Examples: Rank Three

Section 5.7 Examples: Rank Three

Table 5.7.1. The Dynkin diagrams for simple Lie algebras of rank \(3\text{.}\)

\(\aa_3\)	\(\begin{aligned} \bullet\!-\!\bullet\!-\!\bullet \end{aligned}\)
\(\bb_3\)	\(\begin{aligned} \bullet=\!\!\Leftarrow\bullet\!-\!\bullet \end{aligned}\)
\(\cc_3\)	\(\begin{aligned} \bullet\Rightarrow\!\!=\bullet\!-\!\bullet \end{aligned}\)
\(\dd_3\)	\(\begin{aligned} \amp\>\bullet \\[-1ex] \amp\,\,\mid \\[-1ex] \bullet\,\!-\!\amp\>\bullet \end{aligned}\)

With three simple roots, the root diagrams are even more interesting. The corresponding Dynkin diagrams are shown in Table 5.7.1.

We begin with the root diagram of \(\aa_3\cong\su(4)\text{,}\) shown in Figure 5.7.2. You should be able to find several copies of of the root diagram for \(\aa_2\cong\su(3)\) contained in this diagram!

Figure 5.7.2. The root diagram of \(\aa_3\cong\su(4)\text{.}\)

Although we have included “\(\dd_3\)” in Table 5.7.1, its Dynkin diagram is clearly equivalent to the one for \(\aa_3\text{.}\) In other words, from the Dynkin diagram alone we can tell that

\begin{equation} \su(4) \cong \aa_3 \ cong \dd_3 \cong so(6)\tag{5.7.1} \end{equation}

the last remaining “special” isomorphisms between simple Lie algebras of different Cartan–Killing type. So Figure 5.7.2 is also the root diagram for \(\so(6)\text{.}\)

In order to get from \(\aa_3\) to either \(\bb_3\) or \(\cc_3\text{,}\) we must replace one of the simple roots by a simple root with a different magnitude, which can be either shorter (\(\bb_3\)) or longer (\(\cc_3\)), resulting in the root diagrams shown in Figures 5.7.3–5.7.4. These two root diagrams clearly differ from each other, unlike the case in rank \(2\text{,}\) where \(\bb_2\cong\cc_2\text{.}\)

Figure 5.7.3. The root diagrams for \(\bb_3\cong\so(7)\text{.}\)

Figure 5.7.4. The root diagram of \(\cc_3\cong\su(3,\HH)\text{.}\)

Activity 5.7.1. Subalgebras of \(\bb_3\) and \(\cc_3\).

Using the root diagram alone, determine whether \(\aa_3\) is a subalgebra of one or both of \(\bb_3\) and \(\cc_3\text{.}\)

Hint.

Since \(\aa_3\cong\so(6)\) and \(\bb_3\cong\so(7)\text{,}\) it must be the case that \(\aa_3\subset\bb_3\text{.}\)

Solution.

As was the case for Activity 5.6.1, short roots can commute to long roots, but not the other way around. So in order to avoid changing the commutators, we want \(\so(6)\) to be built from long roots. This argument correctly suggests that the “extra” roots should be short. So adding a short root (to obtain \(\bb_3\)) should ensure that \(\so(6)\) remains as a subalgebra, whereas adding a long root (to obtain \(\cc)3\)) does not.