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Why can't compact symplectic groups $Sp(n)\equiv USp(2n)\equiv U(2n)\cap Sp(2n,\mathbb{C})$ be gauge groups in Yang-Mills theory?

+ 3 like - 0 dislike
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The gauge groups in Yang-Mills theory can be things like $O(10)$ or $SU(5)$ but continuing the pattern from real to complex, the next obvious thing would be quaternion matrices. A group like $U(4,H)$ where $H$ is the quaternions. This is another name for $Sp(4)$ (according to Wikipedia!).

A group like $U(4,H)$ I always thought would be interesting since it would be split $U(1,H)\times U(3,H)$ and $U(1,H)=SU(2)$ and $U(3,H)$ would have subgroup $SU(3)$.

But I have never seen a Yang-Mills theory with a compact symplectic gauge group so apparently there must be a good reason for that.

Do you know the reason? Is there a theoretical reason or an experimental reason?


This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user zooby

asked Sep 16, 2016 in Theoretical Physics by zooby (15 points) [ revision history ]
edited Sep 20, 2016 by Dilaton
Isn't SU(2) equivalent to Sp(1)?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user pathintegral
John Baez has thought a lot about how quaternions fit into quantum mechanics. For some enjoyable procrastination: google.com/search?q=baez+quaternion+quantum

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user Ruben Verresen
Well, the groups must come from somewhere. For example, they must accommodate the Standard Model gauge group, or arise from some other physical context naturally. No one just picks random groups and studies their gauge theory (I hope). It might just be that symplectic groups don't arise as gauge groups, except in the cases where they are isomorphic to some $\mathrm{SO}$ or $\mathrm{SU}$.

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user ACuriousMind
I think some people do just pick gauge groups just to study them. Why not?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user zooby

1 Answer

+ 1 like - 0 dislike

Well the answer of your question is not so trivial i guess. Here is my try. I want to give a glimpse why symplectric group is not a good choice for model building from a phenomenological point of view.

Now look at the symplectic group closely.

  • $Sp(2)$ is isomorphic to $SU(2)$
  • $Sp(4)$ is isomorphic to $SO(5)$ (which is due to a deeper connection between $SO(2n+1)$ and $Sp(2n)$)

The standard model gauge group is $SU(3)_{C}\times SU(2)_{L}\times U(1)_{Y}$. If we have a closer look then $SU(3)$ has complex representation (fundamental and anti-fundamental representation don't talk to each other), $SU(2)$ has pseudo real representation. That simply says particles belongs to standard model (also belongs to real world!) gauge group has complex representations.

Most strikingly, the symplectic group does not have complex representations. For example $USp(2n)$ with $n\geq 3$ has only real and pseudo-real representations. So, any gauge theory which is can not accommodate complex representation is not a good choice for model building.

For more rigorous perspective one can consult with, Group theory for unified model building by Slansly.

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user mas
answered Sep 16, 2016 by mas (30 points) [ no revision ]
Why must it have a complex representation? What would happen if you took a real representation and just complexified it? Or Sp(n) has a quaternion representation. Wouldn't that be even better than a complex representation? Or not?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user zooby
Are you asking that what happen if one choose complex symplectic group?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user mas
Its not obvious what you mean by complexification of 'real a representation'.

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user mas
I'm not convinced by this answer - having no truly complex representations doesn't strike me as bad. That all representations are "real" just means that if you take any complex representation of the group, you can find a real structure such that the representation restricts to a real vector space, it doesn't forbid representations on complex spaces.

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user ACuriousMind
Having a (pseudo) real representation of a Lie group means, its fundamental and anti-fundamental representation are connected by a similarity transformation. am i right?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user mas
I read in a different answer that real gauge groups must have mirror fermions. Is this because the real rep transforms both the real part and imaginary part of a complex spinor in the same way. So it is like two separate particles?

This post imported from StackExchange Physics at 2016-09-20 21:55 (UTC), posted by SE-user zooby

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