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Consider $S^2$ with the two stereographic projections obtained by removing respectively the North pole and the South pole, find the gluing function obtained through the definition of Riemannian surface with the two charts given by the two projections.

We call $\varphi_1$ the stereographic projection from $S^2- \{N \}$ to $\mathbb{C}$ and $\varphi_2$ the stereographic projection from $S^2- \{S\}$ to $\mathbb{C}$. We assume that the gluing map is defined by $(\varphi_1)^{-1} \circ \varphi_2: \mathbb{C}^* \to \mathbb{C}^*$. However, considering $z=x+iy \in \mathbb{C}^*$, recalling that $(\varphi_1)^{-1}(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2})$ and $\varphi_2(x,y,z)=(\frac{x}{z+1},\frac{y}{z+1})$, we have $(\varphi_1)^{-1} \circ \varphi_2 (z)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})=\frac{z}{|z|^2}=\frac{1}{\bar z}$. However, as I have seen in other parts, the result should be $\frac{1}{z}$. What is wrong?

Andreadel1988
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1 Answers1

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This is almost a duplicate to this post but I think it is worth going through some details. Not least because of notation.

In this notation $(x,y,z)$ are Cartesian coordinates on the sphere, and $(X,Y)$ on the plane. The projection from the north pole and its inverse are \begin{align} \pi_N(x,y,z)=(X,Y)&=\left(\frac{x}{1-z},\frac{y}{1-z}\right)\,,\\ \pi^{-1}_N(X,Y)=(x,y,z)&=\left(\frac{2X}{1+X^2+Y^2}\,,\frac{2Y}{1+X^2+Y^2}\,,\frac{-1+X^2+Y^2}{1+X^2+Y^2}\right)\,. \end{align} Checking $\pi_N\circ\pi^{-1}_N$: \begin{align} 1-z&=\frac{2}{1+X^2+Y^2}\,,& \frac{x}{1-z}&=X\,,&\frac{y}{1-z}&=Y \end{align} as it should. According to Andrew D. Hwang's hint in his answer we have to take the complex conjugate projection from the south pole. This and its inverse are \begin{align} \pi_S(x,y,z)=(X,Y)&=\left(\frac{x}{1+z},\frac{-y}{1+z}\right)\,,\\ \pi^{-1}_S(X,Y)=(x,y,z)&=\left(\frac{2X}{1+X^2+Y^2}\,,\frac{-2Y}{1+X^2+Y^2}\,,\frac{1-X^2-Y^2}{1+X^2+Y^2}\right)\,. \end{align} Complex numbers should now be denoted as $X+iY$ and not by the Cartesian coordinate $z\,.$ To get $$ X+iY\mapsto \frac{1}{X+iY}=\frac{X-iY}{X^2+Y^2} $$ we have to compose $\pi_N\circ\pi^{-1}_{\color{red}{S}}\,:$ $$ 1-z=1-\frac{1-X^2-Y^2}{1+X^2+Y^2}=\frac{2X^2+2Y^2}{1+X^2+Y^2} $$ leads to the expected complex inverse of $X+iY\,:$ $$ \pi_N\circ\pi^{-1}_{\color{red}{S}}(X,Y)=\left(\frac{X}{X^2+Y^2},\frac{-Y}{X^2+Y^2}\right)\,. $$

Kurt G.
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