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April 9, 2007

Decoupling N=8 Supergravity

There’s been a lot of buzz, lately, about the possibility that maximal (N8N-8) supergravity might be a finite theory in d=4d=4. These supergravity theories arise as the low-energy limits of Type-II string theory compactified on T 10dT^{10-d}. The question is: is there a decoupling limit, in which one can hold M (d)M_{(d)}, the Planck mass in dd dimensions, fixed, while decoupling all of the degrees of freedom except for the massless supergravity multiplet.

As I explained recently, the massless scalar fields of the supergravity multiplet take values on =E˜ n/K n\mathcal{M}=\tilde{E}_n/K_n, where n=11dn=11-d. Type II string theory, compactified on T 10dT^{10-d}, yields not quite that. Because of the massive charged states of the theory, the continuous E˜ n\tilde{E}_n symmetry is broken to E˜ n()\tilde{E}_n(\mathbb{Z}), which is, in fact, a gauge symmetry, and the true moduli space is /E˜ n()\mathcal{M}/\tilde{E}_n(\mathbb{Z}).

Might there be a decoupling limit (as one approaches the boundary of the moduli space), in which all of the massive degrees of freedom decouple, leaving only the supergravity multiplet (whose moduli space, in the limit, looks like \mathcal{M})?

Green, Ooguri and Schwarz say no (at least, for d4d\geq 4). The computation is fairly trivial, and I suspect that the result is well-known to most of you. For variety, let me present it in M-theory language (which has the advantage of being a bit more concise).

We’re interested in M-theory on d×T 11d\mathbb{R}^d\times T^{11-d}. There are BPS pp-branes in the dd-dimensional theory, with tensions (masses, for p=0p=0) given by the following table

BPS pp-branes in the dd-dimensional theory
pp Branes Tension
KK modes p=0p=0 1/R1/R 1/R1/R
Wrapped M2 p2p\leq 2 M (11) 3R 2pM_{(11)}^3 R^{2-p} M (d) (d2)/3R (d53p)/3M_{(d)}^{(d-2)/3}R^{(d-5-3p)/3}
Wrapped M5 p5p\leq 5 M (11) 6R 5pM_{(11)}^6 R^{5-p} M (d) 2(d2)/3R (2d73p)/3M_{(d)}^{2(d-2)/3}R^{(2d-7-3p)/3}
KK Monopole p=d4p=d-4 M (11) 9R 12dM_{(11)}^9 R^{12-d} M (d) d2RM_{(d)}^{d-2}R

Here, M 11M_{11} is the 11-dimensional Planck mass, and RR is a typical radius of the T nT^n. The last row corresponds to d3×(Taub-NUT)×T 10d\mathbb{R}^{d-3}\times (\text{Taub-NUT})\times T^{10-d}, and is the D6-brane in d=10d=10 and the KK monopole in d=4d=4.

We’ll hold the rest of the moduli fixed, and consider varying just the overall size of the torus. To decouple the Kaluza-Klein modes, we want to take R0R\to 0, while holding the dd-dimensional Planck mass, M (d) d2=M (11) 9R 11d M_{(d)}^{d-2} = M_{(11)}^9 R^{11-d} fixed. Rewriting the tensions in terms of M (d)M_{(d)} gives the last column in the table above.

As you can see, for d=2,3d=2,3, all of the particles and branes (p0p\geq 0) have masses that go to infinity in this limit. For d=4d=4, however, the KK monopoles and the wrapped M5-brane go to zero mass, which is to say that decoupling fails. And similarly, for higher dimensions.

One can try approaching the boundary of moduli space differently, scaling different radii to zero at different rates (the limit that Green et al take is, in fact, one such asymmetrical limit), but the conclusion is the same.

The reason should be familiar. Consider a p-brane and its magnetic dual p=d4pp'=d-4-p brane (where KK modes are dual to KK monopoles and M5s are dual to M2s). The product of their tensions is M (d) d2M_{(d)}^{d-2}, which we are holding fixed. So, if you send one to infinity, the other goes to zero.

In d=4d=4, particles are dual to particles. The BPS charges form a 56 of E˜ 7()\tilde{E}_7(\mathbb{Z}) and we’ve chosen a particular way to send 28 “electric” states to infinite mass, while sending the 28 “magnetic” states to zero.

A loophole exists for d=2,3d=2,3 because there isn’t electric-magnetic duality these dimensions (in d=3d=3, particles are dual to instantons1).

What can we conclude about N=8N=8 supergravity in d=4d=4?

That’s not so clear to me. I don’t think we want to take a limit where all the BPS masses go to infinity (even if that were possible). After all, there are charged blackhole solutions of the supergravity theory. And we don’t want to send their masses to infinity. That corresponds to sending the electric gauge coupling to infinity. To the contrary, starting with the supergravity theory and its charged blackhole solutions, we ought to be able to more-or-less bootstrap ourselves up to the full string theory, by demanding that they have a consistent quantum-mechanical interpretation.

Said differently, the relation T pT 4dp=M (d) d2 T_p T_{4-d-p} = M_{(d)}^{d-2} can be derived purely in the supergravity theory. There really wasn’t any stringy input. So, if the supergravity theory really is UV-complete, then all these degrees of freedom must be there. In other words, like it or not, the full quantum-mechanical theory of N=8N=8 supergravity is M-theory, compactified on T 7T^7, whether we put that in from the outset or not.

Of course, that’s something that’s not visible in perturbation theory…


1 In d=3d=3, the KK “monopole” is an instanton (p=1p=-1-brane). That these become unsuppressed is probably the statement that perturbation theory about flat space is not a good quantization scheme, a lesson that seems to be borne-out by the Chern-Simons approach to (super) gravity theory in d=3d=3.

Posted by distler at April 9, 2007 1:49 AM

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16 Comments & 1 Trackback

Re: Decoupling N=8 Supergravity

Hi Jacques, I am wondering if this has any bearing on the question of the finiteness of N=8 SUGRA. In other words, can one use these considerations to demonstrate that the S-matrix of N=8 SUGRA necessarily will have UV divergences non-perturbatively.

Posted by: Moshe on April 9, 2007 12:43 PM | Permalink | Reply to this

Re: Decoupling N=8 Supergravity

I presume this means that there isn’t a (unitary) S-matrix, nonperturbatively, unless we include these charged states as external states (after all, there is some amplitude to pair-produce them).

In that sense, the “pure” N=8 supergravity either

  1. Is not UV-finite, in which case, it must be completed to the full string theory.
  2. Is UV-finite, in which case it already includes all of the asymptotic states of the compactified string theory.

Either way, the “full” quantum theory is the compactified string theory.

Posted by: Jacques Distler on April 9, 2007 1:12 PM | Permalink | PGP Sig | Reply to this

Re: Decoupling N=8 Supergravity

Yeah, it is clear that existence of states which don’t propagate in loops means the theory is incomplete. Also, it is clear that this non-unitarity will not manifest itself in perturbation theory. I was asking specifically if this non-unitarity has to do with UV finiteness, or could it be that the S-matrix is finite but not unitary.

Posted by: Moshe on April 9, 2007 1:36 PM | Permalink | Reply to this

Re: Decoupling N=8 Supergravity

In other words, I am wondering if there is a doable calculation in N=8 SUGRA, probably involving an instanton leading to pair creation of non-perturbative objects, that explicitly demonstrates that it is not UV finite beyond perturbation theory

Posted by: Moshe on April 9, 2007 1:47 PM | Permalink | Reply to this

Re: Decoupling N=8 Supergravity

Yeah, it is clear that existence of states which don’t propagate in loops means the theory is incomplete.

They exist as solitons in the theory.

In the sine-Gordon theory, the existence of solitons does not mean the theory is incomplete. Nor would it be correct to include them as new (fermionic) massive degrees of freedom.

I was asking specifically if this non-unitarity has to do with UV finiteness…

In other words: does it distinguish between possibilities 1&2 above?

I don’t see how. The same argument holds for d5d\geq 5, where the supergravity theory is, surely, UV-incomplete.

Personally, I would find possibility 2 much more satisfying. It would be a beautiful example of emergent geometry, where the full T 7T^7 emerges nonperturbatively from the quantization of this ostensibly 4-dimensional theory.

In other words, I am wondering if there is a doable calculation in N=8 SUGRA, probably involving an instanton leading to pair creation of non-perturbative objects, that explicitly demonstrates that it is not UV finite beyond perturbation theory

I suppose one would want to be convinced that it really is finite to all orders in perturbation theory, before looking for nonperturbative divergences.

Posted by: Jacques Distler on April 9, 2007 3:34 PM | Permalink | PGP Sig | Reply to this

Re: Decoupling N=8 Supergravity

Thanks Jacques, this makes things clear. I agree that possibility 2 would be nicer.

Regarding finiteness, if there is a clean argument that the theory is not finite non-perturbatively, maybe perturbative finiteness would be less interesting. Also, proving the former is probably much easier than proving the latter.

Posted by: Moshe on April 9, 2007 3:51 PM | Permalink | Reply to this

existence in the UV

How is the coupling constant behaving in the UV? (Of course, one should talk about a dimensionless coupling constant here.) Is there something like asymptotic freedom?

Posted by: Lord Sidious on April 9, 2007 2:31 PM | Permalink | Reply to this

Re: existence in the UV

There are no dimensionless coupling constants in the theory.

Posted by: Jacques Distler on April 9, 2007 3:36 PM | Permalink | PGP Sig | Reply to this

Re: existence in the UV

There are no dimensionless coupling constants in the theory.

I know that, the way the action is usually written, there is a dimensionful coupling constant. However, one can define a dimensionless coupling by dividing by a renormalisation scale to a suitable power. I was asking about the running of that coupling constant.

Posted by: Lord Sidious on April 9, 2007 4:20 PM | Permalink | Reply to this

Re: existence in the UV

I’m not sure what that particular sleight-of-hand is supposed to achieve.

The conjecture alluded to above (verified to an increasing number of loops) is that the theory is finite.

Posted by: Jacques Distler on April 10, 2007 12:15 AM | Permalink | PGP Sig | Reply to this

Re: existence in the UV

I’m not sure what that particular sleight-of-hand is supposed to achieve.

I was trying to establish how the beta function behaves in the UV and, in order to do that, I tried to use the dimesionless coupling constant constructed as above.

However, if the theory is UV finite, it seems there’s no coupling constant renormalisation so my worries about what happens to the coupling constant in the UV were unfounded.

Posted by: Lord Sidious on April 10, 2007 7:11 AM | Permalink | Reply to this

Re: Decoupling N=8 Supergravity

I think one should worry that the perturbation series is only an asymptotic series (they usually are). In this sense, the radius of convergence of perturbation theory is zero and non-perturbative effects are required to make sense of the theory to all orders.

Here, supergravity non-perturbative effects can be associated to objects of very small action. This is because there is no “minimum size of a black hole” classically. I’m still trying to figure out how to make sense of that situation in general. I think that the bubling of microscopic geometries being taken care of by perturbation theory was argued by Lin, Lunin and maldacena in some specical cases. I don’t see how it could work here with just supergravity.

Posted by: David Berenstein on April 10, 2007 12:39 PM | Permalink | Reply to this

Leading nonperturbative effects

As you can see from the above table, the strongest nonperturbative effects in d=4d=4 come from M5 instantons (Euclidean M5’s wrapped on the torus), whose action goes to zero like (M (4)R) 4/3{(M_{(4)} R)}^{4/3}. M2 instantons also become unsuppressed; their action going like (M (4)R) 2/3{(M_{(4)} R)}^{2/3}

But, to some extent, the previous paragraph is silly. There’s no point in even trying to go to the boundary of moduli space, since there’s no decoupling there.

Posted by: Jacques Distler on April 10, 2007 2:23 PM | Permalink | PGP Sig | Reply to this

Re: Leading nonperturbative effects

Yes.

Silly me. I was thinking about a situation where one has neutral black holes too, and for the sake of argument, just having supergravity with no string theory. In particular, the BPS black holes do not have to satisfy the minimal Dirac quantization condition, I don’t know any theorem that says that is
inconsistent.

Obviously the minimal Dirac quantization happens in
string theory, so there is no possible decoupling there.

Posted by: David Berenstein on April 10, 2007 4:12 PM | Permalink | Reply to this

Re: Leading nonperturbative effects

In particular, the BPS black holes do not have to satisfy the minimal Dirac quantization condition, I don’t know any theorem that says that is inconsistent.

Yes. That is interesting. Say, instead of satisfying the minimal Dirac quantization condition, they satisfy a quantization condition “NN” times as large. Then the product of the BPS masses is m em m=NM (4) 2 m_e m_m = N M_{(4)}^2

If you change the lattice of allowed electric and magnetic charges, this changes the group E˜ 7()\tilde{E}_7(\mathbb{Z}) to some other discrete group, ΓE˜ 7\Gamma \subset \tilde{E}_7, which preserves the new lattice (a finite-index sublattice of the original one).

I agree that the choice of this (sub)lattice is an additional nonperturbative choice in quantizing the supergravity theory.

Clearly, choices other than the standard one (if they’re nonperturbatively consistent) would not correspond to the usual string theory.

(We’re tacitly assuming perturbative finiteness here; otherwise the theory needs to be UV-completed by embedding in string theory, and only the standard choice is allowed.)

Posted by: Jacques Distler on April 10, 2007 5:31 PM | Permalink | PGP Sig | Reply to this

Re: Leading nonperturbative effects

I’ll think more about it. It would be nice if one can prove some sort of theorem about minimal quantization

Posted by: David Berenstein on April 10, 2007 5:47 PM | Permalink | Reply to this
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Excerpt: More on N=8 supergravity.
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