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For the B2 root system with ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ long and ${\alpha }_2={\epsilon }_2$ short, the spin fundamental weight is ${\Lambda }_2=({\epsilon }_1+{\epsilon }_2)/2$. Its four-dimensional highest-weight representation has the minuscule weight orbit $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$, each of multiplicity one. The crystal basis is the chain $++\stackrel{2}{\to }+-\stackrel{1}{\to }-+\stackrel{2}{\to }--$, where signs denote the coordinates. Each Kashiwara operator subtracts its indicated simple root; the sl2 Lie algebra strings have length one. This fixes every edge and gives a connected highest-weight graph with the correct character.

In an orthonormal basis $({\epsilon }_1,{\epsilon }_2)$, the roots are $\pm ({\epsilon }_1-{\epsilon }_2)$, $\pm ({\epsilon }_1+{\epsilon }_2)$, $\pm 2{\epsilon }_1$, and $\pm 2{\epsilon }_2$. Choose simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$, with ${\alpha }_1$ a short root. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2={\epsilon }_1+{\epsilon }_2$.

The Dynkin labels of a weight $\lambda$ are its simple-coroot coordinates ${\lambda }^i=⟨\lambda ,{\alpha }_i^{\vee }⟩$. Thus $\lambda ={\sum }_i{\lambda }^i{\omega }_i$ in the basis of fundamental weights.

With simple roots ${\alpha }_i={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n={\epsilon }_{n-1}+{\epsilon }_n$, the fundamental weights are ${\omega }_i={\epsilon }_1+\cdots +{\epsilon }_i$ for $i\le n-2$, followed by ${\omega }_{n-1}=({\epsilon }_1+\cdots +{\epsilon }_{n-1}-{\epsilon }_n)/2$ and ${\omega }_n=({\epsilon }_1+\cdots +{\epsilon }_n)/2$. All roots have squared length two, so their pairings with coroots verify the dual-basis identities. The last two weights are the half-spin highest weights; for $n=2$, the defining highest weight is ${\omega }_1+{\omega }_2$.

For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$ with upper triangular positive roots, the simple coroots are $\mathrm{diag}(1,-1,0)$ and $\mathrm{diag}(0,1,-1)$. Their dual fundamental weights are ${\omega }_1(\mathrm{diag}(t_1,t_2,t_3))=t_1$ and ${\omega }_2(\mathrm{diag}(t_1,t_2,t_3))=t_1+t_2=-t_3$. In simple-root coordinates, ${\omega }_1=(2{\alpha }_1+{\alpha }_2)/3$ and ${\omega }_2=({\alpha }_1+2{\alpha }_2)/3$. The primitive vectors $e_1\in {\mathbb{C}}^3$ and $e_1\wedge e_2\in {\Lambda }^2{\mathbb{C}}^3$ realize these Fundamental representations of sl3.

For the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, the fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$ and ${\omega }_2=3{\alpha }_1+2{\alpha }_2$. The positive coroots, relative to the simple coroots, have coordinates $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. The Weyl dimension formula therefore gives, for nonnegative integers $a,b$,The denominator is the product of the pairings of the Weyl vector with those coroots. The dimensions at ${\omega }_1,{\omega }_2,2{\omega }_1$ are seven, fourteen and twenty-seven.

For short and long simple roots $\alpha ,\beta$ of $G_2$, the long roots form an $A_2$ root system with simple roots ${\delta }_1=\beta$ and ${\delta }_2=3\alpha +\beta$. If ${\omega }_1,{\omega }_2$ are the corresponding $A_2$ fundamental weights, then $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$.

Give the real root span an inner product with ${\epsilon }_1,{\epsilon }_2$ orthonormal. The B2 root system consists of four short roots on the coordinate axes and four long roots on the diagonals. Choose the long-first simple rootsThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2={\epsilon }_1$, and ${\alpha }_1+2{\alpha }_2={\epsilon }_1+{\epsilon }_2$. The corresponding simple reflections areThey generate every signed permutation of the two coordinates. Thus the Weyl group has order eight and is the symmetry group of a square; every element sends $(x,y)$ to $(\pm x,\pm y)$ or $(\pm y,\pm x)$, with independent signs.

Since ${\alpha }_1,{\alpha }_2$ form an integral basis of ${\mathbb{Z}}^2$, the root lattice is $Q=\mathbb{Z}{\epsilon }_1\oplus \mathbb{Z}{\epsilon }_2$. The simple coroots are ${\alpha }_1^{\vee }={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2^{\vee }=2{\epsilon }_2$. Therefore the weight lattice isSolving $⟨{\Lambda }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsThe cone of dominant weights isIndeed $a=x-y$ and $b=2y$ are exactly the two simple-coroot pairings. This describes representations of the Lie algebra, or of its simply connected Spin group; half-integral weights are allowed. For representations descending to $\mathrm{SO}(5)$ itself, the weight must lie in ${\mathbb{Z}}^2$, equivalently $b$ must be even.

Use directed color-$i$ edges $u\stackrel{i}{\to }v$ for the Kashiwara operator ${\tilde{f}}_i$, so every edge subtracts ${\alpha }_i$ from the weight. The defining crystal of the defining odd-orthogonal representation isThe defining module has highest weight ${\Lambda }_1$ and precisely these five weights, all of multiplicity one. The color-two string through zero has length two, since $⟨{\epsilon }_2,{\alpha }_2^{\vee }⟩=2$. The other indicated strings have length one. Restriction to each simple-root sl2 Lie algebra and the root-string property of a crystal justify the edges; no other weight difference is the relevant simple root.

For the B2 spin crystal $B({\Lambda }_2)$, denote the four weights $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$ by their sign pairs. Its graph isThe fundamental weight ${\Lambda }_2$ is a minuscule weight: its pairings with all positive coroots are zero or one. Its Weyl orbit has four weights, and the dimension formula gives four, so each occurs once and there are no additional weights. All simple-root strings have length one. Subtracting ${\alpha }_1$ or ${\alpha }_2$ where another weight is present gives exactly this graph.

For the B2 adjoint crystal, write $x_{\beta }$ for the vertex of root weight $\beta$, and use two distinct zero-weight vertices $z_1,z_2$. There are eight root vertices and two zero vertices because the zero-weight space is the two-dimensional Cartan subalgebra. The complete lists of colored strings areEvery arrow in a given row has that row's color; the reverse arrows are the raising operators, and all unlisted lowering operations vanish. In particular the two zero vertices must not be merged.

For a constructive justification, let $V={\mathbb{C}}^5$ with the invariant symmetric form from part(a). The mapis an equivariant isomorphism ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$, identifying the highest weight as ${\epsilon }_1+{\epsilon }_2$. Compute its crystal as the highest-weight component starting at $1\otimes 2$ in $B({\Lambda }_1)\otimes B({\Lambda }_1)$, with labels $1,2,0,\bar{2},\bar{1}$ for the defining chain. The crystal tensor-product rule iswhere ${\epsilon }_i$ counts available raising steps and ${\phi }_i$ counts lowering steps in the color-$i$ string. It generates the following ten vertices:Applying the displayed rule to these labels gives precisely the listed strings, proving the entire diagram rather than only its weight multiplicities. Tensor labels here are crystal vertices in the combinatorial limit, not an assertion that a bare ordinary tensor such as $v_0\otimes v_0$ lies in ${\Lambda }^{2}V$.

Take the diagonal Cartan subalgebra of ${\mathfrak{sl}}_3$ and let ${\epsilon }_i$ select its $i$th diagonal entry, so ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3=0$ on it. Use simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2=-{\epsilon }_3={\epsilon }_1+{\epsilon }_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^{\ast }$. ThusTheir crystal basis diagrams are $B:1\stackrel{1}{\to }2\stackrel{2}{\to }3$ and $B^{\vee }:\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights ${\epsilon }_i$ and $-{\epsilon }_i$ respectively. Use the same crystal tensor-product rule as in2(d): lowering acts on the first factor when ${\phi }_i(u)>{\epsilon }_i(v)$; otherwise on the second. Raising uses the first factor when ${\phi }_i(u)\ge {\epsilon }_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.

For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to 21\to 22$, $31\to 32$, $13\to 23$; the complete color-two list is $21\to 31$, $22\to 32\to 33$, $12\to 13$. There are two connected components:For example ${\tilde{f}}_1(1\otimes 1)=2\otimes 1$ because ${\phi }_1(1)=1>{\epsilon }_1(1)=0$, whereas ${\tilde{f}}_1(1\otimes 2)=0$ because equality selects the second factor and ${\tilde{f}}_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional symmetric square and the three-dimensional exterior square:

For $B\otimes B^{\vee }$, abbreviate $i\otimes \bar{j}$ by $i\bar{j}$. The complete color-one edge list is $1\bar{3}\to 2\bar{3}$, $1\bar{2}\to 2\bar{2}\to 2\bar{1}$, $3\bar{2}\to 3\bar{1}$; the complete color-two list is $1\bar{3}\to 1\bar{2}$, $2\bar{3}\to 3\bar{3}\to 3\bar{2}$, $2\bar{1}\to 3\bar{1}$. The isolated vertex is $1\bar{1}$. Thus the two highest vertices are $1\bar{3}$ of weight ${\epsilon }_1-{\epsilon }_3={\omega }_1+{\omega }_2$, and $1\bar{1}$ of weight zero. Lowering from the former visits all eight other vertices. ConsequentlyThis also follows concretely from $V\otimes V^{\ast }\cong \mathrm{End}(V)={\mathfrak{sl}}_3\oplus \mathbb{C}I$: the first term is the Adjoint representation, with highest weight the highest root, and the second is the trivial Lie algebra representation. The ordinary invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.

Let $E$ be the real span of the roots in ${\mathfrak{h}}^{\ast }$. The weight lattice iswhere $h_{\alpha }$ is the coroot in the root's sl2 subalgebra associated with a root, normalized by $\alpha (h_{\alpha })=2$. It suffices to test the simple coroots. Equivalently, if ${\omega }_i(h_{{\alpha }_j})={\delta }_{ij}$ defines the fundamental weights, then $P={⨁}_i\mathbb{Z}{\omega }_i$.

To prove integrality of representation weights on coroots, restrict a finite-dimensional module $V$ to the root's sl2 Lie algebra. By the classification of finite-dimensional sl2 representations, the eigenvalues of its Cartan element $h_{\alpha }$ are integers. Thus every $\mu \in {\Lambda }_V$ has $\mu (h_{\alpha })\in \mathbb{Z}$. The simple coroots form a basis of the real Cartan space, so these integral values also put $\mu$ in $E$. Therefore $\boxed{{\Lambda }_V\subseteq {\Lambda }_W}$ for every finite-dimensional representation in this setting. The weight lattice need not coincide with the smaller root lattice.

For the chosen simple roots $\Delta$, the open fundamental chamber of a root system isEquivalently $\lambda (h_{\alpha })>0$ for every simple coroot. Since all positive roots are nonnegative combinations of simple roots, requiring the inequality for all positive roots gives the same chamber. It is a connected component of the complement of the root reflection hyperplanes; the Weyl group permutes those chambers simply transitively.

The Closed dominant Weyl chamber is obtained by replacing the inequalities by $\ge 0$. It is the cone generated by the fundamental weights. When the fundamental Weyl chamber is taken to include its walls, this closed convention is intended; both conventions determine the same positive system and boundary rays.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

The weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$. The coroots use the normalization fixed above.

The Killing form defines the complex-linear mapBy nondegeneracy of the Killing form on a Cartan subalgebra, this is an isomorphism. For $H\in {\mathfrak{h}}_{\mathbb{R}}$, expand its image in the dual fundamental-weight basis:The coefficients are real by the preceding positivity calculation, and the ${\omega }_i$ are in ${\Lambda }_W$. ConsequentlyIn fact both spaces have real dimension $r$, so this inclusion is equality.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

Put $\bar{i}=2n+2-i$ and $o=n+1$. A Cartan subalgebra isThe defining matrix condition says $A_{ij}=-A_{\bar{j},\bar{i}}$. Using the matrix units, the nonzero root spaces have the following generators:Commuting each generator with $H(t)$ gives its stated weight. Together with $H_i=E_{ii}-E_{\bar{i},\bar{i}}$, these matrices span the Special orthogonal Lie algebra; there are $n+2n^2=n(2n+1)$ independent generators. Thus its decomposition as a Cartan subalgebra module isThis is the Bn root system. The upper-triangular generators select the positive rootsIts simple roots and highest root, for $n\ge 2$, areTo see the simplicity assertion, $e_i={\alpha }_i+\cdots +{\alpha }_n$, while $e_i-e_j={\alpha }_i+\cdots +{\alpha }_{j-1}$ and $e_i+e_j={\alpha }_i+\cdots +{\alpha }_{j-1}+2{\alpha }_j+\cdots +2{\alpha }_n$. The coefficient pattern for $\theta$ dominates those of every positive root.

Normalize the inner product so the $e_i$ are orthonormal. The simple coroots are ${\alpha }_i^{\vee }={\alpha }_i$ for $i<n$ and ${\alpha }_n^{\vee }=2e_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the half-sum of positive roots, the coordinate $e_i$ occurs with total coefficient $2(n-i)+1$. Hence the Weyl vector is

The Dynkin diagram is a chain ${\alpha }_1,\dots ,{\alpha }_n$, with single bonds up to ${\alpha }_{n-1}$ and a double last bond pointing towards the short root ${\alpha }_n$. In the Extended Dynkin diagram, add ${\alpha }_0=-\theta$. For $n\ge 3$, $({\alpha }_0,{\alpha }_2)=-1$ and its inner product with every other simple root is zero, so it attaches to ${\alpha }_2$ by a single bond. For $n=2$, both long nodes $0,1$ attach by double bonds to short node $2$. These rules give the Bn Dynkin diagram and affine extension drawn below.

For rank one, $R=\{\pm e_1\}$, ${\alpha }_1=\theta =e_1$, ${\omega }_1=e_1/2$, and $\rho =e_1/2$. The finite diagram is one node; its affine extension has two equal-length nodes joined by a double bond, without a short-root arrow.

Number the simple roots of the G2 root system so that ${\alpha }_1$ is short and ${\alpha }_2$ is long. In the Euclidean plane chooseTheir squared lengths are one and three, and their angle is $150$ degrees. The six positive roots areTogether with their negatives these form the twelve-root diagram: two regular hexagons of radii one and $\sqrt{3}$, rotated by $30$ degrees relative to one another. Solving the fundamental weight equations givesIn particular ${\omega }_1$ is the highest short root and ${\omega }_2$ the highest long root. The root diagram below marks both weights.

To evaluate the Weyl dimension formula, put $a=n_1$, $b=n_2$ and $\lambda =a{\omega }_1+b{\omega }_2$. The Weyl vector is $\rho ={\omega }_1+{\omega }_2$. In the order of positive roots listed above, their coroots have simple-coroot coefficient pairsFor example the short root ${\alpha }_1+{\alpha }_2$ has coroot ${\alpha }_1^{\vee }+3{\alpha }_2^{\vee }$, while the long root $3{\alpha }_1+2{\alpha }_2$ has coroot ${\alpha }_1^{\vee }+2{\alpha }_2^{\vee }$. Pairing with $\lambda +\rho$ and dividing by the same factors at $a=b=0$ gives the G2 dimension polynomialThe denominator is $1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5$. As checks, the fundamental representations have dimensions seven and fourteen, and $a=b=0$ gives the trivial representation of dimension one.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.

For the quark model, the relevant approximate flavour symmetry rotates the three light quarks $u,d,s$ in the fundamental $3$. The up quark, down quark, and strange quark basis states have the weight diagram of the defining SU(3) representation. In physical coordinates define isospin $I_3=H_1$ and flavor hypercharge $Y=2H_2/\sqrt{3}$. ThenThe last row follows from the Gell-Mann--Nishijima formula $Q=I_3+Y/2$, in units of the positive elementary electric charge. The antiquarks transform in $\overline{3}$ and have the negatives of these additive quantum numbers. Each quark has baryon number $B=1/3$; the strange quark has strangeness $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing flavor hypercharge from electroweak hypercharge. The Weyl group geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.

A conventional meson has valence content $q\bar{q}$, so its flavour symmetry gives a meson octet and a singlet. For the light pseudoscalar mesons, the octet contains the pions ${\pi }^{+},{\pi }^0,{\pi }^{-}$ at $Y=0$ with isospin $I=1$, the kaons $K^{+}=u\bar{s},K^0=d\bar{s}$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ Eta octet state. The central weight multiplicity two is realized byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The eta and eta prime mesons are mixtures of ${\eta }_8,{\eta }_1$, rather than exact octet and singlet mass eigenstates; this gives the pseudoscalar meson nonet. A different spin coupling of the same flavour product gives the vector mesons, again an octet and singlet, including the $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ mixtures. The lowest orbital states have $J^P=0^{-}$ and $1^{-}$ respectively.

A conventional baryon has three valence quarks, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest baryon octet has spin $1/2$ and positive parity. Its isospin rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the nucleons $p,n$, the three Sigma baryons, the Lambda baryon, and the two Xi baryons. The central states ${\Sigma }^0$ and $\Lambda$ have the same weight $(I_3,Y)=(0,0)$ but different total isospin; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity baryon decuplet is the triangle with rowsThe Delta baryons form its top row, and the Omega baryon is the bottom $sss$ state. All ten weights have multiplicity one. Their electric charges follow directly from the Gell-Mann--Nishijima formula; for example ${\Delta }^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and ${\Omega }^{-}$ has $I_3=0,Y=-2$, so $Q=-1$.

The two algebraic octets do not mean that two independent ground-state octets must occur. A three-quark colour singlet has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the Pauli exclusion principle therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the symmetric spin-flavour SU6 representation hasThe three-quark flavour singlet is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but ${\bigwedge }^3{\mathbb{C}}^2=0$. The Pauli constraint on three-quark flavour multiplets thus excludes it from this ground-state sector; orbital excitation permits other multiplets.

Finally, this flavour symmetry is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The weight diagrams classify flavour states, rather than asserting exact equality of their masses. Colour SU(3) is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the eightfold way. The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.

Write $\bar{i}=2n+1-i$, and let $E_{ab}$ denote a matrix unit. The diagonal Cartan subalgebra consists ofFor a diagonal matrix, $[t,E_{ab}]=(t_a-t_b)E_{ab}$. Substituting into the symplectic relation shows that the following are root vectors:The zero weight space is spanned by $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The symplectic equation pairs the remaining entries exactly as in this table; the displayed vectors and the $H_i$ are independent and span the algebra. This proves the root-space decompositionThere are $2n^2$ roots of a root system, each with one-dimensional root space, giving $\dim \mathfrak{g}=n(2n+1)$.