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The baryon octet contains an isospin triplet ${\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-}$. Each has strangeness $-1$; their valence quark contents are $uus$, $uds$, $dds$.

For three light flavours and two spin states, an approximate $SU(6)$ spin-flavour symmetry acts on the six-dimensional single-quark spin-flavour space. A spatial ground-state three-quark colour singlet requires the symmetric representation of dimension $56$. Under flavour symmetry and ordinary spin rotations, it decomposes as $(10,4)\oplus (8,2)$: the baryon decuplet has spin $3/2$ and the baryon octet spin $1/2$. The product of spin and flavour factors can require a correlated sum, rather than a single unentangled product.

Three light quarks have a six-dimensional single-quark spin-flavour space ${\mathbb{C}}^3\otimes {\mathbb{C}}^2$. A symmetric spatial ground state and the antisymmetric three-quark colour singlet require the symmetric cube of that space. Its dimension is $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$, and restriction to flavour $SU(3)$ and spin $SU(2)$ gives a baryon decuplet of spin $3/2$ and a baryon octet of spin $1/2$. The two mixed-permutation multiplicity spaces pair once in the symmetric sector, explaining why the two flavour-octet copies in the unconstrained tensor cube do not represent two distinct ground-state octets.

An octet operator between baryon octet states has two reduced matrix elements, because the tensor square of the SU(3) adjoint representation contains two octets. The symmetric invariant $d_{ris}$ and antisymmetric invariant $f_{ris}$ give independent component tensors. Equivalently, represent baryons by traceless matrices $B$ and let the operator act as $F[t_i,B]+D(\{t_i,B\}-\frac{2}{3}\mathrm{tr}(t_{i}B)I)$. Normalization choices are absorbed into $F,D$; Hermitian operators permit real coefficients in a Hermitian octet basis.

The baryon octet contains the isospin doublet ${\Xi }^0,{\Xi }^{-}$. Each has strangeness $-2$, with valence quark content $uss$, $dss$.