Understanding SU(5) Subgroups and the SM Choice

[ChrisVer]

Well working with the dynkin diagram of SU(5), one can easily see (by Dynkin's rule) that possible choices of SU(5) spontaneous symmetry breaking could be:
$SU(5)→ SU(4) \times U(1)$ I call the broken group $G$
and
$SU(5)→ SU(3) \times SU(2) \times U(1)$ which I call $SM$

So I have a question. Apart from the natural imposition of the $SM$ subgroup (since we know that this is the gauge symmetry of our below GUT scale physics) is there any particular way the nature could have chosen it to $G$?

If you understood the question so far, don't procceed I'll try to make it clearer. I would expect that both G and SM are equally possible candidates, but something must have been there to choose the Standard Model to G... what is that something?

Thanks

 

[Vanadium 50]

But we don't know that nature is even SU(5) symmetric - and indeed, there is evidence that it is not.

 

[ChrisVer]

SU(5) won't die that easily, will it? XD Well, the minimal SU(5) theories are in fact rulled out by experiment (no proton decay), leaving nonminimal SU(5)s still around in the game.

Nevertheless, this doesn't cancel my question, which I think can be more general. Having a bigger symmetry, containing several subgroups, one of them being the Standard Model, how are the rest subgroups ruled out in the procedure of Spontaneous Symmetry Breaking?

One could be this, I heard of in a talk of Prof.Nielsen:
http://arxiv.org/pdf/1306.2668v1.pdf
But I think it leaves out the spontaneous symmetry breaking formalism?

 

[andrien]

Nevertheless, this doesn't cancel my question, which I think can be more general. Having a bigger symmetry, containing several subgroups, one of them being the Standard Model, how are the rest subgroups ruled out in the procedure of Spontaneous Symmetry Breaking?

I don't think that I am understanding what you are asking correctly.The most general generalization of SU(5) group is SO(10) which apart from SU(5) contains U(1),while SU(5) contains the embedding of standard model.You just introduce one other lepton in case of SO(10) i.e. antineutrino.The embedding of any group structure inside a group is determined from
1-the adjoint representation of larger group should contain the adjoint representation of smaller one.
2-the smallest representation of larger group is the sum of all non-trivial representations of smaller group.
In your case,if you use the embedding $SU(5)→ SU(4) \times U(1)$,then SU(4) actually has an embedding of $SU(3) \times U(1)$,which is not what you want.Apart from this the next larger group is the Exceptional group E₆,which contains SO(10).

 

[ChrisVer]

In your case,if you use the embedding $SU(5)→ SU(4) \times U(1)$,then SU(4) actually has an embedding of $SU(3) \times U(1)$,which is not what you want..

why isn't it what I want?