Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
$iz$ is a rotation by 90 degrees. –  PAD Jan 9 '13 at 0:22

2 Answers 2

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.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.