Dolbeault cohomology explained

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

Hp,q(M,\Complex)

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

\bar{\partial}:\Omegap,q\to\Omegap,q+1

Since

\bar{\partial}2=0

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

Hp,q(M,\Complex)=

\ker(\bar\partial:\Omegap,q\to\Omegap,q+1)
im(\bar\partial:\Omegap,q-1\to\Omegap,q)

.

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

lO(E)

of holomorphic sections of E, using the Dolbeault operator of E. This is therefore a resolution of the sheaf cohomology of

lO(E)

.

In particular associated to the holomorphic structure of

E

is a Dolbeault operator

\bar\partialE:\Gamma(E)\to\Omega0,1(E)

taking sections of

E

to

(0,1)

-forms with values in

E

. This satisfies the characteristic Leibniz rule with respect to the Dolbeault operator

\bar\partial

on differential forms, and is therefore sometimes known as a

(0,1)

-connection
on

E

, Therefore, in the same way that a connection on a vector bundle can be extended to the exterior covariant derivative, the Dolbeault operator of

E

can be extended to an operator

\bar \partial_E : \Omega^(E) \to \Omega^(E)which acts on a section

\alphas\in\Omegap,q(E)

by

\bar \partial_E (\alpha \otimes s) = (\bar \partial \alpha) \otimes s + (-1)^ \alpha \wedge \bar \partial_E sand is extended linearly to any section in

\Omegap,q(E)

. The Dolbeault operator satisfies the integrability condition

\bar

2
\partial
E

=0

and so Dolbeault cohomology with coefficients in

E

can be defined as above:

H^(X,(E,\bar \partial_E)) = \frac .The Dolbeault cohomology groups do not depend on the choice of Dolbeault operator

\bar\partialE

compatible with the holomorphic structure of

E

, so are typically denoted by

Hp,q(X,E)

dropping the dependence on

\bar\partialE

.

Dolbeault–Grothendieck lemma

In order to establish the Dolbeault isomorphism we need to prove the Dolbeault–Grothendieck lemma (or

\bar{\partial}

-Poincaré lemma). First we prove a one-dimensional version of the

\bar{\partial}

-Poincaré lemma; we shall use the following generalised form of the Cauchy integral representation for smooth functions:

Proposition: Let

B\varepsilon(0):=\lbracez\in\Complex\mid|z|<\varepsilon\rbrace

the open ball centered in

0

of radius

\varepsilon\in\R>0,

\overline{B\varepsilon(0)}\subseteqU

open and

f\inl{C}infty(U)

, then

\forallz\inB\varepsilon(0):f(z)=

1
2\pii
\int
\partialB\varepsilon(0)
f(\xi)d\xi+
\xi-z
1
2\pii
\iint
B\varepsilon(0)
\partialf
\partial\bar{\xi
}\frac.

Lemma (

\bar{\partial}

-Poincaré lemma on the complex plane): Let

B\varepsilon(0),U

be as before and

\alpha=f

0,1
d\bar{z}\inl{A}
\Complex

(U)

a smooth form, then

l{C}infty(U)\nig(z):=

1
2\pii
\int
B\varepsilon(0)
f(\xi)
\xi-z

d\xi\wedged\bar{\xi}

satisfies

\alpha=\bar{\partial}g

on

B\varepsilon(0).

Proof. Our claim is that

g

defined above is a well-defined smooth function and

\alpha=fd\bar{z}=\bar{\partial}g

. To show this we choose a point

z\inB\varepsilon(0)

and an open neighbourhood

z\inV\subseteqB\varepsilon(0)

, then we can find a smooth function

\rho:B\varepsilon(0)\to\R

whose support is compact and lies in

B\varepsilon(0)

and

\rho|V\equiv1.

Then we can write

f=f1+f2:=\rhof+(1-\rho)f

and define

g
i:=1
2\pii
\int
B\varepsilon(0)
fi(\xi)
\xi-z

d\xi\wedged\bar{\xi}.

Since

f2\equiv0

in

V

then

g2

is clearly well-defined and smooth; we note that

\begin{align}

g
1&=1
2\pii
\int
B\varepsilon(0)
f1(\xi)
\xi-z

d\xi\wedged\bar{\xi}\\ &=

1
2\pii

\int\Complex

f1(\xi)
\xi-z

d\xi\wedged\bar{\xi}\\ &=\pi-1

2\pi
\int
0
i\theta
f
1(z+re

)e-i\thetad\thetadr, \end{align}

which is indeed well-defined and smooth, therefore the same is true for

g

. Now we show that

\bar{\partial}g=\alpha

on

B\varepsilon(0)

.
\partialg2
\partial\bar{z
}=\frac\int_f_2(\xi)\frac\Big(\frac\Big)d\xi\wedge d\bar=0

since

(\xi-z)-1

is holomorphic in

B\varepsilon(0)\setminusV

.
\begin{align} \partialg1
\partial\bar{z
}=&\pi^\int_\frac e^d\theta\wedge dr\\=& \pi^\int_\Big(\frac\Big)(z+re^) e^d\theta\wedge dr\\=&\frac\iint_\frac\frac\end

applying the generalised Cauchy formula to

f1

we find
f
1(z)=1
2\pii
\int
\partialB\varepsilon(0)
f1(\xi)d\xi+
\xi-z
1
2\pii
\iint
B\varepsilon(0)
\partialf1
\partial\bar{\xi
}\frac =\frac\iint_\frac\frac

since

f1|

\partialB\varepsilon(0)

=0

, but then
f=f
1=\partialg1
\partial\bar{z
}=\frac on

V

. Since

z

was arbitrary, the lemma is now proved.

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

n(0)
\Delta
\varepsilon
the open polydisc centered in

0\in\Complexn

with radius

\varepsilon\in\R>0

.

Lemma (Dolbeault–Grothendieck): Let

p,q
\alpha\inl{A}
\Complexn

(U)

where
n(0)}
\overline{\Delta
\varepsilon

\subseteqU

open and

q>0

such that

\bar{\partial}\alpha=0

, then there exists
p,q-1
\beta\inl{A}
\Complexn

(U)

which satisfies:

\alpha=\bar{\partial}\beta

on
n(0).
\Delta
\varepsilon

Before starting the proof we note that any

(p,q)

-form can be written as

\alpha=\sumIJ\alphaIJdzI\wedged\bar{z}J=\sumJ\left(\sumI\alphaIJdzI\right)J\wedged\bar{z}J

for multi-indices

I,J,|I|=p,|J|=q

, therefore we can reduce the proof to the case
0,q
\alpha\inl{A}
\Complexn

(U)

.

Proof. Let

k>0

be the smallest index such that

\alpha\in(d\bar{z}1,...,d\bar{z}k)

in the sheaf of

l{C}infty

-modules, we proceed by induction on

k

. For

k=0

we have

\alpha\equiv0

since

q>0

; next we suppose that if

\alpha\in(d\bar{z}1,...,d\bar{z}k)

then there exists
0,q-1
\beta\inl{A}
\Complexn

(U)

such that

\alpha=\bar{\partial}\beta

on
n(0)
\Delta
\varepsilon
. Then suppose

\omega\in(d\bar{z}1,...,d\bar{z}k+1)

and observe that we can write

\omega=d\bar{z}k+1\wedge\psi+\mu,    \psi,\mu\in(d\bar{z}1,...,d\bar{z}k).

Since

\omega

is

\bar{\partial}

-closed it follows that

\psi,\mu

are holomorphic in variables

zk+2,...,zn

and smooth in the remaining ones on the polydisc
n(0)
\Delta
\varepsilon
. Moreover we can apply the

\bar{\partial}

-Poincaré lemma to the smooth functions

zk+1\mapsto\psiJ(z1,...,zk+1,...,zn)

on the open ball
B
\varepsilonk+1

(0)

, hence there exist a family of smooth functions

gJ

which satisfy
\psi
J=\partialgJ
\partial\bar{z

k+1

}\quad \text \quad B_(0).

gJ

are also holomorphic in

zk+2,...,zn

. Define

\tilde{\psi}:=\sumJgJd\bar{z}J

then

\begin{align} \omega-\bar{\partial}\tilde{\psi}&=d\bar{z}k+1\wedge\psi

+\mu-\sum
J\partialgJ
\partial\bar{z

k+1

}d\bar_\wedge d\bar_J +\sum_^k\sum_J\fracd\bar_j\wedge d\bar_\\&=d\bar_\wedge\psi+\mu-d\bar_\wedge\psi+\sum_^k\sum_J\fracd\bar_j\wedge d\bar_\\&=\mu+\sum_^k\sum_J\fracd\bar_j\wedge d\bar_ \in (d\bar_1, \dots, d\bar_),\end

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

0,q-1
η\inl{A}
\Complexn

(U)

such that

\omega-\bar{\partial}\tilde{\psi}=\bar{\partial}ηon

n(0)
\Delta
\varepsilon

and

\zeta:=η+\tilde{\psi}

ends the induction step. QED

The previous lemma can be generalised by admitting polydiscs with

\varepsilonk=+infty

for some of the components of the polyradius.

Lemma (extended Dolbeault-Grothendieck). If

n(0)
\Delta
\varepsilon
is an open polydisc with

\varepsilonk\in\R\cup\lbrace+infty\rbrace

and

q>0

, then
p,q
H
\bar{\partial
}(\Delta_\varepsilon^n(0))=0.

Proof. We consider two cases:

p,q+1
\alpha\inl{A}
\Complexn

(U),q>0

and
p,1
\alpha\inl{A}
\Complexn

(U)

.

Case 1. Let

p,q+1
\alpha\inl{A}
\Complexn

(U),q>0

, and we cover
n(0)
\Delta
\varepsilon
with polydiscs

\overline{\Deltai}\subset\Deltai+1

, then by the Dolbeault–Grothendieck lemma we can find forms

\betai

of bidegree

(p,q-1)

on

\overline{\Deltai}\subseteqUi

open such that

\alpha

|
\Deltai

=\bar{\partial}\betai

; we want to show that

\betai+1

|
\Deltai

=\betai.

We proceed by induction on

i

: the case when

i=1

holds by the previous lemma. Let the claim be true for

k>1

and take

\Deltak+1

with
k+1
\Delta
i=1

\Deltaiand\overline{\Deltak}\subset\Deltak+1.

Then we find a

(p,q-1)

-form

\beta'k+1

defined in an open neighbourhood of

\overline{\Deltak+1

} such that
\alpha|
\Deltak+1

=\bar{\partial}\betak+1

. Let

Uk

be an open neighbourhood of

\overline{\Deltak}

then

\bar{\partial}(\betak-\beta'k+1)=0

on

Uk

and we can apply again the Dolbeault-Grothendieck lemma to find a

(p,q-2)

-form

\gammak

such that

\betak-\beta'k+1=\bar{\partial}\gammak

on

\Deltak

. Now, let

Vk

be an open set with

\overline{\Deltak}\subsetVk\subsetneqUk

and

\rhok:

n(0)\to\R
\Delta
\varepsilon
a smooth function such that:

\operatorname{supp}(\rhok)\subsetUk,   

\rho|
Vk

=1,    \rhok|

n(0)\setminus
\DeltaUk
\varepsilon

=0.

Then

\rhok\gammak

is a well-defined smooth form on
n(0)
\Delta
\varepsilon
which satisfies

\betak=\beta'k+1+\bar{\partial}(\gammak\rhok)on\Deltak,

hence the form

\betak+1:=\beta'k+1+\bar{\partial}(\gammak\rhok)

satisfies

\begin{align} \betak+1

|
\Deltak

&=\beta'k+1+\bar{\partial}\gammak=\betak\\ \bar{\partial}\betak+1&=\bar{\partial}\beta'k+1

=\alpha|
\Deltak+1

\end{align}

Case 2. If instead

p,1
\alpha\inl{A}
\Complexn

(U),

we cannot apply the Dolbeault-Grothendieck lemma twice; we take

\betai

and

\Deltai

as before, we want to show that

\left\|\left.\left({\betai}I-{\betai+1

}_I \right) \right |_ \right \|_\infty<2^.

Again, we proceed by induction on

i

: for

i=1

the answer is given by the Dolbeault-Grothendieck lemma. Next we suppose that the claim is true for

k>1

. We take

\Deltak+1\supset\overline{\Deltak}

such that

\Deltak+1\cup\lbrace\Deltai\rbrace

k
i=1
covers
n(0)
\Delta
\varepsilon
, then we can find a

(p,0)

-form

\beta'k+1

such that
\alpha|
\Deltak+1

=\bar{\partial}\beta'k+1,

which also satisfies

\bar{\partial}(\betak-\beta'k+1)=0

on

\Deltak

, i.e.

\betak-\beta'k+1

is a holomorphic

(p,0)

-form wherever defined, hence by the Stone–Weierstrass theorem we can write it as

\betak-\beta'k+1=\sum|I|=p(PI+rI)dzI

where

PI

are polynomials and

\left\|rI|

\Deltak-1

\right

-k
\|
infty<2

,

but then the form

\betak+1:=\beta'k+1+\sum|I|=pPIdzI

satisfies

\begin{align} \bar{\partial}\betak+1&=\bar{\partial}\beta'k+1

=\alpha|
\Deltak+1

\\ \left\|({\betak}I-{\betak+1

}_I)|_ \right \|_\infty&=\| r_I\|_\infty<2^\end

which completes the induction step; therefore we have built a sequence

\lbrace\betai\rbracei\in\N

which uniformly converges to some

(p,0)

-form

\beta

such that
\alpha|
n(0)
\Delta
\varepsilon

=\bar{\partial}\beta

. 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,

Hp,q(M)\congHq(M,\Omegap)

where

\Omegap

is the sheaf of holomorphic p forms on M.

A version of the Dolbeault theorem also holds for Dolbeault cohomology with coefficients in a holomorphic vector bundle

E

. Namely one has an isomorphism

H^(M,E) \cong H^q(M, \Omega^ \otimes E).

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

Proof

Let

l{F}p,q

be the fine sheaf of

Cinfty

forms of type

(p,q)

. Then the

\overline{\partial}

-Poincaré lemma says that the sequence

\Omegap,q\xrightarrow{\overline{\partial}}l{F}p,q+1\xrightarrow{\overline{\partial}}l{F}p,q+2\xrightarrow{\overline{\partial}}

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

n

-dimensional complex projective space is
p,q
H
\bar{\partial
}(P^n_)=\begin\Complex & p=q\\ 0 &\text\end

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

k
H
\rmdR

\left

n
(P
\Complex

,\Complex\right)=oplusp+q=k

p,q
H
\bar{\partial
}(P^n_)

because

n
P
\Complex
is a compact Kähler complex manifold. Then

b2k+1=0

and

b2k=hk,k+\sump+q=2k,p\nehp,q=1.

Furthermore we know that

n
P
\Complex
is Kähler, and

0\ne[\omegak]\in

k,k
H
\bar{\partial
}(P^n_), where

\omega

is the fundamental form associated to the Fubini–Study metric (which is indeed Kähler), therefore

hk,k=1

and

hp,q=0

whenever

p\neq,

which yields the result.

See also

\bar\partial

-exact differential form in the setting of compact Kähler manifolds.

References

. Raymond O. Wells, Jr.. Differential Analysis on Complex Manifolds . Springer-Verlag . 1980 . 978-0-387-90419-1.

. Gunning. Robert C.. Robert C. Gunning. Introduction to Holomorphic Functions of Several Variables, Volume 1. 1990 . Chapman and Hall/CRC. 9780534133085. 198.

Notes and References

  1. Book: 10.1017/CBO9780511629327.002. Calculus on Complex Manifolds . Several Complex Variables and Complex Manifolds II . 1982 . 1–64 . 9780521288880 .
  2. In contrast to de Rham cohomology, Dolbeault cohomology is no longer a topological invariant because it depends closely on complex structure.
  3. , Section 8