Classification of Lie Algebra

Recall that the Lie group representation only depends a linear combination of generators as in its exponential form  ${\cal{D}}(\alpha)= e^{i\,\alpha^a {\cal{T}}_a}$, we are able to choose a particular set of generators for the classification of Lie algebra.

We consider the compact Lie groups whose representation can always be unitary and thus their generators are Hermitian ${\cal{T}}_a={\cal{T}}^{\dagger}_a$.

Cartan generators

Out of all the generators, we can construct a maximal subset of mutually commuting Hermitian generators (Cartan generators), ${\cal{H}}_i$ for $i=1,\cdots, m$, such that \begin{equation}{\cal{H}}_i={\cal{H}}^{\dagger}_i\,,\quad \Big[{\cal{H}}_i\,,{\cal{H}}_j\Big]=0\,,\quad\text{Tr}\left({\cal{H}}_i{\cal{H}}_j\right)=k_D\,\delta_{ij}\,\tag{1}\end{equation} for some positive constant $k_D$.

The construction of such Cartan generators is in the following steps: 

1. From the generators ${\cal{T}}_1, \cdots,{\cal{T}}_N$, we can pick a maximal subset of mutually commuting generators, say ${\cal{T}}_1, \cdots,{\cal{T}}_m$, such that \begin{equation}\Big[{\cal{T}}_i\,,{\cal{T}}_j\Big]=0\end{equation} for $i,\,j=1,\cdots, m$.
2. Compute a $m\times m$ matrix $\mathbf{G}$ whose matrix element is defined by \begin{equation}\mathbf{G}_{ij}\equiv\text{Tr}({\cal{T}}_i\,{\cal{T}}_j)\end{equation} for $i,\,j=1,\cdots, m$.
3. Since $\mathbf{G}$ is a real symmetric matrix, it can be diagonalized \begin{equation}\mathbf{O}\,\mathbf{G}\,\mathbf{O}^T=\text{diag}[\lambda_1,\cdots, \lambda_m]\end{equation} by an orthonormal matrix $\mathbf{O}$.
4. Finally, define a set of new generators \begin{equation}{\cal{H}}_i\equiv \sqrt{\frac{k_D}{\lambda_i}}\sum_{j=1}^m \mathbf{O}_{ij}\,{\cal{T}}_j\,,\end{equation} which satsifies all the properties in Eq. (1).

Remarks:

• Cartan generators can be simultaneously diagonalized as \begin{equation}{\cal{H}}_i |\mu\rangle =\mu_i|\mu\rangle\,,\tag{2}\end{equation} where $\mu$ is a $m$-component vector called weights.
• In Lie algebra $\mathfrak{su}(2)$, $m=1$, $\cal{H}_1=\cal{J}_3$ and $k_D=1/2$.

Raising and lowering operators

Raising operators

Out of the remaining generators, we can construct a set of raising operators $\cal{E}_{\alpha}$ such that \begin{equation}\Big[{\cal{H}}_i\,,{\cal{E}}_{\alpha}\Big]=\alpha_i\,\cal{E}_{\alpha}\,,\tag{3}\end{equation} where $\alpha$ is a $m$-component vector called roots. 

Note that $\cal{E}_{\alpha}$ in Eq. (3) can be up to an arbitrary constant. So we impose the same normalization as Eq. (1) \begin{equation}\text{Tr}\left({\cal{E}}^{\dagger}_{\alpha}\,{\cal{E}}_{\alpha}\right)=k_D\,.\tag{4}\end{equation}

Remarks:

• The generator $\cal{E}_{\alpha}$ in Eq. (3) is called raising operator because the weight increases from $\mu$ to $\mu+\alpha$ when acting $\cal{E}_{\alpha}$ on the state $|\mu\rangle$ in Eq. (2): \begin{equation} {\cal{H}}_i\,\cal{E}_{\alpha}|\mu\rangle =\cal{E}_{\alpha}\,{\cal{H}}_i|\mu\rangle+\left[{\cal{H}}_i\,,\cal{E}_{\alpha}\right]|\mu\rangle = (\mu_i+\alpha_i)\,\cal{E}_{\alpha}|\mu\rangle\,. \end{equation}
• In Lie algebra $\mathfrak{su}(2)$, such $\cal{E}_{\alpha}$ is $\frac{1}{\sqrt{2}}{\cal{J}}_+\equiv \frac{1}{\sqrt{2}}\left(\cal{J}_1+i\cal{J}_2\right)$ and the commutator (3) is $[\cal{J}_3\,,\cal{J}_+]=\cal{J}_+$.

The construction of the raising operator in Eq. (3) is in the following steps:

1. For each Cartan generator $\cal{H}_i$, find its adjoint representation $\cal{A}_i$ in the representation space spanned by ${\cal{T}}_1, \cdots,{\cal{T}}_N$.
2. Find the eigenvalues and eigenvectors of $\cal{A}_i$ \begin{equation}{\cal{A}}_i\,\left| {\cal{E}}_{\alpha}\right\rangle=\alpha_i\,\left| {\cal{E}}_{\alpha}\right\rangle\,.\tag{5}\end{equation}
3. Expand the eigenvectors in the basis \begin{equation}\left| {\cal{E}}_{\alpha}\right\rangle =c_{m+1}\left| {\cal{T}}_{m+1}\right\rangle+c_{m+2}\left| {\cal{T}}_{m+2}\right\rangle+\cdots+c_{N}\left| {\cal{T}}_{N}\right\rangle\,.\tag{6}\end{equation}
4. Finally, the generator $\cal{E}_{\alpha}$ in Eq. (3) is \begin{equation} {\cal{E}}_{\alpha}\propto c_{m+1}\,{\cal{T}}_{m+1}+c_{m+2}\,{\cal{T}}_{m+2}+\cdots+c_{N}\,{\cal{T}}_{N}\,,\end{equation} up to a constant to in order to satisfy the normalization (3).

Remarks:

• The key insight for the classification of Lie algebra is to work with adjoint representation because of the property \begin{equation}{\cal{A}}_i\,\left|{\cal{T}}_j\right\rangle=\left|\left[{\cal{H}}_i, {\cal{T}}_j\right]\right\rangle\end{equation}for the adjoint representation $\cal{A}_i$. This is why the eigenvalue equation of the adjoint representation (4) leads to the commutator (3).
• The expansion (5) does not contain the terms of Cartan generators because ${\cal{A}}_i\,\left|{\cal{H}}_j\right\rangle=\left|\left[{\cal{H}}_i, {\cal{H}}_j\right]\right\rangle=\mathbf{0}$.
• From Eq. (2) and (4), roots are the weights in the adjoint representation.

Lowering operators

By taking the Hermitian conjugate of the commutator (3), we have \begin{equation} \Big[{\cal{H}}_i\,,{\cal{E}}^{\dagger}_{\alpha}\Big]=-\alpha_i\,\cal{E}^{\dagger}_{\alpha}\,,\end{equation} which suggests that we can take \begin{equation}\cal{E}^{\dagger}_{\alpha}=\cal{E}_{-\alpha}\,.\tag{7}\end{equation} $\cal{E}_{-\alpha}$ is the lowering operator that is similar to $\frac{1}{\sqrt{2}}{\cal{J}}_-\equiv \frac{1}{\sqrt{2}}\left(\cal{J}_1-i\cal{J}_2\right)$ in Lie algebra $\mathfrak{su}(2)$.

Finally, in Lie algebra $\mathfrak{su}(2)$, there is also commutator $\left[\cal{J}_+\,,\cal{J}_-\right]=2\,\cal{J}_3$. This motivates to consider the commutator $\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]$. By the Jacobi identity, we have \begin{equation}\Big[{\cal{H}}_i\,,\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]\Big]=\Big[\Big[{\cal{H}}_i\,,{\cal{E}}_{\alpha}\Big]\,, {\cal{E}}_{-\alpha}\Big]+\Big[{\cal{E}}_{\alpha}\,,\Big[{\cal{H}}_i\,,{\cal{E}}_{-\alpha}\Big]\Big]=\left(\alpha_i-\alpha_i\right)\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]=0\,,\end{equation} which suggests that $\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]$ is a linear combination of Cartan generators, i.e. \begin{equation}\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]=\sum_{j=1}^m\beta^j {\cal{H}}_j\,.\end{equation} To determine the coefficients $\beta^i$, we multiply $\cal{H}_i$ on both sides of the above equation, take the trace, apply Eq. (1) and (8), and obtain \begin{equation}\beta^i=\frac{1}{k_D}\text{Tr}\left(\Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]{\cal{H}}_i\right)=\frac{1}{k_D}\text{Tr}\left(\Big[{\cal{H}}_i\,,{\cal{E}}_{\alpha}\Big]{\cal{E}}_{-\alpha}\right)=\frac{\alpha_i}{k_D}\text{Tr}\left({\cal{E}}_{\alpha}{\cal{E}}_{-\alpha}\right)=\alpha_i\,.\end{equation}

Each root forms a $\mathfrak{su}(2)$

In summary, starting with the generators ${\cal{T}}_1, \cdots,{\cal{T}}_N$, we find a different set of generators, $\cal{H}$s and $\cal{E}$s, satisfying the commutation relations \begin{equation} \color{red}{\Big[{\cal{H}}_i\,,{\cal{E}}_{\pm\alpha}\Big]=\pm\alpha_i\,{\cal{E}}_{\pm\alpha}\,,\quad \Big[{\cal{E}}_{\alpha}\,,{\cal{E}}_{-\alpha}\Big]=\sum_{j=1}^m \alpha_j {\cal{H}}_j}\,.\tag{8}\end{equation}

For each root $\alpha$, let \begin{equation}{\cal{E}}_{\pm}\equiv\frac{1}{|\alpha|}{\cal{E}}_{\pm\alpha}\,,\quad{\cal{E}}_{3}\equiv\sum_{j=1}^m \frac{\alpha_j}{|\alpha|^2} {\cal{H}}_j\,,\tag{9}\end{equation} where $|\alpha|\equiv\sqrt{\sum_{j=1}^m\alpha^2_j}$, the commutation relations (8) becomes $\mathfrak{su}(2)$: \begin{equation}\Big[{\cal{E}}_{3}\,,{\cal{E}}_{\pm}\Big] =\pm {\cal{E}}_{\pm}\,,\quad\Big[{\cal{E}}_{+}\,,{\cal{E}}_{-}\Big] = {\cal{E}}_{3}\,,\end{equation} which is identical to $\mathfrak{su}(2)$ with $\cal{E}_3=\cal{J}_3$ and $\cal{E}_{\pm}=\frac{1}{\sqrt{2}}\cal{J}_{\pm}$.

Classification of Lie Algebra

From Eq. (2) and (9), we have \begin{equation}{\cal{E}}_3\,|\mu\rangle=\frac{\alpha\cdot\mu}{|\alpha|^2}\,|\mu\rangle\end{equation} where $\alpha\cdot\mu\equiv \sum_{j=1}^m\alpha_j\mu_j$. Recall that in Lie algebra $\mathfrak{su}(2)$, the eignvalue of $\cal{J}_3$ takes integer or half-integer values, so \begin{equation}2\frac{\alpha\cdot\mu}{|\alpha|^2}\in \mathbb{N}\,.\end{equation}

For finite representations, let $p, q$ be the maximal integer such that ${\cal{E}}^p_+|\mu\rangle\neq \mathbf{0}$ and ${\cal{E}}^q_-|\mu\rangle\neq \mathbf{0}$, then ${\cal{E}}^p_+|\mu\rangle$ and ${\cal{E}}^q_-|\mu\rangle$ are the highest and lowest weight states $|j\rangle$ and $|-j\rangle$ in Lie algebra $\mathfrak{su}(2)$, and thus they should have the equal weight: \begin{equation}\frac{\alpha\cdot\mu}{|\alpha|^2}+p=j\,,\quad\frac{\alpha\cdot\mu}{|\alpha|^2}-q=-j\,.\quad \end{equation} As a result, we have \begin{equation}\color{red}{\frac{\alpha\cdot\mu}{|\alpha|^2}=-\frac{1}{2}(p-q)}\,.\tag{10}\end{equation}

Eq. (10) is the key result for the classification of Lie algebra. 

Recall that roots are a special weights, we can take $\mu$ in Eq. (10) as another root $\beta$, and obtain the constraint \begin{equation}\frac{\alpha\cdot\beta}{|\alpha|^2}=-\frac{1}{2}(p-q)\,.\end{equation} Similarly, \begin{equation}\frac{\beta\cdot\alpha}{|\beta|^2}=-\frac{1}{2}(p'-q')\,.\end{equation} Therefore, \begin{equation}\cos^2\theta_{\alpha\beta}\equiv \frac{(\alpha\cdot \beta)^2}{|\alpha|^2\,|\beta|^2}=\frac{(p-q)(p'-q')}{4}\,.\tag{11}\end{equation} That is, the angle between any two root vectors can only take values among $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$, $120^\circ$, $135^\circ$, $150^\circ$ and $180^\circ$.