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.

  • 7
    $\begingroup$ 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 Nov 23 '10 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 Nov 23 '10 at 14:14
  • $\begingroup$ @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 Nov 23 '10 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 Nov 23 '10 at 16:16

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).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.