Reality check on integrability of almost-complex surface. Let $(M^2,J)$ be an almost-complex surface. I do not assume integrability of $J$ now. I understand that from the Newlander-Nirenberg theorem, one of the several equivalences for the integrability of $J$ is that ${\rm d} = \partial + \overline{\partial}$. But I also know that in dimension $2$ the Nijenhuis tensor always vanishes. So this means that ${\rm d} = \partial + \overline{\partial}$ holds in dimension $2$, no matter which $J$ I begin with. So I should be able to give a direct proof of this, from scratch.
Ok, here's an attempt: assume that $(Z,\overline{Z})$ is a local complex frame for $M$, with $Z$ of type $(1,0)$ and $\overline{Z}$ of type $(0,1)$. If $(\zeta,\overline{\zeta})$ is the dual coframe, we automatically have that $\zeta$ is of type $(1,0)$ and $\overline{\zeta}$ is of type $(0,1)$.


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*Let $f\colon M \to \Bbb C$ be a $0$-form. Then ${\rm d}f = Z(f)\zeta + \overline{Z}(f)\overline{\zeta}$, and hence $\partial f = Z(f)\zeta$ and $\overline{\partial}f = \overline{Z}(f)\overline{\zeta}$. Thus for $0$-forms ${\rm d} = \partial + \overline{\partial}$ holds. Beautiful.

*Let $\alpha = f\zeta + g\overline{\zeta}$ be a $1$-form. We compute $${\rm d}\alpha = {\rm d}f\wedge \zeta + f\,{\rm d}\zeta + {\rm d}g\wedge \overline{\zeta} + g\,{\rm d}\overline{\zeta} = \big(-\overline{Z}(f) + Z(g) + f{\rm d}\zeta(Z,\overline{Z}) + g\,{\rm d}\overline{\zeta}(Z,\overline{Z})\big)\zeta\wedge \overline{\zeta}.$$Since ${\rm d}\alpha$ has no $(2,0)$ or $(0,2)$ component, I conclude that $\partial\alpha = \overline{\partial}\alpha = 0$. But ${\rm d}\alpha = 0$ need not be zero.
Question: what's the screw up?
 A: You can't conclude $\partial\alpha=\bar\partial\alpha=0$.  What you can conclude is $\pi^{2,0}(\partial\alpha)=\pi^{0,2}(\partial\alpha)=0$ and $\pi^{2,0}(\bar\partial\alpha)=\pi^{0,2}(\bar\partial\alpha)=0$.  You can't say much about the $(1,1)$-part (except that this is where your $d\alpha$ has to end up).
But of course that is enough: To get $d=\partial+\bar\partial$ on $\Omega^{1,0}$, you need to show $d(f\zeta)$ to have no $(0,2)$-part, which we know from above.  Similarly, you get $d=\partial+\bar\partial$ on $\Omega^{0,1}$.
A: You have shown that if $\alpha$ is a one-form, then $d\alpha$ is a $(1,1)$-form (since there aren't any $(2,0)$ or $(0,2)$ forms on a one-dimensional almost complex manifold). Now, if $\alpha$ is a $(1,0)$-form then
$$ \partial \alpha = (d \alpha)^{(2,0)} = 0, \,\,\bar{\partial} \alpha = (d \alpha)^{(1,1)} = d\alpha  $$
and in particular
$$ d\alpha = 0 + (d\alpha)^{(1,1)} = \partial \alpha + \bar{\partial} \alpha = (\partial + \bar{\partial})(\alpha). $$
Similarly, if $d\alpha$ is a $(0,1)$-form then
$$ \partial \alpha = (d \alpha)^{(1,1)} = d\alpha, \,\,\,\bar{\partial} \alpha = (d \alpha)^{(0,2)} = 0 $$
and in particular
$$ d\alpha = (d\alpha)^{(1,1)} + 0 = \partial \alpha + \bar{\partial} \alpha = (\partial + \bar{\partial})(\alpha). $$
The point is that if you assume that $\alpha = \alpha^{(1,0)} + \alpha^{(0,1)}$ is a general one-form, you can't conclude that $\partial \alpha$ or $\bar{\partial} \alpha = 0$ since
$$ \partial \alpha = \partial \left( \alpha^{(1,0)} + \alpha^{(0,1)} \right) = (d\alpha)^{(2,0)} + (d\alpha)^{(1,1)}, \\
\bar{\partial} \alpha = \bar{\partial} \left( \alpha^{(1,0)} + \alpha^{(0,1)} \right) = (d\alpha)^{(1,1)} + (d\alpha)^{(0,2)} $$
are neither the $(2,0)$ nor the $(0,2)$ components of $d\alpha$.
