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For the defining representation of the special linear Lie algebra ${\mathfrak{sl}}_3$, the crystal basis is $1\stackrel{1}{\to }2\stackrel{2}{\to }3$, with weights ${\epsilon }_i$ restricted to trace-zero diagonal matrices. Its dual is $\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights $-{\epsilon }_i$. Using the crystal tensor-product rule with lowering on the first factor when ${\phi }_i(a)>{\epsilon }_i(b)$, the tensor square has highest vertices $1\otimes 1$ and $1\otimes 2$, of highest weights $2{\omega }_1,{\omega }_2$, generating components of six and three vertices. The defining tensor dual has highest vertices $1\otimes \bar{3}$ and $1\otimes \bar{1}$, generating components of eight and one vertices and weights ${\omega }_1+{\omega }_2,0$. These encode the symmetric square plus exterior square, and the trace-free endomorphisms plus scalar identity, respectively. A crystal vertex labelled by a tensor pair need not represent that same undeformed ordinary simple tensor; the actual invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$.

In a field of characteristic two, $1+1=0$. The usual splitting of a tensor product into its symmetric square and exterior square by $(I\pm \tau )/2$ is unavailable.

The defining representation of ${\mathfrak{so}}_{2n+1}$ has weights ${\epsilon }_1,\dots ,{\epsilon }_n,0,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Its crystal basis is a chain on these vertices, with lowering-edge colors $1,\dots ,n-1,n,n,n-1,\dots ,1$. The short-root string has length two through zero. With the crystal tensor-product rule, for $n\ge 2$ the tensor square has highest vertices $1\otimes 1$, $1\otimes 2$ and $1\otimes \bar{1}$, corresponding to the traceless symmetric square, the exterior square and the trivial Lie algebra representation. Their dimensions are $(2n+1)(n+1)-1$, $n(2n+1)$ and one.

For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

With 24 physical half-integer open-string oscillators, the first three string level operator eigenvalues are $0,1/2,1$. Their states are the vacuum, $a_{1/2}^{i†}|0⟩$, and $a_{1/2}^{i†}{a}_{1/2}^{j†}|0⟩$. The creation operators commute, so the third level is the symmetric square of the 24-dimensional transverse vector, of dimension $24\cdot 25/2=300$. Its trace is a scalar and its trace-free part has dimension 299. Including the vacuum shift gives ${\alpha }^{\prime }{M}^2=1/2,1,3/2$ when the two longitudinal directions are common directions with Neumann boundary conditions. The representation labels refer to the surviving ND rotation group, not full unbroken target Lorentz symmetry.

The first levels are the oscillator vacuum, ${\alpha }_{-1}^I|0⟩$, and ${\alpha }_{-2}^I|0⟩,{\alpha }_{-1}^I{\alpha }_{-1}^J|0⟩$. At level two the transverse vector representation and symmetric square assemble into the massive symmetric traceless square of one extra rotation dimension.

Take the diagonal Cartan subalgebra of ${\mathfrak{sl}}_3$ and let ${\epsilon }_i$ select its $i$th diagonal entry, so ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3=0$ on it. Use simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2=-{\epsilon }_3={\epsilon }_1+{\epsilon }_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^{\ast }$. ThusTheir crystal basis diagrams are $B:1\stackrel{1}{\to }2\stackrel{2}{\to }3$ and $B^{\vee }:\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights ${\epsilon }_i$ and $-{\epsilon }_i$ respectively. Use the same crystal tensor-product rule as in2(d): lowering acts on the first factor when ${\phi }_i(u)>{\epsilon }_i(v)$; otherwise on the second. Raising uses the first factor when ${\phi }_i(u)\ge {\epsilon }_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.

For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to 21\to 22$, $31\to 32$, $13\to 23$; the complete color-two list is $21\to 31$, $22\to 32\to 33$, $12\to 13$. There are two connected components:For example ${\tilde{f}}_1(1\otimes 1)=2\otimes 1$ because ${\phi }_1(1)=1>{\epsilon }_1(1)=0$, whereas ${\tilde{f}}_1(1\otimes 2)=0$ because equality selects the second factor and ${\tilde{f}}_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional symmetric square and the three-dimensional exterior square:

For $B\otimes B^{\vee }$, abbreviate $i\otimes \bar{j}$ by $i\bar{j}$. The complete color-one edge list is $1\bar{3}\to 2\bar{3}$, $1\bar{2}\to 2\bar{2}\to 2\bar{1}$, $3\bar{2}\to 3\bar{1}$; the complete color-two list is $1\bar{3}\to 1\bar{2}$, $2\bar{3}\to 3\bar{3}\to 3\bar{2}$, $2\bar{1}\to 3\bar{1}$. The isolated vertex is $1\bar{1}$. Thus the two highest vertices are $1\bar{3}$ of weight ${\epsilon }_1-{\epsilon }_3={\omega }_1+{\omega }_2$, and $1\bar{1}$ of weight zero. Lowering from the former visits all eight other vertices. ConsequentlyThis also follows concretely from $V\otimes V^{\ast }\cong \mathrm{End}(V)={\mathfrak{sl}}_3\oplus \mathbb{C}I$: the first term is the Adjoint representation, with highest weight the highest root, and the second is the trivial Lie algebra representation. The ordinary invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.

Let $e_1,e_2,e_3$ be the basis of the defining representation and $y_1,y_2,y_3$ its dual. On a diagonal element of the Cartan subalgebra their weights are $L_i$ and $-L_i$, with $L_1+L_2+L_3=0$. Use the tensor product and symmetric square basisIts weight is $L_i-L_j-L_k$. For the three repeated weights putIf $\{i,l\}=\{1,2,3\}\setminus \{j\}$, the complete weight-space decomposition isThere are six spaces of the first type, three of the second, and three of the third. Their dimensions sum to $6+3+9=18$. The weight diagram below labels every space with all of its basis vectors.

The printed phrase “not reducible” is a misprint: the requested nonzero complementary submodules establish that this tensor square is reducible. The natural reason is that the flip $\tau (v\otimes w)=w\otimes v$ commutes with the Lie algebra representation. Its two eigenspaces are the symmetric square and the exterior square, givingIn the symmetric square, the four weights $\pm 2L_i$ and four weights $\pm L_1\pm L_2$ each occur once, while zero occurs twice. In the exterior square, the four weights $\pm L_1\pm L_2$ each occur once and zero occurs twice; the $\pm 2L_i$ are absent because repeated wedge factors vanish. These are the two requested complementary weight diagrams.

More precisely, the tensor-square decomposition of the defining symplectic representation identifies ${\mathrm{Sym}}^{2}V$ with the irreducible adjoint module $L(2{\omega }_1)$. The identification is an intertwiner sending a symmetric tensor $vw$ to the element of the symplectic Lie algebra $x\mapsto \omega (v,x)w+\omega (w,x)v$. For the exterior square, symplectic contraction of an exterior square has a five-dimensional kernel, the primitive exterior square, and the invariant tensor $\Omega =e_1\wedge e_3+e_2\wedge e_4$ supplies a complementary line. Thus the full irreducible decomposition isHere ${\omega }_1=L_1$ and ${\omega }_2=L_1+L_2$. The two zero weights in the exterior square split into one primitive zero weight and the invariant line.

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.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

Use the Clebsch-Gordan decomposition for SU2, which states that $V_m\otimes V_n\cong {⨁}_{j=0}^{\min (m,n)}{V}_{m+n-2j}$. Here this givesTo determine symmetry, realize the tensor product as polynomials in two pairs of variables. A highest-weight vector in the $j$th summand can be chosen proportional to $(x_1{y}_2-y_1{x}_2{)}^j{x}_1^{3-j}{x}_2^{3-j}$. Exchanging the two factors multiplies it by $(-1{)}^j$, and hence multiplies the entire irreducible summand by that sign. Even $j$ belong to the symmetric square, odd $j$ to the exterior square. ThusTheir dimensions are respectively $7+3=10$ and $5+1=6$, as required for a four-dimensional space.

Label the weight vertices by $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$, with weights $e_1,\dots ,e_n,0,-e_n,\dots ,-e_1$. The crystal of the defining odd-orthogonal representation is the chainEach arrow of color $i$ subtracts the simple root ${\alpha }_i$. The two arrows through zero form the length-two short-root string.

Here is a complete grid description of the tensor product of crystals, valid for arbitrary $n$. Place the vertex $a\otimes b$ in row $a$, column $b$, in the above chain order; there are $(2n+1{)}^2$ vertices. Let ${\epsilon }_i(a)$ and ${\phi }_i(a)$ count incoming and outgoing steps along the color-$i$ string. For $i<n$, the two strings are $i\to i+1$ and $\overline{i+1}\to \bar{i}$, each of length one. For $i=n$, the string is $n\to 0\to \bar{n}$, with pairs $({\epsilon }_n,{\phi }_n)$ equal to $(0,2),(1,1),(2,0)$. Other vertices have both counts zero.

Using the crystal tensor-product rule, draw every color-$i$ edge byThe first case is a downward arrow to the next row and the second a rightward arrow to the next column; omit an arrow when its indicated factor has no outgoing edge. Thus the rule specifies every vertex and every arrow of the general tensor-square diagram, including the boundary nodes. The figure displays the resulting entire grid for $n=3$, alongside the vector crystal; the generator can draw any positive rank.

The raising rule chooses the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. For $n\ge 2$, its only highest vertices areIndeed a highest tensor must have highest first factor, hence $a=1$. For colors $i>1$, the second factor cannot have an incoming edge. For color one, its incoming count must be at most ${\phi }_1(1)=1$. Inspecting the strings leaves precisely $b=1,2,\bar{1}$.

These three connected components encode the Irreducible Lie algebra representationsHere $S_0^{2}V$ is the traceless symmetric square, ${\Lambda }^{2}V$ is the Adjoint representation, and the invariant symmetric form supplies the trivial summand. Their dimensions are $n(2n+3)$, $n(2n+1)$, and one, summing to $(2n+1{)}^2$. In fundamental-weight notation the highest weights are $2{\omega }_1,{\omega }_2,0$ for $n\ge 3$, but $2{\omega }_1,2{\omega }_2,0$ for $n=2$.

For $n=1$ the vector crystal is $1\stackrel{1}{\to }0\stackrel{1}{\to }\bar{1}$. The same tensor rule gives highest vertices $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, of weights $2e_1,e_1,0$, or $4{\omega }_1,2{\omega }_1,0$, respectively. Its summand dimensions are $5,3,1$.

Use the ordered alphabetwith weights $\mathrm{wt}(i)={\epsilon }_i$ and $\mathrm{wt}(\bar{i})=-{\epsilon }_i$. The crystal of the defining symplectic representation is the chainThe label of an arrow is its simple root color; a Kashiwara operator ${\tilde{f}}_i$ follows an arrow of color $i$ and lowers the weight by ${\alpha }_i$.

Here is a grid description of the entire tensor product of crystals, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:Their indicators are ${\phi }_i$ and ${\epsilon }_i$. Use the crystal tensor-product ruleAn arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.

For completeness, highest vertices can be read from the raising rule, which selects the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$. If $a$ has an incoming edge of color $i$, then ${\phi }_i(a)=0$; either the raising operator acts nontrivially on $a$, or ${\epsilon }_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all ${\phi }_i(a)$ vanish except ${\phi }_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge 2$, the possibilities are exactlyTheir weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$, and zero. Each occurs once. Connected highest-weight components of the crystal basis describe the irreducible representations in the tensor product of Lie algebra representations, soThis is the tensor-square decomposition of the defining symplectic representation. The symmetric square is $L(2{\omega }_1)$; the exterior square is the direct sum of its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$. Their dimensions are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim (V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes 1$ and $1\otimes \bar{1}$: the decomposition is $L(2{\omega }_1)\oplus L(0)$, of dimensions three and one.

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.

Solving the massive subsidiary conditions. In light-cone coordinates, use $d{s}^2=-2d{X}^{+}d{X}^{-}+{\delta }_{IJ}d{X}^{I}d{X}^J$ and write $d=D-2$. The divergence condition readsThus, when ${\partial }_{-}$ is invertible,First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The trace condition becomesThe independent components are $h_{--}$, $h_{-I}$ and the trace-free part of $h_{IJ}$. Under transverse rotations they form a scalar representation, a vector representation, and a symmetric traceless rank-two tensor. ThereforeThis is the light-cone decomposition of a massive spin-two field for $D\ge 3$. In $D=2$, the trace and divergence give $h_{+-}=h_{++}=0$ and ${\partial }_{+}{h}_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The massive particle little group is $SO(D-1)=SO(d+1)$, and its symmetric traceless square branches asThe mixed components with the extra direction give the vector; one independent trace combination gives the scalar. These are exactly the polarizations of a massive spin-two field. The remaining independent components retain the massive Klein-Gordon equation.

Transverse bosonic modes and mass levels. The variables ${\alpha }_k^I$ are Fourier amplitudes of the physical transverse open-string mode expansion. Classically, reality requires ${\alpha }_{-k}^I=({\alpha }_k^I{)}^{\ast }$. The symplectic term in the action fixes their quantum commutators:With string tension convention ${\alpha }^{\prime }=(2\pi T{)}^{-1}$, the zero mode of the constraint givesThe longitudinal nonzero modes have already been removed in light-cone gauge in string theory. Quantum normal ordering introduces the string intercept $a$, giving the open bosonic string mass spectrumEach bosonic occupation number is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.

Suppressing the common momentum label, the lowest light-cone levels of an open bosonic string areThe oscillator vacuum is annihilated by every positive ${\alpha }_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its massless particle little group. A massive vector would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus the first bosonic vector level must be massless, fixing $a=1$.

At level two the commuting creation operators give a symmetric square. Its scalar trace and symmetric traceless square, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one massive spin-two field with $m^2=1/{\alpha }^{\prime }$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the symmetric traceless rank-two tensor dimension of $SO(25)$.

Half-integer fermionic modes. The Neveu–Schwarz sector has antiperiodic worldsheet Majorana fermions. Its Neveu–Schwarz fermionic oscillators obeyThe oscillator vacuum satisfies ${\alpha }_k^I|0;p⟩=0$ for $k>0$ and $b_r^I|0;p⟩=0$ for $r>0$. A negative fermion mode is a fermionic creation operator for a transverse worldsheet excitation. The Neveu–Schwarz level operator and mass condition areThe smallest positive frequency is $1/2$, so the only first-excited states areThe same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.

At $N=1$, ${\alpha }_{-1}^I|0;p⟩$ is a vector, while $b_{-1/2}^I{b}_{-1/2}^J|0;p⟩$ is the exterior square: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:Thus the Neveu–Schwarz level-one massive tensor has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2{\alpha }^{\prime })$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric trace-free representation nor its scalar trace. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are before the GSO projection; the usual tachyon-removing GSO projection also removes this integer level.

Write $V_r={\mathrm{Sym}}^r{\mathbb{C}}^2$ for the irreducible SU(2) representation of dimension $r+1$, with character of a representation $z^r+z^{r-2}+\cdots +z^{-r}$. Here $V=V_3$, with weights $3,1,-1,-3$. Squaring its character of a representation and successively removing the highest-weight irreducible characters givesThe dimensions $7+5+3+1=16$ check the decomposition; complete reducibility follows from the invariant inner product.

For the exterior square, add distinct pairs of weights. Its character of a representation is $z^4+z^2+2+z^{-2}+z^{-4}={\chi }_{V_4}+{\chi }_{V_0}$. Equivalently it is $({\chi }_V(z{)}^2-{\chi }_V(z^2))/2$. Hence $\boxed{{\Lambda }^{2}V\cong V_4\oplus V_0}$, of dimensions five and one. The complementary symmetric square is $V_6\oplus V_2$.

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.