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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

For simple-root generators $e_i,f_i,h_i$ with $[h_i,e_j]=a_{ij}{e}_j$, the Serre relations include $[e_i,f_j]={\delta }_{ij}{h}_i$ and $(\mathrm{ad}{e}_i{)}^{1-a_{ij}}{e}_j=(\mathrm{ad}{f}_i{)}^{1-a_{ij}}{f}_j=0$ for $i\ne j$. The off-diagonal Cartan entries are nonpositive integers. These relations restrict which iterated simple-root brackets remain nonzero.