1. Complex projective Space The n-dimensional complex projective space Pn over C is the space Cn+1 n f0g modulo the relation ∼ defined below. Two elements x and y in Cn+1 n f0g are said to be equivalent if there exists λ 2 C n f0g 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 nf0g is denoted by [x] = (x0 : ··· : xn): Let π : C nf0g ! 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 nf0g: Since π is continuous and Cn+1 nf0g is connected (it is path connected and hence connected), Pn = π(Cn+1 n f0g) is connected. (Continuous functions send connected sets to connected sets). The unit sphere n 2n+1 n X 2 S = f(x0; ··· ; xn) 2 C : jxij = 1g i=0 is contained in Cn+1 n f0g: 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 n f0g) and hence Ui is open by the continuity of πi and the openess of C n f0g: Since −1 0 n+1 π (Ui) = Ui \ (C n f0g); n Sn n Ui is open in P : One can easily see that i=0 Ui = Pk : Hence fU0; ··· ;Ung forms an open cover for Pn: For each 0 ≤ i ≤ n; we define n x0 xn 'i : Ui ! C ; (x0 : ··· : xn) = ; ··· ; : xi xi This map is well-defined (left to the reader) and is a bijection. Let us prove that 'i is a homeomor- phism. −1 n To show that 'i is continuous, we show that 'i (V ) is open in Ui for any open set V of C : Since n −1 −1 Ui is open in P ; to show that 'i (V ) is open in Ui; we only need to show that 'i (V ) is open in n −1 n −1 −1 n+1 0 P : To show that 'i (V ) is open in P ; we show that π ('i (V )) is open in C n f0g: Let Ui 0 n be as above. The composition 'i ◦ π : Ui ! C is given by x0 xn ('i ◦ π)(x0; ··· ; xn) = ; ··· ; : xi xi 0 n −1 0 It is easy for us to see that 'i◦π : Ui ! C is continuous. Hence ('i◦π) (V ) is open in Ui and hence n+1 0 n+1 −1 −1 −1 open in C nf0g (we use the fact that Ui is open in C nf0g:) Since π ('i (V )) = ('i ◦π) (V ) n+1 −1 n is open in C n f0g;'i (V ) is open in P : n Let us prove that 'i is an open mapping, i.e. 'i(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 n f0g and hence open in C : We only need to show that 'i(π(W )) is open. In fact, 0 n we can show that 'i ◦ 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 n f0g 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 n f0g: Lemma 1.1. Let f : X ! Y be any function. Suppose fAα : α 2 Λg is a family of subsets of X: Then ! [ [ f Aα = f(Aα): α2Λ α2Λ n Since any union of open sets is open, if we can show that ('i ◦π)(I0 ×I1 ×· · ·×In) is open in C for 0 any open set of the form I0 ×I1 ×· · ·×In with Ii open in Cnf0g; Lemma 1.1 implies that ('i ◦π)(W ) 0 −1 is open for any open subset W of Ui so that W = π (W ): Let Jk = fxk=xi : xi 2 Ii; xk 2 Ikg for any k 6= i: Then ('i ◦ 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 ('i ◦ π)(I0 × · · · × In) is open in C : For each µ 2 Ii; we see that [ [ −1 Jk = fxk/µ : xk 2 Ikg = µ Ik: µ2Ii µ2Ii −1 Since Ik is open in C; µ Ik is open for any µ 2 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 'i : Ui ! C is a homeomorphism for all 0 ≤ i ≤ n: Let 0 ≤ j < i ≤ n: Then n 'i(Ui \ Uj) = f(z1; ··· ; zn) 2 C : zj 6= 0g: Hence 'i(Ui \ Uj) is open. The transition functions are given by −1 n ('j ◦ 'i ): 'i(Ui \ Uj) ! C −1 z1 zj−1 zj+1 zi 1 zi+1 zn ('j ◦ 'i )(z1; ··· ; zn) = ; ··· ; ; ; ··· ; ; ; ; ··· ; zj zj zj zj zj zj zj −1 Hence 'j ◦ 'i is biholomorphic on 'i(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 n K is open. Let us prove that every point of X n K is an interior point of X n K: Let x 2 X n K: For any y 2 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 fVy : y 2 Kg forms an open cover for K: Since K is compact, we can find y1; ··· ; yk so that Tk fVyi : 1 ≤ i ≤ kg covers K: Take U = i=1 Uyk : Then U is an open neighorhood of x: Let us show that U is contained in X n K: Let z 2 U: If z 2 K; then z 2 Vyi for some 1 ≤ i ≤ k: Then z 2 Vyi \ U ⊆ Vyi \ Uyi = ; which is impossible. Therefore z 62 K and hence z 2 X n K: We see that U ⊆ X n K: We find that x is an interior point of X n 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 2 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 2 K and an open neighborhood Ub of b such that Ub \ V = ;: Since B is compact and fUb : b 2 Bg forms an open cover for B; there exist b1; ··· ; bl 2 B so that fUbi : 1 ≤ i ≤ lg 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 2 U \ V: Then z 2 V and z 2 Ubi for some i: Hence z 2 V \ Ubi = ; which is not possible.