Question: Let $\varphi_S: S^2\setminus \{N\} \to \mathbb C$ be given by $\varphi_S (x_1, x_2, x_3) = \left(\frac{x_1}{1 - x_3}, \frac{x_2}{1 - x_3}\right)$, i.e., the stereographic projection. Also let $\varphi_0^{-1}: \mathbb C \to \mathcal U_0$ be given by $\varphi_0^{-1} (z) = [1,z]$, where $\mathcal U_0 = \{[z_0, z_1] \in \mathbb {CP}^1; z_0 \neq 0\}$. Show that $$(\varphi_0^{-1} \circ \varphi_S)^*\omega_{FS} = \frac{1}{4} \omega_{std}$$ where $$\omega_{FS} = \frac{dx \wedge dy}{(1 + x^2 + y^2)^2}\ \ , \ \ \text{on} \ \ \mathcal U_0 \ \ \ \text{and} \ \ \ \omega_{std} = d\theta \wedge dh $$ where $0 \leq \theta \leq 2\pi$ and $-1\leq h \leq 1$ are cylindrical coordinates on the 2-sphere.

Attempt: The composition $\varphi_0^{-1} \circ \varphi_S$ is given by,

$$\varphi_0^{-1} \circ \varphi_S (z) = \left[1, \frac{x_1}{1-x_3} + i\frac{x_2}{1 - x_3}\right] $$

Then, identifying $z = x + i y$,

$$x = \frac{x_1}{1 -x_3} \implies dx = \frac{1}{1-x_3} dx_1 + \frac{x_1}{(1-x_3)^2}dx_3 \\y = \frac{x_2}{1 -x_3} \implies dy = \frac{1}{1-x_3} dx_2 + \frac{x_2}{(1-x_3)^2}dx_3$$ So it follows that

$$dx \wedge dy = \frac{1}{(1-x_3)^2}dx_1 \wedge dx_2 + \frac{x_2}{(1-x_3)^3}dx_3 \wedge dx_2 + \frac{x_1}{(1- x_3)^3} dx_3 \wedge dx_2$$ On the other hand

$$(1 + x^2 + y^2)^2 = \frac{4}{(1 - x_3)^2}$$ Therefore

$$\begin{aligned}(\varphi_0^{-1} \circ \varphi_S)^* \omega_{FS} &= \frac{1}{4}\left(dx_1 \wedge dx_2 + \frac{x_2}{1- x_3}dx_1 \wedge dx_3 + \frac{x_1}{1- x_3}dx_3 \wedge dx_2 \right) \\&= \frac{1}{4}\left(dx_1 \wedge dx_2 + \frac{x_2}{x_3 - 1}dx_3 \wedge dx_1 + \frac{x_1}{x_3 - 1}dx_2 \wedge dx_3 \right) \end{aligned}$$

I know that we may write $\omega_{std}$ as the pullback of the inclusion $\iota : S^2 \to \mathbb R^3$ as

$$\omega_{std} = x_3 dx_1 \wedge dx_2 + x_2dx_3 \wedge dx_1 + x_1dx_2 \wedge dx_3 $$

What am I missing? I couldn`t get any further than this. Any ideas?

  • 1
    $\begingroup$ Note that the formula for $\omega_{FS}$ is actually the formula for $(\varphi_0^{-1})^*\omega_{FS}$. I haven't done this specific computation, but polar coordinates might be easier. Then you only have to relate $r$ and $h$. $\endgroup$ – Ted Shifrin Mar 18 '19 at 17:05
  • $\begingroup$ Thanks @TedShifrin, I still couldn't make any progress. $\endgroup$ – Aaron Maroja Mar 18 '19 at 19:15

Here's one approach. Note that in $(h,\theta)$ coordinates, $\varphi_S$ is given by $$\varphi_S(h,\theta) = \sqrt{\frac{1+h}{1-h}}(\cos\theta,\sin\theta).$$ Pulling back $\omega_{FS}$ is pulling back $\dfrac{r\,dr\wedge d\theta}{(1+r^2)^2}$, and we get $$\frac{\sqrt{\frac{1+h}{1-h}} d\theta\wedge \frac{dh}{\sqrt{1+h}(1-h)^{3/2}}}{\left(1+\frac{1+h}{1-h}\right)^2} = \frac14 d\theta\wedge dh.$$

  • $\begingroup$ Thank you again for your answer. It is really helpful. I got $-\frac{1}{4} d\theta \wedge dh$ though, what does this sign mean? $\endgroup$ – Aaron Maroja Mar 19 '19 at 0:21
  • $\begingroup$ Well, because $r=\sqrt{1-h^2}$, orientation flips and $r\,dr\wedge d\theta = -h\,dh\wedge d\theta= h\,d\theta\wedge dh$. I guess it's interesting that the sign disappeared. You're right that my calculation is off by that sign. Oh, I see ... different $r$. The $r$ for stereographic projection increases as $h$ increases, so now I'm not sure their sign is correct. $\endgroup$ – Ted Shifrin Mar 19 '19 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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