1. Complex Projective Space the N-Dimensional Complex Projective Space Pn Over C Is the Space Cn+1 \{0} Modulo the Relation ∼ Deﬁned Below

Home , Complex projective space

1. Complex projective Space The n-dimensional complex projective space Pn over C is the space Cn+1 \{0} modulo the relation ∼ deﬁned below. Two elements x and y in Cn+1 \{0} are said to be equivalent if there exists λ ∈ C \{0} so that x = λy. If x and y are equivalent, we write x ∼ y. The equivalent class of n+1 n+1 n an element x = (x0, ··· , xn) of C \{0} is denoted by [x] = (x0 : ··· : xn). Let π : C \{0} → P be the quotient map. We equip Pn with the quotient topology, i.e. U ⊆ Pn is open if and only if π−1(U) is open in Cn+1 \{0}. Since π is continuous and Cn+1 \{0} is connected (it is path connected and hence connected), Pn = π(Cn+1 \{0}) is connected. (Continuous functions send connected sets to connected sets). The unit sphere n 2n+1 n X 2 S = {(x0, ··· , xn) ∈ C : |xi| = 1} i=0 is contained in Cn+1 \{0}. Let π0 be the restriction of π to S2n+1. Then π0 : S2n+1 → Pn is a surjective continuous map. Since S2n+1 is closed and bounded in Cn, it is compact by Heine-Borel Theorem. Since continuous functions map compact sets to compact sets and S2n+1 is compact, Pn = π0(S2n+1) is compact. We conclude that Proposition 1.1. The n-dimensional complex projective space Pn is a compact and connected space.

n For each 0 ≤ i ≤ n, let Ui be the set of all points (x0 : ··· : xn) in P with the property that there exists a representative (x0, ··· , xn) of (x0 : ··· : xn) such that xi 6= 0. Let us check that this 0 0 set is well-defined. Let (x0, ··· , xn) be another representative of (x0 : ··· : xn). Then there exists a 0 0 nonzero complex number λ in C so that xj = λxi for 0 ≤ j ≤ n. Since xi 6= 0 and λ 6= 0, xi 6= 0. (C is a field.) Therefore Ui is well-defined. Let us show that Ui is open. n+1 Let 0 ≤ i ≤ n. We define πi : C → C by (x0, ··· , xn) to xi. Then πi is smooth for any 0 n+1 0 ≤ i ≤ n. Let Ui be the subset of C consisting of vectors (x0, ··· , xn) so that xi 6= 0. Then 0 −1 0 Ui = πi (C \{0}) and hence Ui is open by the continuity of πi and the openess of C \{0}. Since −1 0 n+1 π (Ui) = Ui ∩ (C \{0}), n Sn n Ui is open in P . One can easily see that i=0 Ui = Pk . Hence {U0, ··· ,Un} forms an open cover for Pn. For each 0 ≤ i ≤ n, we define n x0 xn ϕi : Ui → C , (x0 : ··· : xn) = , ··· , . xi xi

This map is well-deﬁned (left to the reader) and is a bijection. Let us prove that ϕi is a homeomorphism. −1 n To show that ϕi is continuous, we show that ϕi (V ) is open in Ui for any open set V of C . Since n −1 −1 Ui is open in P , to show that ϕi (V ) is open in Ui, we only need to show that ϕi (V ) is open in n −1 n −1 −1 n+1 0 P . To show that ϕi (V ) is open in P , we show that π (ϕi (V )) is open in C \{0}. Let Ui 0 n be as above. The composition ϕi ◦ π : Ui → C is given by x0 xn (ϕi ◦ π)(x0, ··· , xn) = , ··· , . xi xi 0 n −1 0 It is easy for us to see that ϕi◦π : Ui → C is continuous. Hence (ϕi◦π) (V ) is open in Ui and hence n+1 0 n+1 −1 −1 −1 open in C \{0} (we use the fact that Ui is open in C \{0}.) Since π (ϕi (V )) = (ϕi ◦π) (V ) n+1 −1 n is open in C \{0}, ϕi (V ) is open in P . n Let us prove that ϕi is an open mapping, i.e. ϕi(W ) is open in C for any open set W of Ui. Since n n 0 −1 W is open in Ui and Ui is an open subset of P ,W is also open in P . Therefore W = π (W ) is n+1 n+1 0 open in C \{0} and hence open in C . We only need to show that ϕi(π(W )) is open. In fact, 0 n we can show that ϕi ◦ π : Ui → C is an open mapping. 1 2

n+1 Since open sets of the form I0 × I1 × · · · × In generates the topology on C , where I0, ··· ,In 0 0 are open subsets of C, open sets of the form I0 × I1 × · · · × In ∩ Ui generates the topology of Ui . 0 Hence open sets of the form I0 × I1 × · · · × In with Ii open in C \{0} generates the topology of Ui . 0 More precisely, W is a union of open sets of the form I0 × I1 × · · · × In where Ii is open in C \{0}.

Lemma 1.1. Let f : X → Y be any function. Suppose {Aα : α ∈ Λ} is a family of subsets of X. Then ! [ [ f Aα = f(Aα). α∈Λ α∈Λ n Since any union of open sets is open, if we can show that (ϕi ◦π)(I0 ×I1 ×· · ·×In) is open in C for 0 any open set of the form I0 ×I1 ×· · ·×In with Ii open in C\{0}, Lemma 1.1 implies that (ϕi ◦π)(W ) 0 −1 is open for any open subset W of Ui so that W = π (W ). Let Jk = {xk/xi : xi ∈ Ii, xk ∈ Ik} for any k 6= i. Then

(ϕi ◦ π)(I0 × · · · × In) = J1 × · · · × Ji−1 × Ji+1 × · · · × Jn.

If we can show that all Jk are open in C for any k 6= i, by the fact that the product of open sets is n n open, J1 × · · · × Ji−1 × Ji+1 × · · · × Jn is open in C and hence (ϕi ◦ π)(I0 × · · · × In) is open in C . For each µ ∈ Ii, we see that [ [ −1 Jk = {xk/µ : xk ∈ Ik} = µ Ik.

µ∈Ii µ∈Ii −1 Since Ik is open in C, µ Ik is open for any µ ∈ Ii. Since any union of open sets is open, Jk is open n in C. We complete the proof of our assertion. We conclude that ϕi : Ui → C is a homeomorphism for all 0 ≤ i ≤ n. Let 0 ≤ j < i ≤ n. Then n ϕi(Ui ∩ Uj) = {(z1, ··· , zn) ∈ C : zj 6= 0}.

Hence ϕi(Ui ∩ Uj) is open. The transition functions are given by −1 n (ϕj ◦ ϕi ): ϕi(Ui ∩ Uj) → C −1 z1 zj−1 zj+1 zi 1 zi+1 zn (ϕj ◦ ϕi )(z1, ··· , zn) = , ··· , , , ··· , , , , ··· , zj zj zj zj zj zj zj −1 Hence ϕj ◦ ϕi is biholomorphic on ϕi(Ui ∩ Uj). Before proving Pn is Hausdorff, let us recall some basic facts from topology. Lemma 1.2. Let X be a Hausdorff space and K be a compact subset of X. Then K is closed. Proof. To show that Y is closed, we show that X \ K is open. Let us prove that every point of X \ K is an interior point of X \ K. Let x ∈ X \ K. For any y ∈ K, x 6= y. Since X is Hausdorff, we can find an open neighborhood Uy of x and an open neighborhood Vy of y such that Uy ∩ Vy = ∅. Then {Vy : y ∈ K} forms an open cover for K. Since K is compact, we can find y1, ··· , yk so that Tk {Vyi : 1 ≤ i ≤ k} covers K. Take U = i=1 Uyk . Then U is an open neighorhood of x. Let us show that U is contained in X \ K.

Let z ∈ U. If z ∈ K, then z ∈ Vyi for some 1 ≤ i ≤ k. Then z ∈ Vyi ∩ U ⊆ Vyi ∩ Uyi = ∅ which is impossible. Therefore z 6∈ K and hence z ∈ X \ K. We see that U ⊆ X \ K. We ﬁnd that x is an interior point of X \ K. Corollary 1.1. Let X be a compact space and Y be a Hausdorﬀ space. Suppose f : X → Y is a bijective continuous function. Then f is a homeomorphism. Proof. To show that f is a homeomorphism, we only need to prove that f is a closed mapping. Let A be any closed subset of X. Since X is compact and A is a closed subset of X,A is compact. Since f is continuous and A is compact, f(A) is compact. Since Y is closed and f(A) is compact, f(A) is a closed subset of Y. 3

Definition 1.1. A topological space X is normal if given any disjoint closed subsets E and F, there exist open neighborhood U of E and V of F such that U ∩ V = ∅. Proposition 1.2. Any compact Hausdorff space is normal. Proof. Let X be a compact Hausdorff space. Let B and K be disjoint closed subset of X. Since X is compact, both B and K are also compact. Let b ∈ B. By the proof of Lemma 1.2 and the Sk compactness of K, we can choose an open neighborhood V = i=1 Vyi of K for some y1, ··· , yk ∈ K and an open neighborhood Ub of b such that Ub ∩ V = ∅. Since B is compact and {Ub : b ∈ B} forms an open cover for B, there exist b1, ··· , bl ∈ B so that {Ubi : 1 ≤ i ≤ l} forms an open cover for B. Sl Let U = i=1 Ubi . Then U is an open neighborhood of B. Claim U ∩ V = ∅. Let z ∈ U ∩ V. Then z ∈ V and z ∈ Ubi for some i. Hence z ∈ V ∩ Ubi = ∅ which is not possible. Proposition 1.3. Let X and Y be compact spaces. Then the product space X ×Y is also compact.

To prove that Pn is Hausdorff, we show that Pn is homeomorphic to S2n+1/S1. We will show that S2n+1/S1 is a Hausdorff space. Define an action of S1 on S2n+1 by 1 2n+1 2n+1 S × S → S , (λ, x) 7→ λx. Let us show that this action is continuous. Let α : C × Cn → Cn be the function (λ, x) 7→ λx. Then α is continuous. The action of S1 on S2n+1 is the restriction of α to S1 × S2n+1; hence it is continuous. Two points x and y of S2n+1 are equivalent if there exists λ ∈ S1 so that x = λy. The quotient space of S2n+1 modulo this relation is denoted by S2n+1/S1. The equivalent class of 2n+1 2n+1 2n+1 1 1 x ∈ S modulo this relation is denoted by [x]S . The quotient map S → S /S is denoted by q. Definition 1.2. A group with a topology is called a topological group if the function G × G → G, (a, b) → ab−1 is continuous. Here G × G is equipped with the product topology.

Example 1.1. C∗ = {z ∈ C : z 6= 0} with the subspace topology induced from C is a commutative topological group called a noncompact (one dimensional) complex torus. The subset S1 of C∗ consisting of complex numbers z so that |z| = 1 forms a compact subgroup of C. To show that S2n+1/S1 is Hausdorff, we need the following Lemma. Lemma 1.3. Let X be a compact Hausdorff space and G be a compact topological group. Suppose G acts on X continuously, i.e. the function m : G × X → X, (g, x) 7→ gx is continuous. Then (1) the quotient map π : X → X/G is an open mapping. (2) the quotient map π : X → X/G is a closed mapping. (3) the orbit space X/G is Haudorff. Proof. At first, let us prove that the quotient map π : X → X/G is an open mapping. To show that π is an open mapping, we need to show that π(U) is open in X/G for any open set U of X. To show that π(U) is open in X/G, we need to show that π−1(π(U)) is open in X. Claim that [ (1.1) π−1(π(U)) = g(U). g∈G Since g : X → X is a homeomorphism for any g ∈ G, g(U) is open in X for any open subset U of X. If the above equation is true, then π−1(π(U)) is open (any union of open subsets of X is open). Let y ∈ π−1(π(U)), then π(y) ∈ π(U) and hence π(y) = π(z) for some z ∈ U. Therefore y = gz ∈ g(U) for S −1 S S some g ∈ G. Hence y ∈ g∈G g(U). Hence π (π(U)) ⊆ g∈G g(U). Conversely, if y ∈ g∈G g(U), 4 then y ∈ g(U) for some g ∈ G. Hence y = gz for some z ∈ U. Therefore π(y) = π(z) ∈ π(U) which −1 S −1 implies that y ∈ π (π(U)). Hence g∈G g(U) ⊆ π (π(U)). We conclude that (1.1) holds. Since X is compact, any closed subset of X is also compact. Let A be a closed subset of X. Then A is compact. Since any product of compact spaces is compact, G × A is compact. Since S G(A) = m(G, A) and m is continuous, G(A) = g∈G g(A) is compact. To show that π is a closed mapping, we show that π−1(π(A)) is closed in X. In fact, π−1(π(A)) = G(A) is a compact subset of a Hausdorff space X, it is closed. The equivalence class of a point x ∈ X is the orbit [x] = Gx = {y ∈ X : y = gx, g ∈ G}. Since G × X → X is continuous and G is compact, the orbit Gx is compact. If [x] 6= [y], then Gx and Gy are disjoint compact subsets of X. Since X is Hausdorff, we can find disjoint open sets U and V of X so that Gx ⊆ U and Gy ⊆ V. Since V is a closed subset of X and π is a closed mapping, π(V ) is closed in X/G. Let U 0 = (X/G) \ π(V ). Then U 0 is an open subset of X/G. Since π is an open mapping, V 0 = π(V ) is open in X/G. We obtain disjoint open sets {U 0,V 0}. Since Gy ⊆ V, [y] ∈ π(V ) = V 0. Let us show that [x] ∈ U 0. Since X is compact and V is a closed subset of X, V is compact. Since Gx is compact and V is compact, we may choose disjoint open sets U and W so that Gx ⊆ U and V ⊆ W. Since U ∩ W = ∅, Gx ∩ W = ∅. Hence [x] 6∈ π(W ) which implies that [x] 6∈ π(V ), i.e. [x] ∈ U 0 = (X/G) \ π(V ). We find {U 0,V 0} disjoint open subsets of X/G so that [x] ∈ U 0 and [y] ∈ V 0 for any [x], [y] in X/G with [x] 6= [y]. Therefore X/G is a Hausdorff space.

This lemma implies that S2n+1/S1 is a (compact) Hausdorﬀ space. Lemma 1.4. The complex projective space Pn is homeomorphic to S2n+1/S1. z Proof. Let r : n+1 \{0} → S2n+1 be the function r(z) = for z ∈ n+1 \{0}. Then r is C kzk C continuous. Deﬁne f : Pn → S2n+1/S1 by

f([z]) = [r(z)] 1 S where z is a representative of [z]. Let us check that this map is well-deﬁned. Suppose z and z0 are equivalent. Then z = µz0 for some nonzero complex number µ. Then z µz0 r(z) = = = λr(z0) kzk |µ|kz0k 0 1 1 for λ = µ/|µ|. In other words, [r(z)]S = [r(z )]S . 0 1 1 If [r(z)]S = [r(z )]S , then z z0 λkzk = λ =⇒ z = z0. kzk kz0k kz0k 0 2n+1 1 1 Hence [z] = [z ]. This shows that f is injective. Let [x]S be a point in S /S . Choose a repre- 2n+1 n+1 1 1 1 sentative x of [x]S . Then x ∈ S ⊆ C \{0} and r(x) = x. Then f([x]) = [r(x)]S = [x]S . We prove that f is surjective. By deﬁnition, we have the following commutative diagram: r Cn+1 \{0} −−−−→ S2n+1   π q y y f Pn −−−−→ S2n+1/S1. To show that f is continuous, we show that f −1(V ) is open in Pn for any open subset V of S2n+1/S1. To show that f −1(V ) is open in Pn, we show that π−1(f −1(V )) is open in Cn+1 \{0}. By the above commutative diagram, π−1(f −1(V )) = (f ◦ π)−1(V ) = (q ◦ r)−1(V ) = r−1(q−1(V )). 5

Since V is open in S2n+1/S1, q−1(V ) is open in S2n+1. Since r is continuous and q−1(V ) is open in S2n+1, r−1(q−1(V )) is open in Cn+1 \{0}. This proves that π−1(f −1(V )) is open in Cn+1 \{0}. Hence f −1(V ) is open in Pn. We conclude that f is continuous. Since f is a bijective continuous map and Pn is compact and S2n+1/S1 is Hausdorff, f is a homeomorphism. Lemma 1.5. Let X and Y be two spaces. Suppose f : X → Y is a homeomorphism. Then X is Hausdorff if and only if Y is Hausdorff.

Proof. Let us assume that X is Hausdorff. Suppose y1 and y2 are two points of Y such that y1 6= y2. Since f is surjective, there exist x1 and x2 in X so that f(xi) = yi for i = 1, 2. Since f is a function, x1 6= x2. (If x1 = x2, y1 = f(x1) = f(x2) = y2.) Since X is Hausdorff, there exist open neighborhoods Ui of xi so that U1 ∩ U2 = ∅. Let Vi = f(Ui) for i = 1, 2. Then Vi are open neighborhood of yi for i = 1, 2. Claim V1 ∩ V2 = ∅. Suppose not. Take z ∈ V1 ∩ V2. By surjectivity of f, we can find x ∈ X −1 so that z = f(x). Then x ∈ f (V1 ∩ V2) ⊆ U1 ∩ U2 = ∅ which is impossible. Since Pn is homeomorphic to S2n+1/S1 and S2n+1/S1 is Hausdorff, Pn is also Hausdorff.