1. Complex Torus
Let C be the complex field and Γ = Z ⊕ Zτ be the lattice in C generated by {1, τ }, where Im τ > 0. Consider the quotient group C/Γ. Equip C/Γ with the quotient topology.
Proposition 1.1. The space C/Γ is a complex manifold of dimension one.
To prove this proposition, let us introduce the notion of covering space.
Definition 1.1. Let p : Y → X be a continuous map between topological spaces. We say that p : Y → X is a covering space on X if for each x ∈ X, there exists an open neighborhood U of x and a disjoint family of open subsets {Ui} of Y so that p−1(U ) =`
iUi and p : Ui→ U is a homeomorphism for each i.
Lemma 1.1. Any open subset of a complex manifold is a complex manifold (of the same dimension)
Proof. Let X be an n-dimensional complex manifold and V be an open subset of X. Since X is a complex manifold, we can choose a complex atlas {(Uα, ϕα)} on X. Take Vα = Uα∩V and ψα = ϕα|Vα. Then we find {(Vα, ψα)} is a complex atlas on V. Hence V is a complex
manifold.
Theorem 1.1. Let π : Y → X be a covering space on X. Suppose Y is a complex manifold.
Then X has a structure of a complex manifold1and dim Y = dim X.
Proof. Assume that dim Y = n. For each x ∈ X, we can choose an open neighborhood U of x and a family of disjoint open sets {Vi} so that π−1(U ) = `
i∈IVα and p|Vi : Vi → U is a homeomorphism. Since any open subset of a complex manifold is also a complex manifold, Vi is a complex manifold. Denote V = Vi and q = p|V. Since V is a complex manifold, we can choose a complex atlas {(Uα, ϕα)} on V. We take Wα= p(Uα) for all α. Since q : V → U is a homeomorphism, {Wα} forms an open covering of U. Define ψα= ϕα◦ q−1: Wα→ Cn. We check that
ψβ◦ ψα−1 = (ϕβ◦ q−1) ◦ (ϕα◦ q−1)−1= ϕβ◦ ϕ−1α : ϕα(Uα∩ Uβ) ⊂ Cn→ ϕβ(Uα∩ Uβ) ⊂ Cn. Hence ψβ◦ ψ−1α is biholomorphic for each α, β. We find {(Wα, ψα)} forms a complex atlas on U. This shows that U is a complex manifold. Since every point of X has a neighborhood which is also a complex manifold, X is a complex manifold. dim Y = dim X. Hence if we can prove that π : C → C/Γ is a covering space on C/Γ, then by this theorem, C/Γ is a complex manifold.
Lemma 1.2. The group Γ is a discrete subgroup of C.
Proof. Let ω = m + nτ in Γ. We can choose = min{1, |τ |}/3. Then B(ω, ) ∩ Γ = {ω}.
Hence Γ is discrete.
Let us consider an action of Γ on C by
Γ × C → C defined by (ω, z) 7→ z + ω, where ω ∈ Γ, and z ∈ C.
Lemma 1.3. For every point z ∈ C, there exists an open neighborhood U of x so that (ω + U ) ∩ U = ∅ for ω 6= 0.
1The statement is also true for smooth manifolds 1
2
Proof. Let δ = min{1, |τ |}/3. Take U = B(z, δ). Let us prove that (ω + U ) ∩ U = ∅ for all ω 6= 0. Suppose w ∈ (ω + U ) ∩ U. Then |w − z| < δ and |w − ω − z| < δ. This implies that
|ω| < 2δ which is impossible.
Definition 1.2. Let G be a topological group and X be a topological space. We say that G acts on X evenly (the action is assumed to be continuous) if for each x ∈ X, there is an open neighborhood U of X such that g(U ) ∩ U = ∅ for all g 6= e, where e is the identity element of G.
By definition, Γ acts on C evenly.
Proposition 1.2. Let G acts on X evenly. The the quotient map p : X → X/G is a covering space.
Proof. Let [x] ∈ X/G. Choose a representative x of [x] in X and an open neighborhood V of x such that g(V ) ∩ V = ∅ for all g 6= e. Then U = p(V ) is an open neighborhood of x. (We equip X/G with the quotient topology. Hence p is an open mapping.) Moreover, let us prove that p−1(U ) = `
g∈Gg(V ). Let y be another point in p−1(U ) so that x 6= y. Then we can find g ∈ G so that y = gx. Hence y ∈ g(V ). We see p−1(U ) ⊂`
g∈Gg(V ). If y ∈`
g∈Gg(V ), then y ∈ g(V ) for some V. Hence y = gv for some v ∈ V. Hence p(y) ∈ p(V ) = U. In other words, y ∈ p−1(U ). We conclude that p−1(U ) =`
g∈Gg(V ).
Since gV and V are homeomorphic, we only need to check that p : V → U is a home- omorphism. This map is a continuous surjective open mapping. We only need to check p : V → U is an injection. Assume that p(x1) = p(x2) in U, with x1, x2 ∈ V. Suppose x1 6= x2. Since p(x1) = p(x2), there exists g 6= e ∈ G so that x1 = gx2. This implies that x1 ∈ V ∩ g(V ) which leads to a contradiction that V ∩ g(V ) = ∅. Hence x1 = x2, i.e. p is
injective.
Using this proposition, we immediately find that π : C → C/Γ is a covering map. Simi- larly, we can prove the following:
Theorem 1.2. Let {ω1, · · · , ωn} be a basis in Cn and Λ be the lattice generated by {ω1, · · · , ωn}. The quotient map π : Cn → Cn/Λ is a covering space on Cn/Λ and thus Cn/Λ is an n-dimensional complex manifold.
