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.


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.

  • 1
    $\begingroup$ Hi. I dont quite understand "you can analyze the argument of $(z_1,z_2,z_3,z_4)$ as being the oriented angle ∠$z_4z_1z_2$ minus the angle ∠$z_4z_3z_2$. How does one measure angles in $\mathbb{R}^4$? $\endgroup$ – Swapnil Tripathi Aug 30 '14 at 13:49
  • $\begingroup$ @SwapnilTripathi He is measuring angle in $R^2$ only. considering $z_i$ as vectors in R^2. $\endgroup$ – Sushil Mar 26 '18 at 13:35

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.