Friday, April 04, 2008

N=8 supergravity: Lance Dixon's puzzle

At Lindefest, Lance Dixon gave a talk about the likely perturbative finiteness of N=8 supergravity. Some of the previous articles about this topic:
Finiteness of supergravity theories
Non-decoupling of N=8 d=4 SUGRA
Finite SUGRA and pure spinors
Non-perturbative well-definedness
Gravitons as squared gluons
More...
On page 29 (PDF: 42/49), Dixon presents a "puzzle" that resulted from his conversations with Banks, Berenstein, Shenker, Susskind, and from reading Green, Ooguri, Schwarz. Let me call these string theorists BBSSGOS but there are surely many others who could join the team. :-)

Well, I don't think that BBSSGOS wanted to confirm to Lance Dixon that there was any puzzle. Quite on the contrary, they probably wanted to subtly explain Lance Dixon that he was making a rudimentary conceptual mistake. But with the BBSSGOS' ultrapolite discourse, their point was most likely misinterpreted. ;-)

OK, there is no puzzle about the perturbative finiteness of N=8 supergravity and its non-perturbative completion. Let me discuss the points on Dixon's transparency one by one:
  • N=8 SUGRA has more symmetry than type II string theory on T^6.
Correct. String theory only preserves E_{7(7)} (Z), a discrete subgroup of E_{7(7)} that respects the lattice of allowed quantized charges under the twenty-eight U(1) gauge groups and their magnetic duals.

In a fully consistent theory, these charges must be allowed to be nonzero. Why? For example because classical black hole solutions can carry these U(1) charges and the full quantum theory probably cannot completely eliminate charged objects if they appear classically as black holes. But these charges must also preserve the Dirac quantization rules, encoding the single-valuedness of the wavefunctions.

If you think about these constraints, the only consistent compromise you can get is that all the 2 x 28 charges live in a lattice that satisfies certain properties and that is the quantum approximation of the classical space of allowed charges, namely of R^{56}.
  • It has continuous, noncompact E_{7(7)}; also SU(8) is a proper subgroup of E_{7(7)}.
Correct. Supergravity is the classical limit of the full quantum theory of gravity in this background and it has a more extensive symmetry which happens to be continuous and noncompact. It is an accidental symmetry of an effective field theory in the same sense as the baryon number conservation in the case of the Standard Model.




When I say that supergravity is the classical limit, I don't prevent you from computing loop corrections to scattering amplitudes. However, I do expect you to interpret the result as a classical action corrected by quantum effects. You must realize that you are not doing a quantum theory until you accept that the Dirac quantization rule must hold and the 56 charges therefore cannot be continuous.
  • [This] symmetry is not present in massive string spectrum.
Correct. The massive string spectrum breaks the noncompact continuous symmetry exactly because the 56 allowed charges - the 7 quantized momenta on T^7, (7 choose 2) = 21 wrapping membrane numbers over 2-cycles, (7 choose 5) = 21 wrapping fivebrane numbers over 5-cycles, and 7 Kaluza-Klein monopole wrapping numbers (the type of the monopole, nominally a six-brane, is associated with one of the 7 U(1) directions of the torus) - are required to belong to a certain lattice.
  • Suppose N=8 SUGRA is finite, order by order in perturbation theory.
Probably correct. I personally find the evidence supporting this assumption very strong and Dixon and others - including some string theorists - have made an amazing job in demonstrating details of this statement and various cancellations of diagrams that the assumption implies.

The most convincing moral explanation is that the closed type II superstring seems to be similar to a tensor product of two type I superstrings - the two copies of the type I string arise from the left-moving and right-moving sector of the string, respectively.

Because the maximally supersymmetric SUGRA emerges as the low-energy limit of the closed strings and the maximally supersymmetric Yang-Mills arises from the low-energy limit of open superstrings, it can't be shocking that the gravitational amplitudes with 32 supercharges are related to the square of the gauge amplitudes with 16 supercharges (KLT relations). Because the latter are finite in d=4, the former ones could be finite, too.

I find it completely plausible that someone will eventually - or soon - transform these arguments into a rigorous 1-page proof of the finiteness that will supersede hundreds of pages with diagrams from the literature that only give us circumstantial evidence for the perturbative finiteness.

However, this relationship between closed strings and open strings - closed strings are open strings "squared" - clearly holds perturbatively only. It is a perturbative coincidence. Nonperturbatively, the left-moving and right-moving sectors of a closed string cannot really be defined and isolated because the whole string ceases to be fundamental at a finite coupling.

It pretty much follows that the consequences of this relationship between closed and open strings - including the finiteness of N=8 SUGRA "derived" from the finiteness of N=4 Super Yang-Mills - will be only satisfied at the perturbative level, too. Non-perturbatively, we can see that N=8 SUGRA is ill-defined.
  • True coupling is (s/M_{Pl}^2).
Correct. The dimensionless coupling is the (squared) energy in the Planck units. Consequently, the scattering amplitude of pure N=8 SUGRA become ill-behaved around the Planck scale and above it. We have many ways to see it. For example, it seems inevitable because of general relativistic reasons: at huge, trans-Planckian energies of colliding gravitons, the cross section to produce a very heavy black hole must increase because the gravitons are likely to get close to each other, within the increasing Schwarzschild radius of the would-be black hole. The black hole production thus becomes real.

The black holes can morally appear as intermediate states, too. At any rate, these effects or effects of similar kind imply the presence of non-analyticities (poles or cuts) in the scattering amplitudes at high energies, something that needed for unitarity but incompatible with the perturbative N=8 SUGRA. The N=8 SUGRA therefore cannot be consistent at trans-Planckian energies.

At the level of quantum gravitational standards, pure N=8 SUGRA is junk. Technically speaking, we - and Green, Ooguri, and Schwarz (summary: Jacques Distler) - say that it belongs to the Swampland, following Vafa's terminology.

A fast albeit not rigorous way to see it is that the pure SUGRA limit cannot be decoupled from the rest of the stuff in string theory that makes it consistent. So far it seems to be the case - and there are many examples that have been checked - that every limiting theory of string theory that can be decoupled from the rest (by proving that the interactions between the subset and the rest goes to zero in a proper limit) is UV-consistent and every theory that can't be decoupled is inconsistent if "violently" separated from string theory. But the argument that the N=8 SUGRA is in the swampland doesn't have to rely on this lore.
  • Series is unlikely to converge, rather be asymptotic.
Correct. The only way how a seemingly consistent perturbative expansion of an inconsistent theory can exist is that the perturbative expansion doesn't converge, much like it diverges in pretty much every quantum field theory. These divergences are the remnants of the forgotten physics.
  • To define it nonperturbatively, need new states/instantons/...
Correct. Forgotten poles and cuts in a UV-divergent theory (even if we talk about the divergence of the resummed expansion only) are a testimony of physics we have omitted. For example, the Fermi 4-fermion interaction is ill-behaved in the UV because it omits the W and Z bosons.
  • Can these come from string theory?
Correct (or yes). String theory adds new states - Kaluza-Klein modes, wrapped M2-branes and M5-branes, Kaluza-Klein monopoles, including the proper excitations of all the previous objects, and microstates of black holes in general - that are both necessary and sufficient to cure the problems of the perturbative approximation including gravity.

Even in this maximally supersymmetric case, the otherwise strong constraints of quantum gravity don't offer us just one way to nonperturbatively complete the perturbative N=8 SUGRA but a 133-63 i.e. 70-dimensional continuum of completions, locally isomorphic to the E_{7(7)}/SU(8) quotient and parameterized by the scalar fields. All these solutions to the problem of the completion of N=8 SUGRA may be interpreted - and they are identical to - superselection sectors of the M-theoretical Hilbert space.
  • Coefficient of every term in N=8 SUGRA amplitude is independent of E_{7(7)}/SU(8) moduli, so lack of convergence is also independent.
Correct. There is only one perturbative N=8 SUGRA. So if it is finite, the whole effective action is independent of the 70 moduli. Just like the E_{7(7)} continuous symmetry of N=8 SUGRA is accidental, this independence is accidental as well. In fact, it is the same thing. And the divergent character of the perturbative expansion is a fact, so it must be true that the asymptotic series approximating the full string-theoretical answer is divergent at every point of the moduli space - because it is the same series.
  • How could stringy states/effects remedy the series if they don't have the right symmetries?
Wrong. Note that the only wrong statement on Dixon's transparency was formulated as a question. I don't know whether Dixon is among them, but incorporating a wrong assumption or a wrong assertion into a question is a favorite rhetorical trick of many ideologues because many people tend to think that questions cannot be wrong (or that questions are not even wrong) which is why questions are usually not questioned (and sadly enough, critics are usually not criticized). ;-) But questions can be wrong and this one is an example.

A symmetry of a theory is defined as a transformation rule applied to arbitrary objects that doesn't influence their evolution according to the dynamical laws. If some objects in your theory don't respect the symmetry, the symmetry is broken, sorry. E_{7(7)} (Z) is the right symmetry while its continuous counterpart is an approximation - an accidental symmetry - only.

There exists no fully complete and consistent theory (or S-matrix) with N=8 supersymmetry as well as the continuous E_{7(7)} symmetry. The breaking of most of the exceptional symmetry is necessary for the completion to be consistent at the quantum level.

Analogously, black holes in the normal Universe are probably able to violate the baryon number conservation, proving that the "right symmetry" doesn't include B at the fundamental level. In the same way, the "right symmetry" of the full N=8 supersymmetric gravitational theory and of its exact convergent S-matrix is and must be E_{7(7)} (Z) only; the E_{7(7)} is a "wrong symmetry".

A reason - one that is independent of any "strings" - has already been explained many times in this article. In a quantum theory, the 56 charges of charged objects - that must exist because some of them must represent the charged black hole solutions that exist in general relativity - must satisfy the Dirac quantization rule. Consequently, they must live in a lattice. There is a 70-dimensional space of possible choices for this lattice and the continuous E_{7(7)} symmetry maps one point of this 70-dimensional space into another point and fails to be a symmetry of the superselection sector - unless you happen to pick an element of E_{7(7)} (Z) that preserves the lattice.

There was a way for Dixon to formulate a similar question that would avoid the mistake above:
  • How can the full stringy S-matrix, including the wrapped branes and stringy/M-theoretical instantons, preserve the continuous E_{7(7)} symmetry once you write down its asymptotic series?
Understood. But there is no puzzle behind this question either. The reason is perfectly understood: the perturbative portion of the S-matrix between uncharged states from the graviton supermultiplet - the asymptotic series approximating its elements - is completely unaffected by the stringy stuff. You can imagine that the N=8 SUGRA with a four-dimensional Planck scale is represented by M-theory on a seven-torus whose radii are comparable to the eleven-dimensional Planck length that gets identified with the four-dimensional Planck scale.

The instantons, much like the intermediate pairs of stringy objects, are therefore suppressed by things like exp(-R^3/Lpl^3) or exp(-R^6/Lpl^6). Note that the amplitudes only depend on the third power of the Planck length - it is probably a general feature of M-theory. The asymptotic expansion of all these objects is simply zero and they cannot break the continuous accidental symmetry of the asymptotic series.

One more comment. There exists a seemingly general rule - although it seems somewhat questionable in heterotic string theory which has no D-branes - that the minimal term in the divergent asymptotic series is of the same order as the leading nonperturbative contribution. In the case of N=8 SUGRA, the minimal term is independent of the 70 moduli.

If the rule is correct, then it should be true that the leading nonperturbative contribution to scattering amplitudes in M-theory on T^7 should be independent of the 70 moduli, too, as long as it is expressed in four-dimensional Planck units. It sounds as a nontrivial statement but I don't see any reason to think that it is impossible. It would be more interesting to know what this leading divergence actually is. As explained in my previous text "Finiteness of SUGRA theories", I suppose that it is something like exp(-1/E^3 Lpl^3) where Lpl is the four-dimensional Planck length, a result that could simultaneously agree with the leading M2-brane instantons. Why not? Or what is it?

At any rate, there is no puzzle of the kind Lance Dixon is describing, regardless of the answer to the question whether the N=8 SUGRA is perturbatively finite (and I would bet it is).

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