Is N=8 Supergravity Finite?

Zvi Bern gave a talk yesterday at the KITP in Santa Barbara entitled The S-Matrix Reloaded: Twistors, Unitarity, Gauge Theories and Gravity. He surveyed recent progress on computing perturbative amplitudes in QCD and N=4 supersymmetric Yang-Mills, some of which involves using twistor methods. The most striking thing though were his last few transparencies (here, here, and here). He notes that all previous studies of divergences in supergravity rely only on power-counting and supersymmetry, assuming that if these two principles allow a divergence to occur, it will. Actually doing the full computation to see if the divergences are there is too hard and no one has done it. Bern notes that in these arguments the extra structure seen by the recent twistor methods is not taken into account, and when one does this, so far all complete calculations show that N=8 supergravity has exactly the same degree of divergence as N=4 Yang-Mills, even though one would naively expect the supergravity amplitudes to have worse behavior. He ends by suggesting that “Serious re-examination of the UV properties of multi-loop N=8 supergravity using modern tools is needed.”

If N=8 supergravity turns out to be renormalizable, this raises an interesting question about string theory….

This entry was posted in Uncategorized. Bookmark the permalink.

16 Responses to Is N=8 Supergravity Finite?

  1. Ben Compson says:

    I don’t understand — supergravity has been around for at least 25 years. How can we only now be asking if N=8 supergravity is renormalizable??? What else have theorists been up to all these years?

  2. woit says:

    “What else have theorists been up to all these years?”

    Hmm, have you heard about this idea called string theory?

    More seriously, divergences in N=8 supergravity don’t occur until at least 5 loops. Ever tried to do a 5-loop calculation in N=8 supergravity?

  3. Wolfgang says:

    Since 3 loops is good enough for superstrings, 5 loops should be more than sufficient ?

  4. logopetria says:

    Sorry to inject such a basic question, but this is something I’ve never been able to find out: in these “N=4”, “N=8” etc theories, what does the ‘N’ denote? Is it a variable that can take on arbitrary integer values? Is it like a coupling constant, or what?

  5. Urs says:

    Is it like a coupling constant, or what?

    N is the ‘number of supersymmetries’. Very roughly and schematically, this means there are N different odd-graded quantities that square to the generator of (time) translation.

  6. logopetria says:

    Thanks. I’m not sure I understand much, but at least I know what it is that I’m not understanding! Does “the number of supersymmetries” dictate how many superpartners each ordinary particle has? That is, does “N=1” say that the electron is partnered with the “selectron”, while “N=2” say that there are 2 flavors of selectron? Or am I way off target here?

  7. Field-Theorist says:

    Even if N=8 happen to be finite, it is not clear that we can embed the standard model in it. Moreover, in contrast to string theory there is no deep reason why it should be finite. Historically, until the first superstring revolution it was considered as an interesting theory, but soon after the superstring revolution people abandoned it.

  8. woit says:


    Basically, yes. When you have multiple supersymmetries, particles come in multiplets more complicated than just fermion-boson pairs.

    Field theorist,

    If Bern is right, N=8 supergravity may be renormalizable because of a combination of supersymmetry and twistor geometry, which would be arguably a deeper reason (symmetry) than the reason for the (conjectured) finiteness of the superstring.

    Yes, it’s hard to get the standard model out of N=8 supergravity, which is one reason people gave up on it. But this may be a much more fruitful starting point than string theory, where the problem is not that you can’t get the standard model, but that you can get anything. And working with a well-defined theory instead of a vague hope that a theory exists might also be a good idea.

    The reason people gave up on N=8 supergravity was not just the problems getting the standard model, but everyone believed it was non-renormalizable. At this point, about the only argument you can give for string theory is that it is the only way to combine gravity and quantum mechanics. If that collapses, I suppose string theorists will keep on arguing for string theory unification, but they won’t have any legitimate arguments left.

  9. Moshe Rozali says:

    This is fascinating, I was under the false impression that Bern and company proved that theory to be non-normalizable, but apparently that was for 11dim SUGRA. Zvi has a review in living reviews in physics (which presumably gets updated regularly) on the subject, and conjectures about N=8 SUGRA being more finite than it should be apparently already appeared before the twistor connections.
    In case N=8 turns out to be finite because of some deep relation to twistor space, that would be absolutely spectacular. Right now it is not clear though (to me) what could be this deep structure (maybe a string theory? just kidding…).
    I am wondering though how can one break SUSY and get more or less realistic spectrum (I was in high school when these things were all the rage). If one has to add matter and thus break the symmetry, the magic is likely to be gone.

  10. woit says:

    Hi Moshe,

    I was in grad school at the time, so I did learn a bit about this, but I’ve forgotten most of it and would have to go look this stuff up. From what I remember, unless you start adding other things in, two big problems are that

    1. you get an SO(8) gauge theory, and SO(8) isn’t big enough to include the standard model gauge groups.

    2. the spectrum is vectorlike, not chiral as needed for the standard model.

    As for supersymmetry breaking, that’s always been a huge problem for any supersymmetric theory.

    Even if the theory is renormalizable, you’re right it’s not clear this would survive whatever one did to try and get around the above problems.

  11. John Baez says:

    Thanks for pointing out this talk by Zvi Bern – it’s really interesting!

    For one thing, it shows that you should never trust a physics “folk theorem” until you see how people convinced themselves of it, and know what the loopholes are.

    For another thing, it’s more evidence that there’s a lot left to understand about perturbative quantum field theory. Connes and Kreimer have been doing cool work on this subject for a while, and now comes this twistor business. I wish I understood more about both!

    If you read Predrag Cvitanovic’s autobiographical remarks entitled Search Without a Plan, you’ll see he’s one of the few people who have added up thousands of Feynman diagrams in QED… and you’ll see he discovered something interesting!

    While field theorists typically guess that the sum of all nth-order diagrams grows like

    n^n (alpha/pi)^n

    with the whopping nn term coming from the combinatorial explosion in the number of diagrams, he found empirically that there were lots of cancellations, leaving a result more on the order of

    n (alpha/pi)^n

    If this were actually true, one could sum all Feynman diagrams in QED and get a finite answer! – contrary to the usual folk wisdom.

  12. Moshe Rozali says:


    I am confused, I thought we already know by the existence of non-perturbative effects (in asymptotically free theories) that perturbation theory cannot converge, regardless of any diagram counting. Maybe there is something special about QED, or maybe one cannot really check explicitly any statements about asymptotic values of n…

  13. Robert says:

    I think the world expert regarding counter terms for gravity theories is Anton van de Ven. He worked for several years to show that the two loop divergence for (N=0) gravity is in fact there but was beaten by Goroff and Sagnotti:

    By Marc H. Goroff (Caltech), Augusto Sagnotti (UC, Berkeley & LBL, Berkeley),. CALT-68-1263, UCB-PTH-85/18, LBL-19512, Apr 1985. 13pp.
    Published in Phys.Lett.B160:81,1985


    By A.E.M. van de Ven (Hamburg U. & SUNY, Stony Brook),. DESY-91-115, ITP-SB-91-52, Oct 1991. 51pp.
    Published in Nucl.Phys.B378:309-366,1992

    The first paper is not covariant whereas van de Ven uses covariant methods (background field gauge, heat kernels etc). The last twenty years, he has worked on the N=1 version of this, where the divergence is supposed to appear at three loop level. As an intermediate result, he (and his student Jan Peter Börnsen) did a three loop YM calculation.

    Anton supevised my master thesis (Diplomarbeit) which was about some minor property of heat kernels in superspace. I should not forget to mention that he now lives in Utrecht and has a teaching position at some high school type institution.

  14. John Baez says:

    Moshe Rozali writes:

    I am confused, I thought we already know by the existence of non-perturbative effects (in asymptotically free theories) that perturbation theory cannot converge, regardless of any diagram counting. Maybe there is something special about QED, or maybe one cannot really check explicitly any statements about asymptotic values of n…

    Yes, it’s confusing – as Cvitanovic himself admits, his observation is “heretical”. His calculations only apply to QED, not asymptotically free theories. But even here, an old argument by Dyson suggests that the power series in the coupling constant can’t really converge: if it did, there should be a sensible version of electrodynamics for particles of small imaginary charge, which would attract each other… and Dyson argues that this is implausible.

    Maybe someday someone will compute the magnetic moment of the electron in QED up to a really huge order in the fine structure constant, and we’ll see if Cvitanovic’s observations hold water.

    Or, maybe someday people will discover some math that sheds more light on these issues.

  15. Arun says:

    Why can’t the QED perturbation series sum to something that has a branch cut going from zero to negative infinity?

  16. Moshe Rozali says:


    Such function is not analytic at zero, in other words does not have a convergent power series expansion in small coupling.

Comments are closed.