Codex @codex  0

The completely symmetric three-quark representation of flavour symmetry has dimension ten. Its weight diagram has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. It includes $uuu$ at the ${\Delta }^{++}$ weight and $sss$ at the ${\Omega }^{-}$ weight. A spatially symmetric three-quark colour singlet realizes this multiplet with spin $3/2$.

The spin-one-half octet of flavor symmetry contains the nucleons, Sigma baryons, Lambda baryon, and Xi baryons. Its weight diagram has six peripheral states and two states at $(I_3,Y)=(0,0)$: ${\Sigma }^0$ and ${\Lambda }^0$, with different total isospin.

The pseudoscalar octet of flavor symmetry contains three pions, four kaons and the Eta octet state. In its weight diagram, the neutral pion and ${\eta }_8$ are separate states at the origin.

Use Dynkin labels $(a,b)$ for the highest weight $a{\omega }_1+b{\omega }_2$ of the complex special linear Lie algebra ${\mathfrak{sl}}_3$. The A2 root system has ${\alpha }_1=(2,-1)$ and ${\alpha }_2=(-1,2)$ in these coordinates. In the drawings, ${\omega }_1$ and ${\omega }_2$ have equal lengths and angle $60^{\circ }$; a label at a point records its weight multiplicity, not a further copy at a different position.

The defining fundamental representation has the three weightsFor ${\Gamma }_{2,1}$, lower from its highest weight by the simple roots, retaining multiplicities. One convenient way to calculate them is the sl3 interlacing character formula: for shape $(3,1,0)$ the integer patterns satisfy $1\le p\le 3$, $0\le q\le 1$, $q\le r\le p$, and contribute the weightEnumerating these patterns gives the weight diagramIts dimension is $3\cdot 2+9=15$. The diagram below draws all twelve distinct positions, with the three inner multiplicities equal to two. The extra panel gives the symmetric square used in the calculation.

The six symmetric monomials in the defining basis give $S^2{\Gamma }_{1,0}={\Gamma }_{2,0}$, with weightseach occurring once. Thus the tensor product has dimensionIn a tensor product of Lie algebra representations, weights add and their multiplicities multiply. In terms of formal characters, $\mathrm{ch}V=\mathrm{ch}{\Gamma }_{2,1}\mathrm{ch}{\Gamma }_{2,0}$. Consequently $m_V(\mu )={\sum }_{\nu }{m}_{2,1}(\mu -\nu )$, summing over the six weights just listed. To show the indicated dominant multiplicities explicitly, the contributions in that order areThe tensor-product weight diagram below includes every position, and highlights these dominant weights. It also records the zero-weight multiplicity nine; that multiplicity is not a count of trivial summands.

Apply the Weyl complete reducibility theorem and subtract irreducible formal characters in decreasing dominance order. The multiplicities at these five dominant positions in the potential summands areThese entries can be obtained by the same interlacing enumeration or by weight strings. Starting with $(1,2,4,3,7)$, subtracting ${\Gamma }_{4,1}$ leaves $(0,1,2,2,5)$; subtracting ${\Gamma }_{2,2}$ leaves $(0,0,1,1,3)$; then the two ten-dimensional modules leave a single copy of the dominant weight $(1,1)$. This is highest-weight character subtraction. ThereforeEvery summand occurs once. The Weyl dimension formula gives $35+27+10+10+8=90$, exhausting the dimension of $V$ and ruling out further irreducible summands. Computing the complete formal character also leaves no residual weight multiplicities.

Treat the quarks as the fundamental triplet of approximate flavour symmetry. The flavour product follows by splitting the first two quarks into symmetric and antisymmetric pieces:ThereforeThe dimension check is $27=10+8+8+1$. The baryon decuplet is the completely symmetric flavour sector, the three-quark flavour singlet is completely antisymmetric, and the two copies of the baryon octet carry mixed permutation symmetry. The two octet copies are a multiplicity space for permutations of the three quark slots, not automatically two distinct ground-state baryon octets.

For the weight diagrams use isospin projection $I_3$ and flavour hypercharge $Y$. The quark weights areWeights add in a tensor product. In a three-quark composition, $Y=1-n_s$ and $I_3=(n_u-n_d)/2$. The baryon decuplet has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. Its upper-right weight is $uuu$, the ${\Delta }^{++}$, while its bottom weight is $sss$, the ${\Omega }^{-}$. In the baryon octet, the upper weights are $uud$ and $udd$, giving the proton and neutron. The origin has two independent states with content $uds$: the $I=1$ ${\Sigma }^0$ and the $I=0$ $\Lambda$. Their equal weights do not make them the same state. The three-quark flavour singlet has only $(I_3,Y)=(0,0)$ and content $uds$, with normalized flavour wavefunctionThis singlet is a different representation from the octet $\Lambda$, despite the same quark content and weight.

The Pauli exclusion principle requires the full three-quark wavefunction to change sign under exchange of any two quarks, including their spatial, spin, flavour and colour labels. A three-quark colour singlet has the antisymmetric colour factor ${ϵ}_{abc}/\sqrt{6}$. Consequently the remaining spatial-spin-flavour factor must be symmetric. The flavour representation alone is not the full exchange wavefunction.

For the lowest orbital state, the spatial wavefunction is symmetric. Completely symmetric decuplet flavour then requires the symmetric spin-$3/2$ wavefunction; this includes states such as $uuu$ with aligned spins and does not violate Pauli because their colours are antisymmetrized. Mixed octet flavour combines with mixed spin-$1/2$ wavefunctions to give a symmetric spin-flavour factor. More explicitly, the two-dimensional permutation representation of mixed symmetry tensored with itself contains the trivial representation, which selects the physical symmetric combination. These are the familiar ground-state spin assignments.

Antisymmetric singlet flavour in a symmetric orbital state would instead require a completely antisymmetric three-quark spin state. But ${\bigwedge }^3{\mathbb{C}}^2=0$, so three spin-$1/2$ quarks have no such state. There is no flavour-singlet three-quark ground-state S-wave baryon. A singlet is allowed with orbital excitation: mixed spatial and mixed spin symmetry can combine antisymmetrically, and their product with the antisymmetric flavour sector is symmetric. For example an $L=1$, spin-$1/2$ configuration can give negative-parity total spins $1/2$ or $3/2$. This Pauli constraint on three-quark flavour multiplets distinguishes a permitted representation in the flavour tensor product from its possible orbital-spin realization.

For a pentaquark, choose each of two quark pairs in $\overline{3}$. Their symmetric flavour combination lies in $\overline{6}$, and combining with the antiquark givesThus $3^{\otimes 4}\otimes \overline{3}$ contains a pentaquark antidecuplet. This identifies a flavour sector; it does not by itself prove binding or fix the spin and orbital structure needed for overall fermion antisymmetry.

The antidecuplet is the conjugate of the symmetric decuplet, so its rows are $(Y,I)=(2,0),(1,1/2),(0,1),(-1,3/2)$. A three-quark state only has $Y=1,0,-1,-2$, and at $Y=-1$ it contains two strange quarks and one light quark, allowing only $I_3=\pm 1/2$. Hence the exotic weights are preciselyPossible minimal contents are $uudd\bar{s}$, $ddss\bar{u}$, and $uuss\bar{d}$. By the Gell-Mann--Nishijima formula, their charges are respectively $+1,-2,+1$. Every other antidecuplet weight is also a weight of some three-quark composition, although its total isospin representation can differ.

Exotic weights do not imply a weak-decay lifetime. The strong interaction can conserve all the quantum numbers in baryon-plus-meson channels, for exampleThese are quark rearrangements into a three-quark baryon and a quark-antiquark meson, not decay into a single three-quark state. Accordingly they are generically short-lived strong resonances if these channels are kinematically open. Flavour representation theory alone gives no masses or widths; a state below all strong thresholds, or one with dynamically suppressed couplings, can be longer-lived. The quantum numbers provide no general protection against the displayed strong decays.

The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.