An Exceptionally Technical Discussion of AESToE

[MTd2]

Hello Mr. Garrett!,

I would like you to comment on this. The guy who pointed you some mistakes (Jacques Distler), made several more remarks about your theory, specially after DEC 9TH. He seems to be changing his mind every day, maybe he is confused.

He is still seems to be very hostile to your theory. But if you discuss with him, it would be profitable, as his got some new maths. He is saying you are not even getting the 1st generation right

http://golem.ph.utexas.edu/~distler/blog/archives/001505.html#AlittleU4
***************

Update (11/29/2007):
David Vogan, from MIT, wrote me to point out that I was too fast in saying that G does not embed in F 4×G 2. It is possible to find such an embedding, but it necessarily leads to a completely nonchiral “fermion” representation (and hence contains no copies of R). I simply didn’t bother considering such embeddings, when I was preparing this post. For the record, though
F 4(−20)⊃Spin(8,1)⊃Spin(3,1)×Spin(5)⊃SL(2,ℂ)×SU(2)×U(1)
and
F 4(4)⊃Spin(5,4)⊃Spin(3,1)×Spin(2,3)⊃SL(2,ℂ)×SU(2)×U(1)
In the latter case, one obtains
26=1+9+16 =(1,1) 0+(4,1) 0+(1,3) 0+(1,1) 2+(1,1) −2 +(2,2) 1+(2,2) −1+(2¯,2) 1+(2¯,2) −1 52=36+16 =(Adj,1) 0+(1,3) 0+(1,1) 0+(1,3) 2+(1,3) −2+(4,3) 0+(4,1) 2+(4,1) −2 +(2,2) 1+(2,2) −1+(2¯,2) 1+(2¯,2) −1
In the former case, there are two distinct embeddings of SU(2)×U(1)⊂Spin(5). For the one under which 4=2 1+2 −1, one obtains the same result as above. For the one under which 4=2 0+1 1+1 −1, one obtains
26 =2(1,1) 0+(4,1) 0+(1,2) 1+(1,2) −1 +(2,2) 0+(2,1) 1+(2,1) −1+(2¯,2) 0+(2¯,1) 1+(2¯,1) −1 52 =(Adj,1) 0+(1,3) 0+(1,1) 0+(4,1) 0+(1,1) 2+(1,1) −2+(1,2) 1+(1,2) −1+(4,2) 1+(4,2) −1 +(2,2) 0+(2,1) 1+(2,1) −1+(2¯,2) 0+(2¯,1) 1+(2¯,1) −1
Putting these, together with the embedding of SU(3)⊂G 2,
7 =1+3+3¯ 14 =8+3+3¯
into (3), one obtains a completely nonchiral representation of G.

Update (12/10/2007):
For more, along these lines, see here http://golem.ph.utexas.edu/~distler/blog/archives/001532.html

Correction (12/11/2007):
Above, I asserted that I had found an embedding of G with two generations. To do that, I had optimistically assumed that there is an embedding of SL(2,ℂ) in a suitable noncompact real form of A 4, such that the 5 decomposes as 5=1+2+2. This is incorrect. It is easy to show that only 5=1+2+2¯ arises. Thus, instead of two generations, one obtains a generation and an anti-generation. That is, the spectrum of “fermions” is, again, completely non-chiral. I believe (but haven’t proven) that this is a completely general result: for any embedding of G in either noncompact real form of E 8, the spectrum of “fermions” is always nonchiral. Let’s have a contest, among you, dear readers, to see who can come up with a proof of this statement.I apologize if I’d gotten anyone’s hopes up, with the above example. Not only can one never hope to get 3 generations out of this “Theory of Everything”; it appears that one can’t even get one generation.

*****************

And here is a post apparently claiming a final blow (not his words, but my emotional interpretation). A certain mark refers to Smolin and you almost as crackpots (again, not his words, but my emotional interpretation)

http://golem.ph.utexas.edu/~distler/blog/archives/001532.html#more

******************

There it is Garrett. Would you have some comments about that?

 

[garrett]

Hello MTd2,
This issue of non-compact subalgebras of non-compact real E8 is pretty tricky. Jacques is trying to pass this stuff off as obvious, but having a hard time doing that since he's been making mistakes. I did the calculations using compact real E8, and figured I could change the signature of part of the Killing form to get a non-compact version, by inserting an i in the roots -- but this was probably naive on my part. In the paper, I do use so(7,1)+so(8), and I though this was in E IX, but it isn't. Jacques asserted in a comment to his first post that so(7,1)+so(8) is in split real E8. This was news to me. Then, in his second post, he said so(7,1)+so(8) isn't in split real E8, as if I were the one who initially said it was. Also, in his second post, Jacques asserted that spin(12,4) was in split real E8 -- another mistake -- then he went back this morning and edited that out of the post, without noting his error.

This behavior makes me pretty wary. Despite his hostility and mistakes, I've learned a bit of useful math from the discussion with Jacques, and will see what I can do with it. I may be able to get things to work with so(7,1)+so(1,7), or with so(12,4), or I might have to try something more drastic. I already knew I was going to have to do something significantly different to get the second and third generations to work in this theory, so, really, not much has changed -- there are now just more clues.

 

[Coin]

garrett, a question wrt Distler's comments:

My understanding of lie groups is very limited and Distler's blog is very ranty so I've had a great deal of trouble picking out what exactly Distler is trying to say in his posts. However it does seem there is one specific important criticism he has made which I haven't seen addressed yet, which is his claim in his second post (which MTd2 quotes from above) where Distler claims that, even if you only attempt a single-generation embedding, the fermions one gets out of the E8 connection are nonchiral.

Has Distler found an actual problem with the E8 connection idea here? Or is this a problem which is real but which you had already forseen somewhere? Or would you say there is some reason that Distler's claim about E8 producing nonchiral fermions is either incorrect or misapplied?

 

[garrett]

Coin,
The Pati-Salam GUT I'm embedding in E8 is a http://en.wikipedia.org/wiki/Left-right_model" standard model -- but ways to do this are well established.

 

[Coin]

OK, thanks for the clarification.

 

[jal]

I want to highlight the effort of embending a link in the words that Garrett used ... ie. chiral ... this is not an exercise needed for the "math kids".
Thanks

 

[MTd2]

Nice, so your E(8) naturaly has massive and oscilating neutrinos. Maybe it can shed light on the doubts sorrounding the data from LSND and MiniBooNE experiments, that shows the possibility of sterile neutrinos.

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

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

Notice that are some anomalies detected in the low energy region of neutrinos, showing a high incidence of eneutrinos. That would be a "confirmation" of brane physics, that is, neutrinos "arriving" from other dimensions, the "bulk of the brane", and intersecting our "brane surface". Some string theorists are excited for that ("Bill Louis, of the MiniBooNE project, has emailed the brane theorists saying: "It is indeed startling to see how well your model appears to fit our excess of low energy events!" There remains the possibly that the effect is a spurious statistical or background anomaly and further analysis is underway."). But, maybe your theory can explain that anomaly without appealing to other dimensions.

 

[garrett]

MTd2,
Neutrino oscillation is going to be a large clue for further development of the theory, but right now the second and third generations aren't described well enough in the theory to make any predictions. Also, this is heading into physical speculation, and I'd like to keep this a technical discussion.

 

[Tony Smith]

In keeping with this being "a technical discussion",
here is something that I asked in response to a comment by Thomas Larsson over on Cosmic Variance in Sean's post "Garrett Lisi’s Theory of Everything!":

Could Garrett Lisi’s model be understood in terms of a 7-grading of e8 that was described in a sci physics research thread Re: Structures preserved by e8, in which Thomas Larsson said:

“… … e_8 also seems to admit a 7-grading,
g = g_-3 + g_-2 + g_-1 + g_0 + g_1 + g_2 + g_3,
of the form

e_8 = 8 + 28* + 56 + (sl(8) + 1) + 56* + 28 + 8* .

…[in]… the above god-given 7-grading of e_8 … g_-3 is identified with spacetime translations and one would therefore get that spacetime has 8 dimensions rather than 11. …”.

So, if you used g-3 for an 8-dim Kaluza-Klein spacetime,
could you see the 28* and 28 as the two copies of D4 used by Garrett Lisi to get MacDowell-Mansouri gravity from one and the Standard Model gauge bosons from the other
and
see the central sl(8)+1 being related to transformations of the 8-dim spacetime
(actually being a 64-dim thing that is substantially 8×8* ).

The even part of the grading would then be the 112 elements
28* + 8×8* + 28
and
the odd part of the grading would then be the 128 elements
8 + 56 + 56* + 8*
If the 8 and 8* are used for 8-dim Kaluza-Klein spacetime
so
could the 56 + 56* be used for fermion particles and antiparticles ?

Even if the above assignment needs improvement,
my basic question is

could Thomas Larsson’s 7-grading of e8 be useful in making Garrett Lisi’s model a realistic description of physics ?

Tony Smith

PS - My personal favorite interpretation of the e8 7-grading is a bit different from what I described above, but I altered it to fit Thomas Larsson's explicit idea that the 8 should correspond to a spacetime.

 

[rntsai]

where's the Z?

Some random questions on exchange particles that I hope are not too basic :

- What happens to the Z boson in your (and Pati-Salam) model;
It looks like the W^+ and W^- bosons show up as is, but
the Z is "replaced" by two new bosons : B_1^+,B_1^-;
The photon is burried somehere inside D2_{ew}; is it
W^3 + B_1^3 -sqrt(2/3)B_2 (page 11)

- I think you use circles as a suggestive notation for
"exchange particles". I can identify the purple and yellow
circles (proudly since age 5). I have trouble with the
green ones, do they correspond to anything that might be
more recongnizable?

 

[MTd2]

Garrett,

it seem Lee Smolin admited he is wrong, and admited that your theory do not include Pati -Salam model:

# Lee Smolin on Dec 15th, 2007 at 8:36 pm

Dear HIGGS

I see, if it is then just a terminological mixup that is of course fine for this issue. I don’t mind making mistakes in public-the time spent studying the Pati-Salam papers was my own and in any case worthwhile-but this shows to me the difficulty of arguing technical issues in the blog environment. Perhaps the experts could find a better way, probably off line, to go through the issues with Lisi point by point and reach a conclusion over the main issues. If so I’d be happy to be involved, so long as everyone involved was patient and professional and no one pretended that the representation theory of non-compact forms of E8 is child’s play.

Thanks,

Lee
# H-I-G-G-S on Dec 15th, 2007 at 10:15 pm

Dear Lee,

I’m glad that we cleared this up, and I appreciate that you admitted error,
in line with your earlier posting on the spirit of science requiring such acknowledgment. I don’t quite agree however that it was a “terminological mixup.” This makes it sounds like there was no real content to the debate, whereas in fact there was. The issue at hand was whether or not Lisi’s embedding contains the Pati-Salam model or not. Jacques showed that it does not. All I did was to provide some helpful clarification. In an earlier post you went on about how “Distler was largely wrong” and so forth, while as far as I can tell, everything he has said has either been correct, or when it was in error, the error was admitted and then clarified. Thus it would be much more appropriate for you to address your admission of error to him than to me. Perhaps if you did so his responses to you would in the future be more temperate.

It is true that blogs are far from the best place to argue technical issues. This discussion was one of the happy exceptions where a point was argued and resolved with all parties in agreement. As for Lisi’s proposal, I believe a conclusion has been reached by the experts.

H
http://cosmicvariance.com/2007/11/16/garrett-lisis-theory-of-everything/

 

[moveon]

Quote:
Now it seems clear that the Coleman-Mandula question pertains only to the unbroken theory, or to put it another way, that in the broken theory, although the connection is still E8-valued, the action is no longer E8-symmetric.

That's right.

If your connection would be valued in the algebra, its would be expandable in the generators of E8. But some components of your connection are fermionic and thus anticommute, or? How can these possibly satisfy any E8 commutation relations? And if not, what on Earth has your construction then to do with E8?

As I was writing over at CV, this is completely different to symmetry breaking (where the proper commutation relations are still satisfied, though non-linearly realized).

 

[garrett]

Tony,
There are many gradings of E8, most of them interesting. I haven't thought about 7-gradings much. My favorite grading of E8 is a 13 grading corresponding to weak hypercharge -- which currently only works correctly for the first generation.

rntsai,
The Z and the photon fields are (rotated) combinations of W, B_1, and B_2. Specifically, as 1-form coefficients,
$$\underline{Z} = \sqrt{\frac{5}{8}}\underline{W}^3 - \sqrt{\frac{3}{8}} (\sqrt{\frac{3}{5}} \underline{B}_1^3 + \sqrt{\frac{2}{5}} \underline{B}_2)$$
and
$$\underline{\gamma} = \sqrt{\frac{3}{8}}\underline{W}^3 + \sqrt{\frac{3}{8}}\underline{B}_1^3 + \sqrt{\frac{2}{8}}\underline{B}_2$$
The $$B_1^\pm$$, and a leftover
$$\underline{X} = \sqrt{\frac{2}{5}} \underline{B}_1^3 - \sqrt{\frac{3}{5}} \underline{B}_2$$
are "new" gauge fields, as in Pati-Salam. (I'm pretty sure I have those right, but I haven't confirmed them.)

The circles are all gauge fields: green for gravitational $$\omega_{L/R}^{\wedge/\vee}$$, yellow for weak $$W^{\pm}$$, blue for gluons, and white for $$B_1^\pm$$. The Z, photon, and X are in the Cartan subalgebra at the origin, and are conventionally not plotted.

MTd2,
H-I-G-G-S was twisting Lee's words, as is clear from his reply (which was visible when you posted).

moveon,
The connection starts out as an E8 valued 1-form. The action (with E8 symmetry broken by hand in my paper, but not in Lee's) introduces dynamical terms for the D4+D4 part of E8, but leaves only the BF term for the rest of E8. These pure gauge degrees of freedom may be replaced by Grassmann fields valued in the non D4+D4 part of E8 -- these are fermions. The resulting generalized connection consists of Lie(E8) valued 1-forms and Grassmann numbers, with the same E8 brackets between Lie algebra elements. This is a fairly standard mathematical construction, used in an unusual way. I'd be happy to expand on it for you or provide references.

 

[Tony Smith]

Garrett, exactly what is the 13-grading of e8 that you like to use?

Tony Smith

 

[CarlB]

Let me make a guess for garrett. The 13 is a weak grading, so it's going to correspond to the weak hypercharge quantum numbers of the standard model, that is, it will use the 13 values: (-1, -5/6, -2/3, -1/2, -1/3, -1/6, 0, +1/6, +1/3, +1/2, +2/3, +5/6, +1). To see the assignment, I would start by looking for the weak hypercharge quantum numbers assignment in his paper. Then you assign a particular root to a blade according to its weak hypercharge quantum number.

My recollection of the standard model is that the +- 5/6 quantum numbers are missing. These blades would be particles that don't appear in the standard model. But my concentration has always been on the fermions -- are there some bosons with weak hypercharge +- 5/6?

The peculiar pattern of the weak hypercharge quantum numbers that are actually used in the standard model, that is, leaving off the +- 5/6, has 11 values. Since I'm a density matrix proponent, (which are bilinear rather than the usual state vector formalism which is linear) I'm going to link in a paper which gives those 11 values, rather than all 13, as a solution to a bilinear equation. See chapter 5: http://www.brannenworks.com/dmfound.pdf

 

[garrett]

Tony,
If we rotate the E8 root system until the vertical axis is weak hypercharge, and rotate out the other axes horizontally to separate the roots a bit, it looks like this:
http://deferentialgeometry.org/blog/hyper.jpg
This makes it visible exactly what is meant by "the charge assignments only work correctly for the first generation," with the other two (smaller triangles) related by triality.

 

[Tony Smith]

When I count the 13-grading from that image I get:

5 + 6 + 15 + 20 + 30 + 30 + 26 + 30 + 30 + 20 + 15 + 6 + 5

which only add to 238, so I must be miscounting two of them somewhere ?

Anyhow, modulo my error of two, the even graded structure would be

5 + 15 + 30 + 26 + 30 + 15 + 5 = 126-dimensional
(if the missing 2 are even, then 128-dimensional)

and the odd graded structure would be

6 + 20 + 30 + 30 + 20 + 6 = 112-dimensional

so

it seems to me that the odd gradings correspond to the 112 root vectors of the adjoint Spin(16) (120 generators - 8 Cartan subalgebra generators = 112)

and

that the even grading probably have the two I miscounted and are the 128 root vectors corresponding to the half-Spinor of Spin(16).

What bothers me about that is that the fermionic spinor-type things are in the even grading and the bosonic vector/bivector adjoint-type things are in the odd grading,

whereas in Thomas Larsson's 7-grading

8 + 28* + 56 + (sl(8) + 1) + 56* + 28 + 8* = 8 + 28 + 56 + 64 + 56 + 28 + 8

the even grade part is
28 + 64 + 28 = 112 dimensional corresponding to the root vectors of adjoint Spin(16) which seems to represent bosonic vector/bivector stuff
while
the odd grade part is
8 + 56 + 56 + 8 = 128-dimensional corresponding to half-spinor of Spin(16) which seems to represent fermionic spinor-type stuff.

Do you have any thoughts about that ?

Tony Smith

 

[garrett]

Tony,
Two gluons overlap. There are many such gradings of E8 -- there may be the same kind of 13 grading along a different direction that gives the even/odd 120/128 split you're after.

 

[sambacisse]

no chiral embedding

Hi Garrett,

I would like to have your comments on http://golem.ph.utexas.edu/~distler/blog/archives/001532.html" ).

I personally feel Distler's argument is fundamental, relatively easy to follow, and seems to be correct, at least up to the level of my knowledge (perhaps I'm making a mistake). Lee has been trying to address it on Cosmic Variance, but hasn't succeeded in finding a mistake or a loophole in it yet. Do you have anything to say about it?

Thanks a lot! :-)

 

[moveon]

The resulting generalized connection consists of Lie(E8) valued 1-forms and Grassmann numbers, with the same E8 brackets between Lie algebra elements. This is a fairly standard mathematical construction, used in an unusual way. I'd be happy to expand on it for you or provide references.

Please! To my knowledge the only algebras that contain both bosonic and fermionic generators are superalgebras, and E8 is not one of them. How can the commutation relations close into E8?

 

[Tony Smith]

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]

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]

This from Tony's website might be good to look at (I'm sure Tony can say more if needed):

http://www.valdostamuseum.org/hamsmith/stringbraneStdModel.html

 

[Thomas Larsson]

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]

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]

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, $$\underline{H}_1+\underline{H}_2$$, and three other fields, the first of which is the first generation fermions, $$\Psi$$, which are Grassmann valued E8 Lie algebra elements. Because of the structure of E8, the Lie brackets between these give the fundamental action:
$$[\underline{H}_1+\underline{H}_2,\Psi] = \underline{H}_1 \Psi - \Psi \underline{H}_2$$
The brackets between two $$\Psi$$'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:
$$\underline{H}_1+\underline{H}_2+\Psi$$
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]

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]

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.

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]

moveon, thanks for that; a very very illuminating comment.

 

[John G]

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]

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.