= Quark-model baryon-decuplet construction
{title2=$\dim\operatorname{Sym}^3\mathbb C^3=\binom{5}{3}=10$}
A particularly economical example is the <baryon decuplet>. Restrict to fully symmetric three-<quark> flavor states. The nonnegative occupation numbers satisfy $n_u+n_d+n_s=3$, giving $\binom{3+3-1}{3}=10$ possibilities by <stars and bars>. Their additive labels are
$$
Q=n_u-1,\qquad I_3=\frac{n_u-n_d}{2},\qquad Y=1-n_s.
$$
These formulas follow directly from the constituent table. With fixed $n_s$, there are $4-n_s$ possible values of $n_u$, so the <weight diagram> has rows of four, three, two and one states. Its outer triangle is therefore a consequence of the constituent construction, rather than an arbitrary drawing convention.
\Image[/quark-flavor-weights-and-baryon-decuplet.png]
{title=Three light-flavor weights generate the triangular <baryon decuplet>}
{description=Original exact constituent calculation using $Y=B+S$. The left panel shows the fundamental flavor triplet. The right panel adds three such weights in fully symmetric flavor states; every point is labelled by its constituent content and <electric charge>. No measured masses or event counts are plotted.}
The bottom corner has $n_s=3$, hence $I_3=0$, $Y=-2$ and $Q=-1$: the <Omega baryon> position. The top-right corner has $uuu$ and $Q=2$, corresponding to the doubly charged <Delta baryon>. Every point satisfies the <Gell-Mann--Nishijima formula> $Q=I_3+Y/2$.
The flavor construction is only part of the state. The <Pauli constraint on three-quark flavour multiplets> also involves <spin angular momentum>, the spatial <wavefunction> and <color charge>. In particular, three identical <fermions> cannot be assigned a completely symmetric total state merely because the flavor diagram is symmetric. Nor does the same flavor diagram determine the <hadron> <masses>: an approximate <flavor symmetry> can organize a spectrum without making all its members degenerate.
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