# Pullback of Fubini-Study form on $\mathbb {CP}^1$

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?

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

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 $$\frac{\sqrt{\frac{1+h}{1-h}} d\theta\wedge \frac{dh}{\sqrt{1+h}(1-h)^{3/2}}}{\left(1+\frac{1+h}{1-h}\right)^2} = \frac14 d\theta\wedge dh.$$
• 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? Mar 19, 2019 at 0:21
• 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. Mar 19, 2019 at 1:17