Weight modules
The study of the representations of Lie algebras is facilitated by describing the Lie algebras as weight modules.
Let {$\frak{g}$} be a Lie algebra with Cartan subalgebra {$\frak{h}$} of {$\frak{g}$}. Let {$\sigma:\frak{g}$}{$\rightarrow\textrm{End}(V)$} be a representation of {$\frak{g}$} on a vector space {$V\cong\mathbb{C}^n$}. Let {$\lambda:\frak{h}\rightarrow\mathbb{C}$} be a linear functional on {$\frak{h}$} and let {$\frak{h}^*$} be the subspace of such linear functionals. The weight space of {$V$} with weight {$\lambda$} is the subspace
{$V_\lambda\def \{v\in V\;|\; \forall H \in\frak{h},$}{$(\sigma(H))(v)=\lambda(H)v$}
{$V$} is a weight module if it is the direct sum of its weight spaces.
{$V=\oplus_{\lambda\in\frak{h}^*}$}{$V_\lambda$}
An inspiration: Symmetric group {$S_3$}
Basic example: {$\frak{sl}_2\mathbb(C)$}
These are the {$2\times 2$} traceless complex matrices. We use the basis
{$H=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}, X=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}, Y=\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$}
{$H$} generates the one-dimensional Cartan subalgebra. We calculate
{$ [H,X]=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}-\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \begin{pmatrix} 0 & 1-(-1) \\ 0 & 0 \\ \end{pmatrix} = 2X$}
{$ [H,Y]=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}-\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1-(1) & 0 \\ \end{pmatrix} = -2Y$}
{$ [X,Y]=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}-\begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix} = H$}
If {$V$} is an irreducible finite-dimensional representation of {$\frak{sl}_2\mathbb{C}$}, then the action of {$H$} on {$V$} is diagonalizable. We can write
{$V=\oplus V_\alpha$} where {$\alpha$} are complex numbers such that for any vector {$v\in V_\alpha$} we have {$H(v)=\alpha\cdot v$}.
Orthogonal matrices
In the case {$m=2n$} is even, we define the quadratic form {$Q$} so that {$Q(e_i,e_j)=Q(e_j,e_i)=\delta_{i,i+n}$}.
This bilinear form can be expressed as
{$Q(x,y)= \begin{pmatrix} x_1 \cdots x_m & x_{m+1} \cdots x_n \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} y_1 \\ \vdots \\ y_m \\ y_{m+1} \\ \vdots \\ y_n \\ \end{pmatrix}$}
Then the Lie algebra {$\frak{so}_{2n}\mathbb{C}$} is the space of matrices {$X$} which satisfy {$X^TM+MX=0$}. This means that {$X$} looks like
{$X = \begin{pmatrix} B & A \\ D & C \\ \end{pmatrix}$} where {$B,C$} are skew-symmetric and {$D=-A^T$}.
The Cartan subalgebra {$\frak{h}$} is generated by the matrices {$H_i=1_{ii}-1_{i+n,i+n}$}. The dual subalgebra {$\frak{h}^*$} is given by {$L_j$} such that {$\langle L_j,H_i\rangle=\delta_{i,j}$}.
We can define {$X_{i,j}=1_{1,j}-1_{n+j,n+i}$} and {$Y_{i,j}=1_{i,n+j}-1_{j,n+i}$} and {$Z_{i,j}=1_{n+i,j}-1_{n+j,i}$}.