Identification of complex plane as $R^2$.

Question

If we have following identification: $$(x,y)\to (z,\overline{z})$$

We will have $$\frac{\partial}{\partial x}= \frac{\partial}{\partial z}+\frac{\partial}{\partial \overline{z}}$$ and $$\frac{\partial}{\partial y}= i(\frac{\partial}{\partial z}-\frac{\partial}{\partial \overline{z}})$$ Also $$dx= \frac{dz+d\overline{z}}{2}, dy= \frac{dz-d\overline{z}}{2i}$$

for $f: \mathbb R^2\sim \mathbb C \to \mathbb C$, we have $$df= \frac{\partial f}{\partial z} dz+ \frac{\partial f}{\partial \overline{z}} d\overline{z}$$

Now Question: I was reading an article, There was one remark: Can someone please explain the following remark. What author intention to make this remark.

Remark: The length $\sqrt{2}$ of $dz$ and $d\overline{z}$, which is imposed by the notation, forces the dual system $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \overline{z}}$ to have the unnatural length $\frac{1}{\sqrt{2}}$; this is why the chain rule is preferable to duality in their notation.

This remark is Remark 1.1 on page-3, "Complex analysis and CR geometry" book by Giuseppe Zampieri

Answer 1

You need to be very careful here. If you want $z = x+iy$, $\overline z = x - iy$ to be a change of coordinates on $\mathbf R^2$, then certainly $iy$ should be a real number, which should probably give you pause. Indeed you can only think of this as a change of coordinates if you first think of $\mathbf C$ as an $\mathbf R$-vector space, which you then complexify, to get a space isomorphic to $\mathbf C^2$. (A remark for those who know a little more about this: this notation really starts to shine when you are working with the complexification of the real (co)tangent bundle to a complex manifold, like in Hodge theory. This is why.)

I don't think the remark you cite at the end makes any sense, can you say where it is from? Usually one writes $$ \frac{\partial}{\partial z} = \frac 1 2 ( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$$ and

$$ \frac{\partial}{\partial \overline z} = \frac 1 2 ( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y}).$$ There is nothing unnatural about this, I think, even though there is a sense in which these are "vectors of length $1/\sqrt 2$".

Answer 2

An hint : what I understand is that, by writing $z=x+iy$, $\bar z=x-iy$, your change of basis seen as a change of basis of $\mathbb R^2$ is $(0,1),(1,0)\rightarrow (1,1),(1,-1)$. The new vectors, which are still orthogonal, have a both lengh $\sqrt{2}$, i.e you induced some dilatation.

But then I don't understand what he means by "the chain rule is preferable to duality (?) in their notation" ...