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.

Let $M_{\alpha}, \alpha \in \mathbb{C}$, be the subgroup of $M$ mapping $\alpha$ to itself, that is, the stabilizer of $\alpha$. Given that

$$M_0 = \left \{w = \frac{z}{cz + d}, d \neq 0 \right \}$$

compute the subgroup $M_i, i = \sqrt{-1}$.

$M$ is the Möbius transformation on the extended complex plane.

What I have said is that in order to compute this subgroup, I want some transformation, $L$, which will send map $z \mapsto z + i$. This subgroup can be worked out using the composition

$$M_i = L^{-1} \circ M_0 \circ L$$

So I know that $L = \frac{(z + i)}{c(z + i) + d}$, but I'm not sure how I get $L^{-1}$. I'm guessing from here, it'll be fairly simple to work out the composition.

EDIT: Also, does order matter? I.e is

$$M_i = L^{-1} \circ M_0 \circ L = L \circ M_0 \circ L^{-1}$$

share|improve this question
1  
If $w = z+i$, then $z=w-i$. This should help you to compute $L^{-1}$. –  mrf Jan 8 '13 at 11:44
    
@mrf So would I say my $L^{-1}$ maps $z \mapsto z - i$? –  Kaish Jan 8 '13 at 11:46
    
Yes.${}{}{}{}{}$ –  mrf Jan 8 '13 at 11:49

1 Answer 1

does order matter?

Yes: you want to move $i$ to $0$, then apply a map that stabilizes $0$, then move $0$ back to $i$ -- in this order. The simplest maps to use are translations by $\pm i$ (suggested by mrf). Thus, $$M_i = \left \{w = i+\frac{z-i}{c(z-i) + d}, d \neq 0 \right \}$$ which you can simplify if desired.

share|improve this answer

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.