## Digging back in time for an identity.

November 23, 2010 by dberenstein

I have written many papers in the last 20 years. Recently, I was taking a trip down memory lane and was reading some of my older stuff. I happened on a pretty identity on one of my papers. Something I had forgotten. That particular identity probably has someone elses name attached to it. I don’t believe that I was the first person who discovered this algebraic identity, but I wouldn’t know where to start looking for the correct precedence. If the identity has a name attached to it, I wouldn’t be surprised if the culprit for finding it first is pushing daisies.

Mathematics keeps on getting rediscovered after all.

The identity is as follows.

Consider a set of of s numbers (they can be real, rational, algebraic, or even more messy commutative algebra objects so long as for the most part their multiplicative inverses are well defined). Let us call this set

And consider the set of permutations of the first integers, where is one such permutation

We are then instructed to take the sum

The stipulation is that the sum has no infinities (the numbers are generic).

This sum is equal to

As everyone can see, it’s an obvious identity so the proof is left to the reader ;-)

For some reason, looking at it I can imagine it appearing miraculously in the twistor formulation of Yang Mills scattering amplitudes… well, this is just a random speculation.

### Like this:

Like Loading...

*Related*

on November 25, 2010 at 8:26 amLuboš MotlIt’s very cute and I haven’t seen it in this form yet.

But I agree it’s almost obvious. The sum is, by construction, a rational function of the alpha’s whose dimension is the same as 1/alpha^s – and that is invariant under all permutations of alpha’s.

However, the only singularities appear when at least one of the alpha’s vanishes. To show so, one must demonstrate that the candidate poles containing 1/sum(several alphas) have vanishing numerators/residues if there are at least two alpha’s in the sum.

For two alpha’s, that’s easy to see. For example, the coefficient of 1/(alpha_{s-1}+alpha_{s}) vanishes because in the summation over the permutations, the terms that are proportional to 1/(alpha_{s-1}+alpha_{s}) include the terms from two permutations that differ by the exchange of these two alpha’s – and their terms differ by the factor 1/alpha_{s-1} and 1/alpha_{s}.

However, the sum 1/alpha_{s-1}+1/alpha_{s} is equal to (alpha_{s-1}+alpha_{s})/alpha_{s-1}alpha_{s}, so the sum of the two alpha’s cancels. In the very same way, the summing over permutations cancels the sum of three alpha’s or any other number of alpha’s. That’s because the sum over all permutations can be obtained as the multiple sum over the cyclic permutations of the last K elements where K goes from 2 to s.

In this arrangement, there’s always just one factor by which the terms containing a particular factor 1/(particular sum of alphas) differ, and by summing this factor over all the permutations of the alpha’s that appear in the denominator, one always gets something that is proportional to the sum, thus canceling it.

The normalization is correct as can be seen from the case when all alpha’s coincide. There are s! terms, each of which is 1/s!, so the coefficient in front of 1/product(all alpha) is indeed one.

Such expressions are indeed similar to some of the twistor ones. But I am not sure about the details of this statement. Do you want to prove the twistor formulae using a sum over more ordinary – Feynman-diagrammatic – terms, or do you want to sum expressions of the twistor type fo get an even simpler prescription for a combined amplitude? What is the combined amplitude?

Best wishes

Lubos

on November 25, 2010 at 8:00 pmdberensteinHi Lubos:

Just for your information, my original proof was by induction.

Regarding the twistor stuff, there are many quirky identities that simplify them in the end. I was thinking maybe a sum of Feynman diagrams or something like that could look like the stuff above, but to tell you the truth I only have this `feeling’ that similar identities might be at play, but no problem in particular to work on.

on November 26, 2010 at 7:33 amLuboš MotlRight, I surely share your feeling at some intuitive level.

Have you noticed the similarity of your identity to the ordinary Feynman parametrization?

http://en.wikipedia.org/wiki/Feynman_parametrization

In both cases, you write 1/A_1.A_2…A_s as a sum. In the Feynman parametrization case, it is really an integral and the denominator has the s-th power of a general “convex” linear combinations of the A’s.

Your formula is like putting some delta-functions to the Feynman parametrization integral (generalized so that non-s-th-powers appear as well), but it gives the same result.

Note that in the applications, the A’s in the Feynman parametrization are the usual (inverse) propagators. And the Feynman trick is useful because you may rewrite everything to a single integral over momentum.

That’s not the case of your alternative because the individual terms are as asymmetric as the result – products of different things. But have you tried to think about your formula as a method to replace the Feynman trick? Try to evaluate a generic n-loop QFT diagram using your formula – you will fail – but you may learn something out of it. ;-)

Cheers

LM

on November 27, 2010 at 4:15 pmLuboš MotlBy the way, David, do you agree that cosmic strings or domain walls

http://motls.blogspot.com/2010/11/penroses-ccc-cosmology-is-either.html

that get expanded during inflation could explain the strong concentric circles in WMAP if the latter are real?

Penrose is “explaining” this “6-sigma observation” by his incoherent theory that is meant not to be inflation except that he only has a causal diagram of his theory which is a causal diagram of inflation. ;-)

I need the extended object to give them a negative pressure so that their total mass doesn’t decrease – in fact, it does increase – as the Universe gets bigger.

If you agree, are they cosmic strings or domain walls? Can one embed it to full-fledged stringy models? If one needs domain walls, do they have to separate “inequivalent” vacua – with different vacuum energy to make it work?

on November 29, 2010 at 3:30 pmdberensteinHi Lubos:

After looking at their paper I would not spend more time on it. I don’t think that their data analysis is particularly convincing. However, I’m not sufficiently on top of the data to actually be able to speculate about it.

on November 29, 2010 at 6:07 pmLuboš MotlDear David,

I only had to spend much more time with it because I had already decided to mention it – which also meant some exchanges with an author etc. ;-)

The data analysis algorithm is clearly amateurish – that’s the approach that the laymen (e.g. media) like. Concentric circles, like epicycles, is exactly what some people like to play with or what they like to hype. That’s how people unexposed to advanced maths parameterize the WMAP-like data. That’s why this paper has been discussed in dozens of outlets.

The very concept of the spherical harmonics decomposition simply looks beyond the abilities of the Armenian guy.

It seems impossible to explain that abundant waves moving anywhere on the sphere but separated by 4.5 degrees will look like enhanced spherical harmonics at some value of L (40). Equally impossible for him to see that the radii of the circles may look “chaotic” but the only “signal” they see that goes above the noise is the signal coming from separations of the circles by a particular distance on the skies, namely 4.5°.

And of course, no one in their echo chamber understands that the Big Bang cosmology actually predicts the whole curve as a function of L – not just some vague qualitative heuristic observations about concentric circles that actually correspond to a small bump at L=40 only, leaving the rest of the function totally unexplained…

Cheers

LM

on November 29, 2010 at 11:22 amLuboš MotlAfter looking at the angular size of the concentric circles, it’s very clear that Penrose et al. have “rediscovered” the L=40 bump in the WMAP data.

http://motls.blogspot.com/2010/11/what-penrose-and-gurzadyan-have.html