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Q: Find a mobius transformation T with the following property or show that no such T exists.

There are distinct points $z_1, z_2, z_3, a, b$ in $\mathbb{C} \cup \{\infty\}$ with $T(z_1)=z_2$, $T(z_2)=z_3$, $T(z_3)=z_1$ and $T(a)=b$, $T(b)=a$?

By the theorem, there exists a unique mobius map that maps three distinct points to three distinct points. So there exists a map $T$ such that $T(z_1)=z_2$, $T(z_2)=z_3$, $T(z_3)=z_1$ and actually this $T$ maps a circle(or extended line) to itself.

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  • $\begingroup$ Having mapped the first three points, there is no general reason why two other points $a,b$ should be swapped. $\endgroup$ – hardmath Sep 28 '17 at 21:56
  • $\begingroup$ Yes, you are right. But I think what the question ask is that could there exists five distinct points that such a mobius exists. Or could there never be such a mobius map for any distinct five points. $\endgroup$ – user424674 Sep 28 '17 at 22:04
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    $\begingroup$ The linear fractional transformation has three degrees of freedom, and once you specify the distinct images of three distinct complex arguments, you've completely specified a Moebius map. From your facts, $T$ will have period $3$. If $a,b$ are distinct we cannot make $T$ swap them. $\endgroup$ – hardmath Sep 28 '17 at 22:27
  • $\begingroup$ What is the period of a linear fractional transformation? I can not find any reference online. $\endgroup$ – user424674 Sep 28 '17 at 23:04
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    $\begingroup$ Sorry for being cryptic, I just mean $T^3$ would have to be the identity map, given your facts. Perhaps you mean you want to find a LFT such that five such distinct points exist, as opposed to starting with five points and finding a LFT? $\endgroup$ – hardmath Sep 28 '17 at 23:08
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Let $T$ be a linear fractional transformation such that for distinct $z_1,z_2,z_3$ in the "extended complex plane" $\mathbb{C} \cup \{\infty\}$ (also known as the Riemann sphere):

$$ T(z_1) = z_2\;, T(z_2) = z_3\;, T(z_3) = z_1 $$

Since $T^3(z_i) = z_i$ for $i=1,2,3$, it follows that $T^3$ is the identity since agreement of two linear fractional transformations on three distinct points implies they are the same.

Assume further that $a,b$ are such that $T(a) = b$ and $T(b)=a$. Then:

$$ a = T^3(a) = T^2(b) = T(a) = b $$

so $a,b$ cannot be distinct.

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