= Solution
For the usual colour <SU(3)> fixing the fourth coordinate of the <SU(4)> defining <representation>,
$$
\mathbf4\downarrow_{SU(3)}=\mathbf3\oplus\mathbf1,\qquad
\overline{\mathbf4}\downarrow_{SU(3)}=\overline{\mathbf3}\oplus\mathbf1.
$$
Taking the <exterior square> gives
$$
\Lambda^2(\mathbf3\oplus\mathbf1)=\Lambda^2\mathbf3\oplus\mathbf3
=\overline{\mathbf3}\oplus\mathbf3.
$$
The identification $\Lambda^2\mathbf3\cong\overline{\mathbf3}$ follows by contracting an <antisymmetric tensor> with the invariant three-index <Levi-Civita symbol>. Combining this with <4/c> gives all colour <multiplets>:
$$
\boxed{\mathbf7\downarrow_{SU(3)}=\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1,\qquad
\mathbf8\downarrow_{SU(3)}=\mathbf3\oplus\overline{\mathbf3}\oplus2\mathbf1.}
$$
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.
\b[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 $SU(3)\subset SU(4)\cong\operatorname{Spin}(6)\subset\operatorname{Spin}(7)$. The standard map of <SU(4)> into $SO(7)$ through $\mathbf6\oplus\mathbf1$ has a nontrivial <kernel>: the concrete central element $-I_4$ acts as $+I_6$ on $\Lambda^2\mathbf4$ 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> $\operatorname{Spin}(7)$ 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 $\mathbf4$ and $\overline{\mathbf4}$ 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.
Back to article page
