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How can I show that the cartesian product of the Lie algebras $\mathfrak{so}(5)$ (type $B_2$) with $\mathfrak{sl}(2)$ (type $A_1$) is isomorphic to a subalgebra of the Lie algebra of type $F_4$?

I'm using $[(x,y),(z,w)]=([x,z],[y,w])$.

I've been trying to construct the Dynkin diagram of $\mathfrak{so}(5)\times\mathfrak{sl}(2)$, but I feel there's an easier argument.

Jendrik Stelzner
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Felipe
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1 Answers1

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Well any subdiagram of the Dynkin diagram gives a root subsystem and thus a Lie subalgebra. This is fairly straightforward to prove with a little familiarity with root systems.

In fact we can take this further. You can extend the diagram by adding in a node corresponding to the lowest root (or the lowest short root) before you take a subdiagram. In $F_4$ these extended diagrams are those with an extra node at one of the ends. You can then (with either of these diagrams) find $B_2\times A_1$ as a subdiagram. The $A_1$ part will be the new node and the $B_2$ will be the two nodes in the middle of the original diagram.

Note in one case the $A_1$ part is given by a short root and in the other case by a long root so these won't be conjugate for example.

Callum
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  • Nice. Did you mean to use the highest short root? – Felipe Dec 10 '22 at 15:20
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    No, the lowest is correct. It is the negative of the highest root. If you look up the affine extended diagrams these are exactly the Dynkin diagrams extended by adding a node corresponding to the lowest root. You have to search a little harder to find the alternative extended diagrams for the lowest short root but here are some drawings of them. – Callum Dec 10 '22 at 15:55
  • Cool. That answers my other question too. Thank's a lot for your help. – Felipe Dec 10 '22 at 17:09