-\g^{sk}\,\g_{ij}\nabla_s + c\,\left(\ell(f)^k_{ij} -\frac{1}{n} 
\, \mathrm{Sym}_{i,j}\,\delta_{i}^k \,\ell(f)_{j}\right), 
\label{MultiSchwar1}$$ where $$\label{qua} 
c=2-\delta n,$$ and $\ell(f)_{ij}^k$ are the components of the tensor (\[ell\])

\[main\] [(i)]{} For all $\d \not= 2/n,$ the map $f\mapsto {\cal A}(f^{-1})$ defines a non-trivial 1-cocycle on ${\mathrm{Diff}}(M)$ with values in ${{\mathcal{D}}}({\cal S}_{\d}^2(M), {\cal S}_\d^1(M)).$\
[(ii)]{} The operator (\[MultiSchwar1\]) does not depend on the rescaling of the metric $\g.$ In particular, if $M:={\mathbb{R}}^n$ and $\g$ is the flat metric of signature $p-q$, this operator vanishes on the conformal group ${\mathrm{O}}(p+1,q+1).$

[**Proof.**]{} To prove (i) we have to verify the 1-cocycle condition $${\cal A}(f\circ h)={h^*}^{-1} {\cal A}(f)+ {\cal A}(h), 
\quad \mbox{for all } f,h\in {\mathrm{Diff}}(M),$$ where $h^*$ is the natural action on ${{\mathcal{D}}}({\cal S}^2_{\d}(M),{\cal S}_\d^1(M)).$ This condition holds because the first part of the operator (\[MultiSchwar1\]) is a coboundary and the second part is a 1-cocycle.

Let us proof that this 1-cocycle is not trivial for $\d\not=2/n$. Suppose that there is a first-order differential operator $A^k_{ij}=u^{sk}_{ij}\nabla_s+v^k_{ij}$ such that $$\label{cn}
{\cal A}(f)={f^*}^{-1} A-A.$$ From (\[cn\]), it is easy to see that ${f^*}^{-1} v^k_{ij}-v^k_{ij}=
(2-\delta n)\left(\ell(f)^k_{ij} -\frac{1}{n} 
\, \mathrm{Sym}_{i,j}\,\delta_{i}^k \,\ell(f)_{j}\right)$. The right-hand side of this equation depends on the second jet of the diffeomorphism $f,$ while the left-hand side depends on the 