An Exceptionally Simple Theory of Everything
I'm not sure yet whether all the attention is good, or if all the hype will have a negative impact, but this has certainly been a strange week. The interest in my work among physicists has been building steadily over the past few months. I've been presenting at conferences, getting invited to cool places, and exchanging emails with some of the best
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but the short version is:
It seems to me that Jacques Distler did not refute your E8 model
(only a possible use of triality)
and
that his analysis, along with your comments,
show how the E8 model can be OK.
Here are details:
Distler said in his blog: "...
E8(8) includes Spin(16) as a maximal compact subgroup ...
In E8(8), the 248 decomposes as 248 = 120 + 128
...
We would like to find an embedding of
G = SL(2,C) x SU(3)xSU(2)xU(1)
in ... E8
...
SL(2,C) = Spin(3,1)o is the connected part of the Lorentz Group,
the "gauge group" in the MacDwoell-Mansouri formulation of gravity.
..."
and
you [Garrett] replied "... The G is embedded in a D4 x D4 subgroup of E8.
...
g = so(3,1) + su(2) + u(1) + su(3)
... is in a so(7,1) + so(8) of e8 via the Pati-Salam, left-right symmetric model,
g' = so(3,1) + su(2)_L + su(2)_R + su(4)
The so(3,1) + su(2)_L + su(2)_R is in so(7,1),
the su(4) is in so(8) ..."
and
Distler said "... Spin(7,1) x Spin(8) ... is a subgroup of E8(8) ...".
So, an E8(8) model should be OK with the following interpretation:
e8(8) = 120 + 128
120 = spin(16) = spin(8) + 64 + spin(7,1) = 28 + 64 + 28
128 = 64 + 64
where
gravity comes from spin(3,1) MacDowell-Mansouri
and
the Standard Model comes from Pati-Salam su(4)_c + su(2)_L + su(2)_R
spin(8) includes su(4) that reduces to su(3)_c
spin(7,1) includes so(3,1) + su(2)_L + su(2)_R
where the so(3,1) of MacDowell-Mansouri gravity is the little group,
or local isotropy group, of 4-dim spacetime M4 described
by the symmetric space G / Spin(3,1) where G can be anti-desitter or deSitter
and
where the su(2)_L + su(2)_R reduces to su(2)_L + u(1) which is the little group
of a 4-dim internal symmetry space CP2 = SU(3) / SU(2)xU(1)
NOTE that the Lie groups of the spin(7,1) Lie algebra form the little groups
of an 8-dim M4 x CP2 Kaluza-Klein space (as used by Batakis in his 1986 paper
Class. Quantum Grav. 3 (1986) L99-L105).
As to the global groups of M4 x CP2,
they are in the spin(8) that includes the su(4):
the su(3) gives the global group SU(3) in CP2 = SU(3) / SU(2)xU(1)
(as used by Batakis)
and
the 4-dimensional deSitter or anti-deSitter rotations of G / Spin(3,1)
should be a part of 6-dim twistor-related CP3 = SU(4) / SU(3)xU(1)
What about the one 64 in the 120 and the two 64 + 64 in the 128 of E8(8) ?
Each of the 64 should be of the form 8x8.
Generalizing the spacetime algebra approach of Hestenes, let, in each of
the three 64, one of the 8 represent 8 Dirac Gammas of the 8-dim K-K space.
Denote it by 8_G
Since the 64 in the 120 is in the adjoint of Spin(16),
its 8x8 should be correspond to the 8-dim vector K-K space,
so denote the 64 in 120 by 8_v x 8_G
Since the 64 + 64 in the 128 is in a spinor space of Spin(16),
its 8x8 + 8x8 should be correspond to fermions ( 8 particles and 8 antiparticles)
so denote the 128 64 + 64 in 128 by 8_f+ x 8_G and 8+f- x 8_G
There should be a Spin(8)-type triality among the three 64 things
8_v x 8_G
8_f+ x 8_G
8+f- x 8_G
The above E8(8) structure describes Gravity, the Standard Model,
4-dim physical spacetime, a 4-dim K-K space, and first-generation fermions,
as well as 8 Gammas of an 8-dim Dirac equation.
If the second and third generation fermions come from combinatorics of
fermions living partly in 4-dim physical spacetime and partly in 4-dim K-K space,
then you get all three generations.
My opinion is that:
a spin foam can be constructed by putting the 120 of E8(8) on links
and the 128 of E8(8) on vertices, and using Jordan algebra structure
related to the 27-dim exceptional Jordan algebra J3(O);
and
particle masses and force strengths come from ratios of geometric volumes
in the spirit of Armand Wyer;
neutrinos are tree-level massless, with masses coming from corrections;
Dark Energy : Dark Matter : Ordinary Matter ratio comes from the structure
of the twistor stuff in the spin(8) of the 120.
Tony Smith
PS - Sorry for the long post on mauitian.
If you want me to not put such stuff here, please just let me know.
Reply
I know Jacques didn't refute it, in fact I think it's more interesting if it's in E8(8) than in E IX, because the representations are more involved.
But I am a little confused. My understanding of E8(8) is that the Killing form is such that the 120 have positive signature and the 128 have negative signature. So so(16) is a subalgebra. But Jacques said so(7,1)+so(8) is also a subalgebra... do you know how that works?
Reply
consider the graded structure of Cl(8)
1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1
and
that by real Clifford algebra 8-periodicity tensor product factoring Cl(16) = Cl(8) (x) Cl(8)
so that
the 120 so(16) bivectors of Cl(16) should be made up of three parts:
28 so(8) bivectors of the first Cl(8) in the tensor product
28 so(8) bivectors of the first Cl(8) in the tensor product
64 = 8x8 tensor product of the two 8-dim 1-vectors of each the two Cl(8)s
to get the 28+28+64 = 120-dim SO(16) bivector algebra of Cl(16) = Cl(8) (x) Cl(8).
So, since so(16) is a subalgebra of E8(8),
its two so(8) parts should also be a subalgebra of E8(8),
and
that gives so(8) + so(8) as a subalgebra of E8(8).
As to how to get the signature of one of the so(8) to be (7,1) and therefore so(7,1),
I am not sure, but here are two guesses
(see the book Spinors and Calibrations by Reese Harvey for details about both of them):
First,
consider that the Clifford algebras Cl(8,0) and Cl(7,1) are isomorphic
(both are the real 16x16 matrix algebra)
and
use that isomorphism to map so(7,1) into one of the so(8)
(here I am using a signature notation that I don't usually use - I usually denote
those Clifford algebras by Cl(0,8) and Cl(1,7) - both conventions appear in the literature).
There is a chart (taken from the book by Reese Harvey) on my web site at
www.valdostamuseum.org/hamsmith/clfR.gif
showing the pattern of Clifford algebras of various signatures. In that chart,
I am talking about the green blocks in the lowest right corner (Cl(8,0)
and the green block diagonally up and to its left (Cl(7,1).
In the chart of Clifford algebras,
green = real matrix algebras, yellow = quaternionic, and white = complex.
Second (this one is more conjectural in my mind right now),
consider the way you build a cross-product in 7 dimensions
(cross-products only live in dimensions 1, 3, and 7,
and the way you build the 7-dimensional cross-product,
starting with the Clifford algebra Cl(7) with graded structure
1 7 21 35 35 21 7 1
and
using a coassociative 4-vector PSI and vectors a, b to get
a x b = *((a /\ b) /\ PSI) as the 7 - (2+4) = 1-vector
(see the book Spinors and Calibrations by Reese Harvey).
Now consider Cl(8) with graded structure
1 8 28 56 70 56 28 8 1
and use an associative 4-vector PHI and 2-vector bivector A to get
*(A /\ PHI) as a corresponding 8 - (2+4) = 2-vector bivector.
Maybe a choice of PHI could be taken such that
PHI maps bivectors that act like so(8) into bivectors that act like so(7,1) ?
Tony Smith
PS - Maybe the math can be a warm and fuzzy friend, and not so scary after all ??
Reply
I wrote up a paper about your E8 model and how it might look in terms of E8(8),
and put it on dotMac at
web.mac.com/t0ny5m17h/Site/E8GLTSCl8Cl16.pdf
and mentioned it in blog comments (including Jacques Distler's blog).
Please be aware that, as I said at the end of the abstract
"... Any errors in this paper are not Garrett Lisi's fault. ...".
Tony Smith
Reply
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