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I am a confused about the role of symmetry transformations in a covariant formulation.

Maxwell's equations can be shown to be invariant under conformal transformations. See e.g. here: https://arxiv.org/abs/hep-th/9701064 So, one would say that the symmetry group (the group of operations under which an object is left unchanged) of Maxwell's equations is the conformal group.

However, in the framework of general relativity, any theory that is formulated covariantly is invariant under general coordinate transformations. As far as I understand a general coordinate transformation is any diffeomorphism (invertible smooth map with smooth inverse) between coordinates. Thus, if I understand correctly, a conformal transformation (which is a coordinate transformation that changes the metric only up to an overall factor) should also be a general coordinate transformation.

Wouldn't this mean that the symmetry group of any theory that is formulated covariantly, is the group of all general coordinate transformations? If so, is there anything special left to the fact that Maxwell's equations are invariant under conformal transformations? Or would the only special fact be that this symmetry even holds in a non-covariant formulation?

EDIT: In formulars, my present understanding is that under a general coordinate transformation \begin{equation} x \rightarrow x'(x), \end{equation} Maxwell's equations \begin{equation} \nabla_\mu F^{\mu\nu}=J^\nu,~\nabla_{[\mu}F_{\nu\lambda]}=0 \end{equation} change as \begin{equation} \nabla_\mu' {F'^{\mu\nu}}=J'^\nu,~\nabla_{[\mu}'F'_{\nu\lambda]}=0. \end{equation} And would they change like this under e.g. a conformal transformation or e.g. a Galilean transformation? If so, is there a way to differentiate these transformations from coordinate transformations?

EDIT 2: Unfortunately, the answer below still does not fully answer my question. The reason is that I need a mathematically precise statement why a conformal transformation is or is not different from a general coordinate transformation.

Say we have a coordinate transformation, induced by a diffeomorphism $f:M\rightarrow M$ where $M$ is our spacetime manifold which induces a pullback on the metric and its vector fields. If it is correct what is written in the answer below (that a coordinate transformation does not change expressions like $v^i v^j g_{ij}$), then it should induce the pullback \begin{equation} (f^*)^{(-1)}g(f_* X,f_* Y)|_p = g_p(X_p, Y_p), \end{equation} which would leave $g(X,Y)$ invariant.

Now let's say, we have a conformal transformation (or rotation or Lorentz trf etc) which is defined as a diffeomorphism that leaves the metric invariant up to a an overall factor, then does this mean that this transformation only acts on the metric as \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} If so, I do not know why the diffeomorphism should not affect $X$ and $Y$ directly as well? Is there a reason for the non-action on the vector fields? Or is there an action on the fields but there is yet another different way in which transformations act in general?

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/300172/2451 , physics.stackexchange.com/q/276753/2451 and links therein. $\endgroup$ – Qmechanic Jan 24 '18 at 15:51
  • $\begingroup$ @Qmechanic Thanks. But the answer in the link only refers to the reason for conformal invariance. I know that the scale invariance due to the masslessness of the photon is the reason. I am confused why conformal invariance is more special than the invariance under any other coordinate transformation in the covariant formulation. $\endgroup$ – exchange Jan 24 '18 at 15:58
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Thus, if I understand correctly, a conformal transformation (which is a coordinate transformation that changes the metric only up to an overall factor) should also be a general coordinate transformation.

Not true. Under a general coordinate transformation, the metric stays the same in an abstract sense, but in order to represent the components of the metric in the new coordinate system, we have to change those components. If we follow all the tensor transformation rules, then the metric gives the same results when we use it to measure things, e.g., $u^iv^jg_{ij}$ gives the same results in the new coordinates as in the old ones.

A conformal transformation is a change in the metric itself. It is not equivalent to a change of coordinates.

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  • $\begingroup$ Thanks! My confusion is probably due to this misunderstanding! It would help me a lot if you could make a bit more precise what you mean with "the metric stays the same in an abstract sense". And what does "a change in the metric itself" mean mathematically? Some equations/mathematical terminology would be great! Thanks again! $\endgroup$ – exchange Jan 24 '18 at 18:13
  • $\begingroup$ You can also make use of differential forms, bundles or other concepts of differential geometry if that makes it easier to explain. $\endgroup$ – exchange Jan 24 '18 at 18:17
  • $\begingroup$ I thought that a Weyl transformation is a change in the metric itself, sending $g \rightarrow \Omega g$ while leaving the coordinates fixed but that a conformal transformation is indeed a coordinate transformation that changes the metric by an overall factor as described in this post: physics.stackexchange.com/questions/38138/… $\endgroup$ – exchange Jan 24 '18 at 21:14
  • $\begingroup$ It would help me a lot if you could make a bit more precise what you mean with "the metric stays the same in an abstract sense". A simpler example is rotating your your coordinate system and looking at the effect on a vector. The vector is the same. Only your description of it has changed. However, the components of the vector will be different in the new coordinate system. $\endgroup$ – Ben Crowell Jan 24 '18 at 22:43
  • $\begingroup$ Yes, I can understand this. But it does not resolve my confusion because I still do not see why a conformal transformation is not also a general coordinate transformation (though a coordinate transformation is more general). I know that a rotation for instance is an isometry and thus the metric does not change at all under this transformation but an isometry is not the most general coordinate transformation or do you claim that all coordinate transformations are isometries? What do you mean mathematically with "stays the same" and "change in itself"? $\endgroup$ – exchange Jan 25 '18 at 7:07

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