Projective Geometry Lecture Notes

Home , Homogeneous coordinates, Klein quadric, Line at infinity, Real projective line

Thomas Baird March 26, 2014

Contents

1 Introduction 2

2 Vector Spaces and Projective Spaces 4 2.1 Vector spaces and their duals ...... 4 2.1.1 Fields ...... 4 2.1.2 Vector spaces and subspaces ...... 5 2.1.3 Matrices ...... 7 2.1.4 Dual vector spaces ...... 7 2.2 Projective spaces and homogeneous coordinates ...... 8 2.2.1 Visualizing projective space ...... 8 2.2.2 Homogeneous coordinates ...... 13 2.3 Linear subspaces ...... 13 2.3.1 Two points determine a line ...... 14 2.3.2 Two planar lines intersect at a point ...... 14 2.4 Projective transformations and the Erlangen Program ...... 15 2.4.1 Erlangen Program ...... 16 2.4.2 Projective versus linear ...... 17 2.4.3 Examples of projective transformations ...... 18 2.4.4 Direct sums ...... 19 2.4.5 General position ...... 20 2.4.6 The Cross-Ratio ...... 22 2.5 Classical Theorems ...... 23 2.5.1 Desargues’ Theorem ...... 23 2.5.2 Pappus’ Theorem ...... 24 2.6 Duality ...... 26

3 Quadrics and Conics 28 3.1 Aﬃne algebraic sets ...... 28 3.2 Projective algebraic sets ...... 30 3.3 Bilinear and quadratic forms ...... 31 3.3.1 Quadratic forms ...... 33 3.3.2 Change of basis ...... 33

1 3.3.3 Digression on the Hessian ...... 36 3.4 Quadrics and conics ...... 37 3.5 Parametrization of the conic ...... 40 3.5.1 Rational parametrization of the circle ...... 42 3.6 Polars ...... 44 3.7 Linear subspaces of quadrics and ruled surfaces ...... 46 3.8 Pencils of quadrics and degeneration ...... 47

4 Exterior Algebras 52 4.1 Multilinear algebra ...... 52 4.2 The exterior algebra ...... 55 4.3 The Grassmanian and the Pl¨ucker embedding ...... 59 4.4 Decomposability of 2-vectors ...... 59 4.5 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 perspective in 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 the apparent angle between straight lines depends on the vantage point of the observer. The more familiar Euclidean geometry is not well equipped to make sense of this, because in Euclidean geometry length and angle are well-defined, measurable quantities independent of the observer. Projective geometry provides a better framework for understanding how shapes change as perspective varies. 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 subspaces of R3. This definition is best motivated by a picture. Imagine an observer sitting at the origin in R3 looking out into 3-dimensional space. The 1-dimensional subspaces of R3 can be understood as lines of sight. If we now situate a (Euclidean) plane P that doesn’t contain the 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 transform a 3-dimensional scene into 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

2 Figure 1: A 2D representation of a cube

Figure 2: Projection onto a plane

3 motivate this terminology, consider a family of projective points that rotate from projective points that intersect P to one that is parallel. The projection onto P becomes a family 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 only at infinity with respect to some Euclidean plane P . One of the characteristic features of projective geometry is that every distinct pair of projective lines in the projective plane intersect. This runs contrary to the parallel postulate in Euclidean geometry, which says that lines in the plane intersect except when they are parallel. We will see that two lines that appear parallel in a Euclidean plane will intersect at a point at infinity when they are considered as projective lines. As a general rule, theorems about intersections between geometric sets are easier to prove, and require fewer exceptions when considered in projective geometry. According to Dieudonné’s History of Algebraic Geometry, projective geometry was among 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 richest and 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 classical geometry. 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 of ﬁelds. A ﬁeld is a set F equipped with two binary operations F × F → F called addition and multiplication, denoted + and · respectively, satisfying the following axioms for all a, b, c ∈ R.

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

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

3. (identities) There exists 0, 1 ∈ F such that a + 0 = a and a · 1 = a

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

5. (additive inverse) There exists −a ∈ F such that a + (−a) = 0

1 1 6. (multiplicative inverse) If a 6= 0, there exists a ∈ F such that a · a = 1.

Two other well know operations are best deﬁned in terms of the above properties. For example a − b = a + (−b)

4 and if b 6= 0 then 1 a/b = a · . b We usually illustrate the ideas in this course using the fields R and C, but it is worth remarking that most of the results are independent of the base field F . A perhaps unfamiliar example that is fun to consider is the field of two elements Z/2. This is the field 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 = 1 and arises as modular arithmetic with respect to 2.

2.1.2 Vector spaces and subspaces Let F denote a ﬁeld. 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 → 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 in the usual way. Similarly, F n is a vector space over F . The vector spaces we consider in this 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 all v1, v2 ∈ V and λ1, λ2 ∈ F , we have

φ(λ1v1 + λ2v2) = λ1φ(v1) + λ2φ(v2).

5 Example 2. A linear map φ from Rm to Rn can be uniquely represented by an n × m matrix of real numbers. In this representation, vectors in Rm and Rn are represented by column vectors, and φ is defined using matrix multiplication. Similarly, a linear map from Cm 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 vector spaces are called isomorphic if there exists an isomorphism between them. A vector space V over F is called n-dimensional if it is isomorphic to F n. V is called finite dimensional if it is n-dimensional for some n ∈ {0, 1, 2, ....}. Otherwise we say that V is infinite dimensional. A vector subspace of a vector space V is a subset U ⊂ V containing zero which is closed under addition and scalar multiplication. This makes U into a vector space in it’s own right, and the inclusion map U,→ V in linear. Given any linear map φ : V → W , we can define two vector subspaces: the image φ(V ) := {φ(v)|v ∈ V } ⊆ W and the kernel ker(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 the nullity of φ. The rank-nullity theorem states that the rank plus the nullity equals the dimension 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 an expression of the form λ1v1 + ... + λkvk where λ1, ..., λk ∈ F are scalars. The set of vectors that can be written as a linear combination of vectors in S is called the span of S. The set S is called linearly independent if the only linear combination satisfying

λ1v1 + ... + λkvk = 0 is when λ1 = λ2 = ... = λk = 0. If some non-trivial linear combination equals zero, then we say S is linearly dependent. A basis of V is a linearly independent set S ⊂ V that spans V . Every vector space has 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 of V , then this deﬁnes an isomorphism φ : F n → V by φ((x1, ..., xn)) = x1v1 + ... + xnvn. n Conversely, given an isomorphism φ : F → V , we obtain a basis {v1, ..., vn} ⊂ V by n setting vi = φ(ei) where {e1, ..., en} ∈ F is the standard basis. So choosing a basis for V 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.

6 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.   A1,1 A1,2 ... A1,n  A2,1 A2,2 ......  A =   .  ......  Am,1 ...... Am,n If A is m × n and B is n × p then the product matrix AB is an m × p matrix with entries

n X (AB)i,j = Ai,kBk,j (1) k=1 The identity matrix I is the n × n-matrix satisfying ( 1 i = j Ii,j := . 0 i 6= j

An n × n-matrix A is called invertible if there exists an n × n-matrix B such that AB = I = BA. If B exists, it is unique and is called the inverse matrix B = A−1. The determinant of an n × n-matrix A is the scalar value

X σ (−1) A1,σ(1)...An,σ(n)

σ∈Sn where the sum is indexed by permutations σ of the set {1, ..., n} and (−1)σ = ±1 is the sign of the permutation.

Theorem 2.2. Let A be an n × n-matrix. The following properties are equivalent (i.e. if one holds, they all hold).

• A is invertible.

• A has non-zero determinant.

• A has rank n.

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 to F = F 1 V ∗ := {φ : V → F | φ is linear}.

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

(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 identiﬁed with the set of n × 1 row vectors under matrix multiplication.

∗ ∗ ∗ Suppose that v1, ..., vn ∈ V is a basis. Then the dual basis v1, ..., vn ∈ V is the unique set of linear maps satisfying ∗ j vi (vj) = δi , j i j where δi is the Kronecker delta deﬁned by δi = 1 and δi = 0 if i 6= j. n Example 4. Let v1, ..., vn ∈ F be a basis of column vectors and let A = [v1...vn] be the n × n-matrix made up of these columns. The rules of matrix multiplication imply that ∗ ∗ the dual basis v1, ..., vn are the rows of the inverse matrix  ∗  v1 ∗ −1  v2  A =  :  .  .  ∗ 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 of 1-dimensional subspaces of V . Deﬁnition 3. If V is a vector space of dimension n + 1, then we say that P (V ) is a projective space of dimension n. A 0-dimensional projective space is called a projective point, a 1-dimensional vector space is called a projective line, and a 2-dimensional vector space 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 dimensional examples. If V is one-dimensional, then the only one dimensional subspace is V itself, and consequently 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, each 1-dimensional subspace of R2 intersects L in exactly one point. Thus RP 1 = P (R2) is isomorphic as a set with the union of L with a single point. Traditionally, we identify L with R and call the extra point {∞}, the point at inﬁnity. Thus we have a bijection of sets 1 ∼ RP = R ∪ {∞}

8 Figure 3: The real projective plane

Figure 4: The Riemann Sphere

Another approach is to observe that every 1-dimensional subspace intersects the unit circle S1 in a pair of antipodal points. Thus a point in RP 1 corresponds to a pair of antipodal points in S1. Topologically, we can construct RP 1 by taking a closed half circle and 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 bijection with C ∪ {∞} 1 ∼ CP = C ∪ {∞}. Topologically, CP 1 is a isomorphic to the 2-dimensional sphere S2 and is sometimes called the Riemann Sphere. The correspondence between C ∪ {∞} and S2 can be understood through stereographic projection, pictured in Figure 2.2.1. We turn next to the real projective plane RP 2 = P (R3). Let x, y, z be the standard coordinates on R3. Then every 1-dimensional subspace of R3 either intersects the plane z = 1 or lies in the plane z = 0. The ﬁrst class of 1-dimensional subspaces is in bijection

9 with R2, and the second class is in bijection with RP 1. Thus we get a bijection 2 ∼ 2 1 ∼ 2 RP = R ∪ RP = R ∪ R ∪ {∞}. It is not hard to see that this reasoning extends inductively to every dimension n to determine the following bijection

n ∼ n n−1 ∼ n n−1 0 RP = R ∪ RP = R ∪ R ∪ ... ∪ R (2) and similarly n ∼ n n−1 ∼ n n−1 0 CP = C ∪ CP = C ∪ C ∪ ... ∪ C . (3) Returning now to the projective plane, observe that every 1-dimensional subspace of R3 intersects the unit sphere S2 in a pair of antipodal points.

Figure 5: Antipodal points on the sphere

Thus we can construct RP 2 by taking (say) the northern hemisphere and identifying antipodal points along the equator. This makes the boundary a copy of RP 1. RP 2 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 a “cross cap”. See the website Math Images for more about the projective plane. It is also interesting to consider the ﬁeld of two elements F = Z2. As a set, the standard n n n vector space F is an n-fold Cartesian product of F with itself so Z2 has cardinality 2 . From the bijection FP n ∼= F n ∪ F n−1 ∪ ... ∪ F 1 ∪ F 0, it follows that

n n n−1 n+1 |Z2P | = 2 + 2 + .... + 1 = 2 − 1 so a projective line contains 3 points and an projective plane contains 7 points. The Z2 projective plane is also called the Fano plane.

10 Figure 6: RP 2 as quotient of a disk

Figure 7: Cross Caps

11 Figure 8: Sliced open Cross Caps

Figure 9: The Fano Plane

12 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 subspace spanned by v. If λ ∈ F \{0} is a non-zero scalar, then [λv] = [v].

Now suppose we choose a basis {v0, ..., vn} for V (recall this is basically the same as deﬁning an isomorphism V ∼= F n). Then every vector in v ∈ V can be written

n X v = xivi i=0 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, for λ 6= 0 we have [λx0 : ... : λxn] = [x0 : ... : xn]

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). 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 ∼ n (1 : y1 : ... : yn). It follows that U0 = F . This result has already been demonstrated geometrically in equations (2) and (3) because U0 is the set of 1-dimensional subspaces in V that intersect the “hyperplane” x0 = 1. The complement of U0 is the set of points n with x0 = 0, which form a copy of P (F ).

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 . Observe that 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 projective line in P (V ). Similarly, if W is 3-dimensional then P (W ) ⊆ P (V ) is a projective plane in P (V ). Observe that when each of these linear subspaces are intersected with the ∼ n subset U0 = {x0 = 1} = F then they become points, lines and planes in the traditional Euclidean sense.

13 2.3.1 Two points determine a line

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

Proof. Since [v1] and [v2] represent distinct points in P (V ), then the representative vectors v1, 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]. 0 Now suppose that there exists another projective line P (W ) ⊆ P (V ) containing [v1] 0 0 and [v2]. Then v1, v2 ∈ W , so W ⊆ W . Since both W and W are 2-dimensional it follows that W = W 0 and P (W ) = P (W 0).

Example 5. Consider the model of RP 2 as pairs of antipodal points on the sphere S2.A projective line in this model is represented by a great circle. Proposition 2.3 amounts to the 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 airplanes ﬂying 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 unique point.

Proof. Let P (V ) be a projective plane where dim(V ) = 3. Two distinct projective lines P (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.

14 Thus to prove dim(U ∩ W ) = 1 it is enough to show that U ∩ W contains a non-zero vector. Choose bases u1, u2 ∈ U and w1, w2 ∈ W . Then the subset {u1, u2, w1, w2} ⊂ V must be linearly dependent because V only has dimension 3. Therefore, there exist scalars λ1, ..., 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 6. Using the model of RP 2 as pairs of antipodal points in S2, then Proposition 2.4 is the geometric fact that pairs of great circles always intersect in a pair of antipodal points.

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 V determines a projective space P (V ) and how a vector subspace W ⊆ V determines a linear subspace P (W ). How do linear maps fit into projective geometry? It is not true in general that a linear map T : V → V 0 induces a map between the projective spaces P (V ) and P (V 0).1 This is because if T has a non-zero kernel, then it sends some 1-dimensional subspaces of V to zero in V 0. However, if ker(T ) = 0, or equivalently if T is injective, then this works fine. Definition 5. Let T : V → V 0 be an injective linear map. Then the map τ : P (V ) → P (V 0) defined by τ([v]) = [T (v)] is called the projective morphism or projective transformation induced by T . Projective transformations from a projective space to itself τ : P (V ) → P (V ) are called projective automophisms and are considered to be the group of symmetries of projective space. 1If T is non-zero but not injective, then it does define a map from a subset of P (V ) to P (V 0). This is an example of a rational map which play an important role in algebraic geometry. We will not consider rational maps in this course though.

15 2.4.1 Erlangen Program To better understand the meaning of this phrase, we make a brief digression to consider the more familiar Euclidean geometry in the plane. This kind of geometry deals with certain 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 three line segments joining them; the line segments have definite lengths and their intersections have definite angles that add up to 180 degrees. A circle is a subset of R2 of points lying a fixed distance r from a point p ∈ R2. The length r is called the radius and circumference of the circle is 2πr. The group of symmetries of the Euclidean plane (or Euclidean transformations) consists of translations, rotations, and combinations of these. All of the objects and quantities studied in Euclidean geometry are preserved by these symmetries: lengths of line segments, angles between lines, circles are sent to circles and triangles are sent to triangles. Two objects in the plane are called congruent if they are related by Euclidean transformations, 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 had been introduced and it was unclear how they related to each other. Klein’s philosophy was 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 symmetries and that geometries can be related in a hierarchy according to how much symmetry they possess. With this philosophy in mind, let us explore the relationship between projective and n Euclidean geometry in more depth. Recall from §2.2.2 that we have a subset U0 ⊂ RP consisting of those points with homogeneous coordinates [1 : y1 : ... : yn]. There is an obvious bijection ∼ n U0 = R , [1 : y1 : ... : yn] ↔ (y1, ..., yn) ∼ n Proposition 2.5. Any Euclidean transformation of U0 = R extends uniquely to a projective transformation of RP n. Proof. We focus on the case n = 2 for notational simplicity. It is enough to show that translations and rotations of U0 extend uniquely. 2 Consider first translation in R by a vector (a1, a2). That is the map (y1, y2) 7→ (y1 + a1, y2 + a2). The corresponding projective transformation is associated to the linear transformation

 1   1 0 0   1   1   y1  7→  a1 1 0   y1  =  y1 + a1  y2 a2 0 1 y2 y2 + a2 2 A rotation by angle θ of U0 = R extends to the projective transformation induced by the linear transformation

16  1   1 0 0   1   1   y1  7→  0 cos(θ) −sin(θ)   y1  =  cos(θ)y1 − sin(θ)y2  y2 0 sin(θ) cos(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 2.5 tells us that projective geometry more basic than Euclidean geometry because it has more symmetry. This means that theorems and concepts in projective geometry translate into theorems and concepts in Euclidean geometry, but not vice-versa. For example, we have lines and points and triangles in projective geometry, but no angles or lengths. Furthermore, objects in Euclidean geometry that are not isomorphic (or congruent) to each other can be 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 called conics.

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

Proposition 2.6. Two injective linear maps T : V → W and T 0 : V → W determine the same projective morphism if and only if T = λT 0 for some non-zero scalar λ.

Proof. Suppose that T = λT 0. Then for [v] ∈ P (V ),

[T (v)] = [λT 0(v)] = [T 0(v)], so T and T 0 deﬁne the same projective morphism. Conversely, suppose that for all [v] ∈ P (V ) that [T (v)] = [T 0(v)]. Then for any basis v0, ..., vn ∈ V , there exist non-zero scalars λ0, ..., λn such that

0 T (vi) = λiT (vi).

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

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

Combining these equations and using linearity, we get

n n n n n X 0 0 X X X X 0 λT (vi) = λT ( vi) = T ( vi) = T (vi) = λiT (vi) i=0 i=0 i=0 i=0 i=0

Pn 0 0 0 Subtracting gives i=0(λ − λi)T (vi) = 0. Since T is an injective, {T (vi)} is a linearly 0 independent set, so λ = λi for all i and T = λT .

17 2.4.3 Examples of projective transformations To develop some intuition about projective transformations, let’s consider transformations of the projective line FP 1 = P (F 2). A linear transformation of F 2 is uniquely described by a 2 × 2 matrix, a b x ax + by = c d y cx + dy so for a projective point of the form [1 : y] we have

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

In the case F = C, the function f : C → C ∪ {∞} of the complex numbers c + dy f(y) = a + by is called a Mobius transformation and plays an important role in complex analysis.

Proposition 2.7. Expressed in a Euclidean coordinate y ↔ [1 : y], every projective transformation 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

1 • An inversion: y 7→ y Proof. Recall from linear algebra that any invertible matrix can be obtained from the identity matrix using elementary row operations: multiplying a row by a non-zero scalar, permuting rows, and adding one row to another. Equivalently, every invertible matrix can be written as a product of matrices of the form

1 0 0 1 1 0 , , 0 λ 1 0 1 1 and these correspond to scaling, inversion and translation respectively. For instance,

0 1 1 y 1 = = y 1 0 y 1 1/y

1 so corresponds to the inversion y 7→ y

It is also informative to try and picture these transformations using our model where RP 1 is a circle and CP 1 is a sphere (leave this to class discussion). A delightful video about Moebius transformations is available here.

18 Remark 1. More generally, a projective transformation of FP n takes the form c¯ + Dy¯ y¯ 7→ a + ¯b · y¯ where D is an n × n matrix, ¯b, c¯ are vectors in F n, a is a scalar and · is the dot product. Proving this is an exercise. A more geometrically defined example of a projective transformation is obtained as follows. Let P (V ) be a projective plane, let P (U), P (U 0) be two projective lines in P (V ), and let [w] ∈ P (V ) be a projective points not lying on P (U) or P (U 0). We define a map τ : P (U) → P (U 0) 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 0). Observe that τ is well defined by Theorem 2.3 and 2.4 . Proposition 2.8. The map τ : P (U) → P (U 0) 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 Let V and W be vector spaces over the same field F . The direct sum V ⊕ W is the vector space 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) + (v0, w0) = (v + v0, w + w0) and λ(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) and p : V ⊕ W → V, p(v, w) = v called inclusion and projection maps respectively. Similar maps exist for W as well. Proposition 2.9. Let V be a vector space and let U, W be vector subspaces of V . Then the natural map Φ: U ⊕ W → V, (u, w) 7→ u + w has image im(Φ) = span{U, W } with kernel ker(Φ) = U ∩ W. In particular, if span{U, W } = V and U ∩ W = {0}, then Φ defines an isomorphism U ⊕ W ∼= V and we say that V is the internal direct sum of U and W .

19 Proof. The image of Φ is the set {u + w ∈ V | u ∈ U, w ∈ W }. Since U and W are both closed under linear combinations, it follows easily that {u + w ∈ V | u ∈ U, w ∈ W } = span{U, W }. 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} and span{U, W } = V , then Φ is both injective and surjective, so it is an isomorphism. Proof of Proposition 2.8. Let W = span{w} be the one-dimensional subspace of V corresponding to [w]. Since [w] 6∈ P (U 0), it follows that W ∩ U 0 = {0}. Counting dimensions, we deduce that V ∼= W ⊕ U 0 is an internal direct sum. Denote by p : V → U 0 the resulting projection map. Now let [u] ∈ P (U) and take

[u0] = τ([u]) ∈ P (U 0).

By deﬁnition of τ,[u0] lies in the line determined by [w] and [u], so u0 is a linear combination

u0 = λw + µu for some scalars λ, µ ∈ F . Moreover, [w] ∈ P (U 0) so [w] 6= [u0] and we deduce that µ 6= 0, so λ 1 u = w − u0 µ µ 1 0 so p(u) = − µ u 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 0.

2.4.5 General position We gain a ﬁrmer grasp of the freedom aﬀorded by projective transformations using the following concept.

Deﬁnition 6. Let P (V ) be a projective space of dimension n. A set of n + 2 points in P (V ) are said to lie in general position if the representative vectors of any proper subset are linearly independent in V (A subset S0 ⊂ S is called a proper subset if S0 is non-empty and S0 does not equal S).

Example 7. Any pair of distinct points in a projective line are represented by linearly independent 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 no three of them are collinear (prove this).

20 Theorem 2.10. If P (V ) and P (W ) are projective spaces of the same dimension n and both p1, ..., pn+2 ∈ P (V ) and q1, ..., qn+2 ∈ P (W ) lie in general position, then there is a unique projective transformation τ : P (V ) → P (W ) sending τ(pi) = qi for all i = 1, 2, ..., n + 2.

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

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

n+1 X λivi − vn+2 = 0 i=1 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 vi with the alterative representing vectors λivi, we may assume without loss of generality that n+1 X vi = vn+2. i=1

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

n+1 X wi = wn+2. i=1

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

n+1 n+1 n+1 X X X T (vn+2) = T ( vi) = T (vi) = wi = wn+2 i=1 i=1 i=1

So it induces a projective transformation satisfying τ(pi) = qi for all i. To prove uniqueness, assume that there is another linear map T 0 : V → W satisfying the conditions of the theorem. Then for each i there exists a non-zero scalar µi such that

0 T (vi) = µiwi. But then by linearity

n+1 n+1 0 0 X X µn+2wn+2 = T (vn+2) = T ( vi) = µiwi i=1 i=1

21 so n+1 n+1 X X µi w = w = w i n+2 µ i i=1 i=1 n+2 and we deduce that µi = 1 for all i and thus that T 0 = µT for some non-zero constant µn+2 scalar µ. Example 9. There is a unique projective transformation τ : FP 1 → FP 1 sending any ordered set of three points to any other ordered set of three points. In the context of complex analysis, we traditionally say that a mobius transformation is determined by where it sends 0, 1, ∞.

Remark 2. In the proof of Theorem 2.10, 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 for i = 1, ..., n + 1 and satisfying n+1 X vi = vn+2. i=1 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 least since 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 AC~ BC~ AC~ BD~ R(A, B; C,D) := / = , AD~ BD~ AD~ BC~ where the ratio of vectors is meaningful because one is a scalar multiple of the other. If a, b, c, d ∈ R, then the cross ratio is (a − c)(b − d) R(a, b; c, d) = . (a − d)(b − c) This extends to 4-tuples of points on RP 1 by defining ∞/∞ = 1. For homework you will prove that cross-ratio is preserved under projective transformations of the line. By Proposition 2.8, it follows that in Diagram 14 we have an equality of cross-ratios R(A, B; C,D) = R(A0,B0; C0,D0). We can define the cross-ratio in a different way using Theorem 2.10. For any field F let τ : L → FP 1 = F ∪ {∞} be the unique projective transformation sending τ(A) = ∞, τ(B) = 0 and τ(C) = 1. Let τ(D) = d. Then

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

22 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 one of the founders of projective geometry. His work was forgotten for many years, until it was rediscovered in the 19th century during the golden age of projective geometry. The following 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 them both.

Theorem 2.11 (Desargues’ Theorem). Let A, B, C, A0,B0,C0 be distinct points in a projective space P (V ) such that the lines AA0, BB0, CC0 are distinct and concurrent (meaning the three lines intersect at a common point). Then the three points of intersection X := AB ∩ A0B0, Y := BC ∩ B0C0 and Z := AC ∩ A0C0 are collinear (meaning they lie on a common line).

Proof. Let P be the common point of intersection of the three lines AA0, BB0 and CC0. First suppose that P coincides with one of the points A, B, C, A0,B0,C0. Without loss of generality, let P = A. Then we have equality of the lines AB = BB0 and AC = CC0, so it follows that X = AB ∩ A0B0 = B0, Z = AC ∩ A0C0 = C0, so B0C0 = XZ must be collinear with Y = BC ∩ B0C0 and the result is proven. Now suppose that P is distinct from A, B, C, A0,B0,C0. Then P , A and A0 are three distinct 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, a0 of A, A0 such that p = a + a0.

23 Figure 13: Illustration of Desargues’ Theorem

Similarly, we have representative vectors b, b0 for B,B0 and c, c0 for C,C0 such that

p = b + b0, p = c + c0

It follows that a + a0 = b + b0 so

x := a − b = b0 − a0 must represent the intersection point X = AB ∩ A0B0, because x is a linear combination of both a, b and of a0, b0. Similarly we get

z := b − c = c0 − b0 and y := c − a = a0 − c0 representing the intersection points Y and Z respectively. To see that X, Y and Z are 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 V corresponding to a projective line.

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

Theorem 2.12. Let A, B, C and A0,B0,C0 be two pairs of collinear triples of distinct points in a projective plane. Then the three points X = BC0 ∩ B0C, Y := CA0 ∩ C0A and Z := AB0 ∩ A0B are collinear.

24 Figure 14: Illustration of Pappus’ Theorem

Proof. Without loss of generality, we can assume that A, B, C0,B0 lie in general position. If not, then two of the three required points coincide, so the conclusion is trivial. By Theorem 2.10, we can assume that

A = [1 : 0 : 0],B = [0 : 1 : 0],C0 = [0 : 0 : 1],B0 = [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] for some c 6= 0 (because it is not equal to A or B). Similarly, the line B0C0 corresponds to the vector space spanned by (0, 0, 1) and (1, 1, 1) so

A0 = [1, 1, a] for some a 6= 1. Next compute the intersection points of lines

0 BC = span{(0, 1, 0), (0, 0, 1)} = {(x0, x1, x2) | x0 = 0} B0C = span{(1, 1, 1), (1, c, 0)} so we deduce that BC0 ∩ B0C is represented by (1, 1, 1) − (1, c, 0) = (0, 1 − c, 1).

0 AC = span{(1, 0, 0), (0, 0, 1)} = {(x0, x1, x2) | x1 = 0} A0C = span{(1, 1, a), (1, c, 0)} so we deduce that A0C ∩ AC0 is represented by (1, c, 0) − c(1, 1, a) = (1 − c, 0, −ca).

0 AB = span{(1, 0, 0), (1, 1, 1)} = {(x0, x1, x2) | x1 = x2} A0B = span{(1, 1, a), (0, 1, 0)} so we deduce that AB0 ∩ A0B is represented by (a − 1)(0, 1, 0) + (1, 1, a) = (1, a, a).

25 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 a   1 a a   1 a a   1 − c 0 −ca  ⇒  0 (c − 1)a −a  ⇒  0 0 0  0 1 − c 1 0 1 − c 1 0 1 − c 1 which has rank two, so the span of the rows has dimension two and the vectors are linearly dependent.

2.6 Duality In this section we explore the relationship between the geometry of P (V ) and that of P (V ∗) where V ∗ denotes the dual vector space. Recall that given any vector space V , the dual 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 the unique 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 isomorphism between them. However there is a natural correspondence between subspaces.

Deﬁnition 7. Let U ⊆ V be a vector subspace. The annihilator U 0 ⊆ V ∗ is deﬁned by

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

0 ∗ 0 To check that U is a subspace of V , observe that if φ1, φ2 ∈ U and u ∈ U then

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

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

0 0 Proposition 2.13. If U1 ⊆ U2 ⊂ V then U1 ⊇ U2 .

26 Proof. This is simply because if φ(v) = 0 for all v ∈ U2 then necessarily φ(v) = 0 for all v ∈ U1. We also have the following result. Proposition 2.14. dim(U) + dim(U 0) = dim(V )

Proof. Choose a basis u1, ..., um ∈ U and extend to a basis u1, ..., um, vm+1, ..., vn of V . Let φ1, ..., φn be the dual basis. For a general element

n X ∗ φ = λiφi ∈ V i=1

0 0 ∗ we have φ(ui) = λi. Thus φ ∈ U if and only if λi = 0 for i = 1, ..., m so U ⊆ V is the subspace with basis φm+1, ..., φn which has dimension n − m. The following proposition justiﬁes the term duality.

S Proposition 2.15. There is a natural isomorphism V ∼= (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 Proposition 2.1) it only remains to prove that S : V → (V ∗)∗ is injective. But if v ∈ V is non-zero, we may extend v1 := v to a basis and then in the dual basis φ1(v1) = S(v1)(φ1) = 1 6= 0 so S(v) 6= 0. Thus ker(S) = {0} and S is injective. Remark 3. Proposition 2.15 depends on V being finite dimensional. If V is infinite dimensional, then S is injective but not surjective. In practice, we tend to replace the phrase “naturally isomorphic” with “equals” and write V = (V ∗)∗. Using this abuse of terminology, one may easily check that (U 0)0 = U, so the operation of taking a subspace to its annihilator is reversible. Proposition 2.16. 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 linear subspaces of P (V ∗) of dimension n − m − 1 by the rule

P (U) ↔ P (U 0).

We use notation P (U 0) = P (U)0 and call P (U)0 the dual of P (U). Proof. By preceding remarks there is a one-to-one correspondence

U ↔ U 0 between subspaces of dimension m+1 in V and subspaces of dimension n−m in V ∗. The result follows immediately. It follows from Propositions 2.13 and 2.16 that geometric statements about linear subspaces of P (V ) translate into geometric statements about linear subspaces of P (V ∗).

27 Example 10. Let P (V ) be a projective plane, so P (V ∗) is also a projective plane. There is 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 ∗). Corollary 2.17. Let A, B be distinct points in a projective plane lying on the line L. Then A0,B0 are lines intersecting at the point L0. Proof. If A, B ⊂ L then L0 ⊂ A0 and L0 ⊂ B0, so L0 = A0 ∩ B0 is their single point of intersection. Corollary 2.18. Two distinct projective planes in projective 3-space always intersect in a line. Proof. This statement is dual to statement “two points determine a line” in 3-space (Theorem 2.3). Duality can be used to establish “dual” versions of theorems. For example, simply be exchanging points with lines and collinear with concurrent, we get the following results. Theorem 2.19 (Dual version of Desargues’ Theorem = Converse of Desargue’s Theorem). Let α, β, γ, α0, β0, γ0 be distinct lines in a projective plane such that the points α ∩α0, β ∩β0 and γ ∩ γ0 are distinct and collinear. Then the lines joining α ∩ β to α0 ∩ β0, α ∩ γ to α0 ∩ γ0 and β ∩ γ0 to β0 ∩ γ0 are concurrent. Theorem 2.20 (Dual version of Pappus’ Theorem). Let α, β, γ and α0, β0, γ0 be two sets of concurrent triples of distinct lines in a projective plane. Then the three lines determined by pairs of points β ∩ γ0 and β0 ∩ γ, γ ∩ α0 and γ0 ∩ α, α ∩ β0 and α0 ∩ β are concurrent. In fact, with a little eﬀort one can show that the statement of Pappus’ Theorem is equivalent 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 2.21. The set of lines in R2 is in one-to-one correspondence with the Mobius strip.

Proof. The set of lines in R2 is the set of lines in RP 2 minus the line at inﬁnity. By duality, this is identiﬁed with RP 2 minus a single point, which is the same as a mobius strip.

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

X I f = f(x1, ..., xn) := aI x I where

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

I i1 in • x = x1 ...xn are called monomials, and

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

I The degree of a monomial x is the sum |I| = i1 + ... + in. The degree of a polynomial f equals 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.

2 A polynomial f(x1, ..., xn) in n variables determines a function

f : F n → F

n by simply plugging in the entries n-tuples in F in place of the variables x1, ..., xn. Deﬁnition 8. Given an n variable polynomial f, the set

n Zaff (f) := {(x1, ..., xn) ∈ F | 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) ∩ ... ∩ Zaff (fk)

n is called the zero set of f1, ..., fk. An affine algebraic set X ⊂ F is a set that equals Zaff (f1, ..., fk) for some set of polynomials f1, ..., fk. Example 12. An affine line in F 2 is defined by a single equation ax+by +c = 0, where a and 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 of a collection of degree one polynomials. Example 13. The graph of a one variable polynomial f(x) is defined by the equation y = f(x), thus is equal to the algebraic set Zaff (y − f(x)).

Example 14. Any ﬁnite set of points {r1, ..., rd} ⊂ F is an algebraic set Zaff (f) wheref(x) = (x − r1)(x − r2)...(x − rd).

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

Example 15. Some examples of affine in R2 conics include: circles, ellipses, hyperbolas 2 2 n and parabolas. The unit sphere Zaff (x1 + ... + xn − 1) is a quadric in R . 2For some fields, two different polynomials can define the same function. For example x2 + x defines the zero function over Z2.

29 3.2 Projective algebraic sets What is the right notion of an algebraic set in projective geometry? We begin with a deﬁnition. Deﬁnition 10. A polynomial is called homogeneous if all non-zero monomials have the same degree. Example 16. • 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 + 1 variables, which we normally denote {x0, x1, ..., xn}. The crucial property that makes homogenous polynomials ﬁt into projective geometry is the following.

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

d f(λx0, ..., λxn) = λ f(x0, ..., xn).

I i0 in Proof. For a monomial x = x0 ...xn of degree d = |I| = i0 + ... + in, we get

I i0 in i0+...+in i0 in d I (λx) = (λx0) ...(λxn) = λ x0 ...xn = λ x . The same will hold true for a linear combination of monomials of degree d.

It follows from Proposition 3.1 that for λ a non-zero scalar, f(x0, ..., xn) = 0 if and only 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

n Z(f1, ..., fk) := {[x0 : ... : xn] ∈ FP | 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 subsets of projective space. These are always of the form Z(l1, ..., lk) for some set of homogeneous linear polynomials l1, ..., lk. To understand the correspondence between projective and algebraic sets we recall that n n ∼ n within FP , we have a copy of the aﬃne plane U0 := {[1 : y1, ... : yn] ∈ FP } = F . Under this correspondence, we have an equality

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

30 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 multiplying the monomials of f by appropriate powers of x0, we can construct a homogenous, degree d polynomial h(x0, ..., xn) such that h(1, x1, ..., xn) = f(x1, ..., xn). We call h the homogenization of f. 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 + 1 ⇒ h(x, y, z) = yz − x2 + z2

3 2 3 3 • g(x1, x2) = x1 + x2 + 7 ⇒ h(x0, x1, x2) = x0x1 + x2 + 7x0

Now 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 research today, of which projective geometry is a small part. So far we have only considered linear algebraic sets, where the polynomials all have order one. Our next big topic are algebraic sets of the form Z(f) where f is a degree two polynomial. Such algebraic sets are 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 if for each non-zero v ∈ V , there is a w ∈ V such that B(v, w) 6= 0. Example 18. The standard example of a symmetric bilinear form is the dot product

n n n X B : R × R → R,B((x1, ..., xn), (y1, ..., yn)) = xiyi. i=1 The dot product is nondegenerate because B(v, v) = |v|2 > 0 if v 6= 0.

31 Example 19. 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 special relativity. In contrast with the dot product, the equation B(v, v) = 0 has holds for some non-trivial vectors called null-vectors. The null-vectors form a subset of R4 called the light cone dividing those vectors satisfying B(v, v) < 0 (called time-like vectors) from those satisfying B(v, v) > 0 (called space-like vectors).

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

Lemma 3.2. Let v1, ..., vn ∈ V be a basis. For any pair of vectors u = x1v1 + ...xnvn and w = y1v1 + ... + ynvn we have

n X B(u, w) = xiyjB(vi, vj). i,j=1

This means that the bilinear form is completely determined by the n × n-matrix β with entries βi,j = B(vi, vj). Proof.

B(u, w) = B(x1v1 + ...xnvn, w) n n X X = xiB(vi, w) = xiB(vi, y1v1 + ... + ynvn) i=1 i=1 n n n X X X = xiyjB(vi, vj) = xiyjβi,j i=1 j=1 i,j=1

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

B(v, w) =v ¯T βw¯ wherev, ¯ w¯ denote the vectors v, w expressed as column vectors and T denotes transpose, so thatv ¯T is v as a row vector.

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

Proof. Suppose that det(β) is zero if and only if there exists a nonzero row vectorv ¯T such thatv ¯T β = 0 if and only ifv ¯T βw¯ = 0 for all column vectorsw ¯ if and only if there exists v ∈ V \{0} such that B(v, w) = 0 for all w ∈ V if and only if B is degenerate.

32 3.3.1 Quadratic forms Given a symmetric bilinear form B : V × V → F , we can associate a function called a quadratic form Q : V → F by the rule Q(v) = B(v, v). Proposition 3.4. Let V be a vector space over a ﬁeld F in which 1 + 1 6= 0. Let B be a symmetric bilinear form on V and let Q : V → F be the associated quadratic form. Then for any u, v ∈ V we 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 3.4 does not hold when F is the ﬁeld of two elements because we are not allowed to divide by 2 = 0.

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

X Q(x1v1 + ... + xnvn) = B(x1v1 + ... + xnvn, x1v1 + ... + xnvn) = βi,jxixj i,j is a homogeneous quadratic polynomial. Some examples:

a ↔ ax2

a b ↔ ax2 + 2bxy + cy2 b c  a b d  2 2 2  b c e  ↔ ax + 2bxy + cy + 2dxy + 2exz + fz d e f

3.3.2 Change of basis We’ve seen so far that a symmetric bilinear form B can be described using a symmetric matrix 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 n {Pi,j}i,j=1 such that X wi = Pi,jvj (4) j

The matrix P with entries Pi,j is called the change of basis matrix from w1, ..., wn to v1, ..., vn.

33 Proposition 3.5. The change of basis matrix P is invertible and the inverse P −1 is the change of basis matrix from v1, ..., vn to w1, ..., wn.

Proof. Let Q be the change of basis matrix from v1, ..., vn to w1, ..., wn. Then

n n n n n n X X X X X X vi = Qi,jwj = Qi,j( Pj,kvk) = ( Qi,jPj,k)vk = (QP )i,kvk j=1 j=1 k=1 k=1 j=1 k=1 where we have used (1) for matrix multiplication. We conclude that ( 1 i = k (QP )i,k = 0 i 6= k and QP is the identity matrix.

−1 Observe that P is invertible, with P providing the change of matrix from w1, ..., wn 0 back to v1, ..., vn. In fact, if P is an invertible matrix then it is the change of basis matrix from v1, ..., vn to the new basis w1, ..., wn satisfying (4). Proposition 3.6. Let B be a bilinear form on V , and let β, β0 be symmetric matrices representing B with respect to bases v1, ..., vn and w1, ..., wn respectively. Then β0 = P βP T where P is the change of basis matrix. Proof. There are two approaches. Simply calculate:

0 X X βi,j = B(wi, wj) = B( Pi,kvk, Pj,lvl) k l X X T = Pi,kPj,lβk,l = Pi,kβk,lPl,j k,l k,l T = (P βP )i,j

Alternatively, we can argue more conceptually as follows. Let ei denote the ith standard basis row vector. Then in terms of the basis v1, ..., vn, wi corresponds to the row vector eiP . Then

0 T T T T βi,j = B(wi, wj) = (eiP )β(ejP ) = eiP βP ej = (P βP )i,j.

Our next task is to classify bilinear forms up to change of basis. First we need a lemma. Lemma 3.7. Let V be a vector space of dimension n equipped with a symmetric bilinear form B. For w ∈ V deﬁne w⊥ := {v ∈ V |B(v, w) = 0}. Then w⊥ ⊂ V is a vector space of dimension ≥ n − 1.

34 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 nullity theorem, ( n im(φ) = {0} dim(w⊥) = dim(V ) − dim(im(φ)) = n − 1 im(φ) = F.

Theorem 3.8. Let B be a symmetric, bilinear form on a vector space V over F = C or F = R. Then there exists a basis v1, ..., vn such that

B(vi, vj) = 0 (5) for i 6= j and such that for all i = 1, ..., n ( 0, 1 F = C B(vi, vi) ∈ (6) 0, ±1 F = R. I.e., there is a choice of basis so that β is diagonal with diagonal entries 0, 1 if F = C and 0, 1, −1 if F = R. Proof. As a ﬁrst step construct a basis satisfying (5) using induction on the dimension of V . In the base case dim(V ) = 1, (5) holds vacuously. Now suppose by induction that all n − 1 dimensional symmetric bilinear forms can be diagonalized. Let V be n dimensional. If B is trivial, then β is the zero matrix with respect to any basis (hence diagonal). If B is non-trivial then by Proposition 3.4, there ⊥ ⊥ exists a vector wn ∈ V such that B(wn, wn) = Q(wn) 6= 0. Clearly wn 6∈ vn , so wn is n−1 ⊥ dimensional. By induction we may choose a basis w1, ..., wn−1 ∈ wn satisfying (5) for all ⊥ i, j < n. Furthermore, for all i < n, B(wi, wn) = 0 because wi ∈ wn . Thus w1, ..., wn satisfy (5). Finally, we want to modify w1, ..., wn to satisfy (6). Suppose that B(wi, wi) = c ∈ F . √1 If c = 0, then let vi = wi. If c 6= 0 and F = C, let vi = c wi so that

1 1 1 2 c B(v , v ) = B(√ w , √ w ) = √ B(w , w ) = = 1. i i c i c i c i i c

1 If c 6= 0 and F = then let vi := √ wi so R |c| c B(v , v ) = = ±1. i i |c|

Corollary 3.9. Let V and B be as above and let Q(v) = B(v, v). Then

35 Pn • If F = C then for some m ≤ n there exists a basis such that if v = i=1 zivi

m X 2 Q(v) = zi i=1

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

p p+q X 2 X 2 Q(v) = zi − zi i=1 i=p+1

Proof. Choose the basis v1, ..., vn so that B(vi, vj) satisfy the equations (5) and (6). Then

n n X X 2 Q(z1v1+...+znvn) = B(z1v1+...+znvn, z1v1+...+znvn) = zizjB(vi, vj) = zi B(vi, vi), i,j=1 i=1 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 the choice of basis. It follows that B and Q are non-degenerate if and only if m = n or m = p + q. In the case of a real vector space, we say that B is positive definite if p = n, negative definite if q = 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 function and let (a1, ..., an) be a point at which all partial derivatives vanish: ∂f fxi (a1, ..., an) = |(a1,...,an) = 0, for all i = 1, ..., n ∂xi If the the second order partials are defined and continuous then the n × n-matrix H with entries ∂2f Hi,j = fxixj (a1, ..., an) = |(a1,...,an) ∂xj∂xi is a symmetric matrix (Hi,j = Hj,i) defining a symmetric bilinear form on the set of tangent vectors called the Hessian. The corresponding quadratic form is equal to the second 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 then f has a local maximum at (a1..., an) if H is negative definite, has a local minimum if H is positive definite, and has neither a max or min if H is neither positive nor negative definite.

36 Similarly, there is a notion of a complex valued function of complex variables being diﬀerentiable (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 version of the maximum principle.

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 = Z(B) := {[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 homogeneous degree two polynomial f. We say that B and B0 are projectively equivalent if there exists non-zero scalar λ ∈ F − {0} such that B = λB0. Observe that

Z(B) = Z(λB) so projectively equivalently symmetric bilinear forms determine same quadric. Consider now the implications of Corollary 3.9 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 coordinates/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, we would say that Z(x2) has a root of “multiplicity” two at [0 : 1], but we won’t define this concept in this course. The case Z(x2 + y2) is empty set because there are no solutions with both 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 polynomials 0 = ax2 + bx + c: such a polynomial can have two roots, one root, or no roots. A consequence of Corollary 3.9 is that up to projective transformations, the number of roots completely characterizes the quadric. Over the complex numbers, the classification is even simpler: a quadric has the form Z(f) where f is one of (up to coordinate change)

• x2

• x2 + y2

37 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 the complex numbers 0 = ax2 + bx + c with a 6= 0 has either two distinct roots or a single root 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 quadric is called a conic). We may choose coordinates so that X = Z(f) where f is one of:

• x2,

• x2 + y2,

• x2 + y2 + z2,

• x2 − y2,

• x2 + y2 − z2 , up to multiplication by −1 which doesn’t change Z(f) = Z(−f). The ﬁrst example x2 = xx = 0 ∼ 1 has the same solution set as x = 0, which is projective line {[0 : a : b]} = RP , however this line is counted twice in a sense we won’t make precise (similar two a double root of single variable polynomial). The second example

x2 + y2 = 0 has a single solution [0 : 0 : 1]. The third example x2 + y2 + z2 = 0, has no projective solutions (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 two distinct lines intersecting at one point. The last example Z(x2 + y2 − z2) is (up to coordinate change) the only non-empty and non-degenerate real conic, up to projective transformation. The apparent difference between the affine conics: ellipses, parabolas, hyperbolas, can be understood as how the conic 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 a hyperbola 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 affine coordinates (so the equation z = 0 is the “line at infinity”). Then the polynomial equation x2 + y2 − z2 = 0 becomes x2 + y2 − 1 = 0 or

x2 + y2 = 1 which deﬁnes the unit circle in the x-y plane.

38 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, or z2 − x2 = (z − x)(z + x) = 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 z0 = y + z. In these new coordinates, the equation becomes

x2 + y2 − (z0 − y)2 = x2 + 2z0y − z02 = 0.

Passing to aﬃne coordinates x, y by setting z0 = 1 we get

x2 + 2y − 1 = 0 or y = −x2/2 + 1/2 which defines a parabola. Since ellipses, hyperbolas, and parabolas can all be obtained by intersecting an affine plane with the cone in R3 defined by x2 + y2 − z2, they are sometimes called conic sections.

Figure 15: Conic sections

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 the quadratic polynomial equations

• x2 = 0,

• x2 + y2 = 0,

39 • x2 + y2 + z2 = 0.

The equation x2 = 0 defines a projective line that is counted twice or has multiplicity two (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 and x − 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 the ﬁeld F in which 2 6= 0, and let A be a point on C. Let L ⊂ P (V ) be a projective line not containing A. Then there is a bijection α : L →∼ C such that, for X ∈ L, the points A, X, α(X) are collinear.

Figure 16: Parametrization of the conic

Proof. Suppose the conic is deﬁned by the symmetric bilinear form B. Let a ∈ V represent the ﬁxed point A. Because A ∈ C, we know that B(a, a) = 0. Let x ∈ V represent a point X ∈ L. Because A 6∈ L, it follows that X 6= A so the vectors a, x are linearly independent, 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 the matrix for B associated to the basis a, x, y would be of the form

 0 0 ∗   0 0 ∗  ∗ ∗ ∗

40 which has determinant zero, and this is impossible because B is non-degenerate. It follows that either B(a, x) or B(x, x) is non-zero. 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 solution µ = 0 corresponding the point A and solution to 2λB(a, x) + µB(x, x) = 0 corresponding to 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 : L → C by α(X) = [w]. It remains to prove that α is a bijection. We do this by proving the preimage of any point is a paint. If Y is a point on C distinct from A, then by Theorem 2.4 there is a unique point of intersection AY ∩ L, so α−1(Y ) = AY ∩ L.

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 if x ∈ a⊥ and X ∈ L. Since a⊥ is 2-dimensional subspace by Lemma 3.7, P (a⊥) is a projective line. The intersection of lines P (a⊥) and L is the unique point in L mapping to A.

In the course of proving Theorem 3.10, we proved the following result that will come up again later. Proposition 3.11. Let C = Z(B) be a non-degenerate conic and P ∈ C a point. The polar of P is the unique line L such that C ∩ L = P . We call L the tangent line of C at P .

41 Though Theorem 3.10 only says that α is a bijection of sets, it is in fact an isomorphism of “algebraic varieties”. In particular, this means that a real projective conic is has the topology to a real projective line, which in turn has the topology of a circle - a fact that we’ve already seen. A second consequence is that the complex projective conic is has the topology of CP 1 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 α in coordinates. Let C = Z(x2 + y2 − z2) corresponding to the matrix  1 0 0  β =  0 1 0  . 0 0 −1 Choose A = [1 : 0 : 1] and let L = Z(x). Then points in L have the form [0 : t : 1] or [0 : 1 : 0] and the map α : L → C satisﬁes

α([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

α([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. For t a real number 1 + t2 > 0 is non-zero so we can divide t2 − 1 2t [t2 − 1 : 2t : t2 + 1] = [ : : 1] t2 + 1 t2 + 1 and we gain a parametrization of the unit circle in aﬃne co-ordinatesα ˜ : R → S1 = 2 2 Zaff (x + y − 1) t2 − 1 2t α˜(t) = , t2 + 1 t2 + 1 that hits all points except (1, 0). These formulas have an interesting application in the case F = Q is the ﬁeld of rational numbers (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 to discuss informally linear algebra and projective geometry over F = Q. To keep matters simple, we concentrate on the standard dimension n vector space over Q, Qn consisting of n-tuples of rational numbers. It is helpful geometrically to consider Qn as a subset

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

n λ1v1 + ...λkvk ∈ Q ,

n n where λ1, ..., λk ∈ Q and v1, ..., vk ∈ Q . The span of a set of vectors span(v1, ..., vk) ⊂ Q is the set of all linear combinations of v1, ..., vk. In particular, a one-dimensional subspace of Qn is the span of a single non-zero vector. We deﬁne the projective space

n n+1 QP = P (Q ) to be the set of one dimensional subspaces in Qn+1. Geometrically, we can regard

n n QP ⊂ RP 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 symmetric bilinear forms because the proof depends on taking square roots but the square root of a rational number is not necessarily rational!). In particular the formulas parametrizing the circle remain valid. That is, solutions to the equation x2 + y2 = z2 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 no common 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 have the 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 deﬁne right triangles with integer side lengths. For example, a = c = 1, b = 2 determines x = 3, y = 4, z = 5, while a = 2, b = 3, c = 1 determines x = 5, y = 8, z = 13.

43 3.6 Polars Let B be a non-degenerate bilinear form on a vector space V . Recall that V ∗ = Hom(V,F ) is the dual vector space for V . Lemma 3.12. The map B# : V → V ∗ deﬁned by

B#(v)(w) := B(v, w) is an isomorphism of vector spaces Proof. The map is linear because

# B (λ1v1 + λ2v2)(w) = B(λ1v1 + λ2v2, w)

= λ1B(v1, w) + λ2B(v2, w) # # = λ1B (v1)(w) + λ2B (v2)(w) so # # # B (λ1v1 + λ2v2) = λ1B (v1) + λ2B (v2). Since B is non-degenerate, if v 6= 0 then there exists w ∈ V such that B#(v)(w) = B(v, w) 6= 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 , deﬁne

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

We call U ⊥ orthogonal or perpendicular subspace of U (relative to B). Proposition 3.13. The isomorphism of Lemma 3.12 sends U ⊥ to U 0. In particular, for any subspaces of V • (U ⊥)⊥ = U

⊥ ⊥ • U1 ⊂ U2 ⇒ U2 ⊂ U1 • dim(U) + dim(U ⊥) = dim(V ) Proof.

B#(U ⊥) = {B#(v) ∈ V ∗|v ∈ U ⊥} = {ξ ∈ V ∗|ξ(u) = 0 for all u ∈ U } = U 0.

The properties is inherited from Section 2.6. The isomorphism B# determines a projective isomorphism between P (V ) with its dual P (V ∗) P (V ) ∼= P (V ∗). (7)

44 Deﬁnition 14. Let B be a non-degenerate symmetric bilinear form on V and let P (U) ⊂ P (V ) be a linear subspace. We call P (U ⊥) the polar of P (U). If P (V ) has dimension n and P (U) ⊂ P (V ) has dimension m then P (U ⊥) has dimension n − m − 1, just like for duality. Proposition 3.14. 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 these points intersect at P .

Figure 17: Duality for conics

Proof. Let U ⊂ V be the 2-dimensional subspace deﬁning the projective line P (U) and let u, v ∈ U be a basis for U. A general point in P (U) has the form [λu + µv] for scalars λ, 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). (8) We think of λ and µ as homogeneous coordinates on the projective line, so that (8) deﬁnes the roots of a quadratic polynomial. From §3.4 we know that (8) has one or two solutions, establishing (i). Property (ii) is a direct consequence of Proposition 3.11. 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]. The same argument applies to [w] and this establishes (iii).

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

Proposition 3.15. Suppose that Q ⊂ P (V ) is a non-degenerate quadric over ﬁeld F in which 0 6= 2. Then if some subspace P (U) is a subset of Q, 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 pair u, u0 ∈ U we have by Proposition 3.4 that 1 B(u, u0) = (B(u + u0, u + u0) − B(u, u) − B(u0, u0)) = 0 2 from which it follows that U ⊆ U ⊥. According to Proposition 2.14 this means that

2 dim(U) ≤ dim(U) + dim(U ⊥) = dim(V ) (9) from which the conclusion follows. 3This duality is explored in the suggested project topic “Pascal’s Theorem, Brianchon’s Theorem and duality”

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

2 2 x0 + ... + xn = (x0 + ix1)(x0 − ix1) + ... + (xn−1 + ixn)(xn−1 − ixn)

2 2 n so the non-degenerate quadric Z(x0 + ... + xn) in CP contains the linear space

Z((x0 − ix1), ..., (xn−1 − 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,

2 2 2 2 x1 + ... + xp − xp+1 − ... − x2p = (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) = 0 passing 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

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 lines satisfying

λ(x − z) = µ(y + 1) −µ(x + z) = λ(y − 1) for some homogeneous coordinates [λ : µ].

3.8 Pencils of quadrics and degeneration In this section, we move beyond the study of individual quadrics to the study of families of 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 a vector space under the operation

(λ1B1 + λ2B2)(v, w) = λ1B1(v, w) + λ2B2(v, w).

47 Figure 19: Bi-ruled hyperboloid

Figure 20: Bi-ruled hyperboloid

48 The vector space of symmetric bilinear forms on V is denoted S2(V ∗).

Proposition 3.17. Two symmetric bilinear forms B,B0 on a complex vector space V deﬁne the same quadric if and only if B = λB0 for some non-zero scalar λ ∈ C. Con- sequently, the set of quadrics in P (V ) is in one-to-one correspondence with the points of P (S2(V ∗)).

Proof. The statement is immediate if dim(V ) = 1. If dim(V ) = 2, then by choosing a basis and passing to the associated quadratic form, the statement is equivalent to saying that a homogeneous degree two polynomial in two variables ax2 +bxy+cy2 is determined up to scalar multiplication by its roots. This follows from the fact that homogeneous polynomials in two variables can always be factored over the complex numbers (by dehomogenizing, factoring and rehomogenizing). This is the homogeneous version of the Fundamental Theorem of Algebra. For the general case, choose a basis v1, ..., vn of V and deﬁne n × n-matrices in which 0 0 B(vi, vj) = βi,j and B (vi, vj) = βi,j. For any pair of i, j ∈ 1, ..., n, we may restrict the bilinear forms to the span of vi and vj. The result for two dimensional vector spaces implies that 0 0 βi,i βi,j βi,i βi,j = λ 0 0 , βj,i βj,j βj,i βj,j for some non-zero scalar λ (possibly depending on i, j). This means in particular that

0 βi,j = 0 if and only if βi,j = 0. (10)

According to Theorem 3.8, we may choose the basis v1, ..., vn so that for some 0 ≤ p ≤ n 0 we have β1,1 = β2,2 = ... = βp,p = 1 and all other entries βi,j = 0 . By (10), βi,j = 0 unless (i, j) ∈ {(1, 1), (2, 2), ..., (p, p)} . If p ≤ 1 then we are done. If p ≥ 2, then for any 1 ≤ i ≤ p,

0 β1,1 0 0 1 0 0 . = β1,1 . 0 βi,i 0 1 0 0 0 so β = λβ with λ = β1,1 and thus B = λB . Remark 5. Proposition 3.17 does not hold over general ﬁelds. For example, the quadratic forms x2 + y2 and x2 + 2y2 are not scalar multiples of each other, but Z(x2 + y2) = Z(x2 + 2y2) = ∅ over the real numbers.

Deﬁnition 15. Let B,B0 be two linearly independent symmetric bilinear forms on a vector space V . The pencil of quadrics generated by B and B0 is the set of quadrics associated to bilinear forms of the form λB + µB0, for λ, µ not both zero. For complex V , the pencil forms a projective line in P (S2(V ∗)).

Proposition 3.18. Let V be a vector space of dimension n. A pencil of quadrics contains at most n degenerate quadrics. If F = C, then the pencil contains at least one degenerate quadric.

49 Proof. Choose a basis for V so that B and B0 are represented by matrices β and β0. The bilinear for λB + µB0 is degenerate if and only if the determinant det(λβ + µβ0) equals zero. Expanding out the expression det(λβ + µβ0) is a degree n homogeneous polynomial in the homogeneous variables [λ : µ], so it has at most n roots. Over C, such a polynomial has at least one root by the Fundamental Theorem of Algebra. Theorem 3.19. Two non-degenerate conics in a complex projective plane intersect in at least 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) and Z(B0). If B00 := λB + µB0 is a linear combination then B00(v, v) = λB(v, v) + µB0(v, v) = 0 + 0 = 0 so [v] ∈ Z(B00). It follows that Z(B) ∩ Z(B0) ⊆ Z(B00) so Z(B) ∩ Z(B0) ⊆ Z(B) ∩ Z(B00). If µ 6= 0, then the pencil determined by B and B0 coincides with the one determinded by B and B00. In particular, 1 λ 1 λ B0 = (λB + µB0) − B = B00 − B µ µ µ µ so arguing conversely, we get

Z(B) ∩ Z(B0) = Z(B) ∩ Z(B00). (11) By Proposition 3.18 we know that Z(B00) is degenerate for some choice [λ : µ], so we are reduced to counting the intersection points between degenerate and a non-degenerate conic. From §3.4 we know that degenerate conics can either be a line or a union of two lines. By Proposition 3.14 it follows that Z(B) ∩ Z(B00) = Z(B) ∩ Z(B0) consists of between one and four points.

50 Corollary 3.20. Two conics in a complex projective plane have between one and four tangent lines in common. Proof. Follows from Theorem 3.19 by duality (see Proposition ??).

Remark 6. With an appropriate notion of multiplicity, Theorem 3.19 (resp. Corollary 3.20) can be improved to say there are exactly four intersection points (resp. common tangents) counted with multiplicity. Using equation (11), we are able to produce pairs on non-degenerate conics realizing all 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 be x2 = 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 quadrics Z(f) and Z(x2) have only one intersection point over the complex numbers. To see this, observe that Z(x2) = {[0 : a : b] | [a : b] ∈ CP 1} and that f(0, a, b) = (0 − b)2 + a2 − b2 = a2 vanishes only if [0 : a : b] = [0 : 0 : 1]. According to (11), this means that Z(−x2 + f(x, y, z)) = Z(y2 − 2xz) intersects Z(f) at the unique point [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 passing through those four points. Proof. All conics in the pencil must contain the four intersection points by (11). It remains to 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 these four 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}.

51 Figure 21: which is equal to the pencil spanned by the two (degenerate) conics Z(x(y − z)) and Z(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 in general position, then the pencil they determine contains three degenerate conics (the 4 maximum allowed by Proposition 3.19), corresponding to the 2 /2 = 3 ways to contain four points in general position in a union of two lines:

4 Exterior Algebras

In this chapter we will always assume that ﬁelds F satisfy 2 6= 0 and consequently that 1 6= −1.

4.1 Multilinear algebra Deﬁnition 16. Let V be a vector space over a ﬁeld F .A multilinear form of degree k (k ≥ 1) or k-linear form is a function

M : V k = V × ... × V → F that is linear in each entry. That is, for each i = 1, 2, ..., k we have

0 0 M(u1, u2, ...., λui + µui, ...., uk) = λM(u1, ..., ui, ..., uk) + µM(u1, ..., ui, ..., uk)

0 k ∗ for all u1, ..., uk, ui ∈ V and λ, µ ∈ F . The set of k-linear forms is denoted T (V ). By convention T 0(V ∗) = F .

We’ve seen many example of multilinear forms already.

• A multilinear form of degree 1 is simply a linear map ξ : V → F , so T 1(V ) = V ∗.

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

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

53 • Symmetric bilinear forms are multilinear forms of degree two.

∼ n • If V = F is the set of n × 1 column matrices, then given v1, ..., vn ∈ V we may form the n × n matrix [v1....vn]. The determinant

det : V × ... × V → F, (v1, ...., vn) 7→ det(v1....vn)

is a multilinear form of degree n.

Multilinearity implies that that the form M is completely determined by it’s values on on k-tuples of basis elements. If v1, ..., vn is a basis for V , then

n n n X X X M( λ1,jvj, ..., λk,jvj) = λ1,j1 ...λk,jk M(vj1 , ..., vjk ). (12)

j=1 j=1 j1,...,jk=1 There are two distinguished subsets of T k(V ∗). The symmetric k-linear forms Sk(V ∗) ⊆ T k(V ∗) possess the property that transposing any two entries does not change the value of 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-symmetric or exterior) forms Λk(V ∗) ⊆ T k(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 have

M(u1, ..., ui, ..., uj, ..., uk) = −M(u1, ..., uj, ..., ui, ...uk). (13)

Of the examples above, dual vectors are alternating because the condition is vacuous in degree one. The determinant is alternating, because transposing columns (or rows for that matter) changes the determinant by minus one. Symmetric bilinear forms are of course not alternating, because transposing the entries leaves it unchanged. A consequence of (13) is that an alternating multilinear form will vanish if two entries are equal. From (12) it follows that there are no nonzero multilinear forms of degree k greater than n the dimension of V . More generally, we can say

Proposition 4.1. Let V be a vector space of dimension n over a ﬁeld F in which 0 6= 2. k ∗ n For k a non-negative integer, then Λ (V ) is a vector space of dimension k .

Proof. Let M1 and M2 be k-linear alternating forms. The vector space structure is deﬁned by

(λ1M1 + λ2M2)(u1, ..., uk) = λ1M1(u1, ..., uk) + λ2M2(u1, ..., uk). (14) That this makes the set of k-linear forms into a vector space is a routine veriﬁcation.

54 Let v1, ..., vn ∈ V be a basis. According to (12), a k-linear form is determined by its values on k-tuples of basis vectors

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

for vij ∈ v1, ..., vn. The constants Mi1,...,ik are sometimes called structur constants. If two indices are duplicated, then Mi1,...,ik = 0 and if two indices are transposed Mi1,...,ik changes only by a minus sign. Thus M is completely determined by the list of scalars

(Mi1,...,ik | i1 < i2 < .... < ik) (15) n which is a list of k 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 matrix. Corollary 4.2. The determinant is the only alternating n-linear form on F n, up to scalar multiplication. Proof. The determinant is a non-trivial element of the one-dimensional vector space n- linear alternating forms, so every other element is a scalar multiple of the determinant.

4.2 The exterior algebra Definition 17. The kth exterior power Λk(V ) is the dual of the vector space of k-linear alternating forms. Elements of Λk(V ) are sometimes called k-vectors. It follows immediately from results in the last section that Λ0(V ) = F , that Λ1(V ) = V k n and that dim(Λ (V )) = k . k Definition 18. If u1, ...uk ∈ V , define the exterior product u1 ∧ .... ∧ uk ∈ Λ (V ) to be the dual vector satisfying

(u1 ∧ .... ∧ uk)(M) = M(u1, ..., uk).

k That u1 ∧ ... ∧ uk is in fact an element of Λ (V ) is immediate from (14).

k An element of Λ (V ) is called decomposable if it equals u1 ∧ ... ∧ uk for some choice of k vectors u1, ..., uk. The following proposition tells us that every k-vector in Λ (V ) is equal to a linear combination of decomposable k-vectors.

Proposition 4.3. Let V be a vector space with basis v1, ..., vn. The set

S := {vi1 ∧ ... ∧ vik | 1 ≤ i1 < ... < ik ≤ n} is a basis for Λk(V ). Proof. It follows from the proof of Proposition 4.1 that the vector space of k-linear alternating forms has a basis {MI } indexed by k-subsets I ∈ {{i1 < ... < ik} | i1, ..., ik ∈

{1, 2, ..., n}}, deﬁned by the condition MI (vi1 , ..., vik ) = 1 and MI (vj1 , ..., vjk ) = 0 for all other k-tuples. The set S is simply the dual basis.

55 2 For example, if V has basis v1, v2, v3, v4 then Λ (V ) has a basis of six 2-vectors v1 ∧ v2, v1 ∧ v3, v1 ∧ v4, v2 ∧ v3, v2 ∧ v4, v3 ∧ v4. Consider the function (called the exterior product)

k k V = V × ... × V → Λ (V ), (u1, ..., uk) 7→ u1 ∧ ... ∧ uk. (16) It is a straightforward consequence of the deﬁnition that (16) is linear in each factor (i.e. it is k-linear) and that transposing two factors introduces a minus sign. Consequently, if ui = uj for some i 6= j, then u1 ∧ ... ∧ uk = 0.

Proposition 4.4. A set of vectors u1, ..., uk ∈ V is linearly independent if and only if k u1 ∧ ... ∧ uk ∈ Λ (V ) is non-zero.

Proof. Suppose that u1, ..., uk ∈ V are linearly dependent. Then it is possible to write one of them as a linear combination of the others X ui = λjuj. j6=i It follows by linearity that X u1 ∧ ... ∧ uk = u1 ∧ ... ∧ ( λjuj) ∧ uk j6=i X = λju1 ∧ ... ∧ uj ∧ ... ∧ uk j6=i = 0 which equals zero because it is a sum of terms containing repeated factors. On the other hand, if u1, ..., uk is linearly independent, then it may be extended to a k basis of V . By Proposition 4.3, this means that u1 ∧ ... ∧ uk is part of a basis for Λ (V ), so in particular it is non-zero. Suppose that T : V → V is linear transformation from V to itself. If M is a k-linear form on V , then the pull-back T ∗M is the k-linear form on V deﬁned by the formula

∗ T M(u1, ..., uk) = M(T (u1), ..., T (uk)). If M is alternating, then T ∗M is alternating because

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

= −M(T (u1), ..., T (uj), ...T (ui), ..., T (uk)) ∗ = −T M(u1, ..., uj, ..., ui, ..., uk). The transpose of T ∗ is the linear map (abusing notation) ΛkT :Λk(V ) → Λk(V ) that sends u1 ∧ ... ∧ uk to T (u1) ∧ ... ∧ T (uk). In the case k = n = dim(V ), we get something familiar (the following proof involves material outside of the prerequisites and will not be tested).

56 Proposition 4.5. If dim(V ) = n and T : V → V is a linear map, then ΛnT :Λn(V ) → Λn(V ) is scalar multiplication by the determinant of T .

n n n n n Proof. The vector space Λ (V ) has dimension n = 1, thus Λ T :Λ (V ) → Λ (V ) must be multiplication by some scalar (it is deﬁned by a 1 × 1-matrix). Choose a basis v1, ..., vn is a basis of V . The linear map T determines an n×n matrix [Ti,j] with entries determined by the formuala n X T (vi) = Ti,jvj. j=1 n By Proposition 4.3, the vector v1 ∧ ... ∧ vn spans Λ (V ), and

n Λ T (v1 ∧ ... ∧ vn) = T (v1) ∧ ... ∧ T (vn) X X = ( T1,j1 vj1 ) ∧ ... ∧ ( Tn,jn vjn )

j1=1 j1=1 n X = T1,j1 ...Tn,jn vj1 ∧ ... ∧ vjn

j1,...,jn=1

To proceed further, observe that the product vj1 ∧ ... ∧ vjn equals zero if any of the indices are repeated and equals ±v1 ∧ ... ∧ vn if they are all distinct. The sign is determined by whether it takes an odd or an even number of transpositions get from the list (j1, ..., jn) to the list (1, 2, ..., n), because a minus sign is introduced for each transposition. Generally, a permutation σ is a list of the numbers {1, ..., n} written without repetition and we deﬁne sign(σ) = (−1)t, where t is how many transpositions it takes to reorder σ to (1, ..., n). With this notation, we have

n n X Λ T (v1 ∧ ... ∧ vn) = T1,j1 ...Tn,jn vj1 ∧ ... ∧ vjn

j1,...,jn=1 X = T1,σ(1)...Tn,σ(n)vσ(1) ∧ ... ∧ vσ(n) σ X = sign(σ)T1,σ(1)...Tn,σ(n)v1 ∧ ... ∧ vn σ

n P so Λ T is scalar multiplication by the scalar σ sign(σ)T1,σ(1)...Tn,σ(n), which is a formula for the determinant of T . Proposition 4.6. There exists a unique bilinear map, called the exterior product,

∧ :Λk(V ) × Λl(V ) → Λk+l(V ), satisfying the property that for decomposable elements, we have

(u1 ∧ ... ∧ uk) ∧ (w1 ∧ ... ∧ wl) = u1 ∧ ... ∧ uk ∧ w1 ∧ ... ∧ wl, (17)

57 Proof. By choosing a basis and applying Proposition 4.3, it is clear that bilinearity com- bined with (17) uniquely determines the map. It remains to show the map exists and is independent of any choice of basis. We do this by supplying a deﬁnition of the exterior product that is independent of any basis. Let b ∈ Λl(V ) be an arbitrary element. Deﬁne a map

k+l ∗ k ∗ ιb :Λ (V ) → Λ (V ) by the rule that for vectors u1, ..., uk ∈ V ,

(ιb(M))(u1, ..., uk) = b(M(u1, ..., uk, −, −, ...−)). k ∗ 0 ∗ Observe that when k = 0, then Λ (V ) = Λ (V ) = F and ιb is simply b. It is easy to check that ιb(M) inherits multi-linearity and the alternating property k l from M so ib is well-defined. Given a ∈ Λ (V ) and b ∈ Λ (V ), we define the exterior product a ∧ b ∈ Λk+l(V ) by the formula 0 ∗ (a ∧ b)(M) = ιaιb(M) ∈ Λ (V ) = F, for all alternating (k + l)-forms M. 1 Example 20. Suppose we have vectors (v1 − v2) ∈ V = Λ (V ) and (v2 ∧ v3 + v1 ∧ v3) ∈ Λ2(V ). Then the product satisfies

(v1 − v2) ∧ (v2 ∧ v3 + v1 ∧ v3) = v1 ∧ (v2 ∧ v3 + v1 ∧ v3) − v2 ∧ (v2 ∧ v3 + v1 ∧ v3)

= v1 ∧ v2 ∧ v3 + v1 ∧ v1 ∧ v3 − v2 ∧ v2 ∧ v3 + v2 ∧ v1 ∧ v3

= v1 ∧ v2 ∧ v3 + v2 ∧ v1 ∧ v3

= v1 ∧ v2 ∧ v3 − v1 ∧ v2 ∧ v3 = 0

An important property of the exterior product is that it is graded commutative. Given two decomposable elements we have

(u1 ∧ ... ∧ uk) ∧ (w1 ∧ ... ∧ wl) = u1 ∧ ... ∧ uk ∧ w1 ∧ ... ∧ wl k = (−1) w1 ∧ u1 ∧ ... ∧ uk ∧ w2 ∧ ... ∧ wl 2k = (−1) w1 ∧ w2 ∧ u1 ∧ ... ∧ uk ∧ w3 ∧ ... ∧ wl = ... kl = (−1) w1 ∧ w2 ∧ .... ∧ wl ∧ u1 ∧ ... ∧ uk

because moving each wj through u1 ∧ ... ∧ uk requires k transpositions. Extending this formula by linearity gives us the third of the following list of properties Proposition 4.7. Let a ∈ Λk(V ), b, b0 ∈ Λl(V ) and c ∈ Λp(V ). Then • a ∧ (b + b0) = a ∧ b + a ∧ b0 (linearity) • (a ∧ b) ∧ c = a ∧ (b ∧ c) (associativity) • a ∧ b = (−1)klb ∧ a (graded commutativity).

58 4.3 The Grassmanian and the Pl¨ucker embedding Let V be a vector space and let k be an integer, 0 ≤ k ≤ dim(V ). The Grassmanian of k-planes in V, denoted Grk(V ), is the set of k-dimensional subspaces of V . In terms of projective geometry, Gr1(V ) is equal to the projective space P (V ), Gr2(V ) is the set of projective lines in P (V ), Gr3(V ) is the set of planes in P (V ), etc.. A k-dimensional subspace U ⊆ V , naturally determines a 1-dimensional subspace

Λk(U) ⊆ Λk(V ).

k k In terms of a basis u1, ..., uk ∈ U, the Λ (U) is spanned by u1 ∧ ... ∧ uk ∈ Λ (V ), which we know is non-zero by Proposition 4.4.

Theorem 4.8. The map

k k Φ: Grk(V ) → P (Λ (V )), Φ(U) = Λ (U)

k is injective so we may identify Grk(V ) the subset of P (Λ (V )) represented by decomposable vectors. The map Φ is called the Pl¨uckerembedding.

Proof. Suppose that U 6= W . Construct a basis of V as follows. Begin with a basis v1, v2, ..., vk of U. Since U 6= W , there exists a vector vk+1 ∈ W \U, such that {v1, ..., vk+1} is linearly independent (since its span has dimension strictly larger than dim(U) = k). Now extend to a basis v1, ..., vn of V . With respect to this basis Φ(U) is spanned by v1 ∧ ... ∧ vk, whereas Φ(W ) is spanned k−1 by a vector a ∧ vk+1, for some a ∈ Λ (V ). Since every term in the expansion of a ∧ vk+1 will include a factor of vk+1, it can not equal v1 ∧ ... ∧ vk, even up to scalar multiplication. We conclude that Φ(U) 6= Φ(W ). It can be shown that the image of the Pl¨ucker embedding is not only a set, it is projective algebraic set in the sense of §3.2. Our next task is to describe this algebraic set in the case k = 2 and use this to study the set of lines in projective space as a geometric object.

4.4 Decomposability of 2-vectors

2 Recall that a 2-vector a ∈ Λ (V ) is called decomposable if it can be factored as a = u1 ∧u2 2 for some vectors u1, u2 ∈ Λ (V ). For example

(3v1 −v2)∧(v1 +v2 +v3) = 3v1 ∧v2 +3v1 ∧v3 −v2 ∧v1 −v2 ∧v3 = 4v1 ∧v2 +3v1 ∧v3 −v2 ∧v3 is decomposable. Now imagine simply being presented with the expression 4v1 ∧v2 +3v1 ∧ v3 − v2 ∧ v3. How can you tell that it factors? Given any decomposable 2-vector, a = u1 ∧ u2, we have

a ∧ a = u1 ∧ u2 ∧ u1 ∧ u2 = 0 because it contains repeated factors. It turns out that the converse of this is also true.

59 Proposition 4.9. Let V be a vector space. A 2-vector a ∈ Λ2(V ) is decomposable if and only if a ∧ a = 0.

Proof. We have already proven the only if direction. Now suppose that a ∧ a = 0. We wish to prove that a is decomposable. We proceed by induction on the dimension of V . Base case: Suppose that V has dimension three (the statement is obvious for dimension less than three, because then all 2-vectors are decomposable). Consider the linear map A : V → Λ3(V ),A(v) = a ∧ v. Since dim(Λ3(V )) = 1, the kernel of A has dimension at least 3 − 1 = 2 by the Rank- Nullity Theorem .... Let v1, v2 ∈ V be two linearly independent vectors such that A(v1) = A(v2) = 0 and extend to a basis v1, v2, v3. By ..., there exist scalars such that

a = λv1 ∧ v2 + µv1 ∧ v3 + νv2 ∧ v3.

By construction a ∧ v1 = 0, so

0 = a ∧ v1

= λv1 ∧ v2 ∧ v1 + µv1 ∧ v3 ∧ v1 + νv2 ∧ v3 ∧ v1

= νv2 ∧ v3 ∧ v1

so we conclude that ν = 0. Similarly, because a ∧ v2 = 0 we conclude that µ = 0. It follows that a = λv1 ∧ v2 = (λv1) ∧ v2 is decomposable. Induction step: Let dim(V ) = n and suppose that the property holds for dimensions 2 less than n. Suppose a ∈ Λ (V ) satisﬁes a∧a = 0. Given an arbitrary basis v1, ..., vn ∈ V , express a as a linear combination X a = ai,jvi ∧ vj 1≤i

0 2 where U = Span{v1, ..., vn−1} ⊂ V , u ∈ U and a ∈ Λ (U). Then

0 = a ∧ a 0 0 = (u ∧ vn + a ) ∧ (u ∧ vn + a ) 0 0 0 0 = u ∧ vn ∧ u ∧ vn + a ∧ u ∧ vn + u ∧ vn ∧ a + a ∧ a 0 0 0 = 2a ∧ u ∧ vn + a ∧ a

60 where in the last step we use the fact that products with repeated factors vanish and graded commutativity of exterior multiplication. Because vn does not appear in the expansion of a0 ∧ u or a0 ∧ a0, they must both vanish separately and we get

a0 ∧ u = 0, a0 ∧ a0 = 0.

0 2 0 Since a ∈ Λ (U) and dim(U) = n−1, it must be decomposable by induction, so a = u1∧u2 for some u1, u2 ∈ U. Subbing in we get the equation

u1 ∧ u2 ∧ u = 0, so by Proposition 4.4, the vectors u1, u2, u are linearly dependent. Consequently

2 a = u ∧ vn + u1 ∧ u2 ∈ Λ (W ). where W is the (at most) 3-dimensional subspace spanned by u, u1, u2, vn, so by the base case must a be decomposable. Example 21. Returning to our previous example, it is immediately clear that a := 4v1 ∧ v2 + 3v1 ∧ v3 − v2 ∧ v3 is decomposable because there are only three basis vectors v1, v2, v3 appearing as factors in its terms, thus every term of the product a ∧ a must contain a repeated factor and thus must vanish, so a ∧ a = 0.

4.5 The Klein Quadric

A special case of Theorem 4.8 when k = 2 identiﬁes Gr2(V ), the set of lines in projective space, with the set of points in P (Λ2(V )) represented by decomposable vectors. Let us consider the implications of our algebraic characterization of decomposability (Proposition 4.9) in various dimensions of V . If dim(V ) = 2, then Λ2(V ) is one-dimensional. In this case P (V ) is itself a projective line so it contains only a single line and P (Λ2(V )) is a single point. 2 3 If dim(V ) = 3, then P (V ) is a projective plane. Since dim(Λ (V )) = 2 = 0, it follows from Proposition 4.9 that every 2-vector is decomposable so Theorem 4.8 identiﬁes the set of lines in P (V ) with the set of points in P (Λ2(V )). Of course, we have already know (§2.6) that the set of lines in a projective plane P (V ) is in one-to-one correspondence with the dual projective plane P (V ∗), so this result implies a natural bijection

P (Λ2(V )) ∼= P (V ∗).

To see where this comes from consider the wedge product

∧ : V × Λ2(V ) → Λ3(V ).

Since the product is bilinear, 2-vectors in Λ2(V ) determine linear maps form V to Λ3(V ). Since Λ3(V ) is one-dimensional, there is an isomorphism Λ3(V ) ∼= F , which is uniquely determined up to a non-zero scalar. Combining these allows us to consider 2-vectors as deﬁning linear maps V → F which are determined up to a non-zero scalar, and this leads to the natural bijection P (Λ2(V )) ∼= P (V ∗).

61 Now consider the case when dim(V ) = 4, so that P (V ) is projective 3-space. Choose 2 a basis v1, ..., vn ∈ V . An arbitrary 2-vector in Λ (V ) has the form

a = x1,2v1 ∧ v2 + x1,3v1 ∧ v3 + x1,4v1 ∧ v4 + x2,3v2 ∧ v3 + x2,4v2 ∧ v4 + x3,4v3 ∧ v4

2 for some scalars xi,j which we treat as homogeneous variables for P (Λ (V )). This gives rise to a formula a ∧ a = B(a, a)v1 ∧ v2 ∧ v3 ∧ v4 where B(a, a) is the quadratic form 2(x1,2x3,4 − x1,3x2,4 + x1,4x2,3). Combing with Propo- sition 4.9 we deduce the following.

Corollary 4.10. The set of lines in a three dimensional projective space P (V ), is in one- to-one correspondence with the points on a quadric X (called the Klein Quadric) lying in the ﬁve dimensional projective space P (Λ2(V )). With respect to the standard basis of 2 Λ (V ) coming from a basis v1, v2, v3, v4 we have X = Z(f) where

f = x1,2x3,4 − x1,3x2,4 + x1,4x2,3

Homework exercise: Prove that for dim(V ) = n ≥ 4, that the set of lines in P (V ) can be identiﬁed with an intersection of quadrics in P (Λ2(V )). In the remainder of this section we study the Klein Quadric Q. Recall from §3.7 that a quadric in complex projective 5-space P (Λ2(V )) is a ﬁlled by families of projective planes. Our next task is to characterize these planes in terms of the the geometry of lines in P (V ).

Proposition 4.11. Let L1 and L2 be two lines in a projective 3-space P (V ). Then L1 2 and L2 intersect if and only if they correspond to points Φ(L1), Φ(L2) ∈ P (Λ (V )) that are joined by a line contained in the Klein Quadric Q.

Proof. The lines L1 and L2 correspond to 2-dimensional subspaces U1,U2 ⊂ V . The lines intersect if and only if U1 ∩ U2 contains an non-zero vector u ∈ U1 ∩ U2. Suppose that the lines do intersect. Then we may choose bases u1, u ∈ U1 and u2, u ∈ 2 U2. The lines are represented by points [u1 ∧ u], [u2 ∧ u] ∈ P (Λ (V )). These determine a line represented by λu1 ∧ u + µu2 ∧ u = (λu1 + µu2) ∧ u which are all decomposable and thus are contained in Q. ∼ Suppose that the lines do not intersect. Then U1 ∩ U2 = {0} and thus by ... U1 ⊕ U2 = V . In particular, there is a basis v1, v2, v3, v4 ∈ V such that span{v1, v2} = U1 and span{v3, v4} = U2. The line containing Φ(L1) and Φ(L2) has points represented by

λv1 ∧ v2 + µv3 ∧ v4, so (λv1 ∧ v2 + µv3 ∧ v4) ∧ (λv1 ∧ v2 + µv3 ∧ v4) = 2λµv1 ∧ v2 ∧ v3 ∧ v4 is non-zero except when either λ or µ is zero, hence does not lie in Q.

62 Proposition 4.12. The set of lines passing through a ﬁxed point X ∈ P (V ) corresponds to a projective plane contained in the Klein Quadric Q ⊆ P (Λ2(V )).

Proof. Let x ∈ V represent the point X ∈ P (V ), and extend to a basis x, v1, v2, v3 ∈ V . Then a projective line L containing X corresponds to span{x, u} for some u = µx+λ1v1 + λ2v2 + λ2v2 with at least one of λi non-zero. Then

Φ(L) = [x ∧ (µx + λ1v1 + λ2v2 + λ2v2)] = [λ1x ∧ v1 + λ2x ∧ v2 + λ3x ∧ v3],

2 which lies in the plane in P (Λ (V ) spanned by the points [x ∧ v1], [x ∧ v2], [x ∧ v3]. Con- versely, any point lying in this plane has form λ1x ∧ v1 + λ2x ∧ v2 + λ3x ∧ v3 and thus corresponds to the line span{x, λ1v1 + λ2v2 + λ3v3}. We call a plane in the Klein Quadric of type described above an α-plane. There is another type of plane as well.

Proposition 4.13. The set of lines lying in a ﬁxed plane P ⊂ P (V ) determines a plane in the Klein quadric Q ⊂ P (Λ2(V )). Such a plane in Q is called a β-plane.

Proof. Let [u1], [u2], [u3] ∈ P be three non-collinear points lying on the plane. An arbitrary 0 0 0 line in P is determined by a pair of points [λu1 + µu2 + νu3] and [λ u1 + µ u2 + ν u3], and maps to the point in Q

0 0 0 [(λu1 + µu2 + νu3) ∧ (λ u1 + µ u2 + ν u3)] 0 0 0 0 0 0 = [(λµ − λ µ)u1 ∧ u2 + (λν − λ ν)u1 ∧ u3 + (µν − µ ν)u2 ∧ u3] contained in the plane determined by u1 ∧ u2, u1 ∧ u3 and u2 ∧ u3. Conversely, points in 2 P (Λ (V )) of the form [λ1u1 ∧ u2 + λ2u1 ∧ u3 + λ3u2 ∧ u3] must be decomposable, because 2 they are contained in Λ (span{u1, u2, u3}) and 2-vectors over a 3-dimensional vector space are always decomposable. It is interesting to note that α-planes and β-planes are related by duality. A point X ⊂ P (V ) corresponds by duality to a plane X0 ∈ P (V ∗). The set of lines in P (V ) containing the point X corresponds via duality to the set of lines contained in X0.

Proposition 4.14. Every plane contained in the Klein quadric is either an α-plane or a β-plane.

Proof. Let P ⊂ Q be a plane in the quadric and choose three non-collinear points in P corresponding to three lines L1,L2,L3 ∈ P (V ). By Proposition 4.11, the intersections Li ∩ Lj are non-empty. There are two cases: Case 1: L1 ∩ L2 ∩ L3 is a point X. Then all Φ(Li) lie in the α-plane associated to X which must equal P . Case 2: L1 ∩ L2 ∩ L3 is empty. Then the pairs of lines intersect three distinct points X,Y,Z ∈ P (V ). These three points can’t be collinear, because that would imply L1 = L2 = L3. Thus X,Y,Z determine a plane containing L1,L2,L3 so P must equal the corresponding β-plane.

63 5 Extra details

Characterization of aﬃne algebraic sets in F 1.

Proposition 5.1. If f(x) is a polynomial of degree n, then Zaff (f) consists of no more 1 than n points. Conversely, if X = {r1, ..., rn} ⊂ F is a set of n points, then there exists a polynomial g(x) of degree n such that X = Z(g(x)).

Proof. For the ﬁrst part, we use induction on n. Base case: If n = 1 then f(x) = ax + b for some scalars a, b ∈ F with a 6= 0. The equation ax + b = 0 has unique solution x = −b/a, so Z(f) = {a/b} is a single point. n n−1 Induction step: Let f(x) = anx + an−1x + ... + a0 with an 6= 0. Suppose that f(x) has at least one root r ∈ F . Then using the Long Division Algorithm we have

f(x) = (x − r)h(x) where h(x) has degree n − 1. Thus f(x) = 0 if and only if x = r or h(x) = 0. By induction, h(x) has at most n − 1 roots, so f(x) has at most n roots. Conversely, note that if

g(x) = (x − r1)(x − r2)...(x − rn) then Z(g(x)) = {r1, ..., rn}. Characterization of algebraic sets in FP 1.

Proposition 5.2. If f(x, y) be a homogeneous polynomial of degree n, then Z(f) is a set with at most n points. Conversely, if X ⊆ FP 1 is a set of n points then there exists a degree n homogeneous polynomial g(x, y) such that X = Z(g).

n n−1 n−1 n Proof. Let f(x, y) = anx + an−1x y + ...a1xy + a0y . Then

[1 : 0] ∈ Z(f) ⇐⇒ f(1, 0) = an = 0.

All remaining points in FP 1 have the form [t : 1], and

n n−1 [t : 1] ∈ Z(f) ⇐⇒ f(t, 1) = ant + an−1t + ... + a0 = 0.

If an 6= 0 then f(t, 1) has at most n roots. If an = 0 then f(t, 1) has no more than n − 1 roots. In either case Z(f) consists of at most n points. n Conversely, suppose that [a1 : b1], ...., [an : bn] are a set of points in FP . Then

f(x, y) = (b1x − a1y)...(bnx − any) vanishes if and only if [x : y] = [ai : bi] for some i = 1, ..., n, so

Z(f) = {[a1 : b1], ...., [an : bn]}.

64 Linear sets. ∗ Let V be a vector space with basis v1, ..., vn ∈ V . The dual basis x1, ..., xn ∈ V are the linear maps xi : V → F for which xj(vi) = δi,j. Equivalently, the dual basis is deﬁned by the property that v = x1(v)v1 + .... + xn(v)vn for all v ∈ V . The general element of V ∗ is a linear combination

f(x1, ..., xn) = a1x1 + .... + anxn.

It follows that V ∗ is equal to the set of homogenous polynomials of degree one for V . Let U ⊂ V be a vector subspace of dimension n. Choose a basis u1, ..., uk ∈ U and ∗ extend to a basis u1, ..., uk, vk+1, ..., vn ∈ V and let x1, ..., xn ∈ V be the dual basis. Then U = Zaff (xk+1, ..., xn) ⊆ V and P (U) = Z(xk+1, ..., xn) ⊆ P (V ). If U1 and U2 are a pair of 2-dimensional subspaces of V , then there exist linear polynomials f1 and f2 such that P (U1) = Z(f1) and P (U2) = Z(f2) and thus

Z(f1f2) = P (U1) ∪ P (U2) is a quadric in P (V ).

6 The hyperbolic plane

The material in this section is not to be tested. Recall that the Riemann sphere

1 CP = C ∪ {∞}

1 is a 2-dimensional sphere. The group GL2(C) of 2 × 2 invertible matrices act on CP by projective transformations. Two matrices determine the same transformation if and only if they are scalar multiples of each other. The real projective line 1 RP = R ∪ {∞} looks like a circle, and sits inside of CP 1 as the equator (or can be understood that way). Picture. 1 1 The subgroup of GL2(R) ⊂ GL2(C) of real matrices sends RP ⊆ CP to itself (i.e. sends the equator to the equator) and acts by projective transformations. If we + restrict to the subgroup GL2 (R) ⊆ GL2(R) with positive determinant, the corresponding transformations of CP 1 also sends the northern hemisphere H to the northern hemisphere. In the spirit of Klein’s Erlangen program, the hyperbolic plane is deﬁned to be H and + hyperbolic geometry is deﬁned to be all concepts left invariant invariant by GL2 (R). Hyperbolic geometry is distinguished from projective geometry in the following ways:

• There is a notion of distance and area

• There is a notion of angle for intersecting lines

65 • Distinct lines to not always intersect (but can intersect at most once).

Hyperbolic geometry is distinguished from Euclidean geometry in the following ways:

• If L is a line and P is a point not on L, then there are many lines passing through P that do not intersect L (multiple parallel lines).

• The sum of angles in a triangle is strictly less than π radians.

There are several diﬀerent models of the hyperbolic plane.

1. The Poincar´ehalf plane model. In this case we pass to aﬃne C, where ∞ lying on the equator, the rest of the equator passing to R ⊂ C, and H passing to the upper half plane of C.

Figure 24: Poincar´ehalf plane model (blue lines and a tiling by 7-gons).

In this model, lines are represented by half circles centred on the real line, or by vertical lines (note that distinct lines intersect at most once). Hyperbolic angles are the same as Euclidean angles in this model.

2. The Poincar´edisk model. We can also pass to aﬃne C by choosing the south pole of CP 1 as ∞ and sending the equator RP 1 to the unit circle and H is sent to the unit disk in C. Hyperbolic angles are the same as Euclidean angles in this model. Lines correspond to circular arcs that intersect the boundary at right angles, or straight lines that bisect the disk. You can see why the angles of a triangle add to less than π radians, because triangles are “bent in” in this geometry.

3. The Klein disk model. In this version, H is represented by the unit disk in R2 and lines a represented by straight Euclidean lines. In this model, the hyperbolic angles are not the same as Euclidean angles, but distance has a nice interpretation. Given points P,Q in the interior of the disk, they

66 Figure 25: Poincar´edisk model (lines and a tiling by triangles with vertices on the boundary).

Figure 26: Klein disk model (lines and tiling by 7-gons and triangles).

67 determine a line that intersects the unit circle in two points A, B. The hyperbolic distance of between P and Q is equal to the logarithm of the cross ratio: 1 d(P,Q) = log(R(A, B; P,Q)). 2

Below, we have an arrangement of lines depicted in the three diﬀerent models:

Figure 27: Three models at once.

Recall that any three distinct points in RP 1 lie in general position

7 Ideas for Projects

• The octonions and the Fano plane

• Projective spaces over ﬁnite ﬁelds and the Fano plane

• Cramer’s paradox

• Cubic curves

• Hopf Fibration

• Incidence geometries and non-Desarguesian planes

68 • Elliptic curves and group laws

• Symmetric bilinear forms in calculus and the Hessian

• The Grassmanian and the Plucker embedding for k > 2.

• Pascal’s Theorem, Brianchon’s Theorem and duality

• Rational quadratic forms