MAT-free reflection arrangements

TL;DR

This paper introduces and classifies MAT-free and MAT2-free hyperplane arrangements, expanding understanding of their structure and relation to other free arrangements in complex reflection groups.

Contribution

It defines new classes of free arrangements based on the Multiple Addition Theorem and provides classifications for irreducible complex reflection arrangements.

Findings

Classified irreducible complex reflection arrangements as MAT-free or MAT2-free.

Connected MAT-free arrangements to the structure of complex reflection groups.

Explored relationships between MAT-free arrangements and other free arrangement classes.

Abstract

We introduce the class of MAT-free hyperplane arrangements which is based on the Multiple Addition Theorem by Abe, Barakat, Cuntz, Hoge, and Terao. We also investigate the closely related class of MAT2-free arrangements based on a recent generalization of the Multiple Addition Theorem by Abe and Terao. We give classifications of the irreducible complex reflection arrangements which are MAT-free respectively MAT2-free. Furthermore, we ask some questions concerning relations to other classes of free arrangements.

Equations

$$G_{24},G_{27},G_{29},G_{31},G_{32},G_{33},G_{34}.$$

$${\mathcal{A}}_X:=\{H\in \mathcal{A}\mid X\subseteq H\},$$

$${\mathcal{A}}^X:=\{X\cap H\mid H\in \mathcal{A}\setminus {\mathcal{A}}_X\}.$$

$${\mathcal{A}}_1\times {\mathcal{A}}_2:=\{H_1\oplus V_2\mid H_1\in {\mathcal{A}}_1\}\cup \{V_1\oplus H_2\mid H_2\in {\mathcal{A}}_2\}.$$

$$D(\mathcal{A}):=\{\theta \in \mathrm{Der}(S)\mid \theta ({\alpha }_H)\in {\alpha }_{H}S\text{ for all }H\in \mathcal{A}\}.$$

$$\exp (\mathcal{A})=(d_1^1,\dots ,d_{{\ell }_1}^1,d_1^2,\dots ,d_{{\ell }_2}^2).$$

$$\varnothing ={\mathcal{A}}_0⊊{\mathcal{A}}_1⊊\cdots ⊊{\mathcal{A}}_{n-1}⊊{\mathcal{A}}_n=\mathcal{A}$$

$$d_{\ell -p}<d_{\ell -p+1}=\cdots =d_{\ell }=:d.$$

$${\mathcal{A}}_j^{\prime \prime }:=({\mathcal{A}}^{\prime }\cup \{H_j\}{)}^{H_j}=\{H\cap H_j\mid H\in {\mathcal{A}}^{\prime }\},j=1,\dots ,q.$$

$$t:=\{\begin{array}{ll}\min \{i\mid d_i\ne 0\}& \text{if }{\mathcal{A}}^{\prime }\ne {\Phi }_{\ell }\\ 0& \text{if }{\mathcal{A}}^{\prime }={\Phi }_{\ell }\end{array}.$$

$$s>\{\begin{array}{ll}\min \{i\mid d_i\ne 0\}& \text{if }{\mathcal{A}}^{\prime }\ne {\Phi }_{\ell }\\ 0& \text{if }{\mathcal{A}}^{\prime }={\Phi }_{\ell }\end{array},$$

$$|{\mathcal{A}}^{\prime }|-|({\mathcal{A}}^{\prime }\cup \{H\}{)}^H|=d_{\ell }-1$$

$$|\mathcal{A}|-|{\mathcal{A}}^H|=|{\mathcal{A}}^{\prime }|+\ell -q+1-|({\mathcal{A}}^{\prime }\cup \{H\}{)}^H|-\ell +q=d_{\ell }.$$

$$\mathcal{A}:=\{$$

$$:=\{$$

$$(1,3,0{)}^{\perp },(1,1,1{)}^{\perp },(2,3,0{)}^{\perp },(1,3,1{)}^{\perp }\}.$$

$$\mathfrak{MF}⊊{\mathfrak{MF}}^{\prime }.$$

$$d_{\ell }-1=$$

$$=$$

$$d_{\ell }-1=$$

$$=$$

$${\mathcal{A}}_2:=\{$$

$$:=\{$$

$$(1,-{\zeta }^2,0{)}^{\perp },(1,0,-{\zeta }^2{)}^{\perp },(1,-1,0{)}^{\perp },(1,0,-1{)}^{\perp },(0,1,-\zeta {)}^{\perp }\}.$$

$$|{\mathcal{A}}_1^{\prime }|-|({\mathcal{A}}_1^{\prime }\cup \{H_1^1\}{)}^{H_1^1}|=4$$

$$|{\mathcal{A}}_1^{\prime }|-|({\mathcal{A}}_1^{\prime }\cup \{H_2^1\}{)}^{H_2^1}|=3.$$

$${\alpha }_{ij}({\zeta }^k)=x_i-{\zeta }^k{x}_j$$

$$H_{ij}({\zeta }^k)=\ker ({\alpha }_{ij}({\zeta }^k)).$$

$$\mathcal{A}(G(r,1,\ell ))=\{\ker (x_i)\mid 1\le i\le \ell \}\dot{\cup }\{H_{ij}({\zeta }^k)\mid 1\le i<j\le \ell ,1\le k\le r\}.$$

$${\pi }_{11}:=\{\ker (x_i)\mid 1\le i\le \ell \},$$

$${\pi }_{ij}:=\{H_{(i-1)k}({\zeta }^j)\mid i\le k\le \ell \},$$

$$\pi =$$

Full text

1. Introduction

A hyperplane arrangement $\mathcal{A}$ is a finite set of hyperplanes in a finite dimensional vector space $V\cong\mathbb{K}^{\ell}$. The intersection lattice $L(\mathcal{A})$ of $\mathcal{A}$ encodes its combinatorial properties. It is a main theme in the study of hyperplane arrangements to link algebraic properties of $\mathcal{A}$ with the combinatorics of $L(\mathcal{A})$.

The algebraic property of freeness of a hyperplane arrangement $\mathcal{A}$ was first studied by Saito [Sai80] and Terao [Ter80a]. In fact, it turns out that freeness of $\mathcal{A}$ imposes strong combinatorial constraints on $L(\mathcal{A})$: by Terao’s Factorization Theorem [OT92, Thm. 4.137] its characteristic polynomial factors over the integers. Conversely, sufficiently strong conditions on $L(\mathcal{A})$ imply the freeness of $\mathcal{A}$. One of the main tools to derive such conditions is Terao’s Addition-Deletion Theorem 2.4. It motivates the class of inductively free arrangements (see Definition 2.5). In this class the freeness of $\mathcal{A}$ is combinatorial, i.e. it is completely determined by $L(\mathcal{A})$ (cf. Definition 2.1). Recently, a remarkable generalization of the Addition-Deletion theorem was obtained by Abe. His Division Theorem [Abe16, Thm. 1.1] motivates the class of divisionally free arrangements. In this class freeness is a combinatorial property too.

Despite having these useful tools at hand, it is still a major open problem, known as Terao’s Conjecture, whether in general the freeness of $\mathcal{A}$ actually depends only on $L(\mathcal{A})$, provided the field $\mathbb{K}$ is fixed (see [Zie90] for a counterexample when one fixes $L(\mathcal{A})$ but changes the field). We should also mention at this point the very recent results by Abe further examining Addition-Deletion constructions together with divisional freeness [Abe18b], [Abe18a].

A variation of the addition part of the Addition-Deletion theorem 2.4 was obtained by Abe, Barakat, Cuntz, Hoge, and Terao in [ABC*+*16]: the Multiple Addition Theorem 3.1 (MAT for short). Using this theorem, the authors gave a new uniform proof of the Kostant-Macdonald-Shapiro-Steinberg formula for the exponents of a Weyl group. In the same way the Addition-Theorem defines the class of inductively free arrangements, it is now natural to consider the class ${\mathfrak{MF}}$ of those free arrangements, called MAT-free, which can be build inductively using the MAT (Definition 3.2). It is not hard to see (Lemma 3.7) that MAT-freeness only depends on $L(\mathcal{A})$. In this paper, we investigate classes of MAT-free arrangements beyond the classes considered in [ABC*+*16].

Complex reflection groups (classified by Shephard and Todd [ST54]) play an important role in the study of hyperplane arrangements: many interesting examples and counterexamples are related or derived from the reflection arrangement $\mathcal{A}(W)$ of a complex reflection group $W$. It was proven by Terao [Ter80b] that reflection arrangements are always free. There has been a series of investigations dealing with reflection arrangements and their connection to the aforementioned combinatorial classes of free arrangements (e.g. [BC12], [HR15], [Abe16]). Therefore, it is natural to study reflection arrangements in conjunction with the new class of MAT-free arrangements.

Our main result is the following.

Theorem 1.1**.**

Except for the arrangement $\mathcal{A}(G_{32})$, an irreducible reflection arrangement is MAT-free if and only if it is inductively free. The arrangement $\mathcal{A}(G_{32})$ is inductively free but not MAT-free. Thus every reflection arrangement is MAT-free except the reflection arrangements of the imprimitive reflection groups $G(e,e,\ell)$, $e>2$, $\ell>2$ and of the reflection groups

[TABLE]

A further generalization of the MAT 3.1 was very recently obtained by Abe and Terao [AT18]: the Multiple Addition Theorem 2 3.3 (MAT2 for short). Again, one might consider the inductively defined class of arrangements which can be build from the empty arrangement using this more general tool, i.e. the class ${\mathfrak{MF}}^{\prime}$ of MAT2-free arrangements (Defintion 3.4). By definition, this class contains the class of MAT-free arrangements. Regarding reflection arrangements we have the following:

Theorem 1.2**.**

Let $\mathcal{A}=\mathcal{A}(W)$ be an irreducible reflection arrangement. Then $\mathcal{A}$ is MAT2-free if and only if it is MAT-free.

In contrast to (irreducible) reflection arrangements, in general the class of MAT-free arrangements is properly contained in the class of MAT2-free arrangements (see Proposition 3.17).

Based on our classification of MAT-free (MAT2-free) reflection arrangements and other known examples ([ABC*+*16], [CRS19]) we arrive at the following question:

Question 1.3**.**

Is every MAT-free (MAT2-free) arrangement inductively free?

In [CRS19] the authors proved that all ideal subarrangements of a Weyl arrangement are inductively free by extensive computer calculations. A positive answer to Question 1.3 would directly imply their result and yield a uniform proof (cf. [CRS19, Rem. 1.5(d)]).

Looking at the class of divisionally free arrangements which properly contains the class of inductively free arrangements [Abe16, Thm. 4.4] a further natural question is:

Question 1.4**.**

Is every MAT-free (MAT2-free) arrangement divisionally free?

This article is organized as follows: in Section 2 we briefly recall some notions and results about hyperplane arrangements and free arrangements used throughout our exposition. In Section 3 we give an alternative characterization of MAT-freeness and two easy necessary conditions for MAT/MAT2-freeness. Furthermore, we comment on the relation of the two classes ${\mathfrak{MF}}$ and ${\mathfrak{MF}}^{\prime}$ and on the product construction. Section 4 and Section 5 contain the proofs of Theorem 1.1 and Theorem 1.2. In the last Section 6 we comment on Question 1.3 and further problems connected with MAT-freeness.

Acknowledgments

We thank Gerhard Röhrle for valuable comments on an earlier draft of our manuscript.

2. Hyperplane arrangements and free arrangements

Let $\mathcal{A}$ be a hyperplane arrangement in $V\cong\mathbb{K}^{\ell}$. If $\mathcal{A}$ is empty, then it is denoted by $\Phi_{\ell}$.

The intersection lattice $L(\mathcal{A})$ of $\mathcal{A}$ consists of all intersections of elements of $\mathcal{A}$ including $V$ as the empty intersection. Indeed, with the partial order by reverse inclusion $L(\mathcal{A})$ is a geometric lattice [OT92, Lem. 2.3]. The rank $\operatorname{rk}(\mathcal{A})$ of $\mathcal{A}$ is defined as the codimension of the intersection of all hyperplanes in $\mathcal{A}$.

If $x_{1},\ldots,x_{\ell}$ is a basis of $V^{*}$, to explicitly give a hyperplane we use the notation $(a_{1},\ldots,a_{\ell})^{\perp}:=\ker(a_{1}x_{1}+\ldots+a_{\ell}x_{\ell})$.

Definition 2.1**.**

Let ${\mathfrak{C}}$ be a class of arrangements and let $\mathcal{A}\in{\mathfrak{C}}$. If for all arrangements $\mathcal{B}$ with $L(\mathcal{B})\cong L(\mathcal{A})$, (where $\mathcal{A}$ and $\mathcal{B}$ do not have to be defined over the same field), we have $\mathcal{B}\in{\mathfrak{C}}$, then the class ${\mathfrak{C}}$ is called combinatorial.

If ${\mathfrak{C}}$ is a combinatorial class of arrangements such that every arrangement in ${\mathfrak{C}}$ is free than $\mathcal{A}\in{\mathfrak{C}}$ is called combinatorially free.

For $X\in L(\mathcal{A})$ the localization $\mathcal{A}_{X}$ of $\mathcal{A}$ at X is defined by:

[TABLE]

and the restriction $\mathcal{A}^{X}$ of $\mathcal{A}$ to $X$ is defined by:

[TABLE]

Let $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ be two arrangements in $V_{1}$ respectively $V_{2}$. Then their product $\mathcal{A}_{1}\times\mathcal{A}_{2}$ is defined as the arrangement in $V=V_{1}\oplus V_{2}$ consisting of the following hyperplanes:

[TABLE]

We note the following facts about products (cf. [OT92, Ch. 2]):

• •

$|\mathcal{A}_{1}\times\mathcal{A}_{2}|=|\mathcal{A}_{1}|+|\mathcal{A}_{2}|$.

• •

$L(\mathcal{A}_{1}\times\mathcal{A}_{2})=\{X_{1}\oplus X_{2}\mid X_{1}\in L(\mathcal{A}_{1})$ and $X_{2}\in L(\mathcal{A}_{2})\}$.

• •

$(\mathcal{A}_{1}\times\mathcal{A}_{2})^{X}=\mathcal{A}_{1}^{X_{1}}\times\mathcal{A}_{2}^{X_{2}}$ if $X=X_{1}\oplus X_{2}$ with $X_{i}\in L(\mathcal{A}_{i})$.

Let $S=S(V^{*})$ be the symmetric algebra of the dual space. We fix a basis $x_{1},\ldots,x_{\ell}$ for $V^{*}$ and identify $S$ with the polynomial ring $\mathbb{K}[x_{1},\ldots,x_{\ell}]$. The algebra $S$ is equipped with the grading by polynomial degree: $S=\bigoplus_{p\in\mathbb{Z}}S_{p}$, where $S_{p}$ is the set of homogeneous polynomials of degree $p$ ($S_{p}=\{0\}$ for $p<0$).

A $\mathbb{K}$-linear map $\theta:S\to S$ which satisfies $\theta(fg)=\theta(f)g+f\theta(g)$ is called a $\mathbb{K}$-derivation. Let $\operatorname{Der}(S)$ be the $S$-module of $\mathbb{K}$-derivations of $S$. It is a free $S$-module with basis $D_{1},\ldots,D_{\ell}$ where $D_{i}$ is the partial derivation $\partial/\partial x_{i}$. We say that $\theta\in\operatorname{Der}(S)$ is homogeneous of polynomial degree $p$ provided $\theta=\sum_{i=1}^{\ell}f_{i}D_{i}$ with $f_{i}\in S_{p}$ for each $1\leq i\leq\ell$. In this case we write $\operatorname{pdeg}{\theta}=p$. We obtain a $\mathbb{Z}$-grading for the $S$-module $\operatorname{Der}(S)$: $\operatorname{Der}(S)=\bigoplus_{p\in\mathbb{Z}}\operatorname{Der}(S)_{p}$.

Definition 2.2**.**

For $H\in\mathcal{A}$ we fix $\alpha_{H}\in V^{*}$ with $H=\ker(\alpha_{H})$. The module of $\mathcal{A}$-derivations is defined by

[TABLE]

We say that $\mathcal{A}$ is free if the module of $\mathcal{A}$-derivations is a free $S$-module.

If $\mathcal{A}$ is a free arrangement we may choose a homogeneous basis $\{\theta_{1},\ldots,\theta_{\ell}\}$ for $D(\mathcal{A})$. Then the polynomial degrees of the $\theta_{i}$ are called the exponents of $\mathcal{A}$ and they are uniquely determined by $\mathcal{A}$, [OT92, Def. 4.25]. We write $\exp(\mathcal{A}):=(\operatorname{pdeg}{\theta_{1}},\ldots$, $\operatorname{pdeg}{\theta_{\ell}})$. Note that the empty arrangement $\Phi_{\ell}$ is free with $\exp(\Phi_{\ell})=(0,\ldots,0)\in\mathbb{Z}^{\ell}$. If $d_{1},\ldots,d_{\ell}\in\mathbb{Z}$ with $d_{1}\leq d_{2}\leq\ldots\leq d_{\ell}$ we write $(d_{1},\ldots,d_{\ell})_{\leq}$.

The notion of freeness is compatible with products of arrangements:

Proposition 2.3** ([OT92, Prop. 4.28]).**

Let $\mathcal{A}=\mathcal{A}_{1}\times\mathcal{A}_{2}$ be a product of two arrangements. Then $\mathcal{A}$ is free if and only if both $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are free. In this case if $\exp(\mathcal{A}_{i})=(d^{i}_{1},\ldots,d^{i}_{\ell_{i}})$ for $i=1,2$ then

[TABLE]

The following theorem provides a useful tool to prove the freeness of arrangements.

Theorem 2.4** (Addition-Deletion [OT92, Thm. 4.51]).**

Let $\mathcal{A}$ be a hyperplane arrangement and $H_{0}\in\mathcal{A}$. We call $(\mathcal{A},\mathcal{A}^{\prime}=\mathcal{A}\setminus\{H_{0}\},\mathcal{A}^{\prime\prime}=\mathcal{A}^{H_{0}})$ a triple of arrangements. Any two of the following statements imply the third:

1. (1)

$\mathcal{A}$* is free with $\exp(\mathcal{A})=(b_{1},\ldots,b_{l-1},b_{\ell})$,* 2. (2)

$\mathcal{A}^{\prime}$* is free with $\exp(\mathcal{A}^{\prime})=(b_{1},\ldots,b_{\ell-1},b_{\ell}-1)$,* 3. (3)

$\mathcal{A}^{\prime\prime}$* is free with $\exp(\mathcal{A}^{\prime\prime})=(b_{1},\ldots,b_{\ell-1})$.*

The preceding theorem motivates the following definition.

Definition 2.5** ([OT92, Def. 4.53]).**

The class ${\mathfrak{IF}}$ of inductively free arrangements is the smallest class of arrangements which satisfies

1. (1)

the empty arrangement $\Phi_{\ell}$ of rank $\ell$ is in ${\mathfrak{IF}}$ for $\ell\geq 0$, 2. (2)

if there exists a hyperplane $H_{0}\in\mathcal{A}$ such that $\mathcal{A}^{\prime\prime}\in{\mathfrak{IF}}$, $\mathcal{A}^{\prime}\in{\mathfrak{IF}}$, and $\exp(\mathcal{A}^{\prime\prime})\subset\exp(\mathcal{A}^{\prime})$, then $\mathcal{A}$ also belongs to ${\mathfrak{IF}}$.

Here $(\mathcal{A},\mathcal{A}^{\prime},\mathcal{A}^{\prime\prime})=(\mathcal{A},\mathcal{A}\setminus\{H_{0}\},\mathcal{A}^{H_{0}})$ is a triple as in Theorem 2.4.

The class ${\mathfrak{IF}}$ is easily seen to be combinatorial [CH15, Lem. 2.5].

The following result was a major step in the investigation of freeness properties for reflection arrangements.

Theorem 2.6** ([HR15, Thm. 1.1], [BC12, Thm. 5.14]).**

For W a finite complex reflection group, the reflection arrangement $\mathcal{A}(W)$ is inductively free if and only if $W$ does not admit an irreducible factor isomorphic to a monomial group $G(r,r,\ell)$ for $r,\ell≥3$, $G_{24}$, $G_{27}$, $G_{29}$, $G_{31}$, $G_{33}$, or $G_{34}$.

Definition 2.7** (cf. [AT16]).**

Let $\mathcal{A}$ be an arrangement with $|\mathcal{A}|=n$. We say that $\mathcal{A}$ has a free filtration if there are subarrangements

[TABLE]

such that $|\mathcal{A}_{i}|=i$ and $\mathcal{A}_{i}$ is free for all $1\leq i\leq n$.

Very recently, Abe [Abe18a] introduced the class ${\mathfrak{AF}}$ of additionally free arrangements. Arrangements in ${\mathfrak{AF}}$ are by definition exactly the arrangements admitting a free filtration. Furthermore, it is a direct consequence of [Abe18a, Thm. 1.4] that the class ${\mathfrak{AF}}$ is combinatorial.

3. Multiple Addition Theorem

The following theorem presented in [ABC*+*16] is a variant of the addition part ((2) and (3) imply (1)) of Theorem 2.4.

Theorem 3.1** (Multiple Addition Theorem (MAT)).**

Let $\mathcal{A}^{\prime}$ be a free arrangement with $\exp(\mathcal{A}^{\prime})=(d_{1},\ldots,d_{\ell})_{\leq}$ and $1\leq p\leq\ell$ the multiplicity of the highest exponent, i.e.,

[TABLE]

Let $H_{1},\ldots,H_{q}$ be hyperplanes with $H_{i}\not\in\mathcal{A}^{\prime}$ for $i=1,\ldots,q$. Define

[TABLE]

Assume that the following three conditions are satisfied:

• (1)

$X:=H_{1}\cap\cdots\cap H_{q}$* is $q$-codimensional.*

• (2)

$X\not\subseteq\bigcup_{H\in\mathcal{A}^{\prime}}H$.

• (3)

$|\mathcal{A}^{\prime}|-|\mathcal{A}^{\prime\prime}_{j}|=d$* for $1\leq j\leq q$.*

Then $q\leq p$ and $\mathcal{A}:=\mathcal{A}^{\prime}\cup\{H_{1},\ldots,H_{q}\}$ is free with $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell-q},d+1,$ $\ldots,d+1)_{\leq}$.

Note that in contrast to Theorem 2.4 no freeness condition on the restriction is needed to conclude the freeness of $\mathcal{A}$ in Theorem 3.1. The MAT motivates the following definition.

Definition 3.2**.**

The class ${\mathfrak{MF}}$ of MAT-free arrangements is the smallest class of arrangements subject to

• (i)

$\Phi_{\ell}$ belongs to ${\mathfrak{MF}}$, for every $\ell\geq 0$;

• (ii)

if $\mathcal{A}^{\prime}\in{\mathfrak{MF}}$ with $\exp(\mathcal{A}^{\prime})=(d_{1},\ldots,d_{\ell})_{\leq}$ and $1\leq p\leq\ell$ the multiplicity of the highest exponent $d=d_{\ell}$, and if $H_{1},\ldots,H_{q}$, $q\leq p$ are hyperplanes with $H_{i}\not\in\mathcal{A}^{\prime}$ for $i=1,\ldots,q$ such that:

• (1)

$X:=H_{1}\cap\cdots\cap H_{q}$ is $q$-codimensional,

• (2)

$X\not\subseteq\bigcup_{H\in\mathcal{A}^{\prime}}H$,

• (3)

$|\mathcal{A}^{\prime}|-|(\mathcal{A}^{\prime}\cup\{H_{j}\})^{H_{j}}|=d$, for $1\leq j\leq q$,

then $\mathcal{A}:=\mathcal{A}^{\prime}\cup\{H_{1},\ldots,H_{q}\}$ also belongs to ${\mathfrak{MF}}$ and has exponents $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell-q},d+1,\ldots,d+1)_{\leq}$.

Abe and Terao [AT18] proved the following generalization of Theorem 3.1:

Theorem 3.3** (Multiple Addition Theorem 2 (MAT2), [AT18, Thm. 1.4]).**

Assume that $\mathcal{A}^{\prime}$ is a free arrangement with $\exp(\mathcal{A}^{\prime})=(d_{1},d_{2},\ldots,d_{\ell})_{\leq}$. Let

[TABLE]

For $H_{s},\ldots,H_{\ell}\notin\mathcal{A}$ with $s>t$, define $\mathcal{A}_{j}^{\prime\prime}:=(\mathcal{A}^{\prime}\cup\{H_{j}\})^{H_{j}}$, $\mathcal{A}:=\mathcal{A}^{\prime}\cup\{H_{s},\ldots,H_{\ell}\}$ and assume the following conditions:

• (1)

$X:=\bigcap_{i=s}^{\ell}H_{i}$* is $(\ell-s+1)$-codimensional,*

• (2)

$X\not\subset\bigcup_{K\in\mathcal{A}^{\prime}}K$, and

• (3)

$|\mathcal{A}^{\prime}|-|\mathcal{A}_{j}^{\prime\prime}|=d_{j}$* for $j=s,\ldots,\ell$.*

Then $\mathcal{A}$ is free with exponents $(d_{1},d_{2},\ldots,d_{s-1},d_{s}+1,\ldots,d_{\ell}+1)_{\leq}$. Moreover, there is a basis $\theta_{1},\theta_{2},\ldots,\theta_{s-1},\eta_{s},\ldots,\eta_{\ell}$ for $D(\mathcal{A}^{\prime})$ such that $\deg\theta_{i}=d_{i},\ \deg\eta_{j}=d_{j}$, $\theta_{i}\in D(\mathcal{A})$ and $\eta_{j}\in D(\mathcal{A}\setminus\{H_{j}\})$ for all $i$ and $j$.

This in turn motivates:

Definition 3.4**.**

The class ${\mathfrak{MF}}^{\prime}$ of MAT2-free arrangements is the smallest class of arrangements subject to

• (i)

$\Phi_{\ell}$ belongs to ${\mathfrak{MF}}^{\prime}$, for every $\ell\geq 0$;

• (ii)

if $\mathcal{A}^{\prime}\in{\mathfrak{MF}}^{\prime}$ with $\exp(\mathcal{A}^{\prime})=(d_{1},d_{2},\ldots,d_{\ell})_{\leq}$ and if $H_{s},\ldots,H_{\ell}$ are hyperplanes with $H_{i}\not\in\mathcal{A}^{\prime}$ for $i=s,\ldots,\ell$, where

[TABLE]

and with

• (1)

$X:=H_{s}\cap\cdots\cap H_{\ell}$ is $(\ell-s+1)$-codimensional,

• (2)

$X\not\subseteq\bigcup_{H\in\mathcal{A}^{\prime}}H$,

• (3)

$|\mathcal{A}^{\prime}|-|(\mathcal{A}^{\prime}\cup\{H_{j}\})^{H_{j}}|=d_{j}$ for $s\leq j\leq\ell$,

then $\mathcal{A}:=\mathcal{A}^{\prime}\cup\{H_{s},\ldots,H_{\ell}\}$ also belongs to ${\mathfrak{MF}}^{\prime}$ and has exponents $\exp(\mathcal{A})=(d_{1},\ldots,d_{s-1},d_{s}+1,\ldots,d_{\ell}+1)_{\leq}$.

We note the following:

Remark 3.5**.**

1. (1)

We have ${\mathfrak{MF}}\subseteq{\mathfrak{MF}}^{\prime}$. 2. (2)

If $\mathcal{A}$ is a free arrangement with $\exp(\mathcal{A})=(0,\ldots,0,1,\ldots,1,d,\ldots,d)_{\leq}$, i.e. $\mathcal{A}$ has only two distinct exponents $\neq 0$, then it is clear from the definitions that $\mathcal{A}$ is MAT2-free if and only if $\mathcal{A}$ is MAT-free.

Example 3.6**.**

1. (1)

If $\operatorname{rk}(\mathcal{A})=2$ then $\mathcal{A}$ is MAT-free and therefore MAT2-free too. 2. (2)

Every ideal subarrangement of a Weyl arrangement is MAT-free and therefore also MAT2-free, [ABC*+*16].

Lemma 3.7**.**

The classes ${\mathfrak{MF}}$ and ${\mathfrak{MF}}^{\prime}$ are combinatorial.

Proof.

The class of all empty arrangements is combinatorial and contained in ${\mathfrak{MF}}$. Let $\mathcal{A}\in{\mathfrak{MF}}$ ($\mathcal{A}\in{\mathfrak{MF}}^{\prime}$). Since conditions (1)–(3) in Defintion 3.2 (respectively Defintion 3.4) only depend on $L(\mathcal{A})$ the claim follows. See also [AT18, Thm. 5.1]. ∎

If an arrangement $\mathcal{A}$ is MAT-free, the MAT-steps yield a partition of $\mathcal{A}$ whose dual partition gives the exponents of $\mathcal{A}$. Vice versa, the existence of such a partition suffices for the MAT-freeness of the arrangement:

Lemma 3.8**.**

Let $\mathcal{A}$ be an $\ell$-arrangement. Then $\mathcal{A}$ is MAT-free if and only if there exists a partition $\pi=(\pi_{1}|\cdots|\pi_{n})$ of $\mathcal{A}$ where for all $0\leq k\leq n-1$,

• (1)

$\operatorname{rk}(\pi_{k+1})=|\pi_{k+1}|$,

• (2)

$\cap_{H\in\pi_{k+1}}H=X_{k+1}\nsubseteq\bigcup_{H^{\prime}\in\mathcal{A}_{k}}H^{\prime}$* where $\mathcal{A}_{k}=\bigcup_{i=1}^{k}\pi_{i}$,*

• (3)

$|\mathcal{A}_{k}|-|(\mathcal{A}_{k}\cup\{H\})^{H}|=k$* for all $H\in\pi_{k+1}$.*

In this case $\mathcal{A}$ has exponents $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell})_{\leq}$ with $d_{i}=|\{k\mid|\pi_{k}|\geq\ell-i+1\}|$.

Proof.

This is immediate from the definition. ∎

Definition 3.9**.**

If $\pi$ is a partition as in Lemma 3.8 then $\pi$ is called an MAT-partition for $\mathcal{A}$.

If we have chosen a linear ordering $\mathcal{A}=\{H_{1},\ldots,H_{m}\}$ of the hyperplanes in $\mathcal{A}$, to specify the partition $\pi$, we give the corresponding ordered set partition of $[m]=\{1,\ldots,m\}$.

Example 3.10**.**

Supersolvable arrangements, a proper subclass of inductively free arrangements [OT92, Thm. 4.58], are not necessarily MAT2-free: an easy calculation shows that the arrangement denoted $\mathcal{A}(10,1)$ in [Grü09] is supersolvable but not MAT2-free. In particular $\mathcal{A}(10,1)$ is neither MAT-free.

Restrictions of MAT2-free (MAT-free) arrangements are not necessarily MAT2-free (MAT-free):

Example 3.11**.**

Let $\mathcal{A}=\mathcal{A}(E_{6})$ be the Weyl arrangement of the Weyl group of type $E_{6}$. Then $\mathcal{A}$ is MAT-free by Example 3.6(2). Let $H\in\mathcal{A}$. A simple calculation (with the computer) shows that $\mathcal{A}^{H}$ is not MAT2-free.

We have two simple necessary conditions for MAT-freeness respectively MAT2-freeness. The first one is:

Lemma 3.12**.**

Let $\mathcal{A}$ be a non-empty MAT2-free arrangement with exponents $\exp(\mathcal{A})$ $=(d_{1},\ldots,d_{\ell})_{\leq}$. Then there is an $H\in\mathcal{A}$ such that $|\mathcal{A}|-|\mathcal{A}^{H}|=d_{\ell}$. In particular, the same holds, if $\mathcal{A}$ is MAT-free.

Proof.

By definition there are $H_{q},\ldots,H_{\ell}\in\mathcal{A}$, $2\leq q$ such that $\mathcal{A}^{\prime}:=\mathcal{A}\setminus\{H_{q},\ldots,H_{\ell}\}$ is MAT2-free. Furthermore by condition (1) the hyperplanes $H_{q},\ldots,H_{\ell}$ are linearly independent. Let $H:=H_{\ell}$. By condition (2), we have $X=\cap_{i=q}^{\ell}H_{i}\nsubseteq\cup_{H^{\prime}\in\mathcal{A}^{\prime}}H^{\prime}$ and thus $|\mathcal{A}^{H}|=|(\mathcal{A}^{\prime}\cup\{H\})^{H}|+\ell-q$. Now

[TABLE]

by condition (3) and hence

[TABLE]

∎

The second one is:

Lemma 3.13**.**

Let $\mathcal{A}$ be an MAT2-free arrangement. Then $\mathcal{A}$ has a free filtration, i.e. $\mathcal{A}$ is additionally free. In particular, the same is true, if $\mathcal{A}$ is MAT-free.

Proof.

Let $\mathcal{A}$ be MAT2-free. Then by definition there are $H_{q},\ldots,H_{\ell}\in\mathcal{A}$ such that $\mathcal{A}^{\prime}:=\mathcal{A}\setminus\{H_{q},\ldots,H_{\ell}\}$ is MAT2-free and conditions (1)–(3) are satisfied. Set $\mathcal{B}:=\{H_{q},\ldots,H_{\ell}\}$. By [AT18, Cor. 3.2] for all $\mathcal{C}\subseteq\mathcal{B}$ the arrangement $\mathcal{A}^{\prime}\cup\mathcal{C}$ is free. Hence by induction $\mathcal{A}$ has a free filtration. ∎

An MAT2-free but not MAT-free arrangement

We now provide an example of an arrangement which is MAT2-free but not MAT-free.

Example 3.14**.**

Let $\mathcal{A}$ be the arrangement defined by

[TABLE]

It is not hard to see that $\mathcal{A}$ is inductively free (actually supersolvable) with $\exp(\mathcal{A})=(1,4,5)$.

Proposition 3.15**.**

The arrangement $\mathcal{A}$ from Example 3.14 is MAT2-free.

Proof.

Let $\mathcal{B}_{1}=\{H_{1},H_{2},H_{3}\}$, $\mathcal{B}_{2}=\{H_{4}\}$, $\mathcal{B}_{3}=\{H_{5},H_{6}\}$, $\mathcal{B}_{4}=\{H_{7},H_{8}\}$, $\mathcal{B}_{5}=\{H_{9},H_{10}\}$, and $\mathcal{A}_{k}=\cup_{i=1}^{k}\mathcal{B}_{i}$ for $1\leq k\leq 5$. It is clear that $\mathcal{A}_{1}$ is MAT2-free. A simple linear algebra computation shows that the addition of $\mathcal{B}_{i+1}$ to $\mathcal{A}_{i}$ for $1\leq i\leq 4$ satisfies Condition (1)–(3) of Definition 3.4. Hence $\mathcal{A}=\mathcal{A}_{5}$ is MAT2-free. ∎

Proposition 3.16**.**

The arrangement $\mathcal{A}$ from Example 3.14 is not MAT-free.

Proof.

Suppose $\mathcal{A}$ is MAT-free and $\pi=(\pi_{1},\ldots,\pi_{5})$ is an MAT-partition. Since $\exp(\mathcal{A})=(1,4,5)$ the last block $\pi_{5}$ has to be a singleton, i.e. $\pi_{5}=\{H\}$. By Condition (3) of Lemma 3.8 we have $|\mathcal{A}^{H}|=5$ and the only hyperplane with this property is $H_{9}=(2,3,0)^{\perp}$. Similarly $\pi_{4}$ can only contain one of $H_{3},H_{6},H_{8},H_{10}$. But looking at their intersections we see that all of the latter are contained in another hyperplane of $\mathcal{A}$, e.g. $H_{3}\cap H_{8}\subseteq H_{4}$. This contradicts Condition (2). Hence $\mathcal{A}$ is not MAT-free. ∎

As a direct consequence we get:

Proposition 3.17**.**

We have

[TABLE]

Products of MAT-free and MAT2-free arrangements

As for freeness in general (Proposition 2.3), the product construction is compatible with the notion of MAT-freeness:

Theorem 3.18**.**

Let $\mathcal{A}=\mathcal{A}_{1}\times\mathcal{A}_{2}$ be a product of two arrangements. Then $\mathcal{A}\in{\mathfrak{MF}}$ if and only if $\mathcal{A}_{1}\in{\mathfrak{MF}}$ and $\mathcal{A}_{2}\in{\mathfrak{MF}}$.

Proof.

Assume $\mathcal{A}_{i}$ is an arrangement in the vector space $V_{i}$ of dimension $\ell_{i}$ for $i=1,2$. We argue by induction on $|\mathcal{A}|$. If $|\mathcal{A}|=0$, i.e. $\mathcal{A}_{1}=\Phi_{\ell_{1}}$, and $\mathcal{A}_{2}=\Phi_{\ell_{2}}$ then the statement is clear. Assume $\mathcal{A}_{1}$ is MAT-free with $\exp(\mathcal{A}_{1})=(d^{1}_{1},\ldots,d^{1}_{\ell_{1}})_{\leq}$ and $\mathcal{A}_{2}$ is MAT-free with $\exp(\mathcal{A}_{1})=(d^{2}_{1},\ldots,d^{2}_{\ell_{2}})_{\leq}$. Then without loss of generality $d:=d^{1}_{\ell_{1}}\geq d^{2}_{\ell_{2}}$. Let $q_{i}$ be the multiplicity of the exponent $d$ in $\exp(\mathcal{A}_{i})$ for $i=1,2$ (note that $q_{2}=0$ if $d>d^{2}_{\ell_{2}}$). Then since $\mathcal{A}_{i}$ is MAT-free there are hyperplanes $\{H^{i}_{1},\ldots,H^{i}_{q_{i}}\}\subseteq\mathcal{A}_{i}$ such that $\mathcal{A}_{i}^{\prime}:=\mathcal{A}_{i}\setminus\{H^{i}_{1},\ldots,H^{i}_{q_{i}}\}$ is MAT-free, i.e. they satisfy Conditions (1)–(3) from Definition 3.2. Now by the induction hypothesis $\mathcal{A}^{\prime}=\mathcal{A}_{1}^{\prime}\times\mathcal{A}_{2}^{\prime}$ is MAT-free and clearly $\{H^{1}_{1}\oplus V_{2},\ldots,H^{1}_{q_{1}}\oplus V_{2}\}\cup\{V_{1}\oplus H^{2}_{1},\ldots,V_{1}\oplus H^{2}_{q_{2}}\}$ satisfy Conditions (1)–(3). Hence $\mathcal{A}$ is MAT-free.

Conversely assume $\mathcal{A}$ is MAT-free with $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell})_{\leq}$. By Proposition 2.3 both factors $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are free with $\exp(\mathcal{A}_{i})=(d^{i}_{1},\ldots,d^{i}_{\ell_{i}})_{\leq}$ and without loss of generality $d_{\ell}=d^{1}_{\ell_{1}}\geq d^{2}_{\ell_{2}}$. Assume further that $q_{i}$ is the multiplicity of $d_{\ell}$ in $\exp(\mathcal{A}_{i})$ and $q$ is the multiplicity of $d_{\ell}$ in $\exp(\mathcal{A})$, i.e. $q=q_{1}+q_{2}$. There are hyperplanes $\{H_{1},\ldots,H_{q}\}\subset\mathcal{A}$ such that $\mathcal{A}^{\prime}=\mathcal{A}\setminus\{H_{1},\ldots,H_{q}\}$ is MAT-free with $\exp(\mathcal{A}^{\prime})=(d_{1},\ldots,d_{\ell-q},d_{\ell-q+1}-1,\ldots,d_{\ell}-1)_{\leq}$, and Conditions (1)–(3) are satisfied. We may further assume that $H_{i}=H^{1}_{i}\oplus V_{2}$ for $1\leq i\leq q_{1}$ and $H_{j}=V_{1}\oplus H^{2}_{j-q_{1}}$ for $q_{1}+1\leq j\leq q$. Let $\mathcal{A}_{i}^{\prime}=\mathcal{A}_{i}\setminus\{H^{i}_{1},\ldots,H^{i}_{q_{i}}\}$ for $i=1,2$. Note that if $d_{\ell}>d^{2}_{\ell_{2}}$ we have $q_{2}=0$ and $\mathcal{A}_{2}^{\prime}=\mathcal{A}_{2}$. But at least we have $\mathcal{A}_{1}^{\prime}\subsetneq\mathcal{A}_{1}$. Then $\mathcal{A}^{\prime}=\mathcal{A}_{1}^{\prime}\times\mathcal{A}_{2}^{\prime}$, $|\mathcal{A}^{\prime}|<|\mathcal{A}|$ and by the induction hypothesis $\mathcal{A}_{1}^{\prime}$ and $\mathcal{A}_{2}^{\prime}$ are MAT-free and Conditions (1) and (2) are clearly satified for $\mathcal{A}_{i}^{\prime}$ and $\{H^{i}_{1},\ldots,H^{i}_{q_{i}}\}$. But since

[TABLE]

for $1\leq i\leq q_{1}$ and

[TABLE]

for $q_{1}+1\leq j\leq q_{2}$, Condition (3) is also satisfied for $\mathcal{A}_{1}^{\prime}$ and $\mathcal{A}_{2}^{\prime}$. Hence both $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are MAT-free. ∎

Altenatively, one can prove Theorem 3.18 by observing that MAT-Partitions for $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$ are directly obtained from an MAT-Partition for $\mathcal{A}$: take the non-empty factors of each block in the same order, and vise versa: take the products of the blocks of partitions for $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$.

Remark 3.19**.**

Thanks to the preceding theorem, our classification of MAT-free irreducible reflection arrangements proved in the next 2 sections gives actually a classification of all MAT-free reflection arrangements: a reflection arrangement $\mathcal{A}(W)$ is MAT-free if and only if it has no irreducible factor isomorphic to one of the non-MAT-free irreducible reflection arrangements listed in Theorem 1.1.

In contrast to MAT-freeness, the weaker notion of MAT2-freeness is not compatible with products as the following example shows:

Example 3.20**.**

Let $\mathcal{A}_{1}$ be the MAT2-free but not MAT-free arrangement of Example 3.14 with exponents $\exp(\mathcal{A}_{1})=(1,4,5)$. Let $\zeta=\frac{1}{2}(-1+i\sqrt{3})$ be a primitive cube root of unity, and let $\mathcal{A}_{2}$ be the arrangement defined by the following linear forms:

[TABLE]

A linear algebra computation shows that $\pi=(1,2,3|4,5|6,7|8,9|10)$ is an MAT-partition for $\mathcal{A}_{2}$. In particular $\mathcal{A}_{2}$ is MAT2-free with $\exp(\mathcal{A}_{2})=(1,4,5)$.

Now by Proposition 2.3 the product $\mathcal{A}:=\mathcal{A}_{1}\times\mathcal{A}_{2}$ is free with $\exp(\mathcal{A})=(1,1,$ $4,4,5,5)$. Suppose $\mathcal{A}$ is MAT2-free. Then either there are hyperplanes $H_{1}\in\mathcal{A}_{1}$ and $H_{2}\in\mathcal{A}_{2}$ such that $\mathcal{A}^{\prime}=\mathcal{A}_{1}^{\prime}\times\mathcal{A}_{2}^{\prime}$ is MAT2-free with exponents $\exp(\mathcal{A}^{\prime})=(1,1,4,4,4,4)$ where $\mathcal{A}_{i}^{\prime}=\mathcal{A}_{i}\setminus\{H_{i}\}$. Or there are hyperplanes $H^{1}_{1},H^{1}_{2}\in\mathcal{A}_{1}$, $H^{2}_{1},H^{2}_{2}\in\mathcal{A}_{2}$ such that $\mathcal{A}^{\prime}=\mathcal{A}_{1}^{\prime}\times\mathcal{A}_{2}^{\prime}$ is MAT2-free with exponents $\exp(\mathcal{A}^{\prime})=(1,1,3,3,4,4)$ where $\mathcal{A}_{i}^{\prime}=\mathcal{A}_{i}\setminus\{H^{i}_{1},H^{i}_{2}\}$.

In the first case $\mathcal{A}^{\prime}$ is actually MAT-free by Remark 3.5. But then by Theorem 3.18 $\mathcal{A}_{2}^{\prime}$ is MAT-free and $\mathcal{A}_{2}$ is MAT-free too which is a contradiction.

In the second case $H^{1}_{1}\oplus V_{2},H^{1}_{2}\oplus V_{2},V_{1}\oplus H^{2}_{1},V_{1}\oplus H^{2}_{2}$ satisfy Condition (1)–(3) of Defintion 3.4. But by Condition (3) we have

[TABLE]

and

[TABLE]

But an easy calculation shows that there are no two hyperplanes in $\mathcal{A}_{1}$ with this property and which also satisfy Condition (2)–(3). This is a contradiction and hence $\mathcal{A}=\mathcal{A}_{1}\times\mathcal{A}_{2}$ is not MAT2-free.

4. MAT-free imprimitive reflection groups

Definition 4.1** ([OT92, §6.4]).**

Let $x_{1},\ldots,x_{\ell}$ be a basis of $V^{*}$. Let $\zeta=\exp(\frac{2\pi i}{r})$ ($r\in\mathbb{N}$) be a primitive $r$-th root of unity. Define the linear forms $\alpha_{ij}(\zeta^{k})\in V^{*}$ by

[TABLE]

and the hyperplanes

[TABLE]

for $1\leq i,j\leq\ell$ and $1\leq k\leq r$. Then the reflection arrangement of the imprimitive complex reflection group $G(r,1,\ell)$ can be defined by:

[TABLE]

Proposition 4.2**.**

Let $\mathcal{A}=\mathcal{A}(G(r,1,\ell))$. Let

[TABLE]

and

[TABLE]

for $2\leq i\leq\ell$, $1\leq j\leq r$. Then

[TABLE]

is an MAT-partition of $\mathcal{A}$. In particular $\mathcal{A}\in{\mathfrak{MF}}$ with exponents

[TABLE]

Proof.

We verify Conditions (1)–(3) from Lemma 3.8 in turn.

Let

[TABLE]

and

[TABLE]

For $\pi_{11}$ we clearly have $|\pi_{11}|=\operatorname{rk}(\pi_{11})=\ell$. Similarly for $2\leq i\leq\ell$, $1\leq j\leq r$ we have $|\pi_{ij}|=\operatorname{rk}(\pi_{ij})=\ell-i+1$ since all the defining linear forms $\alpha_{(i-1)k}(\zeta^{j})$ ($i\leq k\leq\ell$) for the hyperplanes in $\pi_{ij}$ are linearly independent. Thus Condition (1) holds.

Furthermore, the forms $\{\alpha_{ac}(\zeta^{b})\}\dot{\cup}\{\alpha_{(i-1)k}(\zeta^{j})\mid i\leq k\leq\ell\}$ are linearly independent for all $1\leq a\leq i-1$, $1\leq b\leq j-1$, and $a+1\leq c\leq\ell$, i.e. $\cap_{H\in\pi_{ij}}H=:X_{ij}\nsubseteq H$ for all $H\in\mathcal{A}_{ij}^{\prime}$. Hence Condition (2) is also satisfied.

To verify Condition (3) let $H=H_{(i-1)k}(\zeta^{j})\in\pi_{ij}$ for a fixed $1\leq k\leq r$. We show

[TABLE]

Let $H_{a}^{\prime}:=H_{(i-1)k}(\zeta^{a})\in\mathcal{A}_{ij}^{\prime}$, $1\leq a\leq j-1$. Then

[TABLE]

and $\operatorname{rk}(\mathcal{B})=2$. So all $H^{\prime}\in\mathcal{B}$ give the same intersection with $H$ and $|\mathcal{B}|=j+1$. For $H^{\prime}=H_{a(i-1)}(\zeta^{b})\in\mathcal{A}_{ij}^{\prime}$ with $a\leq i-2$, and $1\leq b\leq r$ we have

[TABLE]

$|\mathcal{C}|=2$ and there are exactly $(i-2)r$ such $H^{\prime}$. All other $H^{\prime\prime}\in\mathcal{A}_{ij}^{\prime}$ intersect $H$ simply. Hence

[TABLE]

or $|\mathcal{A}_{ij}^{\prime}|-|(\mathcal{A}_{ij}^{\prime})^{H})|=\sum_{a=1}^{i-1}m_{i}+(j-1)$. This finishes the proof. ∎

Proposition 4.3**.**

Let $\mathcal{A}=\mathcal{A}(G(r,r,\ell))$ ($r,\ell\geq 3$). Then $\mathcal{A}$ is not MAT2-free. In particular $\mathcal{A}$ is not MAT-free.

Proof.

By [OT92, Prop. 6.85] the arrangement $\mathcal{A}$ is free with $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell})=(1,r+1,2r+1,\ldots,(\ell-2)r+1,(\ell-1)(r-1))$. In particular we have $(\ell-1)(r-1)=d_{\ell}$ and $|\mathcal{A}|=\frac{\ell(\ell-1)}{2}r$. But for all $H\in\mathcal{A}$ by [OT92, Prop. 6.82, 6.85] we have $|\mathcal{A}^{H}|=\frac{(\ell-1)(\ell-2)}{2}r+1$. Hence $|\mathcal{A}|-|\mathcal{A}^{H}|=(\ell-1)r-1\neq d_{\ell}$ and by Lemma 3.12 the arrangement $\mathcal{A}$ is not MAT2-free. ∎

Theorem 4.4**.**

Let $\mathcal{A}=\mathcal{A}(W)$ be the reflection arrangement of the imprimitive complex reflection group $W=G(r,e,\ell)$ ($r,\ell\geq 3$). Then $\mathcal{A}$ is MAT-free if and only if it is MAT2-free if and only if $e\neq r$.

Proof.

Since $\mathcal{A}=\mathcal{A}(G(r,1,\ell))$ if and only if $r\neq e$, this is Proposition 4.2 and Proposition 4.3. ∎

5. MAT-free exceptional complex reflection groups

To prove the MAT-freeness of one of the following reflection arrangements, we explicitly give a realization by linear forms.

First note that if $W$ is an exceptional Weyl group, or a group of rank $\leq 2$, then by Example 3.6 $\mathcal{A}(W)$ is MAT-free.

Proposition 5.1**.**

Let $\mathcal{A}$ be the reflection arrangement of the reflection group $H_{3}$ (Shephard-Todd: $G_{23}$). Then $\mathcal{A}$ is MAT-free. In particular $\mathcal{A}$ is MAT2-free.

Proof.

Let $\tau=\frac{1+\sqrt{5}}{2}$ be the golden ratio and $\tau^{\prime}=1/\tau$ its reciprocal. The arrangement $\mathcal{A}$ can be defined by the following linear forms:

[TABLE]

With this linear ordering of the hyperplanes the partition

[TABLE]

satisfies Conditions (1)–(3) of Lemma 3.8 as one can verify by an easy linear algebra computation. Hence $\pi$ is an MAT-partition and $\mathcal{A}$ is MAT-free. ∎

Proposition 5.2**.**

Let $\mathcal{A}$ be the reflection arrangement of the complex reflection group $G_{24}$. Then $\mathcal{A}$ is not MAT2-free. In particular $\mathcal{A}$ is not MAT-free.

Proof.

The arrangement $\mathcal{A}$ is free with $\exp(\mathcal{A})=(1,9,11)$ and $|\mathcal{A}|-|\mathcal{A}^{H}|=13$ for all $H\in\mathcal{A}$ by [OT92, Tab. C.5]. Hence by Lemma 3.12 $\mathcal{A}$ is not MAT2-free. ∎

Proposition 5.3**.**

Let $\mathcal{A}$ be the reflection arrangement of the complex reflection group $G_{25}$. Then $\mathcal{A}$ is MAT-free. In particular $\mathcal{A}$ is MAT2-free.

Proof.

Let $\zeta=\frac{1}{2}(-1+i\sqrt{3})$ be a primitive cube root of unity. The reflecting hyperplanes of $\mathcal{A}$ can be defined by the following linear forms (cf. [LT09, Ch. 8, 5.3]):

[TABLE]

With this linear ordering of the hyperplanes the partition

[TABLE]

satisfies the three conditions of Lemma 3.8 as one can easily verify by a linear algebra computation. Hence $\pi$ is an MAT-partition and $\mathcal{A}$ is MAT-free. ∎

Proposition 5.4**.**

Let $\mathcal{A}$ be the reflection arrangement of the complex reflection group $G_{26}$. Then $\mathcal{A}$ is MAT-free. In particular $\mathcal{A}$ is MAT2-free.

Proof.

Let $\zeta=\frac{1}{2}(-1+i\sqrt{3})$ be a primitive cube root of unity. The reflection arrangement $\mathcal{A}$ is the union of the reflecting hyperplanes of $\mathcal{A}(G_{25})$ and $\mathcal{A}(G(3,3,3))$ (cf. [LT09, Ch. 8, 5.5]). In particular the hyperplanes contained in $\mathcal{A}$ can be defined by the following linear forms:

[TABLE]

With this linear ordering of the hyperplanes the partition

[TABLE]

satisfies the three conditions of Lemma 3.8 as one can verify by a standard linear algebra computation. Hence $\pi$ is an MAT-partition and $\mathcal{A}$ is MAT-free. ∎

Proposition 5.5**.**

Let $\mathcal{A}$ be the reflection arrangement of the complex reflection group $G_{27}$. Then $\mathcal{A}$ is not MAT2-free. In particular $\mathcal{A}$ is not MAT-free.

Proof.

The arrangement $\mathcal{A}$ is free with $\exp(\mathcal{A})=(1,19,25)$ and $|\mathcal{A}|-|\mathcal{A}^{H}|=29$ for all $H\in\mathcal{A}$ by [OT92, Tab. C.8]. Hence by Lemma 3.12 $\mathcal{A}$ is not MAT2-free. ∎

Proposition 5.6**.**

Let $\mathcal{A}$ be the reflection arrangement of the reflection group $H_{4}$ (Shephard-Todd: $G_{30}$). Then $\mathcal{A}$ is MAT-free. In particular $\mathcal{A}$ is MAT2-free.

Proof.

Let $\tau=\frac{1+\sqrt{5}}{2}$ be the golden ratio and $\tau^{\prime}=1/\tau$ its reciprocal. The arrangement $\mathcal{A}$ can be defined by the following linear forms:

[TABLE]

With this linear ordering of the hyperplanes the partition

[TABLE]

satisfies Conditions (1)–(3) of Lemma 3.8 as one can verify with a linear algebra computation. Hence $\pi$ is an MAT-partition and $\mathcal{A}$ is MAT-free. In particular $\mathcal{A}$ is MAT2-free. ∎

We recall the following result about free filtration subarrangements of $\mathcal{A}(G_{31})$:

Proposition 5.7** ([Müc17, Pro. 3.8]).**

Let $\mathcal{A}:=\mathcal{A}(G_{31})$ be the reflection arrangement of the finite complex reflection group $G_{31}$. Let $\tilde{\mathcal{A}}$ be a minimal (w.r.t. the number of hyperplanes) free filtration subarrangement. Then $\tilde{\mathcal{A}}\cong\mathcal{A}(G_{29})$.

Corollary 5.8**.**

Let $\mathcal{A}$ be the reflection arrangement of one of the complex reflection groups $G_{29}$ or $G_{31}$. Then $\mathcal{A}$ has no free filtration.

Proposition 5.9**.**

Let $\mathcal{A}$ be the reflection arrangement of one of the complex reflection groups $G_{29}$ or $G_{31}$. Then $\mathcal{A}$ is not MAT2-free. In particular $\mathcal{A}$ is not MAT-free.

Proof.

By Corollary 5.8 both arrangements have no free filtration and hence are not MAT2-free by Lemma 3.13. ∎

Proposition 5.10**.**

Let $\mathcal{A}$ be the reflection arrangement of the complex reflection group $G_{32}$. Then $\mathcal{A}$ is not MAT-free and also not MAT2-free.

Proof.

Up to symmetry of the intersection lattice there are exactly 9 different choices of a basis, where a basis is a subarrangement $\mathcal{B}\subseteq\mathcal{A}$ with $|\mathcal{B}|=r(\mathcal{B})=r(\mathcal{A})=4$. Suppose that $\mathcal{A}$ is MAT-free. Then the first block in an MAT-partition for $\mathcal{A}$ has to be one of these bases. But a computer calculation shows that non of these bases may be extended to an MAT-partition for $\mathcal{A}$. Hence $\mathcal{A}$ is not MAT-free. A similar but more cumbersome calculation shows that $\mathcal{A}$ is also not MAT2-free. ∎

Proposition 5.11**.**

Let $\mathcal{A}$ be the reflection arrangement of one of the complex reflection group $G_{33}$ or $G_{34}$. Then $\mathcal{A}$ is not MAT2-free. In particular $\mathcal{A}$ is not MAT-free.

Proof.

First, let $\mathcal{A}=\mathcal{A}(G_{33})$. Then $\exp(\mathcal{A})=(1,7,9,13,15)$ by [OT92, Tab. C.14]. But $|\mathcal{A}|-|\mathcal{A}^{H}|=17$ for all $H\in\mathcal{A}$ also by [OT92, Tab. C.14]. So $\mathcal{A}$ is not MAT2-free by Lemma 3.12.

Similarly $\mathcal{A}=\mathcal{A}(G_{34})$ is free with $\exp(\mathcal{A})=(1,13,19,25,31,37)$ by [OT92, Tab. C.17] and $|\mathcal{A}|-|\mathcal{A}^{H}|=41$ for all $H\in\mathcal{A}$. Hence $\mathcal{A}$ is not MAT2-free by Lemma 3.12. ∎

Comparing with Theorem 2.6 finishes the proofs of Theorem 1.1 and Theorem 1.2.

6. Further remarks on MAT-freeness

In their very recent note [HR19] Hoge and Röhrle confirmed a conjecture by Abe [Abe18a] by providing two examples $\mathcal{B}$, $\mathcal{D}$ of arrangements, related to the exceptional reflection arrangement $\mathcal{A}(E_{7})$, which are additionally free but not divisionally free and in particular also not inductively free. The arrangements have exponents $\exp(\mathcal{B})=(1,5,5,5,5,5,5)$ and $\exp(\mathcal{D})=(1,5,5,5,5)$. Since both arrangements have only 2 different exponents by Remark 3.5 they are MAT-free if and only if they are MAT2-free. Now a computer calculation shows that both arrangements are not MAT-free and hence also not MAT2-free. In particular they provide no negative answer to Question 1.3 and Question 1.4.

Several computer experiments suggest that similar to the poset obtained from the positive roots of a Weyl group giving rise to an MAT-partition (cf. Example 3.6) MAT-free arrangements might in general satisfy a certain poset structure:

Problem 6.1**.**

Can MAT-freeness be characterized by the existence of a partial order on the hyperplanes, generalizing the classical partial order on the positive roots of a Weyl group?

Recall that by Example 3.11 the restriction $\mathcal{A}^{H}$ is in general not MAT-free (MAT2-free) if the arrangement $\mathcal{A}$ is MAT-free (MAT2-free). But regarding localizations there is the following:

Problem 6.2**.**

Is $\mathcal{A}_{X}$ MAT-free (MAT2-free) for all $X\in L(\mathcal{A})$ provided $\mathcal{A}$ is MAT-free (MAT2-free)?

Last but not least, related to the previous problem, our investigated examples suggest the following:

Problem 6.3**.**

Suppose $\mathcal{A}^{\prime}$ and $\mathcal{A}=\mathcal{A}^{\prime}\cup\{H\}$ are free arrangements such that $\exp(\mathcal{A}^{\prime})=(d_{1},\ldots,d_{\ell})_{\leq}$ and $\exp(\mathcal{A})=(d_{1},\ldots,d_{\ell-1},d_{\ell}+1)_{\leq}$. Let $X\in L(\mathcal{A})$ with $X\subseteq H$. By [OT92, Thm. 4.37] both localizations $\mathcal{A}^{\prime}_{X}$ and $\mathcal{A}_{X}$ are free. If $\exp(\mathcal{A}^{\prime}_{X})=(c_{1},\ldots,c_{r})_{\leq}$ is it true that $\exp(\mathcal{A})=(c_{1},\ldots,c_{r-1},c_{r}+1)_{\leq}$, i.e.  if we only increase the highest exponent is the same true for all localizations?

Note that the answer is yes if we only look at localizations of rank $\leq 2$. Our proceeding investigation of Problem 6.1 suggests that this should be true at least for MAT-free arrangements. Furthermore, a positive answer to Problem 6.3 would imply (with a bit more work) a positive answer to Problem 6.2.