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The appropriate abstract object is a finite reduced crystallographic root system $R$ in a real inner product space $E$. Its axioms are: $R$ is finite, spans $E$, and does not contain zero; for $\alpha \in R$, $R\cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$; each root reflectionpermutes $R$; and every Cartan integer $n_{\alpha \beta }=2(\alpha ,\beta )/(\beta ,\beta )$ is an integer. The restriction to a reduced root system and crystallographic integrality distinguishes roots of complex semisimple Lie algebras from more general reflection configurations.

For any two roots of a root system, the Cauchy-Schwarz inequality givesIf the inner product is zero, both Cartan integers vanish. Otherwise their signs agree, and the absolute value of each is a positive integer. Dividing their product by an integer of absolute value at least one provesFor nonproportional roots the product is strictly less than four. In a reduced root system, proportional roots are just $\pm \alpha$ and have Cartan integers $\pm 2$, so the printed bound is intentionally looser than the resulting bound of three.

A fundamental system of a root system is a basis $\Delta$ of $E$ made of roots, such that each root is an integer combination of $\Delta$ with either all coefficients nonnegative or all nonpositive. Its members are the simple roots. Suppose distinct $\alpha ,\beta \in \Delta$ had $n_{\alpha \beta }>0$. Thenhas a positive coefficient of $\alpha$ and a negative coefficient of $\beta$, contradicting the defining sign condition. Thus $n_{\alpha \beta }\le 0$. Distinct simple roots are linearly independent, so their Cartan-integer product is strictly less than four. Combining integrality and the sign condition givesHere nonpositive is the intended sense of the printed convention that includes zero among “negative” numbers; orthogonal simple roots really do give zero.

To form a Dynkin diagram, place a vertex at each simple root. Join two vertices by $n_{\alpha \beta }{n}_{\beta \alpha }$ bonds, hence zero, one, two or three. A multiple bond has an arrow toward the short root. Indeed $n_{\alpha \beta }/n_{\beta \alpha }=(\alpha ,\alpha )/(\beta ,\beta )$ determines the squared length ratio, and the diagram with the Cartan matrix reconstructs the angles and relative lengths. A single bond joins equal-length roots.

The connected finite Dynkin diagrams are the following. The descriptions include bond multiplicities and arrow directions, so distinguish dual diagrams:

There are no other connected finite Dynkin diagrams. Low-rank conventions also identify $B_1=C_1=A_1$ and $D_3=A_3$; $D_2$ is disconnected, so introduces no further connected type. Affine diagrams are outside this finite classification.

Finally suppose the underlying graph contained a cycle on $m\ge 3$ distinct simple roots ${\alpha }_1,\dots ,{\alpha }_m$. Put $u_i={\alpha }_i/\Vert {\alpha }_i\Vert$. Any bonded pair hasand all other distinct pairs have nonpositive inner products. The cycle contributes at least $m$ bonded pairs, soBut the simple roots, and hence these normalized vectors, are linearly independent, making the displayed sum nonzero. Positive definiteness gives a contradiction. Thus the underlying graph of a finite Dynkin diagram has no cycle. This acyclicity of a finite Dynkin diagram argument also excludes cycles with extra chords or multiple bonds; multiple bonds themselves are not treated as two-edge cycles.

Let ${\alpha }_i^{\vee }=2{\alpha }_i/({\alpha }_i,{\alpha }_i)$ be the coroots. With a chosen set of simple roots, the root lattice and weight lattice areThe fundamental weights ${\omega }_j$ are defined by $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$, and form an integral basis of $P$. We use the Cartan matrix convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$.

For the A2 root system, inversion of its Cartan matrix yields the A2 fundamental weights and weight lattice:Equivalently ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. The root lattice has index three in the weight lattice, since the change-of-basis matrix has determinant three. For a planar realization takeThe weight lattice is triangular. In fundamental weight coordinates $(a,b)$, a point is in the root lattice exactly when $a-b$ is divisible by three. The following sketch marks both bases and the sublattice:

The integers called Dynkin indices here are the Dynkin labels of the highest weight:For finite-dimensional Irreducible Lie algebra representations, they are nonnegative integers. All weight coordinates below are these Dynkin label coordinates, not simple-root coordinates.

A weight-string enumeration algorithm gives the set of weights without their weight multiplicities. Begin with $\Lambda$ and process known weights by increasing height below $\Lambda$. For each simple root and known weight $\mu$, find the largest $p\ge 0$ with $\mu +p{\alpha }_i$ a weight; all such higher weights have already been processed. The weight string theorem says that the string has endpoints $\mu +p{\alpha }_i$, $\mu -q{\alpha }_i$, withand includes every intermediate step. Append $\mu -{\alpha }_i,\dots ,\mu -q{\alpha }_i$ and repeat until no new weights appear. This terminates in finite dimension and supplies all weights; every nonhighest weight can be reached by simple-root lowering. A string can contain contributions from several sl2 summands, so the resulting set does not by itself determine weight multiplicities.

For the explicit calculation we use the following general facts: finite-dimensional representations of a complex semisimple Lie algebra are completely reducible by the Weyl complete reducibility theorem; the symmetric powers of the defining sln representation are irreducible of highest weight $m{\omega }_1$; and weights in a tensor product of Lie algebra representations add with multiplicities multiplied. We also use the highest-weight representation classification: each finite-dimensional irreducible has a unique dominant integral highest weight, and a nonzero vector killed by all simple-root raising operators supplies an irreducible summand with that highest weight in a completely reducible module. For $A_2$, the Weyl dimension formula specializes toFor $A_2\cong {\mathfrak{sl}}_3(\mathbb{C})$, the defining fundamental representation has weights $(1,0),(-1,1),(0,-1)$. Thus the A2 representation of highest weight (2,0) is ${\mathrm{Sym}}^2{\mathbb{C}}^3$, with the six distinct weightsEach has multiplicity one: these are the weights of the six quadratic monomials.

For the tensor square of the A2 representation of highest weight (2,0), add every ordered pair of elements of $W_6$. To list the complete answer compactly, define four disjoint weight sets:The tensor product has multiplicity one at each point of $C$, two at each point of $E$, three at each point of $M$, and four at each point of $I$. These fifteen distinct weights account for $3+12+9+12=36$ states.

We next identify the irreducible summands rather than just their dimensions. Split the tensor square of the six-dimensional space into its symmetric power and exterior power, of dimensions $21$ and $15$. The square of a highest-weight vector in the symmetric part has highest weight $(4,0)$, giving $R(4,0)$ of dimension $15$.

Let $v_a,v_b$ be vectors of weights $a=(2,0)$ and $b=(0,1)=a-{\alpha }_1$. In the exterior part $v_a\wedge v_b$ has weight $(2,1)$ and is killed by both simple-root raising operators: raising $v_b$ along ${\alpha }_1$ gives a multiple of $v_a$, whose wedge with itself is zero, and the other raising actions vanish. Thus it is a highest-weight vector. The Weyl dimension formula gives $\dim R(2,1)=15$, so the entire exterior part is this irreducible summand.

The remaining symmetric part has dimension six. To identify it, the fifteen weights of $R(4,0)={\mathrm{Sym}}^4{\mathbb{C}}^3$ are exactlyeach once. Subtract them from the unordered-pair weights of ${\mathrm{Sym}}^{2}R(2,0)$. The residual weights are $M\cup I$, each once, with highest weight $(0,2)$. They are the negatives of $W_6$, so this six-dimensional summand is $R(0,2)$. ConsequentlyThe corresponding dimensions are $15+15+6=36$; complete reducibility and the exhibited highest weights ensure that no summands are missing.

The full weight multiplicities in the A2 tensor square of highest weight (2,0) are summarized below. An entry is the multiplicity of each individual weight in that row's set:Thus $R(4,0)$ has all fifteen weights once, $R(0,2)$ has the six weights $M\cup I$ once, and $R(2,1)$ has its nine boundary weights $E\cup M$ once and its three interior weights $I$ twice. Only the three interior weights of $R(2,1)$ are degenerate among the irreducible components.

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.

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

For the restriction to a sl2 subalgebra associated with a root, normalize its Cartan element to be $h_{\alpha }={\alpha }^{\vee }$. On a root vector of root $\beta$, its eigenvalue is $⟨\beta ,{\alpha }^{\vee }⟩$. The classification of finite-dimensional sl2 representations says that $R(\Lambda )$ has weights $\Lambda ,\Lambda -2,\dots ,-\Lambda$ and dimension $\Lambda +1$. A root string of $\Lambda +1$ nonparallel roots therefore supplies one such irreducible module: the raising and lowering brackets connect its consecutive root spaces.

For clarity, the positive-side strings for each simple-root direction are listed below. Brackets denote the whole consecutive string; a single listed root is a string of length one. In every row also include each negative string in reverse order, and separately the triple formed by the root ${\alpha }_i$, its negative, and $h_{{\alpha }_i}$.The root's own triple is always $R(2)$. The one-dimensional subspace of $\mathfrak{h}$ annihilated by ${\alpha }_i$ commutes with this subalgebra and gives $R(0)$. Each singleton root in the table, and its negative, contributes another $R(0)$. Reading the other string lengths now gives the adjoint branching to root sl2 subalgebras in rank two:Their dimensions are respectively $8,10,10,14,14$. The last row agrees with G2 adjoint branching to a short-root sl2 subalgebra. All labels in the table are highest weights, rather than dimensions.

A convenient root basis isThe corresponding Cartan matrix isThus the Dynkin diagram is the three-node $C_3$ chain, with a double edge between ${\alpha }_2$ and ${\alpha }_3$ and its arrow pointing toward the shorter root ${\alpha }_2$.

The Weyl reflection formula $s_i(\beta )=\beta -⟨\beta ,{\alpha }_i^{\vee }⟩{\alpha }_i$, together with the Cartan matrix, givesThese are the images of every simple root under each simple reflection.

The root lattice is $Q={\sum }_i\mathbb{Z}{\alpha }^{(i)}$. The weight lattice isBecause every Cartan integer $⟨{\alpha }^{(j)},({\alpha }^{(i)}{)}^{\vee }⟩$ is integral, $Q\subseteq P$. The Dynkin labels of $\lambda$ are

There are three isomorphism classes of complex simple rank-three Lie algebras: types A3 root system$A_3$, B3 root system$B_3$, and C3 root system$C_3$. Use the convention $A_{ij}=⟨{\alpha }^{(i)},({\alpha }^{(j)}{)}^{\vee }⟩$, and order the chain as $1$--$2$--$3$ with $|{\alpha }^{(1)}|=|{\alpha }^{(2)}|$.

The Dynkin labels of a finite-dimensional irreducible representation are those of its highest weight. Thus $[1,0,0]$ means the fundamental representation $V({\omega }_1)$. The weights, written in Dynkin labels and in a lowering order, areFor $A_3$ these are the weights of the defining representation of ${\mathfrak{sl}}_4$; for $B_3$ they are $\{\pm e_1,\pm e_2,\pm e_3,0\}$ in the vector representation of ${\mathfrak{so}}_7$; for $C_3$ they are $\{\pm e_1,\pm e_2,\pm e_3\}$ in the defining representation of ${\mathfrak{sp}}_6$. Hence the requested dimensions are $4$, $7$, and $6$, respectively.

The squared root lengths and inner product areWith $A_{ij}=2{\alpha }_i\cdot {\alpha }_j/|{\alpha }_j{|}^2$, the Cartan matrix isThis is the Cartan matrix of the G2 root system, with ${\alpha }_1$ short and ${\alpha }_2$ long.

A Cartan subalgebra $\mathfrak{h}$ of a complex semisimple Lie algebra is a maximal commuting subalgebra consisting of semisimple elements. A nonzero functional $\alpha \in {\mathfrak{h}}^{\ast }$ is a root when its space in the root-space decompositionis nonzero. A Cartan-Weyl basis combines a basis of $\mathfrak{h}$ with root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$.

A choice of regular hyperplane divides roots into positive and negative roots. The simple roots are the positive roots that are not sums of two positive roots; they form a basis of the real root span. The Cartan matrix isup to the equivalent transposed indexing convention.

For ordered simple roots, the Cartan matrix is $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$. It determines their relative lengths and angles.

For simple roots ${\alpha }_i$, the Cartan matrix isIts diagonal entries are two, while its off-diagonal entries encode the angles and relative lengths in the Dynkin diagram.

With $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$ and roots labeled from left to right, the three Cartan matrices are

A reduced root system of rank one is $\Phi =\{\alpha ,-\alpha \}$. Its fundamental weight is $\alpha /2$, and its Weyl group has two elements. Its ordinary Dynkin diagram has one node; adjoining ${\alpha }_0=-\alpha$ gives affine Cartan matrix $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$, drawn as a double bond between equal-length roots.