Complete expression of special conformal generator in $d\geq 3$ does not satisfy conformal algebra

Question

I am having trouble with the equation (4.31) from DiFrancesco's Conformal Field Theory. \begin{equation} K_\mu = 2 x_\mu \tilde{\Delta}-x^\nu S_{\mu \nu} -2i x_\mu x^\nu \partial _\nu +i x^2 \partial _\mu. \tag{4.31} \end{equation} It seems from his derivation as if there is a factor of $2$ missing in the spin part. In fact if you work out the commutator of $K_\mu$ with $P_\nu=- i\partial_\nu$ it is supposed to give \begin{equation} [K_\mu , P_\nu] = 2i(\eta _{\mu \nu}D-L_{\mu \nu}) \end{equation} but for me that just does not work out unless there really is a factor missing. If this is a typo then it is not yet listed in the errata and has been copied into the lecture notes by J. Qualls. (arXiv:1511.04074 Eq.2.37)

On top of that I tried to find an alternative derivation of the expression of $K_\mu$ to check for myself. I started from the definition of the generator and the transformation rule for primary scalars under special conformal transformations. All terms come out as expected except of course for the spin part. To get that one I wanted to look at the infinitesimal version of the SCT and treat it like a rotation, i.e. \begin{equation} \tilde{x}_\mu = x_\mu +\omega_{\mu \nu}x^\nu \end{equation} where $ \omega_{\mu \nu}=2b_\nu x_\mu-b_\mu x_\nu$. Then disregarding the other parts of the transformation that we already know a field with spin should transform as \begin{equation} \tilde{\phi} = \left(1-\frac{i}{2} \omega_{\rho \nu} S^{\rho \nu}\right )\phi \end{equation} But if we work out the resulting contribution to $K$ we find \begin{equation} i\frac{\delta \tilde{\phi}}{\delta b^\mu}=-\frac{3}{2}x^\nu S_{\mu \nu} \end{equation} which is definitely wrong.

So my question is:

1. What is the correct factor for $S$?

2. Why is my derivation wrong?

Answer 1

1. It does appear, that there is a typo in the references. The correct expression for the full generator is $$K_\mu \ = 2 x_\mu \tilde{ \Delta} - 2 x^\nu S_{\mu \nu} - 2x_\mu x^\nu \partial_\nu + i x^2 \partial_\mu $$ A correct version is given in a slightly different form in the lecture notes by David Simmons-Duffin (arXiv:1602.07982 Eq.51).

2. The derivation was wrong because I naively assumed what $\omega$ was. The correct form follows from $$\omega_{\mu \nu} = \frac{\partial \tilde{x}_\mu}{\partial x^\nu} - \delta_{\mu \nu}$$