I am trying to some computations about the Kähler structure of $(S^2,\omega, J)$, where $\omega_x(u,v)=\langle x,u\times v\rangle $ and $J_x(u)=x\times u$. First I have tried to check that if we have $\phi:S^2\rightarrow \mathbb{R}^2$ to be the stereographic projection we will have that $\phi^*(\frac{4dx\wedge dy}{(1+x^2+y^2)^2})$, but I think I have gotten a sign wrong but I can't find my mistake.
Recall that in cylindrical coordinates $\omega$ is given by $d\theta \wedge dz$. Also in cylindrical coordinates the local representation of the stereographic projection is $(\frac{\sqrt{1-z^2}\cos(\theta)}{1-z},\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})$.Now we will have that
$\phi^* \tau = \frac{4d(\frac{\sqrt{1-z^2}\cos(\theta)}{1-z})\wedge d(\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})}{((\frac{\sqrt{1-z^2}\cos(\theta)}{1-z})^2+(\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})^2+1)^2}$
Now notice that $d(\frac{\sqrt{1-z^2}\cos(\theta)}{1-z})=-\sin(\theta)\frac{\sqrt{1-z^2}}{1-z}d\theta+\frac{\cos(\theta)}{(1-z)\sqrt{1-z^2}}dz=d(\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})=\cos(\theta)\frac{\sqrt{1-z^2}}{1-z}d\theta+\frac{\sin(\theta)}{(1-z)\sqrt{1-z^2}}dz$
And so we get that
$d(\frac{\sqrt{1-z^2}\cos(\theta)}{1-z})\wedge d(\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})=\frac{\cos(\theta)^2}{(1-z)^2}dz\wedge d\theta+\frac{\sin(\theta)^2}{(1-z)^2}dz\wedge d\theta= \frac{1}{(1-z)^2}dz\wedge d\theta$. Note also that $((\frac{\sqrt{1-z^2}\cos(\theta)}{1-z})^2+(\frac{\sqrt{1-z^2}\sin(\theta)}{1-z})^2+1)^2= (\frac{(1-z^2)}{(1-z)^2}+1)^2=\frac{4}{(1-z)^2}$
If anyone can spot a mistake I would appreciate it, since I think there should be a mignus sign somewhere.
The other thing is that I am not understanding is how can I actually represent $J$ in local coordinates, say for example in cyclindrical ones? I think I need to use $d\phi$ and it's inverse but this would lead to alot of computations, and so I am wondering if there is an easier way to see it.
Any help is appreciated. Thanks in advance.
It would probably be easier to pull back $\omega$ by $\phi^{-1}$, but just write $\tau$ in polar coordinates to start with, and this won't be bad at all. In particular, $$\phi(z,\theta) = R(\cos\theta,\sin\theta), \quad\text{with}\quad R=\sqrt{\frac{1+z}{1-z}}.$$ Write $R^2 = \dfrac{1+z}{1-z}$, and so $R\,dR = \dfrac{dz}{(1-z)^2}$. Thus, $$\phi^*\tau = \phi^*\left(\frac{4R\,dR\wedge d\theta}{(1+R^2)^2}\right) = \frac{4\,dz/(1-z)^2}{(2/(1-z))^2}\wedge d\theta = dz\wedge d\theta,$$ as required.