= Solution
For a complex finite-dimensional <simple Lie algebra>, a <Cartan subalgebra> $\mathfrak h$ is a maximal commuting subalgebra of elements whose adjoint maps are semisimple. Equivalently in this setting it is a nilpotent self-normalizing subalgebra. Its dimension is the <rank of a semisimple Lie algebra>. Simultaneous diagonalization of its <Adjoint representation> gives the <root-space decomposition>
$$
\mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha,\qquad \mathfrak g_\alpha=\{E:[H,E]=\alpha(H)E\text{ for all }H\in\mathfrak h\}.
$$
A <root of a root system> is a nonzero linear functional $\alpha$ for which this <root space> is nonzero. For a complex semisimple algebra each <root space> is one-dimensional. A <Cartan-Weyl basis> consists of a basis $H_i$ of $\mathfrak h$ and one nonzero <root vector> $E_\alpha$ for every root.
The general <Lie brackets> have the form
$$
[H_i,H_j]=0,\qquad[H_i,E_\alpha]=\alpha(H_i)E_\alpha,
$$
$$
[E_\alpha,E_\beta]=\begin{cases}N_{\alpha\beta}E_{\alpha+\beta},&\alpha+\beta\in\Phi,\\\text{an element of }\mathfrak h,&\beta=-\alpha,\\0,&\alpha+\beta\notin\Phi\cup\{0\}.\end{cases}
$$
For the opposite-root bracket, use the <Killing form> to define $h_\alpha$ by $\kappa(h_\alpha,H)=\alpha(H)$. Its <invariant bilinear form on a Lie algebra> property gives
$$
[E_\alpha,E_{-\alpha}]=\kappa(E_\alpha,E_{-\alpha})h_\alpha.
$$
One may normalize the <root vectors> so that the pairing is one. If instead one uses a <coroot> as the opposite-root bracket, the root-vector normalization changes accordingly. In particular, root evaluation coordinates cannot simply be used as coefficients in a nonorthonormal Cartan basis.
For the matrix calculation take $N\ge2$. The <complexification of a Lie algebra> of the <special unitary group> <Lie algebra> is the <special linear Lie algebra> $\mathfrak{sl}_N(\mathbb C)$: traceless complex matrices. Its <Cartan subalgebra> consists of traceless diagonal matrices. Write $E_{jk}=\mathcal T^{(j,k)}$ for the <matrix units>. The given Cartan basis is $H_i=E_{ii}-E_{i+1,i+1}$, $1\le i<N$, and the other basis elements are $E_{jk}$ with $j\ne k$.
The matrix-unit identity $E_{jk}E_{lm}=\delta_{kl}E_{jm}$ gives
$$
[H_i,E_{jk}]=\left(\delta_{ij}-\delta_{i+1,j}-\delta_{ik}+\delta_{i+1,k}\right)E_{jk}.
$$
Thus all the roots, expressed as evaluation vectors in this precise Cartan basis, are
$$
\boxed{(\alpha_{jk})_i=\alpha_{jk}(H_i)=\delta_{ij}-\delta_{i+1,j}-\delta_{ik}+\delta_{i+1,k},\quad j\ne k,\quad 1\le i<N.}
$$
They are the functionals $e_j-e_k$ on traceless diagonal matrices; there are $N(N-1)$ of them. The corresponding <root vector> is $E_{jk}$. Together with $N-1$ Cartan generators, they give $N^2-1$ basis elements. The <simple roots> can be chosen as $\alpha_{i,i+1}$, whose evaluation vectors are the rows of the type-$A_{N-1}$ <Cartan matrix>, with $2$ on the diagonal and $-1$ on adjacent entries. These vectors are evaluations on $H_i$, not coordinates in an orthonormal realization of the <root system>.
To express every bracket strictly in the chosen basis, introduce the abbreviation
$$
D_{jk}=E_{jj}-E_{kk}=\begin{cases}\displaystyle\sum_{p=j}^{k-1}H_p,&j<k,\\\displaystyle-\sum_{p=k}^{j-1}H_p,&j>k.\end{cases}
$$
Then all pairs are covered by
$$
\boxed{[H_i,H_l]=0,\qquad[H_i,E_{jk}]=(\alpha_{jk})_iE_{jk},}
$$
$$
\boxed{[E_{jk},E_{lm}]=\begin{cases}D_{jk},&k=l,\ j=m,\\E_{jm},&k=l,\ j\ne m,\\-E_{lk},&j=m,\ k\ne l,\\0,&k\ne l,\ j\ne m.\end{cases}}
$$
Here both input <root vectors> have distinct row and column indices. The first case is the only one producing diagonal <matrix units>, and the displayed sum of $H_p$ resolves them completely into the chosen Cartan basis. Reversing the order gives the negative bracket. This also shows explicitly that the two nonzero non-Cartan cases have <structure constants> $+1$ and $-1$, and verifies the required root-addition rule.
Back to article page
