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For the defining representation of the Special orthogonal Lie algebra, Hermitian Lie algebra generators $t_{ij}=i(E_{ij}-E_{ji})/\sqrt{2}$ have trace normalization $\mathrm{tr}(t_{ij}{t}_{kl})={\delta }_{ik}{\delta }_{jl}$ for ordered pairs. Summing their squares gives ${\sum }_{i<j}{t}_{ij}^2=(N-1)I/2$. Consequently the Casimir eigenvalue is $(N-1)/2$ and the Dynkin index is one. Doubling the trace normalization doubles both numbers; a gauge coupling must be rescaled inversely with the Lie algebra generators. With long roots of squared length two, the first convention is $C_2(\lambda )=(\lambda ,\lambda +2\rho )/2$.

Choose five Hermitian four-by-four matrices obeying the Euclidean Clifford algebra relations ${\gamma }_a{\gamma }_b+{\gamma }_b{\gamma }_a=2{\delta }_{ab}I$. For $M_{ab}=E_{ab}-E_{ba}$, the displayed anti-Hermitian matrices satisfy the Special orthogonal Lie algebra Lie bracket relations. A direct proof first establishes $[\rho (M_{ab}),{\gamma }_c]={\delta }_{bc}{\gamma }_a-{\delta }_{ac}{\gamma }_b$, then applies the commutator product rule to $[{\gamma }_c,{\gamma }_d]$. One explicit choice is ${\gamma }_1={\sigma }_1\otimes I$, ${\gamma }_2={\sigma }_2\otimes I$, ${\gamma }_3={\sigma }_3\otimes {\sigma }_1$, ${\gamma }_4={\sigma }_3\otimes {\sigma }_2$, ${\gamma }_5={\sigma }_3\otimes {\sigma }_3$, with the Pauli matrices. The Cartan subalgebra operators $-i\rho (M_{12})$ and $-i\rho (M_{34})$ have simultaneous eigenvalues $(\pm 1/2,\pm 1/2)$, identifying the four-dimensional spinor representation. A full $2\pi$ rotation acts as $-I_4$, proving it does not descend to $SO(5)$.

For a Lie subalgebra $\mathfrak{g}\subseteq \mathfrak{gl}(V)$, its defining representation is its given action on the vector space $V$. The Lie bracket identity $[X,Y]v=X(Yv)-Y(Xv)$ verifies the representation property. For the special linear Lie algebra, diagonal trace-zero matrices have coordinate weights of a representation $L_i$ on the standard basis; for a Special orthogonal Lie algebra, the same action preserves the defining symmetric bilinear form.

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.

Over $\mathbb{C}$, the symplectic Lie algebra on a four-dimensional symplectic vector space acts faithfully on its five-dimensional primitive exterior square, preserving its symmetric bilinear form. Faithfulness follows because an operator killing all wedges $e_i\wedge e_j$ has all off-diagonal coefficients zero and satisfies $A_{ii}+A_{jj}=0$ for every pair, hence is zero. The source and target Lie algebras both have dimension ten, so this faithful map is an isomorphism to the Special orthogonal Lie algebra ${\mathfrak{so}}_5$.

For a real inner product space $V$, the displayed map identifies ${\Lambda }^{2}V$ with its Special orthogonal Lie algebra. It sends a wedge of distinct orthonormal basis vectors to the corresponding elementary skew-symmetric matrix, so it is an isomorphism. For an orthogonal transformation $g$, invariance of the inner product gives $\Phi (gu\wedge gv)=g\Phi (u\wedge v)g^{-1}$. Thus the exterior square of the defining representation is the Adjoint representation of the orthogonal group. Complexification identifies their weight spaces, allowing the root system to be read from the exterior-square weights and the zero-weight multiplicity.

For an infinitesimal element of the SO(4) group, write $U=I+iϵT+O({ϵ}^2)$. Reality of $U$ makes $T$ purely imaginary; $U^{T}U=I$ gives $T^T=-T$. Together these imply $T^{†}=T$, while antisymmetry makes the trace vanish. An antisymmetric four-by-four matrix has $4(4-1)/2=6$ independent entries. Thus six imaginary, Hermitian, antisymmetric, traceless matrices span the Special orthogonal Lie algebra in this convention. The physicists' Hermitian generators differ by a factor of $i$ from the real antisymmetric matrices used for its real Lie algebra.

A nonzero real four-vector has a length and a direction. Proper rotations act transitively on the unit three-sphere, so a gauge choice aligns the scalar vacuum expectation value along the fourth coordinate: ${\Phi }_0=v{e}_4$. This chooses an orientation of the vacuum without changing its length. Its stabilizer rotates the first three coordinates and is an SO(3) group.

The supplied generators obeyHence the three sums are unbroken generators. The orthogonal field combinationsare respectively the unbroken and broken gauge directions. With $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^A{T}^A$, direct action on the vacuum givesCompare with the vector mass term $m^2{A}_{\mu }{A}^{\mu }/2$. ThereforeThe SO(4) vector-Higgs gauge-boson spectrum uses the scalar kinetic normalization printed in the paper, $|D\Phi {|}^2$. Inserting an additional factor $1/2$ while keeping the same numerical $v$ would instead give $|gv|/\sqrt{2}$. This is a field-normalization convention, not a disagreement about the broken subgroup. The massless fields generate the residual SO(3) gauge symmetry; they are sums, not three of the six original gauge fields individually.

Multiplying the first three supplied matrices givesThey generate a copy of the SU(2) Lie algebra. The other triple generates a second copy, and the two commute. Since they span all six dimensions,This is the two commuting rotation algebras in so4 decomposition. A Lie-algebra isomorphism fixes the local group structure but not its global central identifications. The Spin(4) double cover givesThus SO(4) is not globally the direct product of the two simply connected groups. The unbroken diagonal Lie algebra integrates to SO(3), with the same simultaneous central sign already identified. This distinguishes the gauge algebra, its covering group and the actual group acting faithfully on the four-vector.

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore

Equip the finite-dimensional real vector space $V$ with a positive-definite inner product. A reduced crystallographic root system is a finite set $R\subset V\setminus \{0\}$ spanning $V$ such that, for every $\alpha \in R$,where $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ is the orthogonal reflection in ${\alpha }^{\perp }$. The Weyl group is $W(R)=⟨s_{\alpha }:\alpha \in R⟩$. It permutes the finite set $R$. Its action is faithful because $R$ spans $V$: an element fixing every root fixes every vector. Hence it embeds in the finite symmetric group of $R$ and is finite.

A Cartan subalgebra is a nilpotent self-normalizing Lie subalgebra. For a finite-dimensional complex semisimple Lie algebra it is equivalently a maximal toral subalgebra: a maximal abelian subalgebra whose elements have diagonalizable adjoint actions. The commuting semisimple operators $\mathrm{ad}(h)$ give the root-space decompositionThe nonzero weights $\alpha$ are the roots. Their real span inside the complex dual ${\mathfrak{h}}^{\ast }$, with the positive-definite real inner product induced by the Killing form, is the Euclidean space in which they form a reduced crystallographic root system. In particular one uses this real span, rather than treating all of ${\mathfrak{h}}^{\ast }$ as a real vector space of twice the rank.

For the specified Special orthogonal Lie algebra, put $\bar{i}=7-i$. The matrix relation is $x_{ij}=-x_{\bar{j},\bar{i}}$. Thus its diagonal Cartan subalgebra consists ofLet ${\epsilon }_i(H(t))=t_i$. The matrix units satisfy $[H,E_{ij}]=(d_i-d_j)E_{ij}$, where $d=(t_1,t_2,t_3,-t_3,-t_2,-t_1)$. A basis of root vectors is supplied, for $1\le i<j\le 3$, byThese are twelve independent off-diagonal matrices. Together with the three diagonal generators, they account for the full dimension $6\cdot 5/2=15$, so no further roots occur. ConsequentlyChoose the simple rootsThe six positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$, and every other root is their negative. In the standard orthonormal coordinates all roots have squared length $2$, and their Cartan matrix isIts Dynkin diagram is the connected chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$. A decomposition into mutually orthogonal nonempty root systems would separate this connected set of simple roots, which is impossible. Thus the root system is irreducible. The chain is also the diagram $A_3$, which is the root system of ${\mathfrak{sl}}_4(\mathbb{C})$. The classification of complex semisimple Lie algebras by their root systems yields the Lie algebra isomorphismAn explicit explanation of this coincidence uses the exterior-power Lie algebra representation of ${\mathfrak{sl}}_4$ on ${\bigwedge }^2{\mathbb{C}}^4$, a six-dimensional space. After choosing a volume form, $u\wedge v$ defines a nondegenerate symmetric bilinear form on that exterior square. The induced action preserves it because a traceless endomorphism acts trivially on the top exterior power. This gives a nonzero homomorphism ${\mathfrak{sl}}_4\to {\mathfrak{so}}_6$. Its kernel is an ideal in the simple algebra ${\mathfrak{sl}}_4$, so it is zero; equal dimensions $15$ then make it an isomorphism. Over $\mathbb{C}$ all nondegenerate symmetric forms in dimension six are equivalent, including the form with matrix $J$.

The complex Special orthogonal Lie algebra ${\mathfrak{so}}_5$ consists of endomorphisms preserving a nondegenerate symmetric bilinear form infinitesimally:In an orthonormal basis this is $\{X\in M_5(\mathbb{C}):X^T+X=0\}$, with the commutator bracket and dimension $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$. Its compact real form is obtained by taking real skew-symmetric matrices. For diagonal root calculations it is more convenient to use an isotropic basis $v_1,v_2,v_0,v_{-2},v_{-1}$ with $B(v_i,v_j)={\delta }_{i,-j}$, including $B(v_0,v_0)=1$.

A maximal torus of the complex special orthogonal group isIts Lie algebra is the Cartan subalgebraIn the compact real form the corresponding torus consists of independent rotations in two orthogonal coordinate planes, fixing the remaining direction. Define ${\epsilon }_i(H)=h_i$ for $i=1,2$, and ${\epsilon }_{-i}=-{\epsilon }_i$, ${\epsilon }_0=0$.

Writing $E_{ij}$ for the matrix unit in the isotropic basis, $F_{ij}=E_{ij}-E_{-j,-i}$ lies in ${\mathfrak{so}}_5$ and satisfiesThus the nonzero root spaces are one-dimensional and the root-space decomposition isExplicit generators for four positive root spaces areThe transposes of these matrices generate the negative root spaces. Each transpose still satisfies the bilinear-form condition and reverses the displayed diagonal weight. These eight root vectors and the two independent Cartan matrices exhaust the dimension ten; a matrix commuting with every $H$ is diagonal and the form condition makes it lie in $\mathfrak{h}$. This also proves that the torus has maximal dimension.

For a real inner product space $V={\mathbb{R}}^5$, defineThis map takes values in the Special orthogonal Lie algebra: its matrices are skew-symmetric. It sends $e_i\wedge e_j$ to $E_{ij}-E_{ji}$, so it is an isomorphism between the two ten-dimensional spaces. For $g\in SO(5)$, invariance of the inner product givesThis proves the exterior square realization of the orthogonal adjoint representation, including equivariance, rather than merely matching dimensions.

After complexification, part (a) is therefore the weight decomposition of the Adjoint representation. Its zero weight space is the complex centralizer of $\mathfrak{t}$. The two zero-weight wedges are nonzero multiples of $e_1\wedge e_2$ and $e_3\wedge e_4$, which map under $\Phi$ to the two infinitesimal rotation generators of $T$. HenceIf a larger torus contained $T$, its Lie algebra would commute with $\mathfrak{t}$ and therefore lie in $\mathfrak{t}$. Equality of Lie algebras and connectedness of both tori force equality of the groups. Thus $T$ is a maximal torus.

Let ${\epsilon }_j$ denote the real coordinate functional $x_j$. In the real weight convention, with the common factor $i$ removed from the differentiated torus weights, the root system isThese are the roots of a root system of type B2 root system. In the imaginary-valued compact Lie-algebra convention each displayed functional is multiplied by $i$. Every root space is one-dimensional, as part (a) explicitly shows.

Put $\bar{i}=2n+1-i$ and let $E_{ij}$ denote a matrix unit. The antidiagonal matrix satisfies $J^{-1}=J$, so the orthogonal condition is $A=-J{A}^{T}J$, or entrywise $A_{ij}=-A_{\bar{j}\bar{i}}$. Consequently the diagonal Cartan subalgebra isWe first take $n\ge 2$. Since $[H,E_{ab}]=(H_{aa}-H_{bb})E_{ab}$, the matrix root basis of the even orthogonal Lie algebra givesThe zero weight space is $\mathfrak{t}$; its basis is $H_i=E_{ii}-E_{\bar{i}\bar{i}}$. No vector of weight $2{\epsilon }_i$ occurs, because the corresponding matrix unit is forced to be its own negative. These spaces account for $n+2n(n-1)=n(2n-1)$ dimensions, exhausting the Special orthogonal Lie algebra. Thus its decomposition as a $\mathfrak{t}$-module and its Dn root system areEach nonzero root space has dimension one; the zero weight has multiplicity $n$.

The upper triangular root spaces are precisely those of weights ${\epsilon }_i-{\epsilon }_j$ and ${\epsilon }_i+{\epsilon }_j$ with $i<j$. ThereforeThese $n$ roots form the simple roots. To check that they generate the chosen positive system with nonnegative coefficients, write ${\epsilon }_i-{\epsilon }_j={\sum }_{k=i}^{j-1}{\alpha }_k$. For $j<n$,and for $j=n$ use ${\epsilon }_i+{\epsilon }_n={\sum }_{k=i}^{n-2}{\alpha }_k+{\alpha }_n$. Empty sums are zero. Their linear independence and these expansions identify the base of the root system.

Equip the real span of the roots in ${\mathfrak{t}}^{\ast }$ with the inner product making the ${\epsilon }_i$ orthonormal. All roots then have squared length two. The coroots of ${\epsilon }_i\pm {\epsilon }_j$ are $H_i\pm H_j$, so the fundamental weights of Dn, characterized by ${\omega }_i({\alpha }_j^{\vee })={\delta }_{ij}$, areTaking their pairings with $H_j-H_{j+1}$ and $H_{n-1}+H_n$ verifies the defining identities directly. For each pair $i<j$, the sum of its two positive roots is $2{\epsilon }_i$. Hence the half-sum of positive roots is

The Dynkin diagram has a chain ending at ${\alpha }_{n-2}$, which is joined to both ${\alpha }_{n-1}$ and ${\alpha }_n$. Every edge is a single bond because the root lengths agree; the nonzero off-diagonal simple-root pairings are minus one. The labelled Dn Dynkin diagram and its low-rank versions are shown below.

For $n=2$, ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2={\epsilon }_1+{\epsilon }_2$ are orthogonal, giving two disconnected A1 root systems. For $n=3$, the diagram is the chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$, of type A3 root system; the displayed weight formulas still apply. If $n=1$ is allowed, ${\mathfrak{so}}_2=\mathfrak{t}$ is one-dimensional and abelian: there are no roots or fundamental weights, and $\rho =0$.

Put $\bar{i}=2n+2-i$ and $o=n+1$. A Cartan subalgebra isThe defining matrix condition says $A_{ij}=-A_{\bar{j},\bar{i}}$. Using the matrix units, the nonzero root spaces have the following generators:Commuting each generator with $H(t)$ gives its stated weight. Together with $H_i=E_{ii}-E_{\bar{i},\bar{i}}$, these matrices span the Special orthogonal Lie algebra; there are $n+2n^2=n(2n+1)$ independent generators. Thus its decomposition as a Cartan subalgebra module isThis is the Bn root system. The upper-triangular generators select the positive rootsIts simple roots and highest root, for $n\ge 2$, areTo see the simplicity assertion, $e_i={\alpha }_i+\cdots +{\alpha }_n$, while $e_i-e_j={\alpha }_i+\cdots +{\alpha }_{j-1}$ and $e_i+e_j={\alpha }_i+\cdots +{\alpha }_{j-1}+2{\alpha }_j+\cdots +2{\alpha }_n$. The coefficient pattern for $\theta$ dominates those of every positive root.

Normalize the inner product so the $e_i$ are orthonormal. The simple coroots are ${\alpha }_i^{\vee }={\alpha }_i$ for $i<n$ and ${\alpha }_n^{\vee }=2e_n$. Solving $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$ gives the fundamental weightsIn the half-sum of positive roots, the coordinate $e_i$ occurs with total coefficient $2(n-i)+1$. Hence the Weyl vector is

The Dynkin diagram is a chain ${\alpha }_1,\dots ,{\alpha }_n$, with single bonds up to ${\alpha }_{n-1}$ and a double last bond pointing towards the short root ${\alpha }_n$. In the Extended Dynkin diagram, add ${\alpha }_0=-\theta$. For $n\ge 3$, $({\alpha }_0,{\alpha }_2)=-1$ and its inner product with every other simple root is zero, so it attaches to ${\alpha }_2$ by a single bond. For $n=2$, both long nodes $0,1$ attach by double bonds to short node $2$. These rules give the Bn Dynkin diagram and affine extension drawn below.

For rank one, $R=\{\pm e_1\}$, ${\alpha }_1=\theta =e_1$, ${\omega }_1=e_1/2$, and $\rho =e_1/2$. The finite diagram is one node; its affine extension has two equal-length nodes joined by a double bond, without a short-root arrow.

The diagonal Cartan subalgebra isPut ${\epsilon }_1=L_1$ and ${\epsilon }_2=L_2$; then $L_3=-{\epsilon }_1$, $L_4=-{\epsilon }_2$ and $L_5=0$ on $\mathfrak{h}$. For the matrix unit $E_{ij}$, $[H,E_{ij}]=(L_i-L_j)(H)E_{ij}$. Applying this to the general matrix in part (i) gives the following one-dimensional root spaces and chosen root vectors:Each row satisfies $[H,X_{\alpha }]=\alpha (H)X_{\alpha }$ and belongs to the Special orthogonal Lie algebra. The two diagonal parameters and eight root vectors span the ten-parameter matrix from part (i), giving the explicit root-space decompositionThis is the B2 root system, with short roots on the coordinate axes and long roots on the diagonals.

Preserving the symmetric bilinear form infinitesimally means $Z^{T}M+MZ=0$. In blocks of sizes $2,2,1$, write $Z$ as a general block matrix and multiply by the given $M$. The condition says that $MZ$ is skew-symmetric. It forces the lower-right entry to be zero, the middle diagonal block to be the negative transpose of the upper-left block, and the two off-diagonal two-by-two blocks to be skew-symmetric. Thus the general element of the Special orthogonal Lie algebra iswhere $A$ is an arbitrary two-by-two matrix and $p,q$ are arbitrary two-component columns. More explicitly, using ten independent complex parameters,Conversely, substituting this matrix into $Z^{T}M+MZ$ gives zero, so there are no further restrictions. In particular $\dim {\mathfrak{so}}_5=4+1+1+2+2=10$.

Use the free-boson operator product expansionWrite $E_p=:e^{ip\cdot X}:$ and $ϵ=z-w$. The needed Wick contractions are $\partial X^{\mu }(z)\partial X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }/(2{ϵ}^2)$ and $\partial X^{\mu }(z)E_p(w)\sim -i{\alpha }^{\prime }{p}^{\mu }{E}_p(w)/(2ϵ)$. For $V_{\zeta }={\zeta }_{\mu }:\partial X^{\mu }{E}_p:$, the cross double contraction in the worldsheet stress tensor has a cubic pole: either of its two derivatives can contract with the derivative in $V_{\zeta }$, while the other contracts with $E_p$. Including the factor $-1/{\alpha }^{\prime }$ givesA primary operator has no cubic pole. Hence the first vertex is primary precisely whenThe antiholomorphic conformal weight is $\bar{h}={\alpha }^{\prime }{p}^2/4$. Primarity alone puts no mass-shell condition on $p$.

For $W_{\zeta }={\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, the holomorphic cubic pole is $-i{\alpha }^{\prime }{p}^{\mu }{\zeta }_{\mu \nu }:\bar{\partial }{X}^{\nu }{E}_p:/(2{ϵ}^3)$. The antiholomorphic operator product expansion gives the analogous pole involving $p^{\nu }{\zeta }_{\mu \nu }$. Thus the transversality of derivative-exponential primary vertices requiresThere is no trace-free requirement for primarity. A physical integrated string vertex operator must additionally have conformal weights $(1,1)$; for this vertex that gives $p^2=0$.

Now keep the fermionic signs in the real chiral Majorana fermion calculation. Contracting the external $\psi (w)$ through $:\psi (z)\partial \psi (z):$ givesMultiplying by $-1/2$ and Taylor expanding $\psi (z)$ at $w$ producesThis proves that $\psi$ is a primary operator with weight $h_{\psi }=1/2$. To find the central charge, in $T(z)T(w)$ the two double-contraction pairings contribute, before the overall factor $1/4$,The single Wick contractions give $2T(w)/{ϵ}^2+\partial T(w)/ϵ$, soComparing the leading term with $c/(2{ϵ}^4)$ proves $c_{\psi }=1/2$, as in the free chiral Majorana fermion conformal field theory.

With four embedding bosons and the ordinary reparameterization bc ghost system, cancellation of the worldsheet Weyl anomaly in each chirality requiresThus there are 44 real left-moving and 44 real right-moving fermions, equivalently 44 two-dimensional nonchiral Majorana fermions. These are internal free fermions in a four-dimensional bosonic string. Adding internal matter to the bosonic string does not introduce local worldsheet supersymmetry or its superghosts.

For the requested polynomial string vertex operators, work in the unprojected Neveu–Schwarz sector. A full model also chooses spin structures and a consistent projection, which the central-charge condition alone does not specify. If the holomorphic and antiholomorphic oscillator contributions to the conformal weights are ${\ell }_L,{\ell }_R$, physicality requiresTherefore only equal levels at most one can contribute to the requested nonpositive bosonic string mass spectrum.

With zero fermion insertions, level zero gives the tachyon vertex operator $E_p$, a scalar with $M^2=-4/{\alpha }^{\prime }$. Level one gives the massless closed-string vertex operator ${\zeta }_{\mu \nu }:\partial X^{\mu }\bar{\partial }{X}^{\nu }{E}_p:$, with the transverse conditions above and the string-state gauge redundancy that removes longitudinal polarizations. The physical transverse polarization tensor decomposes into a symmetric trace-free graviton, an antisymmetric Kalb–Ramond field, and a scalar dilaton. In four spacetime dimensions these have respectively two, one, and one physical polarizations; the two-form can be dualized to a scalar.

With two fermion insertions there are two possibilities. One fermion in each chirality givesThese are $44^2=1936$ scalar tachyons before any projection. There is no spacetime derivative or spacetime polarization to impose a vector transversality condition.