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Use a finite-dimensional complex semisimple Lie algebra, or the complexification of its compact real form, as required for the root construction. Choose a Cartan subalgebra with basis $H_1,\dots ,H_r$. Simultaneous diagonalization of its adjoint action gives the root-space decompositionThe nonzero weights are the roots of a root system; each root space is one-dimensional in this setting, and roots occur in opposite pairs. This supplies the Cartan and root-vector basis.

The Jacobi identity givesIf the bracket is nonzero and $\alpha +\beta \ne 0$, it is a nonzero vector in the root space of weight $\alpha +\beta$, so that sum is a root. The printed implication needs the opposite-root exception: for $\beta =-\alpha$, $[E_{\alpha },E_{-\alpha }]$ can be a nonzero Cartan element, while zero is not a root. The supplied simple-root commutator itself gives this counterexample. This is the root bracket and the opposite-root exception.

Choose a real linear functional nonzero on every root; positive roots are those on which it is positive, and negative roots are their negatives. The simple roots are positive roots not decomposable into two positive roots. They form a basis ${\alpha }_1,\dots ,{\alpha }_r$ of the root span, and every root is an integer combinationwith all $m_i\ge 0$ for a positive root and all $m_i\le 0$ for a negative root. Positive roots can be decomposed repeatedly until simple roots remain; the finite root set makes this procedure terminate.

For $i\ne j$, ${\alpha }_j-{\alpha }_i$ has simple-root coefficients of both signs and is not a root. It is also nonzero, so the root-space bracket rule givesThe operators $E_i^{\pm },{\hat{H}}_i$ have the SU(2) Lie algebra normalization ${\hat{H}}_i=2J_3$, $E_i^{\pm }=J_{\pm }$. In a finite-dimensional representation, a vector killed by $E_i^{-}$ is a lowest-weight vector in its SU(2) summands. If its ${\hat{H}}_i$ value is $-\lambda$, those summands have spin $\lambda /2$, soOne can also read the termination from successive raising coefficients $\sqrt{(\lambda -k)(k+1)}$ starting at the lowest weight.

Apply this result in the finite-dimensional Adjoint representation, taking the lowest-weight vector to be $E_j^{+}$. The vanishing bracket just proved is its lowering condition. If $[{\hat{H}}_i,E_j^{+}]=a_{ij}{E}_j^{+}$, the lowest-weight result forces $a_{ij}=-n_{ij}$ with $n_{ij}$ a nonnegative integer. It also yieldsThe last identity is exactly the requested nested commutator. This is the SU(2) proof of the Serre relations, and the identities are the Serre relations in these Chevalley basis conventions.

To construct a highest-weight representation, start from its highest-weight vector and apply negative-root operators. Fix an order on the negative roots. The Poincaré-Birkhoff-Witt theorem gives spanning vectorsof weight $w-{\sum }_i{k}_i{\beta }_i$. The spanning assertion follows by commuting all negative-root operators to the left, Cartan operators to the middle and positive-root operators to the right; commutators supply lower-order words, positive operators kill the highest-weight vector and Cartan operators act there by scalars. In the finite-dimensional highest-weight representation, impose the simple-root string relations $(E_i^{-}{)}^{w_i+1}|w⟩=0$ and the commutator/Serre relations. At each weight retain a linearly independent set of the resulting words, discarding zero words and dependencies. This gives a basis, with separate independent vectors retained when a weight has multiplicity greater than one. The construction is finite in the assumed finite-dimensional representation; treating all lowering words as automatically independent would overcount.

For SU(3), put $e_i=E_i^{+}$, $f_i=E_i^{-}$ and $h_i={\hat{H}}_i$. The off-diagonal Cartan entries are minus one, so the Serre relations give $(\mathrm{ad}{e}_1{)}^2{e}_2=(\mathrm{ad}{e}_2{)}^2{e}_1=0$. With $e_3=[e_1,e_2]$, these implyFor $f_3=-[f_1,f_2]$, expand the commutator using Jacobi identity and the distinct-root brackets:since $[h_1,f_2]=f_2$. Similarly,and the same calculation on $f_3$ gives $[h_1,f_3]=-f_3$. ThusThese signs also agree with the explicit model $e_1=E_{12}$, $e_2=E_{23}$, $e_3=E_{13}$, $f_1=E_{21}$, $f_2=E_{32}$, $f_3=E_{31}$.

For the SU(3) Casimir in Chevalley generators, setUsing $[h_1,e_1]=2e_1$, $[h_2,e_1]=-e_1$, and putting Cartan elements on the left, the four terms of $[P,e_1]$ areTheir sum is $3h_1{e}_1$. The remaining terms giveThe last two cancel, so $[C,e_1]=3h_1{e}_1-3h_1{e}_1=0$. Also $h_1$ commutes with every Cartan polynomial and with each $f_n{e}_n$: their raising and lowering weights cancel in the product. HenceOn $|w_1,w_2⟩$, $e_1$ and $e_2$ vanish and therefore $e_3=[e_1,e_2]$ vanishes as well. All three $f_n{e}_n$ terms consequently give zero, leavingThe paper's normalization is $C=3C_2$ relative to the conventional SU(3) quadratic Casimir eigenvalue. For example it gives $4$ in the fundamental representation, $9$ in the octet and $18$ in the decuplet.

For the usual positive roots, the chamber condition is that the coefficients of $L_1,L_2,L_3$ be weakly decreasing, since their consecutive differences are the simple-coroot pairings. Adding the same constant to all three coefficients changes no weight, because $L_1+L_2+L_3=0$.

The given coefficients are $(-2,3,0)$. Let $s_1$ exchange $L_1,L_2$, and $s_2$ exchange $L_2,L_3$. With the rightmost reflection acting first,ThusIts coefficients $(3,0,-2)$ are strictly decreasing, so it lies even in the open Weyl chamber; its Dynkin labels are $(3,2)$.

Choose a linear functional $\ell :E\to \mathbb{R}$ which is nonzero on every root; one exists because there are only finitely many root hyperplanes to avoid. Declare a root positive when $\ell (\alpha )>0$, and negative when $\ell (\alpha )<0$. This orders the roots by sign, compatibly with addition whenever the sum is a root. One can refine it to a lexicographic total order by completing $\ell$ to a coordinate system. Since negation permutes the roots,A simple root is a positive root not expressible as the sum of two positive roots. If distinct simple roots $\alpha ,\beta$ had $\alpha -\beta$ as a root, it would be either positive or negative. In the first case $\alpha =\beta +(\alpha -\beta )$ decomposes $\alpha$ into two positive roots; in the second $\beta =\alpha +(\beta -\alpha )$ decomposes $\beta$. Both contradict simplicity. If they are equal, their difference is zero and is also not a root. Hence the difference of two simple roots is never a root.

For use in the final subpart, the simple roots form a basis, and every positive root is a nonnegative integer combination of them. Here are the needed reasons. Repeatedly decompose a nonsimple positive root into two positive roots; their $\ell$ values decrease, and the finite root set makes this terminate. This gives the integer combinations. Distinct simple roots have nonpositive inner product: a positive inner product would make $\alpha$ and $-\beta$ obtuse, so the qualified root-sum lemma would make $\alpha -\beta$ a root, just ruled out. Finally, a nontrivial linear relation between simple roots can be split into positive and negative coefficients, giving the same vector $u$ as positive combinations of disjoint sets. Its squared norm, computed using those two expressions, is nonpositive, whereas a nonzero positive combination has positive $\ell$ value. This contradiction proves linear independence. Their span is $E$ because they generate all roots.

The single bonds give equal simple-root lengths and inner products, after normalization,The positive roots of this A3 root system areThe first two sums are roots by the qualified obtuse-root lemma. The last is a root because $(\alpha +\beta ,\gamma )=-1$. All have nonnegative simple-root coordinates, hence are positive.

To prove completeness, first note that every root is Weyl-conjugate to a simple root. For a nonsimple positive root $\delta =\sum c_i{\alpha }_i$, some $(\delta ,{\alpha }_i)>0$, since $(\delta ,\delta )=\sum c_i(\delta ,{\alpha }_i)>0$. Reflection subtracts a positive integer multiple of ${\alpha }_i$, decreasing its integer height. The reflected root stays positive: it leaves the coefficients at all other simple roots unchanged, at least one of which is positive; a root has coordinates all of one sign. Repeating this height decrease reaches a simple root. Thus every root has squared length two in this connected single-bond system.

Write an arbitrary positive root as $a\alpha +b\beta +c\gamma$, with nonnegative integers $a,b,c$. Its squared length condition isThus $a,c\in \{0,1\}$. For $(a,c)=(0,0)$ the equation gives $b=1$; for $(1,0)$ or $(0,1)$ it gives $b=0$ or $1$; for $(1,1)$ it gives $b=1$. These are exactly the six coefficient triples in the displayed list. This proves completeness from the axioms, not merely from recognizing the Dynkin diagram.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

Put $V={\mathbb{C}}^n$. We use complete reducibility of compact-group representations, and the highest-weight classification: the multiplicity of $V_{\lambda }$ in a completely reducible representation is the dimension of its weight-$\lambda$ subspace killed by all positive root vectors. Each irreducible summand contributes just its one-dimensional highest-weight line to that subspace.

Symmetrizing and antisymmetrizing the three tensor factors give invariant subspaces ${\mathrm{Sym}}^{3}V$ and ${\Lambda }^{3}V$ of $V^{\otimes 3}$. The vectors $e_1^{\otimes 3}$ and $e_1\wedge e_2\wedge e_3$ are highest-weight vectors of weights $(3,0,\dots ,0)$ and $(1,1,1,0,\dots ,0)$ respectively. The unitary version of the Weyl dimension formula isIt givesThese are exactly the dimensions of the third symmetric power and exterior power, so each is irreducible and occurs once in those subspaces.

The weight $(2,1,0,\dots ,0)$ in $V^{\otimes 3}$ has basisThe only positive root vector that acts nontrivially on this weight space is $E_{12}$. Acting on the tensor product as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes 3}$. Consequently the highest-weight vectors in this space are preciselyThis is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its dimension isThe dimensions of the summands already found add toThus no other irreducible summand can remain. We have proved the full decompositionThere are four summands counted with multiplicity and three distinct isomorphism types.

For $n=2$, the exterior cube vanishes. The third symmetric power still has highest weight $(3,0)$ and dimension $4$, while the same two-dimensional highest-vector calculation gives two copies of the representation of weight $(2,1)$ and dimension $2$. It is $\det \otimes V$, since a determinant twist shifts the standard highest weight $(1,0)$ to $(2,1)$. Hence

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

The printed identity has a sign error. With the defined Kostant partition function and the exponent $e^{-\nu }$, the cancelling factors must be $1-e^{-\alpha }$. This is already visible for $SU(2)$ with positive root $\alpha$: the formal series ${\sum }_{k\ge 0}{e}^{-k\alpha }$ multiplied by $1-e^{\alpha }$ is $-e^{\alpha }$, not $1$.

For the corrected formula, a partition of $\nu$ is a tuple of nonnegative integers $(k_{\alpha }{)}_{\alpha >0}$ satisfying $\nu ={\sum }_{\alpha >0}{k}_{\alpha }\alpha$. The order of summands is not counted. Multiplying the formal geometric series yieldsThis product is coefficientwise meaningful. Choose a linear functional strictly positive on every positive root. For a fixed $\nu$ it bounds each $k_{\alpha }$, so only finitely many tuples contribute. We are working in the completion supported on the negative positive-root cone, not claiming convergence of an ordinary Fourier series on the torus.

Each geometric series cancels its factor $1-e^{-\alpha }$, provingEquivalently, reversing all the exponential signs gives $({\sum }_{\nu }p(\nu )e^{\nu }){\prod }_{\alpha >0}(1-e^{\alpha })=1$ in the opposite completion. More generally, if $N$ is the number of positive roots, the expression literally printed in the paper equalssince ${\prod }_{\alpha >0}(1-e^{\alpha })=(-1{)}^N{e}^{2\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$. Thus the correction is substantive, not merely a choice of Fourier convention.