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How do I find out the symmetry of a function?

Let $f:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be a Möbius transformation $z\mapsto (az+b)/(cz+d)$ sending $\{0,1,\infty\}$ to $\{0,1,\infty\}$ with $ad-bc = 1$.

I suspect there are only a finite number of such Möbius transformations. What are these?

A non-trivial example is $f(z) = 1/(1-z)$. It sends $0$ to $1$, $1$ to $\infty$ and $\infty$ to $0$. Note that $f(z) = -z+1$ is not an example, because $ad-bc = -1$ in this case.

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marked as duplicate by t.b., Did, J. M., Asaf Karagila, Srivatsan Nov 7 '11 at 19:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The determinant condition is just for normalization; note that $-z+1=(-iz+i)/(0z+i)$, so it is indeed an example. – Hans Lundmark Oct 22 '11 at 20:37
Anyway, except for such normalization, I guess that the Möbius transformations that you're looking for are these six: – Hans Lundmark Oct 22 '11 at 20:45
In general, for any $x_1,x_2,x_3$ distinct, and $y_1,y_2,y_3$ distinct, there is a unique Mobius transform $f$ with $f(x_i)=y_i$. That means that there are as many transformations sending the set $\{0,1,\infty\}$ to itself as there are permutations of that set, or exactly 6. – Thomas Andrews Oct 22 '11 at 22:36
up vote 1 down vote accepted

If $0\mapsto 0$, then $b=0$. This means that $d=\frac{1}{a}$.

  • If $f(1)=1$ and $f(\infty)=\infty$, then we must have $a=c+d$ and $c=0$, so $ad=1$, $a=1$. Hence $a=d=\pm 1$, $b=c=0$. The only transformation is the identity.

  • If $f(1)=\infty$ and $f(\infty)=1$, then we must have $a=c$ and $c+d=0$; since $d=\frac{1}{a}= -c = -a$, we have $-a^2=1$, so $a=c=\pm i$, $d=\mp i$.

If $0\mapsto 1$, then $b=d$, so $(a-c)d=1$.

  • If $1\mapsto 0$ and $\infty\mapsto\infty$, then $a=-b$ to get $1\mapsto 0$, and $c=0$ to get $\infty\mapsto \infty$. Then $(a-c)d = -d^2=1$, so $b=d=\pm i$, $a=\mp i$. (corrected; apologies to the anonymous user who suggested the edit)
  • If $1\mapsto\infty$ and $\infty\mapsto 0$ then we must have $a=0$, and $c+d=0$. So $1 = (a-c)d = -cd = d^2$. Hence $d=b=\pm 1$, $c=\mp 1$, $a=0$.

If $0\mapsto\infty$, then $d=0$, so $bc=-1$.

  • If $1\mapsto 0$ and $\infty\mapsto 1$, then $a=c$ and $a+b=0$. So $-1 = bc = -ac = -c^2$, hence $a=c=\pm 1$, $b=\mp 1$.
  • If $1\mapsto 1$ and $\infty\mapsto 0$, then $a=0$ and $b=c$. Hence $-1=bc=b^2$, so $b=c=\pm i$, $a=d=0$.
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