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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.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger representations containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large representation.

A dual vector must be pulled back by $P^T$, since $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus a G2 root system weight $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet meson retains its values from solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor instance the first three come from the G2 root system weights $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The weight diagram displays the preserved octet and the new states.

The fractional electric charges make these additional states qualitatively different from ordinary quark-antiquark mesons, whose electric charges are integers. This is a consequence of the proposed symmetry extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

Among the representations in c, the $10$ of $SU(5)$ occurs in the $15$ of $SU(6)$, while the simplest source of $\bar{5}$ is $\bar{6}$. Thus the economical direct sum containing the prescribed family isThe extra $SU(5)$ multiplets are one $5$ and one trivial representation. Calling $R$ a representation does not require it to be an irreducible representation; none of the five individual choices in c contains both required $SU(5)$ summands.

There is a further physical qualification if this is to be the entire chiral matter content of a consistent grand unified theory. Merely containing one Standard Model family does not guarantee cancellation of the cubic $SU(6)$ gauge anomaly. In the convention $A(6)=1$, one has $A(\bar{6})=-1$ and $A(15)=2$, so the displayed sum has $A(R)=1$. With these types of representations, the minimal anomaly-free completion is insteadBeyond one family this adds a vectorlike gauge spectrum $5\oplus \bar{5}$ and two trivial representations. The conjugate pair can receive a mass after the appropriate symmetry breaking, without removing the remaining family.^[ref] The first boxed answer addresses representation containment; the second includes the extra consistency condition for a chiral gauge theory.