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Compute the subgroup $M_R \subset M$, preserving the line

$$R = \{x = 0\} \subset \mathbb{R}^2 = \mathbb{C}.$$

Here $M$ is a Möbius transformation on the extended comlex plane.

What I thought was that if I could define some transformation $L$, which shows how the elements map from point to the next, then I could set $M_0 = \frac{az + b}{cz + d}$ and compute the composition $$L \circ M_0 \circ L^{-1}.$$

Now for my $L$, as I want to preserve the line $x = 0$, I want that anything on the $y$-axis can be moved, but nothing on the $x$, so my transformation basically shifts the line up or down. So what I thought was I'll define the the transformation of mapping $0$ to $i$ as my $L$ and then compute the composition. So what I said was let $L$ be the transformation such that $z \mapsto z + i$. This gives me $L^{-1} : z \mapsto z - i$ and I tried this composition, but the answer came out wrong.

What the answer said was that I had to set $L : z \mapsto iz$, why is this? This again relates back to my previous question: Why does my transformation sending $0$ to $w$ change in these Möbius transforms? , which is basically, how do you know when you do say $z \mapsto xz$ or $z \mapsto z + x$? Also, before I made the mistake in defining $L$, was my reasoning correct?

EDIT: The reason way I tried thinking about it was like this. Let's say we pick the line $x = 0$, this is just a vertical line. Now if I shift it to $x + i$, this will just move the line up by $i$ units, and so it still stays going through $x = 0$, doesn't it?

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$iz$ is a rotation by 90 degrees. – PAD Jan 9 '13 at 0:22
up vote 0 down vote accepted

The group that fixes the $x-axis $ are all Möbius of the form $$\phi(z)=\frac{az+b}{cz+d} $$ with $a,b,c,d$ real. Therefore your transformations have the form $T=L^{-1} \phi L $, i.e. $$ T(z)=\frac{az-bi}{ciz+d}$$ with $a,b,c,d$ real.

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Note that $L(z)=iz$. – PAD Jan 9 '13 at 0:16
Why is $L(z) = iz$? – Kaish Jan 9 '13 at 11:40
Because you want to rotate the y-axis 90 degrees onto the x-axis where it is easy to get the answer, and then rotate back. – PAD Jan 9 '13 at 14:01
Hang on, so what is the question exactly asking me to do? I thought it was that they wanted me to find a map such that I can transform the line $x = 0$ however I wanted and it remained as $x = 0$, is this correct? – Kaish Jan 9 '13 at 14:12
Yes. But it has to satisfy two requirements. It has to be a Mobius transformation and also the most general, not a specific example. – PAD Jan 11 '13 at 6:39

You have $M_0$, the group that fixes $\mathbb R$. You want to find $M_R$, the group that fixes $i\mathbb R$. The idea behind writing $L\circ M_0\circ L^{-1}$ is to find $L\colon \mathbb R\to i\mathbb R$. Then $L\circ M_0\circ L^{-1}$ maps from $i\mathbb R$ to $\mathbb R$, moves the point around withn $\mathbb R$, then goes back to $i\mathbb R$ (or $\infty$ at each stage, of course). The obvious bijection $\mathbb R\to i\mathbb R$ is $x\mapsto i x$. On the other hand $x\mapsto x+i$ would send $\mathbb R\to i+\mathbb R$, a line parallel to the $x$-axis.

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How do you know you want the group that fixes $i \mathbb{R}$ and not $i + \mathbb{R}$? – Kaish Jan 8 '13 at 18:21

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