# Why is the Standard Model gauge group a simple product?

The Standard Model gauge group is always given as $\text{SU}(3)\times\text{SU}(2)\times\text{U}(1)$ (possibly quotiented by some subgroup that acts trivially on all konwn fields). However, at the level of (perturbative) Yang-Mills theory, only the local structure of this group is needed, that is, the algebra $\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak{u}(1)$. However, since this is only a local structure, it would still be the Lie algebra of a more complicated bundle of the three gauge groups. So, why do we say that the Standard Model gauge group can be written as a trivial product of three gauge groups, instead of something more topologically interesting? Are there theoretical or experimental considerations that force the trivialization?

Note: I am not asking why the QCD term is $\text{SU}(3)$ instead of some other group with the same Lie algebra (and similarly for the $\text{SU}(2)$ and $U(1)$ parts), since that question was answered here.

• "However, since this is only a local structure, it would still be the Lie algebra of a more complicated bundle of the three gauge groups." What do you mean by this? Why "bundle" (bundle over what base?)? Can you give an example of a different group that you think has the same Lie algebra? Note that twisting the product structure, e.g. as a semi-direct product, would also twist the Lie algebra structure, i.e. the sum of the algebras would not be a direct sum anymore. May 6, 2018 at 19:07
• This article by David Tong might be relevant to your question (he argues that the gauge group of the standard model is actually an undetermined quotient of the direct product) : arxiv.org/abs/1705.01853 May 6, 2018 at 19:26
• @ACuriousMind Construct an arbitrary $\text{SU}(3)$ bundle over $\text{SU}(2)$, then construct a $\text{U}(1)$ bundle over that resulting bundle, as manifolds. This will locally look like the typical standard model gauge group, and so the local structure will be the same, no? May 6, 2018 at 19:43
• How is this question different from the linked one? I don't think you're characterizing the linked question correctly, it seems to be asking the same thing as this one. May 6, 2018 at 19:59
• The local structure as a manifold will be the same, but you don't have any group structure on such bundles - while you could locally in trivial patchs use the group structure from the trivial product, it is not evident that this lifts to a well-defined group structure on the whole bundle. May 6, 2018 at 20:06

Actually while the Lie algebra $$su(3) \oplus su(2) \oplus u(1)$$ of the standard model is a direct sum of three simple Lie algebras, the gauge group itself appears to be the group $$S(U(3) \times U(2))$$.
• $S(U(3)\times U(2))$ is $SU(3)\times SU(2)\times U(1)$ "quotiented by [a] subgroup that acts trivially on all k[no]wn fields", so apparently it isn't what OP was looking for. Mar 15, 2021 at 6:27