Petersen plane arrangements and a surface with multiple 7-secants

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DOI 10.1515 / ADVGEOM.2009.019 de Gruyter 2009

Petersen plane arrangements and a surface with multiple 7-secants

Hirotachi Abo, Holger P. Kley^∗and Chris Peterson^∗

(Communicated by R. Miranda)

Abstract. We study configurations of 2-planes inP⁴ that are combinatorially described by the Petersen graph. We discuss conditions for configurations to be locally Cohen–Macaulay and de- scribe the Hilbert scheme of such arrangements. An analysis of the homogeneous ideals of these configurations leads, via linkage, to a class of smooth, general type surfaces inP⁴. We compute their numerical invariants and show that they have the unusual property that they admit (multiple) 7-secants. Finally, we demonstrate that the construction applied to Petersen arrangements with additional symmetry leads to surfaces with exceptional automorphism groups.

Key words. Plane arrangements, Petersen graph, automorphisms, 7-secants, general type surface, liaison, group actions, linkage.

2000 Mathematics Subject Classification. 14N20, 14J50, 14J25, 14M06, 05E20, 14J10

1 Introduction

Linkage (or liaison) theory can provide a bridge between combinatorially interesting varieties such as`-plane arrangements (equidimensional unions of projective subspaces of Pⁿ) and geometrically interesting ones: smooth projective varieties. This paper studies one such bridge, connecting the class of 2-plane arrangements inP⁴ whose incidence structure may be described by the Petersen graph with a previously unpublished class of smooth surfaces.

We begin, in Section 2, with a review of the relevant combinatorics of the Petersen graph. Next, in Section 3, we associate a graph to an arrangement: the vertices are the linear spaces in the arrangement and two vertices are joined by an edge if the spaces meet in greater than the expected dimension. Our study of the geometry of 2-plane arrangements begins with a necessary combinatorial condition (Corollary 3.10) on the associated graph to insure that the underlying arrangement is locally Cohen–Macaulay (lCM). Since

∗The second and third author were partially supported by NSF grant MSPA-MCS-0434351.

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linkage preserves the lCM property, such a result is essential if we are interested in linking to smooth surfaces.

The discussion leads naturally to the consideration of 2-plane arrangements whose associated graph is the Petersen graph, which begs the following questions: do such arrangements exist, and are they (generally) locally Cohen–Macaulay? In Section 4, we answer both questions in the affirmative (Propositions 4.9) via an explicit and combinatorially interesting construction and show (Proposition 4.10) that in fact, all lCM Petersen arrangements arise in this manner. The modular nature of the construction has immediate benefits. For example, we are able to give detailed information (Theorem 4.11) about the Hilbert scheme of such arrangements.

Furthermore, the explicit construction allows a detailed description of the homogeneous ideal of an lCM Petersen arrangement; this is the subject of Section 5. In particular, we find (Theorem 5.7) that the ideal always contains six quintic forms. Thus, it is not surprising that Petersen arrangements may be linked, via a general pair of quintics, to (smooth) degree fifteen surfaces inP⁴. Ellingsrud and Peskine [8] proved that there exists a constantd0such that every smooth surface inP⁴of degreed > d0 is of general type.

While the smallest possible value ford0is still an open problem, many believe that the answer isd0 =15. In any case, for surfaces inP⁴there seems to be a propensity towards special behavior at this degree. For instance, the lowest degree of a non-arithmetically Cohen–Macaulay smooth quasi-complete intersection surface inP⁴ occurs in degree 15.

It is a surface of non-general type and can be used to construct the Horrocks–Mumford bundle on P⁴. Through linkage (or by considering the zero locus of a section of the Horrocks–Mumford bundle), we obtain another very special object; the abelian surface of degree 10. Furthermore, the largest known degree of an abelian surface and of a bielliptic surface occurs in degree 15. This confluence of special behavior provided motivation to the authors to search in degree 15 (and degree 10) for other interesting phenomena. It should be noted that Ranestad and Aure (independently and unpublished) explored such a construction while considering the classification of low-degree surfaces inP⁴.

The degree fifteen surfaces, constructed through linkage from lCM Petersen arrangements, are the subject of Section 6. In particular, we show that they are smooth and of general type (Theorem 6.2). By exploiting the link between combinatorics and geometry once more, we show (Theorem 6.6) that the surfaces so obtained have the surprising property that they admit a one-dimensional family of 6-secant lines and fifteen 7-secant lines. This is unusual in that surfaces inP⁴ typically have a finite number of 6-secants and no 7-secants. In particular, we know of no other examples of surfaces inP⁴, of such low degree, that have as many 7-secants. Finally, in Theorem 7.1, we exploit the family of 2-planes as a combinatorial object, determine configurations with non-trivial automorphism groups, and show how to use such configurations to generate examples of smooth surfaces with non-trivial automorphism groups.

Note. All computer algebra scripts used in examples and computations throughout the present article may be found at [1].

Notation. For simplicity, fix an algebraically closed fieldK of characteristic zero. (In fact, the reader will note that all results will hold over sufficiently large algebraic exten- sions of the prime field in all but finitely many positive characteristics.)

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Forna non-negative integer, letPⁿ := P(Kⁿ⁺¹)denote the projective space of di- mensionnoverK. By alinear(sub)space(ofPⁿ) we meanP(V)for some vector sub- spaceV ⊂Kⁿ⁺¹. IfdimV =`+1, we also refer toP(V)as an`-planeinPⁿ. Finally, for linear subspacesΛ1=P(V)andΛ2 =P(W)ofPⁿ, letΛ1,Λ2 =P(V +W)denote the linear space they span.

Acknowledgments. The authors would like to thank Rick Miranda, Kristian Ranestad, Bernd Sturmfels and Ravi Vakil for stimulating conversations and questions.

2 The Petersen graph

In this section, we recall a few definitions from graph theory. We then define the Petersen graph and outline some of its standard properties.

Definition 2.1. LetΓbe a graph with vertex setV and edge setE.

• AmatchinginΓis a subsetE⁰ ⊂Esuch that no two edges inE⁰share a vertex.

• Aperfect matchinginΓis a matching that covers every vertex inΓ.

• Anindependent set of verticesofΓis a subsetS⊂V such that no pair of vertices in Sform an edge ofΓ.

• Amaximal independent set of verticesis an independent set of vertices which is not properly contained in any other independent set of vertices.

• Thegirth ofΓ is the length of a shortest, non-trivial cycle inΓ. IfΓcontains no non-trivial cycles, it is said to haveinfinite girth.

By a slight abuse of notation, we write{0,1,2,3,4} =Z/(5)for (the elements of) the integers modulo 5.

ThePetersen graphis the graphΓPetewith vertex set

V(Γ_Pete) :={0,1,2,3,4,0⁰,1⁰,2⁰,3⁰,4⁰}={i, i⁰:i∈Z/(5)}

and edge set

E(Γ_Pete) :={01,12,23,34,40,00⁰,11⁰,22⁰,33⁰,44⁰,0⁰2⁰,1⁰3⁰,2⁰4⁰,3⁰0⁰,4⁰1⁰}

={i(i+1), ii⁰, i⁰(i+2)⁰:i∈Z/(5)}.

One standard way to realizeΓ_Peteis shown in Figure 1. Notice thatΓ_Petehas diameter 2 and girth 5.

Henceforth, let ρ ∈ Aut(Γ_Pete)be the “rotation by 2π/5,” i.e., the automorphism whose action onV(Γ_Pete)is given by

ρ(i) =i+1 andρ(i⁰) = (i+1)⁰fori∈Z/(5),

letτ ∈ Aut(ΓPete)be the automorphism whose action on V(ΓPete) is given (in cycle notation) by

τ:= (00⁰)(13⁰42⁰)(21⁰34⁰)

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0 2 0

⁰

2

⁰

4

⁰

1

⁰

4

3

1

3

⁰

Figure 1. The Petersen graph.

and letω∈Aut(ΓPete)be the automorphism whose action onV(ΓPete)is given (in cycle notation) by

ω:= (21⁰)(34⁰)(2⁰3⁰).

Referring to Figure 1,τmore or less “turnsΓ_Peteinside-out.” Also, observe that σ:=τ²= (14)(23)(1⁰4⁰)(2⁰3⁰)∈Aut(Γ_Pete)

is the “reflection across the vertical.”

We identify the maximal independent vertex set V₀:={1,4,2⁰,3⁰} ⊂V(Γ_Pete)

ofΓ_Peteand observe that any 4-element maximal independent vertex set ofΓ_Peteis one of Vj :=ρ^jV0

forj =0, . . . ,4. Notice that each vertex ofΓPetelies in precisely two of theVjand that any two of theVjintersect in exactly one vertex ofΓPete. Consequently,Aut(ΓPete)acts faithfully on{V0, . . . , V₄}. Furthermore,ρacts as a 5-cycle on{V0, . . . , V₄}whileωacts as the transposition(V₁V₄)on{V0, . . . , V₄}, so we recover the well-known

Proposition 2.2. The automorphism group of the Petersen graph is isomorphic to the symmetric group on5letters.

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Similarly, we identify the perfect matching

E₅ :={00⁰,11⁰,22⁰,33⁰,44⁰} ⊂E(Γ_Pete), (2.1) and observe that every perfect matching ofΓPeteis eitherE5or one of

E_j:=ρ^jωE₅ forj=0, . . . ,4. (2.2) Clearly,Aut(ΓPete)acts transitively on{E0, . . . , E5}, so the group

G:= Stab(E₅)≤Aut(Γ_Pete)

has order 20. One easily checks thatρ, τ∈G, and concludes that

G=hρ, τi ≤Aut(Γ_Pete). (2.3) Finally, we identify

D:=hρ, σi=hρ, τ²i ≤G, (2.4) which is isomorphic to the dihedral group of order 10.

3 Arrangements and their incidence graphs

In this section we define our basic objects of study and prove a simple but fundamental proposition about their algebro-geometric structure.

We begin with a trivial but useful observation about linear spaces:

Remark 3.1. LetP,QandR⊂Pⁿbe linear spaces. If

dim(P∩R) + dim(Q∩R)−dimR≥dim(P∩Q),

then(P ∩Q) ⊂ R. In particular, ifR is a 2-plane meetingP andQin lines and if P∩Q={p}, thenp∈R.

Definition 3.2. For positive integers ` < n, an `-plane arrangement in Pⁿ is a finite collection of`-dimensional linear subspaces ofPⁿ.

Notation 3.3. LetA be an`-plane arrangement inPⁿ, letS = K[x₀, . . . , x_n]be the homogeneous coordinate ring ofPⁿ, letH ⊂Pⁿbe a hyperplane, and letp∈Pⁿ. Then we write

• V_A:=S

Λ∈AΛfor the corresponding projective variety inPⁿ,

• I_A⊂Sfor the homogeneous ideal ofV_AandI_Afor its ideal sheaf inO_Pn,

• A_p := {Λ ∈ A : p ∈ Λ} for the subarrangement of `-planes in Awhich pass throughp,

• H∩ A:={H∩Λ : Λ∈ A}for thehyperplane sectionofAbyH, and

• p,A:={p,Λ : Λ∈ A}for theconeoverAwith vertexp.

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Remark 3.4. We freely use three simple facts about cones and hyperplane sections of

`-plane arrangements:

• IfAis an`-plane arrangement inPⁿand ifH ⊂Pⁿ is a hyperplane not containing any of the members ofA, thenH∩ Ais an(`−1)-plane arrangement inH.

• IfAis an(`−1)-plane arrangement in a hyperplaneH ⊂Pⁿ and ifp /∈H, then p,Ais an`-plane arrangement inPⁿ.

• Forp∈PⁿandH ⊂Pⁿa general hyperplane,Ap=p, H∩ Ap.

In general, we will say that an arrangement has a certain geometric property if its corresponding variety does. In particular, we have:

Definition 3.5. LetAbe an`-plane arrangement inPⁿ and letp ∈ V_A be a point corresponding to the prime idealp⊂S. We say thatAisarithmetically Cohen–Macaulay (aCM) if the homogeneous coordinate ringS/IAis a Cohen–Macaulay ring. We say that Aislocally Cohen–Macaulay(lCM)atpif the local ring(S/IA)p is Cohen–Macaulay.

Finally, we say thatAislocally Cohen–Macaulayif it is lCM at all closed pointsp∈VA. The following proposition is useful in understanding when an arrangement is lCM at a point.

Proposition 3.6. LetAbe an`-plane arrangement inPⁿwith`≥2. Letp∈VA. LetH be a hyperplane not containing any plane ofA. The following are equivalent:

(i) Ais lCM atp.

(ii) Apis lCM atp.

(iii) Apis aCM.

(iv) H∩ Apis aCM.

Proof. The equivalence of (i) and (ii) is clear. The equivalence of (iii) and (iv) is an immediate corollary of a result of Huneke and Ulrich [6]. Finally, the equivalence of (ii) and (iv) follows becauseAp is a cone overH∩ Ap, and a variety is aCM if and only if the cone over the variety is lCM at the vertex of the cone. 2 Example 3.7. Consider distinct 2-planes P₁, P₂, P₃, P₄ ⊂ P⁴ not all contained in a hyperplane. Suppose thatP₁ ∩P₃ = P₂∩P₄ = P₁∩P₄ = {p} consists of a single point whileP₁∩P₂,P₂∩P₃, andP₃∩P₄are lines (necessarily passing throughpby Remark 3.1). Then the arrangement{P₁, P₂, P₃, P₄}is a cone over a projection intoP³ of a degenerate rational normal curve of degree 4. Such curves are not aCM and hence the arrangement{P₁, P₂, P₃, P₄}is not lCM.

It is useful to keep track of those pairs of planes in an`-plane arrangement that are in special position with respect to each other. We represent this data via a graph.

Definition 3.8. If Ais an `-plane arrangement in Pⁿ, theincidence graph ofA is the graphΓ(A), with vertices the planes of A and edges the pairs of planes with special intersection. In other words,Γ(A)has vertex setA, and edge set{ΛΛ⁰ : dim Λ∩Λ⁰ >

n−2`}.

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Remark 3.9. Because it captures non-generic behavior only betweenpairsof`-planes, the graphΓ(A) may not completely characterize the geometry of the arrangementA.

For example, inP⁴, both the arrangement of three 2-planes all containing a common line and the arrangement of three 2-planes meeting pairwise in three distinct lines have the complete graph on three vertices as their incidence graphs. (These arrangements could, however, be distinguished by anincidence complex.)

Corollary 3.10. LetA be a2-plane arrangement inP⁴ and let p ∈ V_A. IfΓ(Ap)is disconnected, thenAis not lCM atp. As a partial converse, ifΓ(Ap)is connected and of order at most three, thenAis lCM atp.

Proof. LetHbe a hyperplane not passing throughp. ThenApis a cone over the arrange- mentL=Ap∩H of lines inH =P³, and by Proposition 3.6,Ais lCM atpif and only ifLis aCM. NowΓ(Ap)is connected if and only ifV_L is connected, and disconnected curves are not aCM. Conversely, supposeVL is connected and has at most three compo- nents. ThenVLis either a complete intersection, a cone over three non-collinear points in a plane, or a degenerate twisted cubic, all of which are aCM. 2 Remark 3.11. LetAbe a 2-plane arrangement inP⁴. Corollary 3.10 suggests that we shouldexpectAto be lCM ifΓ(A)is connected and of diameter at most 2. On the other hand, Example 3.7 suggests that if the diameter ofΓ(A)exceeds 2, thenAmay well fail to be lCM. IfΓ(A)is disconnected,Afails to be lCM in all cases.

4 Petersen arrangements

For the remainder of this paper, aplanemeans a 2-plane. In Remark 3.11, we observed that a plane arrangementAinP⁴, constructed in such a way thatΓ(A)has diameter 2, is expected to be lCM. Among the graphs of finite girth, only those ofgirth ≤ 5 may havediameter≤2. Thus, graphs of girth 5 and diameter 2 are a natural place to look for interesting examples of arrangements.

The simplest such graph is the pentagon, and one can readily construct a plane ar- rangementAinP⁴such thatΓ(A)is isomorphic to the pentagon. This plane arrangement appears as a degenerate case in the family containing the elliptic quintic scrolls. Here elliptic quintic scrolls are ruled surfaces over elliptic curves embedded inP⁴ in such a way that all the fibers have degree 1. These surfaces determine one of the few known families of smooth irregular surfaces inP⁴. The (degenerate) elliptic quintic scrollV_A is cut out by five cubic hypersurfaces and can be linked in the complete intersection of two such cubic hypersurfaces to the Veronese surface (which is known as the only smooth non-degenerate surface inP⁴that is not linearly normal).

As we observed in Section 2, the Petersen graphΓ_Pete likewise has diameter 2 and girth 5.

Definition 4.1. A (two-)plane arrangementAinP⁴is aPetersen arrangementifΓ(A)' ΓPete. Alabelingof a Petersen arrangementAis an isomorphismψ: ΓPete

−∼→Γ(A).

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Several natural questions are: Do Petersen arrangements exist? If so, how big is the family of such arrangements? Do lCM Petersen arrangements exist? If so, can liaison theory be used to produce (smooth) surfaces from Petersen arrangements? Do Petersen arrangements with extra structure (such as a prescribed automorphism group) exist? The remainder of this article addresses these questions.

4.1 Construction of Petersen arrangements. Let L = {Li}_i∈_Z_/(5) be an indexed collection of lines inP⁴. Suppose thatLsatisfies

Condition 1. L0, . . . , L4are pairwise skew.

Then for each pair{i, j} of distinct elements ofZ/(5), Li andLj span a hyperplane Hi,j:=Li, Lj. Clearly,Hi,j=Hj,i.

Suppose thatLalso satisfies

Condition 2. No three lines ofLare contained in a hyperplane.

Lemma 4.2. The following are equivalent conditions on a collectionLsatisfying Condi- tion1:

(i) Condition2.

(ii) For all choices of distincti, j, k∈Z/(5), the hyperplanesHi,jandHi,kare distinct.

(iii) For all choices of distincti, j, k ∈Z/(5), there is a unique “trisecant” lineMi,j,k

meetingL_i,L_jandL_k.

(iv) For all choices of three distincti, j, k ∈Z/(5), there is a unique plane containing L_iwhich meetsL_jandL_k.

Proof. This is immediate. The key observations are that

Mi,j,k=Hi,j∩Hj,k∩Hk,i, (4.1)

and that

H_i,j∩H_i,k=L_i, M_i,j,k (4.2)

is the unique plane containingL_iand meetingL_jandL_k. 2 For eachi∈Z/(5), we define two planes

P_i:=H_i,i+1∩H_i,i−1 and P_i⁰ :=H_i,i+2∩H_i,i−2 (4.3) and set

A=A(L) :={Pi, P_i⁰ :i∈Z/(5)}. (4.4) We would like to show thatAis a Petersen arrangement. There are, however, two potential problems: Γ(A)could containΓ_Pete as a proper subgraph, and some of the planes inA could coincide. To eliminate these possibilities, we require thatLadditionally satisfies Condition 3. No plane containing a line ofLmeets three other lines ofL.

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Lemma 4.3. The following are equivalent conditions on a collectionLsatisfying Condi- tions1and2:

(i) Condition3.

(ii) L_`∩L_i, M_i,j,k=∅for all choices of distincti,j,k,`∈Z/(5).

(iii) H_i,j∩H_i,k∩H_i,`=L_ifor all choices of distincti,j,k,`∈Z/(5).

(iv) M_i,j,k∩M_i,j,`=∅for all choices of distincti,j,k,`∈Z/(5).

Proof. This is immediate in light of Lemma 4.2. 2

We can now establish

Proposition 4.4. LetL ={L_i}_i∈_Z_/(5)be an indexed collection of lines inP⁴satisfying Conditions1–3. ThenA(L)is a Petersen arrangement with labelingψ_L: Γ_Pete →Γ(A) given byψ_L(i) =Piandψ_L(i⁰) =P_i⁰. Moreover,Pi∩P_i⁰ =Li.

Proof. Sinceψ_Lis a bijection on vertices, it suffices to show that it induces a bijection of edges, which we establish as a series of claims. Leti∈Z/(5).

Claim.dim(Pi∩Pi+1) =1.

Proof. By construction,Pi∩Pi+1 =H_i−1,i∩Hi,i+1∩Hi+1,i+2, which is as least one-dimensional. On the other hand,Pi containsLi andPi+1 containsLi+1 and since these lines are skew by Condition 1,Pi6=Pi+1.

Claim.dim(P_i⁰∩P_i+2⁰ ) =1.

Proof.This is essentially identical to the previous case.

Claim.dim(P_i∩P_i+2) =0.

Proof. Combining (4.2) with (ii) of Lemma 4.3 shows thatP_i=L_i, M_i−1,i,i+1does not meetL_i+2⊂P_i+2. It follows thatP_iandP_i+2cannot meet in a line.

Claim.dim(P_i⁰∩P_i+1⁰ ) =0.

Proof.This is essentially identical to the previous case.

Claim.Pi∩P_i⁰=Li.

Proof.Using (iii) of Lemma 4.3,

Pi∩P_i⁰ =Hi−1,i∩Hi,i+1∩Hi−2,i∩Hi,i+2=Li∩Hi,i+2=Li. Claim.dim(Pi∩P_j⁰) =0 for allj6=i.

Proof. We consider two cases: Pi∩P_i+1⁰ andPi∩P_i+2⁰ . The other two cases are essentially identical. Keeping in mind that indices are calculated modulo five, we have

P_i∩P_i+1⁰ =H_i−1,i∩H_i,i+1∩H_i−1,i+1∩H_i+1,i−2.

Notice that three terms involve the indexi+1. Applying (iii) of Lemma 4.3, we reduce to

Pi∩P_i+1⁰ =H_i−1,i∩Li+1

which must be a point by Condition 2.

For the second case, an analogous argument shows that

P_i∩P_i+2⁰ =H_i−1,i∩H_i,i+1∩H_i,i+2∩H_i+2,i−1=H_i+2,i∩L_i

is also a point. 2

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Remark 4.5. The symmetric groupS =S_Z_/(5)acts naturally on the indexing of a col- lectionL = {Li}_i∈_Z_/(5) of lines inP⁴ satisfying Conditions 1–3. LetF ≤ S be the subgroup of order 20 generated by the 5-cycle(01234)and the 4-cycle(1342). Recall the groupG ≤ Aut(ΓPete)of (2.3) and observe that the mapping (01234) 7→ ρ and (1342) 7→ τ extends to a group isomorphism f: F −^∼→ G. Now for s ∈ S, we have A(sL) = A(L)if and only if s ∈ F, in which case the corresponding labelings are related byψ_sL=ψ_L(f(s)).

Of course, labeled Petersen arrangements lead to line arrangements via a reverse construction. Specifically, letAbe a Petersen arrangement andψ: ΓPete

−∼→Γ(A)a labeling.

Then we have the indexed collection of lines

L(A, ψ) ={ψ(i)∩ψ(i⁰)}_i∈Z/(5).

Remark 4.6. The lines of L(A, ψ) correspond to the edges in the perfect matching ψ(E₅) ={ψ(ii⁰)}_i∈_Z_/(5)ofΓ(A). Likewise, forα∈Aut(Γ_Pete), the lines ofL(A, ψα) correspond to the edges in the perfect matchingψα(E₅)ofΓ(A). Keeping in mind the notation of Remark 4.5,L(A, ψα) =L(A, ψ)as setsif and only ifα∈G= Stab(E₅)≤ Aut(ΓPete), in which caseL(A, ψα) =f⁻¹(α)L(A, ψ)as indexed sets.

Unfortunately, not every Petersen arrangement can be obtained as in (4.4):

Example 4.7. LetAbe any Petersen arrangement and letH ⊂ P⁴ be a general hyperplane. Choose a pointp /∈H. Then

B:=p,(A ∩H) is a Petersen arrangement. Choose a labeling ψ: ΓPete

−∼→ Γ(B). Then the lines of L(B, ψ)all pass through the vertexpand thus do not satisfy Condition 1. Sinceψwas arbitrary, this shows that it is impossible to realizeBasA(L)for any indexed collection Lof lines.

4.2 LCM Petersen arrangements. LetL ={Li}_i∈Z/(5)be an indexed collection of lines inP⁴satisfying Conditions 1–3, and letA=A(L)as in (4.4). While Proposition 4.4 shows thatAis a Petersen arrangement,Aneed not be locally Cohen–Macaulay. In light of Proposition 3.6, the study of lCM plane arrangements inP⁴can be reduced to the study of aCM line arrangements inP³. For our purposes, the following will suffice:

Lemma 4.8. LetLbe an arrangement of lines inP³. IfΓ(L)is isomorphic to either (i) the Petersen graphΓ_Pete,

(ii) the pentagonC₅, or

(iii) the complete bipartite graphK_1,3, thenVLis not arithmetically Cohen–Macaulay.

Proof. In case (i),V_Lis a curve of degree 10 and arithmetic genus 6. In case (ii),V_Lis a curve of degree 5 and arithmetic genus 1. Finally, in case (iii),V_Lis a curve of degree 4 and arithmetic genus 0. In each case, Riemann–Roch now implies thatH¹(IVL(1))6=0.

Consequently,VLis not aCM. 2

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For instance, the arrangementBof Example 4.7, being a cone over a non-aCM line arrangement, fails to be lCM by Proposition 3.6.

As in the previous section, letL={Li}_i∈_Z_/(5)be an indexed collection of lines inP⁴ satisfying Conditions 1–3, and letA=A(L)be the resulting Petersen arrangement.

LetP be a plane inA. Then there exist three planes inAthat are adjacent toP in Γ(A). Conditions 1–3 do not imply that the lines along which these planes meetPform a triangle inP. Indeed, if these three lines all pass through a singlep∈P, thenApfails to be lCM atpby case (iii) of Lemma 4.8. Keeping in mind the definition of the “trisecant”

linesM_i,j,kin Lemma 4.2, we therefore consider

Condition 4. For alli ∈ Z/(5), both M_{i−2,i−1,i}∩M_i,i+1,i+2 = ∅ andM_i−2,i,i+1∩ M_i−1,i,i+2=∅.

Notice the similarity of this Condition 4 to Condition 3 via (iv) of Lemma 4.3. Finally, in order to rule out non-lCM points arising as vertices of cones over the line arrangement of case (ii) of Lemma 4.8, we consider:

Condition 5. \

i∈Z/(5)

Hi,i+1= \

i∈Z/(5)

Hi,i+2=∅.

Proposition 4.9. LetLbe an indexed collection of lines inP⁴satisfying Conditions1–5.

ThenA=A(L)is a locally Cohen–Macaulay Petersen arrangement.

Proof. By Corollary 3.10, it suffices to show thatΓ(Ap)is connected of order at most 3 for allp∈V_A. All cases of this assertion are either obvious (i.e., whenplies in exactly one plane ofAor in exactly two planes ofAthat meet in a line) or can be deduced from the

Claim. IfP,Q∈ AwithP ∩Q={p}, then there exists a uniqueR∈ Ameeting bothPandQin lines, in which caseAp={P, Q, R}.

Note thatpcannot lie in three or more planes meeting pairwise in lines as the Petersen graph contains no 3-cycles. Thus, ifApconsists of three or more planes, there must be at least two of them which meet in a point, in which case the claim applies.

Proof of claim. P∩Q={p}means thatP andQare not adjacent inΓ(A). Since the Petersen graph has no 4-cycles and any two of its vertices lie on a 5-cycle, there is a uniqueR ∈ Ameeting bothP andQin lines. These two lines meet at a point ofR which, by Remark 3.1 is necessarilyp, so we are reduced to showing that no otherT ∈ A containsp. Suppose to the contrary, that such aTexists. Then either

(i) Tis adjacent to eitherPorQ,

(ii) Tis adjacent to none ofP,QandR, or (iii) TandRare adjacent.

In case (i), assume without loss of generality thatP andT are adjacent. Then there exists a unique planeU ∈ Asuch thatC:= Γ({T, P, R, Q, U})is a 5-cycle inΓ(A). By Remark 3.1,p∈Uas well, so

P∩R∩Q∩U∩T ={p}. (4.5)

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Now eitherC= Γ({Pi :i∈Z/(5)})orC= Γ({P_i⁰ :i∈Z/(5)})orCcontains two of the edges ofE5. In the first two cases, (4.5) contradicts Condition 5, while in the third, it contradicts Condition 1.

In case (ii), P andQ are distance 2 fromT, and there exist unique planesU and U⁰ ∈ Aadjacent toTsuch thatPandQare adjacent toUandU⁰respectively inΓ(A). In other words,C:= Γ({T, U, P, R, Q, U⁰})is a 6-cycle inΓ(A). By repeated applications of Remark 3.1, all planes ofC must pass through p. Since any 6-cycle in Γ_Pete must contain two of the edges ofE₅, this contradicts Condition 1 as in the previous case.

Finally, in case (iii), eitherR=P_i orR =P_i⁰for somei ∈Z/(5). IfR=P_i, then without loss of generality,P =P_i−1,Q=P_i+1andT =P_i⁰. Then, using (4.3) and (4.1),

{p}=P∩Q∩R∩T =M_{i−2,i−1,i}∩Mi,i+1,i+2,

contradicting Condition 4. IfR=P_i⁰, an analogous argument again leads to a contradic-

tion to Condition 4. 2

As it turns out, every locally Cohen–Macaulay Petersen arrangement arises as a result of our construction:

Proposition 4.10. Let A be a locally Cohen–Macaulay Petersen arrangement and ψ: ΓPete

−∼→Γ(A)a labeling. ThenL(A, ψ)satisfies Conditions1–5,A=A(L(A, ψ)) andψ=ψ_L(A,ψ).

Proof. Recall that we setPi =ψ(i),P_i⁰=ψ(i⁰)andLi=Pi∩P_i⁰for alli∈Z/(5)and thatL={Li}_i∈_Z_/(5).

For any distincti, j ∈Z/(5), there is a uniqueQ∈ Asuch that{Pi, P_i⁰, Pj, P_j⁰, Q}

are the vertices of a 5-cycle inΓ(A). IfLi∩Lj6=∅, then using Remark 3.1,{Pi, P_i⁰, Pj, P_j⁰, Q}is a cone over a line arrangement with incidence graph isomorphic to the pentagon C₅. If any other plane ofAalso passes through the vertexpof this cone, then, by repeated application of Remark 3.1, all planes ofAhavepin common. Thus, eitherApis a cone over a line arrangement with pentagon incidence graph, or it is a cone over a line arrangement with Petersen incidence graph. In either case,Ais not lCM at pby Lemma 4.8.

Thus,L_i∩L_j = ∅, Condition 1 is satisfied and the hyperplanesH_i,j =L_i, L_j are defined.

SincePiandPi+1are adjacent inΓ(A)for alli, they span a hyperplane. This hyperplane must containLiandLi+1so that

Hi,i+1 =Pi, Pi+1 for alli∈Z/(5). (4.6)

Similarly,

Hi,i+2=P_i⁰, P_i+2⁰ for alli∈Z/(5). (4.7)

Now supposeLfails Condition 2. By Lemma 4.2, there exist distincti, j, k∈Z/(5) such thatLi,Lj, andLk lie in some hyperplaneH. Without loss of generality, we may assume that eitheri=j−1 andk=j+1 or thati=j−2 andk=j+2. Using (4.6), the first case yields

Pj−1, Pj =Hj−1,j =H =Hj,j+1=Pj, Pj+1.

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Thus,P_j−1 andPj+1 must meet in a line, contradicting their non-adjacency in Γ(A).

In the second case, a parallel argument using (4.7) shows that P_j−2⁰ meets P_j+2⁰ in a line, contradicting their non-adjacency inΓ(A). Thus, Condition 2 is satisfied andPi = Hi−1,i∩Hi,i+1whileP_i⁰=Hi−2,i∩Hi,i+2for alli.

Suppose Condition 3 fails. Then by Lemma 4.3, there exist distinct indicesi, j, k, `∈ Z/(5)such thatQ=Hi,j∩Hi,k∩Hi,`has dimension 2. Up to obvious reindexing, there are just two cases to consider: eitherj =i−1,k=i+1 and` =i+2, orj =i+1, k=i+2 and`=i+3. In the first of these,P_i =Q⊂H_i,i+2 ⊃P_i+2⁰ , contradicting the non-adjacency ofP_i andP_i+2⁰ . In the second case, Q = P_i⁰ ⊂ H_i,i+1 ⊃ P_i+1, contradicting the non-adjacency ofP_i+1andP_i⁰. Thus, Condition 3 is satisfied. Note that we have also established thatA=A(L(A, ψ))andψ=ψ_L(A,ψ).

Finally, ifL(A, ψ)fails Conditions 4 or 5, thenV_Acontains a pointpsuch thatA_p is a cone over a line arrangement with incidence graph isomorphic toΓ_Pete,C₅, orK_1,3,

contradicting, via Lemma 4.8, thatAis lCM. 2

4.3 Examples. Conditions 1–5 are non-empty open conditions on the set of indexed collectionsL of five lines inP⁴, so lCM Petersen arrangements exist. To construct a specific example, choose random 2×5 matricesA_iof rank 2 to represent the linesL_i∈ L and check by matrix manipulation and rank computations of matrices whetherLsatisfies Conditions 1–5.

Alternatively, use a computer algebra system such asMACAULAY2. Represent the linesL_ivia their ideals, each of which is generated by picking three random linear forms in five variables. In sufficiently large characteristic, Conditions 1–5 can be verified ideal- theoretically. Also, one can check thatAis locally Cohen–Macaulay directly by looking at the second syzygy module ofI_A; see [1].

Once such a sufficiently generic example has been constructed, one can check compu- tationally that at the corresponding point, the Zariski tangent space to the Hilbert scheme ofP⁴has dimension 30. Again, see [1]. Now combining Remarks 4.5 and 4.6 and Propo- sitions 4.9 and 4.10, we see that a parameter space for locally Cohen–Macaulay Petersen arrangements isU/Aut(ΓPete)whereU ⊂Q

Z/(5)G(1,4)is the open set of indexed col- lections of lines satisfying Conditions 1–5 andAut(ΓPete)acts freely, though only the subgroupGacts by permutation of indices.

Given thatdimU =30, we have established:

Theorem 4.11. Let A be a general locally Cohen–Macaulay Petersen arrangement.

ThenV_A lies on a unique component of the Hilbert scheme ofP⁴ which is birational toU/Aut(ΓPete). Furthermore, the Hilbert scheme is smooth atVAof dimension30and VAcannot be smoothed inP⁴.

Example 4.12. Recall the dihedral subgroupD =hρ, σiofAut(Γ_Pete)defined in (2.4).

OverK, construct a 5-dimensional representationV ofDwith basis{ei}_i∈_Z_/(5), by let- tingρact as the 5-cycle(e0e1e2e3e4)andσas the product of transpositions(e1e4)(e2e3).

Thenσhas eigenvalues 1 and−1 with corresponding eigenspaceshe0, e1+e4, e2+e3i, andh−e1+e4,−e2+e3i, respectively. Consider theσ-invariant two-dimensional sub- spaceW ofV spanned byv1=e0+e2+e3andv2=−e1−e2+e3+e4.

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Now in the induced projective representationP(V), the lineL0 := P(W)is alsoσ- invariant. Fori∈Z/(5), setLi :=ρⁱ(L0). Then as a set,L ={Li}_i∈_Z_/(5)is invariant under the action ofD ,→ PGL(V), and it is straightforward to show that L satisfies Conditions 1–5. ThusA(L) is a locally Cohen–Macaulay Petersen arrangement (see [1] for a direct verification), invariant under the action ofD. Note, however, that the corresponding labelings change in the obvious way:ψ_δL=ψ_L◦δforδ∈D.

5 Generators ofIA

LetAbe a locally Cohen–Macaulay Petersen arrangement and fix a labelingψ: ΓPete→ Γ(A). As before, letPi=ψ(i)andP_i⁰ =ψ(i⁰)fori∈Z/(5). The idealIAis obtained as the intersection of the ideals of the planes inAand, by means of an example, one can determine that a minimal set of generators forI_Aconsists of six quintics and five sextics.

The main goal of this section is to provide a combinatorial description of these generators.

Among other things, this description will allow us to show that there are exactly five 6- secant lines toV_A.

Letuandvbe adjacent vertices ofΓ_Pete ande= uvthe corresponding edge. Then the planesψ(u), ψ(v)∈ Aspan a hyperplane which we denote

H^u,v=H^e:=ψ(u), ψ(v).

For each of the six perfect matchingsE_i of Γ_Pete (see Section 2), we define a quintic hypersurface inP⁴by

Qi:= [

e∈Ei

H^e.

Letq_i ∈Sbe a quintic polynomial that definesQ_i. We will prove thatq_i ∈(I_A)₅. For eachi∈Z/(5), define

l_i:=H^i,i⁰∩H⁽ⁱ⁺¹⁾⁰^,(i−1)⁰∩H(i+2),(i−2). (5.1) Lemma 5.1. The subspaceslisatisfy:

(i) l_iis a line.

(ii) l_i6⊂V_A.

(iii) Theliare pairwise disjoint.

Proof. The techniques are similar and rely on knowing thatAis lCM and Petersen. For (i), without loss of generality consider the casei=0. The three hyperplanesH^0,0⁰,H¹⁰^,4⁰ andH^2,3are distinct. (If, e.g.,H^0,0⁰ =H^2,3, thenP₀andP₂meet in a line, contradicting thatAis Petersen.) So ifl₀is not a line, it is a plane. Suppose that were the case. Consider the three pointsP₀∩P₁⁰,P₁⁰∩P₂andP₂∩P₀; by Remark 3.1, all three must lie inP₁. SinceAis lCM, they cannot coincide. (See case (iii) of Lemma 4.8.) They cannot be collinear since no two ofP₁⁰,P2, andP0meet in a line. Thus, they spanP1. Since each of them lies in two of the hyperplanes definingl0, they also spanl0and we concludeP1=l0. But this would give, e.g.,dimP1∩P₀⁰ =1, again contradicting thatAis Petersen.

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To prove (ii), suppose to the contrary, for example, thatl0⊂P0. (The remaining cases are similar.) Then the four pointsP0∩P₁⁰,P0∩P2,P0∩P₄⁰andP0∩P3lie onl0, and they must be distinct sinceAis lCM. Now by Remark 3.1, the first two of these lie on P1 and we deducel0 =P0∩P1. Working with the third and fourth points, we see that l0=P0∩P4as well, contradicting thatdimP1∩P4consists of a point.

Finally, to prove (iii), suppose, for example, thatl_iandl_i+1share a point. ThenP_i⁰, P_i+1⁰ andP_i−2 must intersect, which, as we saw in Section 4.2, contradicts the assump- tion thatAis an lCM Petersen arrangement. The remaining cases follow in a similar

manner. 2

Finally, suppose that the arrangementAsatisfies the following open condition:

Condition 6. For each perfect matchingE_iofΓ_Pete, \

e∈Ei

H^e=∅.

Then for eachi∈Z/(5),

pi:= \

e∈Ei\E5

H^e,

is a point and we have the following lemma:

Lemma 5.2. Suppose thatAsatisfies Condition6. Then the union ofVA, the five lines {li}and the five points{pi}is cut out set theoretically by the five quintic hypersurfaces Q₀, . . . , Q₄.

Proof. LetQ =Q₀∩ · · · ∩Q₄denote the intersection of the five quintic hypersurfaces and letI_Q = (q₀, . . . , q₄)denote the ideal generated by the five quintics. Then a pointp ofP⁴lies inQif and only if for eachi∈Z/(5), there exists at least onee∈E_isuch that H^econtainsp. A pointpofQlies inV_Aif and only if for two distinct elementsiandj ofZ/(5), there existe∈Eiandf ∈Ejsuch thatH^eandH^f containpand intersect in one of the planes inA. (The latter condition is equivalent to the condition that the edges eandfshare a common vertex.)

Thus, a pointplies in the closure of the complement ofV_AinQ(whose ideal is the ideal quotientIQ : I_A) ifp∈T

e∈EH^efor some matchingE(not necessarily perfect) ofΓPetewhich is minimal among matchingsE such thatE∩Ei 6=∅for alli ∈Z/(5).

Any suchEis eitherEi\E5,{ii⁰,(i+1)⁰(i−1)⁰,(i+2)⁰(i−2)⁰}, orE5. However, by Condition 6, the last subset cannot occur. Thus, the residual variety consists precisely of the union of thepiand theli, which completes the proof. 2 Theorem 5.3. The union ofV_Aand the five lines{l_i}is scheme-theoretically cut out by the six quintic hypersurfacesQ₀, . . . , Q₅.

Proof. For the set-theoretic statement, it is enough, by Lemma 5.2, to show that ifi ∈ Z/(5)andpi6∈V_A, thenpi6∈H^efor alle∈E5. Condition 6 implies thatH^i,i⁰ does not containpi. On the other hand, ifj 6=i, then either(j)(j+1)∈Eior(j−1)(j)∈Ei, so thatpi ∈ H^j,j⁰ would imply that pi ∈ Pj = H^j,j⁰ ∩H^j−1,j = H^j,j⁰ ∩H^j,j+1, contradicting thatpi6∈VA. A computation withMACAULAY2shows that the union ofVA

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and the five lines{li}is not only set-theoretically but also scheme-theoretically cut out by the six quintic hypersurfacesQ0, . . . , Q5. Indeed, the degree 5 component of the ideal is the span of the corresponding quintic formsq0, . . . , q5. Since this is an open condition, the vanishing ideal of the union of a general lCM Petersen arrangement and the five lines {li}is minimally generated by the quinticsq0, . . . , q5. 2

As a corollary, we have:

Corollary 5.4. LetVAbe a general lCM Petersen arrangement. Then there are exactly five6-secant lines toVA.

Proof. Let l be a 6-secant line to V_A. Then any quintic hypersurface containing V_A containsl. It follows from Theorem 5.3 thatl =l_i for somei ∈ Z/(5). But for each i ∈Z/(5), the linelimeets the six planesPi,Pi⁰,P_(i+1)0,P_(i−1)0,Pi+2andP_i−2, but does not meet the linesPi∩Pi⁰,P_(i+1)0∩P_(i−1)0 andPi+2∩P_i−2and hence does not lie

in the planes it meets. Soliis a 6-secant toV_A. 2

Remark 5.5. The general lCM Petersen arrangement satisfies Condition 6. One can ver- ify, however, that the arrangement of Example 4.12 does not. In that case, the five quinticsq₀, . . . , q₄ are linearly independent, butq₅can be written as a linear combination of q₀, . . . , q₄. We will discuss this example further in the last section.

We recall a well-known fact:

Lemma 5.6. If four2-planes inP⁴ meet pairwise in distinct points, then there exists a unique cubic hypersurface containing them.

Now for eachi∈Z/(5), letCibe the unique cubic hypersurface inP⁴containing the four planesPi+1, P_(i+2)⁰, P_(i−2)⁰, Pi−1 (which correspond to the maximal independent vertex setVidefined in Section 2) and set

Si:=H^i,i⁰∪H⁽ⁱ⁺¹⁾⁰^,(i−1)⁰∪H(i+2),(i−2)∪Ci.

Observe thatV_A∪li⊂Si.

Theorem 5.7. V_A is scheme-theoretically cut out by the six quintic hypersurfacesQ₀, . . . , Q₅and any two different sextic hypersurfacesSjandSk.

Proof. Since thelj are pairwise skew, it suffices to show thatV_A∪lj is scheme-theoretically cut out byQ0, . . . , Q5andSjfor everyj ∈Z/(5). Without loss of generality, we may assume thatj =0. Leti6=0. Ifli does not lie inS0, then, sinceli is a 6-secant to VA,

(li∩S0)⊂VA

and we are done by Theorem 5.3. Thus, our task is to prove thatlidoes not lie inS0. Here we will only show thatl1does not lie inS0; essentially the same proof applies to the remaining cases. Suppose, to the contrary, thatl1⊂S0. Thenl1lies inC0or in one of the hyperplanesH^0,0⁰,H¹⁰^,4⁰orH^2,3. Given thatl1 =H^1,1⁰∩H²⁰^,0⁰∩H^3,4, we deduce

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thatl1 lies in eitherP₀⁰ = H^0,0⁰ ∩H⁰⁰^,2⁰,P1⁰ = H^1,1⁰ ∩H¹⁰^,4⁰ or P3 = H^2,3∩H^3,4, contradicting Lemma 5.1.

Finally, suppose thatl1 ⊂ C0. Recall thatl1 meets (but is not contained in)P1,P4

andP₂⁰. SinceC0contains these three planes,C0must be the union of the three distinct hyperplanes spanned byl1and these planes. Thus the fourth plane,P₃⁰thatC0contains must lie in one of these hyperplanes and therefore meets eitherP₁,P₄, orP₂⁰ in a line,

contradicting thatAis Petersen. 2

Remark 5.8. LetAbe a Petersen arrangement and letI_Abe its vanishing ideal. One can show thatI_Aispl-generated, i.e., that it is generated by products of linear forms. Recall that every quintic polynomialqi we defined is a product of linear forms. Furthermore, five partitions ofE(ΓPete)arise naturally from the 6-cycles ofΓPete. To each of these set partitions, we can attach a sextic polynomial which is a product of linear forms. One then shows that these quintics and sextics together generate the same ideal as theqiand the forms vanishing on theSj. Arrangements defined by products of linear forms have been studied by Bj¨orner, Peeva and Sidman [2].

6 Smooth general-type surfaces of degree 15

Once again, letA = {Pi, P_i⁰}i∈Z5 be a (labeled) lCM Petersen arrangement. We have shown that the vector space formed by quintic hypersurfaces containingV_Ahas dimension 6, soV_A can be linked in the complete intersection of two quintic hypersurfaces to a surface of degree 15. In this section, we will study the general surface arising in this way.

First, however, we recall (from [8] or [7], for example) the main definition and some basic results of liaison theory.

Definition 6.1. LetX andX⁰ be surfaces inP⁴with no irreducible component in common. ThenX andX⁰ are said to belinkedby a complete intersection of type(m, n)if there exist hypersurfacesY andY⁰of degreesmandnrespectively such thatY ∩Y⁰ = X∪X⁰scheme-theoretically.

In this situation,Xis lCM if and only ifX⁰is lCM,X∩X⁰is a curve, and there are two standardexact sequences of linkage,

0−→ω_X−→ OX∪X⁰(m+n−5)−→ OX⁰(m+n−5)−→0 and

0−→ωX−→ OX(m+n−5)−→ O_X∩X⁰(m+n−5)−→0.

The first sequence yields the relation between the Euler–Poincar´e characteristics:

χ(O_X⁰) =χ(O_X∪X⁰)−χ(ω_X(5−m−n)). (6.1) Lettingdandd⁰denote the degrees ofX andX⁰andπandπ⁰their sectional genera, the corresponding sequence for linkage of curves inP³yields the relation

π−π⁰ =1

2(m+n−4)(d−d⁰). (6.2)

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Theorem 6.2. Let V_A be a general lCM Petersen arrangement. Then the surface X linked toVA by two general quintic hypersurfaces is smooth, non-minimal and of general type, with degreed = 15, sectional genusπ = 21, Euler–Poincar´e characteristic χ = χ(OX) = 5, geometric genus pg = 4, irregularity q = 0, and Euler number e(X) =55. Moreover,Xis embedded inP⁴by the linear system

Hmin−

4

X

i=0

li

,

whereH_minis the hyperplane class of the minimal model ofX and thel_iare the lines of (5.1).

Proof. As we have shown in Section 5,(I_A)₅ is generated by six quintics. Thus,V_Ais linked—via a complete intersection of two general elements of(I_A)₅—to a surface X of degree 15. Furthermore, a computation withMACAULAY2(see [1]) shows thatX is smooth for a particular choice ofAand quintics inI_A. HenceX is smooth in general.

By (6.2),X has sectional genusπ=21 and by (6.1),X has Euler–Poincar´e charac- teristicχ=5.

LetHandKbe a hyperplane section and a canonical divisor ofXrespectively. Recall the double point formula for surfaces inP⁴[5]:

d²−10d−5H.K−2K²+12χ=0.

Using this formula, it follows thatK²=5, showing thatX is of general type. This also proves thate(X) = 55, because χ = ₁₂¹(K²+e). A computation withMACAULAY2 shows thatp_g=4, and henceq=χ−p_g−1=0.

Recall that the linesl₀, . . . , l₄ of (5.1) are 6-secants toV_A. From the first standard linkage sequence, we obtain:

0−→ I_X∪V_A(5)−→ I_A(5)−→ OX(K)−→0.

SinceX∪VAis a complete intersection of two quintic hypersurfaces and sinceIA(5)is globally generated by its sections outsideS

li, the canonical bundleωXis also globally generated outsideS

li. As a consequence,S

li is the fixed part of the canonical linear series, andl0, . . . , l4are the exceptional lines ofX. In particular, the embedding linear

system ofXhas the desired form. 2

Remark 6.3. Letϕ_|H+K|:X →P^N be the adjunction map, whereN =π+p_a−1 = 21+4−1 = 24. Then the minimal modelX_minof X is obtained as the image ofX underϕ_|H+K|. LetH_min be its hyperplane class and letK_minbe its canonical divisor.

ThenXminhas degreeH_min² = (H+K)²=70. SinceHmin.Kmin= (H+K).K=30, the sectional genus ofXminis 51. We note that the canonical model ofXminis a singular surface of degree 10 inP³.

Remark 6.4. For a general surfaceXobtained in this way,H⁰(IX(5)) =2. Therefore, since the family of lCM Petersen arrangements has dimension 30 and since the space

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of quintics containing each Petersen arrangement has dimension 6, the family of smooth general type surfaces of degree 15 obtained from the Petersen arrangements via linkage has dimension 38=30+2(6−2). On the other hand, a standard computation shows that χ(N_X/_P4) = 35. We conclude thatdimH¹(N_X/_P4) ≥3, but at present, we are unable to show that equality holds in general. Consequently, we cannot rule out the possibility that the Hilbert scheme ofP⁴may be obstructed at the general surfaceXobtained via our construction.

Now letP ∈ A. Then exactly three distinct planes ofAmeetPin lines, and sinceA is lCM, these three lines meet pairwise in three distinct pointsr₁,r₂, andr₃ofP.

Lemma 6.5. There exists a quintic plane curveQ⊂P, not passing throughr₁,r₂orr₃, such thatX∩P =Q∪ {r₁, r₂, r₃}scheme-theoretically.

Proof. SetB =A \ {P}and letU =X ∪V_B. ThenP andU are linked (by the same complete intersection of degree(5,5)that linksV_AandX), soC=P∩Uis a curve and the second standard exact sequence of linkage becomes

0−→ωP −→ OP(5)−→ OC(5)−→0. (6.3) Consider the divisorial exact sequence:

0−→ OP(−C)−→ OP −→ OC−→0. (6.4)

LetL ⊂ P be a line so thatωP = OP(−3L). By comparing (6.3) with (6.4) twisted byOP(5), we conclude thatC≡8LinP. Now there are exactly three planes ofBthat intersectPin lines, so the intersectionX∩Pcontains a curveQ≡8L−3L=5L, i.e., a plane curve of degree 5.

Now letK = KX be a canonical divisor forX. Let X·P denote the intersection product ofXandP. Recall that the equivalence ofQfor the intersection productX·P inP⁴is given by a formula of Todd:

(X·P)^Q= (K·Q)X+ (KP ·Q)P−(K_P4·Q)_P4−KQ;

see [3, Example 9.1.7]. SinceQis a quintic plane curve,K_P.Q=−15,K_P4.Q=−25, anddegK_Q=2·6−2=10. The length of the residual scheme is

deg(X·P)−deg(X·P)^Q = degX−(K.Q) +15−25+10

=15−K.Q. (6.5)

Consider the exact sequence of linkage:

0−→ωX−→ OX(5)−→ OD(5)−→0, whereD=X∩V_A. By comparing this sequence with

0−→ OX(−D)−→ OX−→ OD−→0,

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we deduce thatD≡5H−KinXand therefore

K.D=K.(5H−K) =120.

On the other hand,Cis the sum of 10 quintics, one in each plane ofA, and we deduce thatK.Q=120/10 =12. So by (6.5), the residual scheme toQinP∩X has length 15−12=3.

NowV_Bfails to be lCM precisely atr1,r2andr3and sinceV_BandX∪Pare linked (by the same pair of quintics that linkXtoV_A),X∪Pfails to be lCM precisely atr1,r2, r3. It follows that the residual scheme toQinP ∩X is supported at theri, completing

the proof. 2

Theorem 6.6. The surfaceX admits a two-dimensional family of5-secant lines, a one- dimensional family of6-secant lines, and fifteen7-secant lines.

Proof. Keeping the notation, we see from Lemma 6.5 that any line in P ∈ Ameets Q—and henceX—in five points (possibly with multiplicity). Furthermore, any line inP passing throughr₁,r₂, orr₃meetsX in that point in addition to the five points where it meetsQ. Finally, any of the three linesri, rj meetsXin those two points in addition to its five points of intersection withQ. Since there are three such lines in each of the ten planesPofAand since each such line consists of the intersection of two of those planes,

we arrive at the count of fifteen 7-secants. 2

7 A general-type surface with symmetries

LetAbe the plane arrangement constructed in Example 4.12. Recall thatA, and thusV_A hasD=hρ, σ:ρ⁵ =σ² =e, ρσ=σρ⁴ias a group of symmetries.

Consider the perfect matchingE₅ ={ii⁰}_i∈_Z_/(5)⊂E(Γ_Pete). Recall that each of the other perfect matchings are obtained as

E_j =ρ^jτ E₅

forj ∈Z/(5). In Section 5, we attached to eachE_j a quintic polynomialq_j ∈I_A. It is easy to check that the first five quintic polynomials,q₀, . . . , q₄are linearly independent.

LetW be the vector space they span; it is clear thatDacts onW.

We now denote byζ a nontrivial 5th root of unity. Letw₁ andw₂ be eigenvectors for the action ofρwith eigenvalues ζ andζ² respectively. Then, for eachi ∈ {1,2}, the quinticswiandσ(wi)span a two-dimensional subspaceWiofW which is invariant underD. Take two general quintic polynomials in eitherW₁orW₂. Then the surfaceX linked, via these two quintics, toV_Ais smooth. Thus,Xis a smoothD-invariant general type surface inP⁴.

As in Example 4.12, letV be the underlying vector space ofP⁴with basis{ei}_i∈_Z_/(5). The action ofD is given byρ(ei) = ei+1 andσ(ei) = e_5−i. In other words,V is the standard permutation representation ofD. Recall thatV contains the trivial representation ofD, which is spanned by the vectorP

i∈Z/(5)ei. Letpbe the corresponding point of