How to pull back the differential form $\omega = \frac{-ydx+xdy}{\sqrt{x^2+y^2}}$ to $S^2$

Question

Consider the stereographic projection chart on $S^2$ which doesn't include the north pole $$(X,Y)=\varphi(x,y,z)=\left(\frac{x}{1-z}, \frac{y}{1-z}\right).$$

I want to pull back the 1-form $\omega = \frac{-ydx+xdy}{\sqrt{x^2+y^2}}$ to $S^2$ from $\mathbb{R}^2$ to $S^2$ but I am not sure about a step in the calculation. The definition of pullback of a form under a smooth map $\varphi $ is

$$\varphi^* \omega=(f \circ \varphi) d\left(X \circ \varphi\right)+(g \circ \varphi) d\left(Y \circ \varphi\right)$$

Then,

$$ \begin{aligned} &\varphi^*\omega =\frac{\frac{-y}{1-z}}{\sqrt{\frac{x^2}{(1-z)^2}+\frac{y^2}{(1-z)^2}}} d\left(X \circ \varphi\right)+\frac{x}{\sqrt{\frac{x^2}{(1-z)^2}+\frac{y^2}{(1-z)^2}}} d\left(Y \circ \varphi\right) \\ & =\frac{-y}{\sqrt{x^2+y^2}} d(X \circ \varphi)+\frac{x}{\sqrt{x^2+y^2}} d(Y \circ \varphi) \\ & \end{aligned} $$

How do I compute $d(X\circ \varphi)$ and $d(Y\circ \varphi)$. Intuitively, this feels like some sort of product rule would have to occur.

$$d\frac{x}{1-z} = \frac{1}{1-z}dx + \frac{x}{(1-z)^2}dz$$ $$d\frac{y}{1-z} = \frac{1}{1-z}dy + \frac{y}{(1-z)^2}dz$$

Then

$$\phi^*\omega = \frac{-y}{\sqrt{x^2+y^2}}\left[\frac{1}{1-z}dx + \frac{x}{(1-z)^2}dz\right] + \frac{x}{\sqrt{x^2+y^2}}\left[\frac{1}{1-z}dy + \frac{y}{(1-z)^2}dz\right]$$ In the end I should get some 1-form on $S^2$. Is this calculation correct?

Thank you!

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EDIT:

Is it actaully the case that if I wanted to pull this back onto $S^2$, what I shuold really do is just consider the map $F:(X,Y)\mapsto (x,y)$ then the pullback will just be

$$F^*\omega = \frac{-YdX+XdY}{\sqrt{X^2+Y^2}}$$

where the upper case $X$ and $Y$ are in stereographic coordinates? Is it this easy?

Answer 1

To answer your question in coordinates if we have $f:U \longrightarrow \mathbb{R}$, for $U \subset\mathbb{R}^n$ and $U$ open. : $$df=\sum_{i=1}^n\frac{\partial f}{\partial x_i}dx_i$$

Answer 2

Imho the simplest but still a bit tedious approach is to use the inverse stereographic projection: \begin{align} \pmatrix{ x\\y\\z}=\frac{1}{1+X^2+Y^2}\pmatrix{2X\\2Y\\-1+X^2+Y^2} \end{align} where $x,y,z$ are the coordinates in $\mathbb R^3$ and $X,Y$ the coordinates on $S^2\,.$ By ordinary calculus, \begin{align} dx&=2\frac{(1-X^2+Y^2)\,dX-2XY\,dY}{(1+X^2+Y^2)^2}\,,\\[2mm] dy&=2\frac{-2XY\,dX+(1+X^2-Y^2)\,dY}{(1+X^2+Y^2)^2}\,. \end{align} This should give \begin{align} -y\,dx+x\,dy&=4\frac{-Y(1-X^2+Y^2)\,dX+2XY^2\,dY-2X^2Y\,dX+ X(1+X^2-Y^2)\,dY}{(1+X^2+Y^2)^3}\\[2mm] &=4\frac{-Y\,dX-X^2Y\,dX-Y^3\,dX+XY^2\,dY+ X\,dY+X^3\,dY}{(1+X^2+Y^2)^3}\,,\\[2mm] &=4\frac{-Y(1+X^2+Y^2)\,dX+X(1+X^2+Y^2 )\,dY}{(1+X^2+Y^2)^3}\,,\\[2mm] &=4\frac{-Y\,dX+X\,dY}{(1+X^2+Y^2)^2}\,. \end{align} Using \begin{align} x^2+y^2=4\frac{X^2+Y^2}{(1+X^2+Y^2)^2} \end{align} we obtain \begin{align}\boxed{\phantom{\BIGG|} \frac{-y\,dx+x\,dy}{\sqrt{x^2+y^2}}=\frac2{1+X^2+Y^2}\frac{-Y\,dX+X\,dY}{\sqrt{X^2+Y^2}}\,.\quad} \end{align} The RHS of this is the pullback of the LHS to $S^2\,.$