The Automorphism Group of Linear Sections

of the Grassmannians 𝔾(1, N)

J. Piontkowski and A. Van de Ven

Received: April 4, 1999 Revised: October 27, 1999

Communicated by Thomas Peternell

Abstract. The Grassmannians of lines in projective N-space, 𝔾(1, N), are embedded by way of the Plücker embedding in the projective space ℙ(⋀^{2} ℂ^{N+1}). Let H^{l} be a general l-codimensional linear subspace in this projective space.

We examine the geometry of the linear sections 𝔾(1, N)∩H^{l} by studying their automorphism groups and list those which are homogeneous or quasihomogeneous.

1991 Mathematics Subject Classification: 14L27, 14M15, 14J50, 14E09 Keywords and Phrases:

0 Introduction

Complete intersections in projective space have been studied extensively from many points of view. A natural generalisation is the study of complete intersections in Grassmannians. The first case that presents itself is the case of intersections with linear spaces. Indeed, there is an extensive literature on the simplest case, the Grassmannian of lines in the 3-space, where intersections are known as linear complexes and congruences of lines. L. Roth has studied the rationality of linear sections of Grassmannians of lines in general. If they are smooth and if the dimension of the intersection is greater than half the dimension of the Grassmannian, then they are rational. R. Donagi determined the cohomology and the intermediate Jacobian of some linear sections of Grassmannians of lines.

In this paper we study the linear sections from the point of automorphism groups. Let 𝔾(1, N) be the Grassmann variety of lines in projective N-space, canonically embedded in ℙ(⋀^{2}ℂ^{N+1}) and let H^{l} be an l-codimensional linear subspace in this space. For general H^{l} we determine the automorphism groups for 𝔾(1, N) ∩ H, 𝔾(1, N) ∩ H^{2}, 𝔾(1, 4) ∩ H^{3}, and 𝔾(1, 5) ∩ H^{3}. In the second case we find for example:

Theorem 3.5 For N = 2n−1 ≥ 5 the automorphism group of 𝔾(1, N)∩H^{2} has SL(2, ℂ)^{n}/{1, −1} as a normal subgroup and the quotient group is isomorphic to the permutation group S(3) for n = 3, to ℤ/2ℤ × ℤ/2ℤ for n = 4, and trivial otherwise.

We believe that apart from trivial cases these are the only general linear sections where automorphism groups of positive dimension appear. Extensive computer checks seem to confirm this.

In particular we prove that the automorphism groups of 𝔾(1, 2n)∩H, 𝔾(1, 4)∩ H^{2}, 𝔾(1, 5)∩H^{2}, 𝔾(1, 6)∩H^{2}, and 𝔾(1, 4)∩H^{3} are quasihomogeneous – those of 𝔾(1, 2n − 1) ∩ H and 𝔾(1, 3) ∩ H^{2} are even homogenous – whereas all others are not.

As to our methods, in our proofs the rich geometry of the Grassmannian plays a decisive role. Otherwise, we mainly use well known tools like multilinear algebra, Lefschetz theorems, vanishing theorems etc.

We are indebted to E. Opdam and A. Pasquale for useful remarks. The first author also thanks the Stieltjes Institut of Leiden University for financial support.

1 Preliminary

The Grassmannian 𝔾(1, N) of lines in ℙ_{N} is embedded by way of the Plücker embedding into ℙ(⋀^{2} ℂ^{N+1})

\begin{matrix}𝔾(1, N) & ⟶ & ℙ(⋀^{2}ℂ^{N+1}) \\ span \{v, w\} & ⟼ & ℙ(v ∧ w). \end{matrix}

We denote by H^{l} an l-codimensional linear subspace of ℙ(⋀^{2} ℂ^{N+1}). Roth [R] examined the geometry of the general linear sections of the Grassmannians and found

Theorem 1.1 For a general H^{l} with 0 ≤ l ≤ 1/2 dim 𝔾(1, N) = N − 1 the intersection with the Grassmannians, 𝔾(1, N) ∩ H^{l}, is rational.

In this article we continue this study by describing the automorphism groups of these sections. As for the notation, given a subvariety Y of a variety X we define Aut(Y, X) to be the automorphisms of X that induce automorphisms of Y, i.e.

Aut(Y, X) = {φ ∈ Aut(X) | φ(Y) ⊆ Y }.

Automorphism Group of 𝔾(1, N) ∩ H^{l} 625

Recall that the automorphism group of the Grassmannian itself is computed in two steps, see e.g. [H, 10.19]. First one shows that all automorphisms are induced by automorphisms of ℙ(⋀^{2} ℂ^{N+1}), i.e.

Aut(𝔾(1, N)) ≅ Aut(𝔾(1, N), ℙ(⋀^{2}ℂ^{N+1})).

Then one proves that for N ≠ 3 the right hand side group is isomorphic to ℙGL(N + 1, ℂ) via

\begin{matrix}ℙGL(N + 1, ℂ) & ⟶ & Aut(𝔾(1, N), ℙ(⋀^{2}ℂ^{N+1})) \\ ℙ(T) & ⟼ & (ℙ(∑ v_{i} ∧ w_{i}) ↦ ℙ(∑ T v_{i} ∧ T w_{i})) . \end{matrix}

For the linear sections of the Grassmannians we follow the same outline. The first step is the following theorem and its corollary; the second step will be done separately for the different cases in the next sections.

Theorem 1.2 For N ≥ 4 and a general linear subspace H^{l} ⊂ ℙ(⋀^{2} ℂ^{N+1}) of codimension l ≤ 2N − 5 the linear section 𝔾(1, N) ∩ H^{l} spans H^{l} and its automorphisms are induced by automorphisms of H^{l}, i.e.

Aut(𝔾(1, N) ∩ H^{l}) = Aut(𝔾(1, N) ∩ H^{l}, H^{l})

Proof. We will abbreviate 𝔾(1, N) by 𝔾. We want to prove that 𝔾 ∩ H^{l} spans H^{l}, i.e. for 𝔾 ∩ H^{l} ⊂ ℙ(⋀^{2} ℂ^{N+1})

h^{0}(𝔾 ∩ H^{l}, O(H)) = dim^{⋀2}ℂ^{N+1} − l.

This is known for l = 0. For l ≥ 1 we take the long exact sequence associated to the restriction sequence tensored by O(H)

\begin{matrix}0 → H^{0}(𝔾 ∩ H^{l−1}, O) = ℂ → H^{0}(𝔾 ∩ H^{l−1}, O(H)) → H^{0}(𝔾 ∩ H^{l}, O(H)) → \\ → H^{1}(𝔾 ∩ H^{l−1}, O) = 0. \end{matrix}

Looking at the dimensions we get

h^{0}(𝔾 ∩ H^{l}, O(H)) = h^{0}(𝔾 ∩ H^{l−1}, O(H)) − 1,

and the claim follows by induction. Now we show that all automorphisms of 𝔾∩H^{l} are induced by automorphisms of H^{l}. This follows if we can show that all divisors of 𝔾 ∩ H^{l} are induced by divisors of H^{l}, i.e.

Pic(𝔾 ∩ H^{l}) = Pic(H^{l}) = ℤ · H,

because then the projective embedding of 𝔾 ∩ H^{l} given by the sections in the line bundle O(H) is equivariant for all automorphisms of 𝔾 ∩ H^{l} To see this, note that by the Lefschetz hyperplane section theorem

ℤ · H = H^{2}(𝔾, ℤ) = H^{2}(𝔾 ∩ H, ℤ) = … = H^{2}(𝔾 ∩ H^{l}, ℤ)

for 0 ≤ l ≤ 2N − 5. From the exponential sequence

0 → ℤ_{𝔾∩H^{l}} → O_{𝔾∩H^{l}} → O_{𝔾∩H^{l}}^{∗} → 0

we get as a part of the associated long exact sequence

… → H^{1}(𝔾 ∩ H^{l}, O) → H^{1}(𝔾 ∩ H^{l}, O^{∗}) → H^{2}(𝔾 ∩ H^{l}, ℤ) = ℤ · H → 0

and therefore

Pic(𝔾 ∩ H^{l}) = H^{1}(𝔾 ∩ H^{l}, O^{∗}) = ℤ · H

as soon as we know that H^{1}(𝔾 ∩ H^{l}, O) = 0.

This is well known for l = 0. For l ≥ 1 we look at the restriction sequence

0 → O_{𝔾∩H^{l−1}}(−H) → O_{𝔾∩H^{l−1}} → O_{𝔾∩H^{l}} → 0

and take its associated long exact sequence

\begin{matrix}… → H^{1}(𝔾 ∩ H^{l−1}, O(−H)) → H^{1}(𝔾 ∩ H^{l−1}, O) → H^{1}(𝔾 ∩ H^{l}, O) → \\ → H^{2}(𝔾 ∩ H^{l−1}, O(−H)) → … \end{matrix}

The right and left cohomology groups are trivial for l ≤ 2N − 4 by Kodaira’s vanishing theorem, so

0 = H^{1}(𝔾, O) = H^{1}(𝔾 ∩ H, O) = … = H^{1}(𝔾 ∩ H^{l}, O). □

Corollary 1.3 For N ≥ 4 and a general linear subspace H^{l} ⊂ ℙ(⋀^{2} ℂ^{N+1}) of codimension l ≤ N − 2

Aut(𝔾(1, N) ∩ H^{l}) = Aut(𝔾(1, N) ∩ H^{l}, ℙ(⋀^{2}ℂ^{N+1})) ∩ Aut(H^{l}, ℙ(⋀^{2}ℂ^{N+1})).

This is also true for 𝔾(1, 4) ∩ H^{3}.

Proof. The special case of 𝔾(1, 4) ∩ H^{3} will be dealt with in Section 7. In view of the theorem we need only to show that an automorphism of 𝔾(1, N)∩ H^{l} can be extended to an automorphism of 𝔾(1, N), which is always linear and fixes the linear subspace H^{l} because 𝔾(1, N) ∩ H^{l} spans H^{l}. To do so we will study the linear subspaces of 𝔾(1, N) ∩ H^{l}. The linear subspaces of the Grassmannian 𝔾(1, N) are the following Schubert cycles:

1. Let p ∈ ℙ_{N} be a point and K ⊂ ℙ_{N} a linear subspace of dimension k that contains the point p then

{L ∈ 𝔾(1, N) | p ∈ L ⊆ K} ⊂ 𝔾(1, N)

is a (k − 1)-dimensional subspace and these form a

dim ℙ_{N} + dim 𝔾(k − 1, N − 1) = N + k(N − k)

dimensional family F_{k−1}.

2. Let E ⊂ ℙ_{N} be a plane then

{L ∈ 𝔾(1, N) | L ⊆ K} ⊂ 𝔾(1, N)

is a plane in 𝔾(1, N) and these form a

dim 𝔾(1, N) = 3(N − 2) = 3N − 6

dimensional family F′_{2}.

Let d = dim ⋀^{2} ℂ^{N+1} − 1. For the variety of 2-planes F_{2} (and analogously for F′_{2}) we consider the incidence correspondence

{(E, h) ∈ F_{2} × 𝔾(d − l, d) | E ⊂ h} ⊂ F_{2} × 𝔾(d − l, d).

The fibre above any point h ∈ 𝔾(d − l, d) is exactly the set of 2-planes of F_{2} on 𝔾(1, N) ∩ h, which we denote by F̃_{2}. From this correspondence we see immediately that for general H^{l} either F̃_{2} is empty or F̃_{2} consists of some components of the same dimension (Stein factorisation). The same argument applied to F′_{2} yields F̃′_{2}. Since for N ≥ 4 dim F_{2} ≠ dim F′_{2} we thus find that if F̃_{2} and F̃′_{2} are not both empty, then dim F̃_{2} ≠ dim F̃′_{2}.

Now any automorphism of 𝔾(1, N) ∩ H^{l} is linear, so we conclude that any automorphism of 𝔾(1, N)∩H^{l} transforms F̃_{2} resp. F̃′_{2} onto itself. For dimension ≠ 2 there are only linear spaces of type 1, so we can state:

Any automorphism φ of 𝔾(1, N) ∩ H^{l} permutes the linear spaces of type 1.

Next, let p ∈ ℙ_{N} and L_{p} the (N − 1)-dimensional linear subspace of 𝔾(1, n) consisting of the lines through p. From l ≤ N − 2 we see e = dim(L_{p} ∩ H^{l}) ≥ 1. Since H^{l} is general, for almost all points of ℙ_{N} this dimension is exactly e. Now, and this is the crucial remark, φ(L_{p}∩H^{l}) being of type 1, is contained in exactly one L_{q}, q ∈ ℙ_{N}. Attaching q to p we obtain for almost all p ∈ ℙ_{N} a map to ℙ_{N}. If the dimension of L_{p_{0}} ∩ H^{l} is bigger than the minimum, then this map can still be defined at p_{0} in exactly the same way. We claim that this is continuous at p_{0}. This can be seen by considering a general sequence of points on ℙ_{N}, say p_{1}, p_{2}, …, converging to p_{0} with dim(L_{p_{i}} ∩H^{l}) minimal and applying the crucial remark twice. The map from ℙ_{N} to ℙ_{N} thus obtained has an inverse and is holomorphic, so it is a linear automorphism of ℙ_{N}. This automorphism induces an automorphism of 𝔾(1, N) for which it is easily verified that it coincides with φ on 𝔾(1, N) ∩ H^{l}. □

It is tempting to assume that the groups Aut(𝔾(1, N), ℙ(⋀^{2} ℂ^{N+1})) and Aut(H^{l}, ℙ(⋀^{2} ℂ^{N+1})) in Aut(ℙ(⋀^{2} ℂ^{N+1})) intersect transversally. Then the dimension of Aut(𝔾(1, N) ∩ H^{l}) could be computed as

dim Aut(𝔾(1, N) ∩ H^{l}) = dim Aut(𝔾(1, N)) − codim Aut(H^{l}, ℙ(⋀^{2} ℂ^{N+1}))
= (N + 1)^{2} − 1 − l (\binom{N+1}{2} − l) .

And we would find the following non-finite groups:

dim Aut(𝔾(1, N) ∩ H) = (N^{2} + 3N + 2)/2
dim Aut(𝔾(1, N) ∩ H^{2}) = N + 4
dim Aut(𝔾(1, 4) ∩ H^{3}) = 3.

Unfortunately, the intersection is not always transversal. Our computation of the automorphism groups will show the following dimensions for N ≥ 4:

dim Aut(𝔾(1, N) ∩ H) = (N^{2} + 3N + 2)/2
dim Aut(𝔾(1, N) ∩ H^{2}) = \{ \begin{matrix}N + 4 & for N even \\ 3(N + 1)/2 & for N odd \end{matrix}
dim Aut(𝔾(1, 4) ∩ H^{3}) = 3
dim Aut(𝔾(1, 5) ∩ H^{3}) = 1.

We conjecture that these are the only non-finite groups. For N + 2 ≤ l the canonical bundle K = O(−N − 1 + l) is positive on 𝔾(1, N) ∩ H^{l}, and this conjecture can be proved by Serre’s duality theorem and Kodaira’s vanishing theorem:

dim Aut(𝔾 ∩ H^{l}) = h^{0}(𝔾 ∩ H^{l}, Θ) = h^{2N−2−l}(𝔾 ∩ H^{l}, KΩ^{1}) = 0

A proof for the remaining cases 3 ≤ l ≤ N + 1 seems difficult. For N ≤ 10 and all l we verified the conjecture for the automorphisms induced by automorphisms of the Grassmannian by computer computations.

With this Theorem and its Corollary our task of determining the automorphisms of 𝔾(1, N) ∩ H^{l} has been immensely simplified. All we need to do is to find the projective transformations of Aut(𝔾(1, N), ℙ(⋀^{2} ℂ^{N+1})) = ℙGL(N + 1, ℂ) such that their induced action on ℙ(⋀^{2} ℂ^{N+1})^{∗} preserves H^{l}. To express this in algebraic terms we identify (⋀^{2} ℂ^{N+1})^{∗} with ⋀^{2}(ℂ^{N+1})^{∗}. If a particular basis of ℂ^{N+1} is chosen, ⋀^{2}(ℂ^{N+1})^{∗} as antisymmetric forms on ℂ^{N+1} can also be identified with the antisymmetric matrices of size N + 1. In concrete terms, if (e_{0}, …, e_{N}) is a basis of ℂ^{N+1} and E_{ij} ∈ M(N + 1, ℂ) the matrix, which has a 1 in the position (i, j) but is otherwise zero, then

\begin{matrix}(^{⋀2}ℂ^{N+1})^{∗} & ⟶ & Antisym(N + 1, ℂ) \\ ∑_{i,j} λ_{ij}(e_{i} ∧ e_{j})^{∗} & ⟼ & \frac{1}{2} ∑_{i,j} λ_{ij}(E_{ij} − E_{ji}). \end{matrix}

In these terms a line l = p ∧ q ∈ 𝔾(1, N) is in the hyperplane H ∈ ℙ(⋀^{2} ℂ^{N+1}) iff for a corresponding antisymmetric matrix A ∈ Antisym(N + 1, ℂ) with ℙ(A) = H we have ^{t}pAq = 0.

Further, the action of ℙGL(N + 1, ℂ) on ℙ(⋀^{2} ℂ^{N+1}), which was given for ℙ(T) ∈ ℙGL(N + 1, ℂ) by

\begin{matrix}ℙ (^{⋀2}ℂ^{N+1}) & ⟶ & ℙ (^{⋀2}ℂ^{N+1}) \\ ℙ(∑ v_{i} ∧ w_{i}) & ⟼ & ℙ(∑ T v_{i} ∧ T w_{i}), \end{matrix}

induces the following action on the dual space

ℙ(Antisym(N + 1, ℂ)) ⟶ ℙ(Antisym(N + 1, ℂ))
ℙ(A)  ⟼  ℙ(^{t}T^{−1}AT^{−1}).

Hence an l-codimensional linear subspace H^{l} ⊆ ℙ(⋀^{2} ℂ^{N+1}) which is dually given by ℙ(span {A_{1}, …, A_{l}}) is preserved under T iff every hyperplane containing H^{l} is mapped to another hyperplane containing H^{l}, i.e.

tT^{−1} (∑ λ_{i}A_{i}) T^{−1} ∈ span {A_{1}, …, A_{l}} for all λ_{i} ∈ ℂ
⟺ ^{t}T^{−1}A_{i}T^{−1} ∈ span {A_{1}, …, A_{l}}  for i = 1 … l.

We conclude

Corollary 1.4 For N ≥ 4, 0 ≤ l ≤ N − 2 and a general H^{l} ⊂ ℙ(⋀^{2} ℂ^{N+1}) given by ℙ(span {A_{1}, …, A_{l}}) ⊂ ℙ(Antisym(N +1, ℂ)) the automorphism group of 𝔾(1, N) ∩ H^{l} is

{ℙ(T) ∈ ℙGL(N + 1, ℂ) | ^{t}T^{−1}A_{i}T^{−1} ∈ span {A_{1}, …, A_{l}} ∀i}.

In the following sections we will compute the automorphism groups using this Corollary. In the course of the computations we will use geometric arguments for which it is essential to know if a hyperplane H ⊂ ℙ(⋀^{2} ℂ^{N+1}) is tangent to 𝔾(1, N) or not. We recall the basic facts together with their short proofs.

Proposition 1.5 For any line l_{0} ∈ 𝔾(1, N) the Schubert cycle

σ := {l ∈ 𝔾(1, N) | l ∩ l_{0} ≠ ∅} ⊆ 𝔾(1, N)

lies inside the tangent space 𝕋_{l_{0}}𝔾(1, N) ⊆ ℙ(⋀^{2} ℂ^{N+1}) and spans it.

Proof. Let l ∈ σ, p ∈ l ∩ l_{0}, q ∈ l_{0} \ {p} and r ∈ l \ {p} then

\begin{matrix}ℂ & ⟶ & 𝔾(1, N) \\ λ & ⟼ & p ∧ (q + λr) \end{matrix}

is a line in σ ⊂ 𝔾(1, N) through l_{0} and l. Therefore it is contained in the tangent space 𝕋_{l_{0}}𝔾(1, N), in particular l ∈ 𝕋_{l_{0}}𝔾(1, N). We choose a basis (e_{0}, …, e_{N}) of ℂ^{N+1} such that l_{0} = ℙ(e_{0} ∧ e_{1}). The 2N − 1 points ℙ(e_{0} ∧ e_{1}), ℙ(e_{0} ∧ e_{i}), ℙ(e_{1} ∧ e_{i}) for i = 2 … N lie in σ ⊂ 𝕋_{l_{0}}𝔾(1, N) and are projectively independent, hence they span 𝕋_{l_{0}}𝔾(1, N). □

Corollary 1.6 Let H = ℙ(A) ∈ ℙ(⋀^{2} ℂ^{N+1})^{∗} be a hyperplane and l_{0} ∈ 𝔾(1, N) a line then

𝕋_{l_{0}}𝔾(1, N) ⊆ H ⟺ l_{0} ⊆ ker A.

Proof. By the Proposition 𝕋_{l_{0}}𝔾(1, N) ⊆ H is equivalent to σ ⊆ H. If we use the same basis of ℂ^{N+1} as in the proof of the proposition, this means that

ℙ((λe_{0} + μe_{1}) ∧ v) ∈ H for all (λ: μ) ∈ ℙ_{1}, v ∈ ℂ^{N+1}

⟺ ^{t}(λe_{0} + μe_{1})Av = 0 for all (λ: μ) ∈ ℙ_{1}, v ∈ ℂ^{N+1}

⟺ l_{0} ⊆ ker A. □

Corollary 1.7 The dual variety 𝔾(1, N)^{∗} ⊂ ℙ(⋀^{2} ℂ^{N+1})^{∗} of the Grassmannian variety 𝔾(1, N) consists of matrices of corank ≥ 2 for N odd resp. corank ≥ 3 for N even.

For N odd it is an irreducible hypersurface of degree (N + 1)/2; for N even it is a 3-codimensional subvariety.

Proof. By the last corollary H = ℙ(A) ∈ ℙ(⋀^{2} ℂ^{N+1})^{∗} is tangential to 𝔾(1, N) iff corank A ≥ 2. Recall that an antisymmetric matrix has even rank. So, for N odd the matrix A ∈ Antisym(N + 1, ℂ) has corank ≥ 2 iff det A = 0. But again since A is antisymmetric, det A is the square of the irreducible Pfaffian polynomial PfA [B, 5.2], which therefore defines 𝔾(1, N)^{∗}.

For N even corank A ≥ 2 is equivalent to corank A ≥ 3. We compute the dimension of 𝔾(1, N)^{∗} following Mumford [M] and find

dim ( \begin{matrix}space of A with \\ dim ker A = 3 \end{matrix} ) = dim G(3, N + 1) + dim ⋀^{2} ℂ^{N+1}/ℂ^{3}
= 3(N − 2) + (N − 2)(N − 3)/2
= (N^{2} + N − 6)/2

⟹ codim 𝔾(1, N)^{∗} = (N + 1)N/2 − (N^{2} + N − 6)/2 = 3. □

2 𝔾(1, 2n − 1) ∩ H

Let the hyperplane H ∈ ℙ(⋀^{2} ℂ^{2n}) be given by an element A ∈ (⋀^{2} ℂ^{2n})^{∗}, which we identify with its corresponding antisymmetric matrix. If H is general, A will be a matrix of full rank. This may be taken as the definition of a general H. We will assume from now on that H is general.

The line system 𝔾(1, 2n − 1) ∩ H in ℙ_{2n−1} does not lead to obvious special points in the ℙ_{2n−1}. Through every point p ∈ ℙ_{2n−1} passes a ℙ_{2n−3} of lines, namely

p ∧ q ∈ 𝔾(1, 2n − 1) with q ∈ ker^{t}pA.

For n ≥ 3 we can compute the automorphism group of 𝔾(1, 2n − 1) ∩ H with the help of Theorem 1.2 and its Corollaries. It consists of elements ℙ(T) ∈ ℙGL(2n, ℂ) = Aut(𝔾(1, 2n−1)) such that ℙ(T) as an element of ℙGL(⋀^{2} ℂ^{2n}) preserves H, i.e.

tT^{−1}AT^{−1} = λA for suitable λ ∈ ℂ^{∗}.

We may choose coordinates on ℙ_{2n−1} such that

A = ( \begin{matrix}0 & −E_{n} \\ E_{n} & 0 \end{matrix} ) .

Then by definition

Sp(2n, ℂ) = {T ∈ GL(2n, ℂ) | ^{t}T^{−1}AT^{−1} = A},

and we have an isomorphism

\begin{matrix}\{T ∈ GL(2n, ℂ) | ∃λ_{T} ∈ ℂ^{∗} : ^{t}T^{−1}AT^{−1} = λ_{T} A\}/ℂ^{∗} & ⟶ & Sp(2n, ℂ)/\{1, −1\} \\ ℂ^{∗} · T & ⟼ & ±\frac{1}{\sqrt{λ_{T}}} T. \end{matrix}

Therefore we see

Proposition 2.1 The automorphism group of 𝔾(1, 2n − 1) ∩ H for a general H ⊂ ℙ(⋀^{2} ℂ^{N+1}) is Sp(2n, ℂ)/{1, −1}. Its action on 𝔾(1, 2n − 1) ∩ H is homogeneous.

Proof. The missing case of 𝔾(1, 3) ∩ H can be found in [FH, p. 278]. The transitivity of the action follows from Witt’s theorem [Br, 12.31]. □

3 𝔾(1, 2n − 1) ∩ H^{2}

A 2-codimensional linear subspace L = H^{2} of ℙ(⋀^{2} ℂ^{2n}) can be thought of as the pencil of hyperplanes containing it. So it gives a line L^{∗} = ℙ(λA − μB) ⊂ ℙ(⋀^{2} ℂ^{2n})^{∗}. We identify again (⋀^{2} ℂ^{2n})^{∗} with the antisymmetric matrices of size 2n. The line L^{∗} intersects the dual Grassmannian 𝔾(1, 2n − 1)^{∗}, which consists of antisymmetric matrices of rank ≤ 2n − 2 and is a hypersurface of degree n by Corollary 1.7, in at most n points. For the moment a line L^{∗}, and hence L, will be called general if it has n points of intersection, H_{i} = ℙ(λ_{i}A − μ_{i}B) ∈ L^{∗}, i = 1 … n, with the dual Grassmannian. These hyperplanes H_{i} are tangent to the Grassmannian 𝔾(1, 2n − 1) at the points l_{i} := ker(λ_{i}A − μ_{i}B) ∈ 𝔾(1, 2n − 1) by Corollary 1.6. Therefore we get n exceptional lines l_{1}, …, l_{n} in ℙ_{2n−1}.

The intersection of the Grassmannian with its tangent hyperplane H_{i} contains all lines that intersect l_{i}, because these lines are already contained in the intersection 𝔾(1, 2n − 1) ∩ 𝕋_{l_{i}}𝔾(1, 2n − 1) by Proposition 1.5.

So, any line through a point p ∈ l_{i} will be in the subspace L ⊂ ℙ(⋀^{2} ℂ^{2n}) as soon as it is contained in any other hyperplane H ∈ L^{∗} \ {H_{i}}. This gives one linear restriction to lines through p, so that there is at least a ℙ_{2n−3} of lines through the points of the lines l_{i}. In contrast, through a general point of ℙ_{2n−1} \ ⋃ l_{i} there is only a ℙ_{2n−4} of lines. In fact, we have

Proposition 3.1 The points of the lines l_{1}, …, l_{n} are characterized by the property that through each of them passes a ℙ_{2n−3} of lines, i.e.

\{p ∈ ℙ_{2n−1} \begin{matrix}through p passes a ℙ_{2n−3} \\ of lines of 𝔾(1, 2n − 1) ∩ L \end{matrix} \} = ⋃ l_{i}.

Furthermore, the lines l_{1}, …, l_{n} span the whole ℙ_{2n−1}.

This may easily be seen if we write the pencil of hyperplanes L^{∗} in its normal form.

Proposition 3.2 (Donagi[D]) Given a pencil of hyperplanes L^{∗} = ℙ(λA − μB) ⊂ ℙ(⋀^{2} ℂ^{2n})^{∗} such that the line L^{∗} intersects the Pfaffian hypersurface in n different points. Then there is a basis of ℂ^{2n} such that

A = ( \begin{matrix}J & & & & 0 \\ & . & . & . \\ 0 & & & & J \end{matrix} ) and B = ( \begin{matrix}λ_{1}J & & & & 0 \\ & . & . & . \\ 0 & & & & λ_{n}J \end{matrix} ) with J = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ).

The points (λ_{1} : 1), …, (λ_{n}: 1) ∈ ℙ_{1} ≅ L^{∗} are unique up to a projective transformation of ℙ_{1}.

Proof of Proposition 3.1. The hyperplane H_{i} = ℙ(λ_{i}A − μ_{i}B) has, written as an antisymmetric matrix, the kernel l_{i} = span {e_{2i−1}, e_{2i}} which means it is tangent to 𝔾(1, 2n − 1) at l_{i}. All lines of 𝔾(1, 2n − 1) ∩ L through the point p are given by p ∧ q with ^{t}pAq = ^{t}pBq = 0. In order to have a ℙ_{2n−2} of lines through p, the linear forms ^{t}pA and ^{t}pB must be linear dependent, i.e. there are λ, μ ∈ ℂ with

0 = λ^{t}pA − μ^{t}pB = ^{t}p(λA − μB).

Therefore p is in the kernel of a matrix of the pencil, but these kernels are the lines l_{1}, …, l_{n}, so p is contained in one of them. □

Knowing the exceptional lines l_{1}, …, l_{n}, one can immediately give some lines which are in the line system.

Proposition 3.3 Any line in ℙ_{2n−1} which intersects two exceptional lines is an element of the line system 𝔾(1, 2n − 1) ∩ L.

The exceptional lines themselves are not in the line system.

Proof. If a line l intersects l_{i} and l_{j}, it lies – as a point of the Grassmannian 𝔾(1, 2n − 1) – in H_{i} and H_{j}, hence in L = H_{i} ∩ H_{j}.

Assume that the exceptional line l_{1} is an element of 𝔾(1, 2n−1). By Proposition 3.1 the lines through a point p ∈ l_{1} sweep out a hyperplane. This hyperplane contains the line l_{1} by assumption and the other exceptional lines l_{2}, …, l_{n} by the first part of this proposition. But this contradicts the second statement of Proposition 3.1. □

Remark 3.4 From Proposition 3.2 we also see that any position of the n points of the line L^{∗} is possible. In particular, we may call a line general if the position of the points is general in the sense needed below.

Using this geometric description we can determine the automorphisms of 𝔾(1, 2n − 1) ∩ L. For the moment we restrict ourselves to n ≥ 3 in order to be able to use Theorem 1.2. By this Theorem and its Corollaries we can view an automorphism of 𝔾(1, 2n − 1) ∩ L as an element ℙ(T) of ℙGL(2n, ℂ). To make the notation simpler, we will write only T for ℙ(T) if no confusion can result. Since the points of the exceptional lines are characterized by the property of Proposition 3.1, T must map the union of the lines l_{i} ⊂ ℙ_{2n−1} onto itself. Permutations of the lines may occur, but – as we will presently see – not all permutations are possible.

If we view the automorphism T as an element of Aut(L, ℙ(^{⋀2}ℂ^{2n})), it interchanges the hyperplanes containing L, i.e. it induces a projective transformation of the line L^{∗} ⊂ ℙ(⋀^{2} ℂ^{2n}). Naturally, the transformation of L^{∗} must preserve the union of points of intersection of L^{∗} with the dual Grassmannian, which determine the lines l_{i}. Now, if a transformation of ℙ_{2n−1} permutes the lines l_{i}, then the induced transformation of L^{∗} must permute the corresponding points of L^{∗} in the same way.

Since not every permutation of four or more points on a line can be induced by a projective transformation, not all permutations are possible. In fact, if the points are in general position, we get the following subgroups of the permutation groups:

n  subgroup of S(n)
3  S(3)
4 {(1 2 3 4), (2 1 4 3), (3 4 1 2), (4 3 2 1)} ≅ ℤ/2ℤ × ℤ/2ℤ
≥ 5  {id}

On the other hand, any permutation σ ∈ S(n) of the points on L^{∗} that is induced by a projective transformation φ of L^{∗} can be induced by an automorphism of 𝔾(1, 2n − 1) ∩ L. To see this, let us write L^{∗} in its normal form and define T ∈ GL(2n, ℂ) as

T (e_{2i}) := e_{2σ(i)} and T (e_{2i−1}) := e_{2σ(i)−1}.

This transformation permutes the lines in the prescribed way, and as an automorphism of ℙ(⋀^{2} ℂ^{2n}) it fixes L since the transformed line L^{∗} is

\begin{matrix}^{t}T^{−1} (λ ( \begin{matrix}J & & & & 0 \\ & . & . & . \\ 0 & & & & J \end{matrix} ) − μ ( \begin{matrix}λ_{1}J & & & & 0 \\ & . & . & . \\ 0 & & & & λ_{n}J \end{matrix} )) T^{−1} \\ = λ ( \begin{matrix}J & & & & 0 \\ & . & . & . \\ 0 & & & & J \end{matrix} ) − μ ( \begin{matrix}λ_{σ^{−1}(1)}J & & & & 0 \\ & . & . & . \\ 0 & & & & λ_{σ^{−1}(n)}J \end{matrix} ) . \end{matrix}

Changing the parametrisation of the line by φ we get back the old parametrisation of the line L^{∗} by the definition of φ. So this T is an automorphism of 𝔾(1, 2n − 1) ∩ L that induces the permutation of lines we started with.

Now we can restrict our attention to transformations that do not permute the lines since we can obtain every permutation by composing with one of the transformations from above. A transformation leaving all the lines individually fixed has the form

T = ( \begin{matrix}t_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & t_{n} \end{matrix} ) with t_{1}, …, t_{n} ∈ GL(2, ℂ).

This T will fix the line system 𝔾(1, 2n − 1) ∩ L in ℙ_{2n−1} iff it preserves L^{∗}, i.e. for all λ, μ ∈ ℂ there exists α, β ∈ ℂ such that

tT^{−1}(λA − μB)T^{−1} = αA + βB.

It is sufficient to check this for (λ, μ) = (1, 0) and (0, −1). Since

tT^{−1}AT^{−1} = ( \begin{matrix}det t_{1}^{−1}J & & & & 0 \\ & ^{.} & ^{.} & . \\ 0 & & & & det t_{n}^{−1}J \end{matrix} )

tT^{−1}BT^{−1} = ( \begin{matrix}λ_{1} det t_{1}^{−1}J & & & & 0 \\ & ^{.} & ^{.} & . \\ 0 & & & & λ_{n} det t_{n}^{−1}J \end{matrix} )

this is equivalent to the question if there exist α, β, γ, δ ∈ ℂ with

(det t_{1}^{−1}, …, det t_{n}^{−1}) = α(1, …, 1) + β(λ_{1}, …, λ_{n})
(λ_{1} det t_{1}^{−1}, …, λ_{n} det t_{n}^{−1}) = γ(1, …, 1) + δ(λ_{1}, …, λ_{n}).

It follows

−γ(1, …, 1) + (α − δ)(λ_{1}, …, λ_{n}) + β(λ_{1}^{2}, …, λ_{n}^{2}) = 0
⟹ α = δ, β = γ = 0
⟹ det t_{1} = … = det t_{n}.

We normalize by det t_{1} = 1, i.e. t_{1}, …, t_{n} ∈ SL(2, ℂ). Then only T and −T ∈ GL(2n, ℂ) give the same element in ℙGL(2n, ℂ). So that as a group the automorphisms of 𝔾(1, 2n − 1) ∩ L that do not permute the exceptional lines are isomorphic to SL(2, ℂ)^{n}/{1, −1}.

Altogether we get

Theorem 3.5 For n ≥ 3 the automorphism group of the intersection of 𝔾(1, 2n − 1) with a general 2-codimensional linear subspace of ℙ(⋀^{2} ℂ^{2n}) has SL(2, ℂ)^{n}/{1, −1} as a normal subgroup and the quotient group is isomorphic to the permutation group S(3) for n = 3, to ℤ/2ℤ × ℤ/2ℤ for n = 4, and trivial otherwise.

The automorphism group is isomorphic to the subgroup of ℙGL(2n, ℂ) that consists of the elements

P_{σ} · ( \begin{matrix}t_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & t_{n} \end{matrix} ) with t_{1}, …, t_{n} ∈ SL(2, ℂ)

where P_{σ} is the identity for n ≥ 5 and otherwise defined by

\begin{aligned}P_{σ}(e_{2i}) & = & e_{2σ(i)} \\ P_{σ}(e_{2i−1}) & = & e_{2σ(i)−1} \end{aligned}

for σ ∈ \{ \begin{matrix}S(n) & if n = 3 \\ \{(1 2 3 4), (2 1 4 3), (3 4 1 2), (4 3 2 1)\} & if n = 4. \end{matrix}

For the sake of completeness we recall the classical case of 𝔾(1, 3) ∩ H^{2}.

Remark 3.6 The automorphism group of 𝔾(1, 3)∩H^{2} is an extension of ℤ/2ℤ by ℙGL(2, ℂ) × ℙGL(2, ℂ). It acts homogeneously on 𝔾(1, 3) ∩ H^{2}.

Proof. The Grassmannian 𝔾(1, 3) is a smooth quadric in ℙ(⋀^{2} ℂ^{4}) ≅ ℙ_{5}. Therefore 𝔾(1, 3)∩H^{2} is a smooth quadric in ℙ_{3}. Hence it is isomorphic to the Segre variety ℙ_{1}×ℙ_{1} in ℙ_{3}. The automorphism group of ℙ_{1}×ℙ_{1} is generated by ℙGL(2, ℂ)×ℙGL(2, ℂ) together with the automorphism that exchanges the ℙ_{1}s. All the automorphisms extend to ℙ_{3}. Obviously, the group acts transitively on ℙ_{1} × ℙ_{1}. □

For the rest of this section we consider the question if the action of the other automorphism groups is quasihomogeous on the corresponding line system, i.e. if there is an open orbit.

This cannot be the case for n ≥ 7 since then the dimension of the line system 𝔾(1, 2n − 1) ∩ H^{2}, 2(2n − 2) − 2 = 4n − 6, is larger than the dimension of the automorphism group, 3n.

For n = 3 the action is quasihomogeneous. To see that one can adjust the (λ_{1}, λ_{2}, λ_{3}) in the normal form of the line system to (1, 0, −1) by a projective transformation and compute the stabiliser of the line (1 : 0 : 1 : 0 : 1 : 0) ∧ (1 : 1 : 1 : −2 : 1 : 1) by hand or computer and see that it is 3-dimensional. So the dimension of its orbit is 3 · 3 − 3 = 6, which is just the dimension of the line system.

For n = 4, 5, 6 the group does not act quasihomogenously anymore. For this one computes again the dimension of the stabiliser of a general line. Since the group acts transitively on ℙ_{2n−1} \ ⋃ L_{i}, we may restrict our attention to lines through one of those points, e.g. (1 : 0 : …: 1 : 0). Using a computer one sees that the stabilizer of a general line through this point has again dimension 3. Hence the orbit has dimension 3n − 3, which is less then the dimension of the line system, 4n − 6.

4 𝔾(1, 5) ∩ H^{3}

Let L = H^{3} ⊂ ℙ(⋀^{2} ℂ^{6}) be a general 3-codimensional subspace. With our usual identification of (⋀^{2} ℂ^{6})^{∗} with the antisymmetric matrices Antisym(6, ℂ) its dual plane L^{∗} = ℙ(λA + μB + vC) ⊂ ℙ(⋀^{2} ℂ^{6})^{∗} intersects the dual Grassmannian 𝔾(1, 5)^{∗}, which consists of matrices of rank ≤ 4 and is a hypersurface of degree 3 by Corollary 1.7, in an irreducible cubic C^{∗}. By Corollary 1.6 a point (λ: μ: ν) ∈ C^{∗} corresponds to the hyperplane h_{(λ:μ:ν)} = ℙ(λA+μB+νC) that is tangent to the Grassmannian at the point

l_{(λ:μ:ν)} := ker(λA + μB + νC) ⊂ ℙ_{5}.

In analogy to the former case we have

Lemma 4.1

\{p ∈ ℙ_{5} | \begin{matrix}through p passes a ℙ_{2} \\ of lines of 𝔾(1, 5) ∩ L \end{matrix} \} = ⋃_{(λ:μ:ν)∈C^{∗}} l_{(λ:μ:ν)} ⊂ ℙ_{5}

Proof. Since by definition the lines in 𝔾(1, 5) ∩ L that contain p are p ∧ q with tpAq = ^{t}pBq = ^{t}pCq = 0, we see that

through p passes at least a ℙ_{2} of lines of 𝔾(1, 5) ∩ L ⟺ ^{t}pA,^{t}pB,^{t}pC are linear dependent ⟺ ∃(λ: μ: ν) ∈ ℙ_{2} with ^{t}p(λA + μB + νC) = 0 ⟺ p ∈ ker(λA + μB + νC) = l_{(λ:μ:ν)}.

We also note that there cannot be a ℙ_{3} of lines of 𝔾(1, 5) ∩ L through a point p. Because if there were one, then dim span {^{t}pA,^{t}pB,^{t}pC} = 1, i.e. there exist two points (λ: μ: ν), (λ′: μ′: ν′) ∈ ℙ_{2} with

tp(λA + μB + νC) = ^{t}p(λ′A + μ′B + ν′C) = 0.

It follows that all the matrices

(αλ + βλ′)A + (αμ + βμ′)B + (αν + βν′)C for all (α: β) ∈ ℙ_{1}

have a non-trivial kernel. Hence the line (αλ + βλ′: αμ + βμ′: αν + βν′) must lie in L^{∗} ∩ 𝔾(1, 5)^{∗} = C^{∗}. But this is a contradiction since the cubic C^{∗} is irreducible. □

Proposition 4.2 The lines l_{(λ:μ:ν)} ⊂ ℙ_{5} with (λ: μ: ν) ∈ C^{∗} do not intersect each other.

Proof. Assume that the line l_{(λ:μ:ν)} intersects the line l_{(λ′:μ′:ν′)} in the point p, i.e.

p ∈ ker(λA + μB + νC) ∩ ker(λ′A + μ′B + ν′C) ≠ 0.

Then

p ∈ ker((αλ + βλ′)A + (αμ + βμ′)B + (αν + βν′)C) ≠ 0 for all (α: β) ∈ ℙ_{1},

and the line (αλ+βλ′: αμ+βμ′: αν +βν′) must be contained in the irreducible cubic C^{∗}, which is a contradiction. □

Let us again derive a normal form:

Proposition 4.3 For a general plane L^{∗} = ℙ(λA + μB + νC) ⊂ ℙ(⋀^{2} ℂ^{6})^{∗} there exists a choice of bases of L^{∗} and ℂ^{6} such that

A = ( \begin{matrix}0 & −1 & & & 0 \\ 1 & 0 \\ & & 0 & −1 \\ & & 1 & 0 \\ & & & & 0 & 0 \\ & 0 & & & 0 & 0 \end{matrix} ) B = ( \begin{matrix}0 & 0 & & & & 0 \\ 0 & 0 \\ & & 0 & −1 \\ & & 1 & 0 \\ & & & & 0 & −1 \\ & 0 & & & 1 & 0 \end{matrix} )

C = ( \begin{matrix}0 & 0 & −α & 0 & −γ & 0 \\ 0 & 0 & 0 & −α & −δ & −γ \\ α & 0 & 0 & −1 & −β & 0 \\ 0 & α & 1 & 0 & 0 & −β \\ γ & δ & β & 0 & 0 & 0 \\ 0 & γ & 0 & β & 0 & 0 \end{matrix} ) .

Documenta Mathematica 4 (1999) 623–664

Remark 4.4 It is also possible to derive a more symmetric normal form where all three matrices look like C only with the {0 −1 \atop 1 0} block moved along the diagonal, but this is not more useful for our computations.

Proof of proposition 4.3. We may assume that the line ℙ(λA + μB) ⊂ L^{∗} is a general line. By Proposition 3.2 there exists a choice of coordinates (corresponding to λ_{1} = 0, λ_{2} = 1, λ_{3} = ∞) such that A and B are of the required form. Further, if we change the coordinates of ℂ^{6} by transformations of the type

T = ( \begin{matrix}t_{1} & 0 & 0 \\ 0 & t_{2} & 0 \\ 0 & 0 & t_{3} \end{matrix} ) t_{1}, t_{2}, t_{3} ∈ SL(2, ℂ),

then A and B will stay the same by Theorem 3.5.

We write the matrix C as

C = ( \begin{matrix}c_{1}J & −^{t}C_{21} & −^{t}C_{31} \\ C_{21} & c_{2}J & −^{t}C_{32} \\ C_{31} & C_{32} & c_{3}J \end{matrix} ) with \begin{matrix}J = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ); c_{1}, c_{2}, c_{3} ∈ ℂ \\ C_{21}, C_{31}, C_{32} ∈ M(2, ℂ). \end{matrix}

We may assume that c_{1} = c_{3} = 0, c_{2} = 1, otherwise we replace C by the matrix 1/(c_{2} − c_{1} − c_{3})(C − c_{1}A − c_{3}B). This is possible since c_{2} − c_{1} − c_{3} ≠ 0, because C is general. So C looks like

C = ( \begin{matrix}0 & −^{t}C_{21} & −^{t}C_{31} \\ C_{21} & J & −^{t}C_{32} \\ C_{31} & C_{32} & 0 \end{matrix} ) .

The generality of C ensures that the matrices C_{21} and C_{32} are invertible, so

T = ( \begin{matrix}\frac{1}{α}C_{21} & 0 & 0 \\ 0 & E_{2} & 0 \\ 0 & 0 & \frac{1}{β}^{t}C_{32} \end{matrix} ) with \begin{matrix}α = \sqrt{det C_{21}} \\ β = \sqrt{det C_{32}} \end{matrix}

is of the above mentioned type and transforms C into

C′ := ^{t}T^{−1}CT^{−1} = ( \begin{matrix}0 & −αE_{2} & −^{t}\overline{C} \\ αE_{2} & J & −βE_{2} \\ \overline{C} & βE_{2} & 0 \end{matrix} ) with \overline{C} := αβC_{32}^{−1}C_{31}C_{21}^{−1}.

This matrix will be transformed under

T = ( \begin{matrix}t^{−1} & 0 & 0 \\ 0 & ^{t}t & 0 \\ 0 & 0 & t^{−1} \end{matrix} ) with t ∈ SL(2, ℂ)

into

tT^{−1}C′T^{−1} = ( \begin{matrix}0 & −αE_{2} & −^{t}t^{t}\overline{C}t \\ αE_{2} & J & −βE_{2} \\ ^{t}t\overline{C}t & βE_{2} & 0 \end{matrix} ) .

So, all that remains to show is: Given a general matrix \overline{C} ∈ M(2, ℂ) there is a matrix t ∈ SL(2, ℂ) such that

tt\overline{C}t = ( \begin{matrix}γ & δ \\ 0 & γ \end{matrix} ) .

If

\overline{C} = ( \begin{matrix}c_{11} & c_{12} \\ c_{21} & c_{22} \end{matrix} ) and t = ( \begin{matrix}1 & −\frac{c_{21}}{c_{11}} \\ 0 & 1 \end{matrix} )

then

\overline{C}′ = ^{t}t\overline{C}t = ( \begin{matrix}c_{11} & c_{12} − c_{21} \\ 0 & \frac{det \overline{C}}{c_{11}} \end{matrix} )

and an additional transformation by

t = ( \begin{matrix}\begin{matrix}^{4} & \sqrt{det \overline{C}} \\ √ & \overline{c_{11}} \end{matrix} & 0 \\ 0 & \begin{matrix}√ & \overline{c_{11}} \\ ^{4} & \sqrt{det \overline{C}} \end{matrix} \end{matrix} )

takes \overline{C} into the desired form

tt\overline{C}′t = ( \begin{matrix}\sqrt{det \overline{C}} & c_{12} − c_{21} \\ 0 & \sqrt{det \overline{C}} \end{matrix} ) . □

Remark 4.5 In terms of this coordinates the cubic C^{∗} ⊂ L^{∗} is given as

λ^{2}μ + μ^{2}λ + λμν − (γ^{2} + β^{2})λν^{2} − (α^{2} + γ^{2})μν^{2} + (αβδ − γ^{2})ν^{3}.

One checks that the cubic is smooth for general α, β, γ, δ.

Now we start to determine the automorphism group of 𝔾(1, 5) ∩ L. A given automorphism

φ ∈ Aut(𝔾(1, 5) ∩ L) = Aut(L, ℙ(^{⋀2}ℂ^{6})) ∩ Aut(𝔾(1, 5), ℙ(^{⋀2}ℂ^{6}))

induces a dual automorphism φ^{∗} on the dual projective space ℙ(⋀^{2} ℂ^{6}))^{∗} that preserves L^{∗} and the dual Grassmannian 𝔾(1, 5)^{∗}, i.e.

φ^{∗} ∈ Aut(L^{∗}, ℙ(^{⋀2}ℂ^{6})^{∗}) ∩ Aut(𝔾(1, 5)^{∗}, ℙ(^{⋀2}ℂ^{6})^{∗}).

In particular, φ^{∗} induces a projective transformation of L^{∗} preserving C^{∗}. But a smooth cubic has only finitely many automorphisms that are induced by a projective linear transformation [BK, 7.3].

To find all automorphisms of 𝔾(1, 5) ∩ L that induce the identity on L^{∗}, we look for the T ∈ ℙGL(6, ℂ) such that

tT^{−1}(λA + μB + νC)T^{−1} ∈ ℂ · (λA + μB + νC) for all λ, μ, ν ∈ ℂ.

It suffices to check this for (λ, μ, ν) = (1, 0, 0), (0, 1, 0), and (0, 0, 1). If we normalize the representation of T in GL(6, ℂ) by det T = 1, we know from the previous section that ^{t}T^{−1}AT^{−1} = ℂ · A and ^{t}T^{−1}BT = ℂ · B is equivalent to

T = ( \begin{matrix}t_{1} & 0 & 0 \\ 0 & t_{2} & 0 \\ 0 & 0 & t_{3} \end{matrix} ) with t_{1}, t_{2}, t_{3} ∈ SL(2, ℂ).

Furthermore, we compute

tT^{−1}CT^{−1} = ( \begin{matrix}0 & −α^{t}t_{1}^{−1}t_{2}^{−1} & −^{t}t_{1}^{−1}\binom{γ 0}{δ γ}t_{3}^{−1} \\ α^{t}t_{2}^{−1}t_{1}^{−1} & 0 & −β^{t}t_{2}^{−1}t_{3}^{−1} \\ ^{t}t_{3}^{−1}\binom{γ δ}{0 γ}t_{1}^{−1} & β^{t}t_{3}^{−1}t_{2}^{−1} & 0 \end{matrix} ) ,

so that ^{t}T^{−1}CT^{−1} = ϑ · C iff t_{1} = \frac{1}{ϑ}^{t}t_{2}^{−1} = t_{3} =: t and

tt^{−1} ( \begin{matrix}γ & δ \\ 0 & γ \end{matrix} ) t^{−1} = ϑ ( \begin{matrix}γ & δ \\ 0 & γ \end{matrix} ) .

Because of det t_{1} = det t_{2} = 1, ϑ must be either 1 or -1. Setting

t = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ⟹ t^{−1} = ( \begin{matrix}d & −b \\ −c & a \end{matrix} )

the last condition together with det t = 1 requires that the following polynomials vanish:

\begin{matrix}(d^{2} + c^{2} − ϑ)γ − dcδ, (db + ac)γ − (ϑ − ad)δ \\ (db + ac)γ − bcδ, (b^{2} + a^{2} − ϑ)γ − baδ, ad − bc − 1 \end{matrix}

The Gröbner basis of the ideal generated by these polynomials with respect to the lexicographical order γ > δ > a > b > c > d can be computed for ϑ = 1 as

γa + δc − γd, b + c, ad + c^{2} − 1,

so that

t = ( \begin{matrix}a & \frac{γ}{δ} (a − d) \\ \frac{γ}{δ} (d − a) & d \end{matrix} ) with det t = 1.

For ϑ = −1 we get as the Gröbner basis

δ, a + d, −c + b, d^{2} + c^{2} + 1.

Since in the general case δ ≠ 0, this gives no further automorphisms.

The one-dimensional subgroup of ℙGL(2, ℂ) consisting of elements like t above acts on ℙ_{1} with the two fixed points (−δ ± \sqrt{δ^{2} − 4γ^{2}} : 2γ). Hence it is conjugate to the one-dimensional subgroup of ℙGL(2, ℂ) that acts on ℙ_{1} with the fixed points 0 and ∞. Now this subgroup consists of the invertible diagonal matrices of ℙGL(2, ℂ), so it is isomorphic to ℂ^{∗}. Therefore we have shown

Theorem 4.6 The component of the automorphism group of 𝔾(1, 5)∩H^{3} containing the identity is isomorphic to ℂ^{∗}. The quotient of Aut(𝔾(1, 5) ∩ H^{3}) by this component is a subgroup of the finite group of projective automorphisms of a smooth cubic in ℙ_{2}.

5 𝔾(1, 2n) ∩ H

The hyperplane H is given by an element A ∈ (⋀^{2} ℂ^{2n+1})^{∗} which can be thought of as an antisymmetric matrix of size 2n + 1. Since antisymmetric matrices have an even rank, the general H corresponds to an A of rank 2n. The one dimensional kernel of A as a point of ℙ_{2n} is called the center c of H.

The center plays a special role in the geometry of the line system 𝔾(1, 2n) ∩ H in ℙ_{2n}.

Proposition 5.1 Every line through the center of the line system 𝔾(1, 2n)∩H is in the line system. The center is the only point with this property.

Moreover, if the line l ∌ c belongs to the line system, so does every line in the plane spanned by the line l and the center c.

Proof. The line c ∧ p through the center will be in the line system if ^{t}cAp = 0. But c is the kernel of A, so this is true. On the other hand, if \overline{c} is a point such that every line through it belongs to the line system, then ^{t}\overline{c}Ap = 0 for all p ∈ ℙ_{2n}. Hence \overline{c} must be in the kernel of A, and therefore \overline{c} = c.

Let the line l = p ∧ q be in the line system. All the lines in the plane spanned by l and c – except the lines through c itself – can be written as

(αp + βc) ∧ (λq + μc) for (α : β), (λ : μ) ∈ ℙ_{1}.

These will be in the line system since

\begin{matrix}(α^{t}p + β^{t}c)A(λq + μc) = & αλ^{t}pAq + αμ^{t}pAc + βλ^{t}cAq + βμ^{t}cAc \\ = & αλ^{t}pAq = 0. \end{matrix} □

Let us for a moment look at the projection ℙ(ℂ^{2n+1}/c) of ℙ_{2n} from the center c. This projection maps all lines in a plane through the center – except the lines through c itself – to only one line. Hence we get a codimension one line system inside ℙ_{2n−1}. In fact, it is of the form 𝔾(1, 2n − 1) ∩ \overline{H}, which is most easily seen in coordinates. We choose a basis (e_{0}, …, e_{2n}) of ℂ^{2n+1} such that the hyperplane H is given by the matrix

A = ( \begin{matrix}\begin{matrix}0 & −E_{n} \\ E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} ) ∈ Antisym(2n + 1, ℂ).

The center of H is c = ℙ(e_{2n}). So the projected line system is 𝔾(1, 2n−1)∩\overline{H}, where \overline{H} is given by the matrix A with the last row and column deleted.

This description helps to determine the automorphism group of 𝔾(1, 2n) ∩ H.

First of all, any of the automorphism must – as a transformation T ∈ ℙGL(2n+ 1, ℂ) – preserve the center, i.e. T c = c. Therefore it induces a transformation \overline{T} of the projected space ℙ(ℂ^{2n+1}/c). This induced transformation \overline{T} has to preserve the projected line system 𝔾(1, 2n − 1) ∩ \overline{H}. Since this case has been treated in Section 2, we know that if we normalize \overline{T} by det \overline{T} = 1, then \overline{T} ∈ Sp(2n, ℂ). Therefore T must have been of the form

T = ( \begin{matrix}\overline{T} & 0 ⋮ 0 \\ a_{0} ⋯ a_{2n−1} & b \end{matrix} ) with \begin{matrix}\overline{T} ∈ Sp(2n, ℂ) \\ a_{i} ∈ ℂ \\ b ∈ ℂ^{∗}. \end{matrix}

One immediately checks that ^{t}T^{−1}AT = A, so that the automorphism group as a subset of ℙGL(2n + 1, ℂ) consists of all elements of the above type. Since we normalized \overline{T}, we have up to multiplication by −1 an unique representative in the class of ℙGL(2n + 1, ℂ).

A small computation shows that

N := \{( \begin{matrix}E_{2n} & 0 ⋮ 0 \\ a_{0} ⋯ a_{2n−1} & 1 \end{matrix} ) | a_{i} ∈ ℂ\} ⊂ Aut(𝔾(1, 2n) ∩ H)

is a normal subgroup which is isomorphic to (ℂ^{2n}, +).

Collecting everything together we have

Proposition 5.2 The automorphism group of 𝔾(1, 2n) ∩ H for a general hyperplane H ⊂ ℙ(⋀^{2} ℂ^{2n+1}) is an extention of Sp(2n, ℂ) × ℂ^{∗}/{1, −1} by (ℂ^{2n}, +) and is isomorphic to the group

\{( \begin{matrix}T & 0 ⋮ 0 \\ a_{0} ⋯ a_{2n−1} & b \end{matrix} ) \begin{matrix}T ∈ Sp(2n, ℂ) \\ a_{i} ∈ ℂ \\ b ∈ ℂ^{∗} \end{matrix} \} /\{1, −1\}.

The action of the automorphism group on the line system is described by the following

Proposition 5.3 The action of the automorphism group of 𝔾(1, 2n) ∩ H on the lines of 𝔾(1, 2n) ∩ H has two orbits:

1. the lines containing the center c

2. the lines that do not.

Proof. Since all the automorphisms preserve the center any orbit will be contained in these two sets.

First we show that the lines containing c form one orbit. For two lines c ∧ p and c ∧ q, we may assume p, q ∈ ℙ(ℂ^{2n}× 0). Take a \overline{T} ∈ Sp(2n, ℂ) that maps p to q. The trivial extention of \overline{T} to T ∈ SL(2n + 1, ℂ) will take c ∧ p to c ∧ q.

The other lines will form the second orbit since any line not containing the center can be pushed into the hyperplane ℙ(ℂ^{2n}× 0) by a transformation with an element of the normal subgroup N. There one can use the transitive action of the Aut(𝔾(1, 2n − 1) ∩ H) subgroup to show that all these lines can be mapped onto each other. □

6 𝔾(1, 2n) ∩ H^{2}

Let L = H^{2} be a 2-codimensional linear subspace of ℙ(⋀^{2} ℂ^{2n+1}). We want to study the linear line system 𝔾(1, 2n) ∩ L. To L corresponds the line L^{∗} = ℙ(λA − μB) ⊂ ℙ(⋀^{2} ℂ^{2n+1})^{∗} of the hyperplanes H_{(λ:μ)} = ℙ(λA − μB) containing L. We identify as always (⋀^{2} ℂ^{2n+1})^{∗} with the antisymmetric matrices Antisym(2n + 1, ℂ). The locus of antisymmetric matrices of corank 3 in Antisym(2n + 1, ℂ) is 3-codimensional by Corollary 1.7. Therefore a line L^{∗} may be called general if it does not intersect it. Hence for the general line L^{∗} the antisymmetric matrices λA − μB corresponding to the hyperplanes H_{(λ:μ)} have all corank 1. So each of the hyperplane sections 𝔾(1, 2n) ∩ H_{(λ:μ)} has a unique center c_{(λ:μ)} ∈ ℙ_{2n} by Proposition 5.1. These centers play an important role in the geometry of 𝔾(1, 2n) ∩ L.

Proposition 6.1 The centers c_{(λ:μ)} are those points of ℙ_{2n} through which there passes a ℙ_{2n−2} of lines of the line system 𝔾(1, 2n) ∩ L. Through all the other points of ℙ_{2n} passes only a ℙ_{2n−3} of lines.

Proof. The lines of the line system through a point p ∈ ℙ_{2n} are p ∧ q with tpAq = ^{t}pBq = 0. So we need to show that ^{t}pA and ^{t}pB are linear dependent iff p is a center of a hyperplane H_{(λ:μ)}. Now ^{t}pA and ^{t}pB are linear dependent precisely if there exists a (λ: μ) ∈ ℙ_{1} with 0 = λ^{t}pA − μ^{t}pB = ^{t}p(λA − μB), i.e. p is the kernel of λA − μB, which is by definition the center of H_{(λ:μ)}. □

Remark 6.2 Any line that contains two centers is a member of the line system 𝔾(1, N) ∩ L.

Proof. If the line contains the centers c_{(α:β)} and c_{(λ:μ)}, it is contained in the hyperplanes H_{(α:β)} and H_{(λ:μ)} by Proposition 5.1 and therefore in their intersection L = H_{(α:β)} ∩ H_{(λ:μ)}. □

Next we want to know more about the curve c_{(λ:μ)}.

Proposition 6.3 Let A, B be two antisymmetric matrices of size 2n + 1 such that every non-zero linear combination of them has corank 1. Then the map

\begin{aligned}c : & ℙ_{1} & ⟶ & ℙ_{2n} \\ & (λ: μ) & ⟼ & ker(λA − μB) \end{aligned}

is a parametrisation of a rational normal curve of degree n.

Proof. (compare [SR, X,4.3] for n = 2.) First we show that the map is injective. If it is not, there are two points of ℙ_{1} with the same image. We may assume that this is the case for (1 : 0) and (0 : 1), i.e. A and B have the same kernel, say e_{0}. Writing A and B in a basis with e_{0} as first element, we have

A = ( \begin{matrix}0 & ⋯ 0 \\ ⋮ 0 & Ã \end{matrix} ) and B = ( \begin{matrix}0 & ⋯ 0 \\ ⋮ 0 & B̃ \end{matrix} ) with Ã, B̃ ∈ Antisym(2n, ℂ).

Since det(λÃ − μB̃) is a homogeneous polynomial of degree 2n, there exist a (λ′: μ′) ∈ ℙ_{1} with det(λ′Ã− μ′B̃) = 0. But then λ′A − μ′B has corank at least two, which contradicts our assumption.

Secondly, we proof that the map is of maximal rank everywhere. If it is not, we may assume that it is not maximal at (1 : 0). Restricting to the chart λ = 1, this means c′(0) = 0. Now from

(A − μB)c(μ) = 0
⟹ Ac′(μ) − Bc(μ) − μBc′(μ) = 0
⟹ Ac′(0) − Bc(0) = 0

Bc(0) = 0 follows. Therefore A and B have the same kernel c(0), and we are back in the above chain of arguments.

Finally, we have to show that the embedding c is of degree n. For this we give an explicit form of the map. Recall [B, 5.2] that the determinant of an antisymmetric matrix C = (c_{ij}) of size 2n is the square of the irreducible Pfaffian polynomial Pf C,

Pf C := ∑_{σ} sgn (σ)c_{σ(1)σ(2)} … c_{σ(2n−1)σ(2n)},

where σ runs through all permutations S(2n) with σ(2i − 1) < σ(2i) for i = 1 … n and σ(2i) < σ(2i + 2) for i = 1 … n − 1.

Let c_{i}(λ: μ) denote (−1)^{i}-times the Pfaffian of the matrix λA − μB with the i-th row and column deleted. Then the c_{i}(λ: μ) are irreducible polynomials of degree n and by a straightforward but messy computation one can check that (c_{0}(λ: μ) : …: c_{2n}(λ: μ)) is the kernel of λA−μB. Therefore c = (c_{0} : …: c_{2n}), which shows that c is a degree n embedding of ℙ_{1}. □

After we have determined the special points in ℙ_{2n} of the line system 𝔾(1, 2n)∩ L, we are nearly ready to compute its automorphism group. It remains to give a normal form for the line L^{∗} ⊂ ℙ(⋀^{2} ℂ^{2n+1}) to make computations easier. This normal form was found by Donagi [D, 2.2]. But he did not give a proof for it since his main interest was lines in ℙ_{2n−1} and not in ℙ_{2n}. So we give the proof here.

Proposition 6.4 Let L^{∗} be a line in ℙ(⋀^{2} ℂ^{2n+1})^{∗} such that the antisymmetric matrices corresponding to the points of L^{∗} have all corank 1. Then there exists a basis (e_{0}, …, e_{2n}) of ℂ^{2n+1} such that the line can be taken as L^{∗} = ℙ(λA − μB) with the matrices

A = ( \begin{matrix}\begin{matrix}0 & −E_{n} \\ E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} ) and B = ( \begin{matrix}0 & 0 ⋮ & −E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ E_{n} & ⋮ 0 & 0 \end{matrix} ) .

Proof. Let A and B any two matrices of L^{∗} in an arbitrary basis. We will adjust the basis in three steps to achieve the required form for A and B.

1^{st} Step: We know that the map

\begin{aligned}c : & ℙ_{1} & ⟶ & ℙ_{2n} \\ & (λ: μ) & ⟼ & ker(λA − μB) \end{aligned}

is a parametrisation of a rational normal curve of degree n. Modulo projective transformations of ℙ_{1} and ℙ_{2n} such parametrisations are all the same. So we can pick a basis of ℙ_{1} and n+1 linear independent vectors e_{n}, …, e_{2n} of ℂ^{2n+1} such that

\begin{aligned}c : & ℙ_{1} & ⟶ & ℙ_{2n} \\ & (λ: μ) & ⟼ & ℙ (∑_{i=0}^{n} λ^{i}μ^{n−i}e_{n+i}) . \end{aligned}

Extending (e_{n}, …, e_{2n}) to a basis (e_{0}, …, e_{2n}) of ℂ^{2n+1} and denoting by a_{0}, …, a_{2n} resp. b_{0}, …, b_{2n} the columns of A resp. B, the fact that c(λ: μ) is the kernel of λA − μB for all (λ: μ) ∈ ℙ_{1} has the following consequences for A and B:

(λA − μB) (∑_{i=0}^{n} λ^{i}μ^{n−i}e_{n+i}) = 0

⟹ ∑_{i=0}^{n} λ^{i+1}μ^{n−i}a_{n+i} − ∑_{i=0}^{n} λ^{i}μ^{n+1−i}b_{n+i} = 0

⟹ −μ^{n+1}b_{n} + ∑_{i=0}^{n−1} λ^{i+1}μ^{n−i}(a_{n+1} − b_{n+i+1}) + λ^{n+1}a_{2n} = 0

⟹ b_{n} = 0, a_{2n} = 0, a_{n+i} = b_{n+i+1} for i = 0 … n − 1. (∗)

We claim that this implies:

a_{n+i,n+j} = 0 for i, j = 0 … n.

Indeed, for 1 ≤ i ≤ n, 0 ≤ j ≤ n − 1 we have using (∗)

a_{n+i,n+j} = b_{n+i,n+j+1} = −b_{n+j+1,n+i} = −a_{n+j+1,n+i−1} = a_{n+i−1,n+j+1}.

This shows that the a_{n+i,n+j} are all the same for i + j = const, in particular a_{n+i,n+j} = a_{n+j,n+i}. On the other hand, by the antisymmetricity of A we have a_{n+i,n+j} = −a_{n+j,n+i}, and the claim follows.

Using (∗) again we know that A and B look in our basis like

A = ( \begin{matrix}\begin{matrix}Ã & −^{t}M \\ M & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} ) and B = ( \begin{matrix}B̃ & 0 ⋮ & −^{t}M \\ 0 ⋯ 0 ⋯ 0 \\ M & ⋮ 0 & 0 \end{matrix} )

with Ã, B̃ ∈ Antisym(n, ℂ) and M ∈ GL(n, ℂ).

2^{nd} Step: Here we will improve the choice of (e_{0}, …, e_{n−1}) to achieve Ã = 0 and M = E_{n}. We claim:

Let an antisymmetric matrix A ∈ Antisym(2n + 1, ℂ) of rank 2n and linear independent vectors e_{n}, …, e_{2n} ∈ ℂ^{2n+1} with ^{t}e_{i}Ae_{j} = 0 for n ≤ i, j ≤ 2n and Ae_{2n} = 0 be given. Then (e_{n}, …, e_{2n}) can be extended to a basis (e_{0}, …, e_{2n}) of ℂ^{2n+1} such that in this basis A is given as

A = ( \begin{matrix}\begin{matrix}0 & −E_{n} \\ E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} ) .

The proof is by induction. The statement is trivial for n = 0. Assuming the claim for n − 1, we prove it for n. Let W := ⋂_{i=1}^{n} ker^{t}e_{n+i}A, then there exists an e_{0} ∈ W with ^{t}e_{n}Ae_{0} = 1. If not, we would have W = W ∩ ker^{t}e_{n}A and with e_{2n} ∈ ker A

dim span {^{t}e_{n}A, …,^{t}e_{2n}A} = dim span {^{t}e_{n+1}A, …,^{t}e_{2n−1}A} ≤ n − 1,

which contradicts rank A = 2n.

Set V := ker^{t}e_{0}A ∩ ker^{t}e_{n}A, then dim V = 2(n − 1) + 1 and e_{n+1}, …, e_{2n} ∈ V. Therefore the induction hypothesis can be applied to A|_{V}. Together with te_{0}Ae_{n} = 1 and ^{t}e_{0}Av = ^{t}e_{n}Av = 0 for v ∈ V this implies the stated form of the matrix.

So up to now A and B look like

A = ( \begin{matrix}\begin{matrix}0 & −E_{n} \\ E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} ) B = ( \begin{matrix}B̃ & 0 ⋮ & −E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ E_{n} & ⋮ 0 & 0 \end{matrix} ) .

3^{rd} Step: We adjust the vectors (e_{0}, …, e_{n−1}) so that B̃ = 0 and A stays the same.

We note that a transformation of ℂ^{2n+1} by

T = ( \begin{matrix}\begin{matrix}E_{n} & 0 \\ t & E_{n} \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 1 \end{matrix} )^{−1} with t ∈ Sym(n, ℂ)

does not change A since

tT^{−1}AT^{−1} = ^{t}T^{−1}(AT^{−1}) = ( \begin{matrix}\begin{matrix}E_{n} & t \\ 0 & E_{n} \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 1 \end{matrix} )( \begin{matrix}\begin{matrix}−t & −E_{n} \\ E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} )= A.

If we denote by \overline{t} resp. |t ∈ M(n × n, ℂ) the matrix that we obtain by deleting the first row resp. column of t and adding a row of zeroes below resp. a column of zeroes on the right side, we can write down the transformation of B as follows:

\begin{matrix}^{t}T^{−1}BT^{−1} = ^{t}T^{−1}(BT^{−1}) = ( \begin{matrix}\begin{matrix}E_{n} & t \\ 0 & E_{n} \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 1 \end{matrix} ) ( \begin{matrix}B̃ − \overline{t} & 0 ⋮ & −E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ E_{n} & ⋮ 0 & 0 \end{matrix} ) \\ = ( \begin{matrix}B̃ − \overline{t} + |t & 0 ⋮ & −E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ E_{n} & ⋮ 0 & 0 \end{matrix} ) . \end{matrix}

So to finish this step, we need to show that every antisymmetric matrix B̃ = (b_{ij}) ∈ Antisym(n, ℂ) can be written as \overline{t} − |t for a symmetric matrix t ∈ Sym(n, ℂ). The entries of \overline{t} − |t are

(t_{i+1,j} − t_{i,j+1})_{ij}

where t_{n+1,i} := t_{i,n+1} := 0 for all i = 1 … n. Obviously, \overline{t}−|t is antisymmetric. We set t_{1i} := t_{i1} := 0 for i = 1 … n and define recursively for j from n down to 2

t_{i+1,j} := t_{i,j+1} + b_{ij} for i = 1 … j − 1.

Then by the symmetry of t the whole matrix t is defined and \overline{t} − |t = B □

For the linear system in normal form the rational curve of centers has the parametrisation

\begin{matrix}c(λ: μ) = ker(λA − μB) = ker ( \begin{matrix}& & & −λ & μ \\ 0 & ... & ... \\ & & & & −λ & μ \\ λ \\ −μ & ... \\ & ... & λ & & 0 \\ & & −μ \end{matrix} ) \\ = (0 : …: 0 : μ^{n}: μ^{n−1}λ: …: λ^{n}). \end{matrix}

The ℙ_{2n−2} of lines through a center c_{(λ:μ)} is given by

c_{(λ:μ)} ∧ q where q ∈ ℙ_{2n} with ^{t}c_{(λ:μ)}Aq = ^{t}c_{(λ:μ)}Bq = 0,

i.e. q must be an element of the hyperplane h_{(λ:μ)} ∈ ℙ_{2n}^{∗}

h_{(λ:μ)} = ker^{t}c_{(λ:μ)}A ∩^{t}c_{(λ:μ)}B
= ker ( \begin{matrix}μ^{n} & μ^{n−1}λ & ⋯ & μλ^{n−1} & 0 & ⋯ & 0 \\ μ^{n−1}λ & μ^{n−2}λ^{2} & ⋯ & λ^{n} & 0 & ⋯ & 0 \end{matrix} )
= ker ( μ^{n−1} μ^{n−2}λ ⋯ λ^{n−1} 0 ⋯ 0 ) .

So the hyperplanes h_{(λ:μ)}, which are traced out by the ℙ_{2n−2} of lines through the centers, give rise to a rational normal curve of degree n − 1 in the space of hyperplanes containing the center curve. That the hyperplanes h_{(λ:μ)} contain the center curve could already be seen from the Remark 6.2, by which h_{(λ:μ)} must contain any line connecting c_{(λ:μ)} with any other point of the center curve.

Now we are ready to study the automorphism group of 𝔾(1, 2n) ∩ L.

Any automorphism T ∈ Aut(𝔾(1, 2n) ∩ L) ⊆ ℙGL(2n + 1, ℂ) has to map the center curve onto itself and also the projective space P ≅ ℙ_{n} spanned by the center curve onto itself. It is known [H, 10.12] that the group of automorphisms of ℙ_{n} fixing a rational normal curve of degree n is isomorphic to ℙGL(2, ℂ). If the rational normal curve is given by

\begin{aligned}c : & ℙ_{1} & ⟶ & ℙ_{n} \\ & (λ: μ) & ⟼ & (μ^{n}: μ^{n−1}λ: …: λ^{n}), \end{aligned}

this isomorphism ℙGL(2, ℂ) ≅ Aut(c, ℙ_{n}) maps

t = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ ℙGL(2, ℂ)

to t_{n+1} ∈ ℙGL(n + 1, ℂ) where t_{n+1} is the unique matrix such that

t_{n+1} ( μ^{n} ; μ^{n−1}λ ; ⋮ ; λ^{n} ) = ( (dμ + cλ)^{n} ; (dμ + cλ)^{n−1}(bμ + aλ) ; ⋮ ; (bμ + aλ)^{n} );

for example

t_{2} = ( \begin{matrix}d & c \\ b & a \end{matrix} ) t_{3} = ( \begin{matrix}d^{2} & 2cd & c^{2} \\ bd & ad + bc & ac \\ b^{2} & 2ab & a^{2} \end{matrix} ) .

Applying this to the center curve restricts the form of the transformation T to

T = ( \begin{matrix}∗ & 0 \\ ∗ & t_{n+1} \end{matrix} ) .

We know further that if T maps c_{(λ:μ)} to c_{(aλ+bμ:cλ+dμ)}, then it must map the hyperplane h_{(λ:μ)} to h_{(aλ+bμ:cλ+dμ)}. Therefore it induces also an automorphism on the rational curve h of degree n − 1 in the dual projective space (ℙ_{2n}/P)^{∗} of hyperplanes containing P. Hence T must be of the form

T = ( \begin{matrix}α^{t}t_{n}^{−1} & 0 \\ ∗ & t_{n+1} \end{matrix} ) with α ∈ ℂ^{∗}.

We make the following claim:

T = ( \begin{matrix}^{t}t_{n}^{−1} & 0 \\ 0 & t_{n+1} \end{matrix} ) ∈ ℙGL(2n + 1, ℂ)

is an automorphism of the linear system 𝔾(1, 2n) ∩ L.

Proof. We need to check that for every t ∈ ℙGL(2, ℂ)

tT^{−1}(λA − μB)T^{−1} ∈ span {A, B}

for all λ, μ ∈ ℂ. Since ℙGL(2, ℂ) is a group, this is equivalent to the statement that for every t ∈ ℙGL(2, ℂ)

tT (λA − μB)T ∈ span {A, B}

for all λ, μ ∈ ℂ. Because of the linearity it is enough to do this for (λ, μ) = (1, 0) and (0, −1). Denoting by \overline{t_{n+1}} resp. \underline{t_{n+1}} the matrix t_{n+1} with the first resp. last row deleted, we compute:

\begin{matrix}^{t}T (AT) =( \begin{matrix}t_{n}^{−1} & 0 \\ 0 & ^{t}t_{n+1} \end{matrix} )( \begin{matrix}0 & −\underline{t_{n+1}} \\ ^{t}t_{n}^{−1}_{0···0} & 0 \end{matrix} )=( ^{t}(t_{n}^{−1}\underline{t_{n+1}}) 0^{0  −t_{n}^{−1}\underline{t_{n+1}}} ) \\ ^{t}T (BT) =( \begin{matrix}t_{n}^{−1} & 0 \\ 0 & ^{t}t_{n+1} \end{matrix} )( \begin{matrix}0 & −\overline{t_{n+1}} \\ ^{t}t_{n}^{−1}^{0···0} & 0 \end{matrix} )=( ^{t}(t_{n}^{−1}\overline{t_{n+1}}) 0^{0  −t_{n}^{−1}\overline{t_{n+1}}} ). \end{matrix}

So, if we show

\begin{matrix}\underline{t_{n+1}} = d (t_{n}_{0}^{0}) + c (_{0}^{0}t_{n}) ⟹ t_{n}^{−1}\underline{t_{n+1}} = d (E_{n}_{0}^{0}) + c (_{0}^{0}E_{n}) \\ \overline{t_{n+1}} = b (t_{n}_{0}^{0}) + a (_{0}^{0}t_{n}) ⟹ t_{n}^{−1}\overline{t_{n+1}} = b (E_{n}_{0}^{0}) + a (_{0}^{0}E_{n}) , \end{matrix}

Documenta Mathematica 4 (1999) 623–664

where _{0}^{0} stands for adding a column of zeroes, then

\begin{matrix}^{t}T AT = dA + cB \\ ^{t}T BT = bA + aB. \end{matrix}

To show the equality for \underline{t_{n+1}} note that on the one hand \underline{t_{n+1}} is the unique matrix with

\underline{t_{n+1}} ( μ^{n} ; μ^{n−1}λ ; ⋮ ; λ^{n} ) = ( (dμ + cλ)^{n} ; (dμ + cλ)^{n−1}(bμ + aλ) ; ⋮ ; (dμ + cλ)(bμ + aλ)^{n−1} )

and on the other hand

( (dμ + cλ)^{n} ; (dμ + cλ)^{n−1}(bμ + aλ) ; ⋮ ; (dμ + cλ)(bμ + aλ)^{n−1} ) = (dμ + cλ) ( (dμ + cλ)^{n−1} ; (dμ + cλ)^{n−2}(bμ + aλ) ; ⋮ ; (bμ + aλ)^{n−1} )

= (dμ + cλ)t_{n} ( μ^{n−1} ; μ^{n−2}λ ; ⋮ ; λ^{n−1} ) = dt_{n} ( μ^{n} ; μ^{n−1}λ ; ⋮ ; μλ^{n−1} ) + ct_{n} ( μ^{n−1}λ ; μ^{n−2}λ^{2} ; ⋮ ; λ^{n} )

= (d (t_{n}_{0}^{0}) + c (_{0}^{0}t_{n})) ( μ^{n} ; μ^{n−1}λ ; ⋮ ; λ^{n} ) .

Of course, the proof for \overline{t_{n+1}} = b (t_{n}_{0}^{0}) + a (_{0}^{0}t_{n}) is analogous. □

Given any automorphism of the line system 𝔾(1, 2n) ∩ L we can compose it with one of the above automorphisms such that the composition fixes the center curve pointwise. So, we can focus our attention to automorphisms of the last type.

Lemma 6.5 All automorphisms of 𝔾(1, 2n) ∩ L that fix the center curve pointwise are of the form

T = ( \begin{matrix}αE_{n} & 0 \\ S & E_{n+1} \end{matrix} ) with α ∈ ℂ^{∗}, S ∈ M((n + 1) × n, ℂ),

where the matrix S ∈ M((n + 1) × n, ℂ) has the same entries along the minor diagonals, i.e. s_{ij} = s_{kl} for i + j = k + l.

As a group these matrices are isomorphic to the semi direct product ℂ^{2n} ⋉ ℂ^{∗}, (s, α) · (s′, α′) = (α′s + s′, αα′).

Proof. We need only to check the property of S and the group structure. T is an automorphism iff

tT^{−1}AT^{−1},^{t}T^{−1}BT^{−1} ∈ span {A, B}.

The inverse of T is

T^{−1} = ( \begin{matrix}\frac{1}{α}E_{n} & 0 \\ −\frac{1}{α}S & E_{n+1} \end{matrix} ) .

Now if \overline{S} resp. \underline{S} ∈ M(n × n, ℂ) denote the matrix S with the first resp. last row deleted, then

tT^{−1}(AT^{−1}) =( \begin{matrix}\frac{1}{α}E_{n} & −\frac{1}{α}^{t}S \\ 0 & E_{n+1} \end{matrix} )( \begin{matrix}\begin{matrix}\frac{1}{α} \underline{S} & −E_{n} \\ \frac{1}{α}E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} )

= ( \begin{matrix}\begin{matrix}\frac{1}{α^{2}} (\underline{S} −^{t}\underline{S}) & −\frac{1}{α}E_{n} \\ \frac{1}{α}E_{n} & 0 \end{matrix} & 0 ⋮ 0 \\ 0 ⋯ 0 & 0 \end{matrix} )

tT^{−1}(BT^{−1}) = ( \begin{matrix}\frac{1}{α}E_{n} & −\frac{1}{α}^{t}S \\ 0 & E_{n+1} \end{matrix} )( \begin{matrix}\frac{1}{α} \overline{S} & 0 ⋮ & −E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ \frac{1}{α}E_{n} & ⋮ 0 & 0 \end{matrix} )

= ( \begin{matrix}\frac{1}{α^{2}} (\overline{S} −^{t}\overline{S}) & 0 ⋮ & −\frac{1}{α}E_{n} \\ 0 ⋯ 0 ⋯ 0 \\ \frac{1}{α}E_{n} & ⋮ 0 & 0 \end{matrix} ).

Therefore T is an automorphism iff \underline{S} = ^{t}\underline{S} and \overline{S} = ^{t}\overline{S}. In other words

\begin{matrix}s_{ij} = & s_{ji} \\ s_{i+1,j} = & s_{j+1,i} \end{matrix} for 1 ≤ i, j ≤ n,

so

s_{ij} = s_{ji} = s_{(j−1)+1,i} = s_{i+1,j−1}

for j > 1 and i < n, hence s_{ij} = s_{kl} for i + j = k + l.

The statement about the group action follows from

( \begin{matrix}αE_{n} & 0 \\ S & E_{n+1} \end{matrix} ) ( \begin{matrix}α′E_{n} & 0 \\ S′ & E_{n+1} \end{matrix} ) = ( \begin{matrix}αα′E_{n} & 0 \\ α′S + S′ & E_{n+1} \end{matrix} ) . □

Collecting the results we have

Theorem 6.6 The automorphism group of 𝔾(1, 2n) ∩ L is an extention of ℙGL(2, ℂ) by the semi direct product ℂ^{2n} ⋉ ℂ^{∗}.

It is isomorphic to the matrix subgroup of ℙGL(2n + 1, ℂ) given by

( \begin{matrix}αE_{n} & 0 \\ S & E_{n+1} \end{matrix} ) ( \begin{matrix}^{t}t_{n}^{−1} & 0 \\ 0 & t_{n+1} \end{matrix} )

where α ∈ ℂ^{∗}, S ∈ M((n + 1) × n, ℂ) with s_{ij} = s_{kl} for i + j = k + l and t_{n} ∈ Aut(h, ℙ_{n−1}) resp. t_{n+1} ∈ Aut(c, ℙ_{n}) are the transformations that are induced by the ℙGL(2, ℂ) action on the rational normal curve h ⊂ ℙ_{n−1} resp. c ⊂ ℙ_{n}.

Proof. It remains to show that the automorphism fixing the center curve pointwise form a normal subgroup, but that can be easily computed. □

Remark 6.7 An automorphism of 𝔾(1, 2n) ∩ L is determined by its action on the lines intersecting the center curve.

In contrast to that, the line system, i.e. the position of the line L^{∗} ⊂ ℙ(⋀^{2} ℂ^{2n+1})^{∗}, is not determined by these lines, as a simple dimension count shows. Giving these lines is equivalent to giving the two rational curves c ⊂ ℙ_{2n} and h ⊂ ℙ_{2n}/P ≅ ℙ_{n−1} and a correspondence between them, so that we have the following dimension count

(2(2n + 1) − 4) + (2n − 4) + 3 < dim 𝔾(1, ℙ(⋀^{2}ℂ^{2n+1})) = 2 (\binom{2n + 1}{2} − 2) .

Proof of the remark. We need to show that only the identity fixes these lines one by one. First a transformation T that fixes the lines must fix the center curve, hence by the Lemma 6.5 it is of the form

T = ( \begin{matrix}αE_{n} & 0 \\ S & E_{n+1} \end{matrix} ) .

We compute the induced action T̃ of T on {l ∈ 𝔾(1, 2n) ∩ L | c_{(0:1)} ∈ l}, the ℙ_{2n−2} of lines through c_{(0:1)} = e_{n}. A line l ∈ 𝔾(1, 2n) through c_{(0:1)} will be in the line system 𝔾(1, 2n)∩L iff it lies in the hyperplane h_{(0:1)} = ker(1 : 0 : …: 0). Therefore the ℙ_{2n−2} of lines through e_{n} is given by

e_{n} ∧ x with x ∈ ℙ(span {e_{1}, …, e_{n−1}, e_{n+1}, …, e_{2n}}).

Using (e_{1} ∧ e_{n}, …, e_{n−1} ∧ e_{n}, e_{n+1} ∧ e_{n}, …, e_{2n} ∧ e_{n}) as a basis, the induced action T̃ is

T̃ = ( \begin{matrix}αE_{n−1} & 0 \\ |\overline{S} & E_{n} \end{matrix} ) .

Here |\overline{S} denotes the matrix S with the first row and column deleted. In order to have T̃ = E_{2n−1}, we must have α = 1 and |\overline{S} = 0.

The same computation for the lines through c_{(1:0)} = e_{2n} yields α = 1 and \underline{S}| = 0 from which S = 0 and the remark follow. □

For the rest of the section we analyze the action of the automorphism group on the line system 𝔾(1, 2n) ∩ L. We start with 𝔾(1, 4) ∩ L.

Proposition 6.8 The action of Aut(𝔾(1, 4) ∩ L) on the lines has four orbits:

1. tangents of the center conic

2. secants of the center conic

3. lines through the center conic that do not lie in the plane of the center conic

4. lines that do not intersect the plane of the center curve.

Proof. Since any automorphism maps the center conic onto itself, it is clear by the geometric description that all the mentioned lines lie in different orbits.

Any line in the plane P of the center conic intersects the conic twice, so by the Remark 6.2 it is a member of the line system. Since the automorphism group acts like Aut(c, P) ≅ ℙGL(2, ℂ) on the plane P, the first two orbits are obvious.

To see that the lines of 3) form one orbit, we have to exhibit an automorphism that given two lines of 3) maps one onto the other. Since the ℙGL(2, ℂ) part of the automorphism group acts transitively on the center conic, we may assume that both lines pass through the same point of the center conic, say e_{2} = c_{(0:1)}. Now the induced action T̃ of an automorphism T fixing the center conic pointwise on the ℙ_{2} of lines through e_{2} was computed in the proof of the Remark 6.7 as

T̃ = ( \begin{matrix}α & 0 & 0 \\ f & 1 & 0 \\ g & 0 & 1 \end{matrix} ) with α ∈ ℂ^{∗}; f, g ∈ ℂ.

These transformations act transitively on ℙ_{2} \ ℙ(span {ẽ_{1}, ẽ_{2}}), where the line ℙ(span {ẽ_{1}, ẽ_{2}}) corresponds to the lines through e_{2} that lie in the plane of the center conic.

The lines of 4) are all the remaining lines since there are no lines that intersect the plane P of the center conic but not the conic c itself. This is clear because the ℙ_{1} of lines through a point p ∈ P \ c is formed by the lines through p in the plane P, so there can be no other line.

Finally, we have to show that the lines of 4) form one orbit. By a small computation one checks that Aut(𝔾(1, 4)∩L) ⊂ ℙGL(5, ℂ) acts transitively on ℙ_{4} \ P. So, it suffices to show that the line e_{0} ∧ e_{1} can be mapped to any other line through e_{0} by an automorphism. Any of these lines can be written as

e_{0} ∧ (e_{1} + βe_{4}) with β ∈ ℂ,

and the automorphism

T = ( \begin{matrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & β & 0 & 0 & 1 \end{matrix} )

will take e_{0} ∧ e_{1} to it. □

Proposition 6.9 The automorphism group acts quasihomogeneously on 𝔾(1, 6) ∩ H^{2}.

Proof. For this it is enough to show that the stabiliser of the line l = e_{0} ∧ e_{2} is a 2-dimensional subgroup since then

dim Orbit(l) = dim Aut(𝔾(1, 6) ∩ H^{2}) − dim Stab(l) = 10 − 2 = 8
= dim 𝔾(1, 6) ∩ H^{2}.

If we normalize the t ∈ ℙGL(2, ℂ) by det t = 1, every T ∈ Aut(𝔾(1, 6) ∩ H^{2}) can be written by the Theorem 6.6 as

\begin{matrix}T = ( \begin{matrix}α & & 0 \\ & α & & & & 0 \\ 0 & & α \\ e & f & g & 1 & & 0 \\ f & g & h & & & 1 \\ g & h & i & & & 1 \\ h & i & j & & 0 & & 1 \end{matrix} ) ( \begin{matrix}^{t}t_{3}^{−1} & 0 \\ 0 & t_{4} \end{matrix} ) \\ with ^{t}t_{3}^{−1} = ( \begin{matrix}a^{2} & −ab & b^{2} \\ −2ac & ab + cd & −2bd \\ c^{2} & −cd & d^{2} \end{matrix} ) . \end{matrix}

To compute the stabilizer we start by looking only at the first three entries of

\begin{matrix}T e_{0} = (αa^{2}, −2αac, αc^{2}, …) \\ T e_{2} = (αb^{2}, −2αbd, αd^{2}, …). \end{matrix}

Since we must have T e_{0}, T e_{2} ∈ l, ac = bd = 0 follows. By det t = ad − bc = 1 we have the two possibilities b = c = 0, d = a^{−1} and a = d = 0, c = −b^{−1}. We examine only the first case, the second being similar. Now we have

\begin{matrix}T e_{0} = a^{2}(α, 0, 0, e, f, g, h) \\ T e_{2} = a^{2}(0, 0, α, g, h, i, j). \end{matrix}

From T e_{0}, T e_{2} ∈ l we conclude e = f = g = h = 0 resp. g = h = i = j = 0. Therefore, including the case (a = d = 0, c = −b^{−1}), the stabilizer of l is

Stab(l) = \{( \begin{matrix}αa^{2} \\ 0 & & 0 \\ & αa^{−2} \\ & & a^{−3} \\ & & a^{−1} \\ 0 & & a \\ & & a^{3} \end{matrix} ) , ( \begin{matrix}0 & αb^{2} \\ & 0 & & 0 \\ αb^{−2} & 0 \\ & & & −b^{−3} \\ & & & b^{−1} \\ & 0 & −b \\ & & b^{3} & 0 \end{matrix} )\} . □

Proposition 6.10 For n ≥ 4 the action of the automorphism group on 𝔾(1, 2n) ∩ H^{2} is not quasihomogeneous.

Proof. We project the ℙ_{2n} from the space P of the center curve onto ℙ_{2n}/P ≅ ℙ_{n−1}. This projects the lines of 𝔾(1, 2n) ∩ H^{2} not intersecting P surjectively onto the lines 𝔾(1, ℙ_{2n}/P) of ℙ_{2n}/P. The automorphisms of 𝔾(1, 2n) ∩ H^{2} induce automorphisms of ℙ_{2n}/P. As matrices these are the upper left n × n matrices of the matrices of Theorem 6.6, i.e. they are of the form ^{t}t_{n}^{−1}. So, as a group this induced automorphism group is isomorphic to Aut(h^{∗}, ℙ_{2n}/P) ≅ ℙGL(2, ℂ). If Aut(𝔾(1, 2n) ∩ H^{2}) acts quasihomogeneously, then Aut(h^{∗}, ℙ_{2n}/P) would have to act quasihomogeneously on 𝔾(1, ℙ_{2n}/P) ≅ 𝔾(1, n − 1), but this contradicts

dim ℙGL(2, ℂ) = 3 < dim 𝔾(1, n − 1) = 2n − 4. □

7 𝔾(1, 4) ∩ H^{3}

Let L = H^{3} be a general 3-codimensional subspace of ℙ(⋀^{2} ℂ^{5}) ≅ ℙ_{9}. To L corresponds the plane L^{∗} = ℙ(λA+μB+νC) ⊆ ℙ(⋀^{2} ℂ^{5})^{∗} of hyperplanes containing L. Since the locus of antisymmetric matrices of corank 3 is 3-codimensional in ℙ(⋀^{2} ℂ^{5})^{∗} by Corollary 1.7, L^{∗} does not contain any. Therefore to each of the hyperplanes H_{(λ:μ:ν)} ⊂ L corresponds a unique center c_{(λ:μ:ν)} ∈ ℙ_{4}. In complete analogy to the last case we get

Lemma 7.1 The centers c_{(λ:μ:ν)} are those points of ℙ_{4} through which there passes a ℙ_{1} of lines of the line system 𝔾(1, 4) ∩ L. Through all the other points passes a unique line.

Proposition 7.2 The map of centers

\begin{aligned}c : & L^{∗} ≅ ℙ_{2} & ⟶ & ℙ_{4} \\ & (λ : μ : ν) & ⟼ & c_{(λ:μ:ν)} = ker(λA + μB + νC) \end{aligned}

is an embedding of ℙ_{2} in ℙ_{4} of degree 2, i.e. its image is a smooth projected Veronese surface.

Remark 7.3 Any line that contains three centers is in the line system.

Proof. Let c(p_{0}), c(p_{1}) and c(p_{2}) with p_{0}, p_{1}, p_{2} ∈ L^{∗} be the three centers on the line l. By the definition of the centers we have l ∈ H_{p_{i}}. Since c maps lines in L^{∗} onto conics in ℙ_{4}, the three points p_{0}, p_{1}, p_{2} do not lie on a line, hence they span L^{∗}. So l ∈ H_{p_{0}} ∩ H_{p_{1}} ∩ H_{p_{2}} = L. □

From the statements we get a complete picture of the lines of 𝔾(1, 4) ∩ H^{3} in ℙ_{4}. We define the trisecant variety Tri(X) of a variety X ⊆ ℙ_{N} by:

Tri(X) := \overline{\{l ∈ 𝔾(1, N) | #(X ∩ l) ≥ 3\}}

Then we have

Corollary 7.4 (Castelnuovo) 𝔾(1, 4) ∩ L is the trisecant variety of the smooth projected Veronese surface Im c ⊂ ℙ_{4}.

Proof.(see [C] or [SR, X, 4.4]) By the remark above the trisecant variety is contained in the irreducible variety 𝔾(1, 4) ∩ L. So it is enough to show that both varieties have the same dimension. The Lemma 7.1 together with the Proposition 7.2 shows that there is an unique line of 𝔾(1, 4) ∩ L through a general point of ℙ_{4} and that 𝔾(1, 4) ∩ L is the closure of such lines. The same statement for the trisecant variety is classical [SR, VII,3.2]. Hence both varieties have dimension three. □

The general trisecant intersects the projected Veronese surface Im c in three different points. Their inverse image under c are triples of points in ℙ_{2} that have to fulfill some conditions since there is only a 3-dimensional family of these triples. To see what these conditions are, we recall some facts about the Veronese surface [H].

The Veronese surface V is the image of the embedding

\begin{aligned}υ : & ℙ_{2} = ℙ(ℂ^{3}) & ⟶ & ℙ(Sym^{2}ℂ^{3}) \\ & ℙ(v) & ⟼ & ℙ(v · v) . \end{aligned}

Its secant variety consists of the points of ℙ(Sym^{2}ℂ^{3}) that are the product of two vectors of ℂ^{3},

Sec(V) = {ℙ(v · w) | v, w ∈ ℂ^{3} \ {0}} .

The projected Veronese surface will be smooth – like in our case – iff the center of projection P is not in the secant variety.

An intersection of the Veronese surface V with a hyperplane H ∈ ℙ(Sym^{2}ℂ^{3})^{∗} gives the conic υ^{−1}(V ∩ H) ⊂ ℙ_{2} which is described by the equation H if we identify ℙ(Sym^{2}ℂ^{3})^{∗} with the polynomials of degree 2 modulo ℂ^{∗}. The conics that we get as hyperplane sections of the projected Veronese surface are precisely the conics that we get as hyperplane sections of the Veronese surface by hyperplanes that contain the projection center P. So these conics fulfill one linear condition given by P. We view P ∈ ℙ(Sym^{2}ℂ^{3}) as a conic C_{P}^{∗} in ℙ_{2}^{∗}. Since P does not lie in the secant variety of the Veronese surface, P is not the product of two elements of ℂ^{3}, hence C_{P}^{∗} is not the union of two lines. Therefore it is smooth. We denote the dual conic of C_{P}^{∗} by C_{P} ⊂ ℙ_{2}.

Now, three different points of the projected Veronese surface Im c ⊂ ℙ_{4} lie on a line, the trisecant, iff any hyperplane that contains two of them contains all three. Under the inverse of the embedding c that means the following on the ℙ_{2}:

Three different points of ℙ_{2} are the inverse image c^{−1}(l) of a trisecant l of the projected Veronese surface Im c iff all conics that fulfill the linear condition given by P (or equivalently by C_{P}) and pass through two of the points pass through all three of them.

The propositions in the appendix tell us that these triples of points are the vertices of the non-degenerated polar triangles of the conic C_{P}. By a continuity argument the trisecants that are tangent to Im c at one point and intersect it in another correspond to the degenerated polar triangles, and the trisecants that intersect Im c in only one point “with multiplicity three” correspond to a triple point on the conic C_{P}. We also see that there are no 4-secants. Since if there is one, there would be four points in ℙ_{2} such that any three of them build a different polar triangle. But this is impossible because a polar triangle is already determined by two of its vertices.

With this geometric description it is easy to compute the automorphism group of 𝔾(1, 4) ∩ H^{3}. By Theorem 1.2 any automorphism is induced by a projective linear transformation of H^{3}. Hence it maps the ℙ_{1}’s of lines of Lemma 7.1 onto themselves and induces thereby an automorphism of the set of centers, the projected Veronese surface Im c. Since the 𝔾(1, 4) ∩ H^{3} consists of the trisecants of Im c, this automorphism determines the original automorphism of 𝔾(1, 4) ∩ H^{3}.

In order to complete the proof of Corollary 1.3 we have to show that the automorphism of the projected Veronese surface Im c ⊂ ℙ_{4} is induced by one of ℙ_{4}. Since c is the composition of the Veronese map v and the projection π_{P} : ℙ_{5} → ℙ_{4}, we consider the situation in the ℙ_{5}. Since Im c ≅ ℙ_{2} ≅ v(ℙ_{2}) and Aut(ℙ_{2}) ≅ Aut(v(ℙ_{2}), ℙ_{5}) [H, 10.9], the automorphism of Im c induces an automorphism of ℙ_{5}. We know further that the automorphism of Im c maps the trisecants of Im c onto themselves. In the ℙ_{5} this means that it preserves the planes that contain the projection center P and three points of the Veronese surface, so it must fix their common intersection, the projection center P. Hence the automorphism descends onto the ℙ_{4}.

Now under the inverse of the embedding c, the automorphism of Im c preserving the trisecants corresponds to an automorphism of ℙ_{2} preserving the polar triangles of the conic C_{P} ⊂ ℙ_{2}. Such an automorphism of ℙ_{2} maps the degenerated polar triangles onto themselves. In particular, it maps the tangents to the conic C_{P} onto themselves. Therefore it has to fix the conic C_{P}.

So we have seen how an automorphism of 𝔾(1, 4) ∩ L induces an unique automorphism of ℙ_{2} fixing C_{P}, hence an automorphism of C_{P} since Aut(C_{P} , ℙ_{2}) ≅ Aut(C_{P}) ≅ ℙGL(2, ℂ).

On the other hand, any projective linear transformation of ℙ_{2} that fixes the conic C_{P} preserves the polar triangles of C_{P}. Therefore it induces via the embedding c an automorphism of the projected Veronese surface Im c that preserves triples of points that lie on a line. So it defines an automorphism of the trisecants of Im c, which is the same as an automorphism of 𝔾(1, 4) ∩ L.

We summarize:

Theorem 7.5 The automorphism group of 𝔾(1, 4) ∩ H^{3} is isomorphic to ℙGL(2, ℂ).

The description of the orbits of this automorphism group follows immediately.

Proposition 7.6 The action of Aut(𝔾(1, 4) ∩ H^{3}) on the linear system 𝔾(1, 4) ∩ H^{3} has three orbits:

1. trisecants of the projected Veronese surface that intersect it in three points

2. trisecants that are tangent to the projected Veronese surface at one point and intersect it in another

3. trisecants that intersect the projected Veronese surface in only one point “with multiplicity three”.

Proof. By what was said above, this is equivalent to the classical statement that the action of group Aut(C_{P} , ℙ_{2}) on the polar triangles has three orbits: the non-degenerated triangles, the degenerated ones and the triple points on C_{P}. □

8 Appendix: Polar Triangles

Here we prove the needed propositions about polar triangles. The whole appendix may be seen as a modern exposition of [SF, 348]. First we recall the basic definitions.

Let C_{A} be a smooth conic in ℙ_{2}, which is given by the quadratic equation txAx = 0, where A ∈ GL(3, ℂ) is a symmetric, invertible matrix. Then C_{A} induces a polarity P by

\begin{aligned}P : & ℙ_{2} & ⟶ & ℙ_{2}^{∗} \\ & ℙ(x) & ⟼ & ℙ(^{t}xA) . \end{aligned}

For a point p ∈ ℙ_{2} the line P (p) is called the polar of p and p the pole of P (p).

A polar triangle is given by three points, at least two of which are different, such that the polar of each point contains the other two points, i.e. (p, q, r) is a polar triangle if ^{t}pAq = ^{t}qAr = ^{t}rAp = 0. The sides of the triangle are the polars of the points. In the non-degenerated case when all three points are different, the three points cannot lie on a line and therefore span the whole ℙ_{2}. In the degenerated case, (p, p, q), p lies on the conic and q on the tangent to the conic at the point p. The sides are the polar of q and twice the tangent.

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Non-degenerated polar triangle

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Degenerated polar triangle

Proposition 8.1 Let C_{A} = {x ∈ ℙ_{2} | ^{t}xAx = 0} be a smooth conic and C_{A}^{∗} = {x ∈ ℙ_{2}^{∗} | ^{t}xA^{−1}x = 0} its dual conic. Further, let C_{B} = {x ∈ ℙ_{2} | ^{t}xBx = 0} be a conic such that

∑_{i,j=0}^{2} a^{ij}b_{ij} = 0 (∗)

where A^{−1} = (a^{ij}), B = (b_{ij}) ∈ Sym(3, ℂ). Finally, let (p, q, r) be a polar triangle of C_{A} then:

If two of the three points p, q, r lie on the conic C_{B}, then also the third.

In the case of a degenerate polar triangle, (p, p, q), the condition that C_{B} contains p twice means that C_{B} contains p and C_{B} is either singular in p or its tangent in b is the polar of q.

Proof. One can show that all the properties in the statement of the proposition are independent of the choice of coordinates, so we may pick nice ones. We have to distinguish between the two cases of the polar triangle being degenerated or not. We treat the case of the non-degenerated polar triangle first.

By a suitable choice of coordinates we may assume that p = (1 : 0 : 0), q = (0 : 1 : 0) and r = (0 : 0 : 1). Then the assumption that (p, q, r) is a polar triangle of C_{A} translates into

tpAq = ^{t}qAr = ^{t}rAp = 0 ⟺ a_{01} = a_{12} = a_{02} = 0.

By a scaling of the coordinates we can achieve that a_{00} = a_{11} = a_{22} = 1, so that C_{A} = {x ∈ ℙ_{2} | x_{0}^{2} + x_{1}^{2} + x_{2}^{2} = 0}. Then the condition (∗) reads b_{00} + b_{11} + b_{22} = 0. If the two points p and q are on the conic C_{B}, we have tpBp = b_{00} = 0 and ^{t}qBq = b_{11} = 0. By (∗) we see 0 = b_{22} = ^{t}rBr, i.e. the third point r lies also on the conic C_{B}.

Now we treat the case of the degenerated polar triangle (p, p, q). We choose coordinates such that C_{A} = {x ∈ ℙ_{2} | x_{0}^{2} + x_{1}^{2} + x_{2}^{2} = 0} and p = (1 : i: 0). The point q ≠ p must lie on the tangent to C_{A}. So it has coordinates q = (λ: iλ: 1), and its polar is spanned by p and (1 : 0 : −λ). Now using the assumptions

b_{00} + b_{11} + b_{22} = 0 (∗)

p ∈ C_{B} ⟺ ^{t}pBp = 0 ⟺ b_{00} + 2ib_{01} − b_{11} = 0, (∗∗)

we have to show

q ∈ B ⟺ C_{B} singular in p or 𝕋_{p}C_{B} = polar of q.

We rewrite this as

tqBq = ^{t} ( \begin{matrix}λ \\ iλ \\ 1 \end{matrix} )B( \begin{matrix}λ \\ iλ \\ 1 \end{matrix} ) = 0 ⟺ ^{t}pB( \begin{matrix}1 \\ 0 \\ −λ \end{matrix} ) = ^{t} ( \begin{matrix}1 \\ i \\ 0 \end{matrix} )B( \begin{matrix}1 \\ 0 \\ −λ \end{matrix} ) = 0.

But this is true since −2 times the left hand side plus (λ^{2} + 1) times (∗∗) plus (∗) gives the right hand side. □

Now we prove the converse of the last proposition.

Proposition 8.2 Given a smooth conic C_{A} = {x ∈ ℙ_{2} | ^{t}xAx = 0}

B := {C_{B} = {x ∈ ℙ_{2} | ^{t}xBx = 0} | ∑_{i,j=0}^{2} a^{ij}b_{ij} = 0}

is a four dimensional family of conics. Let p, q, r ∈ ℙ_{2} be three points, at least two of which are different, with the property that if two of them lie on a conic C_{B} ∈ B then also the third.

Then (p, q, r) is a polar triangle of C_{A}.

Proof. We will show that if (p, q, r) is not a polar triangle then there exits a C_{A} ∈ B for which this property is violated. We have to treat several cases.

First let the three points be all different, then they cannot lie on a line. Because if they would, the conics in the at least two dimensional family

B_{p,q} := {C_{B} ∈ B | p, q ∈ C_{B}}

of conics of B passing through p and q must split off the line through the three points. If we pick coordinates such that this line is given by {x_{2} = 0}, then B_{p,q} must be

B_{p,q} = {V(x_{2}(λ_{0}x_{0} + λ_{1}x_{1} + λ_{2}x_{2})) | (λ_{0} : λ_{1} : λ_{2}) ∈ ℙ_{2}} .

This means that the conics C_{B} with the matrices

B = ( \begin{matrix}0 & 0 & b_{02} \\ 0 & 0 & b_{12} \\ b_{02} & b_{12} & b_{22} \end{matrix} ) with b_{02}, b_{12}, b_{22} ∈ ℂ

are all in B. Hence the matrix A^{−1} must be of the type

A^{−1} = ( \begin{matrix}a^{00} & a^{01} & 0 \\ a^{01} & a^{11} & 0 \\ 0 & 0 & 0 \end{matrix} ) with a^{00}, a^{01}, a^{11} ∈ ℂ,

but this contradicts the invertibility of A^{−1}.

Now since p, q, r span the ℙ_{2} we may pick coordinates such that p = (1 : 0 : 0), q = (0 : 1 : 0) and r = (0 : 0 : 1). That (p, q, r) is not a polar triangle of C_{A} means that ^{t}pAq = a_{01} ≠ 0, ^{t}qAr = a_{12} ≠ 0 or ^{t}rAp = a_{02} ≠ 0. Assuming det A = 1 we conclude that not all of the a^{01} = a_{02}a_{12} − a_{01}a_{22}, a^{02} = a_{01}a_{12} − a_{02}a_{11} and a^{12} = a_{01}a_{02} − a_{00}a_{12} can be zero. If for example a^{02} ≠ 0, then

B = ( \begin{matrix}0 & 0 & −a^{11} \\ 0 & 2a^{02} & 0 \\ −a^{11} & 0 & 0 \end{matrix} )

gives a conic C_{B} ∈ B that contains the points p and q, but not r.

Now let us look at the case where two of the points p, q, r are the same. The points (p, p, q) will not form a polar triangle if ^{t}pAp ≠ 0 or ^{t}pAq ≠ 0.

For the case ^{t}pAp ≠ 0 we pick coordinates such that A is the identity matrix and p = (1 : 0 : 0). Let

B = \{ \begin{matrix}(\begin{matrix}0 & −q_{2} & q_{1} \\ −q_{2} & 0 & 0 \\ q_{1} & 0 & 0 \end{matrix} ) & for q_{0} ≠ 0 or q_{1}^{2} + q_{2}^{2} ≠ 0 \\ ( \begin{matrix}0 & 0 & 1 \\ 0 & 1 & ±i \\ 1 & ±i & −1 \end{matrix} ) & for q = (0 : 1 : ±i), \end{matrix}

then C_{B} is a conic of B that contains p and q, but is smooth in p, and its tangent in p is not the polar of q, so it does not contain p twice.

Finally, if ^{t}pAp = 0 and ^{t}pAq ≠ 0, we pick coordinates such that A is the identity matrix and p = (1 : i: 0). Let

B = \{ \begin{matrix}( \begin{matrix}−2i & 2 & iq_{0} − 2q_{1} \\ 2 & 2i & −iq_{1} \\ iq_{0} − 2q_{1} & −iq_{1} & 0 \end{matrix} ) & for q = (q_{0} : q_{1} : 1) \\ ( \begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} ) & for q_{2} = 0, \end{matrix}

then we are in the same situation as above. □

References

[B] Bourbaki, N.: Eléments de Mathématique XXIV, Formes Sesquilinéaires et Formes Quadratiques. Herman, Paris 1959.

[BK] Brieskorn, E. & H. Knörrer: Plane Algebraic Curves. Birkhäuser, Basel 1986.

[Br] Brieskorn, E.: Lineare Algebra und analytische Geometrie II. Vieweg, Wiesbaden 1985.

[C] Castelnuovo, G.: Geometria della retta nello spazio a quattro dimensioni. Atti del R. Ist. Veneto (7), 2 (1891), p. 855-901.

[D] Donagi, R.: On the Geometry of Grassmannians. Duke Math. J. 44 (1977), p. 795-837.

[FH] Fulton, W. & J. Harris: Representation Theory. Springer, New York 1991.

[H] Harris, J.: Algebraic Geometry. Springer, New York 1992.

[M] Mumford, D.: Some Footnotes to the Work of C. P. Ramanujam. C. P. Ramanujam - A Tribute. Springer, Berlin 1978.

[R] Roth, L.: Some properties of Grassmannians. Rend. de Mat. e Appl. V 10 (1951), p. 96-114.

[SF] Salmon, G. & W. Fiedler: Analytische Geometrie der Kegelschnitte II. Teubner Verlag, Leipzig 1918.

[SR] Semple, J. & L. Roth: Introduction to Algebraic Geometry. Oxford University Press, Oxford 1985.

Mathematisches Institut IV Heinrich-Heine-Universität Universitätsstr. 1 40239 Düsseldorf Germany piontkow@uni-duesseldorf.de Faculteit der Wiskunde en Natuurwetenschappen Afdeling Wiskunde en Informatica Postbus 9215 2300 RA Leiden Netherlands VEN@rulwinw.leidenuniv.nl