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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.

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    $\begingroup$ "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. $\endgroup$
    – ACuriousMind ♦
    Commented May 6, 2018 at 19:07
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    $\begingroup$ 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 $\endgroup$
    – Antoine
    Commented May 6, 2018 at 19:26
  • $\begingroup$ @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? $\endgroup$
    – Bob Knighton
    Commented May 6, 2018 at 19:43
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    $\begingroup$ 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. $\endgroup$
    – tparker
    Commented May 6, 2018 at 19:59
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    $\begingroup$ 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. $\endgroup$
    – ACuriousMind ♦
    Commented May 6, 2018 at 20:06

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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 $S(U(3) \times U(2))$.

For much more details see the discussion on the web page ''Standard Model Group Structure'' by Tony Smith.

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  • $\begingroup$ $S(U(3)\times U(2))$ 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. $\endgroup$
    – benrg
    Commented Mar 15, 2021 at 6:27

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