# Liftings of curves $u\cdot v$ and $v\cdot u$ with respect to the sine covering map.

I'm trying to work through the exercises in Otto Forster's book on Riemann Surfaces. While most of them seemed not that hard, this one gives me a headache:

Let $X=\mathbb{C}\setminus\{\pm1\}$ and $Y = \mathbb{C}\setminus\{\frac{\pi}{2}+k\pi~|~k\in\mathbb{Z}\}$. Then $\sin: Y\rightarrow X$ is a covering map. This is clear.

Now consider the following curves:

$$u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$$

and

$$v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$$

Let $w_1: [0,1]\rightarrow Y$ be the lifting of $u\cdot v$ and $w_2$ be the lifting of $v\cdot u$ with $w_1(0)=w_2(0)=0$.

Show that $w_1(1) = 2\pi$ while $w_2(1) = -2\pi$.

As of now, I don't have many ideas how to solve this. Of course, the inverse sine is a multi-valued function so the two liftings should use different branches. But I don't see how exactly they come into play.

I'm also not sure whether the solution can be done by calculations alone or if there is some more general principle underlying. Any suggestions would be welcome.

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Is $u\cdot v\,:\, [0,1] \rightarrow X, (u\cdot v)(t) = 1-e^{4\pi i t}\,(0\le t\le \frac{1}{2}), \,(u\cdot v)(t) = -1+e^{4\pi i t}\,(\frac{1}{2}\le t\le 1)$ ? – ts375_zk26 Jan 30 at 23:58
@ts375_zk26: Yes, $u\cdot v$ means the curve obtained by first tracing $u$ and then $v$. – Prism Feb 2 at 17:10

Mapping property of $w=\sin z$.

Let $G=\{z: -\frac{\pi}{2}<\Re z<\frac{\pi}{2}, \Im z>0\}$. This is one of pink-colored vertical stripe regions of Fig.1. When $z$ moves along the half-line $l_1=\{z: z=-\frac{\pi}{2}+it, t:\infty \to 0\}\subset \partial G$, $w=\sin (-\frac{\pi}{2}+it)=-\frac{1}{2}(e^{-t}+e^t)$ moves along $L_1=(-\infty,-1).$ Arrows indicate those directions of moving. When $z$ moves along the segment $l_2=[-\frac{\pi}{2}, \frac{\pi}{2}]\subset \partial G$, $w=\sin z$ moves along $L_2=[-1,1]$. Also when $z$ moves along the half-line $l_3=\{z: z=\frac{\pi}{2}+it, 0<t<\infty\}\subset \partial G$, $w=\sin\left(\frac{\pi}{2}+it\right)=\frac{1}{2}(e^{-t}+e^t)$ moves along $L_3=(1, \infty)$. Therefore it follows from Darboux theorem that $w=\sin z$ mapps $G$ onto the upper half-plane $H^+=\{w:\Im w>0\}$ injectively. The same argument shows that each pink-colored stripe region is mapped onto $H^+$ injectively by $w=\sin z$ and all $l_i$ correspond to $L_i$, $i=1,2,3$.

Similarly all white-colored stripe regions are mapped onto the lower half-plane $H^-=\{w:\Im w<0\}$.

Now we consider four curves $a,b,c$ and $d$ depicted in Fig.2. Of course $a\cdot b\cdot c\cdot d=u\cdot v$ and $c\cdot d\cdot a\cdot b=v\cdot u$.

The lifting of $u\cdot v.$

The curve $a$ starts at the origin, lies in $H^-$ and ends at a point $w=2\in L_3.$ Therefore it's lifting $\tilde{a}$ must start at the origin, lie in a white-colored region, and end at a point $P(=\sin^{-1}2)\in l_3$. Also the lifting of $b$, say $\tilde{b}$, starts at $P$, lies in a pink-colored region and ends at a point $Q(=\pi)\in l_2$, since $b$ starts at the end point of $a$, lies in a pink-colored region and ends at $w=0\in L_2$. These are illustrated in Fig.2.

We continue our arguments.
The lifting of $c$ must start at $Q$, lie in a pink-colored region and end at a point $R\in l_1$, since $c$ starts at the end point of $b$, lies in a pink-colored region and ends at $w=-2\in L_1$.

Finally the lifting of $d$ starts at $R$, lies in a white-colored region and ends at $z_0\in l_2$, since $d$ starts at the end point of $c\, (w=-2),$ lies in a white-colored region and ends at $w=0$.

Since $\sin z_0=0,\, \frac{3\pi}{2}<z_0<\frac{5\pi}{2},$ we know that $w_1(1)=z_0=2\pi.$

We only depict the lifting of $v\cdot u$ in Fig.2. We have $w_2(1)=z_1=-2\pi$.

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Nice graphics, may I ask how you created them? – One small point: Unless I am mistaken, the curve $u(t) = 1-e^{2\pi i t}$ runs clockwise, not counter-clockwise as $v$. As a consequence, the lifted curve should not have those "cusps". – Martin R Feb 3 at 8:57
Both $u: [0,1] \rightarrow X, u(t) = 1-e^{2\pi i t}$ and $v: [0,1] \rightarrow X, v(t) = -1+e^{2\pi i t} = -u(t)$ run clock-wise, I think. If it were the case, both $u\cdot v$ and $v\cdot u$ have a cusp and therefore both liftings of them have a cusp. – ts375_zk26 Feb 3 at 10:09
My previous comment was wrong, sorry. But both run counter-clockwise. – Martin R Feb 3 at 10:12
Sorry, mistaken. Yes, both run just counter-clockwise. So the graphics is correct. – ts375_zk26 Feb 3 at 11:04