Download 1. Projective Space Let X be a topological space and R be an

1. Projective Space
Let X be a topological space and R be an equivalence relation on X. The set of all Requivalence 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
ζ0
ζi−1 ζi+1
ζn
(1.2)
ϕi (ζ0 : · · · : ζn ) =
,··· ,
,
,···
.
ζi
ζi
ζi
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 ) = (ζ1 /ζ0 , · · · , ζn /ζ0 ). 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 Cn whose inverse map ψi : Cn → Ui
is given by
ψi (z1 , · · · , zn ) = (z1 , · · · , zi−1 : 1 : zi : · · · : zn ),
1
(z1 , · · · , zn ) ∈ Cn .
2
Assume that i < j. Then Ui ∩ Uj 6= ∅ and
ϕi (Ui ∩ Uj ) = {(z1 , · · · , zn ) : zj 6= 0},
ϕj (Ui ∩ Uj ) = {(z1 , · · · , zn ) : zi 6= 0}.
Then both of ϕi (Ui ∩ Uj ) and ϕj (Ui ∩ Uj ) are open subsets of Cn . The
ϕj ◦ ϕ−1
i : ϕi (Ui ∩ Uj ) → ϕj (Ui ∩ Uj ) is given by
zj−1 zj+1
zi−1 1 zi+1
z1
−1
ϕj ◦ ϕi (z1 , · · · , zn ) =
,··· ,
, ,
,··· ,
,
,···
zj
zj zj z j
zj
zj
transition map
zn
,
zj
,
for each 0 ≤ i, j ≤ n. We see that ϕj ◦ ϕ−1
i : 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 ).
Sn
n
Suppose V is open in P . Then V = 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 ),
−1
(ϕi ◦ f )−1 (ϕi (V ∩ Ui )) = f −1 (ϕ−1
(V ∩ Ui )
i (ϕi (V ∩ Ui ))) = f
is open in X. Since any union of open sets is open in X, and
f
−1
(V ) =
n
[
f −1 (V ∩ Ui ),
i=0
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 Pn is 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 S 2n+1 in Cn is the set of points z so that |z| = 1. Equip S 2n+1
with the subspace topology of Cn . Then S 2n+1 is closed and bounded in Cn ; by Heine-Borel
theorem S 2n+1 compact. Define a map f from S 2n+1 to Cn by (z0 , · · · , zn ) 7→ (z0 : · · · : zn ).
Then f : S 2n+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 ) = S 2n+1 ∩ π −1 (U ),
where π : Cn+1 \ {0} → Pn is the quotient map. Hence f −1 (U ) is open in S 2n+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 (S 2n+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.
3
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 Pn is 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→ (ζ0 /λ0 , · · · , ζn /λn ). 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
(ζ1 /ζ0 , · · · , ζn /ζ0 ) is a homeomorphism and Cn is a Hausdorff space, we can find V and W
−1
in Cn with V ∩ W = ∅ and ϕ0 (x) ∈ V and ϕ0 (y) ∈ W. We know ϕ−1
0 (V ) and ϕ0 (W ) are
n
disjoint open subsets in P containing x and y respectively. Since f is a bijection and f
−1 (ϕ−1 (W )) are disjoint open subsets of Pn containing p
is continuous, f −1 (ϕ−1
0
0 (V )) and f
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 ) : ζ1 6= 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 ) =
ζ0 /ζ1 and ϕ(ζ0 : ζ1 ) = ζ1 /ζ0 . Then ψ and ϕ are homeomorphisms. Denote z = ζ0 /ζ1 and
w = ζ1 /ζ0 . 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 Hausdorff space.
4
The map h : S 3 → P1 defined by h(ζ1 , ζ2 ) = [ζ1 , ζ2 ] is called the Hopf map. Let U 0 =
∩ S 3 and V 0 = π −1 (V ) ∩ S 3 . Then U 0 , V 0 are open sets in S 3 . Then
z2
z2
(ϕ ◦ h)(z1 , z2 ) = , (ψ ◦ h)(z1 , z2 ) = .
z1
z1
∞
This shows that h is a C -map.
π −1 (U )