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.

## Best Answer

I think your confusion is most concisely caught in this part of your question:

Without getting into the argument that's emerged in comments to your question about whether your particular choice of coordinates makes sense, it is generally true that once you choose a set of coordinates, you do

nothave any gauge freedom left. The gauge freedom in general relativity isexactlythe freedom to choose your coordinates. No more and no less.I think that's most naturally viewed through an ADM decomposition to 3+1 dimensions. (The part of your question about gravitational waves also gets computed in this formalism, or a variant of it.) Assuming for simplicity at this point a vacuum solution, if $g_{\mu\nu}$ is the metric of your spacetime, Arnowitt, Deser, and Misner showed in the 1950s that if you take a foliation of spacelike hypersurfaces, you can decompose this into an induced 3-metric on the surfaces $\gamma_{ij}$ plus and "lapse" function $\alpha$ and a shift vector $\beta^i$. These are related by $$ \left( \begin{array}{cc} g_{00} & g_{0j} \\ g_{i0} & g_{ij} \end{array} \right) = \left( \begin{array}{cc} \beta_k \beta^k - \alpha^2 & \beta_j \\ \beta_i & \gamma_{ij} \end{array} \right) $$ where the index on the shift is lowered by the 3-metric, $\beta_k = \gamma_{ik} \beta^i$. There's also a conjugate field $\pi^{ij}$ in this formalism that allow (coordinate) time derivatives $\partial_t \gamma_{ij}$ to be expressed as first-order-in-time PDEs.

I'm not going to write out all of the equations here because they are readily available, but making this split divides the Einstein equations into two types of equations: There are two "evolution" equations for $\gamma_{ij}$ and for $\pi^{ij}$, and there are two constraint equations. The constraints are a scalar constraint called the Hamiltonian constraint and a 3-vector constraint, known as the momentum constraint. (This is analogous, respectively, in EM to the two Maxwell equations that have time derivatives and the two that do not.)

In both the 3+1 formulation of GR and the Maxwell equations, one can prove that the "constraints propagate". That means that if the constraints are satisfied at any given time, they are satisfied at all times. See, for example, the appendix in Wald, which treats this in great detail.

Now note that

the lapse and the shift are your "gauge" fields in the this formalism. They have no physical meaning. In particular, they don't appear in the Hamiltonian constraint or in the momentum constraint, which is analogous to the fact that the EM gauge fields don't appear in the constraint equations $\nabla \cdot E = 0$ or $\nabla \cdot B = 0$ (again assuming vacuum). Not appearing in the constraint equations means that if you have any ($\gamma_{ij}$, $\pi^{ij}$) pair that satisfy the constraint equations at any given time, i.e. they are "physical" at any time, you have can construct a solution to the full Einstein equationsfor any choice of lapse and shiftby solving the evolution equations as an initial value problem, due to the fact that the constraints propagate, as mentioned above.Secondly,

note that the choice of lapse and shift are essentially equivalent to a choice of coordinates. You can find figures of this in many references, including MTW and Wald, but basically you can show that between two "nearby" surfaces $\Sigma_0$ and $\Sigma_1$, a point $x$ on $\Sigma_0$ will have the same spatial coordinate as a point $y$ on $\Sigma_1$ if $y \approx x + \alpha n + \beta$, where $n$ is normal to $\Sigma_0$ at $x$. You also have that the lapse is proportional to the proper time (rather than coordinate time) between slices.So, in some sense you are right in your complaint. Once you fix the coordinates, you don't have gauge freedom left. When you construct your "personal LIGO" ala the end of your question, you will have effectively made your coordinate choice since the "natural" coordinates for your configuration will be determined by how you layout your detector's arms, giving you a rectangular coordinate system on your "spatial slices". In the weak-field region of Earth, that will be completely compatible with unit lapse and zero shift, i.e. the Minkowski space that you would have linearized around to compute your wave solution at the detector.

Even with this understanding, you can still find the effects of gauge. When you eventually detect a wave on your personal LIGO, it will have a polarization, which you'll probably decompose into "plus" and "cross" polarizations. If you rotate your detector, which is effectively a different gauge choice, you'll mix those differently for a given signal. This probably the simplest of multiple factors that you need to "match" between your theoretical calculation of the wave - where you have to fix the gauge - and your actual configuration of the detector on the ground. This is the other end of your mistake / confusion. When you do the theory, if you want to match to your detector, you not only need to choose a gauge, you need to choose a gauge that's compatible with how you're going to read the results from your detector. You cannot not just choose any gauge that you like and try to use the results from the calculation directly with your detector.