A Cayley-Bacharach theorem and plane configurations
aa r X i v : . [ m a t h . AG ] F e b A CAYLEY–BACHARACH THEOREM AND PLANECONFIGURATIONS
JAKE LEVINSON AND BROOKE ULLERY
Abstract.
In this paper, we examine linear conditions on finite sets of pointsin projective space implied by the Cayley–Bacharach condition. In particular,by bounding the number of points satisfying the Cayley–Bacharach condition,we force them to lie on unions of low-dimensional linear spaces. These resultsare motivated by investigations into degrees of irrationality of complete inter-sections, which are controlled by minimum-degree rational maps to projectivespace. As an application of our main theorem, we describe the fibers of suchmaps for certain complete intersections of codimension two. Introduction
Let Γ ⊂ P n be a finite set of points. We say that Γ satisfies the Cayley–Bacharach condition with respect to degree r polynomials or simply is CB( r ) if, whenever a homogeneous degree r polynomial F vanishes at all but onepoint of Γ, F vanishes at the last point.This notion was familiarized by the classical 19th century theorem of Cayleyand Bacharach which states that if two plane curves of degree d and e , respectively,meet in de points, then any curve of degree d + e − de points isCB( d + e − X is a smooth varietyof dimension n and Γ ⊂ X is a finite set of points, we say Γ satisfies the Cayley–Bacharach condition with respect to a linear system | V | if any section s ∈ V vanishing at all but one point of Γ vanishes at the last point. In the case whereΓ is a general fiber of a generically finite dominant rational map X P n , Bas-tianelli showed that Γ satisfies the Cayley–Bacharach condition with respect to thecanonical linear system | K X | (see [Bas12, Proposition 4.2]).Several recent papers have exploited the Bastianelli result to compute the go-nality, or the degree of irrationality, a higher dimensional analog of gonality, ofseveral classes of smooth complete intersection varieties in projective space (seee.g. [BCD14], [BDE + X is sufficiently positive (as, for example, in the case of high degreecomplete intersections), the fibers are forced to lie in special positions.For instance, to this end, Bastianelli, Cortini, and De Poi prove the following: Date : January 2021.
Theorem 1.1. [BCD14, Lemma 2.4] If Γ ⊂ P n is CB( r ) and (1.1) | Γ | ≤ r + 1 , then Γ lies on a line. This result allows them to show that the fibers of rational maps to P r from ahigh degree hypersurface X ⊂ P r +1 are collinear.In other (non-hypersurface) examples, however, we do not expect the fibers tobe collinear. A natural question, then, is: what can one say about the geometry ofΓ when (1.1) is weakened? For instance, when the bound is raised to | Γ | ≤ r + 1,Stapleton and the second author show that Γ lies on either a plane conic or twoskew lines [SU20] (see Remark 1.5).In this paper, we examine linear conditions on Γ implied by the Cayley–Bacharachcondition. We call a union P = P ∪ · · · ∪ P k ⊆ P n of positive-dimensional linearspaces a plane configuration of dimension dim( P ) = P dim( P i ) and length ℓ ( P ) = k . We have the following conjecture: Conjecture 1.2.
Let Γ ⊂ P n be a finite set of points satisfying CB( r ) .If | Γ | ≤ ( d + 1) r + 1 , then Γ lies on a plane configuration P of dimension d . Note that this bound is easily shown to be tight: a counterexample for r ( d +1)+2is given by points on a rational normal curve (see Example 3.1).Our main result is as follows. Theorem 1.3.
Conjecture 1.2 holds in the following cases: (i)
For r ≤ and all d . Moreover, we may take ℓ ( P ) = 1 . (ii) For all r and for d ≤ . Moreover, for r ≤ , we may take ℓ ( P ) ≤ . (iii) For d = 4 and r = 3 . Moreover, we may take ℓ ( P ) ≤ . We believe a stronger statement is true in at least two respects. First, for each d , we believe the length of the plane configuration can be bounded in terms of r , asin the above theorem. Note that for larger values of r and d , this minimum lengthcan be arbitrarily large (Example 3.5). See also Corollary 4.8, which gives a boundon rd − in the case where P has maximum length, i.e. P is d disjoint lines.Second, we believe Γ lies on certain unions of curves of low degree and arithmeticgenus, not necessarily rational, whose exact description depends on refinements ofthe bound on | Γ | , as in the result of Stapleton and the second author. See Section7 for additional discussion.Theorem 1.3 immediately leads to a result about the geometry of fibers of lowdegree maps to projective space from certain codimension two complete intersec-tions: Theorem 1.4.
Let X ⊂ P n +2 be the complete intersection of a quartic hypersurfaceY and a hypersurface of degree a ≥ n − . (a) If f : X P n is a finite rational map of degree ≤ a and Γ ⊂ X is ageneral fiber of f , then Γ lies on a plane configuration of dimension . (b) In particular, if n ≥ , or more generally if Y contains a line, the conclusionof (a) holds for any dominant rational map f : X P n of minimumdegree. Remark 1.5.
The cases of hypersurfaces of quadrics and cubics were dealt with in[SU20]. The authors showed that in the former case, the fibers must be collinear,
CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 3 and in the latter, the fibers must lie on plane conics or two skew lines – in particular,the fibers lie on plane configurations of dimension two.Applying the conclusion of Conjecture 1.2, we obtain the following more generalconjectural result about codimension two complete intersections of arbitrary degree.
Proposition 1.6.
Let X ⊂ P n +2 be the complete intersection of hypersurfaces Y a and Y b , of degrees a and b , respectively. Suppose that Conjecture 1.2 holds for d = b − and r = a + b − n − and that a + 1 ≥ b (3 + n − b ) . (a) If f : X P n is a finite rational map of degree ≤ a ( b − and Γ ⊂ X isa general fiber of f , then Γ lies on the union of disjoint linear spaces whosedimensions sum to at most b − . (b) In particular, if n ≥ b − , or more generally if Y b contains a line, theconclusion of (a) holds for any dominant rational map f : X P n ofminimum degree. The paper is organized as follows. In Section 2, we introduce the notion of aplane configuration and prove basic results about the Cayley–Bacharach conditionand plane configurations which we make frequent use of throughout the remainderof the paper. In Section 3, we give some concrete examples. Sections 4 and 5 aredevoted to the proof of the main theorem and requisite lemmas. Section 6 gives ashort proof of Theorem 1.4 and Proposition 1.6. Finally, in Section 7, we discussdirections for potential research and leave the reader with several open questions.The authors would like to thank Nic Ford, Joe Harris, Rob Lazarsfeld, JessicaSidman, and David Stapleton. 2.
Preliminaries
Our main object of interest is the following.
Definition 2.1. A plane configuration P is a union of distinct positive-dimensionallinear spaces P = k [ i =1 P i in P n . We say its dimension is dim P = P i dim P i and its length is ℓ ( P ) = k .Thus a 2-dimensional plane configuration is either two lines or one 2-plane. Ina sufficiently large P n , if the planes of P are in general position then span( P )has dimension dim P + ℓ ( P ) −
1. Note that, by definition, a 0-dimensional planeconfiguration is empty.
Definition 2.2.
We say a plane configuration P is skew if P i ∩ P j = ∅ for all i = j .We say it is split if, in addition, P is a projectivized direct sum decomposition ofthe affine cone over span( P ). Equivalently,dim span( P ) = dim( P ) + ℓ ( P ) − . Equivalently, for all i , P i ∩ P j = i P j = ∅ .If there exist i = j for which P i ∩ P j is nonempty, it will frequently be preferablefor us to replace P i and P j by span( P i , P j ). The resulting plane configuration haslower length than P and at most the same dimension, while also being larger (asa variety). Since we wish to cover finite sets Γ ⊆ P n by low-dimensional planeconfigurations, we may often freely reduce to P being skew. JAKE LEVINSON AND BROOKE ULLERY
Note that when ℓ ( P ) = 2, P is skew if and only if P is split. Example 2.3. In P , let P consist of three general lines. Then dim P = 3 and ℓ ( P ) = 3, but span( P ) = P , so P is skew but not split.2.1. Basic facts.
We now prove some basic facts about plane configurations andthe Cayley–Bacharach condition. We will use them throughout the remaining ar-guments of the paper, often without explicit reference.
Proposition 2.4.
Let Γ ⊆ P n be a finite set and P a linear space. Then P can be extended to a hyperplane H containing no additional points of Γ . That is, Γ ∩ H = Γ ∩ P .Proof. Omitted. (cid:3)
Proposition 2.5 (Excision property) . Let Γ ⊆ P n be a finite set satisfying CB( r ) .Let P be a plane configuration of length ℓ . Then Γ \ P is CB( r − ℓ ) . In particular, the complement of Γ by a single linear space is CB( r − Proof.
It suffices to show that, for a single linear space P , the set ˜Γ := Γ \ P isCB( r − P and avoiding ˜Γ, we may replace P by a hyperplane H .Let Z be a hypersurface of degree r − x ∈ ˜Γ. Then Z ∪ H has degree r and contains Γ \ { x } . Since Γ is CB( r ), Z ∪ H contains thefinal point x . By construction H does not contain x , so Z does. (cid:3) Proposition 2.6 (Basic lower bound) . Let Γ ⊆ P n be a finite set satisfying CB( r ) .If Γ is nonempty, | Γ | ≥ r + 2 .Proof. Suppose 1 ≤ | Γ | ≤ r +1. Fix x ∈ Γ and find hyperplanes H i , each containingexactly one of the other points of Γ. The union of the H i ’s has degree | Γ | − ≤ r and misses x ∈ Γ, a contradiction. (cid:3)
In particular, Conjecture 1.2 holds in the case d = 0 (recall that a 0-dimensionalplane configuration is empty).We also record the following stronger lower bound, whose proof is trivial, for usein inductive arguments below. Proposition 2.7 (Inductive lower bound) . Assume Conjecture 1.2 holds for fixed r and dimension up to d − . Let Γ ⊆ P n be a finite set of cardinality | Γ | ≤ r ( d +1)+1 ,satisfying CB( r ) .Then if Γ does not lie on a plane configuration of dimension d − , | Γ | ≥ rd + 2 . Examples
Our conjecture is sharp in certain respects and not others, as the followingexamples demonstrate.
Example 3.1 (Rational normal curves) . Let r ≥ m = r ( d + 1) + 2. LetΓ ⊂ P d +1 be any set of m distinct points on a rational normal curve C ⊂ P d +1 ofdegree d + 1. Then Γ is in general linear position, hence is not contained in a planeconfiguration of dimension d . (Any such configuration contains at most 2 d pointsof Γ, by taking d lines.) Nonetheless, Γ satisfies CB( r ) by B´ezout.This example demonstrates the following. CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 5
Corollary 3.2.
The bound | Γ | ≤ r ( d + 1) + 1 in Conjecture 1.2 is tight. We next see that, under the hypotheses of the conjecture, the points need notlie on a degree- d rational curve. Example 3.3 (Elliptic curve) . Let r = 2 and d = 3. In P , let Γ consist of9 = 2(3 + 1) + 1 general points on a quartic elliptic curve E . Then no 8 points of Γare a complete intersection of E with a quadric surface (such complete intersectionsform only a 7-dimensional family), so Γ is CB(2). The unique 3-dimensional planeconfiguration containing Γ is the P . Moreover, Γ does not lie on any union ofrational curves of total degree 3.Plane configurations of length greater than one are unavoidable: Example 3.4 (Two 2-planes) . Let d = 4 and r = 3. Let P be a split configurationof two 2-planes P , P . Let C i be a smooth conic on P i ( i = 1 ,
2) and let Γ consistof 8 general points from each C i . Then Γ is CB(3) and has r ( d + 1) + 1 = 16.Moreover, P is the unique plane configuration of dimension 4 containing Γ. Example 3.5 ( d general lines) . Suppose r ≥ d −
1. Let P consist of d generallines in P n , where n > d > n . Let Γ consist of r + 2 general points from each line.Then Γ is CB( r ) by B´ezout and | Γ | = ( r + 2) d ≤ r ( d + 1) + 1. The only dimension- d plane configuration containing Γ is the d lines.Notably, this example shows that plane configurations of arbitrary length arisein Conjecture 1.2 for sufficiently large r . See also Corollary 4.8, which shows thatskew plane configurations of maximum length d occur only when 1 ≤ rd − ≤ Key lemmas
We have seen that complements of Γ by hyperplanes must satisfy lower-degreeCayley–Bacharach conditions. These conditions seem too weak to allow inductiveapproaches to Conjecture 1.2. Our core approach is to establish that, under certainconditions, subsets of Γ again satisfy CB( r ), with the same value of r .4.1. Split configurations.Lemma 4.1.
Let Γ ⊂ P n be a finite set of points. Let P , P be disjoint, positive-dimensional planes, and suppose Γ ⊂ P ∪ P .Let r ≥ . Then Γ is CB( r ) if and only if Γ ∩ P and Γ ∩ P are both CB( r ) .Proof. Let p i = dim P i . First, we choose coordinates [ X : · · · : X p : Y : · · · : Y p ]for P + P ⊆ P n so that P = [ X : · · · : X p : 0 : · · · : 0] and P ′ = [0 : · · · : 0 : Y : · · · Y p ]. Write Γ = Γ ∪ Γ where Γ i = Γ ∩ P i .( ⇒ ) Suppose Γ satisfies CB( r ). Let F be a polynomial of degree r vanishingat all but possibly one point of Γ . We claim that F vanishes at the last point.Let G ( X, Y ) = F ( X, Y j ’s. We may assume G is not identically zero. Now G | P = F | P , and G vanishes identically on P , hence on all of Γ . By CB( r ) for Γ, G vanishes at thelast point of Γ , so F does also.( ⇐ ) Suppose Γ and Γ satisfy CB( r ). Let F be a degree- r polynomial vanishingat all but possibly one point of Γ, say p ∈ Γ . We may assume F | P is not identicallyzero, so by CB( r ) for Γ , F vanishes at the last point. (cid:3) JAKE LEVINSON AND BROOKE ULLERY
We will primarily use this for pairs of disjoint planes, but the same argumentapplies to split configurations:
Proposition 4.2.
Let P = S P i be a split plane configuration and let Γ ⊆ P be afinite set. Let r ≥ . Then Γ is CB( r ) if and only if, for each i , Γ ∩ P i is CB( r ) . Proposition 4.3.
Suppose Conjecture 1.2 holds up to d − for fixed r . Let | Γ | ≤ r ( d + 1) + 1 . Let A, B ⊆ P n be disjoint, positive-dimensional linear spaces andsuppose Γ ⊂ A ∪ B . Then Γ lies on a plane configuration of dimension d .Proof. Let Γ A = Γ ∩ A and Γ B = Γ ∩ B . By counting, either | Γ A | ≤ rd + 1 or | Γ B | ≤ rd + 1 . Without loss of generality, assume | Γ A | ≤ rd + 1. Let d A be minimal such that Γ A lies on a plane configuration of dimension d A . By Proposition 4.1, Γ A is CB( r ), sothe earlier cases of the Conjecture imply d A ≤ d −
1. The inductive lower bound(Proposition 2.7) implies | Γ A | ≥ rd A + 2 by minimality of d A , and so | Γ B | = | Γ | − | Γ A | ≤ r ( d − d A + 1) − < r ( d − d A + 1) + 1 , so Γ B lies on a plane configuration of dimension d − d A . Combining these two planeconfigurations gives a plane configuration of dimension d as desired. (cid:3) Unions of two planes meeting at a point.
Let
A, B ⊂ P n be linear spacesmeeting at a point p . Without loss of generality, assume A and B span P n . Wemay choose coordinates of the form[ X : · · · Y i · · · : · · · Z j · · · ]such that p = [1 : 0 : · · · : 0], A = [ X : · · · Y i · · · : 0 · · ·
0] and B = [ X : 0 · · · · · · Z j · · · ]. Lemma 4.4.
Let F A , F B be polynomials of degree r on A and B , such that F A ( p ) = F B ( p ) . Then there exists a polynomial F on P n such that F | A = F A and F | B = F B .Proof. Restricting to p means setting all Y i and Z j to 0, so F A ( p ) = F B ( p ) meansthe coefficient c of X r is the same for both. Then put F ( X, Y, Z ) = cX r + ( F A ( X, Y, − cX r ) + ( F B ( X, , Z ) − cX r ) . Setting all Z j = 0 recovers F A and setting all Y i = 0 recovers F B . (cid:3) Now let Γ ⊂ A ∪ B be a finite set. Let Γ A = (Γ ∩ A ) \ { p } and Γ B = (Γ ∩ B ) \ { p } . Lemma 4.5.
Suppose Γ satisfies CB( r ) . At least one of Γ A and Γ A ∪ { p } is CB( r ) . If p / ∈ Γ , at least one of Γ A and Γ B ∪ { p } is CB( r ) . If p ∈ Γ , at least one of Γ A ∪ { p } and Γ B ∪ { p } is CB( r ) .Note that we may exchange A and B in statements 1 and 2. Statement (1) is the most important as it guarantees information about thepoints on A (and, analogously, the points on B ). We can visualize the other rela-tionships according to a graph: CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 7 Γ A ❍❍❍❍❍❍❍❍❍ Γ A ∪ { p } ✈✈✈✈✈✈✈✈✈ Γ B Γ B ∪ { p } Γ A Γ A ∪ { p } Γ B Γ B ∪ { p } ( p / ∈ Γ) ( p ∈ Γ)Each edge indicates “at least one of these two sets must be CB( r )”. Proof.
We repeatedly use the following observation:( ∗ ) Suppose F A is a polynomial of degree r vanishing at p and at all but onepoint of Γ A . Then extending by F B = 0 via Lemma 4.4 contradicts Γ beingCB( r ). Thus no such F A exists.Effectively, ( ∗ ) says Γ A ∪{ p } is “almost CB( r )”: the only degree r polynomials thatcan leave out a point of Γ A ∪ { p } are those leaving out p (and therefore vanishingat all of Γ A ).1. Suppose Γ A is not CB( r ). Then there exists F A leaving out one point a ∈ Γ A .By observation ( ∗ ), F A ( p ) = 0. Suppose further that Γ A ∪ { p } is not CB( r ). Thenthere exists ˜ F A leaving out one point of Γ A ∪{ p } . Again, ( ∗ ) implies ˜ F A ( p ) = 0, andso ˜ F A vanishes on all of Γ A . But then a linear combination of F A and ˜ F A vanisheson Γ A \ { a } (since both F A and ˜ F A do) and at { p } (by choice of combination) anddoes not vanish at { a } (since F A does not and ˜ F A does). This contradicts ( ∗ ).2. Suppose Γ A is not CB( r ) and let F A leave out a ∈ A . By observation ( ∗ ), F A ( p ) = 0. Suppose Γ B ∪ { p } is also not CB( r ) and let F B be a correspondingpolynomial. By ( ∗ ) applied with A and B exchanged, F B must leave out p andvanish on all of Γ B . By Lemma 4.4, there exists a global polynomial F on P n extending F A and F B (up to scalar). Since p / ∈ Γ, this contradicts Γ being CB( r ).3. Suppose there exist F A and F B leaving out points of Γ A ∪ { p } and Γ B ∪ { p } ,respectively. By ( ∗ ), the left-out point must be p in both cases. By Lemma 4.4, F A and F B extend to a global polynomial on P n vanishing on Γ A ∪ Γ B . Since p ∈ Γ,this contradicts Γ being CB( r ). (cid:3) Proposition 4.6.
Suppose the Conjecture holds up to d − for fixed r . Let | Γ | ≤ r ( d + 1) + 1 . Let A, B ⊆ P n be positive-dimensional linear spaces meeting at a pointand suppose Γ ⊂ A ∪ B . Then Γ lies on a plane configuration of dimension d .Proof. Let p, Γ A , Γ B be as above. By counting, either | Γ A | ≤ rd or | Γ B | ≤ rd .Without loss of generality, assume the first inequality holds.Let S be whichever of the sets Γ A or Γ A ∪ { p } is CB( r ), according to Lemma4.5(1). Let d be minimal such that S lies on a plane configuration of dimension d . Then | S | ≤ rd + 1, so the earlier cases of the conjecture imply d ≤ d − d , it follows | S | ≥ rd +2. Then, the inequality | Γ B | + | S | ≤ | Γ | implies | Γ B | + 1 ≤ | Γ | − | S | + 1 ≤ r ( d − d + 1) , so whichever of the sets Γ B or Γ B ∪ { p } satisfies CB( r ) lies on a plane configurationof dimension d − d .So we are done if the two plane configurations we have constructed cover Γ. Theremaining case is where p ∈ Γ but is not included in either configuration. That is,Γ A and Γ B are both CB( r ), but neither Γ A ∪ { p } nor Γ B ∪ { p } are CB( r ). Thiscontradicts Lemma 4.5(3). (cid:3) JAKE LEVINSON AND BROOKE ULLERY
Bounding the length of the configuration.
The proposition below is notused outside of this section. Its main use is to prove Corollary 4.8, which givesconditions on r and d under which Γ lies on a plane configuration of maximallength, i.e. d disjoint lines. Proposition 4.7.
Suppose Γ is CB( r ) and is contained in a skew plane configura-tion P = S P i . Then one of the following holds: (i) Each plane P i contains at least max( ℓ ( P ) , r + 2) points of Γ , or (ii) Some plane P i contains fewer than ℓ ( P ) points, and also ℓ ( P ) ≥ r + 2 .Proof. Let i be such that γ i = ∩ P i ) is of minimal cardinality. Fix x ∈ Γ ∩ P i .We construct a collection of hyperplanes covering Γ \ { x } as follows.Observe that, given j = i and a point x ′ ∈ Γ ∩ P i \{ x } , the plane P ′ = span( P j , x ′ )intersects P i only in x ′ , by the skewness assumption. We may extend it to ahyperplane H ′ containing no additional points of Γ ∩ P i .We first choose min( γ i − , ℓ ( P ) −
1) hyperplanes H ′ according to this observation,taking a different j and a different x ′ ∈ Γ ∩ P i \ { x } each time.Suppose γ i ≥ ℓ ( P ). Then for each remaining point of Γ ∩ P i \ { x } , we takea hyperplane H ′′ containing that point and no other points of Γ, giving γ i − x ∈ Γ. Since Γ is CB( r ), we see that r < γ i −
1. By minimality of γ i , we concludethat each plane in P contains at least max( r + 2 , ℓ ( P )) points. This gives case (i).Suppose instead γ i < ℓ ( P ). Then we instead extend the remaining P j ’s tohyperplanes H ′′ not containing x , giving a total of ℓ ( P ) − \ { x } . Then r < ℓ ( P ) −
1. This gives case (ii). (cid:3)
Corollary 4.8.
Let Γ ⊆ P n be CB( r ) and | Γ | ≤ r ( d +1)+1 . Suppose the Conjectureholds, so Γ lies on a plane configuration P of dimension d . By summing anyintersecting planes of P , assume P is skew.Suppose ≤ rd − ≤ . Then the length of P is at most d − .Proof. Suppose instead that ℓ ( P ) = d , that is, Γ lies on d skew lines.We are not in case (ii) of Proposition 4.7 since d < r + 2. But then we must bein case (i), so each line contains at least r + 2 points of Γ. This gives | Γ | ≥ d ( r + 2).Combining with | Γ | ≤ r ( d + 1) + 1, we see that r ≥ d −
1, a contradiction. (cid:3) Proof of main theorem
In this section, we prove Theorem 1.3. Throughout, Γ ⊆ P n denotes a finite setsatisfying CB( r ), and such that | Γ | ≤ r ( d + 1) + 1 . We wish to show that Γ lieson a plane configuration P of dimension d and with length ℓ ( P ) bounded as in thestatement of the theorem. Note that the case r = 0 is obvious and the case d = 0is precisely Proposition 2.6.We frequently use the following notation. Remark 5.1 (Setup) . Let α ≥ d + 1 be the maximum number of points of Γ lyingon a d -plane. Let A be such a d -plane and let Γ A = Γ ∩ A . Next, denote thecomplementary set of points as Γ B := Γ \ Γ A , let β = | Γ B | and let B be the linearspan of Γ B . We may assume β > A . (cid:3) CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 9
Fixed r , varying d . The case r = 1 holds essentially by definition: Proposition 5.2. If Γ is CB(1) and | Γ | ≤ d + 2 , then Γ lies on a d -plane.Proof. Take the first d + 1 points. If they don’t span a d -plane, add the last pointand we’re done. And if they do span a d -plane, that plane contains the last pointby CB(1). (cid:3) Remark 5.3.
For r = 1, the weaker statement, allowing a plane configuration ofarbitrary length, is vacuous and does not require CB(1) at all: any d + 2 pointstrivially lie on a ( d − ( d + 2) lines.We next consider the case r = 2. This is the final case in which the planeconfiguration always has length 1. Proposition 5.4. If Γ is CB(2) and | Γ | ≤ d + 3 , then Γ lies on a d -plane.Proof. Let α, Γ A , A and β, Γ B , B be as in Remark 5.1. For contradiction, assume β > d + 1 points Γ ′ ⊆ Γ A lie on a ( d − ′ and extend B by the other α − d − A . By excision Γ B is CB(1), so dim( B ) ≤ β −
2, so the resulting plane P hasdimension at most ( β −
2) + ( α − d − ≤ d. If Γ ′ ⊂ P , then Γ ⊂ P and the proof is complete. Otherwise, Γ \ P = Γ ′ \ P isCB(1) by excision and so lies on a ( | Γ ′ \ P | − ′ ∩ P to the set, we see Γ ′ spans a plane of dimension at most | Γ ′ | − d −
1, asdesired.Now, by maximality, Γ A must span the d -plane A . But then Γ A contains d + 1linearly independent points. This is a contradiction, so β = 0. (cid:3) Notice that as soon as r ≥ , we can no longer conclude that Γ lies on a planeconfiguration of length one, as we see in the following example. Example 5.5.
Let Γ be a set of 10 = 3(2 + 1) + 1 points on two skew lines,five points on each line. Then Γ satisfies CB(3) by B´ezout, but the only planeconfiguration of dimension two that Γ lies on has length two.See Example 3.5 for a generalization to d skew lines, for sufficiently large r .5.2. Fixed d , varying r . We prove the cases d = 1 , ,
3. The case d = 1 is fairlystraightforward. Theorem 5.6 (Case d = 1) . If Γ is CB( r ) and | Γ | ≤ r + 1 , then Γ lies on a line.Proof. Let α, Γ A , A and β, Γ B , B be as in Remark 5.1. Assume for contradictionthat β >
0. By excision, Γ B is CB( r −
1) and, by counting, β ≤ r −
1) + 1 (since α ≥ r , Γ B lies on a line.By Proposition 2.6, we have ( r −
1) + 2 ≤ β . By maximality of α we have β ≤ α ,which gives | Γ | ≥ r + 2, a contradiction. (cid:3) Theorem 5.7 (Case d = 2) . If Γ is CB( r ) and | Γ | ≤ r + 1 , then Γ lies on a2-plane or two skew lines. Proof.
We have established the cases r = 1 and r = 2, so we assume r ≥ r .Let α, Γ A , A and β, Γ B , B be as in Remark 5.1. We may assume that β >
0. Wehave α ≥ d + 1 = 3, and so β = | Γ | − α ≤ (3 r + 1) − r −
1) + 1 . By excision, Γ B is CB( r − B lies on a 2-plane or on two skewlines. In fact, we will show that it lies on a line. We consider both cases. Case 1 : Γ B lies on a 2-plane. By maximality of α and the fact r ≥
3, we have β ≤ | Γ | = r + ≤ r −
1) + 1 . Since Γ B is CB( r − Case 2 : Γ B lies on two skew lines. Suppose each line contains at least one pointof Γ B . One line together with one point from the other line span a 2-plane. So, bymaximality of α , each line contains at most α − B . Since α + 2( α − ≤ α + β = | Γ | ≤ r + 1 , we see that α ≤ r + 1. On the other hand, since the two lines are skew, they areboth CB( r −
1) (since Γ B is) by Lemma 4.1, so each contains at least r + 1 ≥ α points. This is a contradiction, so we again conclude that Γ B lies on a single line.We now finish the proof. If A and B are disjoint, then Γ lies on plane configura-tion of dimension 2 by Proposition 4.3. If A and B meet a point, the same conclusionfollows from Proposition 4.6. Otherwise, A ⊇ B and Γ lies on A itself. (cid:3) Theorem 5.8 (Case d = 3) . If Γ is CB( r ) and | Γ | ≤ r + 1 , then Γ lies on a planeconfiguration of dimension . Note that if r ≤
4, then we may further take ℓ ( P ) ≤ Proof.
The statement holds for r = 1 and 2 by Propositions 5.2 and 5.4. We canthus assume r ≥ r .Let α, Γ A , A and β, Γ B , B be as in Remark 5.1. We may assume β >
0. Byexcision, Γ B is CB( r − β ≥ r + 1. On the otherhand, β ≤ | Γ | − α ≤ r −
1) + 1so by induction, Γ B lies on a plane configuration of dimension 3. We consider thepossibilities below; in fact we will reduce to the case where Γ B lies on a single line. Case 1: Γ B lies on 3 disjoint lines. If Γ B lies on just two lines, then it lies in a3-plane, which is covered by Case 3. Thus, assume it lies in three disjoint lines butnot in a 3-plane. We obtain a contradiction as follows.Choose two of the lines whose union contains at least β points of Γ B . The spanof those lines is a 3-plane, so by maximality of α , β ≤ α . We calculate β + β ≤ α + β ≤ r + 1and so β ≤ (12 r + 3) / . If r ≥ , this gives β ≤ r −
1) + 1 , so by Theorem 5.7, Γ B lies on a dimension 2 plane configuration, which means itmust lie on a 3-plane, contradicting our assumption. CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 11 If r = 3, then by the above inequality, β ≤ (12 r + 3) / / . So β ≤ , which means at least one of the three lines contains ≤ B . The other two lines span a 3-plane, which means the leftover ≤ r = 4, one of the lines has ≤ Case 2: Γ B lies on the union of a disjoint line and a 2-plane. Again, we assumeΓ B does not lie on a 3-plane.Therefore, the plane and line are disjoint, so the subsets of Γ B on the plane andthe line each satisfy CB( r −
1) by Lemma 4.1. Thus there are ≥ r −
1) + 2 = 2 r points of Γ B on the 2-plane and ≥ r − r + 1 on the line, which gives β ≥ r + 1. The span of one point of Γ B on the line along with the 2-plane is a3-plane, which by hypothesis implies that α ≥ r + 1 , and in turn | Γ | ≥ r + 2 , acontradiction.This leaves one final case. Case 3: Γ B lies on a 3-plane. By maximality of α , we have β ≤ α , which implies β ≤ r. We split this up into two cases.
Case 3a: β = 2 r. Then α = 2 r or α = 2 r + 1. Since r > , we have 2 r < r − r −
1) + 1 , which by Proposition 5.7 implies that the β = 2 r points of Γ B lie on a 2-plane ortwo skew lines. If two lines, the set of points of Γ B on each line are CB( r − r −
1) + 2 points of Γ B , a contradiction. Thus, inthis case, Γ B lies on a 2-plane. But then for any point a ∈ Γ A , the span of Γ B and a contains exactly 2 r + 1 points, otherwise contradicting the maximality of α . ThusΓ B − { a } is CB( r − a was chosen arbitrarily, and α >
4, all of Γ A must lie on a 2-plane, acontradiction, which completes this case. Case 3b: β ≤ r − . Since 2 r − r − B lies on a line, that is, B = span(Γ B )is a line. The desired statement now follows by Proposition 4.3 (if A and B aredisjoint), by Proposition 4.6 (if A and B meet at a point) or directly (if A ⊃ B ). (cid:3) The case d = 4 , r = 3 . Having established Conjecture 1.2 for d ≤ r ≤
2, we consider the next case not covered, namely d = 4, r = 3. Ourargument uses Propositions 4.6 and 4.3 to handle nearly all cases. We give an adhoc argument for the remaining case, in which Γ lies on two hyperplanes meetingalong a line. Theorem 5.9 (Case d = 4, r = 3) . Let d = 4 , let Γ be CB(3) and suppose | Γ | ≤ r ( d + 1) + 1 = 16 . Then Γ lies on a plane configuration of dimension .Proof. We may assume | Γ | ≥
14; otherwise we are done by the case r = 3 , d = 3.Let α, Γ A , A and β, Γ B , B be as in Remark 5.1. We assume β >
0, so in particularΓ B is CB(2), β ≥ α ≤ B = 1, we are done by Proposition4.3 if A ∩ B is empty and by Proposition 4.6 if A ∩ B is a point. Since Γ B is CB(2), we may in particular assume β ≥
6, and we also have dim( B ) ≤ ⌈ β − ⌉ . Second,we may extend B to a P by adding points. So, by maximality of α , we get β + 4 − ⌈ β − ⌉ ≤ β + (4 − dim( B )) ≤ α. By numerical consideration of α and β , this covers nearly every case. The remainingones are below; we also assume dim( B ) ≥ A ∩ B ) ≥ α, β ) = ( , ) or ( , ) : By CB(2), dim( B ) = 2. We must have Γ ∩ A ∩ B emptyor we contradict the maximality of α by extending Γ B by two points. But then Γ A is itself CB(2) and so lies on a P , contradicting the choice of A .( α, β ) = ( , ) or ( , ) : By CB(2) and the discussion above, dim B = 2 and A ∩ B is a line L . By construction, any points of Γ lying on L have been countedamong the α points of Γ A .Suppose Γ ∩ L is nonempty. Then | Γ \ B | ≤
9, hence by CB(2) lies on a 3-plane A ′ ⊂ A . Then A ′ intersects L in at least a point, and does not contain L since Γ A does not lie on a P . Thus A ′ and B cover Γ and intersect in a point, so we aredone by Lemma 4.6.Suppose Γ ∩ L is empty. We are done as above if α = 9. For α = 10, we considerall 2-planes A ′ such that L ⊂ A ′ ⊂ A and A ′ contains at least one point of Γ.Then Γ \ span( A ′ , B ) is CB(2), hence lies on some C ∼ = P (since α − ≤ C Γ ∩ A ′ since Γ A does not lie on a P . So, by CB(1) (excising B and C ),Γ ∩ A ′ contains at least 3 points. Since this is true for any choice of A ′ , combining B with two such choices yields a P containing at least β + 3 + 3 > α points, acontradiction. (cid:3) Finally, to check that the plane configuration P can be taken to have length ℓ ( P ) ≤
2: we have ℓ ( P ) ≤ ℓ ( P ) = 3 (i.e., two lines and a 2-plane) does not arise.6. Codimension two complete intersections
In this section, we give a proof of Proposition 1.6 and Theorem 1.4, which quicklyfollow from the main theorem.Let X ⊂ P n +2 be a complete intersection of hypersurfaces Y a and Y a of degrees a and b , respectively.Let f : X P n be a dominant rational map, and let Γ ⊂ X be a generalfiber of f . The result of Bastianelli [Bas12, Proposition 4.2] mentioned in theintroduction then implies that Γ satisfies the Cayley–Bacharach condition withrespect to K X = ( a + b − n − H , where H is a hyperplane section. That is, Γ isCB( a + b − n − b = 4), we just need the following inequality: a ( b − ≤ (( b −
1) + 1)( a + b − n −
3) + 1 . This inequality easily follows from the hypotheses of the proposition. We thenapply Theorem 1.3 and Conjecture 1.2, which completes the proof.For part (b) of the Theorem and Proposition, assume Y b contains a line ℓ . Inparticular, as long as n ≥ b − , any choice of Y b will contain a line.Notice that ℓ ∩ X will have length a, so projection from ℓ yields a dominantrational map X P n CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 13 of degree a ( b − degree of irrationality of X , or irr( X ) , which is defined to be the minimum degree of a dominant rationalmap to P n . This completes the proof.7. Further considerations
As stated in the introduction, an immediate question raised by our Conjectureis to determine how the length of the plane configuration relates to r and d . Question 7.1.
Let Γ ⊂ P n be a finite set satisfying CB( r ) and such that | Γ | ≤ ( d + 1) r + 1. Assuming Conjecture 1.2 holds, what can be said about the minimumlength of a plane configuration P containing Γ?Our main result shows that ℓ ( P ) ≤ r and d ; it’s straightfor-ward to exhibit examples in which ℓ ( P ) ≥ r and d . Corollary4.8 shows, at the other extreme, that the case ℓ ( P ) = d occurs only in the range1 ≤ rd − ≤ . The Cayley–Bacharach locus of the Hilbert scheme.
Let H = Hilb( P n , m )be the Hilbert scheme of m points in P n . The Cayley–Bacharach condition givesa locally-closed subset CB ( r ) ⊂ H corresponding to m -tuples of distinct pointssatisfying CB( r ). Problem 7.2.
Understand the geometry of CB ( r ) and its closure in H , in particularits dimension, smooth locus, irreducible components and general points.We also observe that the conclusion of Conjecture 1.2, of being contained in aplane configuration, is a closed condition, hence applies to CB ( r ) if it applies to CB ( r ). Both loci CB ( r ) and CB ( r ) can be further decomposed into closed subsetsaccording to the combinatorial type of the plane configuration, CB ( r ) = [ ℓ,~d CB ( r ) ℓ,~d , CB ( r ) = [ ℓ,~d CB ( r ) ℓ,~d where ℓ denotes the length, ~d = ( d , . . . , d ℓ ) the tuple of dimensions and the result-ing loci parametrize those Γ for which there exists a plane configuration P of thespecified type containing Γ. Problem 7.3.
Determine for which combinatorial types the corresponding subsets CB ( r ) ℓ,~d and CB ( r ) ℓ,~d are nonempty, and determine their dimensions and geometry.There is also a natural analog of the Cayley–Bacharach condition for a subscheme ξ ⊆ P n of finite length m : we may say ξ is CB( r ) if, whenever a homogeneousdegree r polynomial F vanishes on a closed subscheme ξ ′ ⊆ ξ of length m −
1, then F vanishes on ξ . This condition is again only locally closed and gives another locus CB ′ ( r ) containing CB ( r ). Question 7.4.
Which CB( r ) subschemes can be smoothed to reduced CB( r ) sub-schemes? That is, what is CB ′ ( r ) ∩ CB ( r )? Matroids.
Conjecture 1.2 and several of our arguments are highly suggestiveof the theory of matroids and matroid varieties [GGMS87]. It would be interestingto understand this connection. Variants of the classical Cayley–Bacharach theoremhave also recently arisen in studying ideals of matroid varieties via the Grassmann–Cayley algebra [STW20].
Question 7.5.
Which matroids arise from sets of points Γ ⊆ P n satisfying theassumptions of Conjecture 1.2?We may also approach the Cayley–Bacharach condition itself in purely matroidalterms. We say that a matroid M satisfies the matroid Cayley–Bacharach con-dition MCB( r ) if, whenever a union of r flats contains all but one point of M , theunion contains the last point. (For a representable matroid, this is equivalent torestricting the geometric Cayley–Bacharach condition to fully reducible hypersur-faces.) Question 7.6.
Does the analog of Conjecture 1.2 hold for matroids M satisfyingMCB( r )? That is, must M be covered by a union of flats of specified dimensions?Some of the arguments of this paper are effectively entirely matroid-theoreticand give positive answers to this question in small cases. The arguments that arenot matroidal (notably Lemma 4.5) can sometimes be replaced, at least for lowvalues of d and r , by substantially longer matroidal arguments.7.3. Curves.
Finally, we revisit the question of describing when Γ lies on a unionof curves (of specified degree and arithmetic genus). As shown in [SU20], if Γ isCB( r ) and | Γ | ≤ r + 1, then Γ lies on a (possibly reducible) curve of degree atmost 2.These considerations are closely related to Γ not imposing independent con-ditions on degree r polynomials, a weaker condition than CB( r ). For example,Castelnuovo [Cas89] gave conditions under which, if a finite set Γ fails to imposeindependent conditions on quadrics, then it lies on a rational normal curve. Question 7.7.
Let c > ⊆ P n be a finite set ofpoints satisfying CB( r ) and such that | Γ | ≤ cr + 1. What is the minimum d suchthat every such Γ lies on a curve of degree d ? What is the genus of such a curve?Thus the result in [SU20] is that for c = , the minimum degree is 2 and thecurve has genus 0. If it could be shown that for c = d + 1, Γ lies on a degree- d curve, this would imply Conjecture 1.2 as such a curve lies on a plane configurationof dimension d . Unfortunately, this is false: for r = d = 2, 7 general points on a2-plane satisfy CB(2), but do not lie on a conic or a union of two lines. For r = 2, d = 3, Example 3.3 of nine general points on a quartic elliptic curve E ⊂ P issimilar: E itself, of degree 4 and genus 1, is a minimal degree curve containing Γ.This question can be formulated more generally: it would be interesting to knowwhen Γ lies on special surfaces or varieties of higher dimension, on special planeconfigurations, and so on. Question 7.8.
With c and Γ as in Question 7.7, what can be said about thegeometry of Γ? CAYLEY–BACHARACH THEOREM AND PLANE CONFIGURATIONS 15
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