I'm sorry I know this has been asked before, but I'm still a bit confused. I understand that an active diffeomorphism $\varphi:M\to M$ can be equivalently viewed as a coordinate transformation so that since the equations of general relativity are tensorial $\varphi^*g$ will be a solution to Einstein's equations if $g$ is. However I don't see how that same reasoning doesn't imply that other physical theories are diffeomorphism invariant. What's the difference between general relativity and other physical theories, like classical mechanics? Why can't diffeomorphisms be viewed as coordinate transformations in both (or am I confused?).

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    $\begingroup$ If I take the wave equation $\Box \psi=0$ for example, this is not diff. invariant because it really means $\eta^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\psi=0$, and if I apply a diffeo $\phi$, it messes up $\eta_{\mu\nu}$, which is part of the background. In other words, the wave equation is not "solving for" $\eta_{\mu\nu}$. However with Einstein's equations, all the variables you have which define the physics - $g_{\mu\nu}$ are the things you're solving for - there are no fixed background quantities. $\endgroup$
    – twistor59
    Commented Oct 20, 2013 at 7:59
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    $\begingroup$ Incidentally, there are possibilities to make any theory diffeomorphism invariant by parametrization see here for example. $\endgroup$
    – twistor59
    Commented Oct 20, 2013 at 8:06
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    $\begingroup$ I thought that all theories are invariant under passive diffs., but only gravity is invariant under active diffs. $\endgroup$
    – jinawee
    Commented Oct 20, 2013 at 11:12
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    $\begingroup$ @twistor59 Ok, but $\eta^{\mu\nu}\nabla_\mu\nabla_\nu\phi=0$ isn't a tensor equation, if you do the change of coordinates you get $g^{\mu\nu}\nabla_\mu\nabla_\nu\phi=0$. I still don't see what the problem is, ex: you can solve the wave equation in cartesian or polar coordinates, you just have to do the proper change of variables. The Euler-Lagrange equations are the same in any coordinate system, so what problems do active diffeomorphisms cause? $\endgroup$
    – JLA
    Commented Oct 20, 2013 at 17:17
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    $\begingroup$ @JLA If you take your example of changing to polar coordinates, this is a passive diffeomorphism - the metric components have changed, but the metric tensor $\eta_{\mu\nu}e^{\mu}\otimes e^{\nu}$ hasn't. With an active diffeomorphism, the metric tensor itself changes, so a solution of the wave equation doesn't (necessarily) get mapped to a solution. In the Euler-Lagrange equations for the wave equation, the metric tensor is still fixed, i.e. part of the background. $\endgroup$
    – twistor59
    Commented Oct 21, 2013 at 6:32

3 Answers 3


The diffeomorphism invariance of GR means we're operating in the category of natural fiber bundles, where for any bundle $Y\to X$ of geometric objects that appear in the theory, we have a monomorphism $$ \mathrm{Diff} X \hookrightarrow \mathrm{Aut} Y $$ Any diffeomorphism of space-time $X$ needs to lift to a general covariant transformation of $Y$, which are not mere coordinate transformations.

These transformations play the role of gauge transformations of GR, but are different from the gauge transformations of Yang-Mills theory: The latter are related to the inner automorphisms of the group and are vertical, ie they leave space-time alone.

I'm not sure about the naturalness of the various geometric formulations of classical mechanics - I'd be interested in that as well (but am too lazy to look into it right now).

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    $\begingroup$ Thanks, though it's a bit technical. Do you have a simple example illustrating the difference between a theory that is diffeomorphism invariant and one that is not? $\endgroup$
    – JLA
    Commented Oct 21, 2013 at 3:32
  • $\begingroup$ "needs to lift to a general covariant transformation of Y, which are not mere coordinate transformations." yes, an example of the actual difference between one and the other would certainly help here $\endgroup$ Commented Jun 17, 2014 at 23:25
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    $\begingroup$ @Christoph so this means the structure group of GR is not exactly Diff(M), right? It is indeed $GL(4,\mathbb{R})$. Am I right? $\endgroup$
    – user74106
    Commented Aug 2, 2016 at 15:58
  • $\begingroup$ Sorry, but I don't understand. Could you put your explanation without using the language of bundles? $\endgroup$ Commented May 14, 2020 at 0:41
  • $\begingroup$ But GR only really looks at bundles associated to the first order (tangent) frame bundle over a spacetime manifold $M$ (i.e. bundle of invertible 1-jets of diffeomorphisms on $M$). Would it be more appropriate to say GR is only invariant under the first jet group (or reductions thereof, such as choosing orthonormal basis) In some ways this is more like a first order (Taylor) approximation to the full group of diffeomorphisms fixing a point $x \in M$. $\endgroup$
    – R. Rankin
    Commented Nov 1, 2023 at 19:20

ben's answer is exactly correct. The idea of diffeomorphism invariance (or "general covariance") was extremely important to Einstein for developing GR, but that fact has led to the unfortunately common misconception that it's somehow special to GR. Under the current way of thinking, all physical theories are diffeomorphism invariant - this follows more or less trivially from the definition of a "physical theory."

What is special about GR is this: for most other theories, there is a natural system of coordinates (usually Cartesian coordinates) in which the fundamental equations that define the theory take on a particularly simple form. Indeed, it's so natural to work in those coordinates that we rarely explicitly specify that the forms of the equations that we write down only apply in that coordinate system. In GR, on the other hand, there's no preferred system of coordinates - e.g. writing Einstein's equations out in Cartesian coordinates would decrease rather than increase the physical intuition - so we almost always go ahead and work with manifestly coordinate-independent equations.

(Technical note: in GR, we often consider spacetimes that are topologically different from Minkowski space, so there would be issues with using global Minkowski coordinates anyway.)

(Also, the extent to which you agree with the second paragraph above may depend on whether consider the metric to be a field on spacetime, or "spacetime itself." See here for a discussion.)

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    $\begingroup$ Weinberg seems to emphasize in his GR book that the important thing is not that GR is covariant but that GR identifies the physical phenomena of gravity with the terms that arise out of making an SR theory generally covariant. So, it is a statement about gravity than about form of equations per se. In this context, in relation to the last line of your first paragraph, I am thinking that you should not be able to make a theory invariant under spacetime diffeomorphisms without introducing gravity into the theory. Correct me if I am wrong. Thanks! $\endgroup$
    – user87745
    Commented May 7, 2019 at 17:28
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    $\begingroup$ @DvijMankad I do not agree with your statement that gravity comes from "making an SR theory generally covariant". SR is already perfectly diffeomorphism invariant/generally covariant, as I explained in my answer. $\endgroup$
    – tparker
    Commented May 8, 2019 at 2:07

You are correct that any theory can be written in diffeomorphism invariant language. Diffeomorphism invariance does not have anything to do with GR, except that it is hard (impossible?) to write down the theory in a gauge-fixed way.

What sets GR apart from other theories, aside from the fact that the metric is dynamical, is that it does not allow for "prior spacetime," in the language of Misner, Thorne, and Wheeler. This means it cannot be consistently coupled to non-dynamical background fields. I do not see how this feature is related to diffeomorphism invariance.


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