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In Weinberg's paper on the cosmological constant problem (CCP), he states that diffeomorphism invariance is always broken by the presence of any given metric $g_{\mu\nu}$. He then goes on to say that there is a residual $GL(4)$ symmetry remaining if we choose the metric (and the set of scalar fields introduced as a putative solution to the CCP) to be constant. (For clarity, he discusses these things on page 4, left-hand column, 2nd paragraph onwards).

I'm quite confused by this though. In particular, I don't see why the choice of metric breaks diff. invariance? I thought the whole point was that, if $g_{\mu\nu}$ is a solution to the Einstein field equations, then any other metric $\hat{g}_{\mu\nu}$ related to $g_{\mu\nu}$ by a diffeomorphism, i.e. $\hat{g}_{\mu\nu}=(\phi^{\ast}g)_{\mu\nu}$, is also a solution. What am I missing here?

Also, why is there a residual $GL(4)$ symmetry? Is it precisely because the fields are constant, i.e. the on-shell solutions are translationally invariant?

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  • $\begingroup$ Rather than "choice of metric", I would prefer "emergence of Minkowski Metric", which is akin to the emergence of the Higgs VEV. And as such, it inevitably spoils the local diffeomorphism invariance. $\endgroup$
    – MadMax
    Commented Jun 19, 2019 at 20:30
  • $\begingroup$ @MadMax Weinberg specifically mentions the presence of any metric breaks diff. invariance though. This is a good point, however. I've always been unsure about this discussion that solutions to the Einstein field equations are equivalence classes of metrics. Clearly the Minkowski metric doesn't preserve diff. invariance. Or is this a confusion between (to use the physics parlance) active and passive diffeomorphisms? $\endgroup$
    – Will
    Commented Jun 19, 2019 at 20:37
  • $\begingroup$ Fundamentally, metric is a quantum field. For a vacuum devoid of CC, it's NOT imperative for the metric to acquire a non-zero Minkowski VEV. Nothing prevents you to have a zero VEV $g_{\mu\nu} \equiv 0$, which preserves the diffeomorphism invariance. In other words, Minkowski VEV $<0|g_{\mu\nu}|0> = \eta_{\mu\nu}$ is an emergent phenomenon. $\endgroup$
    – MadMax
    Commented Jun 19, 2019 at 20:44
  • $\begingroup$ Maybe I should make myself more clear. General relativity (or Einstein's gravity equation) does NOT dictate that the vacuum solution for metric should be non-zero. Rather, zero $g_{\mu\nu} =0$ is a more natural solution and zero $g_{\mu\nu}= 0$ enjoys the diffeomorphism invariance. Non-zero metric (e.g. the Minkowski metric) is NOT an unavoidable corollary of tenet of general relativity. $\endgroup$
    – MadMax
    Commented Jun 19, 2019 at 21:30
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    $\begingroup$ @Will: Thanks. When people ask for clarification of a question, normally the best thing to do is to edit the question to provide the clarification. That saves everyone else from having to read the comment thread to understand the question. $\endgroup$
    – user4552
    Commented Jun 19, 2019 at 23:26

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I don't see why the choice of metric breaks diff. invariance? I thought the whole point was that, if $g_{\mu\nu}$ is a solution to the Einstein field equations, then any other metric $\hat{g}_{\mu\nu}$ related to $g_{\mu\nu}$ by a diffeomorphism, i.e. $\hat{g}_{\mu\nu}=(\phi^{\ast}g)_{\mu\nu}$, is also a solution. What am I missing here?

I think what he means is that if you fix the dependence of the metric's components on the coordinates, then that breaks diffeomorphism invariance. The metric will usually look different when you choose different coordinates. It's unusual that for Minkowski space, you can have different coordinates in which the components of the metric look the same.

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  • $\begingroup$ Ok, but won’t it generally be true that two metrics related by a diffeomorphism will have different components in a given set of coordinates? They still describe the same physical spacetime, however. $\endgroup$
    – Will
    Commented Jun 20, 2019 at 7:57
  • $\begingroup$ @Will: won’t it generally be true that two metrics related by a diffeomorphism will have different components in a given set of coordinates? They still describe the same physical spacetime, however. Correct. $\endgroup$
    – user4552
    Commented Jun 20, 2019 at 12:21
  • $\begingroup$ I'm still confused then as to what is meant by his statement that the presence for a given metric breaks diff. invariance, and also his comment about the residual $GL(4,R)$ symmetry? $\endgroup$
    – Will
    Commented Jun 20, 2019 at 14:22

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