How to find a Cartan subalgebra of $so(3)$.

Question

Let $so(3)$ be the Lie algebra given by $$ so(3) = \{X \in \text{Mat}_{3 \times 3}: X^T = - X \}. $$ Here $\text{Mat}_{3 \times 3}$ is the set of all $3 \times 3$ matrices and $X^T$ is the transpose of $X$. How to find a Cartan subalgebra of $so(3)$? Thank you very much.

Answer 1

Using the standard basis of $\mathfrak{so}(3)$, given by $$ e_1=\begin{pmatrix} 0 & 1 & 0 \cr -1 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix},\; e_2=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr -1 & 0 & 0 \end{pmatrix},\; e_3=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & -1 & 0 \end{pmatrix}, $$ the Lie brackets are given by commutator, i.e., $$ [e_1,e_2]=-e_3,\;[e_1,e_3]=e_2,\;[e_2,e_3]=-e_1. $$

A Cartan subalgebra of a simple (complex) Lie algebra is given by a maximal abelian Lie subalgebra $H$ consisting of diagonalizable elements. Now it is easy to see that every subalgebra of dimension $k>1$ of $\mathfrak{so}(3)$ is not abelian. Hence a Cartan subalgebra $H$ must have dimension $1$. The difficulty now is to find a diagonalizable element, generating $H$. This depends on the field. If the field is algebraically closed of characteristic zero, say $\mathbb{C}$, then each basis elements is diagonalizable. Over the real numbers one usually tries to avoid this difficulty and passes to another real Lie algebra, with is isomorphic to $\mathfrak{so}_3(\mathbb{R})$, namely the Lie algebra $\mathfrak{su}(2)$, with the Pauli matrices as basis. Then $H=\langle \sigma_3\rangle$, see here. Using the isomorphism $\phi\colon \mathfrak{su}(2)\rightarrow \mathfrak{so}_3(\mathbb{R})$ we obtain a Cartan subalgebra for $\mathfrak{so}_3(\mathbb{R})$.