# Map that sends a $2\times 2$ matrix to a Mobius transformation is a homomorphism

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.

• Have you tried computing $\phi(A)(\phi(B)(z))$ and $\phi(AB)(z)$ and comparing them? – Robert Israel Mar 18 '15 at 20:30
• 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. – mr eyeglasses Mar 18 '15 at 20:34
• If $B=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, then $\phi(B)(z)=\frac{az+b}{cz+d}$. – celtschk Mar 18 '15 at 20:36
• $\phi(B)=z\mapsto\frac{az+b}{cz+d}$. Or maybe I should write the (equivalent) $\phi(B)=x\mapsto\frac{ax+b}{cx+d}$ to avoid using $z$ again. – celtschk Mar 18 '15 at 20:39
• $B$ 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 $z$. – 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.
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.