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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3$\begingroup$ 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. $\endgroup$– José Figueroa-O'FarrillCommented Jan 16, 2011 at 20:28
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$\begingroup$ The question is: what are the matrices corresponding to the Chevalley generators, i.e. the famous (e_i,f_i,h_i)? $\endgroup$– GuntramCommented Jan 16, 2011 at 21:26
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2$\begingroup$ @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. $\endgroup$– José Figueroa-O'FarrillCommented Jan 17, 2011 at 6:54
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$\begingroup$ 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. $\endgroup$– garretstarCommented Jan 17, 2011 at 18:13
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1 Answer
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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 \begin{pmatrix} 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 \end{pmatrix} and the $e_{x_2-x_1}$ is \begin{pmatrix} 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 \end{pmatrix}
answered Jan 17, 2011 at 1:06