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In a complex semisimple Lie algebra whose rank of a semisimple Lie algebra is $\ell$, a root vector $x$ satisfies $\dim Z_{\mathfrak{g}}(x)\ge 3\ell -2$. To see this, select $\ell -1$ roots whose images form a basis of the real root span modulo $\mathbb{R}\alpha$, where $x\in {\mathfrak{g}}_{\alpha }$. Take the highest endpoints of the root strings through both signs of each selected root. Their $2(\ell -1)$ distinct root spaces commute with $x$. Together with $\ker \alpha \subset \mathfrak{t}$ and $\mathbb{C}x$, these give $2(\ell -1)+(\ell -1)+1$ independent vectors in the Lie algebra centralizer.

A Chevalley basis of a complex semisimple Lie algebra consists of suitable coroot elements and one vector $e_{\alpha }$ in each root space, chosen so all Lie bracket structure constants are integers. In particular $[e_{\alpha },e_{\beta }]=N_{\alpha ,\beta }{e}_{\alpha +\beta }$ when the sum is a root, with $N_{\alpha ,\beta }=\pm (p+1)$ where $\beta -p\alpha$ is the last negative-direction root in the root string. For independent roots with no root sum the bracket is zero. The integral lattice makes reduction to other fields possible.

For the antidiagonal symmetric matrix $J$ in dimension $2n$, put $\bar{i}=2n+1-i$ and $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The diagonal Cartan subalgebra has coordinates ${\epsilon }_i(H)=t_i$. The root spaces of weights ${\epsilon }_i-{\epsilon }_j$ are spanned by $E_{ij}-E_{\bar{j}\bar{i}}$; those of weights ${\epsilon }_i+{\epsilon }_j$ by $E_{i\bar{j}}-E_{j\bar{i}}$, with negative-root spaces given by transposes. The condition $A_{ij}=-A_{\bar{j}\bar{i}}$ gives these generators directly and excludes roots $2{\epsilon }_i$.

For a complex semisimple Lie algebra, the Cartan subalgebra is orthogonal under the Killing form to every nonzero root space. A vector in the Cartan subalgebra orthogonal to that whole subalgebra is therefore orthogonal to the entire Lie algebra and must vanish. On the Cartan subalgebra the form is $\kappa (h,h^{\prime })={\sum }_{\alpha \in \Phi }\alpha (h)\alpha (h^{\prime })$.

Choose an orthonormal basis $e_1,e_2,e_3$ in the real Euclidean space containing the root system, and take simple rootsTheir squared lengths are $2,2,1$. Use the Cartan matrix convention $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, with the denominator indexed by the column. Direct inner products giveThe Dynkin diagram of the B3 root system is a three-node chain with a single bond from node 1 to node 2 and a double bond from node 2 to the short node 3; the arrow points toward node 3. The A3 root system has a chain of three nodes joined by single bonds. A convention putting the row length in the denominator transposes the first displayed Cartan matrix.

For the Extended Dynkin diagram, the highest root is $\theta =e_1+e_2={\alpha }_1+2{\alpha }_2+2{\alpha }_3$. Add ${\alpha }_0=-\theta$. Its only nonzero off-diagonal inner product among these simple roots is $({\alpha }_0,{\alpha }_2)=-1$. Deleting the short node 3 leaves the chain $0-2-1$, with three equal-length roots, which is $A_3$.

More concretely, the long roots $\pm e_i\pm e_j$ form the D3 root system, which is isomorphic to the A3 root system. The corresponding root spaces and Cartan subalgebra generatefor the compact real forms. At the matrix level the inclusion fixes one coordinate in seven-dimensional space. This establishes the required Lie subalgebra inclusion; the associated global groups require the distinction explained in d.

Order the simple roots of the G2 root system with ${\alpha }_1$ long and ${\alpha }_2$ short. This ordering is consistent with the supplied projection in solution; reversing the nodes would interchange the two fundamental Dynkin labels. Choose $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=2/3$, and $({\alpha }_1,{\alpha }_2)=-1$. With the same row convention for the Cartan matrix as in solution,The positive roots, with their Dynkin labels, areFor the highest weight $(0,1)$, a lowering chain isThe middle three terms form the length-three weight string for the short simple root. The Weyl dimension formula, with this long-first convention, isIt gives seven for $(0,1)$, so the chain exhausts the weight system, with every weight multiplicity equal to one:For $(1,0)$, observe that $2{\alpha }_1+3{\alpha }_2=(1,0)$ is the highest root. Thus this is the Adjoint representation; the twelve root spaces contribute the twelve nonzero weights once, and the rank-two Cartan subalgebra contributes two zero weights. ThereforeThe same Weyl dimension formula gives fourteen, confirming the count.

The highest root is $\theta =2{\alpha }_1+3{\alpha }_2$. In the Extended Dynkin diagram, the extra node $-\theta$ is long and is joined by a single bond to the long node ${\alpha }_1$; it is orthogonal to the short node ${\alpha }_2$. Deleting the short node therefore leaves the A2 root system Dynkin diagram. Equivalently, the six long roots are a closed root subsystem, so their root spaces together with the Cartan subalgebra form an A2 root system Lie subalgebra. The corresponding compact subgroup is $SU(3)\subset G_2$.

For the particular projection in the paper, choose the A2 root system simple roots as ${\beta }_1={\alpha }_1+3{\alpha }_2$ and ${\beta }_2={\alpha }_1$. Since these are long, ${\beta }_1^{\vee }={\alpha }_1^{\vee }+{\alpha }_2^{\vee }$ and ${\beta }_2^{\vee }={\alpha }_1^{\vee }$. Consequently the Dynkin labels restrict asProjecting the six short roots gives $\pm (1,0),\pm (0,1),\pm (1,-1)$. Group these into the three weights of $3$ and the three of $\overline{3}$ listed in solution. The zero weight of $7$ is left over, soThe three positive long roots project as $(2,-3)\mapsto (-1,2)$, $(-1,3)\mapsto (2,-1)$ and $(1,0)\mapsto (1,1)$. Together with their negatives and the two zero weights, these are exactly the Adjoint representation $8$. The remaining six short roots again give $3\oplus \overline{3}$. Hence the branching of a representation is

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.

The complex Special orthogonal Lie algebra ${\mathfrak{so}}_5$ consists of endomorphisms preserving a nondegenerate symmetric bilinear form infinitesimally:In an orthonormal basis this is $\{X\in M_5(\mathbb{C}):X^T+X=0\}$, with the commutator bracket and dimension $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Its compact real form is obtained by taking real skew-symmetric matrices. For diagonal root calculations it is more convenient to use an isotropic basis $v_1,v_2,v_0,v_{-2},v_{-1}$ with $B(v_i,v_j)={\delta }_{i,-j}$, including $B(v_0,v_0)=1$.

A maximal torus of the complex special orthogonal group isIts Lie algebra is the Cartan subalgebraIn the compact real form the corresponding torus consists of independent rotations in two orthogonal coordinate planes, fixing the remaining direction. Define ${\epsilon }_i(H)=h_i$ for $i=1,2$, and ${\epsilon }_{-i}=-{\epsilon }_i$, ${\epsilon }_0=0$.

Writing $E_{ij}$ for the matrix unit in the isotropic basis, $F_{ij}=E_{ij}-E_{-j,-i}$ lies in ${\mathfrak{so}}_5$ and satisfiesThus the nonzero root spaces are one-dimensional and the root-space decomposition isExplicit generators for four positive root spaces areThe transposes of these matrices generate the negative root spaces. Each transpose still satisfies the bilinear-form condition and reverses the displayed diagonal weight. These eight root vectors and the two independent Cartan matrices exhaust the dimension ten; a matrix commuting with every $H$ is diagonal and the form condition makes it lie in $\mathfrak{h}$. This also proves that the torus has maximal dimension.

The displayed torus-only Weyl integration formula needs $\phi$ to be a class function, meaning $\phi (xg{x}^{-1})=\phi (g)$. This hypothesis is implicit in the usual formula but is not printed here. We prove the general version for a continuous function, then specialize:Here $dg$ and $dt$ are probability Haar measures; $d\mu$ is the invariant probability measure on the homogeneous space $G/T$; and $W=N_G(T)/T$ is the finite Weyl group.

We use the following standard maximal-torus facts, as permitted: every element of a connected compact Lie group lies in some maximal torus; all maximal tori are conjugate; two elements of a fixed $T$ are conjugate in $G$ exactly when they are in the same $W$-orbit; and the complexified Lie algebra decomposes into ${\mathfrak{t}}_{\mathbb{C}}$ and one-dimensional root spaces. Opposite roots of a root system give real two-dimensional planes in ${\mathfrak{t}}^{\perp }$. At an element $t$ for which no root character has value one, the identity component of its centralizer is $T$.

Choose a positive root system ${\Phi }^{+}$. Regard each root $\alpha$ as a character $\alpha :T\to U(1)$ through the Adjoint representation. A convenient globally defined denominator isThe usual Weyl denominator additionally has a unit-modulus factor $e^{\rho }(t)$, with $\rho$ half the sum of positive roots. Where that half-weight is not a global torus character, it may be interpreted on a covering torus; the squared modulus is nevertheless unambiguous. In either convention,

Average an inner product on $\mathfrak{g}$ over the Adjoint representation to make it invariant. It defines a bi-invariant Riemannian metric on $G$ and the quotient metric on $G/T$. Consider the conjugation mapAt $(T,t)$, identify the source tangent directions with $X\in {\mathfrak{t}}^{\perp }$ and $Y\in \mathfrak{t}$, and translate the target tangent vector to the identity by $t^{-1}$. Differentiating givesThe two summands are orthogonal. On a real root plane, $\mathrm{Ad}(t^{-1})$ is a rotation with complex eigenvalues $\alpha (t{)}^{-1},\alpha (t)$. Thus the absolute real Jacobian determinant on this plane is $|1-\alpha (t){|}^2$. The torus direction contributes one. Therefore the conjugation Jacobian for a compact Lie group isThis is also the Jacobian for the normalized measures: the Riemannian submersion has fibers isometric to $T$, so $\mathrm{vol}(G)=\mathrm{vol}(G/T)\mathrm{vol}(T)$; the probability-volume normalization factors cancel.

Let $T_{\mathrm{reg}}$ be the elements for which $J_q\ne 0$. The restriction of $q$ is a local diffeomorphism onto the regular elements of $G$. It is $|W|$-to-one. Indeed, fixing $t_0\in T_{\mathrm{reg}}$, any pair with $xt{x}^{-1}=t_0$ has $xT{x}^{-1}=T$, because conjugation identifies the identity components of the two centralizers. Thus $x\in N_G(T)$, and there is exactly one pair $(xT,x^{-1}{t}_{0}x)$ for each coset in $N_G(T)/T$. This argument does not require the full centralizer of $t_0$ to be connected.

The nonregular locus has measure zero. In $T$ it is a finite union of sets $\alpha (t)=1$ of smaller dimension; their conjugation images are images of smaller-dimensional manifolds under $q$, and hence have zero volume in $G$. The change-of-variables formula for the regular covering therefore givesFor a class function, the inner integral is $\phi (t)$, proving the required formula with its correct hypothesis.

Without conjugation invariance, the torus-only expression can be false. For $G=SU(2)$ with diagonal $T$, take $\phi (g)=|g_{12}{|}^2$. It vanishes on $T$, whereas its group integral is $1/2$: the two squared coordinates of a Haar-distributed unit column have equal means and sum to one. Thus the conjugacy average cannot simply be dropped.

For a Cartan-Weyl basis, the Jacobi identity givesThus the bracket has joint Cartan weight $\alpha +\beta$. If $\beta =-\alpha$, it belongs to the zero-weight space, which is the Cartan subalgebra, and so is a linear combination ${\tilde{\alpha }}_i{H}_i$. If $\alpha +\beta$ is a nonzero root, its root space is one-dimensional and the bracket is proportional to $E_{\alpha +\beta }$. If that nonzero sum is not a root, the bracket vanishes. This proves the asserted forms without assuming every pair of roots has a root sum.

With the stated normalization, put $H_{\alpha }=2{\alpha }_i{H}_i/{\alpha }^2$. The commutators becomeThe adjoint relations make $J_3=H_{\alpha }/2$, $J_{+}=E_{\alpha }$, $J_{-}=E_{-\alpha }$ the usual SU(2) Lie algebra generators. Hence every finite-dimensional SU(2) irreducible component has $J_3$ eigenvalues $m=-j,-j+1,\dots ,j$, with $2j$ an integer. Therefore $H_{\alpha }$ has integer eigenvalues $2m$, including in the finite-dimensional Adjoint representation. Algebraically this is the sl2 subalgebra associated with a root.

If $[H_{\alpha },X_{\lambda }]=\lambda X_{\lambda }$, another use of the Jacobi identity givesThe bracket is therefore zero or lies in the eigenvalue-$\lambda +2$ space. Within one irreducible SU(2) ladder it is proportional to the next chosen vector $X_{\lambda +2}$. If the full eigenspace has multiplicity, the weight statement is the general one; proportionality to one named vector presupposes the ladder choice. Similarly $E_{-\alpha }$ lowers this eigenvalue by two, and the ladder weights run from $-n$ to $n$ in steps of two.

For a root vector $E_{\beta }$, its $H_{\alpha }$ weight isThe adjoint SU(2) ladder therefore provesFor nonparallel $\alpha ,\beta$, raising and lowering $E_{\beta }$ move its joint Cartan weight through the root string $\beta +k\alpha$. The finite irreducible ladder has weights symmetric about zero. Starting at $m_{\alpha \beta }$, reaching its partner $-m_{\alpha \beta }$ requires $k=-m_{\alpha \beta }$. The corresponding nonzero vector lies in the root space of $\beta -m_{\alpha \beta }\alpha$, proving the Weyl reflection property:There is no zero joint root in this string when $\alpha ,\beta$ are nonparallel. Its one-dimensional root spaces prevent two irreducible ladders with the same weight parity from overlapping, so the ladder argument applies to the entire string. For $\beta =\pm \alpha$, reflection simply interchanges the two opposite roots, which already exist by the adjoint relation.

For angle and length restrictions, set $p=2\alpha \cdot \beta /{\alpha }^2$ and $q=2\alpha \cdot \beta /{\beta }^2$. These Cartan integers have the same sign when nonzero and obeyFor nonparallel, nonorthogonal roots, $pq$ is an integer strictly between zero and four. The possibilities are consequently:

Orthogonal roots have $p=q=0$ and $\theta =\pi /2$. Parallel roots have $pq=4$; the integer pairs would allow only length multiples $1,2,1/2$. The stated reducedness condition excludes the latter two, so parallel roots are equal or opposite, with angles zero or $\pi$ and equal lengths. ThusNot every root system realizes every listed angle.

The orthogonal pair alone gives no ratio through $q/p$. To finish the length restriction for a simple Lie algebra, use irreducibility of its root system. Let $U$ span the Weyl group orbit of $\alpha$. Both $U$ and $U^{\perp }$ are invariant under the orthogonal root reflections. If a root had nonzero projections in both, reflection of its projection in $U$ would have a nonzero $U^{\perp }$ component, violating invariance. Hence every root is entirely in one of these subspaces. A proper nonzero $U$ would split the root system into two orthogonal parts, contradicting irreducibility. This proves that the Weyl orbit spans an irreducible root space.

In particular, some orbit root $\gamma$ is nonorthogonal to any chosen $\beta$, and ${\gamma }^2={\alpha }^2$. Applying the previous integer-pair restriction to $\gamma ,\beta$ therefore also handles orthogonal $\alpha ,\beta$:There can be at most two different root lengths: three lengths would have ratios two and three relative to the shortest, but then the longest-to-middle squared-length ratio would be $3/2$, which is forbidden. A reducible collection of orthogonal rank-one factors could have independently chosen scales; the simple-algebra assumption excludes that exception.

Choose a generic linear functional to define positive roots. The simple roots are those positive roots which cannot be written as sums of two positive roots; they form a basis in which all root coefficients are integers of one sign. Distinct simple roots have nonpositive inner product: if their inner product were positive, their root string would contain their difference, and whichever difference is positive would decompose one of them. In rank two, irreducibility excludes the orthogonal case. The allowed simple-root angles are therefore $120^{\circ },135^{\circ },150^{\circ }$, giving the three irreducible diagrams in the classification of rank-two root systems: A2 root system, B2 root system and G2 root system. Reflections in the two simple roots generate their respective six, eight and twelve roots.

Here are explicit simple-root choices and positive roots; adding their negatives gives each complete diagram.

The nonnegative coefficients in each positive-root list verify the chosen simple roots explicitly. The orthogonal four-root system $A_1\times A_1$ is rank two but reducible, so it is not a diagram for the simple algebra assumed here.

Define the Cartan matrix in the column-coroot convention:Using the ordered roots above givesThe alternative row-coroot convention transposes these matrices. Stating the convention and root ordering avoids a spurious disagreement in the unequal-length cases.

First show that the Cartan subalgebra is orthogonal, for the Killing form, to every nonzero root space. For $H\in \mathfrak{h}$ and $X\in {\mathfrak{g}}_{\alpha }$, choose $H_0\in \mathfrak{h}$ with $\alpha (H_0)\ne 0$. Invariance and the abelian nature of $\mathfrak{h}$ giveHence $B(H,X)=0$.

Now let $H$ be in the radical of the restriction to $\mathfrak{h}$, so $B(H,H^{\prime })=0$ for every $H^{\prime }\in \mathfrak{h}$. The preceding argument also gives $B(H,{\mathfrak{g}}_{\alpha })=0$ for every root. By the root-space decomposition, $H$ is orthogonal to all of $\mathfrak{g}$. The assumed nondegeneracy on $\mathfrak{g}$ forces $H=0$. Therefore the Killing form restricted to the Cartan subalgebra is nondegenerate.

Here the Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra is a maximal toral subalgebra: it is abelian, its adjoint operators are simultaneously diagonalizable, and its centralizer is itself. These standard properties give the root-space decompositionThe roots of a root system $\alpha$ are the nonzero linear functionals in ${\mathfrak{h}}^{\ast }$ for which ${\mathfrak{g}}_{\alpha }\ne 0$; $\Phi$ is their finite set. Each root space is one-dimensional. The zero weight space is the centralizer of $\mathfrak{h}$, hence exactly $\mathfrak{h}$. Thus the displayed sum is a direct sum of simultaneous eigenspaces of the Adjoint representation. It is the Cartan root-space decomposition meant here.

The bracket respects weight addition: $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, where ${\mathfrak{g}}_0=\mathfrak{h}$ and a missing weight means the zero space. This follows by applying the Jacobi identity with $H\in \mathfrak{h}$ to the bracket of two root vectors. It explains how the vector-space decomposition encodes the Lie algebra structure.

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.

Put $\bar{i}=2n+1-i$ and let $E_{ij}$ denote a matrix unit. The antidiagonal matrix satisfies $J^{-1}=J$, so the orthogonal condition is $A=-J{A}^{T}J$, or entrywise $A_{ij}=-A_{\bar{j}\bar{i}}$. Consequently the diagonal Cartan subalgebra isWe first take $n\ge 2$. Since $[H,E_{ab}]=(H_{aa}-H_{bb})E_{ab}$, the matrix root basis of the even orthogonal Lie algebra givesThe zero weight space is $\mathfrak{t}$; its basis is $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. No vector of weight $2{\epsilon }_i$ occurs, because the corresponding matrix unit is forced to be its own negative. These spaces account for $n+2n(n-1)=n(2n-1)$ dimensions, exhausting the Special orthogonal Lie algebra. Thus its decomposition as a $\mathfrak{t}$-module and its Dn root system areEach nonzero root space has dimension one; the zero weight has multiplicity $n$.