It is often said $E_8$ gauge group is forbidden because it has no complex representation. That seems to be true in that it has no fundamental representation in terms of complex vectors. But just taking the adjoint representation and identifying opposite roots with complex conjugates and leaving the 8 chargeless generators as Majorana fermions seems to be a "pseudo-complex" representation. If $\alpha$ are the 240 roots of $E_8$ and $n=1..8$ are the 8 :
$$\overline{\Psi_{\alpha}} = \Psi_{-\alpha}$$ $$\psi_n = \overline{\psi_n}$$
In other words it would have 120 particles, 120 anti-particles and 8 Majorana fermions which are their own anti-particle. (Treating left and right handed fermions as separate particles)
With Lagrangian for the kinetic part $\mathcal{L} = \Psi_{-\alpha} \sigma^\mu \partial_\mu \Psi_{\alpha} + \psi_n \sigma_\mu \partial_\mu \psi_n$
Similarly for the bosons you would have 240 complex bosons $A_{\alpha} = \overline{A}_{-\alpha}$ with 8 chargless bosons $B_n$ (which include combinations of the photon, Z-boson and the 2 chargless gluons plus 4 extra). (The subgroup $O(16)$ of $E_8$ consisting of the roots $(0,...\pm 2,..\pm 2,..0)$ plus the 8 neutral bosons can act entirely on a subset of the complex fermions with roots $(\pm\frac{1}{2},\pm\frac{1}{2},...)$ as this really does have a true complex representation.) Interaction terms would include $\Psi_{\alpha}A_{\beta}\Psi_{-\alpha-\beta}$ as well as those with neutral particles.
In fact this pseudo-complex representation should presumably work for any Lie group in adjoint representation.
I can't see anything particularly wrong with this argument and it doesn't seem to contradict the no-go theorem because it is not a fully complex representation due to the inclusion of the 8 Marjorana fermions.
Presumably this is completely wrong otherwise this would have been done before. So what is the reason? (Is it perhaps a real representation in disguise?)
