Why do the Wirtinger derivatives behave like actual partial derivative operators?

Question

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?

Answer 1

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 $\beta_k$.