side depends on the first jet of $f,$ which is absurd.

For $\d= \frac{n+2}{2n},$ the 1-cocycle (\[MultiSchwar2\]) is trivial: ${\cal B}(f)_{ij}={f^*}^{-1} (\g_{ij}B) -B\,\g_{ij},$ where $B:= \g^{st}\nabla_s\nabla_t -\frac{1}{4}\frac{n-2}{n-1}R$ is the so-called Yamabe-Laplace operator (see, e.g., [@besse]).\
Let us prove (ii). Consider a metric $\tilde \g:=F\cdot \g,$ where $F$ is a non-zero positive function. Denote by $\tilde {\cal B}(f)$ the operator (\[MultiSchwar2\]) written with the metric $\tilde \g.$ We have to prove that $\tilde {\cal B}(f)={\cal B}(f).$\
The proof is similar to the proof of part (ii) of Theorem (\[main\]), by means of the equation (\[lien\]), (\[bebe\]) and $$\begin{aligned}
\tilde \nabla_l \tilde \nabla_k P^{ij}&=&\nabla_l \tilde \nabla_k P^{ij}+
\frac{1}{2F} \left( (1-\d n) F_l\,\tilde\nabla_k P^{ij}- F_k\, 
\tilde \nabla_l P^{ij}+\g^{st}\, \g_{lk}\, F_s \tilde 
\nabla_t P^{ij}\right ) \nonumber \\
&& +\frac{1}{2F} \mathrm{Sym}_{i,j} \tilde \nabla_k P^{mi} \left (F_m \, \delta^j_l -
{\g}^{sj}{\g}_{ml}\, F_s\right),\nonumber\\
\tilde \nabla_l \, \tilde \ell(f)^k_{ij}&=&\nabla_l \,\tilde \ell(f)^k_{ij} - 
\frac{1}{2F}\, F_l\,\,\tilde \ell(f)^k_{ij}+\frac{1}{2F} \left( F_t \, 
\delta^k_l -\g^{sk}\, \g_{tl}\, F_s \right ) \tilde \ell(f)^t_{ij}
\nonumber \\
&&-\frac{1}{2F} \mathrm{Sym}_{i,j} \left (F_i \delta^s_l-F_m \,{\g}^{ms}
{\g}_{il}\right)\tilde \ell(f)^k_{js}\nonumber\\
\tilde R&=& \frac{1}{F} \left ( R-(n-1)\frac{1}{F} 
\left( \g^{ij}\nabla_i F_j+ (n-6)\frac{1}{4 \, F}\g^{ij}F_i 
F_j \right) \right)  \end{aligned}$$ for all $P^{ij}\xi_i\xi_j\in {\cal S}_\d^2(M),$ where $\tilde \nabla,$ $\tilde \ell(f)$ and $\tilde R$ are, the covariant 