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For the defining representation of the Special orthogonal Lie algebra, Hermitian Lie algebra generators $t_{ij}=i(E_{ij}-E_{ji})/\sqrt{2}$ have trace normalization $\mathrm{tr}(t_{ij}{t}_{kl})={\delta }_{ik}{\delta }_{jl}$ for ordered pairs. Summing their squares gives ${\sum }_{i<j}{t}_{ij}^2=(N-1)I/2$. Consequently the Casimir eigenvalue is $(N-1)/2$ and the Dynkin index is one. Doubling the trace normalization doubles both numbers; a gauge coupling must be rescaled inversely with the Lie algebra generators. With long roots of squared length two, the first convention is $C_2(\lambda )=(\lambda ,\lambda +2\rho )/2$.

For group representations $R:G\to GL(V)$ and $S:G\to GL(W)$, their direct sum acts on the direct sum of vector spaces by $(R\oplus S)(g)(v,w)=(R(g)v,S(g)w)$. Its character and Dynkin index, when defined, are the sums of those of the two summands.

Identify the product with endomorphisms of the defining space. Special unitary group conjugation fixes scalar matrices and preserves the complementary traceless subspace. For $N\ge 2$, the latter is the complex Adjoint representation of a Lie algebra, which is irreducible because ${\mathfrak{sl}}_N$ is simple. With fundamental Dynkin index $1/2$, the trace-Casimir identity gives $C_2(\mathbf{N})=(N^2-1)/(2N)$; the tensor-product Casimir trace identity then gives $C_2(\mathbf{Adj})=N$.

For left-handed Weyl fermions in representations $(R_i,Q_i)$ of a simple semisimple factor times $U(1)$, the displayed traces cancel the local cubic non-Abelian, mixed non-Abelian–Abelian, and cubic Abelian gauge anomalies. Include spectator multiplicities in every trace. A triangle with one non-Abelian insertion and two Abelian insertions vanishes by tracelessness. Coupling to gravity also requires ${\sum }_i\dim (R_i)Q_i=0$, and global anomalies must be checked separately. For a conjugate representation the cubic coefficient changes sign but the Dynkin index $T(R)$ does not.

Write $a=a_1$, $b=a_2$ with $a,b\ge 0$, and let $\lambda =a{\omega }_1+b{\omega }_2$ be the highest weight. The positive roots are ${\alpha }_1$, ${\alpha }_2$ and ${\alpha }_1+{\alpha }_2$, and the Weyl vector is $\rho ={\omega }_1+{\omega }_2$. The Weyl dimension formula givesFor the quadratic Casimir operator, use the root-length normalization $C(\lambda )=(\lambda ,\lambda +2\rho )$. Substituting the Gram matrix in solution yieldsTaking the trace of the defining quadratic Casimir operator identity gives $d(a,b)C(a,b)=\dim (\mathfrak{su}(3))c(a,b)=8c(a,b)$, where $c$ is the Dynkin index in the matching normalization. ThereforeIn particular $c(3)=1$ and $c(8)=6$. The common particle-physics convention uses $C_2=C/2$ and $T=c/2$, giving $C_2(3)=4/3$ and $T(3)=1/2$. The subsequent discrete gauge anomaly congruences use the integer Dynkin index $c$, while the sign test for the supplied beta function is independent of a common rescaling of all indices.

The Weyl dimension formula is increasing in each nonnegative Dynkin label. If $a\ge 3$ or $b\ge 3$, the dimension is at least $d(3,0)=10$. Among $0\le a,b\le 2$, the cases with both labels nonzero, other than $(1,1)$, have dimension at least $d(2,1)=15$. Thus the following is the complete list of irreducible representations of dimension strictly less than $10$:Here $6={\mathrm{Sym}}^{2}3$ is the two-index symmetric representation and $8$ is the Adjoint representation. Interchanging $a$ and $b$ gives the conjugate representation, explaining equal dimensions, Casimir eigenvalues and Dynkin indices in each conjugate pair. In the alternative convention of solution, divide the $C$ and $c$ columns by two.

The two-index symmetric representation of SU(3) is $6$, with Dynkin index $5$. Thus $n$ copies have $c(r_S)=5n$, and the supplied beta function isThe strict sign condition for asymptotic freedom is $n<132/5=26.4$, givingIndeed $132-5(26)=2>0$, while $132-5(27)=-3<0$.

The Dynkin index is additive over a direct sum of group representations, so $n$ copies of the fundamental representation have $c(r_S)=n$. With $c(\mathrm{ad})=6$, the supplied beta function becomesFor positive gauge coupling $g$, asymptotic freedom at this order requires $\beta (g)<0$, hence $n<132$. For nonnegative integer $n$,At $n=132$ the one-loop coefficient vanishes, so this formula alone does not establish asymptotic freedom. This count follows the scalar multiplicity convention of the printed beta function; changing how a real or complex scalar field is counted requires changing its coefficient too.

For the fundamental representation and the two-index symmetric representation, the Dynkin indices are $1$ and $5$. Therefore the specified discrete gauge anomaly condition for ${\mathbb{Z}}_4$ isAll ordered solutions areOnly $(1,3)$ and $(3,1)$ give a faithful ${\mathbb{Z}}_4$ action on this pair of Weyl fermions; faithfulness was not imposed. As in solution, cancellation here concerns the stipulated discrete gauge anomaly, not cancellation of the pure cubic SU(3) gauge anomaly. The integer normalization matters: one must not replace $c$ by a half-integer trace index inside an unchanged modular congruence.

The complexification of a Lie algebra gives $\mathfrak{so}(2n,\mathbb{C})$ of type Dn root system and $\mathfrak{so}(2n+1,\mathbb{C})$ of type Bn root system. For $D_n$, the simple roots ${\alpha }_i=e_i-e_{i+1}$ for $1\le i<n$ and ${\alpha }_n=e_{n-1}+e_n$ give a chain $1,\dots ,n-2$ with two terminal nodes $n-1,n$ joined to $n-2$. All edges are single. For $B_n$, use ${\alpha }_i=e_i-e_{i+1}$ for $i<n$ and ${\alpha }_n=e_n$. The Bn Dynkin diagram is a chain with a double final edge and an arrow pointing to the short node $n$.

The usual family diagrams apply to $D_n$ for $n\ge 4$ and $B_n$ for $n\ge 2$. The small cases are $D_3\cong A_3$, $D_2\cong A_1\oplus A_1$, $B_1\cong A_1$; $\mathfrak{so}(2)$ is one-dimensional and Abelian.

To fix Casimir normalization for orthogonal vector representations, use Hermitian Lie algebra generators $t_{ij}=i(E_{ij}-E_{ji})/\sqrt{2}$, so the defining representation has Dynkin index $c(\mathbf{N})=1$. With long roots of squared length two, the quadratic Casimir operator has Casimir eigenvaluewhere $\rho$ is the Weyl vector and $\lambda$ in the final expression is the column of Dynkin labels. The defining fundamental representation has highest weight $(1,0,\dots ,0)$ in the stated family conventions.

For $D_n$, the row supplied in the source has sum $n-1$ and first entry one, soFor $B_n$, the supplied row has sum $n-\frac{1}{2}$, givingTaking the trace of ${\sum }_a{t}_a{t}_a=C_2(R)I_R$ gives the relation between the Dynkin index and Casimir eigenvalue, $c(R)\dim \mathfrak{g}=C_2(R)\dim R$. ThereforeThere is also a direct check: each coordinate appears in $N-1$ planes and $t_{ij}^2=(E_{ii}+E_{jj})/2$, so ${\sum }_{i<j}{t}_{ij}^2=(N-1)I/2$.

If the Casimir operator is instead defined by $C(\lambda )=(\lambda ,\lambda +2\rho )$, every Lie algebra generator is larger by $\sqrt{2}$ and the corresponding answers are $C(\mathbf{N})=N-1$ and $c(\mathbf{N})=2$. These are the same Lie algebra in a different normalization. At the exceptional small ranks the family Dynkin labels of the vector change, but the direct matrix computation remains valid.

In the normalization $c(8)=1$ of c, the Adjoint representation of $D_4$ has highest weight ${\lambda }_{\mathrm{adj}}=(0,1,0,0)$, the highest root. The supplied Gram matrix has second row $(1,2,1,1)$, whose sum is five. HenceThe Adjoint representation has the same dimension as the Lie algebra, so its Dynkin index is also $c(\mathrm{adj})=6$. The $k$ Weyl fermions form a direct sum of $k$ vector representations; additivity of the trace gives $c(r_W)=k$.

Substitution into the given one-loop renormalization-group beta function yieldsFor positive $g$, asymptotic freedom at this order requires a negative beta function, equivalentlyIndeed, if $b_0=22-2k/3>0$, integrating $\mu dg/d\mu =-b_0{g}^3/(16{\pi }^2)$ givesso the gauge coupling decreases towards zero as the energy scale increases. At $k=33$ the one-loop coefficient vanishes and the given equation cannot decide asymptotic freedom; $k>33$ has the opposite one-loop running.

With the alternative convention $C(\lambda )={\lambda }^{T}G(\lambda +21)$ described in c, $c(8)=2$, $c(\mathrm{adj})=12$, and the same source formula reads $\beta (g)=-(44-4k/3)g^3/(16{\pi }^2)$. Rescaling the gauge coupling together with the Lie algebra generators relates the conventions; the bound $k<33$ is identical.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The renormalization-group beta function impliesIntegrating the one-loop running coupling equation givesHere the sign of ${\beta }_i$ is the sign convention in the question: a negative coefficient makes the inverse coupling increase towards high energy.

For gauge coupling unification, let $L=\log (M_{\mathrm{GUT}}/M_Z)/(2\pi )$ and let $a_G$ be the common inverse coupling at the unification scale. The normalized hypercharge coupling is $(5/3){\alpha }_1$, soSubtract the second relation from the first to determine $L$, then subtract it from the third. Provided $3{\beta }_1/5-{\beta }_2\ne 0$, this provesThe factors $3/5$ belong to the inverse normalized coupling; dropping them changes the prediction.

Use the Standard Model gauge group $SU(3{)}_C\times SU(2{)}_L\times U(1{)}_Y$ and convention $Q=T_3+Y$. One fermion family and one complex Higgs doublet have the following Standard Model representations. The left-handed quark and lepton doublets contain two Weyl spinors each; the right-handed fields are single Weyl species.The row $u_R$ denotes the right-handed up quark. Equivalently an all-left-handed table replaces the three right-handed fields by $u_R^c:(\overline{3},1{)}_{-2/3}$, $d_R^c:(\overline{3},1{)}_{1/3}$, and $e_R^c:(1,1{)}_1$; these are alternative descriptions, not extra fermions. The minimal model has no right-handed neutrino. For completeness, the family-independent gauge fields are the spin-one gluons $(8,1{)}_0$, weak gauge bosons $(1,3{)}_0$, and hypercharge gauge boson $(1,1{)}_0$; they are not counted once per family.

For the Standard Model one-loop gauge coefficients, include three fermion families and one Higgs doublet. With $T(\mathbf{N})=1/2$, the colour Dynkin index sum in one family isThe first factor two counts the two members of $Q_L$; the colour trace is already included in the Dynkin index. There is no coloured scalar. HenceFor $SU(2{)}_L$, the three colours of $Q_L$ and the lepton doublet give $3(1/2)+1/2=2$ per family. The Higgs contributes $T_2(\varphi )=1/2$. ThusFor $U(1{)}_Y$, sum squared hypercharges over every colour and doublet component. One family contributesThe two complex Higgs components contribute $2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction term, soThis ${\beta }_1$ is for the unnormalized coupling $g_Y$ used in the question. For $g_1^{\mathrm{GUT}}=\sqrt{5/3}{g}_Y$, the corresponding coefficient is $(3/5){\beta }_1=41/10$. This distinction is required for a consistent unification calculation.

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

The transformation of the holomorphic strong-coupling scale is fixed by the mixed $SU(2{)}^{2}U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\tilde{\Phi }$ cancel, so ${\Lambda }^{b_0}$ is neutral under $U(1{)}_B$. Under $U(1{)}_A$, all $2N_f$ doublets have charge one and Dynkin index one, giving axial charge $2N_f$. For the R-symmetry, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus

Using the supplied Dynkin indices and $p=N+4$, the one-loop beta function of a supersymmetric gauge theory isThus $b_0>0$ for every nontrivial $SU(N)$ in this family, and the electric theory is asymptotically free.

For a field in a tensor-product representation of a product gauge group, its contribution to the beta function of factor $i$ includes the dimensions of all other factors. Those indices label independent copies for that factor. For an Abelian factor, replace its Dynkin index by the squared charge. Thus a quark weak doublet contributes three copies to the weak beta function and two copies to the colour beta function; hypercharge sums include all six components. Chirality conjugation changes a representation and the sign of an Abelian charge, but not its quadratic index.

For three quark generations, their corresponding leptons, and one complex Higgs doublet, use $Q=T_3+Y$ and $D_{\mu }\supset i{g}_{Y}Y{B}_{\mu }$. In $\mu d{g}_i/d\mu =b_i{g}_i^3/(16{\pi }^2)$ the coefficients are the displayed values. Per generation, the fermionic Dynkin index sums are two for both non-Abelian factors and the sum of squared hypercharges is $10/3$. The Higgs index for $SU(2)$ and its sum of squared hypercharges are each $1/2$. Gauge terms are $-22/3,-11$ for $SU(2),SU(3)$, respectively. For the grand-unified normalization $g_1=\sqrt{5/3}{g}_Y$, the Abelian coefficient instead is $41/10$.

The two-index antisymmetric representation of $SU(N)$ is the exterior square of its fundamental representation. For $N\ge 3$, it has dimension $N(N-1)/2$, Dynkin index $N-2$, and cubic anomaly coefficient $N-4$ in the same normalization. For $N=2$, the Dynkin index formula still applies in the stated normalization, but every symmetric cubic trace vanishes, so the stated fundamental-normalized cubic anomaly coefficient is not defined.

The two-index symmetric representation of $SU(N)$ is the symmetric square of its fundamental representation. For $N\ge 3$, it has dimension $N(N+1)/2$, Dynkin index $N+2$, and cubic anomaly coefficient $N+4$ when the fundamental index and anomaly are normalized to one. For $N=2$, the Dynkin index formula still applies in the stated normalization, but every symmetric cubic trace vanishes, so the stated fundamental-normalized cubic anomaly coefficient is not defined.