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An Exceptionally Technical Discussion of AESToE

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Tony Smith
#91
Dec19-07, 09:40 AM
P: 40
moveon said "... the only algebras that contain both bosonic and fermionic generators are superalgebras ...".

No, that is not true.
As Pierre Ramond in hep-th/0112261 said "... exceptional algebras relate tensor and spinor representations of their orthogonal subgroups ...",
and
the exceptional algebra 248-dim E8 contains
120 generators corresponding to the tensor/bosonic part 120-dim adjoint Spin(16)
and
128 generators corresponding to the spinor/fermionic part 128-dim half-spinor Spin(16)
and
their commutation relations do close into E8.

However, Pierre Ramond went on to say in that paper:
"... Spin_Statistics requires them [ the adjoint/bosonic and half-spinor/fermionic ] to be treated differently ...",
so
any model you build with E8 must somehow treat them differently.

For example, you might just construct a Lagrangian into which you put
the 128 half-spinor fermionic generators into a fermion term
and
8 of the 120 bosonic generators into a spacetime base manifold term
and
120-8 = 112 of the 120 bosonic generators into a gauge boson curvature term.

Then you might have disagreement as to how natural (or ad hoc) is such an assignment of parts of E8 to terms in a Lagrangian,
but all should agree that you have "treat[ed] them differently" as required by Spin-Statistics.

However, in Garrett's 13-grading decomposition of the 240 root vectors of E8

5 + 6 + 15 + 20 + 30 + 30 + 28 + 30 + 30 + 20 + 15 + 6 + 5

some of the graded parts contain both bosonic terms and fermionic terms,
for example the central 28 has both circles (bosons) and triangles (leptons and quarks),
which has led Thomas Larsson to complain (on Cosmic Variance):
"... both fermions and bosons belong to the same E8 multiplet. This is surely plain wrong. ...".

I think that the point of Thomas Larsson is that
the model must treat the fermions and bosons differently to satisfy Spin-Statistics
so
the fermionic generators must be put into some part of the model where the bosonic generators are not put
so
if you decompose the generators into multiplets some of which contain both fermionic and bosonic generators (as in Garrett's 13-grading decomposition) then you are not respecting your multiplets when you, from a given multiplet, put some of them into a fermionic part of the model and some of them into a bosonic part of the model.

This is not merely an objection of ad hoc assignments of generators to parts of the model,
it is an objection that the assignments do not respect the chosen decomposition into multiplets.

Tony Smith

PS - It is possible to choose a decomposition that does keep the bosonic and fermionic generators separate, the simplest being 64 + 120 + 64
where the 120 is bosonic and the 64+64 = 128 is fermionic.
moveon
#92
Dec19-07, 10:12 AM
P: 9
Quote Quote by Tony Smith View Post
moveon said "... the only algebras that contain both bosonic and fermionic generators are superalgebras ...".

No, that is not true.
As Pierre Ramond in hep-th/0112261 said "... exceptional algebras relate tensor and spinor representations of their orthogonal subgroups ...",
and
the exceptional algebra 248-dim E8 contains
120 generators corresponding to the tensor/bosonic part 120-dim adjoint Spin(16)
and
128 generators corresponding to the spinor/fermionic part 128-dim half-spinor Spin(16)
and
their commutation relations do close into E8.

Oh yes, this is of course very well known since ages. But those tensor and spinor rep generators are all bosonic, and close into the usual E8 commutator relations. My point is, apparently still not appreciated, that if some of the generators are made fermionic (as it happens for superalgebras), then they cannot produce the E8 commutation relations (and jacobi identities etc) any more. The opposite seems to be claimed here all over, so I'd like to see, how. Please prove this by writing them down!

And if the E8 commutation relations are not there, there is no E8 to talk about. There is "somewhat" more to E8 than a drawing of the projection of its polytope...
John G
#93
Dec19-07, 10:51 AM
P: 19
This from Tony's website might be good to look at (I'm sure Tony can say more if needed):

http://www.valdostamuseum.org/hamsmi...eStdModel.html
Thomas Larsson
#94
Dec19-07, 01:45 PM
P: 102
One needs to distinguish between spin and statistics.

There are two types of statistics: fermions, which anticommute and obey Pauli's exclusion principle, and bosons, which commute.

There are also two types of spin: spinors, which have half-integer spin, and tensors and vectors, which have integer spin.

The spin-statistics theorem asserts that physical fermions always have half-integer spin and physical bosons have integer spin. But this is non-trivial and surprisingly difficult to prove. In contrast, BRST ghosts are fermions with integer spin, and therefore unphysical. Physical and unphysical fermions are not the same.

What is quite easy to prove is that statistics is conserved, i.e.

[boson, boson] = boson
[boson, fermion] = fermion
{fermion, fermion} = boson.

People like Lee, Peter and Bee know this, of course, and it must be obvious that putting both bosons and fermions into the same E8 multiplet violates this fundamental principle. That they don't emphasize this simple fact but instead complain about manners is something that I find surprising and quite disappointing.
Tony Smith
#95
Dec19-07, 02:50 PM
P: 40
Here is what I hope is a concrete example of what I think that Thomas Larsson is saying (please feel free to correct my errors):

If you were to (not what Garrett did) make a physics model by decomposing E8 according to its e17 5-grading:

g(-2) = 14-dim physically being spacetime transformations
g(-1) = 64-dim physically being fermion antiparticles
g(0) = so(7,7)+R = 92-dim physically being gauge bosons
g(+1) = 64-dim physically being fermion particles
g(+2) = 14-dim physically being spacetime transformations

then that would be consistent with spin-statistics because
the products fermion(-1) times fermion(+1) would be gauge bosons(-1+1=0)
the products of gauge bosons(0) times gauge bosons(0) would be gauge bosons(0+0=0)
the products of gauge bosons(0) times fermions(-1) would be fermions(0-1=-1)
the products of gauge bosons(0) times fermions(+1) would be fermions(0+1=+1)

etc

The point is that if you have fermions and bosons mixed up together in the same part of the graded decomposition, you do not get good spin-statistics,
but
it is possible to decompose in a way that you do get good spin-statistics
and
that is something that should be taken into account in model-building.

Tony Smith

PS - Sorry for burying stuff like fermion(-1) times fermion(-1) giving spacetime(-2) into an "etc" (sort of like spinor x spinor = vector) but in this comment I am just trying to make a point and not build a complete model here.
garrett
#96
Dec19-07, 07:21 PM
garrett's Avatar
P: 360
sambacisse,
The issue is a bit more complicated than it appears because of how the real representations are mixed together in exceptional groups into complex representation spaces, relying on an inherent complex structure. This sort of thing is described halfway through John Baez's TWF253 for the case of E6. When describing so(3,1) reps in terms of sl(2,c) this is further complicated, and when swapping in conjugated anti-fermions it's more complicated still -- because one has to be clear in each step which complex structure one is conjugating with respect to. I thought I had this figured out several years ago, but I don't like to make statements about complicated things without having slowly worked through them in detail. So I've stayed out of the arguments. Of course, I can say that the worst case scenario is that one might have to use a complex E8.

moveon,
Tony addressed this a bit, and I'll try to summarize the specific case in the paper. The E8 Lie algebra may be naturally decomposed into a D4+D4 subalgebra, and everything else. In terms of the number of elements, this decomposition is:
(28+28)+64+64+64
which I don't consider a "grading," but it relates to gradings. The important thing is the Lie brackets. If we label the D4+D4 elements "bosons," and the rest "fermions," the brackets are as Thomas Larsson has helpfully described. Now, if the E8 symmetry is broken such that the "fermion" part of the Lie algebra is pure gauge, then that part of the connection may be replaced by Lie algebra valued Grassmann fields. We end up with a D4+D4 valued connection 1-form field, [tex]\underline{H}_1+\underline{H}_2[/tex], and three other fields, the first of which is the first generation fermions, [tex]\Psi[/tex], which are Grassmann valued E8 Lie algebra elements. Because of the structure of E8, the Lie brackets between these give the fundamental action:
[tex][\underline{H}_1+\underline{H}_2,\Psi] = \underline{H}_1 \Psi - \Psi \underline{H}_2[/tex]
The brackets between two [tex]\Psi[/tex]'s are in D4+D4, but these terms vanish in the action. Notice that there is no symmetry here relating the fermions to bosons. That symmetry was destroyed when we broke the E8 symmetry by adding the terms we did to the action. I did that by hand in my paper, and Lee talks about how that can happen dynamically in his. There is a cute trick in the BRST literature whereby these objects can be formally added in a generalized connection:
[tex]\underline{H}_1+\underline{H}_2+\Psi[/tex]
Since I like cute math tricks, I used it -- allowing all fields to be written as parts of this "superconnection," with the dynamics coming from its generalized curvature.
Tony Smith
#97
Dec19-07, 08:47 PM
P: 40
So, it seems to me that:

1 - Garrett has shown that his physical identifications of E8 generators are consistent with spin-statistics;

2 - Garrett is not claiming that any BRST ghost-fermions-with-integer-spin are physical,
but
is just using one of the technical "math tricks" from BRST literature in order to construct his "superconnection" containing both gauge boson curvature terms and curvature terms derived from spinor/fermions;

3 - Garrett has explicitly broken full E8 symmetry so that it is irrelevant whether or not Garrett's physics stuff (whether it is Pati-Salam or not) fits inside E8,
so that Jacques Distler's arguments about it not fitting inside E8 are irrelevant.

4 - However, just as Jacques Distler's comments were useful in seeing that E8(8) might be more useful than E8(-24),
it may be that his comments about Pati-Salam vs. the Standard Model might also be useful indicators that Garrett's model should perhaps be put directly in terms of the minimal Standard Model than in terms of Pati-Salam.

Tony Smith

PS - If I had to guess, I would guess that Garrett used Pati-Salam because he thought that it was an established particle physics model, and its use would make his E8 model more acceptable to conventional physicists.
Since it has turned out otherwise, maybe just using the plain vanilla minimal Standard Model plus MacDowell-Mansouri gravity might be a way to go.

PPS - It is unfortunate that a "food-fight" atmosphere has obscured much of the sensible physics in discussions on some parts of the web, and I would like to say that I very much appreciate the moderate (in more meanings than one) atmosphere here on Physics Forums. Such moderation-in-climate does not come about without moderation-in-the-other-sense, and that takes effort, which I appreciate very much.
moveon
#98
Dec20-07, 05:03 AM
P: 9
Garret,

OK so let me translate this in my language.. your superconnection does not take values in the Lie algebra of E8 as some generators are fermionic (they square to zero, eg).
Therefore the curvature, or field strength does not take values in all of E8, but in D4+D4 only. The full commutation relations of E8 are therefore not non-trivially realized. So in what sense then does E8 play a role? It seems that the purpose of your E8 is to organize, as a bookkeeping device, the fermionic part of the spectrum in terms of the coset E8/(D4+D4), as far as their quantum numbers are concerned.

This is linked to the "breaking" of E8. There are different notions of a symmetry being broken. Usually in particle physics a symmetry is spontaneously broken, which means it is "still there" albeit non-linearly realized. It reflects itself in terms of Ward identities of the low energy effective theory.
There is an energy scale above which the symmetry is restored and the theory is in an "unbroken phase". So one may speak of an "underlying" symmetry.

In contrast, you write a theory where there is no E8 symmetry to begin with (ie, its commutation relations are not fully realized) and there is no energy scale above which it is restored. So calling it "breaking" may be misleading...it is just not there. It is a bit like saying the standard model has monster group symmetry, although most of it is broken.


Quote Quote by Tony Smith View Post
So, it seems to me that:

1 - Garrett has shown that his physical identifications of E8 generators are consistent with spin-statistics;

.....

3 - Garrett has explicitly broken full E8 symmetry so that it is irrelevant whether or not Garrett's physics stuff (whether it is Pati-Salam or not) fits inside E8,
so that Jacques Distler's arguments about it not fitting inside E8 are irrelevant.

To 1- .... they are not the generators of E8. They are the generators of some superalgebra whose bosonic piece is D4+D4.


To 2- .... it seems to me that the claim was that that the standard model spectrum can be organized in terms of E8/(D4+D4) (rather, of the relevant non-compact real forms). That has been shown by Distler not to be the case.


I would thus 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. Also, superalgebras are consistent with Coleman-Mandula (that's why supergravity works).
mitchell porter
#99
Dec20-07, 05:59 AM
P: 751
moveon, thanks for that; a very very illuminating comment.
John G
#100
Dec20-07, 12:15 PM
P: 19
Quote Quote by moveon View Post
In contrast, you write a theory where there is no E8 symmetry to begin with (ie, its commutation relations are not fully realized) and there is no energy scale above which it is restored. So calling it "breaking" may be misleading...it is just not there. It is a bit like saying the standard model has monster group symmetry, although most of it is broken.

To 1- .... they are not the generators of E8. They are the generators of some superalgebra whose bosonic piece is D4+D4.

To 2- .... it seems to me that the claim was that that the standard model spectrum can be organized in terms of E8/(D4+D4) (rather, of the relevant non-compact real forms). That has been shown by Distler not to be the case.
The full E8 symmetry would seem to be E8/D8, I personally am more familiar with E8/E7xSU(2) and so on down the A-D-E series but maybe one can do something with E8/D8. The D4+D4 part seems after symmetry breaking so one should not expect any E8/(D4+D4) physics.
Tony Smith
#101
Dec20-07, 02:46 PM
P: 40
moveon "... advise[s] 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
#102
Dec20-07, 07:01 PM
garrett's Avatar
P: 360
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
#103
Dec21-07, 01:16 AM
Sci Advisor
HW Helper
CarlB's Avatar
P: 1,204
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 [tex]\langle 0 | stuff | 0\rangle[/tex]. Then you take the squared magnitude, that is, you multiply your amplitude by its complex conjugate:

[tex]\langle 0 | stuff | 0\rangle \langle 0 | stuff^* |0\rangle [/tex]

Now suppose you commute two creation operators in "stuff" and get a minus sign. That minus sign is cancelled 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 [tex]\psi[/tex] that you want to give "fermion statistics" to. Add a term to the Hamiltonian of [tex]\kappa\;\psi\psi[/tex]. Let [tex]\kappa \to \infty[/tex] 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 [tex]k^*n^*u^*j^*|0\rangle[/tex]. Then the average of the operator Q over this quantum state is:
[tex]\langle 0 | j\; u\; n\; k\; | Q |\;k^*\;n^*\;u^*\;j^*| 0\rangle[/tex]
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 [tex]j^*\;u^*\; - u^*\;j^*\;|0\rangle[/tex]. 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
#104
Dec21-07, 06:58 PM
P: 587
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
#105
Dec21-07, 08:52 PM
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P: 360
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
#106
Dec21-07, 09:32 PM
P: n/a
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
#107
Dec21-07, 09:47 PM
garrett's Avatar
P: 360
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
#108
Dec21-07, 11:21 PM
P: n/a
Quote Quote by garrett View Post
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


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