most recent 30 from http://mathoverflow.net 2013-05-22T01:04:10Z http://mathoverflow.net/feeds/question/95126 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95126/profiniteness-condition-for-hochschild-serre-spectral-sequence Zuriel 2012-04-25T07:46:32Z 2012-04-27T08:56:01Z

<p>This question may seem elementary to experts but I am quite confused about it:</p> <p>According to the entry of Lyndon–Hochschild–Serre spectral sequence on wikipedia, for a group extension $1\to N\to G\to Q\to1$, there is a Lyndon–Hochschild–Serre spectral sequence if $G$ is a profinite group and $N$ is a closed normal subgroup of $G$:</p> <p><a href="http://en.wikipedia.org/wiki/Lyndon-Hochschild-Serre_spectral_sequence" rel="nofollow">http://en.wikipedia.org/wiki/Lyndon-Hochschild-Serre_spectral_sequence</a></p> <p>Also in the book Neukirch, Jürgen; Schmidt, Alexander; Wingberg, Kay (2000), <em>Cohomology of Number Fields</em>, the same condition was assumed.</p> <p>However, according to some other books, for example Brown, Kenneth S. (1972), <em>Cohomology of Groups</em>, and <em>A User's Guide to Spectral Sequences</em> by John McCleary, the profiniteness condition on the group $G$ was NOT assumed. </p> <p>Why there is such a difference? Do we really need the condition that $G$ is a profinite group and $N$ is a closed normal subgroup of $G$ to construct the Lyndon–Hochschild–Serre spectral sequence?</p>