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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.

If a dominant integral weight $\lambda$ lies in the root lattice, its finite-dimensional irreducible highest-weight representation contains zero as a weight. First $\lambda$ is a nonnegative integral combination of simple roots: a negative part would have negative inner product with $\lambda$, contradicting dominance. For any nonzero weight $\mu =\sum m_i{\alpha }_i$ with $m_i\ge 0$, the identity $(\mu ,\mu )=\sum m_i(\mu ,{\alpha }_i)>0$ supplies an index with $m_i>0$ and $⟨\mu ,{\alpha }_i^{\vee }⟩>0$. By Injectivity of sl2 lowering above weight zero, applying the corresponding lowering operator produces a nonzero vector of weight $\mu -{\alpha }_i$. Repetition reaches zero.

For the irreducible highest-weight representation of the special linear Lie algebra ${\mathfrak{sl}}_3$ with Dynkin labels $(a,b)$, let $(p,q)$ be a dominant integral weight. Put $r=(2a+b-2p-q)/3$ and $s=(a+2b-p-2q)/3$. If these are not integers, the weight multiplicity is zero; otherwise it is $\max \{0,1+\min (a,b,r,s)\}$.

To derive the formula, use the sl3 interlacing character formula with top row $(a+b,b,0)$. The target diagonal exponents are $(s+p+q,s+q,s)$, and an interlacing middle row $(P,Q)$ has $P+Q=2s+p+2q$, with bottom entry $s+p+q$. ThusEvery integral $P$ in this interval gives exactly one Gelfand–Tsetlin basis vector. The upper bound minus the lower bound is the minimum of six differences: $a,s,r,r+p+q,b,s+q$. Since $p,q\ge 0$, this minimum reduces to $\min (a,b,r,s)$. Counting the interval proves the formula, including absent weights of a representation. This converts a two-dimensional weight diagram calculation into four elementary boundary distances.

For a finite-dimensional irreducible highest-weight representation with highest weight $\lambda$, every weight in the Weyl group orbit $W\lambda$ is extremal and has weight multiplicity one.

A highest-weight vector is a nonzero weight vector annihilated by every positive-root raising operator. The module it generates is a highest-weight representation.

A dominant integral weight $\lambda$ is minuscule when its irreducible highest-weight representation is a minuscule representation. Equivalently, $⟨\lambda ,{\alpha }^{\vee }⟩\in \{0,1\}$ for every positive root $\alpha$. This convention includes the zero weight and its trivial Lie algebra representation.

Work over $\mathbb{C}$ with a finite-dimensional semisimple Lie algebra $\mathfrak{g}$. Choose a Cartan subalgebra $\mathfrak{h}$, its root system $R\subset {\mathfrak{h}}^{\ast }$, a positive system $R^{+}$ and its simple roots. Let $W$ be the Weyl group, generated by the corresponding root reflections, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the Weyl vector. The finite-dimensional irreducible highest-weight representation $L(\lambda )$ is indexed by a dominant integral weight $\lambda$.

If $P$ is the weight lattice, its group ring has formal symbols $e^{\mu }$ with multiplication $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. The formal character iswhere $L(\lambda {)}_{\mu }=\{v:Hv=\mu (H)v\text{ for all }H\in \mathfrak{h}\}$ is a weight space. For a weight $\nu$, define the Weyl alternant $A_{\nu }={\sum }_{w\in W}\det (w)e^{w\nu }$; the sign $\det (w)=(-1{)}^{\ell (w)}$ is the determinant of $w$ on the real root span, equivalently the parity of its expression in simple reflections.

The Weyl character formula isThe second denominator is the Weyl denominator formula. The quotient can initially be taken in the fraction field of the group ring; the formula says it is the finite character on the left, so its apparent singularities are removable. The shifts by $\rho$ make $\lambda +\rho$ strictly dominant.

Work first in a finite-dimensional SU(2) representation, with its invariant positive inner product. In the complexified Lie algebra, choose $H^{†}=H$ and $(E^{-}{)}^{†}=E^{+}$. Let $v_0$ be a highest-weight vector with $H{v}_0=m{v}_0$ and $E^{+}{v}_0=0$, and put $v_r=(E^{-}{)}^r{v}_0$. The commutator with $H$ givesInduction also givesFor $r=1$ this is $E^{+}{E}^{-}{v}_0=H{v}_0=m{v}_0$. For the induction step, use $E^{+}{E}^{-}=E^{-}{E}^{+}+H$ on $v_r$: the coefficient becomes $r(m-r+1)+m-2r=(r+1)(m-r)$. In particularwhich proves the assumed formula rather than merely invoking it.

The norm identity $\Vert E^{-}{v}_r{\Vert }^2=(r+1)(m-r)\Vert v_r{\Vert }^2$ shows that, for nonnegative integer $m$, the ladder is nonzero for $0\le r\le m$ and stops with $E^{-}{v}_m=0$. Distinct eigenvalues make these vectors independent. Their span is stable under $H,E^{\pm }$, and in the irreducible module it is the full space. ThusFor normalized states, $E^{-}|n⟩=\frac{1}{2}\sqrt{(m+n)(m-n+2)}|n-2⟩$, with the last state killed. General finite-dimensional SU(2) representations are direct sums of these ladders.

The representation category matters here. The Lie-algebra relations alone allow an infinite Verma module even at integral highest weight: its nonzero vector $v_{m+1}$ is singular. The finite representation can instead be constructed algebraically by taking the basis $v_0,\dots ,v_m$ with the displayed actions and declaring $E^{-}{v}_m=0$; all three Lie brackets then hold, including at the endpoints. Thus the finite-dimensional assertion uses the integrable SU(2) module, or this explicit finite quotient, rather than assuming that a singular vector must vanish in every highest-weight module.

For rank two, take a Cartan subalgebra $\mathfrak{h}$ with commuting basis elements. A root $\alpha$ is a nonzero linear functional on $\mathfrak{h}$ for which a root space contains an element $E_{\alpha }$ satisfyingChoose simple roots ${\alpha }_1,{\alpha }_2$ and their coroots $H_i={\alpha }_i^{\vee }$. In the PDF's index convention, the Cartan matrix isNormalize $E_i^{\pm }=E_{\pm {\alpha }_i}$ so $[E_i^{+},E_i^{-}]=H_i$. The difference of distinct simple roots is not a root, so the mixed positive-negative bracket is zero. This givesThe transpose in the last formula is essential and is printed correctly in the original PDF.

A highest-weight vector of labels $({\overline{n}}_1,{\overline{n}}_2)$ is killed by both $E_i^{+}$ and is an eigenvector of the two $H_i$. Repeated words in $E_1^{-},E_2^{-}$ generate its highest-weight representation; each lowers the weight by the corresponding simple root. Restricting to the sl2 subalgebra associated with a root for either simple root gives the preceding integer ladder, so finite-dimensional highest labels must be nonnegative integers. The finite irreducible module is the highest-weight quotient imposing $(E_i^{-}{)}^{{\overline{n}}_i+1}v=0$ and the submodules generated by these relations, together with the Serre relations. Choosing independent nonzero lowering words gives a simultaneous weight basis; an extra multiplicity index is needed where a weight space has dimension greater than one.

For the specified B2 root system, the simple-root squared lengths are $1,2$ and their inner product is $-1$, givingThe positive root ${\alpha }_1+{\alpha }_2$ is generated by $[E_1^{+},E_2^{+}]\ne 0$, and $2{\alpha }_1+{\alpha }_2$ by $[E_1^{+},[E_1^{+},E_2^{+}]]\ne 0$. No further positive sums occur, so $(\mathrm{ad}{E}_1^{+}{)}^3{E}_2^{+}=0$ and $(\mathrm{ad}{E}_2^{+}{)}^2{E}_1^{+}=0$. These are the relevant Serre relations. Applying the Cartan commutators to a weight vector proveswhenever the image is nonzero.

To enumerate the requested representation without guessing when arrows stop, realize it concretely as the Adjoint representation. For a Euclidean weight $(x,y)$, its labels are $(n_1,n_2)=(2x,-x+y)$. The highest root $2{\alpha }_1+{\alpha }_2=(1,1)$ has labels $(2,0)$, so it generates the irreducible adjoint module of this simple rank-two Lie algebra. Its nonzero weights are exactly its eight roots, and its zero weight space is the two-dimensional Cartan subalgebra. This gives the B2 adjoint representation weight diagram and the nine distinct states labels:Their actual production by lowering can be seen fromAn arrow labelled $i$ denotes a nonzero action of $E_i^{-}$ between the indicated weight spaces. The two paths into zero produce independent Cartan vectors. Hence there are nine distinct pairs, as requested when degeneracy is disregarded, but the full representation has dimension ten, not nine.

Use $[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m){\delta }_{m+n,0}$ and $L_n^{†}=L_{-n}$. A normalized highest-weight vector obeys $L_0|h⟩=h|h⟩$ and $L_n|h⟩=0$ for $n>0$. Its Virasoro descendants are products of negative modes. Their inner products are determined entirely by $c,h$ and the Virasoro algebra; for example $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. An irreducible unitary highest-weight Virasoro module is the Virasoro Verma module quotient by its null submodules, and its surviving inner product must be positive definite.

The classification for $c<1$ is the Unitary Virasoro discrete series:where $m$ is an integer at least $2$, $1\le r\le m-1$, and $1\le s\le m$. The identification $(r,s)\sim (m-r,m+1-s)$ leaves $m(m-1)/2$ distinct modules. The two generating singular-vector levels are $rs$ and $(m-r)(m+1-s)$; all their Virasoro descendants are removed in the irreducible quotient. For $m=2$ the only module is the trivial $c=h=0$ module; for $m\ge 3$ these are the nontrivial unitary discrete-series modules. There are no other irreducible unitary highest-weight representations with $c<1$. This is the classification theorem, rather than a condition inferred only from finitely many descendant norms.

At $c=1/2$, $m=3$ and the possible weights areConstruct them with a single free chiral Majorana fermion:The free chiral Majorana fermion conformal field theory has $T=-:\psi \partial \psi :/2$. In the Neveu–Schwarz sector, $r\in \mathbb{Z}+1/2$, while in the Ramond sector, $r\in \mathbb{Z}$. With positive modes moved to the right and the zero-mode product antisymmetrized, the corresponding generators areThe oscillator relation gives $[L_n,b_r]=-(r+n/2)b_{n+r}$. The stress-tensor operator product expansion, or the reordering of the quadratic generators, then gives the Virasoro algebra with $c=1/2$. One can check both constants directly in the ground states. In the NS sector,In the R sector, $b_0^2=1/2$ and $L_{-1}|R⟩=\frac{1}{2}{b}_{-1}{b}_0|R⟩$, whose norm squared is $1/8$. The relation $[L_1,L_{-1}]=2L_0$ requires $2h_R=1/8$, so $h_R=a=1/16$. Thus the Ramond sector zero-mode shift is essential.

The three constructions in Ising Virasoro modules from a Majorana fermion are explicit. The NS vacuum is a highest-weight vector of weight $0$. The state $b_{-1/2}|0⟩$ is a highest-weight vector of weight $1/2$: applying any positive $L_n$ gives $-(n/2-1/2)b_{n-1/2}|0⟩=0$, including the zero coefficient for $n=1$. Since $L_n$ is quadratic in fermions, the two cyclic modules lie in the even and odd fermion parity sectors, respectively. For R, choose a normalized ground state annihilated by all $b_n$ with $n>0$ and with $b_0|R⟩=\pm |R⟩/\sqrt{2}$. All positive $L_n$ annihilate it and its weight is $1/16$. The two zero-mode sign choices give isomorphic Virasoro modules; one choice suffices to construct the third module.

The oscillator inner products are positive, and each cyclic highest-weight representation just constructed is irreducible. Indeed any nonzero proper invariant graded submodule would have a lowest-energy vector $w$ of weight greater than the cyclic ground weight. Every positive $L_n$ annihilates $w$; thus $w$ is orthogonal to every negative-mode descendant of the original ground vector, by moving the modes across the inner product. It is also orthogonal to that ground vector by its different $L_0$ eigenvalue. Since those Virasoro descendants span the cyclic module, $w$ would have zero norm, a contradiction. This proves irreducibility directly inside the positive fermionic Fock space.

In fact the cyclic modules fill the entire even NS, odd NS and chosen R Fock spaces. Otherwise an orthogonal remaining lowest vector would be the ground vector of a further unitary highest-weight Virasoro module; its weight would lie in ${\mathbb{Z}}_{>0}$, $1/2+{\mathbb{Z}}_{>0}$ or $1/16+{\mathbb{Z}}_{>0}$, respectively. None belongs to the classified list $\{0,1/2,1/16\}$. As a check, their Virasoro characters, defined by ${\chi }_h(q)=\mathrm{Tr}{q}^{L_0-c/24}$, areThe NS signs project onto even and odd occupation number; the R product uses the single chosen ground state. Null vectors of the abstract Virasoro Verma modules are automatically zero in this construction. In particular $L_{-1}|0⟩=0$, and the weight-$1/2$ and weight-$1/16$ ground states obey, respectively, $(L_{-2}-\frac{3}{4}{L}_{-1}^2)|1/2⟩=0$ and $(L_{-2}-\frac{4}{3}{L}_{-1}^2)|1/16⟩=0$, as is also verified by their level-two Gram matrices.

Write the affine modes as $J_n^a=T_n^a$, and put $Q=Q^{\mathfrak{g}}$. The Sugawara construction requires a restricted module, meaning that sufficiently large positive current modes annihilate each fixed vector, and a noncritical level $2k+Q\ne 0$. A positive-energy highest-weight representation has the required restriction. DefineNormal ordering puts positive modes to the right, making each sum finite on a fixed finite-energy vector. The invariant sum over the current index $a$ is the quadratic Casimir operator construction for the affine current algebra.

Here is the normal ordering calculation responsible for its denominator. In commuting $S_m$ with $J_n^b$, the two affine central terms supply $-2kn{J}_{m+n}^b$. The remaining terms contain $i{f}^{abc}$ times the two ways to replace a current by $J_{p+n}^c$. Relabeling the summation index and using antisymmetry of $f^{abc}$ cancels the un-reordered terms. Reordering the modes that cross the creation/annihilation boundary leaves $n$ signed terms, whose two structure constants contract by $f^{abc}{f}^{abd}=Q{\delta }^{cd}$. Their sum is $-Qn{J}_{m+n}^b$. ThusThe adjoint-Casimir contribution is genuinely quantum. For an explicit check of its coefficient, let $|0⟩$ be the affine vacuum, with all $J_n^a|0⟩=0$ for $n\ge 0$. ThenThe first term acts by the adjoint Casimir operator $Q$, and only $p=1$ contributes $2k$ to the second. This also verifies that a current has conformal weight one under the displayed normalization.

Now commute the quadratic generators with one another. Acting on either current factor produces $(m-n)L_{m+n}^{\mathfrak{g}}$; the residual full contractions are scalar and vanish unless $m+n=0$. The Jacobi identity fixes their cubic dependence on $m$, and the plane vacuum convention $L_{-1}|0⟩=L_0|0⟩=L_1|0⟩=0$ fixes the combination $m^3-m$. To determine its coefficient explicitly, noteCommuting a single $J_1^a$ through the numerator gives $(2k+Q)J_{-1}^a|0⟩$: the two central contractions give $2k{J}_{-1}^a$, and the zero-mode commutator in the structure-constant term gives $Q{J}_{-1}^a$. The remaining $J_1^a$ contraction gives $k$ for each adjoint index. ThereforeFor an algebraic module without a positive inner product, the same calculation is its vacuum matrix element and fixes the universal central term. ConsequentlyThis is a representation of the Virasoro algebra, not merely its centerless part.

For a complex semisimple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal abelian subalgebra of semisimple elements; equivalently it is the complexification of a maximal torus algebra in the compact real form. Its dimension is the rank of a semisimple Lie algebra, $r$. Choose an orthonormal Cartan basis with currents $H_n^i$. They have no structure-constant term:For $k\ne 0$, their abelian Sugawara construction isEach Cartan current has weight one under both $L^{\mathfrak{g}}$ and $L^{\mathfrak{h}}$, so $K_m=L_m^{\mathfrak{g}}-L_m^{\mathfrak{h}}$ commutes with all $H_n^i$, hence also with $L_n^{\mathfrak{h}}$. Subtracting the two commuting stress tensors yields the Coset constructionIn the intended nontrivial compact unitary theory at positive admissible level, this coset acts on a positive inner-product space, with $K_m^{†}=K_{-m}$. Its vacuum satisfies $K_n|0⟩=0$ for $n\ge -1$, soThus $c^{\mathfrak{g}}\ge r$. Also $Q>0$ for a compact simple Lie algebra and $k>0$, so $2k/(2k+Q)<1$, which proves the upper bound directly. Hence, under these physical representation hypotheses,The upper inequality is strict at finite positive level; $c^{\mathfrak{g}}$ tends to $\dim \mathfrak{g}$ as the level grows.

The positivity hypothesis is necessary and is not implied by the bare affine commutation relation. For example, at $k=0$ its trivial representation has all currents zero and gives $c^{\mathfrak{g}}=0<\mathrm{rank}\mathfrak{g}$. Even positive arbitrary levels need not satisfy the bound: with $\mathfrak{g}=\mathfrak{su}(2)$, $f^{abc}={ϵ}^{abc}$, $Q=2$ and $k=1/4$, the formula gives $c^{\mathfrak{g}}=3/5<1$. Such a level does not give the intended compact unitary positive-energy theory. Thus the last bound is a statement about nontrivial unitary admissible-level representations; it cannot hold for every representation in the unrestricted algebraic formulation. At $2k+Q=0$ the displayed Sugawara construction itself is undefined.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

Use a finite-dimensional complex semisimple Lie algebra, or the complexification of its compact real form, as required for the root construction. Choose a Cartan subalgebra with basis $H_1,\dots ,H_r$. Simultaneous diagonalization of its adjoint action gives the root-space decompositionThe nonzero weights are the roots of a root system; each root space is one-dimensional in this setting, and roots occur in opposite pairs. This supplies the Cartan and root-vector basis.

The Jacobi identity givesIf the bracket is nonzero and $\alpha +\beta \ne 0$, it is a nonzero vector in the root space of weight $\alpha +\beta$, so that sum is a root. The printed implication needs the opposite-root exception: for $\beta =-\alpha$, $[E_{\alpha },E_{-\alpha }]$ can be a nonzero Cartan element, while zero is not a root. The supplied simple-root commutator itself gives this counterexample. This is the root bracket and the opposite-root exception.

Choose a real linear functional nonzero on every root; positive roots are those on which it is positive, and negative roots are their negatives. The simple roots are positive roots not decomposable into two positive roots. They form a basis ${\alpha }_1,\dots ,{\alpha }_r$ of the root span, and every root is an integer combinationwith all $m_i\ge 0$ for a positive root and all $m_i\le 0$ for a negative root. Positive roots can be decomposed repeatedly until simple roots remain; the finite root set makes this procedure terminate.

For $i\ne j$, ${\alpha }_j-{\alpha }_i$ has simple-root coefficients of both signs and is not a root. It is also nonzero, so the root-space bracket rule givesThe operators $E_i^{\pm },{\hat{H}}_i$ have the SU(2) Lie algebra normalization ${\hat{H}}_i=2J_3$, $E_i^{\pm }=J_{\pm }$. In a finite-dimensional representation, a vector killed by $E_i^{-}$ is a lowest-weight vector in its SU(2) summands. If its ${\hat{H}}_i$ value is $-\lambda$, those summands have spin $\lambda /2$, soOne can also read the termination from successive raising coefficients $\sqrt{(\lambda -k)(k+1)}$ starting at the lowest weight.

Apply this result in the finite-dimensional Adjoint representation, taking the lowest-weight vector to be $E_j^{+}$. The vanishing bracket just proved is its lowering condition. If $[{\hat{H}}_i,E_j^{+}]=a_{ij}{E}_j^{+}$, the lowest-weight result forces $a_{ij}=-n_{ij}$ with $n_{ij}$ a nonnegative integer. It also yieldsThe last identity is exactly the requested nested commutator. This is the SU(2) proof of the Serre relations, and the identities are the Serre relations in these Chevalley basis conventions.

To construct a highest-weight representation, start from its highest-weight vector and apply negative-root operators. Fix an order on the negative roots. The Poincaré-Birkhoff-Witt theorem gives spanning vectorsof weight $w-{\sum }_i{k}_i{\beta }_i$. The spanning assertion follows by commuting all negative-root operators to the left, Cartan operators to the middle and positive-root operators to the right; commutators supply lower-order words, positive operators kill the highest-weight vector and Cartan operators act there by scalars. In the finite-dimensional highest-weight representation, impose the simple-root string relations $(E_i^{-}{)}^{w_i+1}|w⟩=0$ and the commutator/Serre relations. At each weight retain a linearly independent set of the resulting words, discarding zero words and dependencies. This gives a basis, with separate independent vectors retained when a weight has multiplicity greater than one. The construction is finite in the assumed finite-dimensional representation; treating all lowering words as automatically independent would overcount.

For SU(3), put $e_i=E_i^{+}$, $f_i=E_i^{-}$ and $h_i={\hat{H}}_i$. The off-diagonal Cartan entries are minus one, so the Serre relations give $(\mathrm{ad}{e}_1{)}^2{e}_2=(\mathrm{ad}{e}_2{)}^2{e}_1=0$. With $e_3=[e_1,e_2]$, these implyFor $f_3=-[f_1,f_2]$, expand the commutator using Jacobi identity and the distinct-root brackets:since $[h_1,f_2]=f_2$. Similarly,and the same calculation on $f_3$ gives $[h_1,f_3]=-f_3$. ThusThese signs also agree with the explicit model $e_1=E_{12}$, $e_2=E_{23}$, $e_3=E_{13}$, $f_1=E_{21}$, $f_2=E_{32}$, $f_3=E_{31}$.

For the SU(3) Casimir in Chevalley generators, setUsing $[h_1,e_1]=2e_1$, $[h_2,e_1]=-e_1$, and putting Cartan elements on the left, the four terms of $[P,e_1]$ areTheir sum is $3h_1{e}_1$. The remaining terms giveThe last two cancel, so $[C,e_1]=3h_1{e}_1-3h_1{e}_1=0$. Also $h_1$ commutes with every Cartan polynomial and with each $f_n{e}_n$: their raising and lowering weights cancel in the product. HenceOn $|w_1,w_2⟩$, $e_1$ and $e_2$ vanish and therefore $e_3=[e_1,e_2]$ vanishes as well. All three $f_n{e}_n$ terms consequently give zero, leavingThe paper's normalization is $C=3C_2$ relative to the conventional SU(3) quadratic Casimir eigenvalue. For example it gives $4$ in the fundamental representation, $9$ in the octet and $18$ in the decuplet.