where $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in \mathrm{GL} _ 2(\mathbb C)$
Suppose $\iota$ is the function that converts between $\mathbb C _ \infty$ and $\mathbb P^1(\mathbb C)$, and that $\Psi _ g$ and $\tilde{\Psi} _ g$ are corresponding Möbius maps on each. Then what statement is true that intuitively says that these really are the same thing, under the identification of $\mathbb C _ \infty$ with $\mathbb P^1 (\mathbb C)$?
Suppose you have a Möbius map $\Psi _ g$ where $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in \mathrm{GL} _ 2(\mathbb C)$. Can you write $\Psi _ g$ as a composition of translations, dilations and inversions?
(For a mnemonic, consider the phrase: “$a$ccidentally $b$ought (the) $C$hannel, and $D$enmark. $J$ust $d$on’t $c$are” for the values in the numerators).
Consider the case where $z _ 1 = 0$ and $z _ 2 = \infty$. Then $T(z) = \lambda z$, and taking $\phi : T \to \lambda$ gives an isomorphism between $\Gamma(0, \infty)$ and $\mathbb C^\times$. Then note that $\Gamma(0, \infty) \cong \Gamma(z _ 1, z _ 2)$, since we have the isomorphism given by $T \mapsto f \circ T \circ f^{-1}$ where $f$ maps $z _ 1$ to $0$ and $z _ 2$ to $\infty$. Chaining isomorphisms, we see $\Gamma(z _ 1, z _ 2) \cong \mathbb C^\times$.
Since any Möbius map can be decomposed into a composition of translations, dilations and inversions, we can check for each of these cases individually. Translations and dilations are immediate (since scaling or moving a circline must also be a circline). Hence we just need to check inversions.
Suppose we have a circle $ \vert z - a \vert = r$.
If $r \ne \vert a \vert $ and $a \ne 0$, then $0$ is not on the circle and it becomes the set of points satisfying $ \vert 1/z - a \vert = r$, or equivalently $ \vert z - 1/a \vert = r/ \vert a \vert \vert z \vert $. This is the equation of a circle (note $r/ \vert a \vert \ne 1$).
If $r = \vert a \vert $ and $a \ne 0$, then $0$ is on the circle and under inversion it becomes the set of points satisfying $ \vert z - 1/a \vert = r/ \vert a \vert \vert z \vert $ and $\infty$. The first set is a line, so this is a circline.
If $a = 0$, we then have the set of points satisfying $ \vert z \vert = 1/r$, which is still a circle.
Now suppose we have a line: this is the set of points satisfying $ \vert z - a \vert = \vert z-b \vert $, together with $\infty$. We can assume wlog that $a \ne b$ (otherwise pick two different points that defines the same line).
Case $ \vert a \vert \ne \vert b \vert $. Then $0$ is not on the line, and under inversion it becomes the set of $z \ne 0$ satisfying $ \vert a \vert \vert 1/a - z \vert = \vert b \vert \vert 1/b - z \vert $, and $\infty$ maps to $0$. This is a circle.
Case $ \vert a \vert = \vert b \vert $. Then $0$ is on the line, so we have the set of $z \ne 0$ satisfying $ \vert a \vert \vert 1/a - z \vert = \vert b \vert \vert 1/b - z \vert $ (a line), with $\infty$ being mapped to $0$ and $0$ being mapped to $\infty$. Hence this is a line.
This is called the Cayley transform. One way to see its correctness is to consider that $\{\infty, -1, 1\}$ is mapped to $\{1, i, -i\}$, so the real line is mapped to the boundary of the disc, and then $i$ is mapped to $0$.
First map semicircle to upper left quadrant by $z \mapsto \frac{z-1}{z+1}$. This could be deduced by the fact that this map must send both bounding arcs to lines, since both the circular arc and the portion of real axis both have points on them mapped to $0$ and $\infty$, and so must be lines. Then these lines are determined by the image of a single point, so since $0 \mapsto -1$ and $i \mapsto i$, and checking a point inside the semicircle concludes that it must get sent to that quadrant (really?)
Then map upper left quadrant to right quadrant by a rotation of $\pi/2$
Then map upper right quadrant to upper half plane by squaring
Then map upper half plane to disc by $z \mapsto \frac{z - i}{z + i}$
Move the intersection of these lines/contours to the origin, since Möbius maps preserve circlines you then have to end up with two lines, which are much easier to manipulate.
Then this maps to the half-disc on the left-hand side of the complex plane. Then you could rotate, map to a quadrant with an appropriate Möbius map, then rotate to the first quadrant, then square to map to the entire upper half-plane, then map to the unit disc using another appropriate Möbius map.
How is the cross-ratio of four points $a, b, c, d \in \mathbb C _ \infty$ defined, and what useful fact about the cross ratio is related to Möbius maps?
Use the fact that $(z, z _ 1; z _ 2, z _ 3)$ represents the Möbius map
\[\begin{aligned}
z _ 1 &\mapsto 1 \\
z _ 2 &\mapsto 0 \\
z _ 3 &\mapsto \infty
\end{aligned}\]
which is straightforward to derive:
\[(z, z _ 1; z _ 2, z _ 3) = \frac{(z - z _ 2)(z _ 1 - z _ 3)}{(z - z _ 3)(z _ 1 - z _ 2)}\]
since the $(z - z _ 2)$ term in the numerator ensures $z _ 2 \mapsto 0$, the $(z - z _ 3)$ term in the denominator ensures $z _ 3 \mapsto \infty$, and then the other terms make sure $z _ 1 \mapsto 1$.
The cross-ratio is preserved under Möbius maps. Using this fact, derive an expression that can be used to find the Möbius map sending $(z _ 1, z _ 2, z _ 3)$ to $(w _ 1, w _ 2, w _ 3)$.
Suppose you have a Möbius map $\Psi _ g$ where $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in \mathrm{GL} _ 2(\mathbb C)$ and $c \ne 0$. @Justify that you can write $\Psi _ g$ as a composition of translations, dilations and inversions, given by:
Can make an argument about multiplying together the associated matrices of each Möbius transformation and then use the fact that $\Psi _ {g _ 1 g _ 2} = \Psi _ {g _ 1} \circ \Psi _ {g _ 2}$.
