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How do you derive conformal Ward identities for operators with spin? You can see in Penedones's notes (page 6) ( https://arxiv.org/abs/1608.04948 ) a brief derivation of Ward identities for general covariance and for Weyl invariance for scalar operators (both together are imposed in order to have conformal invariance) but how do I include the operators with spin in this discussion, I guess I have to modify the diffeomorphism invariant Ward Identity... but how?

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The left hand side is obviously the same. In the notes you linked Prof. Penedones keeps always explicit the disconnected part. I'll omit it here for brevity, remembering that it's always there. Let me also consider a flat background. $$ I = \int_{\partial B} dS_\mu\,\epsilon_\nu(x)\,\langle T^{\mu\nu}(x)\mathcal{O}^{I_1}_1(x_1)\ldots \mathcal{O}^{I_n}_n(x_n)\rangle\,, $$ where $I_1,\ldots,I_n$ are collective indices of the appropriate representations. In order to generalize it I prefer to use the notation in Simmons-Duffin's notes. The integral above is the correlator that contains a topological operator $Q_\epsilon(\partial B)$ $$ Q_\epsilon(\partial B) = \int_{\partial B} dS_\mu\,\epsilon_\nu(x)\, T^{\mu\nu}(x)\,. $$ Assuming that the interior of $B$ contains all the points $x_1,\ldots, x_n$ we can show that $$ I = - \sum_{x_i \in B}\langle \mathcal{O}^{I_1}_1(x_1)\ldots [\hat{Q}_\epsilon,\mathcal{O}^{I_i}_i(x_i)]\ldots\mathcal{O}^{I_n}_n(x_n)\rangle\,. $$ One may think $\hat{Q}_\epsilon$ as the quantum operator corresponding to the charge that generates the conformal transformation with Killing vector $\epsilon_\mu$. Or, more simply, one can see the commutator $[\hat{Q}_\epsilon,\mathcal{O}^{I_i}_i(x_i)]$ as a shorthand for the transformation law of $\mathcal{O}^{I_i}_i$.

Equation $(17)$ in Penedones' notes assumes that the $\mathcal{O}_i$'s transform as scalars. However, we can modify his derivation by assuming that the transformation law is generic and the logic essentially remains the same. The final result is then obtained by computing the exact expression of the commutator.

This can be found in equation $(54)$ of the other notes (I added explicit indices). $$ [Q_\epsilon,\mathcal{O}^{I_i}_i(x_i)] = \left(\delta^{I_i}_{J_i}\, \epsilon\cdot \partial + \delta^{I_i}_{J_i}\,\frac{\Delta_i}{d}(\partial\cdot\epsilon) - \frac{1}{2}(\partial^\mu\epsilon^\nu){\mathcal{S}_{\mu\nu}}^{I_i}_{\phantom{I_i}J_i}\right)\mathcal{O}^{J_i}_i(x_i)\,. $$ The tensor $\mathcal{S}_{\mu\nu}$ is the generator of the rotation group in the appropriate representation. It depends on the spin of $\mathcal{O}^{I_i}_i$ and it vanishes for scalars (as it should by comparison with the non-spinning case). Some examples (possibly up to factors) are $$ \begin{aligned} &\mbox{Spin $\frac{1}{2}$:}\qquad {\mathcal{S}_{\mu\nu}}^{\alpha}_{\phantom{\alpha}\beta} = \frac{1}{4}[\gamma_\mu,\gamma_\nu]^{\alpha}_{\phantom{\alpha}\beta}\,,\\ &\mbox{Spin $1$:}\qquad {\mathcal{S}_{\mu\nu}}^{\rho}_{\phantom{\rho}\sigma} = \delta^\rho_\mu\,\eta_{\nu\sigma} - \delta^\rho_\nu\,\eta_{\mu\sigma}\,. \end{aligned} $$ Note that Ward identities for spinning operators are far more complicated to work out explicitly because in general the correlators have more structures. I can point you to a reference (Elkhidir-Karateev) where it is worked out for spinors (appendix A) and to another reference (Cordova-Diab) where it is worked out for operators with spin $(0,j)$ and $(1,j)$, both in four dimensions.

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