An Exceptionally Technical Discussion of AESToE

garrett

Gold Member
The media frenzy is dying down a bit, and I'm opening this thread to discuss technical questions from researchers and students as they read this paper.

I'd like the discussion to involve short questions and statements involving equations so it can be quick and conversational. The main purpose of the thread is to elucidate some of the unusual math and notation used, such as vector-form contraction,
$$\vec{v} \underline{f} = v^i \vec{\partial_i} \underline{dx^j} f_j = v^i f_i$$
To help explain things, I will often refer people to specific pages of the Deferential Geometry wiki. (And if the discussion gets really good, I'll add stuff there.)

I'll be directing several new people to this discussion -- math and physics professors as well as students -- and I hope many find it interesting. One advantage here is the ability to typeset TeX by surrounding it in [ t e x ] and [ / t e x ] (without the spaces). If your post does not involve math, it is probably inappropriate for this thread. Please do not start physics debates here -- there are many other places for that; this thread is mostly about the mathematical tools, tricks, and notation used, with connections to physics as appropriate. I hope the techniques discussed will be of use beyond this paper. I expect questions and discussion from all levels, and tangents are OK, so don't be shy.

Best,
Garrett

Related Beyond the Standard Model News on Phys.org

josh1

Hi Garret,

Would you mind posting a description of the problems with your paper?

morbello

Sounds interesting

garrett

Gold Member
Hi Josh,
The physics problems with the theory are all discussed in the paper. And this thread is not the appropriate place for listing or discussing them -- this thread is for discussing the math in the paper. If you are asking for a list of specific errata, then this is the appropriate place, and here is the answer:

The matrix at the bottom of page 18 needs a $$\frac{1}{\sqrt{2}}$$ scaling in front.

The matrix at the bottom of page 22 has a $$\frac{-1}{\sqrt{2}}$$ that should be a $$\frac{-1}{2}$$.

On page 29, in the discussion, the non-compact version of E8 used was misidentified as $$E \, IX$$ when it is in fact the split real form, $$E \, VIII$$.

These were caught thanks to public collaborative peer review, and will be corrected in a revision.

rntsai

g2 <-> su(3) + 3 + 3'

I have a question on the g2-su(3) relation (section 2.1 of paper).
Looking at the diagram I can see how the weights of 3,3',and 8
irreps of su(3) have the same coordinates as the g2 roots. So
there's a correspondance between three irreps of su(3) (maybe
a2 is better here) and one irrep of g2 :

g2 <-> su(3) + 3 + 3'

g2(0,1) <-> a2(1,1) + a2(1,0) + a2(0,1)

(notation : g2(0,1) = 14 dim adjoint rep of g2, g2(1,0) is the 7 dim rep
a2(1,1) = 8 dim ajoint rep of a2, a2(1,0) and a2(0,1) are the 3 dim reps)

My question is how do you define the "+" above between the a2 reps?
I can get explicit 8x8 matrices for a2(1,1), 3x3 mats for a2(1,0) and
3x3 mats for a2(0,1). How do you combine these into 14x14 mats of g2(0,1)?
direct sum isn't it since that just gives you a2. Any other explicit description
of this correspondance woul be helpful.

I hope this question is appropriate for this thread.

garrett

Gold Member
Hello rntsai,
Good question. The Lie algebra and representation spaces here are being treated as vector spaces, and the "+" is a direct sum of vector spaces (which is often written as "$$\oplus$$"). To see how this works explicitly with specific representations, eq(2.3) on p6 shows how the su(3), 3, and 3' subspaces can be represented within a 7x7 matrix representation of g2.

rntsai

Hello rntsai,
Good question. The Lie algebra and representation spaces here are being treated as vector spaces, and the "+" is a direct sum of vector spaces (which is often written as "$$\oplus$$"). To see how this works explicitly with specific representations, eq(2.3) on p6 shows how the su(3), 3, and 3' subspaces can be represented within a 7x7 matrix representation of g2.

Thanks Garry,

I didn't read ahead to the 7 dimensional rep until I understood
the embedding of su(3) in g2 better; I thought the 14 dim adjoint
rep of g2 was more relevant, but it looks like this embedding is
independant of which g2 rep you work with.

Let me collect my understanding of the mapping here.

g2 is generated by 14 elements : g2=<h1,h2,e1,e2,e3,e4,e5,e6,f1,f2,f3,f4,f5,f6>

h1,h2 : cartan algebra generators : g^3,g^8 in paper
e1,e2,e3 : long positive roots : g^{rb'},g^{rg'},g^{bg'}
f1,f2,f3 : long negative roots : g^{r'b},g^{r'g},g^{b'g}
e4,e5,e6 : short positive roots : q^{r},q^{g},q^{b}
f4,f5,f6 : short negative roots : q^{r'},q^{g'},q^{b'}

<e1,e2,e3,f1,f2,f3> generate 8 dim subalgebra; Levi-Malcev decomposition : 8 dim (a2)
<e4,f5,f6> generate 8 dim subalgebra; Levi-Malcev decomposition : 3 dim (a2) + radical
<f4,e5,e6> generate 8 dim subalgebra; Levi-Malcev decomposition : 3 dim (a2) + radical

Things seem to fit, but I don't know how <e4,f5,f6> is associated with rep 3 for
example and <f4,e5,e6> with 3'. I know the root coordinates are very suggestive of
3 and 3', but is there another path?

Cold Winter

Garrett:

One thing about this whole thread, is that if any of this is valid, it took a computer to do it. So I'll ask. Anybody know where the specifications for this particular model might be?

Personally, I'd love to write this up in "C" code. Suspect it might be more useful to all than the original program used to do this.

Y'all realize, that as of your new theory we have arrived at a point where a computer is needed to do the math, and no single human being will ever get a complete handle on all this??? garrett

Gold Member
rntsai,
Everything you've said is correct -- and since these are Lie algebra elements, the decomposition is representation independent. The minimal 7x7 matrix rep of G2 is convenient. Using this rep, we can take any element of G2 corresponding to an element of a2 and compute the Lie bracket (anti-symmetric matrix product) with any element of G2 corresponding to a quark root vector. The result will be a quark root vector, the same as if we acted on the original quark in a 3 with an a2 element. In eq(2.3) the matrix rep has been written using the Cartan-Weyl basis, so these calculations are easier.

You are right that the root system is not the whole story -- since root addition doesn't precisely determine the result of brackets that land in the Cartan subalgebra. To get the whole story we have to work with some representation, such as the 7x7 matrix. This works, but for E8 the matrices are more cumbersome and it's more efficient to say what we can just using the roots. But the main point is that there are many paths, since the quarks as well as the gluons are identified as Lie algebra elements. If we wanted to be really wild, we could even calculate the Lie derivatives between gluons and quarks as vector fields on the G2 group manifold. That would be a purely geometric description. But it's much easier to work with a matrix representation, and easier still to work with the roots.

Hello Cold Winter,
(It's snowing outside my window.) I think computers have been used to do fundamental physics for quite a while now. Personally, I have a very large Mathematica notebook related to this paper.

Last edited:

Cold Winter

... Personally, I have a very large Mathematica notebook related to this paper.
Mathematica is nice, but for this particular excercise, I imagine the entire E8 entity would run ( a lot ) faster in "C". Any ideas?

Snow here too. Good for coding.

garrett

Gold Member
Cold Winter,
Lots of ideas -- but what is it you want to do?

samalkhaiat

Hi,

Have you worked out how the relevant E(8)-Noether currents depend on the new fields $\undeline{x}\Phi$?

Regards

Sam
(an Exceptionally lazy person)

garrett

Gold Member
Hi Sam,
Some others have been playing with current algebra, but I haven't, no.

rntsai

Thanks Garrett. The relation between g2 and a2 is getting clearer.
Letting :

g2=<h1,h2,e1,e2,e3,e4,e5,e6,f1,f2,f3,f4,f5,f6>
a2=<h1,h2,e1,e2,e3,f1,f2,f3>= subalgebra of g2 isomorphic to a2

V8 ={h1,h2,e1,e2,e3,f1,f2,f3} 8 dim vector subspace of g2, this is also a subalgebra
V3 ={e4,f5,f6} 3 dim vector subspace of g2, this is not a subalgebra
V3'={f4,e5,e6} 3 dim vector subspace of g2, this is not a subalgebra

Then g2=V8+V3+V3' as vector space direct sum

a2*V8 = V8 action gives an 8 dim rep of a2
a2*V3 = V3 action gives a 3 dim rep of a2
a2*V3'= V3' action gives a 3 dim rep of a2

I was able to explicitely verify that these three subspaces are indeed closed under
the action of a2. I use GAP software for such calculations; running on a generic PC.
On to the next page!

rntsai

Moving to the decomposition so(6) decomposition (page 7,eq. 2.5) :

so(6)=su(4)=u(1)+su(3)+3+3' -> u(1)+g2

I don't think g2 occurs as a subalgebra of so(6) so it's hard to
interpret the u(1)+g2. Skipping over this, so(6)=su(4)=d3=a3.

a3=<h1,..,h3, e1,...,e6, f1,... f6> (15 generators of a3)
a2=<h1,h2,e1,e2,e1+e2,f1,f2,f1+f2> a2 subalgebra of a3

V8={h1,h2,e1,e2,e1+e2,f1,f2,f1+f2} invariant subspace (a2*V8=V8)
V3={e3,e1+e3,e2+e3} (a2*V3=V3)
V3'={f3,f1+f3,f2+f3} (a2*V3'=V3')

altogether these give a 14 dimensional subspace; there's one more
that corresponds to the u(1). I have trouble identifying this one.
Any suggestions?

rntsai

> a2=<h1,h2,e1,e2,e1+e2,f1,f2,f1+f2> a2 subalgebra of a3

This should be
a2=<h1,h2,e1,e2,e4,f1,f2,f4> a2 subalgebra of a3
(e4 corresponds to root that is the sum of roots 1 and 2,that confused the notation)

> V8={h1,h2,e1,e2,e1+e2,f1,f2,f1+f2} invariant subspace (a2*V8=V8)
> V3={e3,e1+e3,e2+e3} (a2*V3=V3)
> V3'={f3,f1+f3,f2+f3} (a2*V3'=V3')

These should be
V8={h1,h2,e1,e2,e4,f1,f2,f4} invariant subspace (a2*V8=V8)
V3={e3,e5,e6} (a2*V3=V3)
V3'={f3,f5,f6} (a2*V3'=V3')

garrett

Gold Member
rntsai,
You are correct that g2 is not an embedded subalgebra of so(6). This is why I have used the arrow, "$$\rightarrow$$," instead of an equality. I describe this in the paper as a "projected" subalgebra. The relevant result, and what is meant, is exactly what you describe -- within a3, the a2 acts on the V3 to take it into V3, and the a2 acts on the V3' to take it into V3'. This is all we need for gluons acting on quarks. And we can also see this by adding the roots in the tables. The u(1) here is the generator in the Cartan subalgebra of a3 orthogonal to the a2. I'm not certain in your notation, but I think the generator for this u(1) is your h3.

rntsai

Hi Garrett,
I use the Chevalley basis for these calculations :

h1,..,hr : generators of Cartan subalgebra, r=rank of algebra
e1,...er...em : correspond to positive roots, e1...er are the simple positive roots
f1,...fr,..fm : correspond to negative roots, f1...fr are the simple negative roots

so you're right in that h3 is the third generaor of the Cartan subalgebra. The problem
is that this isn't invariant under the a2 subalgebra. Fortunately I found a way to get
a 1 dim subspace that is :

V1={h1+2h2+3h3}; a2*V1=V1

actually a2*V1=0, <V1> is the centralizer of a2 in a3; this is actually how I calculated it.

The next projection looks trickier (Table 2, eq. 2.6). so(7)=b3 has 21 roots: 3 cartan +
9 positive + 9 negative root. Looking at Table 2 : there are 6 g^{xy} 6qI,6qII,and 6qIII
and two l's. I'm guessing that there are three different decompositions of so(7) : all
include g^{xy} and then one of the three qI,qII,qIII sets...Is this going along the right
path?

Cold Winter

Cold Winter,
Lots of ideas -- but what is it you want to do?
I'm looking to mechanize E8 in a manner that anyone can use ( without 16way Opteron based computers ). Then I'm wondering at some method of "posing questions" to the model.

garrett

Gold Member
rntsai,
Yes, this is the right path. The qI relate to the first generation quarks. The qII and qIII relate to second and third generation quarks and new particles -- but these second and third generation assignments are very speculative at this point. Things in this paper are only currently described perfectly for the first generation, with the second and third generation included using handwaving.

Cold Winter,
The most fun you'll be able to have is writing fast code for spinning and displaying the E8 root system with particle labels. Or maybe cranking out lists of allowed interactions.

MTd2

Gold Member
Last edited by a moderator:

rntsai

Garrett,
b3 has a lot of a2 subalgebras.I got 12 just by trying simple combinations of
the roots. I still have a problem with the dimensionality : assume you find an
a2 subalgebra of b3 with desired properties. Then you need to find (simultaneously)
6 subspaces each of dim 3 that correspond to (qI,qI',qII,qII',qIII,qIII'); adding
the 8 subspace that corresponds to a2 itself, you have dim=26. b3 is only 21 dimensional.
Maybe b3 (so(7)) isn't a good intermediate step. F4 has 52 dim to work with so the chances of fitting these there are better.

On the other hand, eq 2.6 : su(3)+3+3'+1+1' only accounts for 16 dimensions. This is somewhat
unsatissfying, but for the sake of moving on to the next sections, is it fair to assume
that the remaining 5 dim subspace can be overlooked.

garrett

Gold Member
MTd2,
Connes is a brilliant guy. He's also trying to describe the standard model via a consistent framework, so I wouldn't be surprised if there are many overlaps. But his approach is based on the spectral action principal, and I haven't looked into it very much.

rntsai,
Yes, section (2.1) is meant as an introductory warmup, good for getting a feel for how things break down, and hints for how the big picture will fit together. The quarks won't separate out and categorize nicely until we work all the way up to E8.

MTd2

Gold Member
It's just that he shows the masses in the end...

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving