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Let G be the Galois group of Q-bar over Q. I have a good idea of what $H^1(G,GL_1)$ is when $G$ acts trivially on $GL_1$: it is the group of homomorphisms from $G$ to $GL_1$. In particular it factors through the abelian quotient of $G$. In fact it is the group Pontrjagin dual to that abelian quotient.

What if $G$ acts nontrivially on $GL_1$? The automorphisms of $GL_1$ are $x$ and $x^{-1}$. To give a nontrivial action of $G$ on $GL_1$, is the same as giving a map from $G$ to that group with two elements. That's the same as giving a quadratic extension of Q, say $Q(\sqrt{z})$. I'll write $GL_{1,\sqrt{z}}$ for $GL_1$ with that Galois module structure.

How can I compute $H^1(G,GL_{1,\sqrt{z}})$? Is it the Pontrjagin dual of the abelian quotient of the Galois group of Q-bar over $Q(\sqrt{z})$?

Class F. Student
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  • I don't know what you mean by $GL_1$ with the trivial $G$-action. Usually $GL_1$ means an algebraic group, namely $\mathbb{G}_m$, and as a $G$-module this should mean $\overline{\mathbb{Q}}^{\times}$ with the obvious $G$-action. In particular the $G$-action is fixed and nontrivial (and $H^1$ with these coefficients vanishes by Hilbert 90). Do you mean a particular group, rather than an algebraic group? If so, is it $\mathbb{Q}^{\times}$ or what? – Qiaochu Yuan Dec 31 '15 at 17:45
  • I mean a particular group, such as GL_1(Q-ell). Actually I would like to know the answer for any value of "Q-ell." But not for the algebraic group that you are describing. – Class F. Student Dec 31 '15 at 18:08
  • if you're referring to a particular group, there's no reason to believe that the automorphism group of that group is $\mathbb{Z}_2$ (which is the automorphism group of $GL_1$ as an algebraic group). For example $\mathbb{Q}^{\times}$ has a much larger (in fact uncountable) automorphism group. – Qiaochu Yuan Dec 31 '15 at 18:20
  • OK, but I am for now only interested in the action I described, that I am calling $GL_{1,\sqrt{z}}$. – Class F. Student Dec 31 '15 at 18:25

1 Answers1

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Let $\rho : G \to \mathbb{Z}_2$ be a nontrivial homomorphism, and let $N$ denote its kernel (an open normal subgroup of index $2$). Let me write $A$ for the coefficient module; I'll assume that in all of the examples you care about this is abelian. In this setting we can apply the inflation-restriction exact sequence, which here reads

$$0 \to H^1(\mathbb{Z}_2, A) \to H^1(G, A) \to H^1(N, A)^{\mathbb{Z}_2} \to H^2(\mathbb{Z}_2, A) \to H^2(G, A).$$

(We have $A = A^N$ because, by hypothesis, the action of $G$ on $A$ factors through $G/N$.) $H^1(N, A)$ is just homomorphisms $N \to A$, or equivalently $N/[N, N] \to A$, since the action of $N$ on $A$ is trivial, but as you can see there are various other things going on.

Qiaochu Yuan
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  • "Z_2" notation considered harmful! – Class F. Student Dec 31 '15 at 18:38
  • I think the first term $H^1(GL_{1,\sqrt{z}})$ vanishes, and the $H^2(GL_{1,\sqrt{z}})$ is of order 2. I feel I understand "H^1(N,A)" a little bit but I don't know how Z/2 acts. And I don't know what the connecting homomorphism is. – Class F. Student Dec 31 '15 at 18:50