get $r_1=\rho$. The next step is to write down an ansatz for $r_1$, linear in $\hat{M}$ and $a_0$, with undetermined coefficients : $-ia_0(c_1\hat{M}_{i0}\otimes p_i
+c_2\hat{M}_{i0}p_i\otimes 1+ d_1\cdot 1\otimes \hat{M}_{i0}p_i
+d_2p_i\otimes \hat{M}_{i0})$. In order to compare this expression with $\rho$, in all the terms from $\rho$ and the ansatz, we move $x_\mu$ to the right side of the tensor product using the relations $\widetilde{\mathcal{R}}$ (relations and for $\lambda=1/2$). We also use the fact that $\hat{M}_{i0}=x_ip_0-x_0p_i+
O(a_0^2)$ ($\hat{M}_{i0}= x_ip_0\frac{\sinh A}{A} -x_0p_i$, see Section \[IV\]). After equating the expressions, we get 7 equations for the 4 coefficients ($c_1,\, c_2,\, d_1,\, d_2$). The system of equations has a unique solution given by $c_1=-d_2=-1,\, c_2=d_1=0$, so that the final result up to the first order in the deformation parameter $a_0$ is $$\label{lukierskicritique}
r_1= -ia_0(p_k\otimes \hat{M}_{k0}-\hat{M}_{k0}\otimes p_k),$$ namely, $r_1 = -ia_0(p_k\otimes \hat{M}_{k0}-\hat{M}_{k0}\otimes p_k)  = \rho + O(a_0^2). $ Using the relations to move $x_\mu$ generates terms of higher order in the deformation parameter, e.g., $\rho=-ia_0( p_kZ^{\frac{1}{2}}\otimes x_kp_0Z^{\frac{1}{2}} -p_0\otimes x_kp_k)= -ia_0\cdot 1\otimes x_k( p_k\otimes p_0 -p_0\otimes p_k) +O(a_0^2)$.

In the second order, $r_2$ is found to be equal to $\frac{1}{2} \rho^{2} -\frac{1}{2}r_1^{2}$ + terms of order $a_0^{2}$ coming from the use of relations $\widetilde{\mathcal{R}}$ ($\lambda=1/2$) in $\rho-r_1$. 