关于通过球极投影将平面微分形式拉回至二维球面的计算疑问

Hey there! Let's walk through your questions step by step to clear things up.

First off, your initial calculation is completely on the right track—great job getting that far! Let's confirm each part:

1. Simplifying the pullback coefficients #

When you substitute (X = \frac{x}{1-z}) and (Y = \frac{y}{1-z}) into (\omega), the denominator (\sqrt{X^2 + Y^2}) simplifies to (\frac{\sqrt{x^2 + y^2}}{1-z}) (since we're excluding the north pole where (z=1), (1-z) is positive so we don't need absolute values). Canceling that with the numerator terms gives you the simplified coefficients (\frac{-y}{\sqrt{x^2+y2}}) and (\frac{x}{\sqrt{x^2+y2}})—that's perfect.

2. Computing the differentials (d(X \circ \varphi)) and (d(Y \circ \varphi)) #

Your use of the quotient rule for differentials here is correct. For (d\left(\frac{x}{1-z}\right)):
[
d\left(\frac{x}{1-z}\right) = \frac{(1-z)dx - x \cdot d(1-z)}{(1-z)^2} = \frac{(1-z)dx + x dz}{(1-z)^2} = \frac{1}{1-z}dx + \frac{x}{(1-z)^2}dz
]
Same logic applies to (d\left(\frac{y}{1-z}\right))—you nailed that step.

3. Simplifying the final pullback form #

If you expand the expression you have, you'll notice a nice simplification: the (dz) terms cancel out entirely! Let's do that:
[
\varphi^* \omega = \frac{-y}{\sqrt{x^2+y2}}\left(\frac{dx}{1-z} + \frac{x dz}{(1-z)^2}\right) + \frac{x}{\sqrt{x^2+y2}}\left(\frac{dy}{1-z} + \frac{y dz}{(1-z)^2}\right)
]
Breaking this into (dx/dy) and (dz) components:

• The (dx/dy) part becomes (\frac{-y dx + x dy}{(1-z)\sqrt{x^2+y2}})
• The (dz) part is (\frac{-xy dz + xy dz}{(1-z)^2\sqrt{x2+y^2}} = 0)

Since we're on (S^2), we can use (x^2 + y^2 = 1 - z^2 = (1-z)(1+z)) to simplify further if needed:
[
\varphi^* \omega = \frac{-y dx + x dy}{(1-z)\sqrt{(1-z)(1+z)}} = \frac{-y dx + x dy}{(1-z)^{3/2}\sqrt{1+z}}
]

4. Addressing your edit question #

Your confusion here comes from mixing up the direction of the map. Let's clarify:

• (\varphi: S^2 \setminus {N} \to \mathbb{R}^2) is the stereographic projection (from sphere to plane)
• (F: \mathbb{R}^2 \to S^2 \setminus {N}) is its inverse (from plane to sphere)

If you want the pullback of (\omega) (a form on (\mathbb{R}^2)) to (S^2), that's exactly what your initial calculation does with (\varphi^* \omega).

If you're thinking about expressing the pullback form in stereographic coordinates ((X,Y)) on (S^2), then yes—it's just (\omega) itself, because (X,Y) are the coordinate system induced by (\varphi). But if you want to write the form in terms of the ambient 3D coordinates ((x,y,z)) of the sphere, your original steps are the correct approach.

Hope that clears up all your questions!

备注：内容来源于stack exchange，提问作者random0620