An Exceptional Supersymmetry

An Exceptional Supersymmetry

It is proposed in this article a way to construct an exceptional supersymmetry distinct from the family of usual supersymmetries usually labelled by N. (N=1, N=2, N=4 and N=8).


It relies on the special properties of the root system of E₈. It is a different kind of super-symmetry in that the superpartners don't have the same charge but together the charges from all the particles in the representation form a group. In a nut-shell we are assigning spins to elements of the E8 algebra and promoting them to super-symmetry operators. The special properties of E8 allows us to do this.


Introduction
The similarities between  N=8 Supersymmetry with 256 generators and the group E8 with 248 generators is suggestive that there might exist a combined structure with properties of both. Noting that E8 can be split into a bosonic part (a representation of O(16) ) and a fermionic part a spinor representation of O(16).

We can assign a "spin" to each element of the E8 roots by taking 1-|Sum(R)|/4. This further divides the bosonic part into spin-1 and spin-0. The spin-1 generators now forming SU(8)xU(1).

The fermionic part split into 16 (spin -1/2) and 112 (spin +1/2).

To form a supersymmetry we need equal numbers of bosons and fermions so we must introduce an extra 8 bosons.

The supersymmetry acts on 248 (256?) fields consisting of 64 spin-1 bosons, 128 spin 1/2 fermions and 56 (64?) spin-0 bosons. (Should there be an additional 8 spin-0 bosons?)

Each generator corresponds to a super-symmetry operator.

The fields form representations of SU(8). For each field we would get a super-symmetry generator.

{Q^A,Q*^B}= t(A+B)P?

It must satisify the Jacobian identity:

For example
Q^{++++---} A(2-2000000) = DY(+-+++---)

Can the Q be represented using superfields?
By taking the converting the roots vectors into a lattice we have E8* (also called E9). Which represents now an infinite number of particles and an infinite supersymmetry algebra.

N=1 with (Q,Q*,P) comes from SU(2)?



Conditions
Why is E8 the only group that can form an exceptional supersymmetry algebra?

• It acts on itself in the adjoint representation
• It can be split into representations of SU(8)

Why wouldn't G2 or O(8) or E7 work?

Phenomenology
Although we are primarily interested in this paper on the mathematical details of this algebra. It is interested to note that if we assign the labels (R,G,B,W,H,F,S,T) to the roots then we can fit a standard-model like structure with 4 generations of fermions with an U(1)xSU(8) GUT gauge group. The usual quantum numbers are found with:

R²+G²+B²+W²+H²+F²+S²+T² = 8

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charge = (R+G+B-3W)/6

isospin = (W-H)/2

color = R+wG+w²B

flavour = F+wS+w²T
|spin| = 1- |R+G+B+W+H+F+S+T|/4 

With Higgs of non-zero values given to (0,0,0,0,2,2,0,0), (0,0,0,0,2,0,2,0), and (0,0,0,0,2,0,0,2) we get masses for some of the particles and mass mixing of the generations. Note that these combinations change the isospin and hence couple left-handed to right-handed particles with the same colour and charge as we would expect.