Can we regard metric as the Higgs field of gravity?

Question

The longer version of the question is: should we regard special relativity just as a spontaneous symmetry breaking phase of general relativity, driven by the non-zero vacuum expectation value (VEV) of metric which is acting as the Higgs field of gravity.

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A side note added after seeing the answer and references provided by @Mitchell Porter: It seems that I am not the first one (too bad!) to investigate this idea of "metric as the Higgs field of gravity", though my version is a bit different since I am the first one (hopefully!) attempted to interpret the global Lorentz symmetry as similar to the global $SU(2)$ custodial symmetry in electroweak theory. A few quotes from some renowned gauge gravity experts:

• Trautman: "The metric tensor is a Higgs field breaking the symmetry from $GL (4,R)$ to the Lorentz group" in THE GEOMETRY OF GAUGE FIELDS
• Sardanashvily: "If gravity is a metric field by Einstein, it is a Higgs field" in Gravity as a Higgs field

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This post is related to another PSE question on Gravity as a gauge theory. When gravity is expressed as a local Lorentz gauge theory (also know as Einstein-Cartan gravity), there are a lot of similarities compared with a typical gauge theory. For example, the gauge group of Lorentz gravity is the local Lorentz group $SO(1,3)$, and the gauge field of Lorentz gravity is the spin connection field $\omega^{ab}_\mu$. However, the metric $g_{\mu\nu}$ itself is not a gauge field in Lorentz gravity. Rather, the metric $g_{\mu\nu}$ is an add-on field in addition to the Lorentz gauge field $\omega^{ab}_\mu$ and there is no such add-on field in a typical gauge theory. Therefore, folks would usually point out that gravity, even in its Lorentz gauge theory format, is not comparable to a typical gauge theory.

But this point of view seems to be inaccurate. Let's take a look at the quintessential gauge theory: the electroweak theory of the standard model. Other than the electroweak gauge fields $W_\mu, A_\mu$, the electroweak theory has also an add-on non-gauge field: the Higgs field $H$ that are responsible for spontaneous symmetry breaking. Can we argue that metric is actually the Higgs field of gravity? Specifically, the symmetry-breaking add-on field of Lorentz gravity is the tetrad/veirbein field $e^a_\mu$ (which could be understood as the square root of metric $g_{\mu\nu} = \eta_{ab}e^a_\mu e^b_\nu$): the non-zero Minkowskian VEV of the tetrad/veirbein field $$<0|e^a_\mu |0> = \delta^a_\mu$$ or equivalently $$<0|g_{\mu\nu} |0> = \eta_{\mu\nu}$$ in Lorentz gravity breaks the gauge symmetry from the local Lorentz gauge symmetry $SO(1,3)$ down to the global Lorentz symmetry of special relativity, similar to the case of the non-zero VEV of Higgs field $$<0|H |0> = v$$ in electroweak theory breaking the gauge symmetry from $SU(2)*U_Y(1)$ down to $U_{EM}(1)$.

So can we really regard metric as the Higgs field of gravity? For more discussions over the similarities and dissimilarities between metric and Higgs field, see here, here, and here. See also here and here for discussions over the weird and peculiar state of global Lorentz symmetry as a combination of the residual elements from both the local Lorentz symmetry and the diffeomorphism symmetry, with tetrad/veirbein acting as the soldering form.

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Added note to address @ACuriousMind comment:

The spontaneous symmetry breaking of the local Lorentz symmetry to global Lorentz symmetry is a complicated SSB process, which needs more explanation. To be exact, the standard model counterpart of the residual global Lorentz symmetry is actually not the electromagnetic $U_{EM}(1)$ symmetry. After spontaneous symmetry breaking, the residual global Lorentz symmetry in Lorentz gravity is more like the residual global $SU(2)$ custodial symmetry in electroweak theory. For how global custodial symmetry can survive SSB, please see details of global $SU(2)$ custodial symmetry in electroweak theory here: P. Sikivie, L. Susskind, M. B. Voloshin and V. I. Zakharov, Nucl. Phys. B 173 (1980) 189. Also see Wikipedia page here explaining custodial symmetry.

Answer 1

We will be using Trautman's unifying formulation of both general relativity and Yang-Mills gauge theories as theories of a $G$-principal bundle $P\to M$ over spacetime $M$. For an exposition of this formalism, see this answer of mine. The Higgs part of this formalism is mostly from Trautman's "The geometry of gauge fields".

tl;dr: The metric is formally analogous to a Higgs field, but there isn't a Higgs mechanism. The Lorentz group is fully analogous to the unbroken subgroup of the gauge group, not to a custodial symmetry.

A Higgs field is a field $\phi$ transforming in some associated bundle $V_\rho$ to some non-trivial representation $\rho$ of the gauge group whose equations of motion are $$ V(\phi) = 0 \quad \land \quad \mathrm{d}_\omega \phi = 0,$$ where $V$ is some potential and $\mathrm{d}_\omega$ is the principal connection on our bundle. The solutions to $V(\phi) = 0$ form the vacuum manifold $\mathcal{V}$ and the group $G$ has a natural action on this space. If the action is transitive but not free, any point $\phi_0\in\mathcal{V}$ has a non-trivial stabilizer group $H:= \{g\in G \mid \rho(g)\phi_0 = \phi_0 \}$ and $\mathcal{V} = G/H$. In general, we need not necessarily assume transitivity and non-freeness and simply talk about $\mathcal{V}_{\phi_0}$ as the orbit of any field configuration $\phi_0$ under $G$.

Any $\phi_0$ allows for the reduction of the $G$-principal bundle to a $H$-principal bundle as the subbundle $P_{\phi_0} = \{ p\in P \mid \rho(p)\phi_0 = \phi_0 \}$. This is the only real mathematical reflection of the idea of symmetry breaking.

For the final discussion to make sense, we need to evade the deeply confusing issue of Elitzur's theorem that gauge symmetries cannot be spontaneously broken. For the "standard" resolution of this issue, see this answer by Dominic Else. We will use instead the idea of formulating the theory entirely in terms of gauge invariant fields from "Higgs Phenomenon without Symmetry Breaking Order Parameter" by Fröhlich, Morchio and Strocchi:

For any $\mathcal{V}_{\phi_0}$ there are two options (see the paper for proofs of the following assertions):

• $H = \{1\}$ is trivial, the group acts transitively on $\mathcal{V}_{\phi_0}$. Then for any other field $\psi$ transforming in another $V_\sigma$, the space of fields $\psi : M\to V_\sigma$ is in bijection to fields $F(\phi)\cdot \psi$ where $F$ is a $\rho$-equivariant polynomial of $\phi$, the $\cdot$ represents a pairing between $V_\rho$ and $V_\sigma$ that results in a $G$-invariant scalar, and we say $F(\phi) = F'(\phi)$ if $F-F'$ vanishes on $\mathcal{V}_{\phi_0}$.

• $H$ is not trivial. Then $V_\sigma = \sum_i V_\sigma^i$ decomposes into representations $V_\sigma^i$ of $H$, and there are polynomials $F^i(\phi)$ valued in projections $V_\sigma \to V_\sigma^i$ in bijection to the projections $V_\rho \to V_\rho^i$. In this case any $\psi$ is in bijection to the set of fields $F^i(\phi)\psi$. $F^i(\phi)\psi$ is either gauge-invariant or transforms in $V_\sigma$ under the original gauge group.

The gauge-invariant ones are straightforwardly meaningful physical fields, while from the covariant ones we can construct invariant operators using Wilson lines. In any case, the masses of the Higgs mechanism just arise from plugging $\langle \phi_0\rangle$ into the terms of these fields that are still invariant or covariant under the full gauge group, there is no symmetry breaking in terms of these fields.

The metric in general relativity is now straightforwardly a Higgs field: It obeys the Einstein Field Equations $V(g) = 0$ with vacuum solution $g = \eta$, and the $\mathrm{d}_\omega g = 0$ is just the condition that the Levi-Civita connection is metric compatible (and in the Palatini formalism this is an equation of motion).

However, while this does produce a reduction of the structure group of the frame bundle to a $\mathrm{SO}(1,3)$-bundle (which purely comes about as the stabilizer of $\eta$), it does not "trigger" the Higgs mechanism: The whole reason that we in the above discussion care about constructing gauge-invariant quantities is that in Yang-Mills theories, they are pure internal gauge symmetries (see my answer from the beginning), while general relativity has no such internal symmetries - there is no pure gauge symmetry here that connects physically indistinguishable states, hence no need to introduce a Higgs-type reasoning.

So it depends very much on what we mean by a "Higgs field" in physical terms whether or not we really should call the metric that: If we just mean the formal equivalence to the Higgs fields of Yang-Mills theories, go ahead. If we are in danger of being misunderstood in the sense that its physical effects (i.e. the Higgs mechanism) are equivalent to that of a Yang-Mills Higgs field, we probably shouldn't say that.

Answer 2

A version of this idea is discussed at the end of section 1.1.1 of

https://inspirehep.net/literature/1823987

which also cites the following as precursors

https://inspirehep.net/literature/150700

https://inspirehep.net/literature/37958