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It is often stated that in quantum gravity only charges coupled to gauge fields can be conserved. This is because of the no hair theorem. If a charge is coupled to a gauge field then when it falls into a black hole the black hole acquires a corresponding field. However if it is not then the black hole doesn't "remember" it. Apparently it implies we can't have exact global symmetries, only gauge symmetries.

How is this expectation realized in string theory? Is it true string vacuum sectors cannot posses global symmetry? Can we prove it?

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up vote 8 down vote accepted

We can prove it in perturbative string theory but it's probably valid beyond it.

In perturbative string theory, any (continuous) global symmetry has to be associated with a conserved charge which, because of the locality of the physics on the world sheet, implies the existence of a world sheet current $j$ or $\bar j$ or both (left movers vs right movers) whose left/right dimension is $(1,0)$ or $(0,1)$ or both (because its integral has to be a conformally invariant charge). Typically, such a symmetry might be something like the isometry of the target (spacetime) manifold or something on equal footing with it.

It follows that one may also construct operators $j\exp(ik\cdot X) \bar\partial X^\mu$ or $\bar j \exp(ik\cdot X) \partial X^\mu$ or both with a null vector $k$ which have the dimension $(1,1)$, transform as spacetime vectors, and therefore belong to the spectrum of vertex operators of physical states which moreover transform as spacetime vectors, i.e. they're the gauge bosons (the $X$ only go over the large spacetime coordinates). One would need to prove that the multiplication of the operators doesn't spoil their tensor character but it usually holds.

Consequently, any would-be global symmetry may automatically be shown to be a gauge symmetry as well.

The argument above only holds for the gauge symmetries that transform things nontrivially in the bulk of the world sheet. But even symmetries acting on the boundary degrees of freedom, i.e. the Chan-Paton factors, obey the same requirement because one may also construct (open string) vertex operators for the corresponding group that transform properly.

We don't have a universal non-perturbative definition of string theory but it's likely that the conclusion holds non-perturbatively, too.

In some moral sense, it holds for discrete symmetries as well even though discrete symmetries don't allow gauge bosons.

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Great answer! Can you add a few more word about the discrete case? –  Squark Nov 5 '11 at 16:53
Thanks, @Squark, but I would prefer not to go to details in the discrete case because it could be inconclusive. One may show that discrete symmetries of various kinds are non-anomalous; and one must figure out what it means for the discrete symmetry to be local. An operational definition is that there has to be a cosmic string (codimension 2 brane in general) such that the monodromy around it is equivalent to any element of the discrete symmetry group. It seems to be satisfied; those cosmic strings exist. –  Luboš Motl Nov 5 '11 at 17:30
Because they do, the domain walls that would separate environments $E$ and $g(E)$ where $g$ is any element of the discrete symmetry group are unstable: holes may start to grow (nucleate) in these domain walls - the boundary of the wall is nothing else than the cosmic string whose existence was advocated above. So ultimately the wall may disappear, proving that $E$ and $g(E)$ are really physically identical environments and the symmetry is a redundancy, not a symmetry acting on distinct objects. Of course, if symmetries change the asymptotic structure of spacetime, they always become physical. –  Luboš Motl Nov 5 '11 at 17:32
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