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The adjoint representation is ${\mathrm{ad}}_x(y)=[x,y]$. For a semisimple Lie algebra, its nonzero weights are the roots and its zero-weight space is the Cartan subalgebra.

The complex Adjoint representation of SU(3) is the action by conjugation on traceless $3\times 3$ matrices. With $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, its highest-weight vector is $E_{13}$ of Dynkin labels $(1,1)$. The six off-diagonal matrix units have weights $(2,-1),(-2,1),(-1,2),(1,-2),(1,1),(-1,-1)$, and the two-dimensional Cartan subalgebra has weight zero. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the weight diagram is a regular hexagon together with a central point of weight multiplicity two. This representation describes both the gluon colour multiplet and the approximate flavour baryon octet, under different physical SU(3) symmetries.

For a choice of positive roots, the corresponding Borel subalgebra is $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$, the sum of a Cartan subalgebra and all positive root spaces.

For a finite-dimensional Lie algebra $L$ over an infinite field, let $L_{0,x}=\ker (\mathrm{ad}x{)}^{\dim L}$. A member of minimum dimension among these generalized zero spaces is a Cartan subalgebra, using the standard fact that each $L_{0,x}$ is a self-normalizing Lie subalgebra. The key minimality argument is more general: if $A\subseteq H=L_{0,x}$ and $H$ minimizes dimension for $x\in A$, then $H\subseteq L_{0,y}$ for every $y\in A$.

Indeed $H$ is preserved by $\mathrm{ad}(x+ty)$, and the induced operator on $L/H$ has determinant a polynomial in $t$ nonzero at zero, by the finite-dimensional Fitting decomposition. For all but finitely many $t$, its generalized zero space lies in $H$ and hence equals $H$ by minimality. The restriction to $H$ is therefore nilpotent for all such $t$. Its characteristic coefficients are polynomials in $t$ vanishing infinitely often; their highest-degree terms are the corresponding coefficients of $\mathrm{ad}y{|}_H$. These vanish, so that restriction is nilpotent. For a global minimum set $A=H$; Engel's theorem makes $H$ nilpotent and self-normalization gives the result.

A Cartan-Weyl basis consists of a basis of a Cartan subalgebra together with root vectors $E_{\alpha }$. It makes weight raising and lowering explicit through $[H,E_{\alpha }]=\alpha (H)E_{\alpha }$.

A Chevalley basis imposes an additional integral normalization on these root vectors; that stricter basis is not interchangeable with an arbitrary Cartan-Weyl basis.

Choose five Hermitian four-by-four matrices obeying the Euclidean Clifford algebra relations ${\gamma }_a{\gamma }_b+{\gamma }_b{\gamma }_a=2{\delta }_{ab}I$. For $M_{ab}=E_{ab}-E_{ba}$, the displayed anti-Hermitian matrices satisfy the Special orthogonal Lie algebra Lie bracket relations. A direct proof first establishes $[\rho (M_{ab}),{\gamma }_c]={\delta }_{bc}{\gamma }_a-{\delta }_{ac}{\gamma }_b$, then applies the commutator product rule to $[{\gamma }_c,{\gamma }_d]$. One explicit choice is ${\gamma }_1={\sigma }_1\otimes I$, ${\gamma }_2={\sigma }_2\otimes I$, ${\gamma }_3={\sigma }_3\otimes {\sigma }_1$, ${\gamma }_4={\sigma }_3\otimes {\sigma }_2$, ${\gamma }_5={\sigma }_3\otimes {\sigma }_3$, with the Pauli matrices. The Cartan subalgebra operators $-i\rho (M_{12})$ and $-i\rho (M_{34})$ have simultaneous eigenvalues $(\pm 1/2,\pm 1/2)$, identifying the four-dimensional spinor representation. A full $2\pi$ rotation acts as $-I_4$, proving it does not descend to $SO(5)$.

Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

For a current subalgebra $\mathfrak{h}\subset \mathfrak{g}$, its currents have conformal weight one under both Sugawara construction stress tensors. Their difference therefore commutes with all $\mathfrak{h}$ currents and with the $\mathfrak{h}$ stress tensor. It obeys the Virasoro algebra with central charge $c_{\mathfrak{g}}-c_{\mathfrak{h}}$. For a positive-level compact unitary vacuum representation, $\Vert L_{-2}^{\mathfrak{g}/\mathfrak{h}}|0⟩{\Vert }^2=(c_{\mathfrak{g}}-c_{\mathfrak{h}})/2\ge 0$. A Cartan subalgebra has $c_{\mathfrak{h}}=\mathrm{rank}\mathfrak{g}$, giving the lower bound on $c_{\mathfrak{g}}$ under these hypotheses.

The real span of the simple coroots inside a complex Cartan subalgebra carries a positive-definite restriction of the Killing form. Equivalently it consists of the $h$ for which all roots have real values. The formula $\kappa (h,h)={\sum }_{\alpha \in \Phi }\alpha (h{)}^2$ is positive for $h\ne 0$, since the roots span the dual. The induced dual inner product makes the root system Euclidean and makes root reflections orthogonal.

For a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$, the shifted Harish-Chandra map identifies the centre of the universal enveloping algebra with the Weyl group-invariant polynomial functions on ${\mathfrak{h}}^{\ast }$. To normalize the shift, let $M^{\chi }$ have highest weight $\chi -\rho$, where $\rho$ is the Weyl vector. A central element $z$ acts by a scalar $j_z(\chi )$, and $z\mapsto j_z$ is the displayed algebra isomorphism. The Poincaré-Birkhoff-Witt theorem gives polynomial dependence on $\chi$ by projecting $z$ onto its pure Cartan component and evaluating at $\chi -\rho$. The associated central characters agree exactly when their shifted parameters lie in the same Weyl orbit. Without incorporating the shift into the parameter, the invariance is expressed using the dot action $w\cdot \lambda =w(\lambda +\rho )-\rho$.

The coordinates of a highest weight record its eigenvalues on a chosen basis of a Cartan subalgebra. They depend on that basis. Dynkin labels instead use the simple coroots; orthonormal Cartesian coordinates and Dynkin labels must be converted before applying a dimension or Casimir eigenvalue formula.

Let $\mathcal{B}(H)=B(H,\cdot )$ be the Killing form identification of a Cartan subalgebra with its dual. The root-space decomposition gives $B(H,K)={\sum }_{\beta \in \Phi }\beta (H)\beta (K)$, and therefore the displayed root expansion. On the real coroot span its coefficients are real, proving $\mathcal{B}({\mathfrak{h}}_{\mathbb{R}})\subseteq \mathbb{R}P$. For ${\mathfrak{sl}}_3$ and $H_{12}=E_{11}-E_{22}$, the formula gives $\mathcal{B}(H_{12})=6(L_1-L_2)$. The normalization is $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$, not a rescaled trace form.

// Target: semisimple-lie-algebra.bigb

For the antidiagonal symmetric matrix $J$ in dimension $2n$, put $\bar{i}=2n+1-i$ and $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The diagonal Cartan subalgebra has coordinates ${\epsilon }_i(H)=t_i$. The root spaces of weights ${\epsilon }_i-{\epsilon }_j$ are spanned by $E_{ij}-E_{\bar{j}\bar{i}}$; those of weights ${\epsilon }_i+{\epsilon }_j$ by $E_{i\bar{j}}-E_{j\bar{i}}$, with negative-root spaces given by transposes. The condition $A_{ij}=-A_{\bar{j}\bar{i}}$ gives these generators directly and excludes roots $2{\epsilon }_i$.

Relative to $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$, a symplectic Lie algebra matrix has blocks $\left(\begin{array}{cc}P& Q\\ R& -P^{\mathsf{T}}\end{array}\right)$ with $Q,R$ symmetric. Its diagonal Cartan subalgebra has basis $E_{ii}-E_{l+i,l+i}$. Write ${\epsilon }_i(h)={\lambda }_i$. A root-space decomposition is obtained fromThe matrix unit rule $[h,E_{ab}]=(h_{aa}-h_{bb})E_{ab}$ proves $[h,e_r]=r(h)e_r$. These root vectors and the diagonal basis span all the permitted blocks and are independent. There are $2l^2$ roots and $l$ diagonal basis elements, giving dimension $l(2l+1)$.

Over a field of characteristic different from two, any Lie algebra ideal in ${\mathfrak{sl}}_n$ containing a nonscalar matrix is the whole special linear Lie algebra. Indeed, if $A_{ji}\ne 0$ for $i\ne j$, then $[E_{ij},[E_{ij},A]]=-2A_{ji}{E}_{ij}$ extracts a matrix unit. For a nonscalar diagonal matrix, $[E_{ij},A]=(A_{jj}-A_{ii})E_{ij}$ extracts one instead. Its bracket with $E_{ji}$ produces $H=E_{ii}-E_{jj}$, and $[H,E_{ji}]=-2E_{ji}$ extracts the reverse unit. Brackets $[E_{ij},E_{jk}]=E_{ik}$ and $[E_{ki},E_{ij}]=E_{kj}$ spread these units to every row and column; opposite pairs span all trace-zero diagonal matrices. Thus only ideals consisting of scalar matrices can be proper. This proof works even when restrictions of distinct roots of a root system to a Cartan subalgebra coincide in positive characteristic.

For a complex semisimple Lie algebra, the Cartan subalgebra is orthogonal under the Killing form to every nonzero root space. A vector in the Cartan subalgebra orthogonal to that whole subalgebra is therefore orthogonal to the entire Lie algebra and must vanish. On the Cartan subalgebra the form is $\kappa (h,h^{\prime })={\sum }_{\alpha \in \Phi }\alpha (h)\alpha (h^{\prime })$.

For a root of a root system of a complex semisimple Lie algebra, define $t_{\alpha }$ by $B(t_{\alpha },h)=\alpha (h)$ using the Killing form on its Cartan subalgebra. Then $\alpha (t_{\alpha })\ne 0$. Otherwise vectors $x\in {\mathfrak{g}}_{\alpha }$, $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$ would generate a solvable Lie algebra with $[x,y]=t_{\alpha }$ central. The Lie theorem would make $\mathrm{ad}{t}_{\alpha }$ nilpotent, while the root-space decomposition makes it diagonalizable, forcing $t_{\alpha }$ into the zero center of a Lie algebra.

Differentiate the defining relation at $\Lambda =I+tX+O(t^2)$. The linear term gives the Lie algebraSolving this equation between matrices yieldswhere a convenient basis isThe first two generate Lorentz boosts and the last a spatial rotation. Direct commutators giveThere are three independent real coefficients, so ${\dim }_{\mathbb{R}}\mathfrak{g}=3$. A Cartan subalgebra is $\mathbb{R}J$, and the rank of a semisimple Lie algebra is one. Explicitly, if $X=a{K}_1+b{K}_2+cJ$ and $Y=a^{\prime }{K}_1+b^{\prime }{K}_2+c^{\prime }J$, their commutator vanishes exactly whenFor nonzero $X$, these equations force $Y$ to be proportional to $X$, so there is no two-dimensional commuting subalgebra. The split real rank is also one, realized by the semisimple Lorentz boost generator $K_1$. ThusFor this Lorentz algebra in three spacetime dimensions, disconnected components of the full Lorentz group do not change its tangent Lie algebra at the identity.

Choose an orthonormal basis $e_1,e_2,e_3$ in the real Euclidean space containing the root system, and take simple rootsTheir squared lengths are $2,2,1$. Use the Cartan matrix convention $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, with the denominator indexed by the column. Direct inner products giveThe Dynkin diagram of the B3 root system is a three-node chain with a single bond from node 1 to node 2 and a double bond from node 2 to the short node 3; the arrow points toward node 3. The A3 root system has a chain of three nodes joined by single bonds. A convention putting the row length in the denominator transposes the first displayed Cartan matrix.

For the Extended Dynkin diagram, the highest root is $\theta =e_1+e_2={\alpha }_1+2{\alpha }_2+2{\alpha }_3$. Add ${\alpha }_0=-\theta$. Its only nonzero off-diagonal inner product among these simple roots is $({\alpha }_0,{\alpha }_2)=-1$. Deleting the short node 3 leaves the chain $0-2-1$, with three equal-length roots, which is $A_3$.

More concretely, the long roots $\pm e_i\pm e_j$ form the D3 root system, which is isomorphic to the A3 root system. The corresponding root spaces and Cartan subalgebra generatefor the compact real forms. At the matrix level the inclusion fixes one coordinate in seven-dimensional space. This establishes the required Lie subalgebra inclusion; the associated global groups require the distinction explained in d.

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore