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Apologies if this question is too naive, but it strikes at the heart of something that's been bothering me for a while.

Under a diffeomorphism $\phi$ we can push forward an arbitrary tensor field $F$ to $\phi_{*}F$. Is the following statement correct?

If $p$ is a point of the manifold then $F$ at $p$ is equal to $\phi_* F$ at $\phi(p)$, since they are related by the tensor transformation law, and tensors are independent of coordinate choice. ()

I have a feeling that I'm missing something crucial here, because this would seem to suggest that diffeomorphisms were isometries in general (which I know is false). (*)

However if the statement isn't true then it menas that physical observables like the electromagnetic tensor $F^{\mu \nu}$ wouldn't be invariant under diffeomorphisms (which they must be because diffeomorphisms are a gauge symmetry of our theory). In fact the proper time $\tau$ won't even be invariant unless we have an isometry!

What am I missing here? Surely it's isometries and not diffeomorphisms that are the gauge symmetries?! Many thanks in advance.

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General diffeomorphisms are not isometries; only the diffeomorphism under which the values of the metric tensor are invariant at every point are isometries, by definition. However, the transformation rules for tensor fields are given by the same universal formula that works pretty much the same whether or not the diffeomorphism is an isommetry or not. The only difference between isometry and non-isometry is that a particular tensor field, the metric tensor field, is or is not invariant. In GR, all diffeomorphisms (or those that are trivial at infinity) form a gauge group (or its part). –  Luboš Motl Dec 22 '12 at 17:45
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Sorry for my confusion. Of course coordinate transformations are local diffeomorphisms of $\bf{R}^n$ so my comment might not be useless... Consider the following. Suppose I have a timelike path $\gamma$ between points $a$ and $b$. Then I can use the diffeomorphism $\phi$ to get a new path $\phi \circ \gamma$. I would think that the proper time of this new path evaluated with $\phi_{\star}g$ would be the same as the original proper time. –  sjasonw Dec 22 '12 at 18:04
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"How on earth can we say that diffeomorphisms are a gauge symmetry if they are changing a physical quantity" they are a gauge symmetry in the sense that they preserve Dirac observables. For the Diff group, though, these observables are rather restricted - things like integrals of various contractions of products of the Riemann tensor with itself. –  twistor59 Dec 22 '12 at 18:25
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@EdwardHughes Actually, if $(M,g)$ is a spacetime, $N$ is a manifold, and $\phi :M\rightarrow N$ is a diffeomorphism, then $\phi$ is an isometry between $(M,g)$ and $(N,\phi_{\star}g)$. So maybe that helps? Also, I don't think it's reasonable that it could be true for geodesics only: arbitrary timelike curves are built out of infinitesimal geodesics. –  sjasonw Dec 22 '12 at 19:00
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@EdwardHughes What you said is correct if we are talking about an isometry between $(M,g)$ and itself. I was talking about an isometry between $(M,g)$ and a different spacetime $(N,\phi_{\star}g)$ (you can take $M=N$ if you want to). –  sjasonw Dec 22 '12 at 19:37
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1 Answer

up vote 4 down vote accepted

If p is a point of the manifold then F at p is equal to ϕ∗F at ϕ(p), since they are related by the tensor transformation law, and tensors are independent of coordinate choice.

This is roughly true. Initially, there is no meaning when one says that tensors at different tangent spaces are equal. However, the diffeomorphism induces an isomorphism between $T_p M$ and $T_{\phi (p)} M$ (the isomorphism is nothing but the vector push forward). The two tensors are equal with respect to this isomorphism.

I have a feeling that I'm missing something crucial here, because this would seem to suggest that diffeomorphisms were isometries in general...

This is actually true in a sense that is relevant. If $(M,g)$ is a spacetime and $\phi \in \text{diff}(M)$, then while there is no reason to think that $\phi$ is an isometry between $(M,g)$ and itself, $\phi$ is always an isometry between $(M,g)$ and $(M,\phi_{\star}g)$.

This last point saves your concern about proper time. If $\gamma$ is a normalized timelike path between two events $a$ and $b$, we can always consider $\phi \circ \gamma$ as a timelike path in $(M,\phi_{\star}g)$. You can check that the new path is normalized with respect to the new metric $\phi_{\star}g$. The domains of the two paths are exactly the same so the proper time between $\phi(a)$ and $\phi(b)$ is the same as the original path's proper time.

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