(M,\g)$, one has

[(i)]{} ${\cal A}^{\tilde \g}(f)=\psi^* {\cal A}^{\g}
(\psi \circ f\circ \psi^{-1}),$

[(ii)]{} ${\cal B}^{\tilde \g}(f)=\psi^* {\cal B}^{\g} 
(\psi \circ f\circ \psi^{-1}).$

[**Proof.**]{} Straightforward computation.

For all conformal map $\psi (M,\tilde \g)
\rightarrow (M,\g)$ one has

[(i)]{} $\tilde {\cal A}^{\tilde \g}(\psi)=-{\cal A}^{\g}(\psi^{-1} ),$

[(ii)]{} $\tilde {\cal B}^{\tilde \g}(\psi)=-{\cal B}^{\g}(\psi^{-1} )\cdot$

[ The Corollary above shows that for a conformal map $\psi,$ the 1-cocycle $\cal B$ is still a second-order differential operator and then does not coincide with the Schwarzian derivative (\[osg\]) defined by Osgood and Stow. ]{}

Appendix
========

We will give a formula for the Schwarzian derivative for the case of surfaces. As explained in section (\[tokyo\]), all surfaces are conformally flat. That means that every metric can be express (locally) as $$\g=F^{-1}\psi^* \g_0,$$ where $\psi$ is a conformal diffeomorphism of $M$, and $F$ is a non-zero positive function, $\g_0$ is a metric of constant curvature.

The explicit formula of the Schwarzian derivative in the case of surfaces is: the following $$\begin{aligned}
{\cal B}'_2(f)_{ij}&=&
{f^*}^{-1} \left( {\g}^{st}\, {\g}_{ij}\nabla_s\nabla_t \right ) 
-\g^{st}\,\g_{ij}\nabla_s \nabla_t +4(1-\delta)\left( \ell(f)_{ij}^s 
-\frac{1}{2}\,Sym_{i,j}\, 
\delta_{i}^s \,\ell(f)_{j} \right)\nabla_s \nonumber \\[2mm]
&&+4 (1-\d)^2 \ell(f)_s \left ( \ell(f)_{ij}^s -\frac{1}{4}
Sym_{i,j}\, \delta_i^s \, \ell(f)_j\right) + 
2(\d-2)(1-\d) \,Sym_{i,j}\,\n_{j}\ell_{i}(f) \nonumber\\[2mm]
&&+8\,(\d-1)^2 \left( 
 {f^{-1}}^*(S(\psi)_{ij})-S(\psi)_{ij}+\frac{1}{2}
