For mobius map why we need a invertible matrix of determinant $1$.

Question

The map $T:\mathbb{C}_\infty\to \mathbb{C}_\infty $ defined by $T(z)=\frac{az+b}{cz+d}$ is called Mobius map . My question is why we may assume $ad-bc=1$ ? If $p\in \mathbb{C}-\{0\},$ then $\frac{paz+pb}{pcz+pd}=\frac{az+b}{cz+d}.$

Answer 1

This is a simple calculation. As you already saw in your question, if you scale $a, b, c, d$ by a common nonzero factor $p$, you don't change the function: $$ \frac{az+b}{cz+d} = \frac{paz+pb}{pcz+pd} $$ for all $z$. But that change alters the nonzero number $ad - bc$ by the square factor $p^2$: $$ (pa)(pd) - (pb)(pc) = p^2(ad - bc). $$ Since all nonzero numbers in $\mathbf C$ have square roots, we can choose $p$ so that $p^2 = 1/(ad - bc)$. Then $p^2(ad - bc) = 1$.