Counting degrees of freedom for gravitational waves as a gauge field

Question

Sean Carroll has a new popularization about the Higgs, The Particle at the End of the Universe. Carroll is a relativist, and I enjoyed seeing how he presented the four forces of nature synoptically, without a lot of math. One thing I'm having trouble puzzling out, however, is his treatment of gravity as just another gauge field. First let me lay out what I understand to be the recipe for introducing a new gauge field, and then I'll try to apply it. I'm not a particle physicist, so I'll probably make lots of mistakes here.

Translating the popularization into a physicist's terminology, the recipe seems to be that we start with some discrete symmetry, expressed by an $m$-dimensional Lie algebra, whose generators are $T^b$, $b=1$ to $m$. Making the symmetry into a local (gauge) symmetry means constructing a unitary matrix $U=\exp[ig \sum \alpha_b T^b]$, where $g$ is a coupling constant and the real $\alpha_b$ are functions of $(t,x)$. If $U$ is to be unitary, then in a matrix representation, the generators must be traceless and hermitian. If you already have an idea of what the matrix representation should look like, you can determine $m$ by figuring out how many degrees of freedom a traceless, hermitian matrix should have in the relevant number of dimensions. For each $b$ from 1 to $m$, you get a vector field $A^{(b)}$, which has two actual d.f. rather than four.

Applying this recipe to electromagnetism, the discrete symmetry is charge conjugation. That's a 1-dimensional Lie algebra, so you get a single field $A$, which is the vector potential from E&M, and its two d.f. correspond to the two helicity states of the photon.

Applying it to the strong force, the discrete symmetry is permutation of colors. That's going to be represented by a 3x3 matrix. The traceless, hermitian 3x3 matrices are an 8-dimensional space, so we get 8 gluon fields.

So far, so good. Now how the heck does this apply to gravity? Carroll identifies the symmetry with Lorentz invariance, so the symmetry group would be SO(1,3), which is a 6-dimensional Lie algebra. That would seem to create six vector fields, which would be 12 d.f. Does this correspond somehow to the classical description of gravitational waves? If you express a gravitational wave as $h^{\alpha\beta}=g^{\alpha\beta}-\eta^{\alpha\beta}$, with $h$ being traceless and symmetric, you get 6 d.f...?

Answer 1

First of all, Sean Carroll is a relativist so his treatment of the diffeomorphism symmetry as a gauge symmetry should be applauded because it's the standard modern view preferred by particle physicists – its origin is linked to names such as Steven Weinberg, it is promoted by physicists like Nima Arkani-Hamed, and naturally incorporated in string theory so seen as "obvious" by all string theorists. In this sense, Carroll throws away the obsolete "culture" of the relativists. There are some other "relativists" who irrationally whine that it shouldn't be allowed to call the metric tensor "just another gauge field" and the diffeomorphism group as "just another gauge symmetry" even though this is exactly what these concepts are.

Second of all, a symmetry expressed by a Lie algebra can't be "discrete", by definition: it is continuous. Lie groups are continuous groups; it is their definition. And only continuous groups are able to make whole polarizations of particles unphysical. It's plausible that a popular book replaces the continuous groups by discrete ones that are easier to imagine by the laymen but this server is not supposed to be "popular" in this sense.

Third, when you say that if $U$ is unitary, the generator has to be Hermitian and traceless, is partly wrong. Unitarity of $U$ means the hermiticity of the generators $T^a$ but the tracelessness of these generators is a different condition, namely the property that $U$ is "special" (having the determinant equal to one). The tracelessness is what reduces $U(N)$ to $SU(N)$, unitary to special unitary.

Fourth, and it is related to the second point above, "charge conjugation" isn't any gauge principle of electromagnetism in any way. Electromagnetism is based on the continuous $U(1)$ gauge group. This group has an outer automorphism – a group of automorphisms is ${\mathbb Z}_2$ – but we're never putting these elements of the discrete group into an exponent.

Fifth, similarly, QCD isn't based on the discrete symmetry of permutations of the colors but on the continuous $SU(3)$ group of special unitary transformations of the 3-dimensional space of colors. Because none of the things you wrote about the non-gravitational case was quite right, it shouldn't be surprising that you have to encounter lots of apparent contradictions in the case of gravity as well because gravity is indeed more difficult in some sense.

Sixth, $SO(3,1)$ isn't related to the diffeomorphism in any direct way. It is surely not the same thing. This group is the Lorentz group and in the GR, you may choose a formalism based on tetrads/vielbeins/vierbeins where it becomes a local symmetry because the orientation of the tetrad may be rotated by a Lorentz transformation independently at each point of the space. But this is just an extra gauge symmetry that one must add if he works with tetrads – it's a symmetry that exists on top of the diffeomorphism symmetry and this symmetry is different and "non-local" because it changes the spacetime coordinates of objects or fields while all the Yang-Mills symmetries above and even the local Lorentz group at the beginning of this paragraph are acting locally, inside the field space associated with a fixed point of the spacetime. (The fact that diffeomorphisms in no way "boil down" to the local Lorentz group is a rudimentary insight that is misunderstood by all the people who talk about the "graviweak unification" and similar physically flawed projects.) I will not use with tetrads in the next paragraph so the gauge symmetry will be just diffeomorphisms and there won't be any local Lorentz group as a part of the gauge symmetry.

The diffeomorphism symmetry is locally generated by the translations, not Lorentz transformations, and the parameters of these 4-translations depend on the position in the 4-dimensional spacetime. This is how a general infinitesimal diffeomorphism may be written down. If there were no gauge symmetries, $g_{\mu\nu}$ would have 10 off-shell degrees of freedom, like 10 scalar fields. However, each generator makes two polarizations unphysical, just like in the case of QED or QCD above (where the 4 polarizations of a vector were reduced down to 2; in QCD, all these numbers were multiplied by 8, the dimension of the adjoint representation of the gauge group, $SU(3)$ etc.). Because the general translation per point has 4 parameters, one removes $2\times 4 = 8$ polarizations and he is left with $10-8=2$ physical polarizations of the gravitational wave (or graviton). The usual bases chosen in this 2-dimensional physical space is a right-handed circular plus left-handed circular polarized wave; or the "linear" polarizations that stretch and shrink the space in the horizontal/vertical direction plus the wave doing the same in directions rotated by 45 degrees:

This counting was actually a bit cheating but it does work in the general dimension. To do the counting properly and controllably, one has to distinguish constraints from dynamical equations and see how many of the modes of a plane wave (gravitational wave) are affected by a diffeomorphism. In the general dimension of $d$, it may be seen that the tensor $\Delta g_{\mu\nu}$ may be described, after making the right diffeomorphism, by $h_{ij}$ in $d-2$ dimensions and moreover the trace $h_{ii}$ may be set to zero. This gives us $(d-2)(d-1)/2-1$ physical polarizations of the graviton. In $d=4$, this yields 2 physical polarizations of the graviton. A gravitational wave moving in the 3rd direction is described by $h_{11}=-h_{22}$ and $h_{12}=h_{21}$ while other components of $h_{\mu\nu}$ may be either made to vanish by a gauge transformation (diffeomorphism), or they're required to vanish by the equations of motion or constraints linked to the same diffeomorphism. Morally speaking, it is true that we eliminate two groups of 4 degrees of freedom, as I indicated in the sloppy calculation that happened to lead to the right result. Note that $$\frac{d(d+1)}2 -2d = \frac{(d-2)(d-1)}2-1 $$ I have to emphasize that these is a standard counting of the "linearized gravity" and it's the same procedure to count as the counting of physical polarizations after the diffeomorphism "gauge symmetry" – just the language involving "gauge symmetries" is more particle-physics-oriented.

Answer 2

To answer your question directly about gravity as a gauge theory, metric-affine gauge gravity helps clarify the role of the metric tensor in relation to gauge symmetry. Metric-affine gravity is a relativistic elasticity theory, which motivates linear diffeomorphisms, as elastic media are fluids with small displacements. The gauge group $GL(4,\mathbb{R})$ relates to linear diffeomorphisms. The affine linear group $AL(4,\mathbb{R})$ adds translational symmetry to the general linear group.

Trautman from 1979-1980 articulated how the metric tensor is a type of Higgs field; Tomboulis has made this point as well and articulated how the metric relates to breaking from $GL(4,\mathbb{R})$ to the Lorentz group $SO(3,1)$. There are different formulations of gauge gravity, but the one from Niederle and Ivanov from 1981 and 1982 is nice, as it introduces a global manifold $M^4$ with a local gauge group $G$. $G$ must minimally contain the Lorentz or Poincare group, while the conformal group was also studied. When looking at the affine linear group $G = AL(4,\mathbb{R})$, we can review the gauge gravitational potentials of metric-affine gravity next.

The Lorentz group relates to the spin connection 1-form:

$\omega = \omega^{ab} J_{ab} = \omega_\mu^{ab}J_{ab} dx^\mu$.

The translations relate to the frame field 1-form:

$e = e^a P_a = e_\mu^a P_a dx^\mu$.

The leftover degrees of freedom are the affine linear group modded out by the Poincare group, which is also $\frac{GL(4,\mathbb{R})}{SO(3,1)}$. Since $GL(4,\mathbb{R})$ has 16 degrees of freedom and $SO(3,1)$ has 6, this quotient space has 10 degrees of freedom. Unsurprisingly, this corresponds precisely to the metric 0-form as a global scalar with respect to $M^4$:

$g = g_{ab}M^{(ab)}$,

where $M_{ab}$ are the generators of $GL(4,\mathbb{R})$ and $J_{ab} = M_{[ab]}$, such that the 10 symmetric generators $M_{(ab)}$ correspond to the inverse metric. General relativity is not a gauge theory, so linearized general relativity is also not a gauge theory. Truly gauging the diffeomorphisms would lead to an infinite number of gauge gravitational field strengths, which I haven't seen any work on. Certainly, Carroll does not discuss this, as he sets torsion and nonmetricity to zero, which is contained in metric-affine gravity.

Technically, $g_{ab}$ is a global scalar with respect to $M^4$, as its indices are associated to $G$. However, the spin-2 graviton is well-known to be recovered from the local metric with the frame fields, which leads to:

$g_{\mu\nu} = e_\mu^a e_\nu^b g_{ab}$.

String theorists have never considered super-metric-affine gravity, but it can be formulated and does not violate the Coleman-Mandula formula, as nonmetricity as the field strength of $g_{ab}$ is a type of internal gravitational symmetry. Since the spin-2 metric $g_{\mu\nu}$ now refers to indices associated to the manifold $M^4$, this corresponds to the metric in general relativity. The equation above clarifies why many claim that translations relate to diffeomorphisms, as they often only consider an affine formulation. An affine formulation of gauge gravity technically has no gauge degrees of freedom for the metric. Supergravity often refers to the metric and the frame field in an interchangeable manner. It is common to set $g_{ab} = \eta_{ab}$ as a local Minkowski metric. Also, most people presume a tetrad postulate $\nabla_\mu e_\nu^a=0$, which is different than claiming that nonmetricity vanishes.

In conclusion, considering $\frac{GL(4,\mathbb{R})}{SO(3,1)}$ rigorously establishes why the metric tensor has 10 degrees of freedom from the perspective of gauge gravity. Coincidentally, the Poincare group has 10 degrees of freedom, but those truly refer to the spin connection and frame field. Asking for the gauge-theoretic origin of $g_{\mu\nu}$ is technically ambiguous, as there are different formulations of gauge gravity. Metric-affine gauge gravity has been increasingly studied recently, but most focus on global manifold coordinate indices, rather than the local gauge indices. It is possible to not consider $M^4 \times G$, but I think this helps clarify until the physics community is ready to establish gauge gravity by gauging the infinite-dimensional diffeomorphism group.