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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?

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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.

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  • $\begingroup$ Thanks, that helps a lot. Is it obvious that any orthogonal group can be broken down into a unitary group (i.e. that $SO(N) \subset SU(N)$)? $\endgroup$ – JeffDror Jun 17 '14 at 2:10
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    $\begingroup$ Probably, but here is how: Restrict the complex unitary matrices of determinant 1 whose complex transpose is their own inverse - which is $\mathrm{SU}(N)$ by definition - to those with real entries. For real matrices, complex transpose and real transpose coincide, so this is the group of real matrices with determinant 1 whose transpose is their inverse, which is by definition $\mathrm{SO}(N)$. $\endgroup$ – ACuriousMind Jun 17 '14 at 5:54
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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.

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    $\begingroup$ This is of course the (only) correct answer. The voting is completely dysfunctional sometimes and seems biased by this reputation points system. $\endgroup$ – xi45 May 27 '16 at 15:23
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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).)

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  • $\begingroup$ Thanks for your response. I do see the many similarities. I tried playing around with classical $SO(N)$ theories and found some strange results. If I have some time I will post a new question with exactly what was strange. $\endgroup$ – JeffDror Jun 16 '14 at 22:40
  • $\begingroup$ Also do you have a comment about the second question, namely whether new physics could exist that is an $SO(N)$ gauge theory but wouldn't show up as a subset of some $SU(N')$ theory? $\endgroup$ – JeffDror Jun 16 '14 at 22:42
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    $\begingroup$ I don't think you get any substantially new phenomena by choosing an SO(N) gauge group. The different representations mean we get different charges for the fermions, but the basic physics of Yang-Mills theory works the same way for any compact gauge group. $\endgroup$ – user1504 Jun 16 '14 at 22:54

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