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Here root systems are finite, reduced and crystallographic, as for complex semisimple Lie algebras. For a base of a root system $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, the Dynkin diagram has one vertex for each simple root. Two distinct vertices have $a_{ij}{a}_{ji}$ bonds, with the arrow on a multiple bond pointing toward the shorter root. The classification of finite crystallographic Dynkin diagrams consists of the following connected diagrams; the integers describing arms count vertices away from the unique branching vertex.

This is classification up to diagram isomorphism and overall rescaling of the root system. It includes the conventional identifications $D_3\cong A_3$ and $B_1\cong A_1$ when those low-rank notations are used. The weighting here encodes root lengths and bonds; it is not the separate theory of vertex labels attached to nilpotent orbits. Dropping the crystallographic or reduced hypothesis changes the classification.

Precisely, a base of a root system is a vector-space basis $\Delta \subset \Phi$ such that each root has a unique expansion $\beta ={\sum }_i{n}_i{\alpha }_i$, with integer coefficients all nonnegative or all nonpositive. This defines the positive and negative roots. We use the column-coroot convention for the Cartan matrix:Some conventions transpose this matrix; specifying which index labels the coroot resolves that difference. The Weyl group is the subgroup of orthogonal transformations generated by the Weyl reflections

For the C3 root system, take an orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{R}}^3$ andThere are six long roots of squared length $4$ and twelve short roots of squared length $2$. A base is ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. The Cartan matrix convention for C3 above givesThus the double bond connects ${\alpha }_2$ to ${\alpha }_3$, with its arrow directed toward ${\alpha }_2$.

An explicit complex Lie algebra realizing this root system is the symplectic Lie algebraIt is closed under commutators, since the two identities $X^{T}J=-JX$, $Y^{T}J=-JY$ imply $[X,Y{]}^{T}J=-J[X,Y]$. Equivalently,so its dimension is $9+6+6=21$. A Cartan subalgebra isWith $E_{ab}$ denoting a matrix unit, the root spaces have generators $E_{ij}-E_{3+j,3+i}$ for $e_i-e_j$ ($i\ne j$), $E_{i,3+j}+E_{j,3+i}$ for $e_i+e_j$ ($i<j$), and $E_{i,3+i}$ for $2e_i$; the analogous lower-left generators give the negative roots. These are all eighteen root spaces, each one-dimensional, and together with $H$ they exhaust the algebra.

For an explicit semisimplicity check, let $I\ne 0$ be an ideal. Simultaneous diagonalization of $\mathrm{ad}H$ makes $I$ a sum of its intersections with $H$ and these root spaces. If it contains a nonzero $h\in H$, some root has $\alpha (h)\ne 0$, and $[h,L_{\alpha }]$ supplies a root vector in $I$. Thus $I$ always contains a root vector. Bracket it with its opposite root vector to obtain a nonzero coroot $h_{\alpha }\in I$. Whenever $(\alpha ,\beta )\ne 0$, bracketing $h_{\alpha }$ with either $L_{\beta }$ or $L_{-\beta }$ puts both root spaces into $I$. The graph of the roots with edges for nonzero inner products is connected: $2e_i$ connects to $e_i+e_j$, which connects to $2e_j$, and every short root connects to an axis root. Repeating the argument yields all root spaces and then all of $H$ from their brackets. Hence $I=L$: the constructed algebra is simple, and therefore semisimple.

The reflections in $e_1-e_2$ and $e_2-e_3$ exchange adjacent coordinates, while reflection in $2e_3$ changes the sign of the third coordinate. They generate every signed permutation. Conversely reflection in any displayed root is a signed permutation. ThusThe reflecting hyperplanes are $x_i=0$ and $x_i=\pm x_j$. A fundamental open Weyl chamber is $x_1>x_2>x_3>0$. Every regular point has three nonzero coordinates of distinct absolute values; its signs and the ordering of those absolute values specify exactly one of the $2^3\cdot 3!=48$ chambers. The closure of this chamber is generated by the rays through $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$. This is the Weyl chamber geometry of C3.

The left panel retains the actual long and short root lengths. The right panel radially projects a hemisphere of the unit sphere, draws every visible reflecting great-circle arc, and highlights the chamber $x_1>x_2>x_3>0$. Each spherical triangle is the section of a three-dimensional chamber by the sphere; opposite triangles on the rear hemisphere supply the remaining chambers. Roots of a root system are normal to the chamber walls, rather than generally lying along their edges.

Here a reduced root system is crystallographic, as appropriate to a semisimple Lie algebra. It is a finite spanning set $\Phi \subset E\setminus \{0\}$ in a real inner product space of dimension $r$, invariant under each Weyl reflectionwith $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for every pair of roots and $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$. A base of a root system $\Delta =({\alpha }_1,\dots ,{\alpha }_r)$ is a basis such that the coefficients of every root are integers that are either all nonnegative or all nonpositive. These define the positive system of a root system. We use the existing Cartan matrix conventionUsing the alternative denominator $({\alpha }_i,{\alpha }_i)$ transposes all the matrices below.

The classification of rank-two root systems gives $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. In orthonormal coordinates we may choose the following simple roots and Cartan matrices:

Each complete root system consists of these positive roots and their negatives. To see why the list is exhaustive, the off-diagonal Cartan matrix entries of two distinct simple roots are nonpositive integers, and their product is $4{\cos }^2\theta <4$. Thus the product is $0,1,2$ or $3$, corresponding to simple-root angles $\pi /2,2\pi /3,3\pi /4$ or $5\pi /6$. These determine the four systems and their length ratios.

The Weyl group of a rank-two root system is generated by the two simple Weyl reflections. Their product is a rotation of order $m=2,3,4,6$, respectively, so the groups are the dihedral groups of order $2m=4,6,8,12$. The first is also the Klein four-group and the second the symmetric group $S_3$. The root-orthogonal lines cut the plane into $2m$ Weyl chambers, each of angle $\pi /m$. A chosen fundamental chamber of a root system is $(x,{\alpha }_1)>0$, $(x,{\alpha }_2)>0$; its walls are the two root-orthogonal lines. The Weyl group acts simply transitively on these open chambers.

The crystallographic hypothesis matters: if the integer-pairing condition is dropped, reduced noncrystallographic systems of type $I_2(m)$ also occur. They are not additional Lie-algebra root systems.

The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.

An abstract root system, in the reduced crystallographic convention, is a finite subset $\Phi$ of a finite-dimensional real Euclidean space $E$ satisfying the following conditions:

The root system is irreducible if it is not a union of two nonempty mutually orthogonal root systems. The integrality condition is essential for the finite angle list below; arbitrary finite reflection systems need not satisfy it.

The possible angles are $0$, $\pi /6$, $\pi /4$, $\pi /3$, $\pi /2$, $2\pi /3$, $3\pi /4$, $5\pi /6$, and $\pi$. For roots $\alpha ,\beta$, putBoth are integers, they have the same sign when nonzero, andIf the roots are linearly independent, strict Cauchy-Schwarz inequality gives $0\le ab<4$. Hence $ab\in \{0,1,2,3\}$. These values give, respectively,For the last column, the possible absolute pairs $(|a|,|b|)$ are $(1,1)$, $(1,2)$, and $(1,3)$, up to interchange, and $a/b=(\beta ,\beta )/(\alpha ,\alpha )$. Parallel roots are equal or opposite by reducedness, giving $0$ or $\pi$. Thus the list is exhaustive, and the rank-two root systems realize each nonparallel possibility.

Now suppose two simple roots $\alpha ,\beta$ have angle $5\pi /6$. Distinct simple roots have nonpositive inner product. Here is a proof from their defining sign property: if $(\delta ,ϵ)>0$, then $s_{\delta }(ϵ)=ϵ-n\delta$ with a positive Cartan integer $n$. This would be a root whose coefficients in the root basis include both a positive and a negative coefficient, which is impossible. Therefore, after normalizing the simple roots to unit length, any nonzero off-diagonal inner product is at most $-1/2$, by the angle list.

Let $u=\alpha /\Vert \alpha \Vert$, $v=\beta /\Vert \beta \Vert$, so $(u,v)=-\sqrt{3}/2$. If there were a third simple root with unit vector $w$, write $r=(u,w)\le 0$ and $s=(v,w)\le 0$. Independence of the simple roots makes their Gram matrix positive definite, soIf either $r$ or $s$ were nonzero, its absolute value would be at least $1/2$. Since $rs\ge 0$, the displayed determinant would then be nonpositive. Thus $r=s=0$. Every other simple root is orthogonal to both $\alpha$ and $\beta$: a triple bond isolates a G2 component.

Indeed, the root reflections in $\alpha ,\beta$ preserve their plane and fix its orthogonal complement; the root reflections in the other simple roots preserve that complement and fix the plane. The given facts that these reflections generate the Weyl group and that every root is Weyl group-conjugate to a simple root imply that every root lies entirely in one of those two orthogonal subspaces. If there were any other simple root, this would make $\Phi$ reducible. Therefore irreducibility forces rank two.

Up to interchanging the simple roots, take $\alpha$ short and $\beta$ long. The squared-length ratio is $3$. Choose a scale and orthonormal coordinates so thatThe Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. In coefficient coordinates $m\alpha +n\beta$, the two root reflections areTheir orbits of $\alpha$ and $\beta$ are, respectively,These sets are closed under both displayed reflections; for example $s_{\beta }(\alpha )=\alpha +\beta$, $s_{\alpha }(\alpha +\beta )=2\alpha +\beta$, $s_{\alpha }(\beta )=3\alpha +\beta$, and $s_{\beta }(3\alpha +\beta )=3\alpha +2\beta$. The assumption that every root is conjugate to a simple root now determines all twelve roots, with no further ones possible.

For completeness, this twelve-element set really is a root system. It is finite, spans the plane, and is reduced. The six short roots have length $1$ and directions spaced by $\pi /3$; the six long roots have length $\sqrt{3}$ and interleaved directions, also spaced by $\pi /3$. Their mutual Cartan integers are integral: within either length class they are $0$, $\pm 1$, or $\pm 2$ as applicable, and between the two length classes they are $0$, $\pm 1$, or $\pm 3$. Closure under all root reflections follows by conjugating the two simple reflections, since every listed root lies in a simple-root orbit. The nonorthogonal simple roots make it irreducible. We have therefore proved existence as well as uniqueness, up to isometry and overall scale:

Choose a positive root system $R^{+}$ for a complex semisimple Lie algebra, with Weyl group $W$ and Weyl vector $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote its finite-dimensional irreducible representation, and use formal symbols with $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. Its formal character of a weight module is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The Weyl character formula saysHere $\det (w)=(-1{)}^{\ell (w)}$, the sign of the product of simple root reflections. The Weyl denominator formula isThe quotient is initially interpreted in the fraction field of the group algebra of the weight lattice; the theorem makes it a finite character.

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 decompositionA 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 formFor 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 givesOne 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 $N\ge 2$. 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 $E_{jk}{E}_{lm}={\delta }_{kl}{E}_{jm}$ givesThus all the roots, expressed as evaluation vectors in this precise Cartan basis, areThey 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 abbreviationThen all pairs are covered byHere 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.

The triple bond in the original Dynkin diagram gives the Cartan integerswhere $\alpha$ is the long root and $\beta$ the short root. Their product is $4{\cos }^2\theta =3$, and the inner product of distinct simple roots is nonpositive. The ratio of the two integers gives the squared-length ratio. ThereforeIt is convenient to normalize $(\alpha ,\alpha )=6$, $(\beta ,\beta )=2$ and $(\alpha ,\beta )=-3$. Scaling the inner product does not affect the root system or the fundamental-weight relations.

For nonproportional roots $\gamma ,\delta$, the root-string theorem states that the root string is consecutive:The endpoints are maximal, and reflection in $\gamma$ reverses the string. The Cartan integer determines $p-q$, not in general the total $p+q+1$ by itself. This result follows by restricting the Adjoint representation to the sl2 subalgebra associated with a root. If the roots are distinct simple roots, $\delta -\gamma$ cannot be a root: its simple-root coefficients have opposite signs. Thus $p=0$ andHere length means number of roots; the number of intervals between them is one less. Distinctness matters. If $\gamma =\delta$, a reduced root system gives the set $\{\gamma ,-\gamma \}$, with a missing zero between them, so the consecutive-string theorem and the displayed simple-root formula do not apply.

In the G2 root system, the initial strings areFor $\delta =\alpha +3\beta$, $⟨\delta ,{\alpha }^{\vee }⟩=-1$. Since $\delta -\alpha =3\beta$ is not a root in a reduced root system, $p=0$ and $q=1$: this generates $2\alpha +3\beta$. The remaining strings explain why the construction stops. The $\alpha$-strings through $\beta$ and $\alpha +\beta$ are the same two-element string; the one through $\alpha +2\beta$ is a singleton since subtracting $\alpha$ gives $2\beta$, and its Cartan integer is zero; the one through $2\alpha +3\beta$ is the string $\{\alpha +3\beta ,2\alpha +3\beta \}$. The $\beta$-strings through $\alpha$, $\alpha +\beta$, $\alpha +2\beta$ and $\alpha +3\beta$ are the initial four-element string. Finally, $2\alpha +3\beta$ is orthogonal to $\beta$; its $\beta$-string is a singleton because $2\alpha +2\beta =2(\alpha +\beta )$ is not a root. Apply the same reasoning to negatives. Using the permitted completeness of this procedure givesThe short positive roots are $\beta ,\alpha +\beta ,\alpha +2\beta$, of squared length two; the other three are long, of squared length six. Each root space is one-dimensional and the Cartan subalgebra has dimension two, soHere the dimension refers to the Lie algebra, with one Cartan generator per rank, not just the number of roots.

Write a prospective weight as $w=a\alpha +b\beta$. The pairings with simple coroots areThe fundamental weights are dual to those coroots. Solving the two linear systems gives, in the long-root-first numbering of this paper,The representation with Dynkin labels $(0,1)$ has highest weight $\lambda ={\omega }_2$, a short root. Numbering the short root first, as some references do, would call this the $(1,0)$ representation instead; the representation itself is unchanged.

The weight set of a finite-dimensional irreducible highest-weight representation is invariant under the Weyl group and lies in the convex hull of the orbit of its highest weight. All weights also differ from the highest weight by an element of the root lattice. The orbit of $\lambda$ comprises the six short roots, so all six are weights. The lowering operators give the chainIn particular zero occurs: at weight $\beta$, its pairing with ${\beta }^{\vee }$ is two, so the lowering operator is nonzero by the finite-dimensional sl2 Lie algebra representation theory. The $\beta$-string through it is the usual three-weight string $\beta ,0,-\beta$; higher weight $2\beta$ would lie outside the highest-weight convex hull.

There can be no further weights. Every point of that convex hull has squared norm at most $\Vert \lambda {\Vert }^2=2$, while a root-lattice point hasThe integer solutions of the bound $\Vert w{\Vert }^2\le 2$ are precisely zero and the six short roots: $a=0$ gives $b=-1,0,1$; $a=1$ gives $b=1,2$; and $a=-1$ gives $b=-1,-2$. ThusThe final dimension uses the stipulated nondegeneracy of the weights. It also agrees with the Weyl dimension formula without that stipulation. The zero weight is not in the Weyl orbit of the nonzero weights, so this is not a minuscule representation.

In the diagram above, a coordinate label $(a,b)$ means $a\alpha +b\beta$. The left panel contains all twelve roots of a root system; the right panel contains the six short-root weights and the zero weight. The long-root-first convention is the same as in the calculations.

Use the coroot normalization ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, with the pairing understood via the root system and its invariant inner product. A dominant integral weight, called a dominant weight here, is a functional $\lambda \in {\mathfrak{t}}^{\ast }$ such thatEquivalently, $\lambda ={\sum }_i{n}_i{\omega }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, where the fundamental weights satisfy $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. If dominance and integrality are defined separately, dominance means the inequalities and membership in the weight lattice means integrality.

A highest-weight vector is a nonzero weight vector $v\in V_{\lambda }$, so $h\cdot v=\lambda (h)v$ for all $h\in \mathfrak{t}$, annihilated by every positive root space:The last condition is equivalently annihilation by the raising operators for the simple roots.

Suppose first that $x\ne 0$, and let $E$ be the real span of the root system, of dimension equal to the rank of a semisimple Lie algebra $\ell$. The images of the roots span $E/\mathbb{R}\alpha$, so select roots ${\beta }_1,\dots ,{\beta }_{\ell -1}$ whose images form a basis of this quotient.

For each $i$, take the highest endpoint ${\gamma }_i^{+}$ of the $\alpha$ root string through ${\beta }_i$, and the highest endpoint ${\gamma }_i^{-}$ of the root string through $-{\beta }_i$. By construction, ${\gamma }_i^{\pm }+\alpha$ is not a root. Moreover ${\gamma }_i^{\pm }\ne -\alpha$, since their images in $E/\mathbb{R}\alpha$ are nonzero. The root-space decomposition and its bracket rule therefore giveThese $2(\ell -1)$ roots are distinct: their quotient images are the two signs of a basis, which are distinct. Their one-dimensional root spaces consequently contribute $2(\ell -1)$ independent vectors to the Lie algebra centralizer.

In addition, the $\ell -1$ dimensional subspace $\ker \alpha \subset \mathfrak{t}$ commutes with $x$, because $[h,x]=\alpha (h)x$. The line $\mathbb{C}x\subset {\mathfrak{g}}_{\alpha }$ also commutes with $x$. These contributions are independent by the root-space decomposition; in particular none of the selected endpoint roots is $\alpha$. Thus the centralizer lower bound for a root vector isIf $x=0$, then $Z_{\mathfrak{g}}(x)=\mathfrak{g}$. The root system contains the $2\ell$ distinct roots $\pm {\alpha }_i$, so $\dim \mathfrak{g}=\ell +|\Phi |\ge 3\ell$, which also proves the desired inequality. The argument includes rank one, where the list of ${\beta }_i$ is empty.

Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are $H(t)\mapsto \pm t_i\pm t_j$, with the two signs independent. This is the D3 root system, with twelve roots.

For a direct check, the defining matrix identity gives the block form $X=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B^T=-B$ and $C^T=-C$. If $E_{ij}$ denotes a matrix unit, the root space for ${\epsilon }_i-{\epsilon }_j$ is spanned by $E_{ij}-E_{j+3,i+3}$; those for ${\epsilon }_i+{\epsilon }_j$ and $-{\epsilon }_i-{\epsilon }_j$ are spanned respectively by $E_{i,j+3}-E_{j,i+3}$ and $E_{i+3,j}-E_{j+3,i}$. Commuting these with $H(t)$ gives exactly the displayed maps.

The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.

Use the wiki's Cartan matrix conventionLabel the long simple root first when the lengths differ. The matrix of an irreducible rank-two root system has the form $A=\left(\begin{array}{cc}2& -a\\ -b& 2\end{array}\right)$, where $a,b$ are positive integers. Integrality and the signs follow from the Cartan integers for distinct simple roots, and irreducibility rules out a zero off-diagonal pair. Its positive-definite symmetrization gives $ab<4$, equivalently $ab=4{\cos }^2\theta <4$. Moreover $a/b=\Vert {\alpha }_1{\Vert }^2/\Vert {\alpha }_2{\Vert }^2\ge 1$. Thus $(a,b)=(1,1),(2,1),(3,1)$, giving the entire classification of rank-two root systems relevant to a simple algebra:These correspond to ${\mathfrak{sl}}_3(\mathbb{C})$, ${\mathfrak{so}}_5(\mathbb{C})\cong {\mathfrak{sp}}_4(\mathbb{C})$, and the exceptional algebra ${\mathfrak{g}}_2$. Here $B_2=C_2$ denotes the Isomorphism between so5 and sp4, not an additional case. The disconnected $A_1\times A_1$ system corresponds to a semisimple but nonsimple algebra and is excluded.

The Dynkin diagrams, with nodes in the same order as the matrices, areThe second and third diagrams have two and three bonds respectively, with the arrowhead pointing to the short root ${\alpha }_2$. In the last diagram $\equiv$ draws the three bonds and $>$ their arrowhead.

To enumerate roots, use the standard root string theorem: for $\beta \ne \pm \alpha$, the roots $\beta +n\alpha$ are consecutive from $\beta -p\alpha$ through $\beta +q\alpha$, with $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. Also use the standard results that the root system is reduced, every root is conjugate under the Weyl group to a simple root, and each root space has dimension one.

Write $m=a\in \{1,2,3\}$. Since a difference of simple roots is not a root, the ${\alpha }_2$-string starting at ${\alpha }_1$ has $p=0$ and $q=m$. It produces ${\alpha }_1,{\alpha }_1+{\alpha }_2,\dots ,{\alpha }_1+m{\alpha }_2$. The ${\alpha }_1$-string starting at ${\alpha }_2$ has $q=1$. In type $G_2$, the further string through $\beta ={\alpha }_1+3{\alpha }_2$ hasand $\beta -{\alpha }_1=3{\alpha }_2$ is not a root, so it also produces $2{\alpha }_1+3{\alpha }_2$. The resulting positive roots areAll roots are these and their negatives. To check completeness, the simple root reflections act on coordinates byEach listed set together with its negatives is stable under both reflections. It contains the simple roots and consists of roots already forced by strings. Since every root lies in a Weyl orbit of a simple root, no other roots can occur. This proves the lists for the A2 root system, B2 root system, and G2 root system. The root-space decomposition then gives