This paper studies the Hilbert scheme component parameterizing pairs of linear spaces in projective space, revealing its smoothness, geometric structure, and birational properties, including its Mori dream space status.
Contribution
It provides a detailed geometric and birational analysis of the Hilbert scheme of pairs of linear spaces, including classifications, blow-up descriptions, and Mori dream space characterization.
Findings
01
The component is smooth and isomorphic to blow-ups of Grassmannians.
02
Classifies subschemes parameterized by the component.
03
Determines the effective and nef cones, and when it is Fano.
Abstract
Let Ha,bn denote the component of the Hilbert scheme whose general point parameterizes an a-plane union a b-plane meeting transversely in Pn. We show that Ha,bn is smooth and isomorphic to successive blow ups of Gr(a,n)×Gr(b,n) or Sym2Gr(a,n) along certain incidence correspondences. We classify the subschemes parameterized by Ha,bn and show that this component has a unique Borel fixed point. We also study the birational geometry of this component. In particular, we describe the effective and nef cones of Ha,bn and determine when the component is Fano. Moreover, we show that Ha,bn is a Mori dream space for all values of a,b,n.
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Let Ha,bn denote the component of the Hilbert scheme whose general point parameterizes an a-plane union a b-plane meeting transversely in Pn. We show that Ha,bn is smooth and isomorphic to successive blow ups of Gr(a,n)×Gr(b,n) or Sym2Gr(a,n) along certain incidence correspondences. We classify the subschemes parameterized by Ha,bn and show that this component has a unique Borel fixed point. We also study the birational geometry of this component. In particular, we describe the effective and nef cones of Ha,bn and determine when the component is Fano. Moreover, we show that Ha,bn is a Mori dream space for all values of a,b,n.
The Hilbert scheme HilbP(t)Pn, which parameterizes closed subschemes of Pn with a fixed Hilbert polynomial P(t), introduced by Grothendieck [G61], has attracted a lot of interest. Although their singularities are known to be arbitrarily complicated [V06], the cases when they are smooth or have smooth components have been extensively studied. Early on these smooth components were used to solve numerous enumerative problems [ES96] and with major advances in the minimal model program [BCHM10], they are also a source of examples with rich birational structure. Fogarty [F68] proved that HilbmP2 is smooth and Arcara, Bertram, Coskun and Huizenga [ABCH13] proved that its a Mori dream space and described the stable base decomposition of its effective cone in numerous cases. Piene and Schlessinger [PS85] showed that Hilb3t+1P3 has two smooth components that meet transversely and described the points of the component corresponding to twisted cubics explicitly. Chen [C09] proved that the component corresponding to the twisted cubics is the flip of \widebarM0,0(P3,3) over the Chow variety. Avritzer and Vainsencher [AV92] proved that the component corresponding to elliptic quartics in Hilb4tP3 is smooth and isomorphic to a double blow up of Gr(1,9); Gallardo, Huerta and Schmidt [GHS18] computed its effective cone. Chen, Coskun and Nollet [CCN11] showed that the component corresponding to a pair of codimension two linear spaces meeting transversely is smooth and isomorphic to a blow of Sym2Gr(n−2,n). They also completely worked out its Mori theory. It is thus very interesting to find components of Hilbert schemes that are smooth and describe their birational geometry.
Let k be an algebraically closed field with chark=2 and let d≥c≥2. Let X be the union of an (n−c)-dimensional plane and an (n−d)-dimensional plane meeting transversely in Pn. The Hilbert polynomial of X is
[TABLE]
There is an integral component of HilbPn−c,n−dn(t)Pn, denoted Hn−c,n−dn or Hn−c,n−d(Pn), whose general point parameterizes X, see Proposition 1.4.
We begin with the natural rational map
[TABLE]
If c=d, the rational map is S2-equivariant where S2 is the group of order 2. It acts on Gr(n−c,n)2 by interchanging the two factors and acts trivially on Hn−c,n−cn.
Definition 0.1**.**
For each 1≤i≤c define an incidence variety
[TABLE]
Note that Ξ is defined on the open subset where the two planes meet transversely. If X spans Pn (when n≥c+d−1) then this open set is precisely the complement of Γc. Moreover, in this case, Ξ is also defined on the complement of Γc−1 (Lemma 1.5). By explicitly resolving Ξ and studying the induced morphism, we obtain
Theorem A.
Let c≥2 and n≥2c−1. The component Hn−c,n−cn is smooth and the map Ξ induces an isomorphism
[TABLE]
where \widebarΓi is the strict transform of Γi/S2.
If n<2c−1, the morphism Hn−c,n−cn⟶Gr(2n−2c+1,n) that sends a scheme to its linear span is smooth; the fiber over a point Λ is Hn−c,n−c(Λ).
Theorem B.
Let d>c≥2 and n≥c+d−1. The component Hn−c,n−dn is smooth and Ξ extends to an isomorphism
[TABLE]
If n<c+d−1, the morphism Hn−c,n−dn⟶Gr(2n−c−d+1,n) that sends a scheme to its linear span is smooth; the fiber over a point Λ is Hn−c,n−d(Λ).
Historically, Harris [H82] suggested that H1,13≃Bl\widebarΓ1Sym2Gr(1,3) and that Hilb2t+2P3 is the union of H1,13 and another smooth component meeting transversely. The authors of [CCN11] generalized this and proved that Hn−2,n−2n≃Bl\widebarΓ1Sym2Gr(n−2,n) is smooth and meets exactly one other component in HilbPn−2,n−2n(t)Pn.
A major step in the proof of these statements was a computation of an analytic neighbourhood of a point in the intersection of the two components using the tangent-obstruction theory for the Hilbert scheme [CCN11, Proposition 2.6].
Unfortunately, for general c,d there are many, sometimes singular, components meeting Hn−c,n−dn (Remark 3.2). Thus a description of a neighbourhood of a point in the intersection of all these components is most likely intractable. Our proof of Theorem A circumvents this by using the explicit construction of Ξ and studying the induced map on tangent spaces.
In [R19] we expounded on the philosophy that the complexity of a Hilbert scheme can be measured by their number of Borel fixed points. In line with this reasoning, we have the following result:
Theorem C, C’.
The component Hn−c,n−dn has a unique Borel fixed point.
We also give a complete description of all the subschemes parameterized by Hn−c,n−dn. In light of Theorem A, B it is enough to consider the case n≥c+d−1. A double structure on an integral subscheme Z⊆Pn is a subscheme Z′⊆Pn such that Zred′=Z and deg(Z′)=2deg(Z). A double structure is said to be pure if it has no embedded components.
Theorem D.
Let n≥2c−1. Let Z be a subscheme parameterized by Hn−c,n−cn. Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1≤i1<⋯<ir≤c and a flag of linear spaces Λ1⊆Λ2⊆⋯⊆Λr⊆Pn with codimPn(Λℓ)=(c+iℓ−1) for each ℓ, such that
(i)
If i1>1 then Z is a union of two planes meeting along Λ1 with embedded pure double structures on Λℓ for each 1≤ℓ≤r.
2. (ii)
If i1=1 then Z is a pure double structure on Λ1 with embedded pure double structures on Λℓ for each 2≤ℓ≤r.
The description when c=d is similar and can be found in Theorem D’.
Corollary E, E’.
Up to projective equivalence, there are exactly 2c schemes parameterized by Hn−c,n−dn.
When chark=0, we use our explicit description of Ξ and the classification of ideals parameterized to study the effective and nef cones of Hn−c,n−dn. As a consequence, we deduce that Hn−c,n−dn is always a Mori dream space.
Definition 0.2**.**
Let Y be a smooth projective variety with Cl(Y) finitely generated. Then Y is a Mori dream space if the Cox Ring of Y is finitely generated over k. The Cox ring of Y is defined to be
⨁m∈ZkH0(Y,OY(∑imiDi))
where D1,…,Dk are chosen to generate Cl(Y).
We also determine the pairs (c,d) for which the component is Fano. For the rest of the introduction Λm will always denote an m-dimensional linear subspace of Pn. We begin with a description of the divisors.
Definition 0.3**.**
Let n≥2c−1. For each 1≤i≤c−1 and a choice of a flag of linear spaces {Λi−1⊆Λ2c−1−i}, let Di denote the divisor class of the locus of subschemes Z∈Hn−c,n−cn, for which the linear span of Λi−1∪(Z∩Λ2c−1−i) has dimension less than 2c−i−1. Let Dc denote the divisor class of the locus of subschemes that meet a fixed Λc−1.
Definition 0.4**.**
Let n≥2c−1. Let N1 denote the divisor class of the locus of generically non-reduced subschemes in Hn−c,n−cn. For each 2≤i≤c−1, let Ni denote the divisor class of the locus of subschemes with an embedded (n−c+1−i)-plane. If n=2c−1 let Nc denote the divisor class of the locus of subschemes with an embedded point. If n>2c−1 let Nc denote the class of the closure of the locus of pairs of planes meeting transversely, where the intersection of the two planes meets a fixed Λ2c−1.
Here are the results when c=d and the pair of planes span Pn.
Theorem F**.**
Let c≥2 and n≥2c−1. The component Hn−c,n−cn is a Mori dream space and we have,
[TABLE]
Moreover, Hn−c,n−cn is Fano if and only if either c=3 and n=5, or c=3 and n∈{2c−1,2c}.
To state the results when the pair of planes do not span Pn, it is more convenient to use dimension instead of codimension to index the component. In particular, the component parameterizing subschemes that do not span Pn are of the form Hc−1,d−1n with n>c+d−1.
Definition 0.5**.**
Let n>2c−1. For each 1≤i≤c−1 and a choice of flag {Λn−2c+i⊆Λn−i}, let Di′ denote the divisor class of the locus of subschemes Z∈Hc−1,c−1n, for which the linear span of Λn−2c+i∪(Λn−i∩Z) has dimension less than n−i. Let Dc′ denote the divisor class of the locus of subschemes meeting a fixed Λn−c. Let F denote the divisor class of the locus of subschemes whose linear span meets a fixed Λn−2c.
Definition 0.6**.**
Let n>2c−1. Let N1′ denote the divisor class of the locus of generically non-reduced subschemes in Hc−1,c−1n. For each 2≤i≤c, let Ni′ denote the divisor class of the locus of subschemes with an embedded (c−i)-plane.
Here are the results when c=d and the pair of planes do not span Pn.
Theorem G**.**
Let c≥2 and n>2c−1. The component Hc−1,c−1n is Fano and thus a Mori dream space. Moreover we have,
[TABLE]
The precise results when c=d can be found in Section 7. We conclude the introduction by describing the components that are Fano in this case; the results mirror the case of c=d.
Theorem H**.**
The component Hc−1,d−1n is Fano. The component Hn−c,n−dn is Fano if and only if either c=2 and n∈{d+1,…,2d−1}, or c≥3 and n∈{c+d−1,c+d}.
Organization**.**
In Section 1 we construct the component Hn−c,n−dn and show that the rational map Ξ is defined away from Γc−1. In Section 2 we thoroughly study the case c=d. We begin by explicitly constructing a morphism, also denoted Ξ, from a sequence of blowups to Hn−c,n−cn (Proposition 2.3, Proposition 2.8). We then construct a Gröbner basis for ideals in the image of Ξ (Lemma 2.5), which is indispensable in showing Ξ is bijective and proving Theorems C, D. By analyzing the differential of Ξ at the Borel fixed point we deduce Theorem A. In Section 3 we explain how to carry out all of the proofs of Section 2 with little to no modification for the case c=d. In Section 4 we study the divisors on Hn−c,n−cn and provide local equations for them. In Sections 5, 6 we study the birational geometry of Hn−c,n−cn,Hc−1,c−1n and prove Theorem F and Theorem G. More precisely, the cones are computed in Proposition 5.12 and Proposition 6.10. The fact that the components are Mori dream spaces is established in Theorem 6.14. In Section 7 we explain how to carry out all of the proofs of Section 5 and 6 for the case c=d.
1. Preliminaries
In this section we fix our notation, verify the existence of a component parameterizing a pair of linear spaces (Proposition 1.4) and describe some of its properties.
Notation: Let k be an algebraically closed field. For Sections 1 - 3 we will assume chark=2 and for Sections 4 - 7 we will assume chark=0. We use S to denote the polynomial ring k[x0,…,xn] and Sd to denote the subspace of monomials of degree d. For a homogenous ideal I⊆S we use Id to denote the subspace of degree d elements of I. We use [I] or [X] to denote the k-point in the Hilbert scheme corresponding to X=Proj(S/I)⊆Pn and we use PX(t) or PS/I(t) to denote its Hilbert polynomial. The ideal associated to a subscheme always refers to its saturated ideal.
We use Gr(r,n) to denote the Grassmannian variety parameterizing r-dimensional linear spaces in Pn. The span of a subscheme X⊆Pn is the linear subspace V(H0(Pn,IX(1)))⊆Pn.
The letters c and d are reserved for the codimension of linear spaces in Pn; throughout the paper, we always assume n≥d≥c≥2. Similarly we reserve the letter k=c=d for the case they are equal.
All the divisors we will consider are assumed to be Cartier. Given a smooth variety Y, we let N1(Y) denote the group of Cartier divisors modulo numerical equivalence. Nef(Y) and Eff(Y) denote the nef and effective cones of Y, respectively. We use ⟨D1,…,Dl⟩ to denote the convex cone in N1(Y)⊗R generated by the divisors Di. For more details we refer to [D01, Chapter 1].
Let X denote the union of an (n−c)-plane and (n−d)-plane meeting transversely in Pn. It is clear that X is parameterized by an open subset of Gr(n−c,n)×Gr(n−d,n) of dimension c(n−c+1)+d(n−d+1). If we show that the tangent space to [X] on its Hilbert scheme has dimension c(n−c+1)+d(n−d+1), it will follow immediately that there is an irreducible component of HilbPn−c,n−dn(t)Pn whose general member parameterizes X and whose natural scheme structure is reduced.
Since X is projectively equivalent to Z=V(x0,…,xc−1)∪V(xn−d+1,…,xn); thus it suffices to compute the tangent space to [Z] on its Hilbert scheme. For the rest of this section we fix Z and P(t)=Pn−c,n−dn(t).
If Z≃Pn−c⊔Pn−d is a disjoint union of linear spaces, it is smooth; this occurs if and only if n≤c+d−1. In this case we have a splitting of normals sheaves
[TABLE]
Thus we obtain, h0(Pn,NZ/Pn)=c(n−c+1)+d(n−d+1) and h1(Pn,NZ/Pn)=0. It follows that [Z] is a smooth point on its Hilbert scheme [H10, Theorem 1.1c]. If n>c+d−1, we will explicitly compute the tangent space to [Z] using the following result:
Let X⊆Pn be a subscheme with ideal IX=(f1,…,fr)⊆S where degfi=ei satisfying, (S/IX)e≃H0(Pn,OX(e)) for e=e1,…,er. Then there is an isomorphism between the universal deformation space of IX and that of X.
In particular, T[X]HilbP(t)Pn=H0(Pn,NX/Pn)=HomS(IX,S/IX)0.
Remark 1.2**.**
With notation as in the above Theorem, consider the following exact sequence in local cohomology [E05, Corollary A1.12],
[TABLE]
If we show that Hmi(S/IX)e=0 for e=e1,…,er and i=0,1, then the Comparison theorem would apply. Here are two instances in which this is true
(i)
The depth of S/IX is at least 2 [E05, Corollary A1.13].
2. (ii)
The Castlenuovo-Mumford regularity of the ideal IX is min{e1,…,er} [E05, Proposition 4.16]. Note that reg(IX)=reg(S/IX)+1.
Since n>c+d−1, the depth of S/IZ is at least 2. It follows from the previous Remark that the comparison theorem applies for Z.
Lemma 1.3**.**
We have dimkT[Z]HilbP(t)Pn=c(n−c+1)+d(n−d+1).
Proof.
We only need to consider the case n>c+d−1. Moreover, it suffices to show that the tangent space dimension is at most c(n−c+1)+d(n−d+1). In particular it is enough to show that any φ∈Hom(IZ,S/IZ)0 can be written as
[TABLE]
for any 0≤i≤c−1 and n−d+1≤j≤n with some constants, aℓi,bℓi∈k.
Let us first show that φ(xixj) is supported on {xix0,…,xixn−d,xjxc,…,xjxn}. Let i,j be any integers satisfying 0≤i≤c−1 and n−d+1≤j≤n. Choose j′ such that n−d+1≤j′≤n and j=j′. Since φ is an S-module homomorphism we have, xj′φ(xixj)=xjφ(xixj′).
This implies that xj divides every non-zero monomial in φ(xixj) that is not annihilated by xj′ in S/IZ. It follows that φ(xixj) is supported on
[TABLE]
Similarly, choose i′ such that 0≤i′≤c−1 and i′=i. Then the equality xi′φ(xixj)=xiφ(xi′xj) implies xi divides every monomial in φ(xixj) that is not annihilated by xi′. Once again we see that φ(xixj) is supported on
[TABLE]
Thus φ(xixj) is supported on
C∩C′={xix0,…,xixn−d,xjxc,…,xjxn}.
For any i,j, write φ(xixj)=∑ℓ=0n−daℓi,jxixℓ+∑ℓ=cnbℓi,jxjxℓ with bℓij,aℓij∈k. Using the relation xj′φ(xixj)=xjφ(xixj′) we see that bℓi,j=bℓi,j′ for each ℓ and all j,j′. Using the relation xi′φ(xixj)=xiφ(xi′xj) we obtain aℓi,j=aℓi′,j for each ℓ and all i,i′. Thus φ is of the form described in (1.1).
∎
We immediately deduce the following.
Proposition 1.4**.**
There is an integral component of HilbP(t)Pn, denoted Hn−c,n−dn or Hn−c,n−d(Pn), whose general point parameterizes an (n−c)-plane and an (n−d)-plane meeting transversely in Pn.
In the introduction we defined a rational map (0.1)
[TABLE]
This map is well defined along the locus where Λ,Λ′ meet transversely, because in this situation IΛIΛ′=IΛ∩IΛ′. In many cases, Ξ is in fact defined on a slightly larger open set.
Lemma 1.5**.**
Let n≥c+d−1. The rational map Ξ extends to the complement of Γc−1.
Proof.
We need to show that Ξ is defined along Γc∖Γc−1. Up to projective equivalence, an element of Γc∖Γc−1 is of the form V(x0,…,xc−1)∪V(x0,xc,…,xc+d−2). It suffices to show that J=(x0,…,xc−1)(x0,xc,…,xc+d−2) has Hilbert polynomial P(t). It follows by inspecting the minimal generators of J that for any t≥1, (S/J)t is spanned by
[TABLE]
Thus the Hilbert polynomial of S/J is
[TABLE]
Using the "Hockey-Stick" identity this simplifies to
[TABLE]
∎
Lemma 1.6**.**
Let n≥c+d−1 and consider the open set
[TABLE]
The morphism Ξ∣V:V⟶Hn−c,n−dn is injective if c=d and two-to-one if c=d.
Proof.
Assume Ξ∣V(Λ,Λ′)=Ξ∣V(Λ~,Λ~′)=[Y] for some scheme Y. Observe that IΛIΛ′ is a saturated ideal. Indeed, up to projective equivalence, Λ∪Λ′=V(x0,…,xc−1)∪V(xc,…,xc−d−2,xi) with i∈{0,c−d−1}. In both cases, IΛIΛ′ is clearly saturated. Thus we have IY=IΛIΛ′ and taking nilradicals we obtain
[TABLE]
Similarly, IΛ~∪Λ~′=IYred. Equating the two expressions we have Λ∪Λ′=Λ~∪Λ~′. The conclusion now follows.
∎
2. Structure of Hn−k,n−kn
This section is devoted to an analysis of Hn−k,n−kn. The first major goal of this section is to prove that Hn−k,n−kn is smooth. We start with the case when the pair of planes parameterized spans Pn. We construct a bijective morphism from a non-singular variety to Hn−k,n−kn and deduce this is an isomorphism by proving its differential is injective (Theorem A). For the case where the pair of planes do not span Pn, we construct a certain fibration to reduce to the case where they do span (Corollary 2.21).
Let n≥2k−1 and X0=Gr(n−k,n)2. For each 1≤v≤k−1, let Xv=BlΓv⋯BlΓ1X0 and let πv:Xv⟶X0 be the blow-up morphism.
The map (0.1) induces a rational map
[TABLE]
defined away from the strict transforms of the exceptional divisors. In order to study the structure of Hn−k,n−kn, we will begin by extending Ξ to a morphism on Xk−1.
For each ordered basis E={e0,…,en} of S1 we obtain an affine neighbourhood UE=Speck[ai,j,bi,j]0≤i≤k−1k≤j≤n of X0 such that the k-points of UE correspond to
[TABLE]
It is clear that as E ranges over all ordered basis of S1, the set of UE cover X0. In particular, it suffices to extend Ξ along each πk−1−1(UE) in a compatible way. For notational convenience we may assume E={x0,…,xn} and let U0=UE. Observe that the locus Γv∩U0 is cut out by the ideal generated by the v×v minors of the matrix
[TABLE]
Thus πk−1−1(U0) is obtained by blowing up U0 along the strict transforms of the ideal generated by the v×v minors of M for v=1,…,k−1, in that order.
Proposition 2.1**.**
For each 1≤v≤k−1, there exists non-singular affine open subsets Uv⊆Xv such that the following hold.
(i)
We have Uv⊆BlΓv∩Uv−1Uv−1⊆Xv.
2. (ii)
On the open set Uv, the matrix πv⋆(M) is row equivalent to the matrix
[TABLE]
where
[TABLE]
3. (iii)
The strict transform of Γv+1 on Uv is cut out by
[TABLE]
4. (iv)
Γv+1∩Uv* is non-singular and the blowup along this locus is given by*
[TABLE]
Proof.
We begin with the definition of U1. Since Γ1 is cut out by (ai,j−bi,j)i,j on U0, it is a non-singular subscheme and we have BlΓ1∩U0U0=Projk[U0][Ti,j(1)]i,j/(Koszul relations). We define U1=D(Tk−1,n(1)).
Let Mv denote the matrix appearing in item (ii). We will prove items (i) - (iv) inductively starting with v=1. Item (i) is true for v=1 by construction. On the open set U1, the Koszul relations simplify to ai,j−bi,j=λ1Ti,j(1); here we have set Tk−1,n(1)=1. Substituting this into the matrix π1⋆(M) and subtracting appropriate multiples of the bottom row from every other row, we obtain the matrix
[TABLE]
This proves item (ii) for v=1. The ideal generated by the 2×2 minors of M1 is λ12(Ti,j(1)−Ti,n(1)Tk−1,j(1))0≤j≤n−10≤i≤k−2. Thus the ideal of the strict transform of Γ2 is (Ti,j(1)−Ti,n(1)Tk−1,j(1))0≤j≤n−10≤i≤k−2. Since this ideal is generated by a regular sequence, the blowup along it is non-singular and equal to BlΓ2∩U1U1:=Projk[U1][Ti,j(2)]i,j/(Koszul relations). This proves item (iii) and (iv) for v=1.
Now assume items (i) - (iv) have been proved for some 1≤v≤k−2. Define Uv+1=D(Tk−v−1,n−v(v+1)); equivalently let Tk−v−1,n−v(v+1)=1. Then the Koszul relations on this open simplify to Ti,j(v)−Ti,n−v+1(v)Tk−v,j(v)=λv+1Ti,j(v+1). Once we substitute this into the matrix Mv, it is straightforward to row reduce the matrix so that it becomes Mv+1. Items (i) - (iv) will follow immediately as explained in the previous paragraph.
∎
Remark 2.2**.**
It follows from Proposition 2.1 that a set of algebraically independent coordinates on Uk−1 is
[TABLE]
with T0,j(k)=T0,j(k−1)−T0,n−k+2(k−1)T1,j(k−1) for all j.
Proposition 2.3**.**
Let n≥2k−1. The rational map Ξ (2.1) extends to a morphism Uk−1⟶Hn−k,n−kn.
Proof.
We will use a to denote the tuple (ai,j)i,j and similarly use b and T(v) to denote their corresponding tuples. Moreover, we will use Λ(a) to denote the (n−k)-plane corresponding to a as in (2.2).
For each 0≤i≤k−1 let yi=xi+∑j=knbi,jxj. At the moment, Ξ maps
[TABLE]
and this is undefined along the strict transforms of the exceptional divisors. Although we may express a in terms of b and {T(v)}v, we will still describe formulas in terms of a as it simplifies the exposition.
Observe that a minimal set of generators for IΛ(a) is given by the rows of [Idk×k∣M]zT where z=[y0⋯yk−1xk⋯xn] is a row vector. Applying row operations to [Idk×k∣M] will produce different minimal sets of generators. In particular, applying the row operations we did to M to get Mk−1 (Proposition 2.1 (ii)) to the matrix [Idk×k∣M]
we obtain a new set of generators α0,…,αk−1 of IΛ(a) where
[TABLE]
and
[TABLE]
with T0,j(k)=T0,j(k−1)−T0,n−k+2(k−1)T1,j(k−1) for all j. By construction, Tk−v,n−v+1(v)=1 for all 1≤v≤k−1.
For 0≤p<q≤k−1 define the following "cross terms"
[TABLE]
where kp=k−1−p for all p and
λp,q={λk−q+1⋯λk−p if p>0λk−q+1⋯λk−1 if p=0.
Note that our convention implies λ0,1=1. Extend Ξ to Uk−1 by mapping
[TABLE]
Note that (2.4) extends the original rational map (2.3). Indeed, (2.3) is defined away from the strict transform of all the the exceptional divisors; this is the locus where λ1,…,λk−1=0. In this case we have
[TABLE]
Thus βp,q∈IΛ(a)(y0,…,yk−1) and (2.3) and (2.4) coincide.
To show that the image of (2.4) is well defined, it is enough to show that the Hilbert polynomial of an ideal J=IΛ(a)IΛ(b)+(βp,q)0≤p<q≤k−1 in this image is Pn−k,n−kn(t). In Lemma 2.5 we define a term order > on S for which
[TABLE]
Since there is a flat degeneration from J to in>J it suffices to show in>J has the desired Hilbert polynomial. It is easy to see that (S/in>J)t is spanned by
[TABLE]
Using this and the Hockey-Stick identity we deduce that Hilbert polynomial of S/in>J is
[TABLE]
∎
Prior to proving Lemma 2.5 we need the following auxiliary result.
Lemma 2.4**.**
The ideal IΛ(a)IΛ(b)+(βp,q)0≤p<q≤k−1 in the image of Equation (2.4) is projectively equivalent to an ideal of the form
[TABLE]
with μi∈k and μp,q=μk−q+1⋯μk−p for any 0≤p<q≤k.
Proof.
Applying the projective transformation that maps xi↦xi−∑j≥kbi,jxj if i≤k−1 and fixes the other xi, we may assume b=0. For each 0≤i≤k−1 let τi denote the map that sends xi↦xi+∑j=1k−i−1Ti,n−j+1(j)xk−j and fixes the other i. It is clear that τk−1∘⋯∘τ0(I) equals,
[TABLE]
For each 0≤i≤k−1 let μi=λi. If T0,j(k)=0 for all j then let μk=0. If not, choose the largest index ℓ for which T0,ℓ(k)=0 and let μk=T0,ℓ(k).
For each 1≤i≤k−1 consider the map τn−ki, that maps xn−ki↦xn−ki−∑j=kn−ki−1Ti,j(k−i)xj and fixes the other xi. As we range over all i, we obtain maps τn,…,τn−(k−2). If μk=0 let τn−(k−1) be the identity; else let τn−(k−1) denote the map that sends xℓ↦xn−k0−μk1∑j=kℓ−1T0,j(k), xn−k0↦xℓ if ℓ<n−k0, and fixes the other xi.
Using the fact that Ti,n−ki(k−i)=1 on the open set Uk−1, it is straightforward to check that τn−(k−1)∘⋯τn∘τk−1∘⋯∘τ0(I) is of the desired form.
∎
Lemma 2.5**.**
Let > denote the lexicographic ordering on S with terms ordered by x0>x1>⋯>xk−1>xn>xn−1>⋯>xk. Let J=IΛ(a)IΛ(b)+(βp,q)0≤p<q≤k−1 denote the ideal in the image of Equation (2.4). Then we have
[TABLE]
Proof.
Let J′ denote the ideal in (2.6). We will first show that
[TABLE]
Let γp,q=(xp+μp,kxn−kp)xq for 0≤p≤q≤k−1 and δp,q=xpxn−kq−μp,qxqxn−kp for 0≤p<q≤k−1. Since in>γp,q=xpxq and in>δp,q=xpxn−kq, to prove (2.7) it is enough to show that G={γp,q,δp,q}p,q is a Gröbner basis for J′. Note that G generates J′ because for p<q we have
[TABLE]
Notice that μp,qμq,k=μp,k and this will be used repeatedly in the rest of the proof.
Given a,b∈S we denote their S-pair by R(a,b)=(hin>b)a−(hin>a)b with h=gcd(in>(a),in>(b)). To show that G forms a Gröbner basis we need to show that there is a standard expression for the S-pairs in terms of elements of G with no remainder [HH11, Section 2.2-2.3].
Case 1. The standard expression of R(γp1,q1,γp2,q2): Let h=gcd(in>γp1,q1,in>γp2,q2) and we may assume p1≤p2. If h=1 then p1<p2 and we have
[TABLE]
This is obviously a standard expression with no remainder. If h=xp1 then p1=p2 or p1=q2; in the latter case we still have p1=p2 as our assumptions imply p1≤p2≤q2. Thus in both the situations we obtain R(γp1,q1,γp2,q2)=xq2γp1,q1−xq1γp1,q2=0. If h=xq1 we have either q1=q2 or q1=p2. If q1=q2 then as shown above we obtain
[TABLE]
Similarly, if q1=p2 we obtain
R(γp1,q1,γp2,q2)=xq2γp1,p2−xp1γp2,q2=−μp2,kxq2δp1,p2
(if p1=p2 this is just [math]). If h=xp1xq1 then we have p1=q1=p2=q2 or p1=p2<q1=q2; in either case R(γp1,q1,γp2,q2)=0.
Case 2. The standard expression of R(δp1,q1,δp2,q2): Let h=gcd(in>δp1,q1,in>δp2,q2) and assume p1≤p2. If h=1 we have p1<p2 and q1=q2. Then we obtain
[TABLE]
Each of the above cases is a standard expression in terms of G with no remainder 222If μp2,q2=0 then in>R(δp1,q1,δp2,q2)=μp2,q2xp1xn−kq1xq2xn−kp2. This is greater or equal to in>(xq2xn−kq1δp1,p2) and in>(xp2xn−kp1δq1,q2).. If h=xn−kq1 we have q1=q2 and p1<p2. Then we obtain
[TABLE]
If h=xp1 we have p1=p2 and wlog we may assume q1<q2. Then we have
[TABLE]
Finally if h=xp1xn−kq1 we have p1=p2<q1=q2 and thus R(δp1,q1,δp2,q2)=0.
Case 3. The standard expression of R(γp1,q1,δp2,q2): Let h=gcd(in>γp1,q1,in>δp2,q2) and note that h∈{1,xp1,xq1}. If h=xp1 we have p1=p2 and using (2.8) we obtain
[TABLE]
Both these cases are standard expressions with no remainder. If h=xq1 then q1=p2 and we obtain,
[TABLE]
Finally consider the case h=1. If we further assume p2<p1 and q2<p1 we have
[TABLE]
This is a standard expression with no remainder. We omit the other cases as their proofs are very similar (use Equation 2.8). We have now shown that G is a Gröbner basis for J′.
Since J′ and in>J′ have the same Hilbert function (as graded S-modules) and J is projectively equivalent to J′, J and in>J′ have the same Hilbert function. On the other hand, (x0,…,xk−1)2⊆in>J and xpxn−kq=in>(βp,q)∈in>J. Thus in>J⊇in>J′. Since these ideals have the same Hilbert function they must be equal, completing the proof.
∎
Remark 2.6**.**
For the rest of the paper, > will always denote the term order from Lemma 2.5 and kp will always denote k−1−p.
The following Lemma sheds some light on the structure of the subschemes in the image of the morphism, Uk−1⟶Hn−k,n−kn.
Lemma 2.7**.**
Let J=IΛ(a)IΛ(b)+(βp,q)0≤p<q≤k−1 denote the ideal in the image of the morphism (2.4). Then the following statements are true
(i)
The ideal J is saturated.
2. (ii)
If all the λi are non-zero and T(k)=0 then J is the ideal of a pair of (n−k)-planes meeting transversely.
3. (iii)
If all the λi are non-zero and T(k)=0 then J is the ideal of a pair of (n−k)-planes meeting along an (n−2k+1)-plane.
4. (iv)
Let ℓ be the smallest index for which λℓ=0. Then we have
[TABLE]
and J is the ideal of a pair of (n−k)-planes meeting along an (n−k+1−ℓ)-plane.
Proof.
Item (i) follows from the fact that depthm(S/J)≥depthm(S/in>J)≥1 where m=(x0,…,xn). The first inequality is [HH11, Theorem 3.3.4] and the second inequality is true because xk is a non-zero divisor on S/in>J.
Notice that Λ(a) and Λ(b) meet along a (n−k+1−ℓ)-plane precisely when the matrix M (Proposition 2.1 (ii)) has rank ℓ−1. As a consequence items (ii), (iii) and the second half of (iv) follow immediately. The other half of item (iv) follows from Equation 2.5 as it shows βp,q∈IΛ(a)IΛ(b) for any q>k−ℓ.
∎
Proposition 2.8**.**
Let n≥2k−1. Then Ξ induces a surjective, GL(n+1)-equivariant morphism
[TABLE]
Moreover, the quotient Xk−1/S2 is non-singular.
Proof.
In Proposition 2.3 we showed that Ξ extends to a map from Uk−1. We will now explain how the same argument gives a morphism on all of πk−1−1(U0). Consider a pair
[TABLE]
with γ1 an ordered k-subset of {0,…,k−1} and γ2 an ordered (k−1)-subset of {k,…,n}. For any such γ we can define a sequence of open sets U1γ,…,Uk−1γ such that
(1)
U1γ=D(Tγ11,γ12(1))⊆BlΓ1∩U0U0 and let Ti,jγ,(1)=Ti,j(1).
2. (2)
For v≥1, the strict transform of Γv+1 on Uvγ is cut out by
[TABLE]
3. (3)
For v≥1, the locus Γv+1∩Uvγ is non-singular and
[TABLE]
4. (4)
For v≥1, we have Uvγ=D(Tγv1,γv2γ,(v))⊆BlΓv∩Uv−1γUv−1γ.
Due to symmetry, the proof of Proposition 2.1 also establishes the above statements (note that Uk−1=Uk−1γ with γ1=(k−1,k−2,…,0) and γ2=(n,n−1,…,n−k+2)). It follows that {Uk−1γ}γ is an affine cover of πk−1−1(U0) with the natural gluing maps. We omit an explicit description of the gluing maps as they will never be used.
To construct the Uvγ and verify statement (2), we would have to row reduce M in a way analogous to Proposition 2.1 (each γ corresponds to a different sequence of row redutions). We will omit an explicit description of the matrix, but the corresponding lambdas are
[TABLE]
As in the proof of Proposition 2.3 we can choose a minimal generating set, α0γ,…,αk−1γ of IΛ(a) where
[TABLE]
for 0<p≤k−1 and
[TABLE]
with Tγk1,jγ,(k)=Tγk1,jγ,(k−1)−Tγk1,γk−12γ,(k−1)Tγk−11,jγ,(k−1).
For 0≤p<q≤k−1 we may define analogous "cross terms"
[TABLE]
Thus we obtain a morphism
[TABLE]
This is well defined as any ideal in the image of ΞUk−1γ is still projectively equivalent to an ideal in (2.6) (the proof of Lemma 2.4 works with straightforward modifications). As explained in Proposition 2.3, ΞUk−1γ will also extend the original rational map (2.3) for each γ. Thus for any γ,γ′, ΞUk−1γ and ΞUk−1γ′ agree on an open subset of Uk−1γ∩Uk−1γ′. By uniqueness of extensions, they will agree on all of Uk−1γ∩Uk−1γ′ . Gluing all these maps gives us a morphism πk−1−1(U0)⟶Hn−k,n−kn.
As mentioned in the beginning of the section, Gr(n−k,n)2 is covered by open sets of the form UE where E ranges over all ordered bases of S1. Since assuming E={x0,…,xn} was purely notational, all the discussion in this section applies verbatim to πk−1−1(UE). In particular, we obtain a morphism on each πk−1−1(UE) that extends the original rational map (2.3). Thus we can glue all these maps to obtain a morphism Ξ:Xk−1⟶Hn−k,n−kn.
Let S2={1,g} be the group on two elements and consider its natural on Gr(n−k,n)2 given by interchanging the two factors. Since each of the Γi are S2 stable, the action extends to the blowup Xk−1. If we consider the trivial action of S2 on Hn−k,n−kn, then our construction shows that Ξ is S2-equivariant. Thus, we get an induced morphism \widebarΞ:Xk−1/S2⟶Hn−k,n−kn.
Since chark=2 and g fixes a divisor (the strict transform of the exceptional divisor of X1), the Chevalley-Shephard-Todd theorem [NS02, Theorem 7.14] implies that the quotient is non-singular. Note that
[TABLE]
Since Ξ is dominant and Xk−1 is projective, \widebarΞ is surjective.
The natural action of GL(n+1) on Pn induces an action on Gr(n−k,n)2 and on Hn−k,n−kn. Since the Γi are stable under this action, it extends to an action on Xk−1. To show that Ξ is GL(n+1)-equivariant we need to show that for any g∈GL(n+1) the two morphisms, Ξ∘g:Xk−1→Hn−k,n−kn given by w↦Ξ(gw) and g∘Ξ:Xk−1→Hn−k,n−kn given by w↦gΞ(w) are identical. For any (Λ,Λ′) in the open set Gr(n−k,n)2∖Γk⊆Xk−1 we have
[TABLE]
Thus Ξ∘g and g∘Ξ must agree on all of Xk−1. It follows that \widebarΞ is also GL(n+1)-equivariant.
∎
Corollary 2.9**.**
Let n≥2k−1. Any subscheme parameterized by Hn−k,n−kn is minimally cut out by k2 quadrics.
Proof.
By the discussion in Proposition 2.8 we may reduce to considering subschemes cut out by ideals in the image of morphism (2.4). Let J denote any such ideal and note that J, as presented, is generated by quadrics. By Lemma 2.7 (i), J is saturated and thus is the ideal of its corresponding subscheme. Therefore it suffices to show that dimkJ2=k2. Since S/J and S/in>J have the same Hilbert function we have dimkJ2=dimk(in>J)2=k2 (Lemma 2.5).
∎
Remark 2.10**.**
The analogue of Lemma 2.7 holds verbatim for ideals in the image of Equation (2.9). The analogue of Lemma 2.5 is as follows: Let J be any ideal in the image of Equation (2.9) and let >γ denote a lexicographic ordering on S for which
[TABLE]
We may choose any hi so that {h1,…,hn−2k+2}={k,…,n}∖{γ12,…,γk−12}. Then we have
[TABLE]
We split the proof of the injectivity of \widebarΞ into two steps. Here is the first step.
Lemma 2.11**.**
For any γ, the restriction \widebarΞ:Uk−1γ/S2⟶Hn−k,n−kn is injective.
Proof.
It is evident from our construction that Uk−1γ is S2-stable and thus the quotient Uk−1γ/S2 is well defined. Without loss of generality we may assume Uk−1γ=Uk−1. To prove the Lemma it suffices to show that for any Z~,Z^∈Uk−1 satisfying Ξ(Z~)=Ξ(Z^), we have Z~=Z^ or g(Z~)=Z^ where where g is the non-identity of S2. Let Z~=(a~,b~,T~(1),…,T~(k)) and Z^=(a^,b^,T^(1),…,T^(k)) be their coordinates on Uk−1. The "betas" and "lambdas" corresponding to Z~ are denoted by β~i,j and λ~i respectively, and the ones corresponding to Z^ are denoted by β^i,j and λ^i.
We have
Λ(a~)∪Λ(b~)=Ξ(Z~)red=Ξ(Z^)red=Λ(a^)∪Λ(b^).
After possibly replacing Z~,Z^ by g(Z~),g(Z^) respectively, we may assume
a~=a^ and b~=b^. Thus to prove that \widebarΞ is injective, we need to now show that Z~=Z^. Since Ξ is GL(n+1)-equivariant we may apply a projective transformation and assume b~=b^=0. For simplicity we let a:=a~=a^.
By Lemma 2.7, Ξ(Z~)red=Ξ(Z^)red is a pair of (n−k)-planes meeting along an (n−k+1−ℓ)-plane for some 1≤ℓ≤k+1. If ℓ∈{k,k+1} then Z,Z lie in an open set along which Ξ was already shown to be two-to-one (Lemma 1.6). Thus we may assume ℓ≤k−1. By Lemma 2.7 it is also the smallest index for which λ~ℓ=0 and, symmetrically, the smallest index for which λ^ℓ=0.
Using Lemma 2.7 (iv) we get Ξ(Z~)=[IΛ(a)IΛ(0)+(β~p,q)0≤p<q≤k−ℓ] and Ξ(Z^)=[IΛ(a)IΛ(0)+(β^p,q)0≤p<q≤k−ℓ]. Using Lemma 2.7 (i) we have the equality
[TABLE]
I claim that (β~p,q)0≤p<q≤k−ℓ=(β^p,q)0≤p<q≤k−ℓ. Assume β~p,q=α+ω with α∈IΛ(a)IΛ(0) and ω∈(β^p,q)0≤p<q≤k−ℓ such that α,ω are linearly independent and homogenous of degree 2. Since λ^ℓ=λ~ℓ=0, the construction in Proposition 2.3 implies
[TABLE]
and
[TABLE]
This implies α=0 and we obtain B=(β~p,q)0≤p<q≤k−ℓ=(β^p,q)0≤p<q≤k−ℓ. The proof will be complete once we the show that the coordinates from Remark 2.2 of Z coincide with those of Z.
It follows from the proof of Proposition 2.1 that the coordinate Ti,j(v) admits a formal expression
[TABLE]
with Ai,j,v a polynomial in a,b,λ1,…,λv and ϵ1,…,ϵv≥1. Similarly, each λv admits a formal expression
[TABLE]
with Bi,j,v a polynomial in a,b,λ1,…,λv−1 and ϵ1,…,ϵv−1≥1.
(i)
λ^i=λ~i for all i≤ℓ: We clearly have λ^1=ak−1,n=λ~1. Since λ^v=0 for all v≤ℓ−1 we can inductively apply (2.11) to obtain
[TABLE]
2. (ii)
T^i,j(v)=T~i,j(v) for all v≤ℓ−1 and all i,j: Analogous to item (i) above, where we instead use (2.10) to conclude
[TABLE]
3. (iii)
T^i,j(v)=T~i,j(v) for all k−1≥v≥ℓ and all relevant i,j (those appearing as coordinates in Remark 2.2): Let r,s be any integers such that 0≤r<s≤k−ℓ and assume β^r,s=∑0≤p<q≤k−ℓcp,qβ~p,q for some constants cp,q∈k.
Let p′=min{p:cp,q=0} and q′=max{q:cp′,q=0}.
Then
[TABLE]
It follows that β~r,s=β^r,s. Equating the terms supported on xr we obtain
[TABLE]
It follows that T^s,j(k−s)=T~s,j(k−s) for all k≤j<n−ks. Similarly, equating the terms supported on xn−ks we obtain T^r,n−j+1(j)=T~r,n−j+1(j) for all 1≤j≤kr.
4. (iv)
T^0,j(k)=T~0,j(k) for all k≤j≤n−k+1: Combining β^0,1=β~0,1 and the equality of coordinates in (iii) we obtain
[TABLE]
Since λ^0,1=1=λ~0,1, equating the coefficients of the monomials containing x1 gives the desired result.
5. (v)
λ^i=λ~i for all i≥ℓ+1: For each ℓ+1≤i≤k−1 we have β~k−i,k−i+1=β^k−i,k−i+1. Note that λ^k−i,k−i+1=λ^i and λ~k−i,k−i+1=λ~i. Using the equality of coordinates in (iii), the expression β~k−i,k−i+1=β^k−i,k−i+1 reduces to
[TABLE]
Equating the coefficients of xk−i+1xn−i+1 gives the desired result.∎
Lemma 2.12**.**
The fiber of Ξ over the point [(x0,…,xk−1)2+(xpxn−kq)0<p<q≤k−1] consists of a single element.
Proof.
Let J denote the ideal (x0,…,xk−1)2+(xpxn−kq)0<p<q≤k−1. Let X∈Uk−1 be the point with all the coordinates of Remark 2.2 equal to [math]. We clearly have Ξ(X)=[J]. Now assume Z∈Xk−1 such that Ξ(Z)=[J]. Since Jred=(x0,…,xk−1), we must have Z∈πk−1−1(U0). In particular, Z∈Uk−1γ for some γ. By Remark 2.10 we have
[TABLE]
Comparing the monomial generators of the two ideals we deduce that γk−p1=p for all 0≤p≤k−2; this forces γ11=k−1. But then we also obtain γk−q2=n−kq=n−(k−q)+1 for all 1≤q≤k−1. Thus Uk−1γ=Uk−1 and by Lemma 2.11, Z=X or g(Z)=X for the non-identity g∈S2. Since Ξ(Z)red=Ξ(X)red=V(x0,…,xk−1) we must have g(Z)=Z; thus Z=X.
∎
Proposition 2.13**.**
Let n≥2k−1. The morphism \widebarΞ:Xk−1/S2⟶Hn−k,n−kn is injective.
Proof.
Let Y,Z∈Xk−1 such that Ξ(Y)=Ξ(Z). Since Ξ(Y)red=Ξ(Z)red we may assume wlog that Y,Z∈πk−1−1(U0). We may also assume wlog that Y∈Uk−1. By Lemma 2.11 we only need to show that Z∈Uk−1. Let ℓ≥1 be the maximal value such that Z∈Uk−1γ with γi1=k−i and γi2=n−i+1 for all i<ℓ. We need to show that ℓ=k (then automatically, γk1=0). For the sake of a contradiction, assume that ℓ<k. Our method is to compare certain initial ideal degenerations of Ξ(Z) and Ξ(Y).
Let w be any integral weight order corresponding to > [E95, Section 15]. For any t∈k⋆ let gt∈GL(n+1) denote the automorphism that maps xi↦t−w(i)xi. Since each gt just scales the coordinates the following facts are immediate
(1)
gt induces an action on X0 and extends to all the blowups Xv.
(2)
gt fixes Uℓγ and also fixes any closed subset of the form V(Ti,jγ,(ℓ)).
(3)
For each ℓ let ψℓ:Xk−1⟶Xℓ denote the blowdown map. Then ψℓ is GL(n+1)-equivariant and thus ψℓ(gt)=gt(ψℓ).
Let Y0=limt→0gt(Y) and Z0=limt→0gt(Z). Using [E95, Theorem 15.17] and Lemma 2.5 we obtain
[TABLE]
Similarly, Ξ(Z0)=(x0,…,xk−1)2+(xpxn−kq)0<p<q≤k−1=Ξ(Y0). By Lemma 2.12, Z0=Y0.
Using the notation in item (3) and our assumption on ℓ, ψℓ(Z) and ψℓ(Y) are k-points of Projk[Uℓ−1][Ti,j(ℓ)]/(Koszul)⊆Xℓ. By maximality of ℓ we have Tk−ℓ,n−ℓ+1(ℓ)(ψℓ(Z))=0 i.e. ψℓ(Z) lies in V(Tk−ℓ,n−ℓ+1(ℓ)). Then by item (2) we still have ψℓ(gt(Z))=gt(ψℓ(Z))∈V(Tk−ℓ,n−ℓ+1(ℓ)). Thus the limit ψℓ(Z0) also lies in there. But this contradicts the fact that Tk−ℓ,n−ℓ+1(ℓ)(ψℓ(Y0))=Tk−ℓ,n−ℓ+1(ℓ)(Y0)=0 (since Y0 lies in Uk−1). Thus ℓ=k and we have Z,Y∈Uk−1, as required.
∎
Remark 2.14**.**
It follows that the preimage Ξ−1(Z) is a single point precisely when Zred is an (n−k)-plane. This occurs precisely when Z is generically non-reduced, c.f. Theorem D 333If the reader is only interested in the classification of subschemes parameterized by Hn−k,n−kn they can directly skip to Lemma 2.22.
The group GL(n+1) acts on S and thus on HilbP(t)Pn by a change of coordinates. An ideal of S or its corresponding point on the Hilbert scheme is said to be Borel fixed if it is fixed by the Borel subgroup of GL(n+1) consisting of upper triangular matrices. Since a Borel fixed ideal is fixed by the subgroup of diagonal matrices, it is generated by monomials. We will now show that Hn−k,n−kn has a unique Borel fixed point. We begin with a combinatorial characterization of the Borel fixed ideals, see [E95, Section 15] for details.
Definition 2.15**.**
Let I⊆S be a monomial ideal and p a prime number. The ideal I is said to be [math]-Borel fixed if for any monomial generator m∈I divisible by xj, we have xjxim∈I for all i<j. The ideal I is said to be p-Borel fixed if for any monomial generator m∈I divisible by xjβ but no higher power of xj, we have (xjxi)αm∈I for all i<j and α⪯pβ (this means that each digit in the p-base expansion of α is less than or equal to each digit in the p-base expansion of β).
Note that a [math]-Borel fixed ideal is always p-Borel fixed for any p.
Proposition 2.16**.**
[E95, Theorem 15.23]* Let chark=p≥0. Then I⊆S is Borel fixed if and only if it I is p-Borel.*
In our situation, chark=p≥0 with p=2. Let I be a saturated p-Borel fixed ideal parameterized by Hn−k,n−kn. Since I is a monomial ideal generated by quadrics (Corollary 2.9) and p=2, the condition α⪯pβ in Definition 2.15 reduces to the condition α≤β. In particular, I is always [math]-Borel.
Proposition 2.17**.**
Let n≥2k−1. Consider the ideal
[TABLE]
Then [In−k,n−kn] is the unique Borel fixed point on Hn−k,n−kn.
Proof.
As noted above, Borel fixed ideals in Hn−k,n−kn are the same as [math]-Borel fixed ideals. Since In−k,n−kn is projectively equivalent to (x0,…,xk−1)2+(xpxn−kq)0≤p<q≤k−1, it lies in Hn−k,n−kn. It also clear that In−k,n−kn is Borel fixed. Let B be any saturated [math]-Borel fixed ideal on Hn−k,n−kn. Then it is of the form B=∑i=0ϵxi(xi,…,xai) with n−1≥a0≥a1≥⋯≥aϵ≥ϵ. Since B=(x0,…,xϵ) has codimension k, we obtain ϵ=k−1.
Arguing as in the end of the proof of Proposition 2.3 we see that the Hilbert polynomial of B is (tn−k+t)+∑i=0k−1(t−1t+n−ai−2). Equating this with the Hilbert polynomial of In−k,n−kn we have
[TABLE]
Since the set {(at−1+a)}a∈N is a Q-basis for Q[t], we obtain ai=2k−i−2 for all i; therefore B=In−k,n−kn.
∎
Lemma 2.18**.**
Let I be a (saturated) ideal parameterized by Hn−k,n−kn. Then the Castelnuovo-Mumford regularity of I is 2 and T[I]HilbPn−k,n−kn(t)Pn=HomS(I,S/I)0.
Proof.
Since I is generated by quadrics, the regularity is at least 2. Up to projective equivalence, we may assume I is of the form (2.6). By [HH11, Theorem 3.3.4] we have also reg(I)≤reg(in>I). Note that in>I is projectively equivalent to In−k,n−kn and the regularity of a [math]-Borel ideal is the highest degree of a minimal monomial generator [HH11, Corollary 7.2.3]. Thus reg(I)≤reg(In−k,n−kn)=2, as required. The description of the tangent space follows from Remark 1.2 and Theorem 1.1.
∎
Definition 2.19**.**
Let ζ denote the pre-image of [In−k,n−kn] in Xk−1 (Remark 2.14) and let ζˉ denote the image of ζ in Xk−1/S2.
By constructing curves passing through ζ and ζˉ we will now show that the differential d\widebarΞζˉ is injective. This is a major portion of the proof of Theorem A.
Lemma 2.20**.**
Let n≥2k−1. The differential d\widebarΞζˉ:Tζˉ(Xk−1/S2)⟶T[In−k,n−kn]Hn−k,n−kn is injective.
Proof.
Note that we have a factorization
[TABLE]
By non-singularity we also have dimkTζXk−1=dimkTζˉ(Xk−1/S2). Thus to show that d\widebarΞζˉ is injective it suffices to establish the following two facts
(1)
dΞζ:TζXk−1⟶T[In−k,n−kn]Hn−k,n−kn has a 1 dimensional kernel
(2)
The exists ω∈Tζˉ(Xk−1/S2) for which d\widebarΞζˉ(ω) does not lie in the image of dΞζ.
We begin with item (1). Let γ1=(k−1,k−2,…,0) and γ2=(k,k+1,…,2k−2). Then ζ is the point 0 on Uk−1γ (Proposition 2.8). As in Remark 2.2 a set of coordinates on Uk−1γ is N=N1∪⋯∪N5 where
[TABLE]
For each η∈N we define a curve Dη:Speck[t]⟶Uk−1γ, passing through 0, by setting η=t and all the other coordinates in N to [math].
Let ι:Speck[t]/(t2)⟶Speck[t] be a first order deformation of the origin. Since Xk−1 is non-singular the set {Dη∘ι}η∈N is a basis for T0Uk−1γ=TζXk−1.
We need to study the dimension of {dΞζ(Dη∘ι)}η. Since dΞζ(Dη∘ι)=(Ξ∘Dη)∘ι we begin with an explicit description of each Ξ∘Dη. The items below follow directly from the construction of the map (2.9).
(i)
If η=bi,j∈N1 then Ξ∘Dη(t) is
[TABLE]
2. (ii)
If η=Ti,k−1+jγ,(j)∈N2 then Ξ∘Dη(t) is
[TABLE]
3. (iii)
If η=Tk−i,jγ,(i)∈N3 then Ξ∘Dη(t) is
[TABLE]
4. (iv)
If η=λiγ with i>1 then Ξ∘Dη(t) is
[TABLE]
5. (v)
If η=λ1γ then Ξ∘Dη(t) is
[TABLE]
6. (vi)
If η=T0,jγ,(k)∈N5 then Ξ∘Dη(t) is
[TABLE]
Let I=In−k,n−kn and under the inclusion Hn−k,n−kn⊆HilbPn−k,n−kn(t)Pn, we may identify T[I]Hn−k,n−kn with a subspace of Hom(I,S/I)0 (Lemma 2.18). We can explicitly describe this identification using [H10, Proposition 2.3]. In particular, by re-indexing, we obtain
[TABLE]
These are the trivial deformations i.e. the ones induced by a change of coordinates. For i∈{1,…,k−2} let Δi be the derivation that maps xix2k−2−i↦xi+1x2k−1−i and other generators to [math]. Let Δk−1 denote the derivation that maps xk−12↦xk−1xk and the other generators to [math].
For i∈{2k−1,…,n} let Δi to the derivation that maps x0x2k−2↦x1xi. Then we have
[TABLE]
Notice that the derivation Δk−1 is a scalar multiple of xk∂xk−1∂. Thus to prove (1) it suffices to show that the set {xj∂xi∂}0≤i≤2k−2i+1≤j≤n∪{Δi}1≤i≤k−2∪{Δi}2k−1≤i≤n is linearly independent.
Assume we had a linear combination
[TABLE]
with some constants ϵi,j,ϵi∈k. Assume ϵp,q=0 for some p<q. Since xpx2k−2−p∈I we may evaluate (2.12) at xpx2k−2−p to obtain
[TABLE]
where
[TABLE]
Observe that the monomial xqx2k−2−p does not appear in the support of Q. Thus, in the left hand side of (2.13), the monomial xqx2k−2−p appears with a coefficient of ϵp,q if p=k−1 and a coefficient of 2ϵp,q if p=k−1. In either case, the coefficient is non-zero. But this is a contradiction as xqx2k−2−p∈/I. Thus we have ϵp,q=0 for all p,q. Evaluating (2.12) at xpx2k−2−p we see that ϵp=0 for every p∈{1,…,k−2}. Finally, evaluating (2.12) at x0x2k−2 we obtain ∑i=2k−1nϵix1xi≡0modI. Since x1xi∈/I for all i≥2k−1, we must have that ϵi=0 for all i. This completes the proof of item (1).
Let Δ∈Hom(I,S/I)0 denote the derivation that maps xk−1xk↦xk2 and all the other generators to [math]. By evaluating at xk−1xk it is easy to see that Δ does not lie in the span of {xj∂xi∂}0≤i≤2k−2i+1≤j≤n∪{Δi}1≤i≤k−2∪{Δi}2k−1≤i≤n. Consider the curve C:Speck[t]→Hn−k,n−kn given by
[TABLE]
This is well defined because for any given s∈k, C(s) is the point in Uk−1γ with λ1γ=−2s, bk−1,k=s and all other coordinates equal [math]. It is also clear that C∘ι corresponds to the derivation Δ. Thus to prove item (2) it suffices to find a curve C′:Speck[t]→Xk−1/S2 passing through ζˉ for which dζˉ\widebarΞ(C′∘ι)=C∘ι.
Let Z denote the image of C and let Z′ denote the pullback \widebarΞ−1(Z)⊆Xk−1/S2. I claim that \widebarΞ∣Z′:Z′→Z is an isomorphism. Since Z is non-singular, Z′ is Cohen-Macaulay and \widebarΞ is bijective, the morphism \widebarΞ∣Z′ is flat. It is clear that a finite flat degree 1 morphism is an isomorphism. Thus C′=\widebarΞ∣Z′−1∘C:Speck[t]→Xk−1/S2 is the desired curve.
∎
We are now ready to prove the main Theorem.
Theorem A**.**
Let n≥2k−1. The component Hn−k,n−kn is smooth and isomorphic to
[TABLE]
Proof.
Proposition 2.8 and 2.13 together show that \widebarΞ is bijective and Xk−1/S2 is non-singular. Since \widebarΞ is GL(n+1)-equivariant, ζˉ (Definition 2.19) is the unique Borel fixed point on Xk−1/S2. By Borel’s fixed point theorem, the closure of the Borel orbit of any point in Xk−1/S2 contains ζˉ. Thus to show that \widebarΞ is an isomorphism, it suffices to show that it is an isomorphism in a neighbourhood of ζˉ. By the proof of [H92, Theorem 14.9], this is equivalent to showing that d\widebarΞζˉ:Tζˉ(Xk−1/S2)⟶T[In−k,n−kn]Hn−k,n−kn is injective. This is precisely the content of Lemma 2.20.
∎
When the pair of planes do not span Pn, we obtain the following fibration
Corollary 2.21**.**
Let n<2k−1. The morphism ρ:Hn−k,n−kn⟶Gr(2n−2k+1,n) that sends a scheme to its linear span is smooth; the fiber over a point Λ is Hn−k,n−k(Λ).
Proof.
Recall that the linear span of a subscheme Z⊆Pn is the linear space V(H0(Pn,IZ(1)))⊆Pn.
Let Y⟶A1 be a flat family such that for t=0, Yt is a disjoint pair of (n−k)-planes. It is clear that for any t=0, the linear span of Yt is a (2n−2k+1)-plane. By upper semicontunity, the limit Y0 also lies in a (2n−2k+1)-plane, which we denote by Λ. Thus Y0 defines a point in Hn−k,n−kn(Λ) and by Corollary 2.9, we see that the linear span of Y0 is all of Λ. It follows that the linear span of any subscheme parameterized by Hn−k,n−k(Pn) is of dimension 2n−2k+1.
For each ordered basis E={e0,…,en} of S1 we obtain an open neighbourhood UE=Speck[fi,j]0≤i≤2k−2−n2k−1−n≤j≤n of ΛE=V(e0,…,e2k−2−n) in Gr(2n−2k+1,n). The k-point f=(fi,j)i,j is identified with
[TABLE]
Let E={ei}i,E′={ei′}i be ordered bases of S1. The isomorphism ΛE→ΛE′ given by mapping ei↦ei′ for all i induces an an isomorphism ψE,E′:Hn−k,n−k(ΛE)⟶Hn−k,n−k(ΛE′). Define the following
•
XE=Hn−k,n−k(ΛE)×UE,
•
XE,E′=Hn−k,n−k(ΛE)×(UE∩UE′)⊆XE,
•
φE,E′=ψE,E′×id:XE,E′⟶XE′,E.
It is clear that
φE,E′−1=φE′,E,
φE′,E′′∘φE,E′=φE,E′′ on XE,E′∩XE,E′′ and
φE,E′(XE,E′∩XE,E′′)=XE′,E∩XE′,E′′.
Thus the set of schemes {XE}E glue to a smooth scheme X (Theorem A).
For each E we obtain a natural morphism gE:UE⟶GL(n+1) such that for any f, gE(f) is the map that sends ei↦ei+∑j=2k−1−nnfi,jej if i≤2k−2−n and fixes the other coordinates. Thus we may define a map
[TABLE]
These maps glue to a morphism Π:X⟶Hn−k,n−kn. By the first paragraph, Π is a bijective morphism. It is also clear that the differential to Π is injective at all points. As noted in Theorem A, this implies that Π is an isomorphism. By construction, there is a smooth fibration ρ:X⟶Gr(2n−2k+1,n) of the desired form.
∎
Theorem C**.**
Hn−k,n−kn* has a unique Borel fixed point.*
Proof.
By Proposition 2.17 we my assume n<2k−1. If X is Borel fixed then its linear span V((IX)1) is also Borel fixed. Thus X lies in the fiber ρ−1(V(x0,…,x2k−2−n))≃Hn−k,n−k2n−2k+1. Moreover, the Borel action on Hn−k,n−kn restricts to the Borel action on this fiber. By Proposition 2.17 this fiber has a unique Borel fixed point; thus X is unique.
∎
We now turn our attention to the subschemes parameterized by Hn−k,n−kn. Since we are going to describe these subschemes up to projective equivalence, we may assume n≥2k−1 (Corollary 2.21). We begin with two Lemmas that will aid in the proof of Theorem D.
Lemma 2.22**.**
Let J=(x0,…,xk−1)2+(xpxn−kq−μp,qxqxn−kp)0≤p<q≤k−1 with μi∈k and μp,q=μk−q+1⋯μk−p for any 0≤p<q≤k. If all the μi are non-zero then the subscheme defined by J is Cohen-Macaulay; in particular, it has no embedded components. Moreover, the subscheme defined by J is double structure on V(x0,…,xk−1).
Proof.
Applying the change of coordinates that maps xp↦μp,kxp for all p≤k−1 and fixing the other coordinates, we may assume μp,q=1 for all p,q. If n>2k−1, the variables xk,…,xn−k form a regular sequence as they do not appear in the support of the generators of J. Thus we may quotient by the ideal (xk,…,xn−k) to reduce to the case n=2k−1; in this case n−kp=k+p. Since Proj(S/J) is supported on V(x0,…,xk−1), it suffices to verify the Cohen-Macaulayness on the open sets D(xk),…,D(x2k−1).
On the open set W=D(xk) we may set xk=1. Then for all j=0 we have xj−x0xk+j=−(x0xk+j−xjxk)∈J∣W and this implies J∣W=(x02,x1−x0xk+1,…,xk−1−x0x2k−1). Since xk,…,x2k−1 forms a regular sequence on (S/J)∣W, Proj(S/J)∣W is a Cohen-Macaulay subscheme of dimension k−1. The argument for the other open sets is the same.
Since the Hilbert polynomial of Proj(S/J) is Pn−k,n−kn(t), its degree is 2; thus it is a double structure on the linear space V(x0,…,xk−1)
∎
Remark 2.23**.**
More generally, (xϵ1,…,xϵ2)2+(xpxn−kq−μp,qxqxn−kp)ϵ1≤p<q≤ϵ2 is Cohen-Macaulay for any 0≤ϵ1≤ϵ2≤k−1, assuming μi=0 for all i.
Lemma 2.24**.**
Let 0≤ϵ1≤ϵ2≤k−1 and let J(ϵ1,ϵ2)=(xϵ1,…,xϵ2)2+(xpxn−kq)ϵ1≤p<q≤ϵ2. Then we have a primary decomposition
[TABLE]
Proof.
For the first statement we proceed by induction on ϵ2. The base case ϵ2=ϵ1 is vacuous and by induction we may assume
[TABLE]
The conclusion now follows from the fact that if I1=(m1,…,mi1),I2=(m1,…,mi2) are monomial ideals then I1∩I2=(lcm(mimj):1≤i≤i1,1≤j≤i2).
∎
Theorem D**.**
Let n≥2k−1. Let Z be a subscheme parameterized by Hn−k,n−kn. Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1≤i1<⋯<ir≤k and a flag of linear spaces Λ1⊆Λ2⊆⋯⊆Λr⊆Pn with codimPn(Λℓ)=(k+iℓ−1) for each ℓ, such that
(i)
If i1>1 then Z is a union of two planes meeting along Λ1 with embedded pure double structures on Λℓ for each 1≤ℓ≤r.
2. (ii)
If i1=1 then Z is a pure double structure on Λ1 with embedded pure double structures on Λℓ for each 2≤ℓ≤r.
Proof.
It suffices to compute a primary decomposition of the ideal
[TABLE]
in (2.6). Let P0=(xp+μp,kxn−kp)0≤p≤k−1, P1=(x0,…,xk−1) and δp,q=xpxn−kq−μp,qxqxn−kp for each 0≤p<q≤k−1. Lemma 2.7 (ii) implies that all the μi are non-zero if and only if J is the ideal of a pair of (n−k)-planes meeting transversely. So we may assume some of the μi are zero. Let i1<⋯<ir be all the indices i for which μi=0. Set i0=0 and ir+1=k+1. Lemma 2.7 (iv) implies J=P0∩P1 and J=P0P1+(δp,q)0≤p<q≤k−i1. For each 2≤ℓ≤r+1 define
[TABLE]
I claim that J=P0∩P1∩⋯∩Pr+1 (note that if μ1=0 then P0=P1). We begin with the inclusion, J⊆P0∩⋯∩Pr+1. It is enough to show that P0P1 and δp,q lie in P0∩⋯∩Pr+1 for 0≤p<q≤k−i1. Observe that
[TABLE]
Clearly, (x0,…,xk−i1)(x0,…,xk−1)⊆Pj for all j. We also have, xp,xn−kp∈Pj for all k−i1+1≤p≤k−1 and all j. Thus P0P1⊆P0∩⋯∩Pr+1. It is clear that δp,q∈P0∩⋯∩Pr+1 if there is some ℓ such that k−iℓ+1≤p<q≤k−iℓ−1. If this was not the case, then there is some ℓ such that p≤k−iℓ<q. This implies δp,q=xpxn−kq and this lies in (x0,…,xk−ij) if j≤ℓ or in (xn−ij−1+2,…,xn) if j>ℓ; in either case, δp,q∈Pj. Thus δp,q∈P0∩⋯∩Pr+1 and we have the desired containment.
To get the other containment it suffices to show that P0∩⋯∩Pr+1 has the same Hilbert function as J. We have
[TABLE]
Our goal is to show all these containments are equalities. Using Equation (2.8) we have
Using Lemma 2.24 we see that in>(P0∩P1)∩in>P2∩⋯∩in>Pr+1 equals
[TABLE]
Applying Lemma 2.24 once again we see that this intersection is just J(0,k−1)∩(x0,…,xk−1). But this ideal is precisely (x0,…,xk−1)2+(xpxn−kq)0<p<q≤k−1=in>J. Thus all the containments in (2.14) are equalities and this shows that J has the same Hilbert function as P0∩⋯∩Pr.
We are left with showing Pℓ is a primary component for all ℓ≥2. Going modulo the linear forms it suffices to show that
(xk−iℓ+1,…,xk−iℓ−1)2+(βp,q)k−iℓ+1≤p<q≤k−iℓ−1
is a primary component. This is the content of Lemma 2.22 and Remark 2.23.
∎
Corollary E**.**
Up to projective equivalence, there are exactly 2k schemes parameterized by Hn−k,n−kn.
Proof.
By Corollary 2.21 we may assume n≥2k−1. It suffices to consider ideals J of the form (2.6). Let φ denote the projective transformation that maps xp↦μp,kxp if μp,k=0 and 0≤p≤k−1 and fixes the other coordinates. For a fixed p, note that if μp,k=0 then μq,k=0 and μp,q=0 for all p<q. Thus after applying φ we may assume that the non-zero μi are equal to 1. In particular, for each subset W⊆{1,…,k} we obtain an ideal parameterized by Hn−k,n−kn by setting μi=0 if i∈W and 1 otherwise; this gives at most 2k distinct ideals. On the other hand, since projective transformations preserve the dimensions of the embedded structures, each of the 2k ideals are projectively inequivalent.
∎
Example 2.25**.**
We can now determine when there is a specialization Z⇝Z′ in Hn−k,n−kn. For any subscheme Z∈Hn−k,n−kn let WZ={ϵ1,…,ϵr} be the set of dimensions of the embedded components of Z; if Z is generically non-reduced include n−k in that set. Then there is a specialization Z⇝Z′ if and only if WZ⊆WZ′
Here is a diagram of specializations for H2,25. The non-reduced structures on points, lines and planes are represented by shadings.
(i)(ii)(iii)(v)(iv)(vi)(vii)(viii)
Remark 2.26**.**
In [V14], Vainsencher uses the map Ξ:BlΓ2BlΓ1Gr(2,5)2→H2,25 to compute the degree of a family of rational cubic fourfolds in P5. However, he does not prove the smoothness of H2,25.
3. Structure of Hn−c,n−dn
In this short section we explain how the proofs of the previous section carry over, almost identically, to the case when the pair of planes are of different dimension. We begin by explaining the special case of c=1 that we have omitted.
Remark 3.1**.**
If c=1 then HilbPn−1,n−dn(t)Pn parameterizes ideals of codimension 1. Using the decomposition in [R19, Proposition 2.4] we obtain
[TABLE]
Thus Hn−1,n−dn is smooth and isomorphic to the full Hilbert scheme. Alternatively, we can deduce this from the proof of Lemma 1.6 and a computation of the tangent space to the unique Borel fixed ideal on HilbPn−1,n−dn(t)Pn.
Let d>c≥2 and assume n≥c+d−1. Let Xc−1=BlΓc−1⋯BlΓ1(Gr(n−c,n)×Gr(n−d,n)) and let πc−1:Xc−1⟶Gr(n−c,n)×Gr(n−d,n) be the blow up.
We have shown in Lemma 1.6 that the rational map Ξ:Xc−1⇢Hn−c,n−dn is defined and one-to-one on the open set Gr(n−c,n)×Gr(n−d,n)∖Γ1∪⋯∪Γc−1. To extend Ξ to Xc−1 we proceed as in Section 2. We first extend Ξ to πc−1−1(U0) where U0=Speck[ai,j,bi,j]i,j is an open subset of Gr(n−c,n)×Gr(n−d,n) such that its k-points correspond to
[TABLE]
We will now perform a few substitutions and obtain a different minimal set of generators for IΛ(a) and IΛ(b). From these new presentations of IΛ(a) and IΛ(b), it will be apparent how one has to mimic the arguments of Section 2 to extend Ξ to πc−1−1(U0), and thus all of Xc−1. For every 0≤i≤c−1, 0≤j≤d−1 and d≤p≤n let
[TABLE]
For any 0≤i≤c−1 we obtain
[TABLE]
Thus we have
[TABLE]
and
[TABLE]
From these descriptions of IΛ(a) and IΛ(b) it follows that Γv∩U0 is cut out by the ideal generated by the v×v minors of the matrix
[TABLE]
We can now prove an analogue of Proposition 2.1. Moreover, using the presentations in (3.1) and (3) and arguing as in Proposition 2.3, 2.8 we can construct a morphism πc−1−1(U0)⟶Hn−c,n−dn extending the rational map Ξ. An argument identical to the one given for Proposition 2.13 will show that this extends to a bijective morphism Ξ:Xc−1⟶Hn−c,n−dn. In a similar manner we may deduce the following results
Theorem C’****.
Let d>c≥2. The component Hn−c,n−dn has a unique Borel fixed point. If n≥c+d−1 the point
[TABLE]
is the unique Borel fixed point on Hn−c,n−dn.
Arguing as in Lemma 2.20, Theorem A and Corollary 2.21 we obtain
Theorem B**.**
Let d>c≥2 and n≥c+d−1. The component Hn−c,n−dn is smooth and there is an isomorphism
[TABLE]
If n<c+d−1, the morphism Hn−c,n−dn⟶Gr(2n−c−d+1,n) that sends a scheme to its linear span is smooth; the fiber over a point Λ is Hn−c,n−d(Λ).
Theorem D’****.
Let n≥c+d−1 and let Z be a subscheme parameterized by Hn−c,n−dn. Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1≤i1<⋯<ir≤c and a flag of linear spaces Λ1⊆Λ2⊆⋯⊆Λr⊆Pn with codimPn(Λℓ)=(d+iℓ−1) for each ℓ, such that
(i)
If i1>1 then Z is a union of two planes meeting along Λ1 with embedded pure double structures on Λℓ for each 1≤ℓ≤r.
2. (ii)
If i1=1 then Z is a codimension c-plane with embedded pure double structures on Λℓ for each 1≤ℓ≤r.
Corollary E’****.
Up to projective equivalence, there are exactly 2c subschemes parameterized by Hn−c,n−dn.
Remark 3.2**.**
In [CCN11] it was shown that Hn−2,n−2n meets exactly one other component in HilbPn−2,n−2n(t)Pn and that this component is smooth. We will give two examples that show these statements are false in general.
The component H2,25 will meet the component whose general member parameterizes a pair of 2-planes meeting at a point union an isolated point. It will also meet the component whose general member parameterizes a quadric union an isolated line.
In [R19, Theorem 3.16] we show that HilbPn−2,1n(t)Pn is a union of Hn−2,1n and a component Y, whose general point parameterizes a line meeting an (n−2)-plane union an isolated point. We show that Y is singular; its singularity is a cone over the Segre embedding of P1×Pn−2↪P2(n−1)−1.
This completes the discussion of the local structure of Hn−c,n−dn. The next four sections will pertain to its global geometry. As we did in Section 2, we begin studying divisors on Hn−c,n−dn with c=d=k and n≥2k−1.
4. Divisors on Hn−k,n−kn
In this section we study the Picard group of Hn−k,n−kn for n≥2k−1. We give an explicit description of the divisors Di,Ni (Remark 4.6, 4.9) and describe equations for their pullback along Ξ∣Uk−1.
Notation 4.1**.**
We will use λk to denote the coordinate T0,n−k+1(k) on Uk−1 from Remark 2.2. This convention will simplify the formulas for the equations we will obtain.
The proofs of Theorem D and Lemma 2.6 give explicit equations for the various loci of embedded structures.
Lemma 4.2**.**
Let n≥2k−1 and let Z be a subscheme parameterized by Ξ(Uk−1). Then
(i)
Z* is a pair of planes meeting transversely if and only if λ1,…,λk−1,T(k)=0.*
2. (ii)
Z* has an embedded (n−2k+1)-plane if and only if T(k)=0.*
3. (iii)
For each 2≤i≤k−1, Z has an embedded (n−k+1−i)-plane if and only if λi=0.
4. (iv)
Z* is generically non-reduced if and only if λ1=0.*
Definition 4.3**.**
Consider the sequence of blowups
Xk−1⟶ψk−1Xk−2⟶ψk−2⋯⟶ψ1X0.
For each i let Ei denote the strict transform in Xk−1 of the exceptional divisor of ψi. Let Ek denote the strict transform of Γk.
Lemma 4.4**.**
Let n≥2k−1. Then N1(Hn−k,n−kn)=Cl(Hn−k,n−kn)=Zk. In particular, linear equivalence and numerical equivalence for divisors coincide.
Proof.
Since Hn−k,n−kn=Xk−1/S2 is a smooth rational variety, its class group is torsion free. In particular, N1(Xk−1/S2)=Cl(Xk−1/S2). Thus it suffices to prove that Cl(Xk−1/S2)Q:=Cl(Xk−1/S2)⊗Q is isomorphic to Qk. By [F98, Example 1.7.6] we have Cl(Xk−1/S2)Q=Cl(Xk−1)QS2. Let E1,0 and E0,1 be the strict transform, in Xk−1, of OX0(1,0) and OX0(0,1), respectively. By [H77, Theorem 8.5], Cl(Xk−1)Q is freely generated by E1,…,Ek−1,E1,0,E0,1. Since S2 fixes Ei and interchanges E1,0 with E0,1, it follows that
[TABLE]
Definition 4.5**.**
Let (X0)trv=X0∖Γk denote the open subset of X0 consisting of pairs of (n−k)-planes such that the two planes in the pair meet transversely. We say that a pair of (n−k)-planes meets another plane Λ transversely, if each plane in the pair meets Λ transversely.
We now describe Di as a scheme theoretic image under Ξ.
Remark 4.6**.**
For each 1≤i≤k−1 consider a flag Fi={Λi−1⊆Λ2k−1−i}. Let Wi⊆(X0)trv be the open subset consisting of pairs of planes that meet Λ2k−1−i transversely. Let D^i denote the (scheme theoretic) closure of
[TABLE]
in X0. Then Di is the image of the strict transform of D^i under the map Ξ.
Similarly, given a plane Λk−1, let D^k be the scheme theoretic closure of
[TABLE]
in X0. Then Dk is the image of the strict transform of D^k under the map Ξ.
Lemma 4.7**.**
The loci Di are divisorial. For 1≤i≤k−1 let Di be defined by the flag
[TABLE]
Then Ξ⋆(Di)∩Uk−1 is cut out by
Ti−1,n−ki(k−i)+Ti−1,n−ki(k−i)Ti,n−ki−1(k−i)+λk−i+1.
Proof.
Assume 1≤i≤k−1 and let Di be defined by the flag (4.1). To show that Di is a divisor, it suffices to show that D^i∩Wi is a divisor in Wi (notation from Remark 4.6). By symmetry, it is enough to show that D^i∩Wi∩U0 is a divisor in Wi∩U0.
Given a point (Λ(a),Λ(b))∈Wi∩U0 we have (Λ(a)∪Λ(b))∩Λ2k−1−i=P∪Q for a pair of (k−1−i)-planes, P and Q. For each n−ki+1≤j≤n let pj (respectively qj) denote the point in P (respectively Q) obtained by setting xj=1 and xℓ=0 for all otherℓ≥k (there are no such points for i=k−1).
Explicitly,
[TABLE]
Let pn−ki (respectively qn−ki) denote the point in P (respectively Q) obtained by setting xn−ki=xn−ki−1=1 and xℓ=0 for all other ℓ≥k. Explicitly,
[TABLE]
For each ℓ∈{0,…,i−2,i} let rℓ=V(x0,…,xℓ−1,xℓ+1,…,xn).
By construction we have, P=span(pn−ki,…,pn), Q=span(qn−ki,…,qn) and Λi−1=span(r0,…,ri−2,ri). It follows that the points in span(Λi−1∪((Λ(a)∪Λ(b))∩Λ2k−1−i)) are in the row span of the matrix
[TABLE]
In particular, D^i∩Wi∩U0 is the locus where the matrix has rank less than 2k−i. Let ϵl,j=al,j−bl,j and apply the row operation
[TABLE]
and swap the i-th column and (i−1)-st column. It follows that the locus
is cut out by the determinant of the submatrix
[TABLE]
Thus D^i∩Wi∩U0 is a divisor and this determinant also cuts out D^i∩U0.
The strict transform of this determinant cuts out Ξ⋆(Di)∩Uk−1. Pulling back this matrix to Uk−1 and column reducing as in Proposition 2.1 we obtain
[TABLE]
The strict transform of its determinant is Ti−1,n−ki(k−i)+Ti−1,n−ki−1(k−i).
•
If i>1 we may use Proposition 2.1 (ii) to rewrite
Ti−1,n−ki−1(k−i)=λk−i+1+Ti−1,n−ki(k−i)Ti,n−ki−1(k−i).
•
If i=1 we may use Remark 2.2 to rewrite
T0,n−k+1(k−1)=λk+T0,n−k+2(k−1)T1,n−k+1(k−1).
In either case, Ξ⋆(Di)∩Uk−1 is cut out by the desired equation. Lastly, Dk is a divisor since D^k is the Weil divisor associated to OX0(1,1)∈PicX0≃Z2.
∎
Corollary 4.8**.**
Let 0≤j<i. For 1≤i≤k−1 let Di be defined by the flag
[TABLE]
55footnotetext: *if j=0 then kj−1=k−1=k is still consistent with our convention, see Remark *2.6
and let Dk be defined by the plane
[TABLE]
Then Ξ⋆(Di)∩Uk−1 is cut out by a polynomial in the coordinates of Remark 2.2 that is linear in λk−j.
Proof.
Assume i≤k−1 and j=0. Imitating the proof of Lemma 4.7 we see that Ξ⋆(Di)∩Uk−1 is cut out by Tj,n−kj(k−i)+Tj,n−kj−1(k−i). To express this in terms of our desired coordinates we will use the relation Tp,q(ℓ)=Tp,n−ℓ+1(ℓ)Tk−ℓ,q(ℓ)+λℓ+1Tp,q(ℓ+1) which is true for any q≤n−kp and any p<k−ℓ and ℓ<k−1 (proof of Proposition 2.1).
Repeatedly applying this relation we obtain the following expressions
[TABLE]
and
[TABLE]
for any q<n−kj. Thus Tj,q(k−i), as a polynomial in the coordinates of Remark 2.2, is linear in λk−j for all q≤n−kj. This implies Ξ⋆(Di)∩Uk−1 is linear in λk−j.
Assume i≤k−1 and j=0. Most of the argument from the previous paragraph still applies in this case. In particular, Ξ⋆(Di)∩Uk−1 is cut out by T0,n−k+1(k−i)+T0,n−k(k−i) and we have
[TABLE]
for all q≤n−k+1=n−k0. Notice that T0,q(k−1)=T0,q(k)+T0,n−k+2(k−1)T1,q(k−1) for all q≤n−k+1 and T0,n−k+1(k)=λk (Remark 2.2). Substituting this into (4.4) we see that T0,n−k+1(k−i)+T0,n−k(k−i) is linear in λk.
Finally assume i=k. The locus of points (Λ(a),Λ(b))∈U0 meeting Λk−1 is clearly cut out by (aj,n−kj−1)(bj,n−kj−1). The pullback of this equation to Uk−1, which coincides with the strict transform, defines Ξ⋆(Dk). If j=0 we can use (4.3) to deduce that
[TABLE]
This expression is linear in λk−j. If j=0 we can argue in the previous paragraph and deduce linearity in λk. This completes the proof.
∎
Here is an alternate description of Ni.
Remark 4.9**.**
For each 1≤i≤k−1, let Ni=Ξ(Ei). If n=2k−1 we let Nk=Ξ(Ek). If n>2k−1, let N^k denote the closure in X0, of the locus of pairs of planes in X0trv where the intersection of the two planes meets a fixed Λ2k−1. Then Nk is the image of the strict transform of N^k under Ξ.
In the next lemma we abuse notation and use "=" to mean equality as divisor classes.
Lemma 4.10**.**
Let n≥2k−1. The loci Ni are divisorial. Moreover, we have
(i)
Ξ⋆(N1)=2E1.
2. (ii)
Ξ⋆(Ni)=Ei* for 2≤i≤k−1.*
3. (iii)
If n=2k−1 then Ξ⋆(Nk)=Ek and Ξ⋆(Nk)∩Uk−1 is cut out by λk.
4. (iv)
If n>2k−1 let Λ2k−1=V(xk,…,xn−k) be the plane defining Nk. Then Ξ⋆(Nk)∩Uk−1 is cut out by λk.
Proof.
Assume 1≤i≤k−1. Remark 4.9 implies that the Ni are divisors. Items (i), (ii) and the first half of (iii) follow from the fact that Ξ is a finite, degree 2 map branched along N1 (although not phrased this way, it is part of the proof of Proposition 2.8), see [F98, Chapter 1.7]. The rest of item (iii) is a consequence of Lemma 4.2 (ii).
Now assume n>2k−1 and let N^k be as in Remark 4.9. To show that Nk is a divisor it is enough to show that N^k∩X0trv∩U0 is a divisor in X0trv∩U0. Given a point (Λ(a),Λ(b))∈X0trv∩U0, the intersection of the two planes is Λ(a)∩Λ(b)=V({∑j=kn(ai,j−bi,j)xj,yi}0≤i≤k−1). Thus the locus of points in X0trv∩U0 satisfying (Λ(a)∩Λ(b))∩Λ2k−1=∅ is cut out by the determinant of
[TABLE]
Column reducing as in Proposition 2.1 (ii) and taking the strict transform gives item (iv).
∎
5. Birational geometry of Hn−k,n−kn for n≥2k−1
This section is devoted to the proof of Proposition 5.12. For the rest of the section we will assume n≥2k−1. We begin by constructing two families of curves and computing their intersection numbers with Di and Ni.
Roughly speaking, the first family of curves will fix a pair of planes and vary the embedded structures while the second family will vary the planes and fix the embedded structures.
Definition 5.1**.**
For each 1≤j≤k−1, define the curve Cj:P1→Hn−k,n−kn by
[TABLE]
with Λ=V(x0,…,xk−1) and Λ′=V(x0,…,xj,xj+1+xn−kj+1,…,xk−1+xn).
Remark 5.2**.**
Theorem D shows that Cj(s:t) is projectively equivalent to (2.6) with
[TABLE]
It also shows that for j≤k−2, the general member of Cj is a pair of (n−k)-planes meeting along a pencil of embedded (n−2k+j+1)-planes and containing fixed embedded (n−2k+ℓ)-planes for all 1≤ℓ≤j−1, while Ck−1 is a pencil of generically non-reduced (n−k)-planes. If (s:t)=(1:0),(0:1), the corresponding subscheme has an embedded (n−2k+j)-plane.
Definition 5.3**.**
Let 0≤j≤k−1. Let Λ=V(x0,…,xk−1) and consider the pencil of (n−k)-planes
Λ′(s:t)=V(x0,…,xj−1,sxj+txn−kj,xj+1+xn−kj+1,…,xk−1+xn).
Define the curve Bj:P1→Hn−k,n−kn by
[TABLE]
Remark 5.4**.**
Theorem D shows that Bj(s:t) is projectively equivalent to (2.6) with
[TABLE]
If (s:t)=(1:0), then B0(s:t) is a pair of (n−k)-planes meeting transversely while Bj(s:t) a pair of (n−k)-planes with a pure embedded (n−2k+j)-plane for j>0. Moreover, the embedded (n−2k+j)-plane is fixed along the curve.
If (s:t)=(1:0), the corresponding subscheme has an embedded (n−2k+j+1)-plane. Note that Bk−1(1:0) is, more precisely, a generically non-reduced (n−k)-plane.
Before we determine the intersection numbers we need to compute a few linear spans. We begin with notation that will be used a great deal in the following Lemmas.
Notation 5.5**.**
We use Cj†(s:t) and Bj†(s:t) to denote the subschemes of Pn cut out by Cj(s:t) and Bj(s:t), respectively. Given an ideal J⊆S, let sat(J) denote its saturation with respect to (x0,…,xn) and let J(1) denote the ideal generated by the linear forms in J.
Lemma 5.6**.**
Let 1≤i≤j≤k−1 and let Λ2k−i−1=V(xk,xk+1,…,xn−ki−2,xn−ki−xn−ki−1). For any (s:t)∈P1, if i=j the linear span of Cj†(s:t)∩Λ2k−i−1 is
[TABLE]
and if i=j the linear span of Ci†(s:t)∩Λ2k−i−1 is
[TABLE]
Proof.
Let Λ=Λ2k−i−1 and note that the linear span of Cj†(s:t)∩Λ is cut out by sat(Cj(s:t)+IΛ)(1). Assume i<j. It is straigthtforward to see that xℓ(x0,…,xn)⊆Cj(s:t)+IΛ for every 0≤ℓ≤i−1. Thus we have
[TABLE]
Moreover, it is clear that Q(d)=(Cj(s:t)+IΛ)(d) for all d≥2. Thus if we show that Q is saturated then Q=sat(Cj(s:t)+IΛ), and this would give the desired linear span. If we write Q=IΛ+(x0,…,xi−1)+Q′, it suffices to show that quadratic portion, Q′, is saturated. But notice that Q′ is projectively equivalent to an ideal of the form (2.6) (for reasons similar to Remark 5.2). It follows from Lemma 2.7 that Q is saturated. The case of i=j is analogous.
∎
Remark 5.7**.**
Here are two simple facts about linear spans:
(i)
If Λp and Λq are disjoint linear spaces in Pn then dimkspan(Λp∪Λq)=p+q+1.
2. (ii)
span(Y1∪Y2)=span(spanY1∪spanY2) for any subschemes Y1,Y2⊆Pn.
The first fact is clear and the second follows from the following chain of equalities,
[TABLE]
Lemma 5.8**.**
Let 1≤i≤k and 1≤j≤k−1. We have the following intersection numbers
(i)
Di⋅Cj=0* whenever i=j,*
2. (ii)
Di⋅Ci=1* for all i≤k−1.*
Proof.
Assume i>j. Since the dimension of any embedded subscheme of Cj†(s:t) is at most n−2k+j+1, a generic (2k−1−i)-plane will not intersect any embedded subscheme of Cj†(s:t). If i<k, the intersection of Cj†(s:t) with a generic Λ2k−1−i is a pair of skew (k−1−i)-planes. Moreover, these skew planes are independent of (s:t) and thus
[TABLE]
is independent of (s:t). As a consequence, we may choose an (i−1)-plane Λi−1⊆Λ2k−1−i that does not meet the P2k−2i−1. It follows from Remark 5.7 that
[TABLE]
If we use the flag {Λi−1⊆Λ2k−1−i} to define Di we see that Di⋅Cj=0. Similarly, if i=k and Λk−1 is generic we have that Cj†(s:t)∩Λk−1=∅. Thus Dk⋅Cj=0.
Assume i<j and let Λ2k−i−1=V(xk,xk+1,…,xn−ki−2,xn−ki−xn−ki−1). By Lemma 5.6 we have that
[TABLE]
is fixed and independent of (s:t). As done in the previous paragraph, if we choose a general Λi−1 inside Λ2k−1−i to define Di, then Di⋅Cj=0. This completes the proof of item (i).
Assume i=j and let the flag {Λi−1⊆Λ2k−1−i} in (4.1) define Di. By Lemma 5.6 we have that
[TABLE]
Thus, if t=0, the linear span of (Ci†(1:t)∩Λ2k−i−1)∪Λi−1 is all of Λ2k−i−1. If t=0, the linear span of (Ci†(1:0)∩Λ2k−i−1)∪Λi−1 is Λ2k−i−1∩V(xi−1). Thus Di∩Ci is supported on the point Z0=Ci(1:0).
Let C~i denote the closure in Xk−1 of the curve, A1↪Uk−1 obtained by setting λ1,…,λk−i−1=1, λk−i+1=t and all the other coordinates of Remark 2.2 to [math]. Since Ξ(C~i)∣Uk−1=Ci(1:t) it follows that Ξ(C~i)=Ci. In particular C~i∩Ξ⋆(Di) is supported at a unique point Z~0∈Ξ−1(Z0). Since Ξ⋆(Di) is linear in λk−i+1 (Lemma 4.7), it follows that Ξ⋆(Di) and C~i intersect transversely at Z~0. Using the push-pull formula we conclude that Ci⋅Di=Ξ⋆C~i⋅Di=Ξ⋆(C~i⋅Ξ⋆(Di))=1.
∎
Lemma 5.9**.**
Let 1≤i≤k and 0≤j≤k−1. We have the following intersection numbers
(i)
Di⋅Bj=0* for all i≤j,*
2. (ii)
Di⋅Bj=1* for all i>j.*
Proof.
Assume i≤j and let Λ2k−1−i=V(xk,…,xn−ki−2,xn−ki−xn−ki−1). Arguing as in Lemma 5.6 we see that
[TABLE]
is independent of (s:t). Arguing as in Lemma 5.8 we deduce item (i).
Assume that j<i≤k−1 and let {Λi−1⊆Λ2k−1−i} be the flag (4.2) defining Di. Then
Bj†(s:t)∩Λ2k−i−1 is a disjoint pair of (k−i−1)-planes defined by
[TABLE]
For t=0, the linear span of (Bj†(s:t)∩Λ2k−i−1)∪Λi−1 is all of Λ2k−i−1. On the other hand if t=0, the linear span of (Bj†(s:t)∩Λ2k−i−1)∪Λi−1 is Λ2k−1−i∩V(xj). Thus Di∩Bj is supported at the point Z0=Bj(1:0).
Let B~j denote the closure in Xk−1 of the curve, A1↪Uk−1 obtained by setting λ1=⋯=λk−j−1=1, λk−j=t, λk−j+2=⋯=λk=1 and all the other coordinates of Remark 2.2 to [math]. Since Ξ(B~j)∣Uk−1=Bj(1:t) we have Ξ(B~j)=Bj. Thus B~j∩Ξ⋆(Di) is supported at a unique point Z~0∈Ξ−1(Z0). Since Ξ⋆(Di) is linear in λk−j (Corollary 4.8), it follows that Ξ⋆(Di) and B~j intersect transversely at Z~0. Using the push-pull formula we conclude that Bj⋅Di=Ξ⋆B~j⋅Di=Ξ⋆(B~j⋅Ξ⋆(Di))=1.
Now assume j<i=k and let Λk−1=V(xj+xn−kj,xk,…,xn−kj−1,xn−kj+1,…,xn) be the plane defining Dk. It is evident that Bj∩Dk is supported at the point Z1,1=Bj(1:1). Once again, B~j (defined in the previous paragraph) and Ξ⋆(Dk) will meet at a unique point Z~1,1∈Ξ−1(Z1,1). Since Ξ⋆(Dk) is linear in λk−j (Corollary 4.8) we see that B~j meets Ξ⋆(Dk) transversely at Z~1,1. Once again we conclude using the push-pull formula.
∎
Lemma 5.10**.**
We have the following intersection numbers,
(i)
Ni⋅Cj=0* for each 1≤i≤k−1 and all 1≤j≤k−i−1,*
2. (ii)
Ni⋅Bj=0* for each 1≤i≤k and all j=k−i,k−i+1,*
3. (iii)
Ni⋅Ck−i+1=2* for each 2≤i≤k,*
4. (iv)
N1⋅Bk−1=2* and Ni⋅Bk−i=1 for 2≤i≤k.*
Proof.
Item (i) and item (ii), except for the case of i=k, follow from the definition of the Ni and the description of the embedded subschemes in Remark 5.2 and Remark 5.4. We will deal with the case of i=k in the last paragraph. For the rest of the proof let Z0=Ck−i+1(1:0) and Z∞=Ck−i+1(0:1). We will also use the curves C~k−i+1 and B~j defined in Lemma 5.8. In particular, let Z~0,Z~∞∈C~k−i+1 be such that Ξ(Z~0)=Z0 and Ξ(Z~∞)=Z∞.
Assume 2≤i≤k−1. Since Ni is the locus of subschemes containing an embedded (n−k+1−i)-plane, it meets the curve Ck−i+1 at Z0 and Z∞. Thus C~k−i+1 meets Ei at Z~0 and Z~∞. Using Lemma 4.10 (ii), we obtain
[TABLE]
Since Z~0∈Uk−1 and Ei is cut out by λi, C~k−i+1 meets Ei transversely at Z~0. Symmetrically, C~k−i+1 will also meet Ei transversally at Z~∞. To see the latter statement, consider the projective transformation g∈GL(n+1) that interchanges xj with xj−1, interchanges xn−kj with xn−kj−1 and fixes the other coordinates. It follows from the definition that g(Ck−i+1)=Ck−i+1 and g interchanges Z0 with Z∞. Since intersection multiplicity is invariant under automorphisms of Hn−k,n−kn we obtain
[TABLE]
This proves item (iii) for i=k.
Since N1 is the locus of generically non-reduced subschemes, it meets the curve Bk−1 at Bk−1(1:0). Using Lemma 4.10 (i) we obtain N1⋅Bk−1=Ξ⋆(B~k−1⋅Ξ⋆(N1))=2B~k−1⋅E1=2. Similarly, using Lemma 4.10 we obtain Ni⋅Bk−i=1 for all 2≤i≤k−1. This finishes item (iv) for i=k
Finally, assume i=k and let Λ2k−1=V(xk,…,xn−k) be the plane defining Nk (if n>2k−1). By Lemma 4.10 (iii), (iv) we see that Ξ⋆(Nk) meets C~1 at Z0 and possibly also at Z∞ (since the latter does not lie in Uk−1). Moreover, Ξ⋆(Nk) meets C~1 transversely at Z~0. We may argue as in the previous paragraph to show that Ξ⋆(Nk) also meets C~1 transversely at Z~∞. Indeed, the projective transformation g fixes Nk. This is clear if n=2k−1 and the case of n>2k−1 follows from the fact that g fixes Λ2k−1. Thus Nk⋅C1=(Nk⋅C1)∣Z0+(Nk⋅C1)∣Z∞=2(Nk⋅C1)∣Z0=2, completing the proof of item (iii). For items (ii) and (iv) we argue similarly using the following projective transformation: g′∈GL(n+1) that maps xn−kj↦xn−kj+xj and fixes the other coordinates. It is straightforward to verify that g′(Bj)=Bj, g′(Bj(0:1))=Bj(1:1) and g′ fixes Nk (since g′ fixes Λ2k−1). This implies
[TABLE]
for j=1. Thus, we may compute Ξ⋆(Nk)⋅B~j along Uk−1 to obtain the desired results.
∎
Proposition 5.11**.**
Let 1≤i≤k. Then we have
•
N1=2Dk−2Dk−1,
•
Ni=2Dk−i+1−Dk−i−Dk−i+2* for all 2≤i≤k−1,*
•
Nk=2D1−D2.
Proof.
By Lemma 4.4, Lemma 5.8 and Lemma 5.9 we see that N1(Hn−k,n−kn) is generated by {D1,…,Dk}. This allows us to write Ni=∑ℓ=1kϵi,ℓDℓ for some ϵi,ℓ∈Z. Using Lemmas 5.8 - 5.10 we obtain
•
N1⋅Cℓ=ϵ1,ℓ=0 for ℓ≤k−2,
•
N1⋅Bk−1=ϵ1,k=2 and N1⋅Bk−2=ϵ1,k−1+ϵ1,k=0.
This immediately implies N1=2Dk−2Dk−1. For each 2≤i≤k we obtain
•
Ni⋅Bj=∑ℓ=j+1kϵi,ℓ=0 for j=k−i,k−i+1
•
Ni⋅Bk−i=∑ℓ=k−i+1kϵi,ℓ=1 and Ni⋅Ck−i+1=ϵi,k−i+1=2.
If i=k, we obtain ϵi,k−i=−1,ϵi,k−i+1=2,ϵi,k−i+2=−1, and ϵi,ℓ=0 for other ℓ. If i=k we obtain ϵk,1=2, ϵk,2=−1 and ϵi,ℓ=0 for other ℓ. This completes the proof.
∎
Proposition 5.12**.**
Let k≥2 and n≥2k−1. Then we have
[TABLE]
Moreover, Hn−k,n−kn is Fano if and only if either k=3 and n=5, or k=3 and n∈{2k−1,2k}.
Proof.
It is clear that the divisors N1,…,Nk are effective and generate N1(Hn−k,n−kn). To conclude that the effective cone is generated by N1,…,Nk, it is enough to show that any R-divisor N=∑i=1kϵiNi, with some ϵj<0, is not effective. Let Aj:P1↪Hn−k,n−kn denote any curve such that for (s:t)=(1:0), Aj(s:t) is a pair of (n−k)-planes meeting transversely while Aj(1:0) it is a pair of (n−k)-planes with a pure embedded (n−k+1−j)-plane if j>1 and generically non-reduced if j=1. Clearly, Aj⋅Ni=0 for i=j and Aj⋅Nj>0. Since N⋅Aj=ϵj<0 and Aj is not contained in the support of N, we see that N cannot be an effective divisor.
By varying the flags it is easy to see that each of the Di is base point free; thus it is also nef. Similar to the previous paragraph, to show that the nef cone gone is generated by D1,…,Dk, it is enough to show that any R-divisor D=∑i=1kϵiDi, with some ϵj<0, is not nef. If j=k, we have D⋅Cj=ϵj<0 and if j=k we have D⋅Bk−1=ϵk<0. Thus D is not nef.
We will now compute the canonical divisor of Hn−k,n−kn using the branched cover Ξ:Xk−1→Hn−k,n−kn. By [H77, Exercise 8.5b] and [E95, Exercise 10.10] we may write
[TABLE]
where D^k is the strict transform of OX0(1,1) (Remark 4.6). Note that the canonical divisor of X0 is OX0(−n−1,−n−1). Let KHn−k,n−kn=ϵ1N1+⋯+ϵk−1Nk−1+ϵkDk for some ϵi∈Q. Hurwitz’s theorem implies that KXk−1=Ξ⋆(KHn−k,n−kn)+E1. Using this and Lemma 4.10 we obtain
[TABLE]
Let ϵ~j=(k−j+1)(n−k−j+2)−1 and using Proposition 5.11 we obtain
[TABLE]
For k=2,3 the above expression simplifies to
[TABLE]
If k≥4 we can rewrite the expression as follows:
[TABLE]
Since 2ϵ~j+1−ϵ~j−ϵ~j+2=−2 for all j we obtain
[TABLE]
Since the ample cone is the interior of the nef cone, we see that −KHn−2,n−2n is ample if and only if n=3,4 and that −KHn−3,n−3n is ample precisely when n=5. If k≥4, −KHn−k,n−kn is ample if and only if n=2k−1,2k.
∎
6. Birational geometry of Hk−1,k−1n for n>2k−1
This section is devoted to the proof of Theorem 6.14. We will show that Hk−1,k−1n is Fano, and thus a Mori dream space. By constructing a contraction from Hk−1,k−1n to Hn−k,n−kn (Proposition 6.11) we will also deduce that Hn−k,n−kn is a Mori dream space.
Notation 6.1**.**
In this section we will primarily be interested in the case when the pair of planes do not span all of Pn. By swapping the roles of codimension and dimension, the components we are interested in are of the form Hk−1,k−1n with n>2k−1.
Corollary 2.21 states that for n>2k−1, the morphism ρ:Hk−1,k−1n⟶Gr(2k−1,n) that sends a scheme to its linear span is smooth; the fiber over a point Λ is Hk−1,k−1(Λ).
Remark 6.2**.**
Let W=Speck[f2k,j,…,fn,j]0≤j≤2k−1 be a neighbourhood of Λ=V(x2k,…,xn)∈Gr(2k−1,n) such that its k-points are identified with
[TABLE]
Then the open subset ρ−1(W) is naturally isomorphic to W×Hk−1,k−1(Λ).
Lemma 6.3**.**
Let n>2k−1. Then N1(Hk−1,k−1n)=Zk+1.
Proof.
As explained in Lemma 4.4, since Hk−1,k−1n is rational and smooth, it suffices to compute N1(Hk−1,k−1n)⊗Q which equals Pic(Hk−1,k−1n)⊗Q=H2(Hk−1,k−1n,Q). By Corollary 2.21 we have a smooth morphism Hk−1,k−1n⟶Gr(2k−1,n) with fibers isomorphic to Hk−1,k−12k−1. Since the base of this morphism is simply connected, we may apply the Leray-Hirsch theorem [V04, Theorem 7.33] and Lemma 4.4 to deduce that H2(Hk−1,k−1n,Q)≃Qk+1.
∎
Using the fibration ρ and Remark 6.2 one can easily verify that the loci Di′,Ni′,F are divisorial. We now define the curves inside Hk−1,k−1n; all but two of them come from curves lying inside Hk−1,k−12k−1.
Definition 6.4**.**
Let Λ=V(x2k,…,xn). For each relevant j, let Aj′,Bj′,Cj′ be the images of Aj,Bj,Cj (Definition 5.1, 5.3, Proposition 5.12) under the inclusion ρ−1(Λ)=Hk−1,k−1(Λ)↪Hk−1,k−1n, respectively.
Definition 6.5**.**
Let Λ′=V(xk,…,xn) and let
Λ(s:t)=V(x0,…,xk−1,sx2k+txk,x2k+1,…,xn)
be a pencil of (k−1)-planes disjoint from Λ′.
Define the curve Y1:P1→Hk−1,k−1n by (s:t)↦Λ(s:t)∪Λ′. Explicitly
[TABLE]
Define the curve Y2:P1→Hk−1,k−1n by
[TABLE]
Remark 6.6**.**
Let Λ=V(x0,…,xk−1,x2k,…,xn) and Λ′=V(x0,xk+1,…,xn) be a pair of (k−1)-planes meeting along a point. Then we have
[TABLE]
In particular, Y2 is a pair of fixed (k−1)-planes with a pencil of embedded points.
Lemma 6.7**.**
Y2* is a moving curve in Nk′ i.e. its deformations span Nk′.*
Proof.
The general subscheme parameterized by Nk′ is a pair of (k−1)-planes meeting along an embedded point. By Corollary 2.21 and Theorem D, up to projectively equivalence, such a subscheme is cut out by
[TABLE]
In particular, the GL(n+1) orbit of Y2 covers a dense subset of Nk′.
∎
Lemma 6.8**.**
For all pairs of relevant indices i,j (the ones appearing in Lemma 5.8, 5.9, 5.10), the intersection numbers of Di′,Ni′ with Bj′,Cj′ are the same as the intersection numbers of Di,Ni with Bj,Cj, respectively.
Proof.
We will only verify Di′⋅Cj′=Di⋅Cj for 1≤i,j≤k−1; the other cases are analogous. Let Λ=V(x2k,…,xn) be a fixed (2k−1)-plane. Let Di′ be defined by a flag Fi′={Λn−2k+i⊆Λn−i}, where the flag is chosen to satisfy the following two properties:
•
Λ is transverse to each element of the flag Fi′,
•
Let Di⊆Hk−1,k−1(Λ) be defined by the flag Fi={Λn−2k+i∩Λ⊆Λn−i∩Λ}. Then either Di∩Cj=∅ if i=j or Di is transverse to Cj if i=j.
Let W be the open neighbourhood of Λ from Remark 6.2. The first bullet point implies that every element of W is transverse to the flag Fi′. It follows that Di′∣ρ−1(W)=W×Di and Cj′={Λ}×Cj. Thus we have Di′⋅Cj′=Di′∣ρ−1(W)⋅Cj′=Di⋅Cj.
∎
Lemma 6.9**.**
We have the following intersection numbers
(i)
Di′⋅Y2=Ni′⋅Y1=0* for all 1≤i≤k,*
2. (ii)
Ni′⋅Y2=0* for all 1≤i≤k−1,*
3. (iii)
Di′⋅Y1=1* for all 1≤i≤k,*
4. (iv)
F⋅Y1=F⋅Y2=1.
Proof.
Items (i) and (ii) are clear from the definition of the divisors.
Let 1≤i≤k, Λ=V(x2k,…,xn) and W be as in Remark 6.2. We may choose a flag Fi′ to define Di′ so that the following properties are satisfied:
•
Λ is transverse to each element of the flag Fi′,
•
Di′∩Y1 is supported at Z0=Y1(1:0).
Let W′=Speck[ϵ1,…,ϵk2]⊆Hk−1,k−1(Λ) be any affine open containing the image of Z0 in Hk−1,k−1(Λ). Then W×W′ is identified with an open neighbourhood of Z0∈Hk−1,k−1n. Along this open set, Y1 is the curve obtained by setting f2k,k=t, fi,j=0 for other i,j, and ϵi=δi for some constants δi∈k. On the other hand, Di′=W×(Di∩W′) where Di is the divisor defined by the flag Fi′∩Λ. It immediately follows that Di′ meets Y1 transversely at Z0 inside W×W′; this proves item (iii).
For item (iv), we will only verify F⋅Y1=1 as the other case is similar. Let F be defined by the (n−2k)-plane, V(x0,…,xk−1,xk+1,…,x2k). It follows that F∩Y1 is also supported at Z0. Moreover, along W×W′, F is cut out by the function f2k,k. Combining this with the equation of Y1 along W×W′ we see that F meets Y1 transversely at Z0.
∎
Proposition 6.10**.**
Let k≥2 and n>2k−1. Then we have,
[TABLE]
Moreover we have,
•
N1′=2Dk′−2Dk−1′,
•
Ni′=2Dk−i+1′−Dk−i′−Dk−i+2′* for all 2≤i≤k−1,*
•
Nk′=2D1′−D2′−F.
Proof.
Using the intersection numbers with the curves {C1′,…,Ck′,Y2} and arguing as in Proposition 5.11, 5.12 we see that N1(Hk−1,k−1n) and Nef(Hk−1,k−1n) are both generated by D1′,…,Dk′,F. Using the curves {A1′,…,Ak′,Y1} and arguing as in Proposition 5.12, we see that N1′,…,Nk′,F generate the effective cone.
By Proposition 5.11 and Remark 6.8 there exists ϵi∈Q such that
•
N1′=2Dk′−2Dk−1′+ϵ1F,
•
Ni′=2Dk−i+1′−Dk−i′−Dk−i+2′+ϵiF for all 2≤i≤k−1,
•
Nk′=2D1′−D2′+ϵkF.
Intersecting these divisors with Y1,Y2 and using Lemma 6.9 we obtain ϵ1,…,ϵk−1=0 and ϵk=−1. ∎
We are now ready to relate Hk−1,k−1n with Hn−k,n−kn.
Proposition 6.11**.**
There is a morphism Ψ:Hk−1,k−1n⟶Hn−k,n−kn with exceptional locus Nk′. Moreover, Nk′ is a Pn−2k+1-fibration over Ψ(Nk′). Geometrically, Ψ "forgets" the embedded points.
Proof.
Given an (n+1)-dimensional vector space V, let
[TABLE]
The Hilbert-Chow morphism induces a birational morphism, Hk−1,k−1(PV)⟶Sym2Gr(k−1,PV) [K96, Theorem 6.3]. Let \widebarΓi(PV) denote the image of Γi(PV) in Sym2Gr(k−1,PV). Since the pullback of each \widebarΓi(PV) is Ni′, we obtain a morphism
[TABLE]
There is an isomorphism Gr(k−1,PV)2≃Gr(n−k,(PV)⋆)2 induced by map Λ↦Λ⋆ that sends a linear space to its dual variety. This isomorphism maps Γi(PV) to Γi (Definition 0.1) and thus maps \widebarΓi(PV) to \widebarΓi after quotienting by S2. Therefore we obtain an isomorphism
[TABLE]
Let Ψ=Ψ2∘Ψ1. One can directly check that Ψ⋆(Di)=Di′ for all i and Ψ⋆(Ni)=Ni′ for 1≤i≤k−1.
To show that Ψ contracts Nk′, it is enough to show that Ψ contracts Y2 (Lemma 6.7). Using Lemma 6.9 we obtain Ψ⋆Y2⋅Di=Ψ⋆(Y2⋅Ψ⋆(Di))=Ψ⋆(Y2⋅Di′)=0 for all i. Since D1,…,Dk generates the nef-cone of Hn−k,n−kn we must have Ψ⋆Y2=0, i.e. Ψ contracts Y2.
Conversely, let C be any curve contracted by Ψ. If C⋅Di′=0 for some i, we would have Ψ⋆C⋅Di=Ψ⋆(C⋅Di′)=0, proving that Ψ does not contract C. Thus we may assume C⋅Di′=0 for all i. Since {Di′}i∪F generates the nef-cone of Hk−1,k−1n we must have F⋅C>0. Using Proposition 6.10 we obtain Nk′⋅C=−F⋅C<0, i.e. C lies inside Nk′.
Lastly, we need to verify that Nk′ is a Pn−2k+1-fibration over Ψ(Nk′). Up to projective equivalence, it is enough to verify that the fiber of Ψ1 over
Z=V(x0,…,xk−1,x2k,…,xn)∪V(x0,xk+1,…,xn)
is isomorphic to Pn−2k+1, c.f. Example 6.12. Let H=spank{x0,x2k,…,xn}. Similar to the proof of Lemma 6.7, any subscheme parameterized by Hk−1,k−1n and supported on Z is cut out by
[TABLE]
where H′∈Gr(n−2k+1,H) and H′′⊆H is chosen so that H′⊕H′′=H. Notice that for a fixed H′, all choices of H′′ give the same ideal as (6.1). It follows that the Ψ1−1(Z) is paramaterized by Gr(n−2k,PH)≃Pn−2k+1.
∎
Example 6.12**.**
Consider X⊆P4 cut out by (x0,x1,x4)∩(x0,x3,x4)∩(x02,x1,x3,x4). This is a pair of lines meeting along an embedded point. Let x0⋆,…,x4⋆ be the dual coordinates on (P4)⋆. We can trace the image of X under the map Ψ:H1,1(P4)→H2,2((P4)⋆) as follows:
[TABLE]
Proposition 6.13**.**
Let k≥2 and n>2k−1. The component Hk−1,k−1n is Fano.
Proof.
Using Proposition 6.11 and the canonical divisor in Proposition 5.12 we deduce that
[TABLE]
The first equality is a modification of [H77, Exercise 8.5] combined with the fact that the codimension of Ψ(Nk′) in Hn−k,n−kn is n−2k+2. It follows from Proposition 6.10 that −KHk−1,k−1n is ample in all cases; thus Hk−1,k−1n is always Fano.
∎
Here is the the main theorem of the paper:
Theorem 6.14**.**
The components Hk−1,k−1n and Hn−k,n−kn are Mori dream spaces.
Proof.
This follows immediately from Proposition 5.12, 6.11 , 6.13 and the subsequent two facts:
(i)
A smooth Fano variety is a Mori dream space [M10, Corollary 4.9],
2. (ii)
Let f:X→Y be a surjective morphism of smooth, projective varieties. If X is a Mori dream space, then so is Y [O16, Theorem 1.1]. ∎
7. Birational geometry of Hn−c,n−dn and Hc−1,d−1n
In this section we explain how the proofs of Section 4, 5 and 6 carry over, almost identically, to the case when the pair of planes are of different dimension. In particular, the definition of the divisors and curves, and computations of their intersection numbers, including transversality, are very similar. Thus we will omit most of the proofs and indicate all the required modifications.
We begin by defining divisors analogous to the ones in Definition 0.3 and 0.4 when the pair of planes span Pn.
Definition 7.1**.**
Let n≥c+d−1. For each 1≤i≤c−1 and a choice of a flag of linear spaces {Λi−1⊆Λc+d−1−i}, let Di denote the divisor class of the locus of subschemes Z∈Hn−c,n−dn, for which the linear span of Λi−1∪(Z∩Λc+d−1−i) has dimension less than c+d−1−i.
Definition 7.2**.**
Let n≥c+d−1. Let Dc(1) denote the closure of the locus of subschemes supported on two distinct planes for which the (n−d)-plane meets a fixed Λd−1. Let Dc(2) denote the closure of the locus of subschemes supported on two distinct planes for which the (n−c)-plane meets a fixed Λc−1.
Remark 7.3**.**
The divisors Dc(1) and Dc(2) are the Weil divisors associated to the strict transforms, under Ξ, of OX0(0,1) and OX0(1,0), respectively. Here X0=Gr(n−c,n)×Gr(n−d,n).
Definition 7.4**.**
Let n≥c+d−1. For each 1≤i≤c−1, let Ni denote the divisor class of the locus of subschemes in Hn−c,n−dn with an embedded (n−d+1−i)-plane. If n=c+d−1 let Nc denote the divisor class of the locus of subschemes with an embedded point. If n>c+d−1 let Nc denote the class of the closure of the locus of pairs of planes meeting transversely, where the intersection of the two planes meets a fixed Λc+d−1.
We can easily modify the curves in Definition 5.1, 5.3 to obtain curves in Hn−c,n−dn. However, this time we can have two variations, depending on whether the (n−c)-plane or (n−d)-plane is fixed along the curve.
Definition 7.5**.**
For each 1≤j≤c−1, define the curve Cj:P1→Hn−c,n−dn by
[TABLE]
with Λ=V(x0,…,xd−1) and Λ′=V(x0,…,xj,xj+1+xn−cj+1,…,xc−1+xn).666Analogous to the notation kj, we define cj=c−1−j and dj=d−1−j.
Definition 7.6**.**
For each 0≤j≤c−1 consider the pencils
[TABLE]
and
[TABLE]
Define the curves Bj(1):P1→Hn−c,n−dn and Bj(2):P1→Hn−c,n−dn by
[TABLE]
and
[TABLE]
Here are the analogues of Lemmas 5.8 - 5.10. To prove these, one first constructs an open set on Hn−c,n−dn analogous to Uk−1 as described in Section 3.
Then we proceed as in Section 4 and describe equations for Di and Ni along this open set.
Lemma 7.7**.**
Let 1≤i≤c−1 and 0≤j≤c−1. We have the following intersection numbers,
(i)
Di⋅Ci=1* and Di⋅Cj=0 for all i=j,*
2. (ii)
Dc(1)⋅Ci=Dc(2)⋅Ci=0* for all i,*
3. (iii)
Di⋅Bj(1)=Di⋅Bj(2)=0* for all i≤j,*
4. (iv)
Di⋅Bj(1)=Di⋅Bj(2)=1* for all i>j,*
5. (v)
Dc(1)⋅Bj(1)=Dc(2)⋅Bj(2)=1* and Dc(1)⋅Bj(2)=Dc(2)⋅Bj(1)=0 for all j.*
Lemma 7.8**.**
We have the following intersection numbers,
(i)
Ni⋅Cj=0* for each 1≤i≤c−1 and all 1≤j≤c−1−i,*
2. (ii)
Ni⋅Bj(1)=Ni⋅Bj(2)=0* for each 1≤i≤c and all j=c−i,c−i+1,*
3. (iii)
Ni=2Dc−i+1−Dc−i−Dc−i+2* for all 3≤i≤c−1,*
4. (iv)
Nc=2D1−D2.
Proposition 7.10**.**
Let d>c≥2 and n≥c+d−1. We have,
[TABLE]
Moreover,
(i)
If c=2, then only Hd−1,1d+1,Hd,2d+2,…,H2d−3,d−12d−1 are Fano,
2. (ii)
If c≥3, then only Hd−1,c−1c+d−1 and Hd,cc+d are Fano.
Proof.
The verification of the effective and nef cone is similar to Proposition 5.12. Using the formula of the canonical divisor of a blowup and arguing as in Proposition 5.11 we obtain, KHn−2,n−dn=(2d−2n−1)D1+(n−2d)(D2(1)+D2(2)) and
[TABLE]
for c≥3. Therefore Hn−c,n−dn is Fano for n∈{c+d−1,c+d} if c≥3, and n∈{d+1,…,2d−1} if c=2.
∎
We move on to the case when the pair of linear spaces do not span Pn.
Definition 7.11**.**
Let n>c+d−1. For each 1≤i≤c−1 and a choice of flag {Λn−c−d+i⊆Λn−i}, let Di′ denote the divisor class of the locus of subschemes Z∈Hc−1,d−1n, for which the linear span of Λn−c−d+i∪(Λn−i∩Z) has dimension less than n−i. Let Dc′(1) denote the class of the closure of the locus of subschemes supported on two distinct planes for which the (d−1)-plane meets a fixed Λn−d. Let Dc′(2) denote class of the closure of the locus of subschemes supported on two distinct planes for which the (c−1)-plane meets a fixed Λn−c.
Let F denote the class of the locus of subschemes Z such that its linear span meets a fixed Λn−c−d.
Definition 7.12**.**
Let n>c+d−1. For each 1≤i≤c, let Ni′ denote the divisor class of the locus of subschemes with an embedded (c−i)-plane.
By lifting the curves Ci,Bi(1),Bi(2) to Hc−1,d−1n (c.f. Lemma 6.8) and defining curves Y1(1),Y1(2),Y2 analogous to Definition 6.5, we obtain the following proposition. Since computations of the intersection numbers are exactly the same as Lemma 6.9 and Proposition 6.10, we omit the proof.
Proposition 7.13**.**
Let c≥2 and n>c+d−1. Then we have,
[TABLE]
Moreover, if c=2 we have N1=D2′(1)+D2′(2)−D1′andN2=2D1′−D2′(1)−D2′(2)−F.
If c≥3 we have
(i)
N1′=Dc′(1)+Dc′(2)−Dc−1′,
2. (ii)
N2′=2Dc−1′−Dc−2′−Dc′(1)−Dc′(2),
3. (iii)
Ni′=2Dc−i+1′−Dc−i′−Dc−i+2′* for all 3≤i≤c−1,*
4. (iv)
Let c≥2 and n>c+d−1. The component Hc−1,d−1n is Fano.
Proof.
Similar to Proposition 6.11, there is a morphism, Ψ:Hc−1,d−1n→Hn−c,n−dn with exceptional locus Nc′. As explained in the Proposition 6.13 we deduce
[TABLE]
Using the expression for KHn−c,n−dn in Proposition 7.10, it follows that −KHc−1,d−1n is ample.
∎
Thus we deduce the main theorem of this section,
Theorem 7.15**.**
The components Hc−1,d−1n and Hn−c,n−dn are Mori dream spaces.
Acknowledgement
We would like to thank Dawei Chen, Michael Christianson, David Eisenbud and Frank-Olaf Schreyer for helpful discussions. We are especially thankful to Michael Christianson for reading a draft and suggesting improvements. The author is partially supported by an NSERC PGSD scholarship.
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