# 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?

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?

• Isn't SU(2) equivalent to Sp(1)? – pathintegral Sep 16 '16 at 17:54
• John Baez has thought a lot about how quaternions fit into quantum mechanics. For some enjoyable procrastination: google.com/search?q=baez+quaternion+quantum – Ruben Verresen Sep 16 '16 at 18:37
• 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 $\mathrm{SU}$. – ACuriousMind Sep 17 '16 at 0:37
• I think some people do just pick gauge groups just to study them. Why not? – zooby Sep 17 '16 at 4:28
• Yang-Mills theory with symplectic groups is sometimes considered in the mathematical physics literature. See, for example, Kapustin & Witten's paper on geometric Langlands or some of Seiberg's early papers on SUSY gauge theory. It's not so relevant to particle physics modeling, but it's a perfectly valid mathematical object. – user1504 Apr 13 '17 at 15:11

The structure of standard model $SU(3)\times SU(2)\times U(1)$ is chiral which basically tells you the necessity of chiral fermions. If left-handed fermions transform under a representation $R$ of the symmetry group then due to charge-conjugation relating left-handed and right-handed fermions as $$\psi_{Right}=C(\bar{\psi^C})^T_{Left}$$ and so, right handed fermions should transform under the complex conjugate representation $R^*$. If $R$ is real or pseudoreal, then left-handed and right-handed fermions transform in same representation of the group and the theory is known to be a vector like theory (QCD). To have chiral structure of fermions, one has to have $R \ne R^*$ which demands $R$ to be complex.

Even though QCD is vector like and $2=2^*$, the whole standard model is chiral as can be seen by writing $R$ for left-handed fermions as, $$R=(3,2)_{\frac{1}{6}}+(3^*,1)_{\frac{-2}{3}}+(3^*,1)_{\frac{1}{3}}+(1,2)_{\frac{-1}{2}}+(1,1)_{1}$$ the complex conjugate to which is not same as $R$.

It is known that $USp(2n)$ for $n>2$ admits real and pseudo-real representations (Weinberg Vol. 2, chapter 22) and $USp(4)$ is not big enough to contain standard model.

Moreover using a $Sp(n)$ like gauge group demands even number of fermion multiplets otherwise the gauge theory will show a non-perturbative anomaly$^1$ involving fourth homotopy group of $Sp(n)$.

1- Ed Witten, nucl. phys. B223 (1983),433-444.

• Perfect. From this I understand that there is nothing theoretical banning the group $Sp(n)$. But it would require mirror fermions that transform the same way as it cannot contain $SU(2)_L$ in a complex representation. So if "mirror fermions" and a fourth generation of fermions were discovered at the LHC then they might have to consider $Sp(n)$. However, that might be unlikely! – zooby Sep 22 '16 at 10:56

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

Now look at the symplectic group closely.

• $Sp(1)$ 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 does not mix with 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 a more rigorous perspective one can consult with, Group theory for unified model building by Slansky.

• 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? – zooby Sep 16 '16 at 19:34
• Are you asking that what happen if one choose complex symplectic group? – AMS Sep 16 '16 at 19:38
• Its not obvious what you mean by complexification of 'real a representation'. – AMS Sep 16 '16 at 19:41
• 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. – ACuriousMind Sep 17 '16 at 0:35
• @AMS: $Sp(1)$ is isomorphic to $SU(2)$, not $Sp(2)$. Plus, the following statement: "For example $USp(2n)$ with $n\geq 3$ has only real and pseudo-real representations. " is incorrect. Every compact real Lie group admits complex representations. – Bilateral Apr 11 '17 at 14:48