Geometric Derivation of the Complex D-Bar Operator

1. Apr 20, 2014

bolbteppa

This picture

from https://www.amazon.com/Visual-Complex-Analysis-Tristan-Needham/dp/0198534469 is all you need to derive the Cauchy-Riemann equations, i.e. from the picture we see $i \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$ should hold so we have

$$i \frac{\partial f}{\partial x} = i \frac{\partial (u+iv)}{\partial x} = \frac{\partial (u+iv)}{\partial y} \rightarrow C \ R \ Eq's$$

Is there a similar picture-derivation of the operators

$$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial }{\partial x} - i\frac{\partial }{\partial y})$$

$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial }{\partial x} + i\frac{\partial }{\partial y})?$$

The fact the differential forms can be visualized in terms of sheets tells me there can be one, any ideas?

Last edited by a moderator: May 6, 2017
2. Apr 20, 2014

homeomorphic

I don't know that you would exactly derive those operators. I understand them as definitions.

The way I think of them is this. Because you are dealing with derivatives, you are linearizing, so it really just boils down to linear algebra. So, what's the corresponding linear algebra result? It is the fact that you can decompose any linear transformation from R^2 to R^2 into one that is conformal plus one that is anti-conformal. So, I suppose I'll leave it as an exercise for the reader to establish that. It is possible to argue geometrically. Visually, the difficulty is that it's hard to visualize the sum of two arbitrary linear transformations. Of course, it's possible there could be a better argument than the one I have in mind.

Anyway, for a smooth mapping of the complex plane to itself, the total derivative at each point decomposes this way with partial z being the conformal part and partial z bar being the anti-conformal part.

This is a little bit post hoc, but a way to make it plausible that those particular formulas extract the conformal and anti-conformal parts in that partial x actually gives you the complex derivative. Partial y gives you i times it, so you need to multiply by -i. Then add and divide by 2. Similarly for partial z bar.

A more algebraic approach is to ask what partial z should do to x or y by writing them in terms of z and z bar. If you hit x with partial z, you ought to get 1/2, and if you hit y with it, you should get -i/2, and that suggests that partial z is what it is.

If you are familiar with the identification of tangent vectors with differential operators, you can interpret them as vector fields, but that doesn't really tell you why you want those particular ones, as far as I know.