Given $f:\Bbb C^{n}\to\Bbb C$ identified with $f:\Bbb R^{2n}\to\Bbb C$, in a book I read that $$ \partial_x f\,dx+\partial_yf\,dy=\partial_zf\,dz+\partial_{\bar z}f\,d\bar z $$ and that this could be rewritten as $$ df=\partial f+\bar{\partial}f $$ and this last equation defines $\partial f$ and $\bar{\partial}f$.

My problem is to write explicitly the last two objects.

Now $\partial_x:=(\partial_{x_1},\dots,\partial_{x_n})$ and $dx:=(dx_1,\dots,dx_n)$ and so on for the other variables.

Hence we can think at $\partial_x f\,dx$ as a scalar product between $\partial_x f$ and the formal vector $dx$. In this way, writing formally $dz=dx+idy$ and $d\bar z=dx-idy$ we deduce easily the first equation. However I would like to write explicitly both its sides.

LHS it's obviously $df$ (just a computation). Now it would be nice to write explicitly even RHS, in order to understand how to define properly $\partial f$ and $\bar{\partial}f$.

We know that $\partial_z=\frac12(\partial_x-i\partial_y)$ and $\partial_\bar z=\frac12(\partial_x+i\partial_y)$. My problem is to understand how $dz$ works.

We can assume wlog $n=1$. As a differential form, $dx$ is a covector: $dx=(1,0)$ and similarly for $dy=(0,1)$. What about $dz$? We know it's equal to $dx+idy=(1,0)+i(0,1)$, but the "ambient" in which $dz$ lives is 1-dim covector.

I think the point is inside the last observation.

The only way to obtain the first equality directly is to write $dz=(1,-1)$ and $d\bar z=(1,1)$, i.e. $i(0,1)=(0,-1)$. How can this be possibile?

I need someone shading lights on this!

Thank you all!


I don't see how you arrive at your conclusion. In $x,y$ coordinates,

$$ \begin{align}\frac{\partial f}{\partial x} \, \mathrm{d}x + \frac{\partial f}{\partial y} \, \mathrm{d}y &= \frac{\partial f}{\partial x} (1,0) + \frac{\partial f}{\partial y} (0,1) \\&= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \end{align}$$

$$ \begin{align}\frac{1}{2}\left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right) \, \mathrm{d}z &+ \frac{1}{2}\left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) \, \mathrm{d}\bar{z} \\&= \frac{1}{2}\left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right) (1,i) + \frac{1}{2}\left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) (1,-i) \\&= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \end{align}$$

  • $\begingroup$ I'm drunk but I'm not aware of being it. It's the only explanation :( thanks a lot $\endgroup$ – Joe Jan 9 '15 at 16:33

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