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When building models people typically gauge $SU(N)$ but rarely try to gauge $SO(N)$ (the only example I know about is $SO(10)$, but even that isn't quite $SO(10)$ but actually its double cover). At least with $SO(2) $ and $SO(3)$ one could choose to gauge these groups or their isomorphisms, $U(1)$ and $SU(2)$.

Is there a good reason why working with the $SO(N)$ is more difficult? Furthermore, could there be some new physics that's hiding in an orthogonal group and not inside a unitary one?

Qmechanic
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JeffDror
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3 Answers3

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If you make a gauge theory about $\mathrm{SU}(N)$, you also have one about $\mathrm{SO}(N)$ since $\mathrm{SO}(N) \subset \mathrm{SU}(N)$.

The representation theory of $\mathrm{SO}(N)$ is more complicated since it has not that many nice properties compared to $\mathrm{SU}(N)$ - the latter preserves orthogonal, complex and symplectic structures while the former only respects orthogonal structures. If you think this should imply that $\mathrm{SO}(N)$ is the larger group, this is because the symplectic and orthogonal structures that $\mathrm{SU}(N)$ preserves are in other dimensions, more precisely $$\mathrm{U}(N) = \mathrm{O}(2N) \cap \mathrm{GL}_\mathbb{C}(N) \cap \mathrm{Sp}_\mathbb{R}(2N)$$

Heuristically, working with less symmetry (i.e $\mathrm{SO}$ vs $\mathrm{SU}$) means more work, thus typical presentations prefer to work with the unitary groups. Also, most quantum theories are inherenty working with complex numbers, so a unitary (gauge) symmetry is more natural than a orthogonal real one.

ACuriousMind
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Mainly because you need complex representations for the fermions such that anomalies cancel. Real representations don't work though these also cancel anomalies since these give large radiative masses. $SO(n)$ has both the tensorial representations(single valued) that are always real and spinorial representations (double valued). For $SO(n)$(or more correctly $Spin(n)$) you only have complex reps when $l\geq 1$ with $n=4l+6$ and n even thus the smallest group that can have these representations is $SO(10)$. In this model you can embed into the spinorial representation all SM fermion fields plus RH neutrino and cancel anomalies and this is one reason why is so used though the algebra has 45 generators. Instead, for $SU(n)$ you can have complex representations since $n \geq 3$ and so the minimal model that contains the gauge group of the SM is the $SU(5)$ with an algebra much more manageable.

Mitor
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SO(N) gauge theory isn't really much different from SU(N) gauge theory. All the calculations are basically the same, and the physics is only different because the representation theory of SO(N) is a little more complicated than the representation theory of SU(N). (If you were lecturing about representation theory, you'd explain SU(N) at the blackboard and make the students solve homework problems about SO(N).)

user1504
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