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Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

Use the basis matrices $M_{ab}=E_{ab}-E_{ba}$ from solution. Define the Pauli matrices and the two-by-two identity byThe following five Hermitian four-by-four matrices, written as Kronecker products, realize a Euclidean Clifford algebra:Define the Lie algebra representation on ${\mathbb{C}}^4$ byThese are anti-Hermitian matrices, appropriate to the real compact Lie algebra. For complete explicitness, all ten representation matrices areEach table entry is the image under $\rho$ of the basis matrix immediately to its left. Since the two-by-two factors are specified above, these Kronecker products specify every entry of each four-by-four matrix.

To verify the representation law, the Clifford algebra relations implyApply this identity to $\rho (M_{cd})=\frac{1}{4}[{\gamma }_c,{\gamma }_d]$ and use the commutator product rule. The result isexactly the Lie bracket relation in solution. Thus $\rho$ is a Lie algebra representation.

It remains to identify its highest weight, rather than just its dimension. The commuting Hermitian Cartan subalgebra operators areTheir simultaneous eigenvalues are $(x,y)=(\pm \frac{1}{2},\pm \frac{1}{2})$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four weights of $V(0,1)$ in solution. A state with $(x,y)=(\frac{1}{2},\frac{1}{2})$ has highest weight $(0,1)$, so this Clifford realization of the so5 spin representation is the requested $r$.

Finally, $\rho (M_{12}{)}^2=-I_4/4$, soThis explicitly exhibits the distinction between the spinor representation of $\mathfrak{so}(5)$, or of $\mathrm{Spin}(5)$, and a single-valued group representation of $SO(5)$.

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.

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.

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

Order the simple roots of the G2 root system with ${\alpha }_1$ long and ${\alpha }_2$ short. This ordering is consistent with the supplied projection in solution; reversing the nodes would interchange the two fundamental Dynkin labels. Choose $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=2/3$, and $({\alpha }_1,{\alpha }_2)=-1$. With the same row convention for the Cartan matrix as in solution,The positive roots, with their Dynkin labels, areFor the highest weight $(0,1)$, a lowering chain isThe middle three terms form the length-three weight string for the short simple root. The Weyl dimension formula, with this long-first convention, isIt gives seven for $(0,1)$, so the chain exhausts the weight system, with every weight multiplicity equal to one:For $(1,0)$, observe that $2{\alpha }_1+3{\alpha }_2=(1,0)$ is the highest root. Thus this is the Adjoint representation; the twelve root spaces contribute the twelve nonzero weights once, and the rank-two Cartan subalgebra contributes two zero weights. ThereforeThe same Weyl dimension formula gives fourteen, confirming the count.

The highest root is $\theta =2{\alpha }_1+3{\alpha }_2$. In the Extended Dynkin diagram, the extra node $-\theta$ is long and is joined by a single bond to the long node ${\alpha }_1$; it is orthogonal to the short node ${\alpha }_2$. Deleting the short node therefore leaves the A2 root system Dynkin diagram. Equivalently, the six long roots are a closed root subsystem, so their root spaces together with the Cartan subalgebra form an A2 root system Lie subalgebra. The corresponding compact subgroup is $SU(3)\subset G_2$.

For the particular projection in the paper, choose the A2 root system simple roots as ${\beta }_1={\alpha }_1+3{\alpha }_2$ and ${\beta }_2={\alpha }_1$. Since these are long, ${\beta }_1^{\vee }={\alpha }_1^{\vee }+{\alpha }_2^{\vee }$ and ${\beta }_2^{\vee }={\alpha }_1^{\vee }$. Consequently the Dynkin labels restrict asProjecting the six short roots gives $\pm (1,0),\pm (0,1),\pm (1,-1)$. Group these into the three weights of $3$ and the three of $\overline{3}$ listed in solution. The zero weight of $7$ is left over, soThe three positive long roots project as $(2,-3)\mapsto (-1,2)$, $(-1,3)\mapsto (2,-1)$ and $(1,0)\mapsto (1,1)$. Together with their negatives and the two zero weights, these are exactly the Adjoint representation $8$. The remaining six short roots again give $3\oplus \overline{3}$. Hence the branching of a representation is

Use the mathematical Lie algebra convention $U(t)=I+tX+O(t^2)$. Differentiating $U^{†}U=I$ and $\det U=1$ at $t=0$ gives $X^{†}=-X$ and $\mathrm{tr}X=0$. Conversely, for such an $X$, the matrix exponential is unitary and has determinant $\det e^{tX}=e^{t\mathrm{tr}X}=1$. ThusThe four diagonal entries are purely imaginary, and the six entries above the diagonal are arbitrary complex numbers determining the entries below it. This gives $4+2\cdot 6=16$ real parameters before imposing the one real trace constraint, so ${\dim }_{\mathbb{R}}\mathfrak{su}(4)=15$.

A Cartan subalgebra of this compact Lie algebra is a maximal commuting Lie subalgebra; chooseA simple basis is $i{H}_1,i{H}_2,i{H}_3$, whereAny matrix commuting with every diagonal element is itself diagonal, which proves maximality. Commuting skew-Hermitian matrices can also be simultaneously diagonalized, so no larger commuting Lie subalgebra is possible. ConsequentlyThe Hermitian quantum generator convention instead uses the Hermitian $H_a$ and $U=\exp (i{\theta }^a{H}_a)$. The two conventions describe the same Lie algebra.

A Cartan subalgebra is a nilpotent Lie subalgebra equal to its normalizer of a Lie subalgebra. Write $H=L_{0,x}$ for a choice $x\in A$ attaining the minimum dimension, and let $h=\dim H$. For $y\in A$, the subalgebra $H$ is invariant under $\mathrm{ad}(x+ty)$ because $A\subseteq H$. On $L/H$, the operator induced by $\mathrm{ad}x$ is invertible: $H$ is its full generalized zero eigenspace, and the finite-dimensional Fitting decomposition separates that from the invertible part.

The determinant of the operator induced by $\mathrm{ad}(x+ty)$ on $L/H$ is a polynomial in $t$, nonzero at zero. Thus it is nonzero for all but finitely many $t\in K$. For these values, the generalized zero eigenspace $L_{0,x+ty}$ lies in $H$. Minimality of $\dim H$ among elements from $A$ forces equality. Hence $\mathrm{ad}(x+ty){|}_H$ is nilpotent for all but finitely many $t$.

Each nonleading coefficient of its characteristic polynomial is a polynomial in $t$. Since $K$ is infinite, a polynomial vanishing at all but finitely many elements is identically zero. The coefficient of degree $i$ in the matrix entries is homogeneous of degree $i$; its highest possible $t^i$ coefficient is the corresponding characteristic coefficient of $\mathrm{ad}y{|}_H$. All these vanish, so that operator has characteristic polynomial $T^h$ and is nilpotent by the Cayley-Hamilton theorem. ThusFor existence, choose an $L_{0,x}=H$ of smallest dimension among all $x\in L$. Since $x\in H$, it also minimizes the same set with $x$ restricted to $A=H$. Applying the proved statement makes every $\mathrm{ad}y{|}_H$ nilpotent. Engel's theorem makes $H$ nilpotent, and the allowed self-idealizer property makes it self-normalizing. Therefore it is a Cartan subalgebra, illustrating Cartan subalgebra existence from a minimal Engel subalgebra.

For the three-dimensional example, direct use of the brackets gives $C_L(z)=Kz$ for every nonzero $z$: the equations $[u,z]=0$ are the vanishing of all two-by-two minors of the two coordinate vectors, so they are proportional. Every nonzero nilpotent Lie algebra has a nonzero center, as seen from its last nonzero lower-central term. A nilpotent subalgebra $H$ with a central nonzero element $z$ must therefore satisfy $H\subseteq C_L(z)=Kz$, and has dimension one. The zero subalgebra is not self-normalizing. Thus all Cartan subalgebras have dimension one. For example $Ka$ is nilpotent and its normalizer is $Ka$, so such subalgebras do occur.

Use the standard split weight-space decomposition in characteristic zero, in particular the usual complex setting: $L={⨁}_{\phi }{L}_{\phi }$, with generalized weights for the nilpotent algebra $H$. The characteristic-zero condition prevents nonzero weight shifts from wrapping around; the unrestricted positive-characteristic version would need additional hypotheses.

For fixed $h\in H$, let $D=\mathrm{ad}h$. On a bracket of vectors from $L_{\phi },L_{\psi }$, the derivation identity givesIterating it gives the binomial expansion of these two commuting actions. At power $2n-1$, each term kills at least one factor, because generalized weight vectors are killed by the $n$th power of the corresponding shifted operator. Therefore the bracket lies in the generalized eigenspace for $(\phi +\psi )(h)$ for every $h$. Stabilization of kernels by dimension $n$ giveswhere $L_{\eta }=0$ when $\eta$ is not a weight.

Nilpotence of $H$ gives $H\subseteq L_0$. If $H$ is a Cartan subalgebra and $L_0/H\ne 0$, the action of $H$ on this quotient consists of nilpotent operators. Engel's theorem supplies a nonzero coset $v+H$ killed by $H$, giving $[H,v]\subseteq H$ with $v\notin H$, a contradiction to self-normalization. Conversely, if $L_0=H$ and $v$ normalizes $H$, then $[h,v]\in H$ for each $h$. Applying further powers of $\mathrm{ad}h$ kills this vector because its restriction to $H$ is nilpotent. Hence $v\in L_0=H$. This proves the equivalence.

The operator $\mathrm{ad}e$ sends $L_{\phi }$ into $L_{\phi +\alpha }$. Repetition follows a consecutive $\alpha$-string; after $r$ steps it must leave the weight set, because $r$ is the largest number of weights in any such string. Thus $\boxed{(\mathrm{ad}e{)}^r=0}$ on every weight space and hence on $L$.

Finally put $E=\mathrm{ad}e$ and $F=\mathrm{ad}f$. Expand $(E+F{)}^n$ as the sum of all words of length $n$, without commuting $E$ and $F$. Acting on $L_{\gamma }$, each word follows weights $\gamma +i\alpha +j\beta$ with $i+j$ increasing by one at each step. Linear independence of $\alpha ,\beta$ in characteristic zero makes the successive weights distinct. A nonzero length-$n$ path would require $n+1$ different weight spaces, but $\dim L=n$. Every word therefore vanishes, giving

By irreducible traceless matrix Lie algebra is semisimple, let $R$ be the solvable radical. Lie's theorem triangularizes its action, with finitely many diagonal characters ${\lambda }_i:R\to \mathbb{C}$, and supplies a nonzero common eigenspace for one of them. For $x\in L$, conjugation by $e^{tx}$ preserves $R$ and gives an equivalent representation of it. Thus the multiset of characters ${\lambda }_i\circ e^{t\mathrm{ad}x}$ is constant. Each character varies continuously in $t$ and has values in that finite set of characters, so it is individually constant. Differentiating at zero gives ${\lambda }_i([x,r])=0$ for all $r\in R$.

Choose a common eigenspace $V_{\lambda }\ne 0$. For $v\in V_{\lambda }$,Thus $V_{\lambda }$ is $L$-invariant, so irreducibility makes it all of ${\mathbb{C}}^n$. Every element of $R$ acts as a scalar. Its trace is zero because $L\subseteq {\mathfrak{sl}}_n$, so that scalar is zero. Faithfulness of the matrix inclusion gives $R=0$. Hence $\boxed{L\text{ is semisimple}}$.

One sufficient diagonal-matrix condition for simplicity is that $L$ contain all traceless diagonal matrices. Indeed invariance under their adjoint actions then decomposes $L$ into that diagonal algebra and some matrix-unit spaces $\mathbb{C}{e}_{ij}$. Put a directed edge $j\to i$ when $e_{ij}\in L$. Irreducibility makes the directed graph strongly connected; otherwise a terminal component spans a proper invariant coordinate subspace. Bracketing matrix units along a simple directed path supplies the matrix unit between its endpoints. Strong connectivity therefore supplies every off-diagonal matrix unit, so $L={\mathfrak{sl}}_n$, which is simple. The mere existence of a diagonal matrix with distinct diagonal entries would not be a sufficient condition.

For the specified construction, $H$ has dimension two. Its diagonal adjoint action and the twelve distinct nonzero weights give a direct sum $L=H\oplus {⨁}_{i=1}^{12}\mathbb{C}{g}_i$, of dimension fourteen. The stated bracket condition and the weight-vector relations make it a Lie algebra. A subspace invariant under it is invariant under all the $g_i$ and hence their associative algebra, so it acts irreducibly on ${\mathbb{C}}^7$. The preceding result makes it semisimple.

Use prime-dimensional faithful irreducible module forces simplicity. To see this, decompose the algebra into simple ideals. Complete reducibility for one ideal and commutation with the others make its isotypic component invariant under the whole algebra, hence all of the representation. Writing that component as $W\otimes \mathrm{Hom}(W,V)$ and repeating produces a tensor product of irreducibles for the simple ideals. Faithfulness makes every factor nontrivial, of dimension at least two. Their dimensions multiply to seven, so only one factor is possible. Thus the constructed $L$ is simple.

The diagonal algebra $H$ is toral, and its centralizer inside $L$ is just $H$, since no $g_i$ has zero weight. A maximal toral subalgebra containing $H$ lies in that centralizer, so equals $H$; it is a Cartan subalgebra. Thus the rank is two and the twelve weights are the roots. Pair them as ${\alpha }_{i+6}=-{\alpha }_i$. The opposite-root brackets have weight zero and belong to $H$. For distinct $i,j\le 6$, the sum cannot be zero, and the supplied root-bracket result says their bracket is nonzero exactly when ${\alpha }_i+{\alpha }_j$ is a root. This is equivalent to $-{\alpha }_i-{\alpha }_j$ being a root, proving the requested equivalence for the paired brackets.

Among rank-two simple root systems, $A_2,B_2,G_2$ have respectively six, eight and twelve roots. ThereforeChoose simple roots $\alpha$ short and $\beta$ long. In $G_2$, $\beta +\alpha$ and $\beta +2\alpha$ are roots. Renumber with $g_1\in L_{\alpha }$, $g_3\in L_{\beta }$, $g_4\in L_{\beta +\alpha }$ and $g_5\in L_{\beta +2\alpha }$, retaining opposite partners in the other six positions. Root spaces are one dimensional, and the allowed bracket equality gives

Use the general definition of a Cartan subalgebra: a nilpotent Lie algebra that is a Lie subalgebra and equals its normalizer of a Lie subalgebra. In a semisimple Lie algebra, the stated structural fact says every Cartan subalgebra $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra. Part (i) therefore givesConversely, suppose $\mathfrak{h}$ is a toral subalgebra and an abelian Lie algebra with $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. An abelian Lie algebra is a nilpotent Lie algebra, because its lower central series of a Lie algebra vanishes at the second term. Part (i) gives $N_{\mathfrak{g}}(\mathfrak{h})=C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. Thus $\mathfrak{h}$ satisfies the defining conditions of a Cartan subalgebra, proving

Use the standard structural fact that $\mathfrak{sp}(V)$ is a semisimple Lie algebra over an algebraically closed field of characteristic zero. If $\mathfrak{h}$ is a Cartan subalgebra, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak{h})=\mathfrak{h}$, and $\mathfrak{h}$ is an abelian Lie algebra. Part (c) therefore shows that its weight spaces occur as $n$ pairs of one-dimensional spaces, with no zero weight space.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_n$ of its nonzero weights of a representation. Pick $p_i\ne 0$ in $V_{{\lambda }_i}$ and $q_i\ne 0$ in $V_{-{\lambda }_i}$, scaling $q_i$ so that $\omega (p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual weight space is one-dimensional and the bilinear form is alternating, so the ordered basisis a symplectic basis. Every $h\in \mathfrak{h}$ is a diagonal matrix in this basis, with entries ${\lambda }_1(h),\dots ,{\lambda }_n(h),-{\lambda }_1(h),\dots ,-{\lambda }_n(h)$.

Let $D_{\mathcal{B}}$ be the Lie subalgebra consisting of all diagonal matricesEach such matrix lies in the symplectic Lie algebra and commutes with $\mathfrak{h}$. Henceand $\mathfrak{h}=D_{\mathcal{B}}$. In particular, every Cartan subalgebra has dimension $n$; the ${\lambda }_i$ are a basis of ${\mathfrak{h}}^{\ast }$.

Carry out the same construction for ${\mathfrak{h}}^{\prime }$, obtaining a symplectic basis ${\mathcal{B}}^{\prime }=(p_1^{\prime },\dots ,p_n^{\prime },q_1^{\prime },\dots ,q_n^{\prime })$ with ${\mathfrak{h}}^{\prime }=D_{{\mathcal{B}}^{\prime }}$. The unique linear map $g$ sending $p_i$ to $p_i^{\prime }$ and $q_i$ to $q_i^{\prime }$ is invertible and preserves all pairings of basis vectors. By bilinearity, it preserves $\omega$ on all of $V$, so $g$ belongs to the symplectic group $\mathrm{Sp}(V)$. The change of basis givesThus conjugacy of Cartan subalgebras of the symplectic Lie algebra follows by explicitly matching their symplectic bases. If $n=0$, the Lie algebra is zero and the same conclusion holds with $g$ the identity on the zero vector space.

Equip the finite-dimensional real vector space $V$ with a positive-definite inner product. A reduced crystallographic root system is a finite set $R\subset V\setminus \{0\}$ spanning $V$ such that, for every $\alpha \in R$,where $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ is the orthogonal reflection in ${\alpha }^{\perp }$. The Weyl group is $W(R)=⟨s_{\alpha }:\alpha \in R⟩$. It permutes the finite set $R$. Its action is faithful because $R$ spans $V$: an element fixing every root fixes every vector. Hence it embeds in the finite symmetric group of $R$ and is finite.

A Cartan subalgebra is a nilpotent self-normalizing Lie subalgebra. For a finite-dimensional complex semisimple Lie algebra it is equivalently a maximal toral subalgebra: a maximal abelian subalgebra whose elements have diagonalizable adjoint actions. The commuting semisimple operators $\mathrm{ad}(h)$ give the root-space decompositionThe nonzero weights $\alpha$ are the roots. Their real span inside the complex dual ${\mathfrak{h}}^{\ast }$, with the positive-definite real inner product induced by the Killing form, is the Euclidean space in which they form a reduced crystallographic root system. In particular one uses this real span, rather than treating all of ${\mathfrak{h}}^{\ast }$ as a real vector space of twice the rank.

For the specified Special orthogonal Lie algebra, put $\bar{i}=7-i$. The matrix relation is $x_{ij}=-x_{\bar{j},\bar{i}}$. Thus its diagonal Cartan subalgebra consists ofLet ${\epsilon }_i(H(t))=t_i$. The matrix units satisfy $[H,E_{ij}]=(d_i-d_j)E_{ij}$, where $d=(t_1,t_2,t_3,-t_3,-t_2,-t_1)$. A basis of root vectors is supplied, for $1\le i<j\le 3$, byThese are twelve independent off-diagonal matrices. Together with the three diagonal generators, they account for the full dimension $6\cdot 5/2=15$, so no further roots occur. ConsequentlyChoose the simple rootsThe six positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$, and every other root is their negative. In the standard orthonormal coordinates all roots have squared length $2$, and their Cartan matrix isIts Dynkin diagram is the connected chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$. A decomposition into mutually orthogonal nonempty root systems would separate this connected set of simple roots, which is impossible. Thus the root system is irreducible. The chain is also the diagram $A_3$, which is the root system of ${\mathfrak{sl}}_4(\mathbb{C})$. The classification of complex semisimple Lie algebras by their root systems yields the Lie algebra isomorphismAn explicit explanation of this coincidence uses the exterior-power Lie algebra representation of ${\mathfrak{sl}}_4$ on ${\bigwedge }^2{\mathbb{C}}^4$, a six-dimensional space. After choosing a volume form, $u\wedge v$ defines a nondegenerate symmetric bilinear form on that exterior square. The induced action preserves it because a traceless endomorphism acts trivially on the top exterior power. This gives a nonzero homomorphism ${\mathfrak{sl}}_4\to {\mathfrak{so}}_6$. Its kernel is an ideal in the simple algebra ${\mathfrak{sl}}_4$, so it is zero; equal dimensions $15$ then make it an isomorphism. Over $\mathbb{C}$ all nondegenerate symmetric forms in dimension six are equivalent, including the form with matrix $J$.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.

Work over $\mathbb{C}$ with a finite-dimensional semisimple Lie algebra $\mathfrak{g}$. Choose a Cartan subalgebra $\mathfrak{h}$, its root system $R\subset {\mathfrak{h}}^{\ast }$, a positive system $R^{+}$ and its simple roots. Let $W$ be the Weyl group, generated by the corresponding root reflections, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the Weyl vector. The finite-dimensional irreducible highest-weight representation $L(\lambda )$ is indexed by a dominant integral weight $\lambda$.

If $P$ is the weight lattice, its group ring has formal symbols $e^{\mu }$ with multiplication $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$. The formal character iswhere $L(\lambda {)}_{\mu }=\{v:Hv=\mu (H)v\text{ for all }H\in \mathfrak{h}\}$ is a weight space. For a weight $\nu$, define the Weyl alternant $A_{\nu }={\sum }_{w\in W}\det (w)e^{w\nu }$; the sign $\det (w)=(-1{)}^{\ell (w)}$ is the determinant of $w$ on the real root span, equivalently the parity of its expression in simple reflections.