equivalence of Grothendieck-style versus Cech-style sheaf cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:26:07Z http://mathoverflow.net/feeds/question/4214 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology equivalence of Grothendieck-style versus Cech-style sheaf cohomology Victoria Flat 2009-11-05T04:47:03Z 2011-03-15T07:00:33Z <p>Given a topological space X we can define the sheaf cohomology of X in </p> <p>I. the Grothendieck style (as the right derived functor of the global sections functor Gamma(X,-))</p> <p>or</p> <p>II. the Cech style (first by defining the Cech cohomology groups subordinate to an open cover, and then taking the direct limit of these groups over all covers).</p> <p>When exactly are these two definitions equivalent? I'm unhappy with the explanation given by Hartshorne. Are they the same for any paracompact Hausdorff space? Or a locally contractible space?</p> <p>And what is the relationship between these two sheaf cohomologies and singular cohomology?</p> <p>Any elaboration on this circle of ideas related to the relationship between all the different cohomology theories would be appreciated.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4217#4217 Answer by Dinakar Muthiah for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Dinakar Muthiah 2009-11-05T05:14:08Z 2009-11-05T05:14:08Z <p>The problem with Cech cohomology is that even if things are acyclic on open sets of your Cech cover, they may not be when you restrict to intersections of those open sets. The usual fix is to make the cover finer so you don't have that problem. Unfortunately there are topological spaces where no cover will be good enough. That's the bad news.</p> <p>The good news is that for a lot of spaces and categories of sheaves you're interested in, there will be such a cover. My favorite example is the category of quasi-coherent sheaves on a separated scheme. Then Cech cohomology computed on any affine cover will compute the derived functor cohomology.</p> <p>The even better news is that there is a way to fix Cech cohomology so that it will work for all situations. This is Verdier's theory of hypercovers, and it computes derived functor cohomology for any category with a Grothendieck topology. I must admit I have not played around much with this, but <a href="http://www.math.uiuc.edu/K-theory/0646/cech.pdf" rel="nofollow">here</a> is a link to a paper that talks about this circle of ideas. </p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4218#4218 Answer by Eric Wofsey for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Eric Wofsey 2009-11-05T05:21:53Z 2009-11-05T14:50:29Z <p>This isn't entirely complete, but here are some results. Exercise III.4.11 in Hartshorne gives that whenever a sheaf is acyclic on any intersection of sets in a cover, Cech cohomology agrees with the derived functors. In particular, if every open cover has a refinement with this property, the derived functors will agree with Cech cohomology defined as the limit over open covers. I don't think you can say any general result involving just the space; you need to know something about the sheaf too. If a sheaf is nowhere locally acyclic, no matter how nice the space is Cech cohomology is probably not going to agree with the derived functors.</p> <p>These are related to singular cohomology by looking only at constant sheaves. On a locally contractible space every open subset of which is paracompact, singular cohomology with coefficients in A is the same as derived functor cohomology of the constant sheaf A (you may be able to get slightly better hypotheses than this; this is what I found I needed when I tried to prove this a while ago). Basically, this is because the (sheafification of the pre)sheaf of singular cochains forms an acyclic resolution of the constant sheaf. On the other hand, for any space having the homotopy type of a CW-complex, Cech cohomology of constant sheaves agrees with singular cohomology (because it satisfies the <a href="http://en.wikipedia.org/wiki/Eilenberg-Steenrod%5Faxioms" rel="nofollow">Eilenberg-Steenrod axioms</a>). It follows, for example, that for any CW-complex or any manifold, singular cohomology agrees with both Cech and derived cohomology of constant sheaves.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4219#4219 Answer by David Zureick-Brown for equivalence of Grothendieck-style versus Cech-style sheaf cohomology David Zureick-Brown 2009-11-05T05:43:10Z 2009-11-05T05:53:36Z <p>I suggest in addition to the other answers checking out Brian Conrad's <a href="http://math.stanford.edu/~conrad/papers/hypercover.pdf" rel="nofollow">notes</a> on cohomological descent. They're a little more to the point for applications to geometry.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4223#4223 Answer by Jonathan Wise for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Jonathan Wise 2009-11-05T07:00:41Z 2011-03-15T06:56:24Z <p>Let $X$ be a topological space, and $T$ its category of open sets with the usual Grothendieck topology. Let $T'$ be any sieve of $T$ (a subcategory of $T$ such that if $U$ is in $T'$ then any subset of $U$ is also in $T'$). For example, $T'$ might be the collection of open subsets subordinate to the open subsets in a cover $\mathcal{U}$. Any sheaf on $T$ induces a functor on $T'$ which can be viewed as a sheaf on $T'$ if $T'$ is given the minimal topology (the only covers are the identity maps). This determines a morphism of topoi $f : T \rightarrow T'$, hence a spectral sequence</p> <p>$H^p(T', R^q f_\ast F) \Rightarrow H^{p+q}(T, F)$ .</p> <p>(One could surely also convince oneself that such a spectral sequence exists without any reference to topoi.)</p> <p>The Cech cohomology of $F$ with respect to some covering family $\mathcal{U}$ is </p> <p>$H^p(\mathcal{U}, F) = H^p(T', f_\ast F)$</p> <p>where $T' = T'(U)$ is the sieve associated to the cover $\mathcal{U}$. The Cech cohomology is then the filtered colimit</p> <p>$\check{H}^p(T, F) = \varinjlim_{(T',f)} H^p(T', f_\ast F)$</p> <p>taken over the projections $f : T \rightarrow T'$ associated as above to covering families $\mathcal{U}$.</p> <p>One evidently has edge homomorphisms</p> <p>$\check{H}^p(T, F) \rightarrow H^p(T, F)$</p> <p>from the spectral sequence, and the question is when these induce an isomorphism. If we could somehow eliminate the $R^p f_\ast F$, $p > 0$, by passing to a "small enough" cover we would have equality. This condition already holds in many cases; the following condition is more general (but I haven't checked carefully that it actually works!):</p> <p>For every cover $\mathcal{U}$ of $X$, every $U_1, \ldots, U_n \in \mathcal{U}$, and every class in $\alpha \in H^p(U_1 \mathop{\times}_X \cdots \mathop{\times}_X U_n, F)$, $p > 0$, there exists a refinement $\mathcal{U}'$ of $\mathcal{U}$ such that the restriction of $\alpha$ under the map</p> <p>$H^p(U_1 \mathop{\times}_X \cdots \mathop{\times}_X U_n, F) \rightarrow H^p(U'_1 \mathop{\times}_X \cdots \mathop{\times}_X U'_n, F)$</p> <p>is zero.</p> <p>To make sense of this, one must use some convention for the covers $\mathcal{U}$ and $\mathcal{U}'$ to ensure there is a map as above. For example, one could work only with covers indexed by the points of $X$ (a cover is then a collection of neighborhoods of each point of $X$).</p> <p>A more refined version of the above condition would say that Cech cohomology equals cohomology in degrees at most $q$ if the above condition holds for $p \leq q$. Since it always holds for $p = 0,1$ this implies that</p> <p>$\check{H}^1(T, F) = H^1(T, F)$ </p> <p>in general.</p> <p><strong>Edit in response to David's comment:</strong></p> <p>The Cech complex always computes cohomology correctly in a presheaf category (i.e., when the topology is "chaotic": an object has no covers by anything except itself). Trying to compute cohomology in an arbitrary site using the Cech complex is (heuristically) something like trying to approximate the site by a presheaf category.</p> <p>Here is how Cech cohomology computes cohomology of presheaves. Consider any category $T'$. If $F$ is a presheaf of groups on $T'$ then the sheaf cohomology groups of $F$ are the derived functors of the inverse limit for diagrams of shape $T'$. They are also computed as</p> <p>$Ext(\mathbf{Z}, F)$</p> <p>where $\mathbf{Z}$ is the constant sheaf associated to the integers. Remarkably, in a presheaf category, $\mathbf{Z}$ has a canonical projective resolution associated to any cover of the final presheaf. A cover of the final presheaf is a collection of objects $U$ of $T'$ such that every object of $T'$ has a map to at least one object of $U$. The $i$-th term of this complex is the direct sum, over all choices of $i$ elements $U_1, ..., U_i$ of $U$, of the groups $\mathbf{Z}_{U_1 \times \cdots \times U_i}$. (You can check this is projective by noting it is the extension by $0$ of $\mathbf{Z}$ from the slice category $T' / U_1 \times \cdots \times U_i$ and extension by $0$ preserves projectives (since it has an exact right adjoint) and $\mathbf{Z}$ is projective on the slice category since all higher cohomology of all sheaves vanishes (since it has a final object). It's also easy to check by a direct calculation.) </p> <p>Denote this complex by $K$. Since this is a projective resolution of $\mathbf{Z}$, $\mathrm{Hom}(K, F)$ computes the cohomology of $F$. But it is also easy to see that this is just the Cech complex of $F$.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/4229#4229 Answer by Georges Elencwajg for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Georges Elencwajg 2009-11-05T08:40:20Z 2011-03-15T07:00:33Z <p>Dear Victoria: here is a summary of the main comparison results I know of between Grothendieck cohomology (which is usually just called cohomology and written $\newcommand{\F}{\mathcal F}H^i(X,\F)$ ) and other cohomologies.</p> <p>1) If $X$ is locally contractible then the cohomology of a constant sheaf coincides with singular cohomology. [This is Eric's answer, but there is no need for his hypothesis that open subsets be acyclic]</p> <p>2) Cartan's theorem: Given a topological space $X$ and a sheaf $\F$, assume there exists a basis of open sets $\mathcal{U}$, stable under finite intersections, such that the CECH cohomology groups for the sheaf $\F$ are trivial (in positive dimension) for every open $U$ in the basis: $H^i(U,\F)=0$ Then the Cech cohomology of $\F$ on $X$ coincides with (Grothendieck) cohomology </p> <p>3) Leray's Theorem: Given a topological space $X$ and a sheaf $\F$, assume that for some covering $(U_i)$ of $X$ we know that the (Grothendieck!) cohomology in positive dimensions of $\F$ vanishes on every finite intersection of the $U_i$'s. Then the cohomology of $\F$ is already calculated by the Cech cohomology OF THE COVERING $(U_i)$: no need to pass to the inductive limit on all covers. This contains Dinakar's favourite example of a quasi-coherent sheaf on a separated scheme covered by affines.</p> <p>4) If $X$ is paracompact, Cech cohomology coincides with Grothendieck cohomology for ALL SHEAVES<br> If you think this is too nice to be true, you can check Théorème 5.10.1 in Godement's book cited below [So Eric's remark that no matter how nice the space is, Cech cohomology would probably not coincide with derived functor cohomology for arbitrary sheaves turns out to be too pessimistic]</p> <p>5) Cohomology can be calculated by taking sections of any acyclic resolution of the studied sheaf: you don't need to take an injective resolution. This contains De Rham's theorem that singular cohomology can be calculated with differential forms on manifolds.</p> <p>6) If you study sheaves of non-abelian groups, Cech cohomology is convenient: for example vector bundles on $X$ ( a topological space or manifold or scheme or...) are parametrized by $H^1(X, GL_r)$. I don't know if there is a description of sheaf cohomology for non-abelian sheaves in the derived functor style.</p> <p>Good references are </p> <p>a) A classic: Godement, Théorie des faisceaux (in French, alas)</p> <p>b) S.Ramanan, Global Calculus,AMS graduate Studies in Mahematics, volume 65. (An amazingly lucid book, in the best Indian tradition.)</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/5814#5814 Answer by Shizhuo Zhang for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Shizhuo Zhang 2009-11-17T13:13:43Z 2009-11-17T13:13:43Z <p>Let f : X-->S be a scheme morphism having a direct image functor f_*(for instance, f_* is the global section functor). Since X is quasicompact, there exists a finite affine cover U of X such that f |U is an affine morphism for any U belongs to a cover. Then the standard complex C(U; f_*) corresponding to the cover U is a resolution of the functor f_*. Therefore it can be used for computing higher direct images (= derived functors) of f_*.</p> <p>If the localizations at different open sets of the cover U commute,the complex C(U; f_*) is homotopically equivalent to the Cech complex,C(U; f_*) of the cover U. One can show that the following conditions are equivalent: (a) For any affine cover U of a scheme X, If the localizations at different open sets of the cover U commute</p> <p>(b) The scheme X is separated.</p> <p>In other words, the ˇ Cech complex is equivalent to the standard complex for any affine cover only if the scheme is separated. If the scheme X is not separated, the higher cohomology of the ˇ Cech complexC(U; f_*) are not isomorphic, for a general affine cover U, to the corresponding derived functors of f_*.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/5815#5815 Answer by Urs Schreiber for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Urs Schreiber 2009-11-17T13:40:54Z 2009-11-17T17:07:06Z <p>I like to say that there is only a single abstract definition of cohomology: in any $(\infty,1)$-topos $\mathbf{H}$ given objects $X$ and $A$, the cohomology of $X$ with coefficients in $A$ is the connected components of the hom-$\infty$-groupoid $H(X,A) := \pi_0 \mathbf{H}(X,A)$.</p> <p>Everything else one sees described as "cohomology" is, i claim, a special case and a special realization of this situation.</p> <p>More on this point of view is at <a href="http://ncatlab.org/nlab/show/cohomology" rel="nofollow">cohomology</a></p> <p>In particular, ordinary abelian sheaf cohomology for sheaves on a cite $C$ is the cohomology in this sense of the <a href="http://ncatlab.org/nlab/show/models+for+infinity-stack+(infinity%2C1)-toposes" rel="nofollow">(oo,1)-topos of oo-stacks on C</a> where the coefficient objects are, moreover, restricted to be objectwise in the image of the <a href="http://ncatlab.org/nlab/show/Dold-Kan+correspondence" rel="nofollow">Dold-Kan map</a> (are "maximally abelian oo-stacks").</p> <p>From this perspective the relation betwen Cech-cohomology and other means to compute sheaf-cohomology become conceptually evident: all of these are just models to model the <a href="http://ncatlab.org/nlab/show/(infinity,1)-categorical+hom-space" rel="nofollow">(oo,1)-cateorical hom-space</a> $\mathbf{H}(X,A)$: Cech cohomology does so by finding cofibrant versions of $X$ (namely <a href="http://ncatlab.org/nlab/show/Cech+nerve" rel="nofollow">Cech nerves</a> of Cech covers), derived-functor-style sheaf cohomology usually does so by finding fibrant versions of $A$ (namely injective resolutions of sheaves).</p> <p>That this is the relation between the two is of course implicitly the old Verdier hypercovering theorem. A particularly clear-sighted description of this is the remarkable old article by Kenneth Brown, <a href="http://ncatlab.org/nlab/show/BrownAHT" rel="nofollow">Abstract homotopy theory and generalized sheaf cohomology</a>.</p> <p>A summary of that in the light of the above comments is at <a href="http://ncatlab.org/nlab/show/abelian+sheaf+cohomology" rel="nofollow">nlab:abelian sheaf cohomology</a>.</p> <p>Technical details are also at <a href="http://ncatlab.org/nlab/show/Cech+cohomology" rel="nofollow">Cech cohomology</a>.</p> http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/18674#18674 Answer by Donu Arapura for equivalence of Grothendieck-style versus Cech-style sheaf cohomology Donu Arapura 2010-03-18T22:26:10Z 2010-03-19T08:52:41Z <p>I don't know if it's bad form to reply to something this old, but I stumbled on this question because I've wondering about the negative result for a couple of days now. That is, for an explicit example where Cech cohomology <em>differs</em> from (derived functor) sheaf cohomology. In case anyone is also curious about this, I did find an example buried in pages 177-179 of Grothendieck's classic "Tohoku" paper "Sur quelque points...". Perhaps I can say a few words about it since it is surprisingly simple. Take X to be the affine plane over a field, and let $Y\subset X$ be the union of two irreducible curves meeting at two distinct points. Let K be the kernel of the restriction map $Z_X\to Z_Y$ of the Z-valued constant sheaves on the Zariski topology. Then he shows that $H^2(X,K)=Z$ but that the Cech group $\check{H}^2(X,K)= 0$. (I wrote this backwards previously, sorry about that.)</p>