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I've tried a few methods but I can't seem to work this one out.

Consider the charts $$f(s) = (\cos s, \sin s) \in \mathbb{R}^2$$ for $-\pi < s < \pi$ and $$g(t)=(\frac{2t}{t^2 + 1}, \frac{t^2 - 1}{t^2 + 1})$$ for $t \in \mathbb{R}$. Are these charts on the circle compatibly oriented?

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2 Answers 2

We will use the identities $$ \sin(2 \theta) = \frac{2 \tan(\theta)}{1 + \tan^2(\theta)}, \;\;\; \cos(2 \theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}. $$ Making the change of variable $t = \tan(\theta)$, we define a chart by $$ \psi(\theta) = g(\tan(\theta)) = (\sin(2\theta), -\cos(2\theta)) : (-\frac{\pi}{2}, \frac{\pi}{2}) \rightarrow \mathbb{R}^2. $$

Drawing the situation, we see that both $f$ and $\psi$ trace the circle counter-clockwise as we go from lower values of $s$ (or $\theta$) to higher values. This implies that the charts are consistently oriented. To show this rigorously, write

$$ \cos(s) = \sin(2\theta) = \cos(\frac{\pi}{2} - 2\theta). $$ Taking into the account the domains of $\theta$ and $s$, we have then $$ 2\theta = \left\{ \begin{array}{lr} s + \frac{\pi}{2}, \;\;\; -\pi < s < \frac{\pi}{2}\\ s - \frac{3\pi}{2},\;\;\;\; \frac{\pi}{2} \leq s < \pi. \end{array} \right.$$

(Why is there a discontinuity in the description?) We see that the coordinate transformation from $\theta$ to $s$ has positive Jacobian, and as this is also true for the transformation from $t$ to $\theta$, the charts are indeed consistently oriented.

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The moving point $f(s)$ traces out the unit circle counterclockwise, starting and ending at $(-1,0)$, which, however is excluded. Put $f\bigl(]{-\pi},\pi[\bigr)=:U$.

In order to better understand $g$ introduce an auxiliary variable $\tau$ that is bijectively and order preserving related to $t$ by $$t=t(\tau)=\tan{\tau\over2}\qquad(-\pi<\tau<\pi)\ .$$ Then we can replace $g$ by its pullback $$\eqalign{\tilde g(\tau):=g\bigl(t(\tau)\bigr)=g\left(\tan{\tau\over2}\right)=\left({2\tan{\tau\over2}\over\tan^2{\tau\over2}+1},{\tan^2{\tau\over2}-1\over \tan^2{\tau\over2}+1}\right)&=(\sin\tau, -\cos\tau)\cr &(-\pi<\tau<\pi)\ .\cr}$$ Wenn $\tau$ goes from $-\pi$ to $+\pi$ the point $\tilde g(\tau)$ traces out the unit circle counterclockwise as well, starting and ending at the point $(0,1)$ which is likewise excluded. Put $g({\mathbb R})=\tilde g\bigl(]{-\pi},\pi[\bigr)=:V$.

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It follows that the intersection $U\cap V$ of the two domains covered by the charts $f$ and $g$ is disconnected, which is definitely undesirable. It implies that we have different transformation formulas between $s$ and $t$ in the two components of $U\cap V$.

For $-\pi<s<{\pi\over2}$ we have $\tau:=s+{\pi\over2}\in\ ]{-\pi},\pi[\ $ and therefore $$f(s)=\bigl(\cos s,\sin s\bigr)=\bigl(\sin\tau,-\cos\tau)=\tilde g(\tau) =g\bigl(\tan{\tau\over2}\bigr)=g\Bigl(\tan\bigl({s\over2}+{\pi\over4}\bigr)\Bigr)\ . $$ So in this part of $U\cap V$ the parameter transformation reads $t=\tan\bigl({s\over2}+{\pi\over4}\bigr)$.

For ${\pi\over2}<s<\pi$ we have $\tau:=s-{3\pi\over2}\in\ ]{-\pi},\pi[\ $ and therefore $$f(s)=\bigl(\cos s,\sin s\bigr)=\bigl(\sin\tau,-\cos\tau)=\tilde g(\tau) =g\bigl(\tan{\tau\over2}\bigr)=g\Bigl(\tan\bigl({s\over2}-{3\pi\over4}\bigr)\Bigr)\ . $$ So in this part of $U\cap V$ the parameter transformation reads $t=\tan\bigl({s\over2}-{3\pi\over4}\bigr)$.

As $\tan'(x)>0$ wherever defined the two charts are everywhere compatibly oriented.

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