High order free hyperplane arrangements in 3-dimensional vector spaces

TL;DR

This paper proves that all 3-dimensional hyperplane arrangements are m-free for sufficiently large m, including specific types like Weyl arrangements of types A and B, extending the understanding of m-freeness.

Contribution

It confirms Holm's conjecture for 3-dimensional arrangements by showing they are m-free for all m beyond a certain bound and computes their m-exponents.

Findings

All 3-arrangements are m-free for m ≥ |A|+2.

Computed m-exponents for these arrangements.

Weyl arrangements of types A and B are m-free for all m ≥ 0.

Abstract

Holm introduced $m$-free $\ell$-arrangements which is a generalization of free arrangements, while he asked whether all $\ell$-arrangements are $m$-free for $m$ large enough. Recently Abe and the author verified that this question is in the negative when $\ell\geq 4$. In this paper we verify that $3$-arrangements $\mathscr{A}$ are $m$-free and compute the $m$-exponents for all $m\geq |\mathscr{A}|+2$, where $|\mathscr{A}|$ is the cardinality of $\mathscr{A}$. Hence Holm's question is in the positive when $\ell=3$. Finally we prove that $3$-dimensional Weyl arrangements of types A and B are $m$-free for all $m\geq 0$.

Equations

$$|\mathbf{a}|=a_1+\cdots +a_{\ell },\mathbf{a}!=a_1!\cdots a_{\ell }!,x^{\mathbf{a}}=x_1^{a_1}\cdots x_{\ell }^{a_{\ell }},\mathrm{and}{\partial }^{\mathbf{a}}={\partial }_1^{a_1}\cdots {\partial }_{\ell }^{a_{\ell }}.$$

$$D^{(m)}(\mathcal{A},V):=\left\{\theta \in D^{(m)}(S)\mid \theta (QS)\subseteq QS\right\},$$

$$s_m(\ell ):=\left(\genfrac{}{}{0ex}{}{m+\ell -1}{m}\right)=\left(\genfrac{}{}{0ex}{}{m+\ell -1}{\ell -1}\right).$$

$${\exp }_m(\mathcal{A},V):=\{\mathrm{deg}({\theta }_1),\dots ,\mathrm{deg}({\theta }_{s_m(\ell )})\}=\{e_1^{i_1},e_2^{i_2},\dots \},$$

$$L(\mathcal{A}):=\left\{\bigcap _{H\in \mathcal{B}}H|\mathcal{B}\subseteq \mathcal{A}\right\}$$

$$L(\mathcal{A}{)}_1:=\left\{X\in L(\mathcal{A})|{\dim }_{\mathbb{K}}(X)=1\right\}$$

$${\delta }_X:=\sum _{i=1}^{\ell }{x}_i(v_X){\partial }_i.$$

$${\mathcal{A}}_X:=\left\{H\in \mathcal{A}|X\subseteq H\right\}.$$

$$\mathbb{K}{\delta }_X=\left\{\delta \in \sum _{i=1}^{\ell }\mathbb{K}{\partial }_i|\delta \left({\alpha }_H\right)=0\mathrm{for}H\in {\mathcal{A}}_X\right\}.$$

$${\delta }_X({\alpha }_H)=\sum _{i=1}^{\ell }{x}_i(v_X){\partial }_i({\alpha }_H)=\sum _{i=1}^{\ell }{\partial }_i({\alpha }_H)x_i(v_X)={\alpha }_H(v_X).$$

$$\left\{(v_1,\dots ,v_{\ell })\in {\mathbb{K}}^{\ell }|\sum _{i=1}^{\ell }{v}_i{\partial }_i\left({\alpha }_H\right)=0\mathrm{for}H\in {\mathcal{A}}_X\right\}$$

$$=$$

$$\begin{array}{ccccccccc}0& \to & \mathrm{Ker}({\delta }_X:V^{\ast }\to \mathbb{K})& \to & V^{\ast }& \stackrel{{\delta }_X}{\to }& \mathbb{K}& \to & 0\end{array}$$

$$y_X:={\iota }_X(1).$$

$$V_X^{\ast }:=\mathrm{Ker}({\delta }_X:V^{\ast }\to \mathbb{K})$$

$${\delta }_X=\sum _{i=1}^{\ell -1}{\delta }_X(y_i){\delta }_i+{\delta }_X(y_X){\delta }_{\ell }={\delta }_X(y_X){\delta }_{\ell }={\delta }_{\ell }.$$

$$X_1:=\{x_1=0,x_2=0\},X_2:=\{x_1=0,x_3=0\},$$

$$X_3:=\{x_2=0,x_3=0\},X_4:=\{x_1=x_2,x_3=0\},$$

$${\mathcal{A}}_{X_1}=\{H_1,H_2,H_4\},{\mathcal{A}}_{X_2}=\{H_1,H_3\},{\mathcal{A}}_{X_3}=\{H_2,H_3\},{\mathcal{A}}_{X_4}=\{H_3,H_4\}.$$

$${\delta }_{X_1}={\partial }_3,{\delta }_{X_2}={\partial }_2,{\delta }_{X_3}={\partial }_1,{\delta }_{X_4}={\partial }_1+{\partial }_2,$$

$$y_{X_1}=x_3,y_{X_2}=x_2,y_{X_3}=x_1,y_{X_4}=\frac{1}{2}(x_1+x_2).$$

$$Q_X:=Q({\mathcal{A}}_X).$$

$$D^{(j)}({\mathcal{A}}_X,V_X)=\left\{\theta \in D^{(j)}(S_X)|\theta (Q_X{S}_X)\subseteq Q_X{S}_X\right\}$$

$$\displaystyle{\rm exp}_{m}(\mathscr{A},V)=\left\{\begin{array}[]{ll}\{m,(|\mathscr{A}|-1)^{m}\}&(0\leq m\leq|\mathscr{A}|-1),\\ \{(|\mathscr{A}|-1)^{|\mathscr{A}|},|\mathscr{A}|^{m-|\mathscr{A}|+1}\}&(m\geq|\mathscr{A}|),\end{array}\right.$$

$$\displaystyle\widetilde{\mathscr{A}}:=\left\{\begin{array}[]{ll}\mathscr{A}&{\rm if}\ m=n-2,\\ \mathscr{A}\cup\{H_{n+1},\dots,H_{m+2}\}&{\rm if}\ m>n-2.\end{array}\right.$$

$$i_X:=|{\tilde{\mathcal{A}}}_X|-2\mathrm{and}{\tilde{P}}_X:=\prod _{H\in \tilde{\mathcal{A}}\setminus {\tilde{\mathcal{A}}}_X}{\alpha }_H=\frac{Q(\tilde{\mathcal{A}})}{Q({\tilde{\mathcal{A}}}_X)}.$$

$$B:=\bigcup _{X\in L(\tilde{\mathcal{A}}{)}_1}\bigcup _{j=0}^{i_X}\left\{{\tilde{P}}_{X}u{y}_X^{i_X-j}|u\in B(X,j)\right\}\subseteq S_m.$$

$$B^{\ast }:=\bigcup _{X\in L(\tilde{\mathcal{A}}{)}_1}\bigcup _{j=0}^{i_X}\left\{\frac{1}{{\delta }_X^{m-j}({\tilde{P}}_X{y}_X^{i_X-j})}{\partial }_u{\delta }_X^{m-j}|u\in B(X,j)\right\}\subseteq \sum _{|\mathbf{a}|=m}\mathbb{K}{\partial }^{\mathbf{a}}.$$

$$B$$

$$B^{\ast }$$

$$B$$

$$B^{\ast }$$

$${\delta }_X^{m-i_X}({\tilde{P}}_{Y}f)$$

Full text

Key Words: hyperplane arrangements, $m$-free arrangements, $m$-exponents, intersection lattices

2010 Mathematics Subject Classification: Primary 32S22, Secondary 52C35.

1 Introduction

Let $\mathbb{K}$ be a field of characteristic zero and let $V$ be an $\ell$-dimensional vector space over $\mathbb{K}$. A (central hyperplane) arrangement $\mathscr{A}$, or $(\mathscr{A},V)$, is a finite set of hyperplanes in $V$ which contain the origin. We call $\mathscr{A}$ an $\ell$-arrangement when we emphasize the dimension of $V$.

For a $k$-dimensional vector space $\Omega$ and for a $\mathbb{K}$-basis $\{\omega_{1},\dots,\omega_{k}\}$ for $\Omega$, a $\mathbb{K}$-basis $\{\omega_{1}^{\ast},\dots,\omega_{k}^{\ast}\}$ for $\Omega^{\ast}:={\rm Hom}_{\mathbb{K}}(\Omega,\mathbb{K})$ is said to be the dual basis for $\{\omega_{1},\dots,\omega_{k}\}$ if \omega_{i}^{\ast}(\omega_{j})=\left\{\begin{array}[]{ll}1&\ (i=j),\\ 0&\ (i\neq j).\end{array}\right. Let $S={\rm Sym}(V^{\ast})$ be the symmetric algebra of $V^{\ast}$, let $\{x_{1},\dots,x_{\ell}\}$ be a $\mathbb{K}$-basis for $V^{\ast}$, and let $\{\partial_{1},\dots,\partial_{\ell}\}$ be the dual basis for $\{x_{1},\dots,x_{\ell}\}$. We can consider $S$ as the polynomial ring $\mathbb{K}[x_{1},\dots,x_{\ell}]$ in variables $x_{1},\dots,x_{\ell}$ and $\partial_{1}=\partial/\partial x_{1},\dots,\partial_{\ell}=\partial/\partial x_{\ell}$ as partial derivatives. For any hyperplane $H\in\mathscr{A}$, there exist a linear form $\alpha_{H}$ in the dual space $V^{\ast}$ such that $\{\alpha_{H}=0\}=H$. We call $Q:=Q(\mathscr{A}):=\prod_{H\in\mathscr{A}}\alpha_{H}$ a defining polynomial of $\mathscr{A}$. The cardinality $n:=|\mathscr{A}|$ of the arrangement $\mathscr{A}$ equals the degree of $Q$.

Let $\mathbb{N}=\{0,1,2,\dots\}$ be the set of nonnegative integers and let $m\in\mathbb{N}$. For $\bm{a}=(a_{1},\dots,a_{\ell})\in\mathbb{N}^{\ell}$, we denote that

[TABLE]

Let $D^{(m)}(S):=\sum_{|\bm{a}|=m}S\partial^{\bm{a}}$ be the $S$-module generated by $m$-th partial derivatives. An $S$-submodule $D^{(m)}(\mathscr{A},V)$ of $D^{(m)}(S)$ is defined by

[TABLE]

which is called the module of $m$-th order $\mathscr{A}$-differential operators. Let

[TABLE]

We say that an arrangement $\mathscr{A}$ is $m$-free if $D^{(m)}(\mathscr{A},V)$ has linearly independent generators $\theta_{1},\dots,\theta_{s_{m}(\ell)}$ over $S$, and the set $\{\theta_{1},\dots,\theta_{s_{m}(\ell)}\}$ of such generators is called a free basis. Let $S_{i}$ be the vector space consisting of homogeneous polynomials of degree $i$ in $S$. For $\theta=\sum_{|\bm{a}|=m}f_{\bm{a}}\partial^{\bm{a}}\in D^{(m)}(S)$, we write $\deg(\theta)=i$ if $f_{\bm{a}}\in S_{i}$ for each $\bm{a}$. A multi-set ${\rm exp}_{m}(\mathscr{A},V)$ of $m$-exponents of an $m$-free arrangement $\mathscr{A}$ is defined by

[TABLE]

where $e^{i}\in{\rm exp}_{m}(\mathscr{A},V)$ means that the integer $e$ occurs $i$ times in the multi-set ${\rm exp}_{m}(\mathscr{A},V)$. We usually call $1$-free arrangements free arrangements. Free arrangements have been studied, relating with combinatorics of hyperplane arrangements. In particular the addition-deletion theorems [10] and Terao’s factorization theorem [11] describe relations of a free arrangement and the intersection poset

[TABLE]

whose ordering is defined by reverse inclusions (see also [8]).

For $m\geq 2$, the $m$-free arrangements are introduced by Holm [2, 3] to study the ring of differential operators on $\mathscr{A}$, which is a quotient ring of the Weyl algebra (see [4]). In particular Holm [2] proved that $2$-arrangements are $m$-free for all $m\geq 0$ by constructing free bases, and these free bases are used in [5] to prove that the ring of differential operators on a $2$-arrangement is a Noetherian ring. It is also an interesting problem to observe the behavior of $m$-freeness. Holm [2] asked whether all arrangements are $m$-free for $m$ large enough. In the case when $\mathscr{A}$ is a generic $\ell$-arrangement (i.e., every $\ell$ hyperplanes of $\mathscr{A}$ intersect only at the origin), $\mathscr{A}$ is $m$-free if and only if $m\geq n-\ell+1$ [2, 7]. This means that the Holm’s question is in the positive for generic arrangements. However for any $\ell\geq 4$, there exist an $\ell$-arrangement $\mathscr{A}$ such that $\mathscr{A}$ is not $m$-free for any $m\geq 1$ (shown by Abe and the author [1]), that is, Holm’s question is in the negative for $\ell\geq 4$. In this paper we verify that Holm’s question is in the positive for $3$-arrangements by proving that all $3$-arrangements are $m$-free for all $m\geq n-2$. After that we compute the $m$-exponents of $3$-arrangements for all $m\geq n-2$ which depend only on the structure of the poset $L(\mathscr{A})$222In other words, let $(\mathscr{A},V)$ and $(\mathscr{A}^{\prime},V)$ be $3$-arrangements with $n=|\mathscr{A}|=|\mathscr{A}^{\prime}|$, and let $m\geq n-2$. If $L(\mathscr{A})\simeq L(\mathscr{A}^{\prime})$ as posets, then ${\rm exp}_{m}(\mathscr{A},V)={\rm exp}_{m}(\mathscr{A}^{\prime},V)$.. In addition we verify that $3$-dimensional Weyl arrangements of types A and B are $m$-free for all $m\geq 0$. Since $3$-dimensional irreducible Weyl arrangements are only of types A and B, all the Weyl $3$-arrangements are $m$-free for all $m\geq 0$.

The paper is structured as follows. In Section 2 we argue a change of variables of $S$ and we define submodules of $D^{(m)}(\mathscr{A},V)$. In Section 3 we prove that if $m\geq n-2$, then $D^{(m)}(\mathscr{A},V)$ is a direct sum of deformed submodules defined in Section 2. By using this direct sum, we prove that $3$-arrangements are $m$-free for all $m\geq n-2$. In Section 4 we compute the $m$-exponents of $3$-arrangements for all $m\geq n-2$. Finally, in Section 5, we construct free bases for $D^{(m)}(\mathscr{A},V)$ for the remaining orders $m<n-2$ when $\mathscr{A}$ is the Weyl $3$-arrangement of type A and of type B.

2 Change of variables

For any arrangement $\mathscr{A}$, let

[TABLE]

be the set of $1$-dimensional elements in $L(\mathscr{A})$. Let $X\in L(\mathscr{A})_{1}$ and let $v_{X}$ be a nonzero vector in $X$. Then $X=\mathbb{K}v_{X}$. We define a derivation $\delta_{X}\in\sum_{i=1}^{\ell}\mathbb{K}\partial_{i}$ by

[TABLE]

For $X\in L(\mathscr{A})$, a localization $\mathscr{A}_{X}$ of $\mathscr{A}$ at $X$ is defined by

[TABLE]

Proposition 2.1**.**

For $H\in\mathscr{A}$, $\delta_{X}(\alpha_{H})=0$ if and only if $H\in\mathscr{A}_{X}$. Moreover

[TABLE]

Proof. For $H\in\mathscr{A}$, we have $\alpha_{H}=\sum_{i=1}^{\ell}\partial_{i}(\alpha_{H})x_{i}$ and

[TABLE]

Therefore $H\in\mathscr{A}_{X}\ \Leftrightarrow\ X\subseteq H\ \Leftrightarrow\ \alpha_{H}(v_{X})=0\ \Leftrightarrow\ \delta_{X}(\alpha_{H})=0$. This implies that $\delta_{X}$ is in the right hand side of (2.4).

It remains to prove that the dimension of the right hand side of (2.4) equals one. Let $\{e_{1},\dots,e_{\ell}\}$ be a basis for $V$ for which $\{x_{1},\dots,x_{\ell}\}$ is the dual basis. Since $\sum_{i=0}^{\ell}\alpha_{H}(e_{i})x_{i}=\alpha_{H}=\sum_{i=0}^{\ell}\partial_{i}(\alpha_{H})x_{i}$, we have $\alpha_{H}(e_{i})=\partial_{i}(\alpha_{H})$ for $1\leq i\leq\ell$. Hence the assertion follows from

[TABLE]

The linear map $\delta_{X}:V^{\ast}\rightarrow\mathbb{K}$ is defined by $\alpha\mapsto\delta_{X}(\alpha)$ for $\alpha\in V^{\ast}$. Since $v_{X}\neq 0$, the map $\delta_{X}$ is not zero. Then the image ${\rm Im}(\delta_{X}:V^{\ast}\rightarrow\mathbb{K})$ is $\mathbb{K}$. We have the split exact sequence

[TABLE]

and there exists a section $\iota_{X}:\mathbb{K}\rightarrow V^{\ast}$ such that $\delta_{X}\circ\iota_{X}=id_{\mathbb{K}}$, where $id_{\mathbb{K}}$ is the identity map of $\mathbb{K}$. We define

[TABLE]

Let

[TABLE]

and let $\{y_{1},\dots,y_{\ell-1}\}$ be a $\mathbb{K}$-basis for $V^{\ast}_{X}$. We note that $\delta_{X}(y_{X})=1$ and $\delta_{X}(y_{i})=0$ for any $1\leq i\leq\ell-1$. Let $\{\delta_{1},\dots,\delta_{\ell-1},\delta_{\ell}\}$ be the dual basis for the basis $\{y_{1},\dots,y_{\ell-1},y_{X}\}$ for $V^{\ast}=V^{\ast}_{X}\oplus\iota_{X}(\mathbb{K})$. Then

[TABLE]

Hence we may consider $\{y_{1},\dots,y_{\ell-1},y_{X}\}$ as variables of $S$ with the partial derivatives $\{\delta_{1},\dots,\delta_{\ell-1},\delta_{X}\}$.

Example 2.2**.**

Let $\ell=3$. Let us consider the $3$-arrangement $\mathscr{A}$ consisting of four hyperplanes $H_{1}:=\{x_{1}=0\},H_{2}:=\{x_{2}=0\},H_{3}:=\{x_{3}=0\},H_{4}:=\{x_{1}-x_{2}=0\}$ (see Figure 1). The elements of $L(\mathscr{A})_{1}$ are

[TABLE]

and

[TABLE]

By Proposition 2.1, we may consider $\delta_{X}$ and $y_{X}$ $(X\in L(\mathscr{A})_{1})$ as

[TABLE]

Let $S_{X}:={\rm Sym}(V^{\ast}_{X})=\mathbb{K}[y_{1},\dots,y_{\ell-1}]$ be the symmetric algebra of $V^{\ast}_{X}$. Then $S=S_{X}\otimes_{\mathbb{K}}\mathbb{K}[y_{X}]$. We note that $\alpha_{H}\in S_{X}$ if $H\in\mathscr{A}_{X}$ and $\alpha_{H}\not\in S_{X}$ if $H\in\mathscr{A}\setminus\mathscr{A}_{X}$. We write

[TABLE]

Let $V_{X}:=V/X$ be the quotient vector space of $V$ by $X$. Then $V^{\ast}_{X}$ is considered as the dual space of $V_{X}$. Moreover, for an integer $j\geq 0$, the module

[TABLE]

is considered as a submodule of $D^{(j)}(\mathscr{A},V)$.

3 Freeness when $\ell=3$ and $m\geq|\mathscr{A}|-2$

In this section we prove that every $3$-arrangement $\mathscr{A}$ is $m$-free when $m\geq n-2$. An arrangement $\mathscr{A}$ is said to be essential if $\bigcap_{H\in\mathscr{A}}H=\{0\}$. We first investigate a nonessential arrangement, which is a direct product of a $2$-arrangement and the empty $1$-arrangement. Holm proved the following.

Proposition 3.1** (Proposition III. 5.2 in [2]).**

Let $(\mathscr{A},V)$ be a $2$-arrangement. Then for any $m\geq 0$, $(\mathscr{A},V)$ is $m$-free with

[TABLE]

where $e^{i}\in{\rm exp}_{m}(\mathscr{A},V)$ means that the integer $e$ occurs $i$ times.

In addition Abe and the author characterised the $m$-freeness of product arrangements as follows.

Theorem 3.2** (Theorem 2.1 in [1]).**

Let $(\mathscr{A}_{1},V_{1})$ and $(\mathscr{A}_{2},V_{2})$ be arrangements with $\dim(V_{1})>0$ and $\dim(V_{2})>0$. Then the product arrangement $(\mathscr{A}_{1}\times\mathscr{A}_{2},V_{1}\oplus V_{2})$ is $m$-free if and only if both $(\mathscr{A}_{1},V_{1})$ and $(\mathscr{A}_{2},V_{2})$ are $i$-free for all $1\leq i\leq m$. Moreover, if $(\mathscr{A}_{1},V_{1})$ and $(\mathscr{A}_{2},V_{2})$ are $m$-free, then ${\rm exp}_{m}(\mathscr{A}_{1}\times\mathscr{A}_{2},V_{1}\oplus V_{2})=\bigcup_{i=0}^{m}\{d+e\mid d\in{\rm exp}_{i}(\mathscr{A}_{1},V_{1}),e\in{\rm exp}_{m-i}(\mathscr{A}_{2},V_{2})\}$.

Proposition 3.1 and Theorem 3.2 imply that a nonessential $3$-arrangement is $m$-free for all $m\geq 0$. If a nonessential $3$-arrangement $(\mathscr{A},V)$ is a product of a $2$-arrangement $(\mathscr{A}^{\prime},V^{\prime})$ and the empty $1$-arrangement, then ${\rm exp}_{m}(\mathscr{A},V)$ equals $\bigcup_{j=0}^{m}{\rm exp}_{j}(\mathscr{A}^{\prime},V^{\prime})$.

We next investigate essential $3$-arrangements. In the rest of this section, we assume that $\mathscr{A}$ is an essential arrangement. Then $L(\mathscr{A})_{1}$ is not empty. We also assume that $\ell=3$ and $m\geq n-2$. If $m>n-2$, then we take distinct hyperplanes $H_{n+1},\dots,H_{m+2}$ in $V$ with $H_{i}\not\in\mathscr{A}$ for $n+1\leq i\leq m+2$. We define an arrangement $\widetilde{\mathscr{A}}$ by

[TABLE]

Then $\widetilde{\mathscr{A}}$ is also an essential arrangement and $L(\widetilde{\mathscr{A}})_{1}$ is not empty. We write $\widetilde{n}:=|\widetilde{\mathscr{A}}|$ and then $m=\widetilde{n}-2$. Let $X\in L(\widetilde{\mathscr{A}})_{1}$. We use the variables $y_{1},y_{2},y_{X}$ and the derivatives $\delta_{1},\delta_{2},\delta_{X}$ defined in Section 2. Let

[TABLE]

Since $\dim(X)=1$, there exist at least two hyperplanes in $\widetilde{\mathscr{A}}_{X}$, that is, $|\widetilde{\mathscr{A}}_{X}|\geq 2$. Then $i_{X}$ is a nonnegative integer. We note that $\deg(\widetilde{P}_{X})=\widetilde{n}-|\widetilde{\mathscr{A}}_{X}|=\widetilde{n}-2-|\widetilde{\mathscr{A}}_{X}|+2=m-i_{X}$. For any $0\leq j\leq i_{X}$, let $B(X,j)$ be a $\mathbb{K}$-basis for the vector space $(S_{X})_{j}$ of homogeneous polynomials of degree $j$ in $S_{X}$. Then we define a set $B$ by

[TABLE]

For any $0\leq j\leq i_{X}$, let $\{\partial_{u}\mid u\in B(X,j)\}$ be the dual basis for $B(X,j)$, that is, for $u,v\in B(X,j)$, \partial_{u}(v)=\left\{\begin{array}[]{ll}1&{\rm if}\ u=v,\\ 0&{\rm if}\ u\neq v.\end{array}\right. We recall that $\delta_{X}(y_{X})=1$ and $\deg(\widetilde{P}_{X}y_{X}^{i_{X}-j})=m-j$. Since $\delta_{X}^{m-j}(\widetilde{P}_{X}y_{X}^{i_{X}-j})=(m-j)!\prod_{H\in\widetilde{\mathscr{A}}\setminus\widetilde{\mathscr{A}}_{X}}\delta_{X}(\alpha_{H})\neq 0$ by Proposition 2.1, we can define a set $B^{\ast}$ by

[TABLE]

Example 3.3**.**

We consider the same $3$-arrangement $\mathscr{A}$ as Example 2.2, i.e., $Q(\mathscr{A})=x_{1}x_{2}x_{3}(x_{1}-x_{2})$.

$(1)$ Let $m=2=|\mathscr{A}|-2$. In this case $\widetilde{\mathscr{A}}=\mathscr{A}$. We recall that $L(\mathscr{A})_{1}=\{X_{1}=\{x_{1}=0,x_{2}=0\},\ X_{2}=\{x_{1}=0,x_{3}=0\},X_{3}=\{x_{2}=0,x_{3}=0\},\ X_{4}=\{x_{1}=x_{2},x_{3}=0\}\}$. Then $\delta_{X}$, $y_{X}$, $S_{X}$, $i_{X}$ and $\widetilde{P}_{X}$ $(X\in L(\mathscr{A})_{1})$ are listed in the following table.

[TABLE]

We can take $B(X_{1},0)=\{1\}$, $B(X_{1},1)=\{x_{1},x_{2}\}$, $B(X_{2},0)=\{1\}$, $B(X_{3},0)=\{1\}$ and $B(X_{4},0)=\{1\}$. Therefore

[TABLE]

$(2)$ Let $m=3$. We take a hyperplane $H_{5}:=\{x_{1}+x_{2}=0\}$ and set $\widetilde{\mathscr{A}}=\mathscr{A}\cup\{H_{5}\}$. The elements of $L(\widetilde{\mathscr{A}})_{1}$ are $X_{1}:=\{x_{1}=0,x_{2}=0\}$, $X_{2}:=\{x_{1}=0,x_{3}=0\}$, $X_{3}:=\{x_{2}=0,x_{3}=0\}$, $X_{4}:=\{x_{1}=x_{2},x_{3}=0\}$, and $X_{5}:=\{x_{1}=-x_{2},x_{3}=0\}$. Then $\delta_{X}$, $y_{X}$, $S_{X}$, $i_{X}$, and $\widetilde{P}_{X}$ $(X\in L(\mathscr{A})_{1})$ are listed in the following table.

[TABLE]

We can take $B(X_{1},0)=\{1\}$, $B(X_{1},1)=\{x_{1},x_{2}\}$, $B(X_{1},2)=\{x_{1}^{2},x_{1}x_{2},x_{2}^{2}\}$, $B(X_{2},0)=\{1\}$, $B(X_{3},0)=\{1\}$, $B(X_{4},0)=\{1\}$  and $B(X_{5},0)=\{1\}$.

[TABLE]

We prepare Lemma 3.4 and Proposition 3.5 to prove that $B$ is a $\mathbb{K}$-basis for the vector space $S_{m}$ and that $B^{\ast}$ is the dual basis for $B$.

Lemma 3.4**.**

Let $\ell=3$ and let $X,Y\in L(\widetilde{\mathscr{A}})_{1}$. If $X\neq Y$, then $\delta_{X}^{m-i_{X}}(\widetilde{P}_{Y}f)=0$ for any $f\in S$ with $\deg(f)=i_{Y}$.

Proof. We suppose that $|\widetilde{\mathscr{A}}_{X}\cap\widetilde{\mathscr{A}}_{Y}|\geq 2$. There exist hyperplanes $H,H^{\prime}\in\widetilde{\mathscr{A}}_{X}\cap\widetilde{\mathscr{A}}_{Y}$ such that $H\neq H^{\prime}$. Then $\dim(H\cap H^{\prime})=3-2=1$. Since $X,Y\subseteq H\cap H^{\prime}$ and $\dim(X)=\dim(Y)=1$, we have $X=H\cap H^{\prime}=Y$. By taking the contraposition, if $X\neq Y$ then $|\widetilde{\mathscr{A}}_{X}\cap\widetilde{\mathscr{A}}_{Y}|<2$.

By Proposition 2.1, for any $f\in S$ with $\deg(f)=i_{Y}$,

[TABLE]

Here if $X\neq Y$, then

[TABLE]

Hence if $X\neq Y$, then $\delta_{X}^{m-i_{X}}(\widetilde{P}_{Y}f)=0$. $\Box$

Proposition 3.5**.**

Let $\mathscr{A}$ be an essential $3$-arrangement.

1. $(1)$

We have the set equation

[TABLE]

where the right hand side is a disjoint union. 2. $(2)$

[TABLE]

Proof. $(1)$ It is obvious that $\left\{\mathscr{B}\subseteq\widetilde{\mathscr{A}}_{X}\,\middle|\,|\mathscr{B}|=2\right\}\subseteq\left\{\mathscr{B}\subseteq\widetilde{\mathscr{A}}\ \middle|\,|\mathscr{B}|=2\right\}$ for any $X\in L(\widetilde{\mathscr{A}})_{1}$. Conversely let $\mathscr{B}=\{H,H^{\prime}\}\subseteq\widetilde{\mathscr{A}}$. We set $X:=H\cap H^{\prime}\in L(\widetilde{\mathscr{A}})_{1}$ and then $\mathscr{B}\subseteq\widetilde{\mathscr{A}}_{X}$. Therefore $\mathscr{B}$ belongs to the right hand side of (3.7).

Next we prove that if $X,Y\in L(\widetilde{\mathscr{A}})_{1}$ with $X\neq Y$, then $\Omega:=\{\mathscr{B}\subseteq\widetilde{\mathscr{A}}_{X}\mid|\mathscr{B}|=2\}\cap\{\mathscr{B}\subseteq\widetilde{\mathscr{A}}_{Y}\mid|\mathscr{B}|=2\}$ is the empty set. We suppose that $\Omega$ is not empty and let $\mathscr{B}=\{H,H^{\prime}\}\in\Omega$. Since $\dim(H\cap H^{\prime})=1$ and $X\subseteq H\cap H^{\prime}$, we have $X=H\cap H^{\prime}$. Similarly $Y=H\cap H^{\prime}$ and hence $X=Y$. The argument above imply that the right hand side of (3.7) is a disjoint union.

$(2)$ We have

[TABLE]

by the equation (3.7). $\Box$

Proposition 3.6**.**

Let $\ell=3$. The set $B$ is a $\mathbb{K}$-basis for $S_{m}$ and $B^{\ast}$ is the dual basis for $B$.

Proof. Let $X\in L(\mathscr{A})_{1}$, let $1\leq j\leq i_{X}$, and let $u\in B(X,j)$. Let $Y\in L(\mathscr{A})_{1}$, let $1\leq k\leq i_{Y}$, and let $v\in B(Y,k)$. If $X\neq Y$, then $\delta_{X}^{m-j}\left(\widetilde{P}_{Y}\,v\,y_{Y}^{i_{Y}-k}\right)=\delta_{X}^{i_{X}-j}\delta_{X}^{m-i_{X}}\left(\widetilde{P}_{Y}\,v\,y_{Y}^{i_{Y}-k}\right)=0$ by Lemma 3.4. If $X=Y$, then $v\in S_{Y}=S_{X}={\rm Sym}\left(V_{X}^{\ast}\right)$ while

[TABLE]

Here since $\partial_{u}\in\sum_{|\bm{a}|=j}\mathbb{K}\partial^{\bm{a}}$, $\deg(v)=k$, and $\deg\left(\widetilde{P}_{X}y_{X}^{i_{X}-k}\right)=m-k$, we have

[TABLE]

Moreover if $X=Y$ and $j=k$, then

[TABLE]

Therefore we have

[TABLE]

It remains to prove that $|B|$ equals the dimension of $S_{m}$. By counting the numbers of monomials of degree $i$ in $\ell^{\prime}$ variables in two ways, we have $s_{i}(\ell^{\prime})=\sum_{j=0}^{i}s_{j}(\ell^{\prime}-1)$ for any $i\geq 0$ and $\ell^{\prime}\geq 2$. The equation (3.10) implies that the right hand side of the equation (3.5) is a disjoint union. Then by Proposition 3.5 (2),

[TABLE]

Therefore $B$ is a $\mathbb{K}$-basis for $S_{m}$ and $B^{\ast}$ is the dual basis for $B$. $\Box$

Let

[TABLE]

Then we have

[TABLE]

Holm [2, 3] proved the following (see also [1, Corollary 2.5]).

Proposition 3.7** (Holm).**

Let $(\mathscr{A},V)$ be an $\ell$-arrangement. Then

[TABLE]

We use Proposition 3.7 in order to prove Theorem 3.8 which is the key to the proof of our main theorem.

Theorem 3.8**.**

Let $\mathscr{A}$ be an essential $3$-arrangement. If $m\geq n-2$, then

[TABLE]

as $S$-modules.

Proof. Let $X\in L(\widetilde{\mathscr{A}})_{1}$, let $0\leq j\leq i_{X}$, and let $\theta\in D^{(j)}(\mathscr{A}_{X},V_{X})$. We recall that $\delta_{X}(Q_{X})=0$ by Proposition 2.1. Then for any $f\in S$,

[TABLE]

This means that $P_{X}\theta\delta_{X}^{m-j}\in D^{(m)}(\mathscr{A},V)$.

Conversely let $\theta\in D^{(m)}(\mathscr{A},V)$. By Proposition 3.6, we have

[TABLE]

Here we fix $X\in L(\widetilde{\mathscr{A}})_{1}$ and $0\leq j\leq i_{X}$. Since $\widetilde{P}_{X}$ is divided by $P_{X}$, $\theta(\widetilde{P}_{X}\,u\,y_{X}^{i_{X}-j})\in P_{X}S$ for any $u\in B(X,j)$. For any $H\in\mathscr{A}_{X}$ and for any $f\in(S_{X})_{j-1}$, since $\partial_{u}(\alpha_{H}f)\in\mathbb{K}$ and $\{\partial_{u}\mid u\in B(X,j)\}$ is the dual basis for $B(X,j)$, we have

[TABLE]

by Proposition 3.7. This implies that $\sum_{u\in B(X,j)}\theta(\widetilde{P}_{X}\,u\,y_{X}^{i_{X}-j})\partial_{u}\in SP_{X}D^{(j)}(\mathscr{A}_{X},V_{X})$ by Proposition 3.7 again. Therefore

[TABLE]

It follows from Proposition 3.6 that the right hand side of (3.12) is a direct sum. $\Box$

Corollary 3.9**.**

An essential $3$-arrangement $\mathscr{A}$ is $m$-free when $m\geq n-2$.

Proof. By Proposition 3.1, $D^{(j)}(\mathscr{A}_{X},V_{X})$ is a free $S_{X}$-module for any $X\in L(\widetilde{\mathscr{A}})_{1}$ and for any $0\leq j\leq i_{X}$. Since $S=S_{X}\otimes_{\mathbb{K}}\mathbb{K}[y_{X}]$, $SP_{X}D^{(j)}(\mathscr{A}_{X},V_{X})$ is a free $S$-module. Therefore by Theorem 3.8, $D^{(m)}(\mathscr{A},V)$ is a free $S$-module. $\Box$

Remark. Let $\mathscr{A}$ be a generic $\ell$-arrangement. It is known that $\mathscr{A}$ is $m$-free if and only if $m\geq n-\ell+1$ [2, 7]. The “if” part of this result coincides with Corollary 3.9 when $\ell=3$. In addition if $\ell=3$ and $n=4$, then $\mathscr{A}$ is not $1$-free. This is an example that a $3$-arrangement is not $m$-free when $m=n-3$.

For operators $\theta_{1},\dots,\theta_{s_{m}(\ell)}\in D^{(m)}(S)$, a coefficient matrix $M_{m}(\theta_{1},\dots,\theta_{s_{m}(\ell)})$ is an $s_{m}(\ell)\times s_{m}(\ell)$ matrix defined by

[TABLE]

In the examples below, we use the following criterion for $D^{(m)}(\mathscr{A},V)$, which is first given by Saito [9] for $D^{(1)}(\mathscr{A},V)$ and which is generalized by Holm [2] for $D^{(m)}(\mathscr{A},V)$ (see also Theorem 3.1 in [1]).

Theorem 3.10** (Saito’s criterion).**

Let $\mathscr{A}$ be an $\ell$-arrangement and let $\theta_{1},\dots,\theta_{s_{m}(\ell)}\in D^{(m)}(\mathscr{A},V)$. Then $\det M_{m}(\theta_{1},\dots,\theta_{s_{m}(\ell)})=cQ^{s_{m-1}(\ell)}$ for some $c\in\mathbb{K}\setminus\{0\}$ if and only if the set $\{\theta_{1},\dots,\theta_{s_{m}(\ell)}\}$ is a free basis for $D^{(m)}(\mathscr{A})$ over $S$.

Example 3.11**.**

We consider the same $3$-arrangement $\mathscr{A}$ as Example 2.2 and Example 3.3, i.e., $Q(\mathscr{A})=x_{1}x_{2}x_{3}(x_{1}-x_{2})$.

$(1)$ Let $m=2$. Then $\widetilde{\mathscr{A}}=\mathscr{A}$ and $\widetilde{P}_{X}=P_{X}$ for any $X\in L(\widetilde{\mathscr{A}})_{1}$. By Saito’s criterion, $\left\{x_{1}\partial_{1}+x_{2}\partial_{2},x_{2}(x_{1}-x_{2})\partial_{2}\right\}$ is a free basis for $D^{(1)}(\mathscr{A}_{X_{1}},V_{X_{1}})$. Hence by Theorem 3.8,

[TABLE]

and ${\rm exp}_{2}(\mathscr{A},V)=\{1,2,3,2,2,2\}=\{1^{1},2^{4},3^{1}\}$.

$(2)$ Let $m=3$, let $H_{5}=\{x_{1}+x_{2}=0\}$, and let $\widetilde{\mathscr{A}}=\mathscr{A}\cup\{H_{5}\}$. Then $P_{X}$ $(X\in L(\widetilde{\mathscr{A}}))_{1}$ are given by

[TABLE]

By Saito’s criterion and Theorem 3.8, we have

[TABLE]

and ${\rm exp}_{3}(\mathscr{A},V)=\{1,2,3,3,3,3,2,2,2,3\}=\{1^{1},2^{4},3^{5}\}$.

We remark that the multi-set of $m$-exponents does not change if we take other hyperplanes $H^{\prime}_{n+1},\dots,H^{\prime}_{m+2}$ instead of $H_{n+1},\dots,H_{m+2}$.

Example 3.12**.**

We consider the same $3$-arrangement $\mathscr{A}$ as Example 2.2, i.e., $Q(\mathscr{A})=x_{1}x_{2}x_{3}(x_{1}-x_{2})$. Let $m=3$. In Example 3.11 we have already constructed a free $S$-basis for $D^{(3)}(\mathscr{A},V)$. In this example we construct another $S$-basis for $D^{(3)}(\mathscr{A},V)$ by taking a hyperplane $H^{\prime}_{5}:=\{x_{1}+x_{2}-x_{3}=0\}$. We set $\widetilde{\mathscr{A}^{\prime}}=\mathscr{A}\cup\{H^{\prime}_{5}\}$. Then the elements of $L(\widetilde{\mathscr{A}^{\prime}})_{1}$ are $X_{1}:=\{x_{1}=0,x_{2}=0\}$, $X_{2}:=\{x_{1}=0,x_{3}=0\}$, $X_{3}:=\{x_{2}=0,x_{3}=0\}$, $X_{4}:=\{x_{1}=x_{2},x_{3}=0\}$, $X_{5}:=\{x_{1}=-x_{2},x_{3}=0\}$, $X_{6}:=\{x_{1}=x_{3},x_{2}=0\}$, $X_{7}:=\{x_{1}=x_{2}=\frac{1}{2}x_{3}\}$, and $X_{8}:=\{x_{2}=x_{3},x_{1}=0\}$. Meanwhile $\delta_{X}$, $i_{X}$, and $P_{X}$ are listed in the following table.

[TABLE]

[TABLE]

By Theorem 3.8 and Saito’s criterion, we have

[TABLE]

and ${\rm exp}_{3}(\mathscr{A},V)=\{1,2,3,2,2,2,3,3,3,3\}=\{1^{1},2^{4},3^{5}\}$. The multi-set of $m$-exponents coincides with that of Example 3.11.

4 Exponents

In this section we determine the $m$-exponents of a $3$-arrangement $\mathscr{A}$ for $m\geq n-2$, which depend only on the poset $L(\mathscr{A})$. The following lemma holds for any $\ell\geq 3$.

Lemma 4.1**.**

Let $\mathscr{A}$ be an essential arrangement. There exist a hyperplane $H_{0}\subseteq V$ with $H_{0}\not\in\mathscr{A}$ such that for any $X\in L(\mathscr{A})_{1}$, $H_{0}\cap X=\{0\}$.

Proof. Let $H_{0}=\{a_{1}x_{1}+\cdots+a_{\ell}x_{\ell}=0\}$ be a hyperplane in $V$. Let $X\in L(\mathscr{A})_{1}$ and let $0\neq v_{X}\in X$. Then $X=\mathbb{K}v_{X}$. We note that $v_{X}=\left(x_{1}(v_{X}),\dots,x_{\ell}(v_{X})\right)$. Here

[TABLE]

Therefore $H_{0}\cap X=\{0\}$ for all $X\in L(\mathscr{A})_{1}$ $\Leftrightarrow$ $(a_{1},\dots,a_{\ell})$ belongs to the complement of $\bigcup_{X\in L(\mathscr{A})_{1}}\{(x_{1},\dots,x_{\ell})\mid x_{1}(v_{X})x_{1}+\cdots+x_{\ell}(v_{X})x_{\ell}=0\}$. Since $\mathbb{K}$ is an infinite field, we can take $(a_{1},\dots,a_{\ell})$ from the complement of $\bigcup_{X\in L(\mathscr{A})_{1}}\{(x_{1},\dots,x_{\ell})\mid x_{1}(v_{X})x_{1}+\cdots+x_{\ell}(v_{X})x_{\ell}=0\}$. Then $H_{0}=\{a_{1}x_{1}+\cdots+a_{\ell}x_{\ell}=0\}$ is a desired hyperplane. $\Box$

Let $\ell=3$, let $m\geq n-2$ and let $\mathscr{A}$ be essential. The argument in Section 3 does not depend on a choice of hyperplanes $H_{n+1},\dots,H_{m+2}$. By Lemma 4.1, if $m>n-2$ then there exist hyperplanes $H^{\prime}_{n+1},\dots,H^{\prime}_{m+2}$ in $V$ such that the next condition (A) is satisfied.

1. (A)

For any $n+1\leq i\leq m+2$ and for any $X\in L(\mathscr{A}\cup\{H^{\prime}_{n+1},\dots,H^{\prime}_{i-1}\})_{1}$, $H^{\prime}_{i}\cap X=\{0\}$, where $\{H^{\prime}_{n+1},\dots,H^{\prime}_{i-1}\}=\emptyset$ if $i=n+1$.

We define $\mathscr{G}:=\mathscr{A}$ if $m=n-2$ and $\mathscr{G}:=\mathscr{A}\cup\{H^{\prime}_{n+1},\dots,H^{\prime}_{m+2}\}$ if $m>n-2$.

Lemma 4.2**.**

$L(\mathscr{G})_{1}=L(\mathscr{A})_{1}\sqcup\{H\cap H^{\prime}\mid H\in\mathscr{G},H^{\prime}\in\mathscr{G}\setminus\mathscr{A}\}$. Moreover if $H\in\mathscr{G}$ and $H^{\prime}\in\mathscr{G}\setminus\mathscr{A}$, then $|\mathscr{G}_{H\cap H^{\prime}}|=2$.

Proof. Let $X\in L(\mathscr{G})_{1}$. If $\mathscr{G}_{X}=\mathscr{A}_{X}$ then $X=\bigcap_{H\in\mathscr{G}_{X}}H=\bigcap_{H\in\mathscr{A}_{X}}H\in L(\mathscr{A})_{1}$. Let $\mathscr{G}_{X}\neq\mathscr{A}_{X}$. We suppose that $|\mathscr{G}_{X}|\geq 3$. Then there exist $i\geq n+1$ and $X^{\prime}\in L(\mathscr{A}\cup\{H^{\prime}_{n+1},\dots,H^{\prime}_{i-1}\})_{1}$ such that $X=X^{\prime}\cap H_{i}$. By the condition (A), we have $X=\{0\}$. This is a contradiction. Therefore $|\mathscr{G}_{X}|=2$ and there exist $H,H^{\prime}\in\mathscr{G}$ such that $X=H\cap H^{\prime}$. Here if $H,H^{\prime}\in\mathscr{A}$ then $\mathscr{G}_{X}=\mathscr{A}_{X}$. So either $H$ or $H^{\prime}$ does not belong to $\mathscr{A}$. $\Box$

In the rest of this section, for a multi-set $\Psi$ and integers $e$ and $i$, $e^{i}\in\Psi$ means that the integer $e$ occurs $i$ times in $\Psi$.

Theorem 4.3**.**

Let $\ell=3$, let $m\geq n-2$, and let $(\mathscr{A},V)$ be an essential arrangement. Then $(\mathscr{A},V)$ is $m$-free with

[TABLE]

Proof. By Corollary 3.9, $(\mathscr{A},V)$ is $m$-free when $\ell=3$ and $m\geq n-2$. By Lemma 4.2,

[TABLE]

Let $X\in L(\mathscr{A})_{1}$. If $\mathscr{G}_{X}\neq\mathscr{A}_{X}$ then $X\not\in L(\mathscr{A})_{1}$. Hence $\mathscr{G}_{X}=\mathscr{A}_{X}$ and $i_{X}=|\mathscr{G}_{X}|-2=|\mathscr{A}_{X}|-2\leq|\mathscr{A}_{X}|-1$. By Proposition 3.1, the multi-set of degrees of the free $S$-basis for $\bigoplus_{X\in L(\mathscr{A})_{1}}\bigoplus_{j=0}^{i_{X}}SP_{X}D^{(j)}(\mathscr{A}_{X},V_{X})\delta_{X}^{m-j}$ is

[TABLE]

The last equation follows from Proposition 3.5 (2).

Let $H,H^{\prime}\in\mathscr{G}\setminus\mathscr{A}$ and $X:=H\cap H^{\prime}$. By the condition (A), we have $\mathscr{G}_{X}=\{H,H^{\prime}\}$, $i_{X}=0$, and $P_{X}=Q$. Therefore

[TABLE]

The multi-set of degrees of the free $S$-basis for $\bigoplus_{H,H^{\prime}\in\mathscr{G}\setminus\mathscr{A}}SQ\delta_{H\cap H^{\prime}}^{m}$ is $\{n^{\binom{\widetilde{n}-n}{2}}\}=\{n^{\binom{m+2-n}{2}}\}$.

Let $H\in\mathscr{A},H^{\prime}\in\mathscr{G}\setminus\mathscr{A}$ and $X:=H\cap H^{\prime}$. Then by the condition (A), we have $\mathscr{G}_{X}=\{H,H^{\prime}\}$, $i_{X}=0$, and $P_{X}=Q/\alpha_{H}$. Therefore

[TABLE]

The multi-set of degrees of the free $S$-basis for $\bigoplus_{H\in\mathscr{A},H^{\prime}\in\mathscr{G}\setminus\mathscr{A}}S(Q/\alpha_{H})\delta_{H\cap H^{\prime}}^{m}$ is $\{(n-1)^{(\widetilde{n}-n)n}\}=\{(n-1)^{(m+2)n-n^{2}}\}$.

By Theorem 3.8,

[TABLE]

Therefore

[TABLE]

as required. $\Box$

5 Weyl arrangements of types A and B

Weyl arrangements $(\mathcal{A}_{\ell},V)$ and $(\mathcal{B}_{\ell},V)$ are defined by

[TABLE]

The author [6] proved that $\mathcal{A}_{\ell}$ and $\mathcal{B}_{\ell}$ are $2$-free for all $\ell\geq 1$. By Corollary 3.9, when $\ell=3$, $\mathcal{A}_{3}$ is $m$-free if $m\geq|\mathcal{A}_{3}|-2=4$, while $\mathcal{B}_{3}$ is $m^{\prime}$-free if $m^{\prime}\geq|\mathcal{B}_{3}|-2=7$. Therefore, to prove that $\mathcal{A}_{3}$ and $\mathcal{B}_{3}$ are $m$-free for all $m\geq 0$, it is enough to prove that $\mathcal{A}_{3}$ is $3$-free and $\mathcal{B}_{3}$ is $i$-free for any $i\in\{3,4,5,6\}$.

We define an operator

[TABLE]

for $k\geq 0$, where we write $\bm{a}=(a_{1},\dots,a_{\ell})$ for any $|\bm{a}|=m$ in (5.1). We note that $\deg(\theta_{k})=m+k$.

Lemma 5.1**.**

Let $\ell\geq 1$. Then the following hold.

• (1)

$\theta_{k}\in D^{(m)}(\mathcal{A}_{\ell})$ for any $k\geq 0$.

• (2)

If $k$ is even, then $\theta_{k}\in D^{(m)}(\mathcal{B}_{\ell})$.

Proof. (1) Let $\bm{b}\in\mathbb{N}^{\ell}$ with $|\bm{b}|=m-1$. For any $1\leq i\leq\ell$, we have

[TABLE]

Meanwhile for any $i,j$ with $1\leq i<j\leq\ell$,

[TABLE]

Therefore $\theta_{k}\in D^{(m)}(\mathcal{A}_{\ell},V)$ by Proposition 3.7.

(2) Let $\bm{b}\in\mathbb{N}^{\ell}$ with $|\bm{b}|=m-1$. Let $i,j$ be integers with $1\leq i<j\leq\ell$. Since $k$ is even, we have $x_{i}^{k+1}+x_{j}^{k+1}=(x_{i}+x_{j})\sum_{t=0}^{k}(-1)^{t}x_{i}^{k-t}x_{j}^{t}$. Then

[TABLE]

Combining the arguments of (1), we have $\theta_{k}\in D^{(m)}(\mathcal{B}_{\ell},V)$ by Proposition 3.7. $\Box$

5.1 The arrangement $\mathcal{A}_{3}$

Let $\ell=3$ and we consider the arrangement $\mathscr{A}=\mathcal{A}_{3}$. The operator $\theta_{k}$ belongs to $D^{(3)}(\mathcal{A}_{3},V)$ by Lemma 5.1, while we can prove similarly to the proof of Theorem 3.8 that $P_{X}\theta\delta_{X}^{3-j}\in D^{(3)}(\mathcal{A}_{3},V)$ for any $X\in L(\mathcal{A}_{3})_{1}$, for any $0\leq j\leq i_{X}$, and for any $\theta\in D^{(j)}\left((\mathcal{A}_{3})_{X},V_{X}\right)$. We construct a free basis for $D^{(3)}(\mathcal{A}_{3},V)$ consisting of these operators. We have exactly seven elements in $L(\mathcal{A}_{3})_{1}$:

[TABLE]

Then

[TABLE]

while

[TABLE]

Proposition 5.2**.**

The set

[TABLE]

is a free basis for $D^{(3)}(\mathcal{A}_{3},V)$, while ${\rm exp}_{3}(\mathcal{A}_{3},V)=\{3^{5},4^{4},5^{1}\}$, where $e^{i}$ means that the integer $e$ occurs $i$ times in the multi-set.

Proof. We prove the assertion by Saito’s criterion. We have that $s_{2}(3)=\binom{2+2}{2}=6$ and $\prod_{i=1}^{7}P_{i}=x_{1}^{4}x_{2}^{4}x_{3}^{4}(x_{1}-x_{2})^{4}(x_{1}-x_{3})^{4}(x_{2}-x_{3})^{4}=Q(\mathcal{A}_{3})^{4}$. It is enough to prove that $M_{3}(\delta_{X_{1}}^{3},\delta_{X_{2}}^{3},\delta_{X_{3}}^{3},\delta_{X_{4}}^{3},\delta_{X_{5}}^{3},\delta_{X_{6}}^{3},\delta_{X_{7}}^{3},\theta_{0},\theta_{1},\theta_{2})=cQ(\mathcal{A}_{3})^{2}$ for some $c\in\mathbb{K}\setminus\{0\}$. Since $\delta_{X_{1}}^{3}=\partial_{1}^{3}$, $\delta_{X_{2}}^{3}=\partial_{2}^{3}$, and $\delta_{X_{3}}^{3}=\partial_{3}^{3}$, we can omit coefficients of $\partial_{1}^{3},\partial_{2}^{3},\partial_{3}^{3}$ in the next determinant calculations. Then

[TABLE]

where rows are arranged in the coefficients of $\partial_{1}^{2}\partial_{2},\partial_{1}^{2}\partial_{3},\partial_{2}^{2}\partial_{1},\partial_{2}^{2}\partial_{3},\partial_{3}^{2}\partial_{1}.\partial_{3}^{2}\partial_{2},\partial_{1}\partial_{2}\partial_{3}$ in the determinant ($\ast$ ‣ 5.1). Therefore the set (5.2) forms a basis for $D^{(3)}(\mathcal{A}_{3},V)$. $\Box$

Corollary 5.3**.**

The arrangement $\mathcal{A}_{3}$ is $m$-free for any $m\geq 0$.

5.2 The arrangement $\mathcal{B}_{3}$

Let $\ell=3$ and we consider the arrangement $\mathscr{A}=\mathcal{B}_{3}$. If $k$ is even, then the operator $\theta_{k}$ belongs to $D^{(m)}(\mathcal{B}_{3},V)$ by Lemma 5.1. Meanwhile $P_{X}\theta\delta_{X}^{m-j}\in D^{(m)}(\mathcal{B}_{3},V)$ for any $X\in L(\mathcal{B}_{3})_{1}$, for any $0\leq j\leq i_{X}$, and for any $\theta\in D^{(j)}\left((\mathcal{B}_{3})_{X},V_{X}\right)$. We have exactly thirteen elements in $L(\mathcal{B}_{3})_{1}$:

[TABLE]

Then

[TABLE]

while

[TABLE]

In addition we define Euler derivations

[TABLE]

with respect to $X_{j}\ (1\leq j\leq 8)$. Then $\varepsilon_{X_{j}}\in D^{(1)}\left(\left(\mathcal{B}_{3}\right)_{X_{j}},V_{X_{j}}\right)$ for any $1\leq j\leq 8$. Although there are thirteen Euler derivations with respect to $X\in L(\mathcal{B}_{3})_{1}$, we only use these eight operators to construct free bases for $D^{(i)}(\mathcal{B}_{3},V)$ for $i\in\{3,4,5,6\}$.

Proposition 5.4**.**

In the following multi-sets of exponents, $e^{i}$ means that the integer $e$ occurs $i$ times.

(1) The set

[TABLE]

is a free basis for $D^{(3)}(\mathcal{B}_{3},V)$, while ${\rm exp}_{3}(\mathcal{B}_{3},V)=\{3^{1},5^{4},6^{4},7^{1}\}$.

(2) The set

[TABLE]

is a free basis for $D^{(4)}(\mathcal{B}_{3},V)$, while ${\rm exp}_{4}(\mathcal{B}_{3},V)=\{4^{1},5^{3},6^{7},7^{3},8^{1}\}$.

(3) The set

[TABLE]

is a free basis for $D^{(5)}(\mathcal{B}_{3},V)$, while ${\rm exp}_{5}(\mathcal{B}_{3},V)=\{5^{4},6^{7},7^{7},8^{3}\}$.

(4) The set

[TABLE]

is a free basis for $D^{(6)}(\mathcal{B}_{3},V)$, while ${\rm exp}_{6}(\mathcal{B}_{3},V)=\{5^{3},6^{8},7^{10},8^{7}\}$.

We can prove Proposition 5.4 by straightforward determinant calculations of coefficient matrices. However it is too long to write them all down. We omit the proofs of Proposition 5.4.

Corollary 5.5**.**

The arrangement $\mathcal{B}_{3}$ is $m$-free for any $m\geq 0$.

Acknowledgments

The author was supported by JSPS Grant-in-Aid for Young Scientists (B) 16K17582.