Dolbeault cohomology

In mathematics, in particular in algebraic geometry and differential geometry, Dolbeault cohomology (named after Pierre Dolbeault) is an analog of de Rham cohomology for complex manifolds. Let M be a complex manifold. Then the Dolbeault cohomology groups depend on a pair of integers p and q and are realized as a subquotient of the space of complex differential forms of degree (p,q).

Construction of the cohomology groups

Let Ωp,q be the vector bundle of complex differential forms of degree (p,q). In the article on complex forms, the Dolbeault operator is defined as a differential operator on smooth sections


this operator has some associated cohomology. Specifically, define the cohomology to be the quotient space

Dolbeault cohomology of vector bundles

If E is a holomorphic vector bundle on a complex manifold X, then one can define likewise a fine resolution of the sheaf of holomorphic sections of E. This is therefore a resolution of the sheaf cohomology of .

Dolbeault–Grothendieck lemma

In order to establish the Dolbeault isomorphism we need to prove the Dolbeault–Grothendieck lemma (or -Poincaré lemma). First we prove a one-dimensional version of the -Poincaré lemma; we shall use the following generalised form of the Cauchy integral representation for smooth functions:

Proposition: Let the open ball centered in of radius open and , then

Lemma ( -Poincaré lemma on the complex plane): Let be as before and a smooth form, then

satisfies on

Proof. Our claim is that defined above is a well-defined smooth function such that is locally -exact. To show this we choose a point and an open neighbourhood , then we can find a smooth function whose support is compact and lies in and Then we can write

and define

Since in then is clearly well-defined and smooth; we note that

which is indeed well-defined and smooth, therefore the same is true for . Now we show that on .

since is holomorphic in .

applying the generalised Cauchy formula to we find

since , but then on . QED

Proof of Dolbeault–Grothendieck lemma

Now are ready to prove the Dolbeault–Grothendieck lemma; the proof presented here is due to Grothendieck[1]. We denote with the open polydisc centered in with radius .

Lemma (Dolbeault–Grothendieck): Let where open and such that , then there exists which satisfies: on

Before starting the proof we note that any -form can be written as

for multi-indices , therefore we can reduce the proof to the case .

Proof. Let be the smallest index such that in the sheaf of -modules, we proceed by induction on . For we have since ; next we suppose that if then there exists such that on . Then suppose and observe that we can write

Since is -closed it follows that are holomorphic in variables and smooth in the remaining ones on the polydisc . Moreover we can apply the -Poincaré lemma to the smooth functions on the open ball , hence there exist a family of smooth functions which satisfy

are also holomorphic in . Define


therefore we can apply the induction hypothesis to it, there exists such that

and ends the induction step. QED

The previous lemma can be generalised by admitting polydiscs with for some of the components of the polyradius.

Lemma (extended Dolbeault-Grothendieck). If is an open polydisc with and , then

Proof. We consider two cases: and .

Case 1. Let , and we cover with polydiscs , then by the Dolbeault–Grothendieck lemma we can find forms of bidegree on open such that ; we want to show that

We proceed by induction on : the case when holds by the previous lemma. Let the claim be true for and take with

Then we find a -form defined in an open neighbourhood of such that . Let be an open neighbourhood of then on and we can apply again the Dolbeault-Grothendieck lemma to find a -form such that on . Now, let be an open set with and a smooth function such that:

Then is a well-defined smooth form on which satisfies

hence the form


Case 2. If instead we cannot apply the Dolbeault-Grothendieck lemma twice; we take and as before, we want to show that

Again, we proceed by induction on : for the answer is given by the Dolbeault-Grothendieck lemma. Next we suppose that the claim is true for . We take such that covers , then we can find a -form such that

which also satisfies on , i.e. is a holomorphic -form wherever defined, hence by the Stone–Weierstrass theorem we can write it as

where are polynomials and

but then the form


which completes the induction step; therefore we have built a sequence which uniformly converges to some -form such that . QED

Dolbeault's theorem

Dolbeault's theorem is a complex analog[2] of de Rham's theorem. It asserts that the Dolbeault cohomology is isomorphic to the sheaf cohomology of the sheaf of holomorphic differential forms. Specifically,

where is the sheaf of holomorphic p forms on M.

A version for logarithmic forms has also been established.[3]


Let be the fine sheaf of forms of type . Then the -Poincaré lemma says that the sequence

is exact. Like any long exact sequence, this sequence breaks up into short exact sequences. The long exact sequences of cohomology corresponding to these give the result, once one uses that the higher cohomologies of a fine sheaf vanish.

Explicit example of calculation

The Dolbeault cohomology of the -dimensional complex projective space is

We apply the following well-known fact from Hodge theory:

because is a compact Kähler complex manifold. Then and

Furthermore we know that is Kähler, and where is the fundamental form associated to the Fubini–Study metric (which is indeed Kähler), therefore and whenever which yields the result.


  1. Serre, Jean-Pierre (1953–1954), "Faisceaux analytiques sur l'espace projectif", Séminaire Henri Cartan, 6 (Talk no. 18): 1–10
  2. In contrast to de Rham cohomology, Dolbeault cohomology is no longer a topological invariant because it depends closely on complex structure.
  3. Navarro Aznar, Vicente (1987), "Sur la théorie de Hodge–Deligne", Inventiones Mathematicae, 90 (1): 11–76, doi:10.1007/bf01389031, Section 8


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