# Intuitive Aproach to Dolbeault Cohomology

I would like to understand an intuitive approach to the definitions of Dolbeault Cohomology (using $\partial$ and $\bar{\partial}$) similar to the one given here.

All suggestions are welcome.

• Yeah, that would be very nice for me too, because Dolbeaut Cohomology seems so mysterious... – juan rojo Jan 22 '15 at 7:42
• If, by intuitive, you mean "geometrical", i.e. related to things like submanifolds and boundaries, I don't think you will find one. $H^{p,q}$, in general, has no interpretation in terms of geometry of submanifolds or such ... – wisefool Jan 22 '15 at 11:34

This question has been asked and answered on MathOverflow. I have replicated David Speyer's accepted answer below.

Here is some nonsense that I find useful: On a complex manifold,

$$\frac{\mbox{locally constant functions}}{\mbox{smooth functions}} \approx \frac{\mbox{locally constant functions}}{\mbox{holomorphic functions}} \cdot \frac{\mbox{holomorphic functions}}{\mbox{smooth functions}}$$

The left hand side is where good geometric intuition lies, and de Rham cohomology measures how complicated it is. For example, $$H^0$$ measures how many locally constant functions there are. On a contractible space, where $$H^{\ast}$$ vanishes, smooth maps are homotopic to locally constant maps.

Dolbeault cohomology measures how complicated the second term on the RHS is. This isn't as geometric, since it is "the square root of geometry". However, it is useful to think about when it is large or small.

On a Stein space, Dolbeault vanishes. This means that smooth functions and holomorphic functions are close to being the same, and all the interesting geometry is in the first fraction. Indeed, on a Stein space, there are lots of holomorphic functions, and every cohomology class has a holomorphic representative.

On a compact Kahler manifold, on the other hand, Dolbeault cohomology is large. This should mean that there are many fewer holomorphic functions than smooth functions, and that only a small part of the geometry can be seen in holomorphic terms. Indeed, in this case, all holomorphic functions are locally constant, and only a small number of cohomology classes have holomorphic representatives.

To actually say something precise, there are three exact sequences of sheaves that come up everywhere in algebraic geometry. Write $$\underline{\mathbb{C}}$$ for the locally constant $$\mathbb{C}$$-valued functions, $$\mathcal{H}^p$$ for the holomorphic $$(p,0)$$ forms and $$\Omega^{(p,q)}$$ for the $$C^{\infty}$$ $$(p,q)$$-forms. Set $$\Omega^n = \bigoplus \Omega^{p, n-p}$$, the smooth $$n$$-forms. Then we have exact sequences:

$$0 \to \underline{\mathbb{C}} \to \Omega^0 \overset{d}{\longrightarrow} \Omega^1 \overset{d}{\longrightarrow} \Omega^2 \overset{d}{\longrightarrow} \cdots$$

$$0 \to \underline{\mathbb{C}} \to \mathcal{H}^0 \overset{\partial}{\longrightarrow} \mathcal{H}^1 \overset{\partial}{\longrightarrow} \mathcal{H}^2 \overset{\partial}{\longrightarrow} \cdots$$

$$0 \to \mathcal{H}^p \to \Omega^{(p,0)} \overset{\bar{\partial}}{\longrightarrow} \Omega^{(p,1)} \overset{\bar{\partial}}{\longrightarrow} \Omega^{(p,2)} \overset{\bar{\partial}}{\longrightarrow} \cdots$$

The LHS of the nonsense equation refers to things related to the first sequence. The two fractions on the RHS refer respectively to the things related to the second and third sequences.