Relationship between mass and twist for operators with spin $s$ in the $AdS/CFT$ correspondence

Question

The relation between scaling dimension and mass of a corresponding field in AdS is given by: $$\Delta=\frac{d}{2}+\sqrt{\frac{d^2}{4}+m^2R^2},$$ Is this formula only valid for scalar fields? I would expect this formula gets modified for spin-$s$ operators and fields so that the minimal twist $\tau=d-1$, where $\tau= \Delta -s$, corresponds to a massless field. Can someone point out the appropriate reference and give a simple explanation?

Answer 1

First of all, you are mixing notations. For the first formula you give to be correct, $d$ needs to be the boundary dimension. But for $\tau = d - 1$ to be the minimal twist, $d$ needs to be the bulk dimension. I will use boundary from now on.

But anyway, that expectation is almost right. The minimal value of twist instead gets mapped to a strictly negative squared mass called the BF bound. The generalization to non-zero spin is \begin{equation} \Delta(\Delta - d) + s(s + d - 2) = m^2 R^2 \end{equation} which is easy to remember because the left hand side is an eigenvalue of the quadratic Casimir of the conformal group. These TASI notes are my favourite source which makes the relation between the bulk Laplacian and boundary Casimir operator manifest.