Example Wanted: When Does Cech Cohomology Fail to be the same as Derived Functor Cohomology?

I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same.

I started worrying about this from this answer to an MO question, and Brian Conrad's comments to another MO question. Let $$\mathcal{F}$$ be a sheaf of abelian groups on a space X. (Here I want to be a little vague about what a "space" means. I'm thinking of either a scheme or a topological space). Then Cech cohomology of X with respect to a cover $$U \to X$$ can be defined as cohomology of the complex

$$\mathcal{F}(U) \to \mathcal{F}(U^{[2]}) \to \mathcal{F}(U^{[3]}) \to \cdots$$

Where $$U^{[ n ]} = U \times_X U \times_X \cdots \times_X U$$. The total Cech cohomology of $$X$$, $$\check H^{ * }(X, \mathcal{F})$$, is then given by taking the colimit over all covers $$U$$ of X. Now if the following condition is satisfied:

Condition 1: For sufficiently many covers $$U$$, the sheaf $$\mathcal{F}|_{U^{[ n ]}}$$ is an acyclic sheaf for each n

then this cohomology will agree with the derived functor version of sheaf cohomology. We have,

$$\check H^{ * }(X, \mathcal{F}) \cong H^*(X; \mathcal{F}).$$

I am told, however, that even if $$\mathcal{F}|_U$$ is acyclic this doesn't imply that it is acyclic on the intersections. It is still okay if this condition fails for some covers as long as it is satisfied for enough covers. However I am also told that there are spaces for which there is no cover satisfying condition 1.

Instead you can replace your covers by hypercovers. Basically this is an augmented simplicial object $$V_\bullet \to X$$ which you use instead of the simplicial object $$U^{[ \bullet +1 ]} \to X$$. There are some conditions which a simplicial object must satisfy in order to be a hypercover, but I don't want to get into it here. You can then define cohomology with respect to a hypercover analogously to Cech cohomology with respect to a cover, and then take a colimit. This seems to always reproduce derived functor sheaf cohomology.

So my question is when is this really necessary?

Question 1: What is the easiest example of a scheme and a sheaf of abelian groups (specifically representable ones such as $$\mathbb{G}_m$$) for which Cech cohomology of that sheaf and derived functor cohomology disagree?

Question 2: What is the easiest example of a (Hausdorff) topological space and a reasonable sheaf for which Cech cohomology and derived functor cohomology disagree?

I also want to be a little flexible about what a "cover" is supposed to be. I definitely want to allow interesting Grothendieck topologies, and would be interested in knowing if passing to a different Grothendieck topology changes the answer. It changes both the notion of sheaf and the notion of Cech cohomology, so I don't really know what to expect.

Also, I edited question 1 slightly from the original version, which just asked about quasi-coherent sheaves. Brian Conrad kindly pointed out to me that for any quasi-coherent sheaf the Cech cohomology and the sheaf cohomology will agree (at least with reasonable assumptions on our scheme, like quasi-compact quasi-separated?) and that the really interesting case is for more general sheaves of groups.

• In the interests of nobody else wasting their time: the Hawaiian earring with coefficients in a locally constant sheaf is not a counterexample to (2). – Tyler Lawson Mar 25 '10 at 18:37
• One comment (which doesn't actually address either question): if you take the limit over all P-hypercovers, with P = set of covering maps, instead of just Cech hypercovers, then this limit cohomology always computes derived functor cohomlogy. Brian Conrad mentions this at the beginning of section 5 of his cohomological descent notes. – David Zureick-Brown Mar 25 '10 at 18:41
• A related phenomenon is exemplified in: [Glen E. Bredon. A Space for Which $H^1(X; Z) \not\approx \lbrack X, S \rbrack$. Proceedings of the American Mathematical Society, Vol. 19, No. 2 (Apr., 1968), pp. 396-398] – Mariano Suárez-Álvarez Mar 29 '10 at 16:06
• @David Brown: That's actually really awesome. – Harry Gindi Mar 30 '10 at 23:27
• An answer to question 2 is at mathoverflow.net/questions/122478/… – David Roberts Feb 21 '13 at 2:25