When is the base change morphism an isomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:00:02Z http://mathoverflow.net/feeds/question/37345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37345/when-is-the-base-change-morphism-an-isomorphism When is the base change morphism an isomorphism? algori 2010-09-01T02:40:00Z 2010-09-20T22:06:26Z <p>This is a rewrite of a previous question, which was in turn a follow up question to <a href="http://mathoverflow.net/questions/29178/leray-hirsch-principle-for-etale-cohomology" rel="nofollow">http://mathoverflow.net/questions/29178/leray-hirsch-principle-for-etale-cohomology</a> The motivation is to clarify some points in Torsten Ekedahl's answer there. (Some time ago I also posted a brief version as a comment there.)</p> <p>Let $X\to Y$ be a surjective morphism of connected algebraic varieties over an algebraically closed field $k$ and let $F$ be a constructible sheaf on $X$ whose stalks are of order prime to $char (k)$. For any closed geometric point $y\in Y$ we have the natural base change map $(R^i f_*F)_y\to H^i(X_y,F)$ where $X_y$ is the fiber over $y$.</p> <blockquote> <p>Can one deduce that all $R^i f_{\ast} F,i>0$ are zero assuming that for each closed $y$ we have $H^i(X_y,F)=0,i>0$? If the answer is yes, I would be interested in knowing whether there is an analog of this for the $l$-adic cohomology with $\mathbf{Q}_l$ coefficients.</p> </blockquote> <p>[upd: nope, as per Dustin's comment below.]</p> <p>Note that in the \'etale case Proposition 4.12, Chapter VI in Milne, \'Etale cohomology gives a stronger conclusion ($H^{\ast}(X,F')\to H^{\ast}(Y,f^{-1}F')$ is an iso for any $F'$) under a stronger hypothesis that the $H^{\ast}(X_y,F'')$ vanishes in positive degrees for any geometric point $y$, which may or may not be closed, and for all $F''$.</p> <blockquote> <p>So here is a side-question: does it suffice to check the vanishing for closed points?</p> </blockquote> <p>At first I thought the answer would be yes, which is why I accepted Torsten's answer to the above question, but then I realized I don't know how to prove this.</p> <p>[upd: .. and for a good reason since it's false.]</p> <p>Remark: if $f$ is proper, the statement is true by the proper base change theorem, but as pointed out by Dustin Clausen in the above thread, assuming $f$ smooth does not help. In fact, if $f:\mathbf{A}^2\setminus 0 \to \mathbf{A}^1$ is the projection to one of the coordinate axes, then (unless I'm mistaken) $R^3 f_*\underline{A}$ is non-zero at the origin (where $A$ is a finite group of order prime to $char(k)$), but the cohomology of the preimage of the origin vanishes in degrees $>1$.</p> <p>Here is some more motivation and one more question. If $G$ is a Lie group that acts on a sufficiently nice topological space $X$, then the stack cohomology of the quotient is simply the cohomology of the Borel construction $X\times EG/G$ where $EG$ is the universal $G$-bundle and the action is diagonal. The Borel construction is mapped to $X/G$ (assuming the topological quotient is reasonable) and the fiber over $[x]\in X/G,x\in X$ is the classifying space of the stabilizer of $x$. So if the action is with finite stabilizers, $H^{\ast}(X/G,\mathbf{Q})$ is isomorphic to the cohomology of the Borel quotient. One can try to mimick this in the \'etale setting, but there are some technical difficulties: in the topological case there are some tools (maximal compact subgroup, the existence of slices for compact group actions, ...) which don't seem to have easy analogs.</p> <p>However, I would still guess that if $G$ is an algebraic group that acts with finite stabilizers on an algebraic variety $X$ (I'd be willing to assume it smooth but not affine), the cohomology of the quotient with $\mathbf{Q}_l$ coefficients should be the same as the cohomology of the Borel construction and it seems plausible that someone has looked into this before. </p> <blockquote> <p>So I would like to ask: is there a reference for that?</p> </blockquote> <p>(Note that although $EG$ does not exist as an algebraic variety, it can be approximated by algebraic varieties: e.g. if $G=GL_n(k)$, one can take the spaces of $n$-frame bundles in $k^N,N\to\infty$.)</p>