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(Duplicated from math.stackexchange)

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. It is not enough to know that it works, but to know what it means.

All suggestions are welcome.

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closed as unclear what you're asking by Fernando Muro, Stefan Kohl, José Figueroa-O'Farrill, Peter Crooks, abx Feb 18 '15 at 15:17

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  • $\begingroup$ Like any derived functor, Dolbeault cohomology measure the nonexactness of something (in this case---global sections), i.e., the extent to which local relations between forms can be extended globally. $\endgroup$ – Alex Degtyarev Jan 21 '15 at 9:10
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    $\begingroup$ @AlexDegtyarev Yes, you are right, cohomology always measures nonexactness. The point is that nonexactness, at least from the definition, is not a concept related to geometry. It works in geometric settings, but it has no geometric meaning, if no further discussion is given (like in the link). So I am looking for this further discussion which reveals the geometric role of nonexactness in the complex setting. $\endgroup$ – Jjm Jan 21 '15 at 9:22
  • $\begingroup$ That's exactly my point :) A posteriori, de Rham cohomology have topological (= geometric) nature, whereas Dolbeault cohomology deal with much more rigid analytic properties. $\endgroup$ – Alex Degtyarev Jan 21 '15 at 10:01
  • $\begingroup$ @AlexDegtyarev But then, has Dolbeault cohomology no geometric meaning? Is it just computing some groups? In general, good analytic or algebraic machinery in geometry has always geometric background and meaning. $\endgroup$ – Jjm Jan 21 '15 at 11:06
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    $\begingroup$ Voting against the closers, I'd love to hear how experts think about this question. I'm not claiming to be an expert, but I put down a few thoughts below. $\endgroup$ – David E Speyer Feb 18 '15 at 14:45
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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.

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    $\begingroup$ that's really cool (but don't you want your first fraction to be the other way round? ie smooth functions/locally constant?) $\endgroup$ – bananastack Feb 18 '15 at 15:26
  • $\begingroup$ @bananastack Well, I either want them all upside down or none of them. I vote for all the way they are. Admittedly, they are "less than one" in the sense that the numerator is a subset of the denominator each time, but greater than one if you think in terms of "how hard are they to work with." $\endgroup$ – David E Speyer Feb 18 '15 at 16:50
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    $\begingroup$ yes, I was bugged by the fact that the fractions were less than one, but fair enough! $\endgroup$ – bananastack Feb 18 '15 at 17:06

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