TITLE: Why do the Wirtinger derivatives behave like actual partial derivative operators?
QUESTION [6 upvotes]: Despite the fact that they're not partial derivative operators, the Wirtinger derivatives obey things like the chain rule.  Of course I can prove such things by manipulating formulas, but this gives no intuition for what's really happening.  Is there a deep reason that everything just seems to work out with these things?

REPLY [4 votes]: I'm not sure it would count as a "deep" reason, but the Wirtinger derivatives are defined as the dual basis to the basis $\{dz,\, d\overline{z}\}$, resp. in higher dimensions to $\{ dz_i,\, d\overline{z}_i : 1 \leqslant i \leqslant n\}$ of the complexified real cotangent space $T_{z}^\ast(U)\otimes \mathbb{C}$.
Now, if we have a basis $\{\omega_i\}$ of the cotangent space of a manifold $M$ around a point $p\in M$ with a coordinate chart $x = (x_1,\dotsc, x_m)$, and the corresponding dual basis $\{\delta^i\}$ of the tangent space (all varying smoothly with the base point), we have
$$df = \sum_{j} \frac{\partial f}{\partial x_j}\cdot dx_j = \sum_i \delta^i(f)\cdot \omega_i$$
for smooth functions $f\colon M \to \mathbb{C}$. If we have another manifold $N$, $q\in N$, and a smooth function $g \colon N\to M$ with $g(q) = p$, and around $y$ a coordinate chart $y = (y_1,\dotsc,y_n)$, a basis $\{\alpha_k\}$ of the cotangent space, and the dual basis $\{\varepsilon^k\}$ of the tangent space, we have for the pull-back by $g$ on the one hand
$$g^\ast(df) = d(g^\ast f) = d(f\circ g) = \sum_k \varepsilon^k(f\circ g)\cdot \alpha_k,$$
and on the other hand, using the representation of $df$ in the basis $\{\omega_i\}$,
$$g^\ast(df) = g^\ast\left(\sum_i \delta^i(f)\cdot\omega_i \right) = \sum_i g^\ast(\delta^i(f))\cdot g^\ast(\omega_i) = \sum_i \delta^i(f)\circ g \cdot g^\ast(\omega_i).$$
Generally, we have no cute formula for the $g^\ast(\omega_i)$, we must use the coefficients of the $\omega_i$ with respect to the basis $dx_j$ (or some other) to compute $g^\ast(\omega_i)$. But when the $\omega_i$ have the nice property of being the differentials of actual functions, as is the case for $dz_r$ and $d\overline{z}_r$, say $\omega_i = d\zeta_i$, then we have the nice pull-back formula for the $g^\ast(\omega_i)$:
$$g^\ast(\omega_i) = g^\ast(d\zeta_i) = d(\zeta_i\circ g) = \sum_k \varepsilon^k(\zeta_i\circ g)\cdot \alpha_k,$$
and hence
$$\begin{align}
g^\ast(df) &= \sum_i \delta^i(f)\circ g\cdot g^\ast(d\zeta_i)\\
&= \sum_i \left(\delta^i(f)\circ g \cdot \sum_k \varepsilon^k(\zeta_i\circ g)\cdot \alpha_k\right)\\
&= \sum_k \left(\sum_i \delta^i(f)\circ g\cdot \varepsilon^k(\zeta_i\circ g)\right)\alpha_k
\end{align}$$
which produces the chain rule
$$\varepsilon^k(f\circ g) = \sum_i \delta^i(f)\circ g \cdot \varepsilon^k(\zeta_i\circ g).$$
Now when we have $\omega_i = d\zeta_i$, it is customary to denote the dual basis by $\dfrac{\partial}{\partial \zeta_i}$ instead of $\delta^i$, and if we have the same situation on $N$, $\alpha_k = d\beta_k$, then the chain rule reads
$$\frac{\partial (f\circ g)}{\partial \beta_k} = \sum_i \frac{\partial f}{\partial \zeta_i}\circ g\cdot \frac{\partial (\zeta_i\circ g)}{\partial \beta_k}.$$
The Wirtinger derivatives are the special case with $\zeta_i = z_i$ for $1 \leqslant i \leqslant m/2$ and $\zeta_i = \overline{z}_{i-m/2}$ for $m/2 < i \leqslant m$ (or some other numeration), and analogous for the $\beta_k$.