1. Holomorphic Functions and their Local Rings Let z = (z1, · · · , zn) be the standard coordinates on Cnwith zk= xk+√
−1yk, 1 ≤ k ≤ n.
In this note, we consider the Euclidean topology on Cn defined by the Euclidean norm kzk2= |z1|2+ · · · + |zn|2.
Define differential operators
(1.1) ∂
∂zi = 1 2
∂
∂xi −√
−1 ∂
∂yi
, ∂
∂zi = 1 2
∂
∂xi +√
−1 ∂
∂yi
.
Let U be an open set of Cn and C∞(U ) be the ring of all complex valued smooth functions on U.
Definition 1.1. A function f ∈ C∞(U, C) is said to be holomorphic on U if
∂f
∂zi = 0, 1 ≤ i ≤ n.
Example 1.1. Let us consider f (z) = |z|2 = zz on C. Then ∂f
∂z = z 6= 0. The function f is not holomorphic on C.
Using quotient rule, we immediate obtain:
Lemma 1.1. Let f be a holomorphic function on U with f 6= 0 on U. Then 1/f is also holomorphic.
The set of holomorphic functions on U is denoted by OCn(U ). It forms an algebra over C.
Remark. The assignment U 7→ OCn(U ) defines a sheaf of complex algebras on Cn. Let P be a point of Cn and OP be the family of pairs (f, U ) where U is an open neigh- borhood of P and f is a holomorphic function on U. Given (f, U ) and (g, V ) in OP, we say that (f, U ) is equivalent to (g, V ) if f = g on U ∩ V. Then ∼ is an equivalent relation on OP. The quotient space OP/ ∼ is denoted by OP whose elements are denoted by f . An element in OP is called a germ at P. Given a germ f at P, we define its value at P by
f = f (P ),
where (f, U ) is a representative of f . Then the value of f at P is well-defined.
On OP, we can define an algebra structure as follows. Let f and g be two elements in OP. Choose representatives (f, U ) of f and (g, V ) of g respectively. On U ∩ V, f + g, f g can be defined. We set f + g to be the equivalent class of (U ∩ V, f + g) and fg to be the equivalent class of (U ∩ V, f g). If a ∈ C, we define af to be the equivalent class of (af, U ).
Lemma 1.2. Let A be a commutative ring. Suppose that the set of all non units in A forms an ideal m. Then A is a local ring with the unique maximal ideal m.
Proof. Let I be a proper ideal of A. Then I contains only non units. (If I contains a unit, then 1 ∈ I. This implies I = A.) Hence I ⊂ m. Let M be a maximal ideal. Since M is a proper ideal, M ⊂ m. Since M is maximal and M ⊂ m, M = m. Thus m is the unique maximal ideal of A.
Corollary 1.1. The set OP is a local ring1 with 1.
1A ring A is local if it has a unique maximal ideal.
1
2
Proof. We leave it to the reader to show that OP is an algebra over C. Now, let us prove that it has a unique maximal ideal.
Let mP = {f ∈ OP : f (P ) = 0}. If f (P ) 6= 0, then we can choose a representative (f, U ) of f such that f 6= 0 on U. Since f is holomorphic on U with f 6= 0, then 1/f is also holomorphic on U. Let g be the equivalent class of (1/f, U ). We see that fg = 1. Hence f is a unit. This is equivalent to say that a non unit is an element of mP. Let f be a unit.
Then there is g so that fg = 1. Hence we can choose a representative (f, U ) of f and a representative (g, V ) of g so that f g = 1 on U ∩ V. In this case, f (P )g(P ) = 1. We find f (P ) 6= 0. Hence f (P ) 6= 0, i.e. f 6∈ mP. We conclude that mP is the set of all non units in OP. This implies that OP is a local ring with the unique maximal ideal mP.
Now, we will think of f as a holomorphic function defined in a neighborhood of P.
2. Complex Differential Forms
A smooth real valued function u(z1, · · · , zn) on an open set U of Cn can be considered as a smooth real valued function u(x1, · · · , xn, y1, · · · , yn), where zi = xi+√
−1yi. From calculus, we have already defined the total differential of u by
du =
n
X
i=1
∂u
∂xidxi+
n
X
j=1
∂u
∂yjdyj.
We know that a smooth complex valued function is give by f = u + iv, where u, v are real value smooth functions on U. Define the total differential df of f by
df = du + idv.
The space of complex-valued differential forms on U is defined to be Ar(U ) = Ωr(U ) ⊗RC,
where Ωr(U ) is the space of smooth r-forms on U when we consider Cn as R2n via the isomorphism (z1, · · · , zn) 7→ (x1, y1, · · · , xn, yn).
Let us use dzi = dxi +√
−1dyi and dzi = dxi −√
−1dyi and the partial differential operators (1.1) to rewrite df as (we leave it to the reader as an exercise)
df =
n
X
i=1
∂f
∂zidzi+
n
X
j=1
∂f
∂zjdzj. This allows us to define two differentials: ∂ and ∂ as follows:
∂f =
n
X
i=1
∂f
∂zidzi, ∂f =
n
X
j=1
∂f
∂zjdzj.
Definition 2.1. A function f ∈ C∞(U ) is holomorphic on U if and only if ∂f = 0.
Let I = (i1, · · · , ip) and J = (j1, · · · , jq) be a p-tuple of integers and q-tuple of integers such that 1 ≤ i1 < · · · < ip ≤ n and 1 ≤ j1 < · · · < jq≤ n. We denote
dzI = dzi1 ∧ · · · dzip, dzJ = dzj1∧ · · · ∧ dzjq. A smooth differential form on U of type (p, q) has the following expression
ω =X
I,J
ωI,JdzI∧ dzJ,
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where ωI,J are smooth functions on U. The set of smooth (p, q)-forms on U is denoted by Ap,q(U ).
It follows from the definition that there is a natural direct sum decomposition:
Ar(U ) = M
p+q=r
Ap,q(U ).
If d is the original de Rham differential on Ωr(U ), we also use d for d ⊗ 1C. The differential dω, ∂ω and ∂ are defined by
dω =X
I,J
dωI,J ∧ dzI∧ dzJ
∂ω =X
I,J
∂ωI,J ∧ dzI∧ dzJ
∂ω =X
I,J
∂ωI,J ∧ dzI∧ dzJ.
Then we obtain linear operators ∂ : Ap,q(U ) → Ap+1,q(U ) and ∂ : Ap,q(U ) → Ap,q+1(U ) and d : Ap,q(U ) → Ap+1,q(U ) ⊕ Ap,q+1(U ) such that d = ∂ + ∂.
Lemma 2.1. ∂2 = 0 and ∂2 = 0.
Proof. Routine check.
In other words, we have two chain complexes:
· · · −−−−→ A∂ p,q−1(U ) −−−−→ A∂ p,q(U ) −−−−→ A∂ p,q+1(U ) −−−−→ · · ·∂
· · · −−−−→ A∂ p−1,q(U ) −−−−→ A∂ p,q(U ) −−−−→ A∂ p+1,(U ) −−−−→ · · ·∂ Corollary 2.1.
∂∂ = −∂∂.
Proof. Since d2= (∂ + ∂)2= 0 and ∂2= 0, ∂2 = 0, we find ∂∂ + ∂∂ = 0. Example 2.1. Let K(z) = log(1 + |z|2) for z ∈ C. Then
∂K = z
1 + |z|2dz, ∂K = z 1 + |z|2dz.
We denote Z(p,q)
∂ (U ) = ker ∂|Ap,q(U ) and Bp,q
∂ (U ) = Im ∂|Ap,q−1(U ). Elements of Z(p,q)
∂ (U ) are called ∂-closed forms on U and elements of Bp,q
∂ (U ) are called ∂-exact forms on U. Since
∂2= 0, Bp,q
∂ (U ) is a vector subspace of Z(p,q)
∂ (U ) over C. We define the (p, q)-th Dolbeault cohomology to be the quotient space
H(p,q)
∂ (U ) = Z(p,q)
∂ (U )/Bp,q
∂ (U ).
Similarly, we can consider H∂p,q(U ).
Definition 2.2. A (p, 0)-form ω on U is called a holomorphic p-form if ∂ω = 0. The set of all holomorphic p-forms on U is denoted by Ωphol(U ).
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3. Dolbeault Lemma and its application
Let B be the open ball {z ∈ Cn: |z| < 1} in Cn. In this section, we would like to prove the following proposition.
Proposition 3.1. Let ω =P
i,jωijdzi∧ dzj be a smooth (1, 1)-form on B. Then dω = 0 if and only if there exists a smooth function f on U such that
ω = ∂∂f.
If ω = ∂∂f for some f ∈ C∞(U ), we know, by Lemma 2.1 and its corollary, dω = (∂ + ∂ω)ω = ∂(∂∂f ) + ∂∂∂f
= 0 − ∂∂2f = 0.
Conversely, assume that dω = 0. Since d = ∂ + ∂, dω = 0 implies that ∂ω = 0 and ∂ω = 0.
Theorem 3.1. (Dolbeault’s Lemma 1) Assume p ≥ 1. A smooth (p, q)-form ω on U satisfies
∂ω = 0 if and only if there exists a (p, q − 1)-form η such that ω = ∂η.
Proof. The proof will be given in another note.
Let us go back to the proof of Proposition 3.1. Since ω is a (1, 1)-form such that ∂ω = 0, by Theorem 3.1, there exists a (1, 0)-form η so that ω = ∂η. Since ∂ω = 0, we find
∂∂η = 0.
Since ∂η is a (2, 0)-form, by definition, ∂η is a holomorphic (2, 0)-form. Moreover, d(∂η) = ∂2η + ∂∂η = 0.
Hence ∂η is a d-closed holomorphic (2, 0)-form.
Theorem 3.2. (Dolbeault Lemma 2) Let η be a holomorphic p-form on U. dη = 0 if and only if there exists a holomorphic p − 1-form τ on U so that η = dτ.
Proof. The proof will be given in another note.
Using Theorem 3.2, there exists a holomorphic one-form τ such that
∂η = dτ.
Since dτ = ∂τ + ∂τ and τ is holomorphic, ∂τ = 0. We obtain ∂η = ∂τ. Hence
∂(η − τ ) = 0.
Theorem 3.3. (Dolbeault Lemma 3). Suppose ω is a (p, q)-form on U. Then ∂ω = 0 if and only if there exists a (p − 1, q)-form η on U such that ω = ∂η.
Proof. The proof will be given in another note.
We can choose a smooth function g on U such that η − τ = ∂g. Since τ is holomorphic, ω = ∂η = ∂(τ + ∂g) = ∂∂f.
Choosing f = −g, we find
ω = ∂∂f.
This completes the proof of Proposition 3.1.
