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Further, a choice of basis of V leads to an identiп¬Ѓcation V = Rn , and thus to a
(II)
coordinate system on An , in which points P в€€ An are represented by coordinates
пЈ«пЈ¶
x1
пЈ¬.пЈ·
P в†” p = пЈ­ . пЈё в€€ Rn where xi в€€ R.
.
xn
в€’ в†’
Two points P, Q в€€ An determine a vector P Q в€€ V as in Figure 4.2. This vector is
(III)
independent of the identiп¬Ѓcations discussed in (I).
Conversely, a vector x в€€ V can be added to a point P в€€ An to get a new point
(IV)
в€’ в†’
Q = P + x в€€ An , and then P Q = x; see again Figure 4.2. This operation is also
independent of the identiп¬Ѓcations discussed in (I).

As with the deп¬Ѓnition of En in 1.3, the deп¬Ѓnition of An involves an identiп¬Ѓcation
An = V or An = Rn , followed by the assurance that any other identiп¬Ѓcation would
do just as well provided that it is related to the п¬Ѓrst by a suitable transformation, in
this case an afп¬Ѓne linear transformation. How to deп¬Ѓne afп¬Ѓne 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 п¬Ѓelds. The most
Remarks
obviously useful case is an n-dimensional vector space over C, giving rise to An , but
C
afп¬Ѓne geometries over п¬Ѓnite п¬Ѓelds F pn , or over other п¬Ѓelds, 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 п¬Ѓelds, and because the course is directed
more towards metric geometries, which are вЂ˜realвЂ™ subjects.
Note also that I work here from the outset in a п¬Ѓnite dimensional space V . However,
in many areas of math, afп¬Ѓne spaces appear as the set of solutions of inhomogeneous
linear equations in inп¬Ѓnite dimensional spaces: there is no preferred solution, but the
differences x в€’ x between any two solutions form a vector space (п¬Ѓnite 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 afп¬Ѓne space of solutions, are often inп¬Ѓnite dimensional.
4.3 THE GEOMETRY OF AFFINE LINEAR SUBSPACES 65

4.3 The geometry of affine linear subspaces
An afп¬Ѓne linear subspace E вЉ‚ An is a nonempty subset of the form

E = P +U = P +v v в€€U ,

with P в€€ An and U вЉ‚ 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 speciп¬Ѓed in E.
Let P, Q в€€ An be two distinct points. The line spanned by P and Q is
в€’в†’
P Q = P + О»P Q О» в€€ R .

The deп¬Ѓnition clearly shows that P Q is an afп¬Ѓne linear subspace, with U the one
в€’в†’
dimensional vector subspace of V generated by P Q в€€ V . As in 1.2, we have the line
segment or interval
в€’ в†’
[P, Q] = P + О» P Q 0 в‰¤ О» в‰¤ 1 .

It is useful to spell this out in vector notation. If P, Q в€€ An correspond to position
vectors p, q, their afп¬Ѓne span is the set

pq = p + О»(q в€’ p) О» в€€ R = (1 в€’ О»)p + О»q О» в€€ R .

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

[p, q] = (1 в€’ О»)p + О»q 0 в‰¤ О» в‰¤ 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 в€’ О»)p + О»q; as we saw in
1.2, there are three subcases here:
пЈј пЈ±
пЈґ пЈґ p в€€ [r, q] so P в€€ [R, Q]
О» в‰¤ 0пЈґ пЈґ
пЈЅ пЈІ
0 в‰¤ О» в‰¤ 1 в‡ђв‡’ r в€€ [p, q] so R в€€ [P, Q]
пЈґ пЈґ
пЈґ пЈґ
1 в‰¤ О»пЈѕ пЈі q в€€ [p, r] so Q в€€ [P, R].

Proposition

Let E = P0 + U be an afп¬Ѓne linear subspace of An . Then the vector space U is
(1)
uniquely deп¬Ѓned by E; explicitly
в€’в†’
U = P Q P, Q в€€ E .

In other words, E = P + U for any P в€€ E.
A necessary and sufп¬Ѓcient condition for a nonempty subset E вЉ‚ An to be an afп¬Ѓne
(2)
subspace is that the line P Q is contained in E for all P, Q в€€ E.
66 AFFINE GEOMETRY

q
1в€’О»

(1 в€’ О» )p + О»q
О»

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 sufп¬Ѓcient condition for E to be an afп¬Ѓne subspace is that it is
nonempty, and deп¬Ѓned 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 в€€ E; informally, any point P в€€ E
can serve as origin. (3) spells out the other easy way of specifying an afп¬Ѓne 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 afп¬Ѓne linear subspace
E. The only n-dimensional afп¬Ѓne linear subspace is An itself; dim E = 0 means
simply that E consists of a single point, whereas a one dimensional afп¬Ѓne linear
subspace is simply a line. The last interesting case with a name of its own is an afп¬Ѓne
linear subspace of dimension n в€’ 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 в€’ 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 вЉ‚ An be any set; an afп¬Ѓne linear combination of is any point
Definition
P в€€ An of the form

k
в€’в†’
в€’
P = P0 + О»i P1 Pi , where Pi в€€ and О»i в€€ R. (2)
i=1

Using position vectors pi of points Pi simpliп¬Ѓes this expression once more; an
afп¬Ѓne linear combination of is any point P в€€ An of the form
4.4 DIMENSION OF INTERSECTION 67

k k
p= Вµi pi , where pi в€€ and Вµi в€€ R with Вµi = 1. (3)
i=0 i=0

This generalises the expression (1 в€’ О»)p + О»q used to parametrise points of the afп¬Ѓne
line P Q. The points Pi appear in the form (3) with (О»0 , . . . , О»k ) = (0, . . . , 1, . . . , 0);
this conп¬Ѓrms that I really mean = 1 in (3) rather than = 0.
The afп¬Ѓne span of any subset is the set of afп¬Ѓne linear combinations of . By
contains all lines spanned by pairs of points in . If P в€€
the previous remark,
= P + U , where U вЉ‚ V is the vector subspace spanned by the vectors
then
в€’ в†’
P Q for Q в€€ . Thus вЉ‚ An is an afп¬Ѓne 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 в€© W + dim(U + W ) (4)

for vector subspaces U, W of a п¬Ѓnite dimensional vector space is familiar from linear
algebra. You remember the proof: pick a basis of U в€© W , extend to two bases of U
and W , and the union is a basis of U + W .

Let E, F вЉ‚ An be afп¬Ѓne subspaces. Then
Theorem

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

provided that E в€© F = в€….
The exceptional case E в€© F = в€… 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 afп¬Ѓne linear subspaces E, F with dim E = a, dim F = b,
E в€© F = в€… and dim E, F = c for any a, b < n and any c with

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

The proof of the п¬Ѓrst statement is almost trivial: if P в€€ E в€© F then the
Proof
four afп¬Ѓne subspaces in question are translates of the four vector subspaces

E ,F ,E в€©F ,E +F вЉ‚V

so that the result follows at once from the linear algebra formula (4).
The counterexamples involve afп¬Ѓne subspaces E, F of An contained in parallel
hyperplanes. To be speciп¬Ѓc, I choose coordinates and put

E вЉ‚ {x1 = 0} F вЉ‚ {x1 = 1}.
and

Then certainly E в€© F = в€…. The converse is proved in Exercise 4.3.
68 AFFINE GEOMETRY

Assume that (0, . . . , 0) в€€ E = E and (1, 0, . . . , 0) в€€ F; then F is the translation
by (1, 0, . . . , 0) of a vector subspace F вЉ‚ V contained in {x1 = 0}. The equality (4)
holds, but the point is that E в€© F = в€… takes no account of dim E в€© 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 в€’ 1. QED

You will п¬Ѓnd 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 в€© 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 deп¬Ѓnition, which I repeat here for completeness.

A map T : An в†’ An is an afп¬Ѓne transformation if it is given in a co-
Definition
ordinate system by T (x) = Ax + b, where A = (ai j ) is an n Г— n matrix with nonzero
determinant and b = (bi ) a vector; in more detail,
пЈ«пЈ¶ пЈ«пЈ¶ пЈ«пЈ¶
x1 x1 b1
n
пЈ¬.пЈ· пЈ¬.пЈ· пЈ¬.пЈ·
x = (xi ) в†’ y = ai j x j + bi , пЈ­ . пЈё в†’ AпЈ­ . пЈё + пЈ­ . пЈё.
or (6)
. . .
j=1
xn xn bn

The set Aff(n) of afп¬Ѓne transformations is the set of вЂ˜allowed symmetriesвЂ™ of
afп¬Ѓne 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 afп¬Ѓne 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 в†’ V are the homogeneous linear maps, all mapping the origin
0 в€€ V to itself.
It is immediate that an afп¬Ѓne transformation takes an afп¬Ѓne linear subspace to an
afп¬Ѓne linear subspace; that is, it preserves the incidence geometry of afп¬Ѓne linear
subspaces. In Proposition 1.9, I proved a converse statement, under the additional
assumption that T restricts to an afп¬Ѓne linear map on each line. In fact, one can prove
that, for n в‰Ґ 2, a bijective map T : An в†’ An that preserves lines and is continuous is
actually afп¬Ѓne 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 afп¬Ѓne linearly independent if the
Definition
в€’в†’
в€’ в€’в†’
в€’
k vectors P0 P1 , . . . , P0 Pk are linearly independent in V . In other words, a set вЉ‚ An
k
is afп¬Ѓne linearly dependent if there exists a nontrivial relation i=0 О»i pi = 0 between
k
position vectors p0 , . . . , pk of points in , with О»i в€€ R and i=0 О»i = 0; is afп¬Ѓne
linearly independent if no such relation exists.
EXERCISES 69

A set вЉ‚ An is an afп¬Ѓne frame of reference if it is afп¬Ѓne linearly independent
and spans An (compare the notion of Euclidean frame in 1.12). This means that every
point P в€€ 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 в€€ is any
в€’в†’
в€’ в€’в†’
в€’
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 afп¬Ѓne frame of reference P0 , . . . , Pn . Then
Proposition

T в†’ T (P0 ), . . . , T (Pn )

deп¬Ѓnes a one-to-one correspondence between afп¬Ѓne transformations and afп¬Ѓne 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 afп¬Ѓne geometry. The other вЂ˜famousвЂ™ centres of a triangle described in 1.16.4 use
notions such as angle or distance that have no meaning in afп¬Ѓne 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 afп¬Ѓne geometry means
that applying any afп¬Ѓne transformation takes Figure 4.7 to a п¬Ѓgure with the same
properties, and in particular takes the centroid of a triangle to the centroid.

Exercises
4.1 Consider the 3 planes

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

in afп¬Ѓne space A3 . Calculate 1 в€© 2 and 1 , 2 and п¬Ѓnd out whether the dimen-
sion of intersection formula works; if not, why not? (Compare Theorem 4.4.) Ditto
for 1 в€© 3 and 1 , 3 .
70 AFFINE GEOMETRY

R

QвЂІ PвЂІ
S

RвЂІ
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 вЉ‚ An are disjoint afп¬Ѓne linear subspaces; prove that there is a
4.3
linear form П• on An such that П•(E) = 0 and П•(F) = 1. [Hint: let P в€€ E, Q в€€ F and
в€’в†’
v = P Q. Then E = P + U for a vector subspace U вЉ‚ V , and v в€€ U . Deduce that
/
there exists a linear form on V that is zero on U but nonzero on v.]
4.4 Write down the afп¬Ѓne transformation taking

(0, 0), (1, 0), (0, 1) в†’ (2, 1), (5, в€’1), (3, 8).

Can you map the same points (0, 0), (1, 0), (0, 1) to (2, 1), (5, в€’2), (3, 0) by an afп¬Ѓne
transformation? Why?
Determine the dimension of the afп¬Ѓne linear subspace E of A5 given by the equations
4.5

x1 + x3 в€’ 2x5 = 1
x2 в€’ 2x4 + x5 = в€’2
x1 + 2x2 + x3 в€’ 4x4 = в€’3.
EXERCISES 71

Find an afп¬Ѓne transformation taking E to an afп¬Ѓne linear subspace given by x1 =
В· В· В· = xk = 0 for some value of k. [Hint: choose a suitable afп¬Ѓne 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 afп¬Ѓne
4.6
linearly dependent (Deп¬Ѓnition 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 afп¬Ѓne geometry over the п¬Ѓnite п¬Ѓeld F p of integers modulo
4.8
the prime p. Count the number of points of afп¬Ѓne 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 afп¬Ѓne 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 afп¬Ѓne
geometry in terms of adding вЂ˜points at inп¬ЃnityвЂ™.
Projective geometry is simple to study as pure homogeneous linear algebra, ignor-
ing the motivation; вЂ˜linear algebra continuedвЂ™ or вЂ˜more things to do with matrixesвЂ™
would be accurate subtitles for this chapter. In Pn , the statement of afп¬Ѓne geometry
analogous to the dimension of intersection formula of Theorem 4.4 holds without
the вЂ˜inhomogeneousвЂ™ conditions of Chapter 4, so that, for example, two distinct lines
L 1 , L 2 вЉ‚ P2 meet in a point P = L 1 в€© L 2 without exception.
Projective geometry has lots of applications in math and other subjects. Projective
transformations include the perspectivities, or projections from a п¬Ѓxed 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 afп¬Ѓne linear subspaces of afп¬Ѓne space An , then
5.1.1
Inhomo- there is a nice formula 4.4 expressing the dimension of their intersection provided
that E в€© F = в€…. 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 deп¬Ѓning an afп¬Ѓne linear subspace
E вЉ‚ 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 вЉ‚ An . But there are also the solutions with x0 = 0, called
вЂ˜points at inп¬ЃnityвЂ™. Including these points adds information to the set of ordinary
solutions; namely, information about all the ways the ratios x1 : В· В· В· : xn can behave
as the xi tend to inп¬Ѓnity. A solution 0, Оѕ1 , . . . , Оѕn (with some Оѕi = 0) corresponds
not to a point of E, but to an (n в€’ 1)-dimensional family of all parallel lines with
slope Оѕ1 : В· В· В· : Оѕn satisfying the homogenised equations ai j Оѕ 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 в€’ 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 п¬Ѓx the origin O в€€ A3 and map points of a plane вЉ‚ A3 , to another
вЉ‚ A3 by taking P в€€ into the point of intersection P = O P в€©
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 п¬Ѓgure on it, and drawing
faithfully the п¬Ѓgure on the glass as you see it (see Figure 5.1b).
I get a map f : в†’ 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
п¬Ѓnite 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 deп¬Ѓned, and the conclusions should be
qualiп¬Ѓed by adding вЂ˜within the domain of deп¬ЃnitionвЂ™; the activity takes place in the
real world, and set theoretic niceties do not cause us undue discomfort.
The map f : P в†’ 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 Г— 2, 2 Г— 1, 1 Г— 2 and 1 Г— 1 matrixes. Note that these can
be assembled into a 3 Г— 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
О
artist's eye drawing
О '

Figure 5.1b Perspective drawing.

parabola y = x2
hyperbola xy = 1
вЂ˜asymptotically x2 = 0вЂ™
вЂ˜asymptotically xy = 0вЂ™

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 вЂ˜the two branches of the
5.2 DEFINITION OF PROJECTIVE SPACE 75

parabola meet at inп¬ЃnityвЂ™, or that the parabola вЂ˜passes through the point at inп¬Ѓnity
corresponding to lines parallel to x = 0.вЂ™
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 inп¬Ѓnity 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-
Afп¬Ѓne 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 вЉ‚ An , you can compactify it
by extending An to Pn ; then X вЉ‚ An вЉ‚ Pn and the closure X вЉ‚ Pn is compact. The
points at inп¬Ѓnity 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 compactiп¬Ѓed to the circle S 1 by
adding the two points (в€ћ, 0) and (0, в€ћ), and the parabola is compactiп¬Ѓed to S 1 by
adding the single point (0, в€ћ) at which the two branches are said to meet.

5.2 Definition of projective space
Provided you forget about the motivation, the deп¬Ѓnition is very simple: introduce the
equivalence relation в€ј on Rn+1 \ 0 deп¬Ѓned by

(x0 , . . . , xn ) = О»(y0 , . . . , yn )
(x0 , . . . , xn ) в€ј (y0 , . . . , yn ) в‡ђв‡’
for some 0 = О» в€€ R.

In other words x в€ј 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 deп¬Ѓne projective space to
be

Pn = Pn = Rn+1 \ 0 в€ј = lines through 0 in Rn+1 .
R

I write (x0 : В· В· В· : 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 = в€ћ.
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 в€ј О»v for О» = 0). Of course, V в€ј Rn+1 (by a choice of basis), so P(V ) в€ј Pn .
= =
76 PROJECTIVE GEOMETRY

A point P в€€ P(V ) is an equivalence class of vectors v в€€ 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 deп¬Ѓning P, for example:

P = Pv , P = [v], v = P, Pv = (x0 : В· В· В· : xn ), etc.
To return to the motivation, Pn contains the subset (x0 = 0) consisting of ratios
that can be written (1 : x1 : В· В· В· : xn ), which is thus naturally identiп¬Ѓed with An . The
language used for motivating projective geometry is quite unsuitable for developing
the theory systematically. For example, the terminology of вЂ˜points at inп¬ЃnityвЂ™ 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 deп¬Ѓnitions 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
в€ј on points of V .
As a п¬Ѓrst example, here is the deп¬Ѓnition of the line P Q through two points P =
(x0 : В· В· В· : xn ) and Q = (y0 : В· В· В· : 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 (О»x0 + Вµy0 : В· В· В· : О»xn + Вµyn ) for all (О», Вµ) = (0, 0) .

The point to notice is that О»P + ВµQ is meaningless as a point of Pn , because the
ratio (О»x0 + Вµy0 : В· В· В· : О»xn + Вµyn ) depends on the choice of P and Q within the
equivalence classes of P and Q. However, the set of all О» P + Вµ Q is a well deп¬Ѓned
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 вЉ‚ V is a vector subspace, P(U )
is the subset (U \ 0)/в€ј вЉ‚ P(V ) of lines through 0 in U . In other words, if U вЉ‚ Rn+1
then P(U ) is the set of ratios (x0 : В· В· В· : xn ) with (x0 , . . . , xn ) в€€ U . The dimension of
P(U ) is deп¬Ѓned to be dim P(U ) = dim U в€’ 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 в€’ 1)-dimensional subspace is a hyperplane.
I sometimes say k-plane to mean k-dimensional projective linear subspace.
Note that the empty set в€… is a projective linear subspace: the trivial vector subspace
0 вЉ‚ Rn+1 has P(0) = в€… вЉ‚ Pn . By convention we write dim в€… = в€’1, to agree with the
5.5 PROJECTIVE LINEAR TRANSFORMATIONS 77

general deп¬Ѓnition 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 в€… = в€’1 has a precise
and useful meaning (in the context of the geometry of linear subspaces only!).

If вЉ‚ P(V ) is a set, write вЉ‚ V for the union of the lines in ; let
Definition
U be the vector subspace of V spanned by , and deп¬Ѓne 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 в‰¤ s; equality holds if
and only if the vectors P0 , . . . , Ps в€€ 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 вЉ‚ Pn be projective linear subspaces. Then
Theorem

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

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

Write E, F вЉ‚ Rn+1 for the vector subspaces overlying E and F. Then
Proof
E в€© F = P( E в€© F) and E, F = P( E + F). By the linear algebra formula 4.4 (4)
we have

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

and since dim P(U ) = dim U в€’ 1 for every vector subspace U вЉ‚ 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 в†’ 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 в€€ Rn+1 . A map T : Pn в†’ 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 в€€ Pn for the point represented by v в€€ Rn+1 , then
T is a projective transformation if there is an invertible matrix A such that

T (Pv ) = PAv for all v в€€ 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 О»v represent the same point of Pn , a scalar matrix О» В· id =
diag(О», . . . , О») with О» = 0 acts as the identity. Moreover, if A is an invertible
78 PROJECTIVE GEOMETRY

matrix and О» в€€ R and О» = 0, then A and the product О»A have exactly the same
effect on every point of Pn . Thus the set of projective transformations is

PGL(n + 1) = invertible (n + 1) Г— (n + 1) matrixes /Rв€—

where Rв€— = О» В· id | 0 = О» в€€ R .
The following deп¬Ѓnition, which may seem unexpected at п¬Ѓrst, 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 п¬Ѓnal point Pn+1 is Pen+1 , where
n
en+1 = О»i ei , with О»i = 0 for every i.
i=0

Indeed, the п¬Ѓrst n + 1 points P0 , . . . , Pn are linearly independent, and the п¬Ѓnal point
Pn+1 is not contained in any of the n + 1 hyperplanes {xi = 0}. The standard frame
of reference is

Pi = (0 : В· В· В· : 1 : В· В· В· : 0) (with 1 in the ith place)
(3)
and Pn+1 = (1 : 1 : В· В· В· : 1).
n
That is, ei for i = 0, . . . , n is the standard basis of Rn+1 and en+1 = i=0 ei . The
п¬Ѓnal point Pn+1 = (1 : В· В· В· : 1) is there to вЂ˜calibrateвЂ™ 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, deп¬Ѓned by T в†’ 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 в€€ 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 в†’ 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 О»i fi of fi (for i = 0, . . . , n), there is a uniquely determined linear
map A : Rn+1 в†’ Rn+1 such that Aei = О»i fi for i = 0, . . . , n, given by the matrix
5.6 P1 AND THE CROSS-RATIO 79

A with columns О»i fi . Using the assumption that f0 , . . . , fn is a basis, I choose the
n
О»i such that fn+1 = i=0 О»i fi . Then, because Q 0 , . . . , Q n+1 is a frame of reference,
О»i = 0 for i = 0, . . . , n, and Aen+1 = fn+1 by choice of A. Since A : Rn+1 в†’ Rn+1
is a linear map with ei в†’ О»i fi and en+1 в†’ fn+1 , it deп¬Ѓnes a projective linear map
T : Pn в†’ Pn taking Pi в†’ Q i for i = 0, . . . , n + 1.
For the uniqueness, let us look back through the construction: п¬Ѓrst, the condition
T (Pi ) = Q i for i = 0, . . . , n determines the columns of A up to multiplying each
column by a scalar О»i ; so far, any О»i will do (possibly different choices for different
columns). Next, the condition T (Pn+1 ) = Q n+1 п¬Ѓxes the О»i up to a common scalar
n
factor: because we must send en+1 = ei into a multiple of fn+1 = i=0 О»i fi , we
have to choose these values of О»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 в€€ 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 Пѓ (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 deп¬Ѓne it, note that any
choices of representatives p, q в€€ R2 \ 0 of P, Q form a basis. Choosing this basis
gives

P = (1 : 0), Q = (0 : 1), R = (1 : О») S = (1 : Вµ)
and (4)

for some О», Вµ. Set {P, Q; R, S} = О»/Вµ.
Changing the representative q в†’ Вµq sets Вµ = 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 deп¬Ѓning
{P, Q; R, S} = О», where R = (1 : О»). 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 deп¬Ѓned,
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 п¬Ѓx the points 0
and 1, then any other point P is п¬Ѓxed by О» = (x в€’ 0)/(1 в€’ 0). In P1 you need also to
п¬Ѓx the point at inп¬Ѓnity.

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
П•
Q R'

R
Q'

The 3-transitive action of PGL(2) on P1 .
Figure 5.6a

x=0
q

L

O
x = О»y
r

s y = Вµx

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
pв€’r qв€’s
{P, Q; R, S} = В· . (5)
pв€’s qв€’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 В±|pв€’r| В· В±|qв€’r| .
В±|qв€’s|
В±|pв€’s|

As in the deп¬Ѓnition 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 О», Вµ are as in (4) then r в€€ R2 is in the
2

equivalence class of (1 : О») and is on L, so that necessarily
(1, О») (1, Вµ)
r= similarly s = .
; (6)
1+О» 1+Вµ
The remaining calculation is very easy:
пЈј
(1, в€’1)пЈЅ
О» в€’1
pв€’r= (1, в€’1), qв€’r= pв€’r qв€’s О»
=в‡’ В· =.
1+О» 1+О»
(7)
(1, в€’1)пЈѕ pв€’s qв€’r Вµ
Вµ в€’1
pв€’s= (1, в€’1), qв€’s=
1+Вµ 1+Вµ

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 : в†’ from O is obtained by mapping P в€€ 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
deп¬Ѓned, since new points have been added to afп¬Ѓne 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 : В· В· В· : xn ) so that = {x0 = 0}, = {x1 = 0} and O =
(1, 1, 0, . . . , 0). Then for a point P = (0 : x1 : В· В· В· : xn ) of , the line O P is the set
of points {(О» : О» + Вµx1 : Вµx2 : В· В· В· : Вµxn )} вЉ‚ Pn (compare the п¬Ѓrst paragraph of 5.3).
is then at (О» : Вµ) = (в€’x1 : 1), so
The intersection point with

f : (0 : x1 : В· В· В· : xn ) в†’ (в€’x1 : 0 : x2 : В· В· В· : xn ).

In particular, you can view the perspectivity f as a projective transformation
= Pnв€’1 with coordinates (x1 : В· В· В· : xn ) to = Pnв€’1 with coordinates
from
(x0 : x2 : В· В· В· : xn ) given by the matrix A = diag(в€’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 в€€ 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 deп¬Ѓned, 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 deп¬Ѓnition of f and the
dimension of intersection formula. Then f : L в†’ 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 вЉ‚ Pn spanned by O and L, and interpret O in Figure 5.6b
as a point of this P2 rather than the afп¬Ѓne origin (0, 0) в€€ R2 . QED

Affine space An as a subset of projective space Pn
5.8
A hyperplane H вЉ‚ Pn corresponds to an n-dimensional subspace W вЉ‚ Rn+1 , the
kernel of a linear form О± : Rn+1 в†’ R. Then Pn \ H can be naturally identiп¬Ѓed with
An , and H = Pnв€’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 вЉ‚ Pn .
Figure 5.8

, I can choose coordinates in Rn+1 so that

О±(x0 , . . . , xn ) = x0

is the п¬Ѓrst coordinate. Then

Pn \ H = ratios (x0 : В· В· В· x1 : В· В· В· : 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 afп¬Ѓne hyperplane An deп¬Ѓned 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 вЂ˜going from inhomogeneous to
homogeneousвЂ™ sketched in 5.1.1; the points of the hyperplane H вЉ‚ Pn are at inп¬Ѓnity
when viewed from the afп¬Ѓne space An deп¬Ѓned by (x0 = 1). However, splitting points
into вЂ˜п¬ЃniteвЂ™ and вЂ˜inп¬ЃniteвЂ™ is not intrinsic to projective space, but depends on the choice
of H (or the linear form О±).

5.9 DesarguesвЂ™ theorem

Let P Q R and P Q R be 2 triangles in Pn
Theorem (DesarguesвЂ™ theorem)
with n в‰Ґ 2. Suppose that P Q R and P Q R are in perspective from some point
O в€€ 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,
пЈј
Q R and Q R meet in A пЈЅ
and A, B, C are collinear (8)
P R and P R meet in B
пЈѕ
P Q and P Q meet in C
5.9 DESARGUESвЂ™ THEOREM 83

A B C

PвЂІ
P
O

Q
QвЂІ
R

RвЂІ

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 п¬Ѓrst 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 в€©
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 в€€ L since A = Q R в€© Q R . The same
applies to B and C, so that also B, C в€€ L and the 3 points are collinear.

We reduce to the п¬Ѓrst 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 в€€ 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 в€© Q S = A, PS в€© P S = B PQ в€© P Q = C
and

are collinear in P3 , so lie on a line L вЉ‚ 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

SвЂІ

О

R RвЂІ O

Lifting the Desargues configuration to P3 .
Figure 5.9b

For proofs of the converse see Exercises 5.14вЂ“5.15 and 5.11. QED
It is interesting to note exactly what is used in the proof of DesarguesвЂ™ theorem
just given. It is pure incidence geometry in Pn with n в‰Ґ 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 п¬Ѓnal 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 в‰Ґ 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 PappusвЂ™ theorem

Let L, L вЉ‚ P2 be two lines and
Theorem (PappusвЂ™ theorem)

P, Q, R вЉ‚ L P ,Q,R вЉ‚ L
and

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

Q R в€© Q R = A, PR в€© P R = B PQ в€© P Q = C
and

are collinear (see Figure 5.10). Notice that the п¬Ѓgure is a conп¬Ѓguration 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 (speciп¬Ѓcally, 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

LвЂІ
PвЂІ

QвЂІ
RвЂІ
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 в€© P Q = (0 : 1 : 1).

Now let R = (1 : ОІ : 0) and R = (1 : 1 : Оі ). Then easy calculations give

P R : {z = Оі y} P R : {y = ОІx}

so that B = (1 : ОІ : (ОІОі )) and

Q R : {z = Оі x} Q R : {y в€’ z = ОІ(x в€’ z)}

so that A = (1 : (ОІ + Оі в€’ ОІОі ) : Оі ). Finally, A, B, C are all on the line

{y в€’ z = ОІ(1 в€’ Оі )x}. QED

5.11 Principle of duality
Projective duality is based on the idea that the space (Rn+1 )в€— of linear forms
О± : Rn+1 в†’ 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
eiв€— : Rn+1 в†’ R deп¬Ѓned by eiв€— (e j ) = Оґi 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 вЉ‚ Rn+1 corresponds to its annihilator (perpendicular)
subspace V вЉҐ , that is, the set of linear forms О± : Rn+1 в†’ R vanishing on V . By
elementary linear algebra, dim V + dim V вЉҐ = 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 )в€— = P(Rn+1 )в€— :

E вЉҐ = Pnв€’dв€’1 = P(V вЉҐ ) вЉ‚ (Pn )в€—
E = P(V ) = Pd вЉ‚ Pn в†ђв†’
вЉ‚ (Pn )в€—
point P = P0 в€€ Pn в†ђв†’ hyperplane Pnв€’1 =
вЉҐ вЉҐ
subspace E 1 вЉ‚ E 2 в†ђв†’ supspace E 1 вЉѓ E 2
вЉҐ вЉҐ
intersection E 1 в€© E 2 в†ђв†’ span E 1 , E 2
вЉҐ вЉҐ
span E 1 , E 2 в†ђв†’ intersection E 1 в€© E 2 .

The case of P2 is special and particularly illustrative: hyperplanes in P2 are simply
lines L = P1 вЉ‚ 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:

в†ђв†’
points P lines L
в†ђв†’
lines L points P
line P1 P2 (= the span P1 , P2 ) в†ђв†’ point of intersection L 1 в€© L 2
intersection L 1 в€© L 2 в†ђв†’ 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 DesarguesвЂ™ theorem in P2 is its converse (which is why
I omitted the proof in 5.9). For the dual of PappusвЂ™ 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( ) вЉ‚ Points( ) Г— Lines( ),

usually called an incidence relation. If (P, L) в€€ Incidence( ), we say that вЂ˜point P
is on line LвЂ™ or вЂ˜line L passes through point PвЂ™; because this is an axiomatic system,
we might as well say with David Hilbert вЂ˜beer mug P is on table LвЂ™.
5.12 AXIOMATIC PROJECTIVE GEOMETRY 87

L
Q

line P Q point L в€© 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 вЉ‚, 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 speciп¬Ѓed. The most important case is the inп¬Ѓnite 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 в€ћ.

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 п¬Ѓeld k, you can take P2 = k 3 \ {0} /в€ј where (x0 , x1 , x2 ) в€ј (О»x0 , О»x1 , О»x2 ) for
k
0 = О» в€€ 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 в€€ 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 ) в€ј (О»x0 , О»x1 , О»x2 ) only left
multiplication by О» в€€ k в€— is allowed, and the linear subspaces of k 3 used to deп¬Ѓne 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 ) в€ј (О»x0 , О»x1 , О»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 .
88 PROJECTIVE GEOMETRY

Qв€ћ Rв€ћ

Pв€ћ

x y x+y L
0

Figure 5.12b Geometric construction of addition.

An axiomatic projective plane gives rise to a
Theorem (HilbertвЂ™s construction)
= P2 .
division ring A such that Moreover,
A

A is an associative ring в‡ђв‡’ DesarguesвЂ™ theorem holds in ;
A is a commutative ring в‡ђв‡’ PappusвЂ™ theorem holds in .

We must make a number of choices in . Pick a line L в€ћ to serve
Flavour of proof
as the line at inп¬Ѓnity, three points Pв€ћ , Q в€ћ and Rв€ћ on it, and a line L through Pв€ћ ,
distinct from L в€ћ . The elements of the division algebra A are the points of L except
for в€ћ = L в€© L в€ћ . Now pick 2 different points of L \ в€ћ, and call them 0 and 1. The
algebraic operation + is constructed in terms of parallels (since we have п¬Ѓxed L в€ћ ,
two lines of are parallel if their intersection is on L в€ћ ) and Г— in terms of similarity.
For example, addition is deп¬Ѓned 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 П• : в†’ from O = (0, 0, 0) in terms of coordinates (x, y)
and (x, z) on . Find and describe the points of where П• is not deп¬Ѓned.
on
Prove that П• takes a line L вЉ‚ to a line L = П•(L) вЉ‚ (with a single exception).
Consider the pencil of parallel lines y = mx + c of (for m п¬Ѓxed and c variable),
and determine how П• maps.
In the notation of the preceding exercise, let S : (x 2 + y 2 = 1) вЉ‚ . Understand the
5.2
effect of the perspectivity П• 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 (О± : 0 : ОІ); write
5.3
down the point of intersection of the 2 lines x + y + z = 0 and О±x + ОІy = 0.
Let i вЉ‚ 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 п¬Ѓrst the span L , M =
P3 .]
Write down all the projective linear maps П• of P2 taking
5.7

(1 : 0 : 0) в†’ (1 : 2 : 3), (0 : 1 : 0) в†’ (2 : 1 : 3), (0 : 0 : 1) в†’ (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 afп¬Ѓne linear map П•0 : A2 в†’ A2 given by
5.8

(x, y) в†’ (3x в€’ 2, 4y в€’ 3).

Prove that П•0 has a unique п¬Ѓxed 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 П• of P2 extending П•0 . Find the locus of п¬Ѓxed
points of П• on P2 . [Hint: either п¬Ѓnd the п¬Ѓxed points вЂ˜by observationвЂ™, or prove that
(x : y : z) is a п¬Ѓxed point of a projective linear map x в†’ 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) в†’ (x в€’ y + 2, x + y + 3).

Suppose z = (1 в€’ О»)x + О»y. Write y = (1 в€’ О» )x + О» z; п¬Ѓnd О» as a function of О».
5.10
Similarly, determine the effect of each permutation of x, y, z on the afп¬Ѓne ratio
О» = (z в€’ x)/(y в€’ x). Thus permuting the 3 points x, y, z deп¬Ѓnes an action of the
symmetric group S3 on the set of values of О».
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 afп¬Ѓne coordinate of a
point R = (1 : О») в€€ P1 ?
(b) Verify that the matrixes 0 1 and 1 в€’1 generate a group under matrix multipli-
0 в€’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 deп¬Ѓned to be

pв€’r qв€’s
{p, q; r, s} = В· .
pв€’s qв€’r

Explain what happens when p, q, r, s are permuted. Prove that there are in general
6 values
О»
1 1 1 1
О», , 1в€’ , О»в€’ , ,
О» О» О» 1в€’О» 1в€’О»
for the cross-ratio, and the group п¬Ѓxing 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
inп¬Ѓnity.]
5.13 DesarguesвЂ™ 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 в€© 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, п¬Ѓnd the coordinates of A, B, C and
prove they are collinear.]
5.14 Modify the argument to prove the converse of DesarguesвЂ™ theorem.
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