the components of the tensor (\[ell\]). This equation implies that $$\begin{array}{lcl}
\n^{\g}_i\phi&=&\n^{\tilde \g}_i\phi-\displaystyle \frac{\l n}{2} 
\displaystyle \frac{F_i}{F}+
\l \,\ell(\psi^{-1})_i \, \phi,\\[2mm]
\n^{\g}_i P^i &=&\n^{\tilde \g}_i P^i+\displaystyle 
\frac{n(1-\d)}{2}\frac{F_i P^i}{F}-(1-\d) \ell(\psi^{-1})_{i}P^i,
\end{array}$$ for all $\phi\in {{\mathcal{F}}}_\l$ and for all $P^i\xi_i\in {{\mathcal{S}}}_{\d}^1(M).$\
Substitute these formul[æ]{} into (\[yah1\]) we see that ${\cal Q}^{\tilde \g}_{\l,\mu}={\cal Q}^{\g}_{\l,\mu}.$

The quantization map ${\cal Q}_{\l,\mu}^{\g}:$ ${\cal D}^2_{\l,\mu}(M)\rightarrow {\cal S}^2_\d(M)$ defined in (\[Tensor1\]) has the property $${\cal Q}_{\l,\mu}^{\tilde \g} (P)={\cal Q}_{\l,\mu}^{\g} (P)+d_1 \, 
{\cal A}^{s}(\psi^{-1})
(P )\nabla_s + d_2 \,\nabla_s \left ( {\cal A}^{s}(\psi^{-1})
(P) \right ) +d_3 \, {\cal B}(\psi^{-1})(P),$$ for all $P\in {\cal S}_{\d}^2(M),$ where the constants $d_1,d_2$ and $d_3$ are given by $$\begin{aligned}
d_1&=&\frac{n(\l+\mu-1)}{(2+n(1-\d))(2-n\d))},  \quad 
\displaystyle d_2=\frac{n\l(\l+\mu-1)}{(2+n(1-\d))(2-n\d)(1-\d)}, \nonumber 
\\[1mm]
d_3&=&\displaystyle \frac{n \l(\mu -1)}{(2+n(1-2\d))(\d -1)(1+n(1-\d))} \cdot 
\nonumber \end{aligned}$$

[**Proof.**]{} The proof involves the calculation of $\nabla_i ^{\tilde \g}\nabla_j ^{\tilde \g} \phi,$ $\nabla_i ^{\tilde \g}\nabla_j ^{\tilde \g} P^{kl}$ and $R^{\tilde \g}$ wich is straightforward but quite complicated.

[ The system $d_1=d_2=d_3=0$ admits a unique solution: $(\l,\mu)=(0,1).$ The value of $\d=1$ is called “resonant”. In this case, the quantization map is not unique; there exists a one-parameter family of such isomorphism (see [@do] for more dtails.) ]{}

For all $f\in {\mathrm{Diff}}(M)$ and for all conformal map $\psi: (M,\tilde \g)
\rightarrow 