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.