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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.

For the defining sl2 Lie algebra module with basis $x,y$, put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$. The raising operator has $e{A}_k=k{A}_{k-1}$ and $e{B}_k=A_k+k{B}_{k-1}$; the lowering operator has $f{A}_k=B_k+(3-k)A_{k+1}$ and $f{B}_k=(3-k)B_{k+1}$. The highest-weight vectors $A_0$ and $B_0-A_1$ generate irreducible submodules of highest weights four and two. Bases are respectively $(A_0,3A_1+B_0,A_2+B_1,A_3+3B_2,B_3)$ and $(B_0-A_1,B_1-A_2,B_2-A_3)$. Their vectors are independent in every weight space, proving the displayed direct sum without only comparing dimensions.

For the defining $n$-dimensional special linear Lie algebra module, with $n\ge 2$, write $T_i(p)=e_i\otimes p(y_1,\dots ,y_n)$. The equivariant contraction is $C(T_i(p))={\partial }_{y_i}p$. The map $\iota (f)={\sum }_i{e}_i\otimes y_{i}f$ is also equivariant, since ${\sum }_i{e}_i\otimes y_i$ is an invariant identity tensor. Direct differentiation gives $C\iota =(n+1)I$, so the displayed direct sum follows. Its trace-free part is an irreducible representation of highest weight ${\omega }_1+2{\omega }_{n-1}$. Its highest-weight vector is $e_1\otimes y_n^2$; the Weyl dimension formula gives dimension $n(n-1)(n+2)/2$, exactly $\dim \ker C$.

// Target: semisimple-lie-algebra.bigb

Write ${\epsilon }_i(b)$ and ${\phi }_i(b)$ for the numbers of raising and lowering steps of color $i$ available from a vertex of a crystal basis. One convention for the tensor product of crystals isThe raising operator uses the first factor for ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. The different strictness makes raising and lowering inverse along every edge. An edge is absent if the chosen Kashiwara operator vanishes. Stating the convention is essential when identifying the highest-weight vectors as ordered tensor vertices.

Let $e_i=x^{4-i}{y}^i$ and $w_{ij}=e_i\wedge e_j$ for $i<j$. For the sl2 Lie algebra, the raising and lowering operators are $E=x{\partial }_y$, $F=y{\partial }_x$, and $w_{ij}$ has weight $8-2(i+j)$. A weight basis of the irreducible representation $L(6)$ isA weight basis of the complementary $L(2)$ isThe first vector in each list is a highest-weight vector. Applying $F$ along the first list gives successive coefficients $3,2,1,12,5,2$; along the second it gives $1,2$. The final vectors are killed by $F$. At the overlapping weights $2,0,-2$, the two listed vectors are independent, proving the direct sum.

// Target: semisimple-lie-algebra.bigb

For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

For a complex semisimple Lie algebra and a choice of positive roots, every finite-dimensional Irreducible Lie algebra representation has a unique highest weight, which is a dominant integral weight. Conversely each dominant integral weight labels exactly one such irreducible module up to isomorphism. Its highest-weight vector spans the highest-weight space and generates the module. This classification and Weyl complete reducibility theorem reduce decomposing any finite-dimensional module to its irreducible highest-weight summands.

For a normalized highest-weight vector with $L_0|h⟩=h|h⟩$, $L_{n>0}|h⟩=0$ and $L_n^{†}=L_{-n}$, the level-two descendants $L_{-2}|h⟩,L_{-1}^2|h⟩$ have the displayed Gram matrix. It follows from the Virasoro algebra: $⟨L_2{L}_{-2}⟩=4h+c/2$, $⟨L_2{L}_{-1}^2⟩=6h$ and $⟨L_1^2{L}_{-1}^2⟩=4h(2h+1)$. At $h=-1,c=26$ its null direction is $(1,3/2)$, giving the level-two scalar Virasoro null state. Nullity of a descendant is a metric statement; satisfying all physical constraints must be checked separately.

For $N$ identical fermions restricted to one angular-momentum multiplet $j$, with all other one-particle labels fixed, the Pauli exclusion principle permits distinct angular momentum projections $j,j-1,\dots ,j-N+1$. Their sum is $Nj-N(N-1)/2$, provided $N\le 2j+1$. The resulting Slater determinant is killed by the total raising angular momentum ladder operator because raising any occupied state either exceeds $j$ or duplicates an occupied state. It is therefore a highest-weight vector of precisely this total angular momentum. The formula is invariant under replacing particles by holes in the full shell.

In an irreducible spin representation, the angular momentum lowering operator has squared norm factor $(I+m)(I-m+1)$. Multiplying these factors from the highest-weight vector gives the factorial normalization displayed above. All factorial arguments are integers even when $I$ is half-integral. This produces a consistent positive-coefficient phase convention for all weight states and fixes signs in subsequent isospin rotations.

The Pauli exclusion principle allows each one-particle magnetic substate only once, since the fermions share the other quantum numbers. To maximize the total angular momentum projection, occupy $m_j=5/2,3/2,1/2,-1/2$, givingThe corresponding Slater determinant is a highest-weight vector: every attempted raising operation either tries to exceed $j$ or duplicates an occupied substate and vanishes. Therefore it belongs to a multiplet with $\boxed{J_{\max }=4}$, not the distinguishable-particle value $4j=10$. Equivalently, four particles in a six-state shell have two holes whose maximum angular momentum projection is $5/2+3/2=4$. The general maximum total angular momentum of equivalent fermions is $J_{\max }=Nj-N(N-1)/2$ for $N\le 2j+1$.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

Choose positive rootsThe corresponding simple roots are ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda ={\epsilon }_1+{\epsilon }_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a highest-weight vector with highest weight $\lambda$.

A representation of a compact Lie group is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the exterior square contains an irreducible summand $L_{\lambda }$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.

Use the Weyl dimension formula, expressly allowed in this part:Here $L_{\lambda }$ is the irreducible highest-weight representation, ${\Phi }^{+}$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with ${\epsilon }_1,{\epsilon }_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,The four factors, in the listed positive-root order, are respectivelyTheir product is $10$. Since $\dim {\Lambda }^2{\mathbb{C}}^5=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, this irreducible summand occupies the entire exterior square. HenceThe proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its B2 adjoint representation weight diagram is exactly the eight roots and the double zero weight obtained above.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

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.

The nine one-dimensional weight spaces from part (i) all lie in $\ker C$, because the dual index differs from every polynomial index. Their bases remain $T_i^{jj}$ at weight $2{ϵ}_j-{ϵ}_i$ for $i\ne j$, and $T_i^{jk}$ at weight $-2{ϵ}_i$ for three distinct indices $i,j,k$.

At weight ${ϵ}_i$, the full space has basis $T_i^{ii},T_j^{ij},T_k^{ik}$, where $j,k$ are the two other indices. Their contractions are $2e_i,e_i,e_i$. Thus the contraction kernel has the independent basis $K_{ij}=2T_j^{ij}-T_i^{ii}$ and $K_{ik}=2T_k^{ik}-T_i^{ii}$. Explicitly the weight basis of the trace-free symmetric-square dual tensor module is completed byTogether with the nine single vectors, these give fifteen independent vectors. They are precisely the $T$ and $K$ labels on the fifteen-dimensional weight diagram.

The defining summand has only the following three weight spaces:Each $S_i=s(e_i)$ is nonzero, has its indicated weight and satisfies $C(S_i)=4e_i$. These are the labels on the three-dimensional diagram. The highest-weight vectors in the two summands areAt each repeated weight, the two kernel vectors together with $S_i$ span the original three-dimensional weight space, since the kernel has dimension two and $C(S_i)\ne 0$. Thus the displayed diagrams specify all submodule weight-space bases.

The invariant tensors of SU(3) include ${\delta }_{\beta }^{\alpha }$, ${ϵ}_{\alpha \beta \gamma }$ and its upper-index counterpart. Consequently an antisymmetric pair of upper fundamental indices can be converted into one lower antifundamental index, and conversely; a completely antisymmetric triple is a determinant singlet. In the Young-symmetrizer tensor decomposition, complete columns of three can therefore be removed. A remaining diagram has rows of lengths $k+l,l$: its two-box columns become $l$ dual indices and its remaining boxes give $k$ defining indices. The irreducible highest-weight component is represented by a mixed SU(3) tensor representation symmetric in all upper indices, symmetric in all lower indices, and with every upper-lower contraction removed. Such contractions produce the lower-weight components. The resulting symmetric traceless SU(3) tensor representation has Dynkin labels $(k,l)$; this is an equivalent canonical realization, not a claim that an arbitrary tensor already has those symmetries.

There are $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)$ symmetric upper-index components and $\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$ symmetric lower-index components. All contractions are equivalent under the separate symmetries. Their image is the separately symmetric tensor space with $(k-1,l-1)$ indices. To subtract its dimension legitimately, we must show that contraction is onto.

Represent such tensors by polynomials of bidegree $(k,l)$ in three variables $x$ and three dual variables $y$. Up to a nonzero normalization, contraction isIf $P$ has bidegree $(k-1,l-1)$, differentiation givesChoose $a_0=1/(k+l+1)$ and $a_j=-a_{j-1}/[(j+1)(k+l+1-j)]$. Thenhas bidegree $(k,l)$ and $\Delta Q=P$: consecutive terms cancel, and the last further derivative vanishes. This constructs a preimage of every contraction target. If $k=0$ or $l=0$, there is no contraction. Therefore the number of trace-free components isThe highest-weight vector $x_1^k{y}_3^l$ lies in this kernel. The Young-symmetry construction identifies the traceless highest-weight component with the irreducible representation of labels $(k,l)$; equivalently its Weyl dimension formula agrees with the independently calculated kernel dimension above.

For traceless matrices $T,S$, the tensor square of the SU(3) adjoint representation decomposes asThe component maps make the decomposition concrete. The singlet is $\mathrm{tr}(TS)$. The two octets are the antisymmetric commutator $[T,S]$ and symmetric traceless anticommutator $\{T,S\}-\frac{2}{3}\mathrm{tr}(TS)I$. Symmetrizing the two upper and two lower indices in $T_{\beta }^{\alpha }{S}_{\delta }^{\gamma }$ and removing contractions gives $R(2,2)$, the 27. Its highest component is already visible in $E_{13}\otimes E_{13}$.

For the 10, contract the two lower indices with ${ϵ}^{\beta \delta \rho }$ and symmetrize the three upper indices $\alpha ,\gamma ,\rho$; for its conjugate, contract the upper pair with ${ϵ}_{\alpha \gamma \rho }$ and symmetrize the three lower indices. These maps are antisymmetric under exchange of $T,S$. They are nonzero: $T=E_{12},S=E_{13}$ produces a nonzero all-upper-1 component, and $T=E_{21},S=E_{31}$ produces the conjugate all-lower-1 component. Thus they reach the irreducible $R(3,0)$ and $R(0,3)$, each of dimension ten. The octet maps are also nonzero, for example on a pair of noncommuting matrix units and on equal nonzero traceless diagonal matrices. These equivariant maps supply distinct irreducible constituents, with the two octets separated by exchange symmetry. Their dimensions exhaust the product:

An octet operator between baryon octet states defines an intertwining operator $8_{\mathrm{operator}}\otimes 8_{\mathrm{state}}\to 8_{\mathrm{state}}$. Its covariance follows from the commutator transformation law in the question. By Schur lemma, the number of independent reduced matrix elements is the multiplicity of the output octet in the product: there are exactly two independent coefficients. To display them, use a Hermitian octet matrix basis $B_s={\lambda }_s$, $t_i={\lambda }_i/2$, normalized by $\mathrm{tr}({\lambda }_r{\lambda }_s)=2{\delta }_{rs}$. The two allowed actions areWith $[{\lambda }_i,{\lambda }_s]=2i{f}_{isk}{\lambda }_k$ and $\{{\lambda }_i,{\lambda }_s\}=\frac{4}{3}{\delta }_{is}I+2d_{isk}{\lambda }_k$, this gives the two reduced octet matrix elements:up to the common normalization absorbed into $F,D$. All component dependence is fixed by the invariant symmetric and antisymmetric tensors. For a Hermitian operator and this Hermitian basis, $F,D$ may be chosen real; neither their values nor a relation between them follows from symmetry alone.

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.