Further, a choice of basis of V leads to an identi¬cation V = Rn , and thus to a

(II)

coordinate system on An , in which points P ∈ An are represented by coordinates

«

x1

¬.·

P ” p = . ∈ Rn where xi ∈ R.

.

xn

’ ’

Two points P, Q ∈ An determine a vector P Q ∈ V as in Figure 4.2. This vector is

(III)

independent of the identi¬cations discussed in (I).

Conversely, a vector x ∈ V can be added to a point P ∈ An to get a new point

(IV)

’ ’

Q = P + x ∈ An , and then P Q = x; see again Figure 4.2. This operation is also

independent of the identi¬cations discussed in (I).

As with the de¬nition of En in 1.3, the de¬nition of An involves an identi¬cation

An = V or An = Rn , followed by the assurance that any other identi¬cation would

do just as well provided that it is related to the ¬rst by a suitable transformation, in

this case an af¬ne linear transformation. How to de¬ne af¬ne space in abstract algebra

(without explicit mention of any origin or coordinates) is a slightly arcane issue, and

is discussed in 9.2.4.

In most of what follows, you can replace R by other ¬elds. The most

Remarks

obviously useful case is an n-dimensional vector space over C, giving rise to An , but

C

af¬ne geometries over ¬nite ¬elds F pn , or over other ¬elds, also have applications in

many areas of math and science. I do not intend to labour this point, because doing it

properly would involve a lot of algebra of ¬elds, and because the course is directed

more towards metric geometries, which are ˜real™ subjects.

Note also that I work here from the outset in a ¬nite dimensional space V . However,

in many areas of math, af¬ne spaces appear as the set of solutions of inhomogeneous

linear equations in in¬nite dimensional spaces: there is no preferred solution, but the

differences x ’ x between any two solutions form a vector space (¬nite dimensional

or otherwise). This happens, for example, in solving Dx(t) = y(t) for functions x =

x(t) in a suitable space of differentiable functions, where D is a linear differential

operator and y(t) a given function. The spaces of functions we work in, and sometimes

also our af¬ne space of solutions, are often in¬nite dimensional.

4.3 THE GEOMETRY OF AFFINE LINEAR SUBSPACES 65

4.3 The geometry of affine linear subspaces

An af¬ne linear subspace E ‚ An is a nonempty subset of the form

E = P +U = P +v v ∈U ,

with P ∈ An and U ‚ V a vector subspace. By Proposition (1) below, any point of

E will do equally well in place of P, so there is no unique origin speci¬ed in E.

Let P, Q ∈ An be two distinct points. The line spanned by P and Q is

’’

P Q = P + »P Q » ∈ R .

The de¬nition clearly shows that P Q is an af¬ne linear subspace, with U the one

’’

dimensional vector subspace of V generated by P Q ∈ V . As in 1.2, we have the line

segment or interval

’ ’

[P, Q] = P + » P Q 0 ¤ » ¤ 1 .

It is useful to spell this out in vector notation. If P, Q ∈ An correspond to position

vectors p, q, their af¬ne span is the set

pq = p + »(q ’ p) » ∈ R = (1 ’ »)p + »q » ∈ R .

The latter is the form of the linear span construction most commonly used. The line

segment now becomes

[p, q] = (1 ’ »)p + »q 0 ¤ » ¤ 1 ,

as shown in Figure 4.3a.

Three points P, Q, R are collinear if they lie on the same line. If I represent the

points by position vectors p, q, r, this means that r = (1 ’ »)p + »q; as we saw in

1.2, there are three subcases here:

±

p ∈ [r, q] so P ∈ [R, Q]

» ¤ 0

0 ¤ » ¤ 1 ⇐’ r ∈ [p, q] so R ∈ [P, Q]

1 ¤ » q ∈ [p, r] so Q ∈ [P, R].

Proposition

Let E = P0 + U be an af¬ne linear subspace of An . Then the vector space U is

(1)

uniquely de¬ned by E; explicitly

’’

U = P Q P, Q ∈ E .

In other words, E = P + U for any P ∈ E.

A necessary and suf¬cient condition for a nonempty subset E ‚ An to be an af¬ne

(2)

subspace is that the line P Q is contained in E for all P, Q ∈ E.

66 AFFINE GEOMETRY

q

1’»

(1 ’ » )p + »q

»

p

Figure 4.3a The affine construction of the line segment [p, q].

P2 + U

P2

P1 + U P1

Figure 4.3b Parallel hyperplanes.

(3) A necessary and suf¬cient condition for E to be an af¬ne subspace is that it is

nonempty, and de¬ned by a set of inhomogeneous linear equations in a coordinate

system.

The proofs are easy exercises in linear algebra. (1) states that E can be translated

back to the vector space U choosing any point P ∈ E; informally, any point P ∈ E

can serve as origin. (3) spells out the other easy way of specifying an af¬ne linear

subspace using coordinates; examples can be found in Exercises 4.1 and 4.5.

Write dim E = dim U for the dimension of a nonempty af¬ne linear subspace

E. The only n-dimensional af¬ne linear subspace is An itself; dim E = 0 means

simply that E consists of a single point, whereas a one dimensional af¬ne linear

subspace is simply a line. The last interesting case with a name of its own is an af¬ne

linear subspace of dimension n ’ 1 (that is, codimension one), a hyperplane. Two

hyperplanes E 1 , E 2 are parallel, if they are translates of the same vector subspace of

V , that is E 1 = P1 + U , E 1 = P2 + U with dim U = n ’ 1, as in Figure 4.3b. An

equivalent condition is to ask that the two hyperplanes should either coincide or have

no common point.

Let ‚ An be any set; an af¬ne linear combination of is any point

Definition

P ∈ An of the form

k

’’

’

P = P0 + »i P1 Pi , where Pi ∈ and »i ∈ R. (2)

i=1

Using position vectors pi of points Pi simpli¬es this expression once more; an

af¬ne linear combination of is any point P ∈ An of the form

4.4 DIMENSION OF INTERSECTION 67

k k

p= µi pi , where pi ∈ and µi ∈ R with µi = 1. (3)

i=0 i=0

This generalises the expression (1 ’ »)p + »q used to parametrise points of the af¬ne

line P Q. The points Pi appear in the form (3) with (»0 , . . . , »k ) = (0, . . . , 1, . . . , 0);

this con¬rms that I really mean = 1 in (3) rather than = 0.

The af¬ne span of any subset is the set of af¬ne linear combinations of . By

contains all lines spanned by pairs of points in . If P ∈

the previous remark,

= P + U , where U ‚ V is the vector subspace spanned by the vectors

then

’ ’

P Q for Q ∈ . Thus ‚ An is an af¬ne linear subspace, in fact the smallest one

containing all the points of .

4.4 Dimension of intersection

The formula

dim U + dim W = dim U © W + dim(U + W ) (4)

for vector subspaces U, W of a ¬nite dimensional vector space is familiar from linear

algebra. You remember the proof: pick a basis of U © W , extend to two bases of U

and W , and the union is a basis of U + W .

Let E, F ‚ An be af¬ne subspaces. Then

Theorem

dim E © F = dim E + dim F ’ dim E, F , (5)

provided that E © F = ….

The exceptional case E © F = … happens if and only if E, F are contained in

parallel hyperplanes. This can happen essentially whatever the dimension of E and F;

more precisely, there exist af¬ne linear subspaces E, F with dim E = a, dim F = b,

E © F = … and dim E, F = c for any a, b < n and any c with

max{a, b} + 1 ¤ c ¤ min{n, a + b + 1}.

The proof of the ¬rst statement is almost trivial: if P ∈ E © F then the

Proof

four af¬ne subspaces in question are translates of the four vector subspaces

E ,F ,E ©F ,E +F ‚V

so that the result follows at once from the linear algebra formula (4).

The counterexamples involve af¬ne subspaces E, F of An contained in parallel

hyperplanes. To be speci¬c, I choose coordinates and put

E ‚ {x1 = 0} F ‚ {x1 = 1}.

and

Then certainly E © F = …. The converse is proved in Exercise 4.3.

68 AFFINE GEOMETRY

Assume that (0, . . . , 0) ∈ E = E and (1, 0, . . . , 0) ∈ F; then F is the translation

by (1, 0, . . . , 0) of a vector subspace F ‚ V contained in {x1 = 0}. The equality (4)

holds, but the point is that E © F = … takes no account of dim E © F . Now E and

F are any vector subspaces contained in the hyperplane given by x1 = 0, so that

dim E , dim F , dim(E + F ) can be anything up to and including n ’ 1. QED

You will ¬nd it instructive to spell out the theorem in a few concrete cases. For

example, if n = 2 and E, F are distinct lines, then dim E, F = 2 and so the conclu-

sion is that E © F is zero dimensional (that is, a point) unless it is empty, the standard

dichotomy of intersecting and parallel lines. For n = 3, see Exercise 4.2.

4.5 Affine transformations

Recall the following de¬nition, which I repeat here for completeness.

A map T : An ’ An is an af¬ne transformation if it is given in a co-

Definition

ordinate system by T (x) = Ax + b, where A = (ai j ) is an n — n matrix with nonzero

determinant and b = (bi ) a vector; in more detail,

« « «

x1 x1 b1

n

¬.· ¬.· ¬.·

x = (xi ) ’ y = ai j x j + bi , . ’ A . + . .

or (6)

. . .

j=1

xn xn bn

The set Aff(n) of af¬ne transformations is the set of ˜allowed symmetries™ of

af¬ne space An . This set consists of invertible maps from An to An (because I require

det A = 0). It acts transitively on An ; that is, a suitable af¬ne transformation maps

any point to any other. In particular, there is no distinguished origin, as I said before:

every point is like every other. Contrast this with the situation in linear algebra, where

the allowed maps V ’ V are the homogeneous linear maps, all mapping the origin

0 ∈ V to itself.

It is immediate that an af¬ne transformation takes an af¬ne linear subspace to an

af¬ne linear subspace; that is, it preserves the incidence geometry of af¬ne linear

subspaces. In Proposition 1.9, I proved a converse statement, under the additional

assumption that T restricts to an af¬ne linear map on each line. In fact, one can prove

that, for n ≥ 2, a bijective map T : An ’ An that preserves lines and is continuous is

actually af¬ne linear. (This is a point where working over R is essential; for a proof,

see Exercise 5.22.)

4.6 Affine frames and affine transformations

A set of points {P0 , . . . , Pk } of An is af¬ne linearly independent if the

Definition

’’

’ ’’

’

k vectors P0 P1 , . . . , P0 Pk are linearly independent in V . In other words, a set ‚ An

k

is af¬ne linearly dependent if there exists a nontrivial relation i=0 »i pi = 0 between

k

position vectors p0 , . . . , pk of points in , with »i ∈ R and i=0 »i = 0; is af¬ne

linearly independent if no such relation exists.

EXERCISES 69

A set ‚ An is an af¬ne frame of reference if it is af¬ne linearly independent

and spans An (compare the notion of Euclidean frame in 1.12). This means that every

point P ∈ An can be written in the form (2) of 4.3 in a unique way; that is, no proper

subset of can span An . Equivalently, = {P0 , P1 , . . . , Pn } where P0 ∈ is any

’’

’ ’’

’

point, and the vectors P0 P1 , . . . , P0 Pn form a basis of V .

In view of the correspondence between bases in a vector space and linear maps,

the last clause gives the following.

Fix one af¬ne frame of reference P0 , . . . , Pn . Then

Proposition

T ’ T (P0 ), . . . , T (Pn )

de¬nes a one-to-one correspondence between af¬ne transformations and af¬ne frames

of reference of An .

4.7 The centroid

The following proposition is usually thought of as part of (plane) Euclidean geometry;

however, it only involves ratios along lines and incidence of lines, so in fact it belongs

to af¬ne geometry. The other ˜famous™ centres of a triangle described in 1.16.4 use

notions such as angle or distance that have no meaning in af¬ne geometry.

Let P, Q, R be three points of An . Then the three medians of P Q R,

Proposition

that is, the three lines connecting each vertex to the midpoint of the opposite side,

meet in a common point S.

Write p, q, r for the position vectors of P, Q, R. Write p = 1 (q + r) for

Proof 2

the midpoint of q and r and s = 2 p + 1 p for the point dividing the segment between

3 3

p and p in ratio one to two. Then s = 1 (p + q + r) is symmetric in p, q, r, so lies

3

on the lines joining q and q = 1 (p + q) and r and r = 1 (p + q). Hence the point S

2 2

with position vector s lies on all medians of P Q R. QED

To reiterate the point: the statement that this is a theorem of af¬ne geometry means

that applying any af¬ne transformation takes Figure 4.7 to a ¬gure with the same

properties, and in particular takes the centroid of a triangle to the centroid.

Exercises

4.1 Consider the 3 planes

: {x ’ 2 = 1 (y ’ z)}, : {x + 2 = y}, : {3(2x + z) = 3y + 1}

1 2 3

2

in af¬ne space A3 . Calculate 1 © 2 and 1 , 2 and ¬nd out whether the dimen-

sion of intersection formula works; if not, why not? (Compare Theorem 4.4.) Ditto

for 1 © 3 and 1 , 3 .

70 AFFINE GEOMETRY

R

Q′ P′

S

R′

P Q

Figure 4.7 The affine centroid.

Q

n 1

m 2

2 1

P R

Figure 4.8 A weighted centroid.

Experiment with 4.4, formula (5) for n = 3 and different E, F. For example, classify

4.2

pairs of lines of A3 into three types, namely intersecting, parallel and skew, drawing

pictures for each case.

Suppose that E, F ‚ An are disjoint af¬ne linear subspaces; prove that there is a

4.3

linear form • on An such that •(E) = 0 and •(F) = 1. [Hint: let P ∈ E, Q ∈ F and

’’

v = P Q. Then E = P + U for a vector subspace U ‚ V , and v ∈ U . Deduce that

/

there exists a linear form on V that is zero on U but nonzero on v.]

4.4 Write down the af¬ne transformation taking

(0, 0), (1, 0), (0, 1) ’ (2, 1), (5, ’1), (3, 8).

Can you map the same points (0, 0), (1, 0), (0, 1) to (2, 1), (5, ’2), (3, 0) by an af¬ne

transformation? Why?

Determine the dimension of the af¬ne linear subspace E of A5 given by the equations

4.5

x1 + x3 ’ 2x5 = 1

x2 ’ 2x4 + x5 = ’2

x1 + 2x2 + x3 ’ 4x4 = ’3.

EXERCISES 71

Find an af¬ne transformation taking E to an af¬ne linear subspace given by x1 =

· · · = xk = 0 for some value of k. [Hint: choose a suitable af¬ne frame consisting of

points on and off the subspaces, compare 4.6.]

Give a determinantal criterion in coordinates for n + 1 points of An to be af¬ne

4.6

linearly dependent (De¬nition 4.6). [Hint: start by saying how you tell whether

3 points of A2 are collinear.]

4.7 In P Q R of Figure 4.8, take points dividing the three sides in the ratios 1 : 2, 1 : 2,

n : m. Assume that the three lines connecting the vertexes to the points on the opposite

sides have a common point. Calculate the value of the ratio m : n. [Hint: follow the

proof of Proposition 4.7. Answer: the ratio is 4 : 1.]

A general project: set up af¬ne geometry over the ¬nite ¬eld F p of integers modulo

4.8

the prime p. Count the number of points of af¬ne space An , and prove analogues of

the theorems of the text. Check that everything remains true, with a single exception

(harder): the statement concerning the centroid fails for one value of p.

5 Projective geometry

The af¬ne geometry studied in Chapter 4 provided one possible solution to the problem

of inhomogeneous linear geometry. However, this turns out not to be the only one.

This chapter treats the alternative: it introduces projective space Pn as another equally

natural linear geometry. The construction of Pn can be motivated starting from af¬ne

geometry in terms of adding ˜points at in¬nity™.

Projective geometry is simple to study as pure homogeneous linear algebra, ignor-

ing the motivation; ˜linear algebra continued™ or ˜more things to do with matrixes™

would be accurate subtitles for this chapter. In Pn , the statement of af¬ne geometry

analogous to the dimension of intersection formula of Theorem 4.4 holds without

the ˜inhomogeneous™ conditions of Chapter 4, so that, for example, two distinct lines

L 1 , L 2 ‚ P2 meet in a point P = L 1 © L 2 without exception.

Projective geometry has lots of applications in math and other subjects. Projective

transformations include the perspectivities, or projections from a ¬xed viewpoint from

one plane to another, that form the foundation of perspective drawing; the fact that

you can readily recognise an object from any angle, or a photograph taken from any

point (and viewed at any angle) indicates that your brain processes perspectivities

automatically and instantaneously.

5.1 Motivation for projective geometry

Recall from Chapter 4 that if E, F are af¬ne linear subspaces of af¬ne space An , then

5.1.1

Inhomo- there is a nice formula 4.4 expressing the dimension of their intersection provided

that E © F = …. One of the points of projective geometry is to get rid of this un-

geneous to

homoge- pleasant condition. The trouble all comes from the inhomogeneity of the equations:

simultaneous inhomogeneous equations include, say, x1 = 0 and x1 = 1, where only

neous

two equations reduce An to the empty set.

ai j x j = bi is a set of in-

The solution is the following formal trick. Suppose

homogeneous equations in n unknowns x1 , . . . , xn de¬ning an af¬ne linear subspace

E ‚ An . Replace these by homogeneous equations ai j x j = bi x0 in n + 1 un-

knowns x0 , x1 , . . . , xn . The solutions with x0 = 0 give ratios x1 /x0 , . . . , xn /x0 that

72

5.1 MOTIVATION FOR PROJECTIVE GEOMETRY 73

give a faithful picture of E ‚ An . But there are also the solutions with x0 = 0, called

˜points at in¬nity™. Including these points adds information to the set of ordinary

solutions; namely, information about all the ways the ratios x1 : · · · : xn can behave

as the xi tend to in¬nity. A solution 0, ξ1 , . . . , ξn (with some ξi = 0) corresponds

not to a point of E, but to an (n ’ 1)-dimensional family of all parallel lines with

slope ξ1 : · · · : ξn satisfying the homogenised equations ai j ξ j = 0, that is, parallel

to some line in E (compare Figure 5.8).

The set E together with these extra solutions is a projective linear subspace of

projective space; the intersection of projective linear subspaces is then governed

by the formula of 4.4 without exception. This does not mean that two projective

linear subspaces cannot have empty intersection; it only means that they have empty

intersection exactly when they have a numerical reason to do so. In modern language,

the quantity dim E + dim F ’ dim E, F on the right-hand side of formula 4.4 is

called the expected dimension of the intersection of E and F; in projective geometry,

linear subspaces always intersect in a subspace whose dimension equals the expected

dimension.

You recognise Figure 5.1a as a plane picture of a cube in R3 . The way it is drawn, the

5.1.2

horizontal parallel edges appear to meet in points of the plane.

Perspective

Suppose I ¬x the origin O ∈ A3 and map points of a plane ‚ A3 , to another

‚ A3 by taking P ∈ into the point of intersection P = O P ©

plane of the

line O P with . A map of this kind is called a perspectivity. It corresponds to putting

your eye at O, with a glass plate, behind it with a ¬gure on it, and drawing

faithfully the ¬gure on the glass as you see it (see Figure 5.1b).

I get a map f : ’ between two planes. It is easy to see that f maps lines

of to lines of , and parallel or concurrent lines L , L , L , on to parallel or

concurrent lines M, M , M on . Here I am ignoring practicalities, such as the

¬nite extent of the plane represented by a physical piece of glass, or the possibility

that some of might poke out in front of rather than behind (see Exercise 5.1 for

details). Strictly speaking, f is only locally de¬ned, and the conclusions should be

quali¬ed by adding ˜within the domain of de¬nition™; the activity takes place in the

real world, and set theoretic niceties do not cause us undue discomfort.

The map f : P ’ P is constructed in linear terms, but is not actually linear

(see Exercise 5.1): choosing coordinates on , = A2 , it can be shown that f is

fractional linear, that is, of the form

Ax + b

f (x) =

Lx + c

where A, b, L and c are 2 — 2, 2 — 1, 1 — 2 and 1 — 1 matrixes. Note that these can

be assembled into a 3 — 3 matrix L b .

A

c

Figure 5.1c depicts the hyperbola x y = 1 and the parabola y = x 2 . Viewed from a

5.1.3

long way off, the hyperbola is very close to the line pair x y = 0. In fact, outside a

Asymptotes

74 PROJECTIVE GEOMETRY

Figure 5.1a A cube in perspective.

P

P'

subject

Π

artist's eye drawing

Π'

Figure 5.1b Perspective drawing.

parabola y = x2

hyperbola xy = 1

˜asymptotically x2 = 0™

˜asymptotically xy = 0™

Figure 5.1c Hyperbola and parabola.

big circle of radius R, either |x| > R and |y| < 1/R or vice versa. One can argue

that, in turn, the parabola is asymptotic to the line x = 0, in the sense that the tangent

line at the point (x0 , x0 ) gets steeper and steeper. This argument is not actually very

2

0, all you can say is y = x 2

convincing: when both x, y x. Nevertheless, in

the theory of conic sections, it is said, for example, that ˜the two branches of the

5.2 DEFINITION OF PROJECTIVE SPACE 75

parabola meet at in¬nity™, or that the parabola ˜passes through the point at in¬nity

corresponding to lines parallel to x = 0.™

The statements on asymptotes are qualitative views of what happens to the curves

when x or y is large (quite vague, even arguable for those in quotes). But we have

not so far said what asymptotic directions or points at in¬nity actually are, which is a

disadvantage in discussing asymptotes formally or in calculating with them. Making

sense of asymptotes (of algebraic plane curves), and providing a simple framework

for calculating with them is one thing that projective geometry does very well.

Here I assume that you know some topology; read this section after Chapter 7 if you

5.1.4

prefer.

Compact-

Af¬ne space An is not compact; in contrast, projective space Pn is compact, as are

ification

its closed subsets, including all projective algebraic varieties. Compact sets are much

more convenient than noncompact ones in many contexts of geometry, topology,

analysis and algebraic geometry. Given a closed set X ‚ An , you can compactify it

by extending An to Pn ; then X ‚ An ‚ Pn and the closure X ‚ Pn is compact. The

points at in¬nity of the closure X correspond in a very precise sense to the asymptotic

lines of X , and are calculated by the same simple trick of adding a homogenising

coordinate x0 . For example, the hyperbola x y = 1 is compacti¬ed to the circle S 1 by

adding the two points (∞, 0) and (0, ∞), and the parabola is compacti¬ed to S 1 by

adding the single point (0, ∞) at which the two branches are said to meet.

5.2 Definition of projective space

Provided you forget about the motivation, the de¬nition is very simple: introduce the

equivalence relation ∼ on Rn+1 \ 0 de¬ned by

(x0 , . . . , xn ) = »(y0 , . . . , yn )

(x0 , . . . , xn ) ∼ (y0 , . . . , yn ) ⇐’

for some 0 = » ∈ R.

In other words x ∼ y if the two vectors x and y are proportional, or span the same line

(1-dimensional vector subspace) through 0 in Rn+1 . Then de¬ne projective space to

be

Pn = Pn = Rn+1 \ 0 ∼ = lines through 0 in Rn+1 .

R

I write (x0 : · · · : xn ) for the equivalence class of (x0 , . . . , xn ); this is the usual notion

of relative ratios of n + 1 real numbers. x0 , . . . , xn are homogeneous coordinates on

Pn . For example, P1 is the set of ratios (x0 : x1 ). If x0 = 0 you might as well just

consider x1 /x0 , but then you are missing one point corresponding to the ratio (0 : 1),

where x1 /x0 = ∞.

In coordinate free language, if V is an (n + 1)-dimensional vector space over R,

write P(V ) for the set of lines of V through 0 (that is, nonzero vectors up to the equiv-

alence v ∼ »v for » = 0). Of course, V ∼ Rn+1 (by a choice of basis), so P(V ) ∼ Pn .

= =

76 PROJECTIVE GEOMETRY

A point P ∈ P(V ) is an equivalence class of vectors v ∈ V , or a line Rv through

0; several kinds of notation are popularly used to indicate that v = (x0 , . . . , xn ) is a

vector in the equivalence class de¬ning P, for example:

P = Pv , P = [v], v = P, Pv = (x0 : · · · : xn ), etc.

To return to the motivation, Pn contains the subset (x0 = 0) consisting of ratios

that can be written (1 : x1 : · · · : xn ), which is thus naturally identi¬ed with An . The

language used for motivating projective geometry is quite unsuitable for developing

the theory systematically. For example, the terminology of ˜points at in¬nity™ is

cumbersome and gives a distorted view of the symmetry of the situation.

The formal language of projective geometry is simply a reinterpretation of the

ideas of linear algebra; the subset with x0 = 0 is not distinguished in Pn , and there

is no discrimination against points of the complement (with x0 = 0). Working with

the de¬nitions of projective geometry and formal calculations in homogeneous co-

ordinates is in many ways easier to understand than how it relates to the motivation

discussed in 5.1.1, and I proceed with this, returning to the motivation in 5.8. So for

the time being, I discuss the geometry of Pn in terms of the vector space Rn+1 , and I

advise you to forget the motivation.

5.3 Projective linear subspaces

The only structures enjoyed by P(V ) are derived from V . Thus all statements or

calculations for P(V ) must reduce to linear algebra in V and the equivalence relation

∼ on points of V .

As a ¬rst example, here is the de¬nition of the line P Q through two points P =

(x0 : · · · : xn ) and Q = (y0 : · · · : yn ) of Pn . First lift to Rn+1 by setting P =

(x0 , . . . , xn ) and Q = (y0 , . . . , yn ) (that is, pick values of xi and yi in the given

ratio), then set

P Q = P, Q = ratios (»x0 + µy0 : · · · : »xn + µyn ) for all (», µ) = (0, 0) .

The point to notice is that »P + µQ is meaningless as a point of Pn , because the

ratio (»x0 + µy0 : · · · : »xn + µyn ) depends on the choice of P and Q within the

equivalence classes of P and Q. However, the set of all » P + µ Q is a well de¬ned

2-dimensional vector subspace of V = Rn+1 , and ratios in it form the line P Q.

Thinking in a purely formal way about vector subspaces of a vector space V gives

the obvious notion of projective linear subspace: if U ‚ V is a vector subspace, P(U )

is the subset (U \ 0)/∼ ‚ P(V ) of lines through 0 in U . In other words, if U ‚ Rn+1

then P(U ) is the set of ratios (x0 : · · · : xn ) with (x0 , . . . , xn ) ∈ U . The dimension of

P(U ) is de¬ned to be dim P(U ) = dim U ’ 1. Thus dim Pn = n.

A 0-dimensional subspace is a single point; a 1- or 2-dimensional projective linear

subspace is called a line or plane; an (n ’ 1)-dimensional subspace is a hyperplane.

I sometimes say k-plane to mean k-dimensional projective linear subspace.

Note that the empty set … is a projective linear subspace: the trivial vector subspace

0 ‚ Rn+1 has P(0) = … ‚ Pn . By convention we write dim … = ’1, to agree with the

5.5 PROJECTIVE LINEAR TRANSFORMATIONS 77

general de¬nition just given. As a rule, prudence might suggest that in mathematical

arguments, we avoid attaching excessive weight to mumbo-jumbo concerning the

empty set or the elements thereof, but here the convention dim … = ’1 has a precise

and useful meaning (in the context of the geometry of linear subspaces only!).

If ‚ P(V ) is a set, write ‚ V for the union of the lines in ; let

Definition

U be the vector subspace of V spanned by , and de¬ne the span or linear span of

= P(U ). This is the smallest projective linear subspace containing .

to be

If P0 , . . . , Ps are (s + 1) points then dim P0 , . . . , Ps ¤ s; equality holds if

and only if the vectors P0 , . . . , Ps ∈ Rn+1 are linearly independent. In this case,

P0 , . . . , Ps are said to be linearly independent in Pn .

5.4 Dimension of intersection

Let E, F ‚ Pn be projective linear subspaces. Then

Theorem

dim E © F = dim E + dim F ’ dim E, F ; (1)

here the convention dim … = ’1 is in use.

Write E, F ‚ Rn+1 for the vector subspaces overlying E and F. Then

Proof

E © F = P( E © F) and E, F = P( E + F). By the linear algebra formula 4.4 (4)

we have

dim( E © F) = dim E + dim F ’ dim( E + F), (2)

and since dim P(U ) = dim U ’ 1 for every vector subspace U ‚ Rn+1 , (1) follows

by subtracting 1 from each term on the left- and right-hand sides of (2). QED

5.5 Projective linear transformations and projective frames of reference

A nonsingular linear map Rn+1 ’ Rn+1 represented by an invertible matrix A acts

in an obvious way on the set of lines of Rn+1 through 0: namely, it takes the line Rv

to R(Av) for every 0 = v ∈ Rn+1 . A map T : Pn ’ Pn is a projective transformation

(also called projectivity or projective linear map) if it arises in this way from a linear

map. In other words, if we write Pv ∈ Pn for the point represented by v ∈ Rn+1 , then

T is a projective transformation if there is an invertible matrix A such that

T (Pv ) = PAv for all v ∈ Rn+1 .

Here Av is the product of A and v, viewed as a column vector. The set of all projective

transformations is written PGL(n + 1).

Because v and »v represent the same point of Pn , a scalar matrix » · id =

diag(», . . . , ») with » = 0 acts as the identity. Moreover, if A is an invertible

78 PROJECTIVE GEOMETRY

matrix and » ∈ R and » = 0, then A and the product »A have exactly the same

effect on every point of Pn . Thus the set of projective transformations is

PGL(n + 1) = invertible (n + 1) — (n + 1) matrixes /R—

where R— = » · id | 0 = » ∈ R .

The following de¬nition, which may seem unexpected at ¬rst, is quite characteristic

of projective geometry.

A projective frame of reference (or simplex of reference) of Pn is a set

Definition

{P0 , . . . , Pn+1 } of n + 2 points such that any n + 1 are linearly independent, that is,

span Pn .

This means

there exists a basis e0 , . . . , en of Rn+1 such that Pi = Pei for i = 0, . . . , n;

1.

2. the ¬nal point Pn+1 is Pen+1 , where

n

en+1 = »i ei , with »i = 0 for every i.

i=0

Indeed, the ¬rst n + 1 points P0 , . . . , Pn are linearly independent, and the ¬nal point

Pn+1 is not contained in any of the n + 1 hyperplanes {xi = 0}. The standard frame

of reference is

Pi = (0 : · · · : 1 : · · · : 0) (with 1 in the ith place)

(3)

and Pn+1 = (1 : 1 : · · · : 1).

n

That is, ei for i = 0, . . . , n is the standard basis of Rn+1 and en+1 = i=0 ei . The

¬nal point Pn+1 = (1 : · · · : 1) is there to ˜calibrate™ the coordinate system.

Let {P0 , . . . , Pn+1 } be the standard frame of reference. Then there

Theorem

is a one-to-one correspondence between projective transformations and frames of

reference, de¬ned by T ’ T (P0 ), . . . , T (Pn+1 ).

n

Write e0 , . . . , en for the standard basis of Rn+1 , and set en+1 = i=0 ei .

Proof

Now let {Q 0 , . . . , Q n+1 } be a different frame of reference, and choose representatives

f0 , . . . , fn , fn+1 ∈ Rn+1 of the points Q 0 , . . . , Q n+1 .

Since e0 , . . . , en and f0 , . . . , fn are two bases of Rn+1 , the usual result of linear

algebra is that there is a uniquely determined linear map A : Rn+1 ’ Rn+1 such that

Aei = fi for i = 0, . . . , n. If f0 , . . . , fn are column vectors, A is the matrix with the

given columns fi . However, that is not what is given, and not what is required! If you

understand that, you have understood the proof.

Indeed, the fi are determined only up to scalar multiples. Start again: for any

nonzero multiples »i fi of fi (for i = 0, . . . , n), there is a uniquely determined linear

map A : Rn+1 ’ Rn+1 such that Aei = »i fi for i = 0, . . . , n, given by the matrix

5.6 P1 AND THE CROSS-RATIO 79

A with columns »i fi . Using the assumption that f0 , . . . , fn is a basis, I choose the

n

»i such that fn+1 = i=0 »i fi . Then, because Q 0 , . . . , Q n+1 is a frame of reference,

»i = 0 for i = 0, . . . , n, and Aen+1 = fn+1 by choice of A. Since A : Rn+1 ’ Rn+1

is a linear map with ei ’ »i fi and en+1 ’ fn+1 , it de¬nes a projective linear map

T : Pn ’ Pn taking Pi ’ Q i for i = 0, . . . , n + 1.

For the uniqueness, let us look back through the construction: ¬rst, the condition

T (Pi ) = Q i for i = 0, . . . , n determines the columns of A up to multiplying each

column by a scalar »i ; so far, any »i will do (possibly different choices for different

columns). Next, the condition T (Pn+1 ) = Q n+1 ¬xes the »i up to a common scalar

n

factor: because we must send en+1 = ei into a multiple of fn+1 = i=0 »i fi , we

have to choose these values of »i . The only remaining choice in A would be to multiply

the whole thing through by a scalar. Thus T is uniquely determined. QED

Projective linear maps of P1 and the cross-ratio

5.6

There exists a unique projective linear transformation of P1 taking

Corollary

any 3 distinct points P, Q, R ∈ P1 to any other 3.

Since any 3 distinct points go into any other 3 points, I can say that projective

linear transformations act 3-transitively on P1 (Figure 5.6a). This means that there

can be no nontrivial function d(P, Q) of 2 points or σ (P, Q, R) of 3 points that is

invariant under these transformations.

However, there is a function of 4 distinct points invariant under projective linear

transformations, namely their cross-ratio {P, Q; R, S}. To de¬ne it, note that any

choices of representatives p, q ∈ R2 \ 0 of P, Q form a basis. Choosing this basis

gives

P = (1 : 0), Q = (0 : 1), R = (1 : ») S = (1 : µ)

and (4)

for some », µ. Set {P, Q; R, S} = »/µ.

Changing the representative q ’ µq sets µ = 1 so that S = (1 : 1). Thus the def-

inition amounts to taking P, Q, S as the frame of reference of P1 , and then de¬ning

{P, Q; R, S} = », where R = (1 : »). Since by Theorem 5.5, the projective transfor-

mation taking P, Q, S to (1 : 0), (0 : 1), (1 : 1) is unique, {P, Q; R, S} is well de¬ned,

and invariant under transformations in PGL(2).

To see the point of cross-ratio, it is useful to compare the invariant

Remark

quantities in A1 and in P1 . In A1 , to be able to measure, you need to ¬x the points 0

and 1, then any other point P is ¬xed by » = (x ’ 0)/(1 ’ 0). In P1 you need also to

¬x the point at in¬nity.

Consider four distinct lines of R2 through O = (0, 0) that are the

Proposition

equivalence classes of P, Q, R, S, and let L be any line of R2 not through the origin

80 PROJECTIVE GEOMETRY

P'

P

•

Q R'

R

Q'

The 3-transitive action of PGL(2) on P1 .

Figure 5.6a

x=0

q

L

O

x = »y

r

s y = µx

p y=0

The cross-ratio {P, Q; R, S}.

Figure 5.6b

intersecting these four lines in p, q, r, s respectively (see Figure 5.6b). Then

p’r q’s

{P, Q; R, S} = · . (5)

p’s q’r

Here the quotients on the right-hand side are ratios of vectors along L. You could

equally take them as ratios of x-coordinates or y-coordinates of the points; or equally,

the ratio of (signed) lengths ±|p’r| · ±|q’r| .

±|q’s|

±|p’s|

As in the de¬nition of {P, Q; R, S}, choose p and q as the standard basis

Proof

of R . Then L is given by x + y = 1. If », µ are as in (4) then r ∈ R2 is in the

2

equivalence class of (1 : ») and is on L, so that necessarily

(1, ») (1, µ)

r= similarly s = .

; (6)

1+» 1+µ

The remaining calculation is very easy:

(1, ’1)

» ’1

p’r= (1, ’1), q’r= p’r q’s »

=’ · =.

1+» 1+»

(7)

(1, ’1) p’s q’r µ

µ ’1

p’s= (1, ’1), q’s=

1+µ 1+µ

This proves the proposition. QED

5.8 AFFINE SPACE An AS A SUBSET OF PROJECTIVE SPACE Pn 81

5.7 Perspectivities

Let , be hyperplanes in Pn and let O be a point outside and . The perspec-

tivity f : ’ from O is obtained by mapping P ∈ to the point of intersection

f (P) of the projective line O P with . Note that since O is not on , the line O P

cannot be contained in , and hence the intersection of O P with is a single point

by the dimension of intersection formula Theorem 5.4. The case n = 3, perspectivity

between two planes in 3-space, was described in 5.1.2 and illustrated on Figure 5.1b.

As opposed to the example in 5.1.2 (compare Exercise 5.1), the map f is everywhere

de¬ned, since new points have been added to af¬ne space to form projective space;

this will be discussed further below.

It is easy to write a perspectivity in terms of suitable coordinates. Choose

coordinates (x0 : x1 : · · · : xn ) so that = {x0 = 0}, = {x1 = 0} and O =

(1, 1, 0, . . . , 0). Then for a point P = (0 : x1 : · · · : xn ) of , the line O P is the set

of points {(» : » + µx1 : µx2 : · · · : µxn )} ‚ Pn (compare the ¬rst paragraph of 5.3).

is then at (» : µ) = (’x1 : 1), so

The intersection point with

f : (0 : x1 : · · · : xn ) ’ (’x1 : 0 : x2 : · · · : xn ).

In particular, you can view the perspectivity f as a projective transformation

= Pn’1 with coordinates (x1 : · · · : xn ) to = Pn’1 with coordinates

from

(x0 : x2 : · · · : xn ) given by the matrix A = diag(’1, 1, . . . , 1).

The cross-ratio of four points on a line is invariant under perspec-

Proposition

tivities; namely, if L is a line in and P, Q, R, S ∈ L are four points on the line,

then

{P, Q; R, S} = { f (P), f (Q); f (R), f (S)}.

First of all, the right-hand side of this expression is de¬ned, since the image

Proof

of L is a line in ; this follows from the fact that f is a projective transformation,

but as an exercise you can check that it also follows from the de¬nition of f and the

dimension of intersection formula. Then f : L ’ f (L) is a projective transformation

between lines; the cross-ratio is preserved under projective transformations of P1 , so

it is preserved under perspectivities also. Note that the equality of cross-ratios also

follows from Figure 5.6b and the discussion of Proposition 5.6, once you restrict the

discussion to the plane P2 ‚ Pn spanned by O and L, and interpret O in Figure 5.6b

as a point of this P2 rather than the af¬ne origin (0, 0) ∈ R2 . QED

Affine space An as a subset of projective space Pn

5.8

A hyperplane H ‚ Pn corresponds to an n-dimensional subspace W ‚ Rn+1 , the

kernel of a linear form ± : Rn+1 ’ R. Then Pn \ H can be naturally identi¬ed with

An , and H = Pn’1 with sets of parallel lines in An . The point is very simple: given

82 PROJECTIVE GEOMETRY

An point at

infinity

P Q = [v]

x0 = 1

v

x0 = 0

The inclusion An ‚ Pn .

Figure 5.8

, I can choose coordinates in Rn+1 so that

±(x0 , . . . , xn ) = x0

is the ¬rst coordinate. Then

Pn \ H = ratios (x0 : · · · x1 : · · · : xn ) x0 = 0

x1 xn

= n-tuples ,..., = An .

x0 x0

In Figure 5.8, P is a point with x0 = 0, so its equivalence class contains a unique

point in the af¬ne hyperplane An de¬ned by (x0 = 1). A point Q with x0 = 0 does

not correspond to any actual point of An ; instead, it corresponds to all the lines of An

parallel to v = Q.

Note that this discussion reverses the process of ˜going from inhomogeneous to

homogeneous™ sketched in 5.1.1; the points of the hyperplane H ‚ Pn are at in¬nity

when viewed from the af¬ne space An de¬ned by (x0 = 1). However, splitting points

into ˜¬nite™ and ˜in¬nite™ is not intrinsic to projective space, but depends on the choice

of H (or the linear form ±).

5.9 Desargues™ theorem

Let P Q R and P Q R be 2 triangles in Pn

Theorem (Desargues™ theorem)

with n ≥ 2. Suppose that P Q R and P Q R are in perspective from some point

O ∈ Pn (that is, O P P , O Q Q and O R R are lines). Then the corresponding sides

of P Q R and P Q R meet in 3 collinear points. In other words,

Q R and Q R meet in A

and A, B, C are collinear (8)

P R and P R meet in B

P Q and P Q meet in C

5.9 DESARGUES™ THEOREM 83

A B C

P′

P

O

Q

Q′

R

R′

The Desargues configuration in P2 or P3 .

Figure 5.9a

(see Figure 5.9a). The converse also holds: condition (8) implies that P Q R and

P Q R are in perspective from some point O.

If the two triangles are in perspective from O, the linear subspaces

Proof

O, P, Q, P , Q and O, P, R, P , R are planes that have at least the line O, P, P

in common. Hence

dim O, P, Q, R, P , Q , R = 2 or 3

by Theorem 5.4. Also, the construction of A, B, C in (8) makes sense: the two lines

P Q and P Q are coplanar (contained in the plane O, P, Q, P , Q ), so meet in a

unique point C, and similarly for the other pairs of sides.

Suppose ¬rst that P, Q, R and P , Q , R span a 3-dimensional space P3

Step 1

so are not in any P2 , and that they are in perspective from O. Set L = P, Q, R ©

P , Q , R . This is the intersection of two distinct planes in P3 , and is therefore a

line by Theorem 5.4. But by construction, A ∈ L since A = Q R © Q R . The same

applies to B and C, so that also B, C ∈ L and the 3 points are collinear.

We reduce to the ¬rst case. Thus suppose that P, Q, R and P , Q , R are

Step 2

in the plane = O P Q R P Q R . Let M ∈ P3 \ be any point, and lift R, R off

the plane: pick S, S as in Figure 5.9b in perspective from O such that S and R are in

perspective from M, and S and R are likewise in perspective from M. Then P Q S

and P Q S are as in Step 1. So the 3 points

Q S © Q S = A, PS © P S = B PQ © P Q = C

and

are collinear in P3 , so lie on a line L ‚ P3 . But it is easy to see from the construction

that A, B lie above A, B in perspective from M, so A, B, C are collinear.

84 PROJECTIVE GEOMETRY

M

S

S′

Π

R R′ O

Lifting the Desargues configuration to P3 .

Figure 5.9b

For proofs of the converse see Exercises 5.14“5.15 and 5.11. QED

It is interesting to note exactly what is used in the proof of Desargues™ theorem

just given. It is pure incidence geometry in Pn with n ≥ 3, in the sense that it uses

nothing beyond particular cases of formula (1) of Theorem 5.4: two distinct points of

Pn span a line, two concurrent lines span a plane, two distinct lines in a plane intersect

in a point, two distinct planes of P3 intersect in a line, etc. The ¬nal part of the proof,

Step 2, assumes also that there exists a point not in the plane (that is, that we are

in Pn with n ≥ 3), and that the two lines M R and M R each have at least one point

in addition to M, R and M, R .

5.10 Pappus™ theorem

Let L, L ‚ P2 be two lines and

Theorem (Pappus™ theorem)

P, Q, R ‚ L P ,Q,R ‚ L

and

two triples of distinct points on L and L (not equal to L © L ). Then the 3 points

Q R © Q R = A, PR © P R = B PQ © P Q = C

and

are collinear (see Figure 5.10). Notice that the ¬gure is a con¬guration of 9 lines and

9 points with 3 lines through each point and 3 points on each line.

This can also be proved via a lifting to P3 , but this requires a bit more

Proof

information about P3 (speci¬cally, quadric surfaces in P3 and properties of lines on

them). I sketch the easy proof in coordinates.

By Theorem 5.5, I can choose homogeneous coordinates (x : y : z) such that

P = (1 : 0 : 0), Q = (0 : 1 : 0), P = (0 : 0 : 1) and Q = (1 : 1 : 1).

5.11 PRINCIPLE OF DUALITY 85

R

Q

P L

C B

A

L′

P′

Q′

R′

Figure 5.10 The Pappus configuration.

Then

L = P Q : {z = 0}, P Q : {x = 0}, L = P Q : {x = y} and P Q : {y = z}.

Therefore

C = P Q © P Q = (0 : 1 : 1).

Now let R = (1 : β : 0) and R = (1 : 1 : γ ). Then easy calculations give

P R : {z = γ y} P R : {y = βx}

so that B = (1 : β : (βγ )) and

Q R : {z = γ x} Q R : {y ’ z = β(x ’ z)}

so that A = (1 : (β + γ ’ βγ ) : γ ). Finally, A, B, C are all on the line

{y ’ z = β(1 ’ γ )x}. QED

5.11 Principle of duality

Projective duality is based on the idea that the space (Rn+1 )— of linear forms

± : Rn+1 ’ R is also isomorphic to Rn+1 . Namely, if e0 , . . . , en+1 is a basis of Rn+1

then the dual basis is given by the linear form

if i = j

1

ei— : Rn+1 ’ R de¬ned by ei— (e j ) = δi j =

if i = j.

0

86 PROJECTIVE GEOMETRY

Further, there is a natural one-to-one correspondence between subspaces of Rn+1

and its dual: a subspace V ‚ Rn+1 corresponds to its annihilator (perpendicular)

subspace V ⊥ , that is, the set of linear forms ± : Rn+1 ’ R vanishing on V . By

elementary linear algebra, dim V + dim V ⊥ = n + 1. Hence we obtain the following

correspondence between elements of the geometry of projective linear subspaces of

Pn = P(Rn+1 ) and those of (Pn )— = P(Rn+1 )— :

E ⊥ = Pn’d’1 = P(V ⊥ ) ‚ (Pn )—

E = P(V ) = Pd ‚ Pn ←’

‚ (Pn )—

point P = P0 ∈ Pn ←’ hyperplane Pn’1 =

⊥ ⊥

subspace E 1 ‚ E 2 ←’ supspace E 1 ⊃ E 2

⊥ ⊥

intersection E 1 © E 2 ←’ span E 1 , E 2

⊥ ⊥

span E 1 , E 2 ←’ intersection E 1 © E 2 .

The case of P2 is special and particularly illustrative: hyperplanes in P2 are simply

lines L = P1 ‚ P2 ; points are dual to lines, and the line through two points is dual to

the intersection of two lines.

Proposition (Principle of duality for P2 ) Every theorem concerning points and

lines in P2 has a dual theorem, obtained from the original one via the following

substitutions:

←’

points P lines L

←’

lines L points P

line P1 P2 (= the span P1 , P2 ) ←’ point of intersection L 1 © L 2

intersection L 1 © L 2 ←’ line P1 P2 .

This means that given a theorem and its proof about points and lines in P2 , you get a

new theorem and its proof by replacing points by lines etc., in a completely automatic

way. For example, the dual of Desargues™ theorem in P2 is its converse (which is why

I omitted the proof in 5.9). For the dual of Pappus™ theorem, see Exercise 5.16.

5.12 Axiomatic projective geometry

An axiomatic projective plane (Figure 5.12a) consists of two sets

Points( ) and Lines( )

and a relation

Incidence( ) ‚ Points( ) — Lines( ),

usually called an incidence relation. If (P, L) ∈ Incidence( ), we say that ˜point P

is on line L™ or ˜line L passes through point P™; because this is an axiomatic system,

we might as well say with David Hilbert ˜beer mug P is on table L™.

5.12 AXIOMATIC PROJECTIVE GEOMETRY 87

L

Q

line P Q point L © M

P

M

Figure 5.12a Axiomatic projective plane.

This data is subject to the following axioms.

1. Every line has at least 3 points.

2. Every point has at least 3 lines through it.

3. Through any 2 distinct points there is a unique line.

4. Any 2 distinct lines meet in a unique point.

Note that these axioms are obviously dual: you can replace the beer mugs on the

tables throughout, and vice versa, and the axioms continue to hold.

More generally an axiomatic projective space has a lattice of projective linear

subspaces, the incidence relation ‚, intersection and linear span, and suitable axioms.

It is best not to insist a priori that the dimension of the space or its projective linear

subspaces is speci¬ed. The most important case is the in¬nite dimensional case,

which von Neumann used to give axiomatic foundations to quantum mechanics,

when dimensions of projective linear subspaces can take values in R or the value ∞.

The real projective plane

Introducing coordinates in axiomatic projective planes

P= P2

2

discussed thus far is certainly not the only axiomatic projective plane: given

R

any ¬eld k, you can take P2 = k 3 \ {0} /∼ where (x0 , x1 , x2 ) ∼ (»x0 , »x1 , »x2 ) for

k

0 = » ∈ k. It is an easy exercise to show that axioms 1 to 4 continue to hold in P2 . For k

example, if k = F2 you get an axiomatic projective plane with 7 points and 7 lines

(see Exercise 5.21).

For this purpose, k has to be a division ring, meaning that ax = b has a solution

for every a, b ∈ A with a = 0, but it is not necessary that k is commutative: you just

have to take care that in the equivalence relation (x0 , x1 , x2 ) ∼ (»x0 , »x1 , »x2 ) only left

multiplication by » ∈ k — is allowed, and the linear subspaces of k 3 used to de¬ne lines

are right k-subspaces. Indeed, even the associative law on k can be weakened, although

some kind of associativity is required in order that (x0 , x1 , x2 ) ∼ (»x0 , »x1 , »x2 ) is an

equivalence relation. For a nontrivial example, do Exercise 8.23. In this course, I do

not have time for a detailed discussion of the following result, one of the most beautiful

contributions of geometry to pure algebra; for details, consult Hartshorne [12].

88 PROJECTIVE GEOMETRY

Q∞ R∞

P∞

x y x+y L

0

Figure 5.12b Geometric construction of addition.

An axiomatic projective plane gives rise to a

Theorem (Hilbert™s construction)

= P2 .

division ring A such that Moreover,

A

A is an associative ring ⇐’ Desargues™ theorem holds in ;

A is a commutative ring ⇐’ Pappus™ theorem holds in .

We must make a number of choices in . Pick a line L ∞ to serve

Flavour of proof

as the line at in¬nity, three points P∞ , Q ∞ and R∞ on it, and a line L through P∞ ,

distinct from L ∞ . The elements of the division algebra A are the points of L except

for ∞ = L © L ∞ . Now pick 2 different points of L \ ∞, and call them 0 and 1. The

algebraic operation + is constructed in terms of parallels (since we have ¬xed L ∞ ,

two lines of are parallel if their intersection is on L ∞ ) and — in terms of similarity.

For example, addition is de¬ned as in Figure 5.12b.

Exercises

Let x, y, z be coordinates in R3 , and : (z = 1), : (y = 1) two hyperplanes. Write

5.1

down the perspectivity • : ’ from O = (0, 0, 0) in terms of coordinates (x, y)

and (x, z) on . Find and describe the points of where • is not de¬ned.

on

Prove that • takes a line L ‚ to a line L = •(L) ‚ (with a single exception).

Consider the pencil of parallel lines y = mx + c of (for m ¬xed and c variable),

and determine how • maps.

In the notation of the preceding exercise, let S : (x 2 + y 2 = 1) ‚ . Understand the

5.2

effect of the perspectivity • on S, both geometrically and in coordinates. Show that

a circle and a hyperbola in R2 correspond to projectively equivalent curves in P2 . R

Account for the 4 asymptotic directions of the hyperbola in terms of S.

In P2 , write down the equation of the line joining P = (1 : 1 : 0) and (± : 0 : β); write

5.3

down the point of intersection of the 2 lines x + y + z = 0 and ±x + βy = 0.

Let i ‚ R3 be the 3 planes of Exercise 4.1. Construct P3 by introducing a fourth

5.4

coordinate t, write down the planes of P3 by homogenising the equations of i , and

calculate again the intersections and spans.

EXERCISES 89

Prove that 3 lines L , M, N of Pn that intersect in pairs are either concurrent (have a

5.5

common point) or coplanar. [Hint: use dimension of intersection.]

Suppose that L , M, N are 3 lines of P4 not all contained in any hyperplane. Prove that

5.6

there exists a unique line meeting all 3 lines. [Hint: consider ¬rst the span L , M =

P3 .]

Write down all the projective linear maps • of P2 taking

5.7

(1 : 0 : 0) ’ (1 : 2 : 3), (0 : 1 : 0) ’ (2 : 1 : 3), (0 : 0 : 1) ’ (3 : 1 : 2).

Now write down the unique projective linear map taking the standard frame of refer-

ence

(1 : 0 : 0), (0 : 1 : 0), (0 : 0 : 1), (1 : 1 : 1)

into

(1 : 2 : 3), (2 : 1 : 3), (3 : 1 : 2), (1 : 2 : 2)

respectively. [Hint: reread the proof of Theorem 5.5.]

Consider the af¬ne linear map •0 : A2 ’ A2 given by

5.8

(x, y) ’ (3x ’ 2, 4y ’ 3).

Prove that •0 has a unique ¬xed point in A2 . [Hint: you can do this by linear algebra,

or by using the contraction mapping theorem from metric spaces.]

Write down the projective linear map • of P2 extending •0 . Find the locus of ¬xed

points of • on P2 . [Hint: either ¬nd the ¬xed points ˜by observation™, or prove that

(x : y : z) is a ¬xed point of a projective linear map x ’ Ax if and only if x = (x, y, z)

is an eigenvector of A.]

5.9 Repeat the previous question for the map

(x, y) ’ (x ’ y + 2, x + y + 3).

Suppose z = (1 ’ »)x + »y. Write y = (1 ’ » )x + » z; ¬nd » as a function of ».

5.10

Similarly, determine the effect of each permutation of x, y, z on the af¬ne ratio

» = (z ’ x)/(y ’ x). Thus permuting the 3 points x, y, z de¬nes an action of the

symmetric group S3 on the set of values of ».

Let P, Q, R = (1 : 0), (0 : 1), (1 : 1) be the standard frame of reference of P1 .

5.11

(a) Find the projective linear map that takes P, Q, S to Q, P, S (in that order); next

P, Q, S to P, S, Q. What is the effect of your map on the af¬ne coordinate of a

point R = (1 : ») ∈ P1 ?

(b) Verify that the matrixes 0 1 and 1 ’1 generate a group under matrix multipli-

0 ’1

10

cation isomorphic to the symmetric group S3 .

90 PROJECTIVE GEOMETRY

(c) The cross-ratio of 4 points p, q, r, s on a line is de¬ned to be

p’r q’s

{p, q; r, s} = · .

p’s q’r

Explain what happens when p, q, r, s are permuted. Prove that there are in general

6 values

»

1 1 1 1

», , 1’ , »’ , ,

» » » 1’» 1’»

for the cross-ratio, and the group ¬xing one value is a 4-group V4 .

5.12 Deduce Proposition 1.16.3 (1) from the invariance of cross-ratio under perspec-

tivity. [Hint: interpret one of the four lines in Proposition 5.6 as the line at

in¬nity.]

5.13 Desargues™ theorem 5.9 states that if P Q R and P Q R are 2 triangles in per-

spective from a point then the 3 points of intersection (e.g., C = P Q © P Q ) of

corresponding sides are collinear. See Figure 5.9a. Give the coordinate proof. [Hint:

as in the proof of Theorem 5.10, take 4 of the points as frame of reference, choose

convenient notation for the 3 remaining points, ¬nd the coordinates of A, B, C and

prove they are collinear.]

5.14 Modify the argument to prove the converse of Desargues™ theorem.