5
$\begingroup$

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?

$\endgroup$
4
  • 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 $h$. $\endgroup$ Mar 18, 2019 at 17:05
  • $\begingroup$ Thanks @TedShifrin, I still couldn't make any progress. $\endgroup$ Mar 18, 2019 at 19:15
  • $\begingroup$ I stuck on the same point as you and I still don't see what went wrong... $\endgroup$
    – user388493
    Jan 20, 2021 at 16:15
  • $\begingroup$ @Lambda8 Have you tried polar coodinates as Ted Shiffrin suggested in his answer. It works. $\endgroup$ Jan 21, 2021 at 18:12

1 Answer 1

2
$\begingroup$

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 $$\frac{\sqrt{\frac{1+h}{1-h}} d\theta\wedge \frac{dh}{\sqrt{1+h}(1-h)^{3/2}}}{\left(1+\frac{1+h}{1-h}\right)^2} = \frac14 d\theta\wedge dh.$$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Mar 19, 2019 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. $\endgroup$ Mar 19, 2019 at 1:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .