An Exceptionally Technical Discussion of AESToE

[Tony Smith]

moveon "... advise to look for superalgebras instead of E8. There exist even exceptional ones; they have been classified by Katz, and a useful ref is hep-th/9607161. Choosing one with D4+D4 as its bosonic piece (and a suitable real form) may be more successful. ...".

hep-th9607161 is indeed a nice reference. Thanks for it. However (please correct me where I am wrong) when I look at it for exceptional Lie superalgebras, I see only three:
F(4) which is 40-dimensional;
G(3) which is 31-dimensional; and
D(2,1;a) which is 17-dimensional,
so
none of them are large enough to contain 28+28=56-dimensional D4+D4.

From Table III on page 13, it seems that the only one with a Dm bosonic part is
D(m,n) which has bosonic part Dm (+) Cn
which the describe on page 37 as being "... osp(2m|2n) ...[ which ]... has as even [ bosonic ] part the Lie algebra so(2m) (+) sp(2n) ...".

osp(2m|2n) is the basis for supergravity and, in his book Supersymmetry (Cambridge 1986 at page 113), Peter G. O. Freund says "... In extended supergravity of type N the largest internal nonabelian gauge group is O(N), corresponding to a gauged osp(N|4) ... The largest nonabelian gauge symmetry is O(8) ...".

So, since the sp(4) in Freund's notation, which is sp(2) in some other notations accounts for gravity and therefore for one of the D4,
you have the O(8) for the other D4,
so
it seems to me that N=8 supergravity is the only superalgebra based model that could reasonably be seen as fitting something like Garrett's D4 + D4 model-making scheme.

As Freund discusses in some detail in chapter 23, N = 8 supergravity and concludes "... all this makes the ultimate absence of a compelling and realistic spectrum all the more frustrating. ...".

In chapter 26, Freund discusses the related 11-dimensonal supergravity, but as far as I know there has been no satisfactory realistic 11-dim supergravity or N=8 supergravity model.

Therefore, to work with D4 + D4 it seems to me that you must abandon superalgebras because they either do not have it or have not been shown to work (despite much effort),
and that ordinary exceptional Lie algebras, which have both bosonic and spinor parts, are a useful place to look for building models,
and
that Garrett has done a good job of seeing how the root vector generators of E8 can be assigned physically realistic roles in constructing a useful physics model, and therefore is worth a substantial amount of research effort (comparable to that spent so far on supergravity).

Tony Smith

 

[garrett]

moveon,
Your translation is interesting, but all fields in the paper are valued in the Lie algebra of E8. I'm not yet certain that the first generation doesn't work in real E8, because of the unusual complex structure employed -- but even if it doesn't work, complex E8 would.

 

[CarlB]

Gosh, it seems to me that in QFT signs are arbitrary and in any observable, fermions always appear in pairs. In that sense, what you really need is to have your fermions square to zero and your bosons not. Zero is not a valid quantum state. To get the equivalent, all you really have to do is make the square of a fermion be "not a valid quantum state", it doesn't actually have to be zero.

What fermions do to each other when you permute them is not a physical observable. Quantum mechanics is a probability theory. To get a probability in QFT you begin with an amplitude, which is a complex number, computed as \langle 0 | stuff | 0\rangle. Then you take the squared magnitude, that is, you multiply your amplitude by its complex conjugate:

\langle 0 | stuff | 0\rangle \langle 0 | stuff^* |0\rangle

Now suppose you commute two creation operators in "stuff" and get a minus sign. That minus sign is canceled by the minus sign that you get when you commute the same two observables in its Hermitian conjugate. No change to the observable whether the result of the commutation is +1 or -1.

So suppose you start with a bosonic QFT and you have a boson \psi that you want to give "fermion statistics" to. Add a term to the Hamiltonian of \kappa\;\psi\psi. Let \kappa \to \infty to prevent it from being energetically possible. The result is a mixed fermion / boson theory by symmetry breaking.

To put the above argument in QM form, consider the ancient physics test problem, "what happens to an electron if you rotate it by 360 degrees?"

Every physicist knows the answer: "it gets multiplied by -1". But that is only true in the spinor representation. In the density matrix representation of a quantum state, spinors appear in pairs and the result of rotating them is to change the density matrix representation by -1 x -1 = 1, or not at all. The act of rotating a fermionic wave function by 360 degrees is related to the act of switching the order of creation operators as is discussed in many QFT textbooks.

To put this in into the operator language, let Q be an operator, we wish to compute the average value of Q for a quantum state produced by the application of say four creation operators on the vacuum to make a four particle state. Label the four particles "k,n,u,j". So the 4-particle state is k^*n^*u^*j^*|0\rangle. Then the average of the operator Q over this quantum state is:
\langle 0 | j\; u\; n\; k\; | Q |\;k^*\;n^*\;u^*\;j^*| 0\rangle
Suppose you've got the above worked out for k, n, u, and j fermion creation and annihilation operators. You might write Q in terms of these creation and annihilation operators, but when you're done writing it, you will have some ordering and you won't have to rearrange them.

Now you can consider the same theory, but with the commutation relations of the k, n, u, j changed (but the operator Q left alone). The ensemble average will be the same as there will be no further need to commute the creation and annihilation operators. You get what you get. And if you want to change the order of the k, n, u, j, then you will be doing it twice and a sign change will cancel.

Another case is when the quantum state is a superposition. For example, consider j^*\;u^*\; - u^*\;j^*\;|0\rangle. If j and u are bosons the result is just zero, no more to say. For fermions, you get twice your choice, of ordering. Choose one of the orderings and relabel your fermions as bosons. No problems. Problems happen when you try to modify your operators (built from creation and annihilation operators with assumed commutation relations) at the same time as you modify the rules you use for how your creation and annihilation operators operate on the vacuum state. But if you do that you will be making a circular argument if you use that to say that the choice of commutation relations is an observable -- what you've done is modified the observable, not the quantum state itself.

 

[Coin]

So this is perhaps a step down in technicalness from the discussion of the last few pages, but this is something I have been wondering for awhile and have only just figured out how to ask correctly:

Something that I keep running across in discussions of symmetry groups is the distinction between local or internal symmetries, and global or spacetime symmetries. In general the idea seems to be that local symmetries, things like quantum phase invariance, apply at a point (or at least to a single structure?); global symmetries, like poincare invariance, apply to "everything".

Are the symmetries of Garrett's E8 construction local, or spacetime symmetries? E8 here contains both things which are usually given as examples of spacetime symmetries, like the Lorentz group, and also things which are usually given as examples of local symmetries, like electroweak SU(2)xU(1). Meanwhile, E8 is here used as a "gauge group", and for some reason I have gotten the impression that all "gauge" symmetries are local symmetries. Are all the E8 symmetries local? Or do they somehow incorporate a mix of local and spacetime symmetries? And if all of the symmetries in the E8 theory are local, then are there assumed to still be any "background" global/spacetime symmetries which exist apart from the symmetries E8 describes?

 

[garrett]

Hello Coin,
All symmetries in this construction are local. The so(3,1) is a local symmetry of the frame, which is a local map from spacetime tangent vectors to a local rest frame, consistent with the equivalence principle. Now, when there are solutions, which give some spacetime, this may or may not have global symmetries.

 

[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 \omegas 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.

 

[patfla]

OK I’m trying to understand the actual procedure here.

Garrett may have given me much (in combination with what I had already) of what I need here

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.

First you pick some simpler subalgebra of E8 – say G2 or F4. This will make the problem vastly more tractable. You can ‘see’ things more clearly and the operations will be simpler and/or faster. Next you need to find a basis within that subalgebra where two of the vectors will use the one-and-only Lie algebra operator – the bracket or commutator – and give a result back at the origin. The origin, in 3D, being x=0,y=0,z=0. What does this mean? Linear algebra I and the dot product (so here we’re talking second yr undergraduate mathematics [imo]). The dot product (in 2D – I guess it’s the wedge product in higher dimensions) will act upon two vectors and give a result of 0 when the two vectors are perpendicular. Perpendicular is the important part. So what does it mean ‘in the real world’ that the vectors are perpendicular? Well that depends. We’ll leave it as an exercise of the reader (but at this point, you are quite close to ‘the real world’).

Back up. In a (probably special, unitary [except that unitary implies group and not algebra]) Lie subalgebra when the commutator is applied to two (root) vectors) and you end up at 0 you have a particle/field. Yowza! A first big success (you found the right basis). So how do you find the right basis? Random doesn’t seem like a good idea (there are a lot of them). Hunches and intuition, if you’re so provided (meaning a professional, practitioner or very talented amateur), can go a long ways. But better yet, some more systematic way of a) try a new (likely) basis b) compute root vectors and see if any computation lands at the origin. Doing this suggests programming.

Doubtless, there’s much more to be said about finding the right basis. What that is that could be said: I don’t know.

But, as Garrett said you still have a (probably) very thorny issue before you. You’ve got a particle, but you don’t yet have enough information to figure out which one. Most particles are ‘known’ (outside) the theory, but, according to this theory, there’s a small number (18 I believe) that are not. As regards identifying the hot particle now in your hand, Garrett said:

He (or she?) [rntsai] 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.

So what procedure is involved here (identifying the particle/field)? With more time, patience (and probably help) I’ll figure it out, but at the moment I don’t know. Although I feel I have, more-or-less, figured out what a representation is. Representation is a keyword and there’s a whole field (of study) corresponding to representation theory. In CS object oriented terms, I’d like to think: you write an OO class in your text editor. It’s a specification - so far it ‘does’ nothing. Then you run a program (containing that class) and the class, as we say, ‘instantiates’. It’s an object in memory and now it’s actually doing something as a part of the program. Which for many people is still somewhat abstract, but is something I’ve been doing for yrs, so for me it’s intuitive. Anyway the instantiation (or object [running in memory]) is the representation. The analogy breaks down quickly though. For a class there’s basically one instantiation (one might use polymorphism to claim that’s there more than one – it’s certainly the case that you can get different ‘behaviors’ out of the object depending on polymorphism). But there are an infinity of representations for E8. Fortunately though, every member of this infinity can be generated from the basic, unitary representations. There’s an enormous number of these, but that number is finite.

The explanation of the last procedure (identifying the particle/field that you’ve just computed) may lie somewhere right before our eyes if we look upthread and know how to recognize what we’re looking at. Certainly somewhere out on the web (again you have to know what you’re looking at).

I’m done (for now) but of course ‘the problem’ isn’t. Is it a boson or fermion (well, you’ve sort of determined that already)? More importantly: computing its actions. Again I’m somewhat guessing here, but an example of an action would be how the W and Z bosons combine to produce the weak interaction.

So what was all that? I’m trying to check my understanding of things (so it’s sort of a big, long question). Berlin told us that he’s using an Excel spreadsheet. Well I know only too well how cumbersome (for me) these become at a point (look up OLAP – extremely cool), so while what I should use popped up as a question in my mind, I think it’s still better that I use the GAP software. Garrett apparently uses Mathematica. And, it goes with saying, these are tools. Then there’s understanding. And (according to Einstein) beyond understanding (knowledge) lies imagination.

Sorry too long. A single, specific question. This depends on some small part of the above being correct. Is it always two roots that combine to produce one particle/field? But wait I’ve probably gotten something (major) wrong. E8 has 240 symmetries, therefore particle/fields. And so 240 roots also. Well if you could only use each root once, then that would give only 120 particles. Maybe you can use a given root more than once?

slipped just saw Garrett’s latest post as I was writing this up.

Pat,
You belong in New Jersey.

Everyone, obviously, is impressed with your IQ. But what’s also struck me from the time I read your first posts (Backreaction) was your EQ as well. And so I’m puzzled: you might have played the academic politics game quite well. There are secondhand descriptions of academic politics all over the place, but I’ve had the opportunity to watch academic politics (and so the application of EQ in this regard) up close in the form of my physicist brother-in-law who made it onto the tenure track a couple of yrs ago. He’s an astrophysicist (currently one project is the polarization of the CMB) and there are web photos of him somewhere both at ESO, high up in the Chilean mountains, as well as at the South Pole. Oh yes, in your neck of the woods, he's also been to the top of Mauna Kea and the Keck installation (or whatever the whole facility is called).

But as you say yourself: “I didn’t want to do string theory”.

I’m from Boston originally and much of my family has lived in or around New York (but not me). I came to California in 1984 to finish up my undergrad at Berkeley. And have stayed in CA ever since, minus 6 yrs in Tokyo that is. We’re in the Bay Area. I’ve always wondered about this. If there’s a SoCal, do we live in NoCal? I finished at Berkeley Phi Beta Kappa with degrees in Japanese and Computer Science but grad school was sort of foreclosed upon by my having lost several yrs to surviving cancer in my early twenties. So all things considered I’m quite happy to be where I am (as opposed to, say, dead [or in New Jersey]).

pat

 

[Toastus]

Could someone describe AESToE in simple terms for me? I don't understand most of this, but I do get that it describes everything with one geometric shape.

 

[rntsai]

OK I’m trying to understand the actual procedure here.

First you pick some simpler subalgebra of E8 – say G2 or F4. This will make the problem vastly more tractable. You can ‘see’ things more clearly and the operations will be simpler and/or faster. Next you need to find a basis within that subalgebra where two of the vectors will use the one-and-only Lie algebra operator – the bracket or commutator – and give a result back at the origin. The origin, in 3D, being x=0,y=0,z=0. What does this mean? Linear algebra I and the dot product (so here we’re talking second yr undergraduate mathematics [imo]). The dot product (in 2D – I guess it’s the wedge product in higher dimensions) will act upon two vectors and give a result of 0 when the two vectors are perpendicular. Perpendicular is the important part. So what does it mean ‘in the real world’ that the vectors are perpendicular? Well that depends. We’ll leave it as an exercise of the reader (but at this point, you are quite close to ‘the real world’).

Hi Pat,
The algebra involved here is actually simpler than this.

There's a lot of jargon in both Lie algebras/groups and in
their application here to elementary particles that make things
even more confusing. There are a lot of constructs that go
by different names; slight variants are sometimes also referred
to by the same name many times sometimes without distinction.
I can try to simplify things to the best of my knowledge (and time).

1 question mark = I know the theory, but don't know what Garrett's saying.
2 question marks = I'm not sure myself because lack of knowledge in the area

- there are no dot products, cross products, ... here at all.
you just have one algebra over the reals(?). You can think of
commutation as its operation or just that it has a product
that satissfies the lie algebra axioms.

- E8 is a group; e8 is an algebra. You can associate several
"E8's" with e8. All the particle assignments and their quantum
can be done working with e8 only. I don't know if e8 over reals
or e8 over complexes is enough; I think reals(?) are enough
in spite of the appearance of complex numbers in some parts.
This importance of this distinction will come up later.

- real and complex e8 is 248 dimensional (over its repective field) and
you can find a basis for it such that 240 of the 248 basis vectors satisfy
certain conditions. You can then call these 240 basis vectors "roots" of the
algebra. These "roots" can be further divided into 120 positive and 120 negative
ones. Again here you're just picking names for basis elements.

I'm beginning to think I'm adding to the jargon rather than clarifying it.
Maybe I'll stop and rethink this.

 

[patfla]

Hello rntsai

The dot product thing was just an analogy. Working my way upwards, as it were, from Linear Algebra I.

The operator in the Lie algebra is the commutator (generic term I think) or Lie bracket. There seem to be several (mathematical) formulations of the Lie bracket on Wikipedia. I think I like this one best

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

pat

 

[Berlin]

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!

Hi Emanual,
Joppe is my family name (Netherlands). I will of course not say to whom your name reminds me of :-)

Jan

 

[rntsai]

The dot product thing was just an analogy. Working my way upwards, as it were, from Linear Algebra I.

OK. There really us just one definition of "Lie bracket" (more jargon);
I prefer x \star y over [x,y] : this is R-bilinear and
x \star y = -y \star x
z \star (x \star y) = (z \star x) \star y + x \star (z \star y)

the last one is "product rule" or "Jacobi Identity" (more jargon);
other "definitions" using differentials, commutation, ... are just
representations of it; the above is more fundemental.

 

[rntsai]

Sorry too long. A single, specific question. This depends on some small part of the above being correct. Is it always two roots that combine to produce one particle/field? But wait I’ve probably gotten something (major) wrong. E8 has 240 symmetries, therefore particle/fields. And so 240 roots also. Well if you could only use each root once, then that would give only 120 particles. Maybe you can use a given root more than once?

It's one root per particle. There are 240 roots, so exactly 240 elementary particles (in this
theory at least); these include some 20 odd new particles not in the standard model
e8 has a lot more symmetries than 240; its reflection
group or "Weyl group" has order 696,729,600. This is a discrete group different
than the Lie group E8...

 

[Tony Smith]

Garrett said that the "... torsion question is especially interesting ...".

Here are some possibly useful things about torsion:
In hep-th/9601066, "Geometric Interpretation of Electromagnetism in a Gravitational Theory with Torsion and Spinorial Matter", Keniche Horie wrote:
"... General relativity is enlarged by allowing for an arbitrary complex linear connection and by constructing an extended spinor derivative based on the complex connection. Thereby the spacetime torsion not only is coupled to the spin of fermions and causes a four-fermion contact interaction, but the non-metric vector-part of torsion is also related to the electromagnetic potential. However, this long-standing relation is shown to be valid only in a special U(1) gauge ...".

Since Horie got U(1) gauge bosons from a complex connection with torsion,
you can speculate about whether:
a quaternionic connection with torsion could have produced SU(2) gauge bosons
and
an octonionic connection with torsion could have produced SU(3) (a subgroup of G2) gauge bosons.

Then you might think of looking at a CxQxO connection (2x4x8 = 64-dim)
and
see that is what Geoffrey Dixon has been using for model-building - see his book "Division algebras ..." to which there is a link at his web page at
http://www.7stones.com/Homepage/AlgebraSite/algebra0.html
which also has links to some of his other works.

As to how that might be related to E8 models,
consider that the 64-dim CxQxO thing might correspond to spinors ( Geoffrey Dixon writes about that in his book ),
and that two copies of it might correspond to the 64+64 = 128-dim Spin(16) half-spinors
inside E8.

Tony Smith

 

[rntsai]

Since Horie got U(1) gauge bosons from a complex connection with torsion,
you can speculate about whether:
a quaternionic connection with torsion could have produced SU(2) gauge bosons
and
an octonionic connection with torsion could have produced SU(3) (a subgroup of G2) gauge bosons.

Then you might think of looking at a CxQxO connection (2x4x8 = 64-dim) ...

As to how that might be related to E8 models,
consider that the 64-dim CxQxO thing might correspond to spinors ( Geoffrey Dixon writes about that in his book ),

Hi Tony,
I don't wish to offend in any way; I think you're pretty sharp guy and have one
of the most interesting websites around. That being said, why one would want to
bother with dealing with octanions for this and other problems completeley
escapes me. There's enough complexity/symmetry/structure with lie algebras
over reals to handle things...I can give up commutativity without problems,
but you lose so much by giving up associativity...is it really worth it?
(I know you lose associativity when you go to Lie algebras, but in that
case it really is worth it : the Jacobi identity is still primitive enough plus
you have the very natural commutation as standard model...)

 

[Tony Smith]

rntsai asked "... why one would want to bother with dealing with octanions ... you lose so much by giving up associativity...is it really worth it? ...".

Sorry that I gave that impression by assuming and not quoting details of how Geoffrey Dixon uses CxQxO which he denotes by T as a spinor space in his book cited in my comment.
As he says ( here I modify some of his notation, such as by denoting the 8x8 real matrix algebra by M(R,8), etc ) (pages 68, 40, 66-67, 84-85, 128)
"... T is not only nonassociative ... but ... also nonalternative ...
The left adjoint algebras are
CL = C
QL = Q
OL = M(R,8)
which imply the Clifford algebra isomorphisms
CL = Cl(0,1)
QL = Cl(0,2)
OL = Cl(0,6)
...
Let T = CxQxO and TL = CLxQLxOL
...
we view TL as the "Pauli" algebra ... from which the "Dirac" algebra ... will be built ...
TL = M(C,16) so we may identify it with the Clifford algebra Cl(0,9) ...
The object space of Cl(0,9) = M(C,16) is the space of 16x1 complex spinors ...
[ 2x2 matrices with entries in TL ]... M(TL,2) = C(32), which is the complexification of Cl(1,9)
So TL is, so-to-speak, the "Pauli" algebra to the ... "Dirac" algebra M(TL,2) ...
The object space of M(TL,2) is T2, 2x1 matrices over T. ...
The spinors of M(TL,2) are elements of the 128-dimensional T2,
the space of 2x1 columns over 64-dimensional T ...".

So, Geoffrey Dixon uses T2, two copies of nonassociative nonalternative T,
as a 128-dimensional spinor space analogous to the 128-dim part of E8
that is the fermionic/spinor half-spinor of Spin(16),
and
he operates on it with an associative Clifford algebra Cl(1,9),
the bivector Lie algebra of which is Spin(1,9) = SL(2,O) as described, for example, in John Baez's paper on Octonions at
http://math.ucr.edu/home/baez/octonions/

By looking at that Clifford algebra, Geoffrey Dixon gets representations of standard model things that may be usefully related to the standard model things that appear in E8 models.

At this point I refer to Geoffrey Dixon's book and papers for more details,
but
the basic point that I want to make is that the nonassociativity of octonions does not render them useless in physics model building because you can work with related associative Clifford algebras.

If you are going to build models based on E8 (or any other exceptional Lie group/algebra), you are effectively using octonions whether you explicitly acknowledge it or not.
For example, the basic structure E8 / Spin(16) = 128-dim half-spinor of Spin(16)
is the symmetric space (OxO)P2
which is the projective plane of the tensor product of two copies of the octonions,
which symmetric space is described by Boris Rosenfeld in his book "Geometry of Lie Groups" (Kluwer 1997).

As to whether or not it is "really worth it",
the worth of any physics model is what you can calculate with it,
and
it permits me to calculate particle masses, force strengths, etc that are substantialy consistent with experimental results, using similar model-building techniques.

Tony Smith

PS - It is not offensive at all to ask such questions. I learned most of what little I know by asking such questions over many years.