Missing minus sign in pullback calculation of a $1$ form on $\mathbb{C}$

Question

Consider $S^2$ with a coordinate chart given by the stereographic projection through the north pole. We identify $\mathbb{C}$ with $\mathbb{R}^2$, then for a point $(\theta,\phi)\in S^2$, the coordinate chart is given by: $$\psi_N:(\theta,\phi)\longmapsto\frac{\cos\theta/2}{\sin\theta/2}e^{i\phi}$$ Now consider the one form on $\mathbb{C}$: $$\omega_N=d\left(\frac{|z|^2-1}{|z|^2+1}\right)$$ The book I am reading from claims that the pullback of this one form to $S^2$ is: $$\psi_N^*(\omega_N)=d\cos\theta$$ However, I am getting an extra minus sign: $$\psi^*(\omega_N)=d\left(\frac{\tan^2\theta/2-1}{\tan^2\theta/2+1}\right)$$ $$\psi^*(\omega_N)=d\left(\frac{\sec^2\theta-2}{\sec^2}\right)$$ $$\psi^*(\omega_N)=d\left(1-2\cos^2\frac{\theta}{2}\right)$$ $$\psi^*(\omega_N)=-2d\left(\cos^2\frac{\theta}{2}\right)$$ $$\psi^*(\omega_N)=2\cos\frac{\theta}{2}\sin\frac{\theta}{2}d\theta$$ $$\psi^*(\omega_N)=\sin\theta d\theta$$ $$\psi^*(\omega_N)=-d\cos\theta$$ For the life of me I can't figure out where this extra minus sign coming from. Where did I mess up?

Answer 1

You begin with $$ \psi_N(\theta, \phi) = \cot \frac{\theta}{2} \, \exp i \phi, $$ but then in your calculation of the pullback, you substitute in for $\lvert z \rvert$ incorrectly. It should be $$ \lvert z \rvert = \biggl\lvert \cot \frac{\theta}{2} \biggr\rvert, $$ which leads to $$ \frac{\lvert z \rvert^2-1}{\lvert z \rvert^2+1} = \frac{\cot^2 \frac{\theta}{2} - 1}{\cot^2 \frac{\theta}{2} + 1} = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = \cos \theta. $$