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This question was prompted by the following passage in Wald (1984)'s chapter on spinors:

Consider, now, a diffeomorphism $\phi:M\to M$ and allow $\phi$ to map the basis fields $(e_\alpha)^a$ into $\phi^*(e_\alpha)^a$ in the manner described in appendix C. In general, the basis $\phi^*(e_\alpha)^a$ will not be orthonormal at each point and thus will not correspond to a physically realizable family of observers, However, when (and only when) $\phi$ is an isometry, $\phi^*(e_\alpha)^a$ will be orthonormal, and we can use $\phi$ to map our original family, $O$, of physical observers associated with the basis field $(e_\alpha)^a$ into a new family $\tilde{O}$ of physical observers associated with $\phi^*(e_\alpha)^a$.

I don't understand how the basis vectors would cease to be orthonormal after the diffeomorphism. Surely the diffeo must also induce a pullback $(\phi^{-1})_*$ (described in the apprendix Wald refers to) such that $(\phi^{-1})_*g(\phi^* e_\alpha,\phi^* e_\beta)=g(e_\alpha,e_\beta)=\eta_{\alpha\beta}$?

This is an issue that has been troubling me lately. I see something similar in the "derivation" of the Lorentz transformation condition from invariance of the line element. It goes: postulate $ds^2$ is invariant under Lorentz transformations, then $$\eta_{\mu\nu}dx^\mu dx^\nu=\eta_{\alpha\beta}(dx^\prime)^\alpha (dx^\prime)^\beta=\eta_{\alpha\beta}\Lambda^\alpha_{\,\,\mu}dx^\mu\Lambda^\beta_{\,\,\nu}dx^\nu=\Lambda^\alpha_{\,\,\mu}\eta_{\alpha\beta}\Lambda^\beta_{\,\,\nu}dx^\mu dx^\nu$$So one concludes that the transformations must satisfy $\Lambda^\alpha_{\,\,\mu}\eta_{\alpha\beta}\Lambda^\beta_{\,\,\nu}=\eta_{\mu\nu}$.

But doesn't $\eta_{\mu\nu}dx^\mu dx^\nu=\eta_{\alpha\beta}(dx^\prime)^\alpha (dx^\prime)^\beta$ already assume that the transformations are isometries of $\eta_{\mu\nu}$? Otherwise the metric components would also change and the invariance of the line element would be a trivial fact. After all, it is a scalar and so is invariant under arbitrary coordinate transformations.

I guess my question is, why do we sometimes treat transformations such as Lorentz transformations as purely acting on vectors, while other times we treat them as diffeomorphisms (or coordinate transformations in the passive viewpoint) that also affect the metric through a pullback? Surely these two cases are very distinct?

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If you are doing a passive transformation where you are merely expressing the same tensors in a new primed basis, then yes the metric changes too in such a way that any scalars don't transform. No matter what metric you have and no matter what the transformation is, this scalar will not change. So it doesn't tell us anything interesting.

If you are instead thinking of an active transformation, it is perfectly reasonable to take only the vectors to transform, not the metric. Then if a dot product is still the same after the transformation it is telling us something interesting, a symmetry of the metric that need not be the case for all metrics and transformations.

To be more specific to your example:

But doesn't $\eta_{\mu\nu}dx^\mu dx^\nu=\eta_{\alpha\beta}(dx^\prime)^\alpha (dx^\prime)^\beta$ already assume that the transformations are isometries of $\eta_{\mu\nu}$?

Yes, by writing this equality we are assuming the transformation is an isometry, and we then use that to restrict the form $\Lambda$ such that this isometry condition is obeyed. If we took the metric to change too, we wouldn't be able to restrict anything at all because the equality is trivially true, as you point out.

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  • $\begingroup$ Thanks, so it looks like my thinking is mostly on the right track. I think the part I'm most interested in is "If you are instead thinking of an active transformation, it is perfectly reasonable to take only the vectors to transform, not the metric.". How does this work? From how I understand it, under an active diffeo we always have $g\to (\phi^{-1})_*g$ such that $(\phi^{-1})_*g(\phi^* X,\phi^* Y)=g(X,Y)$. I know that for $\phi$ an isometry we have $(\phi^{-1})_*g=g$, but that isn't necessary for preserving orthonormality like Wald claims. $\endgroup$ Jan 2 at 1:45
  • $\begingroup$ Perhaps the use of "active transformation" is confusing since some (most?) define it in such a way that it is equivalent to a passive transformation. The point is nothing requires you to transform both the metric and vectors, and only transforming one not the other is what lets you talk about symmetry. $(\phi^{-1})_*g=g$ is absolutely necessary for an isometry since beginning with the relation in your question $(\phi^{-1})_*g(\phi^* e_\alpha,\phi^* e_\beta)=g(e_\alpha,e_\beta)=g(\phi^* e_\alpha,\phi^* e_\beta)$, where the last equality follows from the nontrivial relation on the metric. $\endgroup$
    – octonion
    Jan 2 at 4:25

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