Folding and realization of type B Lie group as a subgroup of $GL_n$.

Question

We can fold the Dynkin diagram of type $A_{2n-1}$ and obtain the Dynkin diagram of type $B_n$. For example, we fold $A_3$ and obtain $B_2$. According to the article, type $B_k$ Lie group is $SO(2k+1)$. So $B_2$ Lie group is $SO(5)$. But type $A_3$ Lie group is $SL_4$. How to realize $SO(5)$ as a subgroup of $SL_4$? Thank you very much.

Answer 1

Your first claim is not true in general: Correctly "folding" a root system of type $A_{2n-1}$ gives a root system of type $C_n$, not $B_n$.

Now in your special case, it happens that $B_2=C_2$ and hence there is an exceptional isomorphism $\mathfrak{so}(5) \simeq \mathfrak{sp}(4)$. On the group level, some centres might intervene, cf. the answers here. But certainly you can realise $Sp(4)$, which would be the "right" group to consider here, as subgroup of $SL_4$.

By the way, are you studying these foldings and their relations to group/Lie algebra morphisms on your own, or do you have a good source? The precise relation has been bugging me for quite a while and I always like to be enlightened about it, cf. my comment to this MO answer.