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I am interested in $H¹$ right now and the cocycle condition $φ_{jk} • φ_{ij} = φ_{ik}$ because of how it is said to relate to automorphic forms. I can't quite see the relationship between factors of automorphy and $H¹(G,M)$ that is mentioned here.

Let $G$ be a group acting on a complex manifold $X$ and let $f$ be a holomorphic function from $X$ to $ℂ$. Recall that the factor of automorphy has $$f(gx) = j_g(x) f(x)$$, which already seems to bear some similarity to a cocycle condition.

The cohomology of a group $G$ with coefficients in $M$ can be calculated from the chain complex whose nth component is the $G$-module of functions from $Gⁿ$ to $M$. See here for the formulas for the differential in group cohomology. I am interested in

$$H¹(G, M) = Z¹(G,M)/B¹(G,M)$$

The wikipedia article on automorphic forms mentions that "The formulation requires the general notion of factor of automorphy j for Γ, which is a type of 1-cocycle in the language of group cohomology."

Can someone spell this out? Is it possible to interpret the automorphy condition as a cocycle condition?

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    $\begingroup$ Yes— here $G$ is acting on the multiplicative group of nonvanishing analytic functions $V$ on $X$. The map $g \mapsto j_g$ defines a 1-cochain which is a cocycle. It’s a good exercise to chase this out from the formulas $\endgroup$
    – Vik78
    Commented Aug 28, 2023 at 5:30
  • $\begingroup$ Automorphic forms have two other conditions making them particular cocylces, perhaps also fulfilling the coboundary condition? Are automorphic forms the same as B¹(G,M)? @Vik78 $\endgroup$
    – user30211
    Commented Aug 28, 2023 at 5:47

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We have $j_{gh}(x)f(x)=f((gh)x)=f(g(hx))=j_g(hx)f(hx)=j_g(hx)j_h(x)f(x)$, and so $j_{gh}(x)=j_g(hx)j_h(x)$, which is the one-cocycle condition.

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