= Solution
Under the <state–operator correspondence>, let $|O_{ij}\rangle$ be the state of the antisymmetric <conformal primary operator> $O_{ij}$. In radial quantization $P_i^\dagger=K_i$, while primarity gives $K_i|O_{jk}\rangle=0$. The norm of the level-one descendant obtained by taking a divergence is therefore determined by the <conformal algebra> commutator
$$
[K_a,P_b]=2i(\delta_{ab}D-M_{ab}).
$$
Using the two-form action of $M_{ab}$ gives, up to a positive normalization,
$$
\|P^i|O_{ij}\rangle\|^2
\propto(\Delta-d+2)\|O\|^2.
$$
Positivity of norm yields the two-form <conformal unitarity bound>
$$
\boxed{\Delta\geq d-2\qquad(d\geq4).}
$$
At saturation the descendant is null, and the operator obeys the conservation equation $\partial^iO_{ij}=0$.
In $d=3$, Hodge duality turns the two-form into the vector primary $V_k=\epsilon_{kij}O^{ij}/2$. The vector divergence descendant has norm proportional to $\Delta-d+1=\Delta-2$. Consequently the stronger bound is
$$
\boxed{\Delta\geq2\qquad(d=3),}
$$
rather than the formal two-form value $d-2=1$.
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