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The root system of the complex special linear Lie algebra ${\mathfrak{sl}}_6(\mathbb{C})$ is $\{e_i-e_j:1\le i\ne j\le 6\}$ in ${\sum }_i{x}_i=0$. Its simple roots $e_i-e_{i+1}$ give a five-node An Dynkin diagram. The defining fundamental representation has six weights $w_i=e_i-\frac{1}{6}{\sum }_j{e}_j$. The dual representation has weights $-w_i$, and the two-index antisymmetric representation has fifteen weights $w_i+w_j$, $i<j$, all with weight multiplicity one.

Write a matrix-valued gauge field ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ in the fundamental representation. Charge conjugation sends it to $-{\mathcal{A}}_{\mu }^T$ because reordering the fermions transposes both the spinor and internal matrices. The identity $[A^T,B^T]=-[A,B{]}^T$ then sends the gauge field strength to $-{\mathcal{F}}_{\mu \nu }^T$. This distinction prevents treating all color components as individually C-odd.

A single complex scalar field in the two-dimensional fundamental representation with potential $\lambda ({\varphi }^{†}\varphi -v^2/2{)}^2$, $\lambda ,v>0$, has a three-sphere vacuum manifold. The stabilizer in $SU(2)$ of a chosen nonzero vector is trivial, so all three gauge bosons acquire the common mass $gv/2$. In unitary gauge, $\varphi =(0,v+h{)}^T/\sqrt{2}$ leaves one physical radial scalar of squared mass $2\lambda v^2$. Its couplings are $g^{2}vh{A}^2/4+g^2{h}^2{A}^2/8-\lambda v{h}^3-\lambda h^4/4$. The three angular fields become longitudinal vector polarizations through the Higgs mechanism.

For the fundamental representation of SU(N), take traceless Hermitian generators with $\mathrm{Tr}(T^a{T}^b)={\delta }^{ab}/2$. Along with the identity matrix, they form an orthogonal basis of all complex matrices. Expanding a matrix in this basis and subtracting its identity component proves the displayed completeness relation. It separates exchanged-index and unchanged-index contributions to color amplitudes.

Set $\epsilon =i{\tau }_2$, whose Pauli matrix identity is $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$. Exponentiation gives $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. Hence complex conjugating a Higgs doublet and multiplying by $\epsilon$ restores the fundamental representation of $SU(2)$, while reversing hypercharge. A Higgs field of hypercharge $1/2$ yields a conjugate doublet of hypercharge $-1/2$. Its product ${\tilde{\varphi }}^{†}L$ with a lepton doublet $L$ of hypercharge $-1/2$ transforms in the trivial representation of the whole electroweak gauge group. At vacuum $(0,v{)}^T/\sqrt{2}$, this composite is $v{\nu }_L/\sqrt{2}$.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.

Choose $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^a{t}^a$ on the Dirac field, $t^a={\sigma }^a/2$ in the fundamental representation of SU(2), and $[t^a,t^b]=i{ϵ}^{abc}{t}^c$. Define the gauge field strength and the Adjoint representation gauge covariant derivative byThe required gauge-fixed Yang-Mills Lagrangian in a covariant gauge, including the matter term, isHere $c^a$ and ${\bar{c}}^a$ are independent Grassmann fields in the Adjoint representation: the Faddeev-Popov ghost fields and antighost fields. Under an infinitesimal gauge transformation, $\delta A_{\mu }^a=(D_{\mu }\omega {)}^a$; hence the Lorenz gauge function $G^a={\partial }^{\mu }{A}_{\mu }^a$ varies by ${\partial }^{\mu }{D}_{\mu }^{ab}{\omega }^b$. Its Faddeev-Popov determinant is represented by the ghost term, with an irrelevant field-independent determinant sign absorbed into normalization. Integration by parts gives ${\mathcal{L}}_{\mathrm{gh}}=({\partial }^{\mu }{\bar{c}}^a)(D_{\mu }c{)}^a$, explicitly showing the ghost-gluon vertex.

There is no scalar vacuum expectation value here, so R-xi gauge reduces to the linear covariant gauge displayed above, with no Goldstone boson term. At $\xi =1$ the Feynman gauge gauge boson propagator is $-i{\delta }^{ab}{g}_{\mu \nu }/(k^2+i0)$. With Fourier convention $e^{-ikx}$ and all three boson four-momenta incoming, this choice gives exactly the source's $+g{ϵ}^{abc}$ three-boson Feynman rule, and the matter interaction vertex is $+ig{\gamma }^{\mu }{t}^a$. The sign of $D_{\mu }$ has been explicitly chosen here; the QED convention in question 2 instead has vertex $-ie{\gamma }^{\mu }$.

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.

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.

Use the integer normalization $c(3)=1$ from solution. For two fundamental representations, the specified discrete gauge anomaly coefficient is $\mathcal{A}=q_1+q_2$. Consequently all solutions in the finite cyclic group of charges areThere are $n$ ordered charge pairs. The trivial pair is included: the problem does not require a faithful action. If a faithful action on these Weyl fermions were additionally required, one would retain precisely the $q$ coprime to $n$.

This cancels the discrete gauge anomaly defined in the paper. Two left-handed fundamental representations by themselves have a nonzero pure cubic SU(3) gauge anomaly; a complete gauge theory would need additional fermions to cancel it. That separate condition does not alter the requested discrete congruence.

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

Choose the first $SU(2)$ as $S{U}_w(2)$ and embed $S{U}_c(3)$ in the upper-left three-by-three block of $SU(4)$. The fundamental representations then restrict as $4\to 3\oplus 1$ and $\overline{4}\to \overline{3}\oplus 1$. Thus the representation in solution, forgetting hypercharge, restricts toThis contains every required $S{U}_w(2)\times S{U}_c(3)$ representation, with one extra singlet.

For the hypercharge embedding, let $X=\mathrm{diag}(1,1,1,-3)$ in the $SU(4)$ fundamental representation, and $T_{3R}=\mathrm{diag}(1/2,-1/2)$ in the second $SU(2)$. These commute with $S{U}_w(2)\times S{U}_c(3)$. The integer normalization in the paper isExplicitly, this circle group embeds in the covering group byBoth nontrivial blocks have determinant one, and the map is injective since the first three entries of the $SU(4)$ block recover $z$.

On $(2,1,4)$, $T_{3R}=0$, so the branching of a representation gives $(2,3{)}_1\oplus (2,1{)}_{-3}$. On $(1,2,\overline{4})$, the U(1) charges under $X$ are $-1$ on $\overline{3}$ and $3$ on $1$, while $6T_{3R}$ takes values $\pm 3$. HenceCombining the two summands givesThis is the Pati-Salam charge embedding, with $X=3(B-L)$. In terms of the common Standard Model convention $Q_{\mathrm{electric}}=T_{3L}+Y_{\mathrm{SM}}$, the paper's integer hypercharge is $Y=6Y_{\mathrm{SM}}$.

The extra neutral singlet can be interpreted as the left-handed charge conjugate ${\nu }_R^c$ of a right-handed neutrino. It is a sterile neutrino under the Standard Model gauge group, and completes the sixteen Weyl fermions of one family in this half-spin representation. Its neutrality permits a Majorana mass term without violating the remaining Standard Model gauge group.

The scalar fields commute, so a degree-$d$ expression in one doublet belongs to a symmetric power, rather than an arbitrary tensor product. Put $x={\varphi }_1$, $y={\varphi }_2$. The generators of a group $r={\tau }_1{\tau }_2$ and $s={\tau }_3$ act byA monomial $x^a{y}^b$ is fixed by $r$ precisely when $a-b$ is divisible by three. Imposing $s$ then pairs the exponents. Through degree four the resulting holomorphic polynomial invariants areIn particular, two explicit invariant terms are ${\varphi }_1{\varphi }_2$ and ${\varphi }_1^3+{\varphi }_2^3$. Their invariance follows directly from ${\alpha }^3=1$ and the interchange of the two components. A quartic example is $({\varphi }_1{\varphi }_2{)}^2$.

For a physical real scalar potential, complex conjugate fields are allowed as well. The Hermitian inner product ${\varphi }^{†}\varphi$ and its square are also invariant, while a holomorphic term can enter with its complex conjugate, for example $b({\varphi }_1^3+{\varphi }_2^3)+\bar{b}({\bar{\varphi }}_1^3+{\bar{\varphi }}_2^3)$. There is still no invariant linear term, because neither the doublet nor its conjugate representation contains a trivial representation. Thus the possible nonconstant total degrees areThe constant term is unrestricted.

For the fundamental representation of the special unitary group $SU(2)$, its central element $-I$ sends $\varphi \mapsto -\varphi$ and $\bar{\varphi }\mapsto -\bar{\varphi }$. Every odd-degree polynomial changes sign, excluding degrees one and three. Further, $SU(2)$ acts transitively on the unit sphere in ${\mathbb{C}}^2$: a unit vector $(u,v{)}^T$ is the first column ofAn invariant polynomial must therefore depend only on ${\varphi }^{†}\varphi$. Through quartic order the general real scalar potential isHence the possible nonconstant degrees are $2,4$; the cubic term permitted by $S_3$ is excluded by $SU(2)$.

If the notation $V(\varphi )$ is instead read as restricting to holomorphic polynomials alone, the distinction matters: a single commuting $SU(2)$ doublet has no nonconstant holomorphic polynomial invariant. The candidate contraction ${ϵ}_{ab}{\varphi }_a{\varphi }_b={\varphi }_1{\varphi }_2-{\varphi }_2{\varphi }_1$ vanishes identically. The quadratic and quartic $SU(2)$ terms above require complex conjugate fields. This also explains why counting a trivial representation in $2\otimes 2$ alone would give an incorrect holomorphic answer.

Write $[r_1,r_2,\dots ]$ for a Young diagram with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the fundamental representation to another single box gives a row or a column:The corresponding symmetric tensors and antisymmetric tensors have dimensions $4\cdot 5/2=10$ and $4\cdot 3/2=6$, soFor the conjugate representation, the invariant volume tensor identifies $\bar{4}\cong {\Lambda }^{3}4$, represented by a column of three boxes. Adding one box givesA column of four boxes is the determinant representation, which is trivial on $SU(4)$. The remaining Young diagram has Dynkin labels $(1,0,1)$ and dimension fifteen. For example the hook-content formula givesEquivalently, a mixed tensor $M^i{}_j$ transforms by $M\mapsto UM{U}^{†}$ and splits into its trace and traceless matrix parts. Thereforewhere $15$ is the Adjoint representation.

For each complex coordinate, exchanging the factors of a wedge product of differential forms reverses its sign, givingConsequently the Kähler form is $\omega =d{x}_{14}+d{x}_{25}+d{x}_{36}$, and $\omega \wedge d{x}_7=d{x}_{147}+d{x}_{257}+d{x}_{367}$. Expanding the holomorphic volume form givesThe sign of $d{x}_{426}$ changes on exchanging $4,2$, while reordering $4,5,3$ to $3,4,5$ requires two exchanges. This verifies every sign inUnder $z\mapsto Uz$, unitarity preserves $\omega$:The determinant is the factor by which a transformation acts on a top exterior power, so ${\Omega }^{\prime }=(\det U)\Omega =\Omega$ for $U\in SU(3)$. Extend the real embedding by fixing $y$. It preserves $\omega$, $dy$ and $\Omega$, and therefore $\varphi$. ThusThe induced seven-dimensional representation is exactly the restriction already computed in c. In the conventional complex notation its branching rule isOver the real numbers, the six-dimensional summand is the underlying real space of the complex fundamental representation, and the last real line is the fixed $y$ axis.

Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\dots ,5$ denote the remaining coordinates. The fundamental representation and its conjugate representation immediately giveFor a symmetric tensor, $S^{00}$ is a trivial representation, $S^{0a}=S^{a0}$ is a $5$, and $S^{ab}$ is the fifteen-dimensional symmetric square of a group representation of $5$. For an antisymmetric tensor, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $5$, and $A^{ab}$ is the ten-dimensional exterior power ${\Lambda }^{2}5$. HenceHere the left-hand $15$ is the antisymmetric tensor of $SU(6)$, while the right-hand $15$ in the first equation is the symmetric tensor of $SU(5)$; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrixThe traceless part of $D$ is the Adjoint representation $24$, the column $c$ is $5$, the row $b$ is $\bar{5}$, and the one remaining diagonal trivial representation is generated by $\mathrm{diag}(-5,1,1,1,1,1)$. ThusAll dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

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.

Let $V={\mathbb{C}}^6$ be the defining fundamental representation of $SU(6)$. Its weights are $w_i=e_i-\frac{1}{6}{\sum }_{j=1}^6{e}_j$. Pairing with the simple coroots gives Dynkin labels $(w_i{)}_a={\delta }_{ia}-{\delta }_{i,a+1}$. Thus the complete list isSuccessive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation $\overline{6}\cong V^{\ast }$ is the dual representation and has the negatives of these weights. Its highest weight is $-w_6=(0,0,0,0,1)$, and its entire weight decomposition isEvery entry has weight multiplicity one.

The two-index antisymmetric representation is ${\Lambda }^{2}V=15$, of dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ and highest weight $w_1+w_2=(0,1,0,0,0)$. A weight basis is $e_i\wedge e_j$ for $i<j$, carrying weight $w_i+w_j$. The fifteen sums, all distinct and all of weight multiplicity one, areThe exclusion of $i=j$ comes from the exterior square, where $e_i\wedge e_i=0$.

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.