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Let $M$ be a complex manifold, and $\omega$ be a $(p,q)$-form. Then $d\omega$ is an element of $\Omega^{p+1,q}(M)\oplus\Omega^{p,q+1}(M)$, so that $d = \partial + \overline{\partial}$, where $\partial$ and $\overline{\partial}$ are the Dolbeault operators.

Now let $M$ be almost complex. It is commonly stated that $d = \partial + \overline{\partial}$ only holds for complex manifolds, and not for almost complex manifolds. But why is this? After extending $d$ to also be complex linear, if $\omega = \sum_i f(z)dz^i$ is a $(0,1)$-form, I'd say that we would have $ d\omega = \sum_i df\wedge dz_i = \sum_{i,j} \frac{\partial f}{\partial z^j}dz^j\wedge dz^i + \frac{\partial f}{\partial \overline{z}^j}d\overline{z}^j\wedge dz^i,$ which clearly does not have a $(0,2)$-part. Why is this wrong?

On the other hand, let $X, Y$ be antiholomorphic tangent vectors, then $d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) = -\omega([X,Y])$. Since $M$ is not nessecarily complex, $[X,Y]$ is not nessecarily also antiholomorphic, so that this term does not nessecarily vanish. But $d\omega$, being a 2-form, can only give a nonzero result if it has a $(0,2)$-part. So from this I can see that it has to have one, but I can't see why this contradicts the calculation of $d\omega$ above.

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    $\begingroup$ In your computation of $d \omega$ you are assuming that $M$ is complex, since you are using holomorphic coordinates $z_i$. For a general almost complex manifold it makes no sense to write $\partial /\partial z$ and $\partial / \partial \bar{z}$, just because no holomorphic coordinates are available. There is just a complex structure on the tangent space, but to write $f(z)$ you need such a structure to be integrable. $\endgroup$
    – Francesco Polizzi
    Commented Nov 23, 2010 at 13:28
  • $\begingroup$ For an explicit example of $d \ne \partial + \bar \partial$ consider $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ with coordinates $x_i, y_i$. The tan. sp. has basis $\partial/\partial x_i, \partial/\partial y_i$. The usual complex structure of $\mathbb{C}^n$ uses the almost complex structure $i(\partial/\partial x_i) = \partial/\partial y_i$ (this determines $i$ on the other basis vectors using $i^2 = -1$. If instead you used another complex structure $J(\partial/\partial x_i) = -\partial/\partial y_i$ and then proceeded to use $J$ to define $(p,q)$ forms then $d \ne \partial + \bar \partial$ $\endgroup$
    – solbap
    Commented Nov 23, 2010 at 14:14
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    $\begingroup$ @solbap: Your example doesn't work. You just replaced the original J by its negative, which is still an integrable complex structure. Francesco and Eric gave the correct reason. $\endgroup$
    – Spiro Karigiannis
    Commented Nov 23, 2010 at 15:01
  • $\begingroup$ hmm yeah it seems I've just reversed the orientation of $\mathbb{C}^n$. I guess I was just thinking that the identity map $(\mathbb{C}^n,i)→(\mathbb{R}^{2n},J)$ doesn't satisfy $i∘D(id)=D(id)∘J$,so this doesn′t give a holomorphic chart for $\mathbb{R}^{2n}$,but I guess $\overline{\mathbb{C}^n}$ does. $\endgroup$
    – solbap
    Commented Nov 23, 2010 at 16:16

2 Answers 2

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In writing $\omega$ you used a symbol $dz$ which doesn't make sense unless there is a holomorphic coordinate. Your $dz$ should really be an element of a frame of (1,0) 1-forms, which need not be closed (as you have assumed).

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Just to follow up on Eric's correct answer: when you have an almost complex structure $J$, you can decompose $1$-forms into type $(1,0)$ and $(0,1)$. Locally, you can find a local basis $e^1, \ldots, e^n$ of $(1,0)$-forms, but these are not of the form $dz^1, \ldots, dz^n$. Indeed, as Eric mentioned, we do not have local holomorphic coordinates. Then $\bar e^1, \ldots, \bar e^n$ are a local basis of $(0,1)$ forms. Now if we compute $de^i$, it is a $2$-form, so it can be written in the form \begin{equation*} de^i = a^i_{jk} e^j \wedge e^k + b^i_{jk} e^j \wedge \bar e^k + c^i_{jk} \bar e^j \wedge \bar e^k. \end{equation*} The almost complex structure $J$ is integrable if and only if all the $c^i_{jk}$'s are zero.

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