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As far as I understand, a general coordinate transformation is induced by a diffeomorphism $f:M\rightarrow M$ where $M$ is a manifold (which can locally be described with coordinates). So if $x:M\rightarrow \mathbb{R}^m$ is a coordinate map, then $y:=x\circ f$ is a new coordinate map after the coordinate transformation. One could locally express the new coordinates in terms of the old ones, writing $y=y(x)$. Is that right so far?

Now if such a diffeomorphism is applied to all points, it should induce an effect on the metric. Let's say $X,Y$ are vector fields (elements of the module of derivatives over the ring of smooth functions $C^\infty(M)$ over $M$) and $g$ is the metric tensor field (an element of the comodule of the module of vector fields which takes two vector fields back to $C^\infty(M)$) and $p$ is a point of $M$. Then, is it right to say that the diffeomorphism induces the change \begin{equation} f^*g(X,Y)|_p = g_p(f_* X_p, f_* Y_p)? \end{equation} I have also seen in another post an action defined on all objects as in \begin{equation} (f^*)^{(-1)}g(f_* X,f_* Y)|_p = g_p(X_p, Y_p), \end{equation} which would mean that any diffeo would leave $g(X,Y)$ invariant? If this is correct, then I am confused here because I don't yet understand why the action should include the inverse of the pull-back?

A conformal transformation is defined as a diffeomorphism that leaves the metric invariant up to a an overall factor, meaning that the diffeomorphism induces a pull-back of the metric that is conformally equivalent (equivalent up to an overall factor) to the old one. Does this mean \begin{equation} f^*g(X,Y)|_p=g_p(f_* X_p, f_* Y_p)=\Omega(p) g_p(X_p,Y_p)? \end{equation}

In that case, a conformal transformation would be a coordinate transformation that changes the metric only by an overall factor and is thus also a coordinate transformation?

But then a conformal invariance of some theory would not be special anymore in a covariant formulation which confuses me. Thus my understanding of the action of a transformation on the metric and the vector fields is probably wrong at some (or multiple) point(s). Would be great if you could help me to clarify that.

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  • $\begingroup$ Does really no one have an answer? I thought it to be pretty clear for people studying differential geometry. $\endgroup$
    – exchange
    Jan 28 '18 at 11:30
  • $\begingroup$ I am also surprised there is not more interest here - I may work on a partial answer to spur activity. But I will point out that I don't think many people would agree with "A conformal transformation is defined as a diffeomorphism that leaves the metric invariant up to a an overall factor". I think the definition of a conformation transformation is simply $\Omega g$, purely a local map. $\endgroup$
    – levitopher
    May 24 at 19:32
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This may not be a complete answer, but it's some perspective.

A conformal transformation should not be considered a diffeomorphism. A diffeomorphism is an isomorphism of manifolds, meaning defined everywhere on each manifold. It may have a local representation in a particular coordinate system, but it will be different at each location and in each coordinate system.

A conformal transformation is a map that takes a metric $g$ (in a particular coordinate system) to $\Omega g$, but it's precise form depends on the local coordinate system. It actually can't be a coordinate system transform, because the scalar curvature is not invariant under a conformal transformation.

Could a conformal transformation be the pullback of a diffeomorphism? I suppose the answer is yes, and one could use the conditions on the invariance of the scalar curvature to find such a thing.

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  • $\begingroup$ Thanks for the answer. Could you elaborate on the sentence "It actually can't be a coordinate system transform, because the scalar curvature is not invariant under a conformal transformation." ? Why should coordinate transformations leave the curvature invariant? I thought only isometries leave the metric and thus curvature invariant. Indeed, there are coordinate transformations for topological manifolds that do not even carry a metric, or am I confusing something here? Thanks! $\endgroup$
    – exchange
    May 31 at 7:33
  • $\begingroup$ @exchange: Changes of the coordinate system should leave the scalar curvature invariant - it's a scalar, so it cannot be different in different coordinate systems. The same would not be true for the Ricci or Riemann curvature tensors. Isometries leave the metric invariant for sure, and therefore also the curvature tensors. If a topological manifold does not carry a metric, there is also no curvature so worrying about what the coordinate transformation does in the smooth category doesn't have any real meaning, right? $\endgroup$
    – levitopher
    Jun 1 at 3:27

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