31

That already fails for $X$ equal to $\text{Spec}\ R$, where $R$ is a DVR with generic point $\eta = \text{Spec}\ K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $\{\eta\}$, there is a straightforward equivalence between the category of $\mathcal{O}_X$-modules and the category of triples $(M,V,\phi)$ of an $R$-module $M$, a $...

22

Since you are asking about computing algebraic de Rham, don't use injective resolutions, since they are not really constructive. You can use a Cech complex: If $\{U_i\}$ is an affine open cover of your variety $X$, form a double complex $C^{\bullet\bullet}=C^\bullet(\{U_i\}, \Omega_X^\bullet)$ with Cech coboundary in one direction, and exterior derivative in ...

19

Functions $(f_0,f_1,f_2)$ as you requested do in fact exist. Your confusion comes from the fact that you are trying to impose the cocycle condition on the intersection $U_0\cap U_1\cap U_2$, which is empty. That is the $f_i$'s are such that
$$ f_2(x) = f_1(x)+1\ \forall x\in U_1\cap U_2$$
$$ f_2(x) = f_0(x)+2\ \forall x\in U_0\cap U_2$$
$$ f_1(x) = f_0(x)+3\ ...

16

This is false even for $\mathrm H^0$: take $X$ to be $\mathbb A^2 \smallsetminus \{0\}$, and as $F$ the structure sheaf of $L \smallsetminus \{0\}$, where $L$ is a line through $0$.

15

I'll have more time to write and provide a more thorough answer later, but I think the most straightforward proof (which I agree is hard to find) comes via sheaf theory: On the one hand, there is a sheaf of locally finite singular chains whose hypercohomology is your $H^{lf}$. I work out the details in the setting of intersection homology on pseudomanifolds ...

15

This is the kind of thing sieves are good for. For an open cover, let $S$ denote the sieve it generates, so $S$ is poset of open subsets $V$ such that $V$ is contained in some element of the cover. A quasi-isomorphic model for the Cech complex is given by the "homotopy limit over the sieve", so the $0$th term in the complex is the product of the $F(U)$ for ...

15

One way of finding a "fully derived" version of Poincaré duality is Atiyah duality. This says that for any closed manifold $M$ there is an equivalence of spectra (in the sense of algebraic topology)
$$M^{-TM}\cong\mathbb{D}(\Sigma^∞_+M)$$
Here $M^{-TM}$ is the Thom spectrum for the (virtual) normal bundle, $\Sigma^\infty_+$ is the suspension spectrum and $\...

14

No. For instance there is a flat, projective morphism $f:X\rightarrow \Bbb{P}^1$ such that $X_t:=f^{-1}(t)$ is a smooth rational curve for $t\neq 0$, but $X_0$ is a nodal plane cubic curve with an embedded point (see Hartshorne, III.9.8.4). Then $H^1(X_t,\mathcal{O}_{X_t})$ is zero for $t\neq 0$, but $\ \dim H^1(X_0,\mathcal{O}_{X_0})=1$ . By base change it ...

14

While I'm sure a hocolim solution also exists (and works in general), one can avoid it, at least for compact topological spaces, by using open covers which are indexed by the points of your space $X$ (this idea goes back at least to Godement). These are called "rigid open covers" by Carlsson–Pedersen, who make the following definition:
Definition: A rigid ...

13

Sure thing! There's an equivalence of categories between cosheaves valued in profinite abelian groups and sheaves valued in torsion abelian groups, by Pontryagin duality one could say. So you can directly define the first etale homology of Z_ell and it will give you the l-adic Tate module. But it's formally the same as defining the l-adic Tate module to ...

13

Here is how I like best to think about this, although I'll point
to a few other proofs. Consider a general space $X$ (say compactly
generated). There is a natural
weak homotopy equivalence $\epsilon\colon |SX|\to X$ from the
geometric realization of the singular simplicial set of $X$ to $X$. Moreover, $|SX|$ is a CW complex whose cellular chains (and ...

13

Taking global sections is the same thing as computing the hom from $O_X$. In other words, there is an isomorphism of functors $\Gamma(X,-)\cong\hom(O_X,-)$, so both functors have the same derived functors.
As for using the derived category to define cohomology: yes, simply because that is what the derived category is for!

answered Jun 1 '17 at 2:21

Mariano Suárez-Álvarez

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11

A Grothendieck topology is called noetherian if every object is quasi-compact (which is defined as usual). On such a topology, sheaf cohomology commutes with filtered colimits (and in particular with arbitrary direct sums). A proof can be found in Tamme's Intoduction to Etale cohomology, Theorem §3.3.11.1. For the etale topology on the spectrum of a field ...

answered Feb 5 '13 at 9:19

Martin Brandenburg

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11

I am not sure if you are only really interested in properly stacky things, but it is perhaps worth pointing out that the result you mentioned from Hartshorne is true in significantly greater generality.
For any quasi-compact quasi-separated scheme $X$ (in fact for any spectral space $X$, or for something even slightly weaker) and any filtered system $(\...

11

Suppose that $\mathcal X$ is an algebraic stack with finite inertia (for example, a separated Deligne-Mumford stack); then, by a well-known result of Keel and Mori, there exist a moduli space $\pi \colon \mathcal X \to M$. The stack $\mathcal X$ is called tame when $\mathrm R^i\pi_* F = 0$ for every quasi-coherent sheaf $F$ on $\mathcal X$ and every $i > ...

11

Counterexample: For $X=\mathbb A^2\setminus\{0\}$, one has $\Gamma(X,O_X)=\mathbb C[u,v]$, but $H^1(X,O_X)\cong\mathbb C[u^{\pm1},v^{\pm1}]/(\mathbb C[u,v^{\pm1}]+\mathbb C[u^{\pm1},v])$ is not finitely generated over $\mathbb C[u,v]$.

11

There is an exact sequence
$$
0 \to \pi^*\Omega_X \to \Omega_{\tilde X} \to i_*\Omega_{E/Z} \to 0,
$$
where $E$ is the exceptional divisor and $i:E \to \tilde X$ is its embedding.
Dualizing it one gets
$$
0 \to T_{\tilde X} \to \pi^*T_X \to i_*T_{E/Z}(E) \to 0.\qquad(*)
$$
On the other hand, since $E = P(N)$ is the projectivization of the normal bundle, one ...

11

For (i), take $X=S^3$, $Y= S^2$, $f$ the Hopf fibration. There is a local $H^1$ everywhere, coming from the fact that the fibers are circles. This local section is defined consistently everywhere (because there is no monodromy since $Y$ is simply-connected), but it does not glue to a global section because $H^1(X,\mathbb Z)=0$.
For (ii), simply take $X=Y$ ...

11

Let $\sigma$ denote the nontrivial automorphism of $\mathbb{F}_4$ and put $C=\{1,\sigma\}$. Let $\mathcal{R}$ denote the category of rings in which $a^4=a$, and let $\mathcal{X}$ denote the category of Stone spaces with an action of $C$. There is a functor $F\colon\mathcal{X}\to\mathcal{R}$ given by $F(X)=\text{Map}_C(X,\mathbb{F}_4)$, and there is a ...

answered Sep 27 '14 at 17:45

Neil Strickland

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11

Suppose $X$ is a connected CW complex with fundamental group $\pi:=\pi_1(X,x)$. Then the cellular chain complex $C_*(\widetilde{X})$ of the universal cover is a chain complex of free $\mathbb{Z}\pi$-modules, where $C_i(\widetilde{X})$ has rank equal to the number of $i$-cells of $X$. We can also regard $L$ as a $\mathbb{Z}\pi$-module. By definition $H^i(X;L)$...

10

I am just posting my comment as an answer. If you want to produce a nontrivial torsor, it suffices to produce a nontrivial torsor after base change to the algebraic closure of a residue field of the base scheme. So now you can use the structure theory of algebraic groups. For multiplicative groups, $\mathbb{A}^{n+1}\setminus\{0\} \to \mathbb{P}^n$ is a ...

9

To answer (1), the idea would be to try and construct a trivialization "by hand". For principal bundles, that means a section. Think of a circle as an interval with the endpoints identified. Constructing the section on an interval is easy (path-lifting property of a bundle) but then you have to check that you can ensure the endpoints meet up. This ...

answered Jun 26 '12 at 19:11

Ryan Budney

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9

(Edit: I answered Q2 initially, ignoring Q1.)
Q1: The spectral sequence is right except that it starts at $E_1$.
Q2: It is true if $\mathcal{F}$ is unitary in the sense that the underlying representation of $\pi_1(M)$ is unitary, then the spectral sequence degenerates at the $E_1$ page. This seems like a folklore fact, so I would need to think to find a ...

8

0) I would guess that the compact spaces you are looking for are extremely rare.
1) For example the extremely simple contractible space $I=[0,1]$ is not suitable:
Consider the inclusion $j\colon U=(0,1)\hookrightarrow I
$ and take on $I$ the sheaf$j_!(\mathbb Z_U)$, the constant sheaf $\mathbb Z_U$ on $U$ extended to $I$ by zero.
Claim: $H^1(I,F)\...

answered Apr 13 '13 at 12:37

Georges Elencwajg

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8

Edit. This follows from the Elkik-Fujita Vanishing Theorem. There is a more general vanishing theorem due to Elkik and Fujita. One version of this theorem (where I read the theorem) is Theorem 1.3.1 of the following article.
MR0946243 (89e:14015)
Kawamata, Yujiro; Matsuda, Katsumi; Matsuki, Kenji
Introduction to the minimal model problem. Algebraic ...

8

For what follows, I recommend SGA 2 (available on Arxiv). There is an exact sequence
$$0\rightarrow H^1(X,\mathcal{O}_X)\rightarrow H^1(X\smallsetminus p,\mathcal{O}_X)\rightarrow H^2_{p}(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X)$$
where $ H^2_{p}(X,\mathcal{O}_X)$ is infinite-dimensional, while $H^i(X,\mathcal{O}_X)$ is finite-dimensional. Therefore $...

7

This is not related to sheafification. The sheaf $\mathbb{C}^{*}$ of locally constant functions on $M$ is already a sheaf, so sheafification will not change it.
This sequence is not an exact sequence of complexes but it is an exact triangle of complexes. That is - it is an exact sequence of complexes, up to quasi-isomorphism. The obvious short exact ...

7

Consider $X = \mathbb{P}^{1}$, $U =\mathbb{A}^{1}$ and j the inclusion of a affine chart in the projective line. The complement of the affine chart is a point P, let i be the inclusion of this point in the projective line.
Consider the sheaf $\mathcal{F} = \mathcal{O}(-2)$ on the projective line. One has $dim H^{1}( \mathbb{P}^{1}, \mathcal{F} ) =1$.
One has ...

7

No. For paracompact spaces sheaf cohomology coincides with Čech cohomology. In particular it applies to the closed topologist's sine curve $C$. There is a map $C \to S^1$ inducing an isomorphism on Čech cohomology, but $C$ is weakly contractible.

7

The example of Alex above is a special case of the Borel-Weil-Bott Theorem applied to $\operatorname{SL}_2/B = \Bbb{P}^1$ with $B$ the standard Borel subgroup in $\operatorname{SL}_2$. The general case is this:
Let $G$ be a semisimple complex Lie group with Weyl Group $W$. If for no $w \in W$ we have $w\ast \lambda$ dominant then $H^i(G/B, L_\lambda)...

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