate.

Case 2. If $b_{j}=b_{i}+1$, then both $s_{i}$ and $s_{j}$ belong to the same
side either above or below the removed line. Hence,
$s^{\prime}_{j}-s_{i}^{\prime}=s_{j}-s_{i}=(a_{j}-a_{i},1)\in\operatorname{{\mathcal{I}}}^{\prime}$.

Case 3. If $b_{j}\geq b_{i}+\ell_{2}-1$, then $s_{i}$ sits below and $s_{j}$
sits above the line at level $\ell_{2}-1$. Thus,
$s^{\prime}_{j}-s_{i}^{\prime}=s_{j}-s_{i}-(0,1)\in\operatorname{{\mathcal{I}}}^{\prime}$.
∎

We say that $s^{\prime}$ is obtained from $s$ by collapsing along $\ell_{2}$.
Similarly, we can collapse along $\ell_{1}$ if $W(s)\leq 2\ell_{1}-1$, cf.
also Theorem 6.7 and Lemma 9.7.

As an immediate consequence of the collapsing procedure we obtain that for a
maximal exceptional sequence where $\\#s=\ell_{1}\ell_{2}$, the inequalities
$\\#s^{\prime}\leq\ell_{1}(\ell_{2}-1)$ and $\\#([y=\ell_{2}-1]\cap
s)\leq\ell_{1}$ imply that

$\\#s^{\prime}=\ell_{1}(\ell_{2}-1)\quad\mbox{and}\quad\\#(s\cap[y=\ell_{2}-1])=\ell_{1}.$

Therefore, $s^{\prime}$ is maximal, too, and we actually removed a horizontal
chain in the sense of Definition 5.9.

This allows to prove fullness in a rather implicit way:

######  Corollary 6.3 (Theorem E, product version).

On $(\ell_{1},\ell_{2};0)$ every maximal exceptional sequence $s$ with
$H(s)\leq 2\ell_{2}-1$ is full.

Of course, the same result holds for $W(s)\leq 2\ell_{1}-1$.

Sketch of the proof. Proceeding by induction we may assume that the collapsed
sequence $s^{\prime}$ generates $\mathbb{Z}^{2}$ by filling horizontal and
vertical lines whenever there are $\ell_{1}$ or $\ell_{2}$ consecutive points,
respectively.

Given $s$ we begin by using the horizontal chain at level $\ell_{2}-1$ to fill
the entire horizontal line on which it lies. Afterwards, we may lift all line
fillings from $s^{\prime}$ to $s$ since any vertical chain of $s^{\prime}$
must reach height $\ell_{2}-1$. $\square$

###  6.3. Maximal exceptional sequences are orderable

######  Theorem 6.4 (Theorem A, product version).

Let $s=(s_{0},\ldots,s_{N})$ be a maximal exceptional sequence on
$(\ell_{1},\ell_{2};0)$. Then $s$ is vertically or horizontally orderable.

###### Proof.

Assume that $s$ is a maximal exceptional sequence which is not vertically not
orderable.

Reordering if necessary, Lemma 5.5 implies that there exists a pair
$s_{i}<s_{i+1}$ with $b_{i}-b_{i+1}\geq\ell_{2}$. Upon applying $i$-times the
helix operator from Subsection (4.4) we may replace $s$ by a sequence where
$i=0$. Choosing suitable coordinates we may therefore suppose without loss of
generality that

$s_{0}=(0,0),\quad s_{1}=(\kappa,\lambda)$

for $\lambda\leq-\ell_{2}$ and $0<\kappa<\ell_{1}$.

Next, there exists precisely one $i_{0}>0$ such that $\Phi(s_{i_{0}})=[1,0]$
by Lemma 4.4. Hence,
$s_{i_{0}}\in\big{(}(1,0)+L\big{)}\cap\operatorname{{\mathcal{I}}}$, where we
recall that

$L=\mathbb{Z}(\ell_{1},0)\oplus\mathbb{Z}(0,\ell_{2}).$

Consequently,

$s_{i_{0}}=(1,m\ell_{2})\in\mathcal{P}$

for some $m\in\mathbb{Z}$, where $\mathcal{P}$ is the parallelogram of the
immaculate locus, cf. Subsection (3.3).

On the other hand, $s_{i_{0}}$ is equal to or a successor of
$s_{1}=(\kappa,\lambda)$, hence

$s_{i_{0}}\in\big{(}(\kappa,\lambda)+\operatorname{{\mathcal{I}}}\big{)}\cup\\{(\kappa,\lambda)\\}.$

But $\lambda\leq-\ell_{2}$ implies
$\mathcal{P}\cap\big{(}(\kappa,\lambda)+{\mathcal{P}}\big{)}=\varnothing$, so
that $s_{i_{0}}\in\operatorname{{\mathcal{I}}}$ yields the contradiction
$0<\kappa<1$ unless $i_{0}=1$, that is, $\kappa=1$ and $m\leq-1$.

Now any point $s_{j}=(a_{j},b_{j})$, $j\geq 2$ must be in

$\operatorname{{\mathcal{I}}}\cap\big{(}(1,m\ell_{2})+\operatorname{{\mathcal{I}}}\big{)}=\\{(a,b)\in\mathbb{Z}^{2}\mid
1<a<\ell_{1}\\}.$

By Remark 5.6 it follows that $s$ must be horizontally orderable. ∎

######  Remark 6.5.

As $s^{1}$ and $s^{2}$ in Example 4.2 show for suitable $a$ and $b$, a maximal
exceptional sequence is not necessarily both vertically and horizontally
orderable. In particular, there are maximal exceptional sequences examples of
height equal to or less than $2\ell_{2}-1$ which are either vertically or
horizontally orderable, but not both. Similarly for constrained width.

###  6.4. The dichotomy of maximal exceptional sequences

As a first step towards the classification of maximal exceptional sequence we
want to combine a lexicographical order with a spatial constraint.

######  Lemma 6.6.

Let $s$ be a maximal exceptional sequence with $H(s)\geq 2\ell_{2}$. Then $s$
is horizontally orderable.

###### Proof.

By Theorem 6.4, it is enough to show that vertical orderability implies
horizontal orderability. So if $<$ denotes the vertical order on $s$, we will
show that $a_{j}-a_{i}>-\ell_{1}$ whenever $s_{i}<s_{j}$, and appeal again to
Remark 5.6.

By definition, $s_{N}-s_{i}=(a_{N}-a_{i},b_{N}-b_{i})$ is in the immaculate
locus for all $i<N$. If $b_{N}-b_{i}\geq\ell_{2}$, then

$0<a_{N}-a_{i}<\ell_{1}$

by the equations defining $\operatorname{{\mathcal{I}}}$. If
$b_{N}-b_{i}<\ell_{2}$, then $H(s)\geq 2\ell_{2}$ implies that $b_{i}$ is at
least at level $\ell_{2}$ whence

$0<a_{i}-a_{0},\,a_{N}-a_{0}<\ell_{1}.$

At any rate, $-\ell_{1}<a_{N}-a_{i}<\ell_{1}$.

Next consider $s_{i}=(a_{i},b_{i})<s_{j}=(a_{j},b_{j})$ for $j<N$. By the
above, $a_{j}-a_{N}>-\ell_{1}$. Since $b_{i}\leq b_{j}$,
$b_{N}-b_{i}<\ell_{2}$ implies $b_{N}-b_{j}<\ell_{2}$ and therefore
$0<a_{i}-a_{0},\,a_{j}-a_{0}<\ell_{1}$. In particular,
$a_{j}-a_{i}>-\ell_{1}$. On the other hand, $b_{N}-b_{i}\geq\ell_{2}$ yields
also

$a_{j}-a_{i}=a_{j}-a_{N}+a_{N}-a_{i}>-\ell_{1}.$

∎

This gives rise to the following dichotomy. If $s$ satisfies $H(s)\geq
2\ell_{2}$, then Lemma 6.6 and Theorem 6.1 imply horizontal orderability and
$W(s)\leq 2\ell_{1}-1$. Similarly, $W(s)\geq 2\ell_{1}$ implies vertical
orderability and $H(s)\leq 2\ell_{2}-1$. Moreover, independently on $H(s)$ or
$W(s)$, $s$ is either vertically or horizontally orderable by Theorem 6.4.
This implies the

######  Theorem 6.7 (Dichotomy of maximal exceptional sequences).

To any maximal exceptional sequence $s$ on $(\ell_{1},\ell_{2};0)$ at least
one of the following two items applies:

  1. (i)

$s$ is vertically orderable with $H(s)\leq 2\ell_{2}-1$

  2. (ii)

$s$ is horizontally orderable with $W(s)\leq 2\ell_{1}-1$.

For sake of concreteness we will concentrate on the first case for the
remainder of this paper since this fits into the twisted case; mutatis
mutandis everything which follows also applies to the second case.

##  7\. The classification for the product case

###  7.1. HeLexing

Recall from Subsection (4.4) the definition of the helix operator $\hslash$
which sends a maximal exceptional sequence $s$ on $(\ell_{1},\ell_{2};0)$ to

$\hslash(s)=\big{(}s_{1},\ldots,s_{N},s_{0}+(\ell_{1},\ell_{2})\big{)}.$

######  Lemma 7.1.

Let $s$ be a vertically ordered maximal exceptional sequence on
$(\ell_{1},\ell_{2};0)$ with $s_{0}=0$ and $\ell_{2}<H(s)\leq 2\ell_{2}-1$.
Then $\hslash(s)$ is vertically orderable with $H(\hslash(s))\leq
2\ell_{2}-1$.

###### Proof.

Let

$s^{\prime}=(s_{0}^{\prime},\ldots,s_{N}^{\prime}):=\hslash(s).$

Since $H(s)>\ell_{2}$ there is a point $s_{i}=(a_{i},b_{i})$ with
$b_{i}\geq\ell_{2}$ for some $i=1,\ldots,N$. Therefore, $H(\hslash(s))\leq
H(s)\leq 2\ell_{2}-1$.

Next assume that $s^{\prime}=\hslash(s)$ is not vertically orderable. By
Proposition 5.3, there exists a pair $s_{i}^{\prime}<s_{j}^{\prime}$ with
$b_{i}^{\prime}-b_{j}^{\prime}\geq\ell_{2}$. Since $s_{i}^{\prime}=s_{i+1}$
and $s_{j}^{\prime}=s_{j+1}$ if $j<N$, vertical orderability of $s$ implies
that $j=N$. However, $b_{i+1}-\ell_{2}\geq\ell_{2}$ implies $b_{i+1}\geq
2\ell_{2}$, contradicting $H(s)\leq 2\ell_{2}-1$. ∎

It follows that we can vertically reorder $\hslash(s)$. Shifting yields an
exceptional sequence $\hslash_{\mbox{\tiny lex}}(s)$ starting at the origin.
We call $\hslash_{\mbox{\tiny lex}}$ the heLex operator.

######  Theorem 7.2 (Theorem C, product version).

On $(\ell_{1},\ell_{2};0)$ every vertically orderable maximal exceptional
sequence $s$ of height $H(s)\leq 2\ell_{2}-1$ can be transformed into a
vertically trivial sequence by vertically reordering and successively applying
$\hslash_{\mbox{\tiny lex}}$ at most $\ell_{1}\ell_{2}$ times.

###### Proof.

If $s$ is vertically ordered with $s_{0}=0$, the helix operator $\hslash$
sends $s_{0}$ to $(\ell_{1},\ell_{2})$. Unless $H(s)\leq\ell_{2}$, that is,
$s$ is vertically trivial, either

  1. (i)

the lowest row becomes empty, whence $H(s)$ decreases at least by one, or

  2. (ii)

the height $H(s)$ remains unchanged, so that we are reducing the number of
elements in $[y=0]\cap s$ by one.

By induction we eventually arrive at a maximal exceptional sequence of height
$H(s)\leq\ell_{2}$, i.e., a vertically trivial one. ∎

###  7.2. The classification of maximal exceptional sequences

Next we discuss the classification of the sets underlying a maximal
exceptional sequence on $(\ell_{1},\ell_{2};0)$ by giving an explicit
algorithm for their construction. We recall that we tacitly assume to work
with vertically orderable sequences $s$ of height $H(s)\leq 2\ell_{2}-1$.

For any subset $X\subseteq\Delta^{\mbox{\tiny
up}}=\\{(a,b)\in\mathbb{Z}^{2}\mid b\geq\ell_{2},\;0<a<\ell_{1}\\}$ or, more
generally, for any $X\subseteq\mathbb{Z}^{2}$ we let

$X_{k}:=[y=k]\cap X=\\{(a,b)\in X\,|\;b=k\\}.$

######  Definition 7.3.

We call a non-empty set $X\subseteq\Delta^{\mbox{\tiny up}}$ admissible if

  1. (Ai)

$X_{k}=\varnothing$ for $k\geq 2\ell_{2}-1$.

  2. (Aii)

the layers $X_{k}\not=\varnothing$ consist of successive points
$(x,k),\,(x+1,k),\ldots,(x+q_{k},k)$.

  3. (Aiii)

for each $k\geq\ell_{2}$ we have

$(0,-1)+X_{k+1}\subseteq X_{k}.$

  4. (Aiv)

the bottom layer $X_{\ell_{2}}$ is right-aligned, i.e.,
$(\ell_{1}-1,\ell_{2})\in X_{\ell_{2}}$.

By convention, the empty set will be admissible, too.

In addition to an admissible $X\subseteq\Delta^{\mbox{\tiny up}}$ we need a
further set to completely classify maximal exceptional sequences.

######  Definition 7.4.

Let $\varnothing\not=X\subseteq\Delta^{\mbox{\tiny up}}$ be admissible. Then
$Y\subseteq\mathbb{Z}^{2}$ is called a complementing partner of $X$ if

  1. (Cv)

$Y_{k}=\varnothing$ for $k\geq 2\ell_{2}$.

  2. (Cvi)

$Y_{k}<X_{k}$, meaning that $(y,k)\in Y_{k}$ and $(x,k)\in X_{k}$ imply $y<x$.

  3. (Cvii)

for each $k\in\\{\ell_{2},\ldots,2\ell_{2}-1\\}$, the union $Y_{k}\cup X_{k}$
forms a horizontal chain.

For $X=\varnothing$ a complementing partner will be any set $Y$ consisting of
$\ell_{2}$ horizontal chains $Y_{\ell_{2}},\ldots,Y_{2\ell_{2}-1}$ such that
$Y_{\ell_{2}}$ starts at $(0,\ell_{2})$.

######  Remark 7.5.

Whenever $X_{k}\neq\varnothing$, the complementary set $Y_{k}$ is uniquely
determined. In contrast, $X_{k}=\varnothing$ for
$k\in\\{\ell_{2},\ldots,2\ell_{2}-1\\}$ implies that $Y_{k}$ defines a
horizontal chain whose horizontal position is unrestricted unless
$k=\ell_{2}$.

######  Example 7.6.

Figure 6 displays a typical maximal exceptional sequence together with its
admissible set and complementing partner.

Figure 6. A maximal exceptional sequence on $(5,4;0)$ with starting point at
the origin in green. The right hand side displays its corresponding admissible
set $X$ in red with complementing partner $Y$ in blue.

######  Theorem 7.7 (Theorem D, product version).

Let $X\subseteq\Delta^{\mbox{\tiny up}}$ be admissible and
$Y\subseteq\mathbb{Z}^{2}$ a complementing partner. Then the union of

$s_{\downarrow}:=Y+(0,-\ell_{2})\subseteq\mathcal{H}\cup[y=0]\quad\mbox{and}\quad
s^{\uparrow}:=X\subseteq\Delta^{\mbox{\tiny up}}$

together with vertical order yields a maximal exceptional sequence $s$ with
$H(s)\leq 2\ell_{2}-1$ and $s_{0}=0$. Moreover, any vertically ordered maximal
exceptional sequence starting at the origin arises this way.

###### Proof.

If $X=\varnothing$, then $Y$ consists of $\ell_{2}$ consecutive horizontal
chains. Shifting down by $(0,-\ell_{2})$ yields a vertically trivial sequence
starting at the origin.

We therefore assume that $X\not=\varnothing$. We order the set
$s_{\downarrow}\cup s^{\uparrow}$ vertically to obtain the sequence $s$. From
Definition 5.9, Definition 7.3 (iv), and Definition 7.4 (Cvii), it is clear
that $Y_{\ell_{2}}\cup X_{\ell_{2}}$ forms a horizontal chain which ends at
$(\ell_{1}-1,\ell_{2})$. Consequently, $Y_{\ell_{2}}$ starts at $(0,\ell_{2})$
and $s_{0}=0$.

For $s_{i}<s_{j}$ we have to show that
$s_{j}-s_{i}\in\operatorname{{\mathcal{I}}}$. If $s_{i}$, $s_{j}\in[y=b]$,
then this follows from $s_{i},s_{j}\in Y_{b+\ell_{2}}+(0,-\ell_{2})$ or
$s_{i},s_{j}\in X_{b}$, (Cvi) and (Cvii). If they are at different levels
$0\leq b_{i}<b_{j}\leq 2\ell_{2}-1$, the only critical case arises from
$b_{j}-b_{i}\geq\ell_{2}$ which implies $0\leq b_{i}\leq\ell_{2}-1$ and
$\ell_{2}\leq b_{j}\leq 2\ell_{2}-1$. This means

$s_{i}\in[y=b_{i}]\cap
s_{\downarrow}=Y_{b_{i}+\ell_{2}}+(0,-\ell_{2})\quad\mbox{and}\quad
s_{j}\in[y=b_{j}]\cap s^{\uparrow}=X_{b_{j}}.$

We proceed via induction over $m:=b_{j}-b_{i}-\ell_{2}\geq 0$.

If $m=0$, then $s_{i}\in Y_{b_{j}}+(0,-\ell_{2})$ whence

$s_{j}-\big{(}s_{i}+(0,\ell_{2})\big{)}\in
X_{b_{j}}-Y_{b_{j}}\subseteq\\{(1,0),\,\ldots,(\ell_{1}-1,0)\\}.$

In particular,
$s_{j}-s_{i}\in(0,\ell_{2})+\\{(1,0),\,\ldots,(\ell_{1}-1,0)\\}\subseteq\Delta^{\mbox{\tiny
up}}\subseteq\operatorname{{\mathcal{I}}}$.

Next let $m\geq 1$. By (Aiii) we know that $s_{j}-(0,1)\in s^{\uparrow}$. On
the other hand, the induction hypothesis implies
$s_{j}-(0,1)-s_{i}\in\operatorname{{\mathcal{I}}}$, and since
$b_{j}-1-b_{i}-\ell_{2}\geq 0$ we even have
$s_{j}-(0,1)-s_{i}\in\Delta^{\mbox{\tiny up}}$. Further,
$s_{j}-s_{i}\in\Delta^{\mbox{\tiny up}}$ by definition of $\Delta^{\mbox{\tiny
up}}$ whence $s_{j}-s_{i}\in\operatorname{{\mathcal{I}}}$.

Finally, we want to show that any vertically ordered maximal exceptional
sequence $s$ with $s_{0}=0$ and $H(s)\leq 2\ell_{2}-1$ arises this way. For
this, we let

$Y:=(0,\ell_{2})+\big{(}s\cap(\mathcal{H}\cup[y=0])\big{)}\quad\mbox{and}\quad
X:=s\cap\Delta^{\mbox{\tiny up}}$

where we identify the sequence $s$ with its underlying set.

If $s$ is vertically trivial, then $Y=(0,\ell_{2})+s$ and $X=\varnothing$. We
therefore assume that $s$ is not trivial and check that $X$ is admissible with
complementing partner $Y$. Properties (Ai) and (Cv) follow from $H(s)\leq
2\ell_{2}-1$ and the definition of $Y$, respectively. Furthermore, (Aii)
follows from Corollary 5.4, while (Cvii) is a consequence of Proposition 5.11.

By Corollary 5.16, $[y=b]\cap s\not=\varnothing$ for $0\leq b\leq H(s)-1$.
Furthermore, we have ${l}=2\ell_{2}-H(s)$ horizontal lines by Corollary 5.13.
For $b=0,\ldots,\ell_{2}-{l}-1$, we put here and in the sequel

(7) $0\leq r_{b}:=\\#[y=b]\cap s-1<\ell_{1}-1$

so that $[y=b]\cap s=\\{(a_{b},b),\ldots,(a_{b}+r_{b},b)\\}$.

In particular, we have
$\\{(r_{0}+1,\ell_{2}),\ldots,(\ell_{1}-1,\ell_{2})\\}=[y=\ell_{2}]\cap s$
which implies (Aiv). For all other $0<b<\ell_{2}-{l}-1$,
$Y_{b+\ell_{2}}=(0,\ell_{2})+[y=b]\cap s$ is either to the left or to the
right of $X_{b+\ell_{2}}$. If $Y_{b+\ell_{2}}$ is to the right, then
$(a_{b}-1,b+\ell_{2})\in X_{b+\ell_{2}}$ whence

$(a_{b}-1,b+\ell_{2})-(a_{b}+r_{b},b)=(-1-r_{b},\ell_{2})\in\operatorname{{\mathcal{I}}},$

contradiction. This implies (Cvi).

Finally, Proposition 5.11 implies $(a_{b}+r_{b}+1,b+\ell_{2})\in
X=s\cap\Delta^{\mbox{\tiny up}}$. It follows that
$(a_{b}+r_{b}+1,b+\ell_{2})-(a_{b-1}+r_{b-1},b-1)\in\operatorname{{\mathcal{I}}}$
whence $a_{b}+r_{b}\geq a_{b-1}+r_{b-1}$. Similarly, we have $a_{b}\leq
a_{b-1}$. In particular, $(0,1)+Y_{b+\ell_{2}}\subseteq Y_{b+\ell_{2}+1}$ and
thus $(0,-1)+X_{b+\ell_{2}+1}\subseteq X_{b+\ell_{2}}$ which yields (Aiii). ∎

###  7.3. Generating the derived category

We can use Theorem 7.7 to prove fullness of any maximal exceptional sequence
on $(\ell_{1},\ell_{2};0)$. By the dichotomy principal, it suffices to
consider the case of a vertically orderable sequence $s$ of height $H(s)\leq
2\ell_{2}-1$. The case of horizontally orderable sequences of width $W(s)\leq
2\ell_{1}-1$ follows analogously.

Let $s$ be a maximal exceptional sequence. Since the standard rectangle
${\mathcal{R}}_{\ell_{1},\ell_{2}}$ generates the Picard group via the
horizontal and vertical Beilinson sequence it suffices to show that
${\mathcal{R}}_{\ell_{1},\ell_{2}}\subseteq\langle s\rangle$.

######  Example 7.8.

We consider again the sequence $s$ from Example 7.6. Figure 7 displays our
strategy to fill all of $\mathbb{Z}^{2}$ starting from $s$. After filling
horizontal and vertical lines in Steps (a)-(e) we see that
${\mathcal{R}}_{5,4}\subseteq\langle s\rangle$.

(a)

(b)

(c)

(d)

(e)

Figure 7. Filling $\operatorname{Pic}(5,4;0,0)$ from $s$. The green point
marks the origin. The red points are generated by $s$ and are used to fill
further lines.

######  Theorem 7.9 (Theorem E, product version).

On $(\ell_{1},\ell_{2};0)$ every maximal exceptional sequence $s$ is full.

###### Proof.

We may assume that $s$ is of the form given in Theorem 7.7. In particular, $s$
is of height $H(s)\leq 2\ell_{2}-1$. Further, we continue to use the notation
introduced there.

We shall describe an explicit filling procedure. First, we fill the
${l}=2\ell_{2}-H(s)$ horizontal chains in $s$ whence $[y=b]\subseteq\langle
s\rangle$, $b=\ell_{2}-1,\ldots,\ell_{2}-{l}$ (cf. also Picture (a) of Figure
7). By decreasing induction on $b$ we will prove that through vertical filling
we can generate sufficiently many points of $[y=b]$ to obtain a horizontal
chain in $[y=b]\cap\langle s\rangle$. Assuming this for a moment, we can fill
the remaining lines $[y=b]$, $b=\ell_{2}-{l}-1,\ldots,0$ from points in
$\langle s\rangle$. In particular,
${\mathcal{R}}_{\ell_{1},\ell_{2}}\subseteq\langle s\rangle$ so that $s$ is
full.

We now start the induction and let $b=\ell_{2}-{l}-1$. By (Cvii) we have

$(a_{b}+r_{b}+1,b+\ell_{2}),\ldots,(a_{b}+\ell_{1}-1,b+\ell_{2})\in
X_{b+\ell_{2}},$

cf. Inequality (7) on Page 7 for the definition of $r_{b}$. From (Aiii) it
follows that

$(a_{b}+r_{b}+1,k+\ell_{2}),\ldots,(a_{b}+\ell_{1}-1,k+\ell_{2})\in
X_{k+\ell_{2}}\quad\mbox{for}\quad k=0,\ldots,b.$

Since $[y=k]\subseteq\langle s\rangle$ for $k=\ell_{2}-1,\ldots,\ell_{2}-{l}$,
we conclude that we have $\ell_{1}-1-r_{b}$ vertical chains

$V_{i}=\\{(a_{b}+r_{b}+i,b+1),\ldots,(a_{b}+r_{b}+i,b+\ell_{2})\\}\quad\mbox{for}\quad
i=1,\ldots,\ell_{1}-1-r_{b}$

in $\langle s\rangle$ (cf. also Picture (b) of Figure 7). Filling the vertical
lines $[x=a_{b}+r_{b}+i]$ starting from $V_{i}$ yields

$(a_{b}+r_{b}+1,b),\ldots,(a_{b}+\ell_{1}-1,b)\in\langle s\rangle$

so that we can fill the line $[y=b]=[y=\ell_{2}-{l}-1]$ (cf. also Picture (c)
of Figure 7).

Next assume that $b<\ell_{2}-{l}-1$. By induction hypothesis,

$[x=a_{b+1}+r_{b+1}+1],\ldots,[x=a_{\ell_{2}-{l}-1}+\ell_{1}-1]\subseteq\langle
s\rangle.$

As shown in Theorem 7.7, $a_{b+1}\leq a_{b}$ and $a_{b}+r_{b}\leq
a_{b+1}+r_{b+1}$. If equality holds, we are done. If not, we first generate
the points $(a_{b}+r_{b}+1,b),\ldots,(a_{b+1}+r_{b+1},b)$ as follows. Namely,
(Cvii) implies

$(a_{b}+r_{b}+1,k+\ell_{2}),\ldots,(a_{b+1}+r_{b+1},k+\ell_{2})\in
X_{k+\ell_{2}}\quad\mbox{for}\quad k=0,\ldots,b$

since otherwise, there would be a pair $(i,j)\in s$ with $j<b$ and thus
$a_{b}+r_{b}\geq a_{j}+r_{j}$ but $i>a_{b}+r_{b}$. This yields a contradiction
for $a_{j}+r_{j}\geq i$.

Combining these points with the lines $[y=k]$, $k=b+1,\ldots,\ell_{2}-1$
inside $\langle s\rangle$ yields vertical chains

$V_{i}^{\prime}=\\{(a_{b}+r_{b}+i,b+1),\ldots,(a_{b}+r_{b}+i,b+\ell_{2})\\},\quad
i=1,\ldots,r_{b+1}-r_{b}.$

contained in $\langle s\rangle$. Therefore,

$[x=a_{b}+r_{b}+1],\ldots,[x=a_{b+1}+r_{b+1}]\subseteq\langle s\rangle$

by filling (cf. also the red line on the left hand side in Picture (d) of
Figure 7).

If $a_{b}+\ell_{1}-1>a_{b+1}+r_{b+1}$, it remains to generate the points

$(a_{b+1}+r_{b+1}+1,b),\ldots,(a_{b}+\ell_{1}-1,b).$

Invoking again (Cvii) and the argument above yields the points

$(a_{b+1}+r_{b+1}+1,k+\ell_{2}),\ldots,(a_{b}+\ell_{1}-1,k+\ell_{2})\in
X_{k+\ell_{2}}\quad\mbox{for }k=0,\ldots,b.$

Together with the horizontal lines $[y=b+1],\ldots,[y=\ell_{2}-1]$ already
generated, we obtain sufficiently many vertical lines to generate
$(a_{b+1}+r_{b+1}+1,b),\ldots,(a_{b}+\ell_{1}-1,b)$ (cf. also the red line on
the right hand side in Picture (d) of Figure 7). Hence we obtain a horizontal
chain at level $b$ in $\langle s\rangle$ which allows us to fill the line
$[y=b]$. ∎

##  8\. Maximal exceptional sequences in the twisted case

For the rest of this article we assume $c\not=0$ and set out to prove the main
theorems A-E in the subsequent sections.

We first address Theorem A. We start with some examples.

###  8.1. Maximal exceptional sequences with $\ell_{1}=2$

The integral depth was defined as

$d_{\mbox{\rm\tiny
int}}=-\Big{\lfloor}{\frac{\ell_{1}-2}{\alpha}}\Big{\rfloor}.$

cf. Subsection (5.2). If $\ell_{1}=2$, then $d_{\mbox{\rm\tiny int}}=0$.
Therefore, the immaculate locus of $(2,\ell_{2};c)$ is given by

$\operatorname{{\mathcal{I}}}(2,\ell_{2};\alpha,\beta)=\mathcal{H}\cup\\{0,\,(1,0),\,(-\beta+1,\ell_{2})\\}$

and the variety is orderable. Hence, the only nontrivial family of maximal
exceptional sequences up to shifts is given by

$s=\\{0,\,(a_{1},1),\,(a_{1}+1,1),\ldots,(a_{\ell_{2}-1},\,\ell_{2}-1),\,(a_{\ell_{2}-1}+1,\ell_{2}-1),(-\beta+1,\ell_{2})\\}$

with $a_{i}\in\mathbb{Z}$. We have $H(s)=\ell_{2}+1$. In the language of
admissible sets and complementing partners which will be developed for the
twisted case in Subsection (10.2), we have

$X=\\{(-\beta+1,\ell_{2})\\}\;\mbox{and}\;Y=\\{(-\beta,\ell_{2}),\,(a_{1}-\beta,1+\ell_{2}),\ldots,(a_{\ell_{2}-1}+1-\beta,\,2\ell_{2}-1)\\}.$

For instance, setting also $\ell_{2}=2$ yields the family of Hirzebruch
surfaces ${\mathcal{H}}_{\alpha}=\big{(}2,2;(0,-\alpha)\big{)}$. Among these,
${\mathcal{H}}_{1}=\mathbb{P}\big{(}{\mathcal{O}}_{\mathbb{P}^{1}}\oplus{\mathcal{O}}_{\mathbb{P}^{1}}(1)\big{)}$
is the only Fano variety. Figure 8 represents
$\operatorname{{\mathcal{I}}}({\mathcal{H}}_{\alpha})=\operatorname{{\mathcal{I}}}(2,2;\alpha,\alpha)$
for $\alpha=1$, $2$ and $3$.

$0$

$0$

$0$

Figure 8. The Hirzebruch surfaces ${\mathcal{H}}_{1}$, ${\mathcal{H}}_{2}$ and
${\mathcal{H}}_{3}$.

###  8.2. Maximal exceptional sequences with $\ell_{1}=\ell_{2}=3$

Since $d_{\mbox{\rm\tiny int}}\geq-1$ any variety $(3,3;c)$ is necessarily
vertically orderable. In order to determine the nontrivial maximal exceptional
sequences $s$ up to shift we may therefore assume that $s$ is vertically
lexicographically ordered. We distinguish two cases:

Case 1: $\alpha\geq 2$. Then $d_{\mbox{\rm\tiny int}}=0$. As in Subsection
(8.1) we find the maximal exceptional sequences

$\displaystyle s^{1}=\big{(}$ $\displaystyle
0,\,(1,0),\,(a_{1},1),\,(a_{1}+1,1),\,(a_{1}+2,1),$
$\displaystyle(a_{2},2),\,(a_{2}+1,2),\,(a_{2}+2,2),\,(-\beta+2,3)\big{)}$
$\displaystyle s^{2}=\big{(}$ $\displaystyle
0,\,(a_{1},1),\,(a_{1}+1,1),\,(a_{1}+2,1),$
$\displaystyle(a_{2},2),\,(a_{2}+1,2),\,(a_{2}+2,2),\,(-\beta+1,3),\,(-\beta+2,3)\big{)}$

for $a_{1}$, $a_{2}\in\mathbb{Z}$. We have $H(s^{1})=H(s^{2})=4$.

Case 2: $\alpha=1$. By the basic inequality (2), $\beta=1$ or $2$. In addition
to the maximal exceptional sequences from the previous case we find

$\displaystyle s^{3}=\big{(}$ $\displaystyle
0,\,(-1,1),\,(0,1),\,(a_{2},2),\,(a_{2}+1,2),\,(a_{2}+2,2),$
$\displaystyle(-\beta+1,3),\,(-\beta+2,3),\,(-\beta+1,4)\big{)}$

for $a_{2}\in\mathbb{Z}$. We have $H(s^{3})=5$.

Figure 9 displays the admissible sets and complementing partners for
$\beta=2$.

(a)

(b)

(c)

Figure 9. The admissible sets (red) and complementing partners (blue) for (a)
$s^{1}$ (b) $s^{2}$ and (c) $s^{3}$ with $a_{1}=-1$, $a_{2}=2$ and $\beta=2$.
The origin is marked in green.

###  8.3. Maximal exceptional sequences are always vertically orderable

General non-extendable sequences might not be vertically orderable.

######  Example 8.1.

On $(4,2;(0,1))$ consider the exceptional sequence

$s=\big{(}0,\,(3,-2),\,(2,-1),\,(3,-1),\,(3,0),\,(3,1),\,(4,1)\big{)},$

cf. Figure 10.

$0$$s_{1}$ Figure 10. The points encircled in red define the sequence $s$ in
$\operatorname{{\mathcal{I}}}(4,2;1,1)$.

Since $s_{1}-s_{0}=(3,-2)$ the order of these two elements cannot be switched,
cf. Proposition 5.3.

On the other hand, the sequence $s$ is not maximal for
$\\#s=7<8=\ell_{1}\ell_{2}$ and eventually not extendable. An easy computation
shows that
$\operatorname{{\mathcal{I}}}\cap\big{(}(3,-2)+\operatorname{{\mathcal{I}}}\big{)}$,
the set of possible common successors of $s_{0}$ and $s_{1}$, is

$\\{(2,-1),\,(3,-1),\,(3,0),\,(3,1),\,(4,1)\\}.$

Alternatively, one may apply Theorem 8.2 below to show that $s$ is the largest
choice for extending $\big{(}0,\,(3,-2)\big{)}$ to an exceptional sequence.

However, we have the

######  Theorem 8.2 (Theorem A, twisted version).

Let $s=(s_{0},\ldots,s_{\ell_{1}\ell_{2}-1})$ be a maximal exceptional
sequence on $(\ell_{1},\ell_{2};c)$, $c\not=0$. Then $s$ is vertically
orderable.

###### Proof.

Assume to the contrary that $s$ is a maximal exceptional sequence which is not
vertically orderable.

By Lemma 5.5 we may assume that there exists a pair $s_{i}<s_{i+1}$ with
$b_{i}-b_{i+1}\geq\ell_{2}$. Furthermore, upon applying $i$-times the helix
operator from Subsection (4.4) we may replace $s$ by a sequence where $i=0$.
Choosing suitable coordinates we may therefore suppose without loss of
generality that

$s_{0}=(0,0),\quad s_{1}=(\kappa,\lambda)$

for $\lambda\leq-\ell_{2}$ and $0<-\lambda\alpha<\kappa<\ell_{1}$.

By Lemma 4.4, there exists precisely one $i_{0}>0$ such that
$\Phi(s_{i_{0}})=[1,0]$. Hence,
$s_{i_{0}}\in\big{(}(1,0)+L\big{)}\cap\operatorname{{\mathcal{I}}}$. Reasoning
as in the proof of Lemma 3.6 we see that

$s_{i_{0}}=(n\ell_{1}-m\beta+1,m\ell_{2})\in\mathcal{P}.$

Further, $m\leq 0$ necessarily implies $n=m=0$ and thus $s_{i_{0}}=(1,0)$.
Finally, $m\geq 1$ implies $n=0$, $m=1$ and thus
$s_{i_{0}}=(-\beta+1,\ell_{2})$.

On the other hand, $s_{i_{0}}$ is equal to or a successor of
$s_{1}=(\kappa,\lambda)$, hence

(8)
$s_{i_{0}}\in\big{(}(\kappa,\lambda)+\operatorname{{\mathcal{I}}}\big{)}\cup\\{(\kappa,\lambda)\\}.$

However, $(1,0)$,
$(-\beta+1,\ell_{2})\not\in(\kappa,\lambda)+\operatorname{{\mathcal{I}}}$,
contradicting (8). Indeed, we have
$\kappa>-\lambda\alpha\geq\alpha\ell_{2}>\beta$. Therefore, a point
$(a,b)\in\mathbb{Z}^{2}$ with $a\leq 1$ and $b\geq 0$ cannot lie in
$(\kappa,\lambda)+\operatorname{{\mathcal{I}}}$ as

$a-\kappa\leq 1-\kappa\leq-\alpha\ell_{2}<-\beta,$

while $b-\lambda\geq\ell_{2}$ implies $-\beta<a-\kappa$. ∎

##  9\. Vertical ensembles

We stay in the twisted case and assume that $c\neq 0$ so that
$\alpha,\beta,{\gamma}\geq 1$.

###  9.1. Replacing vertical chains by ensembles

Rather than to vertical chains, the $V$-sequence from Corollary 3.2 gives rise
to a complicated shape for the involved locus in
$\operatorname{Pic}X=\mathbb{Z}^{2}$. The sheaves ${\mathcal{F}}_{k}$
occurring in Subsection (3.2) suggest the following

######  Definition 9.1.

We denote by

$V=V(\ell_{2};\alpha,\beta)\subseteq\mathbb{Z}^{2}$

the set of lattice points $\sum_{j\in J}(f^{j},1)=(\sum_{j\in J}f^{j},\\#J)$
where $J\subseteq\\{1,\ldots,\ell_{2}\\}$ is an arbitrary subset of
cardinality $1\leq\\#J\leq\ell_{2}-1$ and
$f=(f^{1},f^{2},\ldots,f^{\ell_{2}})$ runs through all integral
$\ell_{2}$-tuples satisfying

$0=f^{1}\geq f^{2}\geq\ldots\geq f^{\ell_{2}-1}\geq
f^{\ell_{2}}=-\alpha\quad\mbox{and}\hskip
5.0pt\sum_{j=1}^{\ell_{2}}f^{j}=-\beta.$

We call the subset $V+(a,b)$ the V-ensemble based at $(a,b)$.

Note that $V$ fits into the rectangular box bounded by $-\beta\leq x\leq 0$
and $0\leq y\leq\ell_{2}$ and whose diagonal of negative slope ends in $(0,0)$
and $(-\beta,\ell_{2})$. Actually, $V\subseteq\mathcal{H}$. Further, in the
extreme case $\alpha=\beta$ which necessarily implies
$c^{1}=c^{2}=\ldots=c^{\ell_{2}-1}=0$ the V-ensemble degenerates to

$V(\ell_{2};\beta,\beta)=\\{(-\beta,b),\,(0,b)\mid b=1,\ldots,\ell_{2}-1\\}.$

See also Figure 11 for an illustration.

$0$$(-\beta,\ell_{2})$$0$$(-\beta,\ell_{2})$ Figure 11. The sets $V(4;2,4)$
and $V(4;4,4)$. On the left hand side the dark red points come from the vector
$c=(0,-1-1,-2)$. Additional points from $c=(0,0,-2,-2)$ which are not dark red
yet are marked in light red.

We turn to existence of V-ensembles inside exceptional sequences next. To
simplify the shape we introduce the larger $W$-ensemble of
$(\ell_{1},\ell_{2};c)$ by

$W=W(\ell_{2};\alpha):=\\{(a,b)\in\mathbb{Z}^{2}\mid
0<b<\ell_{2},\,-b\alpha\leq a\leq 0\\}.$

######  Lemma 9.2.

$\;V(\ell_{2};\alpha,\beta)$ is symmetric under the transformation
$(x,y)\mapsto(-\beta,\ell_{2})-(x,y)$. Moreover,
$V(\ell_{2};\alpha,\beta)\subseteq W(\ell_{2};\alpha)$.

###### Proof.

For the symmetry note that $\sum_{j=1}^{\ell_{2}}(f^{j},1)=(-\beta,\ell_{2})$.
Hence,

$(-\beta,\ell_{2})-\sum_{j\in J}(f^{j},1)=\sum_{i\in I}(f^{i},1)$

with $I=\\{1,\ldots,\ell_{2}\\}\setminus J$.

Further, $(a,b)\in V$ implies $(a,b)=(\sum_{j\in J_{b}}f^{j},b)$ for some
suitable $0\geq f^{j}\geq-\alpha$ and $\\#J_{b}=b$. Hence $a\geq-b\alpha$. ∎

Let us define

$\displaystyle\overline{\mathcal{P}}_{V}$
$\displaystyle\phantom{:}=\\{(a,b)\in\mathbb{Z}^{2}\mid-\beta\leq a\leq
0\hskip 5.0pt\mbox{and}\hskip
5.0pt0\leq\langle(a,b),\,(1,\alpha)\rangle\leq{\gamma}\\}$

as a smaller (and closed) version of the immaculate parallelogram
$\mathcal{P}$ from Subsection (3.3), cf. Figure 12. The
$(-\beta,\ell_{2})$-symmetry of $V$ implies the following observation.

$0$$(-\beta,\ell_{2})$$(\ell_{1}-\beta,\ell_{2})$$(\ell_{1},0)$$(2\ell_{2}-1)$-line$\mathcal{P}$$\overline{\mathcal{P}}_{V}$$W$$(-\beta,\ell_{2})-W$${\mathcal{H}}$
Figure 12. $\overline{\mathcal{P}}_{V}$ and $\mathcal{P}$ and ${\mathcal{H}}$
for
$\operatorname{{\mathcal{I}}}(\ell_{1},\ell_{2},\alpha,\beta)=\operatorname{{\mathcal{I}}}(16,7;2,6)$

######  Lemma 9.3.

We have

  1. (i)

$W(\ell_{2};\alpha)\cap\big{(}(-\beta,\ell_{2})-W(\ell_{2};\alpha)\big{)}=\overline{\mathcal{P}}_{V}\setminus\big{\\{}0,\;(-\beta,\ell_{2})\big{\\}}$.

  2. (ii)

$V(\ell_{2};\alpha,\beta)\subseteq\overline{\mathcal{P}}_{V}\setminus\big{\\{}0,\;(-\beta,\ell_{2})\big{\\}}$.

######  Remark 9.4.

The fact that $f^{1}=0$ and $f^{\ell_{2}}=-\alpha$ for the sequences defining
$V$ allows actually an even more refined description. It turns out that
$V\cup\big{\\{}0,\;(-\beta,\ell_{2})\big{\\}}$ equals the union of four
smaller $\overline{\mathcal{P}}_{V}$-like parallelograms located at the four
corners of the ambient $\overline{\mathcal{P}}_{V}$. However, the rather
coarse relationship $V\subseteq W$ from Lemma 9.2 will be sufficient for our
purposes.

###  9.2. Finding $V$\- and $W$-ensembles inside maximal exceptional sequences

Our goal is to localise sufficiently many $V$-ensembles inside any given
maximal exceptional sequence $s$.

######  Lemma 9.5.

Let $s$ be a vertically ordered exceptional sequence on
$(\ell_{1},\ell_{2};c)$ with $s_{0}=0$ and $H(s)\geq 2\ell_{2}$. Then $s\cup
W(\ell_{2};\alpha)$ is also exceptional with respect to the vertical order.

###### Proof.

As suggested by Figure 13, $(a,b)-W\subseteq\operatorname{{\mathcal{I}}}$ for
all $(a,b)\in\Delta^{\mbox{\tiny up}}$. Indeed, let
$(a^{\prime},b^{\prime})\in W$. By Subsection (3.3),

$\ell_{2}\leq b,\quad-\beta<a<{\ell_{1}+{\gamma}-b\alpha}.$

Since $b^{\prime}<\ell_{2}$ it follows that $0<b-b^{\prime}$. If
$b-b^{\prime}<\ell_{2}$, then $(a-a^{\prime},b-b^{\prime})\in\mathcal{H}$, and
we are done. If $\ell_{2}\leq b-b^{\prime}$, then $0\leq-a^{\prime}\leq
b^{\prime}\alpha$ implies

$-\beta<a-a^{\prime}<{\ell_{1}+{\gamma}-(b-b^{\prime})\alpha}=\ell_{1}-\beta-(b-b^{\prime}-\ell_{2})\alpha<\ell_{1}$

whence $(a-a^{\prime},b-b^{\prime})\in\operatorname{{\mathcal{I}}}$.

$0$ Figure 13. $\operatorname{{\mathcal{I}}}(4,3;1,1)$ with the $W$-ensemble
in light red and the set $(1,4)-W$ in red.

Furthermore, $H(s)\geq 2\ell_{2}$ implies existence of a point $(a,b)\in s$
with ${b\geq 2\ell_{2}-1}$. Therefore, $(a,b)-W$ and
$(a,b)-(s\cap\mathcal{H})$ are contained in $\Delta^{\mbox{\tiny up}}$. In
virtue of Lemma 5.14 it follows that
$|a^{\prime}-a^{\prime\prime}|\leq\ell_{1}-2$ for any
$(a^{\prime},b^{\prime})\in W$ and
$(a^{\prime\prime},b^{\prime})\in(s\cap\mathcal{H})\cup W$. Hence $s\cup W$ is
exceptional. ∎

######  Corollary 9.6.

Let $s$ be a vertically ordered maximal exceptional sequence with $H(s)\geq
2\ell_{2}$. Then $s_{0}+W\subseteq s$. In particular,
$s_{0}+(V\cup\\{(0,b)\mid 0\leq b<\ell_{2}\\})$ is contained in $s$.

###### Proof.

Since $s\cup(s_{0}+W)$ is an exceptional extension of $s$, maximality of $s$
implies $s\cup(s_{0}+W)\subseteq s$. The second claim follows from

$s_{0}+\\{(0,b)\mid 0\leq b<\ell_{2}\\}\subseteq(s_{0}+W)\cup\\{s_{0}\\}$

and Lemma 9.2 which asserts that $V\subseteq W$. ∎

###  9.3. Bounding the height

Our final goal in this section is to establish Theorem B in the twisted case.
We first define

$\mathcal{H}^{-}:=\\{(a,b)\in\mathbb{Z}^{2}\mid
a<0,\,0<b<\ell_{2}\\}\subseteq{\mathcal{H}}.$

######  Lemma 9.7.

Let $s$ be a maximal exceptional sequence on $(\ell_{1},\ell_{2};c)$ with
$c\not=0$, $\ell_{1}\geq 3$ and $H(s)\geq 2\ell_{2}$. If $s$ is vertically
ordered with $s_{0}=0$, then we obtain a sequence $s^{\prime}$ on

$(\ell_{1}^{\prime},\ell_{2}^{\prime};c^{\prime})=(\ell_{1}-1,\ell_{2};c)$

via the following procedure:

  1. (i)

Remove the set $\\{(0,b)\mid 0\leq b<\ell_{2}\\}$ from $s$.

  2. (ii)

For every $s_{i}=(a_{i},b_{i})\in{\mathcal{H}}^{-}$ put
$s^{\prime}_{i}:=s_{i}+(1,0)$.

  3. (iii)

For all remaining $s_{i}$ put $s_{i}^{\prime}:=s_{i}$.

If we endow $s^{\prime}$ with the order induced by $s$, then $s^{\prime}$
defines a maximal exceptional sequence. We say that $s^{\prime}$ is obtained
from $s$ by collapsing along $\ell_{1}$.

###### Proof.

We denote the immaculate locus of
$(\ell_{1}^{\prime},\ell_{2}^{\prime};c^{\prime})$ by
$\operatorname{{\mathcal{I}}}^{\prime}$. Let $s_{i}^{\prime}<s_{j}^{\prime}$
be a pair of elements in $s^{\prime}$ coming from
$s_{i}=(a_{i},b_{i})<s_{j}=(a_{j},b_{j})$ in $s$. We need to show that
$s_{j}^{\prime}-s_{i}^{\prime}\in\operatorname{{\mathcal{I}}}^{\prime}$.
Figure 14 sketches how the immaculate locus adjusts
$\operatorname{{\mathcal{I}}}^{\prime}$ under passing from
$(\ell_{1},\ell_{2};c)$ to $(\ell_{1}-1,\ell_{2};c)$.

$0$ Figure 14. $\operatorname{{\mathcal{I}}}(5,3;1,1)$ and
$\operatorname{{\mathcal{I}}}^{\prime}=\operatorname{{\mathcal{I}}}(4,3;1,1)$.
The additional points in $\operatorname{{\mathcal{I}}}(5,3;1,1)$ are marked in
light blue. The upper left boundary and the upper right boundary are marked in
red.

Case 1: $s_{i},\,s_{j}\not\in{\mathcal{H}}^{-}$. Then
$s_{j}^{\prime}-s_{i}^{\prime}=s_{j}-s_{i}$. If $\ell_{2}>b_{j}-b_{i}$, then
$s_{j}^{\prime}-s_{i}^{\prime}\in{\mathcal{H}}={\mathcal{H}}^{\prime}$ or
$b_{j}=b_{i}$. In the latter case we have $a_{j}-a_{i}<\ell_{1}-1$ for
otherwise, $s$ would have a horizontal chain and thus $H(s)\leq 2\ell_{2}-1$
by Corollary 5.15.

If, on the other hand, $b_{j}-b_{i}\geq\ell_{2}$, then
$s_{j}^{\prime}-s_{i}^{\prime}\in\operatorname{{\mathcal{I}}}^{\prime}$ unless

$a_{j}-a_{i}+(b_{j}-b_{i})\alpha=\ell_{1}-1-\beta+\ell_{2}\alpha=\ell_{1}^{\prime}-\beta+\ell_{2}\alpha,$

that is, $s_{j}-s_{i}$ sits in the upper right boundary of
${\mathcal{P}}^{\prime}$, cf. Figure 14. Since $\ell_{2}\leq b_{j}$ we have
$a_{j}+b_{j}\alpha\leq\ell_{1}-1-\beta+\ell_{2}\alpha$. But
$0<a_{i}+b_{i}\alpha$ – this is clear for $s_{i}\in\mathcal{H}$ and follows
from the defining inequalities of the parallelogram if $s_{i}\in\mathcal{P}$.
Hence

$a_{j}-a_{i}+(b_{j}-b_{i})\alpha<a_{j}+b_{j}\alpha\leq\ell_{1}-1-\beta+\ell_{2}\alpha,$

contradiction.

Case 2: $s_{i},\,s_{j}\in{\mathcal{H}}^{-}$. Again
$s_{j}^{\prime}-s_{i}^{\prime}=s_{j}-s_{i}$, and $s_{j}-s_{i}\in{\mathcal{H}}$
or $b_{j}=b_{i}$. As for Case 1 we find that
$s_{j}^{\prime}-s_{i}^{\prime}\in{\mathcal{H}}^{\prime}\subseteq\operatorname{{\mathcal{I}}}^{\prime}$.

Case 3: $s_{i}\in{\mathcal{H}}^{-}$, $s_{j}\not\in{\mathcal{H}}^{-}$. Then
$s_{j}^{\prime}-s_{i}^{\prime}=s_{j}-s_{i}-(1,0)$. If $\ell_{2}>b_{j}-b_{i}$,
we conclude as in Case 1.

If, on the other hand, $b_{j}-b_{i}\geq\ell_{2}$, then
$s_{j}^{\prime}-s_{i}^{\prime}\in\operatorname{{\mathcal{I}}}^{\prime}$ unless

$s_{j}-s_{i}=(a_{j}-a_{i},b_{j}-b_{i})=(-\beta+1,b_{j}-b_{i}),$

that is, $s_{j}-s_{i}$ sits in the upper left boundary of
${\mathcal{P}}^{\prime}$, cf. Figure 14. Now $\ell_{2}<b_{j}$ implies
$a_{j}-a_{i}=-\beta+1\leq a_{j}$, but $a_{i}<0$.

Case 4: $s_{i}\not\in{\mathcal{H}}^{-}$, $s_{j}\in{\mathcal{H}}^{-}$. Then
$s_{j}^{\prime}-s_{i}^{\prime}=s_{j}+(1,0)-s_{i}$ and
$\ell_{2}>b_{j}>b_{i}\geq 0$, where the middle inequality follows from
$a_{j}<a_{i}$. We conclude again as in Case 1. ∎

######  Corollary 9.8 (Theorem B, twisted version).

Let $s$ be a maximal exceptional sequence. Then $H(s)\leq 2\ell_{2}$.

###### Proof.

We assume that $s$ is vertically ordered and starts at $s_{0}=0$. We proceed
by induction on $\ell_{1}\geq 2$.

If $\ell_{1}=2$, then $H(s)\leq\ell_{2}+1<2\ell_{2}$, cf. Subsection (8.1).

Next assume that $\ell_{1}\geq 3$. Let $s$ be a maximal exceptional sequence
with $H(s)>2\ell_{2}$. By Lemma 9.7 we can collapse $s$ along $\ell_{1}$ and
obtain the maximal exceptional sequence $s^{\prime}$ on
$(\ell_{1}-1,\ell_{2};c)$ with $2\ell_{2}\leq H(s)-1\leq H(s^{\prime})$. Here,
the latter inequality is a consequence of $W\subseteq s$ from Corollary 9.6.
Further, $H(s^{\prime})\leq 2\ell_{2}$ by induction hypothesis so that
$H(s^{\prime})=2\ell_{2}<H(s)$. In particular, the collapsed sequence
$s^{\prime}$ starts at $s_{0}^{\prime}=(a,1)$ with $a\leq 0$.

By Corollary 9.6, $s_{0}^{\prime}+W=(a,1)+W\subseteq s^{\prime}$. Therefore

$(a,1)+\big{(}-(\ell_{2}-1)\alpha,\ell_{2}-1\big{)}=\big{(}-(\ell_{2}-1)\alpha+a,\ell_{2}\big{)}\in
s^{\prime}.$

By design of the collapsing procedure, $(-(\ell_{2}-1)\alpha+a,\ell_{2})$ is
also in $s$ whence

$-\beta<-(\ell_{2}-1)\alpha+a\leq-(\ell_{2}-1)\alpha.$

But this contradicts the basic inequality (2). ∎

##  10\. The classification for the twisted case

We now discuss the twisted analogues of heLexing (cf. Subsection (7.1)) and
the structure of maximal exceptional sequences (cf. Subsection (7.2)).

###  10.1. HeLexing

Let $s$ be a maximal exceptional sequence starting at the origin which by
Theorem 8.2 we may take to be vertically ordered. The helix operator $\hslash$
sends $s_{0}$, the leftmost element of the lowest row, to the point
$(\ell_{1}-\beta,\ell_{2})$ at level $\ell_{2}$, using the terminology of
Subsection (5.3).

In Subsection (7.1) we considered $\hslash_{\mbox{\tiny lex}}$ which was the
helix operator $\hslash$ followed by vertically lexicographic reordering and a
shift sending the resulting $s_{0}$ back to the origin. The proof of
Proposition 7.2 applies verbatim and yields the

######  Theorem 10.1 (Theorem C, twisted version).

Every maximal exceptional sequence on $(\ell_{1},\ell_{2};c)$, $c\not=0$, can
be transformed into a trivial sequence by successively applying
$\hslash_{\mbox{\tiny lex}}$ at most $\ell_{1}\ell_{2}$ times.

###  10.2. The classification

Again we can establish an algorithmic recipe for the construction of maximal
exceptional sequences.

The definition of admissible sets and complementing partners carries over from
Subsection (7.2) except for (Aiii) which gets replaced by

  1. (Aiii’)

for each $k\geq\ell_{2}$ and $(x,k+1)\in X_{k+1}$ the points
$(x,k),\,(x+1,k),\,\ldots,\,(x+\alpha,k)$ belong to $X_{k}$.

See Subsections (8.1) and (8.2) for examples. Note that $X_{\ell_{2}}$ being
right-aligned means now that $(\ell_{1}-1-\beta,\ell_{2})\in X_{\ell_{2}}$.
Then we obtain the

######  Theorem 10.2 (Theorem D, twisted version).

If $X\subseteq\Delta^{\mbox{\tiny up}}$ is admissible and
$Y\subseteq\mathbb{Z}^{2}$ a complementing partner, then the union of

$s_{\downarrow}:=Y+(\beta,-\ell_{2})\subseteq\mathcal{H}\cup[y=0]\quad\mbox{and}\quad
s^{\uparrow}:=X\subseteq\Delta^{\mbox{\tiny up}}$

together with vertical order yields a maximal exceptional sequence $s$ with
$s_{0}=0$. Moreover, any vertically ordered maximal exceptional sequence
starting at the origin arises this way.

###### Proof.

The proof goes along the lines of the proof of Theorem 7.7.

If $X=\varnothing$, then $Y$ consists of $\ell_{2}$ consecutive horizontal
chains. Shifting down by $(\beta,-\ell_{2})$ yields a vertically trivial
sequence starting at the origin.

We therefore assume that $X\not=\varnothing$. We order the set
$s_{\downarrow}\cup s^{\uparrow}$ vertically to obtain the sequence $s$. From
Definition 5.9, Definition 7.3 (iv), and Definition 7.4 (Cvii), it is clear