# Property of Mobius transformations which send $\mathbb{R}$ in $\mathbb{R}$ and that $ad-bc=1$

I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers.

So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see if I get some information about $a,b,c,d$ considering that $T(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ but this doesn't really work. Also I've considered $SL(2,\mathbb{C}) / \{ \pm I\}$ which is isomorphic to the group of Mobius transformations but this doesn't really help either.

Hints: if we assume $c\neq 0$, $T(z)=\frac{a}{c}-\frac{1}{c(cz+d)}$. Letting $z$ go to infinity shows $\frac{a}{c}$ is real, hence $c(cz+d)$ is real for all $z\in\mathbb R$. From there conclude $cd$ and $c^2$ are real. The rest should be easy. The case $c=0$ is also not difficult.

• So you are saying $T(\infty) \in \mathbb{R}$? Why? Sep 10 '17 at 23:55
• @CamiloEscobar By continuity. Sep 11 '17 at 6:55

Prove first that $$T(z)=\frac{a'z+b'}{c'z+d'}$$ for real $a'$, $b'$, $c'$ and $d'$ and that then $a=\pm a'/\sqrt{a'd'-b'c'}$ etc.

Suppose $T$ is as asked for.

• Suppose $c\neq 0$. Then if $\frac{-d}{c} \in \mathbb{R}$ and we know that $T(\frac{-d}{c}) = \infty \notin \mathbb{R}$ unless the numerator is $0$ as well, in which $a(\frac{-d}{c}) = b = 0$ which means $\frac{-ad}{c} = -b$ or $-ad = -bc$ or $ad=bc$ and we have a contradiction with $ad-bc = 1$.

So $c=0$ and so $ad=1$. What more can you do now?

• You are assuming $\frac{-d}{c}\in\mathbb R$. Sep 10 '17 at 6:59
• @Wojowu yes, to get a contradiction. Sep 10 '17 at 7:58
• You are supposed to show that it's not possible to have $T(\mathbb R)=\mathbb R$ if they are neither all real nor all imaginary. That's what (the contrapositive of) the question is. You show that $T(\mathbb R)=\mathbb R$ is impossible if they are all real or all imaginary. Sep 10 '17 at 8:01
• @Wojowu The question says that if we have a $T$ that has $[\mathbb{R}] = \mathbb{R}$ and $ad-bc=1$, then the coefficients are all real or all imaginary.(say, $\phi \implies \psi$). I showed that if $T$ obeys the left hand side, the right hand side cannot hold. So I showed $\lnot (\phi \land \psi) = \lnot \phi \lor \lnot \psi$, while the implication is equivalent to $\lnot \phi \lor \psi)$. So I think the question is wrong. Sep 10 '17 at 8:06
• By the way, it's possible to have $T(\mathbb R)=\mathbb R$ -- take $a=d=1,b=c=0$. You are not taking $c=0$ into consideration. Sep 10 '17 at 8:10