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Let $[z_1,z_2,z_3,z_4]$ denote the cross ratio of the complex numbers $z_1,z_2,z_3,z_4\in \mathbb{C}$. Show that the distinct points $z_1,z_2,z_3,z_4\in\widehat{\mathbb{C}}$ lie on a generalized circle if and only if $[z_1,z_2,z_3,z_4]\in\mathbb{R}$.


I saw this statement claimed in this answer, and wondered how one would go about showing it. I know of some facts we could work with:

  • First, the definition: A generalized circle in $\mathbb{C}$ is either a circle or a line in $\mathbb{C}$.
  • I know that a Mobius transformation maps generalized circles to generalized circles.
  • I know that if $f$ is a Mobius transformation and $z_j'=f(z_j)$, $j=1,2,3,4$, then $[z_1',z_2',z_3',z_4']=[z_1,z_2,z_3,z_4]$.

I have posted a potential proof as an answer below. Please let me know if you think it's complete. Thank you!

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We will show (1) if $z_1,z_2,z_3,z_4$ lie on a generalized circle, then $[z_1,z_2,z_3,z_4]\in\mathbb R$, and (2) if $[z_1,z_2,z_3,z_4]\in\mathbb R$, then $z_1,z_2,z_3,z_4$ lie on a generalized circle.

  1. First, suppose $z_1,z_2,z_3,z_4$ lie on a generalized circle. We know that, given three real numbers $x_1,x_2,x_3 \in \mathbb{R}$, there exists a Mobius transformation \begin{equation*} F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3} \end{equation*} such that $F(z_i)=x_i$, $i=1,2,3$. Also, since Mobius transformations map generalized circles to generalized circles, we know that $z_4$ is also mapped to some $x_4 \in \mathbb{R}$. Therefore, since $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then their cross ratio $[x_1,x_2,x_3,x_4]=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \in \mathbb{R}$. Therefore, since the cross ratio is invariant under Mobius transformation, we have \begin{align*} [z_1,z_2,z_3,z_4] &=[F(z_1),F(z_2),F(z_3),F(z_4)] \\ &=[x_1,x_2,x_3,x_4] \\ &=\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \\ &\in \mathbb{R} \end{align*}
  2. Next, suppose $[z_1,z_2,z_3,z_4] \in \mathbb{R}$. Let \begin{equation*} F=F_{x_1,x_2,x_3}^{-1} \circ F_{z_1,z_2,z_3} \end{equation*} be the Mobius transformation which sends $F(z_i)=x_i$, $i=1,2,3$, and let $z_4'=F(z_4)$. Since the cross ratio is invariant under Mobius transformation, we know that \begin{equation*} [z_1,z_2,z_3,z_4]=[x_1,x_2,x_3,z_4'] \end{equation*} Therefore the cross ratio $[x_1,x_2,x_3,z_4']$ is also a real number, which implies that $z_4'$ is a real number as well. Now consider $F^{-1}$, which is also a Mobius transformation. Since $x_1,x_2,x_3,z_4'$ lie on a generalized circle (the real line), and since a Mobius transformation maps generalized circles to generalized circles, then $z_1,z_2,z_3,z_4 \in F^{-1}(\mathbb{R})$ belong to a generalized circle.
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First, show that every generalized circle can be sent by a Mobius transformation to the real line (transitivity). Then, all you have to do is to show that if the cross ratio of $4$ real points is always real.

Also note that this theorem can be shown with elementary geometry using the inscribed angle theorem.

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  • $\begingroup$ OK, naturally if $x_1,x_2,x_3,x_4 \in \mathbb{R}$, then $\frac{x_1-x_3}{x_2-x_3}\cdot\frac{x_2-x_4}{x_1-x_4} \in \mathbb{R}$. But can you elaborate on the part about every generalized circle can be mapped by a Mobius transformation to the real line please? $\endgroup$ – Mathemanic Apr 29 '15 at 10:11
  • $\begingroup$ Here are precisions about the first point. If you have $\mathcal{C}$ a generalized circle, you have to find a Mobius transformation such that $f(\mathcal{C}) = \mathbb{R}$. This is not trivial and be done in different ways. $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 29 '15 at 10:15
  • $\begingroup$ Extra hint : take $3$ distincts points among the $4$ you have. Is there a Mobius transformation sending the first to 0, the second to $\infty$ and the third to $1$ ? If yes, where is the image if the last point ? $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 29 '15 at 10:20

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