Consistent couplings between spin-2 and spin-3 massless fields

Abstract

We solve the problem of constructing consistent first-order cross-interactions between spin-2 and spin-3 massless fields in flat spacetime of arbitrary dimension n > 3 and in such a way that the deformed gauge algebra is non-Abelian. No assumptions are made on the number of derivatives involved in the Lagrangian, except that it should be finite. Together with locality, we also impose manifest Poincaré invariance, parity invariance and analyticity of the deformations in the coupling constants.1 Chargé de Recherches FNRS (Belgium); nicolas.boulanger@umh.ac.be 2 serge.leclercq@umh.ac.be plying γ to Equation (4.16), dγb 1 = 0 is obtained. Thanks to the Poincaré lemma and Proposition 3, we see that b 1 can be taken inLet us now introduce the differential D defined in Section 3.4, we obtainThis is because the left-hand side is strictly non γ-exact. Let us label the ghosts more preciselyThen, as D raises by 1 the D-degree, the only non zero components of the matrix A are A J i J i+1 and the last equation decomposes into:The first equation means thatthanks to Proposition 5 and Proposition 6. The λ's are constants, because of the Poincaré invariance. We obtainThus, β J 0 A J 0 J 1 depends only on the underivated antifields, which cannot be δ-exact modulo d unless they vanish, because a δ-exact term depends on the derivatives of the antifields or on the equations of motion, and because of the Poincaré invariance. Thus, δα J 1 + dβ J 1 = 0 and β J 0 A J 0 J 1 = 0 independently. By applying the same reasoning recursively, the same decomposition appears to occur at every D-degree, so we finally obtain:(4.18)

Universal limits on massless high-spin particles

“…In particular, together with the mass-shell conditions (19) it makes the matrix element transverse and traceless. In reality, constraints (27)(28)(29)(30)(31) are too weak, because if any of the coefficients A k ,B k , C k and α k A k had a singularity 1/q 2 6 then vertex (25) would imply the existence of another massless spin 2 field (it couples to a transverse-traceless vertex!) which mixes linearly with the graviton.…”

“…Yet, the cubic vertex of [30], or any other vertex that may have been proposed in the literature, cannot be extended beyond cubic order: Weinberg's theorem [5] forbids it 10 . This is seen by applying Weinberg's factorization argument, reviewed in Section 2, to a vertex with two spin s particles and two gravitons.…”

Higher-spin fermionic gauge fields and their electromagnetic coupling

“…(3.7), shows that both the elements in Subset-1 enjoy such a lift, thanks to the relations (B.9)-(B.10) among others. Explicitly, 8) and similarly for the hermitian conjugate terms. Hereã 1 is the usual ambiguity: Γã 1 = 0.…”

“…The latter include one non-abelian (2s − 2)-derivative vertex, a 2s-derivative abelian one that is gauge invariant up to a total derivative and exists in D ≥ 5, and a Born-Infeld type abelian one containing 2s + 2 derivatives. This could be seen by using either the light-cone method [10] or the cohomological methods [8,9].…”

“…Indeed, N = 2 SUGRA [6,7] allows massless gravitini to have dipole and higher-multipole couplings, but forbids a non-zero U(1) charge in flat space. Gravitational and EM multipole interactions also show up, for example, as the 2 − s − s and 1 − s − s trilinear vertices constructed in [8,9] for bosonic fields. 1 These cubic vertices are but special cases of the general form s − s ′ − s ′′ , that involves massless fields of arbitrary spins.…”