The Hilbert scheme of a pair of linear spaces

Let Ha,bn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}_{a,b}^n$$\end{document} denote the component of the Hilbert scheme whose general point parameterizes an a-plane union a b-plane meeting transversely in Pn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbf {P}}^n$$\end{document}. We show that Ha,bn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}_{a,b}^n$$\end{document} is smooth and isomorphic to successive blow ups of Gr(a,n)×Gr(b,n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbf {Gr}(a,n) \times \mathbf {Gr}(b,n)$$\end{document} or Sym2Gr(a,n)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\text {Sym}^2 \mathbf {Gr}(a,n)$$\end{document} along certain incidence correspondences. We classify the subschemes parameterized by Ha,bn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}_{a,b}^n$$\end{document} 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\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}_{a,b}^n$$\end{document} and determine when the component is Fano. Moreover, we show that Ha,bn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}_{a,b}^n$$\end{document} is a Mori dream space for all values of a, b, n.

isomorphic to a double blow up of Gr(1, 9); Gallardo, Huerta and Schmidt [13] computed its effective cone. Chen, Coskun and Nollet [5] showed that the component corresponding to a pair of codimension two linear spaces meeting transversely is smooth and isomorphic to a blow of Sym 2 Gr(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 char k = 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 P n . The Hilbert polynomial of X is There is an integral component of Hilb P n n−c,n−d (t) P n , denoted H n n−c,n−d or H n−c,n−d (P n ), whose general point parameterizes X , see Proposition 1.4.
We begin with the natural rational map meets exactly one other component in Hilb P n n−2,n−2 (t) P n . 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 [5,Proposition 2.6]. Unfortunately, for general c, d there are many, sometimes singular, components meeting H n n−c,n−d (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 and studying the induced map on tangent spaces.
In [24] 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 The component H n n−c,n−d has a unique Borel fixed point.
We also give a complete description of all the subschemes parameterized by H n n−c,n−d . 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 ⊆ P n is a subscheme Z ⊆ P n such that Z red = Z and deg(Z ) = 2 deg(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 H n n−c,n−c . Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1 ≤ i 1 < · · · < i r ≤ c and a flag of linear spaces 1 ⊆ 2 ⊆ · · · ⊆ r ⊆ P n with codim P n ( ) = (c + i − 1) for each , such that When char k = 0, we use our explicit description of and the classification of ideals parameterized to study the effective and nef cones of H n n−c,n−d . As a consequence, we deduce that H n n−c,n−d is always a Mori dream space.
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 Sect. 7 we explain how to carry out all of the proofs of Sects. 5 and 6 for the case c = d.

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 Sects. 1-3 we will assume char k = 2 and for Sects. 4-7 we will assume char k = 0. We use S to denote the polynomial ring k[x 0 , . . . , x n ] and S d to denote the subspace of monomials of degree d. For a homogenous ideal I ⊆ S we use I d 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 ) ⊆ P n and we use P X (t) or P S/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 P n . The span of a subscheme X ⊆ P n is the linear subspace V (H 0 (P n , I X (1))) ⊆ P n . The letters c and d are reserved for the codimension of linear spaces in P n ; 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 N 1 (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 D 1 , . . . , D l to denote the convex cone in N 1 (Y ) ⊗ R generated by the divisors D i . For more details we refer to [6,Chapter 1].
Let X denote the union of an (n − c)-plane and (n − d)-plane meeting transversely in P n . 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 Hilb P n n−c,n−d (t) P n whose general member parameterizes X and whose natural scheme structure is reduced.
Since X is projectively equivalent to Z = V (x 0 , . . . , x c−1 ) ∪ V (x n−d+1 , . . . , x n ); 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) = P n n−c,n−d (t). If Z P n−c P n−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 Thus we obtain, h 0 (P n , N Z /P n ) = c(n − c + 1) + d(n − d + 1) and h 1 (P n , N Z /P n ) = 0. It follows that [Z ] is a smooth point on its Hilbert scheme [15,Theorem 1.1c]. If n > c + d − 1, we will explicitly compute the tangent space to [Z ] using the following result: Theorem 1.1 (Comparison Theorem [23]) Let X ⊆ P n be a subscheme with ideal I X = ( f 1 , . . . , f r ) ⊆ S where deg f i = e i satisfying, (S/I X ) e H 0 (P n , O X (e)) for e = e 1 , . . . , e r . Then there is an isomorphism between the universal deformation space of I X and that of X . In particular, T [X ] Hilb P(t) P n = H 0 (P n , N X /P n ) = Hom S (I X , S/I X ) 0 . Remark 1. 2 With notation as in the above Theorem, consider the following exact sequence in local cohomology [ Since n > c + d − 1, the depth of S/I Z is at least 2. It follows from the previous Remark that the comparison theorem applies for Z . Lemma 1. 3 We have dim k T [Z ] Hilb P(t) P n = 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(I Z , S/I Z ) 0 can be written as for any 0 ≤ i ≤ c − 1 and n − d + 1 ≤ j ≤ n with some constants, a i , b i ∈ k.
Let us first show that ϕ(x i x j ) is supported on {x i x 0 , . . . , x i x n−d , x j x c , . . . , x j x n }. 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, x j ϕ(x i x j ) = x j ϕ(x i x j ). This implies that x j divides every non-zero monomial in ϕ(x i x j ) that is not annihilated by x j in S/I Z . It follows that ϕ(x i x j ) is supported on Similarly, choose i such that 0 ≤ i ≤ c − 1 and i = i. Then the equality x i ϕ(x i x j ) = x i ϕ(x i x j ) implies x i divides every monomial in ϕ(x i x j ) that is not annihilated by x i . Once again we see that ϕ(x i x j ) is supported on Thus ϕ(x i x j ) is supported on C ∩ C = {x i x 0 , . . . , x i x n−d , x j x c , . . . , x j x n }.
For any i, j, write ϕ( Proof We need to show that is defined along c \ c−1 . Up to projective equivalence, an 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 Thus the Hilbert polynomial of S/J is Using the "Hockey-Stick" identity this simplifies to for some scheme Y . Observe that I I is a saturated ideal. Indeed, up to projective equivalence, In both cases, I I is clearly saturated. Thus we have I Y = I I and taking nilradicals we obtain Similarly, I˜ ∪˜ = I Y red . Equating the two expressions we have ∪ =˜ ∪˜ . The conclusion now follows.

Structure of H n n−k,n−k
This section is devoted to an analysis of H n n−k,n−k . The first major goal of this section is to prove that H n n−k,n−k is smooth. We start with the case when the pair of planes parameterized spans P n . We construct a bijective morphism from a non-singular variety to H n n−k,n−k 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 P n , we construct a certain fibration to reduce to the case where they do span (Corollary 2.21).
Let n ≥ 2k −1 and X 0 = Gr(n−k, n) 2 . For each 1 ≤ v ≤ k −1, let X v = Bl v · · · Bl 1 X 0 and let π v : X v −→ X 0 be the blow-up morphism. The map (0.1) induces a rational map defined away from the strict transforms of the exceptional divisors. In order to study the structure of H n n−k,n−k , we will begin by extending to a morphism on X k−1 . For each ordered basis E = {e 0 , . . . , e n } of S 1 we obtain an affine neighbourhood (2. 2) It is clear that as E ranges over all ordered basis of S 1 , the set of U E cover X 0 . In particular, it suffices to extend along each π −1 k−1 (U E ) in a compatible way. For notational convenience we may assume E = {x 0 , . . . , x n } and let U 0 = U E . Observe that the locus v ∩ U 0 is cut out by the ideal generated by the v × v minors of the matrix Thus π −1 k−1 (U 0 ) is obtained by blowing up U 0 along the strict transforms of the ideal generated by the v × v minors of M for v = 1, . . . , k − 1, in that order.
(iv) v+1 ∩ U v is non-singular and the blowup along this locus is given by Proof We begin with the definition of U 1 . Since 1 is cut out by (a i, j − b i, j ) i, j on U 0 , it is a non-singular subscheme and we have Bl Let M v 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 U 1 , the Koszul relations simplify to a i, j − b i, j = λ 1 T (1) i, j ; here we have set T (1) k−1,n = 1. Substituting this into the matrix π 1 (M) and subtracting appropriate multiples of the bottom row from every other row, we obtain the matrix This proves item (ii) for v = 1. The ideal generated by the 2 × 2 minors of M 1 is Since this ideal is generated by a regular sequence, the blowup along it is non-singular and equal to Bl 2 ∩U 1 U 1 := Proj k[U 1 ][T (2) i, j ] i, j /(Koszul relations). This proves item (iii) and (iv) for v = 1. Now assume items (i)-(iv) have been proved for some Then the Koszul relations on this open  simplify . Once we substitute this into the matrix M v , it is straightforward to row reduce the matrix so that it becomes M v+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 U k−1 is Proof We will use a to denote the tuple (a i, j ) i, j and similarly use b and T (v) to denote their corresponding tuples. Moreover, we will use (a) to denote the (n − k)-plane corresponding to a as in (2.2).
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 I (a) is given by the rows of Id k×k | M z T where z = y 0 · · · y k−1 x k · · · x n is a row vector. Applying row operations to Id k×k | M will produce different minimal sets of generators. In particular, applying the row operations we did to M to get M k−1 (Proposition 2.1 (ii)) to the matrix Id k×k | M we obtain a new set of generators α 0 , . . . , α k−1 of I (a) where For 0 ≤ p < q ≤ k − 1 define the following "cross terms" where k p = k − 1 − p for all p and λ p,q = λ k−q+1 · · · λ k− p if p > 0 λ k−q+1 · · · λ k−1 if p = 0. Note that our convention implies λ 0,1 = 1. Extend to U k−1 by mapping (a, b, T (1) , . . . , T (k) ) → I (a) (y 0 , . . . , 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 λ 1 , . . . , λ k−1 = 0. In this case we have (2.5) Thus β p,q ∈ I (a) (y 0 , . . . , y k−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 = I (a) I (b) + (β p,q ) 0≤ p<q≤k−1 in this image is P n n−k,n−k (t). In Lemma 2.5 we define a term order > on S for which 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 Using this and the Hockey-Stick identity we deduce that Hilbert polynomial of S/in > J is Prior to proving Lemma 2.5 we need the following auxiliary result.
Proof Applying the projective transformation that maps and fixes the other x i , we may assume b = 0. For each 0 ≤ i ≤ k −1 let τ i denote the map that sends i,n− j+1 x k− j and fixes the other i. It is clear that τ k−1 • · · · • τ 0 (I ) equals, x j and fixes the other x i . As we range over all i, we obtain maps τ n , . . . , τ n−(k−2) . If μ k = 0 let τ n−(k−1) be the identity; else let τ n−(k−1) denote the map that and fixes the other x i . Using the fact that T (k−i) i,n−k i = 1 on the open set U k−1 , it is straightforward to check that τ n−(k−1) • · · · τ n • τ k−1 • · · · • τ 0 (I ) is of the desired form.

Lemma 2.5 Let > denote the lexicographic ordering on S with terms ordered by x
Proof Let J denote the ideal in (2.6). We will first show that (2.7) Let γ p,q = (x p +μ p,k x n−k p )x q for 0 ≤ p ≤ q ≤ k−1 and δ p,q = x p x n−k q −μ p,q x q x n−k p for 0 ≤ p < q ≤ k − 1. Since in > γ p,q = x p x q and in > δ p,q = x p x n−k q , to prove (2.7) it is enough to show that G = {γ p,q , δ p,q } p,q is a Gröbner basis for J . Note that G generates J because for p < q we have (2.8) Notice that μ p,q μ q,k = μ p,k and this will be used repeatedly in the rest of the proof.
). 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 [18, Section 2.2-2.3].
Case 1. The standard expression of R(γ p 1 ,q 1 , γ p 2 ,q 2 ): Let h = gcd(in > γ p 1 ,q 1 , in > γ p 2 ,q 2 ) and we may assume p 1 ≤ p 2 . If h = 1 then p 1 < p 2 and we have This is obviously a standard expression with no remainder. If h = x p 1 then p 1 = p 2 or p 1 = q 2 ; in the latter case we still have p 1 = p 2 as our assumptions imply p 1 ≤ p 2 ≤ q 2 . Thus in both the situations we obtain R(γ p 1 ,q 1 , γ p 2 ,q 2 ) = x q 2 γ p 1 ,q 1 − x q 1 γ p 1 ,q 2 = 0. If h = x q 1 we have either q 1 = q 2 or q 1 = p 2 . If q 1 = q 2 then as shown above we obtain The standard expression of R(δ p 1 ,q 1 , δ p 2 ,q 2 ): Let h = gcd(in > δ p 1 ,q 1 , in > δ p 2 ,q 2 ) and assume p 1 ≤ p 2 . If h = 1 we have p 1 < p 2 and q 1 = q 2 . Then we obtain Each of the above cases is a standard expression in terms of G with no remainder 2 . If h = x n−k q 1 we have q 1 = q 2 and p 1 < p 2 . Then we obtain If h = x p 1 we have p 1 = p 2 and wlog we may assume q 1 < q 2 . Then we have Finally if h = x p 1 x n−k q 1 we have p 1 = p 2 < q 1 = q 2 and thus R(δ p 1 ,q 1 , δ p 2 ,q 2 ) = 0. Case 3. The standard expression of R(γ p 1 ,q 1 , δ p 2 ,q 2 ): Let h = gcd(in > γ p 1 ,q 1 , in > δ p 2 ,q 2 ) and note that h ∈ {1, x p 1 , x q 1 }. If h = x p 1 we have p 1 = p 2 and using (2.8) we obtain Both these cases are standard expressions with no remainder. If h = x q 1 then q 1 = p 2 and we obtain, Finally consider the case h = 1. If we further assume p 2 < p 1 and q 2 < p 1 we have This is a standard expression with no remainder. We omit the other cases as their proofs are very similar (use Eq. 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, (x 0 , . . . , x k−1 ) 2 ⊆ in > J and x p x n−k q = in > (β p,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 k p will always denote k − 1 − p.
The following Lemma sheds some light on the structure of the subschemes in the image of the morphism, U k−1 −→ H n n−k,n−k .
The first inequality is [18, Theorem 3.3.4] and the second inequality is true because x k is a non-zero divisor on S/in > J . Notice that (a) and (b) meet along a (n − k + 1 − )-plane precisely when the matrix M (Proposition 2.1 (ii)) has rank − 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 β p,q ∈ I (a) I (b) for any q > k − .
Proof In Proposition 2.3 we showed that extends to a map from U k−1 . We will now explain how the same argument gives a morphism on all of π −1 k−1 (U 0 ). Consider a pair For any such γ we can define a sequence of open sets U Due to symmetry, the proof of Proposition 2.1 also establishes the above statements (note that . . , 0) and γ 2 = (n, n − 1, . . . , n − k + 2)). It follows that {U γ k−1 } γ is an affine cover of π −1 k−1 (U 0 ) with the natural gluing maps. We omit an explicit description of the gluing maps as they will never be used.
To construct the U γ v 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 As in the proof of Proposition 2.3 we can choose a minimal generating set, Thus we obtain a morphism This is well defined as any ideal in the image of As mentioned in the beginning of the section, Gr(n − k, n) 2 is covered by open sets of the form U E where E ranges over all ordered bases of S 1 . Since assuming E = {x 0 , . . . , x n } was purely notational, all the discussion in this section applies verbatim to π −1 k−1 (U E ). In particular, we obtain a morphism on each π −1 k−1 (U E ) that extends the original rational map (2.3). Thus we can glue all these maps to obtain a morphism : Let S 2 = {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 i are S 2 stable, the action extends to the blowup X k−1 . If we consider the trivial action of S 2 on H n n−k,n−k , then our construction shows that is S 2 -equivariant. Thus, we get an induced morphism Since char k = 2 and g fixes a divisor (the strict transform of the exceptional divisor of X 1 ), the Chevalley-Shephard-Todd theorem [21,Theorem 7.14] implies that the quotient is non-singular. Note that The natural action of GL(n+1) on P n induces an action on Gr(n−k, n) 2 and on H n n−k,n−k . Since the i are stable under this action, it extends to an action on X k−1 . To show that is GL(n + 1)-equivariant we need to show that for any g ∈ GL(n + 1) the two morphisms, • g : X k−1 → H n n−k,n−k given by w → (gw) and g • : Thus •g and g • must agree on all of X k−1 . It follows that is also GL(n +1)-equivariant.

Corollary 2.9
Let n ≥ 2k − 1. Any subscheme parameterized by H n n−k,n−k is minimally cut out by k 2 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 dim k J 2 = k 2 . Since S/J and S/in > J have the same Hilbert function we have dim k J 2 = dim k (in > J ) 2 = k 2 (Lemma 2.5).

Remark 2.10
The analogue of Lemma 2.7 holds verbatim for ideals in the image of Eq. (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 We may choose any h i so that We split the proof of the injectivity of into two steps. Here is the first step.

Lemma 2.11
For any γ , the restriction : Proof It is evident from our construction that U γ k−1 is S 2 -stable and thus the quotient U γ k−1 /S 2 is well defined. Without loss of generality we may assume U γ k−1 = U k−1 . To prove the Lemma it suffices to show that for anyZ ,Ẑ ∈ U k−1 satisfying (Z ) = (Ẑ ), we have (1) , . . . ,T (k) ) andẐ = (â,b,T (1) , . . . ,T (k) ) be their coordinates on U k−1 . The "betas" and "lambdas" corresponding toZ are denoted byβ i, j andλ i respectively, and the ones corresponding toẐ are denoted byβ i, j andλ i . We After possibly replacing Z ,Ẑ by g(Z ), g(Ẑ ) respectively, we may assumeã =â andb =b. Thus to prove that is injective, we need to now show thatZ =Ẑ . Since is GL(n + 1)-equivariant we may apply a projective transformation and assumeb =b = 0. For simplicity we let a :=ã =â.
By Lemma 2.7, (Z ) red = (Ẑ ) red is a pair of (n − k)-planes meeting along an (n − k + 1 − )-plane for some 1 ≤ ≤ k + 1. If ∈ {k, k + 1} then Z , Z lie in an open set along which was already shown to be two-to-one (Lemma 1.6). Thus we may assume ≤ k − 1. By Lemma 2.7 it is also the smallest index for whichλ = 0 and, symmetrically, the smallest index for whichλ = 0.
Using Lemma 2.7 (iv) we get Using Lemma 2.7 (i) we have the equality and ω ∈ (β p,q ) 0≤ p<q≤k− such that α, ω are linearly independent and homogenous of degree 2. Sinceλ =λ = 0, the construction in Proposition 2.3 implies This implies α = 0 and we obtain B = (β p,q ) 0≤ p<q≤k− = (β p,q ) 0≤ p<q≤k− . 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 T (i)λ i =λ i for all i ≤ : We clearly haveλ 1 = a k−1,n =λ 1 . Sinceλ v = 0 for all v ≤ −1 we can inductively apply (2.11) to obtain i, j for all v ≤ − 1 and all i, j: Analogous to item (i) above, where we instead use (2.10) to concludê i, j for all k − 1 ≥ v ≥ 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 − and assumê β r ,s = 0≤ p<q≤k− c p,qβ p,q for some constants c p,q ∈ k. Let p = min{ p : c p,q = 0} and q = max{q : c p ,q = 0}. Then It follows thatβ r ,s =β r ,s . Equating the terms supported on x r we obtain Sinceλ 0,1 = 1 =λ 0,1 , equating the coefficients of the monomials containing x 1 gives the desired result.
Equating the coefficients of x k−i+1 x n−i+1 gives the desired result.

Lemma 2.12
The fiber of over the point consists of a single element. .
Comparing the monomial generators of the two ideals we deduce that γ 1 We need to show that = k (then automatically, γ 1 k = 0). For the sake of a contradiction, assume that < k. Our method is to compare certain initial ideal degenerations of (Z ) and (Y ).
Let w be any integral weight order corresponding to > [7, Section 15]. For any t ∈ k let g t ∈ GL(n + 1) denote the automorphism that maps x i → t −w(i) x i . Since each g t just scales the coordinates the following facts are immediate (1) g t induces an action on X 0 and extends to all the blowups X v .
(2) g t fixes U γ and also fixes any closed subset of the form For each let ψ : X k−1 −→ X denote the blowdown map. Then ψ is GL(n + 1)equivariant and thus ψ (g t ) = g t (ψ ).
Using the notation in item (3) and our assumption on , ψ (Z ) and ψ (Y ) are k-points Remark 2.14 It follows that the preimage −1 (Z ) is a single point precisely when Z red is an (n − k)-plane. This occurs precisely when Z is generically non-reduced, c.f. Theorem D 3 .
The group GL(n + 1) acts on S and thus on Hilb P(t) P n 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 H n n−k,n−k has a unique Borel fixed point. We begin with a combinatorial characterization of the Borel fixed ideals, see [7,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 0-Borel fixed if for any monomial generator m ∈ I divisible by x j , we have x i x j m ∈ I for all i < j. The ideal I is said to be p-Borel fixed if for any monomial generator m ∈ I divisible by x β j but no higher power of x j , we have ( x i x j ) α m ∈ I for all i < j and α p β (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 0-Borel fixed ideal is always p-Borel fixed for any p. In our situation, char k = p ≥ 0 with p = 2. Let I be a saturated p-Borel fixed ideal parameterized by H n n−k,n−k . Since I is a monomial ideal generated by quadrics (Corollary 2.9) and p = 2, the condition α p β in Definition 2.15 reduces to the condition α ≤ β. In particular, I is always 0-Borel.
Then [I n n−k,n−k ] is the unique Borel fixed point on H n n−k,n−k . Proof As noted above, Borel fixed ideals in H n n−k,n−k are the same as 0-Borel fixed ideals. Since I n n−k,n−k is projectively equivalent to (x 0 , . . . , has codimension k, we obtain = k − 1. Arguing as in the end of the proof of Proposition 2.3 we see that the Hilbert polynomial of B is n−k+t . Equating this with the Hilbert polynomial of I n 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 [18,Theorem 3.3.4] we have also reg(I ) ≤ reg(in > I ). Note that in > I is projectively equivalent to I n n−k,n−k and the regularity of a 0-Borel ideal is the highest degree of a minimal monomial generator [18,Corollary 7.2.3]. Thus reg(I ) ≤ reg(I n n−k,n−k ) = 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 [I n n−k,n−k ] in X k−1 (Remark 2.14) and letζ denote the image of ζ in X k−1 /S 2 .
By constructing curves passing through ζ andζ we will now show that the differential d ζ is injective. This is a major portion of the proof of Theorem A.
Proof Note that we have a factorization . Thus to show that d ζ is injective it suffices to establish the following two facts We begin with item (1). Let Let I = I n n−k,n−k and under the inclusion H n n−k,n−k ⊆ Hilb P n n−k,n−k (t) P n , we may identify T [I ] H n n−k,n−k with a subspace of Hom(I , S/I ) 0 (Lemma 2.18). We can explicitly describe this identification using [15,Proposition 2.3]. In particular, by re-indexing, we obtain These Notice that the derivation k−1 is a scalar multiple of x k ∂ ∂ x k−1 . Thus to prove (1) it suffices to show that the set {x j with some constants i, j , i ∈ k. Assume p,q = 0 for some p < q. Since x p x 2k−2− p ∈ I we may evaluate (2.12) at x p x 2k−2− p to obtain p+1≤ j≤n Observe that the monomial x q x 2k−2− p does not appear in the support of Q. Thus, in the left hand side of (2.13), the monomial x q x 2k−2− p appears with a coefficient of p,q if p = k − 1 and a coefficient of 2 p,q if p = k − 1. In either case, the coefficient is non-zero. But this is a contradiction as x q x 2k−2− p / ∈ I . Thus we have p,q = 0 for all p, q. Evaluating (2.12) at x p x 2k−2− p we see that p = 0 for every p ∈ {1, . . . , k − 2}. Finally, evaluating (2.12) at x 0 x 2k−2 we obtain n i=2k−1 i x 1 x i ≡ 0 mod I . Since x 1 x i / ∈ I for all i ≥ 2k − 1, we must have that i = 0 for all i. This completes the proof of item (1).
Let ∈ Hom(I , S/I ) 0 denote the derivation that maps x k−1 x k → x 2 k and all the other generators to 0. By evaluating at x k−1 x k it is easy to see that does not lie in the span This is well defined because for any given s and all other coordinates equal 0. It is also clear that C • ι corresponds to the derivation . Thus to prove item (2) it suffices to find a curve C : Spec Let Z denote the image of C and let Z denote the pullback −1 (Z ) ⊆ X k−1 /S 2 . I claim that | Z : Z → Z is an isomorphism. Since Z is non-singular, Z is Cohen-Macaulay and is bijective, the morphism | Z is flat. It is clear that a finite flat degree 1 morphism is an isomorphism. Thus We are now ready to prove the main Theorem.
Proof Propositions 2.8 and 2.13 together show that is bijective and X k−1 /S 2 is nonsingular. Since is GL(n + 1)-equivariant,ζ (Definition 2.19) is the unique Borel fixed point on X k−1 /S 2 . By Borel's fixed point theorem, the closure of the Borel orbit of any point in X k−1 /S 2 containsζ . Thus to show that is an isomorphism, it suffices to show that it is an isomorphism in a neighbourhood ofζ . By the proof of [17,Theorem 14.9], this is equivalent to showing that d ζ : Tζ (X k−1 /S 2 ) −→ T [I n n−k,n−k ] H n n−k,n−k is injective. This is precisely the content of Lemma 2.20.
When the pair of planes do not span P n , we obtain the following fibration Corollary 2.21 Let n < 2k − 1. The morphism ρ : H n n−k,n−k −→ Gr(2n − 2k + 1, n) that sends a scheme to its linear span is smooth; the fiber over a point is H n−k,n−k ( ).
Proof Recall that the linear span of a subscheme Z ⊆ P n is the linear space V (H 0 (P n , I Z (1))) ⊆ P n . Let Y −→ A 1 be a flat family such that for t = 0, Y t is a disjoint pair of (n − k)planes. It is clear that for any t = 0, the linear span of Y t is a (2n − 2k + 1)-plane. By upper semicontunity, the limit Y 0 also lies in a (2n − 2k + 1)-plane, which we denote by . Thus Y 0 defines a point in H n n−k,n−k ( ) and by Corollary 2.9, we see that the linear span of Y 0 is all of . It follows that the linear span of any subscheme parameterized by H n−k,n−k (P n ) is of dimension 2n − 2k + 1.
Thus the set of schemes {X E } E glue to a smooth scheme X (Theorem A).
For each E we obtain a natural morphism g E : U E −→ GL(n + 1) such that for any f, g E (f) is the map that sends e i → e i + n j=2k−1−n f i, j e j if i ≤ 2k − 2 − n and fixes the other coordinates. Thus we may define a map These maps glue to a morphism : X −→ H n n−k,n−k . By the first paragraph, is a bijective morphism. It is also clear that the differential to is injective at all points. As noted in Theorem A, this implies that is an isomorphism. By construction, there is a smooth fibration ρ : X −→ Gr(2n − 2k + 1, n) of the desired form.

Theorem C H n n−k,n−k 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 ((I X ) 1 ) is also Borel fixed. Thus X lies in the fiber ρ −1 (V (x 0 , . . . , x 2k−2−n )) H 2n−2k+1 n−k,n−k . Moreover, the Borel action on H n n−k,n−k 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 H n n−k,n−k . 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 = (x 0 , . . . , x k−1 ) 2 + (x p x n−k q − μ p,q x q x n−k p ) 0≤ p<q≤k−1 with μ i ∈ k and μ p,q = μ k−q+1 · · · μ k− p for any 0 ≤ p < q ≤ k. If all the μ i 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 (x 0 , . . . , x k−1 ).
Proof Applying the change of coordinates that maps x p → μ p,k x p for all p ≤ k − 1 and fixing the other coordinates, we may assume μ p,q = 1 for all p, q. If n > 2k −1, the variables x k , . . . , x n−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 (x k , . . . , x n−k ) to reduce to the case n = 2k − 1; in this case n − k p = k + p. Since the Hilbert polynomial of Proj(S/J ) is P n n−k,n−k (t), its degree is 2; thus it is a double structure on the linear space V (x 0 , . . . , x k−1 )
Proof For the first statement we proceed by induction on 2 . The base case 2 = 1 is vacuous and by induction we may assume ). The conclusion now follows from the fact that if I 1 = (m 1 , . . . , m i 1 ), Theorem D Let n ≥ 2k − 1. Let Z be a subscheme parameterized by H n n−k,n−k . Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1 ≤ i 1 < · · · < i r ≤ k and a flag of linear spaces 1 ⊆ 2 ⊆ · · · ⊆ r ⊆ P n with codim P n ( ) = (k + i − 1) for each , such that Proof It suffices to compute a primary decomposition of the ideal implies that all the μ i 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 μ i are zero. Let i 1 < · · · < i r be all the indices i for which μ i = 0. Set i 0 = 0 and i r +1 = k + 1. Lemma 2.7 (iv) implies √ J = P 0 ∩ P 1 and J = P 0 P 1 + (δ p,q ) 0≤ p<q≤k−i 1 . For each 2 ≤ ≤ r + 1 define I claim that J = P 0 ∩ P 1 ∩ · · · ∩ P r +1 (note that if μ 1 = 0 then P 0 = P 1 ). We begin with the inclusion, J ⊆ P 0 ∩ · · · ∩ P r +1 . It is enough to show that P 0 P 1 and δ p,q lie in P 0 ∩ · · · ∩ P r +1 for 0 ≤ p < q ≤ k − i 1 . Observe that Clearly, (x 0 , . . . , x k−i 1 )(x 0 , . . . , x k−1 ) ⊆ P j for all j. We also have, x p , x n−k p ∈ P j for all k − i 1 + 1 ≤ p ≤ k − 1 and all j. Thus P 0 P 1 ⊆ P 0 ∩ · · · ∩ P r +1 . It is clear that δ p,q ∈ P 0 ∩ · · · ∩ P r +1 if there is some such that k − i + 1 ≤ p < q ≤ k − i −1 . If this was not the case, then there is some such that p ≤ k − i < q. This implies δ p,q = x p x n−k q and this lies in (x 0 , . . . , x k−i j ) if j ≤ or in (x n−i j−1 +2 , . . . , x n ) if j > ; in either case, δ p,q ∈ P j . Thus δ p,q ∈ P 0 ∩ · · · ∩ P r +1 and we have the desired containment.
To get the other containment it suffices to show that P 0 ∩ · · · ∩ P r +1 has the same Hilbert function as J . We have Our goal is to show all these containments are equalities. Using Eq. (2.8) we have Then the proof of Lemma 2.5 immediately implies Similarly for ≥ 2 Using Lemma 2.24 we see that in > (P 0 ∩ P 1 ) ∩ in > P 2 ∩ · · · ∩ in > P r +1 equals j , x j+1 , . . . , x k−1 , x n−k j+1 , . . . , x n ). 4 Applying Lemma 2.24 once again we see that this intersection is just J (0, k − 1) ∩ (x 0 , . . . , x k−1 ). But this ideal is precisely (x 0 , . . . , Thus all the containments in (2.14) are equalities and this shows that J has the same Hilbert function as P 0 ∩ · · · ∩ P r . We are left with showing P is a primary component for all ≥ 2. Going modulo the linear forms it suffices to show that ( Corollary E Up to projective equivalence, there are exactly 2 k schemes parameterized by H n n−k,n−k . 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 x p → μ p,k x p if μ p,k = 0 and 0 ≤ p ≤ k − 1 and fixes the other coordinates. For a fixed p, note that if μ p,k = 0 then μ q,k = 0 and μ p,q = 0 for all p < q. Thus after applying ϕ we may assume that the non-zero μ i are equal to 1. In particular, for each subset W ⊆ {1, . . . , k} we obtain an ideal parameterized by H n n−k,n−k by setting μ i = 0 if i ∈ W and 1 otherwise; this gives at most 2 k distinct ideals. On the other hand, since projective transformations preserve the dimensions of the embedded structures, each of the 2 k ideals are projectively inequivalent. Example 2. 25 We can now determine when there is a specialization Z Z in H n n−k,n−k . For any subscheme Z ∈ H n n−k,n−k let W Z = { 1 , . . . , r } 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 W Z ⊆ W Z Here is a diagram of specializations for H 5 2,2 . The non-reduced structures on points, lines and planes are represented by shadings.

Structure of H n n−c,n−d
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 Hilb P n n−1,n−d (t) P n parameterizes ideals of codimension 1. Using the decomposition in [24, Proposition 2.4] we obtain Thus H n n−1,n−d 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 Hilb P n n−1,n−d (t) P n .
We have shown in Lemma 1.6 that the rational map : We will now perform a few substitutions and obtain a different minimal set of generators for I (a) and I (b) . From these new presentations of I (a) and I (b) , it will be apparent how one has to mimic the arguments of Sect. 2 to extend to π −1 c−1 (U 0 ), and thus all of X c−1 .
For any 0 ≤ i ≤ c − 1 we obtain  From these descriptions of I (a) and We can now prove an analogue of Proposition 2.

that sends a scheme to its linear span is smooth; the fiber over a point is H n−c,n−d ( ).
Theorem D' Let n ≥ c + d − 1 and let Z be a subscheme parameterized by H n n−c,n−d . Then Z is a pair of planes meeting transversely, or there exists a sequence of integers 1 ≤ i 1 < · · · < i r ≤ c and a flag of linear spaces 1 ⊆ 2 ⊆ · · · ⊆ r ⊆ P n with codim P n ( ) = (d + i − 1) for each , such that  [5] it was shown that H n n−2,n−2 meets exactly one other component in Hilb P n n−2,n−2 (t) P n and that this component is smooth. We will give two examples that show these statements are false in general.
The component H 5 2,2 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 [24,Theorem 3.16] we show that Hilb P n n−2,1 (t) P n is a union of H n n−2,1 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 P 1 × P n−2 → P 2(n−1)−1 .
This completes the discussion of the local structure of H n n−c,n−d . The next four sections will pertain to its global geometry. As we did in Sect. 2, we begin studying divisors on

Divisors on H n n−k,n−k
In this section we study the Picard group of H n n−k,n−k for n ≥ 2k − 1. We give an explicit description of the divisors D i , N i (Remark 4.6, 4.9) and describe equations for their pullback along | U k−1 .

Notation 4.1
We will use λ k to denote the coordinate T (k) 0,n−k+1 on U k−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.   Proof Since H n n−k,n−k = X k−1 /S 2 is a smooth rational variety, its class group is torsion free. In particular, 0) and O X 0 (0, 1), respectively. By [14,Theorem 8.5], Cl(X k−1 ) Q is freely generated by E 1 , . . . , E k−1 , E 1,0 , E 0,1 . Since S 2 fixes E i and interchanges E 1,0 with E 0,1 , it follows that

Definition 4.5
Let (X 0 ) trv = X 0 \ k denote the open subset of X 0 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 transversely, if each plane in the pair meets transversely.
We now describe D i as a scheme theoretic image under .

Remark 4.6 For each 1
be the open subset consisting of pairs of planes that meet 2k−1−i transversely. Let D i denote the (scheme theoretic) closure of Proof Assume 1 ≤ i ≤ k − 1 and let D i be defined by the flag (4.1). To show that D i is a divisor, it suffices to show thatD i ∩ W i is a divisor in W i (notation from Remark 4.6). By symmetry, it is enough to show thatD i ∩ W i ∩ U 0 is a divisor in W i ∩ U 0 . Given a point ( (a), (b)) ∈ W i ∩ U 0 we have ( (a) ∪ (b)) ∩ 2k−1−i = P ∪ Q for a pair of (k − 1 − i)-planes, P and Q. For each n − k i+1 ≤ j ≤ n let p j (respectively q j ) denote the point in P (respectively Q) obtained by setting x j = 1 and x = 0 for all other ≥ k (there are no such points for i = k − 1). Explicitly, p j = (−a 0, j : · · · : −a k−1, j : 0 : · · · : 0 : 1 : 0 : · · · : 0) q j = (−b 0, j : · · · : −b k−1, j : 0 : · · · : 0 : 1 : 0 : · · · : 0).
In particular,D i ∩ W i ∩ U 0 is the locus where the matrix has rank less than 2k − i. Let l, j = a l, j − b l, j and apply the row operation q n − p n − l l,n r l p n−k i . . . p n r 0 . . .
and swap the ith column and (i − 1)st column. It follows that the locus is cut out by the determinant of the submatrix ⎛ ThusD i ∩ W i ∩ U 0 is a divisor and this determinant also cuts outD i ∩ U 0 . The strict transform of this determinant cuts out (D i ) ∩ U k−1 . Pulling back this matrix to U k−1 and column reducing as in Proposition 2.1 we obtain ⎛ In either case, (D i ) ∩ U k−1 is cut out by the desired equation. Lastly, D k is a divisor sincê D k is the Weil divisor associated to O X 0 (1, 1) ∈ Pic X 0 Z 2 .
and let D k be defined by the plane

is cut out by a polynomial in the coordinates of Remark 2.2 that is linear in λ k− j .
Proof Assume i ≤ k − 1 and j = 0. Imitating the proof of Lemma 4.7 we see that To express this in terms of our desired coordinates we will use the relation T ( ) which is true for any q ≤ n − k p and any p < k − and < k −1 (proof of Proposition 2.1). Repeatedly applying this relation we obtain the following expressions for any q < n − k j . Thus T (k−i) j,q , as a polynomial in the coordinates of Remark 2.2, is linear in λ k− j for all q ≤ n − k j . This implies Assume i ≤ k − 1 and j = 0. Most of the argument from the previous paragraph still applies in this case. In particular, Finally assume i = k. The locus of points ( (a), (b)) ∈ U 0 meeting k−1 is clearly cut out by (a j,n−k j − 1)(b j,n−k j − 1). The pullback of this equation to U k−1 , which coincides with the strict transform, defines (D k ). If j = 0 we can use (4.3) to deduce that This expression is linear in λ k− j . If j = 0 we can argue in the previous paragraph and deduce linearity in λ k . This completes the proof.
Here is an alternate description of N i .

Remark 4.9 For each 1
If n > 2k − 1, letN k denote the closure in X 0 , of the locus of pairs of planes in X trv 0 where the intersection of the two planes meets a fixed 2k−1 . Then N k is the image of the strict transform ofN k under .
In the next lemma we abuse notation and use "=" to mean equality as divisor classes.
Proof Assume 1 ≤ i ≤ k − 1. Remark 4.9 implies that the N i are divisors. Items (i), (ii) and the first half of (iii) follow from the fact that is a finite, degree 2 map branched along N 1 (although not phrased this way, it is part of the proof of Proposition 2.8), see [11,Chapter 1.7]. The rest of item (iii) is a consequence of Lemma 4.2 (ii). Now assume n > 2k − 1 and letN k be as in Remark 4.9. To show that N k is a divisor it is enough to show thatN k ∩ X trv 0 ∩ U 0 is a divisor in X trv 0 ∩ U 0 . Given a point ( (a), (b)) ∈ X trv 0 ∩ U 0 , the intersection of the two planes is (a) Column reducing as in Proposition 2.1 (ii) and taking the strict transform gives item (iv).

Birational geometry of H n n−k,n−k 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 D i and N i . 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.

Remark 5.2
Theorem D shows that C j (s : t) is projectively equivalent to (2.6) with It also shows that for j ≤ k − 2, the general member of C j is a pair of (n − k)-planes meeting along a pencil of embedded (n − 2k + j + 1)-planes and containing fixed embedded (n − 2k + )-planes for all 1 ≤ ≤ j − 1, while C k−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.
and consider the pencil of (n − k)-planes (s :

Remark 5.4
Theorem D shows that B j (s : t) is projectively equivalent to (2.6) with If (s : t) = (1 : 0), then B 0 (s : t) is a pair of (n − k)-planes meeting transversely while B j (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.
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 C † j (s : t) and B † j (s : t) to denote the subschemes of P n cut out by C j (s : t) and B j (s : t), respectively. Given an ideal J ⊆ S, let sat(J ) denote its saturation with respect to (x 0 , . . . , x n ) and let J (1) denote the ideal generated by the linear forms in J .
Proof Let = 2k−i−1 and note that the linear span of C † j (s : t) ∩ is cut out by sat(C j (s : Moreover, it is clear that Q(d) = (C j (s : t) + I )(d) for all d ≥ 2. Thus if we show that Q is saturated then Q = sat(C j (s : t) + I ), and this would give the desired linear span. If we write Q = I + (x 0 , . . . , x i−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 p and q are disjoint linear spaces in P n then dim k span( p ∪ q ) = p + q + 1.
The first fact is clear and the second follows from the following chain of equalities, Lemma 5.8 Let 1 ≤ i ≤ k and 1 ≤ j ≤ k − 1. We have the following intersection numbers Proof Assume i > j. Since the dimension of any embedded subscheme of C † j (s : t) is at most n − 2k + j + 1, a generic (2k − 1 − i)-plane will not intersect any embedded subscheme of C † j (s : t). If i < k, the intersection of C † j (s : t) with a generic 2k−1−i is a pair of skew (k − 1 − i)-planes. Moreover, these skew planes are independent of (s : t) and thus span (C † j (s : t) ∩ 2k−1−i ) P 2k−2i−1 is independent of (s : t). As a consequence, we may choose an (i −1)-plane i−1 ⊆ 2k−1−i that does not meet the P 2k−2i−1 . It follows from Remark 5.7 that is fixed and independent of (s : t). As done in the previous paragraph, if we choose a general i−1 inside 2k−1−i to define D i , then D i · C j = 0. This completes the proof of item (i). Assume i = j and let the flag Thus, if t = 0, the linear span of ( LetC i denote the closure in X k−1 of the curve, A 1 → U k−1 obtained by setting λ 1 , . . . , λ k−i−1 = 1, λ k−i+1 = t and all the other coordinates of Remark 2.2 to 0. Since , it follows that (D i ) andC i intersect transversely atZ 0 . Using the push-pull formula we conclude that C i · D i = C i · D i = (C i · (D i )) = 1.

Lemma 5.9
Let 1 ≤ i ≤ k and 0 ≤ j ≤ k − 1. We have the following intersection numbers . Arguing as in Lemma 5.6 we see that . , x n−k i−1 ) P 2k−2i−1 is independent of (s : t). Arguing as in Lemma 5.8 we deduce item (i).
Assume that j < i ≤ k − 1 and let For t = 0, the linear span of (B † j (s : t) ∩ 2k−i−1 ) ∪ i−1 is all of 2k−i−1 . On the other hand if t = 0, the linear span of (B † j (s : LetB j denote the closure in X k−1 of the curve, A 1 → U k−1 obtained by setting λ 1 = · · · = λ k− j−1 = 1, λ k− j = t, λ k− j+2 = · · · = λ k = 1 and all the other coordinates of Remark 2.2 to 0. Since it follows that (D i ) andB j intersect transversely atZ 0 . Using the push-pull formula we . , x n ) be the plane defining D k . It is evident that B j ∩ D k is supported at the point Z 1,1 = B j (1 : 1). Once again,B j (defined in the previous paragraph) and (D k ) will meet at a unique point is linear in λ k− j (Corollary 4.8) we see thatB j meets (D k ) transversely atZ 1,1 . Once again we conclude using the push-pull formula. Lemma 5. 10 We have the following intersection numbers, Proof Item (i) and item (ii), except for the case of i = k, follow from the definition of the N i 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 Z 0 = C k−i+1 (1 : 0) and Z ∞ = C k−i+1 (0 : 1). We will also use the curvesC k−i+1 andB j defined in Lemma 5.8. In particular, letZ 0 ,Z ∞ ∈C k−i+1 be such that (Z 0 ) = Z 0 and (Z ∞ ) = Z ∞ . Assume 2 ≤ i ≤ k − 1. Since N i is the locus of subschemes containing an embedded (n − k + 1 − i)-plane, it meets the curve C k−i+1 at Z 0 and Z ∞ . ThusC k−i+1 meets E i at Z 0 andZ ∞ . Using Lemma 4.10 (ii), we obtain SinceZ 0 ∈ U k−1 and E i is cut out by λ i ,C k−i+1 meets E i transversely atZ 0 . Symmetrically, C k−i+1 will also meet E i transversally atZ ∞ . To see the latter statement, consider the projective transformation g ∈ GL(n + 1) that interchanges x j with x j−1 , interchanges x n−k j with x n−k j−1 and fixes the other coordinates. It follows from the definition that g(C k−i+1 ) = C k−i+1 and g interchanges Z 0 with Z ∞ . Since intersection multiplicity is invariant under automorphisms of H n n−k,n−k we obtain Since N 1 is the locus of generically non-reduced subschemes, it meets the curve B k−1 at B k−1 (1 : 0). Using Lemma 4.10 (i) we obtain N 1 ·B k−1 = (B k−1 · (N 1 )) = 2B k−1 ·E 1 = 2. Similarly, using Lemma 4.10 we obtain N i · B k−i = 1 for all 2 ≤ i ≤ k − 1. This finishes item (iv) for i = k Finally, assume i = k and let 2k−1 = V (x k , . . . , x n−k ) be the plane defining N k (if n > 2k − 1). By Lemma 4.10 (iii), (iv) we see that (N k ) meetsC 1 at Z 0 and possibly also at Z ∞ (since the latter does not lie in U k−1 ). Moreover, (N k ) meetsC 1 transversely atZ 0 . We may argue as in the previous paragraph to show that (N k ) also meetsC 1 transversely atZ ∞ . Indeed, the projective transformation g fixes N k . This is clear if n = 2k−1 and the case of n > 2k − 1 follows from the fact that g fixes 2k−1 . Thus N k · C 1 = (N k ·C 1 )| Z 0 +(N k ·C 1 )| Z ∞ = 2(N k ·C 1 )| Z 0 = 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 x n−k j → x n−k j + x j and fixes the other coordinates. It is straightforward to verify that g (B j ) = B j , g (B j (0 : 1)) = B j (1 : 1) and g fixes N k (since g fixes 2k−1 ). This implies for j = 1. Thus, we may compute (N k ) ·B j along U k−1 to obtain the desired results. Proposition 5.11 Let 1 ≤ i ≤ k. Then we have Proof By Lemma 4.4, Lemma 5.8 and Lemma 5.9 we see that N 1 (H n n−k,n−k ) is generated by {D 1 , . . . , D k }. This allows us to write N i = k =1 i, D for some i, ∈ Z. Using Lemmas 5.8-5.10 we obtain Proof It is clear that the divisors N 1 , . . . , N k are effective and generate N 1 (H n n−k,n−k ). To conclude that the effective cone is generated by N 1 , . . . , N k , it is enough to show that any denote any curve such that for (s : t) = (1 : 0), A j (s : t) is a pair of (n − k)-planes meeting transversely while A j (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, A j · N i = 0 for i = j and A j · N j > 0. Since N · A j = j < 0 and A j 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 D i is base point free; thus it is also nef. Similar to the previous paragraph, to show that the nef cone gone is generated by D 1 , . . . , D k , it is enough to show that any R-divisor D = k i=1 i D i , with some j < 0, is not nef. If j = k, we have D · C j = j < 0 and if j = k we have D · B k−1 = k < 0. Thus D is not nef.
We will now compute the canonical divisor of H n n−k,n−k using the branched cover : Hurwitz's theorem implies that K X k−1 = (K H n n−k,n−k ) + E 1 . Using this and Lemma 4.10 we obtain Let˜ j = (k − j + 1)(n − k − j + 2) − 1 and using Proposition 5.11 we obtain For k = 2, 3 the above expression simplifies to If k ≥ 4 we can rewrite the expression as follows: Since the ample cone is the interior of the nef cone, we see that −K H n n−2,n−2 is ample if and only if n = 3, 4 and that −K This section is devoted to the proof of Theorem 6.14. We will show that H n k−1,k−1 is Fano, and thus a Mori dream space. By constructing a contraction from H n k−1,k−1 to H n n−k,n−k (Proposition 6.11) we will also deduce that H n n−k,n−k 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 P n . By swapping the roles of codimension and dimension, the components we are interested in are of the form H n k−1,k−1 with n > 2k − 1.
Proof The general subscheme parameterized by N k 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 In particular, the GL(n + 1) orbit of Y 2 covers a dense subset of N k . Proof We will only verify D i · C j = D i · C j for 1 ≤ i, j ≤ k − 1; the other cases are analogous. Let = V (x 2k , . . . , x n ) be a fixed (2k − 1)-plane. Let D i be defined by a flag where the flag is chosen to satisfy the following two properties: • is transverse to each element of the flag F i , Let W be the open neighbourhood of from Remark 6.2. The first bullet point implies that every element of W is transverse to the flag F i . It follows that Lemma 6. 9 We have the following intersection numbers Proof Items (i) and (ii) are clear from the definition of the divisors.
Let 1 ≤ i ≤ k, = V (x 2k , . . . , x n ) and W be as in Remark 6.2. We may choose a flag F i to define D i so that the following properties are satisfied: • is transverse to each element of the flag F i , Along this open set, Y 1 is the curve obtained by setting f 2k,k = t, f i, j = 0 for other i, j, and i = δ i for some constants δ i ∈ k. On the other hand, where D i is the divisor defined by the flag F i ∩ . It immediately follows that D i meets Y 1 transversely at Z 0 inside W × W ; this proves item (iii).
For item (iv), we will only verify F ·Y 1 = 1 as the other case is similar. Let F be defined by the (n − 2k)-plane, V (x 0 , . . . , x k−1 , x k+1 , . . . , x 2k ). It follows that F ∩ Y 1 is also supported at Z 0 . Moreover, along W × W , F is cut out by the function f 2k,k . Combining this with the equation of Y 1 along W × W we see that F meets Y 1 transversely at Z 0 .
Proof Using the intersection numbers with the curves {C 1 , . . . , C k , Y 2 } and arguing as in Proposition 5.11, 5.12 we see that N 1 (H n k−1,k−1 ) and Nef(H n k−1,k−1 ) are both generated by D 1 , . . . , D k , F. Using the curves {A 1 , . . . , A k , Y 1 } and arguing as in Proposition 5.12, we see that N 1 , . . . , N k , F generate the effective cone.
We are now ready to relate H n k−1,k−1 with H n n−k,n−k .

Proposition 6.11
There is a morphism : H n k−1,k−1 −→ H n n−k,n−k with exceptional locus N k . Moreover, N k is a P n−2k+1 -fibration over (N k ). Geometrically, "forgets" the embedded points.
To show that contracts N k , it is enough to show that contracts Y 2 (Lemma 6.7). Using Lemma 6.9 we obtain Y 2 · D i = (Y 2 · (D i )) = (Y 2 · D i ) = 0 for all i. Since D 1 , . . . , D k generates the nef-cone of H n n−k,n−k we must have Y 2 = 0, i.e. contracts Y 2 .
Conversely, let C be any curve contracted by . If C · D i = 0 for some i, we would have C · D i = (C · D i ) = 0, proving that does not contract C. Thus we may assume C · D i = 0 for all i. Since {D i } i ∪ F generates the nef-cone of H n k−1,k−1 we must have F · C > 0. Using Proposition 6.10 we obtain N k · C = −F · C < 0, i.e. C lies inside N k .
Here is the the main theorem of the paper: Theorem 6.14 The components H n k−1,k−1 and H n n−k,n−k are Mori dream spaces. Proof This follows immediately from Proposition 5.12, 6.11 , 6.13 and the subsequent two facts: In this section we explain how the proofs of Sects. 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 P n .  (1) c and D (2) c are the Weil divisors associated to the strict transforms, under , of O X 0 (0, 1) and O X 0 (1, 0), respectively. Here X 0 = Gr(n −c, n)×Gr(n − d, n). Here are the analogues of Lemmas 5. 8-5.10. To prove these, one first constructs an open set on H n n−c,n−d analogous to U k−1 as described in Sect. 3. Then we proceed as in Sect. 4 and describe equations for D i and N i along this open set. Lemma 7.7 Let 1 ≤ i ≤ c − 1 and 0 ≤ j ≤ c − 1. We have the following intersection numbers, (i) D i · C i = 1 and D i · C j = 0 for all i = j, (ii) D (1) c · C i = D (2) c · C i = 0 for all i,