first jet of $f,$ which is absurd.

For $\d=2/n,$ one can easily see that the 1-cocycle (\[MultiSchwar1\]) is a coboundary.

Let us prove (ii). Consider a metric $\tilde \g= F\cdot\g,$ where $F$ is a non-zero positive function. Denote by $\tilde {\cal A}(f)$ the operator (\[MultiSchwar1\]) written with the metric $\tilde \g.$ We have to prove that $\tilde {\cal A}(f)={\cal A}(f).$ The Levi-Civita connections associated with the metrics $\g$ and $\tilde \g$ are related by $$\label{lien}
\tilde \Gamma^k_{ij}=\Gamma^k_{ij}+\frac{1}{2F}\left (F_i \d^k_j +F_j \d^k_i
-F_t \,\g^{tk} \g_{ij}\right),$$ where $F_i=\partial_i F.$\
We need some formul[æ]{}: denote by $\ell(f)$ the tensor (\[ell\]) written with the connection $\tilde \g,$ then we have $$\begin{aligned}
\label{bebe}
\tilde \nabla_k P^{ij}&=&\nabla_k P^{ij}+\frac{1}{2F}\left(  \mathrm{Sym}_{i,j} 
P^{mi} \left(F_m \delta^j_k-F_t \,\g^{tj}\g_{km}\right) +(2-n\delta )\,
P^{ij}F_k\right),\\
\tilde \ell(f)^k_{ij}&=&\ell(f)^k_{ij}+\frac{1}{2 F\circ f}
\left(\mathrm{Sym}_{i,j} \stackrel{*}{F_i}\delta^k_j -\stackrel{*}{F}_t\, 
\stackrel{*}{\g}^{tk}\stackrel{*}{\g}_{ij}\right)
-\frac{1}{2F}\left( \mathrm{Sym}_{i,j} \,F_i\, \delta^k_j -F_t\, 
\g^{tk}\g_{ij}\right), 
\nonumber\end{aligned}$$ where $\stackrel{*}{F}_i={f^{*}}^{-1}F_i$ and $\stackrel{*}{\g}_{ij}=
{f^{*}}^{-1}\g_{ij}$ for all $P^{ij}\xi_i \xi_j\in {\cal S}_{\d}^2 (M).$\
By substituting the formul[æ]{} (\[bebe\]) into (\[MultiSchwar1\]) we get $${\cal A} (f)_{ij}^k=\tilde {\cal A} (f)_{ij}^k+
\frac{1}{2 F} (\delta n +c -2 ) \,{\g}^{sk} \,\g_{ij}\, F_s\, 
+  
\frac{1} {2 F\circ f}(2-c -\delta  n) 
\stackrel{*}{\g}^{tk} \stackrel{*}{\g}_{ij}\, \stackrel{*}{F}_t \, \,
\cdot$$ We see that ${\cal A} (f)=\tilde {\cal A} (f)$ if and only if $c=2-\delta n 