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The triple bond in the original Dynkin diagram gives the Cartan integerswhere $\alpha$ is the long root and $\beta$ the short root. Their product is $4{\cos }^2\theta =3$, and the inner product of distinct simple roots is nonpositive. The ratio of the two integers gives the squared-length ratio. ThereforeIt is convenient to normalize $(\alpha ,\alpha )=6$, $(\beta ,\beta )=2$ and $(\alpha ,\beta )=-3$. Scaling the inner product does not affect the root system or the fundamental-weight relations.

For nonproportional roots $\gamma ,\delta$, the root-string theorem states that the root string is consecutive:The endpoints are maximal, and reflection in $\gamma$ reverses the string. The Cartan integer determines $p-q$, not in general the total $p+q+1$ by itself. This result follows by restricting the Adjoint representation to the sl2 subalgebra associated with a root. If the roots are distinct simple roots, $\delta -\gamma$ cannot be a root: its simple-root coefficients have opposite signs. Thus $p=0$ andHere length means number of roots; the number of intervals between them is one less. Distinctness matters. If $\gamma =\delta$, a reduced root system gives the set $\{\gamma ,-\gamma \}$, with a missing zero between them, so the consecutive-string theorem and the displayed simple-root formula do not apply.

In the G2 root system, the initial strings areFor $\delta =\alpha +3\beta$, $⟨\delta ,{\alpha }^{\vee }⟩=-1$. Since $\delta -\alpha =3\beta$ is not a root in a reduced root system, $p=0$ and $q=1$: this generates $2\alpha +3\beta$. The remaining strings explain why the construction stops. The $\alpha$-strings through $\beta$ and $\alpha +\beta$ are the same two-element string; the one through $\alpha +2\beta$ is a singleton since subtracting $\alpha$ gives $2\beta$, and its Cartan integer is zero; the one through $2\alpha +3\beta$ is the string $\{\alpha +3\beta ,2\alpha +3\beta \}$. The $\beta$-strings through $\alpha$, $\alpha +\beta$, $\alpha +2\beta$ and $\alpha +3\beta$ are the initial four-element string. Finally, $2\alpha +3\beta$ is orthogonal to $\beta$; its $\beta$-string is a singleton because $2\alpha +2\beta =2(\alpha +\beta )$ is not a root. Apply the same reasoning to negatives. Using the permitted completeness of this procedure givesThe short positive roots are $\beta ,\alpha +\beta ,\alpha +2\beta$, of squared length two; the other three are long, of squared length six. Each root space is one-dimensional and the Cartan subalgebra has dimension two, soHere the dimension refers to the Lie algebra, with one Cartan generator per rank, not just the number of roots.

Write a prospective weight as $w=a\alpha +b\beta$. The pairings with simple coroots areThe fundamental weights are dual to those coroots. Solving the two linear systems gives, in the long-root-first numbering of this paper,The representation with Dynkin labels $(0,1)$ has highest weight $\lambda ={\omega }_2$, a short root. Numbering the short root first, as some references do, would call this the $(1,0)$ representation instead; the representation itself is unchanged.

The weight set of a finite-dimensional irreducible highest-weight representation is invariant under the Weyl group and lies in the convex hull of the orbit of its highest weight. All weights also differ from the highest weight by an element of the root lattice. The orbit of $\lambda$ comprises the six short roots, so all six are weights. The lowering operators give the chainIn particular zero occurs: at weight $\beta$, its pairing with ${\beta }^{\vee }$ is two, so the lowering operator is nonzero by the finite-dimensional sl2 Lie algebra representation theory. The $\beta$-string through it is the usual three-weight string $\beta ,0,-\beta$; higher weight $2\beta$ would lie outside the highest-weight convex hull.

There can be no further weights. Every point of that convex hull has squared norm at most $\Vert \lambda {\Vert }^2=2$, while a root-lattice point hasThe integer solutions of the bound $\Vert w{\Vert }^2\le 2$ are precisely zero and the six short roots: $a=0$ gives $b=-1,0,1$; $a=1$ gives $b=1,2$; and $a=-1$ gives $b=-1,-2$. ThusThe final dimension uses the stipulated nondegeneracy of the weights. It also agrees with the Weyl dimension formula without that stipulation. The zero weight is not in the Weyl orbit of the nonzero weights, so this is not a minuscule representation.

In the diagram above, a coordinate label $(a,b)$ means $a\alpha +b\beta$. The left panel contains all twelve roots of a root system; the right panel contains the six short-root weights and the zero weight. The long-root-first convention is the same as in the calculations.

Choose the root basisIn the usual invariant inner product, each simple root has squared length $2$, while $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$ and $({\alpha }_2,{\alpha }_3)=0$. The labeled Dynkin diagram is therefore the following chain, with a single bond between adjacent nodes:It is type $D_3$, equivalently the A3 root system. To check the root basis explicitly, the corresponding positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$; these and their negatives are all twelve roots.

An automorphism of a root system outside the Weyl group is the sign changeIt preserves the set $\{\pm {\epsilon }_i\pm {\epsilon }_j\}$, fixes ${\alpha }_1$, and interchanges ${\alpha }_2$ and ${\alpha }_3$. It therefore realizes the nontrivial symmetry of the Dynkin diagram while preserving the chosen root basis.

To verify that $\tau \notin W$, note that $s_1$ swaps ${\epsilon }_1,{\epsilon }_2$, $s_2$ swaps ${\epsilon }_2,{\epsilon }_3$, and $s_3$ swaps ${\epsilon }_2,{\epsilon }_3$ while changing both their signs. Every product of these generators is a signed permutation with an even number of sign changes, whereas $\tau$ has one sign change. Thus it cannot be a Weyl group element. The parity here concerns sign changes, rather than the determinant of the permutation.

The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.

Work over $\mathbb{R}$ or $\mathbb{C}$, in finite dimension and characteristic zero. The Killing form of a Lie algebra is the trace form of its Adjoint representation:It is bilinear and symmetric by the cyclic property of the trace. It is an invariant bilinear form on a Lie algebra, since the Jacobi identity gives ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and henceEquivalently, $\kappa ([z,x],y)+\kappa (x,[z,y])=0$. A Lie algebra automorphism preserves the form, because it conjugates the adjoint matrices. Its radical of a bilinear form ${\mathfrak{r}}_{\kappa }=\{x:\kappa (x,\mathfrak{g})=0\}$ is an ideal of a Lie algebra by invariance. The center of a Lie algebra lies in this radical, so the form vanishes for an abelian algebra. On a direct sum of ideals the summands are orthogonal and the form restricts to their own Killing forms.

A short argument proves the requested implication without assuming the Cartan criterion for semisimplicity. Let $\mathfrak{a}$ be an abelian ideal of a Lie algebra, $x\in \mathfrak{a}$, and $y\in \mathfrak{g}$. The map ${\mathrm{ad}}_x$ takes $\mathfrak{g}$ into $\mathfrak{a}$ and vanishes on $\mathfrak{a}$, while ${\mathrm{ad}}_y$ preserves $\mathfrak{a}$. Therefore ${\mathrm{ad}}_x{\mathrm{ad}}_y$ has image in $\mathfrak{a}$ and is zero on $\mathfrak{a}$. In a basis extending a basis of $\mathfrak{a}$, both diagonal blocks are zero, so its trace vanishes. This proves that abelian ideals lie in the radical of the Killing form.

If the solvable radical $\mathfrak{r}$ were nonzero, its derived series of a Lie algebra would have a last nonzero term $\mathfrak{a}$. The Jacobi identity makes every derived term an ideal of $\mathfrak{g}$, and the last one is abelian. Thus $0\ne \mathfrak{a}\subseteq {\mathfrak{r}}_{\kappa }$, contradicting nondegeneracy. We obtainThe converse holds as well in characteristic zero; together these implications are the Cartan criterion for semisimplicity.

For a complex simple Lie algebra, that criterion gives nondegeneracy on $\mathfrak{g}$. Let $\mathfrak{h}$ be a Cartan subalgebra. Use the standard root-space decompositionFor $h,h^{\prime }\in \mathfrak{h}$ and a root vector $e_{\alpha }$, choose $t\in \mathfrak{h}$ with $\alpha (t)\ne 0$. Invariance givesThus $\mathfrak{h}$ is orthogonal to every nonzero root space. If $h\in \mathfrak{h}$ is also orthogonal to $\mathfrak{h}$, it is orthogonal to all of $\mathfrak{g}$ and hence is zero. This establishes nondegeneracy of the Killing form on a Cartan subalgebra:The same invariance calculation shows that $\kappa ({\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta })=0$ unless $\alpha +\beta =0$. Opposite root spaces are therefore paired nondegenerately. Tracing the adjoint action on the root-space decomposition yieldssince the root spaces of a complex semisimple algebra are one-dimensional and its adjoint action on $\mathfrak{h}$ is zero.

The relevant Euclidean subspace of a Cartan subalgebra iswhere the $h_{{\alpha }_i}$ are the simple coroots, normalized by ${\alpha }_i(h_{{\alpha }_i})=2$. The standard Euclidean property is that $\kappa$ is real and positive definite on this space. The displayed trace formula explains it: $\kappa (h,h)$ is a sum of real squares, and the roots span ${\mathfrak{h}}^{\ast }$, so all squares vanish only for $h=0$. Via the nondegenerate form, the roots can consequently be regarded as vectors in a real Euclidean normed vector space. The induced dual inner product makes root reflections orthogonal and allows root lengths and angles to be encoded by Cartan integers and the Dynkin diagram.

This positivity is on a specified real subspace; the complex Killing form is a bilinear form, not a positive Hermitian inner product. On a compact real form the Killing form is instead negative definite. For example the Killing form of the special linear Lie algebra gives $\kappa (X,Y)=4\mathrm{tr}(XY)$ for ${\mathfrak{sl}}_2$, so $\kappa (L_0,L_0)=8$ but $\kappa (i{L}_0,i{L}_0)=-8$.

Use the wiki's Cartan matrix conventionLabel the long simple root first when the lengths differ. The matrix of an irreducible rank-two root system has the form $A=\left(\begin{array}{cc}2& -a\\ -b& 2\end{array}\right)$, where $a,b$ are positive integers. Integrality and the signs follow from the Cartan integers for distinct simple roots, and irreducibility rules out a zero off-diagonal pair. Its positive-definite symmetrization gives $ab<4$, equivalently $ab=4{\cos }^2\theta <4$. Moreover $a/b=\Vert {\alpha }_1{\Vert }^2/\Vert {\alpha }_2{\Vert }^2\ge 1$. Thus $(a,b)=(1,1),(2,1),(3,1)$, giving the entire classification of rank-two root systems relevant to a simple algebra:These correspond to ${\mathfrak{sl}}_3(\mathbb{C})$, ${\mathfrak{so}}_5(\mathbb{C})\cong {\mathfrak{sp}}_4(\mathbb{C})$, and the exceptional algebra ${\mathfrak{g}}_2$. Here $B_2=C_2$ denotes the Isomorphism between so5 and sp4, not an additional case. The disconnected $A_1\times A_1$ system corresponds to a semisimple but nonsimple algebra and is excluded.

The Dynkin diagrams, with nodes in the same order as the matrices, areThe second and third diagrams have two and three bonds respectively, with the arrowhead pointing to the short root ${\alpha }_2$. In the last diagram $\equiv$ draws the three bonds and $>$ their arrowhead.

To enumerate roots, use the standard root string theorem: for $\beta \ne \pm \alpha$, the roots $\beta +n\alpha$ are consecutive from $\beta -p\alpha$ through $\beta +q\alpha$, with $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. Also use the standard results that the root system is reduced, every root is conjugate under the Weyl group to a simple root, and each root space has dimension one.

Write $m=a\in \{1,2,3\}$. Since a difference of simple roots is not a root, the ${\alpha }_2$-string starting at ${\alpha }_1$ has $p=0$ and $q=m$. It produces ${\alpha }_1,{\alpha }_1+{\alpha }_2,\dots ,{\alpha }_1+m{\alpha }_2$. The ${\alpha }_1$-string starting at ${\alpha }_2$ has $q=1$. In type $G_2$, the further string through $\beta ={\alpha }_1+3{\alpha }_2$ hasand $\beta -{\alpha }_1=3{\alpha }_2$ is not a root, so it also produces $2{\alpha }_1+3{\alpha }_2$. The resulting positive roots areAll roots are these and their negatives. To check completeness, the simple root reflections act on coordinates byEach listed set together with its negatives is stable under both reflections. It contains the simple roots and consists of roots already forced by strings. Since every root lies in a Weyl orbit of a simple root, no other roots can occur. This proves the lists for the A2 root system, B2 root system, and G2 root system. The root-space decomposition then gives

For the restriction to a sl2 subalgebra associated with a root, normalize its Cartan element to be $h_{\alpha }={\alpha }^{\vee }$. On a root vector of root $\beta$, its eigenvalue is $⟨\beta ,{\alpha }^{\vee }⟩$. The classification of finite-dimensional sl2 representations says that $R(\Lambda )$ has weights $\Lambda ,\Lambda -2,\dots ,-\Lambda$ and dimension $\Lambda +1$. A root string of $\Lambda +1$ nonparallel roots therefore supplies one such irreducible module: the raising and lowering brackets connect its consecutive root spaces.

For clarity, the positive-side strings for each simple-root direction are listed below. Brackets denote the whole consecutive string; a single listed root is a string of length one. In every row also include each negative string in reverse order, and separately the triple formed by the root ${\alpha }_i$, its negative, and $h_{{\alpha }_i}$.The root's own triple is always $R(2)$. The one-dimensional subspace of $\mathfrak{h}$ annihilated by ${\alpha }_i$ commutes with this subalgebra and gives $R(0)$. Each singleton root in the table, and its negative, contributes another $R(0)$. Reading the other string lengths now gives the adjoint branching to root sl2 subalgebras in rank two:Their dimensions are respectively $8,10,10,14,14$. The last row agrees with G2 adjoint branching to a short-root sl2 subalgebra. All labels in the table are highest weights, rather than dimensions.

A convenient root basis isThe corresponding Cartan matrix isThus the Dynkin diagram is the three-node $C_3$ chain, with a double edge between ${\alpha }_2$ and ${\alpha }_3$ and its arrow pointing toward the shorter root ${\alpha }_2$.

With the Euclidean inner product used above, the simple coroots areIndeed, if a positive root is $\beta ={\sum }_i{n}_i{\alpha }_i$, thenwhose coefficients are nonnegative. The negative roots give the negatives of these combinations, so the displayed coroots form a fundamental system of a root system for the dual root system.

Duality reverses root lengths. The dual of $C_3$ is therefore the B3 root system: its Dynkin diagram is again a three-node chain with a double final edge, but its arrow points toward the now-short root ${\alpha }_3^{\vee }$.

Write an element of the diagonal torus asand let ${\epsilon }_i$ extract $t_i$. The root-space decomposition iswith one-dimensional root spaces. Thus the $D_n$ root system is

The upper-triangular choice givesA compatible simple system isThe highest root and Weyl vector areThe fundamental weights are

Using the paper's letters, the root lattice and weight lattice are respectivelyTheir quotient is

The Dynkin diagram is the $D_n$ diagram: a chain ${\alpha }_1-\cdots -{\alpha }_{n-2}$ whose last node is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. The Extended Dynkin diagram adds ${\alpha }_0=-\theta$ joined to ${\alpha }_2$. For $D_4$, the central node ${\alpha }_2$ consequently has the four leaves ${\alpha }_0,{\alpha }_1,{\alpha }_3,{\alpha }_4$.

One root basis isAll roots have the same length. The nonzero inner products are $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$, so the labeled Dynkin diagram isIt is the three-node $A_3$ diagram.

The preceding Dynkin diagram identifies the root system $D_3$ with $A_3$. The classification of finite-dimensional complex simple Lie algebras by connected Dynkin diagrams therefore givesThis is the Isomorphism between so6 and sl4. Both algebras have dimension $15$, consistently with the classification.

An irreducible root system is simply laced when every root has the same length, equivalently when its Dynkin diagram has no multiple edge.

If all roots have the same length, the two Cartan integers for $\alpha$ and $\beta$ are equal. The root-system finiteness lemma then makes their product either zero or one, sofor $\alpha \ne \pm \beta$.

Conversely, when all such Cartan integers lie in $\{0,\pm 1\}$, any two nonorthogonal roots have Cartan integers of absolute value one in both directions. Their squared lengths are therefore equal. Irreducibility makes the graph joining nonorthogonal roots connected, so all roots have the same length. Thus the system is simply laced.

ChooseHere ${\alpha }_1$ is long and ${\alpha }_2$ is short. The Dynkin diagram consists of two vertices joined by a double edge, with its arrow pointing from ${\alpha }_1$ toward the shorter root ${\alpha }_2$:

For simple roots ${\alpha }_i$, the Cartan matrix isIts diagonal entries are two, while its off-diagonal entries encode the angles and relative lengths in the Dynkin diagram.

Disconnected nodes in a Dynkin diagram represent orthogonal roots, so $\angle ({\alpha }_1,{\alpha }_3)=90^{\circ }$ in all three cases. A single edge gives $120^{\circ }$, hence $\angle ({\alpha }_1,{\alpha }_2)=120^{\circ }$ and, for $A_3$, also $\angle ({\alpha }_2,{\alpha }_3)=120^{\circ }$. The double edge in both $B_3$ and $C_3$ gives $\angle ({\alpha }_2,{\alpha }_3)=135^{\circ }$.

After ordering a symplectic basis in two blocks, writeMatrices in the symplectic Lie algebra have block formThe root-space decomposition isFor example, these one-dimensional spaces are spanned respectively byThus this is the Cn root systemThe upper-triangular choice givesIts simple roots, highest root, fundamental weights, and half-sum of positive roots are

Using the notation requested in the paper, the root lattice $P$ and weight lattice $Q$ areso $Q/P\cong \mathbb{Z}/2\mathbb{Z}$. This reverses the common notation in which the root lattice is called $Q$ and the weight lattice is called $P$.

Since a multiple-edge arrow in a Dynkin diagram points toward the shorter root, the finite and extended diagrams areand

A reduced root system of rank one is $\Phi =\{\alpha ,-\alpha \}$. Its fundamental weight is $\alpha /2$, and its Weyl group has two elements. Its ordinary Dynkin diagram has one node; adjoining ${\alpha }_0=-\alpha$ gives affine Cartan matrix $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$, drawn as a double bond between equal-length roots.

A root system is a finite set of nonzero vectors closed under the reflections they define and satisfying the crystallographic integrality condition when it arises from a semisimple Lie algebra.

An irreducible root system is simply laced when all roots have the same length, equivalently when its Dynkin diagram has only single edges. This is also equivalent towhenever $\alpha \ne \pm \beta$.