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September 2015

QUADRATIC FORMS AND INVERSIVE GEOMETRY

Nicholas Phat Nguyen 1

Abstract. This paper develops an inversive geometry for anisotropic quadradic spaces, in

analogy with the classical inversive geometry of a Euclidean plane. Such a generalization,

although relatively straight-forward, does not seem to have been published or generally known,

and therefore the account in this paper may be of interest and utility to some readers, especially as

the theory has certain elegance.

TABLE OF CONTENTS

Section 1. Anisotropic quadratic forms

Section 2. The projective space of circles and lines for an anisotropic quadratic form

Section 3. Zero sets

Section 4. A geometrical interpretation of the scalar product on circles and lines

Section 5. Group action on isotropic points

Section 6. Affine reflections and circle inversions

Section 7. Hyperplane reflections in the projective space of circles and lines

Section 8. Inversive geometry is projective geometry

Section 9. Conjugate points

Section 10. Stereographic projection

Section 11. Inversive geometry in dimension one

Section 12. The involution theorem of Desargues

Section 13. The eleven-point conic

1 E-mail address: nicholas.pn@gmail.com

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Section 1. Anisotropic quadratic forms

Let k be a field of characteristic ≠ 2. In this paper, we will consider anisotropic quadratic forms

defined for finite-dimensional vector spaces over k.

Recall that anisotropic quadratic forms are those quadratic forms h with the property that h(X) =

0 if and only if X = 0. Such an anisotropic quadratic form q can be viewed as the norm

associated with a regular symmetric bilinear form, which we will denote by using the dot product

notation. Specifically, we will write h(X) = X.X, where (X, Y) ↦ X.Y is the associated

symmetric bilinear form.

Anisotropic quadratic forms occur naturally in many mathematical contexts.

(a) Multiplication on k naturally gives us an anisotropic form of dimension 1.

(b) If E is a quadratic field extension of k, then the norm is naturally an anisotropic quadratic

form on E.

(c) If E is quaternion algebra defined over k, then the reduced norm is a quadratic form that

is anisotropic if and only if E is a division algebra. (If the reduced norm is isotropic, then

E is isomorphic to a matrix algebra.)

(d) Given an anisotropic quadratic form, the extension by scalars of that form to a field L

over k is also anisotropic when L is a finite extension of odd degree over k, or when L is

the rational function field k(T). See [Scharlau] at chapter 2, section 5.

(e) Anisotropic quadratic spaces represent all the different elements of the Witt ring W(k) of

the ground field k.

(f) If k is an ordered field, the standard scalar product

x.y = x1y1 + … + xnyn where x = (x1, …, xn) and y = (y1, …, yn)

is always anisotropic.

Recall that an ordered field is a field endowed with a total ordering relation compatible with the

additive and multiplicative operations of the field. An ordered field could be defined by giving

a total ordering ≤ that satisfies the following basic properties associated with the usual ordering

on the real line:

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if a ≤ b then a + c ≤ b + c for any c

if 0 ≤ a and 0 ≤ b then 0 ≤ a b

Examples of ordered fields include the rational numbers Q, any real algebraic number field, the

real numbers R, the field R(X) of rational functions with real coefficients, or the field R((X)) of

formal Laurent series with real coefficients. More generally, given any ordered field F, we can

always extend the ordering on F to an ordering on the following extension fields over F that

make them into ordered fields: (i) any finite extension of odd degree over F, (ii) any field

obtained by adjoining square roots of positive elements in F, (iii) the field F(X) of rational

functions in one variable with coefficients in F, and (iv) the field F((X)) of formal Laurent series

in one variable with coefficients in F. For more background and information on ordered fields,

the reader can refer to [Lang] at chapter XI (Real Fields).

The fact that the standard scalar product x.y = x1y1 + … + xnyn is anisotropic for any kn is a

characteristic property of ordered field, as discovered by Artin and Schreier (a field can be

ordered if and only if it is formally real).

Section 2. The projective space of circles and lines for an anisotropic quadratic form

For an anisotropic quadratic space E of dimension n ≥ 1 over k, consider the set F of all functions

p from E to k of the form:

p(X) = aX.X + b.X + c,

where X and b are vectors in E, and a and c are elements in the field k.2

The set F of all such functions can naturally be endowed with the structure of a k-vector space of

dimension n + 2, being parametrized by the vector b and the elements a and c. We will refer to

2 Given a basis of E, and regard the coefficients of the vector X in that basis as variables, the function p is

at most a second degree polynomial in n variables. Because the field k has 3 or more elements, such a

function p is a zero function if and only if the parameters a, b, and c are all zero.

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a function p where the coefficient a is non-zero as a circle, and a function p where the coefficient

a = 0 but b ≠ 0 as a line, in analogy with the classical case of a real Euclidean plane.

Aside from the natural structure of a vector space over k, we can also endow F with a symmetric

bilinear form < _ , _ > as follows.

Given p = aX.X + b.X + c and p* = a*X.X + b*.X + c*,

define <p,p*> = b.b* – 2ac* – 2a*c.

This scalar product is clearly symmetric and bilinear. Moreover, it is regular, as can be checked

by sight from the defining formula. (It is easy to see that F is isometric to the sum of E and a

hyperbolic plane.)

In case n = 1, the above scalar product has long made an appearance in the study of involutions

on a complex projective line. Such involutions can be described by quadratic polynomials p =

ax2 + 2bx + c where x is a suitable variable parametrizing the given line. See, for example, the

classic 19th

-century Salmon treatise on conics, [Salmon] at Articles 332 through 346. In the

classical case of real Euclidean spaces, this scalar product has also been known for a long time

for the case n = 2 and 3, at least since William Kingdon Clifford. From the references cited by

[Pfiefer & Hook], the earliest paper that described this scalar product for Euclidean spaces

seems to be the paper On the Powers of Spheres by William Kingdon Clifford, containing his

communications to the London Mathematical Society in February 1868.

For our purpose, we consider only functions p ≠ 0, and moreover we regard all non-zero scalar

multiples of the same function as equivalent because they have the same zero set in E (if any). In

other words, let Z(p) = {x ∈ E such that p(x) = 0}. Then Z(p) = Z(αp) where α is any non-zero

element of k.

Accordingly, we are led naturally to the projective space F consisting of all equivalent classes of

such non-zero functions, or what is essentially the same, the set of all 1-dimensional subspaces

of F. As a matter of convenient notation, for each letter denoting an element or subset of F, we

will put a hat over that letter to denote the corresponding element or subset of F under the natural

correspondence that sends elements of F to their projective equivalence classes.

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Since F has dimension n + 2, the projective space F has dimension n + 1. F can be decomposed

into the following disjoint subsets:

(i) A subset of all circles �� where p = aX.X + b.X + c with coefficient a ≠ 0.

By completing the square, we can rewrite each circle as p = aX.X + b.X + + c

= a(X + b/2a). (X + b/2a) – as, where s = (b.b – 4ac)/4a2 = <p, p>/4a

2. By

geometric analogy, we will refer to the point (–b/2a) in E as the center of the

circle p, and the number s as the size of the circle form. In case k = R, the

number s would obviously be equal to the square of the radius r if Z(p) is an

ordinary geometric circle or sphere.

Note that all circles have centers, even though some circles may not have any

zeros in E. We will refer to circles with size s = 0 as circles of size zero or with

zero radius, or even more informally as zero circles.

(ii) A projective subspace of dimension n corresponding to all affine functions p =

b.X + c. The constant functions p(X) = c represent an exceptional point of this

subset, which we will denote by the infinity symbol ∞.

There is a natural bijection between zero circles in the projective space F and the elements of E.

For any point in E, take the set of all circles with zero radius centered at that point. These circles

are all scalar multiples of each other, and map to a unique projective element in F.

Using standard language for bilinear spaces, we say that u is orthogonal to v if <u, v> = 0. Since

orthogonal relations are obviously unchanged when we replace each vector by a scalar multiple,

such relations are consistent with projective equivalence and it makes sense to talk about

orthogonal elements in the projective space F.

For any subset S of F (or F) we denote by S#

the subset of elements of F (or F) that are

orthogonal to all elements in S. If S is a linear subspace of F or a projective subspace of F, then

so is S#.

A vector u of F is said to be isotropic if <u, u> = 0 and non-isotropic or anisotropic otherwise.

This concept is clearly compatible with the projective equivalence, so we can similarly refer to

isotropic and non-isotropic points in the projective space F.

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The proposition below summarizes the basic orthogonal relationships in F.

Proposition 2.1:

(a) Any constant function is orthogonal to itself and all lines. A non-zero constant

function is not orthogonal to any circle.

(b) Two lines p = b.X + c and q = d.X + e have scalar product <p, q> = b.d.

Consequently, p and q are orthogonal in F if and only if b and d are orthogonal in

E. A line is therefore non-isotropic.

(c) A circle p with center C is orthogonal to a line q if and only if q(C) = 0, i.e. the

affine hyperplane defined by the equation q(x) = 0 passes through C.

(d) For circles p = a(X + b/2a). (X + b/2a) – as and q = d(X + e/2d). (X + e/2d) – dt,

the scalar product of p and q are given the following formula:

<p, q> = 2ad(s + t – w), where w = (b/2a – e/2d). (b/2a – e/2d)

(e) The only isotropic vectors in F are the constant functions and the zero circles,

namely functions p(X) of the form a(X + b/2a). (X + b/2a).

Proof. Statements (a) and (b) are clear.

For statement (c), let p = a(X + b/2a). (X + b/2a) – as = aX.X + b.X + c be a circle with center C

= –b/2a. Let q = e.X + f be a line. We then have <p, q> = b.e – 2af = –2a(e.( –b/2a) + f) = –

2a.q(C) = 0 if and only if q(C) = 0.

For statement (d), let:

p = a(X + b/2a). (X + b/2a) – as = aX.X + b.X + c where s = (b.b – 4ac)/4a2

q = d(X + e/2a). (X + e/2a) – dt = dX.X + e.X + f where t = (e.e – 4df)/4d2

then by definition <p, q> = b.e – 2af – 2dc.

Because w = (b/2a – e/2d). (b/2a – e/2d) = b.b/4a2 + e.e/4d

2 – 2b.e/4ad, we have, after

cancelling out the terms b.b/4a2 and e.e/4d

2:

s + t – w = – c/a – f/d + 2b.e/4ad, or

2ad(s + t – w) = b.e – 2af – 2dc = <p, q>

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Statement (e) follows from (a) through (d). Among the affine functions, the only isotropic

vectors are clearly the constant functions. For a circle p, <p, p> = b.b – 4ac = 2a2(2s – 0) = 0 if

and only if s = 0.

Section 3. Zero sets

Let V be the set of isotropic points in F. We just saw that V consists of: (i) the exceptional point

∞ representing the non-zero constant functions, and (ii) all circles of size zero, which can be

naturally identified with E. Accordingly, we can identify V with E ∪ {∞}. With this

identification, we can think of V as an extension of the affine space E, and we will refer to the

exceptional point ∞ of V as the point at infinity of the extended affine space E.

We also have the following description for the zero set Z(p) of any function p in F.

Proposition 3.1: Let p be a non-zero element of F.

(a) The orthogonal complement of (p), written as (p)#, is a linear subspace of co-dimension

1 (i.e., a hyperplane). If p is non-isotropic, then the hyperplane (p)# is regular with

respect to the induced scalar product, and F is isometric to the direct sum of (p) and (p)#.

(b) If p is a line, then the intersection (𝑝) # ∩ V contains the following: (i) the exceptional

point ∞ and (ii) all those points of E that lie on the affine hyperplane Z(p). (Here we are

identifying E with the subset of V consisting of zero circles.)

(c) If p is a circle, then the intersection (𝑝) # ∩ V contains precisely all those points of E that

lie on the zero set Z(p) of p. Z(p) can be empty, in which case (𝑝) #

does not intersect V.

Proof. Statement (a) follows from the basic facts about symmetric bilinear spaces. See,

e.g., [Artin], particularly Chapter III (Symplectic and Orthogonal Geometry), or [Szymiczek],

Chapter 5 (Nonsingular Bilinear Spaces).

Statement (b) is an immediate consequence of Proposition 2.1 (a) and (c). Statement (b) implies

that the exceptional point ∞ belongs to the orthogonal complement (𝑝) #

of any line p. 3

3 In classical inversive geometry, it is greatly convenient to adopt the convention that any line in

𝐑2 passes through the point at infinity. Statement (b) explains why that convention is helpful.

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In most cases, as we will see more clearly later in this paper, the essential set that we look at is

not really the zero set Z(p), but the orthogonal complement (𝑝) #

of p and particularly its

intersection with V. For a line p, such a set includes not just the points on the affine hyperplane

Z(p), but also the exceptional point ∞ of V.

For statement (c), assuming that p is a circle, then the only elements of V that are potentially

orthogonal to p are the zero circles. A zero circle q is orthogonal to p if and only if (using the

same notation as in Proposition 2.1), the expression s + t – w = s – w = 0. But aw = a(b/2a –

e/2d). (b/2a – e/2d) = p(–e/2d) + as. Therefore, s = w if and only if as = aw or p(–e/2d) = 0.

This means the center of the circle form q belongs to the zero set Z(p) of p.

The zero set Z(p) can of course be empty, e.g., consider the case s < 0 when F = R. In that case,

the orthogonal complement (𝑝) #

does not intersect with V.

We can say more about the intersection of V and (𝑝) # for circles p.

Proposition 3.2: Let p be a circle in F. For convenience, we identify the zero set Z(p) in E

with the intersection V ∩ (𝑝) # by virtue of Proposition 3.1.

(a) Z(p) consists of exactly one point if and only if p is a zero circle.

(b) If p is not a zero circle, then Z(p) is non-empty if and only if <p, p> is represented

by the quadratic form on E. If not, then Z(p) is empty.

If the orthogonal complement (p)# is regular then either it has no isotropic vectors, or it contains

a hyperbolic plane (with two linearly independent isotropic vectors). Therefore Z(p) consists of

exactly one point if and only if the orthogonal complement (p)# is singular. That happens if and

only if the circle p is isotropic, that is when p is a circle of size zero.

Assume that p is not a zero circle. To determine whether or not Z(p) is empty we do a

comparison with Z(q) where q is a line.

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If the subspace (p) is isometric to (q), then Witt’s extension theorem says that there is an

orthogonal transformation of F mapping p to q.4 The same transformation would map (p)

# to

(q)#, and the intersection (p)

# ∩ V to the intersection (q)# ∩ V (because the set V of isotropic

elements is stable under such an orthogonal transformation). Accordingly, if (p) is isometric to

such a (q), then Z(p) must be non-empty, since the set (q)# ∩ V consists of the affine hyperplane

Z(q) and ∞.

The subspace (p) is isometric to (q) if and only if the numbers <p, p> and <q, q> differ by a

square factor. For a line q, the scalar product <q, q> is equal to the norm of a vector in E, in

light of Proposition 2.1(b). That means (p) is isometric to (q) if and only if the number <p, p> is

represented by the quadratic form on E.

What we have shown is that if <p, p> is represented by the quadratic form on E, then Z(p) must

be non-empty. The converse is also true. Write the circle p as

p(X) = aX.X + b.X + c = a(X + b/2a). (X + b/2a) – as,

where s = (b.b – 4ac)/4a2 = <p, p>/4a

2.

If Z(p) is empty, then there is no vector x in E satisfying the equation

(x + b/2a). (x + b/2a) = s.

That means s cannot be represented by the quadratic form on E. Because <p, p> = 4a2s, <p, p>

also cannot be represented by the quadratic form on E.

Section 4. A geometrical interpretation of the scalar product on circles and lines

It is worthwhile to note the following geometrical interpretation of the scalar product defined on

the vector space F of circles and lines in the classical case of a Euclidean space Rn.

Proposition 4.1: For E = Rn, let p and q be circles with sizes s and t respectively.

4 For a statement and proof of Witt’s extension theorem (which is also known equivalently as Witt’s

cancellation theorem), the reader can consult a wide range of books on quadratic forms or geometric

algebra, such as [Artin], theorem 3.9, or [Szymiczek], Chapter 7 (Witt’s Cancellation Theorem).

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(a) Assume that both s and t > 0. In other words, Z(p) and Z(q) are ordinary spheres

in Rn. Then <p, q> = 0 if and only if the associated spheres are orthogonal in the

classical sense.

(b) If both s and t are < 0, then <p, q> ≠ 0.

(c) Assume that s > 0 but t <0. Let q* be the circle obtained by replacing t by the

number –t in the expression for q. We have <p, q> = 0 if and only if the

intersection of Z(p) and Z(q*) lies in an affine hyperplane passing through the

center of Z(q*).

Proof. Note that <p, q> = 0 is equivalent to s + t – w = 0 in light of Proposition 2.1(d).

When s and t are both > 0,

s = square of the radius of Z(p)

t = square of the radius of Z(q)

w = square of the distance between the centers of Z(p) and Z(q).

The equation s + t – w = 0 is then a simple restatement of the classical meaning of orthogonal

spheres in Euclidean space Rn.

If both s and t are < 0, then s + t – w < 0 (as w is by definition always > 0, being the square of the

distance between two center points). Accordingly, <p, q> is never zero in this case.

If s > 0 but t < 0, then rewrite s + t – w = 0 as s = – t + w. That means the square of the radius of

Z(p) is equal to the square of the radius of Z(q*) plus the square of the distance between the

center points of Z(p) and Z(q*). (q and q* have the same center points.) That happens if and

only if the hyperplane that passes through the intersection of Z(p) and Z(q*) (and which is

orthogonal to the line connecting the two center points) also passes through the center point of

Z(q*). When n = 2, that is the same as saying that the geometric circle Z(p) cuts the geometric

circle Z(q*) in a diameter of Z(q*).

Section 5. The group of projective orthogonal transformations.

We consider the group O(F) of orthogonal transformations of the vector space F that leaves

invariant the scalar product defined on F. In other words,

O(F) := {transformations L ∈ GL(F) such that <L(p), L(q)> = <p, q> for any p, q in F}.

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We denote by G the corresponding group of projective orthogonal transformations on F, i.e., it is

the subgroup of PGL(F) induced by orthogonal transformations in O(F).

The most basic orthogonal transformations in O(F) are the hyperplane reflections defined by

non-isotropic vectors. Given a non-isotropic vector p in F, we can express F as the direct sum of

(p) and its orthogonal complement (p)#. Any vector x of F is then equal to a unique sum x = u +

v, where u ∈ (p) and v ∈ (p)#. The reflection R across the orthogonal hyperplane (p)

# is the

linear transformation: x = u + v → – u + v.

Expressed in terms of p, the orthogonal reflection R has the following well-known formula:

R(x) = x – 2 <𝒙,𝑝>

<𝑝,𝑝> p

(Note that < 𝑝, 𝑝 > ≠ 0 because p is a non-isotropic vector.)

As the reader can readily verify from the above formula, R sends p to – p, and for any vector v

orthogonal to p, R (v) = v.

The theorem of Cartan – Dieudonne says that the group O(F) is generated by hyperplane

reflections. More precisely, each orthogonal transformation in O(F) can be expressed as product

of at most dimF = n + 2 hyperplane reflections defined by non-isotropic vectors of F. See, e.g.,

[Artin], theorem 3.20.

According to Proposition 2.1, the set V of isotropic elements of F can be naturally identified with

E ∪ {∞}. Each orthogonal transformation in G maps V to itself, as the image of an isotropic

vector is clearly another isotropic vector. We therefore have a natural group action of G on V.

Theorem 5.1: The natural group action of G on V is faithful and doubly transitive.

Proof. To say that the action of G on V is doubly transitive means that we can always

find an orthogonal projective transformation that moves any two distinct points u, w of V to any

other two distinct points u’, w’ of V. Indeed, let M be the 2-dimensional subspace of F spanned

by {u, w} and N the subspace spanned by {u’, w’}. Both M and N are isometric to a hyperbolic

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plane. By Witt’s extension theorem, there is an orthogonal projective transformation in G that

maps u to u’ and w to w’.

To say that the action of G on V is faithful means that the only transformation in G leaving all

elements of V invariant is the identity transformation. Recall that we can identify V with

E ∪ {∞}, where ∞ is identified with the non-zero constants, and each point of E is identified with

the zero circles centered at that point. Let w be the zero vector in E, and let u1, u2, …, un be an

orthogonal basis in E. Then it is easy to see that the images of w, u1, u2, …, un, and ∞ under this

identification come from the following linear basis of the vector space F:

v0 = X.X

v1 = X.X – 2u1.X + 1

………….

vn = X.X – 2un.X + 1

vn+1 = 1

In addition to the (n+2) points w, u1, u2, …, un, and ∞, we can also find another point z in V

whose expression as linear combination of v0, v1, …, vn, and vn+1 has no zero coefficient.

Indeed, consider z = (2u1 + u2 + … + un) in E. As an element of V, it is the point corresponding

to the zero circle X.X – 2(2u1 + u2 + … + un).X + (n + 3) = (– n)v0 + 2v1 + v2 + … + vn + 2vn+1

in F.

From linear algebra, we know that the (n + 3) points of V corresponding to w, u1, u2, …, un, ∞

and z form a projective frame of F. This implies that any projective transformation of F which

leaves these points fixed must be the identity transformation itself. Hence, the action of G on V

is certainly faithful.

In fact, if a non-singular hyperplane H intersects V in a non-empty set W, then a transformation

in G that leaves all points of W fixed must either be the identity transformation or the

orthogonal reflection across the hyperplane H. This is certainly true in the case where H is the

hyperplane orthogonal to u1 (the projective hyperplane defined by the equation v1 = 0). H can

be regarded as a space of lines and circles in (n – 1) variables and the intersection W in that case

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is just the projective set of isotropic elements in that lower space. From the foregoing

discussion, it follows that any orthogonal transformation leaving all of the points of W fixed

must also leave all vectors in H invariant. The only orthogonal transformations with that

property are the identity transformation and the orthogonal reflection across H.

According to the proof of Proposition 3.2, if a regular hyperplane H in F intersects V in a non-

empty set W, then there is a projective orthogonal transformation mapping that hyperplane to the

projective hyperplane orthogonal to u1. So the result also holds in the general case.

Section 6. Affine reflections and circle inversions

Recall that we start with an anisotropic quadratic space E. On the extended affine space E ∪ {∞}

obtained by adjoining an extra point ∞ called the “point at infinity,” we can define certain

transformations that are analogous to the classical inversions of Euclidean plane geometry.

The zero set of any affine function L(x) = b.x + c where b ≠ 𝟎 is an affine hyperplane in E. The

corresponding affine reflection is the mapping that sends ∞ to itself, and any finite point x to a

finite point x’ = x – 2b L(𝐱)

𝐛.𝐛. Note that b.b is a non-zero number because the space is

anisotropic and b ≠ 𝟎 by hypothesis.

Likewise, for any non-zero circle a(x – b).(x – b) – as, where a and s are non-zero elements of F,

we can define an “inversion” mapping the space E ∪ {∞} to itself as follows:

b ↔ ∞,

for any finite point x ≠ b, x → the finite point x’ collinear with b and x, and such that

(x – b).(x’ – b) = s.

The collinear condition and dot product equation are equivalent to x’ – b = (x – b) s

(𝐱−𝐛).(𝐱−𝐛) .

Note that (x – b).(x – b) is a non-zero number because (x – b) ≠ 𝟎 and the quadratic space E is

anisotropic by hypothesis. This shows that inversion is a well-defined transformation

regardless of whether or not the general circle a(x – b).(x – b) – as has any zero in the space E.

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Any non-zero scalar multiple of a line or circle defines the same mapping, and any reflection or

inversion so defined is a transformation of E ∪ {∞} that is its own inverse. If x and x’ are

mapped to each other by such a transformation, we will call these points inverse points or

conjugate points relative to the line or non-zero circle that defines the transformation.

We denote by Inv(E) the group of transformations of E ∪ {∞} generated by such affine

reflections and circle inversions. We will refer to the elements of Inv(E) as inversive

transformations.

The classical case is when E = R2. If we identify 𝐑𝟐 ∪ {∞} with C ∪ {∞} (the extended

complex line), then the group Inv(R2) = Inv(2, R) contains all the Mobius transformations (also

called fractional linear transformations or homographies) and the conjugate Mobius

transformations (obtained by combining the regular Mobius transformations with the action of

complex conjugation). The group of Mobius transformations is a subgroup of index 2 in the

inversive group Inv(2, R), and can be characterized as the subgroup of all inversive

transformations that can be written as product of an even number of the basic inversions or

reflections.

For n > 2, the group Inv(n, R) has long been the subject of study in geometry, often under the

name of Mobius group and Mobius transformations. See, e.g., [Ratcliffe] chapter 4 (Inversive

Geometry). The importance of the inversive groups Inv(n, R) in geometry derives from

conformal properties of such inversive transformations and the fact that the groups are

isomorphic to the isometry groups of certain standard hyperbolic manifolds. Our perspective on

the groups Inv(E) as presented in this paper is different, since we approach them mainly from the

view of algebra and relying on methods that apply to arbitrary anisotropic spaces E. We also

have used the name inversive transformations rather than Mobius transformations in order to

minimize the risk of confusion, since Mobius transformations are generally regarded as

fractional linear transformations with complex coefficients, and that group is a proper subgroup

of the inversive group for n = 2.

For more historical background on the development of inversive geometry, the reader can refer

to the historical notes at the end of chapter 4 (Inversive Geometry) in [Ratcliffe].

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Section 7. Hyperplane reflections in the projective space of lines and circles

Let p be a non-isotropic vector in F and let R be the hyperplane reflection defined by p. We will

use the same letter R to denote the above orthogonal transformation in O(F) and the projective

transformation in G induced by the same reflection. Because the meaning will be clear from the

context, there should be little risk of confusion.

The following theorem describes the action of those reflections on V.

Theorem 7.1:

(a) If p is a line b.X + c, then the action of R on V is the same as the affine reflection

defined by p.

(b) If p is a non-zero circle (X – b). (X – b) – s with s ≠ 0, then the action of R on V is

the same as the inversion with respect to the circle p.

Proof. We will prove (a) and (b) by showing that in each case the action of R on V

coincides with such an affine reflection or circle inversion, which we denote by the letter L.

Recall that the reflection R is the orthogonal transformation with the formula

R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p

We begin with statement (a). When p is the line b.X + c, <p, p> = b.b. We know that the

exceptional point ∞ is orthogonal to p. That means R(∞) = ∞ = L(∞).

For any finite point w in V represented by a zero circle q = X.X – 2w.X + w.w, we note first that

L(w) by definition is the point w’ = (w – b 2𝑝(𝒘)

𝒃.𝒃).

We have <q, p> = –2b.w – 2c = –2p(w). Hence,

R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p = q – 2

–2𝑝(𝒘)

𝒃.𝒃 (b.X + c)

= X.X – 2(w – b 2𝑝(𝒘)

𝒃.𝒃).X + w.w + 4c

𝑝(𝒘)

𝒃.𝒃

16

= X.X – 2 w’.X + w.w + 4c 𝑝(𝒘)

𝒃.𝒃

Now observe that:

w’.w’ = w.w + 4𝑝(𝒘)𝑝(𝒘)

𝒃.𝒃 –

4𝑝(𝒘)𝒃.𝒘

𝒃.𝒃 = w.w + 4

𝑝(𝒘)

𝒃.𝒃 (p(w) – b.w) = w.w + 4

𝑝(𝒘)

𝒃.𝒃c

Substituting the above in the expression for R(q), we have R(q) = X.X – 2w’.X + w’.w’.

What the above expression means is that the finite point w in V is mapped to w’ under the

reflection R. But we already see that w’ = L(w) is the image of w under the reflection across

the affine hyperplane Z(p).

Accordingly, R has the same action on V as the affine reflection L. This proves statement (a).

We now consider statement (b). Let p be the non-zero circle (X – b). (X – b) – s = X.X – 2b.X +

+ b.b – s with center b and size s ≠ 0.

The inversion L with respect to the circle p is by definition the mapping:

b ↔ ∞,

for any finite point x ≠ b, x → the finite point x’ collinear with b and x, and that

satisfies the dot product equation (x – b).(x’ – b) = s.

For any vector q, we have R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p. From the defining formula for the scalar

product, it is easy to verify that:

<p, p> = (– 2b). (– 2b) – 4(b.b – s) = 4b.b – 4b.b + 4s = 4s.

For the constant function q(X) = s, <q, p> = – 2s.

Hence, if q is the constant function q(X) = s,

R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p = s + p = (X – b). (X – b)

But the zero circle (X – b). (X – b) = X.X – 2b.X + b.b is exactly the image of b under the

identification V ↔ E ∪ {∞}. Hence b ↔ ∞ under the action of both R and L.

17

For any finite point m ≠ b, take the zero circle q(X) = X.X – 2m.X + m.m corresponding to m

under the identification V ↔ E ∪ {∞}. We know that the image R(q) of q will be projectively

equivalent to another zero circle q’(X) = X.X – 2m’.X + m’.m’ associated with some point m’.

Our task is to determine m’.

In this case, <q, p> = 4m.b – 2m.m – 2b.b + 2s = 2(s – (m – b). (m – b)). Accordingly,

R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p = q – (1 −

(𝒎−𝒃).(𝒎−𝒃)𝑠

)p

If q’ = X.X – 2m’.X + m’.m’ is projectively equivalent to R(q), then we must have

λq' = q – (1 – (𝒎−𝒃).(𝒎−𝒃)

𝑠 )p for some non-zero scalar λ.

Because we are working with m ≠ b, (m – b). (m – b) is a non-zero number. To simplify our

expression, let α be the number in k such that

1/α = (𝐦−𝐛).(𝐦−𝐛)

s or equivalently α(m – b).(m – b) = s.

We then have λq' = q – (1 – 1/α)p.

Equating the respective coefficients of X.X and X from each side, we have the equations:

λ = 1/α

λm’ = m – b(1 – 1/α)

Because αλ = 1, we have:

m’ = αm – (α – 1)b

or (m’ – b) = α(m – b)

But the last equation implies that (i) m’ is collinear with m and b, and (ii) (m’ – b).(m – b) =

α(m – b).(m – b) = s. These are exactly the conditions for m’ and m to be conjugate or inverse

points with respect to the circle equation p(x) = 0.

To sum up, the orthogonal reflection R defined by the non-zero circle p has the same action on

V as the inversion L with respect to the same non-zero circle p. This proves statement (b), and

our proof of Theorem 7.1 is complete.

18

Section 8. Inversive geometry is projective geometry

Because the lines and non-zero circles p in F for which a reflection or inversion on the space of

E ∪ {∞} can be defined are exactly the non-isotropic vectors for which we can define orthogonal

reflections in F, the above theorem shows that all inversive transformations in Inv(E) come from

orthogonal transformations in in G. Stated differently, the natural action of G on the variety V

is exactly the same of the action of the group Inv(n, E) on the space E ∪ {∞} under the natural

identification of V and of E ∪ {∞}. Because the actions of both groups are faithful, that means

the two groups are naturally isomorphic.

Accordingly, inversive geometry can be regarded as a particular case of projective geometry.

We know that the zero set Z(p) of any line or circle p is just the intersection of (𝑝) # with V.

Because orthogonal transformations map hyperplanes to hyperplanes while leaving V stable, the

above projective description of inverstive transformations means that inversive transformations

map a zero set to another zero set. Moreover, such action is orthogonal in the sense that it maps

orthogonal zero sets to orthogonal zero sets.

The classical study of pencils and bundles of hyperplanes and spheres can best be viewed as the

study of 1-dimentional and 2-dimensional subspaces of F. For example, consider the classical

case of the Euclidean plane 𝐑2 . The corresponding vector space F of lines and circles has

dimension 2 + 2 = 4 and the projective space F has dimension n + 1 = 3. Let H be a 2-

dimensional subspace of F. Then H is a hyperplane in F, and the orthogonal complement of H

has dimension 1. That means we can always describe H as the set of projective points

orthogonal to some �� in F. The traditional classification of such bundles defined by H into

parabolic, hyperbolic and elliptic types follows naturally from the above general consideration

based on the character of the orthogonal vector p (isotropic, non-isotropic with non-empty zero

set, non-isotropic with empty zero set). For a discussion of the classical view, see, e.g.,

[Caratheodory], chapter 1 (Mobius transformations).

19

The traditional classification of pencils into 3 different types also has a natural explanation. For

any n, let L be a 1-dimensional projective subspace in F. Looking at the induced scalar product

on the 2-dimensional vector space L, we have 3 possibilities:

(i) L is singular. In this case, L cannot be totally isotropic because the isotropic variety

V contains no subspace of dimension 1 or more. Hence, its radical is 1-dimensional

and the pencil L is tangent to V at a point representing that radical. That means all

elements in the pencil L have a common zero in V.

(ii) L is regular and anisotropic (the pencil L does not intersect V). In this case its

orthogonal complement L# is also regular and must contain isotropic vectors. (F is

the direct sum of L and L# and F contains isotropic vectors.) All vectors in L are

orthogonal to these isotropic vectors, which means that these points of V are the

common zeros of all the circles and lines in L.

(iii) L is regular and isotropic. In this case, L is a hyperbolic plane, meaning that the

pencil L intersects V at exactly two distinct points).5 As we will see in further

discussion below, the points in E ∪ {∞} corresponding to these isotropic points are

conjugate or inverse points relative to any circle or line in L.

When n = 2, F has dimension 4, and both L and L# have dimension 2. If L has two independent

isotropic vectors (hyperbolic), then its orthogonal complement L# must have none (anisotropic),

and vice versa.

Section 9. Conjugate points

Recall that a pair of points in E ∪ {∞} are said to be conjugate or inverse with respect to a line or

non-zero circle p if they are mapped to each other under the reflection or inversion defined by p.

Theorem 7.1 shows that two points are conjugate with respect to p if and only if (identifying V

with E ∪ {∞}) the zero circles associated to these points are mapped to each other by the

orthogonal reflection R defined by the non-isotropic vector p in F.

5 Recall that over a field of characteristic ≠ 2, a plane with a regular symmetric bilinear form is isotropic

if and only if it is a hyperbolic plane.

20

The following proposition provides another characterization of such conjugate points.

Proposition 9.1: Two distinct points m and m’ are conjugate with respect to p if and only if

the non-isotropic vector p belongs to the linear subspace spanned by the following two zero

circles associated with m and m’.

q(X) = X.X – 2m.X + m.m and q*(X) = X.X – 2m’.X + m’.m’

Proof. For points m and m’ to be conjugate with respect to p, we must have:

λq* = R(q) = q – 2 <𝑞,𝑝>

<𝑝,𝑝> p for some non-zero scalar λ

This means that q* is spanned by q and p, i.e., the projective classes of p, q and q* are collinear.

The point m’ (represented by q*) is in the line through m (represented by q) and the projective

class of p.

Because m and m’ are distinct, the coefficient of p in the above equation must also be non-zero.

Accordingly, to say q* is spanned by q and p is the same as saying p is spanned by q and q*.

Conversely, suppose that a non-isotropic element p is spanned by q and q*. The pencil L

generated by p and q intersects V in at most two points, as discussed in Section 8 above. In this

case, the pencil L intersects V in exactly two distinct points, m and m’. The conjugate of m is in

the same pencil, as evidenced in the first half of the proof. Such a conjugate point of course lies

in V, and hence that conjugate point must be the same as point m’.

Proposition 9.2: Let u and v be two non-isotropic vectors in F. Recall our convention that

the zero set of a line includes also the point at infinity.

(a) If the zero set Z(v) passes through two points m and m’ in E ∪ {∞} that are

conjugate with respect to u, then <u, v> = 0.

(b) Conversely, assume that <u, v> = 0. Then for any point m in the zero set Z(v), its

conjugate point m’ with respect to u also belongs to Z(v).

Proof. Let q and q* be two zero circles associated with the points m and m’ in E ∪ {∞}.

If Z(v) passes through m and m’, then that means v is orthogonal to both q and q* in light of

21

Proposition 2.1. Therefore v is orthogonal to all vectors spanned by q and q*, which includes u

by virtue of Proposition 9.1.

Conversely, if v is orthogonal to u, and v is also orthogonal to the zero circle q, then v must be

orthogonal to all vectors spanned by u and q, which includes q*. That means v(m’) = 0 or that

the conjugate point m’ also lies on the zero set Z(v) just like m does.

In the classical case of R2, the above result tracks the well-known classical theorem about

orthogonal circles and conjugate points.

Section 10. Stereographic projection

In this section, we briefly discuss the well-known stereographic projection mapping and its

generalization to our case.

Recall our basic setting of an anisotropic quadratic space E defined over the field k. Let L be the

vector space of elements of the form (x, y, z), where x and z are numbers in k, and y is a vector

in the space E.

Define a scalar product < _ , _> on L as follows. For t = (x, y, z) and t* = (x*, y*, z*), we

define the scalar product <t, t*> of t and t* as the number:

<t, t*> = y.y* + xx* – zz*

This scalar product on L is clearly non-degenerate, since it makes L isometric to the sum of E

and a hyperbolic plane. We will call this scalar product the Lorentz product by virtue of its

analogy with the classical Lorentz metric.

As with our foregoing study of F and its scalar product, we can look at the projective space L and

the set U of all isotropic elements in the space L. 6

6 In the case of an ordered field k, the set U lies entirely in the affine subset of the projective space L

corresponding to z = 1 because the equation x2 + y.y = 0 has no non-zero solution. If we identify the

vector (x, y, 1) in L with the point (x, y) in k1+n

, then the set U can be identified with the unit sphere Sn of

all points (x, y) in k1+n

satisfying the equation x2 + y.y = 1.

22

Now consider the following linear mapping S from L to F:

S: (x, y, z) → (z − x)

2 X.X – y.X +

(z + x)

2

S is clearly an invertible linear mapping of L onto F. When x ≠ z, it maps (x, y, z) to a circle

centered at y/(z – x). It maps (1, 0, 1) to the constant function p(X) = 1.

Moreover, S is compatible with the inner products of the two spaces in the sense that:

<u, v> (Lorentz product) = < S (u), S (v)> (scalar product on F)

To see this, note that by virtue of bilinear conditions, it is sufficient to verify

<u, u> = <S(u), S(u)> for any u

If u = (x, y, z), we have <u, u> = y.y + x2 – z

2.

On the other side, for S (u) = (z − x)

2 X.X – y.X +

(z + x)

2, we have:

< S (u), S (u)> = y.y – 4 (z − x)

2 (z+ x)

2 = y.y – (z

2 – x

2) = y.y + x

2 – z

2

Consequently, S is an isometry of the vector space L (endowed with the Lorentz product) onto

the vector space F. The isometry S induces a natural bijection between the set U of isotropic

elements in the projective space L and the set V of isotropic elements in the projective space F,

under which the point (1, 0, 1) of U is mapped to the point ∞ of V = E ∪ {∞} and each other

point (x, y, z) (x ≠ z) of U is mapped to the point y/(z – x) in V = E ∪ {∞} .

This is exactly the good old stereographic projection in case k = R. From our current point of

view, this mapping represents the trace of a larger isometry S between the spaces L and T. From

this perspective, intersections of U with hyperplanes in the space L are transformed into

intersections of V = E ∪ {∞} with hyperplanes in the space F in a linear and orthogonal fashion.

23

Section 11. Inversive geometry in dimension one

The results of this paper apply to any field k of characteristic ≠ 2 in case of dimension 1,

because multiplication in the field k is naturally an anisotropic bilinear form.

In this one-dimensional case, the vector space F has linear dimension 1 + 2 = 3 and the projective

space F has dimension n + 1 = 2 (a projective plane). The elements of F are expressions Ax2 +

Bx + C in one variable x, where A, B, C vary in k.

The set V of isotropic points is identified with the extended line k ∪ {∞}. The constant

expressions C represent the exceptional point ∞, while the expressions A(x – u)2 represent the

finite point u for any number u in the field k.

Any line Bx + C has exactly one zero in F, namely the number (– C/B). Geometrically, the

projective hyperplane orthogonal to such a line intersects V in exactly two points, namely at ∞

and at (– C/B).

The affine reflection defined by the line Bx + C is the transformation that sends ∞ to itself, and

any number x to x’ = x – 2 𝐵𝑥+𝐶

𝐵 = – x – 2(C/B), which is a combination of the inversion x → –

x and the translation x → x – 2(C/B).

For a circle p = Ax2 + Bx + C, rewrite it as A(x + B/2A)

2 – As where the number s = (B

2 –

4AC)/4A2. The circle p is non-isotropic when s is non-zero, or equivalently, when the equation

Ax2 + Bx + C = 0 does not have a double root. That means each non-isotropic circle either has

two distinct roots or none at all. Geometrically, the orthogonal complement of �� either cuts the

set V in two distinct points, or not at all.

The inversion relative to such a non-zero circle is the transformation x → x* described by the

equation: (x* + B/2A)(x + B/2A) = (B2 – 4AC)/4A

2. If B = 0 and A = 1, that is simply the

mapping x* = – (C/x).

From the basic background outlined above, it follows that the inversive transformations in this

case are simply all the fractional linear transformations described by:

x → x’ = 𝑎𝑥+𝑏

𝑐𝑥+𝑑 where a, b, c, d are numbers such that ad – bc ≠ 0,

24

because such transformations are generated by the translations and inversions described above.

The group of all fractional linear transformations of V = E ∪ {∞} can be described as the

projective transformation group PGL(1, k) of the projective line if we identify E ∪ {∞} with the

projective space of dimension 1 over k under the standard identification that maps k to the affine

set (x, 1) and ∞ to the projective point corresponding to the line (x, 0) in k2.

Therefore in the one-dimensional case, inversive transformations can either be identified with

fractional linear transformations on a projective line, or with orthogonal projective

transformations in the space F of lines and circles. This double connection with projective

transformations in dimension 1 (which is valid for any field k of characteristic ≠ 2) can give us

some new insights on the classical geometry of the projective line, as we briefly outline below.

A. Involutions are orthogonal reflections. A major focus of the classical geometry of

projective line is the study of involutions, defined as projective transformations of order

two. From our perspective this focus is very natural, since the involutions in G when the

dimension n = 1 are precisely the hyperplane reflections. Indeed, for odd dimension n,

any involution in G must come from an involution in O(F), since we cannot have M2 = –

IdF for an orthogonal transformation M (the determinant of one side is 1, while the

determinant of the other side is (– 1)n+2

= – 1). An involution in O(F) is simply a

reflection across some regular subspace. Because F has dimension 3 in this case, and

because a reflection R and its negative –R represent the same projective transformation in

G, the involutions in G can be represented by hyperplane reflections.

B. Pairs of conjugate points: Observe that any pair of distinct points (u, w) in V = k ∪ {∞}

can be thought of as the intersection of V with a non-singular line in the space F. Given

any pair of distinct points, we can find an expression Ax2 + Bx + C whose orthogonal

complement intersects V in exactly those two points. If one of the points, say w, is ∞,

then we just take the affine expression x – u. If both points are finite, then we take the

quadratic expression (x – u)(x – w).

Suppose we have an involution I. We know that the involution I must be an orthogonal

reflection defined by some non-isotropic vector. Consider any two distinct points u and

25

w that are conjugate under I, i.e., I(u) = w and I(w) = u. Let H be the non-singular

hyperplane in F that projectively intersects V at u and w. These two points u and w are

the roots of a function q orthogonal to H. Because H is generated by u and w, H is

stable under the action of I. Since I is an orthogonal reflection, (q) must also be stable

under I, and I(q) is either q or – q. The second case cannot happen because H by

hypothesis is not invariant under I. Hence the space (q) must be invariant under the

action of I. In other words, given an involution I, pairs of I-conjugate points are roots of

functions in the invariant hyperplane of I (viewed as an orthogonal reflection).

Conversely, if we have a non-singular pencil L in the projective plane F, then the

orthogonal reflection across L induces an involution I on V = k ∪ {∞}. For any non-

zero element q of L, its orthogonal complement H is stable under the reflection. That

means the intersection of H with V (if non-empty) must be a pair of distinct conjugate

points relative to I. Accordingly, given a non-singular hyperplane L in F, the roots of

the functions in L are conjugate pairs under the involution on V = k ∪ {∞} induced by

reflection across L.

Expressed geometrically, we can say that if elements of non-singular pencil in F intersect

the line V = k ∪ {∞}, then they will do so in conjugate points of an involution.

Section 12. The involution theorem of Desargues

As evidenced in the above outline, the double connection between projective geometry and

inversive geometry in dimension n = 1 can give us significant insights on some classical results

of projective geometry. In particular, it can provide significant insight on the involution theorem

of Desargues and related results, valid for geometry over any field k of characteristic ≠ 2.

Girard Desargues, one of the founders of projective geometry, discovered the following

remarkable theorem, which can be stated roughly as follows: A pencil of conics in a projective

plane will generally intersect a line in pairs of points that are conjugate under an involution.

Specifically, let Q be a projective plane with homogeneous coordinates (u, v, w). A conic in Q

can be thought of as a quadratic form in 3 variables u, v, w, i.e., a homogeneous polynomial of

26

degree 2 in u, v, w. The set of all such quadratic expressions is a vector space C of dimension 6

over k. A pencil of conics can be thought of as either a vector subspace of dimension 2 in C or a

line in the projective space of C.

Let D be a line in Q. The Desargues involution theorem says that the conics in a pencil L will

generally intersect D in pairs of points that are conjugate under an involution. We say

“generally” because this theorem is true for most but not all configurations of D and L. If the

pencil in question is the set of all conics passing through 4 points in general position (meaning in

this case that no 3 of them are collinear), then the most commonly stated condition is that the line

D does not pass through any of the 4 given points.

By changing coordinates if necessary, we can assume that D is the projective line w = 0. The

projection onto the first two coordinates, (u, v, w) → (u, v), defines a projective transformation P

of Q to the projective line k ∪ {∞} parametrized by the coordinate x = u/v. Clearly P maps D

bijectively to k ∪ {∞}.

At the same time, consider the truncation map W: C → F whereby each quadratic form f = Au2 +

Buv + Cv2 + (terms with variable w) is sent to W(f) = Ax

2 + Bx + C. W is a linear mapping of C

to F that is compatible with the projection P in the following sense. A conic f has a zero m on D

if and only if its truncation W(f) has P(m) as a zero on the projective line k ∪ {∞}.

Accordingly, any intersections of the line D with a pencil L in C will be reflected faithfully in the

intersections of the line k ∪ {∞} with the pencil W(L). In particular, the conics in the pencil L

intersect with D in pairs of an involution if and only if the forms in the pencil W(L) intersect

k ∪ {∞} in pairs of an involution. That happens if and only if the pencil W(L) is a non-singular

hyperplane in the space F.

So we have a precise condition for when the Desargues involution theorem would hold for the

line D (w = 0). Just consider the scalar product:

f = Au2 + Buv + Cv

2 + (terms with variable w)

g = au2 + buv + cv

2 + (terms with variable w)

< f, g > = Bb – 2Ac – 2Ca

27

for elements f and g of the pencil L, regarded as a 2-dimensional linear subspace of C. The

Desargues involution theorem holds if and only if the scalar product <f, g> is non-degenerate in

L. That is something that can readily be checked in general (by computing discriminant, for

example).

Section 13. The eleven-point conic

The mappings P: D → k ∪ {∞} and W: C → F also preserve some nice symmetry relations

that we will briefly indicate below.

Recall that in the space F, the non-isotropic vectors are the expressions that either have two

distinct roots in k ∪ {∞}, or none at all. In the projective plane E, the conics that map to non-

isotropic vectors in F are precisely the conics that either intersect the line D in two points or none

at all, and the conics that map to isotropic vectors in F are the conics that intersect D in exactly

one point, namely the conics that are tangent to D.

The equation of any conic defines a scalar product in the vector space (x, y, z), which induces a

scalar product on D (z = 0). Conics that intersect the line D in either two points or none at all are

precisely the conics that induce a non-degenerate scalar product on D. This is because a non-

singular inner product space will either have two different isotropic lines, or none at all, and

these isotropic lines correspond exactly to the projective points where D intersect the conic.

For such a conic f, there is a natural involution on D induced by the scalar product, which sends a

point to its orthogonal conjugate. (This is well-defined because the scalar product is non-

degenerate on D.) The transformation P: D → k ∪ {∞} transports that involution to an

involution on k ∪ {∞}. This involution is the same as the reflection defined by the non-isotropic

vector p = W(f), i.e. the orthogonal reflection along (p). In fact, if W(f) is the expression Ax2 +

Bx + C, then in either case the involution or reflection is just the transformation x → x* defined

by the relation Axx* + Bx/2 + Bx*/2 + C = 0.

This nice symmetry allows us to gain some more insight into the following remarkable conic. In

our situation, let K, L, M, N be four points of E in general position and consider the pencil L of

28

all conics passing through the four points K, L, M, and N. Assume that the Desargues

involution theorem holds for that pencil L and the line D. For each non-degenerate conic in the

pencil L, consider the pole of D relative to that conic. It is known that all such poles as the

conics range over the pencil F constitute a conic E that has some remarkable properties. In

particular, that conic E contains certain 11 points that are defined by the configuration of D and

the 4 points K, L, M, N. Accordingly, it is known as the eleven-point conic.

By hypothesis the conics of the pencil L cut the line D in pairs of an involution. If we look at

the corresponding involution on k ∪ {∞}, we know it is just the orthogonal reflection defined by

a non-isotropic vector p orthogonal to the pencil W(L). What we want to show is that the

eleven-point conic E will be mapped (up to projective equivalence) to p under the truncation map

W: C → F.

Observe first that the conics in L that are tangent to D are mapped to isotropic vectors in the

space F. Because the pencil W(L) is non-singular, it contains either two isotropic vectors or

none at all. That means there are either zero or two conics in F that are tangent to D. But those

tangent points are exactly the poles of D relative to the touching conics, i.e., the points of

intersection between E and D. Hence the conic E will intersect D either in two separate points or

none at all.

Accordingly, the conic E induces an involution on D. We have an involution on D defined by the

pairs of intersection with the pencil L, and now we also have an involution defined by this conic

E. Are these two involutions related? The answer is that yes, they are exactly one and the same

involution.

If the involution induced by the pencil F has two fixed points, then these two fixed

points are also the fixed points of the involution induced by the conic E. That is

because they are necessarily the tangent points of D with two conics in the pencil L

and hence also intersection points between D and E, and these intersection points are

invariant under the orthogonal involution induced by E. But two involutions with

the same two fixed points must be the same.

29

If the involution induced by the pencil L has no fixed point, then we look at the same

configuration and equations in the algebraic closure of the base field k. In that

algebraic closure, any non-singular subspace of dimension 2 must have two isotropic

lines, and therefore the involution induced by the pencil must have two fixed points.

Consequently, by extending the base field to its algebraic closure, we see that the two

involutions are the same. But that can only be the case if they are the same

transformations over the base field k.

Suppose that W maps E to a non-isotropic vector p in the space F. What we have seen is that

under the identification of D with k ∪ {∞} by the projection P, the orthogonal involution induced

by E on the line D is the same as: (i) the involution on k ∪ {∞} induced by orthogonal reflection

across the non-singular hyperplane W(L); and (ii) the involution on k ∪ {∞} induced by the

orthogonal reflection relative to p.

Therefore, the isotropic vector p = W(E) must be the unique vector (up to projective equivalence)

orthogonal to the pencil W(L) in the space F. If E = Au2 + Buv + Cv

2 + (terms with variable w),

and if q = au2 + buv + cv

2 + (terms with variable w) is any conic in the pencil L, then we must

have Bb – 2Ac –2Ca = 0, a remarkable relationship.

30

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