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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

Use $h=\mathrm{diag}(1,-1)$, $e=E_{12}$ and $f=E_{21}$ in the sl2 Lie algebra, so $hx=x$, $hy=-y$, $ey=x$ and $fx=y$. On the symmetric power, a monomial $x^{3-k}{y}^k$ has weight $3-2k$. The tensor product of Lie algebra representations adds weights, because $h$ acts as $h\otimes I+I\otimes h$.

Put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$ for $0\le k\le 3$. These eight vectors are a basis of $U$; their weights are $4-2k$ and $2-2k$. Thus the weight-space decomposition and a weight basis for each summand areIn particular $\boxed{U=U_4\oplus U_2\oplus U_0\oplus U_{-2}\oplus U_{-4}}$, with dimensions summing to eight. Here the subscripts are the eigenvalues of the chosen Cartan element $h$.

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 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 $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, and let $f_i$ be the dual basis. The equivariant contraction sends $t_{ij}\otimes f_k$ to ${\delta }_{ik}{e}_j+{\delta }_{jk}{e}_i$. Its kernel has a weight basis consisting of $t_{ii}\otimes f_k$ for $i\ne k$, $t_{ij}\otimes f_k$ for three distinct indices with $i<j$, and $t_{ij}\otimes f_j-\frac{1}{2}{t}_{ii}\otimes f_i$ for $i\ne j$. The three families have sizes six, three and six. This is the irreducible fifteen-dimensional highest-weight representation with highest weight $2L_1-L_3$.