Cartan-Weyl basis for the complexified Lie algebra $L_{\mathbb{C}}(SU(N))$ I'm trying to construct the Cartan-Weyl basis for $L_{\mathbb{C}}(SU(N))$. 
Looking at the basis for the complexified lie algebra for $SU(2)$ (consisting of the cartan element and the step operators), is there a straightforward generalisation here? The basis for the Cartan subalgebra is clear, so do we just shift the Cartan-Weyl basis for $SU(2)$ down the diagonal? 
 A: It's not so clear what you mean but there are very simple ways of obtaining what I think you want.
Define the matrix $N\times N$ matrix
\begin{align}
E_{ij}= \left\{\begin{array}{cl}
1&\hbox{at position}\ (i,j)\\
0&\hbox{elsewhere}
\end{array}\right.
\end{align}
Thus for $N=3$ we have 
\begin{align}
E_{11}= \left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)\, ,\qquad E_{12}=\left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)\, ,\qquad E_{13}=\left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right)
\end{align}
etc.  One can then define a Cartan subalgebra for $\mathfrak{su}(3)$ using
\begin{align}
h_1=E_{11}-E_{22}\, ,\qquad h_2=E_{22}-E_{33}
\end{align}
The $E_{ij}$ with $i<j$ are raising operators and $E_{ij}$ with $i>j$ are lowering operators.  
For instance, in this simple $3$-dimensional representation, the vector
$\vert 1\rangle\to (1,0,0)^\top$ is the highest weight vector, has weight $(1,0)$, and is killed by $E_{12}$, $E_{23}$ and $E_{13}$, i.e. is killed by all the raising operators.  
This obviously generalizes to an $\mathfrak{su}(n)$.
