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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 $...


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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....


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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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