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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.

Use $|m,m^{\prime }⟩$ as shorthand for the tensor product of two total-isospin-one states with third components $m,m^{\prime }$. A total-isospin-zero state must have total third component zero, so write $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$. Applying the total raising operator givesAn isospin singlet must be annihilated by this operator; hence $b=-a$ and $c=a$. Normalization gives the two-isovector singletIt is also annihilated by the total lowering operator. A raising or lowering action changes the third-component label by one; the unchanged output label in the PDF's preliminary ladder formula is a typographical omission, not the action used here.

Exchange commutes with the total isospin operators. The displayed singlet is symmetric. The $I=2$ multiplet has symmetric highest state $|1,1⟩$; applying the symmetric total lowering operator preserves that exchange symmetry. The symmetric square of a three-dimensional space has dimension six, exactly $1+5$, while its antisymmetric square has dimension three. The remaining $I=1$ multiplet is therefore antisymmetric. Thus the Clebsch-Gordan decomposition isPions have zero intrinsic spin and form an isospin triplet. Bose-Einstein statistics requires the complete two-pion state to be symmetric under exchange. Its relative orbital state changes by $(-1{)}^{\ell }$, so odd orbital angular momentum requires antisymmetric isospin: an odd-$\ell$ two-pion state has $I=1$. This statement treats the charge components as states of the same isospin multiplet, rather than ignoring exchange when their charges differ.

For the decay, let $|\chi ⟩=H_I|K^0⟩$. The stated commutator with $I_3$ gives $I_3|\chi ⟩=0$. More decisively, the commuting raising operator givesThe kaon doublet cannot be raised twice. An $I=2,M=0$ component would survive two raisings, so it is excluded. Meanwhile a spinless kaon decaying into two spinless pions has $\ell =0$ by conservation of total angular momentum, independently of whether parity is conserved by the decay. The exchange condition excludes $I=1$. This proves the highest-weight weak-isospin selection in kaon decay: the two-pion final state has $I=0$.

In a consistent pion phase convention, the singlet's charge components have the structureThe normalized symmetric charged channel has coefficient $\sqrt{2/3}$ and the neutral channel coefficient $-1/\sqrt{3}$. Thus, in the isospin limit with equal pion masses and matching two-body phase space,Equivalently, ordered charged and neutral amplitudes have equal magnitudes, but neutral identical particles contribute the phase-space factor $1/2!$. This is the same counting expressed in another basis, and must not be included a second time after using the normalized charge-channel coefficients. Small pion-mass and electromagnetic effects can modify the ideal ratio.

In the notation of part (i), the tensor raising operator and lowering operator act bywith terms beyond the stated index range omitted. Both $v_4=A_0$ and $v_2=B_0-A_1$ are killed by $e$, and have respective weights four and two. They are the desired highest-weight vectors.

For $v_4$, successive lowering givesThus a basis of the five-dimensional summand, in descending weight order, isFor $v_2$, the successive vectors are $f{v}_2=2(B_1-A_2)$, $f^2{v}_2=2(B_2-A_3)$ and $f^3{v}_2=0$. HenceThese spans are invariant under $h,e,f$. The sl2 highest-weight lowering formula gives $e{f}^r{v}_n=r(n-r+1)f^{r-1}{v}_n$, so all interior raising coefficients in either chain are nonzero. Any nonzero submodule therefore contains the highest vector and then the whole chain; each span is irreducible.

At the shared weights $2,0,-2$, the two displayed summand vectors are independent: their coefficient determinants in the original weight bases are $4,2,4$. At weights $4,-4$ only the five-dimensional summand occurs. Consequently the eight displayed vectors form a basis of $U$, proving that these are complementary submodules, not only modules with the right dimensions.

Use the standard sl2 triple $H=x{\partial }_x-y{\partial }_y$, $X=x{\partial }_y$, $Y=y{\partial }_x$. In the tensor product of Lie algebra representations, each operator acts on both factors. PutThese twelve vectors form a weight basis, and $H{u}_{ij}=(5-2(i+j))u_{ij}$. Thus the bases of the weight spaces areThere are no other weights. The weight diagram shows their multiplicities, with the decomposition also displayed for comparison.

The raising operator acts bywith out-of-range terms omitted. At weights five, three and one its kernels are respectively spanned byAt weight one the kernel equations are $2a+b=0$ and $b+2c=0$ for $a{u}_{20}+b{u}_{11}+c{u}_{02}$. At weight minus one they are $3a+b=0$, $2b+2c=0$, $c=0$, which force zero; the maps on the two remaining negative-weight spaces are injective as well. Thus these three lines give all nonzero highest-weight vectors up to scalar, at their respective weights. A linear combination of different lines is killed by $X$ but is not a weight vector, so is not a highest-weight vector.

The lowering operator isIn particular,is an explicit basis for the submodule ${\Gamma }_1$. Indeed $Y^2{h}_1=0$, $X{h}_1=0$, $X(Y{h}_1)=h_1$, and the two vectors have weights $1,-1$. The analogous strings from $h_5,h_3$ have dimensions six and four. Their distinct irreducible highest weights and total dimension twelve give the sl2 tensor product of cubic and quadratic symmetric powers

The anti-Hermitian operators represent the compact real form; its complexification also contains the Hermitian $H_i$ and the raising operators and lowering operators $E_{\pm }^r$, with $(E_{+}^r{)}^{†}=E_{-}^r$. Define the roots in the two-dimensional Cartan plane byEach has squared length one. For each $r=1,2,3$, setThe displayed commutators give $[J_3^{(r)},J_{\pm }^{(r)}]=\pm J_{\pm }^{(r)}$ and $[J_{+}^{(r)},J_{-}^{(r)}]=2J_3^{(r)}$. Thus these are the three root SU(2) subalgebras of SU(3).

The commuting Hermitian operators $H_1,H_2$ can be simultaneously diagonalized. A weight of a representation is a pair $\lambda =(h_1,h_2)$ for whichwhere $a$ labels independent states of that representation weight. The weight multiplicity is the dimension of this simultaneous eigenspace. Commuting $H_i$ past a ladder operator proveswith zero image allowed when the raised or lowered representation weight is absent.

Choose ${\alpha }_1,{\alpha }_2$ as positive simple roots. A highest-weight vector is a nonzero simultaneous eigenvector annihilated by $E_{+}^1,E_{+}^2,E_{+}^3$; the third condition follows from the first two and their commutator. In an irreducible representation, its highest weight $\Lambda$ has multiplicity one and repeated lowering generates the representation. The three root SU(2) subalgebras of SU(3) give nonnegative integer Dynkin labelsEquivalently, with ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$,These are the SU(3) highest-weight coordinates in the normalization used here.

The Weyl group is generated by the reflections $s_r(\lambda )=\lambda -2(\lambda \cdot {\alpha }_r){\alpha }_r$. The SU(2) Lie algebra root-string symmetry makes these reflections preserve both representation weights and their weight multiplicities. They generate the symmetries of an equilateral triangle: rotations through $120^{\circ }$ and three reflections. The outer polygon is the convex hull of the Weyl group orbit of $\Lambda$. When $p,q>0$ it is a hexagon with alternating side lengths $p,q$ in root-step units; when exactly one is zero it is a triangle; when both vanish it is a point. The representation conjugate to $(p,q)$ has labels $(q,p)$ and negative representation weights. For $p=q$, this gives an additional inversion symmetry and a regular hexagon with $60^{\circ }$ rotations. A general unequal-sided hexagon need not have that extra symmetry.

The shell multiplicities in an SU(3) weight diagram can be stated precisely. Let $P_{a,b}$ be the convex hull of the Weyl group orbit of $a{\omega }_1+b{\omega }_2$. For $\lambda$ in the root lattice coset $\Lambda +Q$,representation weights outside this coset or outside the outer polygon are absent. This states that boundary representation weights have multiplicity one, each inward shell adds one, and the innermost triangular region has constant multiplicity $\min (p,q)+1$. When $p=q$ the innermost region is a single point. All representation weights of a triangular diagram have multiplicity one. Counting the diagram gives $\dim (p,q)=(p+1)(q+1)(p+q+2)/2$. The examples below display both triangular orientations, a regular hexagon, an unequal-sided hexagon, a central multiplicity three, and the trivial diagram.

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.

Apply the lowering operator $F$ to the highest-weight vector from part (ii). Since $F{w}_{03}=4w_{13}+w_{04}$ and $F{w}_{12}=2w_{13}$, we obtainThis is nonzero because $w_{04},w_{13}$ are distinct elements of the exterior power basis. Both have weight zero. As a consistency check, the raising operator sends this vector to $2v_{+}$, as required by the sl2 highest-weight lowering formula.

The induced raising operator and lowering operator act bySince $E{w}_{01}=0$ and $w_{01}$ has weight $6$, it generates the seven-dimensional irreducible sl2 Lie algebra module ${\mathrm{Sym}}^{6}V$. By complete reducibility of semisimple Lie algebra representations, subtracting its weights $6,4,2,0,-2,-4,-6$, each of multiplicity one, from the weight diagram leaves weights $2,0,-2$ once each. HenceThese are the only two irreducible isomorphism types occurring as submodules.

For the three-dimensional summand, calculate $E{w}_{03}=3w_{02}$ and $E{w}_{12}=w_{02}$. Thereforeis a nonzero highest-weight vector of weight $2$. The classification of finite-dimensional sl2 representations shows that the submodule $W$ it generates is the irreducible module ${\mathrm{Sym}}^{2}V$, of dimension three.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

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.

Fix a simple root ${\alpha }_i$ and choose generators $e_i,f_i,h_i$ of the sl2 subalgebra associated with a root, with $h_i={\alpha }_i^{\vee }$ andThe highest-weight vector satisfies $e_{i}v=0$ and $h_{i}v=m_{i}v$, where $m_i=⟨\lambda ,{\alpha }_i^{\vee }⟩$. The classification of finite-dimensional sl2 representations already implies that $m_i$ is a nonnegative integer. One can also see the necessary integrality directly from the lowering operator.

Indeed, $f_i^{k}v$, if nonzero, has $h_i$ eigenvalue $m_i-2k$. These distinct eigenvalues cannot occur indefinitely in a finite-dimensional space, so there is a largest $r\ge 0$ with $f_i^{r}v\ne 0$. The sl2 Lie algebra relations imply, by induction on $k$,For $k=r+1$ the left side is zero, hence $(r+1)(m_i-r)f_i^{r}v=0$. Thus $m_i=r\in {\mathbb{Z}}_{\ge 0}$. Since this holds for every simple root,

Choose the given short root $\alpha$ as a simple root, and let $\beta$ be the long simple root. This entails no loss of generality: the Weyl group is transitive on the short roots of the G2 root system. The relevant Cartan integers are $⟨\beta ,{\alpha }^{\vee }⟩=-3$ and $⟨\alpha ,{\beta }^{\vee }⟩=-1$. The positive roots areUnder ${\mathfrak{m}}_{\alpha }\cong {\mathfrak{sl}}_2$, the root space for $m\alpha +n\beta$ has $h_{\alpha }$ eigenvalue $2m-3n$. The subalgebra ${\mathfrak{m}}_{\alpha }$ itself is the three-dimensional irreducible $V(2)$ in the Adjoint representation.

The four root spaces along the root string $\beta ,\alpha +\beta ,2\alpha +\beta ,3\alpha +\beta$ have eigenvalues $-3,-1,1,3$. Consecutive spaces are connected by nonzero raising operators and lowering operators, so their sum is $V(3)$. The negative root string supplies another $V(3)$. The root spaces for $\pm (3\alpha +2\beta )$ have eigenvalue zero, and neither adding nor subtracting $\alpha$ gives a root; they are two copies of $V(0)$. Finally, the one-dimensional space $\ker \alpha \subset \mathfrak{t}$ commutes with ${\mathfrak{m}}_{\alpha }$, giving one more $V(0)$. We have accounted for all $14$ dimensions, and therefore the G2 adjoint branching to a short-root sl2 subalgebra isFor a nonzero root vector $x\in {\mathfrak{g}}_{\alpha }$, its Adjoint representation action is a nonzero scalar multiple of the raising operator on each summand. On each irreducible $V(n)$ this operator has a one-dimensional kernel, including $V(0)$. Since the Lie algebra centralizer is that kernel,