In his lecture notes, C. McMullen discusses "the three great cocycles" in Complex Analysis: the derivative $$f\mapsto\log f',$$ the non-linearity $$f\mapsto (\log f')'dz$$ and the Schwarzian derivative $$f\mapsto \left(\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2\right)(dz)^2$$ I would like to understand what "cocycle" means here precisely, and in which sense these maps are "the only" cocycles.
For diffeomorphisms of the circle, this is very clearly explained, e. g., in the book by Ovsienko and Tabachnikov. Namely, $\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\lambda(S^1)$ of tensor densities of degree $\lambda$. The corresponding (group) 1-cohomology is trivial unless $\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.
Question: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This MO answer suggests that the group is $\mathrm{Aut}$(meromorphic functions), but, at least naively, this seems to coincide with a subgroup of the Mobius group.
Update: Let me restate the question in simpler terms.The maps above satisfy the cocycle equation $$ D(g\circ f)(z)=(f'(z))^\lambda D(g)(f(z))+D(f(z)), $$ for $\lambda=0,1,2$, respectively. For a tensor density $\psi$, one can define the coboundary $$ D_\psi:f\mapsto (f')^\lambda(\psi\circ f)-\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?
