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Momentum cut-off regularisation leads to non-covariant results, i.e., it breaks the Poincaré covariance of the theory. Is there any guarantee that Poincaré covariance is always restored when we remove the cut-off? Or can this symmetry be anomalous? Is there any topological argument about the Poincaré group that would automatically rule out any anomaly?

Assume that the regulator is the only source of non-covariance (for example, if there are gauge fields, the fixing condition is covariant as opposed to, say, Coulomb). Also, assume that Poincaré is a global symmetry of the classical theory (i.e., no gravity).

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  • $\begingroup$ My understanding is that existence of other regulators implies Poincaré covariance. You don’t specify any restrictions on the models you are considering, but if a Pauli-Villars regularizer exists for them all (for which there’s a general proof), then Poincaré is unbroken, and it is safe to use the cut-off regularizer in practical calculations. But I may be wrong. $\endgroup$ – Prof. Legolasov Apr 27 '18 at 20:49
  • $\begingroup$ @SolenodonParadoxus That's actually a more general version of my question: the existence of a regulator that respects some (non-gauge) symmetry implies that any regulator does? I guess you are saying that the answer is "yes", which sounds reasonable. It would be great to have a more or less convincing argument to why that is so. $\endgroup$ – AccidentalFourierTransform Apr 27 '18 at 20:57
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    $\begingroup$ there’s an even more general version — apparently, if multiple regularizers exist and give rise to well defined renormalized theories, then those are equivalent. I don’t know how to prove this, but have heard people making this claim so I just assumed it is true so far ;) $\endgroup$ – Prof. Legolasov Apr 27 '18 at 21:00

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