most recent 30 from http://mathoverflow.net 2013-05-22T06:26:40Z http://mathoverflow.net/feeds/question/34924 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34924/tale-cohomology-of-linear-groups algori 2010-08-08T14:03:44Z 2010-08-08T14:18:06Z

<p>This is in a sense a follow up question to the answer here <a href="http://mathoverflow.net/questions/34717/analytic-tools-in-algebraic-geometry/34743#34743" rel="nofollow">http://mathoverflow.net/questions/34717/analytic-tools-in-algebraic-geometry/34743#34743</a> </p> <p>Let $k$ be an algebraically closed field of positive characteristic and let $R$ be the result of applying the Witt vector procedure to $k$ (see e.g. <a href="http://eom.springer.de/W/w098100.htm" rel="nofollow">http://eom.springer.de/W/w098100.htm</a>) This is a complete local ring with residue field $k$ and fraction field of characteristic 0. Let $L$ be an algebraic closure of the fraction field.</p> <p>Using the procedure explained in SGA 4 1/2, pp. 54-56 one can construct, given a smooth scheme over $R$, a comparison morphism $H^*(X_1,A)\to H^*(X_0,A)$ where $X_0$ and $X_1$ are the fibers of $X$ over $Spec(k)$, respectively, $Spec(L)$, and $A$ is a finite abelian group of order prime to $char(k)$. The construction is as follows: we have the maps $X_1\to X\gets X_0$; due to the smoothness of $X$, the map $H^{\ast}(X,A)\to H^{\ast}(X_1,A)$ is an isomorphism, which we invert and then take the restriction to $X_0$.</p> <p>These comparison maps are in general not isomorphisms, but they are if we take $X=GL_n(R)$ or $SL_n(R)$. I think I know how to prove this, but I also think this should be standard material. So I'd like to ask if somebody knows a reference for that.</p>