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For B3 root system simple roots $e_1-e_2,e_2-e_3,e_3$, choose the D3 root system simple roots in chain order $e_2-e_3,e_1-e_2,e_2+e_3$. The Dynkin labels project by $(a,b,c)\mapsto (b,a,b+c)$. This separates the vector weights $\{0,\pm e_i\}$ into the six-dimensional exterior square of the defining SU(4) representation and a singlet representation. The spinor cube vertices $\frac{1}{2}(\pm e_1\pm e_2\pm e_3)$ split into its defining representation and its dual representation.

Mixing of the flavour singlet and isoscalar octet vector states so that the mass eigenstates separate into nonstrange and strange flavour. In the basis $|1⟩=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$, $|8_0⟩=(2s\bar{s}-u\bar{u}-d\bar{d})/\sqrt{6}$, the determinant-one rotation to $(\omega ,\varphi )$ has angle $-\arctan (1/\sqrt{2})$. Basis phases change the quoted angle but not the flavour content.

An eight-dimensional representation. For SU(3), the adjoint octet is realized by traceless matrices, and decomposing $\bar{3}\otimes 3$ separates this traceless part from the scalar singlet.

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 ${\Lambda }^{3}U\cong U^{\ast }$.

For the usual colour SU(3) fixing the fourth coordinate of the SU(4) defining representation,Taking the exterior square givesThe identification ${\Lambda }^{2}3\cong \overline{3}$ follows by contracting an antisymmetric tensor with the invariant three-index Levi-Civita symbol. Combining this with c gives all colour multiplets:A colour triplet has the color charge transformation law of a quark, an antitriplet that of an antiquark, and a colour singlet is colour-neutral, as a lepton is. Thus both parent representations permit quark-like and antiquark-like colour sectors; the vector has one neutral sector and the spinor two. These are colour assignments only. They do not determine particle spin, statistics, electric charge, weak quantum numbers or a full generation of the Standard Model. In a real parent field the conjugate colour sectors may describe one complex colour field and its conjugate, rather than independent particle species.

Global group qualification. The printed chain is correct at the Lie algebra level, $\mathfrak{su}(3)\subset \mathfrak{su}(4)\cong \mathfrak{so}(6)\subset \mathfrak{so}(7)$. Globally its simply connected form is $SU(3)\subset SU(4)\cong \mathrm{Spin}(6)\subset \mathrm{Spin}(7)$. The standard map of SU(4) into $SO(7)$ through $6\oplus 1$ has a nontrivial kernel: the concrete central element $-I_4$ acts as $+I_6$ on ${\Lambda }^{2}4$ and trivially on the singlet representation. Its image is $SO(6)\cong SU(4)/\{\pm I_4\}$, rather than an injective copy of SU(4) under this map.

The global form obstruction for the Spin7 spinor is explicit: the central element $-1$ of the Spin group $\mathrm{Spin}(7)$ maps to the identity in $SO(7)$ but acts as $-I_8$ on the eight-dimensional spin representation. Equivalently $-I_4$ acts as $-I$ on both $4$ and $\overline{4}$ in its restriction. Therefore the spinor does not descend to a genuine SO(7) group representation. A theory containing that spinor needs the Spin group global form, or must understand the source's notation as specifying only its Lie algebra. The displayed colour branchings remain valid with this correction.

Let $z=\cos {\theta }^{\star }$. Rotating the spin-one intermediate state between initial and final axes gives probability amplitudes proportional to $d_{\pm 1,\pm 1}^1$ for equal fermion helicities and $d_{\mp 1,\pm 1}^1$ for opposite ones. Thus, apart from a common positive normalization,

The common factor $1/4$ from the rotation elements is absorbed into the normalization. Distinct external helicities are orthogonal, so an unpolarized rate adds their squared probability amplitudes incoherently.

The source interchanges the conventional left and right electron coupling labels. With $g_L=T_3-Q{\sin }^2{\theta }_W$ and $g_R=-Q{\sin }^2{\theta }_W$, the physical electron values areAt ${\sin }^2{\theta }_W=0$, for example, the left electron still carries weak isospin $-1/2$, while the right electron is a weak singlet; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.^[ref]

For a muon final state, lepton universality gives the same pair of couplings. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; thenHere $\mathcal{N}$ includes the resonance propagator, initial spin average and common coupling normalization. Equivalently,The source's simultaneous label interchange reverses ${\mathcal{A}}_{\ell }$ but leaves its square unchanged.

Use the orthonormal basis $|u\bar{u}⟩,|d\bar{d}⟩,|s\bar{s}⟩$. The normalized flavour singlet isThe SU(3) invariant contraction pairs each fundamental flavour with its conjugate. Applying any supplied raising operator to the singlet gives zero because its quark and antiquark contributions cancel.

To fix the neutral isospin triplet phase consistently with the source's charged states, apply $I_{+}$ to $-|d\bar{u}⟩$:Thus one missing member of the meson octet isIndeed $I_{+}|T_0⟩=\sqrt{2}|u\bar{d}⟩$, the correct ladder operator strength for the $I=1$ triplet.

The other missing state is orthogonal to both $|1⟩$ and $|T_0⟩$. For coefficients $(a,b,c)$, orthogonality requires $a=b$ and $a+b+c=0$. Choose the convenient overall phase $c>0$ and normalize:It can also be obtained by adding the supplied ladder operations $V_{+}[-|s\bar{u}⟩]=|s\bar{s}⟩-|u\bar{u}⟩$ and $U_{+}[-|s\bar{d}⟩]=|s\bar{s}⟩-|d\bar{d}⟩$. The three displayed states are mutually orthonormal. The meson octet here is the traceless adjoint component of $\bar{3}\otimes 3$; the source's word “antisymmetric” does not impose antisymmetry on all its quark-antiquark flavour wavefunctions.

Only the singlet $|1⟩$ and isoscalar octet member $|8_0⟩$ can mix without violating isospin: they have the same $I=I_3=S=0$ and, for the vector multiplet, the same $J^{PC}=1^{--}$. The neutral triplet member has $I=1$ and is excluded in the isospin limit.

Set $|{\psi }_a⟩=|1⟩$ and $|{\psi }_b⟩=|8_0⟩$, with precisely the phases in a. Define $|n⟩=(|u\bar{u}⟩+|d\bar{d}⟩)/\sqrt{2}$. Taking overlaps givesTherefore the source's rotation convention is satisfied byThis yields $|\omega ⟩=|n⟩$ and $|\varphi ⟩=|s\bar{s}⟩$ exactly. Reversing the octet phase changes the quoted sign, with an irrelevant compensating phase on the physical phi state; exchanging the basis order gives the complementary angle. The physical statement of ideal vector-meson mixing is the pure strange and pure nonstrange content, rather than a convention-independent sign for the angle.

The observed dominance of kaons supports a predominantly $s\bar{s}$ phi meson. In a connected strong decay, the original strange quark and antiquark survive into different kaons, while a gluon produces a light $q\bar{q}$ pair:This gives $K^{-}{K}^{+}$ or ${\bar{K}}^0{K}^0$ and is allowed by the OZI rule. A pionic final state contains no strange valence constituents, so the initial $s\bar{s}$ pair must annihilate. The strong annihilation of a colour-singlet $1^{--}$ state requires at least three gluons: one gluon cannot carry the singlet colour, and two have the wrong charge-conjugation quantum number. This disconnected flavour topology is suppressed by the OZI rule, although it need not vanish.

The leptonic mode proceeds through $s\bar{s}\to {\gamma }^{\ast }\to {\ell }^{+}{\ell }^{-}$ and is suppressed by electromagnetic couplings; its width is proportional to ${\alpha }^2$ times the vector-meson overlap. Unlike a strong hadronic decay, it cannot create its final particles through quark rearrangement. Hence a branching fraction of order $10^{-4}$ is qualitatively consistent with electromagnetic annihilation competing with strong modes.

Kaon dominance is especially suggestive because phase space is limited: using the supplied neutral-kaon and phi masses,The two spin-zero kaons have orbital angular momentum $L=1$ to match the vector parent, so their partial width has the additional near-threshold factor $p_K^3$. Pionic modes have more phase space yet remain subdominant because of their annihilation topology. The observations therefore support, but do not establish exact, ideal vector-meson mixing: a small nonstrange admixture, electromagnetic amplitudes and nonperturbative strong dynamics can all contribute to the pionic rate.

A one-dimensional group representation. For connected semisimple groups such as SU(2) or SU(3), its continuous irreducible singlet is trivial, so all Lie-algebra generators act as zero. A singlet of one subgroup need not be a singlet of a larger group.