U ⊆ Pn is open if and only if π−1(U ) is open in Cn+1\ {0}

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 an element x = (x0, · · · , xn) of Cⁿ⁺¹\ {0} is denoted by [x] = (x0: · · · : xn). Let π : Cⁿ⁺¹\ {0} → Pⁿ 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

S²ⁿ⁺¹= {(x₀, · · · , x_n) ∈ Cⁿ:

n

X

i=0

|x_i|²= 1}

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.

For each 0 ≤ i ≤ n, let Ui be the set of all points (x0 : · · · : xn) in Pⁿ with the property that there exists a representative (x0, · · · , xn) of (x0: · · · : xn) such that xi6= 0. Let us check that this set is well-defined. Let (x⁰₀, · · · , x⁰_n) be another representative of (x0: · · · : xn). Then there exists a nonzero complex number λ in C so that x⁰j = λxifor 0 ≤ j ≤ n. Since xi6= 0 and λ 6= 0, x⁰_i6= 0. (C is a field.) Therefore Ui is well-defined. Let us show that Ui is open.

Let 0 ≤ i ≤ n. We define πi : Cⁿ⁺¹ → C by (x0, · · · , xn) to xi. Then πi is smooth for any 0 ≤ i ≤ n. Let U_i⁰ be the subset of Cⁿ⁺¹ consisting of vectors (x₀, · · · , x_n) so that x_i 6= 0. Then U_i⁰= π_i⁻¹(C \ {0}) and hence Ui⁰ is open by the continuity of πi and the openess of C \ {0}. Since

π⁻¹(Ui) = U_i⁰∩ (Cⁿ⁺¹\ {0}), Ui is open in Pⁿ. One can easily see thatSn

i=0Ui = Pⁿk. Hence {U0, · · · , Un} forms an open cover for Pⁿ.

For each 0 ≤ i ≤ n, we define

ϕ_i: U_i→ Cⁿ, (x₀: · · · : x_n) = x₀ xi

, · · · ,x_n xi

 .

This map is well-defined (left to the reader) and is a bijection. Let us prove that ϕi is a homeomor- phism.

To show that ϕ_iis continuous, we show that ϕ⁻¹_i (V ) is open in U_ifor any open set V of Cⁿ. Since U_i is open in Pⁿ, to show that ϕ⁻¹_i (V ) is open in U_i, we only need to show that ϕ⁻¹_i (V ) is open in Pⁿ. To show that ϕ⁻¹_i (V ) is open in Pⁿ, we show that π⁻¹(ϕ⁻¹_i (V )) is open in Cⁿ⁺¹\ {0}. Let U_i⁰ be as above. The composition ϕi◦ π : U_i⁰ → Cⁿ is given by

(ϕ_i◦ π)(x₀, · · · , x_n) = x₀ xi

, · · · ,x_n xi

 .

It is easy for us to see that ϕ_i◦π : U_i⁰→ Cⁿis continuous. Hence (ϕ_i◦π)⁻¹(V ) is open in U_i⁰and hence open in Cⁿ⁺¹\{0} (we use the fact that U_i⁰is open in Cⁿ⁺¹\{0}.) Since π⁻¹(ϕ⁻¹_i (V )) = (ϕi◦π)⁻¹(V ) is open in Cⁿ⁺¹\ {0}, ϕ⁻¹_i (V ) is open in Pⁿ.

Let us prove that ϕiis an open mapping, i.e. ϕi(W ) is open in Cⁿfor any open set W of Ui. Since W is open in Ui and Ui is an open subset of Pⁿ, W is also open in Pⁿ. Therefore W⁰ = π⁻¹(W ) is open in Cⁿ⁺¹\ {0} and hence open in Cⁿ⁺¹. We only need to show that ϕi(π(W⁰)) is open. In fact, we can show that ϕi◦ π : U_i⁰→ Cⁿ is an open mapping.

1

Since open sets of the form I0× I1× · · · × In generates the topology on Cⁿ⁺¹, where I0, · · · , In

are open subsets of C, open sets of the form I⁰× I1× · · · × In∩ U_i⁰ generates the topology of U_i⁰. Hence open sets of the form I0× I1× · · · × In with Ii open in C \ {0} generates the topology of Ui⁰. More precisely, W⁰ 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 [

α∈Λ

A_α

!

= [

α∈Λ

f (A_α).

Since any union of open sets is open, if we can show that (ϕi◦π)(I0×I1×· · ·×In) is open in Cⁿfor any open set of the form I0×I1×· · ·×Inwith Iiopen in C\{0}, Lemma 1.1 implies that (ϕⁱ◦π)(W⁰) is open for any open subset W of Uiso that W⁰ = π⁻¹(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 open, J₁× · · · × Ji−1× Ji+1× · · · × Jn is open in Cⁿ and hence (ϕ_i◦ π)(I0× · · · × In) is open in Cⁿ.

For each µ ∈ I_i, we see that

Jk = [

µ∈Ii

{xk/µ : xk∈ Ik} = [

µ∈Ii

µ⁻¹Ik.

Since Ik is open in C, µ⁻¹Ik is open for any µ ∈ Ii. Since any union of open sets is open, Jk is open in C. We complete the proof of our assertion. We conclude that ϕⁱ: Ui→ Cⁿ is a homeomorphism for all 0 ≤ i ≤ n.

Let 0 ≤ j < i ≤ n. Then

ϕi(Ui∩ Uj) = {(z1, · · · , zn) ∈ Cⁿ: zj6= 0}.

Hence ϕi(Ui∩ Uj) is open. The transition functions are given by (ϕj◦ ϕ⁻¹_i ) : ϕi(Ui∩ Uj) → Cⁿ

(ϕ_j◦ ϕ⁻¹_i )(z₁, · · · , z_n) = z₁ zj

, · · · ,z_j−1 zj

,z_j+1 zj

, · · · ,z_i zj

, 1 zj

,z_i+1 zj

, · · · ,z_n zj



Hence ϕj◦ ϕ⁻¹_i 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 {Vy_i : 1 ≤ i ≤ k} covers K. Take U =Tk

i=1Uy_k. 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 ∈ V_yi for some 1 ≤ i ≤ k. Then z ∈ V_yi∩ 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. 

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 compactness of K, we can choose an open neighborhood V =Sk

i=1Vy_iof K for some y1, · · · , yk∈ K and an open neighborhood Ubof 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 {Ub_i : 1 ≤ i ≤ l} forms an open cover for B.

Let U =Sl

i=1Ub_i. Then U is an open neighborhood of B. Claim U ∩ V = ∅. Let z ∈ U ∩ V. Then z ∈ V and z ∈ Ub_i for some i. Hence z ∈ V ∩ Ub_i= ∅ 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

S¹× S²ⁿ⁺¹→ S²ⁿ⁺¹, (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 x ∈ S²ⁿ⁺¹modulo this relation is denoted by [x]_S1. The quotient map S²ⁿ⁺¹→ S²ⁿ⁺¹/S¹is 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/S¹is 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∈G

g(U ).

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 some g ∈ G. Hence y ∈S

g∈Gg(U ). Hence π⁻¹(π(U )) ⊆S

g∈Gg(U ). Conversely, if y ∈S

g∈Gg(U ),

then y ∈ g(U ) for some g ∈ G. Hence y = gz for some z ∈ U. Therefore π(y) = π(z) ∈ π(U ) which implies that y ∈ π⁻¹(π(U )). HenceS

g∈Gg(U ) ⊆ π⁻¹(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 G(A) = m(G, A) and m is continuous, G(A) =S

g∈Gg(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 U⁰ = (X/G) \ π(V ). Then U⁰ is an open subset of X/G. Since π is an open mapping, V⁰ = π(V ) is open in X/G. We obtain disjoint open sets {U⁰, V⁰}. Since Gy ⊆ V, [y] ∈ π(V ) = V⁰. Let us show that [x] ∈ U⁰.

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] ∈ U⁰ = (X/G) \ π(V ).

We find {U⁰, V⁰} disjoint open subsets of X/G so that [x] ∈ U⁰ and [y] ∈ V⁰ for any [x], [y] in X/G with [x] 6= [y]. Therefore X/G is a Hausdorff space.

 This lemma implies that S²ⁿ⁺¹/S¹is a (compact) Hausdorff space.

Lemma 1.4. The complex projective space Pⁿ is homeomorphic to S²ⁿ⁺¹/S¹. Proof. Let r : Cⁿ⁺¹\ {0} → S²ⁿ⁺¹ be the function r(z) = z

kzk for z ∈ Cⁿ⁺¹\ {0}. Then r is continuous. Define f : Pⁿ → S²ⁿ⁺¹/S¹ by

f ([z]) = [r(z)]_S1

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

r(z) = z

kzk = µz⁰

|µ|kz⁰k = λr(z⁰) for λ = µ/|µ|. In other words, [r(z)]_S¹= [r(z⁰)]_S¹.

If [r(z)]_S1 = [r(z⁰)]_S1, then

z

kzk= λ z⁰

kz⁰k =⇒ z = λkzk kz⁰kz⁰.

Hence [z] = [z⁰]. This shows that f is injective. Let [x]_S1 be a point in S²ⁿ⁺¹/S¹. Choose a repre- sentative x of [x]_S1. Then x ∈ S²ⁿ⁺¹⊆ Cⁿ⁺¹\ {0} and r(x) = x. Then f ([x]) = [r(x)]_S1 = [x]_S1. We prove that f is surjective.

By definition, we have the following commutative diagram:

Cⁿ⁺¹\ {0} −−−−→^r S²ⁿ⁺¹

π



 y



 y

q

Pⁿ

−−−−→ Sf ²ⁿ⁺¹/S¹.

To show that f is continuous, we show that f⁻¹(V ) is open in Pⁿfor 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 )).

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 y₁and y₂are two points of Y such that y₁6= y₂. Since f is surjective, there exist x₁and x₂in X so that f (x_i) = y_i for i = 1, 2. Since f is a function, x₁6= 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 so that z = f (x). Then x ∈ f⁻¹(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.