4 Projective space and projective varieties

One of the motivations for projective geometry is to understand perspective. It is well-known phenomenon that parallel lines such as railway tracks ap- pear to meet at a point, called the vanishing point. It was not until the Renaissance that painters were able to give a geometrically correct represen- tation of objects. The painting below, The delivery of keys by Pietro Perug- ino in the Sistine Chapel (source https://en.wikipedia.org/wiki/File: Entrega_de_las_llaves_a_San_Pedro_(Perugino).jpg) is a good exam- ple of the use of perspective, the parallel lines on the ground converge towards a point near the centre of the painting.

Projective geometry gives a nicer theory than affine geometry in many ways, for example, two lines in the projective plane always meet, or more generally two curves defined by degree m and degree n equations, resp., have mn intersection points counted with multiplicity. Projective geometry also has important practical applications, images are combined by applying projective transformations in photo stitching, which is used in panoramic photos, Google Street View and virtual reality.

43 4.1 Basic properties of projective space and projective varieties

The definition of projective space is also motivated by the idea that if a camera is at the origin of the co-ordinate system, all points of a ray through the origin will be mapped to the same point of the image, so points of the image correspond to lines through the origin. Definition. The n-dimensional projective space over a field K, denoted by Pn(K) or by Pn if the field is understood, is the set of equivalence classes n+1 of K \{(0, 0,..., 0)} under the equivalence relation (x0, x1, . . . , xn) ∼ n (λx0, λx1, . . . , λxn) for any λ ∈ K \{0}. (The points of P correspond to lines through the origin in Kn+1.)

n+1 The equivalence class of a point (X0,X1,...,Xn) ∈ K \{(0, 0,..., 0)} is denoted by (X0 : X1 : ... : Xn). X0, X1,... Xn are called homogeneous co-ordinates on Pn.

Example: (1 : 2 : 3), (2 : 4 : 6) and (−1 : −2 : −3) are the same point in P2, this is a peculiarity of projective space that the co-ordinates of a point are not unique. Two ways of looking at projective space

n 1. Let U0 = {(X0 : X1 : ... : Xn) ∈ P | X0 6= 0}.(X0 : X1 : ... : Xn) = (1 : X1/X0 : ... : Xn/X0) in U0 and X1/X0, X2/X0,..., Xn/X0 can take arbitrary values in K, so the points of U0 are in bijection with the points of n n n A . The set P \ U0 = {(X0 : X1 : ... : Xn) ∈ P | X0 = 0} is clearly a copy of Pn−1. Therefore Pn = An ∪ Pn−1 as a set. If n = 1, then

1 X1
U0 = {(X0 : X1) ∈ P | X0 6= 0} = { 1 : | X1 ∈ K,X0 6= 0} = {(1 : x) | x ∈ K}, X0

1 1 so we can identify U0 with A via (X0 : X1) ↔ X1/X0 ∈ A .

1 1 P \ U0 = {(X0 : X1) ∈ P | X0 = 0} = {(0 : X1) | X1 ∈ K \{0}} = {(0 : 1)},

because (0 : X1) = (0 : 1) for any K \{0}. This is the decomposition, P1 = A1 ∪ P0, as P0 consists of just a single point. We can identify A1 with K and denote the single point of P0 by ∞, this way we can identify P1 with K ∪ {∞}.

44 If n = 2, then

2 X1 X2
U0 = {(X0 : X1 : X2) ∈ P | X0 6= 0} = { 1 : : | X1,X2 ∈ K,X0 6= 0} X0 X0 = {(1 : x : y) | x, y ∈ K}

2 2 so we can identify U0 with A via (X0 : X1 : X2) ↔ (X1/X0,X2/X0) ∈ A .

The complement of U0 is

2 2 2 P \U0 = {(X0 : X1 : X2) ∈ P | X0 = 0} = {(0 : X1 : X2) | (X1,X2) ∈ K \{(0, 0)}}, this is a copy of P1. To understand these points, consider the parallel lines x+y = 0 and x+y−2 = 2 0 in A = U0. As x = X1/X0 and y = X2/X0, we can rewrite the equations X X X X in terms of homogeneous co-ordinates as 1 + 2 = 0 and 1 + 2 −2 = 0. X0 X0 X0 X0 After multiplying by X0 we get X1 + X2 = 0 and X1 + X2 − 2X0 = 0. The solutions of this system of linear equations are X0 = 0, X1 = −X2, which correspond to the point (0 : 1 : −1) ∈ P1. Therefore these parallel lines in 2 2 A = U0 intersect in P \ U0. Any other line parallel to them also contains (0 : 1 : −1).

y

0:1:-1 4

H L

2 x+y-2=0

x -4 -2 2 4

-2 x+y=0

-4

Similarly, all lines with direction vector (X1,X2) intersect at (0 : X1 : X2) ∈ 2 P \ U0.

45 n n−1 This is true in general for arbitrary n, the points of P \U0 = P correspond n to directions of lines in A or equivalence classes of parallel lines. (0 : X1 : ... : Xn) is the point where all the lines with direction vector (X1,X2,...,Xn) in An meet. (The single point of P0 corresponds to the only line in A1.) For this reason, these points are often called points at infinity, but this does not mean that they are intrinsically different, the disctinction depends on the choice of co-ordinates. The youtube video https://www.youtube.com/watch?v=q3turHmOWq4 (also linked to from the Animations and videos section of the course website) contains a good explanation of the projective plane.

2. The other approach is is to consider the sets

n Ui = {(X0 : X1 : ... : Xn) ∈ P | Xi 6= 0}, 0 ≤ i ≤ n. Each one is a copy of An and their union is Pn, since there is no point (0 : 0 : ... : 0). These n + 1 copies are glued together by identifying their points via rational maps ϕij : Ui 99K Uj so that P ∈ Ui is the same n point of P as ϕij(P ) ∈ Uj.

For example, if n = 1, let t = X1/X0 be the co-ordinate on U0, and u = X0/X1 the co-ordinate on U1. We have the rational maps ϕ01 : U0 99K U1, t 7→ 1/t, and ϕ10 : U1 99K U0, u 7→ 1/u. A point t ∈ U0 is the same point in 1 P as ϕ01(t) ∈ U1, whenever ϕ01(t) is defined and similarly, a point u ∈ U1 is 1 the same point in P as ϕ10(u) ∈ U0, whenever ϕ10(u) is defined.

If n = 2, then let x = X1/X0, y = X2/X0 be the co-ordinates on U0, u = X0/X1, v = X2/X1 the co-ordinates on U1 and s = X0/X2, t = X1/X2 the co-ordinates on U2. Now we have u = 1/x and v = y/x, which give the rational map ϕ01 : U0 99K U1,(x, y) 7→ (1/x, y/x), so that (x, y) ∈ U0 2 and ϕ01(x, y) ∈ U1 are the same point in P whenever the latter is defined. The point (2 : 3 : 4) ∈ P2 is the same as (1 : 3/2 : 2), (2/3 : 1 : 4/3) or (1/2 : 3/4 : 1), therefore (3/2, 2) ∈ U0, (2/3, 4/3) ∈ U1 and (1/2, 3/4) ∈ U2 correspond to the same point of P2 and they get identified in the gluing process.

n In general, if F ∈ K[X0,X1,...,Xn] and P = (X0 : X1 : ... : Xn) ∈ P , F (P ) cannot be defined, since F (λX0, λX1, . . . , λXn) will give different values for different values of λ ∈ K \{0}. However, if F is homogeneous of degree d, d then F (λX0, λX1, . . . , λXn) = λ F (X0,X1,...,Xn), so we can tell whether F (P ) = 0 or not, and this is enough to define projective algebraic varieties. Definition. A polynomial is called homegeneous if and only if all of its terms have the same degree.

46 E. g., x3 + 2y2z − 3xz2 is homogeneous of degree 3, while x3 − y2 is not homogeneous.

Definition. An ideal I/K[X0,X1,...,Xn] is called homogeneous if and only if it can be generated by homogeneous elements.( Warning: This does not mean that all elements of the ideal are homogeneous polynomials.)

Definition. Let I/K[X0,X1,...,Xn] be a homogeneous ideal. The projective algebraic variety defined by I is the set

n V(I) = {(X0 : X1 : ... : Xn) ∈ P | F (X0,X1,...,Xn) = 0, ∀F ∈ I,F homogeneous}.

Two ways of looking at projective varieties 2 2 1. Let’s consider the variety V = V(hX1X2 − X0 i) ⊂ P . If X0 6= 0, we can 2 2 divide X1X2 − X0 = 0 by X0 to get (X1/X0)(X2/X0) − 1 = 0, which can be written as xy − 1 = 0 in terms of the affine co-ordinates x = X1/X0 and 2 y = X2/X0 on U0 = A , so V0 = V ∩U0 is a hyperbola. The points V \V0 are 2 the points of V with X0 = 0. By substituting X0 = 0 into X1X2 − X0 = 0 we obtain X1X2 = 0, therefore the points at infinity are (0 : 1 : 0), which corresponds to lines parallel to the x-axis, and (0 : 0 : 1), which corresponds to lines parallel to the y-axis. The asymptotes of the hyperbola pass through these points at infinity. V consists of the hyperbola with two points at infinity corresponding to the asymptotes.

y 4 0:0:1

H L 2

xy-1=0

x -4 -2 2 4 0:1:0

-2 H L

-4

2 Starting with the equation xy − 1 = 0 we can recover X1X2 − X0 = 0 by substituting x = X1/X0 and y = X2/X0 into xy − 1 = 0 and multiplying it 2 by X0 .

47 2 2 Let’s now consider the variety V = V(hX1 − X0X2i) ⊂ P . If X0 6= 0, we can 2 2 2 divide X1 − X0X2 = 0 by X0 to get (X1/X0) − (X2/X0) = 0, which can 2 be written as x − y = 0 in terms of the affine co-ordinates x = X1/X0 and 2 y = X2/X0 on U0 = A , so V0 = V ∩ U0 is a parabola. The points V \ V0 are 2 the points of V with X0 = 0. By substituting X0 = 0 into X1 − X0X2 = 0 2 we obtain X1 = 0, therefore the only points at infinity is (0 : 0 : 1), which corresponds to lines parallel to the y-axis. Unlike the hyperbola, the parabola has no asympotes, but for large x and y, its tangent direction approaches the direction of the y-axis.

y

(0:0:1)

8

6

4

2

x2-y=0

x -3 -2 -1 1 2 3

2 2 Starting with the equation x − y = 0 we can recover X1 − X0X2 = 0 by 2 substituting x = X1/X0 and y = X2/X0 into x − y = 0 and multiplying it 2 by X0 .

Given a homogeneous polynomial F ∈ K[X0,X1,...,Xn],

F [X0,X1,...,Xn] deg F = F [1, x1, x2, . . . , xn] X0 where xi = Xi/X0, 1 ≤ i ≤ n, is called the dehomogenisation of F with respect to X0. If we dehomogenise all homogeneous elements of an homo- geneous ideal J/K[X0,X1,...,Xn], we obtain an ideal defining the affine n algebraic variety V0 = V ∩ U0 ⊆ A , called the affine piece X0 6= 0 of V = V(J) ⊆ Pn.

If we start with a polynomial f(x1, x2, . . . , xn),

deg f X0 f(X1/X0,X2/X0,...,Xn/X0) ∈ K[X0,X1,...,Xn]

48 is a homogeneous polynomial, the homogenisation of f. Homogenisation can also be done at the level of ideals, the projective algebraic variety in Pn defined by the ideal generated by the homogenisation of the elements of an n ideal J/K[x1, x2, . . . , xn] is called the projective closure of V(J) ⊆ A . The affine piece X0 6= 0 of the projective closure is exactly V(J), while the points at infinity correspond to asymptotic directions of V(J) as we have seen in the example. Homogenisation and dehomogenisation are one-sided inverses of each other. Homogenisation followed by dehomogenisation always yields the same poly- nomial, and similarly for varieties, taking the projective closure of an affine variety and then the affine piece X0 6= 0 gives the original variety. Dehomogenising and then homogenising a homogeneous polynomial will give the polynomial divided by the highest power of X0 dividing it. (For example, the dehomogenisation of X0X1 is x1, whose homogenisation is just X1.) For varieties this means that taking projective closure of the affine piece X0 6= 0 of a projective algebraic variety will give the union of irreducible components of the original variety not contained in the hyperplane X0 = 0. (Irreducibility for projective varieties will be defined later, but it is completely analogous to the affine case.) If a projective variety V has no irreducible components contained in the hyperplane X0 = 0, then V is the projective closure the affine piece V0 = V ∩ U0 and the points of V correspond to the points of V0 and the asympotic directions of V0.

2. The other approach is to consider all the affine pieces Vi = V ∩ Ui, 0 ≤ i ≤ n, of a projective variety V . These are affine varieties and they are glued together by identifying the points via the rational maps ϕij : Ui 99K Uj defined previously. For example, by homogenising the equation y2 −x3 −x2 = 2 2 3 2 0 of the nodal cubic in A , we obtain X0X2 −X1 −X0X1 = 0. The projective curve defined by this equation is the projective closure V , whose affine piece V0 = V ∩ U0 is original affine nodal cubic curve. Substituting X0 = 0 into 3 the equation gives X1 = 0, therefore the only point with X0 = 0 is (0 : 0 : 1), the asymptotic direction of the y-axis. x and y can be expressed as x = X1/X0, y = X2/X0 in terms of the homo- geneous co-ordinates. The affine co-ordinates on the other affine pieces are u = X0/X1, v = X2/X1 on U1 and s = X0/X2, t = X1/X2 on U2. Deho- 2 3 2 2 mogenising X0X2 −X1 −X0X1 = 0 with respect to X1 gives uv −1−u = 0, 3 2 dehomogenising it with respect to X2 gives s − t − st = 0. Therefore the projective curve has the affine pieces shown below.

49 y2 − x3 − x2 = 0 uv2 − 1 − u = 0 s − t3 − st2 = 0

y v t 10 10 10

5 5 5

x u s -10 -5 5 10 -10 -5 5 10 -10 -5 5 10

-5 -5 -5

-10 -10 -10

They are glued together by the maps ϕ01 : U0 99K U1,(x, y) 7→ (1/x, y/x) and ϕ02 : U0 99K U2,(x, y) 7→ (1/y, x/y).

Definition. Let F ∈ K[X0,X1,...,Xn] be a polynomial. The degree i ho- mogeneous part of F , denoted by F[i], is the sum of all the terms of degree i in F . If i < 0 or i > deg F , F[i] is defined to be 0.

It is clear from this definition that F = F[0] + F[1] + ... + F[deg F ]. 2 Example: Let F = X0+X0X2−X1 +X0X1X2 ∈ K[X0,X1,X2], then F[0] = 0, 2 F[1] = X0, F[2] = X0X2 − X1 and F[3] = X0X1X2.

Lemma 4.1 The ideal I/K[X0,X1,...,Xn] is homogeneous if and only if for any F ∈ I, all the homogeneous parts of F are also elements of I.

Proof. Assume that I is homogeneous, let F1, F2,..., Fr be a set of ho- r P mogeneous generators for I. Let F ∈ I, then F = FiGi for some i=1 r P Gi ∈ K[X0,X1,...,Xn], 1 ≤ i ≤ r. Then F[d] = FiG[d−deg Fi] ∈ I for i=1 every d, 0 ≤ d ≤ deg F , so all the homogeneous parts of F are also elements of I.

Assume now that F ∈ I implies that F[0], F[1],..., F[deg F ] are also in I. Let F1, F2,..., Fr be an arbitrary set of generators for I. We claim that the set deg F P i {Fi[j] | 1 ≤ i ≤ r, 0 ≤ j ≤ deg Fi} generates I. On one hand Fi = Fi[j] is j=0 contained in the ideal generated by this set for each i, 1 ≤ i ≤ r, so this ideal contains I. On the other hand, each of the generators is in I, so the ideal generated by them is a subset of I. Therefore the ideal generated by this set is exactly I and since these generators are homogeneous, I is a homogeneous ideal. Proposition 4.2 (Cf. Proposition 1.3)

50 (i) Let V1 = V(I1), V2 = V(I2), . . . , Vk = V(Ik) be projective algebraic varieties in Pn. Then

V1 ∪ V2 ∪ ... ∪ Vk = V(I1 ∩ I2 ∩ ... ∩ Ik) = V(I1I2 ...Ik) is also a projective algebraic variety. n (ii) Let Vα = V(Iα), α ∈ A be projective algebraic varieties in P . Then \ X
Vα = V Iα α∈A α∈A is also a projective algebraic variety. Proof. Just imitate the proof of Proposition 1.3 (i) and (ii). Warning: There is no direct analogue of Proposition 1.3 (iii) for projective varieties because Pm × Pn is very different from Pm+n. Definition. The homogeneous ideal of a set Z ⊆ Pn is the ideal I(Z) / K[X0,X1,...,Xn] generated by the set

{F ∈ K[X0,X1,...,Xn] | F homogeneous, F (X0,...,Xn) = 0 ∀(X0 : ... : Xn) ∈ Z}.

Theorem 4.3 (Projective Nullstellensatz, cf. Theorem 1.7) Let K be an algebraically closed field and let J/K[X0,X1,...,Xn] be a homogeneous ideal. √ (i) V(J) = ∅ if and√ only if √J = K[X0,X1,...,Xn] or J = hX0,X1,...,Xni. (ii) I(V(J)) = J unless J = hX0,X1,...,Xni.

Idea of proof. For any homogeneous ideal J/K[X0,X1,...,Xn] we can consider the projective algebraic variety V(J) defined by in Pn and also the affine algebraic variety defined by J in An+1, called affine cone on V(J). This gives an almost bijective correspondence between projective algebraic varieties in Pn and certain affine algebraic varieties in An+1, the only failure of bijectivity is that K[X0,X1,...,Xn] and hX0,X1,...,Xni both define the empty set as a projective variety. Definition. A projective algebraic variety V is reducible if and only if it can be written as V = V1 ∪V2, where V1, V2 are also projective algebraic varieties, V1 6= V 6= V2. If V is not reducible, it is called irreducible. Proposition 4.4 (Cf. Theorem 1.8) Every projective algebraic variety V can be decomposed into a union V = V1 ∪ V2 ∪ ... ∪ Vk such that every Vi, 1 ≤ i ≤ k, is an irreducible projective algebraic variety and Vi 6⊆ Vj for i 6= j. The decomposition is unique up to the ordering of the components. The Vi, 1 ≤ i ≤ k, are called the irreducible components of V .

51 Proof. Completely analogous to the proof of Theorem 1.8.

Lemma 4.5 A homogeneous ideal I/K[X0,X1,...,Xn] is prime if and only if for any homogeneous polynomials F , G ∈ K[X0,X1,...,Xn], FG ∈ I implies F ∈ I or G ∈ I. Proof. If I is prime then FG ∈ I implies F ∈ I or G ∈ I for all F,G ∈ K[X0,X1,...,Xn] by the definition of a prime ideal. Let’s assume I is a homogeneous ideal which is not prime. Then there exist polynomials P/∈ I, Q/∈ I such that PQ ∈ I. Let d ≥ 0 be the minimal integer such that P[d] ∈/ I. There exists such a d since if P[j] ∈ I for every deg P P j ≥ 0, then P = P[j] ∈ I, too. Similarly, let e ≥ 0 be the minimal integer j=0 d+e P such that Q[e] ∈/ I. Then (PQ)[d+e] = P[j]Q[d+e−j]. If j < d, then P[j] ∈ I, j=0 so P[j]Q[d+e−j] ∈ I, too. If j > d, then then Q[d+e−j] ∈ I as d + e − j < e, so P[j]Q[d+e−j] ∈ I, too. Therefore all the terms in the sum except for P[d]Q[e] are in I,(PQ)[d+e] ∈ I by Lemma 4.1 since PQ ∈ I and I is homogeneous. Hence P[d]Q[e] ∈ I, too, but P[d] ∈/ I and Q[e] ∈/ I. Therefore F = P[d] and G = Q[e] are homogeneous polynomials with the property that F,G/∈ I but FG ∈ I. Proposition 4.6 (Cf. Proposition 1.9) A projective algebraic variety V is irreducible if and only if I(V ) is prime. Proof. Imitate the proof of Proposition 1.9 and use Lemma 4.5.

52 4.2 Tangent spaces, dimension and singularities

There are two possible approaches. 1. One can define tangent lines and imitate the development of the affine theory. The line through the points P 6= Q consists of the points λP + µQ, where (λ : µ) ∈ P1, and this line is tangent to V at P if and only if for any homogeneous polynomial F ∈ I(V ), F (λP + µQ) contains no degree 1 term in µ. 2. Let V ⊆ Pn be an irreducible projective algebraic variety, and let P ∈ V . n Let (X0 : X1 : X2 : ...,Xn) be homogeneous co-ordinates on P . Choose i such that P is not contained in the hyperplane Xi = 0. Let Vi = {(X0 : X1 : ... : Xn) ∈ V | Xi 6= 0} be the affine piece Xi 6= 0 of V . The tangent space TP V is defined as the projective closure of TP Vi, the local dimension of V at P is the local dimension of Vi at P , and the P is a singular point of V if and only if it is a singular point of Vi. (It needs to be proved that these definitions are independent of the choice of i, which can be done by an improved version of Theorem 3.5.) Tangent spaces, dimension and singular points of projective varieties can be calculated by using the Jacobian matrix in the same way as in the affine case. Example: Find the singular points, if any, of the curve C defined by the equation Y 2Z − X3 − X2Z = 0 in P2(C). 2 3 2 2 Let F = Y Z − X − X Z, then FX = 3X − 2XZ, FY = 2YZ and FZ = Y 2 − X2. The rank of the Jacobian J is 0 where all three partial derivatives vanish, and 1 elsewhere.

The points P ∈ C where rank JP = 0 are the solutions of F = FX = FY = FZ = 0. FY = 0 implies Y = 0 or Z = 0. If Y = 0, then FZ = 0 implies X = 0, and we get the point (0 : 0 : 1), which is indeed on C. (In projective space, (0 : 0 : 1) = (0 : 0 : Z) for any Z 6= 0.) If Z = 0, then FX = 0 implies 2 X = 0, and then FZ = 0 implies Y = 0, but (0 : 0 : 0) is not a point in P .

Therefore rank J = 0 and dim TP C = 2 at (0 : 0 : 1), and rank J = 1 and dim TP C = 1 at all other points of C. Hence dim C = 1 and (0 : 0 : 1) is the only singular point of C. If we take the affine piece Z 6= 0 of C, we get the familiar nodal cubic curve y2 − x3 − x2 = 0, which has a singular point at (0, 0), corresponding the singular point of the projective curve.

53