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In several papers (including a recent one by Banks and Seiberg) people mention a "folk-theorem" about the impossibility to have global symmetries in a consistent theory of quantum gravity. I remember having heard one particular argument that seemed quite reasonable (and almost obvious), but I can't remember it.

I have found other arguments in the literature, including (forgive my sloppiness):

  • In string theory global symmetries on the world-sheet become gauge symmetries in the target space, so there is no (known) way to have global symmetries.

  • in AdS/CFT global symmetries on the boundary correspond to gauge symmetries in the bulk so there again there is no way to have global symmetries in the bulk.

  • The argument in the Banks-Seiberg paper about the formation of a black hole charged under the global symmetry.

I find none of these completely satisfactory. Does anybody know of better arguments?

Nikita
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bangnab
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2 Answers2

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Perhaps this is just rephrasing your last explanation, so I am not sure if you consider this as a "better argument", but I'll give you a good reference for further reading.

Quantum gravity may break global symmetries because the global charge can be eaten by virtual black holes or wormholes, see this paper.

Daniel Grumiller
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  • Yes, it's something along these lines that I was looking for, but can you be maybe a bit more precise? – bangnab Feb 10 '11 at 10:30
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    Dear @inovaovao, could you please be more precise about what you find imprecise about those arguments and the paper above? For example, do you want to explain the paper showing that e.g. "black holes destroy the baryon charge unless it's a gauge symmetry" in more elementary terms, or do you want, on the contrary, a more rigorous and technical paper than the papers above? Those things are a huge amount of evidence that global symmetries can't exist in quantum gravity - which is, by itself, natural already in classical GR because "everything is made local" in GR. – Luboš Motl Feb 10 '11 at 11:39
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    For example, one may explain why the black holes destroy the non-gauged baryon number. Take a star with $B=10^{48}$, make it collapse into a black hole. Long-distance approximation - GR - will show that the black hole event horizon is locally independent of the baryon charge because there are no fields that could remember the baryon charge. So the Hawking radiation from the event horizon, by locality, has to be independent of the initial baryon charge, too. It follows that the black hole emits radiation that is independent of the initial $B$, which means $B=0$ radiation in average: B is gone. – Luboš Motl Feb 10 '11 at 11:44
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    @Lubos: I'm a bit puzzled by this. How do you know that you can trust a semiclassical calculation all the way to the end of the black hole evaporation? Wouldn't it be possible that in a truly quantum gravity description the black hole actually "knows" that it contains some nonzero $B$ and that at the end of evaporation you get it back? – bangnab Feb 18 '11 at 08:52
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    @inovaovao: Because there isn't enough mass at the end of evaporation to get $10^{48}$ baryons back. – Ron Maimon Jul 27 '12 at 06:46
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If one builds a QG in a flat space-time, a la Logunov's RTG, then one may have global symmetries. But it is forbidden to say and is punished, beware.

Vladimir Kalitvianski
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