On the freeness of hypersurface arrangements consisting of hyperplanes and spheres

Ruimei Gao; Qun Dai; Zhe Li

doi:10.1515/math-2018-0041

Article Open Access

On the freeness of hypersurface arrangements consisting of hyperplanes and spheres

Ruimei Gao , Qun Dai and Zhe Li

Published/Copyright: April 23, 2018

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$Open Mathematics$

From the journal Open Mathematics Volume 16 Issue 1

Abstract

Let V be a smooth variety. A hypersurface arrangement 𝓜 in V is a union of smooth hypersurfaces, which locally looks like a union of hyperplanes. We say 𝓜 is free if all these local models can be chosen to be free hyperplane arrangements. In this paper, we use Saito’s criterion to study the freeness of hypersurface arrangements consisting of hyperplanes and spheres, and construct the bases for the derivation modules explicitly.

Keywords: Hypersurface arrangement; Freeness; Hyperplane; Sphere

MSC 2010: 52C35; 32S22

1 Introduction

A hypersurface arrangement 𝓜 in a smooth variety V is a reduced divisor D consisting of a union of smooth hypersurfaces, such that at each point D is locally analytically isomorphic to a hyperplane arrangement. For hypersurface arrangements, many researchers made focus on the study of Milnor fibers, higher homotopy groups and Alexander invariants of the hypersurface complements, such as [1, 2, 3, 4]. Besides the topological properties, the freeness of a hypersurface arrangement could also be considered. We say a hypersurface arrangement is free if D is itself a free divisor on V. The study of free hyperplane arrangements was initiated by H. Terao in [5], and has been playing the central role in this area. Recently, there have been several studies to determine when a hyperplane arrangement is free, e.g., [6, 7, 8, 9] and so on. However, it is still very difficult to determine the freeness. Freeness of hyperplane arrangements implies several interesting geometric and combinatorial properties of the arrangements, for example see [6, 10, 11]. Therefore, there were many works on the freeness of hyperplane arrangements, especially on Coxeter arrangements and the cones over Catalan and Shi arrangements[12, 13, 14, 15, 16].

In [17], H. Schenck and S. Tohǎneanu studied the freeness of Conic-Line arrangements in P₂ and their results are the first to give an inductive criterion for freeness of nonlinear arrangements. Until now, the papers about the freeness of hypersurface arrangements are few. In this paper, we will consider the freeness of hypersurface arrangements consisting of hyperplanes and spheres, and will construct bases for the derivation modules of hypersurface arrangements.

The paper is organized as follows: in Section 2, we recall the basic definitions and generalize Saito’s criterion to hypersurface arrangements consisting of hyperplanes and spheres. In Section 3, for the hypersurface arrangement consisting of n spheres, the hypersurface arrangement containing a free hyperplane arrangement and n spheres, we present the constructions of bases for the derivation modules respectively.

2 Preliminaries and Notations

We begin with some basic concepts and notations of arrangements, for more information see P. Orlik and H. Terao [18].

Let V be an ℓ-dimensional vector space on 𝕂 with a coordinate system {x₁,…,x_ℓ} ⊂ V^*. Let S = S(V^*) be the symmetric algebra of V^* and Der_𝕂(S) be the module of derivations

DerK(S)={θ:S→S| θ(fg)=fθ(g)+gθ(f), f,g∈S}.

Define D_i = ∂/∂x_i, 1 ≤ i ≤ ℓ, then D₁,…,D_\ℓ is a basis for Der_𝕂(S) over S.

Definition 2.1

A nonzero elementθ ∈ Der_𝕂(S) is of polynomial degree p if θ = ∑k=1ℓfkDkand the maximum of the degrees of coefficient polynomials f₁,…,f_ℓ (get rid of 0) isp. In this case we write pdegθ = p.

Definition 2.2

For a hypersurface arrangement 𝓜 in V, the derivation module D(𝓜) is defined by

D(M)={θ∈DerK(S)| θ(αX)∈αXS for all X∈M},

where X = ker(α_X), 𝓜 is called free if D(𝓜) is free.

Definition 2.3

Let 𝓜 be a free hypersurface arrangement and let {θ₁,…,θ_ℓ} be a basis for D(𝓜). We call pdegθ₁,…, pdegθ_ℓthe exponents of 𝓜 and write

expM={pdegθ1,…,pdegθℓ}.

Definition 2.4

Given derivationsθ₁,…,θ_ℓ ∈ D(𝓜), define the coefficient matrix M(θ₁,…,θ_ℓ) by M_i,j = θ_j(x_i), thus

M(θ1,…,θℓ)=θ1(x1)...θℓ(x1)...............θ1(xℓ)...θℓ(xℓ),

and θ_j = ∑i=1ℓMi,jDi.

Definition 2.5

Let 𝓜 be a hypersurface arrangement, the product

Q(M)=∏X∈MαX

is called a defining polynomial of 𝓜, where X = ker(α_X).

For hyperplane arrangements, Saito’s criterion provides a wonderful method to prove the freeness. Next, we will prove it also holds for 𝓜, where 𝓜 is a hypersurface arrangement in ℝ^ℓ consisting of linear hyperplanes and spheres.

Lemma 2.6

Ifθ₁,…,θ_ℓ ∈ D(𝓜), then det M(θ₁,…,θ_ℓ) ∈ Q(𝓜)S.

Proof

Let X ∈ 𝓜, and let X = ker(α_X), then

detM(θ1,…,θℓ)=fdetθ1(x1)...θℓ(x1)..........θ1(αX)...θℓ(αX)..........θ1(xℓ)...θℓ(xℓ),

If αX=∑k=1ℓckxk,thenf=ck∈R;IfαX=∑k=1ℓ(xk−ak)2−r, then f = 2(x_k – a_k). For any case, det M(θ₁,…,θ_ℓ) is divisible by α_X. Since X is arbitrary, det M(θ₁,…,θ_ℓ) ∈ Q(𝓜)S. □

Lemma 2.7

Let M_n be an n × n matrix with the (p, q) entry as follows:

Mpq=xpxqif p≠q,xp2−rif p=q,

where 1 ≤ p, q ≤ n, r ∈ ℝ. Therefore,

detMn=(−r)n−1(∑k=1nxk2−r).

Proof

We will prove the lemma by induction on n.

For the case n = 1, M₁ = x12 – r, then det M₁ = x12 – r.
We assume that for the case n the result holds, that is det Mn=(−r)n−1(∑k=1nxk2−r).
For the case n + 1 we have
Mn+1=MnNNTxn+12−r,
where N = (x₁x_n+1,…,x_nx_n+1)^T. Therefore,
detMn+1=detMnNNTxn+12+detMnOn×1NT−r=det−rEnNO1×nxn+12+(−r)detMn=(−r)n(xn+12)+(−r)(−r)n−1(∑i=1nxi2−r)=(−r)n(∑k=1n+1xk2−r),
where O_n×1, O_1×n are the n × 1 and 1 × n null matrices respectively, and E_n is the n × n identity matrix. □

Lemma 2.8

Let

S={(x1,…,xℓ)∣∑k=1ℓ(xk−ak)2=r∈R+}

be the (ℓ – 1)-dimensional sphere in ℝ^ℓwith center (a₁, a₂, …, a_ℓ) and radiusr,define

θq=∑p=1ℓfpqDp,1≤q≤ℓ,

where

fpq=(xp−ap)(xq−aq)if p≠q,(xp−ap)2−rif p=q.

Thenθ₁,…,θ_ℓ ∈ D({𝓢}) and det M(θ₁,…,θ_ℓ) ≐ Q({𝓢}).

Proof

We can see

θq=∑p=1ℓfpqDp=(xq−aq)∑p=1ℓ(xp−ap)Dp−rDq,

and

θq[∑k=1ℓ(xk−ak)2−r]=2(xq−aq)[∑k=1ℓ(xk−ak)2−r]∈[∑k=1ℓ(xk−ak)2−r]S,

Thus θ_q ∈ D({𝓢}) for 1 ≤ q ≤ ℓ.

By Lemma 2.7, we obtain

detM(θ1,…,θℓ)=(−r)ℓ−1[∑k=1ℓ(xk−ak)2−r]≐Q({S}).

□

Next, we will show Saito’s criterion for hypersurface arrangements.

Theorem 2.9

Givenθ₁,…,θ_ℓ ∈ D(𝓜), the following two conditions are equivalent:

det M(θ₁,…,θ_ℓ) ≐ Q(𝓜).
θ₁,…,θ_ℓform a basis forD(𝓜) overS.

Proof

(1)⇒(2) The proof is exactly the same with that of Saito’s criterion in [18].

(2)⇒(1) By Lemma 2.6, we can write det M(θ₁,…,θ_ℓ) = fQ(𝓜) for some f ∈ S. Fix X ∈ 𝓜, if X is a hyperplane, then {X} is a free hyperplane arrangement; if X is a sphere, by Lemma 2.8 and (1)⇒(2), {X} is a free hypersurface arrangement. Assume η₁,…,η_ℓ is the basis of X, then Q_Xη₁,…,Q_Xη_ℓ ∈ D(𝓜), where Q_X = Q(𝓜)/α_X. Since each Q_Xη_i is an S-linear combination of η₁,…,η_ℓ, then there exists an ℓ × ℓ matrix N with entries in S, such that

M(QXη1,…,QXηℓ)=M(θ1,…,θℓ)N.

Thus we have

Q(M)QXℓ−1≐detM(QXη1,…,QXηℓ)∈detM(θ1,…,θℓ)S=fQ(M)S.

Therefore f divides QXℓ−1 for all X ∈ 𝓜. Since the polynomials {QXℓ−1}X∈M have no common factor, f ∈ ℝ^*. □

Corollary 2.10

If 𝓢 is an (ℓ − 1)-dimensional sphere in ℝ^ℓ, then {𝓢} is a free hypersurface arrangement with

exp{S}={2,…,2},

where 2 appears ℓ times.

Proof

The result is obtained directly from Lemma 2.8 and Theorem 2.9. □

3 Main results

In this section, we will consider the freeness for hypersurface arrangements containing hyperplanes and spheres, and give the explicit bases for the derivation modules of the free ones. First, we show that the hypersurface arrangement having n spheres is free.

Theorem 3.1

Let 𝓜_nbe the hypersurface arrangement consisting ofn spheres 𝓢₁,…,𝓢_n, where

Si={(x1,…,xℓ)∣∑k=1ℓ(xk−ak)2=ri, (a1,a2,…,aℓ)∈Rℓ, ri∈R+}.

Define derivationsφ1n,…,φℓnby

M(φ1n,…,φℓn)=AnAn−1⋯A1,

where A_i is anℓ × ℓmatrix and the (p, q) entry of A_i is

(Ai)pq=(xp−ap)(xq−aq) if p≠q,(xp−ap)2−ri if p=q.

Thenφ1n,…,φℓnform a basis for D(𝓜_n) and exp 𝓜_n = {2n,…,2n}, where 2n appearsℓ times.

Proof

We will prove this result by Theorem 2.9: Saito’s criterion. By Lemma 2.7, we obtain

detAi=(−ri)ℓ−1[∑k=1ℓ(xk−ak)2−ri].

Therefore,

detM(φ1n,…,φℓn)=∏i=1ndetAi≐∏i=1n[∑k=1ℓ(xk−ak)2−ri]=Q(Mn).

Next, we will prove φin ∈ D(𝓜_n) and deg φin = 2n for any 1 ≤ i ≤ ℓ by induction on n.

The case n = 1 is clear according to Lemma 2.8 and Corollary 2.10. For the case n + 1, we notice that

φin+1=∑p=1ℓφin+1(xp)Dp=∑p=1ℓ[∑q=1ℓ(An+1)pqφin(xq)]Dp=∑p=1ℓ[∑q≠p(xp−ap)(xq−aq)φin(xq)+[(xp−ap)2−rn+1]φin(xp)]Dp=∑p=1ℓ[(xp−ap)∑q=1ℓ(xq−aq)φin(xq)]Dp−rn+1∑p=1ℓφin(xp)Dp=∑q=1ℓ(xq−aq)φin(xq)∑p=1ℓ(xp−ap)Dp−rn+1φin=12φin[∑q=1ℓ(xq−aq)2]∑p=1ℓ(xp−ap)Dp−rn+1φin.

Therefore, for 1 ≤ i ≤ ℓ and 1 ≤ j ≤ n,

φin+1[∑k=1ℓ(xk−ak)2−rj]=φin[∑q=1ℓ(xq−aq)2]∑p=1ℓ(xp−ap)2−rn+1φin[∑k=1ℓ(xk−ak)2−rj]=φin[∑q=1ℓ(xq−aq)2][∑p=1ℓ(xp−ap)2−rn+1],

by induction hypothesis,

φin[∑q=1ℓ(xq−aq)2]∈∏j=1n[∑k=1ℓ(xk−ak)2−rj]S,

hence,

φin+1[∑k=1ℓ(xk−ak)2−rj]∈∏j=1n+1[∑k=1ℓ(xk−ak)2−rj]S.

This means φin+1∈D(Mn+1) for any 1 ≤ i ≤ ℓ, in addition,

pdegφin+1=pdegφin+2=2n+2=2(n+1).

We complete the induction, so by Saito’s criterion φ1n,…,φℓn form a basis for 𝓜_n, and exp 𝓜_n = {2n,…,2n}. □

Next, we will study the freeness for the hypersurface arrangements consisting of hyperplanes and spheres, where all the spheres are centered at origin.

Theorem 3.2

Assume 𝓐 is a free hyperplane arrangement with a homogeneous basisθ₁,…,θ_ℓ, and exp 𝓐 = {d₁,…,d_ℓ}, Si0is the sphere centered at origin:

Si0={(x1,…,xℓ)∣∑k=1ℓxk2=ri∈R+},1≤i≤n,

and

Mn=A∪{S10,…,Sn0},

Define derivationsφ1n,…,φℓnby

M(φ1n,…,φℓn)=(AnAn−1⋯A1)M(θ1,…,θℓ),

where A_i is anℓ × ℓmatrix and the (p, q) entry of A_i is

(Ai)pq=xpxq if p≠q,xp2−ri if p=q,

thenφ1n,…,φℓnform a basis for D(𝓜_n) and exp 𝓜_n = {d₁ + 2n,…,d_ℓ + 2n}.

Proof

By Lemma 2.7, we obtain

detAi=(−ri)ℓ−1(∑k=1ℓxk2−ri).

Since 𝓐 is a free arrangement with a homogeneous basis θ₁,…,θ_ℓ, by Saito’s criterion,

detM(θ1,…,θℓ)≐Q(A).

Therefore,

detM(φ1n,…,φℓn)=∏i=1n(detAi)detM(θ1,…,θℓ)≐∏i=1n(∑k=1ℓxk2−ri)Q(A)=Q(Mn).

Next, we will prove φin ∈ D(𝓜_n) and deg φin = d_i + 2n for any 1 ≤ i ≤ ℓ by induction on n.

For the case n = 1,

φi1=∑p=1ℓφi1(xp)Dp=∑p=1ℓ[∑q=1ℓ(A1)pqθi(xq)]Dp=∑p=1ℓ[∑q≠pxpxqθi(xq)+(xp2−r1)θi(xp)]Dp=∑p=1ℓ[xp∑q=1ℓxqθi(xq)]Dp−r1∑p=1ℓθi(xp)Dp=∑q=1ℓxqθi(xq)∑p=1ℓxpDp−r1θi=∑q=1ℓxqθi(xq)θE−r1θi.

Since θ_E, θ_i ∈ D(𝓐), we have φi1 ∈ D(𝓐) for any 1 ≤ i ≤ ℓ. And

pdegφi1=pdeg[∑q=1ℓxqθi(xq)θE]=pdegθi+2=di+2.

In addition,

φi1(∑k=1ℓxk2−r1)=[∑q=1ℓxqθi(xq)θE−r1θi](∑k=1ℓxk2−r1)=∑q=1ℓxqθi(xq)(2∑k=1ℓxk2)−2r1∑q=1ℓxqθi(xq)=2(∑k=1ℓxk2−r1)∑q=1ℓxqθi(xq)∈(∑k=1ℓxk2−r1)S

that is, φi1∈D({S10}). Therefore, φi1∈D(A)∩D({S10})=D(M1) for any 1 ≤ i ≤ ℓ.

For the case n + 1, by the similar calculation of φi1, we get

φin+1=∑q=1ℓxqφin(xq)θE−rn+1φin=12φin(∑q=1ℓxq2)θE−rn+1φin.

By induction hypothesis,

φin∈D(Mn)⊆D(⋃i=1n{Si0}),

we obtain

φin(∑q=1ℓxq2)∈∏j=1n(∑k=1ℓxk2−rj)S.

Therefore,

φin(∑q=1ℓxq2)θE∈D(⋃i=1n{Si0}).

Combining θ_E ∈ D(𝓐) with

D(Mn)=D(⋃i=1n{Si0})⋂D(A),

we conclude

φin(∑q=1ℓxq2)θE∈D(Mn).

Hence, φin+1∈D(Mn) since φin∈D(Mn).

In addition, we have

φin+1(∑k=1ℓxk2−rn+1)=φin(∑q=1ℓxq2)∑k=1ℓxk2−rn+1φin(∑k=1ℓxk2)=(∑k=1ℓxk2−rn+1)φin(∑q=1ℓxq2)∈(∑k=1ℓxk2−rn+1)S,

We obtain φin+1∈D({Sn+10}) for any 1 ≤ i ≤ ℓ, therefore

φin+1∈D({Sn+10})∩D(Mn)=D(Mn+1), 1≤i≤ℓ.

Moreover,

pdegφin+1=pdeg[φin(∑q=1ℓxq2)θE]=pdegφin+2=di+2n+2=di+2(n+1), 1≤i≤ℓ.

We complete the induction. □

Corollary 3.3

Let 𝓜_n = A∪{S10,…,Sn0}be the hypersurface arrangement defined in Theorem 3.2. Then 𝓐 is free if and only if 𝓜_nis free.

Proof

If 𝓐 is free we can obtain that 𝓜_n is free directly from Theorem 3.2. Assume 𝓜_n is free, 𝓐 ⊆ 𝓜_n, then D(𝓜_n) ⊆ D(𝓐). Let φ₁,…,φ_ℓ be a basis for D(𝓜_n), then φ_i ∈ D(𝓐) for 1 ≤ i ≤ ℓ. Write φi=∑k≥0φi(k), where φi(k) is zero or homogeneous of degree k ≥ 0. Since Q(𝓐)S is generated by homogeneous polynomial Q(𝓐), each homogeneous component φi(k)(Q(A))ofφi(Q(A)) also lies in Q(𝓐)S. This shows that φi(k)∈D(A) for k ≥ 0. Since [Q(⋃i=1n{Si})](0)≠0, there exist φ1(d1),…,φℓ(dℓ) such that

detM(φ1(d1),…,φℓ(dℓ))≐Q(A).

By Saito’s criterion, φ1(d1),…,φℓ(dℓ) form a basis for D(𝓐). □

Remark 3.4

In Theorem 3.1 and Theorem 3.2 the preconditions did not impose the restrictions on the size relations of r₁, r₂, …, r_n. Hence, if r₁ = r₂ = … = r_n, Theorem 3.1 and Theorem 3.2 also hold. In this case, (𝓜, m) is a hypersurface arrangement with a multiplicity m_i for each hypersurface in 𝓜, we call it hypersurface multiarrangement. As defined by G. Ziegler in [19], the module of derivations consists ofθsuch thatθ(α_i) ∈ αimiS.

Example 3.5

Let 𝓜 be a hypersurface arrangement with the defining polynomial

Q(M)=(x1−x2)(x1−x3)(x2−x3)(x12+x22+x32−1).

In this case, the hyperplane arrangement 𝓐 ⊆ 𝓜 is the Coxeter arrangement of typeA₂, it is a free arrangement with exp 𝓐 = {0, 1, 2}. By Theorem 3.2, 𝓜 is a free hypersurface arrangement andD(𝓜) has the basisφ₁, φ₂, φ₃as follows:

(φ1,φ2,φ3)=(D1,D2,D3)x12−1x1x2x1x3x2x1x22−1x2x3x3x1x3x2x32−11x1x121x2x221x3x32.

That is

φ1=(x12+x1x2+x1x3−1)D1+(x1x2+x22+x2x3−1)D2+(x1x3+x2x3+x32−1)D3,φ2=x1(x12+x22+x32−1)D1+x2(x12+x22+x32−1)D2+x3(x12+x22+x32−1)D3,φ3=x1(x13+x23+x33−x1)D1+x2(x13+x23+x33−x2)D2+x3(x13+x23+x33−x3)D3.

And exp 𝓜 = {pdegφ₁, pdegφ₂, pdegφ₃} = {2, 3, 4}.

Example 3.6

Let 𝓜 be a hypersurface arrangement with the defining polynomial

Q(M)=x1x2x3(x1+x2)(x1+x3)(x2+x3)(x1−x2)(x1−x3)(x2−x3)(x12+x22+x32−1)(x12+x22+x32−2).

In this case, the hyperplane arrangement 𝓐 ⊆ 𝓜 is the Coxeter arrangement of typeB₃, it is a free arrangement with exp 𝓐 = {1, 3, 5}. By Theorem 3.2, 𝓜 is a free hypersurface arrangement andD(𝓜) has the basisφ₁, φ₂, φ₃as follows:

(φ1,φ2,φ3)=(D1,D2,D3)x12−2x1x2x1x3x2x1x22−2x2x3x3x1x3x2x32−2x12−1x1x2x1x3x2x1x22−1x2x3x3x1x3x2x32−1x1x13x15x2x23x25x3x33x35.

And exp 𝓜 = {pdegφ₁, pdegφ₂, pdegφ₃} = {5, 7, 9}.

Acknowledgement

The work was partially supported by NSF of China No. 11501051, No. 11601039 and Science and Technology Development Foundation of Jilin Province (No.20180520025JH and No.20180101345JC).

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Received: 2017-08-29

Accepted: 2018-01-15

Published Online: 2018-04-23

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

Articles in the same Issue

https://doi.org/10.1515/math-2018-0041

Keywords for this article

Hypersurface arrangement; Freeness; Hyperplane; Sphere

Creative Commons

BY-NC-ND 4.0