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Write a matrix-valued gauge field ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ in the fundamental representation. Charge conjugation sends it to $-{\mathcal{A}}_{\mu }^T$ because reordering the fermions transposes both the spinor and internal matrices. The identity $[A^T,B^T]=-[A,B{]}^T$ then sends the gauge field strength to $-{\mathcal{F}}_{\mu \nu }^T$. This distinction prevents treating all color components as individually C-odd.

Set $\epsilon =i{\tau }_2$, whose Pauli matrix identity is $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$. Exponentiation gives $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. Hence complex conjugating a Higgs doublet and multiplying by $\epsilon$ restores the fundamental representation of $SU(2)$, while reversing hypercharge. A Higgs field of hypercharge $1/2$ yields a conjugate doublet of hypercharge $-1/2$. Its product ${\tilde{\varphi }}^{†}L$ with a lepton doublet $L$ of hypercharge $-1/2$ transforms in the trivial representation of the whole electroweak gauge group. At vacuum $(0,v{)}^T/\sqrt{2}$, this composite is $v{\nu }_L/\sqrt{2}$.

Let $\epsilon =i{\sigma }_2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so ${\epsilon }^{-1}=-\epsilon$. Direct multiplication of the Pauli matrices gives $\epsilon {\sigma }_i^{\ast }\epsilon ={\sigma }_i$, equivalently $\epsilon {\sigma }_i^{\ast }{\epsilon }^{-1}=-{\sigma }_i$. For $U=\exp (i{\theta }_i{\sigma }_i/2)\in SU(2)$ this impliesTherefore ${\varphi }^c=\epsilon {\varphi }^{\ast }$ transforms as the same weak fundamental representation doublet as $\varphi$, while complex conjugation reverses its hypercharge to $-1/2$. This is the Higgs conjugate doublet. The Yukawa coupling product $\bar{L}{\varphi }^c{u}_R$ is an $SU(2)$ singlet, and its hypercharges add toIt is consequently gauge-invariant and can supply an up-quark mass.

In unitary gauge, ${\varphi }^c=((v+H)/\sqrt{2},0{)}^T$. Writing $y_u$ for the full arbitrary coefficient of the Yukawa coupling term, its physical form isA one-generation field rephasing makes the mass positive, and its Higgs boson coupling is $m_u/v$. The printed Yukawa expression has an additional trailing $m$ after $R^{+}$ which is never defined. If it denotes a constant, absorb it into $y_u=f^{+}m$; in the conventional dimensionless Yukawa notation without that factor, $y_u=f^{+}$. It does not affect the gauge-transformation argument.

For three families, use a singular value decomposition of the two mass matrices: $U_{uL}^{†}{M}_u{U}_{uR}=D_u$ and $U_{dL}^{†}{M}_d{U}_{dR}=D_d$, with positive diagonal entries. The original left-handed weak current then becomes a current between mass eigenstates withA complex $3\times 3$ matrix has eighteen real parameters, and $U^{†}U=I$ imposes nine independent real conditions, leaving nine. A generic unitary matrix can be built from three complex plane rotations and three diagonal phases: each plane rotation supplies one angle and one phase. Thus these are three rotation angles and six phases. Rephase each of the six Dirac quark mass eigenfields, applying the same phase to its two chiralities so the diagonal masses stay real. A common phase changes no entry of $V$, leaving five effective rephasings. Removing five of the six phases leaves three angles and one physical phase. This CKM parameter counting assumes generic nondegenerate masses; degeneracies give additional basis freedom.

Exactly massless neutrinos have an arbitrary unitary flavor rotation that changes no mass term. After diagonalizing the charged-lepton mass matrix, choose the neutrino basis with the same left-handed rotation. The apparent lepton charged-current mixing matrix becomes the identity. Thus mixing of exactly massless neutrinos is removable; it is not an analogue of physical CKM mixing.

For the requested CP symmetry calculation, write $j_{ij}^{\mu }={\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)d_j$. The supplied charge conjugation transformations, with the fermion-reordering sign, giveFor $\Gamma ={\gamma }^{\mu }(1-{\gamma }^5)$, use $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$ and the given gamma matrix relation to obtainUnder parity, the current acquires the Lorentz parity matrix and $1+{\gamma }^5$ becomes $1-{\gamma }^5$. Under charge conjugation the charged vector also changes sign and becomes its conjugate. The two minus signs cancel, and the parity matrices cancel in the contracted operator. ConsequentlyWith $k=g/(2\sqrt{2})$, the interaction is $-k\sum (V_{ij}{O}_{ij}+V_{ij}^{\ast }{O}_{ij}^{†})$. Thereforewith the transformed density evaluated at $x_P$. This is the CP transformation of a charged quark current.

Physical noninvariance requires a phase that cannot be removed by quark rephasings. The invariant quartetis the Jarlskog invariant, where $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$ in the standard three-angle parametrization. If $J\ne 0$, no allowed row/column phases can make the matrix real, and the charged-current interaction violates CP symmetry. For example all three angles $\pi /4$ with $\delta =\pi /2$ give $J=1/(8\sqrt{2})\ne 0$.

The printed noninvariance claim is conditional on a nonzero physical phase. Arbitrary Yukawa matrices do not force it: $V=I$, or any real CKM matrix, gives a CP-invariant charged-current interaction. The correct statement is generic CP violation when the irreducible phase is nontrivial, as explained by quark rephasing and CP conservation.

Finally a right-handed neutrino has gauge representation $(1,1{)}_0$. Its Lorentz-invariant same-chirality bilinear with the field obtained by charge conjugation is therefore a gauge singlet and permits a Majorana mass term, conventionallyThe factor $1/2$ is the usual convention for a field paired with itself; it may be absorbed into the printed coefficient. A Higgs Yukawa interaction can independently give a Dirac mass term. The Majorana term violates lepton number by two units but preserves the gauge symmetries. Corresponding terms for the other charged Standard Model fermions carry nonzero electric charge, and quarks also carry nontrivial color charge. Thus this gauge-invariant right-handed neutrino Majorana mass is allowed while those charged-fermion analogues are forbidden. Merely introducing a right-handed neutrino permits mass; vanishing mass coefficients would still leave it massless.

The local matrix $U(x)$ is a gauge transformation in the representation acting on the matter field. With Hermitian Lie algebra generators it can be written locally in spacetime, and pointwise for $SU(3)$, asThe real functions ${\omega }^a$ can instead include a factor of $g$ by redefining the transformation parameters. The gauge covariant derivative convention in this paper is $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$. Inserting the stated transformed $A_{\mu }$ shows explicitly that the ${\partial }_{\mu }U$ term cancels, giving $D_{\mu }^{\prime }(U\psi )=U{D}_{\mu }\psi$.

In QCD, each quark field is in the colour fundamental representation $3$, with ${\psi }^i\mapsto U^i{}_j{\psi }^j$. Thus its colour generators are eight $3\times 3$ matrices, conventionally $T^a={\lambda }^a/2$ for the Gell-Mann matrices; their number is not their matrix dimension. The eight gluons themselves transform in the adjoint representation of SU(3).

With $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, the pure Yang-Mills theory has Lagrangian ${\mathcal{L}}_{\mathrm{YM}}=-\frac{1}{4}{F}_{\mu \nu }^a{F}^{a\mu \nu }=-\frac{1}{2}\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$. Put $G_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a$ and $C_{\mu \nu }^a=f_{abc}{A}_{\mu }^b{A}_{\nu }^c$. Since $F=G-gC$,The cross term has two derivative contributions. Interchange $\mu ,\nu$ and then $b,c$ in the second; antisymmetry of $f_{abc}$ makes it equal to the first. The cubic and quartic Yang-Mills self-interactions are thereforeThe positive cubic sign follows from the negative $gfAA$ term in the given Yang-Mills field strength; changing the covariant-derivative convention consistently changes intermediate signs.

For the left-handed quark doublet $Q_L=(u_L,d_L{)}^T$, the hypercharge is $Y=1/6$ and the weak generators are ${\sigma }^{\alpha }/2$. Its electroweak term isThe Pauli matrices give diagonal entries $\pm {\mathcal{A}}^3$ and off-diagonal entries ${\mathcal{A}}^1\mp i{\mathcal{A}}^2$. Expanding the term givesIn particular the off-diagonal weak charged current coupling is $-g(W_{\mu }{\bar{u}}_L{\gamma }^{\mu }{d}_L+W_{\mu }^{\ast }{\bar{d}}_L{\gamma }^{\mu }{u}_L)/\sqrt{2}$.

Write $s_W=\sin {\theta }_W$, $c_W=\cos {\theta }_W$. Invert the neutral-field rotation: ${\mathcal{A}}_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$, $B_{\mu }=c_W{A}_{\mu }-s_W{Z}_{\mu }$. The original PDF gives $\tan {\theta }_W=g^{\prime }/g$, soThe photon coefficients inside the derivative are $g{s}_W/2+g^{\prime }{c}_W/6=2e/3$ for $u_L$, and $-g{s}_W/2+g^{\prime }{c}_W/6=-e/3$ for $d_L$. Multiplication by the $i$ in the kinetic Lagrangian gives the requested electromagnetic coupling of a weak multiplet:Equivalently it is $-e{A}_{\mu }{\overline{Q}}_L{\gamma }^{\mu }(T_3+Y)Q_L$, making the electric charges explicit. The remaining neutral terms are $-(g/c_W)Z_{\mu }\{(1/2-2s_W^2/3){\bar{u}}_L{\gamma }^{\mu }{u}_L+(-1/2+s_W^2/3){\bar{d}}_L{\gamma }^{\mu }{d}_L\}$. The TeX's missing prime in the mixing-angle ratio and truncated first Pauli matrix are conversion errors, not physical assumptions.

Take $T_i={\tau }_i/2$ and the hypercharge generator $Y=I_2/2$ on the scalar. The two commuting factors of the electroweak interaction have independent invariant gauge kinetic terms and therefore independent coupling constants. Non-Abelian gauge invariance requires the same $g$ for all three $SU(2)$ generators, while the $U(1)$ factor has its own $g^{\prime }$; no symmetry of this product group requires $g=g^{\prime }$.

A nonzero Higgs doublet can be rotated by $SU(2)$ toThe fundamental representation acts transitively on normalized complex doublets, so this fixes an internal direction without changing the vacuum norm. A general infinitesimal generator annihilates this vacuum only ifFor real parameters this means ${\alpha }_1={\alpha }_2=0$ and $\beta ={\alpha }_3$. Thus the connected stabilizer subgroup is one-dimensional:This proves the claimed electroweak symmetry breaking and identifies the unbroken electric charge generator.

The gauge covariant derivative isWriting the connection as ${\mathcal{A}}_{\mu }=g{A}_{\mu }^i{T}_i+g^{\prime }{B}_{\mu }Y$, its local transformation is ${\mathcal{A}}_{\mu }^{\prime }=U{\mathcal{A}}_{\mu }{U}^{-1}-i({\partial }_{\mu }U)U^{-1}$. Direct substitution gives $D_{\mu }^{\prime }(U\varphi )=U{D}_{\mu }\varphi$. Unitarity then provesso the scalar kinetic term is gauge-invariant.

At the constant vacuum the gauge covariant derivative becomesConsequentlywhere $\mathcal{V}=(A^1,A^2,A^3,B{)}^T$ andThe factor $1/2$ is the canonical real-vector mass-term normalization; the displayed electroweak doublet gauge-boson mass matrix contains squared masses when the gauge kinetic terms are canonical.

The electric charge generator acts on gauge fields through the Adjoint representation, not through the scalar's two-by-two matrix. With $U=e^{i\alpha Q}$, the brackets $[T_3,T_1]=i{T}_2$ and $[T_3,T_2]=-i{T}_1$ give $\delta A^1=\alpha A^2$, $\delta A^2=-\alpha A^1$, and neutral $A^3,B$. Hence the electromagnetic charge matrix of electroweak gauge fields in the requested basis isChoose positive gauge couplings by the gauge-field sign convention. Set $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The Weinberg angle rotation diagonalizes the neutral block, while $W^{\pm }=(A^1\mp i{A}^2)/\sqrt{2}$ obey $\delta W^{\pm }=\pm i\alpha W^{\pm }$. The physical electroweak gauge bosons areThe charges are in units of $e=g{s}_W=g^{\prime }{c}_W$. Thus the tree-level electroweak gauge-boson masses satisfy $m_W=m_Z{c}_W$, and the unbroken photon is massless.

The electron and neutrino form the left-chiral lepton doublet, while the right-chiral electron is a weak singlet. In four-component notation use chiral projectors $P_{L,R}=(1\mp {\gamma }_5)/2$ and packageHere each entry can be obtained by projecting a Dirac field; only the indicated chirality propagates. The hypercharges follow from $Q=T_3+Y$: $(0,-1)=(1/2,-1/2)+Y_{\Psi }$ gives $Y_{\Psi }=-1/2$, and $T_3=0,Q=-1$ gives $Y_{\psi }=-1$. The minimal theory has no right-handed neutrino.

The gauge-invariant lepton kinetic Lagrangian isIf unprojected Dirac spinors are used as notation instead, insert $P_L$ in the first kinetic term and $P_R$ in the second. The inactive components are not additional physical degrees of freedom. Each term is invariant because the gauge covariant derivative transforms in the same gauge group representation as its field.

No bare Dirac mass term is possible: $\bar{\Psi }\Psi$ and $\bar{\psi }\psi$ vanish by opposite adjoint projectors, while a cross term $\bar{\Psi }\psi$ has an uncontracted weak-doublet index and hypercharge $-1/2$. A neutrino Dirac mass would additionally require the absent right-handed field. A Majorana mass term pairs a chiral field with itself; ${\Psi }^T{C}^{-1}\Psi$ has hypercharge $-1$ and ${\psi }^T{C}^{-1}\psi$ has $-2$, so neither is invariant. These statements concern quadratic masses before electroweak symmetry breaking, not masses involving scalar insertions.

For the Higgs conjugate doublet, put $\epsilon =i{\tau }_2$. The given identity $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$ exponentiates to $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. ThereforeIn particular its weak isospin remains $T=1/2$ while its hypercharge reverses. Since $\Psi \mapsto e^{-i\beta /2}U\Psi$, both ${\varphi }^{\prime †}\Psi$ and $\bar{\Psi }{\varphi }^{\prime }$ transform in the trivial representation of the full gauge group. They remain a Lorentz spinor and its adjoint; gauge singlet does not mean Lorentz scalar.

Finally the Yukawa interactionis gauge invariant: the first term has total hypercharge $+1/2+1/2-1=0$, and its doublet indices are contracted. At the Higgs field vacuum it becomes $-(y_{e}v/\sqrt{2}){\bar{e}}_L{e}_R+h.c.$. Rephase $e_R$ to make the coefficient real and positive and assemble $e=e_L+e_R$. ThenThis explicitly generates the gauge-invariant electron Yukawa mass, without a bare mass in the unbroken theory.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

Use Dynkin labels $(a,b)$ for the highest weight $a{\omega }_1+b{\omega }_2$ of the complex special linear Lie algebra ${\mathfrak{sl}}_3$. The A2 root system has ${\alpha }_1=(2,-1)$ and ${\alpha }_2=(-1,2)$ in these coordinates. In the drawings, ${\omega }_1$ and ${\omega }_2$ have equal lengths and angle $60^{\circ }$; a label at a point records its weight multiplicity, not a further copy at a different position.

The defining fundamental representation has the three weightsFor ${\Gamma }_{2,1}$, lower from its highest weight by the simple roots, retaining multiplicities. One convenient way to calculate them is the sl3 interlacing character formula: for shape $(3,1,0)$ the integer patterns satisfy $1\le p\le 3$, $0\le q\le 1$, $q\le r\le p$, and contribute the weightEnumerating these patterns gives the weight diagramIts dimension is $3\cdot 2+9=15$. The diagram below draws all twelve distinct positions, with the three inner multiplicities equal to two. The extra panel gives the symmetric square used in the calculation.

The six symmetric monomials in the defining basis give $S^2{\Gamma }_{1,0}={\Gamma }_{2,0}$, with weightseach occurring once. Thus the tensor product has dimensionIn a tensor product of Lie algebra representations, weights add and their multiplicities multiply. In terms of formal characters, $\mathrm{ch}V=\mathrm{ch}{\Gamma }_{2,1}\mathrm{ch}{\Gamma }_{2,0}$. Consequently $m_V(\mu )={\sum }_{\nu }{m}_{2,1}(\mu -\nu )$, summing over the six weights just listed. To show the indicated dominant multiplicities explicitly, the contributions in that order areThe tensor-product weight diagram below includes every position, and highlights these dominant weights. It also records the zero-weight multiplicity nine; that multiplicity is not a count of trivial summands.

Apply the Weyl complete reducibility theorem and subtract irreducible formal characters in decreasing dominance order. The multiplicities at these five dominant positions in the potential summands areThese entries can be obtained by the same interlacing enumeration or by weight strings. Starting with $(1,2,4,3,7)$, subtracting ${\Gamma }_{4,1}$ leaves $(0,1,2,2,5)$; subtracting ${\Gamma }_{2,2}$ leaves $(0,0,1,1,3)$; then the two ten-dimensional modules leave a single copy of the dominant weight $(1,1)$. This is highest-weight character subtraction. ThereforeEvery summand occurs once. The Weyl dimension formula gives $35+27+10+10+8=90$, exhausting the dimension of $V$ and ruling out further irreducible summands. Computing the complete formal character also leaves no residual weight multiplicities.

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

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

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

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

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

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

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

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

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