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Open Mathematics

formerly Central European Journal of Mathematics

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Volume 16, Issue 1

Issues

Volume 13 (2015)

On the freeness of hypersurface arrangements consisting of hyperplanes and spheres

Ruimei Gao / Qun Dai
  • Corresponding author
  • Department of Science, Changchun University of Science and Technology, Changchun, 130022, China
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/ Zhe Li
Published Online: 2018-04-23 | DOI: https://doi.org/10.1515/math-2018-0041

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

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

Define Di = /∂xi, 1 ≤ i, then D1,…,D\ℓ is a basis for Der𝕂(S) over S.

Definition 2.1

A nonzero element θ ∈ Der𝕂(S) is of polynomial degree p if θ = k=1fkDk and the maximum of the degrees of coefficient polynomials f1,…,f (get rid of 0) is p. 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)={θDerK(S)|θ(αX)αXSforallXM},

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

Definition 2.3

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

expM={pdegθ1,,pdegθ}.

Definition 2.4

Given derivations θ1,…,θD(𝓜), define the coefficient matrix M(θ1,…,θ) by Mi,j = θj(xi), thus

M(θ1,,θ)=θ1(x1)...θ(x1)...............θ1(x)...θ(x),

and θj = i=1Mi,jDi.

Definition 2.5

Let 𝓜 be a hypersurface arrangement, the product

Q(M)=XMαX

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

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 consisting of linear hyperplanes and spheres.

Lemma 2.6

If θ1,…,θD(𝓜), then det M(θ1,…,θ) ∈ Q(𝓜)S.

Proof

Let X ∈ 𝓜, and let X = ker(αX), then

detM(θ1,,θ)=fdetθ1(x1)...θ(x1)..........θ1(αX)...θ(αX)..........θ1(x)...θ(x),

If αX=k=1ckxk,thenf=ckR;IfαX=k=1(xkak)2r, then f = 2(xkak). For any case, det M(θ1,…,θ) is divisible by αX. Since X is arbitrary, det M(θ1,…,θ) ∈ Q(𝓜)S. □

Lemma 2.7

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

Mpq=xpxqifpq,xp2rifp=q,

where 1 ≤ p, qn, r ∈ ℝ. Therefore,

detMn=(r)n1(k=1nxk2r).

Proof

We will prove the lemma by induction on n.

  1. For the case n = 1, M1 = x12r, then det M1 = x12r.

  2. We assume that for the case n the result holds, that is det Mn=(r)n1(k=1nxk2r).

    For the case n + 1 we have

    Mn+1=MnNNTxn+12r,

    where N = (x1xn+1,…,xnxn+1)T. Therefore,

    detMn+1=detMnNNTxn+12+detMnOn×1NTr=detrEnNO1×nxn+12+(r)detMn=(r)n(xn+12)+(r)(r)n1(i=1nxi2r)=(r)n(k=1n+1xk2r),

    where On×1, On are the n × 1 and 1 × n null matrices respectively, and En is the n × n identity matrix. □

Lemma 2.8

Let

S={(x1,,x)k=1(xkak)2=rR+}

be the ( – 1)-dimensional sphere in with center (a1, a2, …, a) and radius r, define

θq=p=1fpqDp,1q,

where

fpq=(xpap)(xqaq)ifpq,(xpap)2rifp=q.

Then θ1,…,θD({𝓢}) and det M(θ1,…,θ) ≐ Q({𝓢}).

Proof

We can see

θq=p=1fpqDp=(xqaq)p=1(xpap)DprDq,

and

θq[k=1(xkak)2r]=2(xqaq)[k=1(xkak)2r][k=1(xkak)2r]S,

Thus θqD({𝓢}) for 1 ≤ q.

By Lemma 2.7, we obtain

detM(θ1,,θ)=(r)1[k=1(xkak)2r]Q({S}). □

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

Theorem 2.9

Given θ1,…,θD(𝓜), the following two conditions are equivalent:

  1. det M(θ1,…,θ) ≐ Q(𝓜).

  2. θ1,…,θ form a basis for D(𝓜) over S.

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) = fQ(𝓜) for some fS. 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 is the basis of X, then QXη1,…,QXηD(𝓜), where QX = Q(𝓜)/αX. Since each QXηi is an S-linear combination of η1,…,η, then there exists an × matrix N with entries in S, such that

M(QXη1,,QXη)=M(θ1,,θ)N.

Thus we have

Q(M)QX1detM(QXη1,,QXη)detM(θ1,,θ)S=fQ(M)S.

Therefore f divides QX1 for all X ∈ 𝓜. Since the polynomials {QX1}XM have no common factor, f ∈ ℝ*. □

Corollary 2.10

If 𝓢 is an ( − 1)-dimensional sphere in, 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 𝓜n be the hypersurface arrangement consisting of n spheres 𝓢1,…,𝓢n, where

Si={(x1,,x)k=1(xkak)2=ri,(a1,a2,,a)R,riR+}.

Define derivations φ1n,,φn by

M(φ1n,,φn)=AnAn1A1,

where Ai is an × matrix and the (p, q) entry of Ai is

(Ai)pq=(xpap)(xqaq)ifpq,(xpap)2riifp=q.

Then φ1n,,φn form a basis for D(𝓜n) and exp 𝓜n = {2n,…,2n}, where 2n appears ℓ times.

Proof

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

detAi=(ri)1[k=1(xkak)2ri].

Therefore,

detM(φ1n,,φn)=i=1ndetAii=1n[k=1(xkak)2ri]=Q(Mn).

Next, we will prove φinD(𝓜n) and deg φin = 2n for any 1 ≤ i 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=1φin+1(xp)Dp=p=1[q=1(An+1)pqφin(xq)]Dp=p=1[qp(xpap)(xqaq)φin(xq)+[(xpap)2rn+1]φin(xp)]Dp=p=1[(xpap)q=1(xqaq)φin(xq)]Dprn+1p=1φin(xp)Dp=q=1(xqaq)φin(xq)p=1(xpap)Dprn+1φin=12φin[q=1(xqaq)2]p=1(xpap)Dprn+1φin.

Therefore, for 1 ≤ i and 1 ≤ jn,

φin+1[k=1(xkak)2rj]=φin[q=1(xqaq)2]p=1(xpap)2rn+1φin[k=1(xkak)2rj]=φin[q=1(xqaq)2][p=1(xpap)2rn+1],

by induction hypothesis,

φin[q=1(xqaq)2]j=1n[k=1(xkak)2rj]S,

hence,

φin+1[k=1(xkak)2rj]j=1n+1[k=1(xkak)2rj]S.

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

pdegφin+1=pdegφin+2=2n+2=2(n+1).

We complete the induction, so by Saito’s criterion φ1n,,φn form a basis for 𝓜n, and exp 𝓜n = {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 basis θ1,…,θ, and exp 𝓐 = {d1,…,d}, Si0 is the sphere centered at origin:

Si0={(x1,,x)k=1xk2=riR+},1in,

and

Mn=A{S10,,Sn0},

Define derivations φ1n,,φn by

M(φ1n,,φn)=(AnAn1A1)M(θ1,,θ),

where Ai is an × matrix and the (p, q) entry of Ai is

(Ai)pq=xpxqifpq,xp2riifp=q,

then φ1n,,φn form a basis for D(𝓜n) and exp 𝓜n = {d1 + 2n,…,d + 2n}.

Proof

By Lemma 2.7, we obtain

detAi=(ri)1(k=1xk2ri).

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

detM(θ1,,θ)Q(A).

Therefore,

detM(φ1n,,φn)=i=1n(detAi)detM(θ1,,θ)i=1n(k=1xk2ri)Q(A)=Q(Mn).

Next, we will prove φinD(𝓜n) and deg φin = di + 2n for any 1 ≤ i by induction on n.

For the case n = 1,

φi1=p=1φi1(xp)Dp=p=1[q=1(A1)pqθi(xq)]Dp=p=1[qpxpxqθi(xq)+(xp2r1)θi(xp)]Dp=p=1[xpq=1xqθi(xq)]Dpr1p=1θi(xp)Dp=q=1xqθi(xq)p=1xpDpr1θi=q=1xqθi(xq)θEr1θi.

Since θE, θiD(𝓐), we have φi1D(𝓐) for any 1 ≤ i. And

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

In addition,

φi1(k=1xk2r1)=[q=1xqθi(xq)θEr1θi](k=1xk2r1)=q=1xqθi(xq)(2k=1xk2)2r1q=1xqθi(xq)=2(k=1xk2r1)q=1xqθi(xq)(k=1xk2r1)S

that is, φi1D({S10}). Therefore, φi1D(A)D({S10})=D(M1) for any 1 ≤ i.

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

φin+1=q=1xqφin(xq)θErn+1φin=12φin(q=1xq2)θErn+1φin.

By induction hypothesis,

φinD(Mn)D(i=1n{Si0}),

we obtain

φin(q=1xq2)j=1n(k=1xk2rj)S.

Therefore,

φin(q=1xq2)θED(i=1n{Si0}).

Combining θED(𝓐) with

D(Mn)=D(i=1n{Si0})D(A),

we conclude

φin(q=1xq2)θED(Mn).

Hence, φin+1D(Mn) since φinD(Mn).

In addition, we have

φin+1(k=1xk2rn+1)=φin(q=1xq2)k=1xk2rn+1φin(k=1xk2)=(k=1xk2rn+1)φin(q=1xq2)(k=1xk2rn+1)S,

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

φin+1D({Sn+10})D(Mn)=D(Mn+1),1i.

Moreover,

pdegφin+1=pdeg[φin(q=1xq2)θE]=pdegφin+2=di+2n+2=di+2(n+1),1i.

We complete the induction. □

Corollary 3.3

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

Proof

If 𝓐 is free we can obtain that 𝓜n is free directly from Theorem 3.2. Assume 𝓜n is free, 𝓐 ⊆ 𝓜n, then D(𝓜n) ⊆ D(𝓐). Let φ1,…,φ be a basis for D(𝓜n), then φiD(𝓐) for 1 ≤ i. Write φi=k0φi(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),,φ(d) such that

detM(φ1(d1),,φ(d))Q(A).

By Saito’s criterion, φ1(d1),,φ(d) 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, …, rn. Hence, if r1 = r2 = … = rn, Theorem 3.1 and Theorem 3.2 also hold. In this case, (𝓜, m) is a hypersurface arrangement with a multiplicity mi for each hypersurface in 𝓜, we call it hypersurface multiarrangement. As defined by G. Ziegler in [19], the module of derivations consists of θ such that θ(αi) ∈ αimiS.

Example 3.5

Let 𝓜 be a hypersurface arrangement with the defining polynomial

Q(M)=(x1x2)(x1x3)(x2x3)(x12+x22+x321).

In this case, the hyperplane arrangement 𝓐 ⊆ 𝓜 is the Coxeter arrangement of type A2, it is a free arrangement with exp 𝓐 = {0, 1, 2}. By Theorem 3.2, 𝓜 is a free hypersurface arrangement and D(𝓜) has the basis φ1, φ2, φ3 as follows:

(φ1,φ2,φ3)=(D1,D2,D3)x121x1x2x1x3x2x1x221x2x3x3x1x3x2x3211x1x121x2x221x3x32.

That is

φ1=(x12+x1x2+x1x31)D1+(x1x2+x22+x2x31)D2+(x1x3+x2x3+x321)D3,φ2=x1(x12+x22+x321)D1+x2(x12+x22+x321)D2+x3(x12+x22+x321)D3,φ3=x1(x13+x23+x33x1)D1+x2(x13+x23+x33x2)D2+x3(x13+x23+x33x3)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)(x1x2)(x1x3)(x2x3)(x12+x22+x321)(x12+x22+x322).

In this case, the hyperplane arrangement 𝓐 ⊆ 𝓜 is the Coxeter arrangement of type B3, it is a free arrangement with exp 𝓐 = {1, 3, 5}. By Theorem 3.2, 𝓜 is a free hypersurface arrangement and D(𝓜) has the basis φ1, φ2, φ3 as follows:

(φ1,φ2,φ3)=(D1,D2,D3)x122x1x2x1x3x2x1x222x2x3x3x1x3x2x322x121x1x2x1x3x2x1x221x2x3x3x1x3x2x321x1x13x15x2x23x25x3x33x35.

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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About the article

Received: 2017-08-29

Accepted: 2018-01-15

Published Online: 2018-04-23


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 437–446, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0041.

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© 2018 Gao et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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