 Solution

Under the state–operator correspondence, let $|O_{ij}⟩$ be the state of the antisymmetric conformal primary operator $O_{ij}$. In radial quantization $P_i^{†}=K_i$, while primarity gives $K_i|O_{jk}⟩=0$. The norm of the level-one descendant obtained by taking a divergence is therefore determined by the conformal algebra commutatorUsing the two-form action of $M_{ab}$ gives, up to a positive normalization,Positivity of norm yields the two-form conformal unitarity boundAt saturation the descendant is null, and the operator obeys the conservation equation ${\partial }^i{O}_{ij}=0$.

In $d=3$, Hodge duality turns the two-form into the vector primary $V_k={ϵ}_{kij}{O}^{ij}/2$. The vector divergence descendant has norm proportional to $\Delta -d+1=\Delta -2$. Consequently the stronger bound israther than the formal two-form value $d-2=1$.