THE HOLOMORPHIC EXTENSION OF TRIANGLE FUNCTIONS

MARKUS ROST

Introduction

Examples of triangle functions are given by the orthocenter, the circumcenter, the incenter, the excenters, the pedal points, the nine-point center, etc. Each of these functions has a “holomorphic extension” which is a complex function in four variables.

1. Triangle functions

Let E be the Euclidean plane.

Let further G be the group of Euclidean similarities of E. If we identify E with the field of complex numbers C, then G consists of the affine transformations x ↦ ax + b with a ≠ 0.

We let G act on E^{r} diagonally.

Definition. By a triangle function we understand a G-equivariant real analytic function

f: U → E

defined on an open G-stable subset U ⊂ E^{3}.

Remark. One could require additionally that a triangle function is also equivariant with respect to reflections at a line. This is indeed the case for most of our examples.

A basic example of such a function is given by the orthocenter^{1} H(x, y, z) of a triangle x, y, z. The function H(x, y, z) is defined (at least) on the subset U_{H} of triples x, y, z ∈ E not lying on a line and with no perpendicular sides. It is invariant under permutations of the coordinates and one has

H(x, y, H(x, y, z)) = z

Let

Φ: U_{H} → E^{4}
Φ(x, y, z) = (x, y, z, H(x, y, z))

Let X be the image of Φ. X is the set of (non-degenerate) orthocentric quadrangles. It is G-stable and invariant under permutations of the coordinates of E^{4}. It is a smooth 6-dimensional real analytic subvariety of E^{4}.

Date: August 6, 2004.

1The orthocenter is the intersection of the altitudes; see [3, Section 1] for orthocenters, orthocentric quadrangles, pedal triangles, etc.

Lemma 1. Let

f: U → E

be a triangle function with U ⊂ U_{H}. Then there exists an open G-stable subset V ⊂ E^{4} with Φ(U) ⊂ V and a G-equivariant complex analytic function

f̂: V → E

such that

f = f̂∘ Φ

The function f̂ is uniquely determined by f on every component of V containing points of Φ(U).

The function f̂ is called the holomorphic extension of f. In all our examples the function f̂(x, y, z, t) is algebraic over C(x, y, z, t) (in fact, algebraic over Q(x, y, z, t)).

The basic idea is that the subvariety X ⊂ E^{4} = C^{4} has 2 complex coordinates, given by the G-action, and 2 further real coordinates. The complexification of the real coordinates will lead to 4 complex coordinates in total.

Conversely, let f̂: V → E be a G-equivariant complex analytic function. Then f|(X ∩ V) or rather f = f̂∘ Φ is a triangle function. When we consider f̂(x, y, z, t) on points (x, y, z, t) ∈ X, we speak of the “Euclidean case”.

The following proof of the Lemma is also a device to construct f̂.

Proof. We make an identification E = C. Because of the G-equivariance, it suffices to consider the restrictions to the slices defined by x = 0, y = 1. Let g(z) = f(0, 1, z) and ϕ(z) = (z, H(0, 1, z)). It suffices to find a complex analytic function ĝ(z, t) with g = ĝ ∘ ϕ.

Let G(z, s) be a complex analytic function with g(z) = G(z, z̄) where z̄ is the complex conjugate.

The orthocenter H of the triangle 0, 1, z is determined by H ⊥ (1 − z) and (1 − H) ⊥ z. One finds

H(0, 1, z) = \frac{(1 − z)(z + z̄)}{z̄ − z}

The function

t = \frac{(1 − z)(z + s)}{s − z}

is a linear fractional function in s. Its inverse is

s = \frac{zt + (1 − z)z}{t + z − 1}

Thus

ĝ(z, t) = G(z, \frac{zt + (1 − z)z}{t + z − 1} )

does the job. □

2. Examples

Example 1. For the orthocenter H(x, y, z) one has obviously

Ĥ(x, y, z, t) = t

Example 2. For the centroid (center of gravity) G(x, y, z) one has

Ĝ(x, y, z, t) = \frac{x + y + z}{3}

Example 3. For the circumcenter (center of circumcircle) O(x, y, z) one has

3G = 2O + H

(the line through these three points is the Euler line). Hence

Ô(x, y, z, t) = \frac{x + y + z − t}{2}

Example 4. Let x_{0}, x_{1}, x_{2}, x_{3} be an orthocentric quadruple. The nine-point circle (also called Euler circle or Feuerbach circle) is the circumcircle of the pedal triangle (which is formed by the bases of the altitudes). It contains all the midpoints m_{ij} of the sides x_{i}x_{j} and m_{ij}m_{kℓ} are diameters of it (ijkℓ is any permutation of 1234).

The nine-point center F (x, y, z) is the center of the Euler circle. It lies one the Euler line and one has

2F = O + H

Hence

F̂(x, y, z, t) = \frac{x + y + z + t}{4}

Example 5 (Pedal points). For a triangle x, y, z let P (x, y, z) be the base of the altitude through z. Note that

P (x, y, z) = P (y, x, z) = P (x, y, H(x, y, z))

Therefore one has for the holomorphic extension P̂ of f the relations

P̂(x, y, z, t) = P̂(y, x, z, t) = P̂(x, y, t, z) = P̂(y, x, t, z)

It follows that the triple

R(x_{0}, x_{1}, x_{2}, x_{3}) = (P̂(x_{0}, x_{1}, x_{2}, x_{3}), P̂(x_{0}, x_{2}, x_{3}, x_{1}), P̂(x_{0}, x_{3}, x_{1}, x_{2}))

is equivariant with respect to the natural operations of S_{4} and S_{3} and the corresponding homomorphism S_{4} → S_{3}.

As for an explicit computation: One first notes that

P (0, 1, z) = \frac{z + z̄}{2}

Then one can use the proof of the Lemma to compute P̂. One finds

P̂(x, y, z, t) = \frac{xy − zt}{x + y − z − t}

Remark 1. The map R is a triple of rational functions in 4 variables defined over any field F, equivariant with respect to S_{4} → S_{3} and with respect to the action of the affine group Aff(1, F). It is my favorite candidate for the construction of a cubic resolvent of a quartic equation. After the previous remarks, it is justified to say that R is the holomorphic extension of the construction of the pedal triangle of an orthocentric quadrangle.

Example 6. Let S(x, y, z) denote the reflection of z along the side xy. Obviously

S(x, y, z) = z + 2(P (x, y, z) − z)

One finds

Ŝ(x, y, z, t) = z + \frac{(z − x)(z − y)}{Ô(x, y, z, t) − z}

Remark 2. Let us consider the denominators of the pedal points. Let ijkℓ be a permutation of 1234 and let

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

In the Euclidean case (that is, x_{0}, x_{1}, x_{2}, x_{3} is an orthocentric quadrangle) one has

(x_{i} − x_{j}) ⊥ (x_{k} − x_{ℓ})

It follows that in this case

r = |ρ_{ij}(x_{0}, x_{1}, x_{2}, x_{3})|

is independent of ij (| · | is the Euclidean absolute value). In fact,

r = 2r_{O} = 4r_{F}

where r_{O} the radius of the circumcircle of any subtriangle and r_{F} is the radius of the Euler circle. It follows also that the quotient

\frac{ρ_{ij}(x_{0}, x_{1}, x_{2}, x_{3})}{ρ_{ik}(x_{0}, x_{1}, x_{2}, x_{3})}

is a complex number of norm 1. Its angle is twice the angle of the triangle x_{j}, x_{k}, x_{ℓ} at x_{ℓ}.

Example 7. [Incenters and Excenters] For a triangle x, y, z let I(x, y, z) be its incenter (intersection of the angle bisectors). The holomorphic extension Î has globally 4-branches with covering group Z/2 × Z/2. There is of course the distinguished branch which extends I in an open neighborhood of Φ(U). The other three branches, obtained by analytic continuation, give by restriction via Φ the excenters of a triangle. Explicit formulas can be found in [6, End of Section 4].

Remark 3. Let I be the incenter, let J_{1}, J_{2}, J_{3} be the excenters, and let O be the circumcenter of a triangle. One has

I + J_{1} + J_{2} + J_{3} = 4O

I don’t know an explicit reference for this basic fact, however see [3, Section 1.7, Exercise 5].

Example 8. Let M(x, y, z) be the intersection of the angle trisectors at x, y adjacent to the side xy. In this case one finds that M̂ has globally 9-branches with covering group Z/3 × Z/3. Explicit formulas can be found in [6].

Remark 4. Connes’ article [2] was the starting point for the considerations of this text. We get the following interpretation: Morley’s theorem states that M(x, y, z) and its permuted versions form an equilateral triangle, that is

M(x, y, z) + ζM(y, z, x) + ζ^{2}M(z, x, y) = 0

where ζ is an appropriate cube root of unity. Now Connes’ Lemma means essentially the corresponding identity

M̂(x, y, z, t) + ζM̂(y, z, x, t) + ζ^{2}M̂(z, x, y, t) = 0

for the holomorphic extensions. Connes’ Lemma provides a geometric interpretation of M̂(x, y, z, t) in terms of fixed points of certain affine transformations. Namely, M̂(x_{1}, x_{2}, x_{3}, t) is the fixed point of g_{1}g_{2} where

g_{1}(u) = \sqrt[3]{\frac{x_{1} + x_{2} − x_{3} − t}{x_{3} + x_{1} − x_{2} − t}}(u − x_{1}) + x_{1}

g_{2}(u) = \sqrt[3]{\frac{x_{2} + x_{3} − x_{1} − t}{x_{1} + x_{2} − x_{3} − t}}(u − x_{2}) + x_{2}

Perhaps it is worthwhile to consider the holomorphic extensions of other so-called triangle centers and to find geometric interpretations.

Example 9. Consider the rational function

A(x, y, z, t) = \frac{(x − y)(y − z)(z − x)}{x + y − z − t}

In the Euclidean case one has

|A(x, y, z, t)| = 2Δ(x, y, z)

where Δ denotes the area of a triangle. This is easy to deduce from

A(0, 1, z) = \frac{z̄ − z}{2}

Remark 5. Here is another point of view. For non-degenerate triangles x_{0}, x_{1}, x_{2} in some Euclidean plane E one can use the Euler line together with its points O and H as a coordinate system. So let us identify E with C by the conditions O = 0 and H = 1. This way the x_{0}, x_{1}, x_{2} become complex numbers subject to the relation x_{0} + x_{1} + x_{2} = 1 and with circumcenter the origin. (One is tempted to call them the “Euler coordinates” of the triangle.)

For the holomorphic extensions f̂(x_{0}, x_{1}, x_{2}, t) this simply means to restrict to t = 1 and x_{0}+x_{1}+x_{2} = 1. The function f̂is uniquely determined by this restriction, because of the G-equivariance. In other words:

Triangle functions are nothing else than complex analytic functions on the 2simplex

Δ^{2} = { (x_{0}, x_{1}, x_{2}) ∈ A^{3} | x_{0} + x_{1} + x_{2} = 1 }

One may also just restrict to t = x_{0} + x_{1} + x_{2}. This way we see: Triangle functions are essentially complex analytic functions on A^{3} which are homogeneous of degree 1; in other words: complex analytic sections in the bundle O(1) on P^{2}.

Remark 6. Continuing this discussion, we may also take the permutation action of S_{3} into account. Given a triangle function f, we consider also its conjugates f ∘ σ, σ ∈ S_{3}.

In the following let us assume that f̂|{t = x_{0} + x_{1} + x_{2}} is rational and can be defined over some field F.

We start with the following setting: a triangle over F is an element x in a cubic extension K of F. K need not be a field, but let’s say it is an etale ring extension of F of rank 3 (having in mind arbitrary flat ring extensions of F of rank 3). Let D be the discriminant algebra of K and H = K ⊗ D. Then, given a rational function f̂(x_{0}, x_{1}, x_{2}, t), the 6-tuple

(f̂∘ σ)_{σ∈S_{3}}

yields a rational function

f̃: K → H

such that

f̃(x_{0}, x_{1}, x_{2}) = (f̂(x_{σ(0)}, x_{σ(1)}, x_{σ(2)}, x_{0} + x_{1} + x_{2}))_{σ∈S_{3}}

in the case K = F × F × F.

Example 10. Suppose char F ≠ 2. For the pedal point P one finds

P̃(x) = \frac{1}{2} (T_{K/F} (x) − \frac{N_{K/F} (x)}{x^{2}})

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

Example 11. For the point S (the reflection of x at the opposite side of the triangle, see Example 6) one finds

S̃(x) = x + 2(P̃(x) − x) = x − \frac{1}{x} \frac{d}{dt}Φ_{x}(t)|_{t=x}

where Φ_{x}(t) is the characteristic polynomial of x.

3. Points on a circle

Here is a variant of “holomorphic extension”, which is actually much simpler. Let C_{n} = G_{m}^{n} where G_{m} is the multiplicative group. Let further

D_{n} = { (z_{1}, …, z_{n}) ∈ C_{n}(C) | |z_{1}| = ⋯ = |z_{n}| }

be the real subvariety of n-tuples of complex numbers lying on some circle around the origin. Let G = G_{m} act on C_{n} diagonally. Then D_{n} is G-stable.

It is easy to see that G-equivariant real analytic functions f: D_{n} → C extend uniquely to G-equivariant complex analytic functions f̂: C_{n} → C.

Example 12. Let n = 4 and for (z_{1}, z_{2}, z_{3}, z_{4}) ∈ D_{4} let f(z_{1}, z_{2}, z_{3}, z_{4}) be the intersection of the lines z_{1}z_{2} and z_{3}z_{4}. Then (cf. [4])

f̂(z_{1}, z_{2}, z_{3}, z_{4}) = \frac{z_{1}^{−1} + z_{2}^{−1} − z_{3}^{−1} − z_{4}^{−1}}{z_{1}^{−1}z_{2}^{−1} − z_{3}^{−1}z_{4}^{−1}}

  The remaining sections have been appended later and could perhaps
be better merged with the previous text.

4. An Involution

4.1. The map τ. The group G_{a} acts on A^{4} via

(x, a) ↦ x + a = (x_{0} + a, x_{1} + a, x_{2} + a, x_{3} + a)

with x = (x_{0}, x_{1}, x_{2}, x_{3}) ∈ A^{4} and a ∈ G_{a}.

The group S_{4} acts on A^{4} by permutation of the coordinates.

In the following ijkℓ is any permutation of 1234.

Let

ρ_{ij}(x) = x_{i} + x_{j} − x_{k} − x_{ℓ}

The 6 polynomials ρ_{ij}(x) are G_{a}-invariant. One has ρ_{ij} = −ρ_{kℓ}. If 2 is invertible, the functions ρ_{01}, ρ_{02}, ρ_{03} form a coordinate system for A^{4}/G_{a}.

Let further

R(x) = ρ_{ij}(x)ρ_{ik}(x)ρ_{iℓ}(x)

The polynomial R is independent of the choice of ijkℓ, it is S_{4}-invariant and G_{a}- invariant.

Let further

D_{i}(x) = x_{j}^{2} + x_{k}^{2} + x_{ℓ}^{2} − x_{j}x_{k} − x_{k}x_{ℓ} − x_{ℓ}x_{j}
= (x_{j} + ζx_{k} + ζ^{−1}x_{ℓ})(x_{j} + ζ^{−1}x_{k} + ζx_{ℓ})

with 1 + ζ + ζ^{−1} = 0. The polynomial D_{i} is G_{a}-invariant and invariant under permutations of jkℓ.

Finally consider the rational map

τ: A^{4}/G_{a} → A^{4}

τ(x) = −\frac{1}{R(x)}(D_{0}(x), D_{1}(x), D_{2}(x), D_{3}(x))

It is of degree −1.

One finds

ρ_{ij}(τ(x)) = \frac{1}{ρ_{ij}(x)}

Thinking of the ρ_{ij}(x)/ρ_{ik}(x) as angles, maybe τ(x) can be thought of as the “algebraic inversion” of the quadrangle x.

The previous relation shows

(π ∘ τ)^{2} = id_{A^{4}/G_{a}}

where π: A^{4} → A^{4}/G_{a} is the projection. In particular, π ∘ τ is a birational isomorphism of A^{4}/G_{a}.

4.2. Coordinate free setups. One can set up τ a little bit more coordinate free as a map

τ: E^{4}/V → (V^{∗})^{4}

where E = e + V is a 1-dimensional affine space with underlying vector space V. Or, let H be a quartic extension of the ground ring F. Define

D̃: H/F → H
D̃(x) = −2x^{2} + xT_{H/F} (x) + T_{H/F} (x^{2}) − Q_{H/F} (x)

where Q denotes the second coefficient of the characteristic polynomial. If H = F^{4}, then

D̃(x) = (D_{0}(x), D_{1}(x), D_{2}(x), D_{3}(x))

One has

D̃^{2}(x) = R(x)x mod F

for a certain polynomial R in the coefficients of the characteristic polynomial. Hence one can generalize τ as

τ: H/F → H

τ(x) = −\frac{D(x)}{R(x)}

as long as R is not identically 0.

4.3. Sections to A^{n} → A^{n}/G_{a}. It follows also that

τ ∘ π ∘ τ: A^{4}/G_{a} → A^{4}

is a (rational) section to π which is S_{4}-equivariant.

Note that if 2 is not invertible, then π does not have a S_{4}-equivariant linear section.

What about S_{n}-equivariant rational sections to the projection A^{n} → A^{n}/G_{a} (with G_{a} acting diagonally)?

For n = 2 it is easy to see that such a section does not exist if 2 = 0.

For n > 2 such a section does exist over any field, basically because S_{n} acts generically free on A^{n}/G_{a}.

For n = 3 such a section is given by

A^{3}/G_{a} → A^{3}
[x_{0}, x_{1}, x_{2}] ↦

\frac{((2x_{0} − x_{1} − x_{2})(x_{1} − x_{2})^{2}, (2x_{1} − x_{2} − x_{0})(x_{2} − x_{0})^{2}, (2x_{2} − x_{0} − x_{1})(x_{0} − x_{1})^{2})}{x_{0}^{2} + x_{1}^{2} + x_{2}^{2} − x_{0}x_{1} − x_{1}x_{2} − x_{2}x_{0}}

This map is closely related to the “basic line” in [5].

For n = 4 we have the section τ ∘ π ∘ τ, which is remarkably more or less the square of the less complicated function τ.

The cases n = 3, 4 can be brought in a common form as follows. Let L/F be a separable extension of degree n. For x ∈ L let

P_{x}(t) = N_{L/F} (t − x) = t^{n} − T_{L/F} (x) + ⋯ + (−1)^{n}N_{L/F} (x)

be the characteristic polynomial of x. Consider the function

g(x) = \frac{N_{L/F} (T_{L/F} (x) − nx)}{T_{L/F} (\frac{dP_{x}}{dt} |_{t=x})}

One has

g(ax + b) = ag(x)

for a, b ∈ F.

One finds for n = 3, 4 that

g(x) = T_{L/F} (x) mod n

(This does not hold for n = 5.) Let further

f(x) = \frac{T_{L/F} (x) − g(x)}{n}

This has sense: We may divide by n for the universal extension L = Z[x_{1}, …, x_{n}] over F = L^{S_{n}} = Z[σ_{1}, …, σ_{n}] and then specialize.

It is clear that

f(ax + b) = af(x) + b

for a, b ∈ F. It follows that

σ: L/F → L
σ(x + F) = x − f(x)

is a section to the projection L → L/F. In the split case L = F^{n} we get exactly the sections A^{n}/G_{a} → A^{n} considered before.

I did not look at n > 4, but I think that there is a substantial difference to the cases n ≤ 4.

4.4. Relation with triangle functions. The automorphism π ∘ τ has the following interpretation as a sort of algebraic version of the complex conjugation for triangle functions.

Over the field C of complex numbers let f_{i}: C^{4} → C, i = 0, 1, 2 be rational functions which are Aff(1, C) equivariant.

Denote by f̄ the complex conjugate of f, that is f̄(x) = \overline{f(x̄)}.

Let X ⊂ C^{4} be the (real) subvariety of orthocentric quadrangles.

Lemma 2. The following statements are equivalent:

(1) For all x ∈ X the points f_{0}(x), f_{1}(x), f_{2}(x) lie on a real line.

(2) For all x ∈ C^{4} one has

\frac{f_{1} − f_{0}}{f_{2} − f_{0}} (x) = \frac{f̄_{1} − f̄_{0}}{f̄_{2} − f̄_{0}} (τ(x))

□

Suppose the triangle functions f_{i} have real coefficients. Then the last condition amounts to the π ∘ τ-invariance of the cross ratio of the f_{i}.

4.5. More remarks on the pedal point. For an orthocentric quadrangle x = (x_{0}, x_{1}, x_{2}, x_{3}) the intersection of the lines x_{0}x_{1}, x_{2}x_{3} is the pedal point

P (x_{0}, x_{1}, x_{2}, x_{3}) = \frac{x_{0}x_{1} − x_{2}x_{3}}{x_{0} + x_{1} − x_{2} − x_{3}}

see Example 5.

For a quadrangle x = (x_{0}, x_{1}, x_{2}, x_{3}) on a circle around 0, the intersection of the lines x_{0}x_{1}, x_{2}x_{3} is given by

f(x_{0}, x_{1}, x_{2}, x_{3}) = \frac{x_{1}x_{2}x_{3} + x_{0}x_{2}x_{3} − x_{0}x_{1}x_{2} − x_{0}x_{1}x_{3}}{x_{0}x_{1} − x_{2}x_{3}}

see Example 12.

In other words,

f(x_{0}, x_{1}, x_{2}, x_{3}) = P (x_{0}^{−1}, x_{1}^{−1}, x_{2}^{−1}, x_{3}^{−1})^{−1}

Maybe the map τ is closely related with this coincidence (or maybe not).

The formula

P (x_{0}, x_{1}, x_{2}, x_{3}) = \frac{x_{0}x_{1} − x_{2}x_{3}}{x_{0} + x_{1} − x_{2} − x_{3}}

appears also in [1].

Here are further interpretations. Let a, b, c, d be four (general) points in an affine line.

Let f be the affine transformation with f(a) = c and f(d) = b. Let g be the affine transformation with g(a) = d and g(c) = b. Consider the commutator h = f^{−1}g^{−1}fg. As any commutator in the group of affine transformations, h is a translation. Moreover h(a) = a, and therefore h is the identity. Thus f and g commute. Their common fixed point is

P (a, b, c, d) = \frac{ab − cd}{a + b − c − d}

This way P (a, b, c, d) appears as the fixed point of an action of Z × Z on A^{1} by affine transformations.

Lemma 3. Let h be a transformation of a projective line. Suppose there are distinct points a, b with h(a) = b, h(b) = a. Then h^{2} is the identity.

Proof. h^{2} has at least 3 distinct fixed points, namely a, b and a fixed point of h. □

Now let h be the projective transformation with h(a) = b, h(b) = a, h(c) = d. By the Lemma one has h(d) = c. Then one has h^{2} = id and one finds

h(∞) = P (a, b, c, d) = \frac{ab − cd}{a + b − c − d}

One can see directly that the fixed point of f above coincides with h(∞), without an explicit calculation. Namely consider k = hf. Then k(a) = d, k(d) = a. By the Lemma, k^{2} = id. Thus

hfh^{−1} = f^{−1}

Therefore h permutes the fixed points of f, which are ∞ and P (a, b, c, d).

Further, let f, g be the projective transformations which fix c, d and which send ∞ to a resp. b. Then fg = gf and one finds fg(∞) = P (a, b, c, d).

5. Exterior Algebra of a cubic extension

Let K be a cubic extension of a field F.

For x ∈ K one denotes by

N_{K/F} (t − x) = t^{3} − T_{K/F} (x)t^{2} + Q_{K/F} (x)t − N_{K/F} (x)

its characteristic polynomial and by

x^{#} = x^{2} − T_{K/F} (x)x + Q_{K/F} (x) = \frac{N_{K/F} (x)}{x}

its adjoint.

The highest exterior power Λ^{3}K is a 1-dimensional F -vector space. Thus for ω, ω′ ∈ Λ^{3}K we have ω/ω′ ∈ F, provided ω′ ≠ 0.

In the following we assume (sometimes) that K/F is separable and use the canonical identifications

Λ^{2}K = Hom(K, Λ^{3}K) = K ⊗ Λ^{3}K
(Λ^{3}K)^{⊗2} = F

via the trace form. Thus for ω, ω′ ∈ Λ^{2}K we have ω/ω′ ∈ K, provided ω′ is nondegenerate.

Here is a basic formula which connects the ∧-product with the multiplication in K:

ux ∧ y ∧ z + x ∧ uy ∧ z + x ∧ y ∧ uz = T_{K/F} (u)x ∧ y ∧ z

with u, x, y, z ∈ K.

Here are a few more formulas:

x(y ∧ z) + y(z ∧ x) + z(x ∧ y) = x ∧ y ∧ z

(1 ∧ x ∧ y)(1 ∧ z) + (1 ∧ y ∧ z)(1 ∧ x) + (1 ∧ z ∧ x)(1 ∧ y) = 0

x(1 ∧ y) + y(1 ∧ x) = 1 ∧ [xT (y) + yT (x) − 2xy]

\begin{matrix}(x(1 ∧ x ∧ y) − (1 ∧ x ∧ xy))(y(1 ∧ y ∧ x) − (1 ∧ y ∧ xy)) \\ = (1 ∧ x ∧ x^{2})(1 ∧ y ∧ y^{2}) \end{matrix}

Consider the binary cubic form

ψ_{K/F} : K/F → Λ^{3}K
ψ_{K/F} (x + F) = 1 ∧ x ∧ x^{2}

This form is fundamental in understanding cubic extensions, since one can recover from it the algebra K. This will be considered elsewhere.

6. Expressions for the circumcenter, the orthocenter and the Euler center

Now let F = C, K = C^{3} and let x = (x_{1}, x_{2}, x_{3}) ∈ K be a triangle.

As usual, x̄ denotes complex conjugation.

For a triangle we call from now on the nine-point center of a triangle x the Euler center and denote it by E(x_{1}, x_{2}, x_{3}) (the letter F will be reserved for the Fermat points). Recall also other basic triangle points: the center of gravity G (or barycenter, center of mass, centroid), the circumcenter O, and the orthocenter H. (I am tempted to rename O to C and H to O, but that might be too confusing for now.)

For a triangle x consider the triangle Φ(x) geometrically obtained from x by drawing through each vertex the line parallel to the opposite side. The construction leads to the expressions

Φ(x) = T_{K/F} (x) − 2x, Φ^{−1}(x) = \frac{T_{K/F} (x) − x}{2}

One has

G(x) = G(Φ(x))
H(x) = O(Φ(x))
O(x) = E(Φ(x))

The observation H(x) = O(Φ(x)) can be used to show that the altitudes are concurrent.

If we use the normalization G(x) = 0, then x ↦ Φ(x) is just multiplication by −2. This implies immediately the first two of the following relations:

3G = 2O + H
3G = O + 2E
2E = O + H
4E = 3G + H

Clearly one has

3G(x) = T_{K/F} (x)

Lemma 4. One has

E(x) = \frac{1}{2} · \frac{1 ∧ x^{2} ∧ x̄}{1 ∧ x ∧ x̄}
O(x) = \frac{1 ∧ x ∧ xx̄}{1 ∧ x ∧ x̄}
O(x) = −\frac{1 ∧ x^{#} ∧ x̄}{1 ∧ x ∧ x̄}
H(x) = \frac{1 ∧ (x^{2} + x^{#}) ∧ x̄}{1 ∧ x ∧ x̄}

Proof. Left to the reader. □

Note further that the denominator 1 ∧ x ∧ x̄ vanishes exactly when the triangle x lies on a real line.

Here is another way to look at the orthocenter. Let H be a quartic extension of the ground ring F and let

θ: H → H
θ(x) = 2x^{2} − xT_{H/F} (x)

Note that θ factors to a map H/F → H/F. One has:

Lemma 5. Let F = C, H = F^{4} and let x ∈ H be a generic element.

(1) The quadrangles x and θ^{2}(x) are similar.

(2) The quadrangle x is orthocentric if and only if \overline{x} is similar to θ(x).

Proof. Left to the reader. But see subsection 4.2. The map θ is very close to D̃. □

7. The Fermat point

See [3, Section 1.8].

Given a triangle x = (x_{1}, x_{2}, x_{3}) ∈ K = C^{3}, draw at each side the outside equilateral triangle. Let y_{1}, y_{2}, y_{3} be the corresponding new vertices, numbered so that x_{i}, x_{j}, y_{k} are the equilateral triangles. Then the lines x_{i}y_{i} meet in one point, the first Fermat point F_{1}. The second Fermat point F_{2} is obtained analogously by using equilateral triangles pointing inwards.

In the following we describe the holomorphic extensions of F_{1} and F_{2}. It turns out that it is better to use the variables (x_{1}, x_{2}, x_{3}, O) instead of (x_{1}, x_{2}, x_{3}, H).

Let K be a cubic extension of a field F (of characteristic different form 3) and let u ∈ K be a Kummer element, that is, its characteristic polynomial is of the form t^{3} − a with a ∈ F \ {0}. For x ∈ K and t ∈ F we define the rational function

F_{u}(x, t) = \frac{(1 ∧ x ∧ ux)t − (1 ∧ x ∧ ux^{2})}{(1 ∧ x ∧ u)t − (1 ∧ x ∧ ux)}

This function has some remarkable properties.

First note that F_{u} depends on u only up to scalar multiplication. Now there are only two Kummer elements up to scalar multiplication, given in the split case K = F^{3} by

u = (1, ζ, ζ^{−1})

with ζ ∈ μ_{3} \ {1}.

Lemma 6. For a triangle x = (x_{1}, x_{2}, x_{3}) ∈ K = C^{3} the two Fermat points are given by

F_{u}(x_{1}, x_{2}, x_{3}, O(x_{1}, x_{2}, x_{3})), F_{u^{−1}}(x_{1}, x_{2}, x_{3}, O(x_{1}, x_{2}, x_{3}))

Proof. Left to the reader. □

Consider the two functions

f_{u}(x) = F_{u}(x, ∞) = \frac{1 ∧ x ∧ ux}{1 ∧ x ∧ u}

In the Euclidean case the interpretation is clear: The f_{u}(x) are the Fermat points for three points on a real line (since O = ∞ if and only if the triangle x lies on a line).

There is also an algebraic interpretation: For x = (x_{1}, x_{2}, x_{3}) ∈ K = F^{3} the f_{u}(x) are the fixed points of the projective transformation Φ ∈ PGL(2, F) with

Φ [\begin{matrix}x_{i} \\ 1 \end{matrix}] = [\begin{matrix}x_{i+1} \\ 1 \end{matrix}] (i mod 3)

For the mid point of the f_{u}(x) one has

2k(x) = f_{u}(x) + f_{u^{−1}}(x)

= \frac{x_{1}^{2}x_{2} + x_{1}^{2}x_{3} + x_{2}^{2}x_{1} + x_{2}^{2}x_{3} + x_{3}^{2}x_{1} + x_{3}^{2}x_{2} − 6x_{1}x_{2}x_{3}}{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} − x_{1}x_{2} − x_{2}x_{3} − x_{3}x_{1}}

One finds

T_{K/F} (x) = 2k(x) + f(x) = f_{u}(x) + f_{u^{−1}}(x) + f(x)

where f is the function from Section 4.3 (for n = 3) which is also the base point of the “basic line” in [5].

It is perhaps worthwhile to consider the denominator of F_{u} more closely. One has for

K = e_{1}F + e_{2}F + e_{3}F
x = (x_{1}, x_{2}, x_{3}) ∈ K, x_{4} ∈ F
u = (u_{1}, u_{2}, u_{3}) ∈ K

the computation

\frac{(1 ∧ x ∧ ux) − x_{4}(1 ∧ x ∧ u)}{e_{1} ∧ e_{2} ∧ e_{3}} =

u_{1}(x_{1} − x_{4})(x_{2} − x_{3}) + u_{2}(x_{2} − x_{4})(x_{3} − x_{1}) + u_{3}(x_{3} − x_{4})(x_{1} − x_{2})

This function is invariant under S_{4} acting on (x_{1}, x_{2}, x_{3}, x_{4}) in the standard way and on (u_{1}, u_{2}, u_{3}) via S_{4} → S_{3}. I have no interpretation for this.

Further, let us consider F_{u}(x, t) as a linear fractional function in t. One has

F_{u}(x)(t) = [\begin{matrix}β(x) & −α(x) \\ γ(x) & −β(x) \end{matrix}] [\begin{matrix}t \\ 1 \end{matrix}]

where

α(x) = 1 ∧ x ∧ ux^{2}
β(x) = 1 ∧ x ∧ ux
γ(x) = 1 ∧ x ∧ u

Since the matrix has trace 0, it follows that t ↦ F_{u}(x, t) is of order 2 (I have no interpretation of this).

Note further that

α(x + a) = α(x) + 2aβ(x) + a^{2}γ(x)

with a ∈ F. Therefore F_{u} is basically determined by α, and so every mystery about the Fermat points should be encoded in α(x) = α_{u}(x).

Consider further the cubic map

A: K → Hom(K, Λ^{3}K) = Λ^{2}K
x ↦ (y ↦ 1 ∧ x ∧ yx^{2})

If u ∈ K is a Kummer element, then K = F + uF + u^{−1}F and the functions α_{u}(x), α_{u^{−1}}(x) and ψ(x) = 1 ∧ x ∧ x^{2} are just the corresponding components of A.

8. On the cubic form in 6 variables

First, more exterior algebra of a cubic extension. Let K be a separable cubic extension of a field F. The norm for K defines a cubic form

N_{K/F} : Λ^{2}K → Λ^{3}K

Let

f: K × K → Λ^{3}(K/F)
f(x, y) = 1 ∧ (x + y) ∧ (xy)

see [7].

For x, y ∈ K with y generic we define

Z_{y}(x) = \frac{y ∧ x}{y ∧ 1}

One finds

(1)  Z_{cy}(ax + b) = aZ_{y}(x) + b  (a, b, c ∈ F, c ≠ 0)
(2)  f(x, Z_{y}(x)) = 0
(3)  Z_{y}(x) = \frac{y^{#}(1 ∧ y ∧ x) + (y^{2} ∧ y ∧ x)}{N_{K/F} (y ∧ 1)}

Equation (2) provides us with the solution z = Z_{y}(x) of the equation f(x, z) = 0 in z.

We now restrict to the case K = F × F × F. Write

z = Z_{y}(x) ∈ F × F × F

One finds that

(4)  \frac{x_{j} − z_{i}}{x_{k} − z_{i}} = \frac{y_{j}}{y_{k}}

is an alternative defining equation for Z_{y}(x).

We now restrict to the Euclidean case F = C.

Equation (3) means that the triangle Z_{y}(x) is similar to the triangle y^{−1}.

Let us consider equation (4). It means that the sum of the angles of the triangles Δ_{k} = (x_{i}, x_{j}, z_{k}) at z_{k} is 2π. Let us further assume that |y_{1}| = |y_{2}| = |y_{3}|, where | · | is the Euclidean norm. Then equation (4) shows the |x_{j} − z_{i}| = |x_{k} − z_{i}| and we can consider the circles C_{k} around z_{k} through x_{i}, x_{k}. It is an exercise to observe that these circles meet in a common point. In the case when y = (1, ζ, ζ^{−1}), this point is the Fermat point, see [3, Section 1.8, Exercise 3].

We recall Napoleon’s theorem: The barycenters of the (outside) equilateral triangles used to define the Fermat point form itself an equilateral triangle. See [3, Section 1.8, Exercise 3].

Now these barycenters are exactly the points z_{k} in the case when when y = (1, ζ, ζ^{−1}) is a Kummer element with ζ a suitable primitive cube root of unity. Napoleon’s theorem follows now from equation (3) which shows that the triangle z is similar to y^{−1}. It follows also directly from the computation

Z_{(1,ζ,ζ^{−1})}(x_{1}, x_{2}, x_{3}) = \frac{(x_{2} − ζx_{3}, x_{3} − ζx_{1}, x_{1} − ζx_{2})}{1 − ζ}

9. More on the cubic form in 6 variables

What about a geometric interpretation of the equation f(x, y) = 0?

Let K = F × F × F. Then canonically Λ^{3}K = F up to a sign due to a choice of the order of the idempotents of K.

One has

f(x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}) =
(x_{1} − y_{2})(x_{2} − y_{3})(x_{3} − y_{1}) − (x_{1} − y_{3})(x_{2} − y_{1})(x_{3} − y_{2})

Let u = (u_{1}, u_{3}, u_{5}, u_{2}, u_{4}, u_{6}). Then f(u) = 0 means essentially that

∏_{i=1}^{6}(u_{i+1} − u_{i})^{(−1)^{i}} = 1

In other words, the condition f(u) = 0 means that for the hexagon u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6} the alternating product of the sides is = 1. This interpretation however does not reflect the symmetry of f under Z/2 × S_{4} = Z/2 ≀ S_{3}.

Lemma 7. Suppose all x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3} are pairwise distinct. Then f(x, y) = 0 if and only if there exists φ ∈ PGL(2, F) = Aut(P^{1}) such that

φ(x_{i}) = y_{i}, φ(y_{i}) = x_{i} (i = 1, 2, 3)

In this case, φ is uniquely determined and one has φ^{2} = 1. □

Clearly, φ is uniquely determined by φ(x_{i}) = y_{i}, i = 1, 2 and one has φ^{2} = 1 by Lemma 3. As for an explicit description, one finds with v = (x_{1}, y_{1}, x_{2}, y_{2})

φ = [\begin{matrix}β(v) & −α(v) \\ γ(v) & −β(v) \end{matrix}]

where

α(v) = x_{1}y_{1}(x_{2} + y_{2}) − x_{2}y_{2}(x_{1} + y_{1})
β(v) = x_{1}y_{1} − x_{2}y_{2}
γ(v) = x_{1} + y_{1} − x_{2} − y_{2}

Note further that

α(v + a) = α(v) + 2aβ(v) + a^{2}γ(v)
α(v^{−1}) = \frac{−1}{x_{1}y_{1}x_{2}y_{2}} γ(v)

with a ∈ F. Therefore φ can be pulled out of the noses of γ and PGL(2). A special case is the pedal point

φ(∞) = P (x_{1}, y_{1}, x_{2}, y_{2})

This and the interpretation of P (x_{1}, y_{1}, x_{2}, y_{2}) in [1] shows:

Lemma 8. Let

u = (u_{01}, u_{02}, u_{03}, u_{23}, u_{31}, u_{12}) ∈ (CP^{1})^{6}

be a generic 6-tuple of points in the Riemann sphere. Then the following statements are equivalent:

(1) f(u) = 0.

(2) The 4 circumcircles u_{01}u_{02}u_{03}, u_{01}u_{12}u_{31}, u_{02}u_{12}u_{23}, u_{03}u_{23}u_{31} meet in a common point.

(3) The 4 circumcircles u_{23}u_{31}u_{12}, u_{23}u_{03}u_{02}, u_{31}u_{03}u_{01}, u_{12}u_{01}u_{02} meet in a common point.

What about describing these points of intersection?

Here is another way to phrase this. Consider 4 generic circles C_{i} through a point P. Then the 6 points u_{ij} of intersection with C_{i} ∩ C_{j} = {P, u_{ij}} satisfy the relation f(u_{01}, u_{02}, u_{03}, u_{23}, u_{31}, u_{12}) = 0 and any (generic) 6-tuple u with f(u) = 0 is obtained this way.

Finally we mention a modular property of f. Let

g = [\begin{matrix}a & b \\ c & d \end{matrix}] ∈ PGL(2, F)

act on u = (u_{1}, u_{2}, u_{3}, u_{4}, u_{5}, u_{6}) diagonally by

u_{i} ↦ \frac{au_{i} + b}{cu_{i} + d}

Then

(f ∘ g)(u) = \frac{(ad − bc)^{3}}{∏_{i=1}^{6}(cu_{i} + d)}f(u)

10. The Kiepert hyperbola

Lemma 9. Fix t ∈ R and let

b = \frac{1 + it}{2}, c = \frac{1 − it}{2}

where i^{2} = −1. For a triangle x = (x_{0}, x_{1}, x_{2}) ∈ C^{3} define the points

y_{i} = bx_{i+1} + cx_{i+2} (i mod 3)

Then the lines ℓ_{i} through x_{i} and y_{i} meet in a point, denoted by P_{t}(x).

One may rephrase this as follows:

Corollary 1. Fix an angle α. For a triangle erect on each side the isosceles triangle with base angles α. Then the lines through the vertices of the triangle and the new vertex of the opposite isosceles triangle are concurrent (=meet in a point).

For a triangle x let H(x) be the real curve determined by t ↦ P_{t}(x). It turns out that H(x) is in general a hyperbola. It is called the Kiepert Hyperbola.

Here are special points on H(x):

• If t = 0 (or α = 0) the points y_{i} are the midpoints of the sides of the triangle. Therefore P_{0}(x) is the center of gravity.

• If t = ∞ (or α = ±π/2) the points y_{i} lie at infinity in direction orthogonal to the corresponding side. Therefore P_{∞}(x) is the orthocenter.

• If t = ±\sqrt{3} (or α = ±π/3) the points y_{i} are vertices of equilateral triangles erected at each side. Therefore P_{±\sqrt{3}}(x) are the (first and second) Fermat points.

• If the angle α is the angle at x_{i} (with appropriate sign) then y_{j} lies on the line x_{i}x_{k}. In this case the point of concurrence is just x_{i}. Thus, for a general triangle x, there are values t = t_{i} such that P_{t_{i}}(x) = x_{i}.

Proof of Lemma 9. The lines ℓ_{i} have the parameterizations

ℓ_{i}: x_{i} + r_{i}(y_{i} − x_{i}),  y_{i} = bx_{i+1} + \overline{b}x_{i+2}  (r_{i} ∈ R)

The lines have a common point if D = 0 where

D = det (\begin{matrix}x_{1} − x_{0} & y_{0} − x_{0} & y_{1} − x_{1} & 0 \\ x_{2} − x_{0} & y_{0} − x_{0} & 0 & y_{2} − x_{2} \\ \overline{x}_{1} − \overline{x}_{0} & \overline{y}_{0} − \overline{x}_{0} & \overline{y}_{1} − \overline{x}_{1} & 0 \\ \overline{x}_{2} − \overline{x}_{0} & \overline{y}_{0} − \overline{x}_{0} & 0 & \overline{y}_{2} − \overline{x}_{2} \end{matrix})

Now D is a priori a cubic polynomial in t. But there are at least 4 zeros: t = 0 and the t_{i}, corresponding to the center of gravity and the points of the triangle. Thus D = 0. □

I haven’t computed P_{t}(x) explicitly in general, only for three points on a real line. For t ≠ 0 and z ∈ R one finds

P_{t}(0, 1, z) = \frac{z(\overline{b}z + b)}{z^{2} − z + 1}

Note that for t = ±\sqrt{3} this simplifies to

P_{±\sqrt{3}}(0, 1, z) = \frac{z}{bz + \overline{b}}

Here is the equation for the Kiepert hyperbola. One considers the functions (with, say, x ∈ K = C^{3})

β(x) = T (x)T (x\overline{x}^{#})
Φ(x) = β(x) − \overline{β(x)}

Then one has

H(x) = { z ∈ C | Φ(x − z) = 0 }

The Kiepert hyperbola is the hyperbola determined by the points of the triangle x, its barycenter T (x)/3 and its orthocenter H(x). It is relatively easy to see that Φ(x − x_{i}) = 0 and Φ(x − T (x)/3) = 0. To see that Φ(x − H(x)) = 0, one considers the function

α(x) = x(1 ∧ \overline{x}) ∈ Λ^{2}K = Λ^{3}K ⊗ K

Then one observes the following identity:

(5)  T (α(x)^{2} − \overline{α(x)}^{2}) = −2Φ(x)

and uses the fact that H(x) = 0 if and only if α(x) + \overline{α(x)} = 0.

Remark 7. Along the way we have used a bunch of identities like (5). Maybe one can set up the whole topic in terms of an appropriate tensor category.

11. On the radius of the incircle

Let x = (x_{1}, x_{2}, x_{3}) ∈ C^{3} be a (generic) Euclidean triangle with the origin as circum center and with circum radius R, i. e., |x_{1}| = |x_{2}| = |x_{3}| = R. Let further r, r_{1}, r_{2}, r_{3} be the radii of the incircle resp. excircles. The functions x ↦ r/R, x ↦ r_{i}/R are real analytic functions on (C^{×}/R^{×})^{3}, invariant also under the diagonal action of C^{×}/R^{×}. When considering their holomorphic extensions (see Section 3), it turns out that they generate a biquadratic extension of C(x_{1}, x_{2}, x_{3}). This will be described in the following.

Let F be field. For x = (x_{1}, x_{2}, x_{3}) ∈ F^{3} we consider the functions

N(x) = x_{1}x_{2}x_{3}
M(x) = (x_{1} − x_{2})^{2}x_{3} + (x_{2} − x_{3})^{2}x_{1} + (x_{3} − x_{1})^{2}x_{2}
Δ(x) = (x_{1} − x_{2})^{2}(x_{2} − x_{3})^{2}(x_{3} − x_{1})^{2}

We consider the polynomial

P_{x}(t) = t^{4} − 8t^{3} − 2\frac{M(x)}{N(x)}t^{2} + \frac{Δ(x)}{N(x)^{2}}

Note that the functions M/N(x), Δ/N^{2}(x) are invariant under x ↦ ax (a ∈ F^{×}) and x ↦ x^{−1}.

The zeros of P_{x} can be described as follows: Let z_{i} with z_{i}^{2} = x_{i}. Then

ρ = \frac{(z_{1} + z_{2})(z_{2} + z_{3})(z_{3} + z_{1})}{z_{1}z_{2}z_{3}}

and its 3 conjugates under z_{i} ↦ ±z_{i} are the zeros of P_{x}. Therefore it is clear that the zeros lie in the biquadratic extension generated by z_{i}/z_{j}. (This is the biquadratic extension mentioned in Example 7.)

Lemma 10. Let x = (x_{1}, x_{2}, x_{3}) ∈ C^{3} be a generic Euclidean triangle with |x_{1}| = |x_{2}| = |x_{3}| = R and let r, r_{1}, r_{2}, r_{3} be the radii of its incircle resp. excircles. Then

−\frac{2r}{R} , +\frac{2r_{1}}{R} , +\frac{2r_{2}}{R} , +\frac{2r_{3}}{R}

are the zeros of P_{x}. □

Lemma 10 includes some standard relations for the radii R, r, r_{i} (with a_{i} the side lengths of the triangle):

Corollary 2. Then

4R = r_{1} + r_{2} + r_{3} − r
\frac{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}{2} = r_{1}r_{2} + r_{2}r_{3} + r_{3}r_{1} − r(r_{1} + r_{2} + r_{3})
0 = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} − \frac{1}{r}
\frac{(a_{1}a_{2}a_{3})^{2}}{R^{2}} = rr_{1}r_{2}r_{3}

Note also that

4|A|R = \sqrt{|Δ(x)|}

where |A| denote the area of the triangle x (see Example 9). Recall also that |A| = rs = r_{i}s_{i} where s = (a_{1} + a_{2} + a_{3})/2 is the semi-perimeter and where s_{i} = s − a_{i}

Remark 8. Describe the Soddy centers. It seems to me that their holomorphic extensions lie in K = F [z_{i}/z_{j}]⊗F [α]/(α^{2}+Δ(x)) and are conjugate under α ↦ −α.

We remark that the tri-quadratic extension K contains also the incenters of the other subtriangles of the orthocentric quadrangle of the given triangle.

It seems that the Soddy centers, the Gergonne point and the incenter lie on a line.

Remark 9. The following observation might be helpful: Let K/F be a biquadratic extension with intermediate quadratic subextensions K_{i}. Let x be an element of K. If T_{K/F} (x) = 0, then

T_{K/K_{1}}(x)T_{K/K_{2}}(x)T_{K/K_{3}}(x) = T_{K/F} (x^{#})

Proof.

(x + x_{1})(x + x_{2})(x + x_{3}) = x^{2}(x + x_{1} + x_{2} + x_{3}) + x(x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1}) + x_{1}x_{2}x_{3}

□

12. The conjugate (or dual?) quadrangle

In this section E is a 2-dimensional affine space over some field F and

V = E − e_{0}

denotes its underlying vector space. Let x = (x_{1}, x_{2}, x_{3}, x_{4}) ∈ E^{4} with x_{i} ≠ x_{j} for i ≠ j. Let further ℓ_{ij} be the line through x_{i} and x_{j} and let

L_{ij} = L_{ji} = [x_{i} − x_{j}] ∈ P(V)

be the corresponding points in the projective space of V. We suppose that the points L_{ij} are distinct, i. e., no three of the x_{i} lie on a line.

Lemma 11. The 6 points L_{ij} “stand in involution”. More precisely, there exist τ ∈ PGL(V) = Aut(P(V)) with τ^{2} = 1 and such that

τ(L_{ij}) = L_{kℓ}

for any permutation ijkℓ of 1234.

Proof. We give an explicit formula for τ. Let v_{i} = x_{i} − x_{4}, i = 1, 2, 3 and define

Φ = Φ_{x_{1},x_{2},x_{3},x_{4}} : V → V ⊗ (Λ^{2}V)^{⊗2}

Φ(w) = ∑_{i=1}^{3} v_{i}(v_{i+1} ∧ v_{i+2})(v_{i+1} ∧ w)

with the indices reduced mod 3.

Note that dim Λ^{2}V = 1. One finds

Φ(v_{i}) = (v_{i+1} − v_{i+2})(v_{i} ∧ v_{i+1})(v_{i+2} ∧ v_{i})
Φ(v_{i+1} − v_{i+2}) = −v_{i}(v_{i+1} ∧ v_{i+2})((v_{i+1} − v_{i}) ∧ (v_{i+2} − v_{i}))

and therefore

Φ^{2} = −(v_{i} ∧ v_{i+1})(v_{i+1} ∧ v_{i+2})(v_{i+2} ∧ v_{i})((v_{i+1} − v_{i}) ∧ (v_{i+2} − v_{i}))id_{V}

Now let τ = P(Φ) be the induced map between the projective spaces. □

Remark 10. For a permutation σ ∈ S_{4} one finds

Φ_{x_{σ(1)},x_{σ(2)},x_{σ(3)},x_{σ(4)}} = (sgn σ)Φ_{x_{1},x_{2},x_{3},x_{4}}

Remark 11. I don’t know a geometric explanation for the existence of τ. It appeared to me first as a consequence of Lemma 7 above and of Lemma 3 in [7].

Remark 12. The converse of Lemma 11 is also true (see also Lemma 4 in [7]): Given 6 points L_{ij} ∈ P(V), 1 ≤ i < j ≤ 4, there exist 4 points x_{i} ∈ E with L_{ij} = [x_{i} − x_{j}]. This way the involution τ gives rise to a self map on nondegenerate quadrangles in 2-dimensional affine space up to translations and scalar multiplication:

Given x = (x_{1}, x_{2}, x_{3}, x_{4}) ∈ E^{4} one may construct a new quadrupel x′ = (x′_{1}, x′_{2}, x′_{3}, x′_{4}) ∈ E^{4}, well defined up to translations and scalar multiplication, such that x′_{i} lies on lines parallel to ℓ_{jk}, ℓ_{kℓ}, ℓ_{ℓj}.

An explicit formula for x′ is given by x′_{i} = Φ_{x_{1},x_{2},x_{3},0}(x_{i}), i = 1, 2, 3 and x′_{4} = 0. Or, for

x = (0, x_{1}, x_{2}, x_{3}) ∈ V^{4}

one may take

x′ = (0, \frac{x_{2} − x_{3}}{x_{2} ∧ x_{3}} , \frac{x_{3} − x_{1}}{x_{3} ∧ x_{1}} , \frac{x_{1} − x_{2}}{x_{1} ∧ x_{2}} ) ∈ (V^{∗})^{4}

If F = R and E = C one may take for x = (0, 1, u, z) the quadrangle

x′ = (0, u − z, (u − z)h(u, z), (u − z)h(z, u))

with

h(u, z) = \frac{(ūz − uz̄)(1 − z)}{(z − z̄)(u − z)}

One observes that

lim_{t→±∞} h(it, z) = \frac{(1 − z)(z + z̄)}{z̄ − z} = H(0, 1, z)

is the orthocenter of (0, 1, z).

References

arXiv:math.QA/0211199.

nary.html, 2003. /binary.html, 2003. nary.html, 2004.

[1] A. Connes, Symetries Galoisiennes et Renormalisation, Seminaire Poincare Octobre 2002, [2] , A new proof of Morley’s theorem, Les relations entre les mathématiques et la physique

théorique, Inst. Hautes Études Sci., Bures, 1998, pp. 43–46, MR 99m:51027.

[3] H. S. M. Coxeter, Introduction to geometry, Wiley Classics Library, John Wiley & Sons Inc.,

New York, 1989, Reprint of the 1969 edition, MR 90a:51001.

[4] C. Lubin, A proof of Morley’s theorem, Amer. Math. Monthly 62 (1955), 110–112, MR 16,848c.

[5] M. Rost, Notes on cubic equations, Notes, http://www.mathematik.uni-bielefeld.de/˜ rost /bi- [6] , Notes on Morley’s theorem, Notes, http://www.mathematik.uni-bielefeld.de/˜ rost [7] , The variety of angles, Notes, http://www.mathematik.uni-bielefeld.de/˜ rost /bi-

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