Complex chain rule

Question

Suppose $V_1$ and $V_2$ are open sets in $\mathbb{C}^m$ and $\mathbb{C}^n$, repsectively, $f:V_1\mapsto V_2$ and $g:V_2\mapsto \mathbb{C}^k$ are $\mathbb{R}-$ differentiable. We know that the real chain rule for the differential of $g\circ f$ at a point $z$ gives
$$D_z(g\circ f)=D_{f(z)}g\circ D_zf.$$ Can someone explain how can we prove that
$$\partial_z(g\circ f)=\partial_{f(z)}g\circ \partial _z f+\overline{\partial}_{f(z)}g\circ \overline{\partial}_zf, $$and
$$\overline{\partial}_z(g\circ f)=\overline{\partial}_{f(z)}g\circ \partial _z f+\partial_{f(z)}g\circ \overline{\partial}_zf? $$
We use the following notation:
$$\partial _z\varphi=\sum_p\frac{\partial \varphi}{\partial z_p}dz_p\quad \& \quad \overline{\partial}_z\varphi=\sum_p\frac{\partial \varphi}{\partial \overline{z}_p}d\overline{z}_p . $$ Also that would be great if someone knows a reference for that kind of stuff. Thanks.

Answer 1

I am going to avoid your notation and use more standard notation, but you can make the translation. I'm also going to do the case $\Bbb C\to\Bbb C$ and you can easily generalize it. So we start with $$df = \partial f + \bar\partial f = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial\bar z}d\bar z.$$ Now, if $z=g(w)$, then $$dz = \frac{\partial g}{\partial w}dw + \frac{\partial g}{\partial\bar w}d\bar w,$$ and we get $d\bar z$ by conjugating, so $$d\bar z = \overline{\frac{\partial g}{\partial w}}d\bar w + \overline{\frac{\partial g}{\partial\bar w}}dw.$$ The chain rule tells us that to compute $d(f\circ g)$ we just substitute these latter expressions. We obtain \begin{align*} d(f\circ g) &= \frac{\partial f}{\partial z}\left(\frac{\partial g}{\partial w}dw + \frac{\partial g}{\partial\bar w}d\bar w\right) + \frac{\partial f}{\partial\bar z}\left(\overline{\frac{\partial g}{\partial\bar w}}dw+ \overline{\frac{\partial g}{\partial w}}d\bar w\right) \\ &= \left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial\bar w}}\right)dw + \left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial w}}\right)d\bar w\,. \end{align*} From this you read off that \begin{align*} \frac{\partial (f\circ g)}{\partial w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial\bar w}}\,, \quad\text{and} \\ \frac{\partial (f\circ g)}{\partial\bar w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\overline{\frac{\partial g}{\partial w}}\,. \end{align*}

You will notice this is a lot like the Einstein summation convention in tensor analysis: We sum a product of a $w$ derivative and a $dw$ and a product of a $\bar w$ derivative and a $d\bar w$ in every case. It may be useful to keep in mind that $\overline{\dfrac{\partial g}{\partial w}} = \dfrac{\partial\bar g}{\partial\bar w}$ and $\overline{\dfrac{\partial g}{\partial \bar w}} = \dfrac{\partial\bar g}{\partial w}$.

Thus, we can finally rewrite this as: \begin{align*} \frac{\partial (f\circ g)}{\partial w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial w}+\frac{\partial f}{\partial\bar z}\frac{\partial\bar g}{\partial w}\,, \quad\text{and} \\ \frac{\partial (f\circ g)}{\partial\bar w} &= \frac{\partial f}{\partial z}\frac{\partial g}{\partial\bar w}+\frac{\partial f}{\partial\bar z}\frac{\partial\bar g}{\partial\bar w}\,. \end{align*}