# Classification and geometric properties of surfaces with property ${\bf
N}_{3,3}$

Hoang Le Truong  Institute of Mathematics, VAST, 18 Hoang Quoc Viet Road,
10307 Hanoi, Viet Nam<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract.

Let $X$ be a closed subscheme of codimension $e$ in a projective space. One
says that $X$ satisfies property ${\bf N}_{d,p}$, if the $i$-th syzygies of
the homogeneous coordinate ring are generated by elements of degree $<d+i$ for
$0\leq i\leq p$. The geometric and algebraic properties of smooth projective
varieties satisfying property ${\bf N}_{2,e}$ are well understood, and the
complete classification of these varieties is a classical result. The aim of
this paper is to study the next case: projective surfaces in $\mathbb{P}^{5}$
satisfying property ${\bf N}_{3,3}$. In particular, we give a classification
of such varieties using adjunction mappings and we also provide illuminating
examples of our results via calculations done with Macaulay 2. As corollaries,
we study the CI-biliaison equivalence class of smooth projective surfaces of
degree $10$ satisfying property ${\bf N}_{3,3}$ on a cubic fourfold.

###### Key words and phrases:

Adjunction mapping, Liaison, Cubic fourfolds, Matrix factorizations,
Arithmetically Cohen-Macaulay

2020 Mathematics Subject Classification: 14J60(Primary), 14D20, 14M06, 13C14
(Secondary).

##  1\. Introduction

A classical problem in algebraic geometry and commutative algebra is the study
of the equations defining projective varieties and of their syzygies. In
[EGHP06], D. Eisenbud et al. introduced the condition ${\bf N}_{d,p}$ for a
graded ring as an indication of the presence of simple syzygies. Let us recall
the definition. Let $R=k[x_{0},\ldots,x_{n}]$ denote the homogeneous
coordinate ring of the projective space $\mathbb{P}^{n}$ over an algebraically
closed field $k$ of characteristic zero, and let $I_{X}\subset R$ denote the
homogeneous ideal of an algebraic set $X\subset\mathbb{P}^{n}$ of codimension
$e$. We say that $X$ satisfies property ${\bf N}_{d,p}$($p\leq\infty$), for
some $d\geq 2$, if

$\beta_{i,j}(X):=\dim\mathrm{Tor}^{R}_{i}(R/I_{X},k)_{i+j}=0,$

for $i\leq p$ and $j\geq d$. This is the same as property ${\bf N}_{p}$
defined by Green and Lazarsfeld in [GrL85] if $X$ is projectively normal and
$d=2$. Notice that property ${\bf N}_{d,\infty}$ means that $X$ is
$d$-regular. One says that $X$ is arithmetically Cohen–Macaulay (ACM) with a
$d$-linear resolution if the only non-zero Betti numbers are $\beta_{0,0}(X)$
and $\beta_{i,d-1}(X)$ for $i=1,2,\ldots,e$.

One of the main interesting problems is to understand geometric properties of
algebraic sets satisfying property ${\bf N}_{d,p}$. In [AhK15], J. Ahn and S.
Kwak showed that if $X$ satisfies property ${\bf N}_{d,e}$, then

$\deg(X)\leq\binom{e+d-1}{d-1}.$

The set of algebraic sets satisfying the above equality is quite large. It
contains many known examples such as the algebraic set defined by the ideal of
maximal minors of a $1$-generic $d\times(e+d-1)$-matrix of linear forms, or
the $(d-1)$-secant variety of a rational normal curve of degree $(e+2d-3)$.

For $d=2$, $X$ satisfies property ${\bf N}_{2,e}$ if and only if $X$ is ACM
with $2$-linear resolution (see [EGHP05, Corollary1.8] and [EiG84,
Corollary1.11]). It follows that if in addition $\deg(X)=1+e$, then $X$ is a
variety of minimal degree. Recall that a variety of minimal degree is either a
rational normal scroll, a quadric hypersurface $Q_{n}\subset\mathbb{P}^{n+1}$,
or the Veronese surface in $\mathbb{P}^{5}$ (see [EiH87]). In this direction,
the papers [AlR02, AhK11, BeE10, BEL91, EGHP05, GrL88, Kwa98, KwP05, Laz87]
give a small sample of the kinds of investigations that have been carried out.
However, not very much is actually known about algebraic sets satisfying
property ${\bf N}_{d,p}$, for $d\geq 3$. For $d=3$, J. Ahn and S. Kwak showed
in [AhK15] that $X$ satisfies property ${\bf N}_{3,e}$ and
$\deg(X)=\binom{e+2}{2}$ if and only if $X$ is ACM with a $3$-linear
resolution. Along this line, it is natural to study the classification problem
of smooth projective varieties $X$ satisfying property ${\bf N}_{3,e}$ and
$\deg(X)=\binom{e+2}{2}$. For $e=2$, smooth projective varieties $X$ of degree
$6$ satisfying property ${\bf N}_{3,2}$ are defined by the ideal of maximal
minors of a $1$-generic $3\times 4$-matrix of linear forms. Thus the first
step would be a classification of smooth projective varieties $X$ of degree
$10$ and codimension $3$ satisfying property ${\bf N}_{3,3}$. For curves, R.
Hartshorne gave a complete classification of projective complex curves of
degree $10$ satisfying property ${\bf N}_{3,3}$ in $\mathbb{P}^{4}$ (see
[Har01]). For surfaces, a numerical classification of rational surfaces of
degree $10$ was given by Livorni in [Liv90]. But the problem of the existence
of such surfaces remains open. In particular, it is still an open problem to
give a complete geometric classification or characterization of projective
complex surfaces of degree $10$ satisfying property ${\bf N}_{3,3}$ ([Man00,
Section 6], [Man01] and [AHK20, Question 3]). The following theorem which is
the main result of this paper gives an answer to this problem.

######  Theorem 1.1.

Let $X$ be a non-degenerate projective surface in $\mathbb{P}^{5}$. Then the
statements are equivalent.

  * $1)$

$\deg X=10$ and $X$ satisfies property ${\bf N}_{3,3}$.

  * $2)$

$X$ is ACM with the $3$-linear resolution

($*$) |  | $0$ | $1$ | $2$ | $3$
---|---|---|---|---|---
$0$ |  | $1$ | $\cdot$ | $\cdot$ | $\cdot$
$1$ |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
$2$ |  | $\cdot$ | $10$ | $15$ | $6$

  * $3)$

$X$ is an ACM surface of degree $10$ and sectional genus $6$.

In this case, $X$ is smooth and $X$ belongs to one of the seven families as in
Table 1.

$K^{2}_{X}$ | $\mathcal{C}_{\delta}$ | $H$ | Abstract structure | $h^{0}(\mathcal{N}_{X/Y})$
---|---|---|---|---
$-6$ | $\mathcal{C}_{8}$ | $X=\mathbb{P}^{2}(5;1^{15})$ | Veronese surface | $12$
$-5$ | $\mathcal{C}_{14}$ | $X=\mathbb{P}^{2}(6;2^{4},1^{10})$ |  del Pezzo surface of degree $5$ | $10$
$-4$ | $\mathcal{C}_{20}$ | $X=\mathbb{P}^{2}(7;3^{1},2^{6},1^{6})$ |  conic bundle over $\mathbb{P}^{1}$ of degree $6$ | $8$
$-3$ | $\mathcal{C}_{26}$ | $X=\mathbb{P}^{2}(7;2^{9},1^{3})$ | $\mathbb{P}^{2}$ | $6$
$-2$ | $\mathcal{C}_{32}$ | $X=\mathbb{P}^{2}(9;3^{6},2^{4},1^{1})$ |  del Pezzo surface of degree $3$. | $4$
$-1$ | $\mathcal{C}_{38}$ | $X=\mathbb{P}^{2}(10;3^{10})$ | $\mathbb{P}^{2}$ | $2$
$0$ | $\mathcal{C}_{44}$ |  $X$ is a Fano model of |  | $0$
|  | an Enriques surface |  |
Table 1.  Smooth projective surfaces $X\subset Y\subset\mathbb{P}^{5}$ of
degree $10$ with property ${\bf N}_{3,3}$, where $Y$ is a cubic fourfolds and
$\mathcal{C}_{\delta}$ is a locus of special cubic fourfolds $Y$ of a
discriminant $\delta$ (see Section 4).

Let us explain how this paper is organized. This paper is divided into 5
sections. We give a proof on our main result and consider some applications
and examples in Section 3. In our argument, adjunction theory play an
important role. Thus in Section 2 let us briefly note the adjunction theory.
In Section 4 we illustrate briefly a connection between the study of smooth
projective surfaces of degree $10$ with property ${\bf N}_{3,3}$ and special
cubic fourfolds. In Section 5, as corollaries to the main theorem, we study
the CI-biliaison equivalence class of smooth projective surfaces of degree
$10$ satisfying property ${\bf N}_{3,3}$ on a cubic fourfold.

##  2\. preliminaries

In this section, we will give the basic definitions and results that are
needed. For a smooth surface $X$ in a projective space we denote by $H_{X}$
and $K_{X}$ the class of a hyperplane and the canonical divisor, respectively.
The structure sheaf of $X$ is $\mathcal{O}_{X}$, and the Euler-Poincare
characteristic of its structure sheaf is
$\chi_{X}=\chi(\mathcal{O}_{X})=1-q+p_{g}$, where $q$ is irregularity and
$p_{g}$ is geometric genus of $X$. The degree of $X$ is $d=H^{2}$ and the
sectional genus $\pi=\frac{1}{2}H(H+K)+1$ is the genus of its general
hyperplane section. The following lemma is a special case of the Riemann-Roch
Theorem.

######  Proposition 2.1.

Let $X\subseteq\mathbb{P}^{5}$ be a smooth surface of degree $d$, sectional
genus $\pi$, irregularity $q$ and geometric genus $p_{g}$. Then we have

$\chi(\mathcal{I}_{X}(m))=\chi(\mathcal{O}_{\mathbb{P}^{5}}(m))-\binom{m+1}{2}d+m(\pi-1)-1+q-p_{g}.$

In fact, we proceed in a similar fashion as in [DeS98, Section 8 and 9] or
[TrY20a] and give a complete geometric classification or characterization of
projective complex surfaces of degree $10$ satisfying property ${\bf N}_{3,3}$
via the adjunction process. Thus we recall a result of Sommese and Van de Ven
for a surface over $\mathbb{C}$.

######  Theorem 2.2 ([Som79, Van79, SoV87], see also [DeS98, Theorem 8.1]).

Let $X$ be a smooth surface in $\mathbb{P}^{n}$ over $\mathbb{C}$. Then the
adjoint linear system $|H+K|$ is non-special of dimension
$N:=h^{0}(\mathcal{O}_{X}(H+K))-1=\chi+\pi-2$ and it defines a birational
morphism

$\Phi=\Phi_{|H+K|}:X\to\mathbb{P}^{N}$

onto a smooth surface $X_{1}$, which blows down precisely all $(-1)$–curves
$E$ on $X$, unless

  1. (i)

$X$ is a plane, or Veronese surface, or $Y$ is ruled by lines;

  2. (ii)

$X$ is a Del Pezzo surface or a conic bundle;

  3. (iii)

$X$ belongs to one of the following four families:

    1. (a)

$X=\mathbb{P}^{2}(p_{1},\ldots,p_{7})$ embedded by
$H=6L-2(p_{1}+\ldots+p_{7})$

    2. (b)

$X=\mathbb{P}^{2}(p_{1},\ldots,p_{8})$ embedded by
$H=6L-2(p_{1}+\ldots+p_{7})-p_{8}$

    3. (c)

$X=\mathbb{P}^{2}(p_{1},\ldots,p_{8})$ embedded by
$H=9L-3(p_{1}+\ldots+p_{8})$

    4. (d)

$X=\mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is an indecomposable rank $2$
bundle over an elliptic curve and $H=B$, where $B$ is a section $B^{2}=1$ on
$X$.

######  Proposition 2.3 ([DeS98, Proposition 8.3]).

Let $X$ be a surface over a field of arbitrary characteristic. Suppose that
the adjoint linear system $|H+K|$ is base point free and that the image
$X_{1}\subseteq\mathbb{P}^{N}$ under the adjunction map $\Phi=\Phi_{|H+K|}$ is
a surface of expected degree $(H+K)^{2}$, expected sectional genus
$\pi_{1}=\frac{1}{2}(H+K)\cdot(H+2K)+1$, and with
$\chi(\mathcal{O}_{X})=\chi(\mathcal{O}_{X_{1}})$. Then $X_{1}$ is smooth and
$\Phi:X\to X_{1}$ is a simultaneous blow down of ${K_{1}}^{2}-K^{2}$ many
$(-1)$–lines on $X$, where $K_{1}$ is the canonical divisor of $X_{1}$.

We also need the Hodge index theorem.

######  Theorem 2.4.

[BeS95, Lemma 2.5.2] Let $X\subseteq\mathbb{P}^{n}$ be an irreducible surface.
Then the intersection matrix

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot K&K^{2}\end{pmatrix}$

is indefinite, unless $H$ and $K$ are numerical dependent.

To apply Theorem 2.2 repeatedly, we have to control the numerical data of the
first adjoint surface $X_{1}\subseteq\mathbb{P}^{N}$. We have

$H_{1}^{2}=(H+K)^{2}\text{ and }H_{1}\cdot K_{1}=(H+K)\cdot K.$

Then $K_{1}^{2}=K^{2}+a$, where $a$ is the number of exceptional lines on $X$,
i.e. of curves $E\subseteq X$ with $E\cdot K=E^{2}=-1$ and $E\cdot H=1$.
Moreover, $X_{1}$ is a smooth projective surface.

######  Proposition 2.5.

Let $X=\mathbb{P}^{2}(p_{1},\ldots,p_{\ell})$ be the blow–up of
$\mathbb{P}^{2}$ in $\ell$ points in general position. We denote by
$E_{1},\ldots,E_{\ell}$ the corresponding exceptional divisors and by $L$ the
pullback of a general line in $\mathbb{P}^{2}$ to $X$. If
$|b_{0}L-\sum\limits_{i=1}^{\ell}b_{i}E_{i}|$ is a very ample linear system of
dimension $5$, then $X\subseteq\mathbb{P}^{5}_{\mathbb{C}}$ is a rational
surface of degree

$d=b_{0}^{2}-\sum\limits_{i=1}^{\ell}b_{i}^{2}$

and sectional genus

$\pi=\binom{b_{0}-1}{2}-\sum\limits_{i=1}^{\ell}\binom{b_{i}}{2}.$

The self–intersection of the canonical divisor of $X$ is $K^{2}=9-\ell$.

###### Proof.

Set $H=b_{0}L-\sum\limits_{i=1}^{\ell}b_{i}E_{i}$. Note that $L^{2}=1$,
$LE_{i}=0$ and $E_{i}^{2}=-1$ for all $i=1,\ldots,\ell$. Then we have
$d=H^{2}=b_{0}^{2}-\sum\limits_{i=1}^{\ell}b_{i}^{2}.$ The canonical divisor
of $X$ is $-3L+\sum\limits_{i=1}^{l}E_{i}$, so $K^{2}=9-\ell$. The sectional
genus of $X$ can be calculated by adjunction:

$\pi=\frac{H(H+K)}{2}+1=\binom{b_{0}-1}{2}-\sum\limits_{i=1}^{\ell}\binom{b_{i}}{2}.$

∎

##  3\. classification via adjunction mapping

In this section, we prove the first main result, Theorem 1.1, using adjunction
mappings. Let $X$ be a non-degenerate projective surface in $\mathbb{P}^{5}$.
Suppose that $X$ satisfies property ${\bf N}_{3,3}$ and $\deg(X)=10$. Then by
Theorem 1.2 in [AhK15], $X$ is arithmetically Cohen-Macaulay and $I_{X}$ has
$3$-linear resolution. Moreover, $X$ has the minimal free resolution as in $*$
‣ $2)$ and sectional genus of $X$ is $6$.

In general, by Lemma 1.1, Proposition 1.3 and Remarks in [Man01] (see also
[Man00]), we know that, if $Z$ is a non-degenerate projective Cohen-Macaulay
surface in $\mathbb{P}^{n_{0}}$, then

$n_{0}=d-\pi+1+p_{a}-h^{2}(\mathcal{O}_{Z}(1))=d-\pi+1+h^{1}(\mathcal{O}_{Z}(1)),$

where $p_{a}$ is the arithmetic genus of $Z$, and so we have the bounds

$n_{0}-1\leq d\leq\binom{n_{0}}{2}+h^{1}(\mathcal{O}_{Z}(1)).$

Moreover, if $\deg(Z)=\binom{n_{0}}{2}$ and $h^{1}(\mathcal{O}_{Z}(1))=0$ then
the homogeneous ideal of $Z$ can always be generated by forms of degree $3$
([Man00]). So, in $\mathbb{P}^{5}$, projective Cohen-Macaulay surfaces $Z$ of
degree $10$ sectional genus $6$ have $h^{1}(\mathcal{O}_{Z}(1))=0$ and so
$p_{a}=q=p_{g}=0$. In particular, $X$ has $h^{1}(\mathcal{O}_{X}(1))=0$, and
$p_{a}=q=p_{g}=0$.

######  Lemma 3.1.

Let $X$ be a non-degenerate projective surface in $\mathbb{P}^{5}$. Suppose
that $X$ satisfies property ${\bf N}_{3,3}$ and $\deg(X)=10$. Then we have

$-6\leq K_{X}^{2}\leq 0.$

###### Proof.

By the hypothesis, $X$ is a surface of degree $10$ and sectional genus $6$ in
$\mathbb{P}^{5}$. Then we have $HK=2.6-2-10=0$. By the Hodge index theorem, we
obtained that $K^{2}\leq\frac{1}{10}(H\cdot K)^{2}=0$. The adjunction mapping
is defined by the linear system $H+K$. By Riemann-Roch and the Kodaira
Vanishing Theorem, we get that

$h^{0}(\mathcal{O}_{X}(H+K))=\chi+\pi-1=6.$

By Theorem 2.2, the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$. Therefore since a nondegenerate
surface $Z\subset\mathbb{P}^{N}$ has degree $\deg Z\geq N-1$ (see [Xam81]), we
have

$4=5-1\leq\deg X_{1}=H_{1}^{2}=(H+K)^{2}=H^{2}+K^{2}.$

Since $H^{2}=10$, we have $K^{2}\geq-6$. ∎

###  3.1. Surface $X$ with $K^{2}=-6$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-6\end{pmatrix}.$

By Theorem 2.2, we have $h^{0}(\mathcal{O}_{X}(H+K))=\chi+\pi-1=6$. Moreover,
the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}4&-6\\\
-6&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem, we have $K_{1}^{2}\leq 9$. Since
$(H_{1}+K_{1})^{2}\geq 0$, we have $K_{1}^{2}\geq-2H_{1}\cdot
K_{1}-H_{1}^{2}=8$. It follows from the classification of surfaces of degree
$4$ (see [Ion82] or [Liv90]) that we have two cases.

  1. (i)

If $K^{2}_{1}=9$ then $X_{1}$ is a smooth Veronese surface with intersection
matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}4&-6\\\ -6&9\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of $15=K_{1}^{2}-K^{2}$
many $(-1)$–lines on $X$. Since $K^{2}=-6$, the number of exceptional divisors
of surface $X$ is $9-K^{2}=15$. It follows that
$X=\mathbb{P}^{2}(p_{1},\ldots,p_{15})$ is a blowup of $\mathbb{P}^{2}$ in
fifteen points in general position by the complete linear system

$H=(5,1^{15})=5L-\sum\limits_{i=1}^{15}E_{i}.$

Moreover, $X$ is smooth by Proposition 2.3 and $\chi_{top}=3+15=18$.

  2. (ii)

If $K^{2}_{1}=8$ then $X_{1}$ is a surface
$\mathbb{P}^{1}\times\mathbb{P}^{1}\subseteq\mathbb{P}^{5}$ of degree $4$ with
intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}4&-6\\\ -6&8\end{pmatrix}.$

It follows that $X=\mathbb{F}_{0}(p_{1},\ldots,p_{14})$ is a blowup of
Hirzebruch surface $\mathbb{F}_{0}$ in fourteen points in general position by
the complete linear system

$H=(4,3;1^{14})=4C_{0}+3f-\sum\limits_{i=1}^{14}E_{i},$

where $C_{0}$ stands for a section of minimal self-intersection and $f$ for a
fiber. Since $X$ has maximum degree $10=\binom{5}{2}$, to show that the
surface $X$ does not satisfy property ${\bf N}_{3,3}$ it is sufficient to
verify that $h^{0}(\mathcal{I}_{X}(2))=1$. We verified this via Macaulay2 and
we can conclude that $h^{0}(\mathcal{I}_{X}(2))=1$ (see [Tr2019]).

###  3.2. Surface with $K^{2}=-5$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-5\end{pmatrix}.$

By Theorem 2.2, we have $h^{0}(\mathcal{O}_{X}(H+K))=\chi+\pi-1=6$. Moreover,
the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}5&-5\\\
-5&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{1}+K_{1})^{2}\geq 0$, we have
$K_{1}^{2}=5$. It follows from the classification of surfaces of degree $5$
(see [Ion82] or [Liv90]) that $X_{1}$ is a smooth Del Pezzo surface of degree
$5$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}5&-5\\\ -5&5\end{pmatrix}.$

Since $K_{1}^{2}-K^{2}=10$, $\Phi:X\to X_{1}$ is a simultaneous blow down of
ten $(-1)$–lines on $X$. Since $K^{2}=-5$, the number of exceptional divisors
of surface $Y$ is $9-K^{2}=14$. It follows that
$X=\mathbb{P}^{2}(p_{1},\ldots,p_{14})$ is a blowup of $\mathbb{P}^{2}$ in
general fourteen points by the complete linear system

$H=(6;2^{4},1^{10})=6L-\sum\limits_{i=1}^{4}2E_{i}-\sum\limits_{i=5}^{14}E_{i}.$

Moreover, $X$ is smooth by Proposition 2.3 and $\chi_{top}=3+14=17$.

###  3.3. Surface with $K^{2}=-4$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-4\end{pmatrix}.$

By Theorem 2.2, we have $h^{0}(\mathcal{O}_{X}(H+K))=\chi+\pi-1=6$. Moreover,
the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}6&-4\\\
-4&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{1}+K_{1})^{2}\geq 0$, we have
$K_{1}^{2}=2$. Therefore $X_{1}$ is a surface of degree $6$ and sectional
genus $2$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}6&-4\\\ -4&2\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of six $(-1)$–lines on $X$.
It follows from the classification of surfaces of degree $6$ (see [Ion82] or
[Liv90]) that $X_{1}\subseteq\mathbb{P}^{5}$ is a conic bundle
$X_{1}\to\mathbb{P}^{1}$ with $6$ singular fibers. So we have the following
two choices for $X$.

  1. (i)

First, $X_{1}=\mathbb{P}^{2}(p_{1},\ldots,p_{7})$ is a blowup of
$\mathbb{P}^{2}$ in general seven points by the complete linear system
$H_{1}=(4;2^{1},1^{6})$, and so $X=\mathbb{P}^{2}(p_{1},\ldots,p_{13})$ is a
blowup of $\mathbb{P}^{2}$ in general thirteen points by the complete linear
system

$H=(7;3^{1},2^{6},1^{6})=7L-3E_{1}-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}.$

Moreover, $X$ is smooth by Proposition 2.3 and $\chi_{top}=3+13=16$.

  2. (ii)

Second, $X_{1}=\mathbb{F}_{e}(p_{2},\ldots,p_{7})$ is a blowup of Hirzebruch
surface $\mathbb{F}_{e}$ in general six points, and so
$X=\mathbb{F}_{e}(p_{2},\ldots,p_{13})$ is a blowup of Hirzebruch surface
$\mathbb{F}_{e}$ in general twelve points by the complete linear system

$H_{e}=(4,2e+5;2^{6},1^{6})=4C_{0}+(2e+5)f-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}.$

Moreover, $0\leq e\leq 2$, since $H\cdot C_{0}\geq 1$. In the case $e=2$, $H$
is not a very ample linear system of dimension $5$. Therefore, $0\leq e\leq
1$. Note that for $e=0$, $1$, another description of the surface $X$ can be
given, using a plane model, as follows.

    1. a)

${\bf e=0:}$ $\mathbb{F}_{0}(p_{2},\ldots,p_{13})$ is isomorphic to the blowup
of $\mathbb{P}^{2}$ at general thirteen points by the complete linear system
$H=(7;3^{1},2^{6},1^{6})$. In fact, $\mathbb{F}_{0}$ is isomorphic to the
quadric surface $Q\subseteq\mathbb{P}^{3}$ and $Q$ is obtained by the complete
(not very ample) linear system $|2L-E_{1}-E_{2}|$. Since the line
$|L-E_{1}-E_{2}|$ is contracted to a point $P\in Q$, we have that to blowup
$Q$ at the point $P$ and at other general eleven points is equivalent to a
blowup of $\mathbb{P}^{2}$ at general thirteen points, as we said. Taking
$C_{0}=L-E_{2}$ and $f=L-E_{1}$, to the very ample divisor
$H_{0}=4C_{0}+5f-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$ on
$\mathbb{F}_{0}(p_{2},\ldots,p_{13})$ corresponds the very ample divisor

$\displaystyle H_{0}$
$\displaystyle=4(L-E_{2})+5(L-E_{1})-2(L-E_{1}-E_{2})-\sum\limits_{i=3}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$
$\displaystyle=7L-3E_{1}-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$

on $X=\mathbb{P}^{2}(p_{1},\ldots,p_{13})$, which is the same divisor we found
in (i).

    2. b)

${\bf e=1:}$ Let us determine the very ample divisor $H=tL-
mE_{1}-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$ on $X$ which
corresponds to the divisor
$4C_{0}+7f-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$ on
$\mathbb{F}_{1}(p_{2},\ldots,p_{13})$. The integers $t,m>0$ are such that

$\begin{cases}\binom{t+2}{2}-\binom{m+1}{2}-3\cdot 6-6=6\\\
t^{2}-m^{2}-30=10.\end{cases}$

Solving the equations we find $t=7$ and $m=3$. Hence
$H=7L-3E_{1}-\sum\limits_{i=2}^{7}2E_{i}-\sum\limits_{i=8}^{13}E_{i}$, which
is the same divisor we found in (i).

###  3.4. Surface with $K^{2}=-3$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-3\end{pmatrix}.$

By Theorem 2.2, we have $h^{0}(\mathcal{O}_{X}(H+K))=\chi+\pi-1=6$. Moreover,
the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}7&-3\\\
-3&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{1}+K_{1})^{2}\geq 0$, we have $-1\leq
K_{1}^{2}\leq 1$. It follows from the classification of surfaces of degree $7$
(see [Ion82] or [Liv90]) that we have three cases.

  1. (i)

If $K^{2}_{1}=1$ then $X_{1}$ is a smooth surface of degree $7$, sectional
genus $3$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}7&-3\\\ -3&1\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of four $(-1)$–lines on
$X$. Moreover, by Theorem 2.2, $X_{1}=\mathbb{P}^{2}(p_{1},\ldots,p_{8})$ is a
blowup of $\mathbb{P}^{2}$ in general eight points by the complete linear
system $H_{1}=(6;2^{7},1^{1})$. Hence, $X=\mathbb{P}^{2}(p_{1},\ldots,p_{12})$
is a blowup of $\mathbb{P}^{2}$ in twelve points in general position by the
complete linear system

$H=(9;3^{7},2^{1},1^{4})=9L-\sum\limits_{i=1}^{7}3E_{i}-2E_{8}-\sum\limits_{i=9}^{12}E_{i}.$

Since $X$ has maximum degree $10=\binom{5}{2}$, to show that the surface $X$
does not satisfy property ${\bf N}_{3,3}$ it is sufficient to verify that
$h^{0}(\mathcal{I}_{X}(2))=1$. We verified this via Macaulay2 and we can
conclude that $h^{0}(\mathcal{I}_{X}(2))=1$ (see [Tr2019]).

  2. (ii)

If $K^{2}_{1}=0$ then $X_{1}$ is a smooth surface of degree $7$, sectional
genus $3$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}7&-3\\\ -3&0\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of three $(-1)$–lines on
$X$. By Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=3$.
Moreover, the adjunction map

$\Phi^{1}=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to\mathbb{P}^{2}$

is birational onto a plane $\mathbb{P}^{2}$. Since $K^{2}=-3$, the number of
exceptional divisors of surface $X$ is $9-K^{2}=12$. Thus
$\Phi^{1}:X_{1}\to\mathbb{P}^{2}$ is a simultaneous blow down of $12-3=9$ many
$(-1)$–lines on $X_{1}$. Therefore $X=\mathbb{P}^{2}(p_{1},\ldots,p_{12})$ is
a blowup of $\mathbb{P}^{2}$ in general twelve points by the complete linear
system

$H=(7;2^{9},1^{3})=7L-\sum\limits_{i=1}^{9}2E_{i}-\sum\limits_{i=10}^{12}E_{i}.$

  3. (iii)

If $K^{2}_{1}=-1$ then $X_{1}$ is a smooth surface of degree $7$, sectional
genus $3$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}7&-3\\\ -3&-1\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of two $(-1)$–lines on $X$.
Therefore $X=\mathbb{F}_{e}(p_{1},\ldots,p_{11})$ is a blowup of Hirzebruch
surface $\mathbb{F}_{e}$ in general eleven points by the complete linear
system

$H=(4,2e+6;2^{9},1^{2})=4C_{0}+(2e+6)f-\sum\limits_{i=1}^{9}2E_{i}-E_{10}-E_{11}.$

Morveover $0\leq e\leq 2$ since $H\cdot C_{0}\geq 1$. Since $X$ has maximum
degree $10=\binom{5}{2}$, to show that the surface $X$ does not satisfy
property ${\bf N}_{3,3}$ it is sufficient to verify that
$h^{0}(\mathcal{I}_{X}(2))=1$. We verified this via Macaulay2 and we can
conclude that $h^{0}(\mathcal{I}_{X}(2))=1$ (see [Tr2019]).

###  3.5. Surface with $K^{2}=-2$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-2\end{pmatrix}.$

By Theorem 2.2, the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}8&-2\\\
-2&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{1}+K_{1})^{2}\geq 0$, we have $-2\leq
K_{1}^{2}\leq 0$. Therefore we have three cases.

  1. (i)

If $K^{2}_{1}=0$ then $X_{1}$ is a smooth surface of degree $8$, sectional
genus $4$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}8&-2\\\ -2&0\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of two $(-1)$–lines on $X$.
By Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=4$.
Moreover, the adjunction map

$\Phi^{1}=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to X_{2}\subseteq\mathbb{P}^{3}$

is birational onto a smooth surface $X_{2}$ of degree $4$ with intersection
matrix

$\begin{pmatrix}H_{2}^{2}&H_{2}\cdot K_{2}\\\ H_{2}\cdot
K_{2}&K_{2}^{2}\end{pmatrix}=\begin{pmatrix}4&-2\\\
-2&K_{2}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{2}+K_{2})^{2}\geq 0$, we have
$K_{2}^{2}\in\\{0,1\\}$. By Theorem 2.2 and Proposition 2.5, there doesn’t
exist surface $X_{2}$.

  2. (ii)

If $K^{2}_{1}=-1$ then $X_{1}$ is a smooth surface of degree $8$, sectional
genus $4$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}8&-2\\\ -2&-1\end{pmatrix}.$

Thus $\Phi:X\to X_{1}$ is a simultaneous blow down of a $(-1)$–line on $X$. By
Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=4$.
Moreover, the adjunction map

$\Phi^{1}=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to X_{2}\subseteq\mathbb{P}^{3}$

is birational onto a smooth cubic surface $X_{2}$ with intersection matrix

$\begin{pmatrix}H_{2}^{2}&H_{2}\cdot K_{2}\\\ H_{2}\cdot
K_{2}&K_{2}^{2}\end{pmatrix}=\begin{pmatrix}3&-3\\\ -3&3\end{pmatrix}.$

By the Hodge index theorem and $(H_{2}+K_{2})^{2}\geq 0$, we have
$K_{2}^{2}=3$. Thus $\Phi^{1}:X_{1}\to X_{2}$ is a simultaneous blow down of
four $(-1)$–lines on $X_{1}$. Since $K_{1}^{2}=-1$, the number of exceptional
divisors of surface $X_{1}$ is $9-K_{1}^{2}=10$. Therefore
$X=\mathbb{P}^{2}(p_{1},\ldots,p_{11})$ is a blowup of $\mathbb{P}^{2}$ in
general eleven points by the complete linear system

$H=(9;3^{6},2^{4},1^{1})=9L-\sum\limits_{i=1}^{6}3E_{i}-\sum\limits_{i=7}^{10}2E_{i}-E_{11}.$

  3. (iii)

If $K^{2}_{1}=-2$ then $X_{1}$ is a smooth surface of degree $8$, sectional
genus $4$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}8&-2\\\ -2&-2\end{pmatrix}.$

By Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=4$.
Moreover, the adjunction map

$\Phi^{1}=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to X_{2}\subseteq\mathbb{P}^{3}$

is birational onto a smooth quadric surface $X_{2}$ with intersection matrix

$\begin{pmatrix}H_{2}^{2}&H_{2}\cdot K_{2}\\\ H_{2}\cdot
K_{2}&K_{2}^{2}\end{pmatrix}=\begin{pmatrix}2&-4\\\
-4&K_{2}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{2}+K_{2})^{2}\geq 0$, we have
$K_{2}^{2}\in\\{6,7,8\\}$. By Proposition 2.5, we have the following two
choices for $X$.

    1. (a)

First $X=\mathbb{P}^{2}(p_{1},\ldots,p_{11})$ is a blowup of $\mathbb{P}^{2}$
in general eleven points by the complete linear system

$H=(8;3^{2},2^{9},1^{0})=8L-3E_{0}-3E_{1}-\sum\limits_{i=2}^{11}2E_{i}.$

Since $X$ has maximum degree $10=\binom{5}{2}$, to show that the surface $X$
does not satisfy property ${\bf N}_{3,3}$ it is sufficient to verify that
$h^{0}(\mathcal{I}_{X}(2))=1$. We verified this via Macaulay2 and we can
conclude that $h^{0}(\mathcal{I}_{X}(2))=1$ (see [Tr2019]).

    2. (b)

Second $X=\mathbb{F}_{e}(p_{2},\ldots,p_{11})$ is a blowup of Hirzebruch
surface in general ten points by the complete linear system

$H=(5,2e+5;2^{10})=5C_{0}+(2e+5)f-\sum\limits_{i=2}^{11}2E_{i}.$

Morveover $e=0$ since $H\cdot C_{0}\geq 1$. Moreover,
$\mathbb{F}_{0}(p_{2},\ldots,p_{11})$ is isomorphic to the blowup of
$\mathbb{P}^{2}$ at general eleven points by the complete linear system
$H=(8;3^{2},2^{9},1^{0})$. In fact, $\mathbb{F}_{0}$ is isomorphic to the
quadric surface $Q\subseteq\mathbb{P}^{3}$ and $Q$ is obtained by the complete
(not very ample) linear system $|2L-E_{1}-E_{2}|$. Since the line
$|L-E_{1}-E_{2}|$ is contracted to a point $P\in Q$, we have that to blowup
$Q$ at the point $P$ and at other general eleven points is equivalent to a
blowup of $\mathbb{P}^{2}$ at general thirteen points, as we said. Taking
$C_{0}=L-E_{2}$ and $f=L-E_{1}$, to the very ample divisor
$H_{0}=5C_{0}+5f-\sum\limits_{i=2}^{10}2E_{i}$ on
$\mathbb{F}_{0}(p_{2},\ldots,p_{11})$ corresponds the very ample divisor

$\displaystyle H_{0}$
$\displaystyle=5(L-E_{2})+5(L-E_{1})-2(L-E_{1}-E_{2})-\sum\limits_{i=3}^{11}2E_{i}$
$\displaystyle=8L-3E_{1}-3E_{2}-\sum\limits_{i=3}^{11}2E_{i}$

on $X=\mathbb{P}^{2}(p_{1},\ldots,p_{11})$, which is the same divisor we found
in (a). Thus surface $X$ is not Cohen-Macaulay.

###  3.6. Surface with $K^{2}=-1$

The intersection matrix of $X$ is

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&-1\end{pmatrix}.$

By Theorem 2.2, the adjunction map

$\Phi=\Phi_{|H+K|}:X\to X_{1}\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X_{1}$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}9&-1\\\
-1&K_{1}^{2}\end{pmatrix}.$

By the Hodge index theorem and $(H_{1}+K_{1})^{2}\geq 0$, we have
$K_{2}^{2}\in\\{-1,0\\}$. Therefore we have two cases.

  1. (i)

If $K^{2}_{1}=0$ then $X_{1}$ is a smooth surface of degree $9$, sectional
genus $5$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}9&-1\\\ -1&0\end{pmatrix}.$

By Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=5$.
Moreover, the adjunction map

$\Phi=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to X_{2}\subseteq\mathbb{P}^{4}$

is birational onto a smooth surface $X_{2}$ with intersection matrix

$\begin{pmatrix}H_{2}^{2}&H_{2}\cdot K_{2}\\\ H_{2}\cdot
K_{2}&K_{2}^{2}\end{pmatrix}=\begin{pmatrix}7&-1\\\
-1&K^{2}_{2}\end{pmatrix}.$

By the Hodge index theorem and $K_{2}^{2}\geq K_{1}^{2}$, we have
$K_{2}^{2}=0$. Since $X_{2}$ is a smooth surface of degree $7$ in
$\mathbb{P}^{4}$ with normal bundle $\mathcal{N}_{X_{2}}$ the numerical
invariants of $X_{2}$ satisfy the double point formula

$d_{X_{2}}^{2}-10d_{X_{2}}-5H_{2}\cdot K_{2}-2K_{2}^{2}+12\chi_{X_{2}}=0$

(see [Har77, Appendix A, 4.1.3]). Thus we have $3\chi_{X_{2}}=4$, impossible.

  2. (ii)

If $K^{2}_{1}=-1$ then $X_{1}$ is a smooth surface of degree $9$, sectional
genus $5$ with intersection matrix

$\begin{pmatrix}H_{1}^{2}&H_{1}\cdot K_{1}\\\ H_{1}\cdot
K_{1}&K_{1}^{2}\end{pmatrix}=\begin{pmatrix}9&-1\\\ -1&-1\end{pmatrix}.$

By Theorem 2.2, we have
$h^{0}(\mathcal{O}_{X_{1}}(H_{1}+K_{1}))=\chi(X_{1})+\pi_{X_{1}}-1=5$.
Moreover, the adjunction map

$\Phi^{1}=\Phi^{1}_{|H_{1}+K_{1}|}:X_{1}\to X_{2}\subseteq\mathbb{P}^{4}$

is birational onto a smooth cubic surface $X_{2}$ with intersection matrix

$\begin{pmatrix}H_{2}^{2}&H_{2}\cdot K_{2}\\\ H_{2}\cdot
K_{2}&K_{2}^{2}\end{pmatrix}=\begin{pmatrix}6&-2\\\ -2&-1\end{pmatrix}.$

Since $X_{2}$ is a smooth surface of degree $6$ in $\mathbb{P}^{4}$ with
normal bundle $\mathcal{N}_{X_{2}}$ there is the relation

$d_{X_{2}}^{2}-10d_{X_{2}}-5H_{2}\cdot K_{2}-2K_{2}^{2}+12\chi_{X_{2}}=0.$

Thus we have $6\chi_{Y_{1}}=7+K_{2}^{2}$. Thus $K_{2}^{2}=-1$. By Theorem 2.2,
we have $h^{0}(\mathcal{O}_{X_{2}}(H_{2}+K_{2}))=\chi(X_{2})+\pi_{X_{2}}-1=3$.
Moreover, the adjunction map

$\Phi^{2}=\Phi^{2}_{|H_{2}+K_{2}|}:X_{2}\to\mathbb{P}^{2}$

is birational onto a plane $\mathbb{P}^{2}$. Since $K^{2}=-1$, the number of
exceptional divisors of surface $X$ is $9-K^{2}=10$. Thus
$\Phi^{2}:X_{2}\to\mathbb{P}^{2}$ is a simultaneous blow down of ten
$(-1)$–lines on $X_{2}$. Therefore $X=\mathbb{P}^{2}(p_{1},\ldots,p_{10})$ is
a blowup of $\mathbb{P}^{2}$ in general ten points by the complete linear
system

$H=(10;3^{10})=10L-\sum\limits_{i=1}^{10}3E_{i}.$

###  3.7. Surface with $K^{2}=0$

Surface $X$ is a smooth Enriques surface of degree $10$ in $\mathbb{P}^{5}$.
Smooth Enriques are well understood, in fact any Enriques surface has a linear
system of degree $10$ and projective dimension $5$ without base points, for
very ampleness it suffices to require that any elliptic curve on the surface
has degree at least $3$ with respect to the linear system, and that there are
no rational curves on it. These surfaces have exactly $20$ plane cubic curves
whose planes form the union of the trisecants to the surface. Thus projecting
the surface from a point on the surface away from these $20$ planes, we get a
smooth surface $X^{\prime}$ of degree $9$, sectional genus $6$ in
$\mathbb{P}^{4}$ (see [DES93]). The goal of this subsection is to construct a
surface $X^{\prime}$ of degree $9$, sectional genus $6$ in $\mathbb{P}^{4}$.
To do so, we will use work of Rao, who showed that the construction of curves
is equivalent to the creation of its Hartshorne–Rao module.

######  Definition 3.2.

The Hartshorne–Rao module of $X^{\prime}$ is the finite length module

$M=M_{X^{\prime}}:=\bigoplus\limits_{n\in\mathbb{Z}}H^{1}(\mathbb{P}^{4},\mathcal{I}_{X^{\prime}}(n))\subseteq\bigoplus\limits_{n\in\mathbb{Z}}H^{0}(\mathbb{P}^{4},\mathcal{O}(n))\cong
R^{\prime}:=k[x_{0},\ldots,x_{4}].$

Assuming the open condition that $X^{\prime}$ has maximal rank, i.e.
restriction map $H^{0}(\mathcal{O}_{\mathbb{P}^{4}}(m))\to
H^{0}(\mathcal{O}_{X^{\prime}}(m))$ has maximal rank for all $m$. Then Hilbert
series of the Hartshorne-Rao module is $H_{M}(t)=3t^{2}+10t^{3}+6t^{4}$ and
the Hilbert numerator is

$(1-t)^{4}H_{M}(t)=3t^{2}-10t^{3}+6t^{4}+15t^{5}-25t^{6}+12t^{7}-t^{9}.$

If $M$ has a natural resolution, which means that for each degree $j$ at most
one $\beta_{ij}$ is nonzero then $M$ has a Betti table

|  | $0$ | $1$ | $2$ | $3$ | $4$ | $5$
---|---|---|---|---|---|---|---
2 |  | $3$ | $10$ | $6$ | $\cdot$ | $\cdot$ | $\cdot$
3 |  | $\cdot$ | $\cdot$ | $15$ | $25$ | $12$ | $\cdot$
4 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $1$

.

Note that having a natural resolution is an open condition in a family of
modules with constant Hilbert function. If homogeneous coordinate ring
$R^{\prime}_{X^{\prime}}=R^{\prime}/I_{X^{\prime}}$ and the section ring
$\Gamma_{*}(\mathcal{O}_{X^{\prime}}):=\bigoplus\limits_{n\in\mathbb{Z}}H^{i}(\mathcal{O}_{X^{\prime}}(n))$
has natural resolution as well, then their Betti tables will be

|  | $0$ | $1$ | $2$ | $3$ | $4$
---|---|---|---|---|---|---
0 |  | $1$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
1 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
2 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
3 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
4 |  | $\cdot$ | $15$ | $25$ | $12$ | $\cdot$
5 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $1$

and  $0$ $1$ $2$ 0 $1$ $\cdot$ $\cdot$ 1 $\cdot$ $\cdot$ $\cdot$ 2 $3$ $10$
$6$ .

To construct a desired $M$, we start with the matrix $\psi$ defining
$\textstyle{{R^{\prime}}^{12}(-3)}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces{R^{\prime}}(-5)}$$\scriptstyle{\psi}$
which we choose randomly. Then we get a smooth surface $X^{\prime}$ of degree
$9$, sectional genus $6$ in $\mathbb{P}^{4}$. Moreover the intersection matrix
of $X^{\prime}$ is

$\begin{pmatrix}H_{X^{\prime}}^{2}&H_{X^{\prime}}\cdot K_{X^{\prime}}\\\
H_{X^{\prime}}\cdot
K_{X^{\prime}}&K_{X^{\prime}}^{2}\end{pmatrix}=\begin{pmatrix}9&1\\\
1&-1\end{pmatrix}.$

Notice that $X^{\prime}$ is regular and $|2K_{X^{\prime}}|\neq 0$, then
$2K_{X^{\prime}}=2E_{X^{\prime}}$. Thus $X^{\prime}$ is an Enriques surface
blown up in one point. By Theorem 2.2, the adjunction map

$\Phi^{\prime}=\Phi^{\prime}_{|H_{X^{\prime}}+K_{X^{\prime}}|}:X^{\prime}\to
X\subseteq\mathbb{P}^{5}$

is birational onto a smooth surface $X$ of degree $10$ with intersection
matrix

$\begin{pmatrix}H^{2}&H\cdot K\\\ H\cdot
K&K^{2}\end{pmatrix}=\begin{pmatrix}10&0\\\ 0&0\end{pmatrix}.$

We summarize the above discussion in the following proposition.

######  Proposition 3.3.

Let $X$ be a non-degenerate projective surface in $\mathbb{P}^{5}$. Suppose
that $X$ satisfies property ${\bf N}_{3,3}$ and $\deg(X)=10$. Then $X$ is a
projective smooth Cohen-Macaulay surface with the minimal free resolution as
in $*$ ‣ $2)$, and $X$ is an image of $\mathbb{P}^{2}$ under a rational map,
defined by a complete linear series as in Table 1 or a Fano model of an
Enriques surface. Moreover, $X$ defines a $9$-dimensional family of cubic
fourfolds $Y$ with $X\subset Y$.

Note in the proof of above Proposition, we claimed that there exists an
example of surfaces $X$ of degree $10$ and sectional genus $6$ which is also
embedded in $\mathbb{P}^{5}$. In an ancillary file (see [Tr2019]), we provide
the explicit homogeneous ideals of such surfaces. Notice that last statement
of above Proposition is an immediate corollary of
$h^{0}(\mathcal{I}_{X}(3))=10$.

To prove Theorem 1.1, we need to prove the following result.

######  Proposition 3.4.

Let $X$ be a general projective Cohen-Macaulay surface of degree $10$,
sectional genus $6$. Then $X$ is of maximal rank and homogeneous coordinate
ring $R_{Y}=R/I_{X}$ and the section ring $\Gamma_{*}(\mathcal{O}_{X})$ have
minimal free resolutions with the Betti tables as in $*$ ‣ $2)$.

###### Proof.

Assuming that restriction map $H^{0}(\mathcal{O}_{\mathbb{P}^{5}}(m))\to
H^{0}(\mathcal{O}_{X}(m))$ has maximal rank. Then we have the following
statements.

  * •

The Hilbert series of the homogeneous coordinate ring of $X$ is

$H_{X}(t)=1+6t+21t^{2}+46t^{3}+81t^{4}+126t^{5}+181t^{6}+246t^{7}+\ldots.$

  * •

The Hartshorne–Rao module

$H^{1}_{*}(\mathcal{I}_{X})=H^{2}_{*}(\mathcal{I}_{X})=0,$

where
$H^{i}_{*}(\mathcal{I}_{X})=\bigoplus\limits_{n\in\mathbb{Z}}H^{i}(\mathcal{I}_{X}(n))$
for all $i\in\mathbb{Z}$.

  * •

The Hilbert numerator has shape

$(1-t)^{5}H_{X}(t)=1-10t^{3}+15t^{4}-6t^{5}.$

Since $H_{X}(t)=H_{\Gamma_{*}(\mathcal{O}_{X})}(t)$, the Betti table of the
resolution $F^{\bullet}$ of the section ring $\Gamma_{*}(\mathcal{O}_{X})$ as
an $R$-module is as in $*$ ‣ $2)$. To show that the Betti tables are indeed
the expected ones and that a general surface $Y$ is of maximal rank, we only
need to exhibit a concrete example, which we construct in Proposition 3.3. ∎

Let us give a proof of Theorem 1.1.

######  Proof of Theorem 1.1.

$1)\Rightarrow 2)$ follows from Theorem 1.2 in [AhK15].
$2)\Rightarrow 3)$ is trivial
$3)\Rightarrow 1)$ follows from Proposition 3.4.
The next result follows immediately from the Proposition 3.3.

######  Corollary 3.5.

The moduli space of projective smooth Cohen-Macaulay surfaces of degree $10$,
sectional genus $6$ have exactly $7$ irreducible components.

##  4\. Connection with cubic fourfolds

The goal of this section is to give the relation between projective surfaces
of degree $10$ satisfying property ${\bf N}_{3,3}$ and special cubic
fourfolds. Recall that a smooth cubic fourfold $Y$, the vanishing locus of a
degree $3$ homogeneous polynomial in $6$ variables, is called special, if the
existence of an embedding of a saturated rank-$2$ lattice

$L_{\delta}:=\langle h^{2},X\rangle\hookrightarrow A(Y),$

where $A(Y)$ is the lattice of middle Hodge classes, $h\in\mathrm{Pic}(Y)$ is
the hyperplane class, $X$ is an algebraic surface not homologous to a complete
intersection, and $\delta$ is the determinant of the intersection matrix of
$\langle h^{2},X\rangle$. In 2000, Hassett showed in [Has00] that the locus
$\mathcal{C}_{\delta}$ of special cubic fourfolds $Y$ of a discriminant
$\delta$ is an irreducible divisor which is nonempty if and only if $\delta>6$
and $\delta\equiv 0,2\pmod{6}$. Recently, the study of divisors
$\mathcal{C}_{\delta}$ for small $\delta$ has received considerable attention.
In particular, we systematically study concrete descriptions of special cubic
fourfolds of small discriminants. For $\delta=8$, $12$, $14$, $20$, Hassett
showed that a generic member $Y\in\mathcal{C}_{\delta}$ could be described
explicitly. The surfaces $X$ in these cases were given by planes, cubic
scrolls, quintic del Pezzos, and Veronese surfaces, respectively. For some
choices of $\delta$, the generic $Y\in\mathcal{C}_{\delta}$ admits an
alternative description. For example, the generic $Y\in\mathcal{C}_{8}$ can be
described as containing an octic K3 surface, and the generic
$Y\in\mathcal{C}_{14}$ containing a quartic scroll. In [Nue17], Nuer proved
that the generic element $Y$ of $\mathcal{C}_{\delta}$ for $12\leq\delta\leq
38$ contains a smooth rational surface obtained as the blowup of
$\mathbb{P}^{2}$ at generic points $p$, and the generic $Y\in\mathcal{C}_{44}$
contains a Fano model of an Enriques surface. Method of Nuer certainly
indicates that the method presented here has been exhausted for the most part.
On the other hand, it is worth noting that Voisin has shown that for a special
cubic fourfold $Y$, the lattice $A(Y)$ can be generated either by smooth
surfaces [Voi17, proof of Theorem 5.6] or by possibly singular rational
surfaces [Voi07], and many have wondered if $A(Y)$ is in fact generated by
smooth rational surfaces. This has certainly been true for the few low-
discriminant cases previously known, and the work of Nuer provides a lot more
evidence for this possibility. Therefore, Nuer gave the following question.

######  Question 4.1.

What does the condition that the generic $Y\in\mathcal{C}_{\delta}$ contains a
smooth rational surface not homologous to a complete intersection say, if
anything, about the geometry of $Y$? and about the geometry of
$\mathcal{C}_{\delta}$?

The following theorem gives a partial answer to this question.

######  Theorem 4.2.

Let $Y$ be a smooth cubic fourfold. Then the statements are equivalent.

  * $1)$

$Y$ contains a non-degenerate projective surface $X$ of degree $10$ satisfies
property ${\bf N}_{3,3}$.

  * $2)$

$Y\in\mathcal{C}_{\delta}$ for $6<\delta\leq 44$ and $\delta\equiv 2\pmod{6}$.

###### Proof.

$1)\Rightarrow 2)$ Assume that the cubic fourfold $Y$ contains a projective
surface $X\subset\mathbb{P}^{5}$ of degree $10$ satisfies property ${\bf
N}_{3,3}$. Let $t=K_{X}^{2}$. By Lemma 3.1, we have $-6\leq t\leq 0$. Then by
computing the self-intersection of $X\subseteq Y$ using the formula from
[Has00] and Proposition 3.3, we have

$X^{2}=c_{2}(\mathcal{N}_{X/Y})=6H^{2}+3H\cdot K+K^{2}-\chi_{top}=48+2t.$

Therefore, $\delta=3X^{2}-10^{2}=44+6t$.

$2)\Rightarrow 1)$ Assume that $X$ is a projective surface of degree $10$
satisfies property ${\bf N}_{3,3}$. Set $t=K_{X}^{2}$. Let $\mathcal{H}_{t}$
be the irreducible component of the Hilbert scheme
$\mathrm{Hilb}_{\mathbb{P}^{5}}^{\chi(\mathcal{O}_{X}(t))}$ containing $[X]$.
Then $\dim\mathcal{H}_{t}=2\cdot(9-t)+27=45-2t$, and $\mathcal{H}_{t}$ is
generically smooth. Indeed, one can verify that
$h^{1}(\mathcal{N}_{X/\mathbb{P}^{5}})=0$. Let
$U\subseteq|H^{0}(\mathbb{P}^{5}(3))|\cong\mathbb{P}^{55}$ be the open set
corresponding to smooth cubic fourfolds. Since $h^{0}(\mathcal{I}_{X}(3))=10$,
the locus $\mathcal{D}_{t}=\\{([X],[Y]):X\subseteq
Y\\}\subseteq\mathcal{H}_{t}\times U$ has dimension $45-2t+10-1=54-2t$. The
image of $\pi_{2}:\mathcal{D}_{t}\to U$ has dimension at most $54$, because
the general cubic fourfold does not contain any $X$ belonging to
$\mathcal{H}_{t}$. For every $[Y]\in\pi_{2}(\mathcal{D}_{t})$, we have

$\dim(\pi^{-1}_{2}([Y]))\geq\dim(\mathcal{D}_{t})-\dim(\pi_{2}(\mathcal{D}_{t}))=54-2t-\dim(\pi_{2}(\mathcal{C}_{\delta}))\geq
54-2t-54=-2t,$

for $-6\leq t\leq 0$. Since $h^{0}(\mathcal{N}_{X/Y})\geq\dim_{[X\subset
Y]}(\pi^{-1}_{2}([Y]))$, for every $[X\subset Y]\in\pi^{-1}_{2}([Y])$, in
order to show that a general $Y\in\mathcal{C}_{\delta}$ contains a surface $X$
it is sufficient to verify that $h^{0}(\mathcal{N}_{X/Y})=-2t$, for a general
smooth cubic fourfolds $Y\in|H^{0}(\mathcal{I}_{X}(3))|$, see also [Nue17] for
a similar argument. We verified this via Macaulay 2 ([GSM2]), and we can
conclude that $h^{0}(\mathcal{N}_{X/Y})=-2t$ for $-6\leq t\leq 0$. ∎

##  5\. ACM bundles

Our goal in this section is to study the relationship between ACM bundles and
projective surfaces of degree $10$ satisfying property ${\bf N}_{3,3}$. A
vector bundle $\mathcal{F}$ on a projective variety $Y$ is ACM if all its
intermediate cohomology groups $H^{i}(Y;\mathcal{F}(m))$ are zero for
$0<i<\dim Y$ and all $m\in\mathbb{Z}$. To associate ACM bundles to surfaces we
work with $\mathcal{F}$-type resolutions (or so-called Bourbaki sequences,
c.f. [BHU87]). We recall the definition here.

######  Definition 5.1.

Let $Y\subset\mathbb{P}^{n}$ be an equidimensional scheme, and $X\subset Y$ be
a codimension $2$ subscheme without embedded components. An $\mathcal{F}$-type
resolution of $X$ on $Y$ is an exact sequence

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
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29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
65.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
65.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
102.2778pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
102.2778pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{I}_{X,Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
146.7078pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
146.7078pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

with $\mathcal{L}$ is dissocié, and $\mathcal{F}$ is a coherent sheaf
satisfying
$H_{\bullet}^{1}(\mathcal{F}^{\vee}):=\bigoplus\limits_{n\in\mathbb{Z}}H^{1}(\mathcal{F}^{\vee}(n))=0$,
and $\mathrm{Ext}^{1}(\mathcal{F},\mathcal{O}_{Y})=0.$

When $Y$ satisfies Serre’s condition $S_{2}$, and
$H_{\bullet}^{1}(\mathcal{O}_{Y})=0$, then an $\mathcal{F}$-type resolution of
$X$ exists (see [Har03, 2.12]). An $\mathcal{F}$-type resolution on $Y$ is not
unique but it is well known that any two $\mathcal{F}$-type resolutions of the
same subscheme are stably equivalent (see [Har03, 1.10]). In other words, if
$\mathcal{F}$ and $\mathcal{F}^{\prime}$ are two sheaves appearing in the
middle of an $\mathcal{F}$-type resolution of a subscheme $X$, then there
exist dissocié sheaves $\mathcal{L}_{1}$, $\mathcal{L}_{2}$, and an integer
$a$ such that

$\mathcal{F}\oplus\mathcal{L}_{1}\cong\mathcal{F}^{\prime}(a)\oplus\mathcal{L}_{2}.$

The following theorem and its converse involve the relationship between ACM
bundles and projective surfaces of degree $10$ satisfying property ${\bf
N}_{3,3}$.

######  Theorem 5.2 (cf. [TrY20, Theorem 3.2]).

Let $X$ be a smooth projective surface of degree $10$ satisfies property ${\bf
N}_{3,3}$ and $Y$ a cubic fourfold containing $X$. Then $X$ has an
$\mathcal{F}$-type resolution on $Y$

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{Y}^{6}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{I}_{X/Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

To prove Theorem 5.2, we use matrix factorizations. Matrix factorizations were
introduced by Eisenbud in his seminal paper ([Eis80]. We recall here some
basic facts and properties for matrix factorizations over the special case of
a polynomial ring $R=K[x_{0},\ldots,x_{n}]$, which is the case of interest for
the paper. Let $Y=V(f)\subseteq\mathbb{P}^{n}$ be a hypersurface cut out by a
homogeneous form $f$ of degree $s$. It is well-known that a matrix
factorization $(A,B)$ of $f$ induces a maximal Cohen-Macaulay module supported
on $Y$ by $\mathrm{coker}A$. Conversely, if we have a maximal Cohen-Macaulay
$R_{Y}=R/(f)$-module, one has a matrix $A$ by reading off its length $1$
resolution. Indeed, it forms a part of a matrix factorization of $f$, so there
is a unique matrix $B$ such that $AB=BA=f\cdot\mathrm{Id}$. As a conclusion,
there is a bijection between the maximal Cohen-Macaulay modules and the
equivalence classes of matrix factorizations of $f$.

Now let us briefly recall Shamash’s construction (see [Sha69], [KiS20, Remark
2.6]). Suppose that $X$ is a subscheme contained in hypersurface
$Y=V(f)\subseteq\mathbb{P}^{n}$. Let $R_{X}$ be the coordinate ring of $X$.
Let $F^{\bullet}$ be the minimal free $R$-resolution of $R_{X}$. Since
$X\subseteq Y$, we have a right exact sequence

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30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
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30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{F_{1}\otimes_{R}R_{Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
95.60825pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
95.60825pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{F_{0}\otimes_{R}R_{Y}\cong
R_{Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
182.90991pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
182.90991pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
225.65894pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
225.65894pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

and hence there is an $R_{Y}$-free resolution of $R_{X}$ (possibly non-
minimal)

$\textstyle{\ldots\to G_{4}\oplus G_{2}(-s)\oplus G_{0}(-2s)\to G_{3}\oplus
G_{1}(-s)\to G_{2}\oplus G_{0}(-s)\to G_{1}\to G_{0}\to R_{X}\to 0}$

where $G_{i}=F_{i}\otimes_{R}R_{Y}$. It becomes eventually $2$-periodic, and
hence induces a matrix factorization of $f$. Such a matrix factorization
provides a presentation of an ACM sheaf on $Y$. Now, we are ready to prove
Theorem 5.2.

######  Proof of Theorem 5.2.

We can consider $\Gamma_{*}(\mathcal{O}_{X})$ as a $R_{Y}$-module. Since the
section ring $\Gamma_{*}(\mathcal{O}_{X})$ have minimal free resolutions with
the Betti tables as in $*$ ‣ $2)$, the minimal resolution of
$\Gamma_{*}(\mathcal{O}_{X})$ as a $R_{Y}$-module is eventually $2$-periodic
with Betti numbers

|  | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$
---|---|---|---|---|---|---|---|---
0 |  | $1$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
1 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$
2 |  | $\cdot$ | $10$ | $15$ | $6$ | $\cdot$ | $\cdot$ | $\cdot$
3 |  | $\cdot$ | $\cdot$ | $\cdot$ | $9$ | $15$ | $6$ | $\cdot$
4 |  | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $\cdot$ | $9$ | $15$

.

Then the periodic part of its minimal free resolution yields, up to twist,
matrix factorization of the form

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
18.24951pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
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0.0pt\hbox{$\textstyle{R_{Y}^{15}(-3)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
26.8902pt\raise 5.1875pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\varphi\
\quad\quad}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 42.24951pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
42.24951pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{R_{Y}^{6}(-1)\oplus
R_{Y}^{9}(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
102.42618pt\raise 6.1111pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\quad\quad\quad\psi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 145.4697pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
145.4697pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R_{Y}^{15}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

The Betti numbers of the minimal periodic resolution of $R_{X}$ as a
$R_{Y}$-module is called the shape of the matrix factorization of $Y$. Then
the shape of a matrix factorization is determined by the Betti numbers of
$\psi$. In the current case they are

$15$ | $6$
---|---
$0$ | $9$

.

Let $F^{\bullet}$ and $\overline{G^{\bullet}}$ be minimal free resolutions of
the section ring $\Gamma_{*}(\mathcal{O}_{X})$ as an $R$-module and
$R_{Y}$-module respectively. Then $\varphi$ is the syzygy map
$\overline{G_{4}}\to\overline{G_{3}}$. Let
$\mathcal{F}=\widetilde{\mathrm{coker}\varphi}(-3)$. Then the Shamash
resolution starts with the Betti numbers

|  | $0$ | $1$ | $2$ | $3$ |
---|---|---|---|---|---|---
0 |  | $1$ | $\cdot$ | $\cdot$ | $\cdot$ |
1 |  | $\cdot$ | $\cdot$ | $1$ | $\cdot$ |
2 |  | $\cdot$ | $10$ | $15$ | $6$ |
3 |  | $\cdot$ | $\cdot$ | $\cdot$ | $10$ |

Thus the Shamash resolution is always non-minimal, and in a minimal resolution
a cancellation occurs, causing $\beta_{1,3}=10$ to decrease by one. Such
cancellation corresponds to the equation $f$ of $Y$ in $R$ becoming
superfluous in $R_{Y}$. By definition, the map
$\overline{G_{3}}\to\overline{G_{2}}$ factors through $\mathcal{F}$. It
follows from $F_{0}=R$, $F_{3}=R(-5)^{6}$ and Shamash’s construction that we
obtain a map $\alpha:\mathcal{O}_{Y}^{6}(-2)\to\mathcal{F}$. Thus we get the
following.

######  Claim 5.3 (cf.[ScT18, Theorem 2.4]).

The sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{Y}^{6}(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\quad\alpha}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{I}_{X/Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

is exact.

###### Proof.

Being the sheafification of a MCM module over $Y$, the sheaf $\mathcal{F}$ is
a vector bundle. We apply the functor $\mathrm{Hom}(\bullet,\omega_{Y})$ to
$\alpha$ and obtain
$\alpha^{*}(-2):\mathcal{F}^{*}(-2)\to\mathcal{O}_{Y}^{6}$. The cokernel of
this map coincides by construction with the cokernel of the dual of the
sheafification of the last map of $F^{\bullet}$

$\mathcal{O}_{Y}^{15}(-1)\to\mathcal{O}_{Y}^{6}$

which is a presentation of $\omega_{X}$ by duality on $\mathbb{P}^{5}$. Since
$\mathrm{rank}\mathcal{F}=7=6+1$, both $\alpha^{*}(-2)$ and $\alpha$ drop rank
in expected codimension $2$. Applying again $\mathrm{Hom}(\bullet,\omega_{Y})$
to $\alpha^{*}(-2)$ we get that $\alpha$ is injective and by the Hilbert–Burch
Theorem, we have an exact complex

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{Y}^{6}(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\mathcal{O}_{Y}(\ell)}$$\textstyle{\mathcal{O}_{X}(\ell)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

for some $\ell$. By applying again $\mathrm{Hom}(\bullet,\omega_{Y})$ to this
last exact sequence one gets that $\alpha^{*}(-2)$ is a presentation of
$\omega_{X}(-\ell)$. Hence $\ell=0$, as required.

∎

The main result of last section is a converse to the Theorem 5.2. To state it,
first of all let us fix our notation and terminology. Liaison has become an
established technique in algebraic geometry. The greatest activity, however,
has been in the last quarter century, beginning with the work of Peskine and
Szpiró in 1974 (see [PeS74]). Liaison is a powerful tool for constructing
examples.

######  Definition 5.4.

Let $X_{1}$ and $X_{2}$ be equidimensional closed subschemes of dimension $r$
of $\mathbb{P}^{n}_{k}$. We say that $X_{2}$ is obtained by an elementary
biliaison of height $m$ from $X_{1}$, if there exists an ACM scheme
$Y\subset\mathbb{P}^{n}$, of dimension $r+1$ containing $X_{1}$ and $X_{2}$,
such that $X_{2}\thicksim X_{1}+mH$ on $Y$, where $H$ denotes the hyperplane
class. (Here $\thicksim$ means linear equivalence of divisors on $Y$ in the
sense of [Har94].) The equivalence relation generated by elementary biliaisons
is called simply biliaison. If $Y$ is a complete intersection scheme in
$\mathbb{P}^{n}$, we speak of CI-biliaison. If $X_{1}$, $X_{2}$, $Y$ are
contained in some projective scheme $Z\subseteq\mathbb{P}^{n}$, we speak of
biliaison (CI-biliaison) on $Z$.

Casanellas and Hartshorne gave a criterion for checking when two codimension
$2$ subschemes $Y_{1}$, $Y_{2}$ of a normal ACM projective scheme $X$ are in
the same biliaison class (see [CaH04, Theorem 3.1]). Using such criterion, we
will prove a converse to the Theorem 5.2.

######  Theorem 5.5.

Let
$(\psi:\mathcal{O}^{6}_{Y}(-3)\oplus\mathcal{O}^{9}_{Y}(-4)\to\mathcal{O}^{15}_{Y}(-2);\varphi:\mathcal{O}^{15}_{Y}(-2)\to\mathcal{O}^{6}_{Y}\oplus\mathcal{O}^{9}_{Y}(-1))$
be a matrix factorization of a cubic fourfold $Y$. Put
$\mathcal{F}=\mathrm{coker}\varphi(-1)$. Then, for every $a\geq 2$, there is
an ACM smooth surface $X^{\prime}$ such that $X^{\prime}$ has an
$\mathcal{F}$-type resolution

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{X^{\prime}}^{6}(-a)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{I}_{X^{\prime}/Y}(a^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

for some $a^{\prime}\in\mathbb{Z}$ and $X^{\prime}$ belongs to the
$CI$-biliaison equivalence class of smooth projective surfaces $X\subset Y$ of
degree $10$ satisfying property ${\bf N}_{3,3}$.

###### Proof.

We denoted by $M=\bigoplus\limits_{n\in\mathbb{Z}}H^{0}(\mathcal{F}(n))$. Then
$M$ is Cohen-Macaulay as $R_{Y}$-module and so
$\mathrm{depth}_{R}M=\mathrm{depth}_{R_{Y}}M=\dim R-1=5$. It follows from the
Auslander-Buchsbaum Formula and the shape of the matrix factorization of the
cubic fourfold $Y$ that we have a short exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{15}(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R^{6}\oplus
R^{9}(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

of $R$-modules. Therefore, the Castelnuovo-Mumford regularity of $M$ is $1$.
Let $N$ be a graded submodule of $M$ generated by homogeneous elements of $M$
all of the same degree $a-1$. Since the Castelnuovo-Mumford regularity of $M$
is $1$, we have $M/N$ is of finite length over $R$. We now observe that $N$
has a presentation

$\textstyle{R^{c}(-a)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{R^{b}(-a+1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

in which the presenting matrix $\phi$ has homogeneous entries with all the
entries have degree $1$. We then adjoin indeterminates $y_{ij}$ to $R$ for
$1\leq i\leq b$, and $1\leq j\leq 6$ to obtain a new ring $A=R[y_{ij}]$, where
the $y_{ij}$’s are considered as degree $1$ elements in the graded ring $A$.
We put $M_{1}=A\otimes_{R}M$, and $N_{1}=A\otimes_{R}N$. We may consider the
matrix $\phi$ as an $A$-homomorphism $A^{c}(-a)\to A^{b}(-a+1)$ presenting
$N_{1}$, that is, we are identifying $1_{A}\otimes\phi$ with $\phi$. Letting
$\Phi$ be the matrix with elements $y_{ij}$, we obtain an $A$-homomorphism
$A^{c}(-a)\oplus A^{6}(-a)\to A^{b}(-a+1)$ via the matrix $\Psi=[\phi|\Phi]$,
where the vertical line represents a matrix partitioning corresponding to the
two direct summands of $A^{c}(-a)\oplus A^{6}(-a)$. Let $E$ be the image of
$A^{6}(-a)$ in $N_{1}$ under the composition of the maps

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
37.84198pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A^{c}(-a)\oplus
A^{6}(-a)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
29.38492pt\raise 5.39166pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.39166pt\hbox{$\scriptstyle{\quad\quad\ \Psi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 61.84198pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
61.84198pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A^{b}(-a+1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
137.58667pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
137.58667pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{N_{1}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Then, $E$ is a free submodule of $N_{1}$ of rank $6$, and also $E$ is a graded
submodule of $N_{1}$. Since $M$ is Cohen-Macaulay as $R_{Y}$-module, $M_{1}$
is Cohen-Macaulay as $A$-module. It follows from $M_{1}/E$ is a graded
$A$-module that its associated prime ideals are homogeneous. Since $\dim
A\geq\dim R=6$, and the $(x_{0},\ldots,x_{5})$-depth of $M$ over $R$ is equal
to the $(x_{0},\ldots,x_{5})$-depth of $M_{1}$ over $A$, we get that
$\mathrm{depth}_{(x_{0},\ldots,x_{5})}(M_{1}/E)=4$. Hence, because our field
is infinite, there must be a form $g$ of degree one in $x_{0},...,x_{5}$ that
is not a zero divisor on $M_{1}/E$.

Next we show that $A_{g}\otimes_{A}(N_{1}/E)$ is isomorphic to a normal prime
ideal $I_{g}$ of $A_{g}$. Indeed, since degree of $g$ is one in
$x_{0},\ldots,x_{5}$, we have that $M_{g}$ at $N_{g}$ is a free
$A_{g}$-module. Hence, $(N_{1})_{g}=A_{g}\otimes_{A}N_{g}$ is also a free
$A_{g}$-module. Thus the sequence

$\textstyle{A_{g}^{c}(-a)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\textstyle{A_{g}^{b}(-a+1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(N_{1})_{g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

splits in which the presenting matrix $\phi$ is as a matrix of homogeneous
entries with all the entries belong to $A_{g}$. After performing invertible
row and column operations on $\phi$, we obtain a $b\times c$\- matrix of the
form

$\bf 0$ | $\bf{I}_{t}$
---|---
$\bf 0$ | $\bf 0$

where $\bf I_{t}$ is the $t$ by $t$ identity matrix for some $t$. The effect
of the row operations on the matrix $\Psi=[\phi|\Phi]$, is to obtain a matrix
$\Psi^{\prime}=[\phi^{\prime}|\Phi^{\prime}]$, where $\Phi^{\prime}$ consists
of entries $y^{\prime}_{ij}$ which are a new set of $6b$ polynomial
indeterminates over $R_{g}$. After applying further invertible column
operations on $\Psi$, we obtain the

$\Psi^{\prime}$=  $\bf 0$ $\bf{I}_{t}$ $\bf 0$ $\bf 0$ $\bf 0$
$y^{\prime\prime}_{ij}$

where the $y^{\prime\prime}_{ij}$ in the lower right-hand corner is a subset
of the $y^{\prime}_{ij}$. Let $I_{g}$ be the $A_{g}$-ideal of projective
dimension one, generated by the maximal minors of the matrix
$(y^{\prime\prime}_{ij})$ in the lower right-hand corner of $\Psi^{\prime}$.
Then, $A_{g}\otimes_{A}(N_{1}/E)\cong I_{g}$ is a normal prime ideal of
$A_{g}$.

Since $g$ is regular on $M_{1}/E$ and $A_{g}\otimes_{A}(N_{1}/E)\cong I_{g}$,
there exists a graded ideal $I$ of $A$ such that $M_{1}/E\cong I$ and
$I_{g}=A_{g}\otimes I$. Since $A$ is factorial and $M_{1}$ is a non free
reflexive module we may assume that $I$ has height exactly two. Since this
matrix consists of indeterminates over $R_{g}$, it follows that $I_{g}$ is a
normal prime ideal of $A_{g}$, and so $I$ a normal prime ideal of $A$.
Therefore, we have an exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A(-a)^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

in which $I$ is a homogeneous prime ideal of $A$ of height two, and $A/I$ is
Cohen-Macaulay. Then, it follows from Flenner’s version of Bertini’s Theorem
that there exists a linear regular sequence $h_{1},\ldots,h_{6b}$ on $A/I$.
Hence the sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(A/(h_{1},\ldots,h_{6b})A)(-a)^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M_{1}/(h_{1},\ldots,h_{6b})M_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I/(h_{1},\ldots,h_{6b})I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

is exact. Since $M_{1}/(h_{1},\ldots,h_{6b})M_{1}\cong M$ and
$A/(h_{1},\ldots,h_{6b})A\cong R$, there is a codimension $2$ subscheme
$X^{\prime}\subset Y$ without embedded components such that $X^{\prime}$ has
an $\mathcal{F}$-type resolution

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{Y}^{6}(-a)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\quad\alpha}$$\textstyle{\mathcal{F}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{I}_{X^{\prime}/Y}(a^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

for some $a^{\prime}\in\mathbb{Z}$. By Theorem 3.1 in [CaH04] and Theorem 5.2,
$X^{\prime}$ and smooth projective surfaces $X\subset Y$ of degree $10$
satisfying property ${\bf N}_{3,3}$ are in the $CI$-biliaison equivalence
class on $Y$. By theorem 1.1, $X$ is a projective Cohen-Macaulay smooth
surface. Therefore $X^{\prime}$ is a projective Cohen-Macaulay smooth surface,
as required.

∎

Here is an immediate corollary of Theorem 5.2 and 5.5 as follows:

######  Corollary 5.6.

Let $Y$ be a smooth cubic fourfold. Then the statements are equivalent.

  * $1)$

$Y\in\mathcal{C}_{\delta}$ for $6<\delta\leq 44$ and $\delta\equiv 2\pmod{6}$.

  * $2)$

$Y$ admits the shape of the matrix factorization

$15$ | $6$ | $\cdot$
---|---|---
$\cdot$ | $9$ | $15$

Acknowledgement The author wishes to thank Prof. F.-O. Schreyer for his
indispensable suggestions about the topics in this paper.

## References


# Preventing RNN from Using Sequence Length as a Feature

Jean-Thomas Baillargeon<EMAIL_ADDRESS>Université
LavalQuébecCanada ,  Hélène Cossette<EMAIL_ADDRESS>Université
LavalQuébecCanada and  Luc Lamontagne<EMAIL_ADDRESS>Université
LavalQuébecCanada

(2022)

###### Abstract.

Recurrent neural networks are deep learning topologies that can be trained to
classify long documents. However, in our recent work, we found a critical
problem with these cells: they can use the length differences between texts of
different classes as a prominent classification feature. This has the effect
of producing models that are brittle and fragile to concept drift, can provide
misleading performances and are trivially explainable regardless of text
content. This paper illustrates the problem using synthetic and real-world
data and provides a simple solution using weight decay regularization.

Recurrent Neural Network, Text Classification, Explainability, Regularization,
Overfitting

††ccs: Information systems Content analysis and feature selection††copyright:
acmcopyright††journalyear: 2022††conference: ; December 16–18, 2022; Bangkok,
Thailand††isbn: 978-1-4503-9763-6

##  1\. Introduction

Neural networks powered by recurrent cells, such as Long Short-Term Memory
(LSTM), have been used to construct state-of-the-art document classification
models. Their theoretical capability of ingesting variable-length inputs into
a machine learning pipeline has made them attractive to the scientific
community. Even though models based on the Transformer architecture presented
in (Vaswani et al.,, 2017) tend to offer improved performance for many natural
language processing tasks, Recurrent Neural Networks (RNN) cells still act as
a solid baseline for many tasks and should be treated accordingly.

In our most recent work, we used RNN to classify insurance claims composed of
many notes (i.e., textual documents) into benign or catastrophic. As we tried
to extract the risk factors explaining the catastrophic claims, we discovered
that the recurrent cells use input length as a predominant feature to classify
claims. Upon investigation, we found that catastrophic and benign claims were
composed of significantly different size entries: catastrophic claims
contained an average of 300 documents, while benign claims contained an
average of 50 documents. Although the length difference was observed during
our preliminary data exploration, no previous work in the literature seems to
warn against this potential behavior.

This feature may be considered legitimate for classification and performance
purposes since the generalization over test sets with similar length
distributions is good. However, we believe that using input lengths to fit a
model makes it susceptible to concept drift, since a change in document length
invalidates its classifications. This behavior complicates knowledge
extraction using an attention or explainability mechanism, since the models’
decision can be trivially explained by the length of the inputs. Moreover this
is a real problem that can go undetected in many real-world applications.

In this paper, we investigate the behavior of recurrent cells learning that
tend to use sequence lengths as a feature and we explore some potential
solutions. We explain in Section 2 that this problem has not yet been properly
addressed in the literature. We illustrate the problem in Section 3 and 4
using synthetic and real-world datasets. Finally, experimental results on how
to deal with this problem are presented and analyzed in Section 5.

Our contributions are twofold. First, we present the statistical red flags
observed during our data investigation, and second, we present a solution to
this problem by using adequate regularization.

##  2\. Related Work

Text document classification is a fundamental task in natural language
processing (NLP). As more complex corpora became available, this generated a
need to provide more flexible algorithms to handle the newly discovered
problems. A good example is the classification of long documents which
requires an algorithm to model the long-term dependency between text elements
in a sequence.

An initial approach is to use an RNN Cell. As the name suggests, this cell
recursively ingests the sequence of tokens from a document to generate a
compact representation that can be used for different NLP tasks. These models
have proven to be complex to train, as the long sequences cause the vanishing
gradient problem, as discussed in (Pascanu et al.,, 2013). Later
implementations, such as LSTM (Hochreiter and Schmidhuber,, 1997), allow
better handling of gradients for modeling very long documents.

RNN’s ability to count and maintain memory of long sequences is reasonably
well documented. LSTMs can explicitly count the length of a sequence (Gers and
Schmidhuber,, 2000), provide the next element in a counting sequence
(Rodriguez et al.,, 1999) or count items in a picture (Fang et al.,, 2018).
However, the literature has not addressed the problem that RNN can converge to
a local minimum based on the sequence length difference in the classes.

Finally, most NLP state-of-the-art models now use the Transformer architecture
(Vaswani et al.,, 2017). To classify long documents, models that belong to the
Longformer family (see (Beltagy et al.,, 2020)) are used. As demonstrated by
(Variš and Bojar,, 2021) for a sequence-to-sequence task, these models face an
input length problem similar to the one we are studying. However, the problem
remains to be investigated for classification tasks.

##  3\. Experimentation with Synthetic Data

We start by exploring the problem using synthetic data. To conduct this
experiment, we generate datasets containing two classes with different
sequence length distributions. By varying the gap between the distributions,
we can evaluate its impact on the decisions of an RNN classification model. In
this section, we present how we generate the input and how we evaluate the
effect of the problem.

###  3.1. Generation of the Synthetic Sequences

The creation of random datasets allows us to create uninformative inputs while
controlling the sequence length meta-feature. A binary classification model
trained on an uninformative training set should, on average, not perform
better than 50 % accuracy when evaluated on an uninformative test set. We
create training and testing sets according to various scenarios. Each of the
two classes are associated to a normal distribution, specifying the mean and
variance of the sequence lengths. To generate an instance of a class, a
sequence length is first drawn from its class distribution. Then each sequence
element of this instance corresponds to a 300 dimensions vector (similar to
GLoVe or Word2Vec word embeddings) filled with random values drawn from a
$[0,1]$ uniform distribution. By selecting different means and variances, we
can control the extent to which the sequence length of both classes overlaps.
Overlap values used for our experiments are presented in Table 1.

###  3.2. Training Procedure

Unless otherwise specified, we train each classifier the same way: we generate
a balanced training set of 10 000 instances (as described in Section 3.1) that
we feed into a vanilla LSTM with a hidden state size of 300. Classification is
made by a feedforward layer taking the last hidden state of the LSTM as input.
Both LSTM and classification head are end-to-end trained together. We train
the network using the Adam optimizer for ten epochs with a learning rate of
0.001. We batch the generated input in groups of 32 observations. We evaluate
our models on a holdout dataset also containing 10 000 randomly generated
instances.

###  3.3. Evaluation of the Problem

This first experimentation aims to evaluate the importance of the problem by
having a total control over the length distribution of the classes. We
selected five sets of parameters such that the theoretical distribution of
lengths overlaps by 100 %, 80%, 50 %, 10%, and 0 %.

We use the length distribution’s overlap to quantify the source of the
problem: the more similar the distributions, the less possible the input
length to be used as a feature. Hence the phenomenon should be absent when the
two class distributions are the same (100% overlap) and appear gradually as
the distributions move apart (0 % overlap). We evaluate how this change in
overlap affects the performance of a vanilla LSTM cell.

###  3.4. Results

The first four columns of Table 1 contain the means and variances of the
length distributions used for the scenarios of this experimentation. The fifth
column contains the theoretical overlap % estimated from the normal
distribution p.d.f. and the scenario parameters. The last column presents the
accuracy obtained with an LSTM model and random inputs. Since the sequences
contain non-informative elements, all accuracy values should ideally tend to
be 50 %.

Table 1. Classifier Accuracy for Synthetic Input Scenarios $\mu_{0}$ | $\sigma_{0}$ | $\mu_{1}$ | $\sigma_{1}$ | overlap % | Accuracy
---|---|---|---|---|---
10 | 2 | 10 | 2 | 100 % | 0.50
10 | 2 | 11 | 2 | 80 % | 0.51
10 | 2 | 13 | 3 | 50 % | 0.71
10 | 2 | 20 | 4 | 10 % | 0.95
10 | 2 | 100 | 10 | 0 % | 1.00

The results clearly indicate that, as the distribution overlap diminishes
(i.e. as the gap between the sequence length distributions of the classes
augments), the accuracy of the LSTM classifier artificially increases. We find
that the problem may appear as soon as the overlapping of the two sequence
length distributions is lower than 80% and becomes overwhelming when the
overlapping is 0 %. These are statistical red flags.

##  4\. The Problem in Real-World Datasets

The next experimentation aims to determine if the sequence length problem is
also encountered in real-world datasets containing informative texts. For this
purpose, we chose the amazon polarity corpus (Zhang et al.,, 2015), a balanced
dataset containing 3.6M training examples that is often used to benchmark
binary classification algorithms.

###  4.1. Training Dataset Alteration

To conduct our experiments, we modify the original training dataset to create
an artificial gap between the class length distributions. In amazon polarity,
the median lengths of negative and positive examples are 90 and 79 tokens
respectively. To create a training dataset with a length gap, we remove
negative examples having a length smaller than 90. Likewise, we remove
positive examples with more than 79 tokens. Therefore, we reduce the length
overlap to 0% between the 2 classes, a gap that could affect the behavior of
the LSTM models. With this modified training dataset, we construct LSTM models
like those described in Section 3.2.

###  4.2. Evaluation Test Datasets

To evaluate if the model uses the unwanted length meta-feature, we alter the
original test dataset with a procedure similar to what we did for the training
set. However we use the different class partitions to create 2 test sets. The
Gap-test dataset contains items with the same length distribution as the
training set, i.e., $[0,79]$ for positive and $[90,\infty]$ for negative
examples. The second test set, the Reverse side, is the complement of the Gap-
test with length distributions of $[80,\infty]$ and $[0,89]$ for positive and
negatives examples respectively. We also create a third test dataset Reverse*
by modifying the length of the examples from the Reverse side dataset so they
correspond to the Gap-test’s range. To achieve this, we truncate the long
positive examples and extend the short negative examples by concatenating a
duplicate of their texts. These manipulations modify the length of the
examples without injecting new textual information. We present the resulting
length distributions for each test dataset in Figure 1.

(a) Original Test Set

(b) Gap Test Set

(c) Reverse Test Set

Figure 1. Class length distributions

###  4.3. Results

Experimental results are presented in Table 2. The first column indicates what
dataset is used to train the LSTM. Original-train is the complete 3.6M
examples of the original amazon polarity training set, whereas the Gap-train
corresponds to the modified training set described in Section 4.1. The four
last columns present the accuracy score obtained on different test sets.
Original-test refers to the original amazon polarity test set while the other
three columns refer the modified test sets presented in Section 4.2.

Table 2. Accuracy for each Model on the Different Test Datasets Training Set | Original-test | Gap-test | Reverse | Reverse*
---|---|---|---|---
Original-train | 94.7% | 95.1 % | 94.1 % | 92.8 %
Gap-train |  53.2 % | 100.0 % |  7.0 % | 90.4 %

We first notice that the classifier trained with the original Original-train
dataset performs well on all the modified test datasets with results similar
to the benchmarks provided in (Mabrouk et al.,, 2020). However, when using the
Gap-train for training purposes, results are significantly different from
those of Original-train for the first three test sets.

A important difference is observed for the Original-test set with a
significant drop in accuracy from 94.7 % to 53.2 %. Even if the reviews remain
unchanged, changes in class length distributions largely deteriorate the
performance. It suggests that, during the training of the Gap-train model,
some of the predictive power is moved from the textual content to the class
length meta-feature.

The Gap-test results indicate that the model learns extremely well when both
training and test sets have the same length distribution as the accuracy
reaches a perfect 100.0 %. In contrast, for the Reverse Side where length
distributions are reversed, performance dramatically drops to an accuracy of
7.0 %. Two factors could contribute to this deterioration: either the Reverse
Side examples are from a different content distribution, or the model
mistakenly learned to use the input length as its prominent feature and
nothing learned from other content features.

The results on the third test set Reverse* help us to determine the cause of
this behavior. The same examples that had an accuracy of 7.0 % now tally up to
90.4 %. This significant improvement clearly shows us that whatever
information these examples contain, the model can classify them correctly as
long as training and test examples share the same length distribution. These
results again support the hypothesis that recurrent cells will use input
length as a prominent feature to classify documents.

A necessary conclusion is that the sequence length problem in classification
tasks may result in LSTM models misperforming its actual value (100.0% vs.
94.7%). This performance bias poses a critical problem: comparing LSTMs with
other classification algorithms could hardly be fair because the local optimum
reached with sequence length at training time could provide a better
performance than models relying on informative content. Finally, as our
experiment shows, if the Original dataset had had an important gap in its
class length distribution, this issue might have gone unnoticed due to the
very good performance with a gap test set (as for Gap-test).

##  5\. Possible Solutions

An intuitive approach to the sequence length problem is to address it as a
regularization problem. As the network is trained to predict which class the
observation belongs to, the model’s ability to adjust its parameters weights
is hampered by regularization and is forced to use variation-robust
information. In this section, we evaluate the impact of traditional
regularization techniques such as weight decay and dropout, as presented in
works like (Merity et al.,, 2017), to mitigate our problem with the synthetic
and amazon_polarity datasets.

###  5.1. Informative Synthetic Dataset

The regularization technique should prevent the model from learning the
sequence length meta-feature while allowing salient features to be selected.
In our current framework, the later part of the behavior cannot be evaluated
for synthetic data as there is no content information in our completely
randomly generated synthetic dataset. To overcome this limitation, we created
an additional synthetic dataset containing information that could help models
to discriminate between the classes.

We generate the informative training dataset with a method similar to that
used for the uninformative synthetic dataset, as explained in Section 3.1. To
add some informative content to a purely random dataset, we modify 10% of the
positive class sequences so they contain a filled vector of 1s. Such a feature
is straightforward to learn and should lead to 100 % classification accuracy.

###  5.2. Dropout

To evaluate if dropout can solve the sequence length problem, we add a 25 %
dropout to the synthetic inputs during training. Dropout randomly sets a
percentage of inputs to zero and forces a model to use multiple features
instead of a few salient ones.

Table 3. Classification Accuracy of LSTMs Trained with Dropout using the Synthetic Datasets Overlap % | Dataset | Dropout | Accuracy
---|---|---|---
0 % | Synthetic (random) | 0 % | 1.00
0 % | Synthetic (informative) | 0 % | 1.00
0 % | Synthetic (random) | 25 % | 1.00
0 % | Synthetic (informative) | 25 % | 1.00

The results presented in Table 3 clearly indicate that the problem is not
solved because the LSTM models can make perfect predictions from random
content (while it should be 50%). For both scenarios, dropout does not lower
the performance of either models as compared to their performance from Table
1. According to our experiments, using a higher dropout percentage (50 % and
75%) has a similar impact on both models’ performance and thus should not be
considered a possible solution. A possible explanation for this behavior is
that dropout is applied to the observations of both classes. As the length
distributions of both classes are modified proportionally by the dropout
regularization, a class length overlap remains present in the inputs.

###  5.3. Weight Decay on Synthetic Data

To assess the impact of weight decay, we add a penalty proportional to the
parameter values to the gradient during backpropagation. This operation forces
the optimizer to converge toward a solution that uses smaller weights.

In our problem, the random nature of our synthetic inputs requires capturing
weights on all dimensions, since there is no pattern in any of the input
dimensions. An optimizer regularized by weight decay prefers models with a
limited number of smaller weights and should be forced to exploit the salient
informative part of the vector instead of sequence lengths.

Table 4. Classification Accuracy of LSTMs Trained with Weight Decay using the Synthetic Datasets Overlap % | Dataset | Weight Decay | Accuracy
---|---|---|---
0 % | Synthetic (random) | 0 | 1.00
0 % | Synthetic (informative) | 0 | 1.00
0 % | Synthetic (random) | 0.1 | 0.50
0 % | Synthetic (informative) | 0.1 | 1.00

As predicted, weight decay significantly reduces the problem of using the
input sequence length as a feature. As presented in Table 4, using a weight
decay of 0.1 prevents LSTM models to learn the sequence length meta-feature,
as the uninformative model accuracy of 50% is equivalent to random class
selection (as it should be). It is also worth noting that weight decay has no
impact on the performance of the model trained with the informative synthetic
dataset as the accuracy remains 100% as it should also be. We conclude that
the informative model accuracy of 100 % (using weight decay) comes from the
salient feature selection and not from the sequence length meta-feature.

We also evaluated the impact of another implicit regularization scheme, the
reduction of the LSTM hidden state size, and even with a very low amount of
parameters, the LSTM learned the unwanted feature.

###  5.4. Weight Decay Impact on Real-World dataset

To verify that our conclusion from the previous section still holds for real-
world data, we evaluate how weight decay modifies the performance of an LSTM
model trained using the amazon_polarity Gap-train dataset.

Table 5. F1 Score for Regularized and Unregularized Models Trained with the Gap-train Dataset Weight Decay | Original-test | Gap-test | Reverse | Reverse*
---|---|---|---|---
0 | 53.2 % | 100.0 % | 7.0 % | 90.4 %
0.045 |  76.6 % | 89.6 % |  63.9 % | 82.3 %

As one can see, the model trained with a thoroughly searched weight decay
parameter of 0.045 has a very different performance profile from its
unregularized counterpart. The classification accuracy of the Original-test
improves by 22 %. An improvement is also observed for the Reverse test set,
going from 7.0 % to 63.9 %. One should also note that the Gap-test performance
has decreased from 100.0 % to 89.6 %. As we hinder the capacity of the model
to learn to use the input length as features, we expect the Gap-test value to
converge to its true value of 95.1 % (from Table 2), obtained by training and
evaluating the model on the original datasets.

These results show us that the conclusion drawn with synthetic inputs still
holds for a real-world dataset, even if weight decay cannot completely negate
the effect of the sequence length problem.

###  5.5. Weight Decay Effect on Representations

As the model learns its prediction task, parameters are fit to the data to
produce document representations the classification module can separate. To
demonstrate the effect of weight decay on the model, we present and analyze 2d
projections (using the TSNE algorithm - see (Van der Maaten and Hinton,,
2008)) of embeddings generated by uninformative data in models trained with
and without weight decay.

The scatter plots from Figure 2 are the embeddings (i.e. the hidden state used
for classification purposes) generated by an LSTM model trained with random
data.

The first subplot 2(a) presents the projection of the random sequences using
the initialization weights of the LSTM model. As there is not much knowledge
stored in the model weights, the embeddings generated for each class
completely overlap in the middle of the vector space.

The next two scatter plots are the same random sequences passed in models
trained without and with weight decay. From 2(b), it can be seen that the
model without weight decay can easily separate the two classes even though no
informational content can support this result except the sequence length meta-
feature. However, as shown in 2(c), applying weight decay removes this class
separation and causes the generated embeddings to overlap once again.

This figure support our hypothesis that LSTM models can learn meta features
such as sequence lengths and that weight decay helps to alleviate this
problem.

(a) Initialisation Weights

(b) No Weight Decay

(c) With Weigth Decay

Figure 2. TSNE Projections of Embeddings Generated by Models Trained with
Uninformative Synthetic Data

##  6\. Conclusion

In conclusion, this paper describes a problem encountered in our current
research work: recurrent networks tend to use the difference in class length
distributions as a prominent classification feature instead of learning the
important textual features. This problem is not trivial because it makes
recurrent models fragile to concept drift and can lead to useless explanations
as for their decisions. Moreover, although using the sequence length
difference helps the model converge to a local optimum, it provides misleading
performance that could lead to an unfair comparison between classification
algorithms of different types.

To isolate the problem, we evaluated its significance using engineered
synthetic examples. Using LSTM models, sequence length issues appear as soon
as the overlap between the class length distribution is lower than 80% and
would trivialize the classification task whenever it approaches 0 %. We also
demonstrated that the undesired length meta-feature can also become
predominant in a real-world dataset affected by class length imbalance.

We then showed and explained that weight decay regularization can mitigate the
impact of the problem on both synthetic and real-world training datasets. Even
if weight decay could fully not hinder the effect of the sequence length
problem, it opens research opportunities in artificial example augmentation,
where training sets could be enhanced to include examples that would balance
the input length distribution.

Preliminary results assert that more advanced classification models, such as
hierarchical attention networks (Yang et al.,, 2016) and transformers (Vaswani
et al.,, 2017), are also impacted by this problem. Future work is required to
evaluate the extent of the problem on these architectures, assess its impact
on classification tasks and determine whether weight decay regularization also
solves the problem as much as it does for LSTM architectures.

## References

  * Beltagy et al., (2020) Beltagy, I., Peters, M. E., and Cohan, A. (2020).  Longformer: The long-document transformer.  arXiv preprint arXiv:2004.05150.
  * Fang et al., (2018) Fang, M., Zhou, Z., Chen, S., and McClelland, J. (2018).  Can a recurrent neural network learn to count things?  In CogSci, pages 360–365.
  * Gers and Schmidhuber, (2000) Gers, F. A. and Schmidhuber, J. (2000).  Recurrent nets that time and count.  In Proceedings of the IEEE-INNS-ENNS International Joint Conference on Neural Networks. IJCNN 2000. Neural Computing: New Challenges and Perspectives for the New Millennium, volume 3, pages 189–194. IEEE.
  * Hochreiter and Schmidhuber, (1997) Hochreiter, S. and Schmidhuber, J. (1997).  Long short-term memory.  Neural computation, 9(8):1735–1780.
  * Mabrouk et al., (2020) Mabrouk, A., Redondo, R. P. D., and Kayed, M. (2020).  Deep learning-based sentiment classification: a comparative survey.  IEEE Access, 8:85616–85638.
  * Merity et al., (2017) Merity, S., Keskar, N. S., and Socher, R. (2017).  Regularizing and optimizing lstm language models.  arXiv preprint arXiv:1708.02182.
  * Pascanu et al., (2013) Pascanu, R., Mikolov, T., and Bengio, Y. (2013).  On the difficulty of training recurrent neural networks.  In International conference on machine learning, pages 1310–1318. PMLR.
  * Rodriguez et al., (1999) Rodriguez, P., Wiles, J., and Elman, J. L. (1999).  A recurrent neural network that learns to count.  Connection Science, 11(1):5–40.
  * Van der Maaten and Hinton, (2008) Van der Maaten, L. and Hinton, G. (2008).  Visualizing data using t-sne.  Journal of machine learning research, 9(11).
  * Variš and Bojar, (2021) Variš, D. and Bojar, O. (2021).  Sequence length is a domain: Length-based overfitting in transformer models.  arXiv preprint arXiv:2109.07276.
  * Vaswani et al., (2017) Vaswani, A., Shazeer, N., Parmar, N., Uszkoreit, J., Jones, L., Gomez, A. N., Kaiser, Ł., and Polosukhin, I. (2017).  Attention is all you need.  Advances in neural information processing systems, 30.
  * Yang et al., (2016) Yang, Z., Yang, D., Dyer, C., He, X., Smola, A., and Hovy, E. (2016).  Hierarchical attention networks for document classification.  In Proceedings of the 2016 conference of the North American chapter of the association for computational linguistics: human language technologies, pages 1480–1489.
  * Zhang et al., (2015) Zhang, X., Zhao, J., and LeCun, Y. (2015).  Character-level convolutional networks for text classification.  Advances in neural information processing systems, 28.