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Let me justify my question before I go on. In string theory, gravitons are strings extended over space. Longitudinal gravitons are pure gauge modes of the diffeomorphism group. However, in string theory, longitudinal gravitons are also extended objects. A condensate of longitudinal gravitons is equivalent to a diffeomorphism, but this diffeomorphism has to be smeared out over the string scale. Is the diffeomorphism group in string theory a quantum deformation smeared out over space?

In the weak field small string coupling limit, the string theory algebra and the classical diffeomorphism algebra ought to coincide, but in this limit, all such algebras over a Poincare invariant background are isomorphic. Away from this limit, what is the form of this quantum deformation?

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A related question is whether the diffeomorphism group gets quantum corrections in conventional quantum gravity (or SuGra). I see no reason that it should not. Compare to the conformal group in finite N=2 (or N=4) SYM theories which receive quantum corrections / deformations. (e.g. – Simon May 29 '11 at 2:32

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nope, there is no deformation of the gauge symmetries in string theory; there is just an infinite extension because aside from the usual massless gauge symmetries we know from quantum field theory, are are - at least in some formulations such as string field theory - a whole infinite tower of additional gauge symmetries that combine into a stringy gauge symmetry (with a whole string field being the parameter of the gauge transformation). The string field theory only works properly for the open strings which can't gravitate but the discussion is analogous for closed strings as well.

However, the massless gauge symmetries form a subgroup of the stringy gauge symmetry algebra and it is completely undeformed. The internal structure of a string is extended in space but it's still true that the unphysical model of the particle that emerges out of the strings knows about a particular point in spacetime - essentially the center-of-mass of the string - so the spatial extension of the string doesn't make the theory nonlocal in this simple sense.

There's a sense in which much of the nonlocality that you would expect from the extended character of the strings is spurious. For example, perturbative string theory's amplitudes still obey (but saturate) the inequalities one may derive for local quantum field theories. Also, you may imagine that the pieces of the string that are "very far from its center of mass" are associated with high-frequency vibrations of the string, so all their physical effects get rapidly averaged out over the time.

However, for many calculations, this is not just approximate: some "derived fields" and "derived gauge symmetries" from string theory behave exactly locally. So string theory, although it may superficially look like a theory of "generic nonlocal objects", is actually much closer to a strictly local quantum field theory than how it looks like.

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If the gauge group isn't deformed, how do we explain T-dualities and mirror symmetry? Is there a much larger gauge group which contains as subgroups two distinct, but overlapping diffeomorphism groups? – user3798 May 31 '11 at 6:31
Yes, that would be the symmetric group of permutations of space-time events. – Philip Gibbs - inactive Jun 30 '11 at 9:24
Lubos: which inequalities are saturated? Can you help with a reference--- I am interested in such things. – Ron Maimon Aug 30 '11 at 7:42
Dear Ron, it's the upper limit on cross section as a function of energy for very high energies and some particular scaling of the impact angle. The inequality may be derived from locality and analyticity in QFT but no ordinary QFT saturates it. Open strings saturate it. But I wouldn't be sure about the inequality on the top of my head, sorry. – Luboš Motl Jan 26 '12 at 19:44
I figured it was the Froissart bound, but I have seen it derived only for massive theories, with a pomeron. The massless analog is going to be different, and interesting--- the original argument used mass-gap in an essential way. I remember vaguely some famous people writing about this recently, but I am not sure that the actual inequalities are, or if they are saturated, or if they exist! This is interesting, because the saturation is obviously through BH states in quantum gravity, so it is different looking than pomeron saturation. – Ron Maimon Jan 27 '12 at 0:57

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