Projective Geometry Lecture Notes

Thomas Baird

March 26, 2012

Contents

1 Introduction 2

2 Vector Spaces and Projective Spaces 32.1 Vector spaces and their duals . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Projective spaces and homogeneous coordinates . . . . . . . . . . . . . . . 5

2.2.1 Visualizing projective space . . . . . . . . . . . . . . . . . . . . . . 52.2.2 Homogeneous coordinates . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Linear subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3.1 Two points determine a line . . . . . . . . . . . . . . . . . . . . . . 72.3.2 Two planar lines intersect at a point . . . . . . . . . . . . . . . . . 8

2.4 Projective transformations and the Erlangen Program . . . . . . . . . . . . 82.4.1 Erlangen Program . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4.2 Projective versus linear . . . . . . . . . . . . . . . . . . . . . . . . . 102.4.3 Examples of projective transformations . . . . . . . . . . . . . . . . 112.4.4 Direct sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.4.5 General position . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5 Classical Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.1 Desargues’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.2 Pappus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Quadrics and Conics 203.1 Affine algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2 Projective algebraic sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Bilinear and quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3.1 Quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3.2 Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3.3 Digression on the Hessian . . . . . . . . . . . . . . . . . . . . . . . 27

3.4 Quadrics and Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.5 The rational parametrization of the circle . . . . . . . . . . . . . . . . . . . 303.6 Polars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.7 Linear subspaces of quadrics and ruled surfaces . . . . . . . . . . . . . . . 35

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4 Ideas for Projects 37

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 shifts.

The projective geometry most relevant to painting is called the real projective plane,and is denoted RP 2 or P (R3).

Definition 1. The real projective plane, RP 2 = P (R3) is the set of 1-dimensional sub-spaces of R3.

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 of R3 can beunderstood as lines of sight. If we now situate a (Euclidean) plane P that doesn’t containthe origin, then each point in P determines unique sight line. Objects in (subsets of) R3

can now be “projected” onto the plane P , enabling us to translate a 3-dimensional sceneonto a 2-dimensional scene. 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. Tomotivate this terminology, consider a family of projective points that rotate from pro-jective points that intersect P to one that is parallel. The projection onto P becomes afamily of points that diverges to infinity and then disappears. But as projective pointsthey 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 pairof projective lines in the projective plane intersect. This runs contrary to the parallelpostulate in 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.

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2 Vector Spaces and Projective Spaces

2.1 Vector spaces and their duals

Let F denote a field (for our purposes F is either the real numbers R or the complexnumbers C). A vector space V over F is set equipped with two operations:

• Addition: V × V → V , (v, w) 7→ v + w.

• Scalar multiplication: F × V , (λ, v) 7→ λv

satisfying the following list of axioms for all u, v, w ∈ V and λ, µ ∈ 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 all v ∈ V(additive identity)

• For every v ∈ V there exists −v ∈ V such that v +−v = 0 (additive inverses)

• (λ+ µ)v = λv + µv (distributivity)

• λ(u+ v) = λu+ λv (distributivity)

• λ(µv) = (λµ)v (associativity of scalar multiplication)

• 1v = v ( identity of scalar multiplication)

I will leave it as an exercise to prove that 0v = 0 and that −1v = −v.

Example 1. The set V = Rn of n-tuples of real numbers is a vector space over F = R inthe usual way. Similarly, Cn is a vector space over C. The vector spaces we consider inthis course will always be isomorphic to one of these examples.

Given two vector spaces V,W over F , a map φ : V → W is called linear if for allv1, v2 ∈ V and λ1, λ2 ∈ F , we have

φ(λ1v1 + λ2v2) = λ1φ(v1) + λ2φ(v2).

Example 2. A linear map φ 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 φ is defined using matrix multiplication. Similarly, a linear map fromCm to Cn can be uniquely represented by an n×m matrix of complex numbers.

We say that a linear map φ : V → W is an isomorphism if φ is a one-to-one and onto.In this case, the inverse map φ−1 is also linear, hence is also a isomorphism. Two vectorspaces are called isomorphic if there exists an isomorphism between them. A vector spaceV over F is called n-dimensional if it is isomorphic to F n. V is called finite dimensional

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if 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 which is closed under additionand scalar multiplication. This makes U into a vector space in it’s own right, and theinclusion map U ↪→ V in linear. Given any linear map φ : V → W , we can define twovector subspaces: the image

φ(V ) := {φ(v)|v ∈ V } ⊆ W

and the kernelker(φ) := {v ∈ V |φ(v) = 0} ⊆ V

The dimension of φ(V ) is called the rank of φ, and the dimension of ker(φ) is called thenullity of φ. The rank-nullity theorem states that the rank plus the nullity equals thedimension of V . I.e., given a linear map φ : V → W

dim(V ) = dim(ker(φ)) + dim(φ(V )).

Given a set of vectors S := {v1, ..., vk} ⊂ V a linear combination is a expression of theform

λ1v1 + ...+ λkvk

where λ1, ..., λk ∈ F are scalars. The set of vectors that can be written as a linearcombination of vectors in S is called the span of S. The set S is called linearly independentif the only linear combination satisfying

λ1v1 + ...+ λkvk = 0

is when λ1 = λ2 = ... = λk = 0. If some non-trivial linear combination equals zero, thenwe say S is linearly dependent.

A basis of V is linearly independent set S ⊂ V that spans V . Every vector space hasa basis, and the dimension of a vector space is equal to the size of any basis of that space.If {v1, ..., vn} is a basis of V , then this defines an isomorphism Φ : F n → V by

Φ((x1, ..., xn)) = x1v1 + ...+ xnvn.

Conversely, given an isomorphism Φ : F n → V , we obtain a basis {v1, ..., vn} ⊂ V bysetting vi = Φ(ei) where {e1, ..., en} ∈ F n is the standard basis. So choosing a basis forV amounts to the same thing as defining an isomorphism from F n to V .

Proposition 2.1. Suppose that W ⊆ V is a vector subspace. Then dim(W ) ≤ dim(V ).Furthermore, we have equality dim(W ) = dim(V ) if and only if W = V .

Proof. Exercise.

Given any vector space V , the dual vector space V ∗ is the set of linear maps from Vto F

V ∗ := {φ : V → F}.

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The dual V ∗ is naturally a vector space under the operations

(φ1 + φ2)(v) = φ1(v) + φ2(v)

and(λφ)(v) = λ(φ(v)).

Example 3. Suppose V ∼= F n is the set of 1 × n column vectors. Then V ∗ is identifiedwith the set of n× 1 row vectors under matrix multiplication.

Suppose that v1, ..., vn ∈ V is a basis. Then the dual basis v∗1, ..., v∗n ∈ V ∗ is the unique

set of linear maps satisfyingv∗i (vj) = δji ,

where δji is the Dirac delta defined by δii = 1 and δji = 0 if i 6= j.

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 of V .

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. If V is one of the standard vector spaces Rn+1 or Cn+1

we also use notation RP n := P (Rn+1) and CP n := P (Cn+1).

2.2.1 Visualizing projective space

To develop some intuition about projective spaces, we consider some low dimensionalexamples.

If V is 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 coordinatesx, y and let L ⊆ R2 be the line defined x = 1. With the exception of the y-axis, each1-dimensional subspace of R2 intersects L in exactly one point. Thus RP 1 = P (R2) isisomorphic as a set with the union of L with a single point. Traditionally, we identify Lwith R and call the extra point {∞}, the point at infinity. Thus we have a bijection ofsets

RP 1 ∼= R ∪ {∞}

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

By similar reasoning we can see that the complex projective line CP 1 is in bijectionwith C ∪ {∞}

CP 1 ∼= C ∪ {∞}.

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Topologically, CP 1 is a isomorphic to the 2-dimensional sphere S2 and is sometimes calledthe Riemann Sphere.

We turn next to the real projective plane RP 2 = P (R3). Let x, y, z be the standardcoordinates on R3. Then every 1-dimensional subspace of R3 either intersects the planez = 1 or lies in the plane z = 0. The first class of 1-dimensional subspaces is in bijectionwith R2, and the second class is in bijection with RP 1. Thus we get a bijection

RP 2 ∼= R2 ∪ RP 1 ∼= R2 ∪ R ∪ {∞}.

It is not hard to see that this reasoning works in every dimension n to determine thefollowing bijection

RP n ∼= Rn ∪ RP n−1 ∼= Rn ∪ Rn−1 ∪ ... ∪ R0 (1)

and similarlyCP n ∼= Cn ∪ CP n−1 ∼= Cn ∪ Cn−1 ∪ ... ∪ C0. (2)

Returning now to the projective plane, observe that every 1-dimensional subspace ofR3 intersects the unit sphere S2 in a pair of antipodal points. Thus we can constructRP 2 by taking (say) the northern hemisphere and identifying antipodal points along theequator.

2.2.2 Homogeneous coordinates

To go further it is convenient to introduce some notation. Given a non-zero vector v ∈V \ {0}, define [v] ∈ P (V ) to be the 1-dimensional spanned by v. If λ ∈ F \ {0} is anon-zero scalar, then

[λv] = [v].

Now suppose we choose a basis {v0, ..., vn} for V (recall this is basically the same asdefining an isomorphism V ∼= F n). Then every vector in v ∈ V can 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 as homogeneous coordinates. Once again, forλ 6= 0 we have

[λx0, ..., λxn] = [x0, ..., xn]

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Consider now the subset U0 ⊂ P (V ) consisting of points [x0, ..., xn] with x0 6= 0(Observe that this subset is well-defined independent of the representative (x0, ..., xn)because for λ 6= 0 we have x0 6= 0 if and only if λx0 6= 0). The for points in U0

[x0, ..., xn] = [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 theform (1, y1, ..., yn). It follows that U0 = F n. This result has already been demonstratedgeometrically in equations (1) and (2) because U0 is the set of 1-dimensional subspacesin V that intersect the “hyperplane” x0 = 1. The complement of U0 is the set of pointswith x0 = 0, which form a copy of P (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 subspace W ⊆ 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. If W ⊆ V is 2-dimensional, we call P (W ) ⊆ P (V ) a projectiveline in P (V ). Similarly, if W 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 thesubset U0 = {x0 = 1} ∼= F n then they become points, lines and planes in the traditionalEuclidean sense.

2.3.1 Two points determine a line

Theorem 2.2. Given two distinct points [v1], [v2] ∈ P (V ), there is a unique projectiveline containing them both.

Proof. Since [v1] and [v2] represent distinct points in P (V ), then the representative vectorsv1, v2 ∈ V must 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]. Then v1, v2 ∈ W , so W ⊆ W ′. Since both W and W ′ are 2-dimensional it followsthat W = W ′ and P (W ) = P (W ′).

Example 4. Consider the model of RP 2 as pairs of antipodal points on the sphere S2. Aprojective line in this model is represented by a great circle. Proposition 2.2 amounts tothe result that any pair of distinct and non-antipodal points in S2 lie on a unique greatcircle (by the way, this great circle also describes the optimal flight route for airplanesflying between two points on the planet).

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2.3.2 Two planar lines intersect at a point

Theorem 2.3. 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 ⊂ V is 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 = W which contradicts the condition that U,W are distinct.

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 because V only has dimension 3. Therefore, there exist scalarsλ1, ..., λ4 not all zero such that

λ1u1 + λ2u2 + λ3w1 + λ4w2 = 0

from which it follows that λ1u1+λ2u2 = −λ3w1−λ4w2 is a non-zero element in U∩W .

Example 5. Using the model of RP 2 as pairs of antipodal points in S2, then Proposition2.3 is the geometric fact that pairs of great circles always intersect in a pair of antipodalpoints.

2.4 Projective transformations and the Erlangen Program

So far we have explored how a vectors space V determines a projective space P (V ) andhow a vector subspace W ⊆ V determines a linear subspace P (W ). What about linearmaps?

It is not true in general that a linear map T : V → V ′ induces a map between theprojective spaces P (V ) and P (V ′). This is because if T has a non-zero kernel, then it sendssome 1-dimensional subspaces of V to zero in V ′. However, if ker(T ) = 0, or equivalentlyif T is injective, then this works fine.

Definition 5. Let T : V → V ′ be an injective linear bap. Then the map

τ : P (V )→ P (V ′)

defined byτ([v]) = [T (v)]

is called the projective morphism induced by T .If T is an isomorphism, then τ is bijective and we call τ a projective transformation.

Projective transformations from a projective space to itself τ : P (V ) → P (V ) areparticularly interesting. These are considered to be group of symmetries of projectivespace.

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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 withcertain subsets of R2 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 of R2 of points lying afixed distance r from a point p ∈ R2. The length r is called the radius and circumferenceof the circle is 2πr.

The group of symmetries of 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 “isomorphic” in this context.

In 1872, Felix Klein initiated a highly influential re-examination of geometry called theErlangen Program. At the time, several different types of “non-Euclidean” geometries hadbeen introduced, and it was unclear how they related to each other. Klein’s 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 §2.2.2 that me have a subset U0 ⊂ RP n

consisting of those points with homogeneous coordinates [1, y1, ..., yn]. There is an obviousbijection

U0∼= Rn, [1, y1, ..., yn]↔ (y1, ..., yn)

Proposition 2.4. Any Euclidean transformation of U0∼= Rn extends uniquely to a pro-

jective transformation of RP n.

Proof. We focus on the case n = 2 for notational simplicity. It is enough to show thattranslations and rotations of U0 extend uniquely.

Consider first translation in R2 by a vector (a1, a2). That is the map (y1, y2) 7→(y1 + a1, y2 + a2). The corresponding projective transformation is associated to the lineartransformation 1

y1

y2

7→ 1 0 0

a1 1 0a2 0 1

1y1

y2

=

1y1 + a1

y2 + a2

A rotation by angle θ of U0 = R2 extends to the projective transformation induced by thelinear transformation

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1y1

y2

7→ 1 0 0

0 cos(θ) −sin(θ)0 sin(θ) cos(θ)

1y1

y2

=

1cos(θ)y1 + sin(θ)y2

cos(θ)y2 − sin(θ)y1

Uniqueness will be clear after we learn about general position.

From the point of view of the Erlangen Program, Proposition ?? 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 allprojective 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 2.5. Two injective linear maps T : V → W and T ′ : V → W determine thesame projective morphism if and only if T = λT ′ for some non-zero scalar λ.

Proof. Suppose that T = λT ′. Then for [v] ∈ P (V ),

[T (v)] = [λT ′(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 scalars λ0, ..., λn such that

T (vi) = λiT′(vi).

However it is also true that [T (v0 + ... + vn)] = [T ′(v0 + ... + vn)] so for some non-zeroscalar λ we must have

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

Combining these equations and using linearity, we get

n∑i=0

λT ′(vi) = λT ′(n∑i=0

vi) = T (n∑i=0

vi) =n∑i=0

T (vi) =n∑i=0

λiT′(vi)

Subtracting gives∑n

i=0(λ − λi)T ′(vi) = 0. Since T ′ is an injective, {T ′(vi)} is a linearlyindependent set, so λ = λi for all i and T = λT ′.

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2.4.3 Examples of projective transformations

To develop some intuition about projective transformations, let’s consider transformationsof the projective line FP 1 = P (F 2) for F = R,C. A linear transformation of F 2 isuniquely 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

τ([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 transformation and plays an important role in complex analysis.

Proposition 2.6. Expressed in a Euclidean coordinate y ↔ [1 : y], every projectivetransformation of the line FP 1 is equal to a combination of the following transformations:

• A translation: y 7→ y + a

• A scaling: y 7→ λy, λ 6= 0

• An inversion: y 7→ 1y

Proof. Recall from linear algebra that any invertible matrix can be obtained from theidentity matrix using elementary 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 λ

],

[0 11 0

],

[1 01 1

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

0 11 0

] [1y

]=

[y1

]= y

[1

1/y

]so corresponds to the inversion y 7→ 1

y

It is also informative to try and picture these transformations using our model whereRP 1 is a circle and CP 1 is a sphere (leave this to class discussion).

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Remark 1. More generally, a projective transformation of RP n takes the form

y 7→ Ay + c

λ+ b · y

where A is an n×n matrix, b, c are vectors in Rn and λ is scalar and · is the dot product.Proving this is an exercise.

A more geometrically defined transformation is obtained as follows. Let P (V ) be aprojective plane, let P (U), P (U ′) be two projective lines in P (V ), and let [w] ∈ P (V ) bea projective points not lying on P (U) or P (U ′). We define a map

τ : P (U)→ P (U ′)

as follows: Given [u] ∈ P (U), there is a unique line L ⊂ P (V ) containing both [w] and[u]. Define τ([u]) to be the unique intersection point L ∩ P (U ′). Observe that τ is welldefined by Theorem 2.2 and 2.3 .

Proposition 2.7. The map τ : P (U)→ P (U ′) is a projective transformation.

Before proving this proposition, it is convenient to introduce a concept from linearalgebra: the direct sum.

2.4.4 Direct sums

Let V and W be vector spaces over the same field F . The direct sum V ⊕W is 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′) andλ(v, w) = (λv, λw). Exercise: show that

dim(V ⊕W ) = dim(V ) + dim(W ).

There are natural linear maps

i : V → V ⊕W, i(v) = (v, 0)

andp : V ⊕W → V, p(v, w) = v

called inclusion and projection maps respectively. Similar maps exist for W as well.

Proposition 2.8. Let V be a vector space and let U,W be vector subspaces such thatspan{U,W} = V . Then the natural map

Φ : U ⊕W → V, (u,w) 7→ u+ w

is surjective with kernelker(Φ) ∼= U ∩W.

In particular, if U ∩W = {0}, then Φ defines an isomorphism U ⊕W ∼= V and wesay that V is the internal direct sum of U and W .

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Proof. Because span{U,W} = V every vector in V is a sum of vectors in U and W , so itfollows that Φ is surjective. To find the kernel:

ker(Φ) = {(u,w) ∈ U ×W | u+ w = 0} = {(u,−u) ∈ U ×W} ∼= U ∩W.

Finally, if ker(Φ) = U ∩ W = {0} then Φ is both injective and surjective, so it is anisomorphism.

Proof of Proposition 2.7. Let W = span{w} be the one-dimensional subspace of V cor-responding to [w]. Then since [w] 6∈ P (U ′), it follows that W ∩ U ′ = {0}. Countingdimensions, we deduce that

V ∼= W ⊕ U ′

is an internal direct sum. Denote by p : V → U ′ the resulting projection map.Now let [u] ∈ P (U) and take [u′] = τ([u]) ∈ P (U ′). By definition of τ , [u′] lies in the

line determined by [w] and [u], so u′ is a linear combination

u′ = λw + µu

for some scalars λ, µ ∈ F . Moreover, [w] ∈ P (U ′) so [w] 6= [u′] and we deduce that µ 6= 0,so

u =λ

µw − 1

µu′

so p(u) = − 1µu′ and τ([u]) = [p(u)]. It follows that τ is the projective transformation

induced 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 of n + 2 points inP (V ) are said to lie in general position if the representative vectors of any proper subsetare linearly independent in V (A subset S ′ ⊂ S is called a proper subset if S ′ is non-emptyand S ′ does not equal S).

Example 6. 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 7. Four points in a projective plane lie in general position if and only if nothree of them is collinear (prove this).

Theorem 2.9. If P (V ) and P (W ) are projective spaces of the same dimension n andboth p1, ..., pn+2 ∈ P (V ) and q1, ..., qn+2 ∈ P (W ) lie in general position, then there is aunique projective transformation

τ : P (V )→ P (W )

sending τ(pi) = qi for all i = 1, 2, ..., n+ 2.

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Proof. Choose representative vectors v1, ..., vn+2 ∈ V such that [vi] = pi for all i. Becausethe points pi lie in general position, it follows that the first n + 1 vectors v1, ..., vn+1 arelinearly independent and thus form a basis of V . It follows that vn+2 can be written as alinear combination

n+1∑i=1

λivi = vn+2

Moreover, all of the scalars λi are non-zero, for otherwise the non-trivial linear expression

n+1∑i=1

λivi − vn+2 = 0

would imply that some proper subset of {v1, ..., vn+2} is linearly dependent, in contradic-tion with the points lying in general position. By replacing our representative vectors viwith the alterative representing vectors λivi, we may assume without loss of generalitythat

n+1∑i=1

vi = vn+2.

Similarly, we may choose representing vectors wi ∈ W for qi ∈ P (W ) such that

n+1∑i=1

wi = wn+2.

Since v1, ..., vn+1 ∈ V are linearly independent, they form a basis for V and we can definea linear map T : V → W by sending T (vi) = wi for i = 1, ..., n+ 1 and extending linearly.By linearity we have

T (vn+2) = T (n+1∑i=1

vi) =n+1∑i=1

T (vi) =n+1∑i=1

wi = wn+2

So it induces a projective transformation satisfying τ(pi) = qi for all i.To prove uniqueness, assume that there is another linear map T ′ : V → W satisfying

the conditions of the theorem. Then for each i there exists a non-zero scalar µi such that

T ′(vi) = µiwi.

But then by linearity

µn+2wn+2 = T ′(vn+2) = T ′(n+1∑i=1

vi) =n+1∑i=1

µiwi

son+1∑i=1

wi = wn+2 =n+1∑i=1

µiµn+2

wi

and we deduce that µi

µn+2= 1 for all i and thus that T ′ = µT for some non-zero constant

scalar µ.

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Example 8. There is a unique projective transformation τ : FP 1 → FP 1 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 Theorem 2.9, 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 for i = 1, ..., n+ 1 and satisfying

n+1∑i=1

vi = vn+2.

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

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

Theorem 2.10 (Desargues’ Theorem). Let A,B,C,A′, B′, C ′ be distinct points in a pro-jective space P (V ) such that the lines AA′, BB′, CC ′ are distinct and concurrent (mean-ing the three lines intersect at a common point). Then the three points of intersectionX := AB ∩ A′B′, Y := BC ∩ B′C ′ and Z := AC ∩ A′C ′ are collinear (meaning they lieon a common line).

Proof. Let P be the common point of intersection of the three lines AA′, BB′ and CC ′.First suppose that P coincides with one of the points A,B,C,A′, B′, C ′. Without loss

of generality, let P = A. Then we have equality of the lines AB = BB′ and AC = CC ′,so it follows that X = AB ∩ A′B′ = B′, Z = AC ∩ A′C ′ = C ′, so B′C ′ = XZ must becollinear with Y = BC ∩B′C ′ 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 Remark 2,given a representative p ∈ V of P , we can find representative vectors a, a′ of A,A′ suchthat

p = a+ a′.

Similarly, we have representative vectors b, b′ for B,B′ and c, c′ for C,C ′ such that

p = b+ b′, p = c+ c′

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It follows that a+ a′ = b+ b′ so

x := a− b = b′ − a′

must represent the intersection point X = AB ∩ A′B′, because x is a linear combinationof both a, b and of a′, b′. Similarly we get

y := b− c = c′ − b′ and z := c− a = a′ − c′

representing the intersection points Y and Z respectively.To see that X, Y and Z are collinear, observe that

x+ y + z = (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.

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.

2.5.2 Pappus’ Theorem

Theorem 2.11. Let A,B,C and A′, B′, C ′ be two pairs of collinear triples of distinctpoints in a projective plane. Then the three points X = BC ′ ∩B′C, Y := CA′ ∩C ′A andZ := AB′ ∩ A′B are collinear.

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. ByTheorem 2.9, 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) in F 3, so the point C ∈ AB must have the form

C = [1 : c : 0]

(because it is not equal to A) for some c 6= 0. Similarly, the line B′C ′ corresponds to thevector space spanned by (0, 0, 1) and (1, 1, 1) so

A′ = [1, 1, a]

for some a 6= 1.Next compute the intersection points of lines

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

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so we deduce that BC ′ ∩B′C is represented by (1, 1, 1)− (1, c, 0) = (0, 1− c, 1).

AC ′ = span{(1, 0, 0), (0, 0, 1)} = {x1 = 0}A′C = span{(1, 1, a), (1, c, 0)}

so we deduce that A′C ∩ AC ′ is represented by (1, c, 0)− c(1, 1, a) = (1− c, 0,−ca).

AB′ = span{(1, 0, 0), (1, 1, 1)} = {x1 = x2}A′B = span{(1, 1, a), (0, 1, 0)}

so we deduce that AB′ ∩ A′B is represented by (a− 1)(0, 1, 0) + (1, 1, a) = (1, a, a).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 a

0 (c− 1)a −a0 1− c 1

⇒

1 a a0 0 00 1− c 1

which has rank two, so span{(0, 1− c, 1), (1− c, 0,−ca), (1, a, a)} has dimension two andthe vectors are linearly dependent.

2.6 Duality

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

V ∗ := {φ : V → F} = Hom(V, F ).

• The dual V ∗ is naturally a vector space under the operations

(φ1 + φ2)(v) = φ1(v) + φ2(v)

(λφ)(v) = λ(φ(v)).

• Suppose that v1, ..., vn ∈ V is a basis. Then the dual basis φ1, ..., φn ∈ V ∗ is theunique set of linear maps satisfying φi(vj) = 1, if i = j and φi(vj) = 0 otherwise.

• If T : 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.

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Definition 7. Let U ⊆ V be a vector subspace. The annihilator U0 ⊆ V ∗ is defined by

U0 := {φ ∈ V ∗ | φ(u) = 0, for all u ∈ U}.

To check that U0 is a subspace of V ∗, observe that if φ1, φ2 ∈ U0 and u ∈ U then

(λ1φ1 + λ2φ2)(u) = λ1φ1(u) + λ2φ2(u) = 0 + 0 = 0

so (λ1φ1 + λ2φ2) ∈ U0, so U0 is closed under linear combinations.

Proposition 2.12. If U1 ⊂ U2 ⊂ V then U01 ⊃ U0

2 .

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

We also have the following result.

Proposition 2.13. dim(U) + dim(U0) = dim(V )

Proof. Choose a basis u1, , um ∈ U and extend to a basis u1, , um, vm+1, , vn of V . Letφ1, , φn be the dual basis. For a general element

φ =n∑i=1

λiφi ∈ V ∗

we have φ(ui) = λi. Thus φ ∈ U0 if and only if λi = 0 for i = 1, ,m so U0 ⊆ V ∗ is thesubspace with basis φm+1, , φn which has dimension n−m.

Next, we explain the use of the word duality.

Proposition 2.14. There is a natural isomorphism VS∼= (V ∗)∗, defined by v 7→ S(v)

where S(v)(φ) = φ(v).

Proof. We already know that dim(V ) = dim(V ∗) = dim((V ∗)∗), so (by ) it only remainsto prove that S : V → (V ∗)∗ is injective. But if v ∈ V is non-zero, we may extend v1 := vto a basis and then in the dual basis φ1(v1) = S(v1)(φ1) = 1 6= 0 so S(v) 6= 0. Thusker(S) = {0} and S is injective.

Remark 3. Proposition 2.14 depends on V being finite dimensional. If V is infinitedimensional, then S is injective but not surjective.

In practice, we tend to replace the phrase “naturally isomorphic” with “equals” andwrite V = (V ∗)∗. Using this abuse of terminology, one may easily check that (U0)0 = U ,so the operation of taking a vector space to it’s dual is reversible.

Proposition 2.15. Let P (V ) be a projective space of dimension n. Then there is a one-to-one correspondence between linear subspaces P (U) ⊂ P (V ) of dimension m and linearsubspaces of P (V ∗) of dimension n−m− 1 by the rule

P (U)↔ P (U0).

We use notation P (U0) = P (U)0.

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

U ↔ U0

between subspaces of dimension m+1 in V and subspaces of dimension n−m in V ∗. Theresult follows immediately.

It follows from Proposition 2.15 that geometric statements about linear subspaces ofP (V ) translate into geometric statements about linear subspaces of P (V ∗).

Example 9. Let P (V ) be a projective plane, so P (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 in P (V ∗).

Proposition 2.16. Let A,B,C be collinear points in a projective plane lying on the lineL. Then A0, B0, C0 are concurrent lines intersecting at the point L0.

Proof. If A,B,C ⊂ L then L0 ⊂ A0, L0 ⊂ B0 and L0 ⊂ C0, so L0 = A0 ∩B0 ∩ C0.

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 2.17 (Dual version of Desargues’ Theorem). Let α, β, γ, α′, β′, γ′ be distinctlines in a projective plane such that the points α ∩ α′, β ∩ β′ and γ ∩ γ′ are distinct andcollinear. Then the lines joining α∩ β to α′ ∩ β′, α∩ γ to α′ ∩ γ′ and β ∩ γ′ to β′ ∩ γ′ areconcurrent.

Theorem 2.18 (Dual version of Pappus’ Theorem). Let α, β, γ and α′, β′, γ′ be two setsof concurrent triples of distinct lines in a projective plane. Then the three lines determinedby pairs of points β ∩ γ′ and β′ ∩ γ, γ ∩ α′ and γ′ ∩ α, α ∩ β′ and α′ ∩ β 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 Hitchin’s notes for details). Thus Pappus’Theorem may be considered “self-dual”.

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

Proposition 2.19. The set of lines in R2 is in one-to-one correspondence with the Mobiusstrip.

Proof. The set of lines in R2 is the set of lines in RP 2 minus the line at infinity. Byduality, this is identified with RP 2 minus a single point, which is the same as a mobiusstrip. Alternatively,

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3 Quadrics and Conics

3.1 Affine algebraic sets

Given a set {x1, ..., xn} of variables, a polynomial f(x1, ..., xn) is an expression of the form

f = f(x1, ..., xn) :=∑I

aIxI

where

• the sum is over (n+ 1)-tuples of non-negative integers I = (i1, ..., in),

• xI = xi11 ...xinn are called monomials, and

• aI ∈ F are scalars called coefficients, of which all but finitely many are zero.

The degree of 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 10. • 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 polynomial f(x1, ..., xn) in n variables may be interpreted as a function

f : F n → F

by simply plugging in the entries n-tuples in F n in place of the variables x1, ..., xn.

Definition 8. Given a n variable polynomial f , the set

Zaff (f) := {(x1, ..., xn) ∈ F n | f(x1, ..., xn) = 0}

is called the (affine) zero set of f . More generally, if f1, ..., fk are polynomials then

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

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

Example 11. An affine line in F 2 is defined by a single equation ax+by+c = 0, where aand b are not both zero. Thus a line in F 2 is the zero set of a single degree one polynomial.More generally, an affine linear subset of F n (points, lines, planes, etc.) is the zero set ofa collection of degree one polynomials.

Example 12. 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)).

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Example 13. Any finite set of points {r1, ..., rd} ⊂ F is an algebraic set Z(f) wheref(x) =(x− r1)(x− r2)...(x− rd).

Definition 9. An affine quadric X ⊂ F n is a subset of the form X = Zaff (f) where fis a single polynomial of degree 2. In the special case that X ⊂ F 2, we call X a (affine)conic.

Example 14. Some examples of affine in R2 conics include: circles, ellipses, hyperbolasand parabolas. The unit sphere Zaff (x

21 + ...+ x2

n − 1) is a quadric in Rn.

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 homogeneous if all non-zero monomials have thesame degree.

Example 15. • x− y is a homogeneous polynomial of degree one.

• x2 + y + 7 is a non-homogeneous polynomial.

• yz2 + 2x3 − xyz is a homogeneous polynomial of degree three.

Projective algebraic sets in FP n are defined using homogeneous polynomials in n+ 1variables, 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 λ ∈ F , we have

f(λx0, ..., λxn) = λdf(x0, ..., xn).

It follows from Proposition 3.1 that for λ a non-zero scalar, f(x0, ..., xn) = 0 if andonly if f(λx0, ..., λxn) = 0. This makes the following definition well-defined.

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

Z(f1, ..., fk) := {[x0 : ... : xn] ∈ FP n | fi(x0, ..., xn) = 0 for all i=1,...k}

is a called a projective algebraic set.

The main example we have encountered so far are linear subsets of 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 FP n, we have a copy of the affine plane U0 := {[1 : y1, ... : yn] ∈ FP n} ∼= F n.Under this correspondence, we have an equality

21

Z(f1, ..., fk) ∩ U0 = Zaff (g1, ..., gk)

where gi(x1, ..., xn) := f(1, x1, ..., xn).Conversely, given an affine algebraic set we construct a projective algebraic set as

follows. Let f(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 of f .

Example 16. Here are some examples of homogenizing polynomials

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

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

• g(x1, x2) = x1 + x32 + 7 ⇒ h(x0, x1, x2) = x2

0x1 + x32 + 7x3

0

Now suppose that g1, ..., gk are non-homogeneous polynomials with homogenizationsh1, ..., 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 called quadrics.

3.3 Bilinear and quadratic forms

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

B : V × V → F

such that

• B(u, v) = B(v, u), (Symmetric)

• B(λ1u1 + λ2u2, v) = λ1B(u1, v) + λ2B(u2, v), (Linear)

Observe that by combining symmetry and linearity, a bilinear form also satisfies

B(u, λ1v1 + λ2v2) = λ1B(u, v1) + λ2B(u, v2).

We say that B is linear in both entries, or bilinear. We say that B is nondegenerate iffor each non-zero v ∈ V , there is a w ∈ V such that B(v, w) 6= 0.

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Example 17. 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 if v 6= 0.

Example 18. The bilinear form B : R4 × R4 → R defined by

B((x0, x1, x2, x3), (y0, y1, y2, y3)) = −x0y0 + x1y1 + x2y2 + x3y3

is called the Minkowski metric that 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 called null-vectors. The null-vectors form a subset of R4 called the lightcone dividing those vectors satisfying B(v, v) < 0 (called time-like vectors) from thosesatisfying B(v, v) > 0 (called space-like vectors).

Now let v1, ..., vn ∈ V be a basis. For any pair of vectors u = x1v1 + ...xnvn andw = y1v1 + ...+ ynvn we have

Lemma 3.2. Let v1, ..., vn ∈ V be a basis. For any pair of vectors u = 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 the n × n-matrix β withentries βi,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

n∑j=1

xiyjB(vi, vj) =n∑

i,j=1

xiyjβi,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× n matrices β, according to the rule

B(v, w) = vTβw

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.

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

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

3.3.1 Quadratic forms

Given a symmetric bilinear form B : V × V → F , we can associate a function called aquadratic form Q : V → F by the rule Q(v) = B(v, v).

Proposition 3.4. Let V be a vector space over F = R or C, let B be a symmetric bilinearform on V and let Q : V → F be the associated quadratic form. Then for any u, v ∈ Vwe have

B(u, v) = (Q(u+ v)−Q(u)−Q(v))/2.

In particular, B is completely determined by its quadratic form Q.

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. Proposition ... does not hold when F is the field of Boolean numbers becausewe are not allowed to divide Boolean numbers by two!

We can understand these relationship better using a basis. Let v1, ..., vn for V be abasis so B can be expressed in terms of the n× n matrix β. According to ...

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

βi,jxixj

is a homogeneous quadratic polynomial.Some examples: [

a]↔ ax2

[a bb c

]↔ ax2 + 2bxy + cy2

a b db c ed e f

↔ ax2 + 2bxy + cy2 + 2dxz + 2exz + fz2

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3.3.2 Change of basis

We’ve 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 kind find scalars{Pi,j}ni,j=1 such that

wj =∑i

Pi,jvi (3)

The matrix P with entries Pi,j is called the change of basis matrix. Observe that P isinvertible, with P−1 providing the change of matrix from w1, ..., wn back to v1, ..., vn. Infact, if P ′ is an invertible matrix then it is the change of basis matrix from v1, ..., vn tothe new basis w1, ..., wn satisfying (3).

Proposition 3.5. Let B be a bilinear form on V , and let β, β′ be symmetric matricesrepresenting B with respect to bases v1, ..., vn and w1, ..., wn respectively. Then

β′ = P TβP

where P is the change of basis matrix.

Proof. There are two approaches. Simply calculate:

β′i,j = B(wi, wj) = B(∑k

Pk,ivk,∑l

Pl,jvl)

=∑k,l

Pk,iPl,jβk,l =∑k,l

P Ti,kβk,lPl,j

= (P TβP )i,j

Alternatively, we can argue more conceptually as follows. Let ei denote the ith stan-dard basis column vector. Then in terms of the basis v1, ..., vn, wi corresponds to thecolumn vector Pei. Then

β′i,j = B(wi, wj) = (Pei)Tβ(Pej) = eTi P

TβPej = (P TβP )i,j.

Our next task is to classify bilinear forms up to change of basis. First we need alemma.

Lemma 3.6. Let V be a vector space of dimension n equipped with a symmetric bilinearform B. For w ∈ V define

w⊥ := {v ∈ V |B(v, w) = 0}.

Then w⊥ ⊂ V is a vector space of dimension ≥ n− 1.

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

φ(λ1v1 + λ2v2) = B(λ1v1 + λ2v2, w) = λ1B(v1, w) + λ2B(v2, w) = λ1φ(v1) + λ2φ(v2)

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

dim(w⊥) = dim(V )− dim(im(φ)) =

{n im(φ) = {0}n− 1 im(φ) = F.

Theorem 3.7. Let B be a symmetric, bilinear form over a vector space V . Then thereexists a basis v1, ..., vn such that

B(vi, vj) = 0 (4)

for i 6= j and such that for all i = 1, ..., n

B(vi, vi) ∈

{0, 1 F = C0,±1 F = R.

(5)

Proof. As a first step construct a basis satisfying (4) using induction on the dimension ofV . In the base case dim(V ) = 1, (4) holds vacuously.

Now suppose by induction that all n − 1 dimensional symmetric bilinear forms canbe diagonalized. Let V be n dimensional. If B is trivial, then β is the zero matrix withrespect to any basis (hence diagonal).

If B is non-trivial then by Proposition 3.4, there exists a vector wn ∈ V such thatB(wn, wn) 6= 0. Clearly wn 6∈ v⊥n , so w⊥n is n − 1 dimensional. By induction we maychoose a basis w1, ..., wn−1 ∈ w⊥n satisfying (4) for all i, j < n. Furthermore, for all i < n,B(wi, wn) = 0 because wi ∈ w⊥n . Thus w1, ..., wn satisfy (4).

Finally, we want to modify w1, ..., wn to satisfy (5). Suppose that B(wi, wi) = c ∈ F .If c = 0, then let vi = wi. If c 6= 0 and F = C, let vi = 1√

cwi so that

B(vi, vi) = B(1√cwi,

1√cwi) =

( 1√c

)2

B(wi, wi) =c

c= 1.

If c 6= 0 and F = R then let vi := 1√|c|wi so

B(vi, vi) =c

|c|= ±1.

Corollary 3.8. Let V and B be as above and let Q(v) = B(v, v). Then

• If F = C then for some m ≤ n there exists a basis such that if v =∑n

i=1 zivi

Q(v) =m∑i=1

z2i

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• If F = R then for some p, q with p+ q ≤ n there exists a basis such that

Q(v) =

p∑i=1

z2i −

p+q∑i=p+1

z2i

Proof. Choose the basis v1, ..., vn so that B(vi, vj) satisfy the equations (4) and (5). Then

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

i,j=1

zizjB(vi, vj) =n∑i=1

z2iB(vi, vi),

where the B(vi, vi) are 0, 1 or −1 as appropriate.

Exercise: Show that the numbers m, p, q occurring above are independent of thechoice of basis.

It follows that B and Q are non-degenerate if and only if m = n or m = p+ q. In thecase of a real vector space, we say that B is positive definite if p = n, negative definite ifq = n and indefinite otherwise.

3.3.3 Digression on the Hessian

This discussion will not be on the test or problem sets.An important place where symmetric bilinear forms occur in mathematics is the study

of critical points of differentiable functions. Let f(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 all i = 1, ..., n

If the the second order partials are defined and continuous then the n× n-matrix H withentries

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

∂2f

∂xj∂xi|(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, if H is non-degenerate thenf has a local maximum at (a1..., an) if H is negative definite, has a local minimum if His positive definite, and has neither a max or min if H is neither positive nor negativedefinite.

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 function f , |f | has no local maximums where f is differentiable. This is a versionof the maximum principle.

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3.4 Quadrics and Conics

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

Definition 13. A quadric X ⊂ P (V ) is a subset defined by

X = {[v] ∈ V |B(v, v) = 0}.

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

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

Consider now the implications of Corollary 3.8 for quadrics in a projective line. Overthe reals we have such a quadric has the form Z(f) = Z(−f) where (up to a change ofcoordinates) f equals

• x2

• x2 + y2

• x2 − y2

In the first case, Z(x2) consists of the single point [0 : 1] which is counted with“multiplicity” two. The case Z(x2 + y2) is empty set because there are no solutions withboth x and y non zero. Finally Z(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 Corollary 3.8 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 f is one of (up to coordinate change)

• x2

• x2 + y2

The first case Z(x2) is a single point of multiplicity 2, while Z(x2+y2) = Z((x−iy)(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 6= 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 and P (V ) is a projective plane (in this case a quadricis called a conic). We may choose coordinates so that X = Z(f) where f is one of:

• x,

• x2 + y2,

• x2 + y2 + z2,

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• x2 − y2,

• x2 + y2 − z2 ,

up to multiplication by −1 which doesn’t change Z(f) = Z(−f).The first example

x2 = xx = 0

has the same solution set as x = 0, which is projective line {[0 : a : b]} ∼= RP 1, howeverthis line is counted twice in a sense we won’t 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 (a degenerate conic).

The last example Z(x2 +y2−z2) is (up to coordinate change) the only non-empty andnon-degenerate conic, up to projective transformation. The apparent difference betweenthe affine conics: ellipses, parabolas, hyperbolas, can be understood as how the conicrelates to the line at infinity. A conic looks like an ellipse if it is disjoint form the line atinfinity, looks like parabola if it is tangent to the line at infinity, and looks like a hyperbolaif 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 equation z = 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.Now instead, set y = 1 and consider x, z as affine coordinates (so y = 0 defines the

line at infinity). Then we get the equation

x2 + 1− z2 = 0,

orx2 − z2 = (x− z)(x+ z) = 1

which is the equation of a hyperbola asymptotic to the lines x = z and x = −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 + 2z′y − z′2 = 0.

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Passing to affine coordinates x, y by setting z′ = 1 we get

x2 + 2y − 1 = 0

ory = −x2/2 + 1/2

which defines a parabola.In case F = C then the possible conics is even more restricted. Up to change of

coordinates, the possible quadratics in the complex projective plane are defined by thequadratic polynomial equations

• x2 = 0,

• x2 + y2 = 0,

• x2 + y2 + z2 = 0.

The equation x2 = 0 defines a projective line that is counted twice or has multiplicitytwo (in a sense we won’t make precise). The equation

x2 + y2 = (x+ iy)(x− iy) = 0

defines 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 The rational parametrization of the circle

Theorem 3.9. Let C be a non-degenerate conic in a projective plane P (V ) over the fieldF , and let A be a point on C. Let P (U) ⊂ P (V ) be a projective line not containing A.Then there is a bijection

α : P (U)→ C

such that, for X ∈ P (U), the points A, X, α(X) are collinear.

Proof. Suppose the conic is defined by the symmetric bilinear form B. Let a ∈ V representthe fixed point A. Because A ∈ X, we know that B(a, a) = 0. Let x ∈ V represent apoint X ∈ P (U). Because A 6∈ P (U), it follows that X 6= 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 for B associated to the basis a, x, y would be of the form 0 0 ∗

0 0 ∗∗ ∗ ∗

which has determinant zero, which is impossible because B is non-degenerate. It followsthat either B(a, x) or B(x, x) is non-zero.

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An arbitrary point on the line AX has the form [λa + µx] for some scalars λ and µ.Such a point lies on the conic C if and only if

B(λa+ µx, λa+ µx) = λ2B(a, a) + 2λµB(a, x) + µ2B(x, x)

= µ(2λB(a, x) + µB(x, x)) = 0

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

w := 2B(a, x)x−B(x, x)a

which is non-zero because one of the coefficients 2B(a, x) or −B(x, x) is non-zero.We define the map α : P (U)→ C by

α(X) = [w].

It remains to prove that α is a bijection. If Y is a point on C distinct from A, thenby Theorem 2.3 there is a unique point of intersection between AY and P (U) so

α−1(Y ) = AY ∩ P (U)

so the function is invertible on the set C \ {A}. To determine α−1(A), observe that

α(X) = [2B(a, x)x−B(x, x)a] = A

if and only if B(a, x) = 0, if and only x ∈ a⊥ ∩ U . Since a⊥ is 2-dimensional (Lemma3.6), and a⊥ 6= U ( a ∈ a⊥ but not in U), then a⊥ ∩ U is an intersection of distinct 2-dimensional subspaces of a 3-dimensional space and hence is 1-dimensional and representsa single point X = P (a⊥ ∩ U).

A consequence of Theorem 3.9 is that a real projective conic is “homeomorphic” to areal projective line, which in turn is homeomorphic to a circle - a fact that we’ve alreadyseen. A second consequence is that the complex projective conic is homeomorphic to CP 1

which in turn is homeomorphic to the two sphere S2.Let us now consider what the map α looks like in coordinates in a special case. Let

C = Z(x2 + y2 − z2) corresponding to the matrix

β =

1 0 00 1 00 0 −1

.

Choose A = [1 : 0 : 1] and let P (U) = Z(x). Then points X ∈ P (U) have the form[0 : 1 : t] or [0 : 0 : 1] and the map α : P (U)→ C satisfies

α([0 : 1 : t]) = [2B((1, 0, 1), (0, 1, t))(0, 1, t)−B((0, 1, t), (0, 1, t))(1, 0, 1)]

= [−2t(0, 1, t)− (1− t2)(1, 0, 1)]

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

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and

α([0 : 0 : 1]) = [2B((1, 0, 1), (0, 0, 1))(0, 0, 1)−B((0, 0, 1), (0, 0, 1))(1, 0, 1)]

= [−2(0, 0, 1) + 1(1, 0, 1)]

= [1 : 0 : −1]

These formulas can be used to parametrization of the circle using rational functions.For t a real number 1 + t2 > 0 is non-zero so we can divide

[1− t2 : 2t : 1 + t2] = [1− t2

1 + t2:

2t

1 + t2: 1]

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

2 + y2 − 1)

α′(t) =(1− t2

1 + t2,

2t

1 + t2

)That hits all points except (−1, 0).

These formulas have an interesting application in the case F = 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 subsetof Rn 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

λ1v1 + ...λkvk ∈ Qn,

where λ1, ..., λk ∈ Q and v1, ..., vk ∈ Qn. The span of a set of vectors span(v1, ..., vk) ⊂ Qn

is the set of all linear combinations of v1, ..., vk. In particular, a one-dimensional subspaceof Qn is the span of a single non-zero vector. We define the projective space

QP n = P (Qn+1)

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

QP n ⊂ RP n

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

immediately to geometry in QP n (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

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for x, y, z ∈ Q must 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 integer solution to the equation x2 + y2 = z2 must havethe form

x = c(b2 − a2),

y = c(2ab),

z = c(b2 + a2)

for some integers a, b, c. Such solutions are called Pythagorean triples because 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 determines x = 5, y = 8, z = 13.

3.6 Polars

Let V be a vector space and B a symmetric, bilinear form on V . Given a subset S ⊂ V ,define

S⊥ := {v ∈ V |B(s, v) = 0, for all v ∈ V }.

Observe that S⊥ must be a vector subspaces, because if v1 ∈ S⊥ and v2 ∈ S⊥ then forany scalars λ1, λ2 ∈ F and s ∈ S we have

B(s, λ1v1 + λ2v2) = λ1B(s, v1) + λ2B(s, v2) = 0

so λ1v1 + λ2v2 ∈ S. Indeed, by a similar linearity argument it can be shown that

S⊥ = span(S)⊥

so the operation of applying ⊥ can be thought of as assigning to a vector space U ⊆ Vits orthogonal or perpendicular subspace U⊥ ( relative to B of course).

Proposition 3.10. If B is a non-degenerate bilinear form, then for any subspace U ⊆ V

• (U⊥)⊥ = U

• U1 ⊂ U2 ⇒ U⊥2 ⊂ U⊥1

• dim(U) + dim(U⊥) = dim(V )

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Proof. Let V ∗ = Hom(V, F ) denote the dual vector space and consider the map

φ : V → V ∗

defined byφ(v)(w) = B(v, w).

This map is linear because

φ(λ1v1 + λ2v2)(w) = B(λ1v1 + λ2v2, w)

= λ1B(v1, w) + λ2B(v2, w)

= λ1φ(v1)(w) + λ2φ(v2)(w)

soφ(λ1v1 + λ2v2) = λ1φ(v1) + λ2φ(v2).

Because B is non-degenerate, if v 6= 0 then there exists w ∈ V such that φ(v)(w) =B(v, w) 6= 0. Thus ker(φ) = 0 and φ is injective. By ... dim(V ) = dim(V ∗) so φ is alsosurjective and hence an isomorphism between V and V ∗.

Finallyφ(U⊥) = {ξ ∈ V ∗|ξ(u) = 0 for all u ∈ U } = U0.

Thus U⊥ is simply the bijective image of the annihilator, so all the conclusions followdirectly from Section 2.6.

Definition 14. Let B be a non-degenerate symmetric bilinear form on V and let P (U) ⊂P (V ) be a linear subspace. Then the polar of P (U) is the linear subspace P (U⊥).

It follows from Section 2.6 that if P (V ) has dimension m and P (U) ⊂ P (V ) hasdimension m then P (U⊥) has dimension n −m − 1. If P (V ) is a projective plane, thenthe polar of point is a line and the polar of a line is a point.

Proposition 3.11. Let C be a non-degenerate conic in a complex projective plane P (V ).Then

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

(ii) If P ∈ C, its polar line is the unique tangent to C passing through P

(iii) If P 6∈ C, the polar line of P meets C in two points, and the tangents to C at thesepoints intersect at P .

Proof. Let U ⊂ V be the 2-dimensional subspace defining the projective line P (U) andlet u, v ∈ U be a basis for U . A general point in P (U) has the form [λu+ µv] for scalarsλ, µ ∈ C not both zero. A point [λu+ µv] lies on the conic C if

0 = B(λu+ µv, λu+ µv) = λ2B(u, u) + 2λµB(u, v) + µ2B(v, v). (6)

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We think of λ and µ as homogeneous coordinates on the projective line, so that (6) definesthe roots of a quadratic polynomial. Over the complex numbers, (6) can be factored

0 = (aλ− bµ)(a′λ+ b′µ)

determining two (possibly coincident) points of intersection between the line and the conic

[bu+ av], [b′u+ a′v],

establishing (i).To simplify equation (6), we choose u so that [u] is one of the intersection points in

C ∩ P (U). Equivalently, B(u, u) = 0, so (6) becomes

0 = 2λµB(u, v) + µ2B(v, v) = µ(2λB(u, v) + µB(v, v)).

This equation has two (possibly coincident) solutions: µ = 0 corresponding to [u] and2λB(u, v) + µB(v, v) corresponding to

[w] = [B(v, v)u+ 2B(u, v)v].

Observe that [u] = [w] if and only if:

[u] = [w]⇔ B(u, v) = 0

⇔ B(u, λu+ µv) = 0 for all points λu+ µv ∈ U⇔ P (U) is the polar of [u].

establishing (ii).Now let [a] denote a point not lying on C, and let P (U) denote the polar of [a]. From

the calculation above, it follows that P (U) meets C in two distinct points [u] and [w] with[u] 6= [w]. Because [u] lies on C we 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, λu+ µa) = λB(u, u) + µB(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).

3.7 Linear subspaces of quadrics and ruled surfaces

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

Proposition 3.12. Suppose that Q ⊂ P (V ) is a non-degenerate quadric. Then if somesubspace P (U) is a subset of Q, then

dim(P (U)) ≤ (dim(P (V ))− 1)/2

,

35

Proof. Suppose that P (U) ⊆ Q. Then for all u ∈ U , B(u, u) = 0. Furthermore, for anypair u, u′ ∈ U we have by Proposition 3.4 that

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 Proposition 2.13 this means that

2 dim(U) ≤ dim(U) + dim(U⊥) = dim(V ) (7)

from which the conclusion follows.

Indeed, over the complex numbers the maximum value of (7) is always realized. If nis odd then we can factor

x20 + ...+ x2

n = (x0 + ix1)(x0 − ix1) + ...+ (xn−1 + ixn)(xn−1 − ixn)

so the non-degenerate quadric Z(x20 + ...+ x2

n) in CP n contains the linear space

Z((x0 − ix1), ..., (xn−1 − ixn))

of dimension (n− 1)/2.In the real case, the maximum value of (7) is achieved when n is odd and n + 1 =

2p = 2q (we say B has split signature) . In this case,

x20 + ...+ x2

p−1 − x2p − ...− x2

2p−1 = (x0 + xp)(x0 − xp) + ...+ (xp−1 + x2p−1)(xp−1 − x2p−1)

so the quadric contains the linear subspace Z(x0 − xp, ..., xp−1 − xn).This is best pictured in the case n = 3:

x2 + y2 − z2 − w2 = (x− z)(x+ z) + (y − w)(y + w) = 0

passing to affine coordinates by setting w = 1 gives

(x− z)(x+ z) + (y − 1)(y + 1) = 0

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

λ(x− z) = µ(y − 1)

µ(x+ z) = λ(y + 1)

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

λ(x− z) = µ(y + 1)

µ(x+ z) = λ(y − 1)

for some homogeneous coordinates [λ : µ].

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4 Ideas for Projects

• Projective spaces over finite fields and the Fano plane

• Cramer’s paradox

• Cross-ratio

• Cubic curves

• Hopf Fibration

• Incidence geometries and non-Desarguesian planes

• Elliptic curves and group laws

• Symmetric bilinear forms in calculus and the Hessian

• The Grassmanian

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