The argument in the preprint assumed the SM and particularly that the turtles were allowed to bind only if they had colour. But the final particle content is very reminiscent of the Georgi-Glashow model.
Consider the usual formulation of the model $$ \begin{eqnarray} 5 =& (1,2)_{-3} + (3,1)_2 \\ 10 =& (1,1)_{-6} + (\bar 3,1)_4 + (3,2)_{-1} \end{eqnarray} $$ and use \(Q=T_3 – Y/6\) to study the electric charge of the colour singlets and colour triplets in each representation of \(SU(5)\)
$$ \begin{array}{r|cccc} repr & singlets & triplets & antitriplets & other \\ 1 & 0 & & &\\ 5 & 0, +1 & -1/3 & & \\ 24 & 0,+1,0,-1 & -4/3, -1/3 & +4/3, +1/3 & (8)_{q=0}\\ 10 & +1 & +2/3, -1/3 & -2/3 &\\ \bar {10} & -1 & +2/3 & -2/3, +1/3 &\\ 15 & +2,+1,0 & +2/3, -1/3 & & (6)_{-2/3}\\ \bar{15} & -2,-1,0 & & -2/3, +1/3 & (\bar 6)_{+2/3}\\ \end{array} $$
It can be argued that the union of \(10 + \bar {10} + 24 \) is the one generation version of our previous construction, albeit with only a left neutrino. See how the X boson of the GUT model is here just the q=+4/3 scalar, and how we have the condition of two states for each fermion. In this case our turtles are all the members of 5 and the elephants are the members of 10. They bypass the previous uniqueness proof because we are allowing leptons to join the game.
It could be interesting to consider variants of this game, such as flipped \(SU(5)\) with some rules for the \(U(1)_X\) charges of the 24 and 15, or even an anomalous «deflipped \(SU(5)\)» with the up quark in the fundamental but the standard hypercharge assignments, so that the \(+4/3\) triplet would appear in the decuplet.
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