The Standard Model of particle physics can be constructed by specifying its gauge group $G$ and the representations of the fields (plus some extra information: Lorentz invariance, values of the coupling constants, etc.). Using this model we can predict many experimental facts, such as the cross sections measured in collider or confinement/asymptotic freedom of QCD.

However, we could've started by just giving the Lie algebra $\mathfrak{g}$ of $G$, the representations of the fields under $\mathfrak{g}$, and so on. For the purpose of making many of these predictions, it suffices to work at the infinitesimal level.

On the other hand, the actual group $G$ is important in gauge theory, as we can see, for example, when dealing with instantons. So it actually matters, for some applications to be able differentiate between different gauge groups even if they have the same Lie algebra.

Is there any experimental evidence or theoretical reason for $G=SU(3)\times SU(2)\times U(1)$ instead of some other group with algebra $\mathfrak{su}(3)\oplus \mathfrak{su}(2)\oplus \mathbb{R}$?

This Phys.SE question is clearly related, but it's only asking about a concrete modification of $SU(3)\times SU(2) \times U(1)$ to take into account a relation between the representations of the matter fields, not about exploring other possibilities for the global structure of the group.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Dec 25 '17 at 12:45

When a compact symmetry is required to hold only locally, i.e., to be a Lie algebra symmetry but without being a corresponding Lie group symmetry, it means that it is allowed to use a non-integrable representation, i.e., a representation of the Lie algebra that do not integrate into a representation of a corresponding Lie group.

An example of such a representation is the representation $J = \mu$ with $2 \mu$ non-integeral of the Lie algebra $\mathfrak{su}(2)$. This is a legitimate infinite dimensional representation of $\mathfrak{su}(2)$ that does not integrate to a representation of the group $SU(2)$, because every unitary irreducible representation of a compact Lie group is finite dimensional. The lack of integrability is reflected in the fact that, in this representation, the finite rotations corresponding to the Lie algebra elements become multivalued.

If you do not allow such representations, then you are automatically assuming that there is a Lie group symmetry. For example, if you attribute quarks and gauge fields, in the standard model, to finite dimensional representations of a Lie algebra of a compact group; then the theory is necessarily invariant under the corresponding Lie group.

Non-integrable representations are known to cause several difficulties. For example, the algebra of the angular momenta of a particle moving in the background of a magnetic monopole constitutes of this type of representation. This representation becomes integrable only if the Dirac's quantization condition is satisfied. Otherwise, several problems will arise: The Dirac string which should be just a coordinate singularity becomes a real singularity, some of the operators become nonassociative and the path integral becomes ill defined. Please see the following article by Bakas and Lüst and references therein.

Similar things can happen in field theory when the Chern-Simons level becomes non-integral.

Nevertheless, there is now a great interest in these cases. Please see the following two articles by Witten who explains a method based on analytic continuation designated to quantize the theories without requiring the Dirac's condition or the integrality of the level.


An important piece of evidence comes from the comparison between predictions made from calculations of Feynman diagrams and experimental results. The group theory factors that appear in the calculations of Feynman diagram depend on the representations of the Lie algebras. A favourable comparison thus fixes these representations.

Examples of such diagrams are those that one would need to compute to determine the beta function for the running coupling of a gauge theory. The one-loop beta function for an SU($n$) gauge theory (which I copied from here) is given by $$ \beta(g) = - \frac{g^3}{16\pi^2} \left[ \frac{11}{3} C_2(G) - \frac{n_s}{3} T(R_s) - \frac{4 n_f}{3} T(R_f) \right] , $$ where $C_2(G)$, $T(R_s)$ and $T(R_f)$ are the group theory factors associated with the group SU($n$).

For QCD $n=3$. One can use the beta function to compute the running of the coupling constant as a function of the energy scale. This has been confirmed experimentally. See for instance Figure 9.3 in: http://pdg.lbl.gov/2015/reviews/rpp2015-rev-qcd.pdf

So, now that we have the representations of the Lie algebra, how does it fix the group structure? Here we look at the specific case of the standard model. The pertinent parts are the SU(2) and SU(3) groups. The fermion content that give agreement with experimental result, fixes the representations of their Lie algebras.

For SU(2), one can have the same Lie algebra with SO(3), but the latter does not have a two-dimensional representation. Therefore, we know that the only compact Lie group with a two-dimensional representation of the Pauli matrices as Lie algebra is SU(2).

For SU(3), the representations are similarly fixed by the matter content of the theory that gives agreement between calculations and observations. The fermions are in the three-dimensional fundmanental representation. This fixes the structure of the compact Lie group to be SU(3).

  • 2
    $\begingroup$ Perhaps you can write down the formula that you have in mind for the QCD beta function that contain effects from the global structure of the gauge group. I don’t know much about this, but the ones I have seen involved only Lie algebra properties $\endgroup$ – coconut Dec 24 '17 at 13:58
  • $\begingroup$ A representation of the Lie algebra doesn't fix in general the group. For example, the adjoint representation of the Lie algebra gives a representation (the adjoint) for any group with that algebra $\endgroup$ – coconut Dec 29 '17 at 14:39
  • $\begingroup$ Edited the answer to give a better explanation $\endgroup$ – flippiefanus Dec 30 '17 at 6:52

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