# Why is the Yang-Mills gauge group assumed compact and semi-simple?

What is the motivation for including the compactness and semi-simplicity assumptions on the groups that one gauges to obtain Yang-Mills theories? I'd think that these hypotheses lead to physically "nice" theories in some way, but I've never, even from a computational perspective. really given these assumptions much thought.

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Compactness is needed for the bilinear form on the adjoint representation to be positively definite. For example, $SO(2,1)$ would be no good because the signature on the adjoint is ${+}{-}{-}$. If we had an indefinite form, the norm of the different colorful polarizations of the gauge bosons would have different signs (ghosts, negative probabilities). In a similar way, some Lie algebras (not semisimple etc.) have "zero norm" directions. Ultimately, we decompose the gauge group to simple compact pieces - the factors behave independently and decouple. – Luboš Motl Jan 29 at 7:34

As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often chosen to be (proportional to) the Killing form, but that need not be the case.

If $\kappa$ is degenerate, this will induce additional zeromodes/gauge-symmetries, which will have to be gauge-fixed, thereby effectively diminishing the gauge group $G$ to a smaller subgroup, where the corresponding (restriction of) $\kappa$ is non-degenerate.

When $G$ is semi-simple, the corresponding Killing form is non-degenerate. But $G$ does not have to be semi-simple. Recall e.g. that $U(1)$ by definition is not a simple Lie group. Its Killing form is identically zero. Nevertheless, we have the following YM-type theories:

1. QED with $G=U(1)$.

2. the Glashow-Weinberg-Salam model for electroweak interaction with $G=U(1)\times SU(2)$.

Also the gauge group $G$ does in principle not have to be compact.

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It's because you want the kinetic part of the Yang Mills action $$\int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's only if G is compact and semi simple though. Maybe someone else could fill in this detail).