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For a representation $R$ of $SU(N)$ with $N\ge 3$, choose the Hermitian quantum generator convention and normalize the cubic anomaly coefficient byHere the anticommutator symmetrizes the last two generators; cyclic property of the trace makes the whole tensor symmetric. This coefficient multiplies the local four-dimensional gauge anomaly from a chiral fermion in $R$. It adds under direct sums.

The conjugate representation has generators $-(T^a{)}^T$, so transposition and the three minus signs give $A(\overline{R})=-A(R)$. Hence a real representation or pseudoreal representation has zero cubic anomaly coefficient. For $SU(3)$, the two-index antisymmetric representation is the conjugate triplet and has coefficient $-1$. For $SU(2)$ the symmetric cubic invariant itself vanishes, so normalization to a nonzero fundamental cubic coefficient is unavailable; this statement concerns the local cubic anomaly and does not settle global anomalies.^[ref]

Differentiating a unitary operator family at the identity gives an anti-Hermitian tangent. In the convention $U(ϵ)=I+iϵF+O({ϵ}^2)$ for real $ϵ$, the generator $F$ is Hermitian. This is the one-parameter version of the Hermitian quantum generator convention. A finite linear truncation need not itself be exactly unitary.

A set $S$ generates a Lie algebra $\mathfrak{g}$ when the smallest Lie subalgebra containing $S$ is $\mathfrak{g}$. Equivalently, every element is a finite linear combination of iterated Lie brackets of elements of $S$. A basis is always a generating set, but a generating set need not span $\mathfrak{g}$ by linear combinations alone: in the Heisenberg Lie algebra, $a,b$ generate because $[a,b]=c$, although their linear span has only two dimensions.

In a Matrix Lie group, physicists also call the matrices specifying infinitesimal transformations generators. Under the Hermitian quantum generator convention, a transformation has the form $I+i{\alpha }_a{t}_a+O(\Vert \alpha {\Vert }^2)$; the corresponding genuine Lie algebra elements are $i{t}_a$. A list of such generators may be a basis or a smaller generating set, so the distinction matters.

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.

Use the metric in the question and fix the Levi-Civita symbol convention ${ϵ}_{0123}=+1$. A consistent choice of rotation Lie algebra generators isSubstituting the indices into the printed Poincare algebra givesThe other cyclic brackets follow the same way. The negative spatial metric and antisymmetry of $M^{\rho \sigma }$ both enter this sign.

The printed brackets are real Lie algebra brackets, without the factor $i$ used for ordinary commutators of Hermitian quantum observables. We use those brackets for the algebraic verifications. For physical eigenvalues below, angular momentum is Hermitian, its spin projection is the real number $j_3$, and $\hslash =1$. In that quantum convention the ordinary operator commutators are $i$ times the displayed brackets. This distinction prevents identifying a real spin projection with an anti-Hermitian matrix eigenvalue.

In the universal enveloping algebra, translations commute. The Pauli-Lubanski pseudovector therefore satisfiesFor fixed $\nu ,\rho$, the product of momenta is symmetric in $\mu ,\tau$ while the epsilon coefficient is antisymmetric in them. There is no need to commute the Lorentz generator through the momenta.

Using the commutator derivation identity and the mixed bracket,The two terms become equal after swapping $\nu ,\rho$, and the last expression vanishes by antisymmetry in $\nu ,\tau$. In the Hermitian observable convention, the calculation has one overall extra $i$ and still vanishes. Thus $W_{\mu }$ preserves each momentum eigenspace.

For the specified epsilon orientation, two useful component identities areThe order displayed matters: the momentum operator is on the right, so it can act first on the momentum eigenstate. At rest, $p^{\mu }=(m,0,0,0)$, and on a spin state with $J_3|j,j_3⟩=j_3|j,j_3⟩$ these giveThese are massive rest-frame Pauli-Lubanski eigenvalues. The raised component would be $W^3=+m{j}_3$; confusing $W_3$ with $W^3$ reverses the answer.

Helicity is the projection of spin angular momentum, or equivalently the rotation generator acting internally, along the momentum direction:For the momentum $p^{\mu }=(k,0,0,k)$ with $k>0$, the helicity operator is $J_3$. Hence a helicity-$j_3$ state hasThe result uses only the two longitudinal components and does not need a separate assumption about the transverse little-group generators. These massless longitudinal Pauli-Lubanski eigenvalues agree with $W^{\mu }=j_3{P}^{\mu }$ for ordinary finite-helicity representations.

Finally, at rest the contraction is $m{W}_0=0$. For the chosen null momentum it is $k(W_0+W_3)$, whose two eigenvalues cancel. Thus both results obey $W_{\mu }{P}^{\mu }=0$. Reversing the epsilon orientation reverses all $W$ eigenvalues together; it leaves both identities and both consistency checks intact. For a literal anti-Hermitian derived representation $\rho$ of the printed real algebra, write $H_X=i\rho (X)$ for the Hermitian observable. Since $W_{\mu }$ is bilinear in generators, its abstract enveloping-algebra image is $\rho (W_{\mu })=-W_{\mu ,\mathrm{phys}}$. Thus the corresponding formal-image eigenvalues, if that convention is intended, areHere $\rho (P^{\mu })$ has eigenvalue $-i{p}^{\mu }$, while $p^{\mu }$ and $j_3$ themselves remain real physical labels. This is the same result after the Hermitian quantum generator convention is applied to both factors, not an inconsistent choice of spin sign. Both the Hermitian-observable and literal anti-Hermitian interpretations are consequently specified.