Casimir conformal generator of $SO(d+1,1)$ The purpose of this post is to ask the help of derivation of equation 2.8 of https://arxiv.org/abs/2106.10822
Let $P_i$ be a point on the conformal boundary of $AdS_{d+1}$ and $Z_i$ be a polarization vector.
The conformal generator acting on the $i$-th particle of $SO(d+1,1)$ is given by
\begin{align}
  D_i^{AB} = P_i^A \frac{\partial}{\partial P_{i, B}} - P_i^{B} \frac{\partial}{\partial P_{i, A}} + Z_i^A \frac{\partial}{\partial Z_{i, B}} - Z_i^B \frac{\partial}{\partial Z_{i, A}}.
\end{align}
I want to obtain
\begin{align}
  -\frac{1}{2} D_i^2 &= \left(P_i \cdot \frac{\partial}{\partial P_i} \right) \left( d + P_i \cdot \frac{\partial}{\partial P_i} \right) + \left(Z_i \cdot \frac{\partial}{\partial Z_i} \right) \left( d-2 + Z_i \cdot \frac{\partial}{\partial Z_i} \right)  \\
  & + 2 \left( Z_i \cdot \frac{\partial}{\partial P_i} \right) \left( P_i \cdot \frac{\partial}{\partial Z_i} \right). 
\end{align}
Naively I can decompose $D_i^2$ into three parts
\begin{align}
\left(P_i^A \frac{\partial}{\partial P_{i, B}} - P_i^{B} \frac{\partial}{\partial P_{i, A}}\right)^2 
 +\left( Z_i^A \frac{\partial}{\partial Z_{i, B}} - Z_i^B \frac{\partial}{\partial Z_{i, A}} \right)^2 
+ 2\left(P_i^A \frac{\partial}{\partial P_{i, B}} - P_i^{B} \frac{\partial}{\partial P_{i, A}} \right)\left( Z_i^A \frac{\partial}{\partial Z_{i, B}} - Z_i^B \frac{\partial}{\partial Z_{i, A}} \right).
\end{align}
Treating independent objects of $Z, P$ from the last term I obtain $2 \left( Z_i \cdot \frac{\partial}{\partial P_i} \right) \left( P_i \cdot \frac{\partial}{\partial Z_i} \right)$, But I have trouble obtaining $d$ and $d-2$
First, expand the first $P$ term I have
\begin{align}
2P_{i,A} \frac{\partial}{\partial P_i^A} +2 P_{i,A}^2 \frac{\partial}{\partial P_{B,i}}\frac{\partial}{\partial P_{B,i}} - 2 \delta_{A}^{A} P_{i,A} \frac{\partial}{\partial P_{i,A}} - 2 P_{i,A} P_{i,B} \frac{\partial}{\partial P_{i,A}} \frac{\partial}{\partial P_{i,B}}.
\end{align}
In the process I identify $\frac{\partial  P_{i,A}}{\partial \partial P_{i,B}} = \delta_{AB}$
and compare the expression
\begin{align}
\left(P_i \cdot \frac{\partial}{\partial P_i} \right) \left( d + P_i \cdot \frac{\partial}{\partial P_i} \right) 
= (d+1) P_i \frac{\partial}{\partial P_i} + P_i^A P_i^B \frac{\partial }{\partial P_{i,A}} \frac{\partial}{\partial P_{i,B}} 
\end{align}
Imposing $P_i^2=0$ and $\delta_{A}^{A} = d+2$ it seems work. But in this case, there seems something wrong with $Z_{i,A}$ parts... To obtain the desired answer I need to impose again $Z^2=0$ and...... Am I missing somewhere?
I know from the appendix of 2106.10822, both $P$ and $Z$ are null i.e., $P^2=0, Z^2=0$ so the crucial part of my question is how $d-2$ appear in $Z$ computations.
 A: This comes from crossing term
Now consider the cross term
\begin{align}
  &2\left(P_i^A \frac{\partial}{\partial P_{i, B}} - P_i^{B} \frac{\partial}{\partial P_{i, A}} \right)\left( Z_i^A \frac{\partial}{\partial Z_{i, B}} - Z_i^B \frac{\partial}{\partial Z_{i, A}} \right)   \\
  & =4 \left( P_i^A  \frac{\partial Z_i^A}{\partial P_{i, B}} \frac{\partial}{\partial Z_{i,B}} + P_i^A Z_i^A \frac{\partial}{\partial P_{i,B}} \frac{\partial}{\partial Z_{i,B}} 
 - P_i^A \frac{\partial Z_i^B}{\partial P_{i, B}} \frac{\partial}{\partial Z_{i,A}} - P_i^A Z_i^B \frac{\partial}{\partial P_{i, B}} \frac{\partial}{\partial Z_{i,A}}  \right)  \\
 & = 4 \left( -2 Z \cdot \frac{\partial}{\partial Z} - P_i^A Z_i^B \frac{\partial}{\partial P_{i, B}} \frac{\partial}{\partial Z_{i,A}} \right) 
 = 4 \left( -\left( Z_i \cdot \frac{\partial}{\partial P_i} \right) \left( P_i \cdot \frac{\partial}{\partial Z_i} \right) - Z \cdot \frac{\partial}{\partial Z} \right)
\end{align}
Imposing $P_i \cdot Z_i =0$ and $Z_{iB} = - P_i^A \frac{\partial Z_{i, A}}{\partial P_{i, B}}$, $P_i^{B} = - \frac{\partial P_i^A}{\partial Z_{i,B}} Z_{i, A}$ one have desired result.
