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\title[Generalization of the addition and restriction theorems]
{Generalization of the addition and restriction theorems from free arrangements
to
the class of projective dimension one}
\author[\initial{T.} Abe]{\firstname{Takuro} \lastname{Abe}}
\address{Rikkyo University\\
Department of Mathematics\\
3-34-1 Nishi-Ikebukuro, Toshima-Ku\\
1718501, Tokyo\\
Japan}
\email{abetaku@rikkyo.ac.jp}
\thanks{
The author is
partially supported by JSPS KAKENHI Grant Number JP21H00975.
}
\keywords{hyperplane arrangements, logarithmic
derivation modules, free arrangements, SPOG arrangements, the addition-deletion theorems}
\subjclass{32S22, 52C35}
\datepublished{2024-04-29}
\begin{document}
\newtheorem{define}[cdrthm]{Definition}
\newtheorem{cor}[cdrthm]{Corollary}
\newtheorem{problem}[cdrthm]{Problem}
\begin{abstract}
We study a generalized version of Terao's addition theorem for free arrangements to
the category of those with projective dimension one.
Namely, we give a formulation to determine the algebraic structure of the logarithmic derivation
module of a hyperplane arrangement obtained by adding one hyperplane to a free arrangement under the assumption that the arrangement obtained by restricting onto that hyperplane is free too.
Also, we introduce a class of stair-strictly plus-one generated (SPOG) arrangements whose SPOGness depends only on the intersection lattice similar to the class
of stair-free arrangements which satisfies Terao's conjecture.
\end{abstract}
\maketitle
\section{Introduction}
Let $\K$ be a field, $V=\K^\ell$, and $S=\mbox{Sym}^*(V^*) =\K[x_1,\ldots,x_\ell]$ be the coordinate ring of
$V$. For the derivation module $\Der S:=\oplus_{i=1}^\ell S\partial_{x_i}$ and a hyperplane arrangement
$\A=\{H_1,\ldots,H_n\}$, where $H_i$ is defined as the zero locus of a non-zero linear form $\alpha_{H_i} \in V^*$, the \textbf{logarithmic derivation module} $D(\A)$ of $\A$ is defined by
$$
D(\A):=\{\theta \in \Der S \mid \theta(\alpha_H) \in S \alpha_H\ (\forall H \in \A)\}.
$$
The module $D(\A)$ is an $S$-graded reflexive module of rank $\ell$, but not free in general. So we say that $\A$ is \textbf{free} with \textbf{exponents} $\exp(\A)=(d_1,\ldots,d_\ell)$ if $D(\A) \simeq \oplus_{i=1}^\ell
S[-d_i]$. In this article $\exp(\A)$ is a \textbf{multiset}.
Also in this article, we assume that all arrangements are \textbf{essential} unless otherwise specified, i.e.,
$\cap_{H \in \A} H=\{0\}$.
If $\A \neq \emptyset$, then the submodule generated by the Euler derivation $\theta_E \in D(\A)$ forms a direct summand of~$D(\A)$. Explicitly,
$D(\A) =S\theta_E \oplus D_H(\A)$, where
$D_H(\A):=\{\theta \in D(\A) \mid \theta (\alpha_H)=0\}$.
So we may assume that $d_1=1=\deg \theta_E \le d_i$ for $i \ge 2$ when an essential arrangement~$\A$ is free.
Free arrangements have been a central topic in the research of hyperplane arrangements. Among them, the most important problem is so called \textbf{Terao's conjecture} asking whether the freeness of $\A$ depends only on the intersection lattice
$$
L(\A):=\{\cap_{H \in \B} H \mid
\B \subset \A\}.
$$
In other words, Terao's conjecture asks whether the freeness is
combinatorial. This is completely open, but it was shown in \cite{Z2} that
the minimal free resolution of $D(\A)$ is not combinatorial. To approach
Terao's conjecture, one of the main tools is Terao's addition-deletion theorem.
For the purpose of this article, we exhibit it in a slightly different way compared to its usual formulation:
\begin{theorem}[Addition and restriction theorems, \cite{T1}]
Let $H \in \A$, $\A':=\A \setminus \{H\}$ and let $\A^H:=\{H \cap L \mid
L \in \A'\}$. Assume that $\A'$ is free with
$\exp(\A')=(1,d_2,\ldots,d_\ell)$. Then the following two conditions are equivalent:
\begin{itemize}
\item [(1)]
$\A$ is free.
\item[(2)]
$\A^H$ is free and $|\exp(\A') \cap \exp(\A^H)|=\ell-1$.
\end{itemize}
If one of these two conditions holds, then for $d_i:=\exp(\A') \setminus \exp(\A^H)$,
it holds that
\begin{eqnarray*}
\exp(\A)&=&(1,d_2,\ldots,d_{i-1},d_i+1,d_{i+1},\ldots,d_\ell),\\
\exp(\A^H)&=&(1,d_2,\ldots,d_{i-1},d_{i+1},\ldots,d_\ell).
\end{eqnarray*}
\label{addition}
\end{theorem}
In general the condition $|\exp(\A') \cap \exp(\A^H)|=\ell-1$ above is
described as $\exp(\A') \supset \exp(\A^H)$. However, note that $|\exp(\A') \cap \exp(\A^H)|<\ell-1$ often occurs even
when both $\A'$ and $\A^H$ are free. So we have the first question in this article:
\begin{problem}
Assume that $\A'$ is free. Then which condition of $D(\A)$ makes $\A^H$ free? More
precisely, are there any explicit condition for $D(\A)$ to make $\A^H$ free
when $\A'$ is free in terms of freeness, projective dimension, free resolution and so on?
\label{problem1}
\end{problem}
Also, recent developments show the following projective dimensional version
of the addition theorem. Note that $\A$ is free
if and only if $\pd \A=0$, where $\pd \A$ denotes the projective dimension of the $S$-module $D(\A)$.
\begin{theorem}[{\cite[Theorem 1.11]{A9}}]
(1)\,\,
Assume that $\pd \A'=\pd \A^H=0$. Then $\pd \A \le 1$.
(2)\,\,
Assume that $\pd \A'=0$ and $\pd \A \le 1$. Then $\pd \A^H = 0$.
\label{pdaddition}
\end{theorem}
Now we have the second question in this article which is related to Problem \ref{problem1}:
\begin{problem}
Can we describe the algebraic structure of $D(\A)$ when $\A'$ and $\A^H$ are both free, but
$|\exp(\A') \cap \exp(\A^H)|<\ell-1$?
\label{problem2}
\end{problem}
Explicitly, we want to know the minimal free resolution of~$D(\A)$ under the above conditions.
Contrary to these problems, when $\A$ is free,
we can describe $D(\A')$, which was proved in \cite{A5}. To see this result, let us recall the definition of
strictly plus-one generated (SPOG).
\begin{define}[\cite{A5}]
We say that $\A$ is
\textbf{strictly plus-one generated (SPOG)} with $\POexp(\A)=(1,d_2,\ldots,d_\ell)$ and \textbf{level} $d$ if there is a
minimal free resolution of the following form:
$$
0 \rightarrow S[-d-1]
\stackrel{(f_1,\ldots,f_\ell,\alpha)}{\rightarrow} \oplus_{i=1}^\ell S[-d_i] \oplus S[-d]
\rightarrow D(\A) \rightarrow 0.
$$
Here $d_1=1, f_i \in S$ and $0\neq \alpha \in V^*$. For the set of generators $\theta_E,\theta_2,\ldots,\theta_\ell,\theta$ with
$\deg \theta_i=d_i$ and $\deg \theta=d$ for the SPOG module $D(\A)$, $\theta_E,\theta_2,\ldots,\theta_\ell$ is called \textbf{the
set of SPOG generators} and $\theta$ the
\textbf{level element}.
\label{SPOGdef}
\end{define}
It was proved in \cite{A5} (see Theorem \ref{SPOG}) that
$\A'$ is SPOG if $\A$ is free and $\A'$ is not free. Interestingly, in this case the structure of~$D(\A')$ is independent of that of~$D(\A^H)$.
However, in general $\A$ is neither free nor SPOG even if $\A'$ is free.
The typical example is the case when
$
\A':\prod_{i=1}^4 x_i=0
$
in
$V=\R^4$.
Then $\A$ is free with exponents~$(1,1,1,1)$. If you add $H:x_1+x_2+x_3+x_4=0$ to $\A'$ to
get $\A$,
then it is well-known that $\A$ is neither free nor SPOG. In fact
$\pd_S D(\A)=2$ in this case.
When $\pd \A'=\pd \A^H=0$, then $\pd \A \le 1$ by Theorem \ref{pdaddition}. Also Theorem \ref{addition} shows that one additional condition for exponents confirms that $\pd \A=0$.
So a weaker condition for exponents when $\pd \A'=\pd \A^H=0$ could determine the
minimal free resolution of $D(\A)$. Namely, we can show the following, which answers Problems \ref{problem1} and \ref{problem2} partially:
\begin{theorem}
Let $\A'$ be free with $\exp(\A')=(1,d_2,\ldots,d_\ell)_\le$. Here for the set of integers $(a_1,\ldots,a_s)$, the notation
$(a_1,\ldots,a_s)_\le $ means that $a_1 \le \ldots \le a_s$.
Let $d_j < d:=d_i+d_j+|\A^H|-|\A'| \le d_{j+1}$ for some $i0}$. Multiarrangements were defined by Ziegler in \cite{Z} and used
in several research of arrangements. We can define their logarithmic derivation module $D(\A,m)$ as follows:
$$
D(\A,m):=\{\theta \in \Der S \mid \theta(\alpha_H)\in S\alpha_H^{m(H)}\ (\forall H \in \A)\}.
$$
Then their freeness and exponents can be defined in the same manner as for $D(\A)$. For details, see \cite{Z}.
We have a canonical way to construct a multiarrangement from~$\A$. Let $H \in \A$. Then the \textbf{Ziegler multiplicity}
$m^H:\A^H \rightarrow \Z$ is defined by
$$
m^H(X):=|\{L \in \A \setminus \{H\} \mid H \cap
L=X\}|.
$$
The pair $(\A^H,m^H)$
is called
the \textbf{Ziegler restriction} of
$\A$ onto $H$. Also recall that for $H \in \A$, the submodule $D_H(\A)$ of $D(\A)$ is defined by
$$
D_H(\A):=\{\theta \in D(\A) \mid \theta(\alpha_H)=0\}.
$$
Since $D(\A)=S\theta_E\oplus D_H(\A)$ by the splitting exact sequence
$$
0 \rightarrow S\theta_E \rightarrow D(\A) \rightarrow D_H(\A) \rightarrow 0
$$
with the map $D(\A) \ni \theta \mapsto \theta-\frac{\theta(\alpha_H)}{\alpha_H} \theta_E \in D_H(\A)$ and
the canonical inclusion as a section, we know that $\A$ is free if and only if $D_H(\A)$ is free.
Then $D_H(\A)$ is closely related to $D(\A^H,m^H)$ as we can see in the following several results:
\begin{theorem}[\cite{Z}]
There is an exact sequence
\begin{equation}
0 \rightarrow D_H(\A) \stackrel{\cdot \alpha_H}{\rightarrow} D_H(\A) \stackrel{\pi}{\rightarrow} D(\A^H,m^H).
\label{eqZexact}
\end{equation}
Here $\pi:=\rho|_{D_H(\A)}$.
This is called the \textbf{Ziegler exact sequence}. Moreover, if $\A$ is free with exponents $(1,d_2,\ldots,d_\ell)$, then
$(\A^H,m^H)$ is also free with exponents $(d_2,\ldots,d_\ell)$.
\label{Zexact}
\end{theorem}
\begin{theorem}[Theorem 5.1, \cite{ADi}]
Let $\pi:D_H(\A) \rightarrow D(\A^H,m^H)$ be the
Ziegler restriction of $\A$ onto $(\A^H,m^H)$. Then
the preimages of a set of generators for $\mbox{Im}(\pi)$ by $\pi$ generate $D_H(\A)$.
\label{preimage}
\end{theorem}
\begin{theorem}[Yoshinaga's criterion, Theorem 2.2, \cite{Y1}]
$\A$ is free if and only if $\A$ is locally free along $H$ (i.e.,
$\A_X$ is free for all $0 \neq X \in L(\A^H))$, and $(\A^H,m^H)$ is free.
\label{Ycriterion}
\end{theorem}
For a multiarrangement we can introduce the concept of SPOG multiarrangements as follows.
\begin{define}[\cite{AD}]
We say that $(\A,m)$ is
\textbf{SPOG} with $\POexp(\A,m)=(d_1,d_2,\ldots,d_\ell)$ and \textbf{level} $d$ if there is a
minimal free resolution of the following form:
$$
0 \rightarrow S[-d-1]
\stackrel{(f_1,\ldots,f_\ell,\alpha)}{\rightarrow} \oplus_{i=1}^\ell S[-d_i] \oplus S[-d]
\rightarrow D(\A,m) \rightarrow 0
$$
with $0 \neq \alpha \in V^*$.
\label{multiSPOGdef}
\end{define}
\section{Cardinality of minimal sets of generators}
In this section we show some new results on the cardinality of
a minimal set of generators for $D(\A)$, which will play a key role
to prove our main theorem.
\begin{define}
For an arrangement $\A$, let $g(\A)$ denote
the cardinality of
a minimal set of generators for $D(\A)$. Clearly it is independent of the choice of the set of minimal
generators.
\label{ga}
\end{define}
Moreover for $\A=\A' \cup \{H\}$ we define an integer for a free arrangement $\A'$ that measures how far
$D(\A')$ is from being tangent to $H$.
\begin{define}
Let $\A=\A' \cup \{H\}$ and assume that $\A'$ is free. Let
$FB(\A')$ be the set of all homogeneous basis for $D(\A')$ and for each
${\textbf{B}}:=\{\theta_1,\ldots,\theta_\ell\} \in FB(\A')$ define
$$
NT({\textbf{B}}):=|\{i\mid 1\le i \le \ell,\
\theta_i \not \in D(\A)\}|,
$$
and define
$$
SNT(\A'):=\min\{NT({\textbf{B}}) \mid{\textbf{B}} \in FB(\A')\}.
$$
\label{MNT}
\end{define}
First we record the following easy facts.
\begin{lemma}
Assume that $\A'$ and $\A^H$ are both free. Then
$g(\A) \le 2\ell-2$.
\label{genless}
\end{lemma}
\noindent
\textbf{Proof}.
By Theorem \ref{FST},
we can choose $\theta_E,\theta_2,\ldots,\theta_{\ell-1} \in D(\A)$ as
preimages of the basis for $D(\A^H)$ by $\rho$.
Let $\theta_E,\varphi_2,\ldots,\varphi_\ell$ be
a basis for $D(\A')$. Then the Euler exact sequence (\ref{EE}) shows that
$$
\theta_E,\theta_2,\ldots,\theta_{\ell-1},\alpha_H\varphi_2,\ldots,\alpha_H \varphi_\ell$$
generate $D(\A)$, hence $g(\A) \le \ell-1+\ell-1
=2\ell-2$. \owari
\medskip
On $g(\A)$ the following proposition is fundamental.
\begin{prop}
Let $\A'$ be free with $SNT(\A')=s$. Let
$\theta_E,\theta_2,\ldots,\theta_\ell$ form a basis for $D(\A')$ such that
$\theta_i \not \in D(\A)\ (2 \le i \le s+1)$ and $\theta_i
\in D(\A)\ (i \ge s+2)$. Then $g(\A) \ge \ell+s-1$.
\label{genlowbdd}
\end{prop}
\noindent
\textbf{Proof}.
Let $\alpha_H=x_1$.
By the assumption on $\theta_E,\theta_2,\ldots,\theta_\ell$, it is clear that we can choose derivations
$\varphi_j\in D(\A)\ (j=1,\ldots,k) $ of the form $\sum_{i=2}^{s+1} f_i^j \theta_i$ such that $f_i^j \in S':=\K[x_2,\ldots,x_\ell]$ are of positive degrees, and
$$
\theta_E,\alpha_H \theta_2,\ldots,\alpha_H \theta_{s+1},\theta_{s+2},\ldots,\theta_\ell$$
together with the derivations $\varphi_j\ (j=1,\ldots,k)$ form a minimal set of
generators for $D(\A)$.
Since their images by $\rho$ generate the rank $(\ell-1)$-module $D(\A^H)$ due to
Theorem \ref{FST}, we can compute
$$
|\{\theta_E,\theta_{s+2},\ldots,\theta_\ell,\varphi_1,\ldots,\varphi_k\}|=k+1
+\ell-s-1=\ell+k-s\ge \ell-1.
$$
So $g(\A) = \ell+k\ge \ell+s-1$.\owari
\medskip
Next let us show a key result to prove Theorem \ref{mainanother}.
\begin{prop}
Let $1 \le id_1$. Then
$$
\theta_1=\sum_{i=1}^s c_i \varphi_i
$$
for constants $c_1,\ldots,c_s \in \K$. We may assume that $c_1 \neq 0$. Then we can choose $
\theta_1,\varphi_2,\ldots,\varphi_{n+t}$ as a basis for $M$.
Now assume that $\theta_i=\varphi_i$ for $1 \le i \le k-1$. Let us prove that we may choose
$\theta_k=\varphi_k$. Again by $\theta_k \in N \subset M$, we can express
$$
\theta_k=\sum_{i=1}^{k-1} f_i \theta_i+c_k\varphi_k+\cdots+c_s \varphi_s,
$$
where $d_k=\cdots=d_s d_\ell$ for some $i$.
Then the following two conditions are equivalent:
\begin{itemize}
\item [(1)]
$\A^H$ is free with $\exp(\A^H)=(1,d_2,\ldots,\hat{d}_i,\ldots,d_{\ell-1}) \cup (d)$
\item[(2)]
$\A$ is SPOG with
$\POexp(\A)=(1,d_2,\ldots,d_{i-1},d_i+1,d_{i+1},\ldots,d_{\ell-1},d_\ell+1)$ and level $d$.
\end{itemize}
\label{onedirection}
\end{cor}
\noindent
\textbf{Proof}.
Clear by the proof of Theorem \ref{mainanother}.
\owari
\medskip
Let us apply Theorem \ref{mainanother} to some examples.
\begin{example}
Let $\A$ be the Weyl arrangement of the type $A_4$ defined by
$$
Q(\A)=\prod_{i=1}^4 x_i \prod_{1 \le i < j \le 4} (x_i-x_j)=0.
$$
$\A$ is well-known to be free with $\exp(\A)=(1,2,3,4)$. Let $\A \not \ni
H:x_1-x_2+2x_3-2x_4=0$ and let $
\B:= \A
\cup \{H\}$. Then $|\B^H|=9 <10=|\A|$.
It is easy to show that
$\B^H$ is free with $\exp(\B^H)=(1,4,4)$. Note that
$$
d:=2+3-|\A|+|\B^H|=5-1=4.
$$
In this setup, from
$\exp(\A) = (1, 2, 3, 4)$, the integers $2$ and $3$
are removed and $d = 4$ coincides with the remaining
integer $4$.
Hence we can apply
Theorem \ref{mainanother} to obtain that
$\B$ is SPOG with $\POexp(\B)=(1,3,4,4)$ and level $4$.
Note that $Q(\B^H)$ is
$$
x_2x_3x_4(x_2-x_3)(x_2-x_4)(x_3-x_4)(x_2-2x_3+2x_4)(x_2-3x_3+2x_4)(x_2-2x_3+x_4).
$$
Let $L:x_2=0$ and let $\C:=\B^{H\cap L}$. Then it is easy to show that
$\chi(\C;t)=(t-1)(t-4)$ and $\chi(\B^H;t)=(t-1)(t-4)^2$. Thus $\B^H$ is divisionally free as in
Theorem \ref{division}. Since
$\A$ is divisionally free too, by Theorem \ref{TF}, the freeness and
exponents of $\A$ and $\B^H$ are both combinatorial.
Thus Theorem \ref{mainanother} shows that the SPOGness of
$\B$ is combinatorially determined.
\label{ex1}
\end{example}
\noindent
\textbf{Proof of Theorem \ref{divSPOG}}.
Clear by Theorems \ref{mainanother} and \ref{SFcombin}.\owari
\medskip
We can use Theorem \ref{mainanother} to show the combinatorial freeness of
arrangements by using a non-free but SPOG arrangements. Let us check it by the following example:
\begin{example}
Let $\A$ be the Weyl arrangement of the type $B_4$ defined by
$$
Q(\A)=\prod_{i=1}^4 x_i \prod_{1 \le i4$,
the condition $d\le d_{j+1}$ in Theorem \ref{mainanother} is necessary.
\label{eximportant}
\end{example}
If we remove the assumption on $\exp(\A')$ and
$\exp(\A^H)$ in Theorem \ref{mainanother},
we have an example related to Problem \ref{mainanotherdiff}.
\begin{example}
Let
$$
Q(\A')=\prod_{i=1}^4 x_i \prod_{i=1}^3 (x_i^2-x_4^2)(x_i^2-4x_4^2)
\prod_{i=2}^3 (x_i^2-9x_4^2)
(x_3^2-16x_4^2).
$$
Then $\A'$ is free with $\exp(\A')=(1,5,7,9)$. Let
$H_1:x_2+x_3+7x_4=0$, $H_2:x_1+x_2+x_3=0$,
and let
$\A_i:=\A' \cup\{H_i\}$. Then $\A_1$ is SPOG with $\POexp(\A_1)=(1,5,8,10)$ and level $15$. So Theorem \ref{mainanother} shows that
$\A^{H_1}_1$ is free with exponents $(1,5,15)$ and vice versa. On the other hand, $\A_2^{H_2}$ is free with exponents
$(1,10,11)$, and $D(\A_2)$ has a minimal free resolution
$$
0 \rightarrow S[-11]\oplus S[-12]
\rightarrow
S[-6]\oplus S[-8] \oplus S[-10]^2 \oplus S[-11]
\rightarrow D_0(\A_2) \rightarrow 0.
$$
So in general, it can happen that $\A$ is of projective dimension one, is not SPOG, but $\A'$ and $\A^H$ are both free.
Note that the freeness of $\A^H$ follows from $g(\A) \le 6$ and the
freeness of $\A'$ by Theorem \ref{freegeneratoringeneral}.
\end{example}
\section{Ziegler restrictions and SPOG arrangements}
Let us study a method to check whether $\A$ is SPOG or not by using Ziegler restrictions, i.e., a theory of multiarrangements.
First recall the following two results which we will use later.
\begin{theorem}[Theorem 2.3, \cite{Y1}]
Let $E$ be a reflexive
sheaf on $\P^n\ (n \ge 3)$ and assume that $E$ is locally free
except for a finite number of points in $\P^n$.
Then $H^1(E(e))=0$ for all $e<<0$.
\label{Y0}
\end{theorem}
\begin{prop}[Proposition 2.5, \cite{AD} and the equation (1.5), \cite{Y2}]
Let $\A$ be an arrangement in $V$ and $m$ be a multiplicity on $\A$. Then
$$
\oplus_{e \in \Z} H^0(\widetilde{D(\A,m)}(e))=D(\A,m),
$$
where $\widetilde{D(\A,m)}$ is a sheaf on ${\bf{Proj}}(V)$ obtained as the coherent sheaf associated to
the module of $D(\A,m)$.
\label{gs}
\end{prop}
Next, we prove the characterization of SPOG arrangements in terms of
that of the Ziegler restrictions as follows:
\begin{prop}
Assume that $\pi:D_H(\A) \rightarrow D(\A^H,m^H)$ is surjective, $\A$ is not free and
$D(\A^H,m^H)$ is SPOG with $\POexp(\A^H,m^H)=(d_2,\ldots,d_{\ell-1},d_\ell)$ and level $d$. Then
$\A$ is SPOG with $\POexp(\A)=(1,d_2,\ldots,d_{\ell})$ and level $d$.
\label{surjSPOG}
\end{prop}
\noindent
\textbf{Proof}.
Since $\pi$ is surjective, there are $\theta_2,\ldots,\theta_\ell,\theta \in D_H(\A)$ such that
$\pi(\theta_2),\ldots,\pi(\theta_\ell)$ form a set of SPOG generators
for $D(\A^H,m^H)$ with a level element $\pi(\theta)$.
For $\varphi \in D_H(\A)$ let $\overline{\varphi}$
denote its image by the Ziegler restriction map $\pi$. Since $\pi$
is surjective, Theorem \ref{preimage} shows that $\theta_2,\ldots,\theta_{\ell},\theta$
together with $\theta_E$ generate $D(\A)$. Let
$$
\overline{\alpha}\overline{\theta}=\sum_{i=2}^{\ell} \overline{f}_i \overline{\theta}_i
$$
be the unique relation in the SPOG module $D(\A^H,m^H)$, where $\alpha_H \neq \alpha \in V^*$ and $ f_i \in S$. Then its preimages are of the form
$$
\alpha \theta-\sum_{i=2}^{\ell}f_i \theta_i \in \alpha_H D_H(\A)
$$
by the Ziegler exact sequence (\ref{eqZexact}).
Since $D_H(\A)=\langle \theta_2,\ldots,\theta_\ell,\theta\rangle_S$ which is a minimal set of generators because of the non-freeness of $D_H(\A)$ and $\mbox{rank}_S D_H(\A)=\ell-1$, it holds that
$$
\alpha \theta-\sum_{i=2}^{\ell}f_i \theta_i =
\alpha_H(\sum_{i=2}^\ell g_i \theta_i+c\theta)
$$
for some $g_i \in S$ and $c \in \K$. Since $\overline{\alpha} \neq 0$,
we have a relation
\begin{equation}
(\alpha-c\alpha_H)\theta-\sum_{i=2}^\ell(f_i+\alpha_Hg_i)\theta_i=0
\label{eq3}
\end{equation}
in $D(\A)$.
On the other hand, assume that there is a relation
$$
h_1\theta_E+\sum_{i=2}^\ell h_i \theta_i+h\theta=0
$$
in $D(\A)$. Since we have a decomposition $D(\A) =S\theta_E \oplus D_H(\A)$, we may assume that $h_1=0$.
Since we have to determine the second syzygy, take a free module
$$
M:=Se+\oplus_{i=2}^\ell S e_i
$$
such that by the map $G:M \rightarrow D_H(\A)$ defined by
$$
G(e_i)=\theta_i\ (i=2,\ldots,\ell),\ G(e)=\theta,
$$
$M$ becomes the first syzygy of $D_H(\A)$. Since $\theta_2,\ldots,\theta_\ell,\theta$ form a minimal set of generators for
$D_H(\A)$ and their images by $\pi$ form a minimal set of generators for $D(\A^H,m^H)$, by the nine-lemma, we have an exact commutative diagram as follows:
$$
\xymatrix{
&0 \ar[d]& 0 \ar[d]& 0 \ar[d]& \\
0 \ar[r] & K \ar[r] \ar[d]^{\cdot \alpha_H}& M \ar[r]^G \ar[d]^{\cdot \alpha_H}& D_H(\A) \ar[r] \ar[d]^{\cdot \alpha_H}&0\\
0 \ar[r] & K \ar[r] \ar[d]& M \ar[r]^G \ar[d]^{\pi}& D_H(\A) \ar[r] \ar[d]^{\pi}&0\\
0 \ar[r] & \overline{S}[-d-1] \ar[r] \ar[d]& \overline{M} \ar[r]^{\overline{G}} \ar[d]& D(\A^H,m^H) \ar[r] \ar[d]&0\\
&0 & 0 & 0 &
}
$$
So what we are assuming is that
\begin{equation}
\sum_{i=2}^\ell h_ie_i +he \in \mbox{ker}(G)=K.
\label{eq1000}
\end{equation}
Sending this by $\pi$, the commutativity shows that
$$
\sum_{i=2}^\ell \overline{h_i} \overline{e_i}+
\overline{h}\overline{e} \in \ker{\overline{G}}.
$$
Since $D(\A^H,m^H)$ is SPOG, $\ker{\overline{G}}$ is generate by the unique relation
$$
\overline{\alpha}
\overline{e}-\sum_{i=2}^\ell \overline{f_i}\overline{e_i}
$$
of degree $d+1$.
Thus the exactness of the middle column in the diagram shows that
\begin{equation}
\sum_{i=2}^\ell h_i e_i+he=
F(\alpha e-\sum_{i=2}^\ell f_i e_i)+\alpha_H \varphi
\label{eq500}
\end{equation}
for some $F \in S$ and $\varphi \in M$.
Rewrite (\ref{eq500}) into the following way:
$$
\sum_{i=2}^\ell h_i e_i+he-
F\{(\alpha-c\alpha_H)e-\sum_{i=2}^\ell(f_i+\alpha_Hg_i)e_i\}
=\alpha_H(Fc e +F\sum_{i=2}^\ell g_i e_i+\varphi).
$$
By (\ref{eq3}), the left hand side of the above is in $\ker(G)$. So is the right hand side.
Since $\alpha_H \neq 0$ in $S$ and $D_H(\A)$ is torsion free, we know that
$$
Fc e +F\sum_{i=2}^\ell g_i e_i+\varphi \in \ker (G).
$$
So we have a new relation among $\theta_2,\ldots,\theta_\ell,\theta$ but the degrees of this relation is lower
than the original relation (\ref{eq1000}). Since the lowest degree relation in $D(\A^H,m^H)$ is at degree
$d+1$ by the assumption,
the lowest degree relation among $\theta_2,\ldots,\theta_\ell,\theta$ in $D_H(\A)$ is (\ref{eq3}), which is of degree $d+1$.
Hence applying the same argument to this new relation continuously,
we can show that
all the relations among $\theta_2,\ldots,\theta_\ell,\theta$ are generated by the unique relation
(\ref{eq3}), i.e., $K \simeq S[-d-1]$. Therefore,
$\A$ is SPOG with the desired exponents and level. \owari
\medskip
To introduce the main result in this section let us recall some definitions and facts on the freeness.
\begin{define}[Proposition 3.6, \cite{A2}]
For $H \in \A$, \textbf{the $b_2$-inequality} for $(\A,H)$ is the inequality
$$
b_2(\A)\ge b_2(\A^H)+|\A^H|(|\A|-|\A^H|).
$$
Moreover, we say that the \textbf{$b_2$-equality} holds for $(\A,H)$ if
$$
b_2(\A)= b_2(\A^H)+|\A^H|(|\A|-|\A^H|).
$$
\label{b2}
\end{define}
\begin{theorem}[Theorem 3.6, \cite{A4}]
\begin{itemize}
\item[(1)]
The $b_2$-inequality holds for all $\A$ and $H \in \A$.
\item[(2)]
$\A$ is free if the $b_2$-equality holds for $(\A,H)$ and
$\A^H$ is free.
\end{itemize}
\end{theorem}
\begin{theorem}[Theorem 3.1, \cite{A4}]
Let $H\in \A$ and assume that the $b_2$-equality holds for $(\A,H)$.
If $\theta_E,\theta_2,\ldots,\theta_\ell$ form a minimal set of generators for $D(\A^H)$, then we may assume that $\theta_2,\ldots,\theta_\ell \in D(\A^H,m^H)$ and
$$
\displaystyle \frac{Q(\A^H,m^H)}{Q(\A^H)}\theta_E,
\theta_2,\ldots,\theta_\ell
$$
form a set of generators for $D(\A^H,m^H)$. Moreover,
they form a minimal set of generators unless $(\A^H,m^H)$ is free.
\label{lowerb2}
\end{theorem}
Now we have the following result for SPOGness.
\begin{theorem}
Let $\ell \ge 5$.
Let $\A \ni H$ and assume that $\A$ is not free, $\A^H$ is SPOG, the $b_2$-equality holds for $(\A,H)$ and $\A$ is locally free along $H$, i.e.,
$\A_X$ is free for all $X \in L(\A^H) \setminus \{0\}$. Then $\A$ is SPOG with
$\POexp(\A)=\POexp(\A^H) \cup (|\A|-|\A^H|)$ and the same level as $\A^H$.
\label{YSPOG}
\end{theorem}
\noindent
\textbf{Proof}.
By Proposition \ref{surjSPOG}, it suffices to show that
$\pi$ is surjective and $(\A^H,m^H)$ is SPOG.
First,
let us show that $\pd_{\overline{S}} D(\A^H,m^H) \le 1$.
Since the $b_2$-equality holds, Theorem \ref{lowerb2} shows that, for a set of SPOG generator and the level element $\theta_E,\theta_2,\ldots,\theta_{\ell-1},\theta$ for $D(\A^H)$ with
$\theta_2,\ldots,\theta_{\ell-1},\theta \in D(\A^H,m^H)$, we know that
$Q'\theta_E,\theta_2,\ldots,\theta_{\ell-1},\theta$ form a set of generators for $D(\A^H,m^H)$.
Here $Q':=Q(\A^H,m^H)/Q(\A^H)$.
If $(\A^H,m^H)$ is free, then clearly $\pd_{\overline{S}} D(\A^H,m^H)=0$. So assume that $(\A^H,m^H)$ is not
free. Since $\mbox{rank}_{\overline{S}} D(\A^H,m^H)=\ell-1$, it holds that
$Q'\theta_E,\theta_2,\ldots,\theta_{\ell-1},\theta$ form a minimal
set of generators by Theorem \ref{lowerb2}. Since $\A^H $ is SPOG, letting $\theta_\ell:=\theta$ as a level element, there are a non-zero $\alpha
\in V^*$ and $\overline{f_i} \in \overline{S}$ such that
\begin{equation}
\overline{f_1} \theta_E+\sum_{i=2}^{\ell-1} \overline{f_i} \theta_i+
\overline{\alpha}\theta_\ell=0
\label{eq55}
\end{equation}
which is the unique relation in $D(\A^H)$.
Again by Theorem \ref{lowerb2}, $\theta_i$
are in $D(\A^H,m^H)$. Thus $Q' \mid \overline{f_1}$. Hence
(\ref{eq55}) is also a relation among a minimal set of generators
in $D(\A^H,m^H)$
obtained above. Since every relation among this minimal set of generators for $D(\A^H,m^H)$ is also
a relation in $D(\A^H)$, the
fact that $\A^H$ is SPOG shows that $D(\A^H,m^H)$ also has the unique relation (\ref{eq55}). Hence in this case $(\A^H,m^H)$ is SPOG.
So in each case, $\pd_{\overline{S}} D(\A^H,m^H) \le 1$. In particular, since $H \simeq \P^{\ell-2}$ and $\ell-2 \ge 3$, it holds that
$H^1(\widetilde{D(\A^H,m^H)}(e))=0$ for all $e \in \Z$.
Second, let us prove the surjectivity of $\pi$. Since $\pi$ is locally free along $H$,
Theorem~\ref{FST} shows that $\pi$ is locally surjective. So we have the sheaf exact sequence
$$
0 \rightarrow
\widetilde{D_H(\A)}
\stackrel{\cdot \alpha_H}{\rightarrow }
\widetilde{D_H(\A)}
\stackrel{\pi}{\rightarrow }
\widetilde{D(\A^H,m^H)} \rightarrow
0.
$$
Since $H^1(\widetilde{D(\A^H,m^H)}(e))=0$ for all $e \in \Z$ as above, the map
\[H^1(\widetilde{D_H(\A)}(e-1)) \stackrel{\cdot
\alpha_H}{\rightarrow }H^1(\widetilde{D_H(\A)}(e))\]
is surjective.
Note that there are at most finite number of non-local free points of $\widetilde{D_H(\A)}$. Assume not, then there is $X \in L(\A)$ such that $\A_X$ is not free and $\dim X\ge 2$. Then it has the intersection
with $H$ of dimension at least one, contradicting the local
freeness of $\A$ along $H$.
Thus Theorem \ref{Y0} shows that $H^1(\widetilde{D_H(\A)}(e))=0$ for all $e \in \Z$.
By using Proposition \ref{gs}, it holds
that $\pi$ is surjective.
Finally let us show that $D(\A^H,m^H)$ is SPOG. By Yoshinaga's criterion (Theorem~\ref{Ycriterion}) and the surjectivity of $\pi$, it holds that $(\A^H,m^H)$ is not free.
Thus the first investigation of the generators for $D(\A^H,m^H)$ shows that $(\A^H,m^H)$ is SPOG. \owari
\medskip
\begin{example}
Let
$$
\A_1:=\prod_{i=1}^5 x_i \prod_{1 \le i < j \le 5} (x_i-x_j)=0.
$$
Then define
$$
\A:=\A_1 \setminus \{x_1=0,x_5=0,x_2=x_3,x_1=x_2\}.
$$
Let $\{x_1=x_5\} =H \in \A$. Then by choosing appropriate coordinates $x,y,z$ for $H^*$,
$\A^H$ is isomorphic to
$$
xyz(x-w)(y-w)(z-w)(x-z)(y-z)=0.
$$
Let $\{y=w\} =X \in \A^H$. Then $\A^H \setminus \{X\}=:\B$ is easily checked to be
divisionally free by Theorem \ref{division},
with exponents $(1,2,2,2)$ and $\A^X$ is free with exponents~$(1,2,3)$, which is also divisionally free. Thus Theorem \ref{mainanother} shows that $\A^H$ is SPOG with $\POexp(\A^H)=(1,2,3,3)$
and level $3$, which is combinatorial by Theorem \ref{divSPOG}. Now we can show by case-by-case argument that $\A$ is locally free
along $H$ and these local freeness depends only on $L(\A)$. Also, since $b_2(\A)=48$ and $b_2(\A^H)=24$, the $b_2$-equality holds for $(\A,H)$.
Thus the SPOGness of $\A^H$ combined with local freeness
along $H$ and Theorem \ref{YSPOG} shows that $\A$ is SPOG with
$\POexp(\A)=(1,2,3,3,3)$ and level $3$, here $3=|\A|-|\A^H|=12-9$. Also
the SPOGness of $\A$ is combinatorial by this
argument.
\end{example}
\longthanks{The author is grateful to Torsten Hoge for his comments to this article.
The author is grateful to the anonymous referees for the careful reading, and in particular, the suggestion to use Lemma \ref{replace}.
The author is
partially supported by JSPS KAKENHI Grant Number JP21H00975.
}
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