An Exceptionally Technical Discussion of AESToE

**garrett** · Dec4-07, 11:28 AM

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,
$LaTeX Code: <BR>\\vec{v} \\underline{f} = v^i \\vec{\\partial_i} \\underline{dx^j} f_j = v^i f_i<BR>$
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

Dec4-07, 11:53 AM

Hi Garret,

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

**morbello** · Dec4-07, 12:00 PM

Sounds interesting

**garrett** · Dec4-07, 12:14 PM

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 $LaTeX Code: \\frac{1}{\\sqrt{2}}$ scaling in front.

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

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

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

**rntsai** · Dec4-07, 02:20 PM

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** · Dec4-07, 02:53 PM

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 " $LaTeX Code: \\oplus$ "). 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** · Dec4-07, 04:22 PM

Originally Posted by garrett

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 " $LaTeX Code: \\oplus$ "). 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?

**rntsai** · Dec4-07, 04:23 PM

[quote=rntsai;1528340]Thanks Garry,
Garrett, Sorry for calling you Garry!

**Cold Winter** · Dec4-07, 05:02 PM

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** · Dec4-07, 05:14 PM

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.

**Cold Winter** · Dec4-07, 05:24 PM

Originally Posted by garrett

... 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** · Dec4-07, 05:35 PM

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

**samalkhaiat** · Dec4-07, 06:40 PM

Hi,

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

Regards

Sam
(an Exceptionally lazy person)

**garrett** · Dec4-07, 07:06 PM

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

**rntsai** · Dec4-07, 07:43 PM

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** · Dec4-07, 10:43 PM

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?