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.

Suppose $\varphi$ is a smooth map on $\mathbb{C}$. For a function $f$, we define a $0$-form $\varphi^*f$ as $\varphi^*f=f\circ\varphi$. Also, $$ \varphi^*\,dx=\frac{\partial\varphi_1}{\partial x}\,dx+\frac{\partial\varphi_1}{\partial y}\,dy, \qquad \varphi^*dy=\frac{\partial\varphi_2}{\partial x}\,dx+\frac{\partial\varphi_2}{\partial y}\,dy, $$ where $\varphi_1$ is the $x$ component of $\varphi$ and $\varphi_2$ is the $y$ component of $\varphi$.

I'm curious, are there sensible analogous definitions for $\varphi^*dz$ and $\varphi^*d\bar{z}$ for the complex case?

I know $dz=dx+idy$, so considering it as a $1$ form I thought maybe \begin{align*} \varphi^*dz &= \varphi^*(dz+idy)\\ &= (\varphi^* 1)\varphi^* dx+(\varphi^* i)\varphi^*dy\\ &= \varphi^*dx+i\varphi^* dy \end{align*} but I think it's not right to equate a constant with a constant function. Likewise in the case of $dz=dx-idy$. What is the proper definition for the complex differential?

share|improve this question
    
In your equations $\phi^*dx=\ldots$, $\phi^*dy=\ldots$ the $dx$, $dy$ on the left side and the $dx$, $dy$ on the right side do not mean the same thing. –  Christian Blatter Apr 23 '12 at 8:23
add comment

1 Answer 1

up vote 1 down vote accepted

One can use the Wirtinger derivatives to define $\phi^*dz$. These derivatives are the operators defined by $$ \frac{\partial}{\partial z} = \frac 12 \Bigl( \frac{\partial}{\partial x} - i\frac{\partial}{\partial y} \Bigr) \quad \text{and} \quad \frac{\partial}{\partial \overline z} = \frac 12 \Bigl( \frac{\partial}{\partial x} + i\frac{\partial}{\partial y} \Bigr). $$ These are defined as to make $\partial z/\partial z = 1$, $\partial z /\partial \overline z = 0$, and so on. Then we can set $$ \phi^* dz = \frac{\partial \phi}{\partial z} dz \quad \text{and} \quad \phi^* dz = \frac{\partial \phi}{\partial \overline z} d\overline z. $$ If you work through the algebra, this should give the same result as your attempt.

This is a very ad-hoc way of defining these things, but it can be justified by saying that it makes things work like we want. A more conceptial way to define pullbacks would be to consider a complex structure on the underlying smooth space and see how the induced splittings of the tangent bundle into holomorphic and antiholomorphic forms force these definitions on us. You could look at Daniel Huybrecht's "Complex geometry" for some discussion of this, but you might have to work out the details yourself.

share|improve this answer
add comment

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.