Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

up vote 0 down vote accepted

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 $1/\sqrt 2$".

share|improve this answer
    
Remark, which i stated, is taken from the book.. I mentioned in the question. –  zapkm Apr 6 '12 at 9:13
    
i understood your concern... But does this answer my question?? –  zapkm Apr 6 '12 at 12:53
    
No, I guess it does not answer your question. All I can say is that I don't understand Zampieri's remark at all. –  Dan Petersen Apr 6 '12 at 12:59

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" ...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.