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Conformal map

  1. Dec 12, 2013 #1
    Conformal transformations as far as I knew are defined as [itex]g_{mn}\rightarrow g'_{mn}=\Omega g_{mn}[/itex].

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

    I can't really make sense of this. Why do we need the derivative of the map to define the transformation?
  2. jcsd
  3. Dec 12, 2013 #2

    Ben Niehoff

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    Gold Member

    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.
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