Is the stress-energy tensor symmetric for a classical action that explicitly breaks rotational symmetry? Suppose you have a classical theory of real or complex scalar fields on flat spacetime, with a Lagrangian density that is space- and time-translationally invariant but explicitly breaks rotational symmetry. Then I believe the following logic is correct:

*

*As derived in the bottom half of Srednicki pg. 137, the stress-energy tensor $T_{\mu \nu}$ - defined as the Noether current corresponding to the translational symmetry - is conserved: $\partial_\mu T^{\mu \nu} = 0$. (I don't believe that this derivation assumes full Lorentz-invariance of the Lagrangian density, but I may be wrong.)


*From the conservation of $T_{\mu \nu}$, it follows that the divergence of the angular momentum tensor
$$M^{\mu \nu \rho} := x^\nu T^{\mu \rho} - x^\rho T^{\mu \nu}$$
equals the antisymmetric part of the stress-energy tensor:
$$\partial_\mu M^{\mu \nu \rho} := T^{[\nu \rho]}.$$
We're led to one of two possible conclusions: either (a) the stress-energy tensor is not symmetric, or (b) angular momentum is conserved even though the action is not rotationally invariant. Both of these conclusions seems kind of weird to me; I was under the impression that you can always choose to symmetrize the stress-energy tensor,*  but I don't see why angular momentum (the Noether current corresponding to rotational symmetry) would be conserved in a system that isn't rotationally symmetic.
Which conclusion is correct? Or is there a mistake in my logic and neither of them is correct?

*I know that the canonical stress-energy tensor in classical EM is not symmetric, but I want to exclude gauge theories from this question because the question of gauge redundany adds an additional complication that I don't think is relevant to my question. My impression had been that while the canonical stress-energy tensor for a general translationally-invariant theory is not always symmetric, you can always add a total derivative to it that renders it symmetric without changing any of the conserved quantities, and that this symmetrized version is what goes into the angular momentum tensor. But now I'm starting to suspect that that may not be the case for a non-rotationally-invariant action.
 A: I like to answer these kinds of questions with a simple working example. So let's write a simple scalar field theory that breaks rotation but not translation invariance:
\begin{equation}
\mathcal{L}= -\frac{1}{2} \left( \eta^{\mu\nu} + \lambda e^\mu e^\nu \right) \partial_\mu \phi \partial_\nu \phi 
\end{equation}
where $\lambda$ is a constant and $e^\mu$ is some fixed unit norm ($\eta_{\mu\nu} e^\mu e^\nu=1$) spacelike 4-vector that is constant over spacetime.
Now let's do a variation with a spacetime dependent shift $\epsilon(x)$; we will then use the trick that because translations with $\epsilon={\rm const}$ are a symmetry we can always write the variation as $\delta \mathcal{L}=-\partial_\mu \epsilon_\nu \tilde{T}^{\mu\nu}$, which after an integration by parts is $\epsilon_\mu \partial_\nu \delta T^{\mu\nu}$ (where $\tilde{T}$ and $T$ potentially differ by a term whose divergence is identically zero). Then since $\partial_\mu T^{\mu\nu}=0$ on shell, we can identify $T^{\mu\nu}$ as the stress-energy tensor.
More concretely, we write $x^\mu \rightarrow x^\mu - \epsilon^\mu(x)$, so $\phi \rightarrow \phi + \epsilon^\mu \partial_\mu \phi$ and $\partial_\mu \phi \rightarrow \partial_\mu \phi + \epsilon^\nu \partial_\mu \partial_\nu \phi + \partial_\nu \phi \partial_\mu \epsilon^\nu$, leading to
\begin{eqnarray}
%
\delta \mathcal{L} 
&=& 
- \frac{1}{2} \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] \left[ \delta(\partial_\rho \phi) \partial_\sigma \phi + \partial_\rho \phi \delta (\partial_\sigma \phi) \right] \\
%
&=& 
- \frac{1}{2}  \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] 
\left[ \epsilon^\nu \partial_\rho \partial_\nu \phi \partial_\sigma \phi 
+  \epsilon^\nu \partial_\sigma \partial_\nu \phi \partial_\rho \phi
+ \partial_\rho \phi \partial_\nu \phi  \partial_\sigma \epsilon^\nu  
+ \partial_\sigma \phi \partial_\nu \phi  \partial_\rho \epsilon^\nu \right] \\
%
&=& 
- \frac{1}{2}  \left[ \eta^{\rho\sigma} + \lambda e^\rho e^\sigma \right] 
\left[ \epsilon^\nu \partial_\nu \left( \partial_\rho  \phi \partial_\sigma \phi \right)  
- \epsilon_\mu \partial_\sigma \left(\partial^\mu \phi \partial_\rho \phi \right)
- \epsilon_\mu \partial_\rho \left(\partial^\mu \phi \partial_\sigma \phi \right) \right] \\
%
&=& 
\epsilon_\mu \partial_\nu \left[ \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu} (\partial \phi)^2 + \lambda \left(  e^\mu  \partial^\nu \phi e^\rho \partial_\rho \phi  -  \frac{1}{2} e^\mu e^\nu (\partial \phi)^2\right) \right]\\
%
&=& \epsilon_\mu \partial_\nu \left[T_{\rm K.G.}^{\mu\nu} + \lambda T_{\rm S.B.}^{\mu\nu} \right]
\end{eqnarray}
where $T_{\rm K.G.}^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi - \frac{1}{2} \eta^{\mu\nu}(\partial \phi)^2$ is the stress tensor for a massless Klein-Gordon field, and the term $T^{\mu\nu}_{\rm S.B.}$ is the stress-energy tensor due to the breaking of rotational symmetry (S.B.=symmetry breaking).
Indeed, the expression for $T^{\mu\nu}_{\rm S.B.}$
\begin{equation}
T^{\mu\nu}_{\rm S.B.} =  e^\mu  \partial^\nu \phi e^\rho \partial_\rho \phi -  \frac{1}{2} e^\mu e^\nu (\partial \phi)^2
\end{equation}
has a non-vanishing anti-symmetric part $\sim (e \cdot \partial)\phi e^{[\mu}\partial^{\nu]} \phi $. One way to explain in words what is happening is that $\epsilon_\mu$ is never directly contracted with $e^\rho$ or $e^\sigma$, and unlike the metric tensor there is no way to "generate more" $e^\rho$ or $e^\sigma$ by raising and lowering indices. Another way is to explain the difference between the rotationally invariant and symmetry breaking terms is to point out that if we replace $e^\mu e^\rho$ with $\eta^{\mu\rho}$ in the first term of $T^{\mu\nu}_{\rm S.B.}$, the first term becomes $\partial^\mu\phi \partial^\nu \phi$, which is symmetric; the "symmetry breaking" term $\sim e^\mu e^\rho$ picks out a special direction on which the final answer can depend, which allows for an anti-symmetric part. To be honest, this isn't the result I was expecting when I started working this out (although I don't know why since there is no way angular momentum can be conserved and I agree with the logic in your question), but as far as I can tell it is correct.
It's always possible to define a symmetric stress-energy tensor by defining it as a variation of the action with respect to the metric, but I take it from the question that this is not what you are interested in.
Another question is whether it is possible to add piece that is identically divergence free to $T^{\mu\nu}_{\rm S.B.}$ to cancel the anti-symmetric part... I suspect (but am not 100% sure) the answer is yes, since I would think you should be able to derive the Einstein-Hilbert version of the stress-energy tensor in this way. (This isn't always possible but I suspect it probably is in this example)
