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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.

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 $n$.

The upper triangular root spaces are precisely those of weights ${\epsilon }_i-{\epsilon }_j$ and ${\epsilon }_i+{\epsilon }_j$ with $i<j$. ThereforeThese $n$ roots form the simple roots. To check that they generate the chosen positive system with nonnegative coefficients, write ${\epsilon }_i-{\epsilon }_j={\sum }_{k=i}^{j-1}{\alpha }_k$. For $j<n$,and for $j=n$ use ${\epsilon }_i+{\epsilon }_n={\sum }_{k=i}^{n-2}{\alpha }_k+{\alpha }_n$. Empty sums are zero. Their linear independence and these expansions identify the base of the root system.

Equip the real span of the roots in ${\mathfrak{t}}^{\ast }$ with the inner product making the ${\epsilon }_i$ orthonormal. All roots then have squared length two. The coroots of ${\epsilon }_i\pm {\epsilon }_j$ are $H_i\pm H_j$, so the fundamental weights of Dn, characterized by ${\omega }_i({\alpha }_j^{\vee })={\delta }_{ij}$, areTaking their pairings with $H_j-H_{j+1}$ and $H_{n-1}+H_n$ verifies the defining identities directly. For each pair $i<j$, the sum of its two positive roots is $2{\epsilon }_i$. Hence the half-sum of positive roots is

The Dynkin diagram has a chain ending at ${\alpha }_{n-2}$, which is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. Every edge is a single bond because the root lengths agree; the nonzero off-diagonal simple-root pairings are minus one. The labelled Dn Dynkin diagram and its low-rank versions are shown below.

For $n=2$, ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_1+{\epsilon }_2$ are orthogonal, giving two disconnected A1 root systems. For $n=3$, the diagram is the chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$, of type A3 root system; the displayed weight formulas still apply. If $n=1$ is allowed, ${\mathfrak{so}}_2=\mathfrak{t}$ is one-dimensional and abelian: there are no roots or fundamental weights, and $\rho =0$.

Name the four vertices $1,2,\bar{2},\bar{1}$ in their displayed order. This is the crystal of the defining symplectic representation in rank two, of highest weight ${\omega }_1$ for the C2 root system. Its weights are respectively ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_2,-{\epsilon }_1$, with simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$. Each lowering arrow subtracts the corresponding simple root. Thus ${\omega }_1={\epsilon }_1$ and ${\omega }_2={\epsilon }_1+{\epsilon }_2$.

Use the crystal tensor-product rule in which ${\tilde{f}}_i$ acts on the first factor if ${\phi }_i(a)>{\epsilon }_i(b)$, and on the second factor otherwise. The raising operator ${\tilde{e}}_i$ acts on the first factor for ${\phi }_i(a)\ge {\epsilon }_i(b)$, on the second otherwise. Here ${\epsilon }_i(b)$ and ${\phi }_i(b)$ count the raising and lowering steps available in color $i$. The data for the four vertices areThe strictness difference between raising and lowering ensures that they are inverse along each colored edge. Testing the sixteen ordered tensor vertices gives exactly three vertices killed by both raising Kashiwara operators:Their highest weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$ and zero.

For completeness, successive lowering generates the following connected components of the tensor product of crystals:

They have sizes ten, five and one, are disjoint, and exhaust all sixteen vertices. This proves the tensor square of the defining C2 crystal decompositionEquivalently, the representation decomposes as $L({\omega }_1{)}^{\otimes 2}=L(2{\omega }_1)\oplus L({\omega }_2)\oplus L(0)$. The dimensions agree with the Weyl dimension formula for C2, $\dim L(a{\omega }_1+b{\omega }_2)=(a+1)(b+1)(a+b+2)(a+2b+3)/6$. The ordered highest vertices would change under the opposite tensor convention, but the irreducible decomposition would be the same.

Work with a finite-dimensional complex semisimple Lie algebra. The Dynkin diagram is obtained from its intrinsic root system, with several structure theorems entering the construction.

First choose a Cartan subalgebra $\mathfrak{h}$. The semisimple structure theory gives the root-space decompositionThe nonzero roots span ${\mathfrak{h}}^{\ast }$, root spaces have dimension one, and every root has its negative. The rank is $r=\dim \mathfrak{h}$.

Next use the Killing form. Its nondegeneracy on $\mathfrak{h}$ and its positive-definite restriction to the real coroot span identify the real root span $E=\mathbb{R}\Phi$ with a Euclidean space. Explicitly, if $t_{\lambda }={\mathcal{B}}^{-1}(\lambda )$, put $(\lambda ,\mu )=B(t_{\lambda },t_{\mu })$. The sl2 subalgebra associated with a root supplies its normalized coroot $H_{\alpha }$ and the integral pairingThe root-string theorem makes the reflection $s_{\alpha }(\beta )=\beta -\beta (H_{\alpha })\alpha$ permute $\Phi$. Together with the absence of multiples other than $\pm \alpha$, this shows that $\Phi$ is a finite reduced root system and a crystallographic root system. These properties come from semisimple Lie algebra theory, not from the diagram's definition.

Choose a vector in $E$ that is not perpendicular to any root. Define the positive roots by positive inner product with this vector. The simple roots are those positive roots that cannot be expressed as sums of two positive roots. Root-system theory shows that they form a basis $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, and each root is an integer combination of them with coefficients all of one sign. Distinct simple roots have nonpositive inner product.

Use the standard Cartan matrix conventionThere is one diagram vertex for each simple root. For $i\ne j$, the integral off-diagonal entries are nonpositive andConnect the two vertices with respectively zero, one, two or three edges. The corresponding angles are $90^{\circ },120^{\circ },135^{\circ },150^{\circ }$. A multiple edge carries an arrow toward the shorter root. For joined vertices, the length ratio is determined by $A_{ij}/A_{ji}=({\alpha }_i,{\alpha }_i)/({\alpha }_j,{\alpha }_j)$; hence the diagram records both angle and relative length.

For example, ${\mathfrak{sl}}_3$ has simple roots $L_1-L_2$ and $L_2-L_3$, giving two equal-length vertices joined by one edge: type $A_2$. For ${\mathfrak{sp}}_4$, take ${\alpha }_1=L_1-L_2$ and ${\alpha }_2=2L_2$. Thenso the two vertices have a double edge directed toward ${\alpha }_1$, the shorter root: type $C_2$.

Finally, orthogonal irreducible components of the root system correspond to the simple ideals of a Lie algebra in the semisimple Lie algebra, so its diagram is the disjoint union of their connected diagrams. Conjugacy of Cartan subalgebras and the Weyl group action on choices of positive roots make the resulting diagram independent of these choices up to isomorphism. Thus the vertices, edge multiplicities, arrows and connected components encode the simple-root geometry and the simple-ideal decomposition.

Choose a fundamental system of a root system $\{\alpha ,\beta \}$. Distinct simple roots have nonpositive inner product. Their Cartan matrix has diagonal entries two and off-diagonal entries $-p,-q$, where either $p=q=0$, or $p,q$ are positive integers. The root-system finiteness lemma gives $pq<4$. After exchanging the simple roots, the possibilities areTheir angles and relative lengths determine the two simple roots up to an orthogonal transformation and common scaling, except that the two orthogonal components can be scaled separately.

To see that no extra roots remain unspecified, write a positive root $\gamma =m\alpha +n\beta$ with nonnegative integer coefficients. If it is not simple, some simple root $\delta$ has $(\gamma ,\delta )>0$: otherwise $\Vert \gamma {\Vert }^2=m(\gamma ,\alpha )+n(\gamma ,\beta )\le 0$. The root reflection in $\delta$ subtracts a positive integer multiple of $\delta$. The other coordinate stays nonnegative; since $\gamma$ is not proportional to $\delta$, it is positive, so the reflected root must still be positive by the defining sign property of a fundamental system of a root system. Its height decreases. Repeating reduces $\gamma$ to a simple root. Thus every root is in a Weyl group orbit of a simple root.

For the four matrices, applying their simple reflections gives, respectively, the following positive roots:where in the last two rows $\alpha$ is long, $\beta$ short. For example, $s_{\beta }(\alpha )=\alpha +q\beta$ and $s_{\alpha }(\beta )=\beta +\alpha$ for the displayed matrices when the convention is $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. The sets comprising these positive roots and their negatives are invariant under both reflections, as direct substitution verifies. The height argument therefore shows they are the whole root system. This proves the classification of rank-two root systems:

A Dynkin diagram has one vertex for each simple root. Vertices $i,j$ are joined by $a_{ij}{a}_{ji}$ bonds, where $a_{ij}$ are the off-diagonal Cartan integers. No bond means orthogonality; one, two and three bonds mean angles $2\pi /3,3\pi /4,5\pi /6$. A multiple-bond arrow points to the shorter root. Thus the Dynkin diagram determines the Cartan matrix, including the relative lengths inside each connected component.

An irreducible root system corresponds to a connected Dynkin diagram. The classification of finite crystallographic Dynkin diagrams givesHere $C_2$ is already represented by $B_2$, and $B_1,C_1$ by $A_1$. General systems are orthogonal direct sums of these, with separate component scales. The following diagrams specify all the types; dots denote a continued chain, not additional vertices.

For orientation, the An Dynkin diagram is a single chain, the Bn Dynkin diagram has a terminal double bond pointing to its terminal short root, and the Cn Dynkin diagram reverses that arrow. The Dn Dynkin diagram has arm lengths $1,1,l-3$. The En Dynkin diagram has arm lengths $1,2,2$; $1,2,3$; or $1,2,4$. The F4 Dynkin diagram has a central double bond and the G2 Dynkin diagram a triple bond.

The positivity behind this finite list is the positive-definite Gram matrix of the simple roots. In particular the underlying graph is a tree: if it contained a cycle, the sum of its normalized roots would have squared norm at most zero, since each bonded inner product is at most $-1/2$. This contradicts linear independence. Restricting the possible bonds and branches by positive definiteness gives precisely the listed chains and trees. Conversely their Cartan matrices are positive definite after symmetrization and their reflection-generated root sets give the indicated finite root systems.

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.

Choose simple roots ${\alpha }_1,\dots ,{\alpha }_r$. For the orbit statement one must use the Closed dominant Weyl chamberIts interior, defined using strict inequalities, is the open Weyl chamber. Weights on reflecting walls, including zero, cannot be moved into that interior, so closure is necessary here.

Choose $\rho$ strictly inside the chamber, for example the sum of the fundamental weights. The finite Weyl group orbit of $\beta$ contains an element $\lambda$ maximizing $(\lambda ,\rho )$. If $(\lambda ,{\alpha }_i)<0$ for some simple root, thencontradicting the maximum. Thus every simple-root pairing of $\lambda$ is nonnegative, and $\lambda \in \mathcal{W}$. Consequently some $w\in W$ sends every integral weight $\beta$ into the closed fundamental chamber. The argument in fact works for every vector in the real weight space.

Choose a linear functional $\ell :E\to \mathbb{R}$ which is nonzero on every root; one exists because there are only finitely many root hyperplanes to avoid. Declare a root positive when $\ell (\alpha )>0$, and negative when $\ell (\alpha )<0$. This orders the roots by sign, compatibly with addition whenever the sum is a root. One can refine it to a lexicographic total order by completing $\ell$ to a coordinate system. Since negation permutes the roots,A simple root is a positive root not expressible as the sum of two positive roots. If distinct simple roots $\alpha ,\beta$ had $\alpha -\beta$ as a root, it would be either positive or negative. In the first case $\alpha =\beta +(\alpha -\beta )$ decomposes $\alpha$ into two positive roots; in the second $\beta =\alpha +(\beta -\alpha )$ decomposes $\beta$. Both contradict simplicity. If they are equal, their difference is zero and is also not a root. Hence the difference of two simple roots is never a root.

For use in the final subpart, the simple roots form a basis, and every positive root is a nonnegative integer combination of them. Here are the needed reasons. Repeatedly decompose a nonsimple positive root into two positive roots; their $\ell$ values decrease, and the finite root set makes this terminate. This gives the integer combinations. Distinct simple roots have nonpositive inner product: a positive inner product would make $\alpha$ and $-\beta$ obtuse, so the qualified root-sum lemma would make $\alpha -\beta$ a root, just ruled out. Finally, a nontrivial linear relation between simple roots can be split into positive and negative coefficients, giving the same vector $u$ as positive combinations of disjoint sets. Its squared norm, computed using those two expressions, is nonpositive, whereas a nonzero positive combination has positive $\ell$ value. This contradiction proves linear independence. Their span is $E$ because they generate all roots.