# A The failure to booststrap SU(3).

1. Jul 12, 2017

### arivero

One of the encouraging points of string theory is the ability to single out specific gauge groups, a feat that Chew himself thougth impossible back in 1970. But it extracts groups as SO(32) or E8xE8.... elegant it is, but not simple.

So lets ask, is there really no way single out SU(3) from some consistency argument. Here the definite property is that $$\bf 3 \times 3 = 6 + \bar 3$$
and that the representations with size $n (n\pm 1) /2$ are seen to happen in string theory when orientifolds are involved. Here we could look to some 14 of Sp(6), or to SO(6), with a 15 what recovers back all the important game of SU(3), via
$$\bf 15 = >> 1_0 + 3_4 + \bar 3_{-4} + 8_0$$
while for higher Sp(2n) or SO(2n) groups using this same decomposition we still get the adjoint but not the defining and conjugate irreps of SU(n).

Last edited: Jul 12, 2017
2. Jul 12, 2017

### arivero

By the way, a pretty nice trick to get 3-quark baryons in the large N expansion is to add a new set of quarks from the above (anti)symmetrisation. But lnot sure of its value... large N is a planar limit, isn't it? Or is it possible to get nonorientable planar large N? Even if so, something as SU(3)xSO(large N) could be more interesting than a swarm of new quarks.

3. Jul 12, 2017

### arivero

Hmm, SU(3) is of course visible as holonomy requirement, and also relatively from decomposition of SO(8) into 1 +3 + 3 +1. But all of this is consequence of D=10, while the former argument in the OP could be used in any dimension. Also, if we are into extra dimensional origins, we have more tools: complex manifolds, compactification enhancements, and even Kaluza Klein on symmetric spaces.