# An Exceptionally Technical Discussion of AESToE

As I told elsewhere, in my case the hexagonal number pattern appears when you ask for the sBootstrap conditions, a coincidence between bosonic and fermionic degrees of freedom that happens in the QCD string. Half of this number (ie 3, 14, 33, 60, ... ) is the number of generations needed for the sBootstrap to exist.

I suspect that some quantisation of flavour will produce at least SO(32), if not E8xE8. This accounts for the 496. But no hint about Leech lattice.

I am not sure what sBoostrap is, and I suppose that the hexagonal number pattern is related to this. These generations, such as the 60, is half the number of spinor on the $M_{12}$ Mathieu group on the 120-cell or icosahedron. I am not sure if degeneracies are considered here, but 196560 is divisible by 32760 with 6 as the answer. I am not sure if there is any significance to this

I think a small verision of quantum gravity is the trio group $S^3\times SL_2(7)~ \subset~M_{24}$. The Leech lattice contains $M_{24}$ as a quantum error correction code, which embeds three $E_8's$ --- an $E_8\times E_8$ for the graded heterotic supergravity field theory and the third for this configuration of all possible spacetimes. In the restricted $S^3\times SL_2(7)$ this is a thee dimensional Bloch sphere where each point on it is a "vector" in a three space spanned by the Fano planes associated with these three $E_8$'s. $S^3\times SL_2(7)$ has 1440 roots and is itself a formidable challenge, but this represents a best first approach. $M_{24}$ has 196560 roots and clearly an explicit calculation of those is not possible at this time.

I am not sure if there is any magical numerology here, but with the 8 additional weights for each of the $E_8$'s, which is then doubled to a total of 48 weights due to the double covering on the Bloch sphere, this gives 1488 as the size of the trio group, which when divided by 6 gives 248, or when divided by 3 gives 496. I am not sure, but this might have some relevancy to Carl Brennan's triality approach to things.

Lawrence B. Crowell

CarlB
Homework Helper
This thread is getting old, so I'll type in the latest Koide fits. These are supposed to be practical applications of triality to quarks interacting to form mesons. As such, they might give a clue on how to fit the quarks better in with the leptons and quarks.

The first observation is that the (handed) quarks and anti quarks are 2/3 or 1/3 of the way between leptons and anti leptons in quantum numbers, and they come in 3 colors, so one ends up with a (1,3,3,1) multiplet with the 1's two types of leptons, say left handed positron and right handed electron, while the 3's two types of quarks, say right handed up quark and left handed anti-down quark. The implication is that the quarks and leptons could be built from three preons each of eight types, charged or neutral, left handed or right handed, and preon or antipreon (with positive charge). Then the electron is composed of three preons with charge +1/3 each, the up quark is made from two +1/3 preons and a neutral preon, etc.

Koide's formula for the masses of the charged leptons reads as follows (ignoring an overall mass scale factor of 25.054 sqrt(MeV)):
$$\sqrt{m_{en}} = \sqrt{0.5} + \cos(2/9 + 0\pi/12 + 2n\pi/3)$$
A similar formula fits the neutrino oscillation mass differenes (with a mass scale of 0.1414 sqrt(eV) also not included):
$$\sqrt{m_{\nu n}} = \sqrt{0.5} + \cos(2/9 + 1\pi/12 + 2n\pi/3)$$
The formula for the charged leptons is quite old and famous. I found the neutrino mass formula a couple years ago and it's now in the literature in various places, for example, Mod. Phys. Lett. A, Vol. 22, No. 4 (2007) 283-288:
http://www.worldscinet.com/mpla/22/2204/S0217732307022621.html

If the quarks are to be composed of a mixture of preons, which form should they follow?

Thinking of the above formulas as resonance conditions for the preons, perhaps a meson made from two quarks could resonate either way. The simplest place to test this is on the mesons that are most carefully and exactly studied, the b-bbar (Upsilon) and c-cbar (J/psi) mesons.

In the particle data group information on the b-bbar and c-cbar mesons:
http://pdg.lbl.gov/2007/listings/contents_listings.html
there are six of each type given:

The Upsilon b-bbar mesons are:
Name, quarks, I^G(J^PC) mass(error) koide_type
\Upsilon(1S) b/b 0^-(1^{--}) 9460.30(26) 1
\Upsilon(2S) b/b 0^-(1^{--}) 10023.26(31) 1
\Upsilon(3S) b/b 0^-(1^{--}) 10355.20(50) 0
\Upsilon(4S) b/b 0^-(1^{--}) 10579.40(120) 1
\Upsilon(10860) b/b 0^-(1^{--}) 10865.00(800) 0
\Upsilon(11020) b/b 0^-(1^{--}) 11019.00(800) 0

The Koide_type is 0 or 1 according as that mass is part of a triplet of states that follow the Koide formula with 0 or 1 copies of pi/12 in the angle. The resulting equations for the Upsilon masses are (leaving off the factor of 25.054 again):
$$\sqrt{m_{\Upsilon 1 n}} = 3.994433 - 0.128815\cos(2/9 + 1\pi/12 + 2n\pi/3)$$
$$\sqrt{m_{\Upsilon 0 n}} = 4.137251 - 0.077550\cos(2/9 + 0\pi/12 + 2n\pi/3)$$

The measured and calculated masses are as follows (MeV):
9460.3(3) ~= 9451.8
10023.2(3) ~= 10041.0
10355.2(5) ~= 10355.1
10579.4(12) ~= 10569.1
10865.0(80) ~= 10864.3
11019.0(80) ~= 11019.5

which is considerably more accurate than random chance would suggest. To put this into perspective, the mass difference between the charged and neutral pions is about 5 MeV.

Fitting the Koide formula to six masses like this is similar to how one would fit a spin-1/2 splitting to six masses. In that case one looks for how one can put the six masses into three pairs of masses with the same difference between the masses. If that were the case for the Upsilons, you can be sure that there would be papers showing why a quark interaction causes this kind of splitting. After you divided the six particles up into three pairs, you need only four degrees of freedom to describe the particles, say the three averages of the pairs, and the split amount. Similarly, with the above Koide fit, you end up removing two degrees of freedom from the six masses. The new four degrees of freedom are 3.994433, -0.128815, 4.137251, and -0.077550.

It would be easy to suppose that this is random chance, but the c-cbar mesons also come in exactly six masses, and these also are very closely fit by four Koide parameters. In this case the mass formulas are:
$$\sqrt{m_{\Psi 1 n}} = 2.442070 - 0.249554\cos(2/9 + 1\pi/12 + 2n\pi/3)$$
$$\sqrt{m_{\Psi 0 n}} = 2.510507 - 0.089433\cos(2/9 + 0\pi/12 + 2n\pi/3)$$
and the mass fits are unnaturally accurate:

J\psi(1S) c/c 0^-(1^{--}) 3096.916(11) 1 ~= 3096.9
\psi(2S) c/c 0^-(1^{--}) 3686.093(34) 0 ~= 3686.1
\psi(3770) c/c 0^-(1^{--}) 3771.1(2.4) 1 ~= 3773.8
\psi(4040) c/c 0^-(1^{--}) 4039(1) 0 ~= 4040.4
\psi(4160) c/c 0^-(1^{--}) 4153(3) 0 ~= 4149.8
\psi(4415) c/c 0^-(1^{--}) 4421(4) 1 ~= 4418.4

I've not yet figured out how to derive this from the assumption that the quarks are composites made from the same things that make up the leptons. I think it has to be doen with perturbation theory. My instinct is that the quarks are acting as a body that has two possible resonances, the type 0 (electron - muon - tau) and the type 1 (neutrinos). Either of these two resonances shows up in threes just like the generations of particles do. But only one resonance can be excited at a time. The result is six resonances that satisfy a Koide relationship.

Among the 8 numbers that give the four Koide fits here, some of them are rather close to rational numbers, or square roots, or what have you. For example, the first of the 8 Koide fit numbers, 3.994433, is very close to 4. But I don't see an overall pattern to the numbers.

My suspicion is that if this can be put into a perturbation expansion, we will see how to derive the 8 Koide fits from simpler assumptions. But I haven't figured out how to do this yet. This may be the reader's opportunity to score a quick paper. Like I mention above, my suspicion is that one should model this as a system that has two available, but mutually exclusive, resonances. And I'm working on this, but I haven't yet got anything worth writing up.

It may or may not help to read the incomplete paper I'm writing that is driving the search for these kinds of coincidences:
http://www.brannenworks.com/qbs.pdf

CarlB
Homework Helper
New paper out on E8:
http://www.iop.org/EJ/abstract/1751-8121/41/33/332001/

Meanwhile, Kea and I are working on the CKM (quark mixing) and MNS (neutrino mixing) matrices. Kea has pointed out the usefulness of 1-circulant and 2-circulant 3x3 matrices for these things.

1-circulant are what I used to call "circulant" while 2-circulants are the reverse order. The even permutations on 3 elements uses the 1-circulants while the odd permutations on 3 elements uses the 2-circulants. The matrix sum given below is of a real 1-circulant and an imaginary 2-circulant. The MNS (neutrino mixing) matrix can be written in a peculiarly simple 3x3 form as the sum of a 1-circulant and a 2-circulant matrix:
$$\sqrt{1/6}\left(\begin{array}{ccc} +\sqrt{2}&1&0\\0&+\sqrt{2}&1\\1&0&+\sqrt{2}\end{array}\right)\;\;\pm i\sqrt{1/6}\left(\begin{array}{ccc} -\sqrt{2}&1&0\\1&0&-\sqrt{2}\\0&-\sqrt{2}&1\end{array}\right)$$

That is, when you take the squared magnitude of the entries of the above 3x3 complex matrix, you get the MNS matrix under the "tribimaximal" form which is a good approximation of current experimental measurement:
$$\left(\begin{array}{c|ccc} &\nu_1&\nu_2&\nu_3\\ \hline e&2/3&1/3&0\\ \mu&1/6&1/3&1/2\\ \tau&1/6&1/3&1/2\end{array}\right)$$

The above matrix has all rows and columns sum to 1. In addition, the complex matrix (i.e. the sum matrix given at the top), has all rows and columns sum to
$$(\sqrt{2} +1 \mp i\sqrt{2}\pm i)/\sqrt{6}$$
which has magnitude 1. So you can multiply by the complex conjugate of this sum and convert the complex matrix given above into a form which is "doubly magic" in that its rows and columns all sum to 1, and the sums of the squared magnitudes of its rows and columns also sum to 1.

Matrices that have this property (double magic) are kind of unusual. It's interesting in that the "double magic" is a linear and bilinear property. It is linear in that the sum of the rows and columns all add to 1; therefore we can take two such matrices and sum them to obtain a new matrix that also has the property that its rows and columns add to a constant. If the two objects being summed are scaled, then we can arrange for the sum matrix to have all rows and columns add to 1. On the other hand, the sum of the squared magnitudes is a bilinear property, a requirement of normalization.

Of course we're looking at how to do the CKM matrix, preferably as a function of the MNS matrix.

Last edited:

Thanks for the link on E8. I see he refers to Bars, Gunaydin, Duff, Witten and other interesting authors, so I look forward to reading it.

Note that another way to write the tribimaximal matrix is as a product of quantum Fourier operators $F_3 F_2$, one associated to three dimensions (mass quantum numbers) and one to two (spin operators). Carl has used a 6x6 matrix version of related operators to derive idempotents for all the standard model fermions. The construction is essentially unique.

Last edited:
Berlin
Gold Member
Just seen the new version of Smolins paper on the Plebanski action. It says that a new paper is in preparation together with Lisi and Speziale on gravity+EW unification, building on Smolins framework.

http://arxiv.org/PS_cache/arxiv/pdf/0712/0712.0977v2.pdf

I would call this "exceptionally" good news! E8 is still alive.

berlin

arivero
Gold Member
Is there some news about Konstant's E8? I was kept thinking about the product of two copies of SU(5)... after all, remember that I suspect that the content of this E8 group (and its cousin SU(32) ) in string theory is to hold the quantum gauge version of flavour.

CarlB could perhaps relate it to mesons, because mesons do actually have global SU(5) as an approximate symmetry.

Gold Member
Hey PF folk,

Things are indeed going very well. The most recent gauntlet for this E8 theory was subjection to formal peer review by the FQXi scientific panel, which was passed with flying colors:

http://www.fqxi.org/

I'm officially taking one hour away from email to celebrate.

I'm also very happy to see many grants awarded to quantum gravity research, and even two others awarded for investigation into exceptional structures in particle physics.

Also, FQXi has announced an essay contest on "The Nature of Time," open to all.

arivero
Gold Member
When one glances the comments of Dixmier, it seems as if the limit of "two generations at most" were coming because E8 containts only two copies of SU(5)... but when one decomposes the infamous 248 into SU(5)xSU(5) one gets a pair of 24, being (24,1) or (1,24) in each group, plus four of 50, combined as (5, \bar 10) and so on.

My bet should be that at least one sum 24+50+50 is the standard model content plus 28 extra degrees of freedom, so that in reality it is 24+(36+9+5)+(36+9+5) and the content of the standard model comes from 24+36+36=96. Does it coincide with Lisi's scheme? I have not looked at it in detail.

The most recent gauntlet for this E8 theory was subjection to formal peer review by the FQXi scientific panel, which was passed with flying colors....

To be honest, I am REALLY not at all impressed. Obviously, it's who you know that counts. Two token women, only one of them a physicist, working on a theory that no longer agrees with observational data. Paolo Bertozzini was also rejected, probably for the crime of living in Thailand. There are only three names on the list that I haven't come across before. These people are already very comfortable. It must be nice to be a happy, smiling, well fed North American male.

Last edited:
marcus
Gold Member
Dearly Missed
... passed with flying colors:

http://www.fqxi.org/

I'm officially taking one hour away from email to celebrate.
...

Wonderful news! Very very glad to hear this!

MTd2
Gold Member
Hey PF folk,

Things are indeed going very well. The most recent gauntlet for this E8 theory was subjection to formal peer review by the FQXi scientific panel, which was passed with flying colors.

Congratulations! :) Is there any restrictions on how to use the money? Can you buy 20,000 Big Macs with it?

Also, in how much in %, aproximately, is your new paper with Simone and Lee away from completion?

Gold Member
Hi MTd2,
I've budgeted for 1 Big iMac. And I'm afraid I can't talk about the paper yet, since it's not just my work.

Kea,
I'm also disappointed there aren't more "outsiders" being funded. But I'm a pretty good counterexample to your "who you know" comment, since a few years ago I knew approximately no one, and wasn't particularly comfortable or well fed. Heck, I'm still technically homeless.

arivero,
Your analysis is correct. The trick is how gravity and generations may or may not work with the "other" su(5). If it can work, it's certainly not obvious how, since the 5's and 10's don't appear to match up desirably. Fun to play with though.

arivero
arivero
Gold Member
Your analysis is correct. The trick is how gravity and generations may or may not work with the "other" su(5). If it can work, it's certainly not obvious how, since the 5's and 10's don't appear to match up desirably. Fun to play with though.

Yep, it is not obvius. Witten himself has been thinking a lot on SU(5)xSU(5), with a couple of papers four or five years ago.

I think the problem is in the 10. As I told elsewhere, it is pretty easy to get generations from the 5 of SU(5) only, via
$$(\bar 5 \otimes 5)_s = 24$$ for all the leptons and
$$3* ((\bar 5 \otimes \bar 5) \oplus ( 5 \otimes 5))_s = 3*(15 + \bar {15})$$ for quarks
but in this case, the representations (10,5) &c., which we need to reach E8, do not appear. A disturbing point is that stringers have really E8xE8, so even not a 248 irrep but a 496 one. For two years now, I have been wondering if the SU(5) could be more straightly related to the SO(32) of the 496 irrep, via $$2^5=32$$.

Heck, I'm still technically homeless.

Garrett, you can no longer claim to be an outsider. And I would happily be homeless with such a fortune, which could sustain me for years.

MTd2
Gold Member
Garrett, would mind putting a different representation instead of colors. I am color blind, and sometimes I can't tell apart heads and tails.

Also, what is your "my standard model"?

arivero
Gold Member
related some grouping of 6 and the projections having sort of hexagonal symmetry in your gadget: I was reading now a series of lectures on Duality Symmetries, by A. Sen in Les Houches in the 2001 session, and for the 4 dimensional duality of SO(32) heterotic, compacted with T^6, some groups of 6 appear:
- a symmetric 6x6 matrix, giving 21 scalars
- an antisymmetric 6x6, giving 15 scalars
- a 6x16 object giving 96 scalars, related to "the gauge fields in the Cartan subalgebra of the gauge group".
Sen refers to his fundational paper http://arxiv.org/abs/hep-th/9402002" [Broken] for more info. It is a dense paper, but one can search for the number 16 :-)

I can not bet for an interpretation of the 21 and 15. An interesting observation is that if one reduces the standard model to SU(3)xU(1) then both the neutrinos and the top are lost (assume the mass of the top is related to the W and Z), and then the number of degrees of freedom of the fermionic sector is 2*36.

Last edited by a moderator:
MTd2
Gold Member
Also, what is your "my standard model"?
Sorry, I just noticed that it is the default name for any moviment I can make.

so...how was lunch with distler?
what did you guys eat?
who paid?
how much of a tip did you leave?

a year later...any progress on finding a way to fit three generations into e8?

arivero
Berlin
Gold Member
Maybe the best strategy for E8 fans would be to accept the fact that only one generation fits into the full E8! That is, I think, not as strange as it looks. In fact, the three generations are only distinguishable in mass (and could so, be identical in the mass to zero limit). Having three doublets of Higgs could cause three different mass levels only after breakdown of some of the E8 symmetries. See it like the only true generation is stuck in three different potential wells (maybe even combined with a Pauli principle). I am trying to use a kind of trinification model based on SU(2)^3, breaking down to three left SU(2)xSU(2)xU(1) groups with a left-right symmetry breaking a la Senjanovic. One of these left SU(2) groups should be part of gravity, the rest is EW. Only the 128 spinor "1/2" states of E8 seems to be required for the fermions. Whether this means that even gravity only exists in its current form after a symmetry breaking I do not know (or that the very notion of left-right comes along at that moment). All crazy ideas and the maths turns out too difficult for me anyway. Fun though.

berlin

Last edited:
CarlB
Homework Helper
On the generation structure, Marni Sheppeard and I have been working on a method of describing it based on (my view of things may differ from hers) the discrete Fourier transform. Partly to justify this, I've applied the same theory to the hadrons and have found 39 equations relating their masses.

This is similar to Regge trajectories (which were used as the basis for string theory). Regge trajectories are equations that relate the masses of the same hadrons but with different angular momenta. The equations I've got relate the same hadrons but with different radial excitations. But it attributes the different radial excitations to color phase effects.

I'm submitting this to Phys Math Central next week and would appreciate comments on it:

As far as E8 theory goes, this would mean something like E8 x discrete Fourier transform for the generations.

MTd2
Gold Member
Hi Carl,

Have you contacted Garrett Lisi about this? He could surely help you. Also, he colaborated intesively with Lee Smolin. I would like to know what you would talk to them, if you are willing to. You should also ask Tommaso Dorigo.

Last edited:
MTd2
Gold Member
Another question,

I really really expected to see the top mass calculated with you method... I know, everything else is really awesome, but I am disapointed with this one...