10

Could anyone tell me what the difference is between a map which is conformal, bi-holomorphic and an automorphism from $D\rightarrow D$ or $D$ to the upper half plane (in that case I know that is not automorphism)?

Maybe I am getting confused about terminology? Please someone explain with examples.

Myshkin
  • 35,974
  • 27
  • 154
  • 332

1 Answers1

12

A conformal map is a holomorphic map whose derivative does not vanish. So it must be locally injective, but not necessarily surjective or injective.

A biholomorphism is a map which is bijective and holomorphic (then its inverse is also holomorphic). An automorphism is a biholomorphism $U \rightarrow U$ where $U$ is a complex domain (or a Riemann surface). In your example they are the same thing.

Example of a conformal map which is not injective : $z \mapsto e^z$. Its derivative does not vanish but it is not injective (and not surjective since its range doesn't contain $0$).

About the automorphism of the unit disk : it can be shown that they are exactly the : $B_\alpha(z)=\frac{\alpha -z}{1-\overline{\alpha}z}$, where $|\alpha|<1$.

Albert
  • 9,170
  • so a biholomorphism need not be conformal right? – Myshkin Mar 03 '13 at 15:19
  • 1
    actually it does. this is a special property of holomorphic functions, wich says that if the derivative vanishes at a point $z_0$ then $f$ cannot be locally injective near $z_0$ (this is of course false in general if $f$ is not holomorphic) – Albert Mar 04 '13 at 14:12