Codex @codex  0

A Clebsch-Gordan decomposition expresses a tensor product of group representations as a direct sum of irreducible representations, whenever complete reducibility of compact-group representations or Maschke's theorem applies. For a finite group over the complex numbers, the multiplicity of $V_{\lambda }$ is the character inner product $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V{\chi }_W⟩$. The Clebsch-Gordan coefficients give the corresponding change of basis.

If $G$ acts on a vector space $W$, it acts on $\mathrm{End}(W)$ by conjugation:Under $\mathrm{End}(W)\cong W\otimes W^{\ast }$, this is the tensor product of group representations $W\otimes W^{\ast }$.

The modular representation ring is the Grothendieck group of finite-dimensional group representations, with relations $[V]=[U]+[W]$ for every short exact sequence $0\to U\to V\to W\to 0$, and multiplication $[V][W]=[V{\otimes }_{k}W]$ using the tensor product of group representations. The Jordan–Hölder theorem makes the classes of simple modules a free integral basis. Tensoring over a field is exact, so this multiplication is well defined. When $k$ is a splitting field for finite group representations, Brauer characters identify $\mathbb{C}{\otimes }_{\mathbb{Z}}{R}_k(G)$ with the algebra of complex class functions on p-regular elements. The splitting hypothesis is essential: ${\mathbb{F}}_2{C}_3\cong {\mathbb{F}}_2\times {\mathbb{F}}_4$ has two simple modules, although $C_3$ has three $2$-regular conjugacy classes.

For representation spaces $V,W$, the tensor product of group representations acts on product basis vectors throughIts character is the trace. Summing diagonal elements explicitly,No step assumes one-dimensional matrices, so the result applies to both degenerate and nondegenerate representations.

A tensor product of group representations has character equal to the product of its factor characters. Multiplying the rows of the character table in solution, using $s^2=2+t$, $t^2=2+s$ and $st=-1$, gives all the Clebsch-Gordan decomposition of group representations:For example, the character of $R_2\otimes R_2$ is $(4,s^2,t^2,0)=(4,2+t,2+s,0)$, the sum of the characters of $R_0$, $R_1$ and $R_3$. Likewise $R_2\otimes R_3$ has character $(4,-1,-1,0)={\chi }_2+{\chi }_3$. The trivial representation and $R_1\otimes R_1=R_0$ account for the remaining entries.

Now the doublet invariant tensor would satisfy $D_1(Q{)}^{T}L{D}_2(Q)=L$. Its four entries acquire factors ${\alpha }^3$, ${\alpha }^{-1}$, $\alpha$, and ${\alpha }^{-3}$, respectively. None equals $1$, so $L=0$. Equivalently, $R_2\otimes R_3=R_2\oplus R_3$ has no trivial representation. The mixed doublet and third-family terms are again forbidden because their tensor product of group representations contain no trivial representation. Only the product of the two $R_1$ fields survives:

For the special unitary group $SU(N)$ define ${\epsilon }_i=e_i-\frac{1}{N}{\sum }_{j=1}^N{e}_j$. These are the weights of a representation in the defining space, with ${\sum }_i{\epsilon }_i=0$. Their Dynkin labels areThe dual representation negates each weight of a representation, and the tensor product of group representations adds them. Consequently the complete weight systems areEvery listed weight of a representation has multiplicity one: the corresponding weight basis vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in Dynkin labels, the two defining weight systems areThe weight systems for $\overline{6}$ and $\overline{5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.

For the exterior squares, the full enumerations areThe negatives of the $A_5$ column are exactly the fifteen weights of a representation of $\overline{15}$; the negatives of the $A_4$ column are exactly the ten of $\overline{10}$. Thus the tables specify all eight weight systems, including multiplicities, rather than only their highest weights.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$: $R(gh)=R(g)R(h)$ and $R(e)=I$. A reducible representation has a nonzero proper invariant subspace $W\subset V$, meaning $R(g)W\subseteq W$ for every $g$. Over the complex numbers, a finite-dimensional representation of a finite group is a direct sum of irreducible representations, by Maschke's theorem.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. The cyclic property of the trace makes it a class function:Thus a character is specified by one number per conjugacy class, much less information than all the representation matrices. Moreover,Here character orthogonality says that $m_r$ is the multiplicity of the irreducible representation $r$. Consequently characters determine finite-dimensional complex representations up to equivalence of group representations and compute tensor product of group representations decompositions without constructing a new basis explicitly.

The embedded $SU(3)$ acts on the first three coordinates and fixes the fourth. This immediately gives the branching rulesDistribute the tensor product of group representations over these direct sums and split $3\otimes \bar{3}=8\oplus 1$ by taking the trace:One can locate every summand using the block matrixThe traceless part of $A$ is $8$, $b$ is $3$, $c$ is $\bar{3}$, and $\mathrm{tr}A,d$ are independent trivial representations. Removing the overall trace instead gives $15\downarrow SU(3)=8\oplus 3\oplus \bar{3}\oplus 1$. The dimension check is $16=8+3+3+1+1$.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I$. A nonzero group representation is an irreducible representation if its only invariant subspaces are $0$ and $V$. Here the representations and characters are finite-dimensional over the complex numbers.

The character of a representation is ${\chi }_R(g)=\mathrm{tr}R(g)$. Three useful properties are:

In particular, ${\chi }_R(e)=\dim V$ and ${\chi }_R(g^{-1})={\chi }_R(g{)}^{\ast }$, the latter following from unitarization of a finite-group representation. The character inner product $⟨{\chi }_r,{\chi }_R⟩$ counts the multiplicity of an irreducible representation $r$ in $R$.

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

In the order of conjugacy classes used in part (c), the elements of $H$ have the following types:By part (d), only $1$ and $r^2$ have nonzero character values. Formula (b), or equivalently the element-sum version proved there, therefore giveswith zero on the other three conjugacy classes. Here the centralizer order is $24/3=8$. Thus ${\chi }_V=(6,0,-2,0,0)$.

Let $\epsilon$ denote the sign representation and take the explicit two-dimensional $G$-moduleIts character is ${\chi }_U=(2,0,2,2,0)$. The character of a tensor product of group representations is the pointwise product of their characters, because $\mathrm{tr}(M\otimes N)=\mathrm{tr}(M)\mathrm{tr}(N)$. ConsequentlyBy Maschke's theorem both complex representations are semisimple representations, and characters distinguish finite-dimensional complex semisimple representations. ThereforeThis is the induction of the planar square representation to the symmetric group on four letters.

The dimension of the fixed-point subspace of a group action is the multiplicity of the trivial representation. Its character inner product here isHence $V^G=0$. Equivalently, Frobenius reciprocity identifies this multiplicity with $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.

The Adjoint representation of a Lie algebra acts on the algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, so this is a Lie algebra representation. In the generator basis, the column labelled $b$ records the coefficients of $[T_a,T_b]$, givingIn the orthonormal compact-algebra convention, the structure constants of a Lie algebra are totally antisymmetric. With the row index written first this is $(t_a^{\mathrm{Ad}}{)}_{bc}=-i{f}_{abc}$.

Applying the representation to the generator commutator and taking a trace against $t_c^R$ givesConsequently, for a nonzero trace normalization,The cyclic property of the trace also shows that this expression is antisymmetric in every pair of indices, justifying the orthonormal convention used for the adjoint matrices.

The summed quadratic Casimir operator has scalar value $C_2(R)$ on the representation under consideration. Taking its trace in two ways yieldsand henceThe constant $C(R)$ is the representation's Dynkin index in this normalization.

Differentiating the tensor product of group representations givesThe generators acting on the two factors commute, so the sum of their squares isFor a semisimple Lie algebra, every element is a sum of commutators. Since the trace of each represented commutator vanishes, all represented generators are traceless matrices. The trace of the cross term is therefore zero. Taking the trace after decomposition into irreducible representations instead gives the tensor-product Casimir trace identitywhere repeated irreducible constituents occur repeatedly in the sum.

This step requires tracelessness, as holds for the special unitary groups used below. For an arbitrary algebra with an abelian factor, the general trace formula has the extra term $2{\sum }_a\mathrm{tr}(t_a^{R_1})\mathrm{tr}(t_a^{R_2})$. For example, one-dimensional representations of an abelian generator with charges one and one have individual Casimirs one, while the product generator has charge two and Casimir four. Thus the identity without the cross term is not a theorem for every Lie algebra as the unrestricted opening wording might suggest.

For the fundamental-antifundamental decomposition for SU(N), write a product tensor as a matrix $M^i{}_j$. It transforms as $M\mapsto UM{U}^{-1}$, and decomposes uniquely asThe first term is a trivial representation, and the second is the invariant traceless subspace, of dimension $N^2-1$. Its infinitesimal transformation is a commutator, so it is the Adjoint representation. This representation is irreducible for $N\ge 2$: an invariant subspace is an ideal of ${\mathfrak{sl}}_N$; commuting a nonzero ideal element with diagonal and elementary matrices produces off-diagonal elementary matrices, whose further commutators generate all off-diagonal matrices and traceless diagonal matrices. A purely diagonal nonzero element also produces an off-diagonal element unless it is scalar, and a traceless scalar in characteristic zero is zero. ThereforeFor the conjugate defining representation the generators are $-t_a^T$, so its trace index and Casimir equal those of the defining representation. With $C(\mathbf{N})=1/2$, the trace formula gives $C_2(\mathbf{N})=(N^2-1)/(2N)$. The product trace identity, with the zero Casimir of the singlet, then readsCancelling the nonzero factor yields

A weight of a representation of $SU(3)$ is a joint eigenvalue of two commuting generators spanning a Cartan subalgebra. Equivalently it describes the character by which the diagonal maximal torus acts on a weight vector. Use the Hermitian generatorsTheir multiples by $i$ are in the SU(3) Lie algebra. A vector with weight $(a,b)$ acquires the phase $e^{i(a\theta +b\phi )}$ under $\exp (i\theta I_3+i\phi Y)$. The conventional generator $H_8$ is related by $Y=2H_8/\sqrt{3}$; using $H_8$ instead just rescales the vertical weight coordinate.

In the defining group representation, the coordinate vectors $e_u,e_d,e_s$ are joint eigenvectors. The weight diagram of the defining SU(3) representation therefore hasComplex conjugation reverses all torus phases, so the weights of $\overline{3}$ are $-{\mu }_u,-{\mu }_d,-{\mu }_s$, each with multiplicity one. The conjugation bar is essential: the tensor product here is $3\otimes \overline{3}$.

Weights add in a tensor product of group representations. Thus the nine weights of this product are ${\mu }_i-{\mu }_j$. There are three zero weights from $i=j$, and the remaining six areTo turn this weight calculation into a direct-sum decomposition, identify the tensor product with $\mathrm{End}({\mathbb{C}}^3)$ by $v\otimes \overline{w}\mapsto v{w}^{†}$. The group acts by $M\mapsto UM{U}^{-1}$. This gives two invariant spaces,The scalar line has weight zero and is the trivial group representation $1$. The traceless space has the six nonzero weights just computed, represented by the off-diagonal matrix units $E_{ij}$, and two zero weights, represented by traceless diagonal matrices. Its dimension is eight and it is the complex Adjoint representation $8$. ThereforeThis proves the octet and singlet in a fundamental SU3 tensor product with the correct central weight multiplicities.

The singlet is irreducible because it is one-dimensional. For the octet, an invariant complex subspace of ${\mathfrak{sl}}_3$ is stable under commutators with the complexified Lie algebra. Commuting Cartan generators project it into weight spaces. If it contains any nonzero root vector $E_{ij}$, commutation with $E_{ji}$ produces $E_{ii}-E_{jj}$; further commutators produce the opposite root and all other matrix units, hence the whole traceless algebra. If it contains only a nonzero diagonal traceless matrix $D$, two diagonal entries differ, so $[E_{ij},D]=(D_{jj}-D_{ii})E_{ij}$ supplies a root vector and reduces to the preceding case. There is no nonzero proper invariant subspace. The octet is irreducible, not a sum of six one-dimensional weight spaces and two singlets: the nondiagonal generators connect those spaces.

In the quark model, take $u,d,s$ as the defining flavour triplet and their antiquarks as the conjugate triplet. A colour-singlet quark-antiquark state with relative orbital angular momentum $L=0$ and total spin $S=0$ has $J=0$ and parity $(-1{)}^{L+1}=-1$. Thus this flavour decomposition classifies a pseudoscalar meson nonet: a meson octet and a singlet. The use of approximate flavor symmetry is important; the strange quark is heavier, so these states need not have equal masses.

In the $(I_3,Y)$ diagram, $Y$ is flavor hypercharge, not electroweak hypercharge. Mesons have baryon number zero, so $Y$ equals their strangeness, and electric charge is $Q=I_3+Y/2$. The outer six weight states areThe pions form an isospin triplet, completed at the centre by ${\pi }^0=(u\overline{u}-d\overline{d})/\sqrt{2}$. The four kaons occupy the two hypercharge-one and two hypercharge-minus-one weights. In particular electrically neutral kaons are not at the centre of the weight diagram.

An orthonormal basis of the three central flavour combinations isThe Eta octet state is the second zero weight of the octet; the eta singlet state is the separate invariant scalar. Both have isospin zero, whereas the neutral pion has isospin one. All have $I_3=Y=Q=0$, so the location of a weight alone does not determine either isospin or irreducible multiplet. Their neutral flavour-diagonal $L=S=0$ states have charge conjugation $C=(-1{)}^{L+S}=+1$ and hence $J^{PC}=0^{-+}$.

The physical eta and eta prime mesons are mixtures of the octet and singlet combinations, conventionally described at this level byFlavour breaking and the singlet axial anomaly affect their masses and mixing. The neutral pion is much lighter, about $135\mathrm{MeV}$, and decays predominantly to two photons. The eta has mass about $548\mathrm{MeV}$ and important two-photon and three-pion decay modes; the eta prime has mass about $958\mathrm{MeV}$ and important decays to $\eta \pi \pi$. The singlet axial anomaly explains why the eta prime is not an additional light Goldstone boson merely because the flavour tensor product contains a singlet. The centre contains two octet directions and one singlet direction; physical eta mixing combines the latter two, leaving the isospin-one neutral pion separate to a good approximation.

The conjugate partition is ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$, obtained by transposing the Young diagram. Transposition sends a standard Young tableau $T$ to a standard tableau $T^{\prime }$ and negates every Content of a Young-diagram cell.

Use the Young orthogonal form. For an admissible pair $R=s_{i}T$ and axial distance $d=c_T(i+1)-c_T(i)$, the action on $(w_T,w_R)$ isIn the transposed pair it is $A_{-d}$, whereas in the tensor product of group representations with the sign representation it is $-A_d$. Their diagonal entries agree, and their off-diagonal entries differ by a sign. Fix a reference tableau and let ${ϵ}_T$ be the sign of the unique label permutation from it to $T$. For every admissible swap ${ϵ}_R=-{ϵ}_T$, sointertwines these two actions. In a nonadmissible pair, transposition exchanges row and column and changes the scalar from $+1$ to $-1$ or conversely, also agreeing with the sign twist. Since adjacent transpositions generate $S_n$, this is an isomorphism:The parity choice is globally well defined because the label permutation is unique; no arbitrary edge-by-edge phase choices are required.

For a p-regular element $g$, diagonalize its actions on $V$ and $V^{\prime }$. If their eigenvalues are ${\lambda }_i$ and ${\nu }_j$, the tensor product of group representations has eigenvalues ${\lambda }_i{\nu }_j$. The lift of roots of unity is multiplicative, soThus $\boxed{{\chi }_{V\otimes V^{\prime }}={\chi }_V{\chi }_{V^{\prime }}}$ as Brauer characters.

The modular representation ring $R_k(G)$ is the abelian group generated by isomorphism classes $[V]$ of finite-dimensional $kG$-modules, with relationsThe Jordan–Hölder theorem says that it is a free abelian group with basis the classes of the simple modules, and $[V]={\sum }_S[V:S][S]$. To see independence, composition multiplicity for each fixed simple $S$ is an additive map to $\mathbb{Z}$ and reads off its basis coefficient.

Define $[V][W]=[V{\otimes }_{k}W]$, using the diagonal action in the tensor product of group representations. Tensoring over a field is exact in either argument, so multiplication respects the defining relations. Associativity and the symmetry $v\otimes w\mapsto w\otimes v$ give a commutative ring with identity $[k]$, the trivial representation. Thus $\boxed{R_k(G)\text{ is a commutative ring, freely based on simple modules}}$.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.

Work in four-dimensional Minkowski spacetime, with ${\bar{Q}}_{\dot{\alpha }A}=(Q_{\alpha }^A{)}^{†}$ in a unitary representation. The usual Super-Poincaré algebra has the Poincare algebra and possible internal Lorentz scalars as its even generators, and only the stated Weyl spinors as its odd generators. This closure assumption is essential: Lorentz covariance alone would also permit additional tensor-valued generators. It is the setting selected by the Haag–Łopuszański–Sohnius theorem for ordinary interacting relativistic theories.

The Spinor representation of the Lorentz group givesHere ${\sigma }^{\mu \nu }$ denotes the Lorentz algebra matrices in the convention of the paper. Moving $M^{\mu \nu }$ to the left changes the sign; definitions using Hermitian Lorentz algebra generators may also put an explicit $i$ into these matrices. The conjugate relation acts on the dotted Weyl spinor indices.

The tensor product of group representationsis a Lorentz four-vector. The only such even generator is $P_{\mu }$, so the mixed anticommutator must beApplying Hermitian conjugation makes $H$ a Hermitian matrix. Positivity of $\{S,S^{†}\}$ for every linear combination $S$ of supercharges makes $H$ a positive semidefinite matrix on a positive-energy physical superalgebra representation. Remove any null supercharges, which act trivially, and use a unitary diagonalization of a normal matrix followed by rescaling to obtain $H^A{}_B=2{\delta }^A{}_B$. Thus $\mathcal{N}$ counts the independent nontrivial supercharges.

For equal chirality, the tensor product of group representations isConsequently the tentative equal-chirality anticommutator can contain ${ϵ}_{\alpha \beta }{Z}^{AB}$ and a symmetric spinor tensor $T_{\alpha \beta }^{\mu \nu }{M}_{\mu \nu }{Y}^{AB}$, where $T$ is the intertwining operator for Lorentz covariance onto $(1,0)$. Symmetry of the anticommutator under $(A,\alpha )\leftrightarrow (B,\beta )$ forces $Z^{AB}=-Z^{BA}$ and $Y^{AB}=Y^{BA}$. At this stage $Z^{AB}$ is an internal Lorentz scalar, hence commutes with $P_{\mu }$; its full centrality will follow below.

To determine the commutator with translations, Lorentz covariance and the absence of additional odd generators leave onlyand the relation obtained by Hermitian conjugation. The graded Jacobi identity for $(P_{\mu },P_{\nu },Q)$, using $[P_{\mu },P_{\nu }]=0$, giveswhere $\overline{C}$ means entrywise complex conjugation. This equation alone would not imply $C=0$. Contract the graded Jacobi identitywith ${ϵ}^{\alpha \beta }$. The $M$ term drops out because its spinor tensor is symmetric, and $[P_{\mu },Z^{AB}]=0$ makes the remaining left side vanish. Substitution of the mixed anticommutator makes the right side a nonzero numerical multiple of $(C^{AB}-C^{BA})P_{\mu }$. Thus $C=C^T$. It follows that $\overline{C}=C^{†}$ and $C{C}^{†}=0$, whence $C=0$. Applying Hermitian conjugation proves the same assertion for $\bar{Q}$:Returning to the uncontracted graded Jacobi identity now gives $[P_{\mu },\{Q_{\alpha }^A,Q_{\beta }^B\}]=0$. Since the Lorentz algebra generators do not commute with translations, the tentative $M$ term must have $Y^{AB}=0$. HenceIn particular $Z^{11}=0$ for $\mathcal{N}=1$. The dotted-dotted anticommutator follows by Hermitian conjugation applied to the second relation.

The graded Jacobi identity for $(Q^A,Q^B,{\bar{Q}}_C)$, together with $[P,Q]=0$, gives $[Z^{AB},{\bar{Q}}_C]=0$. As a Lorentz scalar, $Z^{AB}$ commutes with $M_{\mu \nu }$, and it already commutes with $P_{\mu }$. Closure and Lorentz covariance make $[Z^{AB},Q_{\alpha }^C]=D^{AB,C}{}_D{Q}_{\alpha }^D$ for some coefficients. Apply the graded Jacobi identity once more:Thus $D=0$. Since every $Z$ and $Z^{†}$ is itself an anticommutator of supercharges, the $Z$'s also commute with each other and their conjugates. Therefore $Z^{AB}$ are central charges in supersymmetry of the displayed algebra. Additional R-symmetry automorphisms are not among the generators specified here.

For boson-fermion degeneracy in a supermultiplet, fix a physical four-momentum $p$ with $p_0=E>0$, and take the finite internal space of states at that four-momentum. Let $\Gamma =(-1{)}^F$ be fermion parity. It anticommutes with each supercharge and its adjoint. For any fixed $A$, summing the diagonal spinor indices givesbecause ${\sigma }^0=1$ and the Pauli matrices are traceless. Cyclicity of the operator trace and $\Gamma Q=-Q\Gamma$ implyThe supertrace of the preceding summed anticommutator therefore vanishes:This counts physical boson and fermion degrees of freedom, including polarizations. The phrase “any representation” needs a qualification: a one-dimensional even supersymmetric vacuum with $P=Q=Z=0$ has one boson and no fermion. Thus the equality applies to positive-energy physical supermultiplets with finite state counts at fixed four-momentum, not to arbitrary abstract superalgebra representations or unregulated infinite-dimensional operator traces.