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For the B2 root system with ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ long and ${\alpha }_2={\epsilon }_2$ short, the spin fundamental weight is ${\Lambda }_2=({\epsilon }_1+{\epsilon }_2)/2$. Its four-dimensional highest-weight representation has the minuscule weight orbit $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$, each of multiplicity one. The crystal basis is the chain $++\stackrel{2}{\to }+-\stackrel{1}{\to }-+\stackrel{2}{\to }--$, where signs denote the coordinates. Each Kashiwara operator subtracts its indicated simple root; the sl2 Lie algebra strings have length one. This fixes every edge and gives a connected highest-weight graph with the correct character.

In an orthonormal basis $({\epsilon }_1,{\epsilon }_2)$, the roots are $\pm ({\epsilon }_1-{\epsilon }_2)$, $\pm ({\epsilon }_1+{\epsilon }_2)$, $\pm 2{\epsilon }_1$, and $\pm 2{\epsilon }_2$. Choose simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$, with ${\alpha }_1$ a short root. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2={\epsilon }_1+{\epsilon }_2$.

The Dynkin labels of a weight $\lambda$ are its simple-coroot coordinates ${\lambda }^i=⟨\lambda ,{\alpha }_i^{\vee }⟩$. Thus $\lambda ={\sum }_i{\lambda }^i{\omega }_i$ in the basis of fundamental weights.

With simple roots ${\alpha }_i={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n={\epsilon }_{n-1}+{\epsilon }_n$, the fundamental weights are ${\omega }_i={\epsilon }_1+\cdots +{\epsilon }_i$ for $i\le n-2$, followed by ${\omega }_{n-1}=({\epsilon }_1+\cdots +{\epsilon }_{n-1}-{\epsilon }_n)/2$ and ${\omega }_n=({\epsilon }_1+\cdots +{\epsilon }_n)/2$. All roots have squared length two, so their pairings with coroots verify the dual-basis identities. The last two weights are the half-spin highest weights; for $n=2$, the defining highest weight is ${\omega }_1+{\omega }_2$.

For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$ with upper triangular positive roots, the simple coroots are $\mathrm{diag}(1,-1,0)$ and $\mathrm{diag}(0,1,-1)$. Their dual fundamental weights are ${\omega }_1(\mathrm{diag}(t_1,t_2,t_3))=t_1$ and ${\omega }_2(\mathrm{diag}(t_1,t_2,t_3))=t_1+t_2=-t_3$. In simple-root coordinates, ${\omega }_1=(2{\alpha }_1+{\alpha }_2)/3$ and ${\omega }_2=({\alpha }_1+2{\alpha }_2)/3$. The primitive vectors $e_1\in {\mathbb{C}}^3$ and $e_1\wedge e_2\in {\Lambda }^2{\mathbb{C}}^3$ realize these Fundamental representations of sl3.

For the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, the fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$ and ${\omega }_2=3{\alpha }_1+2{\alpha }_2$. The positive coroots, relative to the simple coroots, have coordinates $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. The Weyl dimension formula therefore gives, for nonnegative integers $a,b$,The denominator is the product of the pairings of the Weyl vector with those coroots. The dimensions at ${\omega }_1,{\omega }_2,2{\omega }_1$ are seven, fourteen and twenty-seven.

For short and long simple roots $\alpha ,\beta$ of $G_2$, the long roots form an $A_2$ root system with simple roots ${\delta }_1=\beta$ and ${\delta }_2=3\alpha +\beta$. If ${\omega }_1,{\omega }_2$ are the corresponding $A_2$ fundamental weights, then $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$.

In the convention of a, the fundamental weights areFor a weight $w=(w_1,w_2,w_3)$ in this orthonormal coordinate system, its Dynkin labels areThese labels, rather than the orthonormal coordinates, are used below and in the projection in c.

The highest weight $(1,0,0)$ is $e_1$. Successive subtraction of the simple roots gives the complete stringThe two consecutive short-root steps are allowed because $e_3$ has coroot pairing two with ${\alpha }_3$. The weight system of this seven-dimensional vector representation is thereforeAll seven weight multiplicities are one, including the zero weight.

The highest weight $(0,0,1)$ is ${\omega }_3$. The Weyl group of the B3 root system acts by signed coordinate permutations. Its orbit is the eight cube vertices $w=(s_1{e}_1+s_2{e}_2+s_3{e}_3)/2$, with independent $s_i=\pm 1$. Starting from $(+,+,+)$, lowering by ${\alpha }_3$, then ${\alpha }_2$, and then either ${\alpha }_1$ or ${\alpha }_3$ generates the middle branches; further lowering reaches all remaining sign choices. Thus the eight-dimensional spin representation hasThe stipulated unit weight multiplicities give dimensions seven and eight by counting these lists.

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.

Use Dynkin labels $w_i=⟨w,{\alpha }_i^{\vee }⟩$, so the rows of the Cartan matrix are the simple roots expressed in fundamental weight coordinates. For the A2 root system,Starting at the highest weight $(1,0)$ and subtracting successive allowed simple roots gives $(1,0)\to (-1,1)\to (0,-1)$. Negating these weights gives the dual representation. Hence the two fundamental weight systems areEach listed weight has weight multiplicity one. The highest weight $(1,1)={\alpha }_1+{\alpha }_2$ is the highest root, so its representation is the Adjoint representation. Its nonzero weights are the six roots, while the Cartan subalgebra supplies the zero weight twice:The superscript $[2]$ indicates weight multiplicity. This also follows from $3\otimes \overline{3}=8\oplus 1$: the nine tensor product weights comprise six differences of distinct fundamental weights and three zeros, one zero belonging to the trivial representation.

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.

For the D5 root system, take an orthonormal basis $e_1,\dots ,e_5$ and simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3,4$, together with ${\alpha }_5=e_4+e_5$. The Dynkin diagram has single bonds $1-2$, $2-3$, $3-4$, $3-5$, with the fork at node $3$. Thus its Cartan matrix isThe fundamental weight ${\omega }_5=\frac{1}{2}(e_1+e_2+e_3+e_4+e_5)$ has Dynkin labels $(0,0,0,0,1)$. Its pairing with every positive coroot $e_i\pm e_j$, $i<j$, is zero or one. Hence its half-spin representation is a minuscule representation, so its weight system is a single Weyl group orbit with multiplicity one. The Weyl group of the D5 root system permutes coordinates and changes an even number of signs. Consequently its complete weight system isThere are $1+\left(\genfrac{}{}{0ex}{}{5}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{4}\right)=16$ such weights. Their Dynkin labels are $((s_1-s_2)/2,(s_2-s_3)/2,(s_3-s_4)/2,(s_4-s_5)/2,(s_4+s_5)/2)$, explicitlyHere the Lie algebra is $\mathfrak{so}(10,\mathbb{C})$, or its compact real form when discussing compact groups. Globally the half-spin representation is a representation of $\mathrm{Spin}(10)$, and does not descend to $SO(10)$; the question only specifies the Lie algebra.

Give the real root span an inner product with ${\epsilon }_1,{\epsilon }_2$ orthonormal. The B2 root system consists of four short roots on the coordinate axes and four long roots on the diagonals. Choose the long-first simple rootsThe positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2={\epsilon }_1$, and ${\alpha }_1+2{\alpha }_2={\epsilon }_1+{\epsilon }_2$. The corresponding simple reflections areThey generate every signed permutation of the two coordinates. Thus the Weyl group has order eight and is the symmetry group of a square; every element sends $(x,y)$ to $(\pm x,\pm y)$ or $(\pm y,\pm x)$, with independent signs.

Since ${\alpha }_1,{\alpha }_2$ form an integral basis of ${\mathbb{Z}}^2$, the root lattice is $Q=\mathbb{Z}{\epsilon }_1\oplus \mathbb{Z}{\epsilon }_2$. The simple coroots are ${\alpha }_1^{\vee }={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2^{\vee }=2{\epsilon }_2$. Therefore the weight lattice isSolving $⟨{\Lambda }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsThe cone of dominant weights isIndeed $a=x-y$ and $b=2y$ are exactly the two simple-coroot pairings. This describes representations of the Lie algebra, or of its simply connected Spin group; half-integral weights are allowed. For representations descending to $\mathrm{SO}(5)$ itself, the weight must lie in ${\mathbb{Z}}^2$, equivalently $b$ must be even.

Use directed color-$i$ edges $u\stackrel{i}{\to }v$ for the Kashiwara operator ${\tilde{f}}_i$, so every edge subtracts ${\alpha }_i$ from the weight. The defining crystal of the defining odd-orthogonal representation isThe defining module has highest weight ${\Lambda }_1$ and precisely these five weights, all of multiplicity one. The color-two string through zero has length two, since $⟨{\epsilon }_2,{\alpha }_2^{\vee }⟩=2$. The other indicated strings have length one. Restriction to each simple-root sl2 Lie algebra and the root-string property of a crystal justify the edges; no other weight difference is the relevant simple root.

For the B2 spin crystal $B({\Lambda }_2)$, denote the four weights $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$ by their sign pairs. Its graph isThe fundamental weight ${\Lambda }_2$ is a minuscule weight: its pairings with all positive coroots are zero or one. Its Weyl orbit has four weights, and the dimension formula gives four, so each occurs once and there are no additional weights. All simple-root strings have length one. Subtracting ${\alpha }_1$ or ${\alpha }_2$ where another weight is present gives exactly this graph.

For the B2 adjoint crystal, write $x_{\beta }$ for the vertex of root weight $\beta$, and use two distinct zero-weight vertices $z_1,z_2$. There are eight root vertices and two zero vertices because the zero-weight space is the two-dimensional Cartan subalgebra. The complete lists of colored strings areEvery arrow in a given row has that row's color; the reverse arrows are the raising operators, and all unlisted lowering operations vanish. In particular the two zero vertices must not be merged.

For a constructive justification, let $V={\mathbb{C}}^5$ with the invariant symmetric form from part(a). The mapis an equivariant isomorphism ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$, identifying the highest weight as ${\epsilon }_1+{\epsilon }_2$. Compute its crystal as the highest-weight component starting at $1\otimes 2$ in $B({\Lambda }_1)\otimes B({\Lambda }_1)$, with labels $1,2,0,\bar{2},\bar{1}$ for the defining chain. The crystal tensor-product rule iswhere ${\epsilon }_i$ counts available raising steps and ${\phi }_i$ counts lowering steps in the color-$i$ string. It generates the following ten vertices:Applying the displayed rule to these labels gives precisely the listed strings, proving the entire diagram rather than only its weight multiplicities. Tensor labels here are crystal vertices in the combinatorial limit, not an assertion that a bare ordinary tensor such as $v_0\otimes v_0$ lies in ${\Lambda }^{2}V$.

Take the diagonal Cartan subalgebra of ${\mathfrak{sl}}_3$ and let ${\epsilon }_i$ select its $i$th diagonal entry, so ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3=0$ on it. Use simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2=-{\epsilon }_3={\epsilon }_1+{\epsilon }_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^{\ast }$. ThusTheir crystal basis diagrams are $B:1\stackrel{1}{\to }2\stackrel{2}{\to }3$ and $B^{\vee }:\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights ${\epsilon }_i$ and $-{\epsilon }_i$ respectively. Use the same crystal tensor-product rule as in2(d): lowering acts on the first factor when ${\phi }_i(u)>{\epsilon }_i(v)$; otherwise on the second. Raising uses the first factor when ${\phi }_i(u)\ge {\epsilon }_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.

For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to 21\to 22$, $31\to 32$, $13\to 23$; the complete color-two list is $21\to 31$, $22\to 32\to 33$, $12\to 13$. There are two connected components:For example ${\tilde{f}}_1(1\otimes 1)=2\otimes 1$ because ${\phi }_1(1)=1>{\epsilon }_1(1)=0$, whereas ${\tilde{f}}_1(1\otimes 2)=0$ because equality selects the second factor and ${\tilde{f}}_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional symmetric square and the three-dimensional exterior square:

For $B\otimes B^{\vee }$, abbreviate $i\otimes \bar{j}$ by $i\bar{j}$. The complete color-one edge list is $1\bar{3}\to 2\bar{3}$, $1\bar{2}\to 2\bar{2}\to 2\bar{1}$, $3\bar{2}\to 3\bar{1}$; the complete color-two list is $1\bar{3}\to 1\bar{2}$, $2\bar{3}\to 3\bar{3}\to 3\bar{2}$, $2\bar{1}\to 3\bar{1}$. The isolated vertex is $1\bar{1}$. Thus the two highest vertices are $1\bar{3}$ of weight ${\epsilon }_1-{\epsilon }_3={\omega }_1+{\omega }_2$, and $1\bar{1}$ of weight zero. Lowering from the former visits all eight other vertices. ConsequentlyThis also follows concretely from $V\otimes V^{\ast }\cong \mathrm{End}(V)={\mathfrak{sl}}_3\oplus \mathbb{C}I$: the first term is the Adjoint representation, with highest weight the highest root, and the second is the trivial Lie algebra representation. The ordinary invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.

Let $E$ be the real span of the roots in ${\mathfrak{h}}^{\ast }$. The weight lattice iswhere $h_{\alpha }$ is the coroot in the root's sl2 subalgebra associated with a root, normalized by $\alpha (h_{\alpha })=2$. It suffices to test the simple coroots. Equivalently, if ${\omega }_i(h_{{\alpha }_j})={\delta }_{ij}$ defines the fundamental weights, then $P={⨁}_i\mathbb{Z}{\omega }_i$.

To prove integrality of representation weights on coroots, restrict a finite-dimensional module $V$ to the root's sl2 Lie algebra. By the classification of finite-dimensional sl2 representations, the eigenvalues of its Cartan element $h_{\alpha }$ are integers. Thus every $\mu \in {\Lambda }_V$ has $\mu (h_{\alpha })\in \mathbb{Z}$. The simple coroots form a basis of the real Cartan space, so these integral values also put $\mu$ in $E$. Therefore $\boxed{{\Lambda }_V\subseteq {\Lambda }_W}$ for every finite-dimensional representation in this setting. The weight lattice need not coincide with the smaller root lattice.

For the chosen simple roots $\Delta$, the open fundamental chamber of a root system isEquivalently $\lambda (h_{\alpha })>0$ for every simple coroot. Since all positive roots are nonnegative combinations of simple roots, requiring the inequality for all positive roots gives the same chamber. It is a connected component of the complement of the root reflection hyperplanes; the Weyl group permutes those chambers simply transitively.

The Closed dominant Weyl chamber is obtained by replacing the inequalities by $\ge 0$. It is the cone generated by the fundamental weights. When the fundamental Weyl chamber is taken to include its walls, this closed convention is intended; both conventions determine the same positive system and boundary rays.

The Weyl group acts on the weight lattice by its root reflections and extends to ring automorphisms of the group ring of a weight lattice:Weights and their multiplicities in a finite-dimensional Lie algebra representation of a semisimple Lie algebra are permuted by this action, so its formal character is Weyl-invariant.

Choose a set ${\Phi }^{+}$ of positive roots and putHere ${\omega }_i$ are the fundamental weights, and $\det (w)=(-1{)}^{\ell (w)}$, where the Coxeter length $\ell (w)$ counts positive roots carried to negative ones. For a dominant integral weight $\lambda$, let $L(\lambda )$ denote the finite-dimensional Irreducible Lie algebra representation of highest weight $\lambda$. The Weyl character formula isThe second denominator expression is the Weyl denominator formula. Both alternants change sign under each root reflection; their quotient is invariant. Although written as a fraction, the theorem asserts that it is a finite element of $\mathbb{Z}[{\Lambda }_W]$ with the nonnegative coefficients given by weight multiplicities. The highest-weight classification specifies which $\lambda$ label the irreducibles, while the root system and choice of positive roots determine every other term.

The weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$. The coroots use the normalization fixed above.

The Killing form defines the complex-linear mapBy nondegeneracy of the Killing form on a Cartan subalgebra, this is an isomorphism. For $H\in {\mathfrak{h}}_{\mathbb{R}}$, expand its image in the dual fundamental-weight basis:The coefficients are real by the preceding positivity calculation, and the ${\omega }_i$ are in ${\Lambda }_W$. ConsequentlyIn fact both spaces have real dimension $r$, so this inclusion is equality.

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.