Let $\Omega,\Omega'\subseteq\Bbb C^n$ open, $F:\Omega\to\Omega'$ holomorphic invertible function; it's a variable change, so let's call $F(z)=\tilde z$. Let $r:\Omega\to\Bbb R$ twice differentiable, and $\tilde r:\Omega'\to\Bbb R$ such that $r=\tilde r\circ F$.

Now we want to differentiate $r=\tilde r\circ F$. To be precise we want to compute $$ \partial_{z_h}\partial_{\bar{z_k}}(\tilde r\circ F)\;\;. $$

First of all I should compute $\partial_{\bar{z_k}}(\tilde r\circ F)$.

Now I'd use the chain rule to get $$ \partial_{\bar{z_k}}(\tilde r\circ F)(z)= \nabla\tilde r(\tilde z)\cdot\partial_{\bar{z_k}}\tilde z= \sum_{i=1}^{n}\partial_{\widetilde{z_i}}\tilde{r}(\tilde z)\partial_{\bar{z_k}}\tilde z_i $$ but the book says that

$$ \partial_{\bar{z_k}}(\tilde r\circ F)(z)= \sum_{i=1}^{n}\partial_{\widetilde{z_i}}\tilde{r}(\tilde z)\partial_{\bar{z_k}}\tilde z_i+ \sum_{i=1}^{n}\partial_{\bar{\widetilde{z_i}}}\tilde{r}(\tilde z)\overline{\partial_{{z_k}}\tilde z_i}\;\; $$ so I missed the second sum.

I suspect the fact that $df=\partial f+\bar{\partial}f$ and $df=2\Re\partial f$ are taken in account but I can't put all the pieces togheter.

(from this, I think it should be easy to compute the second derivative).

EDIT As suggested by the user mrf, this computation is simply a Wirtinger derivative. However I don't know nothing about this and I wasn't able to find anything on internet or on the Ahlfors.

I really want (and need) to understand how these basic real/complex differential operators behaves, how do they work.

Can someone suggest me an appropriate reference?

Many thanks!

  • 2
    $\begingroup$ See Wikipedia for the chain rule expressed with Wirtinger derivatives. $\endgroup$ – mrf Mar 29 '15 at 10:26
  • $\begingroup$ EXACTLY!! That's the point! These rules are the ones I searched for! But my book wants me to deduce them by a previous discussion on complex differentiabilty; and here's I'm stuck. Can you give a reference on which I can find that proof? Many many thanks $\endgroup$ – Joe Mar 29 '15 at 15:04
  • $\begingroup$ This might be what you're looking for: wcherry.math.unt.edu/math5410/lecture9_24.pdf The derivation is not very simple. Very, very tedious. There are 64 terms to keep track of.. $\endgroup$ – Cameron Williams Mar 29 '15 at 20:43

Probably, the best way to do this is via "best linear approximation". A linear operator $L\in GL_{2n}(\mathbb{R})$ is the differential of $F$ at $p$ if and only if $$F(p+h)=F(p)+Lh+o(\vert h\vert)$$ as $\vert h\vert\to0$. Let us suppose that we fixed the basis of $\mathbb{R}^{2n}$ as $\{x_1,\ldots, x_n, y_1, \ldots, y_n\}$ so that $\{x_1+iy_1,\ldots, x_n+iy_n\}$ is a basis for $\mathbb{C}^n$. Then, writing $h=(h_1,\ldots, h_{2n})$ and thus interpreting $\overline{h}$ as $(h_1,\ldots, h_n, -h_{n+1},\ldots, -h_{2n})$, we obtain a $\mathbb{C}$-linear operator $L_1$ and a $\mathbb{C}$-antilinear operator $L_2$ such that $$F(p+h)=F(p)+L_1h+L_2\overline{h}+o(|h|)\;.$$ We call $L_1=\partial F\vert_p$ and $L_2=\overline{\partial}F\vert_p$.

Also, we have that there is a vector $u\in\mathbb{C}^n$ such that $$r(q+h)=r(q)+\langle u,h\rangle+\langle \overline{u}, \overline{h}\rangle + o(|h|)$$ and we call $u=\partial r\vert_p$. (As $r$ is real valued, $\dfrac{\partial r}{\partial \overline{z_j}}$ and $\dfrac{\partial \overline{r}}{\partial \overline{z_j}}$ are the same).

So, plugging $F$ in, we have $$r(F(p+h))=r(F(p)+L_1h+L_2\overline{h}+o(|h|))=r(F(p))+\langle u, L_1h\rangle + \langle u, L_2\overline{h}\rangle + \langle \overline{u}, \overline{L_1h}\rangle + \langle\overline{u},\overline{L_2\overline{h}}\rangle+ o(|h|)$$ i.e. $$r(F(p+h))=r(F(p))+(u^tL_1 + \overline{u^tL_2})h+(u^tL_2+\overline{u^tL_1})\overline{h}+o(|h|)$$ This means that $$\partial (r\circ F)=(\partial r)^t\partial F + (\overline{\partial}r)^t\partial\overline{F}$$

  • $\begingroup$ Hey Wisefool! Too nice! You are helping me, I appreciate this! Many thanks! :-) $\endgroup$ – Joe Apr 9 '15 at 4:48

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