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The Lie algebra $so_5$ has 10 elements and its root structure is given by the Dynkin diagram B2. I have been having trouble creating an explicit $5 \times 5$ complex matrix representation of its 10 elements from its Cartan matrix. I would greatly appreciate help with this.

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closed as no longer relevant by S. Carnahan Jan 31 '11 at 3:19

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I am not sure I understand this question. Why would you expect the Cartan matrix to give you a particular matrix representation? The Cartan matrix gives you the structure of the Lie algebra using the Serre relations. This is explained in a variety of places, e.g., Humphrey's book on Lie algebras and representation theory. $$ $$ The compact real form of the Lie algebra of type B2 is the algebra of skewsymmetric endomorphisms of a five-dimensional euclidean space. It doesn't get any more explicit than that: just take the 5x5 skewsymmetric matrices. – José Figueroa-O'Farrill Jan 16 '11 at 20:28
The question is: what are the matrices corresponding to the Chevalley generators, i.e. the famous (e_i,f_i,h_i)? – Guntram Jan 16 '11 at 21:26
@Guntram: that's also not a research-level question, Allen's answer notwithstanding. See his answer for how to go about it. In any case, you don't get this from the Cartan matrix. – José Figueroa-O'Farrill Jan 17 '11 at 6:54
Thanks, I have found this answer helpful. Also, I have just discovered the book, "Notes on Lie Algebras" by Hans Samelson, which has clarified the issue. – garretstar Jan 17 '11 at 18:13

While I'm not sure this question is appropriate for this site, here goes.

First, you need a maximal torus. Inside SO(5) we have SO(4) and then SO(2) x SO(2).

Write your antisymmetric matrices as

$$\begin{matrix} aJ & C & v \end{matrix} $$ $$\begin{matrix} -C^T & bJ & w \end{matrix} $$ $$\begin{matrix} -v^T & -w^T & 0 \end{matrix} $$ where $J = \left( {0\atop -1}{1\atop 0} \right) $, $C$ is square, and $v$ and $w$ are columns. Then the $a$ and $b$ parts are the torus, the $v$ gets you the $\pm x_1$ weights, the $w$ gets you the $\pm x_2$, and the $C$ gets you the $\pm x_1\pm x_2$.

Taking $x_1$ and $x_2 - x_1$ as simple roots, the $e_{x_1}$ is $$ 0 0 0 0 1 $$ $$ 0 0 0 0 i $$ $$ 0 0 0 0 0 $$ $$ 0 0 0 0 0 $$ $$ -1 -i 0 0 0 $$ and the $e_{x_2-x_1}$ is $$ 0 0 1 i 0 $$ $$ 0 0 i -1 0 $$ $$ -1 -i 0 0 0 $$ $$ -i 1 0 0 0 $$ $$ 0 0 0 0 0 $$ Sorry for the ugly matrices -- I'm having trouble getting the matrix environment working here.

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