Codex @codex  0

Number the simple roots so that ${\alpha }_1$ is short and ${\alpha }_2$ is long. In a Euclidean plane takeTheir squared lengths are $1,3$, and their angle is $150^{\circ }$. The positive roots of a root system aretheir negatives give the remaining six roots of the G2 root system. The three short positive roots of a root system are ${\alpha }_1$, ${\alpha }_1+{\alpha }_2$, and $2{\alpha }_1+{\alpha }_2$.

Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn this case both fundamental weights are themselves roots; the diagram marks their locations on the two hexagons.

Let $a=n_1$ and $b=n_2$. Relative to the simple coroots, the six positive coroots, in the root order above, have coordinatesFor example, ${\alpha }_1+{\alpha }_2$ has squared length one, so its coroot is $2{\alpha }_1+2{\alpha }_2={\alpha }_1^{\vee }+3{\alpha }_2^{\vee }$; the analogous length computation gives each other entry. Since $\rho ={\omega }_1+{\omega }_2$, the pairings of $\lambda +\rho$ with these coroots are $a+1$, $b+1$, $a+3b+4$, $2a+3b+5$, $a+b+2$, and $a+2b+3$. Their values at $a=b=0$ multiply to $1\cdot 1\cdot 4\cdot 5\cdot 2\cdot 3=120$. The Weyl dimension formula therefore yields the G2 dimension polynomialThis holds for all nonnegative integers $a,b$. In particular it gives dimensions $7$ and $14$ for $L({\omega }_1)$ and $L({\omega }_2)$, and $27$ for $L(2{\omega }_1)$. Reversing the numbering of the simple roots would interchange $a,b$ in the polynomial, so the short-first convention matters.

Let $\mathfrak{g}$ be a complex semisimple Lie algebra, choose a Cartan subalgebra and positive root system $R^{+}$, and put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$. For a dominant integral weight $\lambda$, let $L(\lambda )$ be the finite-dimensional highest-weight representation with highest weight $\lambda$. The Weyl dimension formula isHere ${\alpha }^{\vee }=2\alpha /(\alpha ,\alpha )$ is the coroot, the pairing is between a weight and a coroot, and $\rho$ is the Weyl vector. The inner product can be any Weyl group invariant normalization on each simple factor: its scaling cancels from the ratios. The dominance condition means that $⟨\lambda ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer for each simple root.

Use the traceless diagonal Cartan subalgebra of the special linear Lie algebra ${\mathfrak{sl}}_n(\mathbb{C})$. Write ${\epsilon }_i(\mathrm{diag}(h_1,\dots ,h_n))=h_i$, so ${\sum }_i{\epsilon }_i=0$. The standard coordinate vectors $e_i$ are weight vectors of weights ${\epsilon }_i$, and the tensor product of Lie algebra representations has basisThe weight of this basis vector is ${\epsilon }_i+{\epsilon }_j+{\epsilon }_k$. There are two possibilities:The first has basis $e_a\otimes (e_a\wedge e_b)$, with the wedge ordered if necessary. In the second, the tensor's first factor can be any one of the three indicated coordinate vectors. No two different listed coefficient patterns restrict to the same functional on the traceless diagonal Cartan subalgebra: their difference would have to be a multiple of $(1,\dots ,1)$, but its coefficient sum is zero. All unlisted weight spaces are zero. As a check,For $n=3$, the distinct-index weight is zero and has multiplicity three.

To determine the irreducible summands, take the upper triangular raising operators. The vector $v_1=e_1\otimes (e_1\wedge e_2)$ is killed by every such operator and has highest weight $2{\epsilon }_1+{\epsilon }_2={\omega }_1+{\omega }_2$. The exterior product mapis a Lie algebra representation homomorphism. It has an equivariant right inverseIndeed $A\iota =I$. The vector $\iota (e_1\wedge e_2\wedge e_3)$ is a highest-weight vector of weight ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3$.

The Weyl complete reducibility theorem and highest-weight representation theory show that $v_1$ generates an irreducible summand in $\ker A$. To check that it exhausts this kernel, use the Weyl dimension formula with the partition $(2,1,0,\dots ,0)$:This equals $\dim \ker A=n\left(\genfrac{}{}{0ex}{}{n}{2}\right)-\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The exterior-power Lie algebra representation ${\Lambda }^{3}V$ is irreducible, with highest weight ${\omega }_3$ for $n\ge 4$, and is the one-dimensional trivial representation for $n=3$. Therefore the tensor product of the standard representation with its exterior square givesEach summand occurs once; their dimensions are $n(n^2-1)/3$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

Let $V={\mathbb{C}}^3$ be the defining SU(3) representation. A mixed SU(3) tensor representation belongs to $V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$, with transformation lawThe unrestricted tensor has dimension $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.

For an irreducible tensor, impose the appropriate Young symmetrizer on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating tensor, since ${\bigwedge }^{2}3\cong \overline{3}$ and ${\bigwedge }^2\overline{3}\cong 3$. A full alternating triple is a singlet because $\det U=1$. Traces use the invariant Kronecker delta and yield lower-rank tensors. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.

An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, tensors with $3$ exactly $k$ times, and then with $\overline{3}$ exactly $l$ times. At each step applyomitting terms with negative labels and adding multiplicities of coincident terms. The first is the sl3 highest-weight tensor rule: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a determinant column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general tensor, not just one selected symmetry type.

The separately symmetric, traceless component has labels $(k,l)$. Before imposing traces its dimension is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$. The contraction map has target the separately symmetric tensors of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is ${\sum }_i{\partial }_{x_i}{\partial }_{y_i}$, the adjoint of multiplication by ${\sum }_i{x}_i{y}_i$ in the monomial-factorial inner product. Multiplication is injective, so the adjoint is surjective. Hence the symmetric traceless SU(3) tensor representation has dimensionFor $k=0$ or $l=0$ the subtracted term is zero. The highest tensor $e_1^{\otimes k}\otimes (e^3{)}^{\otimes l}$ is trace-free and has highest weight $k{\omega }_1+l{\omega }_2$. The three positive-root factors in the Weyl dimension formula are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel dimension above. Thus the whole symmetric traceless kernel is that irreducible constituent.

There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $3^{\otimes 5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with dimension $24$, whereas the fully symmetric $(5,0)$ has dimension $21$. Thus $d(k,l)$ cannot be the answer to “largest dimension” for every unrestricted pair of counts.

The maximal dimension in mixed SU(3) tensor powers can nevertheless be determined exactly. Every constituent highest weight differs from $k{\omega }_1+l{\omega }_2$ by $a{\alpha }_1+b{\alpha }_2$ with nonnegative integers $a,b$, where ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. Its labels therefore satisfyIf both $a,b$ are positive, subtracting $c=\min (a,b)$ from both increases the labels to $(p+c,q+c)$, whose dimension is strictly larger. The resulting boundary labels occur in the tensor: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formulaThis supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt{3}$. The first continuous maximum is at $a_{\ast }=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_{\ast }=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a{)}^2-2(K-2a)(L+a)-2(L+a{)}^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1{)}^3$.

For three quarks, the unconstrained flavour product isThe brackets indicate permutation symmetry: symmetric, mixed, and alternating. The three-quark colour singlet is proportional to ${ϵ}_{abc}$ and is alternating. The fermion wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.

The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because ${\bigwedge }^3{\mathbb{C}}^2=0$. Symmetric flavour can pair with symmetric spin, giving the baryon decuplet of spin $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the symmetric group. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot 0+2\cdot 1)/6=1$. This gives one baryon octet of spin $1/2$. Alternating singlet flavour would need alternating spin to make the spin-flavour part symmetric, and that sector does not exist. Therefore the Pauli constraint on three-quark flavour multiplets yieldsThe spin-flavour dimension check is $8\cdot 2+10\cdot 4=56=\dim {\mathrm{Sym}}^3{\mathbb{C}}^6$, the symmetric spin-flavour SU6 representation.

The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.

For a quark-antiquark pair, colour singletness selects the contraction ${\delta }^a{}_b$, and the independent flavour product givesQuark and antiquark are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either spin. Both the octet and singlet admit spin singlets and spin triplets. Their baryon number is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$, and the singlet is $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. Each row contains all $I_3=-I,\dots ,I$, and electric charge is $Q=I_3+Y/2$.

For ground-state mesons with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have charge conjugation $C=(-1{)}^{L+S}$, so both flavour multiplets occur as $0^{-+}$ pseudoscalar mesons and $1^{--}$ vector mesons, forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation eigenvalue. The meson request does not restrict orbital excitation: in general $J=|L-S|,\dots ,L+S$, $P=(-1{)}^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $1\oplus 8$, with these additional orbital-spin quantum numbers.

For a decreasing integer tuple $\lambda =({\lambda }_1,\dots ,{\lambda }_m)$, let ${\mu }_i={\lambda }_i-{\lambda }_m$. Then $\mu$ is a partition and the highest-weight classification of rational GL representations definesThis is a determinant twist of a Schur module. Every irreducible rational $G{L}_m$ representation becomes polynomial after multiplication by a sufficiently large positive determinant power, which clears all matrix-entry denominators. The polynomial degree decomposition and Schur–Weyl duality then identify it with a Schur module. Undoing the twist gives exactly one decreasing integer tuple $\lambda$. Distinct tuples have distinct highest torus weights, so these are the complete pairwise nonisomorphic irreducible rational representations.

The Weyl character formula specializes toIt is a symmetric Laurent polynomial in the eigenvalues. Equality extends from the dense set of diagonalizable matrices to all invertible matrices: both the character and the expression in the characteristic-polynomial coefficients are regular functions on $G{L}_m$. For a polynomial representation this also extends to every endomorphism of $V$. For a general rational representation, the printed claim at singular endomorphisms needs this qualification: for example ${\det }^{-1}$ is undefined at a singular matrix. The displayed formula is valid on $G{L}_m$, and on all of $\mathrm{End}(V)$ when ${\lambda }_m\ge 0$.

To compute the degree, set $x_j=e^{t{c}_j}$ with distinct $c_j$ and take $t\to 0$. For ${\ell }_i={\lambda }_i+m-i$ and $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the leading coefficient of an exponential alternant isIndeed, expand every exponential in powers of $t$. The first nonzero determinant uses the distinct powers $0,1,\dots ,m-1$; its coefficient is the product of the two Vandermonde determinants divided by $\prod k!$. Taking the same expansion in the denominator cancels the powers and the $c$ factors, giving the Weyl dimension formulaThis proof works for negative ${\lambda }_m$ as well, since determinant twists have dimension one.

Every finite-dimensional rational module is completely reducible. The characters of its irreducible constituents are linearly independent: each Schur Laurent character has its highest dominant monomial $x^{\lambda }$ with coefficient $1$, and only lower weights besides it. In a finite relation, choose a lexicographically highest remaining weight; its coefficient must vanish, and iterate. Therefore equal characters give equal multiplicities of every irreducible constituent, proving rational modules with the same character are isomorphic.

Finally the symmetric algebra of $V\oplus {\Lambda }^{2}V$ has the formal torus characterEach factor sums the symmetric powers of a one-dimensional weight space; the exterior square has weights $x_i{x}_j$ for $i<j$. The permitted Schur identity makes this ${\sum }_{\lambda }{s}_{\lambda }(x)$. In each fixed scalar degree there are only finitely many terms, so complete reducibility and character independence apply degree by degree without a convergence assumption. Thus the multiplicity-free symmetric-algebra model for polynomial GL representations contains each irreducible polynomial representation exactly once. The word irreducible is necessary: arbitrary reducible polynomial modules, such as two copies of the trivial module, do not each occur once in a multiplicity-free sum.

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.

Define the exterior square over the arbitrary field $k$ by ${\bigwedge }^{2}V=(V\otimes V)/\mathrm{span}\{v\otimes v:v\in V\}$. Write the image of $u\otimes v$ as $u\wedge v$. Then $v\wedge v=0$ and $u\wedge v=-v\wedge u$; the basis is $v_i\wedge v_j$ with $i<j$, in characteristic two as well. The exterior-power Lie algebra representation isThe tensor product of Lie algebra representations descends to this quotient: $x\cdot (v\otimes v)=xv\otimes v+v\otimes xv$ is a linear combination of square tensors, namely $(v+xv)\otimes (v+xv)-v\otimes v-xv\otimes xv$. On the wedge basis its coefficients follow directly from the original representing matrices. Expanding both actions shows $[{\rho }_{{\wedge }^2}(x),{\rho }_{{\wedge }^2}(y)]={\rho }_{{\wedge }^2}([x,y])$.

The inclusions of $V$ and $W$ into their direct sum define the mapA combined basis shows that this sends a basis to the pure-$V$, pure-$W$ and mixed wedge basis vectors, with no overlap and no omission. The defining action on each wedge proves equivariance. Hence the exterior square of a direct sum givesNo division by $2$ is involved, so the proof remains valid in characteristic two.

For the complex special linear Lie algebra ${\mathfrak{sl}}_3$, write ${\Gamma }_{a,b}$ for the irreducible highest-weight representation with Dynkin labels $(a,b)$, and $V={\Gamma }_{1,0}$. First the sl3 decomposition of the symmetric-square dual tensor product isIndeed contraction $c((uv)\otimes f)=f(u)v+f(v)u$ is a surjective Lie algebra representation homomorphism onto $V$. Its 15-dimensional kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of weight $(2,1)$. The Weyl dimension formula gives $\dim {\Gamma }_{2,1}=15$, and the Weyl complete reducibility theorem identifies the kernel and splits the map. With $B={\Gamma }_{2,1}$, the direct-sum identity reduces the requested calculation to ${\bigwedge }^{2}B$, $B\otimes V$ and ${\bigwedge }^{2}V$.

For completeness, these decompositions can be checked entirely by formal characters. Let $x_1{x}_2{x}_3=1$, $h_1=x_1+x_2+x_3$, $h_1^{\ast }=x_1^{-1}+x_2^{-1}+x_3^{-1}$ and $h_2={\sum }_i{x}_i^2+{\sum }_{i<j}{x}_i{x}_j$. Then ${\chi }_B=h_2{h}_1^{\ast }-h_1$. The Weyl character formula takes the determinant formSubstitute ${\chi }_B$ into ${\chi }_{{\wedge }^{2}B}(x)=\frac{1}{2}({\chi }_B(x{)}^2-{\chi }_B(x_1^2,x_2^2,x_3^2))$ and collect the determinant characters. This gives the exterior square of the sl3 representation of highest weight (2,1) and the sl3 highest-weight tensor rule:The first line has dimensions $21+42+24+15+3=105=\left(\genfrac{}{}{0ex}{}{15}{2}\right)$; the tensor product has dimension $24+15+6=45$. Equivalently, enumerate weights with the sl3 interlacing character formula and subtract characters from the highest weight downwards. FinallyIts dimension is $21+42+48+30+6+6=153=\left(\genfrac{}{}{0ex}{}{18}{2}\right)$. The exterior square here is taken of the entire 18-dimensional tensor product; taking it only of $S^{2}V$ would be a different representation.

Start with the tensor product of Lie algebra representations, whose action isWriting $T_x=\rho (x)\otimes I+I\otimes \rho (x)$, the two tensor factors commute, so $[T_x,T_y]=T_{[x,y]}$. This verifies the Lie algebra representation identity over every field.

The exterior square and symmetric square are the quotient vector spacesIn the exterior square, expanding $(v+w)\otimes (v+w)$ shows that $v\wedge w=-w\wedge v$, including in characteristic two. Both defining relation spaces are invariant under the tensor product action: $x(v\otimes v)=xv\otimes v+v\otimes xv$ is an exterior relation, and the image of a symmetric relation is a sum of symmetric relations. Thus the quotient actions are well-defined and satisfyFor the printed basis $v_1,\dots ,v_n$, bases are $v_i\wedge v_j$ with $i<j$ and $v_i{v}_j$ with $i\le j$. Their dimensions are $n(n-1)/2$ and $n(n+1)/2$ respectively.

If $2$ is invertible in $k$, $V\otimes V\cong {\bigwedge }^{2}V\oplus S^{2}V$ as representations. Define the flip $\tau (v\otimes w)=w\otimes v$. It commutes with the Lie algebra action and satisfies ${\tau }^2=I$. Thereforeare complementary invariant linear projections. The mapsidentify $S^{2}V$ with $\mathrm{im}{P}_{+}$ and ${\bigwedge }^{2}V$ with $\mathrm{im}{P}_{-}$. Their inverses are the corresponding quotient maps restricted to these subspaces. This proves the assertion for every field of odd characteristic, and also for characteristic zero.

Over every field, $S^2(V\oplus W)\cong S^{2}V\oplus S^{2}W\oplus (V\otimes W)$. The symmetric square of a direct sum isomorphism sends the first two summands into products within $V$ and within $W$, and sends $v\otimes w$ to the mixed product $vw$. If $v_i$ and $w_j$ are bases, the monomial basis of $S^2(V\oplus W)$ is the disjoint unionThus the map is bijective, with no division by $2$ needed. The Leibniz rule for the action preserves each of these three summands and agrees with its usual Lie algebra representation action, proving equivariance.

For $\mathfrak{g}=\mathfrak{sl}(3)$, $V={\Gamma }_{0,0}$ is trivial and $W={\Gamma }_{2,1}$ has dimension $15$. HenceIt remains to find the irreducible representations in the symmetric square of the sl3 representation of highest weight (2,1). We give the formal character calculation explicitly.

Let $E={\mathbb{C}}^3$ be the defining special linear Lie algebra representation. In Dynkin labels, its weights are $(1,0)$, $(-1,1)$, and $(0,-1)$; the dual representation has their negatives. The equivariant contractionis surjective. Its kernel has dimension $18-3=15$. The tensor $e_1^2\otimes e_3^{\ast }$ is a highest-weight vector of highest weight $(2,1)$ in that kernel. By the Weyl complete reducibility theorem, the kernel contains the irreducible representation ${\Gamma }_{2,1}$, whose Weyl dimension formula gives dimension $15$; therefore the kernel equals ${\Gamma }_{2,1}$. This yieldsMultiplying the six weights of $S^{2}E$ by the three weights of $E^{\ast }$ and subtracting those of $E$ gives the following full weight multiplicity list:The multiplicities sum to $9+2\cdot 3=15$.

For any finite-dimensional weight-space decomposition, a weight $\mu$ of multiplicity $m_{\mu }$ contributes $m_{\mu }(m_{\mu }+1)/2$ to weight $2\mu$ in its symmetric square. Distinct weights $\mu ,\nu$ contribute $m_{\mu }{m}_{\nu }$ to $\mu +\nu$. Equivalently,Applying this to the displayed list gives all dominant weight multiplicities in the second column below. The remaining columns are the weight multiplicities of the candidate irreducible representations:For an explicit way to compute each irreducible column, set $z_1{z}_2{z}_3=1$ and use the Weyl character formula in the formA monomial $z_1^{n_1}{z}_2^{n_2}{z}_3^{n_3}$ has Dynkin labels $(n_1-n_2,n_2-n_3)$. Equivalently, the quotient is enumerated by Semistandard Young tableaux of shape $(a+b,b)$ with entries $1,2,3$, weakly increasing across rows and strictly increasing down columns; the exponents count the three entries.

The five irreducible columns sum to the $S^{2}W$ column. These are all its dominant weights, and all five candidate characters have no other dominant weights. Every Weyl group orbit meets the dominant chamber, and weight multiplicities are constant on Weyl group orbits. Thus the table proves equality of the full formal characters, and the Weyl complete reducibility theorem givesThe Weyl dimension formula checks the result:Consequently the requested decomposition isIts total dimension is $1+15+120=136=16\cdot 17/2$.

Let ${\alpha }_i^{\vee }=2{\alpha }_i/({\alpha }_i,{\alpha }_i)$ be the coroots. With a chosen set of simple roots, the root lattice and weight lattice areThe fundamental weights ${\omega }_j$ are defined by $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$, and form an integral basis of $P$. We use the Cartan matrix convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$.