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In Georgi, 1999; pg254 the following question is posed:

Find the regular maximal subalgebras of $SO(12)$. To find them all, you will have to apply the extended Dynkin diagram algorithm several times, because some of the regular maximal subalgebras themselves have nontrival regular maximal subalgebras.

I am confused by this last statement. Let $\mathcal{L}$ be a Lie-algebra and $\mathcal{A}$ be a regular maximal subalgebra (RMS) of $\cal L$. Then if $\cal B$ is a RMS of $\cal A$ then surely it can't be a RMS of $\cal L$ since it is not maximal in $\cal L$. Is this correct and if it is how do I make sense of the above question?

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  • $\begingroup$ I suggest you add a precise reference to this exercise. As written, it is indeed incorrect. $\endgroup$
    – Moishe Kohan
    Commented Jan 4, 2018 at 23:16
  • $\begingroup$ @MoisheCohen Georgi, H., 1999. Lie algebras in particle physics: from isospin to unified theories (Vol. 54). 2ed. Westview press. pg254 $\endgroup$
    – Quantum spaghettification
    Commented Jan 5, 2018 at 6:54

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When you apply the Dynkin diagram algorithm, you typically end up with a product of algebras that make up the regular maximal subalgebra, i.e. $$ SO(12) \to SU(2)\times SU(2)\times SO(8) $$ The $SO(8)$ component here has a regular maximal subalgebra $SU(2)\times SU(2)\times SU(2)\times SU(2)$, so the product of this with the other $SU(2)$s will also be a regular maximal subalgebra. That is $SU(2)^6$.

In other words, if $\mathcal{A}$ is maximal in $\mathcal{L}$ and $\mathcal{B}$ is maximal in $\mathcal{A}$, then $\mathcal{B}$ is maximal in $\mathcal{L}$. Maximality here means the rank of the subalgebra is the same as the rank of the containing algebra.

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