THE VARIETY OF ANGLES

MARKUS ROST

                         preliminary version

Introduction

This text grew out of an attempt to understand the derived triangle construction, see Section 5. We consider the variety of (algebraic) angles of an Euclidean quadrangle and an action of S_{6} on it. This action is induced from an action on a 5-dimensional torus.

1. An action of S_{6}

Let T be the torus with the 6 characters

u_{ij} = u_{ji}: T → G_{m}, 0 ≤ i < j ≤ 3

as a base of Hom(T, G_{m}). Let U = T/d(G_{m}) where

d: G_{m} → T, d^{∗}(u_{ij}) = id

is the diagonal.

In the following ijkℓ stands for a permutation of 0123. The lattice Hom(U, G_{m}) is generated by the elements

α_{ijk} = u_{ij}^{−1}u_{jk}

One has

α_{ij0} = α_{0ji}^{−1}
α_{i0j} = α_{0ij}α_{ij0}
α_{ijk} = α_{0jk}α_{ij0}

It follows that Hom(U, G_{m}) is generated by the 6 elements

α_{0ij}, 1 ≤ i, j ≤ 3

They are subject to the relation

(1)  α_{012}α_{023}α_{031} = α_{013}α_{032}α_{021}

Since dim U = 5, this gives a complete presentation of U.

We consider the following automorphisms of U. Here the first three types are defined on T and factor to U.

• The elements g ∈ S_{4} which act by permutations of the indices 0123.

• The inverse ι with ι(u_{ij}) = u_{ij}^{−1}. One has

ι(α_{ijk}) = α_{kji}

• The involution τ with τ(u_{ij}) = u_{kℓ}. One has

τ(α_{ijk}) = α_{kℓi}

Date: March 23, 2004; last changes on January 18, 2007.

• The automorphism Φ of U defined on the generators α_{0ij} by

Φ(α_{0ij}) = α_{0jk}

It is easy to see that this assignment preserves the relation (1). It is clear that Φ^{3} = 1. In Section 5 we will describe a geometric interpretation of Φ.

Let G be the subgroup of Aut(U) generated by these elements.

One considers the sets of characters

M_{a} = {α_{ijk}},  α_{ijk} = u_{ij}^{−1}u_{jk}
M_{c} = {γ_{ijkℓ}},  γ_{ijkℓ} = u_{ij}^{−1}u_{kℓ}^{−1}u_{iℓ}u_{jk}
M = M_{a} ∪ M_{c}

One has |M_{a}| = 24 and |M_{c}| = 6 and therefore |M| = 30. (Later the sets M_{a}, M_{c} will be interpreted in terms of angles and cross ratios in a quadrangle, respectively.)

It is clear that the automorphisms ι, τ and g ∈ S_{4} preserve M_{a} and M_{c}. The element Φ preserves M. Indeed, one has

Φ(α_{0ij}) = α_{0jk}
Φ(α_{ij0}) = α_{ki0}
Φ(α_{ijk}) = Φ(α_{0jk}α_{ij0}) = α_{0ki}α_{ki0} = α_{k0i}
Φ(α_{i0j}) = Φ(α_{0ij}α_{ij0}) = α_{0jk}α_{ki0} = γ_{0jki}
Φ(γ_{0ijk}) = Φ(α_{0ij}α_{jk0}) = α_{0jk}α_{ij0} = α_{ijk}

In the following we denote by S_{3} ⊂ S_{4} the isotropy group of the index 0.

Lemma 1. One has the following relations in Aut(U):

• The element ι lies in the center of Aut(U). One has ι^{2} = 1.

• The element τ commutes with S_{4}. One has τ^{2} = 1.

• The element Φ commutes with S_{3}. One has Φ^{3} = 1.

• One has (τΦ)^{2} = 1.

• Let h = (01) ∈ S_{4} (the transposition). One has (hΦ^{−1}τΦ)^{3} = ι.

Proof. To see (τΦ)^{2} = 1 one observes

τΦ(α_{0ij}) = τ(α_{0jk}) = α_{ki0}
τΦ(α_{ki0}) = τ(α_{jk0}) = α_{0ij}

with ijk a permutation of 123 To see (hΦ^{−1}τΦ)^{3} = ι one observes first (with ij a permutation of 23)

hΦ^{−1}τΦ(α_{01i}) = hΦ^{−1}(α_{j10}) = h(α_{1i0}) = α_{0i1}
hΦ^{−1}τΦ(α_{0i1}) = hΦ^{−1}(α_{ji0}) = h(α_{i10}) = α_{i01}

Next note that τ leaves every γ_{ijkℓ} fixed and therefore Φ^{−1}τΦ leaves all elements α_{i0j} fixed. Hence

hΦ^{−1}τΦ(α_{i01}) = h(α_{i01}) = α_{i10} = α_{01i}^{−1}

Moreover

hΦ^{−1}τΦ(α_{0ij}) = hΦ^{−1}(α_{1i0}) = h(α_{ij0}) = α_{ij1}
hΦ^{−1}τΦ(α_{ij1}) = hΦ^{−1}τ(α_{10i}) = hΦ^{−1}(α_{ij1}) = hΦ^{−1}(α_{ij0}α_{0j1})
= h(α_{j10}α_{0ij}) = α_{j01}α_{1ij}
hΦ^{−1}τΦ(α_{j01}α_{1ij}) = hΦ^{−1}τ(α_{01i}α_{ij0}α_{j01}) = hΦ^{−1}τ(α_{ji1})
= hΦ^{−1}(α_{10j}) = h(α_{ji1}) = α_{ji0} = α_{0ij}^{−1}

(There might be a simpler way to present this using the relation with the root system A_{5}, see below). □

Corollary 1. The group G is isomorphic to S_{6} × Z/2, with S_{6} generated by S_{4}, τι and Φ and with Z/2 generated by ι.

Proof. This follows from Lemma 1 and, for instance, the known presentation of S_{6} as the Weyl group of A_{5}. □

It seems that the action of S_{6} on M is the standard action of S_{6} on the 2-element subsets of a 6-element set followed by a non-inner automorphism of S_{6}.

2. Relation with the root system A_{5}

Consider the set of characters on U:

A = \{\frac{u_{ij}}{u_{kl}} \} ∪ \{(\frac{u_{ij}u_{kl}}{u_{ik}u_{il}} )^{±1}\}

Here ijkℓ runs through the permutations of 0123.

The set A is also invariant under the group G, it has 6 + 24 = 30 elements and is in fact the root system A_{5}. A simple base of positive roots in the order of the Dynkin diagram is given by

\frac{u_{01}}{u_{23}} , \frac{u_{03}u_{12}}{u_{01}u_{02}} , \frac{u_{02}}{u_{13}} , \frac{u_{01}u_{23}}{u_{02}u_{03}} , \frac{u_{03}}{u_{12}}

The lattice generated by A has index 3 in the lattice generated by M. Thus U is the torus of SL(6)/μ_{2}.

Let us describe U ⊂ SL(6)/μ_{2} directly. Let

X = ker(G_{m}^{6} → G_{m}) ⊂ SL(6)

be the standard diagonal torus in SL(6) with coordinates (z_{1}, z_{2}, z_{3}, z_{4}, z_{5}, z_{6}) subject to z_{1}z_{2}z_{3}z_{4}z_{5}z_{6} = 1, z_{i} ≠ 0. One finds that

ρ: X → U
(z_{1}, z_{2}, z_{3}, z_{4}, z_{5}, z_{6}) ↦
[u_{01}, u_{02}, u_{03}, u_{23}, u_{31}, u_{12}] = [z_{2}, z_{1}z_{2}z_{4}, z_{5}^{−1}, z_{1}, z_{1}z_{2}z_{3}, z_{6}^{−1}]

is the corresponding map of tori. It is equivariant with respect to S_{6} ×Z/2 = G. (If I had known this earlier, I could have saved a lot of typing in the proof of Lemma 1.)

3. A cubic form in 6 variables

Let F be a ring, let K be a flat F -algebra of rank 3 and let L be a flat K-algebra of rank 2. Then Λ^{3}(K/F) is an invertible F -module. We have the cubic map

f = f_{L/K/F} : L → Λ^{3}(K/F)
f(x) = 1 ∧ T_{L/K}(x) ∧ N_{L/K}(x)

where T_{L/K}, N_{L/K}: L → K is the trace resp. norm.

Lemma 2.

f(x^{−1}) = \frac{−f(x)}{N_{L/F} (x)}

Proof. This can be seen from the explicit formulas below. □

For our later purposes it suffices to consider case K = F ×F ×F and L = K ×K. In this case we get a cubic form f in 6 variables, well defined up to a sign (depending on the choice of a generator of Λ^{3}(K/F)). It is obviously invariant under

Aut(L/K/F) = Z/2 × S_{4} = Z/2 ≀ S_{3}

One finds

f(x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}) = ∑_{i=1,2,3} x_{i}x_{i+1}(y_{i} − y_{i+1}) + y_{i}y_{i+1}(x_{i} − x_{i+1})

or

f(u_{01}, u_{02}, u_{03}, u_{23}, u_{31}, u_{12}) = − ∑_{g∈S_{4}/{1,(01)(23)}} sgn(g)g(u_{01}u_{02}u_{13})

If we compose this with the map ρ, then a direct calculation gives the following amazing identity:

f(z_{2}, z_{1}z_{2}z_{4}, z_{5}^{−1}, z_{1}, z_{1}z_{2}z_{3}, z_{6}^{−1}) =

−\frac{z_{1}z_{2}}{z_{5}z_{6}} (z_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} − z_{1}^{−1} − z_{2}^{−1} − z_{3}^{−1} − z_{4}^{−1} − z_{5}^{−1} − z_{6}^{−1})

Here (z_{1}, z_{2}, z_{3}, z_{4}, z_{5}, z_{6}) are subject to z_{1}z_{2}z_{3}z_{4}z_{5}z_{6} = 1, z_{i} ≠ 0.

This shows that f has a (S_{6} × Z/2)-symmetry. This will find an explanation later on.

Here is another way of writing the map ρ: The collection of the 12 elements z_{r}^{±1} is the collection of the 12 elements

\frac{u_{ij}u_{ik}u_{jℓ}}{t}

where t is subject to

t^{2} = ∏_{ij} u_{ij} (6 factors)

Since z_{r} ∈ S^{1}, one has z_{r}^{−1} = \overline{z}_{r}. Hence, if we write

z_{r} = cos φ_{r} + i sin φ_{r}

then

z_{r} − z_{r}^{−1} = 2i sin φ_{r}

and we get the following fundamental relation:

∑_{r=1}^{6} sin φ_{r} = 0

Here the φ_{r} are certain angles (well defined up to 2πZ and a common constant in πZ) associated to an Euclidean quadrangle. It would be desirable to have a geometric interpretation of this relation.

4. The fundamental relation for angles in a quadrangle

Let x_{0}, x_{1}, x_{2}, x_{3} ∈ C be four distinct points in the Euclidean plane. Define the quantities (i ≠ j)

u_{ij}(x) = u_{ji}(x) = \frac{x_{i} − x_{j}}{x̄_{i} − x̄_{j}}

where x̄ is the complex conjugate of x.

We call u_{ij}(x) the “algebraic side” given by x_{i}, x_{j}. We wish to understand which values u_{ij}(x) are possible. Note that the u_{ij}(x) are 6 complex numbers of norm 1. They are also invariant under translations of the quadrangle and under multiplication of the quadrangle with a real number. Therefore the (real) dimension of the space of possible u_{ij} is at most 8 − 2 − 1 = 5. It is in fact 5 and here is the corresponding basic equation in (S^{1})^{6}:

Lemma 3. Let L = C^{6} with coordinates u_{ij}, let τ be the involution of L with τ(u_{ij}) = u_{kℓ} and let K = L^{τ} be the invariant subalgebra. Then for every x ∈ C^{4} outside the general diagonal one has

f_{L/K/F} (u_{ij}(x)) = 0

Proof. Omitted. □

Lemma 4. Given 6 (generic) complex numbers u_{ij} ∈ S^{1} of norm 1 with

f_{L/K/F} (u_{ij}) = 0

Then there exists a quadrangle x ∈ C^{4} with u_{ij} = u_{ij}(x).

Proof. Omitted. □

These observations translate to similar considerations for the “algebraic angles”

α_{ijk}(x) = u_{ij}(x)^{−1}u_{jk}(x)

Corollary 2. Let U′ = (S^{1})^{6}/S^{1} be the quotient by the diagonal. Let further

V_{4} ⊂ U′

be the image of

{f = 0} ∩ (S^{1})^{6}

Then for a generic quadrangle x ∈ C^{4} the elements α_{ijk}(x) define a point α(x) ∈ V_{4} and x is uniquely determined by α(x) up to Euclidean similarities.

In other words, V_{4} is the variety of algebraic angles of an Euclidean quadrangle. It is also the moduli space of (general) Euclidean quadrangles with ordered vertices up to Euclidean similarities.

Or: An Euclidean quadrangle has 24 (algebraic) angles. The relations among them are given by the straightforward multiplicative relations and the form f.

For instance, suppose that for a quadrangle ABCD the vertices A, B, and the angles CAB, ABC, DAB, ABD are known. Then C, D are known as well. Using the form f one can express the angle ACD in terms of the given 4 angles. All the other angles of ABCD are then easy to compute.

To be more algebraic, let W_{4} → Spec Z be the subvariety of the torus U from Section 1 given by f = 0. The variety W_{4} is invariant under the involution ι. Now V_{4} is the set of real points of the twisted form of W_{4} with Gal(C/R) acting via ι.

5. The derived triangle

Cf. [1, p. 16, Section 1.5, Exercise 12]

For a generic quadrangle x = (x_{0}, x_{1}, x_{2}, x_{3}) ∈ C^{4} define

ϕ(x) = y = (y_{0}, y_{1}, y_{2}, y_{3}) ∈ C^{4}

as follows: y_{0} = x_{0} and y_{i} is the reflection of x_{0} at the side x_{j}x_{k} where ijk stands for a permutation of 123.

Then the angles x_{0}x_{i}x_{j} and y_{0}y_{j}y_{k} are equal (this is Coxeter’s exercise and left to the reader). It follows that the quadrangles x and ϕ^{3}(x) are similar.

Therefore ϕ defines an automorphism (possibly rational) of V_{4}. It turns out that ϕ is the restriction of Φ to V_{4}. The considerations of Section 1 yield therefore an action of G = Z/2 × S_{6} on V_{4}.

References

[1] H. S. M. Coxeter, Introduction to geometry, Wiley Classics Library, John Wiley & Sons Inc., New York, 1989, Reprint of the 1969 edition, 90a:51001.

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