= Solution
The <SU(3) Lie algebra> is the compact real algebra of traceless <skew-Hermitian matrices>; its complexification is $\mathfrak{sl}_3(\mathbb C)$, a <special linear Lie algebra>. In an antihermitian <Lie algebra representation>, the two commuting Cartan operators $H_1,H_2$ are <Hermitian matrices>, and the root operators satisfy $(E_+^a)^\dagger=E_-^a$. Thus the <spectral theorem> gives an orthogonal <weight-space decomposition>
$$
V=\bigoplus_\mu V_\mu,\qquad H_rv=\mu_rv\quad(v\in V_\mu),\qquad \mu=(\mu_1,\mu_2)\in\mathbb R^2.
$$
A <weight diagram> plots the pairs $\mu$, recording the <weight multiplicity> $\dim V_\mu$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the <roots of a root system>
$$
\alpha_1=(1,0),\qquad \alpha_2=\left(-\frac12,\frac{\sqrt3}2\right),\qquad \alpha_3=\alpha_1+\alpha_2.
$$
The <raising operators> $E_+^a$ move a nonzero <weight vector> by $+\alpha_a$ and the <lowering operators> move it by $-\alpha_a$. These six root directions form the <A2 root system>. Each root and its negative generate an <SU(2) representation> ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt3H_2/2$ constrain the two simple-root strings.
Choose a linear functional positive on the three positive roots. A maximal <weight> for this ordering is killed by every <raising operator>. In an <Irreducible Lie algebra representation> it generates the entire representation by lowering, and its <weight space> has dimension one. The two <SU(2) representation> strings show that its <Dynkin labels> are nonnegative integers:
$$
p=2\lambda_1,\qquad q=-\lambda_1+\sqrt3\lambda_2,\qquad p,q\in\mathbb Z_{\ge0}.
$$
Conversely, each such pair specifies exactly one finite-dimensional irreducible <highest-weight representation>. Write it as $R(p,q)$. The <fundamental weights> in these Cartesian coordinates are $\omega_1=(1/2,1/(2\sqrt3))$ and $\omega_2=(0,1/\sqrt3)$, so
$$
\lambda=p\omega_1+q\omega_2=\left(\frac p2,\frac{p+2q}{2\sqrt3}\right),\qquad
\boxed{\dim R(p,q)=\frac{(p+1)(q+1)(p+q+2)}2.}
$$
The dimension follows from the three positive-root factors in the <Weyl dimension formula>: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every <weight> negated. The <SU(3) highest-weight coordinates> here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.
The <weight diagram> is invariant under the <Weyl group>, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the <highest weight>. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda+$ root lattice. A point in a different <weight lattice> coset is not a weight merely because it lies inside the polygon.
The <shell multiplicities in an SU(3) weight diagram> can be stated geometrically. The outer boundary has <weight multiplicity> one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual <weight multiplicity> by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min(p,q)+1$; when $p=q$ it is just the central point. A triangular <weight diagram> has multiplicity one everywhere. Thus $R(1,1)=\mathbf8$, the <adjoint representation of SU(3)>, has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=\mathbf{15}$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=\mathbf{27}$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot2+3=27$. The <dominant weight multiplicity formula for sl3> is an algebraic expression of the same rule.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-52-su3-weights.png]
{title=Triangular and hexagonal SU(3) weight diagrams, with each weight-space dimension shown at its point}
For a <tensor product of Lie algebra representations>, the Cartan operators add. Consequently its <tensor-product weight diagram> has <weights> $\mu+\nu$ and multiplicities
$$
m_{V\otimes W}(\rho)=\sum_{\mu+\nu=\rho}m_V(\mu)m_W(\nu).
$$
Geometrically, place a copy of the second <weight diagram> at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining <dominant weight>, identify its irreducible <highest-weight representation>, subtract that representation's whole diagram with its <weight multiplicities>, and repeat. This <highest-weight character subtraction> must exhaust every weight, not merely the corners. Unlike the <Clebsch-Gordan decomposition for SU2>, an SU(3) product may contain repeated copies of an irreducible representation.
The two three-dimensional <Fundamental representations of sl3> are $\mathbf3=R(1,0)$ and $\overline{\mathbf3}=R(0,1)$. In $\mathbf3\otimes\mathbf3$, the largest weight is $2\omega_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $\mathbf3\otimes\overline{\mathbf3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the <octet and singlet in a fundamental SU3 tensor product> and the two-quark product are
$$
\boxed{\mathbf3\otimes\mathbf3=\mathbf6\oplus\overline{\mathbf3},\qquad
\mathbf3\otimes\overline{\mathbf3}=\mathbf8\oplus\mathbf1.}
$$
The first dimensions are $9=6+3$, and the second are $9=8+1$. Next $\mathbf6\otimes\mathbf3$ has highest weight $3\omega_1$, giving $R(3,0)=\mathbf{10}$. Subtraction leaves one $R(1,1)=\mathbf8$. Together with $\overline{\mathbf3}\otimes\mathbf3=\mathbf8\oplus\mathbf1$, this proves the <tensor cube of the defining SU(3) representation>:
$$
\boxed{\mathbf3^{\otimes3}=\mathbf{10}\oplus\mathbf8\oplus\mathbf8\oplus\mathbf1,\qquad 27=10+8+8+1.}
$$
The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.
For the <quark model>, the relevant approximate <flavour symmetry> rotates the three light <quarks> $u,d,s$ in the fundamental $\mathbf3$. The <up quark>, <down quark>, and <strange quark> basis states have the <weight diagram of the defining SU(3) representation>. In physical coordinates define <isospin> $I_3=H_1$ and <flavor hypercharge> $Y=2H_2/\sqrt3$. Then
$$
\begin{array}{c|ccc}
&u&d&s\\\hline
I_3&1/2&-1/2&0\\
Y&1/3&1/3&-2/3\\
Q&2/3&-1/3&-1/3
\end{array}
$$
The last row follows from the <Gell-Mann--Nishijima formula> $Q=I_3+Y/2$, in units of the positive elementary <electric charge>. The <antiquarks> transform in $\overline{\mathbf3}$ and have the negatives of these additive quantum numbers. Each <quark> has <baryon number> $B=1/3$; the <strange quark> has <strangeness> $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing <flavor hypercharge> from electroweak hypercharge. The <Weyl group> geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.
A conventional <meson> has valence content $q\bar q$, so its <flavour symmetry> gives a <meson octet> and a singlet. For the light <pseudoscalar mesons>, the octet contains the <pions> $\pi^+,\pi^0,\pi^-$ at $Y=0$ with <isospin> $I=1$, the <kaons> $K^+=u\bar s,K^0=d\bar s$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ <Eta octet state>. The central <weight multiplicity> two is realized by
$$
\pi^0=\frac{u\bar u-d\bar d}{\sqrt2},\qquad
\eta_8=\frac{u\bar u+d\bar d-2s\bar s}{\sqrt6}.
$$
The orthogonal <eta singlet state> is $\eta_1=(u\bar u+d\bar d+s\bar s)/\sqrt3$. The <eta and eta prime mesons> are mixtures of $\eta_8,\eta_1$, rather than exact octet and singlet mass eigenstates; this gives the <pseudoscalar meson nonet>. A different spin coupling of the same flavour product gives the <vector mesons>, again an octet and singlet, including the $\rho$, $K^*$, and neutral $\omega,\phi$ mixtures. The lowest orbital states have $J^P=0^-$ and $1^-$ respectively.
A conventional <baryon> has three valence <quarks>, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest <baryon octet> has spin $1/2$ and positive <parity>. Its <isospin> rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the <nucleons> $p,n$, the three <Sigma baryons>, the <Lambda baryon>, and the two <Xi baryons>. The central states $\Sigma^0$ and $\Lambda$ have the same <weight> $(I_3,Y)=(0,0)$ but different total <isospin>; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity <baryon decuplet> is the triangle with rows
$$
\begin{array}{c|c|c}
Y&I&\text{states}\\\hline
1&3/2&\Delta^{++},\Delta^+,\Delta^0,\Delta^-\\
0&1&\Sigma^{*+},\Sigma^{*0},\Sigma^{*-}\\
-1&1/2&\Xi^{*0},\Xi^{*-}\\
-2&0&\Omega^-
\end{array}
$$
The <Delta baryons> form its top row, and the <Omega baryon> is the bottom $sss$ state. All ten <weights> have multiplicity one. Their <electric charges> follow directly from the <Gell-Mann--Nishijima formula>; for example $\Delta^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and $\Omega^-$ has $I_3=0,Y=-2$, so $Q=-1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-52-hadron-multiplets.png]
{title=Light pseudoscalar meson octet, spin-one-half baryon octet, and spin-three-halves baryon decuplet in isospin and flavour-hypercharge coordinates}
The two algebraic octets do not mean that two independent ground-state octets must occur. A <three-quark colour singlet> has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the <Pauli exclusion principle> therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the <symmetric spin-flavour SU6 representation> has
$$
\operatorname{Sym}^3(\mathbb C^3\otimes\mathbb C^2)
\cong(\mathbf{10},\mathbf4)\oplus(\mathbf8,\mathbf2),\qquad 56=10\cdot4+8\cdot2.
$$
The <three-quark flavour singlet> is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but $\bigwedge^3\mathbb C^2=0$. The <Pauli constraint on three-quark flavour multiplets> thus excludes it from this ground-state sector; orbital excitation permits other multiplets.
Finally, this <flavour symmetry> is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The <weight diagrams> classify flavour states, rather than asserting exact equality of their masses. Colour <SU(3)> is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the <eightfold way>. \b[The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.]
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