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The underlying graph of a finite Dynkin diagram is a forest. Normalize every simple root to length one. Each bonded pair has inner product at most $-1/2$, and every other distinct pair has nonpositive inner product. A cycle with $m$ vertices would make the squared norm of their sum at most $m-m=0$, contradicting their linear independence in an inner product space. Multiple bonds denote root-length information, not graph-theoretic parallel-edge cycles.

For $n\ge 1$, the finite Dynkin diagram $A_n$ is a chain of $n$ vertices joined by single bonds. It represents the root system of ${\mathfrak{sl}}_{n+1}$.

An automorphism of a root system $\Phi \subset E$ is an orthogonal transformation of $E$ mapping $\Phi$ onto itself. The Weyl group is a subgroup of this automorphism group; additional automorphisms can act nontrivially on a Dynkin diagram while preserving a root basis.

For $n\ge 2$, the Bn root system has simple roots ${\alpha }_i={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n={\epsilon }_n$. Its Dynkin diagram is a chain with a double terminal bond pointing toward short ${\alpha }_n$. The highest root is $\theta ={\epsilon }_1+{\epsilon }_2$. In the Extended Dynkin diagram, ${\alpha }_0=-\theta$ attaches by a single bond to ${\alpha }_2$ for $n\ge 3$. For $n=2$, both long nodes $0,1$ attach by double bonds to short node $2$. Rank one is the rank-one root system; its affine diagram has two equal-length nodes joined by a double bond without a short-root arrow.

The connected Dynkin diagrams of finite reduced crystallographic root systems are $A_r$, $B_r$, $C_r$, $D_r$, $E_6$, $E_7$, $E_8$, $F_4$ and $G_2$. Bonds encode products of off-diagonal Cartan integers; multiple-bond arrows point toward shorter roots. The simply laced branching diagrams have arm lengths $(1,1,r-3)$ for $D_r$ and $(1,2,2)$, $(1,2,3)$, $(1,2,4)$ for the three exceptional $E$ types. The $B$ and $C$ chains have an end double bond, $F_4$ a middle double bond, and $G_2$ a triple bond. This edge weighting is distinct from the vertex labels used to classify nilpotent orbits.

For $n\ge 2$, the finite Dynkin diagram $C_n$ is a chain with a terminal double bond. Its arrow points from the terminal long root toward the adjacent short root. It represents the Cn root system; $C_2$ is isomorphic to $B_2$ after relabelling.

For $n\ge 2$, the finite Dynkin diagram has a chain of short roots ${\alpha }_1,\dots ,{\alpha }_{n-1}$ with single bonds, followed by a double bond to the long root ${\alpha }_n$, pointing toward ${\alpha }_{n-1}$. The Extended Dynkin diagram adds the long root ${\alpha }_0=-2{\epsilon }_1$, doubly bonded to ${\alpha }_1$ with arrow toward ${\alpha }_1$. All other bonds remain. For $n=1$, the finite diagram is a single vertex; the affine diagram has two equal-length vertices joined by a double bond without an arrow, with Cartan matrix $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$.

The root system $D_5$ consists of the forty roots $\pm e_i\pm e_j$, $1\le i<j\le 5$, in an orthonormal basis. One choice of simple roots is $e_1-e_2,e_2-e_3,e_3-e_4,e_4-e_5,e_4+e_5$. The Dynkin diagram has a fork at node $3$ with ends $4$ and $5$. Its Weyl group permutes coordinates and changes an even number of signs. The associated complex simple Lie algebra is $\mathfrak{so}(10,\mathbb{C})$. Its two half-spin representations have dimension $16$, with highest weights $\frac{1}{2}(e_1+e_2+e_3+e_4\mp e_5)$.

For $n\ge 4$, the finite Dynkin diagram $D_n$ is a simply laced tree with one trivalent vertex and three arms containing $1,1,n-3$ edges. All its roots have the same length.

The dual root system of $\Phi$ is ${\Phi }^{\vee }=\{{\alpha }^{\vee }:\alpha \in \Phi \}$. It has the transposed Cartan matrix, so its Dynkin diagram reverses every multiple-edge arrow; in particular, $B_n$ and $C_n$ are dual.

The finite Dynkin diagrams $E_6,E_7,E_8$ are simply laced trees with one trivalent vertex. Their three arm lengths, counting edges, are respectively $(1,2,2)$, $(1,2,3)$ and $(1,2,4)$.

The finite Dynkin diagram $F_4$ is a four-vertex chain whose central bond is double. The two consecutive long roots lie on one side, the two consecutive short roots on the other, and the arrow points toward the short pair. The remaining two bonds are single.

The finite Dynkin diagram $G_2$ has two vertices joined by a triple bond, with the arrow toward the short root. The squared root-length ratio is three. It represents the G2 root system.

A root system spanning a real inner product space is irreducible when it cannot be partitioned into two nonempty mutually orthogonal root systems. Its Dynkin diagram is connected. This condition concerns the roots themselves, rather than an arbitrary ambient space larger than their span.

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.

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.

Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

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

Normalize the simple roots to $({\alpha }_i,{\alpha }_i)=2$. The A2 root system has $({\alpha }_1,{\alpha }_2)=-1$, so its Dynkin diagram is two nodes joined by one bond:The fundamental weights satisfy $({\omega }_i,{\alpha }_j)={\delta }_{ij}$. Hence the Gram matrix in the fundamental weight basis, the metric for Dynkin label coordinates, isFor weights of a representation with Dynkin labels $a$ and $b$, their inner product is $a^{T}Gb$. In the simple root basis the metric is instead $A$ itself; stating the basis removes this apparent convention ambiguity.