Symmetry breaking to a special subalgebra?

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This is a follow-up to my question here.

• For regular subalgebras of some group's Lie algebra the root system of the subalgebra is a subset of the root system of the original's group algebra. In other words, the generators of the subalgebra are a subset of the generators of the original group.
• Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators.

Using the Higgs mechanism, as explained nicely by @Heterotic in the question I linked to above, we simply check which generators remain unbroken after one or more Higgs fields get a vev. Then:

"The remaining subgroup after the symmetry breaking is simply the group generated by the unbroken generators."

This is certainly correct for regular subalgebras, but how does this work for special subalgebras? How can we determine that we broke to a special subalgebra by giving a vev to a Higgs field, if the generators of the subalgebra are not a subset of the original's group generators?

This post imported from StackExchange Physics at 2015-11-01 19:26 (UTC), posted by SE-user JakobH

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Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators.

That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is meant by a root system not being a subset of another root system. It does not imply that the subalgebra generators are not a subset of the algebra generators.

Example: Consider the three dimensional representation of $\mathfrak g=\mathfrak{su}(3)$, whose generators $T_a=\lambda_a/2$, where $\lambda_a$ are the Gell-Mann matrices. The step operators are
\begin{align*}
E_{\pm\alpha_1}=T_1\pm iT_2,\\
E_{\pm\alpha_2}=T_6\pm iT_7,\\
E_{\pm(\alpha_1+\alpha_2)}=T_4\pm iT_5,
\end{align*}

Regular embedding: The generators $T_1$, $T_2$ and $T_3$ form a subalgebra $\mathfrak{su}(2)$. This subalgebra step operators are $t_\pm=T_1\pm iT_2=E_{\pm\alpha_1}$. The subalgebra step operators form a subset of the algebra step operators.

Special embedding: Another $\mathfrak{su}(2)$ subalgebra is given by $T_2$, $T_5$ and $T_7$. In this case the step operators are $t_\pm=T_5\pm iT_7$ which cannot be written as a subset of the step operators of $\mathfrak{su}(3)$.

Notice the branching rules for the first embedding is $3=2\oplus 1$ while for the second is $3=3$.

answered Jul 12, 2016 by (90 points)

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