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I was reading a paper in which a part of it they want to classify the closed connect subgroups of $SO(5)$. What they write is this:

Let $G^0$ be a closed connected subgroup of $SO(5)$. Let $T$ be a maximal Torus of $G^0$, then it is contained in the 2-dimensional maximal torus of $SO(5)$. Let $\mathfrak{h}$ be the Lie algebra of $G^0$. By the classification of Dynkin diagrams $\mathfrak{h}_\mathbb{C}$ must be isomorphic to

$$\mathfrak{t}_1, \mathfrak{sl}_2 = \mathfrak{so}_3 \mbox{ if } \dim(T) = 1$$ $$\mathfrak{t}_2, \mathfrak{t}_1\times\mathfrak{sl}_2, \mathfrak{sl}_2\times\mathfrak{sl}_2 = \mathfrak{so}_4, \mathfrak{sl}_3, \mathfrak{so}_5, \mathfrak{g}_2 \mbox{ if } \dim(T) =2 $$

My question is how did they came up with these Lie algebras. I think I know where they get most of them from.

$\mathfrak{t}_n$ is the abelian Lie algebra of dimension $n$

$\mathfrak{sl}_2$ corresponds to the Dynkin diagram $A_1$

$\mathfrak{so}_5$ corresponds to the Dynkin diagram $B_2$

$\mathfrak{g}_2$ correspond to the exceptional Dynkin diagram $G_2$

The other three I am not so sure about. I would believe that $\mathfrak{t}_1 \times \mathfrak{sl}_2$ and $\mathfrak{sl}_2 \times \mathfrak{sl}_2$ just come from the fact that if we have two Lie algebras $\mathfrak{g}, \mathfrak{h}$ with maximal Torus of dimension $m$ and $n$ respectively then $\mathfrak{g}\times\mathfrak{h}$ has a maximal Torus with dimension $m+n$. I am not sure if this is true though.

However, I can't see where they are getting $\mathfrak{sl}_3$ from. The only other Dynkin diagram missing is $A_2$ but doesn't this correspond with $\mathfrak{su}_2$ and not $\mathfrak{sl}_3$.

I would greatly appreciate confirmation that my analysis in all the non $\mathfrak{sl}_3$ were correct as well as clarification as to where $\mathfrak{sl}_3$ comes from.

Mastrel
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  • This just comes from the classification of complex semisimple Lie algebras. In fact, if they knew better they would have known that it suffices to search for maximal semisimple Lie subalgebras. Dynkin proved in 1950s that these are obtained by removing one node of the extended Dynkin diagram of the original Lie algebra. This works for general complex Lie semisimple algebras. – Moishe Kohan Aug 29 '14 at 01:59
  • Note that for the $A_2$ case, $\mathfrak{su}(3)_{\mathbb{C}} = \mathfrak{sl}_3(\mathbb{C})$ (and $\mathfrak{su}(2) = \mathfrak{so}(3)$ has complexification $A_1$). – Travis Willse Aug 29 '14 at 07:29

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