# Global symmetries in quantum gravity

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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?

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user inovaovao

edited Aug 16, 2014

The black hole argument is completely satisfactory, there's no way for Hawking radiation to be sensitive to a global charge, so that it must be true in any theory with gravity. The other arguments depend on a specific model.

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To summarize the complete argument here. Let's pretend Baryon number is an exact symmetry. A neutron star has 10^40-something baryons. It collapses into a black hole which then evaporates into photons and gravitons. During the semi-classical phase, when the black hole is enormous, the emissions cannot be of massive particles, only massless particles can come out, because the mass-scale for emission is the inverse black-hole radius in natural units, which is tiny. At the end of evaporation, when this mass scale is large enough for massive particles to come out, the black hole is very small, and doesn't have enough mass to emit 10^48 baryons. Therefore baryon number is violated in black hole emissions.

This argument shows that baryon number is not exact for sure, and similarly for any global symmetry where all charged particles are massive. It's a complete physical argument, and it can't be sidestepped.

But it leaves a loophole when there is a massless globally charged particle. In this case, you need to think a little more about the details of black hole decay. The decay of black holes is thermal, into photons and gravitons, and any massless mode equally. You can consider forming a large black hole of large mass and global charge from the collision of a large number of such massless particles converging from opposite directions, and this black hole will decay into an equal proportion of these particles and gravitons, which you should always be able to arrange to violate the global symmetry.

The reason it seems this way is that the black hole emissions are thermal, and are produced at thermal equilibrium at the Hawking temperature, regardless of global charge, just by counting field mode degrees of freedom. Choosing a bunch of particles of some global charge, these are not in global equilibrium, because the state is specially chosen to have large value of the charge, and the process of infalling into the black hole removes the charge without a trace. This argument is not airtight, unlike the case with massive globally charged particles.

Black holes preserve gauge symmetries, because these produce a field outside the black hole which keeps track of the total charge inside. But in order to do so, this places a bound on the mass of the lightest charged particle, so that the black hole can properly decay. The bound is that the lightest charged particle must be lighter than it's mass in natural units, in other words, that gravitational attraction is necessarily weaker than the gauge-repulsion.

This means that black holes also can conserve discrete global symmetries, like $Z_n$ symmetries, as these can be the residuals gauge symmetries at low-energies after breaking. When $n$ is not enormously large, these are symmetries that only restrict the very end of the black hole evaporation, and are not inconsistent with the arguments above. When $n$ becomes large, this becomes more and more like a global symmetry, but the same argument gives a similar inequality on the mass of the $Z_n$ charged particles as $n$ gets larger. This suggests that perhaps you can find a situation where there is a global symmetry or near-global with massless or near-massless charged particles, but then it is not about the real world, where all possibly globally charged particles in the standard model are massive (like the baryon).

The inequalities from complete black-hole decay are the most restrictive swampland constraints we know about, and are the most predictive statements string theory can make in general today.

The string theory arguments, in AdS/CFT and worldsheet string theory, seem superficially different from the Hawking semiclassical black-hole argument, but once you understand the holographic principle, they are simply the exact same thing stated mathematically precisely. The string itself is a kind of quantum-sized black hole, and the vertex operators for absorption/emission are representing  the amplitude of a particular history of infalling stuff and then Hawking emission, the fact that these are described by simple motion modes of the string is required by the holographic principle.

So the statement that this reproduces the swampland constraints from semiclassical black hole decay is nonperturbative completely persuasive evidence that string theory is consistent as a theory of gravity. The consistency of the two pictures is the content of the 1990s string revolution, which arrived at the string dualities by considerations of BPS states, which are, in string theory, classical extremal black holes, with gauge-charge equal to mass. This is the only thing that was strong enough to persuaded me, personally, that string theory is a consistent theory of quantum gravity, as the constraint of being consistent with semiclassical black-hole decay pretty much uniquely fixes the properties of the mathematical description which string theory has.

answered Aug 16, 2014 by (7,535 points)
edited Aug 16, 2014

About the argument of the first paragraph: "the black hole is very small, and doesn't have enough mass to emit 10^48 baryons", how to exclude the possibility to have a particle of baryonic charge 10^48 as final state of the black hole decay?

That's the remnant idea, you're right. It's not logically ruled out, it's ruled out by what I consider to be the common sense axiom that black holes that are small don't have a constant enormous entropy that can store all the detailed history of how they formed. The idea is that the Bekenstein bound actually is a real thing, that small objects can't have an enormous arbitrarily large information (since the Baryon number can be made as large as you like, by throwing more baryons into the black hole when it's big, and waiting for other stuff to come out)

So the argument is: if one creates a black hole of baryonic charge N, and if the baryonic number is conserved during the decay, at the end of the decay we have a small object, of size around the Planck scale, with baryonic charge N. As the size of the small object is independent of N and  N can be arbitrary large, there is a contradiction with the entropy bound for N large enough. This argument is convincing in general but I am puzzled by a special (maybe unrealistic) case. The argument uses the fact that the decay of the black hole is almost complete. It is certainly the case if the black hole is neutral under all the gauge fields but not necessarely otherwise: the final state could be a stable extremal black hole.

More precisely, I assume there is an electromagnetic gauge field. If there exists a neutral baryon or more generally if there exists baryons of positive electric charge and baryons of negative electric charge, I can still produce a electrically neutral black hole and the argument applies. But if the baryons are all positively electrically charged, the black hole is electrically  charged and can decay to an extremal black hole. The argument would still work if the size of the final extremal black hole is independent of N but I don't know how to show this. Of course, one has to use the hypothesis that the global symmetry is not the gauge symmetry, i.e. that there exists non-baryonic electrically charged particles but I don't know how to use it.

"Extremal" requires charge under a gauge field, that's the way the metric is modified, by the mass-energy in the surrounding field. If there is no field sourced by the charge, as for a global symmetry, there can be no extremality, and the black hole has to decay all the way down to some fixed size.

I understand that. In what I consider, there is a true gauge field, which is different from the global symmetry. In a generic case, if there is no special relation between the gauge field and the global symmetry, the gauge field is irrelevant for the argument. I consider a special case where there is a special relation between the gauge field and the global symmetry, for example if "all the baryons are positively electrically charged" (where baryon refers to the global symmetry and electrically refers to the gauge symmetry). In this case, a black hole made of baryons is electrically charged and can be extremal. But I guess I was stupid in my preceding comment: to solve my problem, it is enough to add some electrically charged non-baryonic matter (which exists by the hypothesis that the global symmetry is not the gauge symmetry) with the correct electric charge to cancel the electric charge of baryons and produce a neutral black hole.

But there is a critical conclusion you can come to from this line of reasoning--- that the field polarization at the black hole horizon has to be strong enough to polarize the outgoing emissions of charged particles to prevent black holes from getting constipated with charge. If you throw charge in, the charge has to be able to come out somehow. What this means is that there has to be a charged state in the theory where separating out the charged particle gains you energy rather than loses energy, and this requires that the mass of at least one particle, naturally this is the lightest particle, should obey the opposite inequality than the black hole. You can avoid this using conspiratorial assignments of charges (for example, having a collection of particles with charges powers of two, which are emitted at the end state of any black hole emissions), but this type of conspiracy is not particularly physical. The natural state of affairs requires a particle with mass<charge in the spectrum, i.e. whose net attractions are repulsive, and this is the mass<charge inequality for the lightest charged particle in a quantum gravity theory.

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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.

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user Daniel Grumiller
answered Feb 10, 2011 by (70 points)
Yes, it's something along these lines that I was looking for, but can you be maybe a bit more precise?

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user inovaovao
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.

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user Luboš Motl
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.

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user Luboš Motl
@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?

This post imported from StackExchange Physics at 2014-08-16 09:12 (UCT), posted by SE-user inovaovao

@inovaovao: Because there isn't enough mass at the end of evaporation to get $10^{48}$ baryons back.

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