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The complex Adjoint representation of SU(3) is the action by conjugation on traceless $3\times 3$ matrices. With $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, its highest-weight vector is $E_{13}$ of Dynkin labels $(1,1)$. The six off-diagonal matrix units have weights $(2,-1),(-2,1),(-1,2),(1,-2),(1,1),(-1,-1)$, and the two-dimensional Cartan subalgebra has weight zero. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the weight diagram is a regular hexagon together with a central point of weight multiplicity two. This representation describes both the gluon colour multiplet and the approximate flavour baryon octet, under different physical SU(3) symmetries.

The completely symmetric three-quark representation of flavour symmetry has dimension ten. Its weight diagram has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. It includes $uuu$ at the ${\Delta }^{++}$ weight and $sss$ at the ${\Omega }^{-}$ weight. A spatially symmetric three-quark colour singlet realizes this multiplet with spin $3/2$.

The spin-one-half octet of flavor symmetry contains the nucleons, Sigma baryons, Lambda baryon, and Xi baryons. Its weight diagram has six peripheral states and two states at $(I_3,Y)=(0,0)$: ${\Sigma }^0$ and ${\Lambda }^0$, with different total isospin.

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.

The pseudoscalar octet of flavor symmetry contains three pions, four kaons and the Eta octet state. In its weight diagram, the neutral pion and ${\eta }_8$ are separate states at the origin.

The light-quark $sss$ member of the spin-three-halves baryon decuplet has strangeness $S=-3$, baryon number $B=1$, flavor hypercharge $Y=-2$, and isospin zero. The Gell-Mann--Nishijima formula gives electric charge minus one. Its position at the bottom of the decuplet's triangular weight diagram follows by replacing all three up or down quarks with strange quarks.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger representations containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large representation.

A dual vector must be pulled back by $P^T$, since $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus a G2 root system weight $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet meson retains its values from solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor instance the first three come from the G2 root system weights $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The weight diagram displays the preserved octet and the new states.

The fractional electric charges make these additional states qualitatively different from ordinary quark-antiquark mesons, whose electric charges are integers. This is a consequence of the proposed symmetry extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

Write $L(n)$ for the irreducible representation of ${\mathfrak{sl}}_2$ with highest weight $n$, dimension $n+1$ and weights $n,n-2,\dots ,-n$. The weight diagram below records multiplicities $1,2,2,2,1$ at weights $4,2,0,-2,-4$.

The unique top weight four forces one copy of $L(4)$, by complete reducibility of semisimple Lie algebra representations. Removing its five weights, each with multiplicity one, leaves weights $2,0,-2$ with multiplicity one, exactly the character of $L(2)$. HenceThe two lower rows draw their separate weight diagrams. The explicit invariant copies constructed next establish the Clebsch-Gordan splitting of a defining module and its cubic symmetric power within $U$ itself.

Let $e_1=x,e_2=y,e_3=z$, and let ${ϵ}_i$ be the weight of a representation defined by ${ϵ}_i(\mathrm{diag}(h_1,h_2,h_3))=h_i$. Since the trace is zero, ${ϵ}_1+{ϵ}_2+{ϵ}_3=0$. The dual representation gives weight $-{ϵ}_i$ to $e_i^{\ast }$. PutTaking $j\le k$ gives an eighteen-vector tensor product basis. Its weight is ${ϵ}_j+{ϵ}_k-{ϵ}_i$.

For explicit coordinates use the usual simple coroots $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, and write each weight as $(\mu (H_1),\mu (H_2))$. Then ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1,1)$ and ${ϵ}_3=(0,-1)$. Collecting equal weights gives every weight space:The first six weights are $2{ϵ}_j-{ϵ}_i$ with $i\ne j$; the next three are $-2{ϵ}_i$; the last three are the defining weights ${ϵ}_i$. Thus there are nine multiplicity-one weights and three multiplicity-three weights, giving total dimension $9+9=18$.

The planar coordinates are $a{\omega }_1+b{\omega }_2$ with ${\omega }_1=(1,0)$ and ${\omega }_2=(1/2,\sqrt{3}/2)$, so the weight diagram uses an equilateral root geometry rather than orthogonal Dynkin-coordinate axes. Every point is labelled with its complete weight-space basis.

For a finite-dimensional Lie algebra representation $V$, a weight of a representation relative to $\mathfrak{h}$ is a linear functional $\mu \in {\mathfrak{h}}^{\ast }$ for which the simultaneous eigenspaceis nonzero. Its elements are weight vectors, and $\dim V_{\mu }$ is the weight multiplicity. The set of weights is $\boxed{{\Lambda }_V=\{\mu :V_{\mu }\ne 0\}}$. For a complex semisimple Lie algebra, the Cartan action on a finite-dimensional module is simultaneously diagonalizable, giving $V={⨁}_{\mu \in {\Lambda }_V}{V}_{\mu }$. The set of weights and their multiplicities are distinct information: several basis vectors can occupy the same point of a weight diagram.

The weight diagram has weights $6,4,2,0,-2,-4,-6$ with multiplicities $1,1,2,2,2,1,1$. An Irreducible Lie algebra representation of the sl2 Lie algebra with highest weight $n$ has weights $n,n-2,\dots ,-n$, all of multiplicity one. The top weight therefore forces one $L(6)$ summand. Subtracting its formal character leaves weights $2,0,-2$, each once, which is $L(2)$. By Weyl complete reducibility theorem,Their dimensions $7+3=10$ account for the entire module.

Use the basis labels in the preceding diagrams. For each $j$, write $\{i,l\}=\{1,2,3\}\setminus \{j\}$ with $i<l$, and defineThe contraction satisfies $C(A_j)=2y_j$ and $C(B_{i;j})=y_j$, so both $K$ vectors lie in $\ker C$. They are independent and form a basis of its weight space at $-L_j$.

Thus the bases marked on the weight diagram of ${\Gamma }_{1,2}$ areand the diagram of ${\Gamma }_{0,1}$ uses the basis $Q_j=A_j+B_{i;j}+B_{l;j}$ at $-L_j$. Each $Q_j$ is complementary to the two kernel vectors because its contraction is $4y_j\ne 0$. All vectors shown are therefore genuine bases, not merely weight counts. The highest-weight vectors areThe latter is highest because $\iota$ intertwines the dual Lie algebra representation and $y_3$ is its highest-weight vector.

Let $e_1,e_2,e_3$ be the basis of the defining representation and $y_1,y_2,y_3$ its dual. On a diagonal element of the Cartan subalgebra their weights are $L_i$ and $-L_i$, with $L_1+L_2+L_3=0$. Use the tensor product and symmetric square basisIts weight is $L_i-L_j-L_k$. For the three repeated weights putIf $\{i,l\}=\{1,2,3\}\setminus \{j\}$, the complete weight-space decomposition isThere are six spaces of the first type, three of the second, and three of the third. Their dimensions sum to $6+3+9=18$. The weight diagram below labels every space with all of its basis vectors.

For the Adjoint representation, the root-space decomposition gives the eight nonzero weightseach of multiplicity one, and weight zero of multiplicity two from the Cartan subalgebra. Thus its weight diagram is the root diagram with a double zero weight, and its total dimension is $8+2=10$.

In the defining representation, the diagonal element of the Cartan subalgebra acts on the four standard basis vectors with eigenvalues $r,s,-r,-s$. ThereforeTheir weight diagram is the four-point cross below.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.

When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by SU(3) leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate flavour symmetry, distinct from the exact colour gauge symmetry. The quark transforms as $3$ and the antiquark as $\overline{3}$.

For mesons, a quark-antiquark tensor is a $3\times 3$ matrix. Its trace and traceless parts are invariant subspaces, givingThe trace is the eta singlet state; the traceless part is the meson octet. For baryons, first separate a pair of quarks into symmetric and antisymmetric tensors:The antisymmetric pair transforms as $\overline{3}$ by contraction with the invariant epsilon tensor. In $6\otimes 3$, complete symmetrization gives ${\mathrm{Sym}}^{3}3$ of dimension $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{3}\otimes 3$, trace and traceless parts give a singlet and another octet. Thus the tensor cube of the defining SU(3) representation isThe two octets carry the two mixed permutation-symmetry components. Their weight diagrams are identical; the extra copy is a multiplicity, not a new arrangement of weights.

The printed charge relation fixes the half-hypercharge convention, rather than the usual Flavor hypercharge normalization:The quark weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the baryon weight diagrams in the half-hypercharge convention. The baryon octet has $(p,n)$ at $Y=1/2$, the isotriplet $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$ and the isosinglet ${\Lambda }^0$ at $Y=0$, and $({\Xi }^0,{\Xi }^{-})$ at $Y=-1/2$. The baryon decuplet has rows $\Delta$ at $Y=1/2$, ${\Sigma }^{\ast }$ at zero, ${\Xi }^{\ast }$ at $-1/2$, and ${\Omega }^{-}$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.

The fully symmetric flavour representation is precisely the baryon decuplet, $10={\mathrm{Sym}}^{3}3$. For a spatial ground state, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ quarks, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the Pauli exclusion principle for the three fermions.

An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state isIt is completely antisymmetric and an SU(3) colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third exterior power: ${\bigwedge }^3{\mathbb{C}}^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the quark model. This argument motivates three colours rather than deriving their exact number without further physical input.

The Pauli constraint on three-quark flavour multiplets also explains the absence of a flavour-singlet baryon in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. Butso three spin-$1/2$ quarks have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state baryon octet with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full Pauli constraint on three-quark flavour multiplets is applied. Excited orbital wavefunctions can permit three-quark flavour singlet; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every baryon at every energy.

A flavour-singlet meson is allowed: a quark and an antiquark are not three identical fermions, and the invariant flavour combination $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no meson exclusion.

Finally, ${\pi }^{+}p$ has baryon number one, strangeness zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ Delta baryon ${\Delta }^{++}$, the low-energy resonance with spin $3/2$. For $K^{+}p$, the kaon $u\bar{s}$ contributes strangeness plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ baryon can have positive strangeness, since every $s$ quark contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic quark content, for example $uuud\bar{s}$, and is not mandated by this spectrum. This is strangeness selection in pion-proton and kaon-proton resonances.

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

Let $f_1,f_2,f_3$ be the dual basis of $W^{\ast }$, and use $L_1+L_2+L_3=0$. The elementary tensors $e_i\otimes e_j\otimes f_k$ have weight $L_i+L_j-L_k$. Grouping them gives every weight space explicitly:The counts are $6+3\cdot 2+3\cdot 5=27$, so this table exhausts the tensor product. The diagrams below use equilateral coordinates $L_1=(1,0)$, $L_2=(-1/2,\sqrt{3}/2)$, $L_3=(-1/2,-\sqrt{3}/2)$; numbers at points are weight multiplicities.

To decompose the module, first split the first two factors into their symmetric and alternating parts. Put $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, including $t_{ii}=e_i\otimes e_i$. The equivariant contractionis onto, since $C(t_{ii}\otimes f_i)=2e_i$. Its kernel $K$ has dimension fifteen and contains the highest-weight vector $t_{11}\otimes f_3$ of weight $2L_1-L_3$.

Here is an explicit weight basis of the trace-free symmetric-square dual tensor module, and thus the requested fifteen-vector basis:Every displayed vector contracts to zero. The first two families occupy distinct one-dimensional weights; at each weight $L_i$ the two vectors in the last family are independent because their respective $t_{ij}\otimes f_j$ terms are distinct. Thus all fifteen are independent and span $K$.

For completeness, $K$ is irreducible, not just an invariant kernel of the right dimension. On elementary tensors the raising action isAt the six weights $2L_i-L_k$, both $E_{12}$ and $E_{23}$ kill the vector only for $(i,k)=(1,3)$. At each weight $-2L_k$, at least one of those operators is nonzero. At weight $L_1$, write a vector as $a(t_{12}\otimes f_2-t_{11}\otimes f_1/2)+b(t_{13}\otimes f_3-t_{11}\otimes f_1/2)$. The two raising equations are $3a+b=0$ and $-a+b=0$, forcing zero. On the two-dimensional weight-$L_2$ space $E_{12}$ is injective, and on the weight-$L_3$ space $E_{23}$ is injective, as substitution in the same action formula shows. Therefore the common raising kernel in $K$ is the single line $\mathbb{C}(t_{11}\otimes f_3)$. By Weyl complete reducibility theorem, every irreducible summand supplies a highest-weight line, so $K$ has only one summand and is irreducible.

The symmetric-factor contraction splits off a defining copy $W$; explicitly $e_i\mapsto {\sum }_j{t}_{ij}\otimes f_j$ is equivariant and its contraction is $4e_i$. For the alternating factor, the invariant volume form identifies ${\Lambda }^{2}W\cong W^{\ast }$, sending $e_1\wedge e_2$ to $f_3$, $e_2\wedge e_3$ to $f_1$ and $e_3\wedge e_1$ to $f_2$. ThereforeThe symmetric dual module is irreducible of dimension six and highest weight $-2L_3$; its symmetric-monomial basis has weights $-2L_i$ and $-L_i-L_j=L_k$, all of multiplicity one. In the original tensor product its highest-weight vector isThe raising action above annihilates it, and its weight is $L_1+L_2-L_3=-2L_3$.

Altogether the Triple tensor decomposition for the defining sl3 representation isThe fifteen-dimensional module has multiplicity one at each $2L_i-L_k$ and $-2L_i$, and multiplicity two at each $L_i$. The six-dimensional module has multiplicity one at $-2L_i$ and $L_i$. Each defining module has the three weights $L_i$, once each. These add to the original table of multiplicities.

The two defining summands are not intrinsically distinguished: other diagonal copies inside their isotypic component have the same three-point weight diagram. Thus the displayed diagrams account for every irreducible submodule type, as well as showing both copies in this decomposition.

For a root $\alpha$, choose normalized root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ and the coroot element $h_{\alpha }\in \mathfrak{h}$ withThe requested sl2 subalgebra associated with a root isThese bracket relations identify its basis with the standard sl2 triple. The notation here denotes this subalgebra, not the root reflection $s_{\alpha }$ used for the Weyl group.

The weight diagram of the Adjoint representation of ${\mathfrak{sl}}_3$ has its six roots $L_i-L_j$, $i\ne j$, each of multiplicity one, and zero of multiplicity two, from the Cartan subalgebra.

Restrict to the root subalgebra generated by $X=E_{12}$, $Y=E_{21}$ and $H=H_{12}$. The explicit decomposition into irreducible modules isThe first summand has $H$ weights $2,0,-2$. The second has weights $1,-1$, since $[Y,E_{13}]=E_{23}$. The third likewise has weights $1,-1$, since $[Y,E_{32}]=-E_{31}$. The last commutes with all three generators and is trivial. These subspaces are stable under the sl2 triple, are independent and have dimensions totaling eight, so they exhaust the Adjoint representation.

The Killing form value is the trace of $\mathrm{ad}{E}_{12}\mathrm{ad}{E}_{21}$. By the formula just proved, the four summands contribute $4,1,1,0$. Therefore