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Let $X$ be an algebraic variety over an algebraically closed field $k$ and let a finite group $G$ act on it so that it acts freely on the generic fibre of the projection $X \to X/G$, so $[k(X):k(X/G)]$ is Galois. Consider group extension $1 \to Gal(k(X)) \to Gal(k(X/G)) \to G \to 1$. I want to understand what the differential on the second page of the corresponding Lyndon-Hochschild-Serre spectral sequence (the transgression map) does: $$ tr: H^1(k(X), M)^G \to H^2(G, M) $$ (say, $M=\mu_n$ to speak about something concrete). Here, a class in $H^1(k(X), M)^G$ is a class corresponding to some $\mu_n$-torsor, and the transgression map puts into correspondence to this class a class of some (central) group extension of $G$ by $\mu_n$. Can this group extension be described in some direct way, without going through the computation of the transgression (which is described, for example, in a reply to this question)?

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May be it will be usefull to add that spectral sequence is coming from $0\to Gal(k(x))\to Gal(k(X/G))\to G\to 0$. More generally, are there derivation of Hochshild-Serre s.s. without abstract factorization of functor of middle term invariants, but directly throgh some good resolvents? May be this is resonable for lie algebra cohomology. –  Bad English Mar 12 '13 at 7:18
    
I have found a (tentative) answer to the question: let k(T) be the field extension of k(X) corresponding to the torsor described by a class $h \in H^1(k(X), \mu_n)$. Then the group $\tilde G$ that is given by the cocycle $tr(h) \in H^2(G,\mu_n)$ is the Galois group $Gal(k(T)/k(X/G))$. I will write a proper answer as soon I get all the details right. –  Dima Sustretov Mar 13 '13 at 23:06

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