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For the special unitary group $SU(N)$ define ${\epsilon }_i=e_i-\frac{1}{N}{\sum }_{j=1}^N{e}_j$. These are the weights of a representation in the defining space, with ${\sum }_i{\epsilon }_i=0$. Their Dynkin labels areThe dual representation negates each weight of a representation, and the tensor product of group representations adds them. Consequently the complete weight systems areEvery listed weight of a representation has multiplicity one: the corresponding weight basis vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in Dynkin labels, the two defining weight systems areThe weight systems for $\overline{6}$ and $\overline{5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.

For the exterior squares, the full enumerations areThe negatives of the $A_5$ column are exactly the fifteen weights of a representation of $\overline{15}$; the negatives of the $A_4$ column are exactly the ten of $\overline{10}$. Thus the tables specify all eight weight systems, including multiplicities, rather than only their highest weights.

Number the simple roots of the $A_4$ root system consecutively along its Dynkin diagram. For a Young diagram with row lengths $r_i$, its highest weight has Dynkin labels $a_i=r_i-r_{i+1}$. The three irreducible representations have the following Young diagrams, which can be filled increasingly down the column to give Young tableaux:Here $i=1,\dots ,5$, $v^i\mapsto U^i{}_j{v}^j$, and $T^{ij}\mapsto U^i{}_k{U}^j{}_l{T}^{kl}$. Contraction with the invariant alternating tensor identifies ${\tilde{v}}_i=(1/4!){ϵ}_{i{i}_1{i}_2{i}_3{i}_4}{w}^{i_1{i}_2{i}_3{i}_4}$, giving the dual representation of $5$. The dimensions follow from $\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.

Use the centralizer and circle group normalization of solution. The first three components have U(1) charge $-2$ and the last two have U(1) charge $3$, so the branching of a representation isFor the dual representation, both U(1) charges reverse and the $SU(3)$ representation is conjugated; the $SU(2)$ doublet is equivalent to its conjugate. Taking the exterior square of the first decomposition gives three types of antisymmetric tensor components: $T^{ab}$, $T^{a\alpha }$, and $T^{\alpha \beta }$, where $a,b=1,2,3$ and $\alpha ,\beta =4,5$. Their U(1) charges add, and ${\Lambda }^{2}3=\overline{3}$, ${\Lambda }^{2}2=1$. HenceThe dimensions are $3+6+1=10$. These product labels describe genuine representations of the quotient group in solution: every element of its ${\mathbb{Z}}_6$ kernel acts trivially on each summand.

For the A3 root system, take simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3$ in the hyperplane ${\sum }_{i=1}^4{x}_i=0$ of an orthonormal ${\mathbb{R}}^4$. Its Dynkin diagram is the single-bond chain $1-2-3$. The Cartan matrix has entries $C_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, givingFor a weight $\mu$, use Dynkin labels $w_i=2(\mu ,{\alpha }_i)/({\alpha }_i,{\alpha }_i)$. The four weights ${\mu }_j=e_j-\frac{1}{4}{\sum }_k{e}_k$ of the defining representation have labelsThey form the weight system with highest weight $(1,0,0)$. Equivalently, subtract successive rows of $C_{A_3}$ from that highest weight. The dual representation negates every weight, so the weight system with highest weight $(0,0,1)$ isAll these weights have multiplicity one. The numbered Dynkin diagrams used in this solution and in solution are shown below.

Choose the doublet bases so that the column of barred fields $B=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $R=({\psi }_R^1,{\psi }_R^2{)}^T$ both transform by the doublet matrices in d. This choice follows the source's assignment of the barred doublet itself. Since $H$ transforms in $\epsilon$, its Yukawa coupling must multiply a fermion bilinear in $\epsilon$.

There is exactly one such bilinear in $2\otimes 2$: the Clebsch-Gordan coefficients in d give $B_1{R}_2-B_2{R}_1$. Equivalently, writing the upper block of the Yukawa matrix as $A$, invariance requiresThe first equation sets $A_{11}=A_{22}=0$, because ${\alpha }^2\ne 1$; the second sets $A_{21}=-A_{12}$.

A mixed doublet-singlet bilinear remains a doublet after multiplication by $H$, so all four mixed entries vanish. Finally ${\bar{\psi }}_L^3{\psi }_R^3$ transforms in $\epsilon \otimes 1=\epsilon$, and multiplication by $H$ makes it invariant. ThereforeThe allowed Yukawa interaction is $H\{a({\bar{\psi }}_L^1{\psi }_R^2-{\bar{\psi }}_L^2{\psi }_R^1)+b{\bar{\psi }}_L^3{\psi }_R^3\}$, together with its Hermitian conjugate when forming a real Lagrangian density.

The displayed texture depends on basis. If instead the unbarred left doublet is assigned $D$ in the same ordering as the right doublet, its barred doublet has the dual representation $D^{\ast }$ and the equivalent upper block is $a\mathrm{diag}(1,-1)$. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex Yukawa couplings.

Let $V={\mathbb{C}}^6$ be the defining fundamental representation of $SU(6)$. Its weights are $w_i=e_i-\frac{1}{6}{\sum }_{j=1}^6{e}_j$. Pairing with the simple coroots gives Dynkin labels $(w_i{)}_a={\delta }_{ia}-{\delta }_{i,a+1}$. Thus the complete list isSuccessive lowering operators subtract the corresponding rows of the Cartan matrix in a. Each weight multiplicity is one.

The complex-conjugate representation $\overline{6}\cong V^{\ast }$ is the dual representation and has the negatives of these weights. Its highest weight is $-w_6=(0,0,0,0,1)$, and its entire weight decomposition isEvery entry has weight multiplicity one.

The two-index antisymmetric representation is ${\Lambda }^{2}V=15$, of dimension $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ and highest weight $w_1+w_2=(0,1,0,0,0)$. A weight basis is $e_i\wedge e_j$ for $i<j$, carrying weight $w_i+w_j$. The fifteen sums, all distinct and all of weight multiplicity one, areThe exclusion of $i=j$ comes from the exterior square, where $e_i\wedge e_i=0$.

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on $L^2$; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in $L^2$ for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.

Choose a continuous nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the left translations $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of $T$. Since $Tf$ lies in the closure of the range, the spectral theorem gives $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional unitary representation $\rho (g)=L_g{|}_M$. It is continuous, since translating each of its finitely many continuous basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.

Let $\ell$ be evaluation at the identity on $M$, and let the dual representation on $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.

To deduce a faithful finite-dimensional representation, first note that these representations separate points. For $g\ne e$, choose a continuous function taking different values at $g$ and $e$, and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy $\theta (g)\ne I$.

There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp (B_{\delta })$ with $0<\delta <r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, so its power $\exp (mX)$ lies outside $U$, a contradiction. For a zero-dimensional group, take $U=\{e\}$.

For every $g\notin U$, choose a representation nontrivial at $g$. The sets on which these representations are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many representations, whose direct sum has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; $G\setminus \{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.

Let $e_1=x,e_2=y,e_3=z$, and let ${ϵ}_i$ be the weight of a representation defined by ${ϵ}_i(\mathrm{diag}(h_1,h_2,h_3))=h_i$. Since the trace is zero, ${ϵ}_1+{ϵ}_2+{ϵ}_3=0$. The dual representation gives weight $-{ϵ}_i$ to $e_i^{\ast }$. PutTaking $j\le k$ gives an eighteen-vector tensor product basis. Its weight is ${ϵ}_j+{ϵ}_k-{ϵ}_i$.

For explicit coordinates use the usual simple coroots $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, and write each weight as $(\mu (H_1),\mu (H_2))$. Then ${ϵ}_1=(1,0)$, ${ϵ}_2=(-1,1)$ and ${ϵ}_3=(0,-1)$. Collecting equal weights gives every weight space:The first six weights are $2{ϵ}_j-{ϵ}_i$ with $i\ne j$; the next three are $-2{ϵ}_i$; the last three are the defining weights ${ϵ}_i$. Thus there are nine multiplicity-one weights and three multiplicity-three weights, giving total dimension $9+9=18$.

The planar coordinates are $a{\omega }_1+b{\omega }_2$ with ${\omega }_1=(1,0)$ and ${\omega }_2=(1/2,\sqrt{3}/2)$, so the weight diagram uses an equilateral root geometry rather than orthogonal Dynkin-coordinate axes. Every point is labelled with its complete weight-space basis.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

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.

The dual representation on $V^{\ast }$ is $(g\ell )(v)=\ell (g^{-1}v)$. The bilinear form $\epsilon (z,w)=\det (z,w)$ on ${\mathbb{C}}^2$ is nondegenerate and invariant under SU(2), because every group matrix has determinant one. Thus it gives an equivariant isomorphism ${\mathbb{C}}^2\to ({\mathbb{C}}^2{)}^{\ast }$. Taking the $n$th symmetric power gives an equivariant isomorphism between its $n$th symmetric power and its dual. Since homogeneous polynomials realize $V_n={\mathrm{Sym}}^n(({\mathbb{C}}^2{)}^{\ast })$, this proves

The character of the dual representation is ${\chi }_{V_n^{\ast }}(g)={\chi }_n(g^{-1})=\overline{{\chi }_n(g)}$, using unitarizability. The displayed sine quotient is real, so this character equals ${\chi }_n$. The theorem that characters determine finite-dimensional semisimple complex representations of a compact group therefore gives $\boxed{V_n\cong V_n^{\ast }}$. This is complex-linear self-duality; it does not assert that every $V_n$ has an invariant real structure.

For a partition $\nu$ with at most $m$ parts, let $D_{\nu }(V)$ be the Schur module constructed in Question 3, equivalently the image of a Young symmetrizer on $V^{\otimes |\nu |}$. For a weakly decreasing integer tuple $\lambda$, put $k=\max (0,-{\lambda }_m)$ and $\nu =\lambda +k(1,\dots ,1)$. Define the rational Schur module byHere ${\det }^{-k}$ is the one-dimensional representation $g\mapsto \det (g{)}^{-k}$. This is rational, and it is irreducible under the permitted irreducibility assumption; for nonnegative $\lambda$ it is the original polynomial Schur module. Larger shifts give the same module, as will also follow from the character formula below.

Write $s_r(x)={\sum }_{i=1}^m{x}_i^r$ and $p_{\alpha }(x)={\prod }_{r=1}^n{s}_r(x{)}^{{\alpha }_r}$. For a permutation $\sigma$ of the given cycle type, the trace of a permuted tensor power isIn a tensor basis, the trace contracts the matrix entries of $\xi$ around each cycle of $\sigma$; a cycle of length $r$ contributes $\mathrm{tr}({\xi }^r)$. This proof applies to nondiagonalizable endomorphisms as well. The eigenvalues give $\mathrm{tr}({\xi }^r)=s_r(x)$.

The Schur–Weyl duality decomposition, on which $\sigma$ and ${\xi }^{\otimes n}$ act on the two respective factors, gives the same trace asFor a partition, the character extends polynomially to all endomorphisms because the tensor-power action does. This extension is not asserted for determinant-twisted modules at singular matrices.

We now derive the alternant character formula for the general linear group. Put $\delta =(m-1,m-2,\dots ,0)$ andUse the permitted symmetric-group character result, the Frobenius alternant character formula:It concerns characters of $S_n$, rather than assuming the character formula we seek for the general linear group. Substitute the proved trace identity and setThen ${\chi }^{\lambda }(\alpha )={\sum }_{\mu }{c}_{\lambda \mu }{\chi }^{\mu }(\alpha )$ for every conjugacy class. Independence of the irreducible symmetric-group characters forces $c_{\lambda \mu }={\delta }_{\lambda \mu }$.