An Exceptionally Technical Discussion of AESToE

[Rade]

Dear Garrett,

Have you contacted CERN directly to be sure they include the predictions of your E8 model in the particle collision data they will capture when they start the LHC experiments soon in 2008 ? As I understand the situation, only a small fraction of the LHC collision data will be captured and stored, the rest is lost forever (CERN does not have enough computer memory storage). What data they do capture is what is predicted by current Standard Model, perhaps some new physics stuff--but, are you 100 % sure they will capture data that can be used to test (e.g., falsify) the predictions of your E8 model ?

 

[garrett]

Hello Rade,
This E8 theory isn't developed well enough to produce such predictions with sufficient confidence. There are pretty clearly twenty or so new particles predicted, but until the problems with the theory are worked out, their properties are kind of up in the air. But, there are plenty of LHC observations which wouldn't be compatible with this theory, so it does have some predictive power in that sense. In any case, the theory needs to be developed further before specific predictions can be made with any confidence. There's a long way to go.

 

[Rade]

Hello Rade, This E8 theory isn't developed well enough to produce such predictions with sufficient confidence. There are pretty clearly twenty or so new particles predicted, but until the problems with the theory are worked out, their properties are kind of up in the air. But, there are plenty of LHC observations which wouldn't be compatible with this theory, so it does have some predictive power in that sense. In any case, the theory needs to be developed further before specific predictions can be made with any confidence. There's a long way to go.

Thank you very much for your clarification. It just seems that it would be a such a great lost to science if the "properties" (by this I mean the LHC collision patterns) of these 20 new particles predicted by your version of E8 are the types of patterns that CERN will never capture and store. Could you please provide some input on nature of the:

plenty of LHC observations which wouldn't be compatible with this [E8] theory​

 

[garrett]

Ha! Umm, no, I don't think this is worth worrying about. I'll do my best to work on this theory and get some precise predictions. But keep in mind that this theory is still developing, and it's a long shot. What isn't a long shot is that the folks at CERN will do an excellent job of ferreting out every bit of new physics they can from their new data, regardless of any predictions I might make.

 

[rntsai]

Garrett,
A quick question on Table 9. Looking at row 4 for example :
second column lists 4 recognizable particles; the last column
lists 8. What's the discrepency?

 

[garrett]

Spin up and down components for each.

 

[Cold Winter]

Hello Rade,
This E8 theory isn't developed well enough to produce such predictions with sufficient confidence.
... There's a long way to go.

Over in another thread, someone asked "what about time?"... to which I ask you, have you looked at this? As well as inertia, gravity, and you did mention vacuum energy.

A simplification of your focus, ( perhaps on one or more of the above items ) might get something that we can lab test far more immediately than what CERN can deliver.

Nothing speaks to physics like experimental proof.

You already seem in danger of getting buried in mathematical techniques. E8 is afterall, far more complex than anyone or group of humans can hope to deal with in a lifetime. And E8 no doubt will be a wonderful proving ground for mathematicians.

Merry Christmas!

 

[rntsai]

Hi Garrett,
A few more question on Table 9. I have an explicit basis that I can
work with now; it's a little different than any of the ones in your
paper, but I don't think that should matter for now. I found these
two (complementary) subalgebra series in e8 useful in identifying
subspaces,... :

e8 > e6p > f4p > d4p > g2p > a2p > 0
0 > a2q < g2q < d4q < f4q < e6q < e8

taking centralizer in e8 moves you from one row to the other. The
"p" and "q" postfix are arbitraty ("a" and "b" are already used).
so(7,1) should correspond to d4p; g2q should correspond to the
strong g2 which shows up as the next to last column in Table 9.
It seems that column really goes more with a2q rather than g2q
since the reps are a2 reps. Similarly for the column before it,
the 8S+,8S-,8V are d4 reps,...

I was able to explicitly decompose 8S+ under a2q, I do get
3+3+1+1; same for 8S- and 8V, these decompose as 3+3+1+1.
I don't understand why sometimes you have l and \bar l;
There's a u(1) that enters the picture here, but I haven't
identified it satisfactorily yet.

The last 4 rows should correspond to the breakup of d4q
under a2q. There are 4 6's in the last column. This does
seem to match what I'm getting : d4q = 8 + 6x3 + 2x1. The
8 corresponds to "A2" in the column with 2 going into the
cartan subalgebra, but what about the 2x1? These are two
1 dimensional subspaces; they're not in the cartan algebra
of d4q. Where do these go?

 

[rntsai]

Hi Garrett,
The last 4 rows should correspond to the breakup of d4q
under a2q. There are 4 6's in the last column. This does
seem to match what I'm getting : d4q = 8 + 6x3 + 2x1. The
8 corresponds to "A2" in the column with 2 going into the
cartan subalgebra, but what about the 2x1? These are two
1 dimensional subspaces; they're not in the cartan algebra
of d4q. Where do these go?

I found a mistake in my calculations; the 2 1-dim subpaces
are in fact in the cartan of d4q, so they're accounted for.

 

[MTd2]

Jacques Distler has posted a final answer to Garrett Lisi:

Final Update (Christmas Edition)Still no responses to my challenge. I suppose that the overlap between the set of people who know some group theory and those who are (still) interested in giving Lisi’s “Theory of Everything” a passing thought is empty.But, since it’s Christmas, I guess it’s time to give the answer.First, I will prove the assertion above, that there can be at most 2 generations in the decomposition of the 248. Then I will proceed to show that even that is impossible.What we seek is an involution of the Lie algebra, e 8. The “bosons” correspond to the subalgebra, on which the involution acts as +1; the “fermions” correspond to generators on which the involution acts as −1. Note that we are not replacing commutators by anti-commutators for the “fermions.” While that would make physical sense, it would correspond to an “e 8 Lie superalgebra.” Victor Kač classified simple Lie superalgebras, and this isn’t one of them. Nope, the “fermions” will have commutators, just like the “bosons.”We would like an involution which maximizes the number of “fermions.” Marcel Berger classified such involutions, and the maximum number of −1 eigenvalues is 128. The “bosonic” subalgebra is a certain real form of d 8, and the 128 is the spinor representation.We’re interested in embedding G in the group generated by the “bosonic” subalgebra, which is Spin(8,8) in the case of E 8(8) or Spin(12,4), in the case of E 8(−24). And we’d like to count the number of generations we can find among the “fermions.” With a maximum of 128 fermions, we can, at best find(6)128=?2(2,ℜ+(1,1) 0)+2(2¯,ℜ¯+(1,1) 0)where
ℜ=(3,2) 1/6+(3¯,1) −2/3+(3¯,1) 1/3+(1,2) −1/2+(1,1) 1
That is, we can, at best, find two generations.Lisi claimed to have found an involution which acted as +1 on 56 generators and as −1 on 192 generators. This, by Berger’s classification, is impossible.In the first version of this post, I mistakenly asserted that I had found a realization of (6). This was wrong, and I had to sheepishly retract the statement. Instead, it — and Lisi’s embedding (after one corrects various mistakes in his paper) — is nonchiral(7)128=(2,ℜ+(1,1) 0+ℜ¯+(1,1) 0)+(2¯,ℜ+(1,1) 0+ℜ¯+(1,1) 0)The reason why (6) cannot occur is very simple. Since we are talking about the spinor representation of Spin(16−4k,4k), we should have
∧ 2128⊃120
In particular, we should find the adjoint representation of G in the decomposition of the antisymmetric square. This does not happen for (6); in particular, you won’t find the (1,8,1) 0 in the decomposition of the antisymmetric square of (6). But it does happen for (7). So (6) can never occur. It doesn’t matter which noncompact real form of E 8 you use, or how you attempt to embed G.Quod Erat Demonstratum. Merry Christmas, y’all!

http://golem.ph.utexas.edu/~distler/blog/archives/001505.html#comments

 

[patfla]

Another educated layperson here trying to dig into Garrett's work as well as the necessary context (huge). Fat chance - right?

Maybe not. Heh.

I was posting, asking questions, on FQXI and got directed here to physicsforums. Very useful place. I first posted the below in “Layman’s Explanation” – it got no response. Then, back at FQXI, Garrett gave me the go-ahead to put it here. All rotten tomatoes should, of course, be directed at my head.

*****

I’ve read the whole topic [in that case, “Layman’s explanation”] and know where that puts me: at the bottom of the totem pole. Which is just fine since then there’s nowhere to go but up.

Have read several of Lee Smolin’s books; Peter Woit’s Not Even Wrong and others. John Baez's blog (the sophistication of which is not to be confused with the aforementioned books) is wonderful. So my ears perked up when I first learned of Garrett and his latest paper. I know that the holy grail (at least at the moment) is the unification of gravity and the standard model. Interesting task, even in mathematical, um, terms alone, since you're trying to reconcile one thing expressed in differential geometry using the tensor calculus with another (SM - not to be confused with S&M) expressed in group theory.

Well there are things that I knew already; things that I’ve learned over the last mth or whatever reading around; and now I have a whole new set of questions. I’ll limit myself to just one of those here. (although as you can see below, it'll hardly be a single sentence).

The components (observables?) of the 8-vectors which are the objects that inhabit the E8 Lie algebra (its operator being the ‘bracket’ or commutator). The components would be the quantum numbers. I’m trying to figure out just what they are.

This topic [“Layman’s”] pointed me to Tbl. 9 on p. 15 of Garrett’s paper. The 8 components seem to be columns 2-9 and they read something as follows (my first stab at TeX):

$$\frac{1}{2i}\omega ^{3} _{T} \;\;\; \frac{1}{2}\omega ^{3} _{S}\;\;\; U^{3} \;\; V^{3} \;\; w \;\; x \;\; y \;\; z \;\;$$

You should see 8 terms above.

Scroll up just slightly from Tbl. 9 in Garrett’s paper where he explains what these are.

The first four are from F4. 2 are associated with so(3,1) gravity and the other 2 are the 2 fields associated with the electroweak. I’m guessing that the omegas on the left are so(3,1) gravity and $$U^{3}$$ and $$V^{3}$$ are the electroweak’s 2 fields?

That’s the first half of my question. The other half consists of the remaining 4.

Here, Garrett explains, one has 3 and 1. 3 are the fields associated with the electrostrong and the remaining 1 is something associated with $$u(1)_{B-L}$$ (whatever that is).

The division of labor here would seem a little clearer: the 3 are $$x\;\;y\;\;z$$ . And the final one ( $$u(1)_{B-L}$$ ) is $$w$$.

Is that right?

(OK I'm a programmer - but I've never used TeX before. How do I force the w, x, y and z above back 'onto the line'? That is, so that they're not floating halfway up.)

All for now – pat

 

[garrett]

Hi Pat,

The mathematical description of tensor calculus and group theory using differential geometry is really neat, and it would make an enjoyable tangent if you wish to discuss it.

For your question: the $$U^3$$ and $$V^3$$ root coordinates are a rotation of the weak $$W^3$$ and $$B_1$$ root coordinates, described on page 10 of the paper. The $$W^3$$ is the weak $$su(2)_L$$ quantum number, but the $$B_1$$ is only part of the weak hypercharge. (This is described a bit in an earlier post in this thread.) The $$x \;\; y \;\; z$$ are rotated into $$B_2$$, which is the $$u(1)_{B-L}$$ baryon minus lepton number, and the $$g^3$$ and $$g^8$$ quantum numbers for the strong interaction. The $$B_1$$ and $$B_2$$ are rotated to give the weak hypercharge, $$Y$$, and something else, $$X$$. The $$w$$ and $$X$$ are two new quantum numbers, not currently part of the standard model.

Hope that helps.

 

[rntsai]

The components (observables?) of the 8-vectors which are the objects that inhabit the E8 Lie algebra (its operator being the ‘bracket’ or commutator). The components would be the quantum numbers. I’m trying to figure out just what they are.

This topic [“Layman’s”] pointed me to Tbl. 9 on p. 15 of Garrett’s paper. The 8 components seem to be columns 2-9 and they read something as follows (my first stab at TeX):

$$\frac{1}{2i}\omega ^{3} _{T} \;\;\; \frac{1}{2}\omega ^{3} _{S}\;\;\; U^{3} \;\; V^{3} \;\; w \;\; x \;\; y \;\; z \;\;$$

It might also help to see these 8 elements as a basis for
the cartan algebra of e8. So these 8 + the 240 in the
second column complete the e8 description.

The cartan algebra is commutative, so these 8 can have
simultaneous eigenvectors. Also note that this is just a
basis, so any other linear combination can be used to define
other (dependant) quantum numbers; electromagnetic charge,
weak hypercharge,...are linear combinations of these).

Looking at the rows and columns this way, table 9 is just the
multiplication table for e8 in the chosen basis. Actually it's
a partial table (cartan x non-cartan). So if you have for
example 1/2 in column c and row v, this means c*v=(1/2)v; c*v
here is commutator "[c,v]"; if you multiply the row elements
(non-cartan x non-cartan) you get the rest of the multiplication
table which tells you how all things interact.

 

[patfla]

Thanx garrett and rntsai – very useful. It goes without saying (but I’ll spell it out anyway): if I go silent for a while is because I’m off processing. And you’ve given me specific leads which make the roadmap more comprehensible.

That is, I’m off processing to the extent that I’m not exercising my duties as a husband or dad. Or bent (not always unwillingly) to my employer’s grindstone. Or, as was the case last night, watching the Patriots-Giants game (good game).

I essentially did a kind of one-to-one mapping, garrett, and if things are more subtle than that, I’m not a bit surprised. As best I understand what you wrote, or rather by omission, it would appear that the 2 $$\omega$$ terms do refer to the gravo part of gravoweak. I confess, I haven’t nearly read the paper to the extent that I should have by now, but if one goes to, say, section 2.2.1, the $$\omega$$s are pretty clearly associated with gravity. A question though: to what extent is the so(3,1) formulation of gravity fully accepted (ahem: clearing of the throat)? Even if it’s not, mathematicians have been using the result of the Riemann hypothesis for yrs (read: centuries). And productively. (hopefully it won’t be disproven). As an aside and as regards differential geometry: I think it’s pretty well accepted that Bernard Riemann is differential geomtery's father or whatever (?). This would have been the first hlf (or thereabouts) of the 19th century. So where does de Brange‘s proof stands these days? It made a splash (plunk?) several yrs ago but I’ve heard nothing since (not that I’ve been that closely tuned in). While on the one hand it would appear that others in the mathematical community, hopefully genially, consider be Branges a bit of a nut, on the other he did solve the Bierbach conjecture. A colleague of mine was actually, for a time, at Purdue (w de Branges) and my colleague has related various interesting ‘stories’.

A second technical point will suffice for the time being. It looks like the weak interaction ‘bleeds’ between $$V^{3}$$ (aka $$B_{1}$$ [reversing the rotation]) and $$B_{2}$$ which arises from $$x\;\;y\;\;z$$. So that’s a specific instance of where my ‘one-to-one’ mapping breaks down. I’ve no doubt put this poorly, but hopefully you can disentangle what I’ve said and confirm or not whether I’ve gotten this one, at a second approximation, correct. I assume that with some kind of rotation or transformation or whatever (that is, using techniques from group theory) that one could ‘reunite’ $$B_{1}$$ and $$B_{2}$$. But then, presumably, that wouldn’t be a part of E8 any longer.

At some point soon (meaning now), I need to retreat; print out; and read carefully both a) Garrett’s paper and b) the whole of this topic. I read “Layman’s” which, while I did learn some things, was kind of picaresque. AESToE is a wholly different matter.

And rntsai, yes - I should definitely try rethinking things as a cartan algebra. I'd like to say more in that regard, but my post is already too long.

In my role as a lurker, I’m curious to see what kind of response M. Distler receives. If I’ve understand him correctly he claims (a proof the merits of which I can’t judge) that one can’t get the three generations of fermions from out of E8.

pat

 

[Berlin]

Talking about my generation..

Hi Garrett,

It took me my holidays to figure out your table 9. It helped to use the “E8 polytope” item of wikipedia to choose the right numbers.

What bothers me is all the attention to squeeze in all three observed generations within E8. As far as I know all the currently observed quantum numbers for the three generations are equal, except for their masses. We all (?) expect the mass to emerge in a higgs-like mechanism, not to be a fundamental property. So, why bother if we have a new quantum number (w) within E8 to play with?

I would guess that the new fields x.phi will do the trick. 18 new fields cannot be a coincidence. 18=3x6=3x3x2. 3 generations, 6 leptons.quarks, 3 colors, 2 catagories of particles…

I would investigate two possibilities:
- first: get one generation right with the right quantum numbers within E8 (mass 0), and use the ‘wrong’ roots mimicing the other two generations. --> use x.phi to turn this into the three observed generations by changing the wrong q-numbers and split the ‘susceptibility’ for higgs between the generations --> standard higgs --> observed generations with observed mass differences.
- second: use three ‘generations’ with wrong q-numbers within E8 (mass 0) --> use x.phi to change their q-numbers and ‘susceptibility’ for higgs (maybe even by mixing between the E8 generations) --> standard higgs mechanism --> observed generations with mass.

Maybe the chosen mechanism is different for leptons an quarks. I will try to be more specific next post.

Jan Leendert

 

[rntsai]

- first: get one generation right with the right quantum numbers within E8 (mass 0), and use the ‘wrong’ roots mimicing the other two generations.

Were you able to confirm the quantm numbers for any of the generations? If you
were, which ones did you check? (charge,spin,...). I've been trying to setup this
step just to see where the correspondance works and where and how it breaks.

 

[CarlB]

I would investigate two possibilities:
- first: get one generation right with the right quantum numbers within E8 (mass 0), and use the ‘wrong’ roots mimicing the other two generations. --> use x.phi to turn this into the three observed generations by changing the wrong q-numbers and split the ‘susceptibility’ for higgs between the generations --> standard higgs --> observed generations with observed mass differences.

Since the utility of Koide's formula here has been discussed before, let me chime in with a quick arithmetic note on it.

The Koide formula for the charged lepton masses is sort of like this, but sort of not. Let $$m_n$$ be the masses of the electron, muon, and tau for n the generation number n=1,2,3. Then, for $$\mu_1$$ a mass scale,

$$m_{1 n} = \mu_1\;\left(1 + \sqrt{2}\cos(2/9 + 2n\pi/3)\right)^2$$

is a close approximation of the charged lepton masses. To get them within the current experimental error bars you have to change 2/9 to 0.22222204717(48), uh, last time I looked anyway. To make the above formula give an electron which is massless, you have to change the 2/9 to $$\pi/4 -\pi/6 = \pi/12$$.

The neutrinos use a different mass scale, $$\mu_0$$, and the 2/9 angle has to be adjusted. Somewhat surprisingly, the adjustment to the angle is the same $$\pi/12$$ that would have made the electron massless, but one still keeps the 2/9 factor:

$$m_{0 n} = \mu_0\;\left(1 + \sqrt{2}\cos(2/9 + \pi/12 + 2n\pi/3)\right)^2,$$

In my mind, the angle adjustment, $$\pi/12$$ comes from Berry or Pancharatnam or geometric phase. Berry phase is half the spherical area (i.e. measured in ster radians) of a path on the sphere. Therefore $$\pi/12$$ corresponds to a spherical surface of$$\pi/6$$ ster radians. This is 1/3 the area of the spherical triangle with corners given (in Cartesian coordinates) of (1,0,0), (0,1,0) and (0,0,1). The factor of 1/3 goes away if you put the Berry phase where it belongs, so the $$2n\pi/3$$ factor becomes $$(2n\pi + \pi/4)/3$$.

 

[Tony Smith]

I just ran across a paper "The octic E8 invariant" by Martin Cederwall and Jakob Palmkvist at
http://arxiv.org/abs/hep-th/0702024
that might be relevant to issues like chirality etc. They say:
"... Spin(16)/Z2 is the maximal compact subgroup of the split form G = E8(8) ...
the adjoint representation of E8 ... decomposes into the adjoint 120 and a chiral spinor 128 ...
The spinorial generator acts similarly to a supersymmetry generator on a superfield ...".

Tony Smith

 

[Berlin]

dazzling numbers

Numbers are dazzling before my eyes. Have been buzy all day understanding and correcting tabel 9. Corrections of my own mistakes. Playing with the x.phi fields (it took me some time to see that they carry two colors) you see that all quarks and leptons are effected by one of the x 1/2/3 . phi fields, except for the third generation of leptons. The third generation of quarks has only a quark-anti quark transition. It seems therefor much more logical to swap the first and third generations in table 9. Is that possible or is Murphy around the corner?

Jan Leendert

 

[garrett]

Pat,
The $$\omega$$'s do correspond to the gravitational spin connection fields, and this $$so(3,1)$$ formulation of gravity is very well accepted. In fact, it's the classical starting point for all Loop Quantum Gravity approaches. Although, sadly, I couldn't find a good description of the spin connection formulation on Wikipedia. I do have a description available on my wiki though -- you might find it helpful:
http://www.deferentialgeometry.org/#spacetime
I think this formulation originally came from Cartan, and better fits the mathematical theory of fiber bundles.

We can rotate the coordinate axes of the root system however we wish, describing the same algebra. This just corresponds to a different choice of basis elements for the same Lie algebra -- still E8. I think rntsai has done a good job of explaining this in his previous post. He (or she?) is also correct that the roots alone aren't enough to tell you which Cartan subalgebra particle/field we get when two roots add to give one at the origin. To describe this, we would have to work in a specific representation, or at least write down these structure constants.

Jan (aka Berlin),
I hope Table 9 made your holidays more rather than less enjoyable. :) I'm open to any interesting way to get the other two generations, so I'll look forward to your more detailed description. The guess I'm following in the paper is that the second and third generations are different because they have different "new" quantum numbers. The first try at doing this in E8 doesn't work very well, as described in the paper ( and emphasized by Distler. ;) ). I have a couple other tricks to try though, and you're right that they'll involve $$x \Phi$$.

rntsai,
It will probably help to rotate the coordinates in Table 9 by the matrix on page 18 in order to identify the "physical" quantum numbers. This gives good quantum numbers for the first generation fermions and gauge fields. The second and third generation quantum numbers are only equivalent to these under triality -- which is a description that needs improvement.

Carl,
I find it very interesting how this relates to tribimaximal mixing. But I haven't had time to play with it yet.

Tony,
Their equation (2.3) is scary... and I'm not sure how these invariant tensors are supposed to work.

Jan,
I've played around with swapping generation elements, succeeding in getting a better set of fields with respect to hypercharge. But the main problem with this kind of swapping is the so(3,1) quantum numbers. I think the gravity part of the theory will have to be formulated in a slightly different way if the three generations are going to work. I certainly encourage you to play with it! There's a good chance that someone else will figure it out before I do.

 

[rntsai]

rntsai,
It will probably help to rotate the coordinates in Table 9 by the matrix on page 18 in order to identify the "physical" quantum numbers. This gives good quantum numbers for the first generation fermions and gauge fields. The second and third generation quantum numbers are only equivalent to these under triality -- which is a description that needs improvement.

Hi Garrett,
I've been working with this rotation and using Table 4 as the reference for quantum numbers.
The table gives $$Y,Q$$ in terms of $$W^3,B_1^3,B_2$$. My question is : what
other quantum numbers are involved; shouldn't $$g^3,g^8$$ (strong force) enter the
picture somehow? Also how do you interpret quantum numbers involving the gravitational part
$$\omega ^{3} _{T}/2i,\omega ^{3} _{S}/2$$? These two are in the
cartan of $$d_2=a_1 + a_1$$ with the familiar 6 rotations/boosts, anything
"physical" in there quantum values?

 

[patfla]

http://www.deferentialgeometry.org/#spacetime

Thanx Garrett. Another resource I should be using. Your site: deferentialgeometry.org. Nice ergonomics (open,close,move around topic boxes – and all linked together). Looked at the About page to see what packages you might have used. Interesting. I imagine this has been discussed elsewhere but can Fractured Atlas add Paypal?

I have, as I intended, printed out the whole of this topic (AESToE) and your paper. It may seem backwards, but I'm reading the topic and referring to your paper (would be hard to do the other way round).

Also on so(3,1) gravity, there appear to be some interesting LQG papers here:

http://cgpg.gravity.psu.edu/people/Ashtekar/articles.html

I take it that Abhay Ashtekar is well known in LQG circles (might as well drop any attempts to flag the unintended puns). Penn State - that's where Lee Smolin used to be and I've come to understand it as a LQG outpost.

Rntsai: thanks for the pointer to the GAP software. Downloaded; installed; playing around. Would it be possible for you to send me some of your GAP code circa, say, post #19? In spite of claims to the contrary, it’s my belief that science almost never works purely either deductively or inductively. It’s a combination of both. And so playing with some computational machine (at the right level of abstraction) can greatly add to one’s intuitions. I have Matlab and that’s my usual platform. Googling around it seems it can be purposed in the direction of subalgebras, Lie groups, etc. But I always like to try out new pieces of software and GAP appears to be your platform of choice.

Garrett, yes I realized quickly enough after my last post that a part of my confusion was simply a matter of changing bases.

Back to $$U^{3}$$ and $$V^{3}$$. You already translated these for me (via rotation) to $$W^{3}$$ and $$B_{1}$$. OK I’m kind of flying by syntactic controls here. The weak interaction is mediated by the W and Z bosons. I’m guessing that the W boson and $$W^{3}$$ are related. Which would leave $$B_{1}$$ ( adding $$B_{2}$$ ) possibly related to the Z boson?

I do see here http://en.wikipedia.org/wiki/Electro-weak that one of the terms in the Lagrangian for the electroweak before symmetry breaking - (the first term), $$\mathcal{L}_g$$ - expands into “three W particles and one B particle”.

 

[Tony Smith]

Sorry about the scary complexity of eq. 2.3 in hep-th/0702024 by Cederwall and Palmkvist.
Since they decompose E8 into the adjoint 120 + spinorial 128 as your model does,
and since they say that "... The spinorial generator acts similarly to a supersymmetry generator on a superfield ...",
their work may be closely related to your use of the spinorial 128 for fermions.
Note, with respect to fermion chirality, that they refer to "... so(16) with chiral spinors ...".

Perhaps another paper by Cederwall and Preitschopf at hep-th/9309030 might be helpful, particularly since your model is based on constructing a connection on E8. They discuss the natural torsion of the 7-sphere S7 and its relation to BRST operators.

Since S7 is the unit sphere in the octonions,
and since the 128-dim spinor space E8 / Spin(16) = (OxO)P2
which is Rosenfeld's octo-octonionic projective plane (i.e., a projective plane based on the product of two copies of the octonions, each of which has an S7 with natural torsion)
then
maybe your connection could have torsion for its fermionic part
and curvature for its bosonic part.

I saw where you mentioned "gravitational torsion" in your paper at 0711.0770 but I did not see a discussion of torsion (which is naturally related to spinors) with respect to the physical fermions of your model.
Is it there implicitly ?

Tony Smith

PS - Cederwall has another paper at hep-th/9310115 that might be more introductory to ideas such as how the natural torsion of S7 is related to exceptional stuff such as parallelizability etc.

 

[Berlin]

Some progress

It seems I made some progress. I have two generations of leptons with exactly the right quantum numbers g3, g8, W3, B13, ½Y, Q. A third generation needs some tuning work, presumably the third generation.

Postulation: a third generation of leptons have ‘tau’ particles made up of original root within E8 plus B1+ or B1- boson. Made just by adding up the 8-dim quantum numbers..
- right tau= E8 root tau + (B1-) (for up and down)
- anti-left tau = E8 root tau + (B1+) (for up and down)

All other particles are their original E8 roots.

This presumably the third generation has all the right quntum numbers for: g3=0, g8=0,
W3, Y and Q !

The B13 number is different for this generation than for the others, but I don’t think it matters.

For getting this all right I would you strongly advise you to re-read the section of Abraham Pais' book 'Inward bound' about Samual Goudsmit and Uhlenbeck (also from here in Leiden). The first had took a course as a detective and was a wizard in cryptograms and hieroglyphs. The second was trained in theoretical physics. Together they solved the fact that the electron should have spin from spectra. I think we need both talents here... Of course those guys had their critics: Lorentz calculated it could not be, Pauli called it heresy.

Jan

 

[rntsai]

Rntsai: thanks for the pointer to the GAP software. Downloaded; installed; playing around. Would it be possible for you to send me some of your GAP code circa, say, post #19? In spite of claims to the contrary, it’s my belief that science almost never works purely either deductively or inductively. It’s a combination of both. And so playing with some computational machine (at the right level of abstraction) can greatly add to one’s intuitions. I have Matlab and that’s my usual platform. Googling around it seems it can be purposed in the direction of subalgebras, Lie groups, etc. But I always like to try out new pieces of software and GAP appears to be your platform of choice.

Hi Pat,
The code is small enough that I'm posting here. The real heavy work is
done by the GAP internals, so this is a simple "application layer" that
sits on top of that. Matlab is a great tool and I use it fairly heavily;
I've gone back and forth between GAP and Matlab on many occasions. For
algebraic computations, GAP is really as good as they get. Magma is also
supposed to be very good, but I've never used it and it's not free.

Just read in this (cut and paste into a text file "Test.gap" and from
the GAP shell type Read("Test.gap");

CheckInSpan:=function(A,V)return ForAll(A,a->ForAll(V,v->IsContainedInSpan(MutableBasis(Rationals,V),a*v)));end;

A3:=SimpleLieAlgebra("A",3,Rationals);
T:=ChevalleyBasis(A3);e:=T[1];f:=T[2];h:=T[3];

A2:=Subalgebra(A3,Concatenation(e{[1,2,4]},f{[1,2,4]}));
Print(SemiSimpleType(A2),"\n");

V8:=Concatenation(ChevalleyBasis(A2));
V3p:=e{[3,5,6]};
V3q:=f{[3,5,6]};
V1:=BasisVectors(Basis(LieCentralizer(A3,A2)));

Print("A3/A2, V8 ",CheckInSpan(V8,V8),"\n");
Print("A3/A2, V3p ",CheckInSpan(V8,V3p),"\n");
Print("A3/A2, V3q ",CheckInSpan(V8,V3q),"\n");
Print("A3/A2, V1 ",CheckInSpan(V8,V1),"\n");

In seconds, this should print to the console :
A2
A3/A2, V8 true
A3/A2, V3p true
A3/A2, V3q true
A3/A2, V1 true

 

[yoyoq]

just a suggestion, if you use complex E8 to fix things up, you could call the next paper
"An exceptionally simply complex theory of everything"

 

[Berlin]

quarks as well..

It seems that I have all the generations of quarks right as well (not checked all the anti's yet). Cannot believe it.

The third generation seems to require an adding of additional root-particles as well, just like the last generation of leptons (see last post)

The quarks require at least the adding of two x.phi fields to the E8 root for the quark. Adding three chosen fields can work as well. The quantum numbers for g3, g8, W3 and Q are all OK! Just like the leptons the B1-3 number is different from the first two generations, but I don't think this matters. The x.phi fields are essential to get the charges right.

Complete list of 'composed' quarks(anti's not checked):
- t-left, for all three colors, all spin-down
- b-left, for all three colors, all spin down
- t-right, for all three colors and both spins

Example:
- t-left-red = E8 root (t-left-red) + x2.phi (BG)+x3.phi(gb)

I seems also possible to use three x.phi fields:
Example:
- t-left-red = E8 root (t-left-red) + x3.phi (rg)+x3.phi(gb)+x3.phi (rb)

This last thing reminds me of a kind of dual to the proton.. Would it be wonderfull physics if we can have a second E8, with other assignments, fully dual to this one? Get a kind of supersymmetry in an unexpected way? This idea is not fully stupid. After all I found out that in total 16 (4 leptons and 12 quarks, all in +/- roots) were in trouble and had to be fixed. This number is just the difference of the 128 and 112 roots of the E8 polytope.. Could you assign all the bosons to the (+/- half) roots and all the leptons to the (+/- 1) 's?? And also fixing 16 at the end?? Could this turn AdS gravity into a CFT? First calc: the 6x3x2= 36 quarks could be dual to the 6 gluons and 18 x.phi's, with twelve fixes. 6x3 leptons could be dual to the 20 (EW bosons+frame higgs) + 2 wR/L=22, fixing 4? May be we could borrow a lot of maths from the string guys after all. Call it the joppe conjecture for now. Forgive me this speculation, it is late and time for a drink. Fun though.

Garrett: would you like to write this all down together? After all, I am not a professional physicist either. I do not surf but play tennis.

Jan

 

[Berlin]

Some more on necessary corrections

I have now fully checkes all roots. It looks like a few leptons and 24 quarks need correction (=observed particle is no bare root), corrected with several B's and x.phi's.

I am sure that this will change the required symmetry groups. This could -maybe- silence some critics...

The 'self dual' character of the E8 could indeed be through 16 of these above. I will work that out, takes some time.

Jan

 

[rntsai]

Hi Garrett,
I've been working with this rotation and using Table 4 as the reference for quantum numbers.
The table gives $$Y,Q$$ in terms of $$W^3,B_1^3,B_2$$. My question is : what
other quantum numbers are involved; shouldn't $$g^3,g^8$$ (strong force) enter the
picture somehow? Also how do you interpret quantum numbers involving the gravitational part
$$\omega ^{3} _{T}/2i,\omega ^{3} _{S}/2$$? These two are in the
cartan of $$d_2=a_1 + a_1$$ with the familiar 6 rotations/boosts, anything
"physical" in there quantum values?

After spinning around in the notations and conventions for
a bit I think I understand this a little better, but not
completely yet. The problem is that there are too many
spins involved : $$\omega_L^3,\omega_R^3,W^3,B_1^3,g_3$$ are all "spins".

$$B_1^3$$ gives weak spin;
$$g_3$$ gives strong spin;
$$\omega_L^3,\omega_R^3$$ are a mix of "regular" spin and chirality;

so for example the "left-chiral spin-up quark", $$u_L^\wedge$$ :

the "left-chiral" part of its name means $$\omega_R^3 u_L^\wedge = 0$$;
the "up" part means $$\omega_L^3 u_L^\wedge = +(1/2)u_L^\wedge$$
Table 4 gives the eigenvalues of $$W^3,B_1^3$$ as 1/2 and 0.
it's not specified what color quark this is, so Table 2 gives the
$$g_3$$ eigenvalue as 1/2 if it's red, -1/2 if green and 0 if blue.

I think I got this right, but not sure.

Also it looks like the new quantum number $$w$$ distinguishes particles
from their antiparticles. This would make $$x_2\Phi$$ anti $$x_1 \Phi$$

 

[patfla]

Berlin: if you’ve succeeded in generating the 2nd and 3rd generation fermions, that would indeed be extraordinary. Finally went by Jacques Distler’s blog Mutterings. Er no, Musings. Topic = ‘A Little More Group Theory’. Whoa dude – that was one hairy ride. But I guess one should always get a second opinion.

Not that I needed it, but the Distler topic certainly confirms that properly generating the 2nd and 3rd generation fermions constitute Outstanding Item #1.

Berlin: May I ask? What computational tool(s) do you use?

rtnsai: thanks – that worked (your code). Now I have to figure out what it’s doing
(I think one inserts a smiley here). There must be a debugger somewhere in GAP – have to find it. And yes the data structures are probably quite large, but if, in perl, I can poke around enormous data structures using perl references (yuk), I’m willing to give GAP a go. I’m of course way behind you guys (meaning men and women both), but the exercise is interesting and engaging.

Have been eying the ‘Is String Theory Correct’ topic. I’ve had to restrain myself from posting there (so I’ll post it here). Something to the effect of:

No.
Everyone knows that the Calabi-Yau manifold is a part of the inner ear. And how and why string theorists have been abusing this, and for as long as they have, is both a scandal and outrage.

It would then be in the spirit of things to sign the above with Roseanne Roseannadanna. Except I should ask first: is levity permitted on physicsforums?

pat

 

[Berlin]

Hi Patfla,

My computational tools are just figuring thinks out by one giant excel sheet with all the roots, Q-#'s etc. Yes, I figured the third generations out, and it is slightly different from my earlier post:

- all the leptons are right, except four of them, who need to be composed particles with a B -/+.
- one full generation of quarks has 'nothing' to do with the other two generations, except that they produce the same quantum numbers. They are all combinations of two other roots.

This reduces the 'strain' on the group theory guys because you only incorporate two identical generations in E8. One generation of leptons as well as one generation of quarks come out only after some kind of symmetry breaking I image.

jan

 

[MTd2]

Hi Berlin!

Could you be more specific with your results? What leptons are wrong? What's up with two generations not having nothing to do with the third? What combination of roots? And what cornders are these?

Daniel

 

[Berlin]

My lepton numbers for the third generation are in ingreement with the other one's except that the B1-3 number is off by a factor of two. This could imply two heavier M, Z particles but I am not sure. If these would be 120 and 182 GeV would anyone knows if they would have been seen by current colliders? So, the third generations only turn up after some symmetry breaking of E8. So, you don't need the group within E8.

Garrett: what we should do is make a second group G2, just like the the strong interaction, put in the third generations of leptons in (only the singlet sector, tau, neutrino (right) and anti tau, neutrino (left) and combine them with the gravitational sector W-L/R. See what turns out! Let this break in a higgs like way using the W/B +/-sector.

My best guess: the third generation of leptons is a kind of hoax. Just because the quantum numbers turns out almost equal does not mean that it is governed by the same symmetry. Forget about the graviweak D4. Go to something like G2xG2xSu(2)ew. I guess we end up with so(6,1) insteed of so(7,1), but my group theory is shaky.

By the way: the third generation of quarks is a hoax too. They are just some symmetry breaking product of the new x.phi fields and the gravi/higgs sector.

If have been working on my dual theory also. I stopped because I needed to many new particles in the w-t, w-s, w sector (+/- 1, 0), but now I see that they are the original roots of the third gen leptons! So, the third generation of leptons uses that part of E8. I will work further on my "joppe conjecture" dual E8. An example of a duality I discovered is that the four w-L, w-r from the gravitational sector are dual to right-hand neutrino's of the first and second gen. Left hand neutrino's are dual to the frame-higgs fields. I hope to complete this program soon.

jan

 

[Emanuel]

Hello Jan, I've been lurking in this topic for a while now and am sorry to pollute it with an off-topic post, but every time you mention your "Joppe conjecture" it makes me smile, as the only Joppe I know is my taijiquan teacher :) So I was wondering, how did you arrive at the name?

PS: good luck with your work on both theories!

 

[garrett]

(sorry I've been away -- I'll respond to these posts in order as a I read them)

rntsai,
The $$x \;\; y \;\; z$$ values of Table 2 are the same as in Table 9. The rotation matrix takes these Table 9 values into the $$B_2 \;\; g^3 \;\; g^8$$ of the strong (and part of the electroweak) force, as in Table 2. The $$\omega_T \;\; \omega_S$$ are temporal and spatial parts of the spin connection. These can be rotated into the left and right-chiral parts of the spin connection, $$\omega_L \;\; \omega_R$$. The quantum numbers with respect to these are "spin."

Pat,
I'm glad you like deferentialgeometry.org. The wiki is all open source, and can be downloaded to your own machine if you wish to play with it. Fractured Atlas doesn't take PayPal, but they do accept credit cards. Abhay Ashtekar is more than known -- he's considered the "founding father" of LQG. Changing basis can be very confusing -- because the Cartan subalgebra basis elements are also used as root space coordinate labels. Your understanding of $$W$$ and $$Z$$ in terms of $$W$$, $$B_1^3$$, and $$B_2$$ is correct.

Tony,
The torsion question is especially interesting. For any theory (such as this one) in which the spin connection and frame are independent variable, the coupling of the spin connection to spinors in curved spacetime will produce nonzero torsion. In this theory, the torsion is also intimately connected with the Higgs. This interaction comes out of the curvature calculation, and I'm not sure if there's a precedent for this.

Jan,
I'm glad you're having fun playing with other possible assignments for the second and third generation. I'm also playing with this, to see what I can cook up. If one takes some liberties with the top quark, and mixes up the right-chiral neutrinos in an interesting way, one can get good weak hypercharge numbers for all three generations. It sounds like you're on to something similar.

yoyoq,
For the next paper I may choose a more conservative title.

Jan,
Yes, I think you've found the same reassignment of generations that I'm playing with, exchanging some of the x.Phi for quarks, and exchanging the $$B_1^\pm$$ for nuetrinos or tau. I don't have time at the moment to co-author anything, but if you'd like to write it up as a paper, I'd be happy to look at it and offer suggestions.

rntsai,
Yep. You need $$g^3$$ and $$g^8$$ for the color quantum numbers though.

Pat,
You belong in New Jersey.

Jan (and MTd2 and Emanuel),
These generation assignments Jan (and I) are now playing with aren't necessarily triality related any more. I think it's great that Jan is playing around with his own related model.