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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 $i<j$. 
Then the following two conditions are equivalent:
\begin{itemize}
\item[(1)]
$\A^H$ is free with $\exp(\A^H)=(1,d_2,\ldots,\hat{d}_i,\ldots,\hat{d}_j,\ldots,d_{\ell}) \cup (d)$
\item[(2)]
$\A$ is SPOG with 
$\POexp(\A)=(1,d_2,\ldots,d_i+1,\ldots,{d}_j+1,\ldots,d_\ell)$ and level~$d$. 
\end{itemize}
\label{mainanother}
\end{theorem}

Theorem \ref{mainanother} can be regarded as an extension of the addition 
and restriction theorems (Theorem \ref{addition}). Namely, Theorem \ref{mainanother} determines a minimal free resolution of $D(\A)$ as an SPOG-module when 
$|\exp(\A') \cap \exp(\A^H)|=\ell-2$. 
The condition $d\le d_{j+1}$ is necessary, see Example \ref{eximportant} for details. 


Now go back to Terao's conjecture. 
As we have seen, SPOG arrangements can be regarded as a close analogue of free arrangements.
Thus to study Terao's conjecture by using 
an inductive approach, 
it is important to study 
combinatorial dependency of SPOG arrangements. 
For that purpose, 
let us introduce the following class of arrangements.

\begin{define}
We say that $\A$ is \textbf{stair-SPOG} if 
there is $H \in \A$ such that both $\A':= \A \setminus \{H\}$ and $\A^H$ are stair-free (see Definition \ref{SF} and Theorem \ref{SFcombin}), and $\exp(\A'),\ \exp(\A^H)$ and $|\A'|-|\A^H|$ 
satisfy the conditions in Theorem \ref{mainanother}. 
Let $\mathcal{S}_\ell$ denote the set of stair-SPOG arrangements in an $\ell$-dimensional vector space and let 
$$
\mathcal{S}:=\bigcup_{\ell \ge 2} \mathcal{S}_\ell.
$$
\label{SPOGcombin}
\end{define}


\begin{theorem}
$\A$ is SPOG if $\A \in \mathcal{S}$. Moreover, if there are $\A,\B$ such that $\A \in \mathcal{S}$ and $L(\A) \simeq L(\B)$, then $\B$ is SPOG too.
\label{divSPOG}
\end{theorem}


The organization of this article is as follows. In \S2 we introduce several results and definitions 
for the proof of the main results in this article. 
In \S3 we prove some useful results on the cardinality of the set of minimal generators. 
In \S4 we prove the main results of this article. Several 
examples are also exhibited in \S4. 
\S5 is devoted to investigate 
the relation between Ziegler restriction of an arrangement and their SPOGness by using the methods introduced in the previous sections.
\medskip






\section{Preliminaries}


In this section let us introduce several definitions and results for the proof of the 
main results in this article. First recall some combinatorics of arrangements. 
For the intersection lattice $L(\A)$, we can define the 
\textbf{M\"{o}bius function} 
$\mu:L(\A) \rightarrow \Z$ by $\mu(V)=1$ and by 
$$
\mu(X):=-\sum_{X \subsetneq Y \subset V,\ Y \in L(\A)} \mu(Y)
$$
for $X \in L(\A) \setminus \{V\}$. The generating function of $\mu$ is called the \textbf{characteristic polynomial} of $\A$ defined by 
$$
\chi(\A;t):=\sum_{X 
\in L(\A)}  \mu (X) t^{\dim X},
$$
which is a combinatorial invariant. 
The absolute value of the coefficient of $t^{\ell-i}$ in $\chi(\A;t)$ is called the \textbf{$i$-th Betti number} of $\A$ and denoted by $b_i(\A)$. 
Next let us recall several useful results on $D(\A)$. 

\begin{theorem}[Terao's polynomial $B$, \cite{T1}]
Let $\C \setminus \{H\}=\C'$. Then there is a homogeneous polynomial $B 
\in S$ of degree $|\C|-1-|\C^H|$ such that 
$$
D(\C')(\alpha_H):=\{\theta(\alpha_H) \mid \theta 
\in D(\C')\} \subset (\alpha_H,B).
$$
We call such $B$ a \textbf{polynomial $B$} of $(\C,H)$. 
\label{B}
\end{theorem}

\begin{theorem}[Terao's factorization theorem, \cite{T2}]
Assume that $\A$ is free with $\exp(\A)=(d_1,d_2,\ldots,d_\ell)$, then 
$$
\chi(\A;t)=\prod_{i=1}^\ell (t-d_i).
$$
\label{TF}
\end{theorem}

For $H \in \C$ and $\C':=\C \setminus \{H\}$ we have the following 
\textbf{Euler exact sequence} 
\begin{equation}
0 \rightarrow D(\C') \stackrel{\cdot \alpha_H}{\rightarrow} D(\C) 
\stackrel{\rho}{\rightarrow} D(\C^H).
\label{EE}
\end{equation}

Here for $\theta \in D(\A)$ and the image $\overline{f} \in S/\alpha_H S$ of a polynomial $f \in S$ by the 
canonical surjection $S \rightarrow S/\alpha_H S$, $\rho(\theta)$ is defined by 
$$
\rho(\theta)(\overline{f}):=\overline{\theta(f)}.
$$
The Euler exact sequence is not right exact in general, but it is so when $\C'$ is free as follows.

\begin{theorem}[Free surjection theorem (FST), Theorem 1.13, \cite{A9}]
Let $\C=\C' \cup \{H\}$ and assume that $\C'$ is free. Then 
$\rho=\rho^H:D(\C) \rightarrow D(\C^H)$ is surjective.
\label{FST}
\end{theorem}

Next let us introduce the results on freeness and SPOGness.

\begin{theorem}[Theorem 1.4, \cite{A5}]
Let $\A$ be free with $\exp(\A)=(d_1,\ldots,d_\ell)$, $H \in \A$ and 
assume that $\A':=
\A \setminus \{H\}$ is not free. Then $\A'$ is SPOG with $\POexp(\A')=(d_1,\ldots,d_\ell)$ and level $d:=|\A'|-|\A^H|$.
\label{SPOG}
\end{theorem}

\begin{theorem}[Division theorem, Theorem 1.1, \cite{A2}]
Assume that $\A^H$ is free and $\chi(\A^H;t) \mid \chi(\A;t)$. Then $\A$ is free. Thus 
if we can show the freeness of $\A$ by using the division theorem several times, then 
the freeness of $\A$ is combinatorial, and such a free arrangement is called a  \textbf{divisionally free arrangement}. 
\label{division}
\end{theorem}

\begin{define}[Definition 4.2, \cite{A6}]
We say that $\A$ is \textbf{stair-free} if the freeness of $\A$ can be 
proved by using the addition and division theorems.
\label{SF}
\end{define}

\begin{theorem}[Theorem 4.3, \cite{A6}]
If $\A$ is stair-free, then 
its freeness depends only on~$L(\A)$.
\label{SFcombin}
\end{theorem}


Finally let us recall the fundamentals of the multiarrangement theory introduced by Ziegler in \cite{Z}. 
A pair $(\A,m)$ is a \textbf{multiarrangement} if $\A$ is an arrangement and 
$m:\A \rightarrow \Z_{>0}$. 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 $1 \le i<j \le \ell$, and let 
$\A'$ be free with $SNT(\A')=2$. Let~$\theta_E,\theta_2,\ldots,\theta_\ell$ be a basis for 
$D(\A')$ such that $\theta_k \in D(\A)\ (k \neq i,j)$. 
Let $\theta_i(\alpha_H) =f_i\alpha_H+g_i B,\ 
\theta_j(\alpha_H)=f_j \alpha_H+g_j B$ by 
Theorem \ref{B}.  Then $(g_i,g_j)=1$ and 
$D(\A)$ is generated by 
$\{\theta_k\}_{k \neq i,j} \cup\{\alpha_H \theta_i,\alpha_H \theta_j,g_j\theta_i-g_i\theta_j\}$. In particular, 
$\A$ is SPOG. 
\label{fund2}
\end{prop}

\noindent
\textbf{Proof}.
The proof is essentially the same as that of \cite[Theorem~1.9]{A5} (see \cite[Proposition~3.6]{Sak} too). 
For the completeness, let us give a sketch of the proof. We may assume that $\alpha_H=x_1$ and let 
$(g_i,g_j)=g\in S$. Let $g_i=gh_i,\ g_j=gh_j$ with $(h_i,h_j)=1$. Then clearly $x_1 \theta_i,x_1 \theta_j,
\varphi:=h_j \theta_i-h_i \theta_j \in D(\A)$. By definition, we can choose $g_i,g_j \in \K[x_2,\ldots,x_\ell]=:S'$ so 
we may assume that $h_i,h_j,g \in S'$. First let us prove that $\{\theta_k\}_{k \neq i,j} \cup\{x_1 \theta_i,x_1 \theta_j,\varphi\}$ generate $D(\A)$. Let $\theta \in D(\A)$. 
Since $D(\A) \subset D(\A')$, there are $a_k \in S$ such that $$
\theta=\sum_{k=1}^\ell a_k \theta_k.
$$
Since $\{\theta_k\}_{k \neq i,j} \subset D(\A)$, it suffices to show that 
$\theta-\sum_{k \neq i,j} a_k
\theta_k=a_i\theta_i+a_j \theta_j$ is expressed as a linear combination of 
$x_1 \theta_i,x_1\theta_j,\varphi$. Let us replace $\theta-\sum_{k \neq i,j} a_k$ by $\theta$. Then we can express 
$$
\theta=b_i(x_1 \theta_i)+b_j (x_1 \theta_j)+c_i\theta_i+c_j \theta_j
$$
for some $b_i,b_j\in S,\ c_i,c_j \in S'$. 
Thus 
$$
\theta(x_1)=b_ix_1 \theta_i(x_1)+b_j x_1 \theta_j(x_1)+c_i\theta_i(x_1)+c_j \theta_j(x_1).
$$
Taking the modulo $x_1=\alpha_H$ 
combined with Theorem \ref{B} 
(see the proof of \cite[Theorem~1.9]{A5} for details), we know that 
$$
c_i\theta_i+c_j \theta_j=c(h_j\theta_i-h_i \theta_j)+c'x_1
$$
for some $c,c' \in S$. As a conclusion, it holds that $\theta \in S\theta_i+S\theta_j+S\varphi$, and 
$$
D(\A')=\langle\{\theta_k\}_{k \neq i,j} \cup\{x_1 \theta_i,x_1 \theta_j,\varphi\}\rangle_S.
$$
By comparing the second Betti number of $\A$ calculated combinatorially and 
algebro-geometrically (see the proof of \cite[Theorem~1.9]{A5} for details), we can see that $h$ is a unit. Thus in fact 
$$
D(\A')=\langle\{\theta_k\}_{k \neq i,j} \cup\{x_1 \theta_i,x_1 \theta_j,
g_j\theta_i-g_i\theta_j\}\rangle_S.
$$
A minimal free resolution of $D(\A)$ is easily obtained by the form of this minimal set of generators, 
which completes the proof.\owari
\medskip


An immediate corollary of Proposition \ref{genlowbdd} is as follows:

\begin{cor}
Let $\A'$ be free and assume that $g(\A) =\ell+1$. Then $\A$ is SPOG.
\label{ell+1SPOG}
\end{cor}

\noindent
\textbf{Proof}.
By Proposition \ref{genlowbdd}, $g(\A) =\ell+1$ only when 
$s\le 2$. $s=0$ cannot occur, and $s=1$ implies that $\A$ is free by the addition theorem, thus 
$g(\A)=\ell$. So the rest case is when $s=2$. In this case, Proposition \ref{fund2} 
shows that $\A$ is SPOG.\owari
\medskip


Since the freeness of $\A^H$ and $\A'$ implies that $\pd \A \le 1$ by Theorem \ref{pdaddition},  
it is natural to study which condition on $g(\A)$ makes the arrangement $\A^H$ free. 



\begin{theorem}
Let $\ell \ge 3$ and $\A'$ be free. 
Then $\A^H$ is free if $g(\A) \le \ell+2$.
\label{freegeneratoringeneral}
\end{theorem}



\noindent
\textbf{Proof}. By the addition and restriction theorems, Theorem \ref{FST}, 
Corollary \ref{ell+1SPOG} and the explicit form of the set of SPOG generators as in Proposition 
\ref{fund2}, the statement follows if 
$g(\A) \le \ell+1$. Assume that $g(\A)=\ell+2$. 
Then by Proposition \ref{genlowbdd}, we have a basis $\theta_E,\theta_2,\ldots,\theta_\ell$ for $D(\A')$  such that 
$\theta_i \not \in D(\A)$ for~$i 
\ge \ell-2$ and $\theta_i \in D(\A)$ for~$i \le \ell-3$. We may 
assume that $\alpha_H=x_1$. 
Then clearly $D(\A)$ has a minimal set of generators with cardinality $\ell+2$ of the form 
$$
\theta_E,\theta_2,\ldots,\theta_{\ell-3},x_1\theta_{\ell-2},
x_1\theta_{\ell-1},x_1\theta_{\ell},\varphi_1,\varphi_2,
$$
by the same argument as in the proof of Proposition \ref{genlowbdd} or \ref{fund2}, 
where $\varphi_j$ is a linear combination of $\theta_{\ell-2},\theta_{\ell-1},\theta_\ell$ over $\K[x_2,\ldots,x_\ell]$. By 
Theorem \ref{FST}, the images of $\theta_E,
\theta_2,\ldots,\theta_{\ell-3},\varphi_1,\varphi_2$ by $\rho$ have to generate $D(\A^H)$. Since $\mbox{rank}_{S/\alpha_H S} D(\A^H)=\ell-1$, 
it holds that $\A^H$ is free. \owari
\medskip



By using results above, we can show the following 
proposition which is fundamental on the relation between free and SPOG arrangements.

\begin{prop}
Let $\A$ be SPOG, $H \in \A$ and $\A':=
\A \setminus \{H\}$. 
If $\A'$ is free, then there are
a set of SPOG generators $\theta_1=\theta_E,\theta_2,\ldots,\theta_\ell$, a level element  $\varphi$ and two distinct integers 
$1 < s < t \le \ell$ such that 
$$
\theta_E,\theta_2,\ldots,\theta_{s-1},
\theta_s/\alpha_H,\theta_{s+1},\ldots,
\theta_{t-1},
\theta_t/\alpha_H,\theta_{t+1},\ldots,\theta_\ell
$$
form a free basis for $D(\A')$.
\label{fund}
\end{prop}

\noindent
\textbf{Proof}.
By Proposition \ref{genlowbdd} and the 
assumption that $g(\A)=\ell+1$, for the basis~$\theta_E,\varphi_2,\ldots,\varphi_\ell$ for $D(\A')$, we may assume that 
$\varphi_i \in D(\A)\ (i \le \ell-2)$ and $\varphi_{\ell-1},\varphi_\ell 
\not \in D(\A)$. Then 
$$
\theta_E,\varphi_2,\ldots,\varphi_{\ell-2},\alpha_H \varphi_{\ell-1}, 
\alpha_H\varphi_\ell,f_{\ell}\varphi_{\ell-1}-f_{\ell-1}
\varphi_\ell
$$
form a minimal set of generators for $D(\A)$, where 
$\varphi_i(\alpha_H)=f_i B$ modulo $\alpha_H$ for $i=\ell-1,\ell$ and 
$B$ is Terao's polynomial, which give the required set of generators for~$D(\A)$.\owari
\medskip






\section{Proof of the main results}


In this section let us prove Theorems \ref{mainanother} and \ref{divSPOG}. For that, let us introduce 
the following two results.
\medskip

\begin{lemma}
    Let $N \subset M$ be $S$-graded free modules. Let $\theta_1,\ldots,\theta_n$ be a homogeneous basis for 
    $N$ with $\deg \theta_1 \le \cdots \le \deg \theta_n$ and $\varphi_1,\ldots,\varphi_{n+t}$ be a homogeneous basis for~$M$ 
    with $\deg \varphi_1 \le \cdots \le \deg \varphi_{n+t}$. If $\deg \theta_i=\deg \varphi_i$ for 
    $1 \le i \le n$, then we may choose 
    $\theta_1,\ldots,\theta_n,\varphi_{n+1},\ldots,\varphi_{n+t}$ 
    as a basis for $M$.
    \label{replace}
\end{lemma}

\noindent
\textbf{Proof}. This is essentially the same as Theorem 4.42 in \cite{OT}, but we give a proof 
for the completeness. 
Let $d_i:=\deg\theta_i=\deg 
\varphi_i$. We prove by induction on $1 \le i \le n$. Let~$\theta_1 \in N \subset M$.
Let $\varphi_1,\ldots,\varphi_s$ be of degree $d_1$ and $d_{s+1}>d_1$. 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 $d_k=\cdots=d_s <d_{s+1}$ or $s=n+t$. If $c_i=0$ for all $i$, then $\theta_1,\ldots,\theta_k$ are not independent over $S$. So we may assume that $c_k=1$, and we can choose 
$\theta_1,\ldots,\theta_k,\varphi_{k+1},\ldots,\varphi_{n+t}$ as a basis for $M$. \owari
\medskip


\begin{lemma}
Let $H \in \A$, $\A':=\A \setminus \{H\}$ and assume that $\A'$ and $\A^H$ are both free. Let $\varphi_1=\theta_E,\varphi_2,\ldots,\varphi_\ell$ form a homogeneous basis for $D(\A')$ with $\deg \varphi_i=:d_i \le d_{i+1}=\deg \varphi_{i+1}$ for all $i$. Assume that there are 
an integer $1 \le k \le \ell$, 
finite subsets $I \subset \{1,\ldots,k-1\}=:[k-1]$, 
$T$, and derivations 
$$
G:=\{\varphi_i\}_{i \in I} \cup \{\phi_j:=\alpha_H \varphi_j\}_{j \in [k-1] \setminus I} 
\cup \{\psi_{t}\}_{t \in 
T} \cup \{\eta_u\}_{u=k}^\ell
$$
in $D(\A)$ which satisfy the following conditions:
\begin{itemize}
\item[(1)]
$\psi_t$ is a linear combination of $\{\varphi_j\}_{j \in 
[k-1]\setminus I}$ over S for all $t \in T$, and 
$\deg \eta_u=\deg \varphi_u=d_u$ for all $k \le u \le \ell$,
    \item[(2)]
$$
\deg \eta_u \ge \max\{\deg \varphi_i,\deg \phi_j,\deg \psi_{t}\mid 
i \in I,j\in [k-1] \setminus I, t \in T\}
$$
for all $k \le u \le \ell$, 
\item[(3)]
$$
(\bigoplus_{ j\in [k-1]\setminus I}S \varphi_j) \cap D(\A)
=\sum_{j \in [k-1]\setminus I}S \phi_j+
\sum_{t \in T} S \psi_{t},
$$
and 
\item[(4)]
the image of $G \setminus 
\{\phi_j\}_{j \in [k-1] \setminus I} 
$ by the Euler restriction map $\rho$ form a basis for $D(\A^H)$.
\end{itemize}
Then the $S$-module generated by 
$$
G':=\{\varphi_i\}_{i \in I} \cup \{\phi_j\}_{j \in [k-1] \setminus I} 
\cup \{\psi_{t}\}_{t \in T} \cup \{\varphi_u\}_{u=k}^\ell
$$
coincides with the $S$-module generated by $G$.
 \label{replace2}
\end{lemma}

\noindent
\textbf{Proof}.
Let $J:=[k-1] \setminus I$. 
We prove by induction on $k \le u\le \ell$. Assume that $\eta_{a}=\theta_{a}$ for $k \le a \le u-1$, and let us show that we can choose $\eta_u=\varphi_u$. Since 
$\eta_u \in D(\A) \subset D(\A')$, we can express
$$
\eta_u=\sum_{a=1}^\ell g_a \varphi_a
$$
for some $g_a \in S$. Since we are interested in sets of generators, we may replace $\eta_u - \sum_{i \in I} g_i \varphi_i$ by $\eta_u$ to get an expression 
$$
\eta_u=\sum_{j \in J} g_j \varphi_j +\sum_{a=k}^{u-1} g_a \eta_a+ 
\sum_{a=u}^{\ell} g_a \varphi_a.
$$
Let $d_u=\cdots=d_b <d_{b+1}$ or $b=\ell$. Then replacing $\eta_u-
\sum_{a=k}^{u-1} g_a \eta_a$ by $\eta_u$, we obtain 
$$
\eta_u=\sum_{j \in J} g_j \varphi_j +
\sum_{a=u}^{b} g_a \varphi_a.
$$
Assume that $\sum_{a=u}^{b} g_a \varphi_a\neq 0$, say $g_u=1$ by the reason of degrees in the conditions~(1) 
and~(2). Then replacing $\sum_{j \in J} g_j \varphi_j +
\sum_{a=u}^{b} g_a \varphi_a$ by $\varphi_u$, we can choose $\eta_u$ as $\varphi_u$. So assume that all $g_a=0$ for $u
\le a 
\le b$. Then 
$$
\eta_u=\sum_{j \in J} g_j \varphi_j. 
$$
Since $\eta_u \in D(\A)$, the condition (3) shows that 
$$
\eta_u=\sum_{ j\in J} h_j \phi_j+\sum_{t \in T}
h_{t} \psi_{t}
$$
for some $h_j,h_{t} \in S$. Sending it by $\rho$, 
the Euler exact sequence shows that 
$$
\rho(\eta_u)-\sum_{t \in T}
h_{t} \rho(\psi_{t})=0,
$$
contradicting the independency of the basis for $D(\A^H)$ in the condition (4), which completes the proof.\owari
\medskip



\noindent
\textbf{Proof of Theorem \ref{mainanother}}. 
First we prove $(2) \Rightarrow (1)$. 
It is easy to see that, for the basis~$\theta_E,\varphi_2,\ldots,\varphi_\ell$ for 
$D(\A')$ with 
$\deg \varphi_k=d_k$, the set of SPOG generators for $D(\A)$ is of the form 
$\{\theta_k\}_{k \neq i,j} \cup \{\alpha_H \theta_k\}_{k=i,j} \cup \{f_j\theta_i-f_i\theta_j\}$ by 
Proposition \ref{fund}. 
Thus Theorem~\ref{FST} shows that $D(\A^H)$ is generated by the image of 
$\{\theta_k\}_{k \neq i,j} \cup\{f_j\theta_i-f_i\theta_j\}$. Since $\mbox{rank}_S D(\A^H)=\ell-1$, it follows that $\A^H$ is free with the 
given 
exponents.

Next we prove $(1) \Rightarrow (2)$. 
Assume that 
$\A^H$ is free with the given exponents above. 
In this assumption, Terao's addition theorem shows that $\A$ is not free since 
$\exp(\A^H) \not 
\subset \exp(\A')$.
By Theorem \ref{FST}, the Euler restriction map 
$\rho^H:D(\A) \rightarrow D(\A^H)$ is surjective. 
Thus there are $S$-independent 
derivations $\theta_E,\theta_2,\ldots,\hat{\theta}_i,\ldots,\hat{\theta}_j,\ldots,\theta_\ell,\theta 
\in D(\A)$ such that $\deg \theta_k=d_k,\ \deg \theta=d$, and their images by $\rho^H$ form a basis for 
$D(\A^H)$. Let $\theta_E=\varphi_1,\varphi_2,\ldots,\varphi_\ell$ be a basis for $D(\A')$ with 
$\deg \varphi_k=d_k$. We may assume that~$d_i<d_{i+1}$ and 
$d_j<d_{j+1}$ or $j=\ell$. 
Since $D(\A) \subset D(\A')$, it holds that $$
\bigoplus_{k=1}^{i-1} S\theta_k \subset
\bigoplus_{k=1}^{i-1} S\varphi_k.
$$
So by Lemma \ref{replace}, 
we can choose $\varphi_k=\theta_k$ for 
$k<i$. 

Next let us show that $\varphi_i \not \in D(\A)$. Assume that $\varphi_i \in D(\A)$. Recall that $d_i<d_{i+1}$. 
So $\rho(\varphi_i) \in D(\A^H)_{d_i}$ is a linear combination of $\rho(\theta_1),\ldots,\rho(\theta_{i-1})$. Hence 
$$
\varphi_i-\sum_{s=1}^{i-1} f_s \theta_s  \in\alpha_H D(\A')
$$ 
for some $f_s \in S$.
Thus replacing $\varphi_i-\sum_{s=1}^{i-1} f_s \theta_s$ by $\varphi_i$, it holds that 
$\varphi_i/\alpha_H \in D(\A')$, contradicting the minimality of the basis 
$\theta_E,\varphi_2,\ldots,\varphi_\ell$ for $D(\A')$. So $\varphi_i \not \in D(\A)$. Hence $D(\A) \cap (S\varphi_i )=S\alpha_H \varphi_i$.
Now apply Lemma \ref{replace2} to obtain that $\theta_s=\varphi_s$ for 
$1 \le s \le i-1,\ i+1 \le s \le j-1$.
Moreover, the same proof as that of $\varphi_i \not \in D(\A)$ shows that $\varphi_j \not \in D(\A)$. 

Now express $\theta$ in the following form by using the fact that 
$d =\deg \theta \le d_{j+1}$:
\begin{equation}
\theta=\sum_{i\neq k=1}^{j-1} f_k \theta_k+
\sum_{k=j+1}^p a_k \varphi_k+f_i \varphi_i+f_j \varphi_j.
\label{eq0}
\end{equation}
Here $d_{j+1}=\cdots=d_p=d < d_{p+1}$ and $a_k \in \K$. First assume that $a_k=0$ for all $k$. Then 
replacing $\theta-\sum_{i\neq k=1}^{j-1} f_k \theta_k$ by $\theta$, we have 
\begin{equation}
\theta=f_i \varphi_i+f_j \varphi_j.
\label{eq1}
\end{equation}
By the independency of images of $\theta_k$ and $\theta$ by $\rho$, at least one of $f_i,f_j$ is not zero. Assume that only one of 
them is not zero, say $f_i\neq 0$ and $f_j=0$. 
Then (\ref{eq1}) combined with the fact that $\varphi_i \not \in D(\A)$ 
shows that $\alpha_H \mid f_i$. 
So 
$\rho^H(\theta)=0$, a contradiction. 
Thus both $f_i$ and $f_j$ are not zero.
Recall 
that 
$\varphi_i,\varphi_j \not \in D(\A)$. 
Thus the same proof as Proposition \ref{fund2} shows that, 
letting $\varphi_i(\alpha_H)=g_i B$, 
$\varphi_j(\alpha_H)=g_j B$ modulo $\alpha_H$, 
$(g_i,g_j)=h$, $g_i=hh_i,\ g_j=hh_j$ and $(h_i,h_j)=1$, it holds that 
$\varphi:=h_j\varphi_i-h_i\varphi_j \in D(\A)$. Note that 
$g_i,g_j,h_i,h_j,h \in S':=\K[x_2,\ldots,x_\ell]$. 
By the construction, $\deg \varphi\le d=\deg \theta=d_i+d_j-|\A'|+|\A^H|=d_i+d_j-\deg B$. 
Assume that $\deg 
\varphi < d$. 
Send $\varphi$ by $\rho$, then we have 
$$
\varphi=h_j\varphi_i-h_i\varphi_j=\sum_{i \neq s=1}^{j-1}b_s \varphi_s+\alpha_H(b_i \varphi_i+b_j \varphi_j)
$$
for some $b_s \in S$. Thus $\alpha_H \mid h_i$ and $\alpha_H \mid b_j$, contradicting $h_i,h_j \in S'$.
Thus we may assume that $h=1$ and 
$\deg (g_i\varphi_j-g_j \varphi_i)=d=\deg \theta$. Hence the equation (\ref{eq1}) shows that 
$$
\theta=g_i\varphi_j-g_j \varphi_i
$$
modulo $\alpha_H$. 

Second assume that $a_k\neq 0$ for some $k$ in the equation (\ref{eq0}). Then we may assume that $a_{j+1}=1$ and the equation (\ref{eq1}) in this case is as follows:
\begin{equation}
\theta=f_i \varphi_i+f_j \varphi_j+\varphi_{j+1}+\sum_{k=j+2}^p a_k \varphi_k.
\label{eq2}
\end{equation}
Replacing 
$f_i\varphi_i+f_j \varphi_j+\varphi_{j+1}+\sum_{k=j+2}^p a_k \varphi_k$ by $\varphi_{j+1}$, 
we may assume that $\theta=\varphi_{j+1}$. Continue this for 
$\theta_{j+1},\ldots,\theta_{p}$, then we obtain either $\theta_k=
\varphi_{k+1}$, or $\theta_k=g_i\varphi_j-g_j \varphi_i$ modulo $\alpha_H$ for 
$j+1\le k \le p$.
So exchanging an appropriate $\theta_k$ by $\theta$, 
we obtain that $\theta=g_j\theta_i-g_i\theta_j$ modulo $\alpha_H$. Hence in both cases,$$
D(\A) \cap(S \varphi_i\oplus S\varphi_j)=
S\alpha_H \varphi_i+S\alpha_H \varphi_j+S\theta.
$$
Thus applying Lemma \ref{replace2}, we obtain that $\varphi_s=\theta_s$ for all $1 \le s \le \ell$ with $s\neq i, j$ and 
$\varphi_i,\varphi_j \not \in D(\A)$. Therefore, Proposition \ref{fund2} shows that 
$\A$ is SPOG with $\POexp(\A)=(1,d_2,\ldots,d_i+1,\ldots,d_j+1,\ldots,d_\ell)$ and level $d=d_i+d_j-|\A'|+|\A^H|$. \owari
\medskip


The following case is the most practical to apply Theorem \ref{mainanother}.

\begin{cor}
Let $\A'$ be free with $\exp(\A')=(1,d_2,\ldots,d_\ell)_\le$ and $d:
=d_i+d_\ell+|\A^H|-|\A'| > d_\ell$ 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 
$$
Q(\A)=\prod_{i=1}^4 x_i \prod_{1 \le i<j \le 4} (x_i^2-x_j^2)=0.
$$
Let  
$\A':=\A \setminus \{H\}$, where $H:x_1=0$. Let $L:x_1+x_2+x_3=0$. 
We know that~$\A'$ is divisionally free with $\exp(\A')=(1,3,5,6)$, and 
$(\A'\cup\{L\})^L$ is also divisionally free with $\exp(\A'\cup\{L\})^L)=(1,5,7)$. Thus Theorem \ref{mainanother} 
confirms that $\B:=\A'\cup\{L\}$ is SPOG with $\POexp(\B)=(1,4,5,7)$ and level $7$. 
Next let $\C:=\B \cup \{H\}$. Then $|\C^H|=9$, so $|\B|-|\C^H|=16-9=7$. 
Thus the set of generators of degrees $1,4,5$ for $D(\B)$ are in $D(\C)$ too, and we may assume that one of 
two generators of degree $7$ is in $D(\C)$ by Theorem \ref{B}. Since the relations in $D(\B)$ are among three derivations of degrees~$4,7,7$ by the explicit construction of the set of 
SPOG generators and a level element in Proposition \ref{fund2}, we know that these 
$4$-basis elements in $D(\C)$ are $S$-independent. So~$\C$ is combinatorially free with $\exp(\C)=(1,4,5,7)$ since the SPOGness of $\B$ and $|\C|$ are both combinatorial.
\label{deletedB}
\end{example}

Unfortunately, there are cases in which $\A'$ and $\A^H$ are free, $\A$ is SPOG but Theorem \ref{mainanother} cannot be applied.


\begin{example}


Let $\A$ be the Weyl arrangement of the type $\A_4$ and let 
$H:x_1+x_2+x_3=0$. Say that $\B:=\A \cup \{H\}$. Then $\B^H$ is free with 
$\exp(\B^H)=(1,4,5) \not \subset (1,2,3,4)=\exp(\A)$. We can check that 
$\B$ is SPOG with $\POexp(\B)=(1,3,4,4)$ and level $5$ by using Macaulay2 in \cite{GS}, but we cannot apply Theorem \ref{mainanother}.

\label{ex2}
 \end{example}


\begin{problem}
Generalize Theorem \ref{mainanother} to all cases when $\A'$ and $\A^H$ are free, 
$\ell-2 \le |\exp(\A') \cap \exp(\A^H)| \le \ell-1$ and $\A$ 
is not free. 
\label{mainanotherdiff}
\end{problem}

In fact, to apply Theorem \ref{mainanother} the condition $d \le d_{j+1}$ is necessary. Let us see the following example.

\begin{example}
Let $\A'$ be an arrangement in $\R^4$ 
defined by 
$$
Q(\A')=(x_1+x_2+x_3+x_4)\prod_{i=1}^4 x_i \prod_{i=2}^4 (x_1+x_i) \prod_{i=2}^4
(x_1+x_2+x_3+x_4-x_i).
$$
Then $\A'$ is free with $\exp(\A')=(1,3,3,4)$. 
Let $H=x_2+x_3+x_4$ and let $\A:=
\A' \cup\{H\}$. Then $\A^H$ is free with $\exp(\A^H)=(1,4,5)$. So 
$\exp(\A') \cap \exp(\A^H)=\{1,4\}$. However, Macaulay2 in 
\cite{GS} shows that $D_0(\A):=D(\A)/S\theta_E$ is not SPOG but has a following minimal 
free resolution:
$$
0 \rightarrow 
S[-5]\oplus S[-6] 
\rightarrow 
S[-4]^3 \oplus S[-5]^2 \rightarrow D_0(\A) 
\rightarrow 0.
$$
Since $d=5=3+3-|\A'|+|\A^H|>4$, 
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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