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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.

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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.

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    $\begingroup$ Let me add that an excellent reference is Serre Local fields, Part III. $\endgroup$ – abx Oct 12 '18 at 13:11
  • $\begingroup$ Ah, brilliant! So the question is in fact quite straightforward with the correct definition. Thank you very much for your help! $\endgroup$ – Sam Streeter Oct 12 '18 at 13:13
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    $\begingroup$ I would also recommend reading Serre, Galois cohomology, Ch. I, Section 5. $\endgroup$ – Mikhail Borovoi Oct 13 '18 at 11:20

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