# Regular maximal subalgebras of $D_6 \cong so(12)$

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?

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

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.