# Conformal map

1. Dec 12, 2013

### gentsagree

Conformal transformations as far as I knew are defined as $g_{mn}\rightarrow g'_{mn}=\Omega g_{mn}$.

Now I come across a new definition, such that a smooth mapping $\phi:U\rightarrow V$ is called a conformal transformation if there exist a smooth function $\Omega:U\rightarrow R_{+}$ such that $\phi^{*}g'=\Omega^{2}g$ where $\phi^{*}g'(X,Y):=g'(T\phi(X),T\phi(Y))$ and $T\phi :TU\rightarrow TV$ denotes the tangent map of $\phi$.

I can't really make sense of this. Why do we need the derivative of the map to define the transformation?

2. Dec 12, 2013

### Ben Niehoff

You can only compare objects that exist on the same space (or region U in this case). So the pullback is needed to accomplish that.