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For the defining $n$-dimensional special linear Lie algebra module, with $n\ge 2$, write $T_i(p)=e_i\otimes p(y_1,\dots ,y_n)$. The equivariant contraction is $C(T_i(p))={\partial }_{y_i}p$. The map $\iota (f)={\sum }_i{e}_i\otimes y_{i}f$ is also equivariant, since ${\sum }_i{e}_i\otimes y_i$ is an invariant identity tensor. Direct differentiation gives $C\iota =(n+1)I$, so the displayed direct sum follows. Its trace-free part is an irreducible representation of highest weight ${\omega }_1+2{\omega }_{n-1}$. Its highest-weight vector is $e_1\otimes y_n^2$; the Weyl dimension formula gives dimension $n(n-1)(n+2)/2$, exactly $\dim \ker C$.

// Target: semisimple-lie-algebra.bigb

For the complexified Special orthogonal Lie algebra ${\mathfrak{so}}_{2n}$, $n\ge 2$, choose positive roots of a root system $e_i-e_j$ and $e_i+e_j$ for $i<j$. Its Weyl vector is $\rho =(n-1,n-2,\dots ,0)$. Dominant coordinates satisfy ${\lambda }_1\ge \cdots \ge {\lambda }_{n-1}\ge |{\lambda }_n|$. The Weyl dimension formula becomes ${\prod }_{i<j}(({\lambda }_i+{\rho }_i{)}^2-({\lambda }_j+{\rho }_j{)}^2)/({\rho }_i^2-{\rho }_j^2)$. The case $D_2$ is two independent A1 root systems, while $D_3$ is isomorphic to the A3 root system.

For the G2 root system with ${\alpha }_1$ short and ${\alpha }_2$ long, the fundamental weights are ${\omega }_1=2{\alpha }_1+{\alpha }_2$ and ${\omega }_2=3{\alpha }_1+2{\alpha }_2$. The positive coroots, relative to the simple coroots, have coordinates $(1,0),(0,1),(1,3),(2,3),(1,1),(1,2)$. The Weyl dimension formula therefore gives, for nonnegative integers $a,b$,The denominator is the product of the pairings of the Weyl vector with those coroots. The dimensions at ${\omega }_1,{\omega }_2,2{\omega }_1$ are seven, fourteen and twenty-seven.

For distinct ${\ell }_i$ and $c_j$, expand the exponential entries in powers of $t$. Nonzero determinant terms first occur at the distinct powers $0,1,\dots ,m-1$. The coefficient is the product of their two Vandermonde determinants divided by the factorial product. Ratios of such alternants yield the Weyl dimension formula by taking a torus character to the identity.

For a root system with $N$ positive roots, put $A_{\nu }(H)={\sum }_{w\in W}\det (w)e^{(w\nu ,H)}$. Its degree-$m$ Taylor coefficient is alternating separately in $\nu$ and $H$, and hence divisible in each set of variables by the product of positive-root linear forms. Thus it vanishes for $m<N$ and its degree-$N$ term is $C{\prod }_{\alpha >0}(\nu ,\alpha ){\prod }_{\alpha >0}(H,\alpha )$. The Weyl denominator formula at $\nu =\rho$ determines the nonzero constant $C$. Taking the ratio $A_{\lambda +\rho }(tH)/A_{\rho }(tH)$ for regular $H$ and letting $t\to 0$ gives the Weyl dimension formula $\dim L(\lambda )={\prod }_{\alpha >0}(\lambda +\rho ,\alpha )/(\rho ,\alpha )$. Factors converting roots to coroots cancel.

Let $N=|R^{+}|$ and identify the real weight and Cartan spaces using a Weyl group-invariant positive inner product. For a regular $H$, meaning $(\alpha ,H)\ne 0$ for every root, specialize $e^{\mu }$ to $e^{t(\mu ,H)}$. Then the value of the formal character at $t=0$ is $\dim L(\lambda )$, but both Weyl alternants in the Weyl character formula vanish there. We compute their leading terms instead of substituting into that zero-over-zero expression.

ExpandFor every root reflection $s$, replacing $\nu$ by $s\nu$ changes the sign of $P_m$, by replacing $w$ with $ws$ in the sum. Replacing $H$ by $sH$ also changes its sign, by replacing $w$ with $sw$. Thus the polynomial vanishes whenever $\nu$ or $H$ lies on any reflecting hyperplane. It is divisible by ${\prod }_{\alpha >0}(\nu ,\alpha )$ in its first variables and by ${\prod }_{\alpha >0}(H,\alpha )$ in its second variables. Distinct positive roots give distinct linear factors; each product has degree $N$.

The polynomial $P_m$ has degree $m$ in each group of variables. Therefore $P_m=0$ for $m<N$, and for $m=N$ it has the formwith a constant $C$ independent of both variables. At $\nu =\rho$, the Weyl denominator formula givesConsequently $C{\prod }_{\alpha >0}(\rho ,\alpha )=1$. For strictly dominant $\nu =\lambda +\rho$, its leading coefficient is nonzero, and the ratio of the two Weyl alternants tends toThe coroot is ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$, so the root-length factors cancel in each ratio. This proves the Weyl dimension formula by the lowest-degree alternant proof of the Weyl dimension formula, including the limiting step.

Apply the Weyl character formula with $\lambda =\rho$. In the Weyl denominator formula, replace every formal exponential $e^{\mu }$ by $e^{2\mu }$. This givesDividing by $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$ and cancelling each factor yieldsThe last expression is a compact way of writing the same character; individual half-root exponentials may need a larger formal lattice, but their full product lies in the group ring of the weight lattice. Expanding the first expression, a subset $S\subseteq R^{+}$ contributes the weight $\rho -{\sum }_{\alpha \in S}\alpha$. Different subsets with the same sum give that weight's multiplicity. In particular $\dim L(\rho )=2^{|R^{+}|}$, consistent with the Weyl dimension formula. This is the case $k=1$ of the character of a Weyl-vector multiple.

Every finite-dimensional irreducible Lie algebra representation has highest weight $\lambda =a{\Lambda }_1+b{\Lambda }_2$ for unique integers $a,b\ge 0$. The Weyl vector isso $\lambda +\rho =(a+b/2+3/2,b/2+1/2)$. In the Weyl dimension formula, the factors for the four positive roots, ordered as ${\epsilon }_1-{\epsilon }_2,{\epsilon }_2,{\epsilon }_1,{\epsilon }_1+{\epsilon }_2$, are respectivelyHenceIn particular the two fundamental representations have dimensions five and four. The highest root is $\theta ={\epsilon }_1+{\epsilon }_2=2{\Lambda }_2$, so the Adjoint representation has dimension ten. These also fix the dimensions of the crystals below. Our numbering places the long root first; interchanging a B2/C2 numbering without interchanging the associated fundamental-weight labels would give the wrong dimension formula.

Let $V={\mathbb{C}}^n$ and $m\ge 0$. Identify the symmetric power ${\mathrm{Sym}}^{m}V$ with the degree-$m$ part of the symmetric algebra on the basis vectors $e_1,\dots ,e_n$, using formal variables $x_i$ for these vectors. Its monomial basis isThe diagonal maximal torus of the unitary group acts on this monomial by the character $z_1^{a_1}\cdots z_n^{a_n}$. Distinct tuples give distinct characters, so each weight space is one-dimensional.

Suppose $M$ is a nonzero invariant complex subspace. For a nonzero vector $w\in M$, select a monomial with nonzero coefficient and extract it by torus averaging:Fourier orthogonality makes this a nonzero scalar multiple of $x^a$. It belongs to $M$, so $M$ contains a monomial.

Differentiating the unitary group action makes $M$ invariant under $\mathfrak{u}(n)$ and hence under its complexification ${\mathfrak{gl}}_n(\mathbb{C})$. The matrix unit $E_{ij}$ replaces an occurrence of $e_j$ by $e_i$, so its action on the symmetric algebra isWhenever $a_j>0$, this transfers one unit of degree from coordinate $j$ to coordinate $i$ with a nonzero coefficient. Starting with any monomial, such transfers can produce every other tuple of nonnegative integers summing to $m$. Thus $M$ contains every monomial and equals ${\mathrm{Sym}}^{m}V$.

ConsequentlyFor $m=0$ this is the one-dimensional trivial representation, and for $n=1$ every symmetric power is also one-dimensional. This weight-transfer proof of irreducibility of symmetric powers uses neither the Weyl character formula nor the Weyl dimension formula, so no unproved dimension formula enters the argument.

Choose positive rootsThe corresponding simple roots are ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda ={\epsilon }_1+{\epsilon }_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a highest-weight vector with highest weight $\lambda$.

A representation of a compact Lie group is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the exterior square contains an irreducible summand $L_{\lambda }$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.

Use the Weyl dimension formula, expressly allowed in this part:Here $L_{\lambda }$ is the irreducible highest-weight representation, ${\Phi }^{+}$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with ${\epsilon }_1,{\epsilon }_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,The four factors, in the listed positive-root order, are respectivelyTheir product is $10$. Since $\dim {\Lambda }^2{\mathbb{C}}^5=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, this irreducible summand occupies the entire exterior square. HenceThe proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its B2 adjoint representation weight diagram is exactly the eight roots and the double zero weight obtained above.

Define an equivariant contraction $C:V^{\ast }\otimes {\mathrm{Sym}}^{2}V\to V$ byThis uses the natural dual pairing, so it commutes with the Lie algebra representation. Also define $s(v)={\sum }_i{e}_i^{\ast }\otimes e_{i}v$. The identity tensor ${\sum }_i{e}_i^{\ast }\otimes e_i$ is invariant, making $s$ equivariant. Direct contraction gives $C(s(v))=3v+v=4v$. Thus $s$ is injective and $C$ is surjective, andis a direct decomposition into $\ker C$ and $s(V)$. In particular $\dim \ker C=18-3=15$ and $s(V)$ is a defining irreducible module of highest weight ${\omega }_1$.

For the usual upper-triangular choice of positive roots, $T_3^{11}=z^{\ast }\otimes x^2$ lies in the kernel and is killed by both simple raising operators $E_{12},E_{23}$. Its highest weight is $2{ϵ}_1-{ϵ}_3=2{\omega }_1+{\omega }_2$, with Dynkin labels $(2,1)$. The highest-weight classification and complete reducibility give an irreducible copy $L(2,1)$ generated by this vector. The Weyl dimension formula for ${\mathfrak{sl}}_3$ isso $\dim L(2,1)=15$. It fills the entire kernel. Therefore the sl3 decomposition of the symmetric-square dual tensor product, with its tensor factors exchanged, isThe weight diagrams below record both summands. Their labels use $K_{ij}=2T_j^{ij}-T_i^{ii}$ for $j\ne i$ and $S_i={\sum }_j{T}_j^{ij}$; these are the weight bases detailed in part (iii).

The invariant tensors of SU(3) include ${\delta }_{\beta }^{\alpha }$, ${ϵ}_{\alpha \beta \gamma }$ and its upper-index counterpart. Consequently an antisymmetric pair of upper fundamental indices can be converted into one lower antifundamental index, and conversely; a completely antisymmetric triple is a determinant singlet. In the Young-symmetrizer tensor decomposition, complete columns of three can therefore be removed. A remaining diagram has rows of lengths $k+l,l$: its two-box columns become $l$ dual indices and its remaining boxes give $k$ defining indices. The irreducible highest-weight component is represented by a mixed SU(3) tensor representation symmetric in all upper indices, symmetric in all lower indices, and with every upper-lower contraction removed. Such contractions produce the lower-weight components. The resulting symmetric traceless SU(3) tensor representation has Dynkin labels $(k,l)$; this is an equivalent canonical realization, not a claim that an arbitrary tensor already has those symmetries.

There are $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)$ symmetric upper-index components and $\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$ symmetric lower-index components. All contractions are equivalent under the separate symmetries. Their image is the separately symmetric tensor space with $(k-1,l-1)$ indices. To subtract its dimension legitimately, we must show that contraction is onto.

Represent such tensors by polynomials of bidegree $(k,l)$ in three variables $x$ and three dual variables $y$. Up to a nonzero normalization, contraction isIf $P$ has bidegree $(k-1,l-1)$, differentiation givesChoose $a_0=1/(k+l+1)$ and $a_j=-a_{j-1}/[(j+1)(k+l+1-j)]$. Thenhas bidegree $(k,l)$ and $\Delta Q=P$: consecutive terms cancel, and the last further derivative vanishes. This constructs a preimage of every contraction target. If $k=0$ or $l=0$, there is no contraction. Therefore the number of trace-free components isThe highest-weight vector $x_1^k{y}_3^l$ lies in this kernel. The Young-symmetry construction identifies the traceless highest-weight component with the irreducible representation of labels $(k,l)$; equivalently its Weyl dimension formula agrees with the independently calculated kernel dimension above.

For traceless matrices $T,S$, the tensor square of the SU(3) adjoint representation decomposes asThe component maps make the decomposition concrete. The singlet is $\mathrm{tr}(TS)$. The two octets are the antisymmetric commutator $[T,S]$ and symmetric traceless anticommutator $\{T,S\}-\frac{2}{3}\mathrm{tr}(TS)I$. Symmetrizing the two upper and two lower indices in $T_{\beta }^{\alpha }{S}_{\delta }^{\gamma }$ and removing contractions gives $R(2,2)$, the 27. Its highest component is already visible in $E_{13}\otimes E_{13}$.

For the 10, contract the two lower indices with ${ϵ}^{\beta \delta \rho }$ and symmetrize the three upper indices $\alpha ,\gamma ,\rho$; for its conjugate, contract the upper pair with ${ϵ}_{\alpha \gamma \rho }$ and symmetrize the three lower indices. These maps are antisymmetric under exchange of $T,S$. They are nonzero: $T=E_{12},S=E_{13}$ produces a nonzero all-upper-1 component, and $T=E_{21},S=E_{31}$ produces the conjugate all-lower-1 component. Thus they reach the irreducible $R(3,0)$ and $R(0,3)$, each of dimension ten. The octet maps are also nonzero, for example on a pair of noncommuting matrix units and on equal nonzero traceless diagonal matrices. These equivariant maps supply distinct irreducible constituents, with the two octets separated by exchange symmetry. Their dimensions exhaust the product:

An octet operator between baryon octet states defines an intertwining operator $8_{\mathrm{operator}}\otimes 8_{\mathrm{state}}\to 8_{\mathrm{state}}$. Its covariance follows from the commutator transformation law in the question. By Schur lemma, the number of independent reduced matrix elements is the multiplicity of the output octet in the product: there are exactly two independent coefficients. To display them, use a Hermitian octet matrix basis $B_s={\lambda }_s$, $t_i={\lambda }_i/2$, normalized by $\mathrm{tr}({\lambda }_r{\lambda }_s)=2{\delta }_{rs}$. The two allowed actions areWith $[{\lambda }_i,{\lambda }_s]=2i{f}_{isk}{\lambda }_k$ and $\{{\lambda }_i,{\lambda }_s\}=\frac{4}{3}{\delta }_{is}I+2d_{isk}{\lambda }_k$, this gives the two reduced octet matrix elements:up to the common normalization absorbed into $F,D$. All component dependence is fixed by the invariant symmetric and antisymmetric tensors. For a Hermitian operator and this Hermitian basis, $F,D$ may be chosen real; neither their values nor a relation between them follows from symmetry alone.

In the defining representation $V={\mathbb{C}}^{2n}$, the basis vectors $v_i$ and $v_{\bar{i}}$ have weights ${\epsilon }_i$ and $-{\epsilon }_i$, respectively, each with multiplicity one. Thus its formal character isThe vector $v_1$ is killed by every positive root space, since all positive-root matrices are upper triangular, and is a highest-weight vector. Any nonzero invariant subspace splits into weight spaces and therefore contains one of the natural weight vectors, since those weight spaces are one-dimensional. The root matrices connect that vector to all the others, so this representation is irreducible for $n\ge 2$. In terms of the fundamental weights of Dn, its highest weight is ${\omega }_1$ for $n\ge 3$ and ${\omega }_1+{\omega }_2$ for $n=2$.

The flip of the two tensor factors is equivariant, henceUse the invariant symmetric bilinear form $Q(u,v)=u^{T}Jv$. Contraction $C(u\otimes v)=Q(u,v)$ is an equivariant map to the trivial Lie algebra representation. Its inverse-form tensoris invariant and satisfies $C(\tau )=2n$. ConsequentlyThe traceless summand contains the highest-weight vector $v_1\otimes v_1$ of weight $2{\epsilon }_1$. The Weyl complete reducibility theorem states that every finite-dimensional complex semisimple Lie algebra representation is a direct sum of irreducible representations, so the submodule generated by this highest vector is $L(2{\epsilon }_1)$. The Weyl dimension formula verifies that it is the entire traceless space: only positive roots involving ${\epsilon }_1$ change their factors, andThis equals $\dim {\mathrm{Sym}}^{2}V-1=n(2n+1)-1$. Thus the traceless symmetric square of the defining even orthogonal representation is irreducible, including $n=2$.

For the other summand, the exterior square of the defining orthogonal representation has the equivariant identificationThe image is skew relative to $Q$, and in a basis making $Q$ the identity the images are the independent elementary skew matrices. The map is therefore an isomorphism onto the Adjoint representation of a Lie algebra. For $n\ge 3$, the connected Dn Dynkin diagram gives a simple Lie algebra (with $D_3=A_3$); its adjoint module is irreducible because its invariant subspaces are exactly its ideals. Its highest weight is ${\epsilon }_1+{\epsilon }_2$, and $v_1\wedge v_2$ is a highest vector. The tensor-square decomposition of the defining even orthogonal representation is consequentlyThe three dimensions are $(n+1)(2n-1)$, $n(2n-1)$ and one, adding to $4n^2$. The highest weights are $2{\omega }_1$, ${\omega }_2$ and zero for $n\ge 4$; for $n=3$ the middle weight is ${\omega }_2+{\omega }_3$.

For $n=2$, the Chiral decomposition of the complexified so4 Lie algebra splits the adjoint module into its two three-dimensional ideals. ThusThe weights are $2{\omega }_1+2{\omega }_2$, $2{\omega }_1$, $2{\omega }_2$ and zero. Equivalently, labelling highest weights by the two sl2 Lie algebra factors, this is $(2,2)\oplus (2,0)\oplus (0,2)\oplus (0,0)$. If $n=1$ is allowed, the abelian algebra has $V=L({\epsilon }_1)\oplus L(-{\epsilon }_1)$ and $V\otimes V=L(2{\epsilon }_1)\oplus 2L(0)\oplus L(-2{\epsilon }_1)$; there is then no single highest weight for an irreducible natural module.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

Use the contraction splitting of a defining module tensor a dual symmetric square. For a quadratic polynomial $p$ in the dual variables, defineBoth maps are intertwiners. The first is the natural tensor contraction; the second uses the invariant tensor ${\sum }_i{e}_i\otimes y_i$. Directly, $C(Q_j)=4y_j$, so $C\iota =4I$ andThe vector $T_{1;33}=e_1\otimes y_3^2$ lies in $\ker C$ and is killed by both positive simple-root operators $E_{12},E_{23}$. It is a highest-weight vector of weight $L_1-2L_3={\omega }_1+2{\omega }_2$. By Weyl complete reducibility theorem, it generates the corresponding irreducible summand. The Weyl dimension formula for ${\mathfrak{sl}}_3$ gives $\dim {\Gamma }_{a,b}=(a+1)(b+1)(a+b+2)/2$, hence $\dim {\Gamma }_{1,2}=15$, exhausting the kernel. ThereforeThe kernel has the nine outer weights of multiplicity one and each inner weight $-L_j$ of multiplicity two; the dual summand has just the three weights $-L_j$, each once.

Put $V={\mathbb{C}}^n$. We use complete reducibility of compact-group representations, and the highest-weight classification: the multiplicity of $V_{\lambda }$ in a completely reducible representation is the dimension of its weight-$\lambda$ subspace killed by all positive root vectors. Each irreducible summand contributes just its one-dimensional highest-weight line to that subspace.

Symmetrizing and antisymmetrizing the three tensor factors give invariant subspaces ${\mathrm{Sym}}^{3}V$ and ${\Lambda }^{3}V$ of $V^{\otimes 3}$. The vectors $e_1^{\otimes 3}$ and $e_1\wedge e_2\wedge e_3$ are highest-weight vectors of weights $(3,0,\dots ,0)$ and $(1,1,1,0,\dots ,0)$ respectively. The unitary version of the Weyl dimension formula isIt givesThese are exactly the dimensions of the third symmetric power and exterior power, so each is irreducible and occurs once in those subspaces.

The weight $(2,1,0,\dots ,0)$ in $V^{\otimes 3}$ has basisThe only positive root vector that acts nontrivially on this weight space is $E_{12}$. Acting on the tensor product as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes 3}$. Consequently the highest-weight vectors in this space are preciselyThis is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its dimension isThe dimensions of the summands already found add toThus no other irreducible summand can remain. We have proved the full decompositionThere are four summands counted with multiplicity and three distinct isomorphism types.

For $n=2$, the exterior cube vanishes. The third symmetric power still has highest weight $(3,0)$ and dimension $4$, while the same two-dimensional highest-vector calculation gives two copies of the representation of weight $(2,1)$ and dimension $2$. It is $\det \otimes V$, since a determinant twist shifts the standard highest weight $(1,0)$ to $(2,1)$. Hence

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.