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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.

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

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

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

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

Let $V={\mathbb{C}}^6$ be the defining fundamental representation of $SU(6)$. Its weights are $w_i=e_i-\frac{1}{6}{\sum }_{j=1}^6{e}_j$. Pairing with the simple coroots gives Dynkin labels $(w_i{)}_a={\delta }_{ia}-{\delta }_{i,a+1}$. Thus the complete list isSuccessive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation $\overline{6}\cong V^{\ast }$ is the dual representation and has the negatives of these weights. Its highest weight is $-w_6=(0,0,0,0,1)$, and its entire weight decomposition isEvery entry has weight multiplicity one.

The two-index antisymmetric representation is ${\Lambda }^{2}V=15$, of dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ and highest weight $w_1+w_2=(0,1,0,0,0)$. A weight basis is $e_i\wedge e_j$ for $i<j$, carrying weight $w_i+w_j$. The fifteen sums, all distinct and all of weight multiplicity one, areThe exclusion of $i=j$ comes from the exterior square, where $e_i\wedge e_i=0$.

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.

Put $\bar{i}=2n+2-i$ and $o=n+1$. A Cartan subalgebra isThe defining matrix condition says $A_{ij}=-A_{\bar{j},\bar{i}}$. Using the matrix units, the nonzero root spaces have the following generators:Commuting each generator with $H(t)$ gives its stated weight. Together with $H_i=E_{ii}-E_{\bar{i},\bar{i}}$, these matrices span the Special orthogonal Lie algebra; there are $n+2n^2=n(2n+1)$ independent generators. Thus its decomposition as a Cartan subalgebra module isThis is the Bn root system. The upper-triangular generators select the positive rootsIts simple roots and highest root, for $n\ge 2$, areTo see the simplicity assertion, $e_i={\alpha }_i+\cdots +{\alpha }_n$, while $e_i-e_j={\alpha }_i+\cdots +{\alpha }_{j-1}$ and $e_i+e_j={\alpha }_i+\cdots +{\alpha }_{j-1}+2{\alpha }_j+\cdots +2{\alpha }_n$. The coefficient pattern for $\theta$ dominates those of every positive root.

Normalize the inner product so the $e_i$ are orthonormal. The simple coroots are ${\alpha }_i^{\vee }={\alpha }_i$ for $i<n$ and ${\alpha }_n^{\vee }=2e_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the half-sum of positive roots, the coordinate $e_i$ occurs with total coefficient $2(n-i)+1$. Hence the Weyl vector is

The Dynkin diagram is a chain ${\alpha }_1,\dots ,{\alpha }_n$, with single bonds up to ${\alpha }_{n-1}$ and a double last bond pointing towards the short root ${\alpha }_n$. In the Extended Dynkin diagram, add ${\alpha }_0=-\theta$. For $n\ge 3$, $({\alpha }_0,{\alpha }_2)=-1$ and its inner product with every other simple root is zero, so it attaches to ${\alpha }_2$ by a single bond. For $n=2$, both long nodes $0,1$ attach by double bonds to short node $2$. These rules give the Bn Dynkin diagram and affine extension drawn below.

For rank one, $R=\{\pm e_1\}$, ${\alpha }_1=\theta =e_1$, ${\omega }_1=e_1/2$, and $\rho =e_1/2$. The finite diagram is one node; its affine extension has two equal-length nodes joined by a double bond, without a short-root arrow.

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.

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 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.