- #1
gentsagree
- 93
- 1
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?
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?
