The set of all Mobius transformations is a group of homeomorphisms of $\mathbb{C} \cup {\infty}$ onto itself. All Mobius transformations have inverses. How can I show the map that sends the $2\times 2$ matrix $$ \left[ \begin{array}{ c c } a & b \\ c & d \end{array} \right] $$ to the Mobius transformation $$\frac{az+b}{cz+d}$$ is a homomorphism of $GL(2, \mathbb{C})$ of invertible $2x2$ complex matrices onto the set of all Mobius transformations? I know we have a map $\phi : GL(2, \mathbb{C}) \to \textrm{set of all Mobius transformations}$ and I need to check that $\phi (AB) = \phi (A) \circ \phi (B)$ ($\circ$ since Mobius transformations are compositions).

Most of my issue in understanding this problem is the notation. We did not learn in this class what a homomorphism is, and I didn't know what $GL(2, \mathbb{C})$ meant until I looked it up. I'm not really that comfortable with groups or abstract algebra, but in this case it doesn't really seem all that necessary.

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    $\begingroup$ Have you tried computing $\phi(A)(\phi(B)(z))$ and $\phi(AB)(z)$ and comparing them? $\endgroup$ – Robert Israel Mar 18 '15 at 20:30
  • $\begingroup$ I don't really understand how to compute such a thing. $\phi$ as I understand it is a mapping from a matrix to an expression, am I correct? So I'm not sure what $\phi (B)(z)$ means. $\endgroup$ – mr eyeglasses Mar 18 '15 at 20:34
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    $\begingroup$ If $B=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $\phi(B)(z)=\frac{az+b}{cz+d}$. $\endgroup$ – celtschk Mar 18 '15 at 20:36
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    $\begingroup$ $\phi(B)=z\mapsto\frac{az+b}{cz+d}$. Or maybe I should write the (equivalent) $\phi(B)=x\mapsto\frac{ax+b}{cx+d}$ to avoid using $z$ again. $\endgroup$ – celtschk Mar 18 '15 at 20:39
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    $\begingroup$ $B$ is a matrix. $\phi(B)$ is a Mobius transformation. So for $z \in \mathbb C \cup \infty$, $\phi(B)(z) \in \mathbb C \cup \infty$ is the result of doing that transformation to $z$. $\endgroup$ – Robert Israel Mar 19 '15 at 2:58

As you say, all you need to do is check that $\phi(A) \circ \phi(B) = \phi(AB)$. As the comments above have explained, $\phi$ is a function that takes a matrix $M$ and gives you a function (in particular, a Möbius transformation) $\phi(M) :\Bbb C \to \Bbb C$.

We write $$ A = \pmatrix{a&b\\c&d}, \quad B = \pmatrix{a'&b'\\c'&d'} $$ We can describe $\phi(B)$ as $$ [\phi(B)](z) = \frac{a'z + b'}{c'z + d'} $$ and so $$ [\phi(A) \circ \phi(B)](z) = [\phi(A)]([\phi(B)](z)) = \\ \frac{a\frac{a'z + b'}{c'z + d'} + b}{c\frac{a'z + b'}{c'z + d'} + d} $$ you want to show that this is the same as $\phi(AB)$. In particular, $$ AB = \pmatrix{ aa' + bc' & ab' + bd'\\ ca' + dc' & cb' + dd'} $$ so that $$ [\phi(AB)](z) = \frac{(aa' + bc')z + (ab' + bd')}{(ca' + dc')z + (cb' + dd')} $$ your goal, then, is to show that these two functions, $[\phi(AB)](z)$ and $[\phi(A)\circ \phi(B)](z)$, are equal.

As for "why this is necessary": nothing in math is strictly necessary. However, this homomorphism is useful since it allows us to say everything about the composition of Möbius functions in terms of matrices, and there are a lot of things we already know how to do with matrices.

For example, we can use this to derive a neat formula for the inverse of a transformation, and we can use this to evaluate things like $f \circ \cdots \circ f$ where $f$ is a transformation.


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