Invariance of cross ratio under Möbius transformation and another problem related to cross ratio.

Problem statement:

Given $z_1,z_2,z_3,z_4$ different points of $\overline {\mathbb C}$, we define the cross ratio $(z_1,z_2,z_3,z_4)$ as $(z_1,z_2,z_3,z_4)=\dfrac{z_1-z_2}{z_1-z_4}\dfrac{z_3-z_4}{z_3-z_2}$.

Note that $(z_1,z_2,z_3,z_4)$ is the image of $z_1$ under the Möbius transformation $T$ such that $T(z_2)=0$, $T(z_3)=1$, $T(z_4)=\infty$.

a) Prove that if $T \in \mathcal H$ then $(T(z_1),T(z_2),T(z_3),T(z_4))=(z_1,z_2,z_3,z_4)$.

b) Show that $z_1,z_2,z_3,z_4$ lie in a line or circle if and only if $(z_1,z_2,z_3,z_4) \in \mathbb R$

My attempt at a solution:

For a), using the "hint" they give, if $T \in \mathcal H$, I did the following:

If I call $H=(z,,z_2,z_3,z_4)$, I can consider $H \circ T^{-1} (z)$. Note that $H \circ T^{-1}(T(z_2))=0$, $H \circ T^{-1} (T(z_3))=1$ and $H \circ T^{-1} (T(z_4))=\infty$. This means that $(z_1,z_2,z_3,z_4)=H \circ T^{-1}(T(z_1))=(T(z_1),T(z_2),T(z_3),T(z_4))$.

I don't know if my answer is correct, I would like to check it, and if anyone has a better or different answer, he/she is very welcome to post it.

For b) I am lost, for the forward implication, I've tried to show that $(z_1,z_2,z_3,z_4)=\overline {(z_1,z_2,z_3,z_4)}$ or that $arg((z_1,z_2,z_3,z_4))$ is a multiple of $\pi$ but I couldn't conclude anything. I would appreciate some help with this point.

Solution 1:

A nice way (and not that difficult) way to prove that a mobius transformation is cross ratio preserving is simply calculating it out.

Let $T$ be the mobius transformation $T(x) = \frac {ax +b}{cx +d}$

Then $$T(x) -T(y) = \frac {a x +b}{c x +d} -\frac {a y +b}{c y +d}= $$

$$ = \frac {(a x +b)(c y +d) - (a y +b)(c x +d) }{(c x +d)(c y + d)} = $$

$$ = \frac {ax(c y +d) +b(c y +d) - a y(c x +d) - b(c x +d)}{(c x) +d)(c y +d)} = $$

$$ = \frac { ac x y + adx +bc y + bd - ac xy -ady - bc x -bd}{(c x +d)(c y +d)} = $$

$$ = \frac { ad x +bc y - ad y - bc x }{(c x +d)(c y +d)} = $$

$$ = \frac{(ad-bc)(x - y)}{(c x + d)(c y +d)} $$

then taking the cross ratio

$$\frac {Tz_1-Tz_2 }{Tz_1-Tz_4} \frac {Tz_3-Tz_4 }{Tz_3-Tz_2} = $$

$$\frac {(Tz_1-Tz_2)(Tz_3-Tz_4) }{(Tz_1-Tz_4)(Tz_3-Tz_2)} = $$

$$ = \frac {( \frac{(ad-bc)(z_1-z_2 )}{(c z_1 + d)(c z_2 +d)})( \frac{(ad-bc)(z_3-z_4)}{(c z_3 + d)(c z_4 +d)}) }{( \frac{(ad-bc)(z_1-z_4)}{(c z_1 + d)(c z_4 +d)})( \frac{(ad-bc)(z_3-z_2)}{(c z_3 + d)(c z_2 +d)})} = $$

$$ = \frac {(z_1-z_2)(z_3-z_4) }{(z_1-z_4) (z_3-z_2)} = $$

$$=\frac {z_1-z_2 }{z_1-z_4} \frac {z_3-z_4 }{z_3-z_2} $$

where you started with.

Solution 2:

Your proof is correct, except that you probably mean $H \circ T^{-1}(T(z_1))$.

For part (b), do you already know that the image of a line (along with $\infty$) or circle under a Moebius transformation is a line or circle?

If so, then part (b) follows from part (a) can be proved as follows. The image under $H$ of the line or circle $C$ through $z_2$, $z_3$, $z_4$ is the line or circle through 0, 1, $\infty$, which is of course $\mathbf{R} \cup \{\infty\}$. So asking whether $z_1$ is on $C$ is the same as asking whether $H(z_1)$ is in its image, which is $\mathbf{R} \cup \{\infty\}$.

If you don't know this fact, then you can analyze the argument of $(z_1, z_2, z_3, z_4)$ as being the oriented angle $\angle z_4 z_1 z_2$ minus the angle $\angle z_4 z_3 z_2$. By the inscribed angle theorem of elementary geometry, the four points $z_1$, $z_2$, $z_3$, $z_4$ are then concyclic or aligned if and only if these two angles differ by 0 modulo $\pi$, i.e., if and only if the argument of their cross ratio is 0 or $\pi$.

Edit: Your proof depends on the uniqueness of the Moebius transformation sending three given points to 0, 1 and $\infty$, but the statement seems to view this fact as already known.