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Use signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the Proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

Use a mostly-plus target Minkowski spacetime metric and restore the string tension by $T=(2\pi {\alpha }^{\prime }{)}^{-1}$. After Wick rotation, the Polyakov action on a Riemann surface isThe Polyakov path integral integrates over embeddings and worldsheet metrics, dividing by worldsheet diffeomorphisms and Weyl transformations. For a consistent flat bosonic string theory, take the critical dimension of the bosonic string $d=26$. Conformal gauge turns the embedding fields into free scalars; its Faddeev-Popov determinant is represented by the bc system. The remaining integrations are over worldsheet moduli and external insertion points.

A tachyon vertex operator isThe worldsheet Green function is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. The operator product expansion with the holomorphic stress-energy tensor gives conformal weights $({\alpha }^{\prime }{k}^2/4,{\alpha }^{\prime }{k}^2/4)$. An integrated string vertex operator must have weights $(1,1)$, so the mass-shell condition is $k^2=4/{\alpha }^{\prime }$, or tachyonic mass parameter $m^2=-4/{\alpha }^{\prime }$.

At lowest order in the string coupling, the worldsheet is a Riemann sphere. For four external states, three insertion points can be fixed by a Möbius transformation. In the worldsheet ghost field description these three vertices carry $c\bar{c}$, while the fourth is integrated. The three-point worldsheet ghost correlator gives the squared product of the three fixed-point separations, precisely the sphere gauge fixing for four string vertices factor. Thus fixing three points is a gauge choice, rather than deleting three integrations without their Jacobian.

The integral over the constant embedding worldsheet zero mode gives $(2\pi {)}^d{\delta }^{(d)}({\sum }_i{k}_i)$. The remaining Gaussian integral, using normal ordering to remove self-contractions, gives the Koba-Nielsen factorSet $(z_1,z_2,z_3,z_4)=(0,z,1,\infty )$; the vertex at infinity is defined with its conformal-weight factor. Let all momenta be incoming and defineThe mass-shell condition and momentum conservation imply $s+t+u=-16/{\alpha }^{\prime }$. Since ${\alpha }^{\prime }{k}_1\cdot k_2=-4-{\alpha }^{\prime }s/2$ and similarly for the pair $(2,3)$, the sphere tachyon position integral becomesHere $\mathcal{N}$ depends on the normalization of external string vertex operators; it does not affect the kinematic dependence.

To evaluate the remaining worldsheet modulus, put $a=-1-{\alpha }^{\prime }s/4$ and $b=-1-{\alpha }^{\prime }t/4$. The complex beta integral isFor example, this identity follows by writing the two powers as Schwinger-parameter Gamma integrals, doing the two-dimensional Gaussian integral, and changing the two positive parameters to their sum and ratio. The complex beta integral first holds for $\mathrm{Re}a,\mathrm{Re}b>0$ and $\mathrm{Re}(a+b)<1$; the scattering result is its analytic continuation. The third gamma-function argument is $1-a-b=-1-{\alpha }^{\prime }u/4$. Absorbing $\pi$ into $\mathcal{N}$, the result is the tree-level four-tachyon amplitude, the Virasoro–Shapiro amplitude:It is symmetric in the three Mandelstam variables and has generic poles at $q=4(N-1)/{\alpha }^{\prime }$, matching the bosonic string mass spectrum. The formula is the leading term of string perturbation theory. Higher orders are constructed by the same Polyakov path integral prescription on higher-genus worldsheets, with four marked points and the corresponding ghost and worldsheet moduli measure; genus $g$ has coupling power $g_s^{2g+2}$.

The harmonic equation, including critical points. All derivatives in this argument use the three-dimensional Levi-Civita connection of $\gamma$. Write $U_i={\nabla }_{i}U$, $|\nabla U{|}^2={\gamma }^{ij}{U}_i{U}_j$ and ${\Delta }_{\gamma }U={\gamma }^{ij}{\nabla }_i{\nabla }_{j}U$. The curvature relation for the static vacuum conformal spatial metric gives $R=2|\nabla U{|}^2$. Substituting it into the contracted Bianchi identity yieldsThe Hessian of a scalar is symmetric, so the last terms cancel:Where $\nabla U\ne 0$, this implies ${\Delta }_{\gamma }U=0$. On the interior of the critical set $\{\nabla U=0\}$, $U$ is locally constant and its Laplacian is also zero. Every other critical point is a limit of noncritical points, so continuity of the Laplacian givesThis avoids incorrectly dividing by a gradient at its zeros.

Integration by parts and rigidity. The usual whole-space energy identity isWith standard static asymptotic falloff $U=O(R^{-1})$ and $\nabla U=O(R^{-2})$, the last term is $O(R^{-1})$ and vanishes. Positivity of the spatial Riemannian metric then makes $U$ constant, and its limiting value fixes that constant to zero.

There is also a direct integration by parts proof requiring only the stated vanishing of $U$ at infinity, rather than an assumed flux decay rate. For a regular value $\epsilon >0$, the region ${\Omega }_{\epsilon }=\{U>\epsilon \}$ has compact closure. Completeness and the absence of an inner boundary ensure no missing boundary pieces; standard asymptotic flatness and $U\to 0$ on all ends keep this positive level set away from infinity. Its outward unit normal is $n=-\nabla U/|\nabla U|$. Multiplying the harmonic equation by $U$ and integrating givesBoth terms are nonpositive, so both vanish. A nonempty component with $U>\epsilon$ cannot have zero gradient throughout and boundary value $\epsilon$. Thus ${\Omega }_{\epsilon }$ is empty. Choose arbitrarily small regular values and apply the same argument to $-U$; it follows thatThis is vanishing harmonic function by level-set integration. It also makes explicit why an inner boundary would invalidate the conclusion.

From zero Ricci curvature to Minkowski spacetime. In dimension three the Schouten tensor is $S_{ij}=R_{ij}-R{\gamma }_{ij}/4$, so it vanishes here. The three-dimensional curvature reconstruction then gives $R_{ijpq}=0$: $\gamma$ is a flat Riemannian manifold. This use of three-dimensional curvature from the Ricci tensor is essential; zero Ricci tensor would not by itself imply flatness in four dimensions.

A complete connected flat spatial manifold has Euclidean universal cover. Standard asymptotic flatness, with an ordinary Euclidean end, excludes a nontrivial free Euclidean quotient: a nontrivial fixed-point-free Euclidean isometry contains a translational or screw component, and its cyclic quotient has at most quadratic volume growth, incompatible with a three-dimensional Euclidean end. Extra identifications cannot restore cubic growth. Thus the complete spatial metric is globally Euclidean, not just locally flat. With $U=0$ and the standard global time coordinate the four-metric becomesTherefore horizonless static vacuum rigidity gives Minkowski spacetime as the sole solution under these hypotheses.

Allowing horizons. A black hole exterior can be static and asymptotically flat without being Minkowski. The Schwarzschild spacetime is the basic example. Its lapse function vanishes at the event horizon, so $U$ is unbounded below there; the horizon introduces an inner end or boundary in the spatial reduction, and the previous energy argument no longer has its hypotheses. In fact for Schwarzschild,so this conformal quotient terminates at $r=2M$ and is not the complete nonsingular quotient used above.

With the usual regularity, connected-horizon and global hypotheses of the static vacuum black-hole uniqueness theorem, the nontrivial black-hole exterior is Schwarzschild. Rotating Kerr black holes are stationary but not static and therefore are not alternatives in this question. This statement concerns a regular domain outside the horizon; the Schwarzschild interior contains a curvature singularity. If “globally static” and “nonsingular everywhere” are retained literally, Schwarzschild does not meet them. Allowing horizons means relaxing the horizonless global hypotheses to the corresponding static exterior problem.

Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

Use Minkowski spacetime signature $(-,+,+,+)$ and a closed spatial parameter of period $2\pi$. Choose the future-directed branch with positive lapse $e$.

Canonical dynamics and gauge freedom. Variation of momentum and embedding in the Nambu-Goto phase-space action givesThe Lagrange multipliers impose the Nambu–Goto phase-space constraints $P^2+T^2{X}^{\prime 2}=0$ and $P\cdot X^{\prime }=0$. These two first-class constraints reflect freedom to relabel time and space on the same string worldsheet. The canonical Hamiltonian is a linear combination of constraints, with arbitrary multiplier functions. These functions specify a coordinate description rather than additional propagating fields. No explicit gauge transformation or Poisson-bracket calculation is required.

Eliminating auxiliary variables. Let $V=\dot{X}-u{X}^{\prime }$. Eliminating momentum using $P=V/e$ leavesOn a patch with spacelike spatial tangent, variation of $u$ gives $V\cdot X^{\prime }=0$, hence $u=(\dot{X}\cdot X^{\prime })/X^{\prime 2}$. For the induced worldsheet metric $g_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$, this gives $V^2=\det g/X^{\prime 2}$. Variation of $e$ then gives $e=\sqrt{-\det g}/(T{X}^{\prime 2})$ on the positive branch. Substitution yieldsThis is minus string tension times Lorentzian worldsheet area. Vary the area using $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{ab}\delta g_{ab}$ and integrate by parts. The Nambu–Goto equations of motion areThe auxiliary-variable elimination and this metric form apply on nondegenerate timelike patches.

Circular motion, length and energy. For the circular embedding, direct differentiation givesAway from collapse, $\sqrt{-g}{g}^{ab}=\mathrm{diag}(-1,1)$, so the equations become ${\ddot{X}}^m-X^{\prime \prime m}=0$. The time and out-of-plane coordinates satisfy them immediately. For $Z=X^1+i{X}^2$, both $\ddot{Z}$ and $Z^{\prime \prime }$ are $-Z$. The relations $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$ verify the Virasoro constraints. Equivalently, $e=1/T$, $u=0$, $P=T\dot{X}$ solve the phase-space equations and constraints at all times.

The ring stays in a fixed plane, with radius $R|\cos t|$. It contracts to a point and re-expands. Each labeled point moves radially with speed $|\sin t|$ in target time $X^0=Rt$. The geometric ring repeats after target-time interval $\pi R$, although the labels have then shifted by half a circumference. It is not rigidly rotating. Velocity is perpendicular to the tangent, so simultaneous spatial arclength is also the local proper length along the string. Its length and conserved energy areAway from collapse, the same energy follows from $T\int dl/\sqrt{1-v^2}$. Increasing kinetic energy compensates the shrinking length. At collapse the induced metric degenerates; the area-form equation alone is undefined there. The regular phase-space solution supplies the continuation and the limiting constant energy. This is a pulsating circular string.

Endpoint variation. For an open string, integration by parts produces the boundary termIts coefficient is the open-string endpoint momentum flux. Evaluate this ungauge-fixed variation in a boundary-adapted temporal gauge for a string $X^0=t$. Then $X^{\prime 0}=0$ and the time equation gives $P^0=1/e\ne 0$. The free variation of time forces $u{P}_0=0$, hence $u=0$ at each end. Since $e$ is nonzero, the remaining spatial term requiresFree variation in every spatial direction gives ${\vec{X}}^{\prime }=0$, and together with $X^{\prime 0}=0$ gives the free-end string boundary condition $X^{\prime m}=0$. The general starting condition is the flux condition; the shift term should not be silently discarded before choosing the gauge.

Target-space charges. The Noether charges for translations and Lorentz transformations are the target-space Noether charges of a stringWriting $F_m=e{T}^2{X}_m^{\prime }+u{P}_m$, the canonical equations implyFor the second identity, antisymmetry cancels the $e{P}_m{P}_n$ terms. Integration by parts then cancels the $u{X}^{\prime }P$ terms and the symmetric $e{T}^2{X}_m^{\prime }{X}_n^{\prime }$ terms. For free ends $F_m=0$ individually, so both charges are constant. In conformal gauge, the same result follows from $u=0$ and $X^{\prime m}=0$. Target-space energy is $-{\mathcal{P}}_0$.

Other possibilities fix selected spatial directions by Dirichlet boundary conditions, while leaving Neumann boundary conditions in the permitted tangent directions. These mixed conditions describe endpoints on a D-brane; the two ends can lie on different branes. Time stays free under the given assumption. Fixed supports may absorb momentum in their Dirichlet directions and break corresponding translations or Lorentz symmetries. More general force-law conditions require an additional boundary interaction.

Independent metric and Hamiltonian parametrization. The Lorentzian Polyakov action isThe worldsheet metric is independent of the embedding. Its equation sets the traceless worldsheet stress tensor to zero, making the metric locally conformal to the induced worldsheet metric; substitution recovers the Nambu–Goto action. Eliminating momentum in the Nambu-Goto phase-space action instead gives $(\dot{X}-u{X}^{\prime }{)}^2/(2e)-e{T}^2{X}^{\prime 2}/2$. On the positive-lapse branch, choose the independent metric, up to a positive Weyl transformation, asIts determinant is $-e^2{T}^2$. Multiplication by $-T/2$ recovers the reduced density, including the mixed term $-u\dot{X}\cdot X^{\prime }/e$. This proves the phase-space identification of the Polyakov metric.

Conformal gauge and residual coordinate freedom. In the Hamiltonian formulation, conformal gauge is $e=1/T,u=0$. The equations become $\ddot{X}-X^{\prime \prime }=0$, while ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$ remain as constraints. In the metric formulation, conformal gauge is ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$, or simply ${\eta }_{ab}$ after fixing Weyl invariance. The metric parametrization proves these choices equivalent. Gauge-fixing the metric does not discard the Virasoro constraints.

With ${\sigma }^{\pm }=t\pm \sigma$, independent reparameterizations ${\sigma }^{\pm }\mapsto f_{\pm }({\sigma }^{\pm })$ multiply the flat metric by a conformal factor. A compensating Weyl transformation restores its chosen representative. Infinitesimally the Conformal Killing equation is ${\partial }_{-}{c}^{+}=0,{\partial }_{+}{c}^{-}=0$. These are the residual conformal transformations of two-dimensional Minkowski spacetime, with the periodicity conditions appropriate to a closed string.

Ghost determinant and anomaly. For gauge conditions $G_A(\varphi )=0$, the Faddeev-Popov determinant is the determinant of the linearized variationInsert this determinant in the gauge-fixed path integral and exponentiate it with anticommuting ghosts and antighosts: the additional action is proportional to $\int b^A{M}_{AB}{c}^B$. In metric gauge fixing, the trace is removed by Weyl invariance, and a worldsheet diffeomorphism changes the trace-free metric throughConsequently $c^a$ is a vector ghost and $b^{ab}$ a symmetric trace-free antighost. A conventional normalization of the worldsheet ghost action isA rescaling of the antighost changes only this overall kinetic normalization. In conformal gauge, after normalizing the chiral fields, the kinetic terms areThe closed-string ghosts are periodic, with independent left and right systems. Their conformal weights are $2$ for $b$ and $-1$ for $c$, so the supplied bc ghost system expression gives $c_{\mathrm{gh}}=-2(6\cdot 2^2-6\cdot 2+1)=-26$. Each free embedding boson has central charge one per chirality. HenceThis central charge of reparameterization ghosts cancels the string conformal anomaly in each chirality separately. Having two sectors does not double the required target dimension.

Channel variables, poles and the vector state. With particles 1,2 incoming and 3,4 outgoing, the Mandelstam variables are $s=-(p_1+p_2{)}^2$ and $t=-(p_1-p_3{)}^2$. They measure squared center-of-mass energy and momentum transfer. In the specified units ${\alpha }^{\prime }=1$, the external tachyon mass squared is $-1$, so $s+t+u=-4$.

The Gamma function residue at $-n$ is $(-1{)}^n/n!$. At $s=n-1$, its argument $-1-s$ varies with the opposite sign to $s$. The Gamma function recurrence givesThus the Veneziano amplitude pole residues areFor generic fixed $t$ the simple poles are $-1,0,1,\dots$. A zero of the residue product at an exceptional negative integer $t$ makes the corresponding pole removable. Fixed $t$ must be away from its own channel poles; fixing it at a channel singularity does not give an ordinary finite meromorphic function of $s$.

At $s=0$, the residue is $-(t+2)=-(t-u)/2$. This is proportional to $(p_1-p_2)\cdot (p_3-p_4)$, the contraction of conserved scalar-pair vector currents. Each current is orthogonal to the exchanged momentum because the external masses in its pair are equal. Alternatively, with scattering-angle cosine $z$, one has $t=-2(s/4+1)(1-z)$; at this pole $t+2=2z$, a pure spin-one angular dependence. ThereforeThe massless vector pole of the Veneziano amplitude is the massless open-string vector seen in the two-tachyon exchange channel.