THE METRIC OF A (n + 2)-GON IN AFFINE n-SPACE

MARKUS ROST

Contents

Summary 1 1. Preliminaries 2 1.1. Affine spaces 2 1.2. Symmetric bilinear forms 2 1.3. The orientation module of a finite set 2 1.4. M-gons 3 1.5. (n + 2)-gons 3 2. The main statements 4 2.1. The (dual) metric of a (n + 2)-gon 4 2.2. The metric of a (n + 2)-gon 5 2.3. Second presentation of Φ_{x} 7 2.4. Third presentation of Φ_{x} 8 2.5. The determinant of Φ_{x} 10 2.6. The dual (n + 2)-gon 10 2.7. The involution of a (n + 2)-gon 10 3. The case of a plane quadrangle 11 3.1. On the lines of a plane quadrangle 11 3.2. On the dual quadrangle 11 3.3. Selfdual quadrangles? 11 3.4. The determinant 11 3.5. Orthocentric quadrangles 12

Summary

For n + 2 points in affine n-space in general position there is a canonical metric (unique up to a similarity factor) such that complementary faces of the (n + 2)-gon are orthogonal.

We describe this metric in terms of a sum over all (n − 1)-dimensional faces (see Proposition 2) and discuss some of its properties.

Date: August 25, 2004.

1. Preliminaries

1.1. Affine spaces. An affine space over a field F consists of a set A, a vector space V over F and an operation

A × V → A
(a, v) ↦ a + v

which makes A into a principal homogeneous V -space.

By a dilation of A we understand an automorphism of A whose linear part (in GL(V)) is a scalar multiple of the identity.

Affine spaces can be presented as follows. Let W be a vector space over F and let

ε: W → F

be an epimorphism. Then A = ε^{−1}(1) is an affine space with underlying vector space V = ε^{−1}(0). The pair (W, ε) is uniquely determined by (A, V, +) up to unique isomorphism.

There are natural exact sequences

0 → Λ^{i+1}V ⟶^{ι} Λ^{i+1}W ⟶^{ε_{i}} Λ^{i}V → 0

where ε_{i} is characterized by

ε_{i}(a ∧ ω) = ω

with a ∈ A and ω ∈ Λ^{i}V.

1.2. Symmetric bilinear forms. Let V be a finite-dimensional vector space and let L be a 1-dimensional vector space. We consider symmetric bilinear maps

Φ: V × V → L

For dual vector spaces we use the notation V^{∨} = Hom(V, F).

The map

Φ̂: V → Hom(V, L) = V^{∨} ⊗ L
Φ̂(v)(w) = Φ(v, w)

is called the duality associated to Φ.

Let n = dim V. The determinant of Φ is defined as

det(Φ) = Λ^{n}Φ̂ ∈ Hom(Λ^{n}V, Λ^{n}(V^{∨} ⊗ L)) = (Λ^{n}V)^{⊗−2} ⊗ L^{⊗n}

1.3. The orientation module of a finite set. Let M be a finite set. The orientation module of M is defined as

O_{M} = Λ^{|M|}Z[M]

where Z[M] is the free abelian group on M. The group O_{M} is free of rank 1 and the natural action of the group of permutations of M on O_{M} is given by the signum. Clearly O_{M} ⊗ O_{M} ≡ Z.

1.4. M-gons. Let A be an affine space (with notations V, ε: W → F as above) and let M be a finite set. By an M-gon in A we understand a map

x: M → A
i ↦ x_{i}

By a r-gon we understand a M-gon for some set M with |M| = r, usually M = {1, …, r}.

For v ∈ V we denote by x + v the translated M-gon, defined by

(x + v)_{i} = x_{i} + v

with i ∈ M.

For a M-gon x ∈ A^{M} and a subset I ⊂ M we denote by

A_{I}(x) ⊂ A

the affine span of the points x_{i} with i ∈ I. Its underlying vector space

V_{I}(x) ⊂ V

is generated by the elements x_{i} − x_{j} with i, j ∈ I.

We denote by

W^{M} → Λ^{M}W
x ↦ ∧ x

be the universal alternating map. Note that

Λ^{M}W ≡ Λ^{|M|}W ⊗ O_{M}

1.5. (n + 2)-gons. Let n be an integer, let A be an affine space with dim A = n, let M be a set with |M| = n + 2 and let x ∈ A^{M} be a M-gon in A. For i ∈ M we define the element

θ_{i}(x) ∈ Λ^{n}V ⊗ O_{M}

by

θ_{i}(x) = ε_{n} (x_{σ(1)} ∧ ⋯ ∧ x_{σ(n+1)}) ⊗ (i ∧ σ(1) ∧ ⋯ ∧ σ(n + 1))

where

σ: {1, …, n + 1} → M \ {i}

is any bijection.

The element θ_{i}(x) has the standard interpretation of a volume element for the i-th face of x. It is invariant under translations:

θ_{i}(x + v) = θ_{i}(x)

This can be easily deduced for instance from

x_{1} ∧ ⋯ ∧ x_{r} = x_{1} ∧ (x_{2} − x_{1}) ∧ ⋯ ∧ (x_{r} − x_{1})

with x_{i} ∈ A.

We say that x is nondegenerate if θ_{i}(x) ≠ 0 for all i ∈ M.

Lemma 1. Let dim A = n, let M be a set with |M| = n + 2 and let x ∈ A^{M} be a M-gon in A. Then

∑_{i∈M} θ_{i}(x)x_{i} = 0

in Λ^{n}V ⊗ O_{M} ⊗ W.

In particular, by applying ε, one gets

∑_{i∈M} θ_{i}(x) = 0

in Λ^{n}V ⊗ O_{M}.

Proof. Basic multilinear algebra: For x_{i} ∈ W the expression

∑_{i=1}^{n+2}(−1)^{i}(x_{1} ∧ ⋯ x̂_{i} ⋯ ∧ x_{n+2})x_{i}

is alternating in the x_{i}. Since n + 2 > dim W, it vanishes. □

For a subset I ⊂ M we write

θ_{I}(x) = ∏_{i∈I} θ_{i}(x) ∈ (Λ^{n}V ⊗ O_{M})^{⊗|I|}

Moreover for i ∈ M we write

ρ_{i}(x) = θ_{M\{i}}(x)

2. The main statements

Let n = dim A, let M be a set with |M| = n + 2 and let x ∈ A^{M} be a M-gon in A.

Notations V, ε: W → F are as in the previous section.

2.1. The (dual) metric of a (n + 2)-gon. This subsection contains a simple definition of the metric. I found it only after typing the other parts of the text, which turned out to be much more complicated than necessary.

Let

L = Λ^{n}V ⊗ O_{M}

Let a, b ∈ A. One considers the tensor

Ω_{x} = ∑_{i∈M}(x_{i} − a) ⊗ (x_{i} − b) ⊗ θ_{i}(x) ∈ V ⊗ V ⊗ L

Proposition 1. (1) The element Ω_{x} does not depend on the choice of a, b ∈ A. (2) The element Ω_{x} is invariant under switch involution on V ⊗ V.

(3) Let I ⊂ M and let f ∈ V^{∨}. If f(V_{I}(x)) = 0, then

(f ⊗ id_{V} ⊗ id_{L})(Ω_{x}) ∈ V_{M\I}(x) ⊗ L

Proof. (1) follows from ∑_{i} θ_{i}(x)x_{i} = 0 (cf. Lemma 1). (2) is obvious. As for (3), we may assume I ≠ ∅, M. Choose a ∈ A_{I}(x) and b ∈ A_{M\I}(x). Then

(f ⊗ id_{V} ⊗ id_{L})(Ω_{x}) = ∑_{i∈M\I} f(x_{i} − a)(x_{i} − b) ⊗ θ_{i}(x)

□

The tensor Ω_{x} defines a symmetric duality

Ω̂_{x}: V^{∨} → V ⊗ L

Consider its (n − 1)-fold exterior power

Λ^{n−1}Ω̂_{x}: Λ^{n−1}V^{∨} → Λ^{n−1}V ⊗ L^{⊗(n−1)}

Since Λ^{n−1}V = V^{∨} ⊗ Λ^{n}V, it defines a symmetric bilinear form

Φ_{x}: V × V → L^{⊗(n+1)}

We call the form Φ_{x} the metric of x.

In the following we give some other descriptions.

2.2. The metric of a (n + 2)-gon. We consider the following symmetric bilinear map on Λ^{2}W with values in an appropriate 1-dimensional vector space:

φ_{x}: Λ^{2}W × Λ^{2}W → (Λ^{n}V ⊗ O_{M})^{⊗(n+1)}

φ_{x}(α, β) = ∑_{{I⊂M \atop |I|=n−1}} θ_{I}(x)ε_{n}(α ∧ (∧ x|_{I}))ε_{n}(β ∧ (∧ x|_{I}))

This is to be read as follows. The product α ∧ (∧ x|_{I}) is an element of

Λ^{2}W ∧ Λ^{I}W = Λ^{2}W ∧ Λ^{|I|}W ⊗ O_{I} = Λ^{n+1}W ⊗ O_{I}

Thus ε_{n}(α ∧ (∧ x|_{I})) is an element of Λ^{n}V ⊗ O_{I} and since O_{I} ⊗ O_{I} ≡ Z we have

ε_{n}(α ∧ (∧ x|_{I}))ε_{n}(β ∧ (∧ x|_{I})) ∈ (Λ^{n}V )^{⊗2} = (Λ^{n}V ⊗ O_{M})^{⊗2}

Lemma 2. If α ∈ Λ^{2}V or β ∈ Λ^{2}V, then φ_{x}(α, β) = 0.

Proof. Suppose α ∈ Λ^{2}V. Fix a ∈ A.

For z ∈ A^{n−1} one has

ε_{n}(α ∧ (∧ z)) = ε_{n}(α ∧ z_{1} ∧ ⋯ ∧ z_{n−1})

= ε_{n}(α ∧ z_{1} ∧ (z_{2} − z_{1}) ∧ ⋯ ∧ (z_{n−1} − z_{1}))

= ε_{n}(α ∧ a ∧ (z_{2} − z_{1}) ∧ ⋯ ∧ (z_{n−1} − z_{1}))

= ∑_{i=1}^{n−1}(−1)^{i+1}ε_{n}(α ∧ a ∧ z_{1} ∧ ⋯ ẑ_{i} ⋯ ∧ z_{n−1})

Moreover

ε_{n}(β ∧ (∧ z)) = (−1)^{i+1}ε_{n}(β ∧ z_{i} ∧ z_{1} ∧ ⋯ ẑ_{i} ⋯ ∧ z_{n−1})

Hence

ε_{n}(α ∧ (∧ z))ε_{n}(β ∧ (∧ z)) =

∑_{i=1}^{n−1} ε_{n}(α ∧ a ∧ z_{1} ∧ ⋯ ẑ_{i} ⋯ ∧ z_{n−1})ε_{n}(β ∧ z_{i} ∧ z_{1} ∧ ⋯ ẑ_{i} ⋯ ∧ z_{n−1})

This shows that

ε_{n}(α ∧ (∧ x|_{I}))ε_{n}(β ∧ (∧ x|_{I})) =

∑_{{i∈I \atop K=I\\{i\}}} ε_{n}(α ∧ a ∧ (∧ x|_{K}))ε_{n}(β ∧ x_{i} ∧ (∧ x|_{K}))

We get

φ_{x}(α, β) = ∑_{{I⊂M \atop |I|=n−1}} θ_{I}(x)ε_{n}(α ∧ (∧ x|_{I}))ε_{n}(β ∧ (∧ x|_{I}))

= ∑_{{K⊂M \atop |K|=n−2}} ∑_{i∈M\K} θ_{I}(x)ε_{n}(α ∧ a ∧ (∧ x|_{K}))ε_{n}(β ∧ x_{i} ∧ (∧ x|_{K}))

For i ∈ K one has x_{i} ∧ (∧ x|_{K}) = 0. Hence we may extend the range of i to all of M and get

φ_{x}(α, β) = ∑_{{K⊂M \atop |K|=n−2}} ∑_{i∈M} θ_{K}(x)θ_{i}(x)ε_{n}(α ∧ a ∧ (∧ x|_{K}))ε_{n}(β ∧ x_{i} ∧ (∧ x|_{K}))

This vanishes because of ∑_{i} θ_{i}(x)x_{i} = 0 (cf. Lemma 1). □

By Lemma 2, the form φ_{x} is essentially a form on Λ^{2}W/Λ^{2}V ≃ V. We describe this as follows:

Proposition 2. Let a, b ∈ A. The form

Φ_{x}: V × V → (Λ^{n}V ⊗ O_{M})^{⊗(n+1)}

Φ_{x}(v, w) = ∑_{{I⊂M \atop |I|=n−1}} θ_{I}(x)ε_{n}(v ∧ a ∧ (∧ x|_{I}))ε_{n}(w ∧ b ∧ (∧ x|_{I}))

does not depend on the choices of a and b.

We call the form Φ_{x} the metric of x.

Lemma 3. Let i, j, k, ℓ ∈ M be distinct elements. Then

(1)  Φ_{x}(x_{i} − x_{j}, x_{k} − x_{ℓ}) = 0
(2)  Φ_{x}(x_{i} − x_{j}, x_{i} − x_{k}) = −ρ_{i}(x)
(3)  Φ_{x}(x_{i} − x_{j}, x_{i} − x_{j}) = −ρ_{i}(x) − ρ_{j}(x)

Proof. It is easy to see that (1) and (3) follow from (2).

As for (2) we choose a = b = x_{i} in the definition of Φ_{x}. Then

Φ_{x}(x_{i} − x_{j}, x_{i} − x_{k}) = ∑_{{I⊂M \atop |I|=n−1}} θ_{I}(x)ε_{n}(−x_{j} ∧ x_{i} ∧ (∧ x|_{I}))ε_{n}(−x_{k} ∧ x_{i} ∧ (∧ x|_{I}))

Every summand vanishes except for I = M \ {i, j, k} and for this term the last two factors amount to −θ_{j}(x)θ_{k}(x). This shows (2). □

Remark 1. One may check (3) also as follows. We have with a = b = x_{j}

Φ_{x}(x_{i} − x_{j}, x_{i} − x_{j}) = ∑_{{I⊂M \atop |I|=n−1}} θ_{I}(x)ε_{n}(x_{i} ∧ x_{j} ∧ (∧ x|_{I}))^{2}

THE METRIC OF A (n + 2)-GON IN AFFINE n-SPACE 7

Here every summand vanishes except for I = M \ {i, j, h} with h ≠ i, j. Hence

Φ_{x}(x_{i} − x_{j}, x_{i} − x_{j}) = ∑_{{h∈M\\{i,j\} \atop I=M\\{i,j,h\}}} θ_{I}(x)ε_{n}(x_{i} ∧ x_{j} ∧ (∧ x|_{I}))^{2}

= ∑_{{h∈M\\{i,j\} \atop I=M\\{i,j,h\}}} θ_{I}(x)θ_{h}(x)^{2}

= θ_{M\{i,j}}(x) ∑_{h∈M\{i,j}} θ_{h}(x)

= θ_{M\{i,j}}(x)(−θ_{i}(x) − θ_{j}(x))

Here we have used ∑_{i} θ_{i}(x) = 0 (cf. Lemma 1). Claim (3) is now immediate. Remark 2. Condition (1) in Lemma 3 is equivalent to

V_{I}(x) ⊥_{Φ_{x}} V_{M\I}(x)

for all subsets I ⊂ M.

Suppose x is nondegenerate. Then

V = V_{I}(x) + V_{M\I}(x)

for all subsets I ⊂ M. Moreover the form Φ_{x} is determined by (1) in Lemma 3 up to multiplication by a scalar. Remark 3. It is clear (for nondegenerate x) that if two lines of the M-gon are parallel, then its metric is isotropic. 2.3. Second presentation of Φ_{x}. Fix h ∈ M and let N = M \ {h}. Then |N| = n + 1. We assume that the family (x_{i})_{i∈N} is a basis for W.

Then there exists a symmetric bilinear map

Ψ_{h}: W × W → (Λ^{n}V ⊗ O_{M})^{⊗(n+1)}

with

Ψ_{h}(x_{i}, x_{j}) = 0
Ψ_{h}(x_{i}, x_{i}) = −ρ_{i}(x)

for i, j ∈ N, i ≠ j. Lemma 4. One has

Ψ_{h}(x_{i}, x_{h}) = ρ_{h}(x)

for i ∈ M. Proof. Indeed,

Ψ_{h}(x_{i}, x_{h}) = Ψ_{h}(x_{i}, −θ_{h}(x)^{−1} ∑_{j∈N} θ_{j}(x)x_{j})
= −θ_{h}(x)^{−1}θ_{i}(x)(−ρ_{i}(x)) = ρ_{h}(x)

and

Ψ_{h}(x_{h}, x_{h}) = Ψ_{h}(x_{h}, −θ_{h}(x)^{−1} ∑_{j∈N} θ_{j}(x)x_{j})

= −θ_{h}(x)^{−1}ρ_{h}(x) ∑_{j∈N} θ_{j}(x) = ρ_{h}(x)

□

Corollary. The form Φ_{x} is the restriction of Ψ_{h} to V. Moreover

Ψ_{h}(V, x_{h}) = 0

and

(4)  W = V ⊕ x_{h}F

is an orthogonal decomposition with respect to Ψ_{h}.

2.4. Third presentation of Φ_{x}. Let |M| = n + 2 and let

a ∈ (F^{×})^{M}

be a M-family of invertible elements in F with

∑_{i∈M} a_{i} = 0

Let U = F [M] be the vector space with basis e_{i}, i ∈ M. Let

Ψ_{a}: U × U → F

be the symmetric bilinear form with

Ψ_{a}(e_{i}, e_{j}) = 0
Ψ_{a}(e_{i}, e_{i}) = a_{i}^{−1}

for i, j ∈ M, i ≠ j.

The vector

z = ∑_{i∈M} a_{i}e_{i} ∈ U

is isotropic. Let us denote by [z] ⊂ U the subspace generated by z. Note that Ψ_{a}(e_{i}, z) = 1 for i ∈ M. Hence for i, j ∈ M one has e_{i} − e_{j} ∈ [z]^{⊥}. Since z ≠ 0, these elements generate [z]^{⊥}.

Now put

W_{a} = U/[z]
V_{a} = [z]^{⊥}/[z]

Further let

ε: W_{a} → F
ε(u + [z])) = Ψ_{a}(u, z)

and

A_{a} = ε^{−1}(1)

Then A_{a} is a n-dimensional affine space with underlying vector space V_{a}. Moreover,

x: M → A_{a}
x_{i} = e_{i} + [z]

defines a M-gon x in A_{a}.

Let

Φ_{a}: V_{a} × V_{a} → F
Φ_{a}(u + [z], u′ + [z]) = Ψ_{a}(u, u′)

be the canonical form associated with Ψ_{a} and the isotropic vector z.

There is a canonical identification

Λ^{n}V_{a} ⊗ O_{M} ≡ [z] ⊗ U/[z]^{⊥} ⊗ Λ^{n}V_{a} ⊗ O_{M} ≡ Λ^{n+2}U ⊗ O_{M} ≡ F

given by

\overline{α} ⊗ (∧σ) ↦ z ⊗ e_{i} ⊗ \overline{α} ⊗ (∧σ) ↦ (z ∧ e_{i} ∧ α) ⊗ (∧σ), (∧σ) ⊗ (∧σ) ↦ 1

with α ∈ Λ^{n}([z]^{⊥}) and σ: {1, …, n + 2} → M a bijection.

With respect to this identification, one has

θ_{i}(x) = a_{i}

and

(5)  (−a_{1} ⋯ a_{n+2})^{−1}Φ_{x} ⊥ H = Ψ_{a}

where H is a hyperbolic plane.

It is easy to see that every nondegenerate M-gon x in a n-dimensional affine space appears in this way from some

a ∈ (F^{×})^{M}

with

∑_{i∈M} a_{i} = 0

Remark 4. For nondegenerate x this gives a very simple way to define Φ_{x}. The first definition of Φ_{x} via a sum over all (n − 1)-dimensional faces works smoothly for all x and has its own appeal anyway. I don’t know an urgent reason to consider the description of Φ_{x} via the form Ψ_{h}, h ∈ M—I used it at first to compute the determinant of Φ_{x}.

Remark 5. Suppose char F ≠ 2. Then a n-dimensional quadratic form Φ appears as Φ_{x} for some x if and only if

Φ ⊥ ⟨1, −1⟩ ≃ −a_{1} ⋯ a_{n+2}⟨a_{1}, …, a_{n+2}⟩

for some a_{i} ∈ F^{×} with ∑_{i=1}^{n+2} a_{i} = 0. From this one sees that every similarity class of a n-dimensional quadratic form appears as the similarity class of the metric of a (n + 2)-gon.

Remark 6. One may also consider twisted forms of (n+2)-gons. The setup would be to consider an etale algebra H of rank n + 2 and a point x: Spec H → A. For nondegenerate x the quadratic form Φ_{x} would be of the form

Φ_{x} ⊥ ⟨1, −1⟩ ≃ −N_{H/F} (a)T_{H/F} (⟨a⟩)

for some a ∈ H^{×} with T_{H/F} (a) = 0.

2.5. The determinant of Φ_{x}. Here is the computation:

Lemma 5.

det(Φ_{x}) = (−θ_{M}(x))^{(n−1)} ∈ (Λ^{n}V)^{⊗(n+2)(n−1)}

In particular we see that Φ_{x} is nondegenerate if and only if x is nondegenerate.

Proof. Since we have to check a polynomial identity in x, we may assume that x is nondegenerate. Then one may use the description of Φ_{x} in (5). But one may also use the orthogonal decomposition (4) which shows

det(Ψ_{h}) = det(Φ_{x})Ψ_{h}(x_{h}, x_{h})

One has

det(Ψ_{h}) = θ_{h}(x)^{−2} ∏_{i∈N}(−ρ_{i}(x)) = (−θ_{M}(x))^{n−1}ρ_{h}(x)

and we are done by Ψ_{h}(x_{h}, x_{h}) = ρ_{h}(x) (cf. Lemma 4). □

2.6. The dual (n + 2)-gon. We assume that x is nondegenerate. Fix a ∈ A and a basis element λ for (Λ^{n}V ⊗ O_{M})^{⊗(n+1)}. Then we get a M-gon

y: M → V^{∨}
y_{i} = Φ̂_{x}(x_{i} − a)/λ

in the dual space V^{∨}. We call it a dual M-gon of x. Dual M-gons of x are determined by x up to dilations (translations and scalar multiplications). They are characterized by

⟨y_{i} − y_{j}, x_{k} − x_{ℓ}⟩ = 0

where i, j, k, ℓ ∈ M are distinct elements and where ⟨ , ⟩ denotes the natural pairing V^{∨} × V → F.

Dual M-gons determine the metric up to multiplication by a scalar.

2.7. The involution of a (n + 2)-gon.

Proposition 3. Let n = dim A, let M be a set with |M| = n + 2 and let x ∈ A^{M} be a M-gon in A. Suppose that x is nondegenerate. Then there exists a unique involution τ_{x} of orthogonal type on End(V) such that

τ_{x}(Hom(V, V_{I}(x))) = Hom(V/V_{M\I}(x), V))

for each subset I ⊂ M.

It is clear that τ_{x} = τ_{y} if x, y differ only by a dilation of A.

Proof. τ_{x} is the involution associated with the symmetric bilinear form Φ_{x}, i. e.,

τ_{x}(Φ_{x}(v) ⊗ v′) = Φ_{x}(v′) ⊗ v

□

Remark 7. Suppose char F = 2, that n is even and that θ_{i}(x) = θ_{j}(x) ≠ 0 for all i, j ∈ M. Then

∑_{i∈M} x_{i} = 0

One finds that there is a unique alternating bilinear form Ω: V × V → F with

Ω(x_{i} − x_{j}, x_{i} − x_{k}) = 1

for distinct elements i, j, k ∈ M.

This is the only case where an involution τ on End(V) with

τ_{x}(Hom(V, V_{I}(x))) = Hom(V/V_{M\I}(x), V))

for each subset I ⊂ M is possibly symplectic.

3. The case of a plane quadrangle

We now look at the case dim A = 2 and |M| = 4.

In this case V = V^{∨} ⊗ Λ^{2}V and the duality Φ̂_{x} becomes a map

Φ̂: V → V^{∨} ⊗ (Λ^{2}V ⊗ O_{M})^{⊗3} = V ⊗ (Λ^{2}V)^{⊗2} ⊗ O_{M}

3.1. On the lines of a plane quadrangle. We assume that x is nondegenerate. Then Φ̂ is an isomorphism and induces an involution

σ_{x}: P(V) → P(V)

on the projective space of lines in V. It has the following interpretation: Let M = {i, j, k, h}. Then

σ_{x}([x_{i} − x_{j}]) = [x_{k} − x_{h}]

One may phrase this by saying that “the 6 lines of a plane quadrangle stand in involution”. The converse is also true: If 6 points in P^{1} stand in involution, they are given by the lines of a quadrangle. (Note: An involution of P^{1} is determined by two pairs of points.)

3.2. On the dual quadrangle. Since V = V^{∨} ⊗ Λ^{2}V we may also speak about dual M-gons in V. They can be described as follows:

Given four general points x_{1}, x_{2}, x_{3}, x_{4} in a 2-dimensional affine space, there exists another sequence of four points y_{1}, y_{2}, y_{3}, y_{4} such that the line x_{i} − x_{j} is parallel to the line y_{k}−y_{h} for any permutation ijkh of 1234. The y-tuple is uniquely determined by the x-tuple up to translation and scalar multiplication.

3.3. Selfdual quadrangles? It turns out that a nondegenerate quadrangle is never dual to itself (in characteristic different from 2).

What about nondegenerate quadrangles which become dual to itself after a permutation? It turns out that then there exists one side x_{i} − x_{j} which is parallel to its opposite side x_{k} − x_{h} and the permutation is (ij)(kh). More specifically, let A = V, let v, w ∈ V be linearly independent and let c ∈ F^{×}. Then the quadrangle (0, v, w, w + cv) is dual to (v, 0, w + cv, w).

3.4. The determinant. Let x_{1}, x_{2}, x_{3}, x_{4} be a nondegenerate plane quadrangle and let a_{i} ∈ F^{×} with

a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3} + a_{4}x_{4} = 0

Then

det(Φ_{x}) = −a_{1}a_{2}a_{3}a_{4}

up to multiplication by a square.

Consider the case of a parallelogram. (This amounts to a_{1} = a_{2} = −a_{3} = −a_{4}.) In this case the metric is hyperbolic; the two isotropic lines are given by the pairs of parallel sides.

Suppose that F = R (real numbers). The Φ_{x} is definite if and only if one of the points x_{i} lies inside the triangle formed by the other points x_{j}, x_{k}, x_{ℓ}.

3.5. Orthocentric quadrangles. Let us consider the case F = R. Let x be a nondegenerate plane quadrangle and suppose that the metric Φ_{x} is definite. Then we have an Euclidean structure on A. With respect to this Euclidean structure, the quadrangle x is orthocentric, i. e., each point x_{i} is the orthocenter of the opposite triangle x_{j}, x_{k}, x_{ℓ}. A dual quadrangle is obtained from x by a rotation of 90^{∘}.

Fakultät für Mathematik, Universität Bielefeld, Postfach 100131, 33501 Bielefeld, Germany

   E-mail address: rost@mathematik.uni-bielefeld.de
   URL: http://www.mathematik.uni-bielefeld.de/~rost