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In the finite-dimensional complex semisimple Lie algebra setting implicit in this root-system construction, the given Cartan subalgebra and root spaces determine a reduced crystallographic root system $R$. The ordering divides $R$ into positive roots and their negatives. The simple roots are the positive roots that cannot be written as a sum of two positive roots; a standard root-system fact is that they form a basis and that every positive root is a nonnegative integral combination of them.

Index the simple roots as ${\alpha }_1,\dots ,{\alpha }_{\ell }$ and form the Cartan matrixFor $i\ne j$, these entries are nonpositive integers and $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Construct one vertex per simple root and join distinct vertices by $A_{ij}{A}_{ji}$ lines. If root lengths differ, direct the arrow toward the shorter simple root. This labelled and oriented graph is the Dynkin diagram. Its unoriented Coxeter graph is admissible: normalizing the simple roots to unit length gives $4(v_i,v_j{)}^2=A_{ij}{A}_{ji}$. A general Lie algebra without the semisimple root-system hypotheses need not produce such a diagram.

Let ${\alpha }_1,{\alpha }_2$ be distinct simple roots. Their difference is not a root, since its simple-root coefficients have opposite signs. The root string through ${\alpha }_2$ in the direction ${\alpha }_1$ therefore starts at ${\alpha }_2$ and has the formHere we use the root-string theorem: if the full string runs from $\beta -p\alpha$ through $\beta +q\alpha$, then $p-q=⟨\beta ,{\alpha }^{\vee }⟩$; in this case $p=0$. Thus ${\alpha }_1+{\alpha }_2$ is a root exactly when $A_{12}<0$, equivalently when the corresponding Dynkin diagram nodes are connected.

The root-space bracket facts are $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, with a zero space when the sum is not a root, and $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]={\mathfrak{g}}_{\alpha +\beta }\ne 0$ when the sum is a root and $\beta \ne -\alpha$. Applying these stated structural facts proves

If the two names denote the same simple root, its one-dimensional root space has zero bracket with itself, and a Dynkin diagram has no self-edge. Thus the equivalence also holds in that case.

If the three simple roots are distinct and every pair of root spaces had nonzero Lie bracket, part (iv) would put all three connections in the Dynkin diagram. Their underlying Coxeter graph would contain a triangle, contradicting part (ii). Hence at least one pair satisfies $\boxed{[{\mathfrak{g}}_{{\alpha }_i},{\mathfrak{g}}_{{\alpha }_j}]=0}$. If two of the listed roots coincide, their common root space is one-dimensional, so its bracket with itself is already zero. Thus the conclusion also covers that case.

Identify the Cartan subalgebra with the coordinate space $(t_1,\dots ,t_n)$ used above. The root reflection in any root of a root system fixes the perpendicular hyperplane. For every root of the Cn root system, its complete coordinate action isEvery unlisted coordinate is fixed, and changing the sign of the root leaves its reflection unchanged. These formulas follow directly from $s_{\alpha }(t)=t-\alpha (t){\alpha }^{\vee }$, with the coroot regarded as a vector of the Cartan subalgebra.

The Cn Weyl group is the signed symmetric group: its elements permute the $n$ coordinates and independently change their signs. The Weyl reflections associated with the simple roots interchange consecutive coordinates for $i<n$ and negate the last coordinate for $i=n$. Thus

Write $\bar{i}=2n+1-i$, and let $E_{ab}$ denote a matrix unit. The diagonal Cartan subalgebra consists ofFor a diagonal matrix, $[t,E_{ab}]=(t_a-t_b)E_{ab}$. Substituting into the symplectic relation shows that the following are root vectors:The zero weight space is spanned by $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. The symplectic equation pairs the remaining entries exactly as in this table; the displayed vectors and the $H_i$ are independent and span the algebra. This proves the root-space decompositionThere are $2n^2$ roots of a root system, each with one-dimensional root space, giving $\dim \mathfrak{g}=n(2n+1)$.

The upper triangular root vectors select preciselyThe simple roots of this Cn root system areIndeed, the positive roots of a root system have expansionsThese expansions also identify the highest root:Normalize the coordinate inner product by $({\epsilon }_i,{\epsilon }_j)={\delta }_{ij}$. The simple coroots are ${\alpha }_i^{\vee }={\epsilon }_i-{\epsilon }_{i+1}$ for $i<n$ and ${\alpha }_n^{\vee }={\epsilon }_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the sum of the positive roots of a root system, the pair $({\epsilon }_i-{\epsilon }_j)+({\epsilon }_i+{\epsilon }_j)$ contributes $2{\epsilon }_i$, and the long root contributes another $2{\epsilon }_i$. Therefore the Weyl vector isThese are the Positive-root data for Cn.

For $n\ge 2$, the finite Dynkin diagram is a chain of $n$ vertices. The bonds between ${\alpha }_i$ and ${\alpha }_{i+1}$ for $i<n-1$ are single. The last bond is double, with its arrow pointing from the long root ${\alpha }_n$ toward the short root ${\alpha }_{n-1}$. For the Extended Dynkin diagram, add ${\alpha }_0=-\theta =-2{\epsilon }_1$ at the other end, doubly bonded to ${\alpha }_1$, with arrow toward ${\alpha }_1$. The following Cn Dynkin diagrams include both ends and the low-rank cases.

At $n=1$, there is just the root ${\alpha }_1=2{\epsilon }_1$ and one finite vertex. The affine Cartan matrix is $\left(\begin{array}{cc}2& -2\\ -2& 2\end{array}\right)$; its two vertices have equal root length, so the double bond has no arrow.

Use the ordered alphabetwith weights $\mathrm{wt}(i)={\epsilon }_i$ and $\mathrm{wt}(\bar{i})=-{\epsilon }_i$. The crystal of the defining symplectic representation is the chainThe label of an arrow is its simple root color; a Kashiwara operator ${\tilde{f}}_i$ follows an arrow of color $i$ and lowers the weight by ${\alpha }_i$.

Here is a grid description of the entire tensor product of crystals, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:Their indicators are ${\phi }_i$ and ${\epsilon }_i$. Use the crystal tensor-product ruleAn arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.

For completeness, highest vertices can be read from the raising rule, which selects the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$. If $a$ has an incoming edge of color $i$, then ${\phi }_i(a)=0$; either the raising operator acts nontrivially on $a$, or ${\epsilon }_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all ${\phi }_i(a)$ vanish except ${\phi }_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge 2$, the possibilities are exactlyTheir weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$, and zero. Each occurs once. Connected highest-weight components of the crystal basis describe the irreducible representations in the tensor product of Lie algebra representations, soThis is the tensor-square decomposition of the defining symplectic representation. The symmetric square is $L(2{\omega }_1)$; the exterior square is the direct sum of its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$. Their dimensions are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim (V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes 1$ and $1\otimes \bar{1}$: the decomposition is $L(2{\omega }_1)\oplus L(0)$, of dimensions three and one.

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.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra, choose a Cartan subalgebra and positive root system $R^{+}$, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ be the finite-dimensional highest-weight representation with highest weight $\lambda$. The Weyl dimension formula isHere ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$ is the coroot, the pairing is between a weight and a coroot, and $\rho$ is the Weyl vector. The inner product can be any Weyl group invariant normalization on each simple factor: its scaling cancels from the ratios. The dominance condition means that $⟨\lambda ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer for each simple root.

Choose a Cartan subalgebra, simple roots ${\alpha }_1,\dots ,{\alpha }_r$, and root vectors $e_i,f_i$ normalized by $[e_i,f_i]=h_i={\alpha }_i^{\vee }$. LetHere the coroots lie in the Cartan subalgebra, and ${\alpha }_i({\rho }^{\vee })=1$. This last identity is the Weyl vector identity applied to the dual root system. For distinct simple roots, ${\alpha }_i-{\alpha }_j$ is not a root, hence $[e_i,f_j]=0$. ConsequentlyThese are exactly the sl2 Lie algebra relations. For nonzero $\mathfrak{g}$ the resulting homomorphism from ${\mathfrak{sl}}_2$ is nonzero and therefore injective, since ${\mathfrak{sl}}_2$ is simple. Its image is a Principal sl2 subalgebra. Equivalently its raising element is a principal nilpotent element, a nilpotent element with Lie algebra centralizer of dimension equal to the rank. The construction works for a semisimple Lie algebra with several simple factors by taking the triple in every factor, giving one diagonally embedded ${\mathfrak{sl}}_2$. This is the construction of a principal sl2 triple.

For a representation $V$, restrict to this subalgebra and take its formal character. If $V_{\mu }$ is the weight space for the original Cartan subalgebra, the resulting q-character of a highest-weight representation, or the analogous q-character for any finite-dimensional representation, isIt is independent of conjugating the chosen principal triple. In particular a root $\beta ={\sum }_i{k}_i{\alpha }_i$ has grading $\beta (h)=2{\sum }_i{k}_i$. Some conventions use $q^{h/2}$ instead; that convention divides all exponents here by two. The normalization above agrees with the ordinary sl2 Lie algebra character in the preceding part.

Write an element of the symplectic Lie algebra in $n\times n$ blocks. Direct multiplication in $AJ+J{A}^T=0$ givesThe diagonal Cartan subalgebra consists of $H=\mathrm{diag}(h_1,\dots ,h_n,-h_1,\dots ,-h_n)$. Define ${\epsilon }_i(H)=h_i$. Under the Adjoint representation, the off-diagonal matrix units in $P$ have weights ${\epsilon }_i-{\epsilon }_j$; the symmetric entries in $Q$ have weights ${\epsilon }_i+{\epsilon }_j$, including $2{\epsilon }_i$ on the diagonal; those in $R$ have the negatives of these weights. All nonzero root spaces are one-dimensional, while the zero weight space is the diagonal Cartan subalgebra. Thus the Cn root system isChoose the positive roots and simple roots asTheir highest root and Weyl vector areFor the last identity, each pair $i<j$ contributes $2{\epsilon }_i$ before division by two, and the long positive root contributes another ${\epsilon }_i$ afterward. Successive simple roots have equal length except for ${\alpha }_n$, whose squared length is twice that of ${\alpha }_{n-1}$. The Cn Dynkin diagram is a chain with a double last bond, whose arrow points toward ${\alpha }_{n-1}$, the short root:

The displayed schematic is for $n\ge 4$; omit intermediate nodes for $n=2,3$. For $n=1$ the diagram has just the single node ${\alpha }_1=2{\epsilon }_1$.

Here root systems are finite, reduced and crystallographic, as for complex semisimple Lie algebras. For a base of a root system $\Delta =\{{\alpha }_1,\dots ,{\alpha }_r\}$, the Dynkin diagram has one vertex for each simple root. Two distinct vertices have $a_{ij}{a}_{ji}$ bonds, with the arrow on a multiple bond pointing toward the shorter root. The classification of finite crystallographic Dynkin diagrams consists of the following connected diagrams; the integers describing arms count vertices away from the unique branching vertex.

This is classification up to diagram isomorphism and overall rescaling of the root system. It includes the conventional identifications $D_3\cong A_3$ and $B_1\cong A_1$ when those low-rank notations are used. The weighting here encodes root lengths and bonds; it is not the separate theory of vertex labels attached to nilpotent orbits. Dropping the crystallographic or reduced hypothesis changes the classification.

Precisely, a base of a root system is a vector-space basis $\Delta \subset \Phi$ such that each root has a unique expansion $\beta ={\sum }_i{n}_i{\alpha }_i$, with integer coefficients all nonnegative or all nonpositive. This defines the positive and negative roots. We use the column-coroot convention for the Cartan matrix:Some conventions transpose this matrix; specifying which index labels the coroot resolves that difference. The Weyl group is the subgroup of orthogonal transformations generated by the Weyl reflections

For the C3 root system, take an orthonormal basis $e_1,e_2,e_3$ of ${\mathbb{R}}^3$ andThere are six long roots of squared length $4$ and twelve short roots of squared length $2$. A base is ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=2e_3$. The Cartan matrix convention for C3 above givesThus the double bond connects ${\alpha }_2$ to ${\alpha }_3$, with its arrow directed toward ${\alpha }_2$.

An explicit complex Lie algebra realizing this root system is the symplectic Lie algebraIt is closed under commutators, since the two identities $X^{T}J=-JX$, $Y^{T}J=-JY$ imply $[X,Y{]}^{T}J=-J[X,Y]$. Equivalently,so its dimension is $9+6+6=21$. A Cartan subalgebra isWith $E_{ab}$ denoting a matrix unit, the root spaces have generators $E_{ij}-E_{3+j,3+i}$ for $e_i-e_j$ ($i\ne j$), $E_{i,3+j}+E_{j,3+i}$ for $e_i+e_j$ ($i<j$), and $E_{i,3+i}$ for $2e_i$; the analogous lower-left generators give the negative roots. These are all eighteen root spaces, each one-dimensional, and together with $H$ they exhaust the algebra.

For an explicit semisimplicity check, let $I\ne 0$ be an ideal. Simultaneous diagonalization of $\mathrm{ad}H$ makes $I$ a sum of its intersections with $H$ and these root spaces. If it contains a nonzero $h\in H$, some root has $\alpha (h)\ne 0$, and $[h,L_{\alpha }]$ supplies a root vector in $I$. Thus $I$ always contains a root vector. Bracket it with its opposite root vector to obtain a nonzero coroot $h_{\alpha }\in I$. Whenever $(\alpha ,\beta )\ne 0$, bracketing $h_{\alpha }$ with either $L_{\beta }$ or $L_{-\beta }$ puts both root spaces into $I$. The graph of the roots with edges for nonzero inner products is connected: $2e_i$ connects to $e_i+e_j$, which connects to $2e_j$, and every short root connects to an axis root. Repeating the argument yields all root spaces and then all of $H$ from their brackets. Hence $I=L$: the constructed algebra is simple, and therefore semisimple.

The reflections in $e_1-e_2$ and $e_2-e_3$ exchange adjacent coordinates, while reflection in $2e_3$ changes the sign of the third coordinate. They generate every signed permutation. Conversely reflection in any displayed root is a signed permutation. ThusThe reflecting hyperplanes are $x_i=0$ and $x_i=\pm x_j$. A fundamental open Weyl chamber is $x_1>x_2>x_3>0$. Every regular point has three nonzero coordinates of distinct absolute values; its signs and the ordering of those absolute values specify exactly one of the $2^3\cdot 3!=48$ chambers. The closure of this chamber is generated by the rays through $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$. This is the Weyl chamber geometry of C3.

The left panel retains the actual long and short root lengths. The right panel radially projects a hemisphere of the unit sphere, draws every visible reflecting great-circle arc, and highlights the chamber $x_1>x_2>x_3>0$. Each spherical triangle is the section of a three-dimensional chamber by the sphere; opposite triangles on the rear hemisphere supply the remaining chambers. Roots of a root system are normal to the chamber walls, rather than generally lying along their edges.

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 $a,b$.

Here a reduced root system is crystallographic, as appropriate to a semisimple Lie algebra. It is a finite spanning set $\Phi \subset E\setminus \{0\}$ in a real inner product space of dimension $r$, invariant under each Weyl reflectionwith $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for every pair of roots and $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$. A base of a root system $\Delta =({\alpha }_1,\dots ,{\alpha }_r)$ is a basis such that the coefficients of every root are integers that are either all nonnegative or all nonpositive. These define the positive system of a root system. We use the existing Cartan matrix conventionUsing the alternative denominator $({\alpha }_i,{\alpha }_i)$ transposes all the matrices below.

The classification of rank-two root systems gives $A_1\times A_1$, $A_2$, $B_2=C_2$, and $G_2$. In orthonormal coordinates we may choose the following simple roots and Cartan matrices:

Each complete root system consists of these positive roots and their negatives. To see why the list is exhaustive, the off-diagonal Cartan matrix entries of two distinct simple roots are nonpositive integers, and their product is $4{\cos }^2\theta <4$. Thus the product is $0,1,2$ or $3$, corresponding to simple-root angles $\pi /2,2\pi /3,3\pi /4$ or $5\pi /6$. These determine the four systems and their length ratios.

The Weyl group of a rank-two root system is generated by the two simple Weyl reflections. Their product is a rotation of order $m=2,3,4,6$, respectively, so the groups are the dihedral groups of order $2m=4,6,8,12$. The first is also the Klein four-group and the second the symmetric group $S_3$. The root-orthogonal lines cut the plane into $2m$ Weyl chambers, each of angle $\pi /m$. A chosen fundamental chamber of a root system is $(x,{\alpha }_1)>0$, $(x,{\alpha }_2)>0$; its walls are the two root-orthogonal lines. The Weyl group acts simply transitively on these open chambers.

The crystallographic hypothesis matters: if the integer-pairing condition is dropped, reduced noncrystallographic systems of type $I_2(m)$ also occur. They are not additional Lie-algebra root systems.

Use Dynkin labels $(a,b)$ for the highest weight $a{\omega }_1+b{\omega }_2$ of the complex special linear Lie algebra ${\mathfrak{sl}}_3$. The A2 root system has ${\alpha }_1=(2,-1)$ and ${\alpha }_2=(-1,2)$ in these coordinates. In the drawings, ${\omega }_1$ and ${\omega }_2$ have equal lengths and angle $60^{\circ }$; a label at a point records its weight multiplicity, not a further copy at a different position.

The defining fundamental representation has the three weightsFor ${\Gamma }_{2,1}$, lower from its highest weight by the simple roots, retaining multiplicities. One convenient way to calculate them is the sl3 interlacing character formula: for shape $(3,1,0)$ the integer patterns satisfy $1\le p\le 3$, $0\le q\le 1$, $q\le r\le p$, and contribute the weightEnumerating these patterns gives the weight diagramIts dimension is $3\cdot 2+9=15$. The diagram below draws all twelve distinct positions, with the three inner multiplicities equal to two. The extra panel gives the symmetric square used in the calculation.

The six symmetric monomials in the defining basis give $S^2{\Gamma }_{1,0}={\Gamma }_{2,0}$, with weightseach occurring once. Thus the tensor product has dimensionIn a tensor product of Lie algebra representations, weights add and their multiplicities multiply. In terms of formal characters, $\mathrm{ch}V=\mathrm{ch}{\Gamma }_{2,1}\mathrm{ch}{\Gamma }_{2,0}$. Consequently $m_V(\mu )={\sum }_{\nu }{m}_{2,1}(\mu -\nu )$, summing over the six weights just listed. To show the indicated dominant multiplicities explicitly, the contributions in that order areThe tensor-product weight diagram below includes every position, and highlights these dominant weights. It also records the zero-weight multiplicity nine; that multiplicity is not a count of trivial summands.