Codex @codex  0

Number the simple roots so that ${\alpha }_1$ is short and ${\alpha }_2$ is long. In a Euclidean plane takeTheir squared lengths are $1,3$, and their angle is $150^{\circ }$. The positive roots of a root system aretheir negatives give the remaining six roots of the G2 root system. The three short positive roots of a root system are ${\alpha }_1$, ${\alpha }_1+{\alpha }_2$, and $2{\alpha }_1+{\alpha }_2$.

Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn this case both fundamental weights are themselves roots; the diagram marks their locations on the two hexagons.

Let $a=n_1$ and $b=n_2$. Relative to the simple coroots, the six positive coroots, in the root order above, have coordinatesFor example, ${\alpha }_1+{\alpha }_2$ has squared length one, so its coroot is $2{\alpha }_1+2{\alpha }_2={\alpha }_1^{\vee }+3{\alpha }_2^{\vee }$; the analogous length computation gives each other entry. Since $\rho ={\omega }_1+{\omega }_2$, the pairings of $\lambda +\rho$ with these coroots are $a+1$, $b+1$, $a+3b+4$, $2a+3b+5$, $a+b+2$, and $a+2b+3$. Their values at $a=b=0$ multiply to $1\cdot 1\cdot 4\cdot 5\cdot 2\cdot 3=120$. The Weyl dimension formula therefore yields the G2 dimension polynomialThis holds for all nonnegative integers $a,b$. In particular it gives dimensions $7$ and $14$ for $L({\omega }_1)$ and $L({\omega }_2)$, and $27$ for $L(2{\omega }_1)$. Reversing the numbering of the simple roots would interchange $a,b$ in the polynomial, so the short-first convention matters.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincaré-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.

Let ${\alpha }_i^{\vee }=2{\alpha }_i/({\alpha }_i,{\alpha }_i)$ be the coroots. With a chosen set of simple roots, the root lattice and weight lattice areThe fundamental weights ${\omega }_j$ are defined by $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$, and form an integral basis of $P$. We use the Cartan matrix convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$.

For the A2 root system, inversion of its Cartan matrix yields the A2 fundamental weights and weight lattice:Equivalently ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. The root lattice has index three in the weight lattice, since the change-of-basis matrix has determinant three. For a planar realization takeThe weight lattice is triangular. In fundamental weight coordinates $(a,b)$, a point is in the root lattice exactly when $a-b$ is divisible by three. The following sketch marks both bases and the sublattice:

The integers called Dynkin indices here are the Dynkin labels of the highest weight:For finite-dimensional Irreducible Lie algebra representations, they are nonnegative integers. All weight coordinates below are these Dynkin label coordinates, not simple-root coordinates.

A weight-string enumeration algorithm gives the set of weights without their weight multiplicities. Begin with $\Lambda$ and process known weights by increasing height below $\Lambda$. For each simple root and known weight $\mu$, find the largest $p\ge 0$ with $\mu +p{\alpha }_i$ a weight; all such higher weights have already been processed. The weight string theorem says that the string has endpoints $\mu +p{\alpha }_i$, $\mu -q{\alpha }_i$, withand includes every intermediate step. Append $\mu -{\alpha }_i,\dots ,\mu -q{\alpha }_i$ and repeat until no new weights appear. This terminates in finite dimension and supplies all weights; every nonhighest weight can be reached by simple-root lowering. A string can contain contributions from several sl2 summands, so the resulting set does not by itself determine weight multiplicities.

For the explicit calculation we use the following general facts: finite-dimensional representations of a complex semisimple Lie algebra are completely reducible by the Weyl complete reducibility theorem; the symmetric powers of the defining sln representation are irreducible of highest weight $m{\omega }_1$; and weights in a tensor product of Lie algebra representations add with multiplicities multiplied. We also use the highest-weight representation classification: each finite-dimensional irreducible has a unique dominant integral highest weight, and a nonzero vector killed by all simple-root raising operators supplies an irreducible summand with that highest weight in a completely reducible module. For $A_2$, the Weyl dimension formula specializes toFor $A_2\cong {\mathfrak{sl}}_3(\mathbb{C})$, the defining fundamental representation has weights $(1,0),(-1,1),(0,-1)$. Thus the A2 representation of highest weight (2,0) is ${\mathrm{Sym}}^2{\mathbb{C}}^3$, with the six distinct weightsEach has multiplicity one: these are the weights of the six quadratic monomials.

For the tensor square of the A2 representation of highest weight (2,0), add every ordered pair of elements of $W_6$. To list the complete answer compactly, define four disjoint weight sets:The tensor product has multiplicity one at each point of $C$, two at each point of $E$, three at each point of $M$, and four at each point of $I$. These fifteen distinct weights account for $3+12+9+12=36$ states.

We next identify the irreducible summands rather than just their dimensions. Split the tensor square of the six-dimensional space into its symmetric power and exterior power, of dimensions $21$ and $15$. The square of a highest-weight vector in the symmetric part has highest weight $(4,0)$, giving $R(4,0)$ of dimension $15$.

Let $v_a,v_b$ be vectors of weights $a=(2,0)$ and $b=(0,1)=a-{\alpha }_1$. In the exterior part $v_a\wedge v_b$ has weight $(2,1)$ and is killed by both simple-root raising operators: raising $v_b$ along ${\alpha }_1$ gives a multiple of $v_a$, whose wedge with itself is zero, and the other raising actions vanish. Thus it is a highest-weight vector. The Weyl dimension formula gives $\dim R(2,1)=15$, so the entire exterior part is this irreducible summand.

The remaining symmetric part has dimension six. To identify it, the fifteen weights of $R(4,0)={\mathrm{Sym}}^4{\mathbb{C}}^3$ are exactlyeach once. Subtract them from the unordered-pair weights of ${\mathrm{Sym}}^{2}R(2,0)$. The residual weights are $M\cup I$, each once, with highest weight $(0,2)$. They are the negatives of $W_6$, so this six-dimensional summand is $R(0,2)$. ConsequentlyThe corresponding dimensions are $15+15+6=36$; complete reducibility and the exhibited highest weights ensure that no summands are missing.

The full weight multiplicities in the A2 tensor square of highest weight (2,0) are summarized below. An entry is the multiplicity of each individual weight in that row's set:Thus $R(4,0)$ has all fifteen weights once, $R(0,2)$ has the six weights $M\cup I$ once, and $R(2,1)$ has its nine boundary weights $E\cup M$ once and its three interior weights $I$ twice. Only the three interior weights of $R(2,1)$ are degenerate among the irreducible components.

The PDF's $J$ is the anti-diagonal identity matrix. Write $\bar{i}=2n+2-i$ and $o=n+1$. Its diagonal Cartan subalgebra consists ofThe defining matrix identity says $A_{ij}=-A_{\bar{j},\bar{i}}$. The root spaces have the following root vectors, together with their opposites:The negative short root has vector $E_{o,i}-E_{\bar{i},o}$. Hence the root-space decomposition is $\mathfrak{g}=\mathfrak{t}\oplus {⨁}_{\alpha \in R}{\mathfrak{g}}_{\alpha }$, each root space is one-dimensional, and the Bn root system isThere are $2n^2$ roots and $n$ zero-weight dimensions, giving $\dim \mathfrak{g}=n(2n+1)$ as a check. Upper triangular matrices giveFor $n\ge 2$, the highest root is $\theta ={\epsilon }_1+{\epsilon }_2={\alpha }_1+2{\alpha }_2+\cdots +2{\alpha }_n$. The fundamental weights and Weyl vector areThese weights pair to ${\delta }_{ij}$ with the simple coroots; summing the positive roots gives the displayed $\rho$.

The Dynkin diagram is the chain $1,\dots ,n$, with single edges except a double edge between $n-1$ and $n$, whose arrow points to the short root ${\alpha }_n$. In the Extended Dynkin diagram, ${\alpha }_0=-\theta$ attaches by a single edge to ${\alpha }_2$ when $n\ge 3$. For $n=2$, both long nodes ${\alpha }_0,{\alpha }_1$ have a double edge to the short node ${\alpha }_2$. The Bn Dynkin diagram and affine extension below distinguishes these small-rank cases.

For $n=1$, the system is $A_1$: $R^{+}=\{{\epsilon }_1\}$, $\theta ={\epsilon }_1$, ${\omega }_1=\rho ={\epsilon }_1/2$. Its ordinary diagram is one node; its affine diagram has two equal-length nodes with the usual affine $A_1$ double bond.

The G2 dimension polynomial gives dimension seven at the short-root fundamental weight, soThe crystal of the seven-dimensional G2 representation is a chain whose weights, in order, areIts successive arrow colors are $1,2,1,1,2,1$, as drawn above. Each arrow subtracts its indicated simple root; in particular the central three vertices form a length-two string of color one. The six nonzero weights are exactly the six short roots, each with multiplicity one, and the zero weight also has multiplicity one.

Number the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, normalized by $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=6$ and $({\alpha }_1,{\alpha }_2)=-3$. Its positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,2{\alpha }_1+{\alpha }_2,3{\alpha }_1+{\alpha }_2,3{\alpha }_1+2{\alpha }_2$, and the negatives complete the twelve-root system. The short roots form one hexagon, and the long roots form a rotated hexagon whose radius is $\sqrt{3}$ times larger. The picture also supplies the crystal requested in the next part.

The fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$, ${\omega }_2=3{\alpha }_1+2{\alpha }_2$, and $\rho ={\omega }_1+{\omega }_2$. The six positive coroots, in coordinates relative to ${\alpha }_1^{\vee },{\alpha }_2^{\vee }$, are $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ therefore gives the G2 dimension polynomialThe denominator is the product $1\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5$ of the corresponding pairings with $\rho$. As checks, the weights ${\omega }_1,{\omega }_2,2{\omega }_1$ give dimensions $7,14,27$ respectively.

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.

Use the coroot normalization ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, with the pairing understood via the root system and its invariant inner product. A dominant integral weight, called a dominant weight here, is a functional $\lambda \in {\mathfrak{t}}^{\ast }$ such thatEquivalently, $\lambda ={\sum }_i{n}_i{\omega }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, where the fundamental weights satisfy $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. If dominance and integrality are defined separately, dominance means the inequalities and membership in the weight lattice means integrality.

A highest-weight vector is a nonzero weight vector $v\in V_{\lambda }$, so $h\cdot v=\lambda (h)v$ for all $h\in \mathfrak{t}$, annihilated by every positive root space:The last condition is equivalently annihilation by the raising operators for the simple roots.

For a dominant integral weight $\lambda$, let $V(\lambda )$ be the finite-dimensional irreducible highest-weight representation with highest weight $\lambda$. Set $\rho =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha$, the half-sum of positive roots. With ${\alpha }^{\vee }$ the coroot of $\alpha$, the Weyl dimension formula isHere ${\Phi }^{+}$ is the positive system of a root system specified by the chosen root basis. The fundamental weights are the basis dual to the simple coroots: $⟨{\omega }_i,{\alpha }_j^{\vee }⟩={\delta }_{ij}$. Thus $\lambda ={\sum }_i{a}_i{\omega }_i$ with nonnegative integers $a_i$. The denominators are positive; equivalently one can use $(\lambda +\rho ,\alpha )/(\rho ,\alpha )$ in each factor, since the coroot normalization cancels.

Use the C2 root system realization in an orthonormal basis ${\epsilon }_1,{\epsilon }_2$ with simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$. This numbers the short root first, as required. The fundamental weights and half-sum of positive roots areso $\lambda +\rho =(a+b+2){\epsilon }_1+(b+1){\epsilon }_2$. For the four positive roots, the factors of the Weyl dimension formula areMultiplying yields the Weyl dimension formula for C2:As checks, $V({\omega }_1)$ has dimension $4$, $V({\omega }_2)$ dimension $5$, and $V(2{\omega }_1)$ dimension $10$.

Write the simple roots of the G2 root system as $\alpha$ short and $\beta$ long. The Long-root A2 subsystem of G2 has simple rootsIf ${\omega }_1,{\omega }_2$ are its fundamental weights, then the $A_2$ root-weight relations giveFor the $A_2$-dominant weight $\lambda =a{\omega }_1+b{\omega }_2$, therefore,The dominant weight inequalities for $G_2$ are thus equivalent tobecause $A_2$-dominance already gives $a,b\ge 0$.