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Although users35736's answer is certainly correct, and the question is old, I think it should be noted that each Killing vector, $\xi^i$, also gives rise to a Noether current: $J^i = T_j{}^i\xi^j$. First note that $$ J^i{}_{;i} = T^{ji}\xi_{i;j} + \xi_jT^{ji}{}_{;i} = 0, $$ by the Killing equation $\xi_{(i;j)} = 0$, the symmetry of the stress-energy tensor $...


Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them are singlets and there are no nontrivial representations, (and 3....


The corresponding symmetry group is the Lorentz group and yes we can use Noether to derive conserved quantities: Invariance under translations $\rightarrow$ momentum conservation Invariance under rotations $\rightarrow$ spin and angular momentum conservation Invariance under boost $\rightarrow$ some strange, not really useful, conserved quantity

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