MAT-free reﬂection arrangements

We introduce the class of MAT-free hyperplane arrangements which is based on the Multiple Addition Theorem by Abe, Barakat, Cuntz, Hoge


Introduction
A hyperplane arrangement A is a finite set of hyperplanes in a finite dimensional vector space V ∼ = K where K is some field.The intersection lattice L(A) of A encodes its combinatorial properties.It is a main theme in the study of hyperplane arrangements to link algebraic properties of A with the combinatorics of L(A).
The algebraic property of freeness of a hyperplane arrangement A was first studied by Saito [Sai80] and Terao [Ter80a].In fact, it turns out that freeness of A imposes strong combinatorial constraints on L(A): by Terao's Factorization Theorem [OT92,Thm. 4.137] its characteristic polynomial factors over the integers.Conversely, sufficiently strong conditions on L(A) imply the freeness of A. One of the main tools to derive such conditions the electronic journal of combinatorics 27(1) (2020), #P1.28 is Terao's Addition-Deletion Theorem 8.It motivates the class of inductively free arrangements (see Definition 9).In this class the freeness of A is combinatorial, i.e. it is completely determined by L(A) (cf.Definition 5).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 A actually depends only on L(A), provided the field K is fixed (see [Zie90] for a counterexample when one fixes L(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 8 was obtained by Abe, Barakat, Cuntz, Hoge, and Terao in [ABC + 16]: the Multiple Addition Theorem 12 (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 MF of those free arrangements, called MAT-free, which can be build inductively using the MAT (Definition 13).It is not hard to see (Lemma 18) that MAT-freeness only depends on L(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 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. Except for the arrangement A(G 32 ), an irreducible reflection arrangement is MAT-free if and only if it is inductively free.The arrangement 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, ), e > 2, > 2 and of the reflection groups A further generalization of the MAT 12 was very recently obtained by Abe and Terao [AT19]: the Multiple Addition Theorem 2 14 (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 MF of MAT2-free arrangements (Definition 15).By definition, this class contains the class of MAT-free arrangements.Regarding reflection arrangements we have the following: Theorem 2. Let A = A(W ) be an irreducible reflection arrangement.Then A is MAT2free 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 28).
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 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 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 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 MF and MF and on the product construction.Section 4 and Section 5 contain the proofs of Theorem 1 and Theorem 2. In the last Section 6 we comment on Question 3 and further problems connected with MAT-freeness.

Hyperplane arrangements and free arrangements
Let A be a hyperplane arrangement in V ∼ = K where K is some field.If A is empty, then it is denoted by Φ .
The intersection lattice L(A) of A consists of all intersections of elements of A including V as the empty intersection.Indeed, with the partial order by reverse inclusion L(A) is a geometric lattice [OT92, Lem.2.3].The rank rk(A) of A is defined as the codimension of the intersection of all hyperplanes in A.
If x 1 , . . ., x is a basis of V * , to explicitly give a hyperplane we use the notation (a 1 , . . ., a ) ⊥ := ker(a Definition 5. Let C be a class of arrangements and let A ∈ C. If for all arrangements B with L(B) ∼ = L(A), (where A and B do not have to be defined over the same field), we have B ∈ C, then the class C is called combinatorial.
If C is a combinatorial class of arrangements such that every arrangement in C is free than A ∈ C is called combinatorially free.
For X ∈ L(A) the localization A X of A at X is defined by: and the restriction A X of A to X is defined by: Let A 1 and A 2 be two arrangements in V 1 respectively V 2 .Then their product A 1 × A 2 is defined as the arrangement in V = V 1 ⊕ V 2 consisting of the following hyperplanes: We note the following facts about products (cf.[OT92, Ch. 2]): Let S = S(V * ) be the symmetric algebra of the dual space.We fix a basis x 1 , . . ., x for V * and identify S with the polynomial ring K[x 1 , . . ., x ].The algebra S is equipped with the grading by polynomial degree: S = p∈Z S p , where S p is the set of homogeneous polynomials of degree p (S p = {0} for p < 0).
A K-linear map θ : S → S which satisfies θ(f g) = θ(f )g + f θ(g) is called a Kderivation.Let Der(S) be the S-module of K-derivations of S. It is a free S-module with basis D 1 , . . ., D where D i is the partial derivation ∂/∂x i .We say that θ ∈ Der(S) is homogeneous of polynomial degree p provided θ = i=1 f i D i with f i ∈ S p for each 1 i .In this case we write pdeg θ = p.We obtain a Z-grading for the S-module Der(S): Der(S) = p∈Z Der(S) p .Definition 6.For H ∈ A we fix α H ∈ V * with H = ker(α H ). The module of Aderivations is defined by We say that A is free if the module of A-derivations is a free S-module.
Then A is free if and only if both A 1 and A 2 are free.In this case if exp The following theorem provides a useful tool to prove the freeness of arrangements.
Theorem 8 (Addition-Deletion [OT92, Thm.4.51]).Let A be a hyperplane arrangement and H 0 ∈ A. We call (A, A = A \ {H 0 }, A = A H 0 ) a triple of arrangements.Any two of the following statements imply the third: The preceding theorem motivates the following definition.

Here (A,
The class 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 10 ([HR15, Thm.1.1], [BC12, Thm.5.14]).For W a finite complex reflection group, the reflection arrangement A(W ) is inductively free if and only if W does not admit an irreducible factor isomorphic to a monomial group G(r, r, ) for r, Definition 11 (cf.[AT16]).Let A be an arrangement with |A| = n.We say that A has a free filtration if there are subarrangements Very recently, Abe [Abe18a] introduced the class AF of additionally free arrangements.Arrangements in 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 AF is combinatorial.

Multiple Addition Theorem
The following theorem presented in [ABC + 16] is a variant of the addition part ((2) and (3) imply (1)) of Theorem 8.
Theorem 12 (Multiple Addition Theorem (MAT)).Let A be a free arrangement with exp(A ) = (d 1 , . . ., d ) and 1 p the multiplicity of the highest exponent, i.e., Let H 1 , . . ., H q be hyperplanes with H i ∈ A for i = 1, . . ., q. Define Assume that the following three conditions are satisfied: (1) (2) X ⊆ H∈A H. ( Then q p and A : Note that in contrast to Theorem 8 no freeness condition on the restriction is needed to conclude the freeness of A in Theorem 12.The MAT motivates the following definition. Definition 13.The class MF of MAT-free arrangements is the smallest class of arrangements subject to (i) Φ belongs to MF, for every 0; (ii) if A ∈ MF with exp(A ) = (d 1 , . . ., d ) and 1 p the multiplicity of the highest exponent d = d , and if H 1 , . . ., H q , q p are hyperplanes with H i ∈ A for i = 1, . . ., q such that: (1) then A := A ∪ {H 1 , . . ., H q } also belongs to MF and has exponents exp(A) = (d 1 , . . ., d −q , d + 1, . . ., d + 1) .

If
A is a free arrangement with exp(A) = (0, . . ., 0, 1, . . ., 1, d, . . ., d) , i.e.A has only two distinct exponents = 0, then it is clear from the definitions that A is MAT2-free if and only if A is MAT-free.
1.If rk(A) = 2 then A is MAT-free and therefore MAT2-free too.If an arrangement A is MAT-free, the MAT-steps yield a partition of A whose dual partition gives the exponents of A. Vice versa, the existence of such a partition suffices for the MAT-freeness of the arrangement: Lemma 19.Let A be an -arrangement.Then A is MAT-free if and only if there exists a partition π = (π

Every ideal subarrangement of a
In this case A has exponents exp(A) = (d 1 , . . ., d ) with Proof.This is immediate from the definition.Example 21.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 A(10, 1) in [Grü09] is supersolvable but not MAT2-free.In particular A(10, 1) is neither MAT-free.
Restrictions of MAT2-free (MAT-free) arrangements are not necessarily MAT2-free (MAT-free): Example 22.Let A = A(E 6 ) be the Weyl arrangement of the Weyl group of type E 6 .Then A is MAT-free by Example 17(2).Let H ∈ A. A simple calculation (with the computer) shows that A H is not MAT2-free.
We have two simple necessary conditions for MAT-freeness respectively MAT2-freeness.The first one is: the electronic journal of combinatorics 27(1) (2020), #P1.28 Lemma 23.Let A be a non-empty MAT2-free arrangement with exponents exp(A) = (d 1 , . . ., d ) .Then there is an H ∈ A such that |A| − |A H | = d .In particular, the same holds, if A is MAT-free.
Proof.By definition there are H q , . . ., H ∈ A, 2 q such that A := A \ {H q , . . ., H } is MAT2-free.Furthermore by condition (1) the hyperplanes H q , . . ., H are linearly independent.Let H := H .By condition (2), we have X = ∩ i=q H i ∪ H ∈A H and thus The second one is: Lemma 24.Let A be an MAT2-free arrangement.Then A has a free filtration, i.e.A is additionally free.In particular, the same is true, if A is MAT-free.
Proof.Let A be MAT2-free.Then by definition there are H q , . . ., H ∈ A such that A := A\{H q , . . ., H } is MAT2-free and conditions (1)-(3) are satisfied.Set B := {H q , . . ., H }. By [AT19, Cor.3.2] for all C ⊆ B the arrangement A ∪ C is free.Hence by induction 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.

Proposition 26. The arrangement
As a direct consequence we get: Proposition 28.We have MF MF .

Products of MAT-free and MAT2-free arrangements
As for freeness in general (Proposition 7), the product construction is compatible with the notion of MAT-freeness: Theorem 29.Let A = A 1 × A 2 be a product of two arrangements.Then A ∈ MF if and only if A 1 ∈ MF and A 2 ∈ MF.
Proof.Assume A i is an arrangement in the vector space V i of dimension i for i = 1, 2. We argue by induction on |A|.If |A| = 0, i.e.A 1 = Φ 1 , and 2 ) .Then without loss of generality

they satisfy Conditions (1)-(3) from Definition 13. Now by the induction hypothesis
Conversely assume A is MAT-free with exp(A) = (d 1 , . . ., d ) .By Proposition 7 both factors A 1 and A 2 are free with exp(A i ) = (d i 1 , . . ., d i i ) and without loss of generality 2 .Assume further that q i is the multiplicity of d in exp(A i ) and q is the multiplicity of d in exp(A), i.e. q = q 1 +q 2 .There are hyperplanes {H 1 , . . ., H q } ⊂ A such that A = A \ {H 1 , . . ., H q } is MAT-free with exp(A ) = (d 1 , . . ., d −q , d −q+1 − 1, . . ., d − 1) , and Conditions (1)-(3) are satisfied.We may further assume that 2 we have q 2 = 0 and A 2 = A 2 .But at least we have and by the induction hypothesis A 1 and A 2 are MAT-free and Conditions (1) and (2) are clearly satified for A i and {H i 1 , . . ., H i q i }.But since the electronic journal of combinatorics 27(1) (2020), #P1.28 for 1 i q 1 and for q 1 + 1 j q 2 , Condition (3) is also satisfied for A 1 and A 2 .Hence both A 1 and A 2 are MAT-free.
Altenatively, one can prove Theorem 29 by observing that MAT-Partitions for A 1 and A 2 are directly obtained from an MAT-Partition for 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 A 1 and A 2 .
Remark 30.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 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.
In contrast to MAT-freeness, the weaker notion of MAT2-freeness is not compatible with products as the following example shows: Example 31.Let A 1 be the MAT2-free but not MAT-free arrangement of Example 25 with exponents exp(A 1 ) = (1, 4, 5).Let ζ = 1 2 (−1 + i √ 3) be a primitive cube root of unity, and let A 2 be the arrangement defined by the following linear forms: A linear algebra computation shows that π = (1, 2, 3|4, 5|6, 7|8, 9|10) is an MAT-partition for A 2 .In particular A 2 is MAT2-free with exp(A 2 ) = (1, 4, 5).Now by Proposition 7 the product A := A 1 × A 2 is free with exp(A) = (1, 1, 4, 4, 5, 5).Suppose A is MAT2-free.Then either there are hyperplanes In the first case A is actually MAT-free by Remark 16.But then by Theorem 29 A 2 is MAT-free and A 2 is MAT-free too which is a contradiction.
In the second case the electronic journal of combinatorics 27(1) (2020), #P1.28 and But an easy calculation shows that there are no two hyperplanes in A 1 with this property and which also satisfy Condition (2)-(3).This is a contradiction and hence A = A 1 × A 2 is not MAT2-free.
To verify Condition (3) let H = H (i−1)k (ζ j ) ∈ π ij for a fixed 1 k r.We show . This finishes the proof.
Proposition 34.Let A = A(G(r, r, )) (r, 3).Then A is not MAT2-free.In particular A is not MAT-free.

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 2, then by Example 17 A(W ) is MAT-free.
Proposition 36.Let A be the reflection arrangement of the reflection group H 3 (G 23 ).Then A is MAT-free.In particular A is MAT2-free.
Proposition 37. Let A be the reflection arrangement of the complex reflection group G 24 .Then A is not MAT2-free.In particular A is not MAT-free.Proposition 38.Let A be the reflection arrangement of the complex reflection group G 25 .Then A is MAT-free.In particular A is MAT2-free.
3) be a primitive cube root of unity.The reflecting hyperplanes of A can be defined by the following linear forms (cf.[LT09, Ch. 8, 5.3]): With this linear ordering of the hyperplanes the partition π = (7, 4, 3|8, 5|9, 6|2, 1|10|11|12) satisfies the three conditions of Lemma 19 as one can easily verify by a linear algebra computation.Hence π is an MAT-partition and A is MAT-free.
Proof.By Corollary 43 both arrangements have no free filtration and hence are not MAT2free by Lemma 24.
Proposition 45.Let A be the reflection arrangement of the complex reflection group G 32 .Then 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 B ⊆ A with |B| = r(B) = r(A) = 4. Suppose that A is MAT-free.Then the first block in an MAT-partition for 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 A. Hence A is not MAT-free.A similar but more cumbersome calculation shows that A is also not MAT2-free.
Proposition 46.Let A be the reflection arrangement of one of the complex reflection group G 33 or G 34 .Then A is not MAT2-free.In particular A is not MAT-free.Comparing with Theorem 10 finishes the proofs of Theorem 1 and Theorem 2.

Further remarks on MAT-freeness
In their very recent note [HR19] Hoge and Röhrle confirmed a conjecture by Abe [Abe18a] by providing two examples B, D of arrangements, related to the exceptional reflection arrangement A(E 7 ), which are additionally free but not divisionally free and in particular also not inductively free.The arrangements have exponents exp(B) = (1, 5, 5, 5, 5, 5, 5) and exp(D) = (1, 5, 5, 5, 5).Since both arrangements have only 2 different exponents by Remark 16 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 MAT2free.In particular they provide no negative answer to Question 3 and Question 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 17) MATfree arrangements might in general satisfy a certain poset structure: Problem 47. 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 22 the restriction A H is in general not MAT-free (MAT2-free) if the arrangement A is MAT-free (MAT2-free).But regarding localizations there is the following: Problem 48.Is A X MAT-free (MAT2-free) for all X ∈ L(A) provided A is MAT-free (MAT2-free)?
Last but not least, related to the previous problem, our investigated examples suggest the following: Problem 49.Suppose A and A = A ∪ {H} are free arrangements such that exp(A ) = (d 1 , . . ., d ) and exp(A) = (d 1 , . . ., d −1 , d +1) .Let X ∈ L(A) with X ⊆ H.By [OT92, Thm.4.37] both localizations A X and A X are free.If exp(A X ) = (c 1 , . . ., c r ) is it true that exp(A) = (c 1 , . . ., c r−1 , c r + 1) , 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 2. Our proceeding investigation of Problem 47 suggests that this should be true at least for MAT-free arrangements.Furthermore, a positive answer to Problem 49 would imply (with a bit more work) a positive answer to Problem 48.
Definition 9 ([OT92, Def.4.53]).The class IF of inductively free arrangements is the smallest class of arrangements which satisfies 1. the empty arrangement Φ of rank is in IF for 0, 2. if there exists a hyperplane H 0 ∈ A such that A ∈ IF, A ∈ IF, and exp(A ) ⊂ exp(A ), then A also belongs to IF.
Weyl arrangement is MAT-free and therefore also MAT2-free, [ABC + 16].Lemma 18.The classes MF and MF are combinatorial.Proof.The class of all empty arrangements is combinatorial and contained in MF.Let A ∈ MF (A ∈ MF ).Since conditions (1)-(3) in Definition 13 (respectively Definition 15) only depend on L(A) the claim follows.See also [AT19, Thm.5.1].

Definition 20 .
If π is a partition as in Lemma 19 then π is called an MAT-partition for A.If we have chosen a linear ordering A = {H 1 , . . ., H m } of the hyperplanes in A, to specify the partition π, we give the corresponding ordered set partition of [m] = {1, . . ., m}.
for all i and j.Definition 15.The class MF of MAT2-free arrangements is the smallest class of ar- (ii) if A ∈ MF with exp(A ) = (d 1 , d 2 , . ..,d ) and if H s , . . ., H are hyperplanes with H i ∈ A for i = s, . . ., , where Lemma 19 as one can verify with a linear algebra computation.Hence π is an MAT-partition and A is MAT-free.In particular A is MAT2-free.We recall the following result about free filtration subarrangements of A(G 31 ):Proposition 42 ([Müc17, Pro.3.8]).Let A := A(G 31 ) be the reflection arrangement of the finite complex reflection group G 31 .Let Ã be a minimal (w.r.t. the number of hyperplanes) free filtration subarrangement.Then Ã ∼ = A(G 29 ).Corollary 43.Let A be the reflection arrangement of one of the complex reflection groups G 29 or G 31 .Then A has no free filtration.