= Solution
Work first in a finite-dimensional <SU(2) representation>, with its invariant positive <inner product>. In the complexified <Lie algebra>, choose $H^\dagger=H$ and $(E^-)^\dagger=E^+$. Let $v_0$ be a <highest-weight vector> with $Hv_0=mv_0$ and $E^+v_0=0$, and put $v_r=(E^-)^rv_0$. The <commutator> with $H$ gives
$$
Hv_r=(m-2r)v_r.
$$
Induction also gives
$$
E^+v_r=r(m-r+1)v_{r-1}.
$$
For $r=1$ this is $E^+E^-v_0=Hv_0=mv_0$. 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 particular
$$
E^+E^-v_r=(r+1)(m-r)v_r=\frac14(m+n)(m-n+2)v_r,\qquad n=m-2r,
$$
which proves the assumed formula rather than merely invoking it.
The norm identity $\|E^-v_r\|^2=(r+1)(m-r)\|v_r\|^2$ 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. Thus
$$
\boxed{n=m,m-2,\ldots,-m,\qquad\dim V(m)=m+1,\qquad j=m/2.}
$$
For normalized states, $E^-|n\rangle=\tfrac12\sqrt{(m+n)(m-n+2)}\,|n-2\rangle$, 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 $v_0,\ldots,v_m$ 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 $E_\alpha$ satisfying
$$
[h,E_\alpha]=\alpha(h)E_\alpha\qquad(h\in\mathfrak h).
$$
Choose <simple roots> $\alpha_1,\alpha_2$ and their <coroots> $H_i=\alpha_i^\vee$. In the PDF's index convention, the <Cartan matrix> is
$$
K_{ij}=\alpha_i(H_j)=\frac{2(\alpha_i,\alpha_j)}{(\alpha_j,\alpha_j)}.
$$
Normalize $E_i^\pm=E_{\pm\alpha_i}$ 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 gives
$$
[H_i,H_j]=0,\qquad [E_i^+,E_j^-]=\delta_{ij}H_j,\qquad[H_i,E_j^\pm]=\pm K_{ji}E_j^\pm.
$$
The 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 $-1$, giving
$$
K=\begin{pmatrix}2&-1\\-2&2\end{pmatrix}.
$$
The positive root $\alpha_1+\alpha_2$ is generated by $[E_1^+,E_2^+]\ne0$, and $2\alpha_1+\alpha_2$ by $[E_1^+,[E_1^+,E_2^+]]\ne0$. No further positive sums occur, so $(\operatorname{ad}E_1^+)^3E_2^+=0$ and $(\operatorname{ad}E_2^+)^2E_1^+=0$. These are the relevant <Serre relations>. Applying the Cartan commutators to a weight vector proves
$$
\boxed{E_1^-:\ (n_1,n_2)\mapsto(n_1-2,n_2+1),\qquad E_2^-:\ (n_1,n_2)\mapsto(n_1+2,n_2-2),}
$$
whenever 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:
$$
\boxed{(2,0),\ (0,1),\ (-2,2),\ (2,-1),\ (0,0),\ (-2,1),\ (2,-2),\ (0,-1),\ (-2,0).}
$$
Their actual production by lowering can be seen from
$$
\begin{aligned}
(2,0)&\xrightarrow{1}(0,1),\\
(0,1)&\xrightarrow{1}(-2,2),\qquad (0,1)\xrightarrow{2}(2,-1),\\
(-2,2)&\xrightarrow{2}(0,0),\qquad (2,-1)\xrightarrow{1}(0,0),\\
(0,0)&\xrightarrow{1}(-2,1),\qquad(0,0)\xrightarrow{2}(2,-2),\\
(-2,1)&\xrightarrow{2}(0,-1),\qquad(2,-2)\xrightarrow{1}(0,-1),\\
(0,-1)&\xrightarrow{1}(-2,0).
\end{aligned}
$$
An 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 \b[the full representation has dimension ten], not nine.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-45-b2-weights.png]
{title=B2 adjoint weights labelled by the PDF Cartan convention, with simple-root lowering arrows}
Back to article page
