# de Rham vs Dolbeault Cohomology

For a complex manifold $M$, one can consider (A) its de Rham cohomology, or (B) its Dolbeault cohomology. I'm looking for some motivation as to why one would bother introducing Dolbeault cohomology. To be more specific, here are some straight questions.

1. What can Dolbeault tell us that de Rham can't?
2. Does there exist some simple relationship between these two cohomologies?
3. When are they equal?
4. Do things become simpler for the Kahler case?
5. What happens for the projective spaces?
6. Why does nobody talk about the holomorphic cohomology?

## 4 Answers

Let $\Omega^{p,q}(M)$ be the $C^{\infty}$ $(p,q)$-forms. One always has a double complex with $\Omega^{p,q}(M)$ in position $(p,q)$. The cohomology in the $q$ direction is Dolbeault cohomology, the cohomology of the total complex is deRham cohomology. (In each case, essentially by definition.) Whenever you have a double complex, you get a spectral sequence. On the first page of the spectral sequence is Dolbeault cohomology; the spectral sequence converges to deRham cohomology. This is sometimes called the Hodge-de Rham spectral sequence, and sometimes called the Frolicher spectral sequence.

If $M$ is compact Kahler (in particular projective) then the spectral sequence collapses at the first page; all maps between Dolbeault groups are zero. So $H^k(M) \cong \bigoplus_{p+q=k} H^{p,q}(M)$ in this case.

If $M$ is Stein (in particular, affine) then the only nonzero Dolbeault groups are the $H^{p,0}(M)$, corresponding to holmorphic $p$-forms. The next page takes the cohomology of the complex of holomorphic $p$-forms; I'll term this holomorphic deRham". After that, there are no further maps, so holomorphic deRham equals deRham.

In general, the spectral sequence can be arbitrarily nondegenerate.

• Can you update this link, if you are able? – Steven Gubkin Sep 9 '15 at 5:13
• @StevenGubkin Done! – David E Speyer Sep 9 '15 at 14:36
• Thanks! A quick question about this answer though: $\frac{1}{\bar{z}} d\bar{z}$ is closed, but not exact, so represents an element of $H^1(M)$, where $M$ is the punctured unit disk. The unit disc is an open Riemann surface, and hence Stein. Does this contradict "deRham = holomorphic deRham", or am I missing something obvious? I feel like I probably am... – Steven Gubkin Sep 9 '15 at 14:40
• @StevenGubkin No contradiction. The class represented by $\frac{d \bar{z}}{\bar{z}}$ is also represented holomorphically by $- \frac{dz}{z}$. (Note that the difference of these classes is $d(\log |z|^2)$, hence exact.) – David E Speyer Sep 9 '15 at 14:44
• Thanks! I was actually thinking along these lines, but didn't recognize $\frac{d\bar{z}}{\bar{z}}+\frac{dz}{z}$ as $d\log(|z|^2)$, although I knew it should be exact. Thanks! – Steven Gubkin Sep 9 '15 at 14:58

Although the main questions have been answered quite well, I would like to say a few words about the first question "why would one bother...".

De Rham and Dolbeault cohomology are measuring different things. The first, which measures the failure of the necessary condition $d\alpha=0$ to guarantee a potential $\alpha=d\beta$, is a topological invariant. The second gives the obstruction to solving the similar problem for the Cauchy-Riemann operator; it measures the holomorphic complexity and it has a priori nothing to do with the topology. That they turn out to be the "same" in good (e.g. compact Kähler) cases is sort of a miracle. Even in such cases, these spaces are not literally the same, and this can be exploited in interesting ways. There is an isomorphism which can be represented by a matrix called a period matrix, which is sensitive to the complex structure. To see how this works in the simplest interesting case, let $X= \mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$ be an elliptic curve. As a topological space it is just a torus, and independent of the choice of $\tau$, but as a Riemann surface it is sensitive to this choice. The natural basis of the first Dolbeault cohomology $H^{10}(X)\oplus H^{01}(X)$is $\lbrace dz,d\bar z\rbrace$. This maps to $\lbrace (1,\tau), (1,\bar \tau)\rbrace$ under the basis of $H^1_{DR}(X)$ dual to loops given by projecting $[0,1],[0,\tau]\subset \mathbb{C}$ to $X$. Thus $X$ can be recovered from its period matrix.

Dolbeaut cohomology is a way to compute the sheaf cohomology of the sheaf of holomorphic $p$-forms. That is,

$$H^{p,q}(X) \cong H^q(X, \Omega_X^p)$$

Now, in the case that $X$ is Kahler, we have that decomposition

$$H^k(X,\mathbb{C}) \cong \bigoplus_{p + q = k} H^{p,q}(X)$$

So in that case, it follows that the Dolbeaut cohomology is a refinement of the de Rham cohomology of $X$. In such a case, you could argue that they are "the same", but it seems a little weird to do so, since the Dolbeaut cohomology has more information than the de Rham cohomology.

As for the cohomology of the holomorphic complex, that does arise; there are often reasons to be interested in the cohomology groups $H^q(X, \mathcal{O}_X) = H^{0,q}(X)$.

Lastly, projective spaces are Kahler, so the above decomposition still holds. In particular, since the cohomology of $\mathbb{P}^n$ is simple enough to describe, you can relatively easily see that

$$H^{p,q}(\mathbb{P}^n) = \mathbb{C}$$

if $0 \leq p = q \leq n$, and is 0 otherwise.

• – roy smith Apr 27 '12 at 17:42
• more specifically, all Riemann surfaces have the same deRham cohomology, but if the decompositions of H^1 by the dolbeault groups are the same, then the surfaces are holomorphically isomorphic. – roy smith Apr 27 '12 at 19:12
• @roy smith: How do the Riemann sphere and a torus have the same deRham cohomology? – YangMills May 1 '12 at 13:28
• @YangMills: I think he means that all Riemann surfaces of a given genus have the same deRham cohomology. – Simon Rose May 1 '12 at 16:03

1.De Rham cohomology is a cohomology of k-forms on complex manifold。 Dolbeault Cohomology is a cohomology of smooth sections of the vector bundle of complex differential forms of degree (p,q) on complex manifold。so Dolbeault Cohomology is subcohomology of the De Rham cohomology.2.when p+q=0