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For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$ with upper triangular positive roots, the simple coroots are $\mathrm{diag}(1,-1,0)$ and $\mathrm{diag}(0,1,-1)$. Their dual fundamental weights are ${\omega }_1(\mathrm{diag}(t_1,t_2,t_3))=t_1$ and ${\omega }_2(\mathrm{diag}(t_1,t_2,t_3))=t_1+t_2=-t_3$. In simple-root coordinates, ${\omega }_1=(2{\alpha }_1+{\alpha }_2)/3$ and ${\omega }_2=({\alpha }_1+2{\alpha }_2)/3$. The primitive vectors $e_1\in {\mathbb{C}}^3$ and $e_1\wedge e_2\in {\Lambda }^2{\mathbb{C}}^3$ realize these Fundamental representations of sl3.

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 $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA 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 systemThe 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 $-H_1/2+\sqrt{3}{H}_2/2$ 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: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 ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe 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 $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, 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 $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, 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 $3\otimes \overline{3}$, 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 areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation: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 $3$. 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 $Y=2H_2/\sqrt{3}$. ThenThe 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 $\overline{3}$ 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 byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. 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 $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ 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 rowsThe 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 $Q=-1$.

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 hasThe 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. 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.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincaré-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.

The weights of the seven-dimensional irreducible $G_2$ representation are zero and the six short roots. In the $A_2$ weight lattice, the three positive short roots areTogether with their negatives, they split into the weight sets of the two dual three-dimensional Fundamental representations of sl3; the zero weight supplies a trivial representation. By the Restriction of the seven-dimensional G2 representation to long-root A2,

For ${\mathfrak{sl}}_3(\mathbb{C})$ there are two fundamental weights. The corresponding Fundamental representations of sl3 are