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First complexify the Clifford algebra in $D=2n$ spacetime dimensions. Multiplying appropriate generators by $i$ converts the signature to the Euclidean form $\{{\Gamma }_a,{\Gamma }_b\}=2{\delta }_{ab}$ without changing the complex module dimension. Pair the generators intoThe Clifford algebra relations giveThe commuting occupation operators $N_j=b_j^{†}{b}_j$ have eigenvalues zero or one. Choose a nonzero state killed by every $b_j$; repeated annihilation reaches such a state in any finite irreducible module. Acting with distinct creation operators produces the $2^n$ occupation statesTheir distinct occupation patterns make them linearly independent, and their span is invariant under every Clifford generator. Irreducibility therefore makes it the entire Dirac module. This derives the even-dimensional Clifford spinor dimension:The chirality matrix is a phase-adjusted product of all $2n$ generators. It squares to one, anticommutes with each generator and acts as occupation parity up to an overall sign. Its two eigenspaces each have $2^{n-1}$ states. The even Clifford algebra, and hence the connected spin group, preserves them, giving irreducible Weyl spinors of complex dimension $2^{D/2-1}$. A Majorana spinor imposes a reality condition only in signatures where it exists; simultaneous Majorana and Weyl conditions are not available in every even dimension.

In Lorentzian ten dimensions, the complex Dirac module has dimension $32$, and a Weyl module dimension $16$. A Majorana-Weyl spinor is allowed and has $16$ real components before its equation of motion. The massless Dirac equation leaves eight physical gaugino polarizations. A ten-dimensional massless gauge field also has $10-2=8$ physical polarizations. Thus ten-dimensional super Yang-Mills theory has eight bosonic and eight fermionic states per gauge generator, with sixteen real supercharges.

For dimensional reduction, take a flat six-torus with periodic fermions and retain fields independent of its coordinates. Split $A_M$ into $A_{\mu }$ and six internal components $A_m$, $m=4,\dots ,9$. The latter are four-dimensional real adjoint scalar fields. Under the spin groups of the two factors, a ten-dimensional chiral spinor decomposes asfor $\mathrm{Spin}(1,3)\times \mathrm{Spin}(6)$, with $\mathrm{Spin}(6)\simeq SU(4)$. The Majorana condition relates these conjugate pieces, leaving four independent four-dimensional Weyl spinors. The supersymmetry parameters reduce in exactly the same fashion: sixteen real charges are four four-dimensional supersymmetries. The resulting field content isThe on-shell counts are $2+6=8$ bosonic and $4\times 2=8$ fermionic states, as required. The internal spin group supplies the $SU(4)$ R-symmetry of the reduced action.

To derive its scalar potential, use Hermitian gauge matrices and a positive trace inner product, withIf the internal volume is $V_6$, integration gives $g_4^2=g_{10}^2/V_6$. Since internal derivatives vanish,With mostly-minus spacetime signature, $F_{\mu m}{F}^{\mu m}$ has an extra minus sign and the double sum over mixed indices gives a factor two. Rescale $A_{\mu }=g_4{a}_{\mu }$ and $A_m=g_4{X}_m$ to canonical fields. The resulting bosonic Lagrangian iswhere $D_{\mu }{X}_m={\partial }_{\mu }{X}_m-i{g}_4[a_{\mu },X_m]$. A commutator of Hermitian matrices is anti-Hermitian, so the last term is minus a nonnegative scalar potential:The sum is over ordered internal pairs; a sum over $m<n$ would instead have coefficient $g_4^2/2$. Trace normalization merely changes the matched gauge-coupling convention.

The origin attains zero potential, so the global minima are exactlyFor a compact connected gauge group, a commuting tuple is simultaneously conjugate into a Cartan subalgebra. Write $X_m={\sum }_{a=1}^r{x}_m^a{H}_a$, where $r$ is the rank. Every choice of the $6r$ real numbers $x_m^a$ has zero potential. After the residual Weyl group identifications, the classical commuting-scalar vacua of four-dimensional N=4 Yang-Mills theory form $({\mathbb{R}}^6{)}^r/\mathcal{W}$ in the four-dimensional theory. For $SU(N)$ one may describe them as six simultaneous diagonal matrices, with tracelessness and permutations of their common eigenvalue labels. Varying these eigenvalues changes gauge-invariant configurations while leaving $V=0$, proving the existence of genuine flat directions of a scalar potential, not just gauge-equivalent directions. Generic values break the gauge group to its maximal torus; coincident eigenvalue vectors produce enhanced unbroken gauge symmetry. If one keeps a finite compact torus rather than the four-dimensional limit, large gauge transformations add periodic identifications of the Wilson-line coordinates.

Identify the Cartan subalgebra with its coordinate vector $t=(t_1,\dots ,t_n)$ using the invariant inner product. A root reflection isFor every root of the Bn root system, the explicit formulas areAll unlisted coordinates stay fixed, and opposite roots define the same reflection. These reflections give all coordinate permutations and independent sign changes. Therefore the Weyl group isIts action on $\mathfrak{t}$ is the group of signed permutation matrices.

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.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra, choose a Cartan subalgebra and a positive root system $R^{+}$, and let $\lambda$ be a dominant integral highest weight. Write $L(\lambda )$ for the corresponding finite-dimensional irreducible representation, ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$ for a coroot, and $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$ for the Weyl vector. The Weyl dimension formula isThe bracket denotes the natural pairing of weights with coroots. The normalization of an invariant inner product cancels from each ratio, so the formula can equivalently use roots instead of coroots in both numerator and denominator.

A highest-weight vector is a nonzero weight vector $v$ such that ${\mathfrak{g}}_{\alpha }v=0$ for every positive root $\alpha$. Its weight of a representation is a highest weight. For an irreducible finite-dimensional Lie algebra representation, this is the unique highest weight of a representation, and $v$ generates the module; a reducible module can have several highest weights of different summands.

Here is an existence argument for every nonzero finite-dimensional module $V$. For each simple root, its root operators and coroot form an sl2 subalgebra associated with a root. The classification of finite-dimensional sl2 representations shows that this coroot acts diagonalizably with integral eigenvalues. These commuting coroots span the Cartan subalgebra, so simultaneous diagonalization gives a finite weight-space decompositionChoose a real linear functional $\ell$ on the weight lattice span that is positive on every simple root, hence on every positive root, and choose a weight $\lambda$ maximizing $\ell$ among the finitely many weights of $V$. If $v\in V_{\lambda }$ and $X_{\alpha }\in {\mathfrak{g}}_{\alpha }$, thenso $X_{\alpha }v$ belongs to $V_{\lambda +\alpha }$ if nonzero. For positive $\alpha$, that would contradict maximality because $\ell (\lambda +\alpha )>\ell (\lambda )$. Thus every nonzero $v\in V_{\lambda }$ is a highest-weight vector. Every nonzero finite-dimensional module therefore has a highest weight. The zero module has no nonzero vector or weight, so it is the vacuous exception to the literal unrestricted wording.

Work with a complex semisimple Lie algebra and its root system $R$ in the real span $E$ of the roots of a root system. The Killing form induces a positive definite inner product on this real root space, after the usual identification with the dual of the real span of the coroots. For each root $\alpha$, its Weyl reflection isThe Weyl group is $\boxed{W=⟨s_{\alpha }:\alpha \in R⟩}$. The structural facts being used are that these reflections preserve the finite root system, that the roots span $E$, and that the reflections in the simple roots already generate $W$. Thus $W$ is finite: it acts faithfully as permutations of $R$. Equivalently, it is the normalizer of the Cartan subalgebra modulo its centralizer in the associated adjoint group.

For the defining representation, the standard coordinate vectors $e_1,e_2,e_3,e_4,e_5$ are weight vectors of weights ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_1,-{\epsilon }_2,0$, respectively. Thus the weight diagram has four axial points and the origin, each of multiplicity one.

For the Adjoint representation, the eight nonzero weights of a representation are the roots from part (ii), each of multiplicity one, with the displayed $X_{\alpha }$ as basis. The zero weight space is the Cartan subalgebra, with basisThe Adjoint representation diagram therefore has the four axial points, four diagonal points, and the origin of multiplicity two. The multiplicities sum to ten, as they must.

The diagonal Cartan subalgebra isPut ${\epsilon }_1=L_1$ and ${\epsilon }_2=L_2$; then $L_3=-{\epsilon }_1$, $L_4=-{\epsilon }_2$ and $L_5=0$ on $\mathfrak{h}$. For the matrix unit $E_{ij}$, $[H,E_{ij}]=(L_i-L_j)(H)E_{ij}$. Applying this to the general matrix in part (i) gives the following one-dimensional root spaces and chosen root vectors:Each row satisfies $[H,X_{\alpha }]=\alpha (H)X_{\alpha }$ and belongs to the Special orthogonal Lie algebra. The two diagonal parameters and eight root vectors span the ten-parameter matrix from part (i), giving the explicit root-space decompositionThis is the B2 root system, with short roots on the coordinate axes and long roots on the diagonals.

For a finite-dimensional Lie algebra, the Killing form is the symmetric bilinear formIn the root-space decomposition of this semisimple Lie algebra, $\mathrm{ad}H$ acts as zero on the abelian Cartan subalgebra and as multiplication by $\alpha (H)$ on ${\mathfrak{g}}_{\alpha }$. For $H,K\in \mathfrak{h}$, their product thus acts as multiplication by $\alpha (H)\alpha (K)$ on each root space. Every root space here is one-dimensional. Taking the trace over the two-dimensional Cartan subalgebra and the eight root spaces provesFor a general weight decomposition the trace would include the corresponding weight multiplicities; the root multiplicities are one in the semisimple setting here.

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.

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.

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.

Under the Principal sl2 subalgebra, a root space of root $\beta$ has $h$-eigenvalue $\beta (2{\rho }^{\vee })=2\mathrm{ht}(\beta )$, where the height is the sum of its simple-root coefficients. The six positive roots of a root system of the G2 root system have heightsEach positive and negative root space has dimension one. The two-dimensional Cartan subalgebra contributes two copies of the zero weight. Therefore the Principal q-character of the G2 adjoint representation isEquivalently,The classification of finite-dimensional sl2 representations therefore gives the actual restricted-module decomposition $\mathfrak{g}\cong L_{10}\oplus L_2$, of dimensions $11+3=14$. In the convention using $q^{h/2}$, the same result is $q^5+q^4+q^3+q^2+2q+2+2q^{-1}+q^{-2}+q^{-3}+q^{-4}+q^{-5}$. Both forms encode the same two weight strings.

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.

Use the traceless diagonal Cartan subalgebra of the special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$. Write ${\epsilon }_i(\mathrm{diag}(h_1,\dots ,h_n))=h_i$, so ${\sum }_i{\epsilon }_i=0$. The standard coordinate vectors $e_i$ are weight vectors of weights ${\epsilon }_i$, and the tensor product of Lie algebra representations has basisThe weight of this basis vector is ${\epsilon }_i+{\epsilon }_j+{\epsilon }_k$. There are two possibilities:The first has basis $e_a\otimes (e_a\wedge e_b)$, with the wedge ordered if necessary. In the second, the tensor's first factor can be any one of the three indicated coordinate vectors. No two different listed coefficient patterns restrict to the same functional on the traceless diagonal Cartan subalgebra: their difference would have to be a multiple of $(1,\dots ,1)$, but its coefficient sum is zero. All unlisted weight spaces are zero. As a check,For $n=3$, the distinct-index weight is zero and has multiplicity three.

To determine the irreducible summands, take the upper triangular raising operators. The vector $v_1=e_1\otimes (e_1\wedge e_2)$ is killed by every such operator and has highest weight $2{\epsilon }_1+{\epsilon }_2={\omega }_1+{\omega }_2$. The exterior product mapis a Lie algebra representation homomorphism. It has an equivariant right inverseIndeed $A\iota =I$. The vector $\iota (e_1\wedge e_2\wedge e_3)$ is a highest-weight vector of weight ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3$.

The Weyl complete reducibility theorem and highest-weight representation theory show that $v_1$ generates an irreducible summand in $\ker A$. To check that it exhausts this kernel, use the Weyl dimension formula with the partition $(2,1,0,\dots ,0)$:This equals $\dim \ker A=n\left(\genfrac{}{}{0ex}{}{n}{2}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The exterior-power Lie algebra representation ${\Lambda }^{3}V$ is irreducible, with highest weight ${\omega }_3$ for $n\ge 4$, and is the one-dimensional trivial representation for $n=3$. Therefore the tensor product of the standard representation with its exterior square givesEach summand occurs once; their dimensions are $n(n^2-1)/3$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

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 ${\alpha }_1=2{\epsilon }_1$.

Identify the diagonal Cartan subalgebra with ${\mathbb{C}}^n$ through its coordinates $(h_1,\dots ,h_n)$, and identify real roots with coordinate vectors using the invariant Euclidean inner product. A Weyl reflection is $s_{\alpha }(h)=h-2(h,\alpha )\alpha /(\alpha ,\alpha )$. It does not change when $\alpha$ is replaced by $-\alpha$. For the Cn root system, its explicit actions areEvery coordinate not displayed stays fixed. Thus each generating Weyl reflection is a signed permutation, so the Weyl group is contained in the signed symmetric group. Conversely, the first row supplies each independent sign change and the second supplies every coordinate transposition; these generate the entire signed symmetric group. This proves the Cn Weyl group identification, including its size: