Wednesday, 31 January 2007

N=8 supergravity according to Lance Dixon

Is N=8 supergravity finite? The first answer that comes to mind is who cares. That's a legitimate approach, since N=8 supergravity is probably not relevant in any way for the real world. More generally speaking, renormalizability or finiteness are no longer considered a holy grail. We now understand that infinities just signal a limited applicability range of a given quantum field theory. I would not expect the ultimate theory of nature to be a finite quantum field theory. More likely, physics at the Planck scale requires a radically new framework (string theory, some may say, but probably that is not radical enough).

Having said all this, I hasten to admit that the opening question is an interesting problem of mathematical physics and there are good reasons to study it. One is just curiosity. N=8 supergravity would be the first example of a finite four-dimensional QFT containing gravity. The second reason is that the supposed finiteness would indicate the presence of some powerful hidden symmetry that is able to control the entire perturbative expansion.

Last week Lance Dixon argued here at CERN that the answer to the opening question might be positive. This conjecture follows from a complicated technical analysis, details of which are not available to mortal men. Still, Lance did a good job explaining what are the tools he is using.

Here is the story so far. The 4D Einstein gravity coupled to generic matter fields is divergent already at one loop. Pure gravity, however, is one-loop finite. This follows from fairly simple symmetry arguments: any possible R^2 counterterm either vanishes by the Einstein equations or is a total derivative. However the divergence strikes back at two loops, as there are R^3 counterterms available. Adding supersymmetry allows to postpone the sentence because the R^3 term haunting the ordinary gravity cannot be supersymmetrized. But there exist supersymmetric R^4 counterterms, so that in supergravity we expect divergences at three loops. The more symmetries we have, the better are our chances. N=8 supergravity has the maximal number of symmetries for a local QFT. But, naively, they are not enough to ensure finiteness of the perturbative expansion. On the other hand, no supergravity amplitude has ever been explicitly shown to be divergent. Hope lives on.

The weapons employed by Lance are the KLT relations and the unitarity cuts method. The KLT relations allow to express supergravity tree-level amplitudes in terms of gauge theory tree -level amplitudes. For N=8 supergravity the corresponding gauge theory is N=4 super-Yang-Mills, a finite one. Sounds promising. The unitarity cuts relate tree-level amplitudes to multi-loop ones. These tools allowed Lance and collaborators to argue in an earlier paper that divergences do not appear at 3 loops. More precisely, they concluded that D-dimensional N= 8 supergravity at L loops is finite if D < 2 + 10/L. This would imply that in D=4 dimensions there could be divergences at L = 5 loops. Lance, however, suspects that this bound is too conservative and the true formula is D < 4+ 6/L, just like in N=4 super-Yang-Mills. The hints are provided by unexpected cancellations that have been observed in one- and two-loop supergravity amplitudes. His work concentrates now on evaluating divergent contributions to higher-loop amplitudes in order to see whether the cancellations persist.

So, is N=8 supergravity finite? I don't know. Go ask Lance. Unfortunately, the transparencies of the CERN talk are not available. You can look at his most recent paper instead. The KLT relations and unitarity cuts are reviewed in a nice article by Zvi Bern on Living Reviews. See also this this post at Not Even Wrong and another one on Reference Frame.


Anonymous said...

"At present, we know only one finite QFT in 4 dimensions: the celebrated N=4 supersymmetric Yang-Mills theory."

Under what definition of "finite" is this true? There are plenty of known conformal QFTs in 4 dimensions, which I would call "finite"....

Anonymous said...

Great wrap up, good service! Good to have you here.

Jester said...

Anonymous, I had in mind relativistic gauge theories (for particle theorists nothing else exists ;-). Let me know if i'm missing something.

Anonymous said...

All the perturbative Banks-Zaks fixed points that exist in relativistic gauge theories?

Jester said...

I WAS missing something. Thanks, I'll correct the statement.