finding an operation for a semicircle Consider the following set
$$S = \{(x, y)\mid x, y \in\mathbb{R}, x^2 + y^2 = 1, y < 0\}$$
Can you think about an operation defined on the elements of $S$ such that $S$ forms a group under that operation?
 A: Given any real $a$, consider the ray $OA$ from $(0,0)$ through $(a,-1)$. It intersects $S$ at exactly one point $C(a)$, and conversely, the ray $OP$ from $(0,0)$ through a point $P$ on $S$ intersects the line $y=-1$ at exactly one point $(t(P),-1)$. So the map $C:\mathbb{R}\mapsto S$ is the inverse of the map $t:S\mapsto\mathbb{R}$. We now use the map $t$ as an isomorphism into the additive group of real numbers. Accordingly, we define addition of points $P,Q\in S$ as $$P+Q = C(t(P)+t(Q))$$
and (sign) inversion as
$$-P = C(-t(P))$$
and we find the neutral element as
$$N = C(0)$$
I leave it to you to work out the details of the maps $C(a)$ and $t(P)$.
Christian Blatter's construction is related; in a sense, it uses $\sinh(a)$ instead of $a$.
A: A hint: The map
$$\phi:\quad {\mathbb R}\to S,\qquad u\mapsto {\bf z}=\left(\tanh u,\ -{1\over\cosh u}\right)$$
is bijective. Now transport the additive group structure on ${\mathbb R}$ to $S$. The sum ${\bf z}_1\oplus{\bf z}_2$ then appears as a complicated formula.
A: The map $(x,y)\mapsto x$ is a bijection from $S$ to $(-1,1)$ so then you need a bijection from $(-1,1)$ to $\mathbb{R}$. The $\tan$ function suitably scaled will work for that. Now, for two elements $s,t\in S$, just define $s\ast t = f^{-1}(f(s)+f(t))$ where $f\colon S\to\mathbb{R}$ is the bijection we created. Show that this operation satisfies the group axioms by using the fact that $\mathbb{R}$ with addition is a group.
