# Understanding symmetry and super symmetry

We have fields associated with the U(1) symmetry; we have other fields associated with the SU(2) symmetry, and we have still other fields associated with the SU(3) symmetry. IIRC we have a particular algebra of multiplication between the fields of the U(1) symmetry; we have a different algebra of multiplication between the fields of the SU(2) symmetry, and we have still another algebra of multiplication between the fields of the SU(3) symmetry. Is it true that the algebra of U(1) is complex, the algebra of SU(2) is quaternions, and the algebra of SU(3) is octonions? I think I heard something like this at one time in my studies.
That's an interesting question, and I doubt I'm in a position to answer it, but Geoffry Dixon does work on this very concept. A little numerology that's always fascinated me is
\begin{align} S^1 \hookrightarrow S^3 \rightarrow S^2 \\ S^3 \hookrightarrow S^7 \rightarrow S^4 \\ S^7 \hookrightarrow S^{15} \rightarrow S^8 \end{align}

notice that the fiber of the complex hopf fibration is $$S^1 \cong U(1)$$ and the fiber of the quatronic hopf fibration is $$S^3 \cong SU(2)$$ however the fiber of the octonic hopf fibration $$S^7$$ isn't even a group. Technically it's a Moufang loop. I think what Dixon does is take the automorphism group of the octonions which is G_2 and get's SU(3) out of that. That's according to something I read on John Baez's this weeks finds.

We have fields associated with the U(1) symmetry; we have other fields associated with the SU(2) symmetry, and we have still other fields associated with the SU(3) symmetry. IIRC we have a particular algebra of multiplication between the fields of the U(1) symmetry; we have a different algebra of multiplication between the fields of the SU(2) symmetry, and we have still another algebra of multiplication between the fields of the SU(3) symmetry. Is it true that the algebra of U(1) is complex, the algebra of SU(2) is quaternions, and the algebra of SU(3) is octonions? I think I heard something like this at one time in my studies.
That's an interesting question, and I doubt I'm in a position to answer it, but Geoffry Dixon does work on this very concept.
I'm sorry, but I thought this would be a straightforward question. Could it still be that we don't know how the fields of the SU(3) symmetry multiply each other? It sounds like a yes or no question. I don't know what could be so complicated about that.

The theory of the SU(3) part is quantum chromodynamics or QCD.

There isn't any way to mark out a red quark as distinct from a green one or a blue one. They all look alike from outside. The same is true of the gluon color states. So one has to use more indirect methods, like comparing quark-quark-gluon and gluon-gluon-gluon overall interaction strengths. But the results are consistent with the QCD symmetry group being SU(3).

One can approach the question in another way. What symmetry groups can there be? We look among the "Lie groups", those whose elements are some differentiable function of some parameters. These groups get very complicated very quickly, so we look at their generators, the "Lie algebras", which are often much simpler to work with. If you've ever done operator algebra with quantum-mechanical angular momentum, you've worked with a Lie algebra.

Also, the particle multiplets are in "representations" of the group and its algebra. Each representation or rep has a matrix realization, and if the matrices cannot all be turned into the same block-diagonal form, then the rep is irreducible, making an irrep.

There's a Lie-algebra rep that's formed from the algebra's structure and that uses the generators as its basis space: the adjoint rep.

The interactions in QCD must all be intertwined and not separable, otherwise it could be decomposed into some subsets that are independent or more-or-less independent of each other. Quarks and gluons must all be in irreps, because otherwise, they'd split into different sets.

As is true of gauge theories in general, the gluons must be in the adjoint rep of the QCD symmetry group, because each gluon mode is associated with a generator of that group.

So we look in the simple Lie algebras. Which ones have 3D reps among all their reps? Not very many. Of the four infinite families and five exceptional ones, only SU(3) and SO(3) have 3D reps. SO(3) is the 3D rotation group, and it and SU(2) have isomorphic algebras. However, the 3D rep of SO(3) is a real rep, and thus not very suited. So we are left with SU(3). Its 3D rep is complex, and it has a conjugate one. Just what one needs for quarks and antiquarks. One can create a colorless state from 3 quarks in both SU(3) and SO(3), which is further support for these solutions.

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Hello Jim,

Thanks for the wonderful reply. It helped me to clear up the concepts.

What I got, is that each internal symmetry is responsible for conservation of charges. Conservation is a fundamental property of Nature.