Projective Geometry Lecture Notes 3

8/10/2019 Projective Geometry Lecture Notes 3

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Projective Geometry Lecture Notes

Thomas Baird

March 26, 2014

Contents

1 Introduction 2

2 Vector Spaces and Projective Spaces 4

2.1 Vector spaces and their duals . . . . . . . . . . . . . . . . . . . . . . . . . 42.1.1 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1.2 Vector spaces and subspaces . . . . . . . . . . . . . . . . . . . . . . 52.1.3 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.4 Dual vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Projective spaces and homogeneous coordinates . . . . . . . . . . . . . . . 82.2.1 Visualizing projective space . . . . . . . . . . . . . . . . . . . . . . 82.2.2 Homogeneous coordinates . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Linear subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.1 Two points determine a line . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Two planar lines intersect at a point . . . . . . . . . . . . . . . . . 14

2.4 Projective transformations and the Erlangen Program. . . . . . . . . . . . 152.4.1 Erlangen Program . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4.2 Projective versus linear. . . . . . . . . . . . . . . . . . . . . . . . . 172.4.3 Examples of projective transformations . . . . . . . . . . . . . . . . 182.4.4 Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.4.5 General position . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.4.6 The Cross-Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.5 Classical Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.1 Desargues Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 232.5.2 Pappus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.6 Duality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Quadrics and Conics 28

3.1 Affine algebraic sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.2 Projective algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.3 Bilinear and quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3.1 Quadratic forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.3.2 Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

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3.3.3 Digression on the Hessian . . . . . . . . . . . . . . . . . . . . . . . 363.4 Quadrics and conics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.5 Parametrization of the conic . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.5.1 Rational parametrization of the circle . . . . . . . . . . . . . . . . . 423.6 Polars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.7 Linear subspaces of quadrics and ruled surfaces . . . . . . . . . . . . . . . 463.8 Pencils of quadrics and degeneration . . . . . . . . . . . . . . . . . . . . . 47

4 Exterior Algebras 52

4.1 Multilinear algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.2 The exterior algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.3 The Grassmanian and the Plucker embedding . . . . . . . . . . . . . . . . 594.4 Decomposability of 2-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 594.5 The Klein Quadric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5 Extra details 64

6 The hyperbolic plane 65

7 Ideas for Projects 68

1 Introduction

Projective geometry has its origins in Renaissance Italy, in the development of perspectivein painting: the problem of capturing a 3-dimensional image on a 2-dimensional canvas.It is a familiar fact that objects appear smaller as the they get farther away, and that theapparent angle between straight lines depends on the vantage point of the observer. Themore familiar Euclidean geometry is not well equipped to make sense of this, because inEuclidean geometry length and angle are well-defined, measurable quantities independentof the observer. Projective geometry provides a better framework for understanding howshapes change as perspective varies.

The projective geometry most relevant to painting is called the real projective plane,and is denoted RP2 or P(R3).

Definition 1. The real projective plane, RP2 = P(R3) is the set of 1-dimensional sub-spaces ofR3.

This definition is best motivated by a picture. Imagine an observer sitting at the originin R3 looking out into 3-dimensional space. The 1-dimensional subspaces ofR3 can beunderstood as lines of sight. If we now situate a (Euclidean) planePthat doesnt containthe origin, then each point in Pdetermines unique sight line. Objects in (subsets of) R3

can now be projected onto the plane P, enabling us to transform a 3-dimensional sceneinto a 2-dimensional image. Since 1-dimensional lines translate into points on the plane,we call the 1-dimensional lines projective points.

Of course, not every projective point corresponds to a point in P, because some 1-dimensional subspaces are parallel to P. Such points are called points at infinity. To

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Figure 1: A 2D representation of a cube

Figure 2: Projection onto a plane

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motivate this terminology, consider a family of projective points that rotate from pro-jective points that intersectP to one that is parallel. The projection ontoP becomes afamily of points that diverges to infinity and then disappears. But as projective points

they converge to a point at infinity. It is important to note that a projective point is onlyat infinity with respect to some Euclidean plane P.One of the characteristic features of projective geometry is that every distinct pair

ofprojective lines in the projective plane intersect. This runs contrary to the parallelpostulatein Euclidean geometry, which says that lines in the plane intersect except whenthey are parallel. We will see that two lines that appear parallel in a Euclidean plane willintersect at a point at infinity when they are considered as projective lines. As a generalrule, theorems about intersections between geometric sets are easier to prove, and requirefewer exceptions when considered in projective geometry.

According to Dieudonnes History of Algebraic Geometry, projective geometry wasamong the most popular fields of mathematical research in the late 19th century. Pro-

jective geometry today is a fundamental part of algebraic geometry, arguably the richestand deepest field in mathematics, of which we will gain a glimpse in this course.

2 Vector Spaces and Projective Spaces

2.1 Vector spaces and their duals

Projective geometry can be understood as the application of linear algebra to classicalgeometry. We begin with a review of basics of linear algebra.

2.1.1 Fields

The rational numbers Q, the real numbers R and the complex numbers C are examples offields. A field is a setFequipped with two binary operationsF F F called additionand multiplication, denoted + and respectively, satisfying the following axioms for alla,b,c R.

1. (commutativity)a+b= b+a, and ab = ba.

2. (associativity) (a+b) +c= a+ (b+c) anda(bc) = (ab)c

3. (identities) There exists 0, 1

F such thata + 0 =a and a

1 =a

4. (distribution) a(b+c) =ab+acand (a+b)c= ac+bc

5. (additive inverse) There existsa F such that a+ (a) = 06. (multiplicative inverse) Ifa = 0, there exists 1

a F such thata 1

a= 1.

Two other well know operations are best defined in terms of the above properties. Forexample

a b= a+ (b)

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and ifb = 0 thena/b= a 1

b.

We usually illustrate the ideas in this course using the fields R and C, but it isworth remarking that most of the results are independent of the base field F. A perhapsunfamiliar example that is fun to consider is the field of two elements Z/2. This is thefield containing two elements Z/2 = {0, 1} subject to addition and multiplication rules

0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 0

0 0 = 0, 0 1 = 1 0 = 0, 1 1 = 1and arises as modular arithmetic with respect to 2.

2.1.2 Vector spaces and subspaces

LetFdenote a field. A vector space V overFis set equipped with two operations:

Addition: V V V, (v, w) v+w. Scalar multiplication: F V V, (, v) v

satisfying the following list of axioms for all u, v, w V and , F. (u+v) +w= u+ (v+w) (additive associativity) u+v= v+u (additive commutativity)

There exists a vector 0 V called the zero vector such that, 0 + v= v for allv V(additive identity)

For everyv V there existsv V such thatv + v= 0 (additive inverses) (+)v= v+v (distributivity) (u+v) =u+v (distributivity) (v) = ()v (associativity of scalar multiplication)

1v= v ( identity of scalar multiplication)

I will leave it as an exercise to prove that 0v= 0 and that1v= v.Example 1. The setV = Rn of n-tuples of real numbers is a vector space overF = Rinthe usual way. Similarly, Fn is a vector space over F. The vector spaces we consider inthis course will always be isomorphicto one of these examples.

Given two vector spaces V, W over F, a map : V W is called linear if for allv1, v2 V and1, 2 F, we have

(1v1+2v2) =1(v1) +2(v2).

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Example 2. A linear map from Rm to Rn can be uniquely represented by an n mmatrix of real numbers. In this representation, vectors in Rm and Rn are represented bycolumn vectors, and is defined using matrix multiplication. Similarly, a linear map from

C

m

to Cn

can be uniquely represented by an n mmatrix of complex numbers.We say that a linear map : V Wis anisomorphism if is a one-to-one and onto.

In this case, the inverse map 1 is also linear, hence is also a isomorphism. Two vectorspaces are calledisomorphicif there exists an isomorphism between them. A vector spaceV over F is called n-dimensional if it is isomorphic to Fn. Vis called finite dimensionalif it is n-dimensional for some n {0, 1, 2,....}. Otherwise we say that V is infinitedimensional.

A vector subspace of a vector space V is a subset U V containing zero which isclosed under addition and scalar multiplication. This makesUinto a vector space in itsown right, and the inclusion map U V in linear. Given any linear map : V W,we can define two vector subspaces: the image

(V) := {(v)|v V} Wand the kernel

ker() := {v V|(v) = 0} V.The dimension of(V) is called the rank of, and the dimension of ker() is called thenullity of . The rank-nullity theoremstates that the rank plus the nullity equals thedimension ofV. I.e., given a linear map : V W

dim(V) = dim(ker()) + dim((V)).

Given a set of vectors S :={v1,...,vk} V a linear combination is an expression ofthe form 1v1+...+kvk

where 1,...,k F are scalars. The set of vectors that can be written as a linearcombination of vectors inSis called thespanofS. The setSis called linearly independentif the only linear combination satisfying

1v1+...+kvk = 0

is when 1 =2 =... = k = 0. If some non-trivial linear combination equals zero, thenwe sayS is linearly dependent.

A basis ofV is a linearly independent set S V that spans V. Every vector spacehas a basis, and the dimension of a vector space is equal to the number of vectors in any

basis of that vector space. If{v1,...,vn} is a basis ofV, then this defines an isomorphism: Fn V by

((x1,...,xn)) = x1v1+...+xnvn.

Conversely, given an isomorphism : Fn V, we obtain a basis{v1,...,vn} V bysettingvi= (ei) where {e1,...,en} Fn is the standard basis. So choosing a basis for Vamounts to the same thing as defining an isomorphism from Fn toV.

Proposition 2.1. Suppose thatW V is a vector subspace. Thendim(W)dim(V).Furthermore, we have equalitydim(W) = dim(V) if and only ifW =V.

Proof. Exercise.

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2.1.3 Matrices

A m n-matrix A over the field F is a set of scalars Ai,j F parametrized by i

{1,...,m

}and j

{1,...,n

}, normally presented as a 2 dimensional array for which i and

j parametrize rows and columns respectively.

A=

A1,1 A1,2 ... A1,nA2,1 A2,2 ... ...

... ... ... ...Am,1 ... ... Am,n

.

IfA is m n and B is n p then the product matrix AB is an m p matrix withentries

(AB)i,j =n

k=1

Ai,kBk,j (1)

The identity matrixI is the n n-matrix satisfying

Ii,j :=

1 i= j

0 i =j .

Ann n-matrixA is called invertible if there exists an n n-matrixB such thatAB =I=BA. IfB exists, it is unique and is called the inverse matrix B=A1.

The determinant of an n n-matrixA is the scalar value

Sn

(1)

A1,(1)...An,(n)

where the sum is indexed by permutations of the set{1,...,n} and (1) =1 is thesign of the permutation.

Theorem 2.2. LetAbe ann n-matrix. The following properties are equivalent (i.e. ifone holds, they all hold).

A is invertible. A has non-zero determinant. A has rankn.The simplest way to check that A is invertible is usually to check if the rank is n by

row reducing.

2.1.4 Dual vector spaces

Given any vector space V, the dual vector space V is the set of linear maps from V toF =F1

V:= {: V F| is linear}.

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The dual V is naturally a vector space under point-wise addition and multiplication

(1+2)(v) =1(v) +2(v)

and()(v) =((v)).

Example 3. Suppose V=Fn is the set of 1 ncolumn vectors. ThenV is identifiedwith the set ofn 1 row vectors under matrix multiplication.

Suppose thatv1,...,vn V is a basis. Then the dual basisv1 ,...,vn V is the uniqueset of linear maps satisfying

vi (vj) =ji ,

whereji is the Kronecker deltadefined by ii = 1 and

ji = 0 ifi =j .

Example 4. Letv1,...,v

nFn be a basis of column vectors and let A = [v

1...v

n] be the

n n-matrix made up of these columns. The rules of matrix multiplication imply thatthe dual basis v1,...,v

n are the rows of the inverse matrix

A1 =

v1v2:.

vn

.

2.2 Projective spaces and homogeneous coordinates

Definition 2. Let V be a vector space. The projective space P(V) of V is the set of1-dimensional subspaces ofV.

Definition 3. If V is a vector space of dimension n+ 1, then we say that P(V) is aprojective space of dimension n. A 0-dimensional projective space is called a projectivepoint, a 1-dimensional vector space is called a projective line, and a 2-dimensional vectorspace is called a projective plane. IfVis one of the standard vector spaces Rn+1 or Cn+1

we also use notation RPn :=P(Rn+1) and CPn :=P(Cn+1).

2.2.1 Visualizing projective space

To develop some intuition about projective spaces, we consider some low dimensional

examples.IfVis one-dimensional, then the only one dimensional subspace is V itself, and con-

sequently P(V) is simply a point (a one element set).Now consider the case of the real projective line. Let V = R2 with standard coordinates

x, y and let L R2 be the line defined x = 1. With the exception of the y-axis, each1-dimensional subspace ofR2 intersects L in exactly one point. Thus RP1 = P(R2) isisomorphic as a set with the union ofL with a single point. Traditionally, we identifyLwith R and call the extra point{}, the point at infinity. Thus we have a bijection ofsets

RP1= R {}

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Figure 3: The real projective plane

Figure 4: The Riemann Sphere

Another approach is to observe that every 1-dimensional subspace intersects the unitcircle S1 in a pair ofantipodal points. Thus a point in RP1 corresponds to a pair ofantipodal points inS1. Topologically, we can construct RP1 by taking a closed half circleand identifying the endpoints. Thus RP1 is itself a circle.

By similar reasoning we can see that the complex projective line CP1 is in bijectionwith C {}

CP1= C {}.Topologically, CP1 is a isomorphic to the 2-dimensional sphereS2 and is sometimes calledthe Riemann Sphere. The correspondence between C {} and S2 can be understoodthrough stereographic projection, pictured in Figure2.2.1.

We turn next to the real projective plane RP2 = P(R3). Let x, y,zbe the standardcoordinates on R3. Then every 1-dimensional subspace ofR3 either intersects the planez= 1 or lies in the plane z= 0. The first class of 1-dimensional subspaces is in bijection

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with R2, and the second class is in bijection with RP1. Thus we get a bijection

RP2= R2 RP1= R2 R {}.

It is not hard to see that this reasoning extends inductively to every dimension n todetermine the following bijection

RPn= Rn RPn1= Rn Rn1 ... R0 (2)

and similarlyCPn= Cn CPn1= Cn Cn1 ... C0. (3)

Returning now to the projective plane, observe that every 1-dimensional subspace ofR3 intersects the unit sphere S2 in a pair of antipodal points.

Figure 5: Antipodal points on the sphere

Thus we can construct RP2 by taking (say) the northern hemisphere and identifyingantipodal points along the equator. This makes the boundary a copy ofRP1.

RP2 is difficult to visual directly because it can not be topologically embedded in R3,though it can be in R4. However, it can be embedded if we allow self intersections as across cap.

See the websiteMath Imagesfor more about the projective plane.It is also interesting to consider the field of two elementsF = Z2. As a set, the standard

vector space Fn is an n-fold Cartesian product ofFwith itself so Zn2 has cardinality 2n.

From the bijectionF Pn=Fn Fn1 ... F1 F0, it follows that

|Z2Pn| = 2n + 2n1 +....+ 1 = 2n+1 1so a projective line contains 3 points and an projective plane contains 7 points. TheZ2projective plane is also called the Fano plane.

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Figure 6: RP2 as quotient of a disk

Figure 7: Cross Caps

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Figure 8: Sliced open Cross Caps

Figure 9: The Fano Plane

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2.2.2 Homogeneous coordinates

To go further it is convenient to introduce some notation. Given a non-zero vector vV

\ {0

}, define [v]

P(V) to be the 1-dimensional subspace spanned by v. If

F

\ {0

}is a non-zero scalar, then[v] = [v].

Now suppose we choose a basis{v0,...,vn} for V (recall this is basically the same asdefining an isomorphismV=Fn). Then every vector in v Vcan be written

v=n

i=0

xivi

for some set of scalars xi and we can write

[v] = [x0: ... : xn]with respect to this basis. These are known ashomogeneous coordinates. Once again, for = 0 we have

[x0 : ... : xn] = [x0: ... : xn]

Consider now the subset U0 P(V) consisting of points [x0 : ... : xn] with x0= 0(Observe that this subset is well-defined independent of the representative (x0 : ... : xn)because for = 0 we have x0= 0 if and only ifx0= 0). For points in U0

[x0 : ... : xn] = [x0(1/x0) :x0(x1/x0) :... : x0(xn/x0)] = [1 : x1/x0: ... : xn/x0]

Thus we can uniquely represent any point in U0 by a representative vector of the form(1 : y1 : ... : yn). It follows that U0= Fn. This result has already been demonstratedgeometrically in equations (2) and (3) because U0 is the set of 1-dimensional subspacesin V that intersect the hyperplane x0 = 1. The complement ofU0 is the set of pointswithx0= 0, which form a copy ofP(F

n).

2.3 Linear subspaces

Definition 4. A linear subspace of a projective space P(V) is a subset

P(W) P(V)consisting of all 1-dimensional subspaces lying in some vector subspaceW V. Observethat any linear subspace P(W) is a projective space in its own right.

For example, each point in P(V) is a linear subspace because it corresponds to a 1-dimensional subspace. IfW V is 2-dimensional, we call P(W) P(V) a projectiveline in P(V). Similarly, ifW is 3-dimensional then P(W)P(V) is a projective planein P(V). Observe that when each of these linear subspaces are intersected with thesubsetU0 = {x0 = 1} =Fn then they become points, lines and planes in the traditionalEuclidean sense.

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2.3.1 Two points determine a line

Theorem 2.3. Given two distinct points [v1], [v2] P(V), there is a unique projectiveline containing them both.

Proof. Since [v1] and [v2] represent distinct points inP(V), then the representative vectorsv1, v2 Vmust be linearly independent, and thus span a 2-dimensional subspace W V.Thus P(W) is a projective line containing both [v1] and [v2].

Now suppose that there exists another projective line P(W)P(V) containing [v1]and [v2]. Thenv1, v2 W, soW W. Since bothW andWare 2-dimensional it followsthatW =W and P(W) =P(W).

Example 5. Consider the model ofRP2 as pairs of antipodal points on the sphereS2. Aprojective line in this model is represented by a great circle. Proposition2.3amounts tothe result that any pair of distinct and non-antipodal points in S2 lie on a unique great

circle (by the way, this great circle also describes the shortest flight route for airplanesflying between two points on the planet).

Figure 10: Two points determine a line

2.3.2 Two planar lines intersect at a point

Theorem 2.4. In a projective plane, two distinct projective lines intersect in a uniquepoint.

Proof. Let P(V) be a projective plane where dim(V) = 3. Two distinct projective linesP(U) and P(W) correspond to two distinct subspaces U, W V each of dimension 2.Then we have an equality

P(U) P(W) =P(U W)so proposition amounts to showing that U W Vis a vector subspace of dimension 1.Certainly

dim(U W) dim(U) = 2.Indeed we see that dim(U W) 1 for otherwise dim(U W) = 2 = dim(U) = dim(W)so U W =U=Wwhich contradicts the condition that U, W are distinct.

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Thus to prove dim(U W) = 1 it is enough to show that U W contains a non-zerovector. Choose bases u1, u2 U and w1, w2 W. Then the subset{u1, u2, w1, w2} Vmust be linearly dependent becauseVonly has dimension 3. Therefore, there exist scalars

1,...,4 not all zero such that1u1+2u2+3w1+4w2 = 0

from which it follows that1u1+2u2 = 3w14w2is a non-zero element inUW.Example 6. Using the model ofRP2 as pairs of antipodal points in S2, then Proposition2.4is the geometric fact that pairs of great circles always intersect in a pair of antipodalpoints.

Figure 11: Two lines in a plane determine a point

2.4 Projective transformations and the Erlangen Program

So far we have explored how a vector space Vdetermines a projective space P(V) andhow a vector subspace W Vdetermines a linear subspace P(W). How do linear mapsfit into projective geometry?

It is nottrue in general that a linear map T : V V induces a map between theprojective spaces P(V) and P(V).1 This is because if T has a non-zero kernel, thenit sends some 1-dimensional subspaces ofV to zero in V. However, if ker(T) = 0, orequivalently ifTis injective, then this works fine.

Definition 5. LetT :V V be an injective linear map. Then the map :P(V)

P(V)

defined by([v]) = [T(v)]

is called the projective morphism or projective transformation induced by T.

Projective transformations from a projective space to itself :P(V) P(V) are calledprojective automophismsand are considered to be the group of symmetriesof projectivespace.

1IfT is non-zero but not injective, then it does define a map from a subset ofP(V) to P(V). This isan example of a rational map which play an important role in algebraic geometry. We will not considerrational maps in this course though.

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2.4.1 Erlangen Program

To better understand the meaning of this phrase, we make a brief digression to considerthe more familiar Euclidean geometry in the plane. This kind of geometry deals with

certain subsets ofR2 and studies their properties. Especially important are points, lines,lengths and angles. For example, a triangle is an object consisting of three points and threeline segments joining them; the line segments have definite lengths and their intersectionshave definite angles that add up to 180 degrees. A circle is a subset ofR2 of points lying afixed distance r from a pointp R2. The lengthr is called the radius and circumferenceof the circle is 2r.

Thegroup of symmetriesof the Euclidean plane (or Euclidean transformations) con-sists of translations, rotations, and combinations of these. All of the objects and quan-tities studied in Euclidean geometry are preserved by these symmetries: lengths of linesegments, angles between lines, circles are sent to circles and triangles are sent to tri-

angles. Two objects in the plane are called congruent if they are related by Euclideantransformations, and this is the concept of isomorphism in this context.In 1872, Felix Klein initiated a highly influential re-examination of geometry called the

Erlangen Program. At the time, several different types of non-Euclidean geometries hadbeen introduced and it was unclear how they related to each other. Kleins philosophywas that a fundamental role was played by the group of symmetries in the geometry;that the meaningful concepts in a geometry are those that preserved by the symmetriesand that geometries can be related in a hierarchy according to how much symmetry theypossess.

With this philosophy in mind, let us explore the relationship between projective andEuclidean geometry in more depth. Recall from

2.2.2that we have a subset U0

RPn

consisting of those points with homogeneous coordinates [1 : y1 : ... : yn]. There is anobvious bijection

U0= Rn, [1 :y1 : ... : yn] (y1,...,yn)Proposition 2.5. Any Euclidean transformation ofU0= Rn extends uniquely to a pro-

jective transformation ofRPn.

Proof. We focus on the case n = 2 for notational simplicity. It is enough to show thattranslations and rotations ofU0 extend uniquely.

Consider first translation in R2 by a vector (a1, a2). That is the map (y1, y2) (y1+ a1, y2+ a2). The corresponding projective transformation is associated to the linear

transformation 1y1

y2

1 0 0a1 1 0

a2 0 1

1y1

y2

=

1y1+a1

y2+a2

A rotation by angle ofU0 = R2 extends to the projective transformation induced by the

linear transformation

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1y1y2

1 0 00 cos()

sin()

0 sin() cos()

1y1y2=

1cos()y1

sin()y2

cos()y2+sin()y1

Uniqueness will be clear after we learn about general position.

From the point of view of the Erlangen Program, Proposition 2.5 tells us that pro-jective geometry more basic than Euclidean geometry because it has more symmetry.This means that theorems and concepts in projective geometry translate into theoremsand concepts in Euclidean geometry, but not vice-versa. For example, we have lines andpoints and triangles in projective geometry, but no angles or lengths. Furthermore, ob-

jects in Euclidean geometry that are not isomorphic (or congruent) to each other canbe isomorphic to each other in projective geometry. We will show for example that all

projective triangles are isomorphic to one another. We will also learn that circles, ellipses,parabolas and hyperbolas all look the same in projective geometry, where they are calledconics.

2.4.2 Projective versus linear

Next, we should clarify the relationship between linear maps and projective morphisms.

Proposition 2.6. Two injective linear mapsT :V W andT: V Wdetermine thesame projective morphism if and only ifT =T for some non-zero scalar.

Proof. Suppose thatT =T. Then for [v] P(V),[T(v)] = [T(v)] = [T(v)],

so T and T define the same projective morphism.Conversely, suppose that for all [v] P(V) that [T(v)] = [T(v)]. Then for any basis

v0,...,vn V, there exist non-zero scalars0,...,n such that

T(vi) =iT(vi).

However it is also true that [T(v0+ ...+vn)] = [T(v0+ ...+ vn)] so for some non-zero

scalar we must have

T(v0+...+vn) =T(v0+...+vn).

Combining these equations and using linearity, we get

ni=0

T(vi) =T(n

i=0

vi) =T(n

i=0

vi) =n

i=0

T(vi) =n

i=0

iT(vi)

Subtracting givesn

i=0( i)T(vi) = 0. Since T is an injective,{T(vi)} is a linearlyindependent set, so= i for alli andT =T

.

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2.4.3 Examples of projective transformations

To develop some intuition about projective transformations, lets consider transformationsof the projective line F P1 =P(F2). A linear transformation ofF2 is uniquely described

by a 2 2 matrix, a bc d

xy

=

ax+bycx+dy

so for a projective point of the form [1 :y] we have

([1 :y]) = [a+by: c+dy] = [1 :c+dy

a+by].

In the case F = C, the function f : C C {} of the complex numbers

f(y) =

c+dy

a+by

is called a Mobius transformationand plays an important role in complex analysis.

Proposition 2.7. Expressed in a Euclidean coordinate y [1 : y], every projectivetransformation of the lineF P1 is equal to a combination of the following transformations:

A translation: y y+a A scaling: y y, = 0

An inversion: y

1y

Proof. Recall from linear algebra that any invertible matrix can be obtained from theidentity matrix usingelementary row operations: multiplying a row by a non-zero scalar,permuting rows, and adding one row to another. Equivalently, every invertible matrixcan be written as a product of matrices of the form

1 00

,

0 11 0

,

1 01 1

and these correspond to scaling, inversion and translation respectively. For instance,

0 11 0

1y

=

y1

= y

11/y

so corresponds to the inversion y 1y

It is also informative to try and picture these transformations using our model whereRP1 is a circle and CP1 is a sphere (leave this to class discussion). A delightful videoabout Moebius transformations is availablehere.

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Remark 1. More generally, a projective transformation ofF Pn takes the form

y c+Dya+ b

y

whereD is an n nmatrix, b, c are vectors in Fn,a is a scalar andis the dot product.Proving this is an exercise.

A more geometrically defined example of a projective transformation is obtained asfollows. LetP(V) be a projective plane, letP(U),P(U) be two projective lines in P(V),and let [w] P(V) be a projective points not lying on P(U) or P(U). We define a map

:P(U) P(U)as follows: Given [u] P(U), there is a unique line L P(V) containing both [w] and[u]. Define ([u]) to be the unique intersection point L P(U). Observe that is welldefined by Theorem2.3 and 2.4 .

Proposition 2.8.The map :P(U) P(U) is a projective transformation.Before proving this proposition, it is convenient to introduce a concept from linear

algebra: the direct sum.

2.4.4 Direct sums

LetV andWbe vector spaces over the same field F. The direct sum V Wis the vectorspace with vector set equal to the cartesian product

V W :=V W := {(v, w)|v V, w W}with addition and scalar multiplication defined by (v, w) + (v, w) = (v+ v, w+ w) and(v, w) = (v,w). Exercise: show that

dim(V W) = dim(V) + dim(W).There are natural linear maps

i: V V W, i(v) = (v, 0)and

p: V W V, p(v, w) =vcalledinclusion andprojectionmaps respectively. Similar maps exist for W as well.

Proposition 2.9. LetVbe a vector space and letU, Wbe vector subspaces ofV. Thenthe natural map

:U W V, (u, w) u+whas image

im() =span{U, W}with kernel

ker() =U W.In particular, ifspan{U, W} =V andU W = {0}, then defines an isomorphism

U W=Vand we say thatV is the internal direct sum ofU and W.

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Proof. The image of is the set{u + w V |u U, w W}. SinceU andWare bothclosed under linear combinations, it follows easily that{u+wV | uU, wW} =span{U, W}. To find the kernel:

ker() = {(u, w) U W| u+w= 0} = {(u, u) U W} =U W.

Finally, if ker() = U W ={0} and span{U, W} = V, then is both injective andsurjective, so it is an isomorphism.

Proof of Proposition2.8. LetW =span{w} be the one-dimensional subspace ofV corre-sponding to [w]. Since [w] P(U), it follows that W U = {0}. Counting dimensions,we deduce that

V=W Uis an internal direct sum. Denote by p : V U the resulting projection map.

Now let [u] P(U) and take[u] =([u]) P(U).

By definition of, [u] lies in the line determined by [w] and [u], so uis a linear combination

u= w+u

for some scalars, F. Moreover, [w] P(U) so [w] = [u] and we deduce that = 0,so

u=

w 1

u

so p(u) = 1

u and ([u]) = [p(u)]. It follows that is the projective transformationinduced by the composition of linear maps i : U V and p : V U.

2.4.5 General position

We gain a firmer grasp of the freedom afforded by projective transformations using thefollowing concept.

Definition 6. Let P(V) be a projective space of dimension n. A set ofn+ 2 points inP(V) are said to lie in general position if the representative vectors of any proper subsetare linearly independent inV (A subsetS

Sis called aproper subset ifSis non-empty

and S does not equal S).

Example 7. Any pair of distinct points in a projective line are represented by linearlyindependent vectors, so any three distinct points on a projective line lie in general position.

Example 8. Four points in a projective plane lie in general position if and only if nothree of them are collinear (prove this).

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son+1

i=1wi= wn+2=

n+1

i=1i

n+2wi

and we deduce that in+2 = 1 for all i and thus thatT = T for some non-zero constant

scalar.

Example 9. There is a unique projective transformation : F P1 F P1 sending anyordered set of three points to any other ordered set of three points. In the context ofcomplex analysis, we traditionally say that a mobius transformation is determined bywhere it sends 0, 1, .Remark 2. In the proof of Theorem2.10,we proved that for points p1,...,pn+2 P(V)in general position, and given a representative vector vn+2 V such that [vn+2] = pn+2,we may choose vectors v1,...,vn+1 V with [vi] =pi fori = 1,...,n+ 1 and satisfying

n+1i=1

vi = vn+2.

This is a useful fact that we will apply repeatedly in the rest of the course.

2.4.6 The Cross-Ratio

The concept of general position enables us to understand an invariant known at leastsince Pappus of Alexandria (c. 290 - c. 350), called the cross-ratio. Consider four points

A,B,C,D lying on a Euclidean line. We denote by AB the vector pointing from A to B .The cross ratio of this ordered 4-tuple of points is

R(A, B; C, D) :=AC

AD/

BC

BD=

AC

AD

BD

BC,

where the ratio of vectors is meaningful because one is a scalar multiple of the other. Ifa,b,c,d R, then the cross ratio is

R(a, b; c, d) =(a c)(b d)(a d)(b c) .

This extends to 4-tuples of points on RP1 by defining/ = 1.For homework you will prove that cross-ratio is preserved under projective transfor-

mations of the line. By Proposition2.8,it follows that in Diagram14we have an equalityof cross-ratiosR(A, B; C, D) =R(A, B; C, D).

We can define the cross-ratio in a different way using Theorem 2.10. For any fieldFlet

:L F P1 =F {}be the unique projective transformation sending (A) = , (B) = 0 and(C) = 1. Let(D) =d. Then

R(A, B; C, D) =R(, 0; 1, d) =( 1)(0 d)( d)(0 1)=

d

1

= d.

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Figure 12: The cross-ratio is preserved under projective transformations

2.5 Classical Theorems

2.5.1 Desargues Theorem

Desargues (1591-1661) a french mathematician, architect and engineer, is considered oneof the founders of projective geometry. His work was forgotten for many years, until itwas rediscovered in the 19th century during the golden age of projective geometry. Thefollowing result is named in his honour.

First, we adopt some notation commonly used in Euclidean geometry. Given distinct

points A, B in a projective space, let AB denote the unique line passing through themboth.

Theorem 2.11 (Desargues Theorem). LetA, B,C,A, B, C be distinct points in a pro-jective spaceP(V) such that the linesAA, BB, C C are distinct and concurrent (mean-ing the three lines intersect at a common point). Then the three points of intersectionX :=AB AB, Y :=BC BC andZ :=AC AC are collinear (meaning they lieon a common line).

Proof. LetPbe the common point of intersection of the three lines AA,BB andCC.

First suppose thatPcoincides with one of the pointsA, B,C,A, B, C. Without lossof generality, let P =A. Then we have equality of the lines AB =BB and AC=C C,so it follows that X = AB AB = B , Z = AC AC = C, so BC = XZ must becollinear with Y =BC BC and the result is proven.

Now suppose that P is distinct from A, B,C,A, B, C. Then P, A and A are threedistinct points lying on a projective line, so they are in general position. By Remark2,given a representative pV ofP, we can find representative vectors a, a ofA, A suchthat

p= a+a.

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Figure 13: Illustration of Desargues Theorem

Similarly, we have representative vectors b, b forB, B and c, c forC, C such that

p= b+b, p= c+c

It follows that a+a= b+b so

x:= a b= b a

must represent the intersection point X=AB AB, because xis a linear combinationof both a, band ofa , b. Similarly we get

z:= b c= c b and y:= c a= a crepresenting the intersection points Y andZ respectively.

To see thatX, Y and Zare collinear, observe that

x+z+y = (a b) + (b c) + (c a) = 0

so x,y,z are linearly dependent and thus must lie in a 2-dimensional subspace of Vcorresponding to a projective line.

2.5.2 Pappus Theorem

Desargues Theorem is our first example of a result that is much easier to prove usingprojective geometry than using Euclidean geometry. Our next result, named after Pappusof Alexandria (290-350) is similar.

Theorem 2.12. Let A,B,C and A, B, C be two pairs of collinear triples of distinctpoints in a projective plane. Then the three pointsX=BC BC, Y :=CA CA andZ:=AB AB are collinear.

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Figure 14: Illustration of Pappus Theorem

Proof. Without loss of generality, we can assume that A, B,C, B lie in general position.If not, then two of the three required points coincide, so the conclusion is trivial. ByTheorem2.10, we can assume that

A= [1 : 0 : 0], B= [0 : 1 : 0], C= [0 : 0 : 1], B= [1 : 1 : 1]

The line AB corresponds to the two dimensional vector space spanned be (1, 0, 0) and(0, 1, 0) inF3, so the point C AB must have the form

C= [1 :c : 0]

for some c= 0 (because it is not equal to A or B). Similarly, the line BC correspondsto the vector space spanned by (0, 0, 1) and (1, 1, 1) soA= [1, 1, a]

for some a = 1.Next compute the intersection points of lines

BC= span{(0, 1, 0), (0, 0, 1)} = {(x0, x1, x2)|x0= 0}BC=span{(1, 1, 1), (1, c, 0)}

so we deduce that BC

BCis represented by (1, 1, 1)

(1, c, 0) = (0, 1

c, 1).

AC = span{(1, 0, 0), (0, 0, 1)} = {(x0, x1, x2)|x1 = 0}AC=span{(1, 1, a), (1, c, 0)}

so we deduce that AC AC is represented by (1, c, 0) c(1, 1, a) = (1 c, 0, ca).AB= span{(1, 0, 0), (1, 1, 1)} = {(x0, x1, x2)|x1= x2}AB= span{(1, 1, a), (0, 1, 0)}

so we deduce that AB AB is represented by (a 1)(0, 1, 0) + (1, 1, a) = (1, a , a).

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Finally, we must check that the intersection points [0 : 1 c: 1], [1 c: 0 : ca] and[1 :a : a] are collinear, which is equivalent to showing that (0, 1 c, 1), (1 c, 0, ca) and(1,a ,a) are linearly dependent. This can be accomplished by row reducing the matrix,

1 a a1 c 0 ca

0 1 c 1

1 a a0 (c 1)a a

0 1 c 1

1 a a0 0 0

0 1 c 1

which has rank two, so the span of the rows has dimension two and the vectors are linearlydependent.

2.6 Duality

In this section we explore the relationship between the geometry of P(V) and that ofP(V) whereV denotes the dual vector space. Recall that given any vector spaceV, thedualvector space V is the set of linear maps from V to F

V:= {: V F} =Hom(V, F).

The dual V is naturally a vector space under the operations

(1+2)(v) =1(v) +2(v)

()(v) =((v)).

Suppose that v1,...,vn V is a basis. Then the dual basis 1,...,n V is theunique set of linear maps satisfying i(vj) = 1, ifi = j andi(vj) = 0 otherwise.

IfT : V W is a linear map, then there is a natural linear map T : W V(called the transpose) defined by

T(f)(v) =f(T(v)).

Although the vector spaces V and V have the same dimension, there is no natural iso-morphism between them. However there is a natural correspondence between subspaces.

Definition 7. LetU Vbe a vector subspace. The annihilator U0

V is defined byU0 := { V| (u) = 0, for all u U}.

To check thatU0 is a subspace ofV, observe that if1, 2 U0 and u U then

(11+22)(u) =11(u) +22(u) = 0 + 0 = 0

so (11+22) U0, soU0 is closed under linear combinations.Proposition 2.13. IfU1 U2 V thenU01 U02 .

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Proof. This is simply because if(v) = 0 for all v U2 then necessarily (v) = 0 for allv U1.

We also have the following result.

Proposition 2.14. dim(U) + dim(U0) = dim(V)

Proof. Choose a basis u1,...,um U and extend to a basis u1,...,um, vm+1,...,vn ofV.Let1,...,n be the dual basis. For a general element

=n

i=1

ii V

we have (ui) =i. Thus U0 if and only ifi = 0 for i= 1,...,mso U0 V is thesubspace with basis m+1,...,n which has dimensionn m.

The following proposition justifies the term duality.

Proposition 2.15. There is a natural isomorphism VS= (V), defined by v S(v)

whereS(v)() =(v).

Proof. We already know that dim(V) = dim(V) = dim((V)), so (by Proposition2.1)it only remains to prove that S : V (V) is injective. But ifv V is non-zero, wemay extend v1 :=v to a basis and then in the dual basis 1(v1) =S(v1)(1) = 1= 0 soS(v) = 0. Thus ker(S) = {0}andSis injective.Remark 3. Proposition 2.15 depends on V being finite dimensional. If V is infinite

dimensional, thenS is injective but not surjective.

In practice, we tend to replace the phrase naturally isomorphic with equals andwriteV = (V). Using this abuse of terminology, one may easily check that (U0)0 =U,so the operation of taking a subspace to its annihilator is reversible.

Proposition 2.16. LetP(V) be a projective space of dimensionn. Then there is a one-to-one correspondence between linear subspacesP(U) P(V)of dimensionmand linearsubspaces ofP(V) of dimensionn m 1 by the rule

P(U) P(U0).

We use notationP(U0) =P(U)0 and callP(U)0 thedualofP(U).

Proof. By preceding remarks there is a one-to-one correspondence

U U0

between subspaces of dimension m +1 inVand subspaces of dimensionn min V. Theresult follows immediately.

It follows from Propositions 2.13 and 2.16 that geometric statements about linearsubspaces ofP(V) translate into geometric statements about linear subspaces ofP(V).

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Example 10. LetP(V) be a projective plane, soP(V) is also a projective plane. Thereis a one-to-one correspondence between points in P(V) and lines in P(V). Similarly,there is one-to-one correspondence between lines in P(V) and points inP(V).

Corollary 2.17. Let A, B be distinct points in a projective plane lying on the line L.ThenA0, B0 are lines intersecting at the pointL0.

Proof. IfA, BL then L0 A0 and L0 B0, so L0 = A0 B0 is their single point ofintersection.

Corollary 2.18. Two distinct projective planes in projective3-space always intersect ina line.

Proof. This statement is dual to statement two points determine a line in 3-space(Theorem2.3).

Duality can be used to establish dual versions of theorems. For example, simply beexchanging points with lines and collinear with concurrent, we get the following results.

Theorem 2.19(Dual version of Desargues Theorem = Converse of Desargues Theorem).Let, ,,, , be distinct lines in a projective plane such that the points , and are distinct and collinear. Then the lines joining to , to and to are concurrent.Theorem 2.20 (Dual version of Pappus Theorem). Let,, and, , be two setsof concurrent triples of distinct lines in a projective plane. Then the three lines determinedby pairs of points and , and , and are concurrent.

In fact, with a little effort one can show that the statement of Pappus Theorem isequivalent to the Dual of Pappus Theorem (see Hitchins notes for details). Thus PappusTheorem may be considered self-dual.

As a final result, we describe the set of lines in R2.

Proposition 2.21. The set of lines inR2 is in one-to-one correspondence with the Mobiusstrip.

Proof. The set of lines in R2 is the set of lines in RP2 minus the line at infinity. Byduality, this is identified with RP2 minus a single point, which is the same as a mobiusstrip.

3 Quadrics and Conics

3.1 Affine algebraic sets

Given a set {x1,...,xn} of variables, a polynomialf(x1,...,xn) is an expression of the formf=f(x1,...,xn) :=

I

aIxI

where

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the sum is over n-tuples of non-negative integersI= (i1,...,in), xI =xi11...xinn are called monomials, and

aI Fare scalars called coefficients, of which all but finitely many are zero.Thedegreeof a monomial xI is the sum|I| =i1+...+in. The degree of a polynomial fequals the largest degree of a monomial occuring with non-zero coefficient.

Example 11. x1 x2 is a polynomial of degree one. x2 +y+ 7 is a polynomial of degree two. xy2 + 2x3 xyz+yz 11 is a polynomial of degree three.A polynomialf(x1,...,xn) in n variables determines a function

2

f :Fn Fby simply plugging in the entries n-tuples inFn in place of the variables x1,...,xn.

Definition 8. Given an n variable polynomialf, the set

Zaff(f) := {(x1,...,xn) Fn | f(x1,...,xn) = 0}is called the (affine) zero set off. More generally, iff1,...,fk are polynomials then

Zaff(f1,...,fk) :=Zaff(f1) ... Zaff(fk)

is called the zero set of f1,...,fk. An affine algebraic set X Fn is a set that equalsZaff(f1,...,fk) for some set of polynomials f1,...,fk.

Example 12. An affine line in F2 is defined by a single equationax + by + c= 0, whereaandbare not both zero. Thus a line inF2 is the zero set of a single degree one polynomial.More generally, an affine linear subset ofFn (points, lines, planes, etc.) is the zero set ofa collection of degree one polynomials.

Example 13. The graph of a one variable polynomial f(x) is defined by the equationy= f(x), thus is equal to the algebraic set Zaff(y f(x)).Example 14. Any finite set of points

{r1,...,rd

} Fis an algebraic set Zaff(f) wheref(x) =

(x r1)(x r2)...(x rd).Definition 9. An affine quadric X Fn is a subset of the form X= Zaff(f) where fis a single polynomial of degree 2. In the special case that X F2, we call X a (affine)conic.

Example 15. Some examples of affine in R2 conics include: circles, ellipses, hyperbolasand parabolas. The unit sphere Zaff(x

21+...+x

2n 1) is a quadric in Rn.

2For some fields, two different polynomials can define the same function. For example x2 + x definesthe zero function over Z2.

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3.2 Projective algebraic sets

What is the right notion of an algebraic set in projective geometry? We begin with adefinition.

Definition 10. A polynomial is called homogeneousif all non-zero monomials have thesame degree.

Example 16. x y is a homogeneous polynomial of degree one. x2 +y+ 7 is a non-homogeneous polynomial. yz2 + 2x3 xyzis a homogeneous polynomial of degree three.Projective algebraic sets in F Pn are defined using homogeneous polynomials in n + 1

variables, which we normally denote{x0, x1,...,xn}. The crucial property that makeshomogenous polynomials fit into projective geometry is the following.

Proposition 3.1. Let f(x0,...,xn) be a homogeneous polynomial of degree d. Then forany scalar F, we have

f(x0,...,xn) =df(x0,...,xn).

Proof. For a monomial xI =xi00...xinn of degree d = |I| =i0+...+in, we get

(x)I = (x0)i0 ...(xn)

in =i0+...+inxi00...xinn =

dxI.

The same will hold true for a linear combination of monomials of degree d.

It follows from Proposition3.1 that for a non-zero scalar, f(x0,...,xn) = 0 if andonly iff(x0,...,xn) = 0. This makes the following definition well-defined.

Definition 11. Let f1,...,fkbe a set of homogeneous polynomials in the variables x0,...,xn.The zero set

Z(f1,...,fk) := {[x0 : ... : xn] F Pn | fi(x0,...,xn) = 0 for all i=1,...k}is a called a projective algebraic set. Notice that

Z(f1,...,fk) =Z(f1) ... Z(fk).The main example we have encountered so far are linear subsetsof projective space.

These are always of the form Z(l1,...,lk) for some set of homogeneous linear polynomialsl1,...,lk.

To understand the correspondence between projective and algebraic sets we recall thatwithin F Pn, we have a copy of the affine plane U0 :={[1 : y1,... : yn] F Pn}= Fn.Under this correspondence, we have an equality

Z(f1,...,fk) U0= Zaff(g1,...,gk)wheregi(x1,...,xn) :=f(1, x1,...,xn).

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Conversely, given an affine algebraic set we construct a projective algebraic set asfollows. Letf(x1,..,xn) be a non-homogeneous polynomial of degree d. By multiplyingthe monomials of f by appropriate powers of x0, we can construct a homogenous, de-

gree d polynomial h(x0,...,xn) such that h(1, x1,...,xn) = f(x1,...,xn). We call h thehomogenization off.

Example 17. Here are some examples of homogenizing polynomials

g(x) = 1 x+ 2x2 h(x, y) =y2 xy+ 2x2

g(x, y) =y x2 + 1h(x,y,z) =yz x2 +z2

g(x1, x2) =x1+x32+ 7h(x0, x1, x2) =x20x1+x32+ 7x30Now suppose that g1,...,gk are non-homogeneous polynomials with homogenizations

h1,...,hk. Then we have equality

Z(h1,...,hk) =U0 Zaff(g1,...,gk).The study of algebraic sets is called algebraic geometry. It is a very active field of re-

search today, of which projective geometry is a small part. So far we have only consideredlinear algebraic sets, where the polynomials all have order one. Our next big topic arealgebraic sets of the form Z(f) where f is a degree two polynomial. Such algebraic setsare calledquadrics.

3.3 Bilinear and quadratic forms

Definition 12. A symmetric bilinear form B on a vector space V is a map

B:V V Fsuch that

B(u, v) =B(v, u), (Symmetric) B(1u1+2u2, v) =1B(u1, v) +2B(u2, v), (Linear)Observe that by combining symmetry and linearity, a bilinear form also satisfies

B(u, 1v1+2v2) =1B(u, v1) +2B(u, v2).

We say that B is linear in both entries, orbilinear. We say that B is nondegenerate iffor each non-zero v V, there is a w V such thatB(v, w) = 0.Example 18. The standard example of a symmetric bilinear form is the dot product

B: Rn Rn R, B((x1,...,xn), (y1,...,yn)) =n

i=1

xiyi.

The dot product is nondegenerate because B(v, v) = |v|2 >0 ifv= 0.

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Example 19. The bilinear form B : R4 R4 Rdefined by

B((x0, x1, x2, x3), (y0, y1, y2, y3)) = x0y0+x1y1+x2y2+x3y3is called the Minkowski metricthat measures lengths in space-time according to specialrelativity. In contrast with the dot product, the equation B(v, v) = 0 has holds for somenon-trivial vectors callednull-vectors. The null-vectors form a subset ofR4 called thelightconedividing those vectors satisfying B(v, v) < 0 (called time-like vectors) from thosesatisfyingB(v, v)> 0 (called space-like vectors).

The bilinear form is completely determined by its values on a basis.

Lemma 3.2. Letv1,...,vn Vbe a basis. For any pair of vectorsu= x1v1+ ...xnvn andw= y1v1+...+ynvn we have

B(u, w) =n

i,j=1

xiyjB(vi, vj).

This means that the bilinear form is completely determined by then n-matrix withentriesi,j =B(vi, vj).

Proof.

B(u, w) =B(x1v1+...xnvn, w)

=n

i=1

xiB(vi, w) =n

i=1

xiB(vi, y1v1+...+ynvn)

=n

i=1

nj=1

xiyjB(vi, vj) =n

i,j=1

xiyji,j

Lemma 3.2 provides a nice classification of bilinear forms: Given a vector space Vequipped with a basis v1,...,vn, then there is a one-to-one correspondence between sym-metric bilinear forms B on V and symmetric n nmatrices , according to the rule

B(v, w) = vTw

where v, w denote the vectors v, w expressed as column vectors and T denotes transpose,so that vT is v as a row vector.

Proposition 3.3. The bilinear formB is non-degenerate if and only if the matrix hasnon-zero determinant.

Proof. Suppose that det() is zero if and only if there exists a nonzero row vector vT suchthat vT= 0 if and only if vTw= 0 for all column vectors w if and only if there existsv V\ {0}such that B(v, w) = 0 for all w Vif and only ifB is degenerate.

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3.3.1 Quadratic forms

Given a symmetric bilinear form B : V V F, we can associate a function called aquadratic formQ: V

Fby the rule Q(v) =B(v, v).

Proposition 3.4. LetVbe a vector space over a fieldF in which1 + 1 = 0. LetB be asymmetric bilinear form onV and letQ : V Fbe the associated quadratic form. Then

for anyu, v V we haveB(u, v) = (Q(u+v) Q(u) Q(v))/2.

In particular, B is completely determined by its quadratic formQ.

Proof. Using bilinearity and symmetry, we have

Q(u+v) =B(u+v, u+v) =B(u, u)+B(u, v)+B(v, u)+B(v, v) =Q(u)+Q(v)+2B(u, v)

from which the result follows.

Remark 4. Proposition3.4does not hold when F is the field of two elements becausewe are not allowed to divide by 2 = 0.

We can understand these relationship better using a basis. Letv1,...,vn for V be abasis so B can be expressed in terms of the n nmatrix. According to ...

Q(x1v1+...+xnvn) =B(x1v1+...+xnvn, x1v1+...+xnvn) =i,j

i,jxixj

is a homogeneous quadratic polynomial.Some examples:

a ax2

a bb c

ax2 + 2bxy+cy2

a b db c e

d e f

ax2 + 2bxy+cy2 + 2dxy+ 2exz+f z2

3.3.2 Change of basis

Weve seen so far that a symmetric bilinear form B can be described using a symmetricmatrix with respect to a basis. What happens if we change the basis?

Let v1,...,vn and w1,...,wn be two different bases of V. Then we can find scalars{Pi,j}ni,j=1 such that

wi=j

Pi,jvj (4)

The matrixPwith entries Pi,j is called the change of basis matrix from w1,...,wn tov1,...,vn.

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Proof. Consider the map : V F, defined by(v) =B(v, w). Then is linear because

(1v1+2v2) =B(1v1+2v2, w) =1B(v1, w) +2B(v2, w) =1(v1) +2(v2)

so w is equal to the kernel of , hence it is a vector subspace. By the rank nullitytheorem,

dim(w) = dim(V) dim(im()) =

n im() = {0}n 1 im() =F.

Theorem 3.8. LetB be a symmetric, bilinear form on a vector spaceV overF =C orF = R. Then there exists a basisv1,...,vn such that

B(vi, vj) = 0 (5)

fori =j and such that for alli= 1,...,n

B(vi, vi)

0, 1 F = C

0, 1 F = R. (6)

I.e., there is a choice of basis so that is diagonal with diagonal entries 0, 1 if F = Cand0, 1, 1 ifF = R.Proof. As a first step construct a basis satisfying (5) using induction on the dimension ofV. In the base case dim(V) = 1, (5) holds vacuously.

Now suppose by induction that all n 1 dimensional symmetric bilinear forms canbe diagonalized. LetV be n dimensional. IfB is trivial, then is the zero matrix withrespect to any basis (hence diagonal). If B is non-trivial then by Proposition3.4, thereexists a vectorwn Vsuch thatB(wn, wn) =Q(wn) = 0. Clearlywn vn, sown isn1dimensional. By induction we may choose a basis w1,...,wn1 wn satisfying (5) for alli ,j < n. Furthermore, for all i < n, B(wi, wn) = 0 because wi wn . Thusw1,...,wnsatisfy (5).

Finally, we want to modify w1,...,wn to satisfy (6). Suppose that B(wi, wi) =c F.Ifc = 0, then let vi = wi. Ifc = 0 and F = C, letvi= 1cwi so that

B(vi, vi) =B( 1cwi, 1cwi) =

1c2

B(wi, wi) = cc = 1.

Ifc = 0 and F = Rthen letvi := 1|c|wi so

B(vi, vi) = c

|c| = 1.

Corollary 3.9. LetV andB be as above and letQ(v) =B(v, v). Then

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IfF = C then for somem n there exists a basis such that ifv=ni=1 ziviQ(v) =

m

i=1

z2i

IfF = R then for somep, qwithp+q n there exists a basis such that

Q(v) =

pi=1

z2ip+qi=p+1

z2i

Proof. Choose the basis v1,...,vnso thatB(vi, vj) satisfy the equations (5) and (6). Then

Q(z1v1+...+znvn) =B(z1v1+...+znvn, z1v1+...+znvn) =

ni,j=1

zizjB(vi, vj) =

ni=1

z2i B(vi, vi),

where the B(vi, vi) are 0, 1 or1 as appropriate.Exercise: Show that the numbers m, p, q occurring above are independent of the

choice of basis.It follows thatB andQ are non-degenerate if and only ifm = n or m = p + q. In the

case of a real vector space, we say that B is positive definite ifp= n, negative definite ifq= n and indefiniteotherwise.

3.3.3 Digression on the HessianThis discussion will not be on the test or problem sets.

An important place where symmetric bilinear forms occur in mathematics is the studyof critical points of differentiable functions. Letf(x1,...,xn) be a differentiable functionand let (a1,...,an) be a point at which all partial derivatives vanish:

fxi(a1,...,an) = f

xi|(a1,...,an)= 0, for alli = 1,...,n

If the the second order partials are defined and continuous then the n n-matrixHwithentries

Hi,j =fxixj (a1,...,an) = 2f

xjxi |(a1,...,an)is a symmetric matrix (Hi,j = Hj,i) defining a symmetric bilinear form on the set oftangent vectors called the Hessian. The corresponding quadratic form is equal to thesecond order Taylor polynomial of the function.

The Hessian can be used to determine if f has a local maximum or minimum at(a1,...,an) (using the Second Derivative Test). Specifically, ifHis non-degenerate thenfhas a local maximum at (a1...,an) ifH is negative definite, has a local minimum ifHis positive definite, and has neither a max or min if H is neither positive nor negativedefinite.

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Similarly, there is a notion of a complex valued function of complex variables beingdifferentiable (also called analytic). Reasoning as above one can show that for a complex-valued functionf,|f|has no local maximums where fis differentiable. This is a versionof the maximum principle.

3.4 Quadrics and conics

We now can state a coordinate independent definition of a quadric.

Definition 13. AquadricX P(V) is a subset defined byX=Z(B) := {[v] V|B(v, v) = 0}.

By convention, we exclude the trivial case B = 0.

Is the case P(V) =F Pn, this definition is equivalent to X=Z(f) for some homoge-neous degree two polynomial f.

We say that B and B are projectively equivalent if there exists non-zero scalar F {0}such that B =B . Observe that

Z(B) =Z(B)

so projectively equivalently symmetric bilinear forms determine same quadric.Consider now the implications of Corollary3.9 for quadrics in a projective line. Over

the reals such a quadric has the form Z(f) = Z(f) where (up to a change of coordi-nates/linear change of variables) f equals

x2

x2 +y2

x2 y2

In the first case, Z(x2) consists of the single point [0 : 1]. In algebraic geometry, wewould say thatZ(x2) has a root of multiplicity two at [0 : 1], but we wont define thisconcept in this course. The caseZ(x2 +y2) is empty set because there are no solutionswith both x and y non zero. FinallyZ(x2 y2) = Z((x y)(x+y)) has two solutions:[1 : 1] and [1 :

1].

These three possibilities correspond to a familiar fact about roots of quadratic poly-nomials 0 = ax2 +bx+c: such a polynomial can have two roots, one root, or no roots.A consequence of Corollary3.9 is that up to projective transformations, the number ofroots completely characterizes the quadric.

Over the complex numbers, the classification is even simpler: a quadric has the formZ(f) where fis one of (up to coordinate change)

x2

x2 +y2

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The first caseZ(x2) is a single point of multiplicity 2, while Z(x2+y2) =Z((xiy)(x+iy))consists of the two points [i: 1] and [i:1]. This corresponds to the fact that over thecomplex numbers 0 = ax2 +bx+c with a= 0 has either two distinct roots or a singleroot of order two (complex polynomials in one variable can always be factored!).We now consider the case F = R andP(V) is a projective plane (in this case a quadricis called a conic). We may choose coordinates so that X=Z(f) where fis one of:

x2, x2 +y2, x2 +y2 +z2, x2 y2,

x2

+y

2

z2

,up to multiplication by1 which doesnt change Z(f) =Z(f).

The first examplex2 =xx = 0

has the same solution set as x= 0, which is projective line{[0 :a : b]}= RP1, howeverthis line is counted twice in a sense we wont make precise (similar two a double root ofsingle variable polynomial). The second example

x2 +y2 = 0

has a single solution [0 : 0 : 1]. The third example x2

+y2

+z2

= 0, has no projectivesolutions (remember (0, 0, 0) does not represent a projective point!). The example

x2 y2 = (x y)(x+y)

factors as a product of linear polynomials. It follows that Z(x2 y2) is a union of twodistinct lines intersecting at one point.

The last example Z(x2 +y2 z2) is (up to coordinate change) the only non-emptyand non-degenerate real conic, up to projective transformation. The apparent differencebetween the affine conics: ellipses, parabolas, hyperbolas, can be understood as how theconic relates to the line at infinity. A conic looks like an ellipse if it is disjoint form the

line at infinity, looks like parabola if it is tangent to the line at infinity, and looks like ahyperbola if it intersects the line at infinity at two points.

To illustrate, consider what happens when we set z= 1 and treat x and y like affinecoordinates (so the equationz= 0 is the line at infinity). Then the polynomial equationx2 +y2 z2 = 0 becomes x2 +y2 1 = 0 or

x2 +y2 = 1

which defines the unit circle in the x-y plane.

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Now instead, set y = 1 and consider x, z as affine coordinates (so y = 0 defines theline at infinity). Then we get the equation

x2 + 1

z2 = 0,

orz2 x2 = (z x)(z+x) = 1

which is the equation of a hyperbola asymptotic to the lines x = zandx= z.Finally, consider performing the linear change of variables z = y+ z. In these new

coordinates, the equation becomes

x2 +y2 (z y)2 =x2 + 2zy z2 = 0.Passing to affine coordinates x, y by setting z= 1 we get

x2 + 2y 1 = 0or

y= x2/2 + 1/2which defines a parabola.

Since ellipses, hyperbolas, and parabolas can all be obtained by intersecting an affineplane with the cone in R3 defined by x2 +y2 z2, they are sometimes called conicsections.

Figure 15: Conic sections

In case F = C then the possible conics is even more restricted. Up to change ofcoordinates, the possible quadratics in the complex projective plane are defined by thequadratic polynomial equations

x2 = 0, x2 +y2 = 0,

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x2 +y2 +z2 = 0.The equationx2 = 0 defines a projective line that is counted twice or has multiplicity

two (in a sense we wont make precise). The equation

x2 +y2 = (x+iy)(x iy) = 0defines the union of two distinct lines determined by the equations x + iy = 0 andx iy= 0. The last equation x2 + y2 + z2 = 0 defines the non-degenerate quadric curve.We study what this curve looks like in the next section.

3.5 Parametrization of the conic

Theorem 3.10. Let C be a non-degenerate conic in a projective plane P(V) over thefieldF in which2

= 0, and letA be a point onC. LetL

P(V) be a projective line not

containingA. Then there is a bijection

: L C

such that, forX L, the pointsA, X, (X) are collinear.

Figure 16: Parametrization of the conic

Proof. Suppose the conic is defined by the symmetric bilinear formB. Leta V representthe fixed point A. Because A C, we know that B(a, a) = 0. Let x V represent apoint X L. Because A L, it follows that X= A so the vectors a, x are linearlyindependent, so we can extend them to a basis a, x, y V.

The restriction of B to the span of a, x is not identically zero. If it were then thematrix forB associated to the basisa, x, y would be of the form

0 0 0 0

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which has determinant zero, and this is impossible because B is non-degenerate. It followsthat eitherB(a, x) orB(x, x) is non-zero.

An arbitrary point on the line AXhas the form [a+x] for some scalars and .

Such a point lies on the conicCif and only ifB(a+x, a+x) =2B(a, a) + 2B(a, x) +2B(x, x)

=(2B(a, x) +B(x, x)) = 0

for which there are two solutions in the homogeneous coordinates and: The solution= 0 corresponding the point A and solution to 2B(a, x) + B(x, x) = 0 correspondingto the point represented by the vector

w:= 2B(a, x)x B(x, x)awhich is non-zero because one of the coefficients 2B(a, x) orB(x, x) is non-zero.

We define the map: L Cby(X) = [w].

It remains to prove that is a bijection. We do this by proving the preimage of anypoint is a paint. IfY is a point on Cdistinct from A, then by Theorem 2.4there is aunique point of intersection AY L, so 1(Y) =AY L.

To determine 1(A), observe that

(X) = [2B(a, x)x B(x, x)a] =A

if and only ifB(a, x) = 0, if and only ifx a and X L. Since a is 2-dimensionalsubspace by Lemma3.7, P(a) is a projective line. The intersection of lines P(a) andLis the unique point in L mapping to A.

In the course of proving Theorem3.10, we proved the following result that will comeup again later.

Proposition 3.11. LetC = Z(B) be a non-degenerate conic and P C a point. Thepolar ofP is the unique lineL such thatC L= P. We callL the tangent line ofCatP.

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Though Theorem3.10only says thatis a bijection of sets, it is in fact an isomorphismof algebraic varieties. In particular, this means that a real projective conic is has thetopology to a real projective line, which in turn has the topology of a circle - a fact that

weve already seen. A second consequence is that the complex projective conic is has thetopology ofCP1 which in turn has the topology of the two sphere S2.

3.5.1 Rational parametrization of the circle

Let us now describe a special case of the map in coordinates. Let C=Z(x2 + y2 z2)corresponding to the matrix

=

1 0 00 1 0

0 0 1

.

Choose A = [1 : 0 : 1] and let L = Z(x). Then points in Lhave the form [0 : t : 1] or

[0 : 1 : 0] and the map : L Csatisfies([0 :t : 1]) = [2B((1, 0, 1), (0, t, 1))(0, t, 1) B((0, t, 1), (0, t, 1))(1, 0, 1)]

= [2(0, t, 1) (t2 1)(1, 0, 1)]= [(t2 + 1, 2t, t2 1)]= [t2 1 : 2t: t2 + 1]

and

([0 : 1 : 0]) = [2B((1, 0, 1), (0, 1, 0))(0, 0, 1) B((0, 1, 0), (0, 1, 0))(1, 0, 1)]= [0(0, 0, 1) + 1(1, 0, 1)]

= [1 : 0 : 1] = A

These formulas can be used to parametrize the circle using rational functions. Fort areal number 1 +t2 >0 is non-zero so we can divide

[t2 1 : 2t: t2 + 1] = [ t2 1

t2 + 1:

2t

t2 + 1: 1]

and we gain a parametrization of the unit circle in affine co-ordinates : R S1 =Zaff(x

2

+y2

1)(t) =

t2 1t2 + 1

, 2t

t2 + 1

that hits all points except (1, 0).

These formulas have an interesting application in the caseF = Q is the field of rationalnumbers (those of the form a/b where a and b are integers).

Since we have focused on the cases F = R and F = C so far, we take a moment todiscuss informally linear algebra and projective geometry over F = Q. To keep matterssimple, we concentrate on the standard dimension n vector space over Q, Qn consistingof n-tuples of rational numbers. It is helpful geometrically to consider Qn as a subset

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ofRn in the standard way. Elements in Qn can be added in the usual way and can bemultiplied by scalars from Q, so it makes sense to form linear combinations

1v1+...

kvk

Qn,

where1,...,k Q andv1,...,vk Qn. The span of a set of vectorsspan(v1,...,vk) Qnis the set of all linear combinations ofv1,...,vk. In particular, a one-dimensional subspaceofQn is the span of a single non-zero vector. We define the projective space

QPn =P(Qn+1)

to be the set of one dimensional subspaces in Qn+1. Geometrically, we can regard

QPn RPn

consisting of those projective points that can be represented by [v] RPn wherev Qn+1.Many of the results about projective geometry we have developed so far generalize

immediately to geometry in QPn (one notable exception is the classification of symmetricbilinear forms because the proof depends on taking square roots but the square root of arational number is not necessarily rational!).

In particular the formulas parametrizing the circle remain valid. That is, solutions tothe equation

x2 +y2 =z2

forx, y,z Qmust be of the form

[x: y : z] = [1 t2

: 2t: 1 +t2

]

for some rational number t Q. If we let t= a/b where a and b are integers sharing nocommon factors, then

[x: y : z] = [1 (a/b)2 : 2(a/b) : 1 + (a/b)2] = [b2 a2 : 2ab: b2 +a2].

It follows in particular that every integersolution to the equationx2 + y2 =z2 must havethe form

x= c(b2 a2),

y= c(2ab),

z= c(b2 +a2)

for some integersa,b,c. Such solutions are calledPythagorean triplesbecause they defineright triangles with integer side lengths. For example, a = c = 1, b = 2 determinesx= 3, y= 4, z= 5, while a = 2, b= 3, c= 1 determinesx = 5, y= 8, z= 13.

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3.6 Polars

LetBbe a non-degenerate bilinear form on a vector space V. Recall thatV = Hom(V, F)is the dual vector space for V.

Lemma 3.12. The map B# :V V defined byB#(v)(w) :=B(v, w)

is an isomorphism of vector spaces

Proof. The map is linear because

B#(1v1+2v2)(w) =B(1v1+2v2, w)

=1B(v1, w) +2B(v2, w)

=1B#(v1)(w) +2B#(v2)(w)

soB#(1v1+2v2) =1B

#(v1) +2B#(v2).

Since B is non-degenerate, ifv= 0 then there exists w V such that B#(v)(w) =B(v, w)= 0. Thus ker(B#) = 0 and B# is injective. By ... dim(V) = dim(V) so B#must also be surjective and thus is an isomorphism between V and V.

Given a vector subspace U V, defineU :=

{v

V

|B(u, v) = 0, for allu

V

}.

We callU orthogonal orperpendicular subspace ofU(relative to B).

Proposition 3.13. The isomorphism of Lemma3.12sendsU to U0. In particular, forany subspaces ofV

(U) = U U1 U2 U2 U1 dim(U) + dim(U) = dim(V)

Proof.

B#(U) = {B#(v) V|v U}= { V|(u) = 0 for all u U}= U0.

The properties is inherited from Section2.6.

The isomorphismB# determines a projective isomorphism betweenP(V) with its dualP(V)

P(V) =P(V). (7)

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Definition 14. LetB be a non-degenerate symmetric bilinear form on Vand letP(U) P(V) be a linear subspace. We call P(U) the polar ofP(U).

IfP(V) has dimensionn and P(U)

P(V) has dimensionm thenP(U) has dimen-

sion n m 1, just like for duality.Proposition 3.14. LetCbe a non-degenerate conic in a complex projective planeP(V).Then

(i) Each line in the plane meets the conic in one or two points

(ii) IfP C, its polar line is the unique tangent to C passing throughP(iii) IfP C, the polar line ofP meetsCin two points, and the tangents to Cat these

points intersect atP.

Figure 17: Duality for conics

Proof. Let U Vbe the 2-dimensional subspace defining the projective line P(U) andletu, v Ube a basis for U. A general point in P(U) has the form [u+v] for scalars, Cnot both zero. A point [u+v] lies on the conic C if

0 =B(u+v, u+v) =2B(u, u) + 2B(u, v) +2B(v, v). (8)

We think ofandas homogeneous coordinates on the projective line, so that (8) defines

the roots of a quadratic polynomial. From 3.4we know that (8) has one or two solutions,establishing (i). Property (ii) is a direct consequence of Proposition3.11.

Now let [a] denote a point notlying onC, and letP(U) denote the polar of [a]. Fromthe calculation above, it follows thatP(U) meetsCin two distinct points [u] and [w] with[u] = [w]. Because [u] lies on Cwe know B(u, u) = 0. Because [u] lies on the polar of [a]we know B(u, a) = 0. It follows that for all points on the line joining [u] and [a] that

B(u,u+a) =B(u, u) +B(u, a) = 0

so the line joining [u] and [a] is the polar of [u] hence also the tangent to C at [u]. Thesame argument applies to [w] and this establishes (iii).

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Proposition 3.14 establishes a duality between circumscribed polygons and in-scribed polygons for complex conics, as depicted in the real plane below.3

Figure 18: Duality for conics

3.7 Linear subspaces of quadrics and ruled surfaces

Quadrics in projective spacesP(V) of dimension greater than 2 often contain linear spaces.However, such a linear space can only be so big.

Proposition 3.15. Suppose thatQ P(V) is a non-degenerate quadric over fieldF inwhich0 = 2. Then if some subspaceP(U) is a subset ofQ, then

dim(P(U)) (dim(P(V)) 1)/2.

Proof. Because of dimension shifts, the statement is equivalent to

dim(U) dim(V)/2

Suppose that P(U) Q. Then for all u U, B(u, u) = 0. Furthermore, for any pairu, u Uwe have by Proposition3.4that

B(u, u) =1

2(B(u+u, u+u)

B(u, u)

B(u, u)) = 0

from which it follows that U U. According to Proposition2.14this means that

2dim(U) dim(U) + dim(U) = dim(V) (9)from which the conclusion follows.

3This duality is explored in the suggested project topic Pascals Theorem, Brianchons Theorem andduality

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Indeed, over the complex numbers the maximum value of (9) is always realized. Ifnis odd then we can factor

x20+...+x2

n

= (x0+ix

1)(x

0 ix

1) +...+ (xn

1+ixn)(xn

1 ixn)

so the non-degenerate quadric Z(x20+...+x2n) in CP

n contains the linear space

Z((x0 ix1),..., (xn1 ixn))of dimension (n 1)/2.

In the real case, the maximum value of (9) is achieved when n is odd and n+ 1 =2p= 2q(we say B has split signature) . In this case,

x21+...+x2p x2p+1 ... x22p= (x1+xp+1)(x1 xp+1) +...+ (xp+x2p)(xp x2p)

so the quadric contains the linear subspace Z(x1 xp+1,...,xp x2p).This is best pictured in the case n = 3:

x2 +y2 z2 w2 = (x z)(x+z) + (y w)(y+w) = 0passing to affine coordinates by setting w = 1 gives

x2 +y2 =z2 + 1

which defines a surface called a hyperboloid or cooling tower.Indeed, the hyperboloid can be filled or ruled by a family of lines satisfying the

equations

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

for some homogeneous coordinates [ : ]. It can also by ruled by the family of linessatisfying

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

for some homogeneous coordinates [: ].

3.8 Pencils of quadrics and degeneration

In this section, we move beyond the study of individual quadrics to the study of familiesof quadrics. The proof of the following proposition is straightforward and will be omitted.

Proposition 3.16. The set of symmetric bilinear forms on a vector space V forms avector space under the operation

(1B1+2B2)(v, w) =1B1(v, w) +2B2(v, w).

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Figure 19: Bi-ruled hyperboloid

Figure 20: Bi-ruled hyperboloid

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Proof. Choose a basis for V so thatB andB are represented by matrices and. Thebilinear for B+ B is degenerate if and only if the determinant det(+ ) equalszero. Expanding out the expression det(+ ) is a degreen homogeneous polynomial

in the homogeneous variables [: ], so it has at mostn roots. Over C, such a polynomialhas at least one root by the Fundamental Theorem of Algebra.

Theorem 3.19. Two non-degenerate conics in a complex projective plane intersect in atleast one and no more than four points.

Proof. First we show that all pairs of conics on the pencil intersect at the same points.Let [v] P(V) be an intersection point of two non-degenerate conics Z(B) andZ(B). IfB:= B+B is a linear combination then

B(v, v) =B(v, v) +B(v, v) = 0 + 0 = 0

so [v] Z(B). It follows that Z(B) Z(B) Z(B) soZ(B) Z(B) Z(B) Z(B).

If = 0, then the pencil determined by B andB coincides with the one determinded byB andB. In particular,

B= 1

(B+B)

B=

1

B

B

so arguing conversely, we get

Z(B) Z(B) =Z(B) Z(B). (11)By Proposition3.18we know that Z(B) is degenerate for some choice [: ], so we

are reduced to counting the intersection points between degenerate and a non-degenerateconic. From3.4we know that degenerate conics can either be a line or a union of twolines. By Proposition3.14it follows that

Z(B) Z(B) =Z(B) Z(B)consists of between one and four points.

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Corollary 3.20. Two conics in a complex projective plane have between one and fourtangent lines in common.

Proof. Follows from Theorem3.19by duality (see Proposition ??).

Remark 6. With an appropriate notion of multiplicity, Theorem3.19 (resp. Corollary3.20) can be improved to say there are exactly four intersection points (resp. commontangents) counted with multiplicity.

Using equation (11), we are able to produce pairs on non-degenerate conics realizingall possible intersections 1 to 4.

For example, consider the affine circle defined by the equation (x 1)2 +y2 1 = 0.This circle intersects the y-axis at one point, and so it intersects the degenerate conic

defined bex2 = 0 exactly once. Homogenize to get polynomials

f(x,y,z) := (x z)2 +y2 z2 =x2 2xz+y2

and x2. I claim that the quadricsZ(f) and Z(x2) have only one intersection point overthe complex numbers. To see this, observe that Z(x2) ={[0 :a : b]| [a: b]CP1} andthatf(0, a , b) = (0 b)2 +a2 b2 =a2 vanishes only if [0 :a : b] = [0 : 0 : 1]. Accordingto (11), this means that Z(x2 + f(x,y,z)) = Z(y2 2xz) intersectsZ(f) at the uniquepoint [0 : 0 : 1].

How do we interpret pencils of conics geometrically? Here is the generic answer.

Proposition 3.21. Suppose that two complex conics intersect in four points lying in

general position. Then the pencil of conics they determine consists of all conics passingthrough those four points.

Proof. All conics in the pencil must contain the four intersection points by (11). It remainsto prove that the set of conics passing through four points in general position is a pencil.Up to a projective transformation, we may assume that the four intersection points are[1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1] and [1 : 1 : 1]. The set of quadratic forms vanishing at thesefour points is

{axy+byz+cxz| a, b, c C, s.t. a+b+c= 0}= {axy+byz+ (b a)xz= ax(y z) +by(x z)|a, b C}.

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Figure 21:

which is equal to the pencil spanned by the two (degenerate) conics Z(x(y z)) andZ(y(x z)). This contains also a third degenerate conic Z(z(x y)).

From the above argument, we see that if two conics intersect in four points lying ingeneral position, then the pencil they determine contains three degenerate conics (themaximum allowed by Proposition3.19), corresponding to the

42

/2 = 3 ways to contain

four points in general position in a union of two lines:

4 Exterior AlgebrasIn this chapter we will always assume that fields F satisfy 2= 0 and consequently that1 = 1.

4.1 Multilinear algebra

Definition 16. Let Vbe a vector space over a field F. A multilinear form of degreek(k 1) ork-linear formis a function

M :Vk =V

...

V

F

that is linear in each entry. That is, for each i = 1, 2,...,k we have

M(u1, u2,....,ui+ui,....,uk) =M(u1,...,ui,...,uk) +M(u1,...,u

i,...,uk)

for all u1,...,uk, ui V and , F. The set ofk-linear forms is denoted Tk(V). By

conventionT0(V) =F.

Weve seen many example of multilinear forms already.

A multilinear form of degree 1 is simply a linear map : V F, so T1(V) =V.

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Figure 22: Degenerate Conics determining pencil, viewed in the z= 1 affine plane.

Figure 23: Degenerate Conics lying on four points in general position.

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Symmetric bilinear forms are multilinear forms of degree two. IfV= Fn is the set ofn 1 column matrices, then given v1,...,vn V we may

form the n

nmatrix [v1....vn]. The determinant

det :V ... V F, (v1,....,vn) det(v1....vn)

is a multilinear form of degree n.

Multilinearity implies that that the form M is completely determined by its valueson on k-tuples of basis elements. Ifv1,...,vn is a basis for V, then

M(n

j=1

1,jvj,...,n

j=1

k,jvj) =n

j1,...,jk=1

1,j1...k,jkM(vj1,...,vjk). (12)

There are two distinguished subsets ofTk(V). Thesymmetric k-linear formsSk(V) Tk(V) possess the property that transposing any two entries does not change the valueof the form. That is, if for any u1,...uk V, we have

M(u1,...,ui,...,uj,...,uk) =M(u1,...,uj ,...,ui,...uk).

In this chapter we will concentrate on what are called alternating (or skew-symmetricor exterior) forms

k(V) Tk(V).A multilinear form M is called alternating if transposing two entries changes M by a

minus sign. That is, if for any u1,...uk V, we haveM(u1,...,ui,...,uj,...,uk) = M(u1,...,uj,...,ui,...uk). (13)

Of the examples above, dual vectors are alternating because the condition is vacuousin degree one. The determinant is alternating, because transposing columns (or rows forthat matter) changes the determinant by minus one. Symmetric bilinear forms are ofcourse not alternating, because transposing the entries leaves it unchanged.

A consequence of (13) is that an alternating multilinear form will vanish if two entriesare equal. From (12) it follows that there are no nonzero multilinear forms of degree kgreater than n the dimension ofV. More generally, we can say

Proposition 4.1. LetVbe a vector space of dimensionn over a fieldF in which0 = 2.Fork a non-negative integer, thenk(V) is a vector space of dimension

nk

.

Proof. LetM1and M2bek-linear alternating forms. The vector space structure is definedby

(1M1+2M2)(u1,...,uk) =1M1(u1,...,uk) +2M2(u1,...,uk). (14)

That this makes the set ofk-linear forms into a vector space is a routine verification.

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Letv1,...,vnV be a basis. According to (12), a k-linear form is determined by itsvalues on k-tuples of basis vectors

Mi1,...,ik :=M(vi1,...,vik)

forvij v1,...,vn. The constantsMi1,...,ik are sometimes called structur constants. If twoindices are duplicated, thenMi1,...,ik = 0 and if two indices are transposed Mi1,...,ik changesonly by a minus sign. Thus Mis completely determined by the list of scalars

(Mi1,...,ik| i1 < i2< .... < ik) (15)which is a list of

nk

scalars. Conversely, it is straightforward to show that any list of

scalars (15) defines a k-linear alternating form.

The following corollary helps establish the significance of the determinant of a

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Projective Geometry Lecture Notes 3