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We say that U ⊂ X/R is open if π−1(U ) is open in X

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1. Projective Space

Let X be a topological space and R be an equivalence relation on X. The set of all R- equivalence classes in X is denoted by X/R. The natural map π : X → X/R sending x to its equivalent class [x] is called the quotient map. We say that U ⊂ X/R is open if π−1(U ) is open in X. This topology on X/R is called the quotient topology.

1.1. Complex Projective Space. On Cn+1\ {0}, we define a relation R as follows. We say that xRy if there is λ 6= 0 so that x = λy. Then R is an equivalence relation. The quotient space of Cn+1 \ {0} modulo R is denoted by Pn and called the n-dimensional complex projective space. We equip Pn with the quotient topology. The quotient map is denoted by π : Cn+1\ {0} → Pn.

Lemma 1.1. Let X be a topological space and f : Pn → X be a function. Then f is continuous if and only if f ◦ π is continuous.

Proof. Let us assume that f : Pn → X is continuous. Since π is continuous and f is continuous, f ◦ π is continuous.

Assume that f ◦ π is continuous. Let U be an open set in X. Then we want to show that f−1(U ) is open in Pn. We only need to show that π−1(f−1(U )) is open in Cn+1\ {0}. Since f ◦ π is continuous, (f ◦ π)−1(U ) is open in Cn+1\ {0}. Since

(f ◦ π)−1(U ) = π−1(f−1(U ))

for any open set U in X, π−1(f−1(U )) is open in Cn+1\ {0} and hence f−1(U ) is open in

Pn. 

If (ζ0, · · · , ζn) ∈ Cn+1\ {0}, we denote (ζ0: · · · : ζn) to be its equivalent class. Let (1.1) Ui = {(ζ0 : · · · : ζn) : ζi 6= 0}.

Then {Ui : 0 ≤ i ≤ n} forms an open cover of Pn. We define ϕi: Ui→ Cn by (1.2) ϕi0 : · · · : ζn) = ζ0

ζi, · · · ,ζi−1

ζii+1

ζi , · · ·ζn

ζi

 . Lemma 1.2. ϕi : Ui → Cn are homeomorphisms for all 0 ≤ i ≤ n.

Proof. Let us prove the case when i = 0. (The proof for others are the same) We denote ϕ = ϕ0, and U = U0, i.e. ϕ(ζ0 : · · · : ζn) = (ζ10, · · · , ζn0). The inverse map ψ = ϕ−1 : Cn → U is given by ψ(z1, · · · , zn) = (1 : z1 : · · · : zn). To show that ϕ : U → Cn is continuous, we only need to show that ϕ ◦ π : π−1(U ) → Cn is continuous. (Use (similar ideas in) Lemma 1.1.) Note that π−1(U ) = {(ζ0, · · · , ζn) : ζ0 6= 0} is open in Cn+1 and (ϕ ◦ π)(ζ0, · · · , ζn) = (z1/z0, · · · , zn/z0). We see that ϕ ◦ π is continuous on π−1(U ).

Let g : Cn → Cn+1 be the function defined by g(z1, · · · , zn) = (1, z1, · · · , zn). Then g is a continuous function on Cn and π ◦ g = ψ. Since both π and g are continuous, ψ is also continuous. (The composition of continuous functions is continuous.)

We conclude that both ϕ and ϕ−1 are continuous and hence ϕ is a homeomorphism from U to Cn.

 By Lemma 1.2, ϕi is a homeomorphism from Ui onto Cnwhose inverse map ψi : Cn→ Ui is given by

ψi(z1, · · · , zn) = (z1, · · · , zi−1: 1 : zi: · · · : zn), (z1, · · · , zn) ∈ Cn.

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Assume that i < j. Then Ui∩ Uj 6= ∅ and

ϕi(Ui∩ Uj) = {(z1, · · · , zn) : zj 6= 0}, ϕj(Ui∩ Uj) = {(z1, · · · , zn) : zi6= 0}.

Then both of ϕi(Ui ∩ Uj) and ϕj(Ui ∩ Uj) are open subsets of Cn. The transition map ϕj◦ ϕ−1i : ϕi(Ui∩ Uj) → ϕj(Ui∩ Uj) is given by

ϕj◦ ϕ−1i (z1, · · · , zn) = z1

zj, · · · ,zi−1 zj , 1

zj,zi+1

zj , · · · ,zj−1

zj ,zj+1

zj , · · · ,zn zj

 ,

for each 0 ≤ i, j ≤ n. We see that ϕj◦ ϕ−1i : ϕi(Ui∩ Uj) → ϕj(Ui∩ Uj) is bi-holomorphic.

By definition, Pn is an n-dimensional complex manifold.

Corollary 1.1. A function f : X → Pn from a topological space X into Pn is continuous if and only iff−1(Uj) is open in X and ϕ ◦ f : f−1(Uj) → Cn is continuous for all j.

Proof. If f is continuous, f−1(Uj) is open for each j (since Uj are open in Pn.) Since the composition of continuous functions are continuous, ϕj◦ f are all continuous on f−1(Uj).

Suppose V is open in Pn. Then V = Sn

i=0(V ∩ Ui). We know that V ∩ Ui is an open subset in the domain Ui of ϕi. Since ϕi is a homeomorphism from Ui to Cn, ϕi(V ∩ Ui) is open in Cn. Since ϕi◦ f is continuous on f−1(Ui),

i◦ f )−1i(V ∩ Ui)) = f−1−1ii(V ∩ Ui))) = f−1(V ∩ Ui) is open in X. Since any union of open sets is open in X, and

f−1(V ) =

n

[

i=0

f−1(V ∩ Ui), we find f−1(V ) is open in X. Hence f is continuous.

 Proposition 1.1. The n-dimensional complex projective space with the quotient topology is an n-dimensional compact complex manifold.

Proof. We have shown that Pnis a complex manifold of dimension n. We only need to show that it is a compact space. For each z = (z0, · · · , zn) ∈ Cn+1, we set

|z|2 = |z0|2+ · · · + |zn|2.

The the closed unit sphere S2n+1 in Cn is the set of points z so that |z| = 1. Equip S2n+1 with the subspace topology of Cn. Then S2n+1 is closed and bounded in Cn; by Heine-Borel theorem S2n+1 compact. Define a map f from S2n+1 to Cn by (z0, · · · , zn) 7→ (z0: · · · : zn).

Then f : S2n+1 → Pn is surjective. (For any point (ζ0 : · · · : ζn) ∈ Pn, we choose a representative ζ = (ζ0, · · · , ζn) of (ζ0 : · · · : ζn). Consider x = ζ/|ζ|. Then f (x) = π(ζ).) For any open subset U of Pn, we have

f−1(U ) = S2n+1∩ π−1(U ),

where π : Cn+1\ {0} → Pnis the quotient map. Hence f−1(U ) is open in S2n+1. This shows that f is continuous. Since the image of a compact space under a continuous map is also compact, by the continuity of f, Pn= f (S2n+1) is also compact.  Definition 1.1. A projective transformation of Pn is a bijection f : Pn → Pn such that there exists an invertible linear operator A : Cn+1 → Cn+1 with f ◦ π = π ◦ A.

Lemma 1.3. A projective transformation f on Pn is continuous.

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Proof. There exists an invertible linear operator A on Cn+1 so that f ◦ π = π ◦ A. Since a linear operator on Cn+1 is continuous, A is continuous. We find π ◦ A is continuous and hence f ◦ π is continuous. By Lemma 1.1, f is continuous.  Definition 1.2. A subset H of Pnis said to be a hyperplane if there exists a n-dimensional linear subspace V of Cn+1 such that H = π(V \ {0}).

Lemma 1.4. Let p0, · · · , pn, q be n + 2 distinct points on Pn so that no n + 1 of which lie on a hyperplane. Then there is a unique projective transformation taking pi to [ei] and q to (1 : · · · : 1). Here {ei : 0 ≤ i ≤ n} is the standard basis in Cn+1.

Proof. Let u0, · · · , un and v be vectors in Cn+1 such that π(ui) = pi and π(v) = q. Then {u0, · · · , un} forms a basis for Cn+1 following from the assumption. Hence we can choose a linear operator A on Cn+1 taking ui to ei for 0 ≤ i ≤ n. Let A(v) = (λ0, · · · , λn). Since no n + 1 of {p0, · · · , pn, q} lies on a hyperplane, all λi are nonzero. (If one of them are zero, then q would lie on a hyper plane spanned by n of {e0, · · · , en}.) Let B : Cn+1→ Cn+1 be the operator defined by ζ 7→ (ζ00, · · · , ζnn). Then B is an invertible linear operator.

Define C = B ◦ A. Then C is invertible. We define f : Pn→ Pn by f ◦ π = π ◦ C. Then f is

the required projective transformation. 

Proposition 1.2. The projective space Pn is a Hausdorff space.

Proof. Assume that p, q are distinct points on Pn. Then we choose p0 = p and p1, · · · , pn so that {p0, · · · , pn, q} are distinct point so that no n + 1 point lie on a hyperplane. We can choose a projective transformation f : Pn → Pn taking pi to [ei] and q to (1 : · · · : 1).

Hence p0 is taken to (1 : 0 : · · · : 0). We know that x = (1 : 0 : · · · : 0) and y = (1 : · · · : 1) both lie in U0 = {(ζ0 : · · · : ζn) : ζ0 6= 0}. Since ϕ0 : U0 → Cn sending (ζ0 : · · · : ζn) to (ζ10, · · · , ζn0) is a homeomorphism and Cnis a Hausdorff space, we can find V and W in Cn with V ∩ W = ∅ and ϕ0(x) ∈ V and ϕ0(y) ∈ W. We know ϕ−10 (V ) and ϕ−10 (W ) are disjoint open subsets in Pn containing x and y respectively. Since f is a bijection and f is continuous, f−1−10 (V )) and f−1−10 (W )) are disjoint open subsets of Pn containing p and q respectively. We proved that Pn is a Hausdorff space.

 1.2. The case P1. Assume that (ζ0, ζ1) is the rectangular coordinate on C2. Let

U = {(ζ0 : ζ1) : ζ16= 0}, V = {(ζ0 : ζ1) : ζ0 6= 0}.

Then {U, V } forms an open cover of P1. Define ψ : U → C and ϕ : V → C by ψ(ζ0 : ζ1) = ζ01 and ϕ(ζ0 : ζ1) = ζ10. Then ψ and ϕ are homeomorphisms. Denote z = ζ01 and w = ζ10. Then the transition functions ϕ ◦ ψ−1 : ψ(U ∩ V ) → ϕ(U ∩ V ) and ψ ◦ ϕ−1 : ϕ(U ∩ V ) → ψ(U ∩ V ) are given by

ϕ ◦ ψ−1(z) = 1

z, ψ ◦ ϕ−1(w) = 1 w,

i.e. w = 1/z on U ∩ V. Note that U ∩ V = {(ζ0 : ζ1) : ζ0, ζ1 6= 0}. Hence ψ(U ∩ V ) = {z ∈ C : z 6= 0} and ϕ(U ∩ V ) = {w ∈ C : w 6= 0}. Therefore the transition function ϕ ◦ ψ−1 is biholomorphic on C \ {0}. Notice that P1 = U ∪ {(1 : 0)}. If we denote ∞ = (1 : 0) and identify U with C via ψ, then we obtain the following identification P1= C ∪ {∞}.

Remark. This is also an example of one-point compactification of locally compact Haus- dorff space.

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The map h : S3 → P1 defined by h(ζ1, ζ2) = [ζ1, ζ2] is called the Hopf map. Let U0 = π−1(U ) ∩ S3 and V0 = π−1(V ) ∩ S3. Then U0, V0 are open sets in S3. Then

(ϕ ◦ h)(z1, z2) = z2

z1, (ψ ◦ h)(z1, z2) = z2

z1. This shows that h is a C-map.

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