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Definition: If $z_1\in \mathbb{C}_{\infty}$ then $(z_1,z_2,z_3,z_4)$ (the cross ratio of $z_1,z_2,z_3$ and $z_4$) is the image of $z_1$ under the unique Mobius transformation which takes $z_2$ to $1$, $z_3$ to $0$ and $z_4$ to $\infty$.

Proposition: Let $z_1,z_2,z_3,z_4$ be four distinct points on $\mathbb{C}_{\infty}$. Then $(z_1,z_2,z_3,z_4)$ is a real number iff all four points lie on a circle.

Proof: Let $S:\mathbb{C}_{\infty}\to \mathbb{C}_{\infty}$ be defined by $Sz=(z,z_2,z_3,z_4) $; then $S^{-1}(\mathbb{R})=$the set of $z$ such that $(z,z_2,z_3,z_4)$ is real. So it is enough to show that the image of $\mathbb{R}_{\infty}$ under a Mobius transformation is a circle.

This is the excerpt from Conway's book on Complex Analysis of one variable.

Can anyone please explain why it is enough to show that $S(\mathbb{R}_{\infty})=\Gamma$ - circle? I have spent about one hour trying to understand it and write down something but I failed to understand it.

Would be very grateful for detailed help, please!

RFZ
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2 Answers2

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@AmanKushwaha pointed out that my original answer involved circular reasoning.

Here is a short proof based on Proposition 3.6 in Conway's book that shows that the image of $\mathbb{R}_\infty$ under a Möbius transformation is a circle.

By "circle" I mean a circle or line. By a line, I mean a set of the form $L= \{ z_0+th \}_{t \in \mathbb{R}_\infty}$, where $z_0,h \in \mathbb{C},h \neq 0$.

By way of notation, if $A \subset \mathbb{C}$ and $h \in \mathbb{C}$, let $hA = \{ ha \mid a \in A \}$.

To simplify subsequent computations, note that we can write $L = h\{ t + {z_0 \over h } \}_{t \in \mathbb{R}_\infty} = h \{ t + i \operatorname{im} {z_0 \over h } \}_{t \in \mathbb{R}_\infty}$, so we can write $L = h \{ t+ i \sigma \}_{t \in \mathbb{R}_\infty}$, where $\sigma \in \mathbb{R}$.

The following shows that a Möbius transformation transforms lines into a "circle". Note that a line is a "circle".

Proposition 3.6 shows that any Möbius transformation can be written as the composition of translations, dilations and inversion (not all need be present). It is straightforward to see that translations & dilations transform lines into lines and "circles" into "circles".

More specifically, the proposition shows that the transformation can be written as the composition of a translation and a dilation, or the composition of a translation, a dilation, and inversion and a translation.

In particular, it is sufficient to show that $S(z) = {1 \over z}$ maps lines into a "circle".

Let $L$ be a line of the form above.

If $\sigma=0$ then it is clear that $S(L)=S(h\mathbb{R}_\infty) = {1 \over h} \mathbb{R}_\infty$, so assume $\sigma \neq 0$. A quick computation shows that $S(h(t+ i \sigma)) -{1 \over 2 h i \sigma} = -{1 \over 2 hi \sigma} \overline{t+i \sigma \over t+i \sigma}$ and it is straightforward to check that $\{ \overline{t+i \sigma \over t+i \sigma} \}_{t \in \mathbb{R}_\infty } = \{ z \}_{|z|=1}$ and so $S(L)$ is a circle of radius ${1 \over 2|\sigma h|}$ and centre ${1 \over 2 hi \sigma}$.

Old answer: (Not correct as it involved circular reasoning.)

Note that $\mathbb{R}_\infty$ is a "circle", and a "circle" is uniquely defined by any three distinct points in $\mathbb{C}_\infty$. In particular, the three points $1,0,\infty$ uniquely define the "circle" $\mathbb{R}_\infty$.

Suppose any Möbius transformation transforms $\mathbb{R}_\infty$ into a "circle".

Define $S(z) = (z,z_2,z_3,z_4)$. Note that $S^{-1}$ is also a Möbius transformation.

Suppose $S(z_1) \in \mathbb{R}$. Then $S(z_1), 1,0,\infty \in \mathbb{R}_\infty$, and $S^{-1}$ transforms $\mathbb{R}_\infty$ into a "circle", and $S^{-1}(S(z_1))=z_1, S^{-1}(1)=z_2, S^{-1}(0)=z_3, S^{-1}(\infty)=z_4$, hence the $z_k$ lie on a "circle".

Now suppose the $z_k$ lie on a "circle" $\Gamma$. Since $S^{-1}(z_k) \in \mathbb{R}_\infty$ for $k=2,3,4$ we see that $S(\mathbb{R}_\infty) = \Gamma$ and since $z_1 \in \Gamma$, we see that $S^{-1}(z_1) \in \mathbb{R}_\infty$. Since $S$ is invertible on $\mathbb{C}_\infty$, and $z_1 \neq z_4$ we see that $S^{-1}(z_1) \neq \infty $ and hence $S^{-1}(z_1) \in \mathbb{R}$.

copper.hat
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  • I think that it should be $S(z_3)= 0$ and $S(z_4)=\infty.$ – Koro Jun 18 '23 at 18:38
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    @Koro Thanks for bringing this to my attention. My previous answer was very sloppy, I have simplified it a bit. – copper.hat Jun 18 '23 at 22:06
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    Thanks a lot, I've understood the spirit of the answer anyways. I found it very helpful as I was also stuck at the same question as OP's. – Koro Jun 19 '23 at 05:30
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    @AmanKushwaha I'm not sure I see the problem? – copper.hat Nov 21 '23 at 15:32
  • How to show $S(z_1) \in \mathbb R$ from $S^{-1}(z_1) \in \mathbb R$? – Aman Kushwaha Nov 21 '23 at 16:16
  • @AmanKushwaha It is straightforward, but complicated to write clearly. Gather the facts first: Any Möbius transformation maps circles to circles and is bijective. A circle is uniquely specified by $3$ points. $S$ is defined as the Möbius transformation that maps $z_2,z_3,z_4$ to $1,0,\infty$. The points $1,0,\infty$ define the 'circle' $\mathbb{R}\infty$ uniquely and the points $z_2, z_3, z_4$ define the circle $\Gamma$ uniquely. Hence $S(\Gamma) = \mathbb{R}\infty$ (this is the main point). Last fact is that $z_1$ is distinct from $z_2,z_3,z_4$. – copper.hat Nov 21 '23 at 18:08
  • @AmanKushwaha ($\Rightarrow$) Now suppose $S(z_1)$ is real. Since $S(z_1)$ is in the circle $\mathbb{R}\infty$, we must have $z_1 \in \Gamma$. ($\Leftarrow$) Now suppose $z_1$ is in $\Gamma$. Then we must have $S(z_1) \in \mathbb{R}\infty$, and since $S(z_4) = \infty$, we have $z_1 \in \mathbb{R}$. – copper.hat Nov 21 '23 at 18:08
  • @copper.hat To claim that $S(\Gamma) = \mathbb{R}\infty$ you used the fact that a mobius transformation maps circles onto circles. However, in the book (from where the excerpt is taken) this fact (circles onto circles) is proved using the quoted proposition. On another note I don't see any use of circles onto circles- property in your answer, neither in $(\Rightarrow)$ nor in $(\Leftarrow)$ (only $S(\mathbb{R}{\infty})=\Gamma$ for some circle $\Gamma$ and the fact that mt is unique given its action on three distinct points is used). But in your comment you use this property for both... – Aman Kushwaha Nov 22 '23 at 13:45
  • ..$(\Rightarrow )$ and $(\Leftarrow)$. I had no problem with $(\Rightarrow)$ in your answer but in $(\Leftarrow)$ I'm still stuck at $S^{-1}(z_1) \in \mathbb R$. – Aman Kushwaha Nov 22 '23 at 13:48
  • @AmanKushwaha I see what you are saying. I missed that completely. I need to rethink. – copper.hat Nov 22 '23 at 18:54
  • @AmanKushwaha I have added a simpler more direct proof of the OP's question that does not involve circular reasoning and is (in my opinion) a bit clearer that Conway's proof. – copper.hat Nov 24 '23 at 22:21
  • @copper.hat Your recent edit shows that a mobius transformation maps lines into circles (and so it follows that "$S(\mathbb{R}{\infty})=\Gamma$"). I don't think there was need of that, looking at OP's question. The question was this: Show that if every mobius transformation maps $\mathbb R{\infty}$ (even "shifted lines" now that you showed$^{\dagger}$) onto a circle then $(z_1,z_2,z_3,z_4)$ is a real number iff $z_1,z_2,z_3,z_4$ lie on a circle. $^{\dagger}$ - if we follow your proof to reach "$S(\mathbb{R}_{\infty})=\Gamma$" it is obvious that the hypothesis be made stronger ... – Aman Kushwaha Nov 25 '23 at 10:41
  • ..but I would still like to see a proof with the weaker (original) hypothesis. – Aman Kushwaha Nov 25 '23 at 10:46
  • @AmanKushwaha I'm missing your point completely. I have not added any hypothesis and have only used results in Conway prior to the OP's question. I have proved a slightly stronger result than necessary but with no additional hypothesis. – copper.hat Nov 25 '23 at 15:20
  • @copper.hat The question was not to prove the result which you proved in your last edit. That was given (not your slightly stronger result but "$S(\mathbb{R}{\infty})=\Gamma$" is given). If you see the proof of proposition $3.10$ in Conway's book. After the first three lines all he does is prove "$S(\mathbb{R}{\infty})=\Gamma$". The problem is only in the first three lines. The latter part i.e "$S(\mathbb{R}_{\infty})=\Gamma$" is perfectly clear and OP doesn't ask how to show this part. This is asked: Once you prove the latter part how do you reach the proposition as claimed in first 3 lines – Aman Kushwaha Nov 25 '23 at 17:37
  • I know you have not added any hypothesis. I was just saying if you are still trying to show the equivalence claimed in proposition with your proof of "$S(\mathbb{R}_{\infty})=\Gamma$" then you can surely use the stronger fact "lines onto circles". But it turns out you weren't trying to show the equivalence "cross ratio of four points is real iff the points lie on a circle" in your last edit. – Aman Kushwaha Nov 25 '23 at 17:44
  • Your old answer is clearly aimed at showing the equivalence claimed in proposition using "$S(\mathbb{R}_{\infty})=\Gamma$". And that was pretty much the question. And also it didn't involve any circulation upto the point it was written (assuming you do not complete the proof from $S^{-1}(z_1) \in \mathbb R$ tacitly by circular arguments). – Aman Kushwaha Nov 25 '23 at 17:52
  • Suggested correction in your old anwer $(\Leftarrow)$ part: Suppose the $z_k$ lie on a "circle" $\Gamma_0 $. By definition of $ S, ; S^{-1}(1)=z_2, S^{-1}(0)=z_3, S^{-1}(\infty)=z_4 $. Since $ S^{-1} $ is a mobius transformation, $S^{-1}(\mathbb R_{\infty})= \Gamma $ for some circle $\Gamma $. As $-1,0,\infty \in \mathbb R_{\infty}$ and $z_2,z_3,z_4$ determine the circle $\Gamma_0$, we must have $\Gamma=\Gamma_0$. Thus $ S^{-1}(\mathbb R_{\infty})= \Gamma_0 $. As $z_1 \in \Gamma_0$, i.e., $z_1 \in S^{-1}({\mathbb R}{\infty})$, there exists $x \in \mathbb R{\infty}$ such that $S^{-1}(x)=z_1$ – Aman Kushwaha Nov 25 '23 at 18:54
  • This implies $S(z_1)=x \in \mathbb R_{\infty}$. As $S(z_4)=\infty$ and $S$ is injective, we must have $S(z_1) \in \mathbb R$ – Aman Kushwaha Nov 25 '23 at 18:57
  • The OP's question was "Can anyone please explain...". This is what I was addressing in both cases. I found Conway's proof a little unsatisfactory and prefer the proof above, it also clarifies how an inversion maps lines into a circle. – copper.hat Nov 25 '23 at 19:22
  • @copper.hat I'd prefer your proof too. But I don't understand how "why it is enough to show that" and "how to show that" can mean the same thing! – Aman Kushwaha Nov 26 '23 at 04:18
  • I'm sorry English is not my first language but is there any chance the word 'it' in OP's question "..why it is enough to show.." is interpreted differently by us. Statement (1): If you want xyz award then it is enough to climb abc mountain. Statement (2): You have a magic rope. It is enough to climb the abc mountain. In statement (1) 'it' refers to an unspecified action or achievement required to receive the XYZ award. In statement (2) 'it' refers to the action of using the magic rope to climb the ABC mountain or just the magic rope. – Aman Kushwaha Nov 26 '23 at 05:27
  • If someone is questioning how (1) holds. To answer, you have to assume you climbed the abc mountain and prove that you'll get the xyz award. If it is questioned how (2) holds. To answer, you actually take the magic rope and climb the abc mountain. 4 years ago, you went with interpretation as in statement (1) but now you're going with (2)? – Aman Kushwaha Nov 26 '23 at 05:33
  • Just to make the analogy clear. XYZ award : Proposition 3.10. Climbing the abc mountain : proving "$S(\mathbb{R}{\infty})=\Gamma$". The magic rope : I don't know, the first two lines of proposition $3.10$'s proof (not according to me) or proposition $3.6$. Surely the magic doesn't work otherwise not this much time(: these many lines) would be required to climb the abc mountain(: to prove "$S(\mathbb{R}{\infty})=\Gamma$"). – Aman Kushwaha Nov 26 '23 at 05:49
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I think to catch every degenerate case you have to view $\infty$ as a real number.

By the first remark of the proof, one has to show that $S^{-1}(\mathbb{R_\infty})$ is exactly the set of points $z$ such that $z, z_2, z_3, z_4$ lie on a circle. We know that $z_2, z_3, z_4 \in S^{-1}(\mathbb{R_\infty})$ since their images are $1, 0, \infty$ respectively and we also know that $S^{-1}$ is a Möbius transform, too. So if $S^{-1}(\mathbb{R_\infty})$ is a circle it is the unique circle through $z_2, z_3$ and $z_4$, completing the proof.

Carlos Esparza
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