I've been stuck for a while on Exercise 1.9 of Bjorn Poonen's "Rational Points on Varieties". We start with $L/K$ a finite Galois extension with Galois group $G$, some $r \in \mathbb{Z}\_{\geq 0}$, and for each 1-cochain $\xi: G \rightarrow GL\_r\left(L\right)$, we define $W\_{\xi}$ to be $L^r$ with equipped with the function $G \times L^r \rightarrow L^r$, $\sigma \mapsto \xi\_{\sigma}\left(\sigma w\right)$.


In the first part of the question, we are asked to show that this function describes a semilinear $G$-action on $W\_{\xi}$ iff $\xi$ is a 1-cocycle, i.e., $\xi\_{\sigma \tau} = \xi\_\sigma {}^{\sigma}\xi\_{\tau}$. I have managed this.


The second part is to show that, for two $1$-cocycles $\xi$ and $\xi'$, the spaces $W\_\xi$ and $W\_{\xi'}$ are isomorphic as $L$-vector spaces with semilinear $G$-action iff $\xi$ and $\xi'$ are cohomologous. This is where I am stuck.


I am new to group cohomology, but I have studied other types of cohomology before and have read a few things on group cohomology in order to try and solve this problem. As I understand it, $\xi$ and $\xi'$ are cohomologous iff their difference is a $1$-coboundary, i.e., a principal crossed homomorphism. Translating the standard definitions into multiplicative language, I believe this amounts to showing that there is some $f \in GL\_r\left(L\right)$ such that, for all $w \in W = L^r$ and all $\sigma \in G$, we have $$\xi'\_{\sigma} \xi\_{\sigma}^{-1}\left(w\right) = \left({}^{\sigma}f\right) f^{-1}\left(w\right),$$ i.e.,



> 
> \begin{equation}
> \xi'\_{\sigma} \xi\_{\sigma}^{-1} = \left({}^{\sigma}f\right) f^{-1}.
> \end{equation}
> 
> 
> 


Now, from what I can gather, $W\_{\xi}$ and $W\_{\xi'}$ are isomorphic as $L$-vector spaces with semilinear $G$-action iff there is $g \in GL\_r\left(L\right)$ such that for all $w \in W = L^r$ and all $\sigma \in G$, we have $$g\left(\xi\_\sigma \left(\sigma w\right)\right) = \xi'\_{\sigma}\left(\sigma g\left(w\right)\right),$$ i.e., $$g\left(\xi\_\sigma \left(\sigma w\right)\right) = \xi'\_{\sigma}\left( {}^{\sigma}g\left(\sigma w\right)\right),$$ meaning precisely that



> 
> \begin{equation}
> g \xi{\_\sigma} = \xi'\_{\sigma}{}^{\sigma}g.
> \end{equation}
> 
> 
> 


My idea for tackling this is to show that we can manipulate the second expression until it is in the form of the first one.


From $\xi$ being a 1-cocycle, we can deduce that $\xi\_{id\_G} = id\_{GL\_r\left(L\right)}$ and then that $\xi\_{\sigma}^{-1} = {}^{\sigma}\xi\_{\sigma^{-1}}$, so, assuming the latter highlighted equation (involving $g$), we get $$\xi'\_{\sigma} \xi\_{\sigma}^{-1} = \xi'\_{\sigma} {}^{\sigma}\xi\_{\sigma^{-1}} = g\xi\_\sigma {}^{\sigma}\left(g^{-1}\right) {}^{\sigma}\xi\_{\sigma^{-1}} = g\xi\_\sigma {}^{\sigma}\left(g^{-1}\xi\_{\sigma^{-1}}\right),$$ but I can't work that into the form I want by a suitable choice of $f$. I have tried other manipulations, but with no success.


My conclusion is that I have probably misunderstood or mistranslated some concept. In particular, I fear I might have the wrong notion of isomorphism for $L$-vector spaces with semilinear $G$-action. The definition I have here is one I found for isomorphism of $G$-sets on Wikipedia. Initially, I had wondered whether an isomorphism should be a pair $\left(\phi,h\right)$ where $\phi: G \rightarrow G$ is an automorphism, $h \in Gl\_r\left(L\right)$ and, for all $w \in W$ and $\sigma \in G$, $$h\left(\xi\_\sigma \left(\sigma w\right)\right) = \xi'\_{\phi\left(\sigma\right)}\left(\phi\left(\sigma\right)h\left(w\right)\right),$$


so in particular I get the first definition back if $\phi$ is the identity map on $G$, but I'm not sure.


If anybody can point out the erroneous definitions/where I am going wrong, then I would be very grateful. 



Correspondence Between First Galois Cohomology and Semilinear Actions Up to Isomorphism
Your mistake is in the meaning of $\xi$ and $\xi'$ being cohomologous. The condition should be 
$\xi\_\sigma =f^\sigma \xi'\_\sigma f^{-1}$, for some $f\in GL\_r(L)$. This is the correct definition of cohomologous cocycles in the non-abelian cituation, and then $f$ just give you the isomorphism. 


In general one should be careful about order and directions when dealing with non-abelian cohomologies. The correct formulas are usually not the first one you would guess. 


