I am reading Milne's *Étale cohomology*, III.4.


A twisted form of an object $Y$ (a scheme, a sheaf of modules, of algebras...) over a scheme $X$ is an object $Y'$ such that there exists a covering in some topology (let's say étale to fix the ideas), $(U\_i \to X)$ such that $Y \times\_X U\_i \cong Y' \times\_X U\_i$ for all $i$. Then, the book says, any twisted form of $Y$ defines a cocycle in $\check{H}^1(X,\mathrm{Aut}(Y))$ as follows: let $f\_i$ be the isomorpisms $Y \times\_X U\_i \to Y' \times\_X U\_i$, then the cocycle is given by $(\alpha\_{ij}): Y \times\_X U\_i \times\_X U\_j \to Y \times\_X U\_i \times\_X U\_j$ where $\alpha\_{ij} = f\_i^{-1} \circ f\_j$, which, in turn, constitutes the descent data that eventually allows to recover $Y'$. As far as I understand, the descent data does not have to be always effective in general, but in some cases it is. The book gives two examples when this is the case: Severi-Brauer varieties and vector bundles.


I would like to understand when the Cech cohomology classes are in bijective correspondnence with the (isomorphism classes of) twisted forms. So I have two questions:


1. what are the general criteria that the descent data as described above is effective?
2. what happens if the trivialising cover of a twisted form $Y'$ contains just one étale morphism $U\_0 \to X$? It seems like the cocycle defined by $Y'$ is always trivial then ($\alpha\_{00} = f\_0^{-1} \circ f\_0$), yet the form $Y'$ might be not isomorphic to $Y$.



Twisted forms and $\check{H}^1$
For your question 2, note that the fiber product $U\_0\times\_XU\_0$ can be non-trivial (unlike the case of an open sub-set $U\_0\subset X$, where it would be $U\_0$ again) with two different projections $p\_1$ and $p\_2$ onto $U\_0$ giving two different structures to it as a $U\_0$-scheme. The isomorphism $\alpha\_{0,0}$ is an isomorphism between these two different $U\_0$-structures on $Y\times\_XU\_0\times\_XU\_0$ and is also a non-trivial piece of information


As an example, if $X=Spec\;k$ and $U\_0=Spec\;K$ where $K/k$ is a Galois extension with Galois group $G$, then we get an isomorphism $G\times U\_0\rightarrow U\_0\times\_XU\_0$ using the group action. The two structures of $G\times U\_0$ as a $U\_0$-scheme correspond to the two maps $(g,u)\mapsto u$ and $(g,u)\mapsto gu$. The isomorphism $\alpha\_{0,0}$ satisfying the co-cycle condition now equates precisely to giving an action of $G$ on $Y'$ compatible with its action on $U\_0$. This is Galois descent. See Serre's Local fields, for example.


EDIT: It occurs to me that I didn't directly talk about twisted objects. The idea is the same. Let us start with an object $Y'$ over $U\_0$ equipped with an isomorphism $f\_0:Y'\rightarrow Y\times\_XU\_0$. This isomorphism has to satisfy the co-cycle condition, which means the following: 


We have two projections $p\_1,p\_2:U\_0\times\_XU\_0\rightarrow U\_0$. So we have two different ways to pull-back $Y'$ to over $U\_0\times\_XU\_0$, giving us $p\_1^\*Y'$ and $p\_2^\*Y'$. The two pull-backs of $Y\times\_XU\_0$ under these projections are canonically isomorphic, since the two projections $p\_1,p\_2$, when composed with the map $U\_0\rightarrow X$ agree with the structure map for $U\_0\times\_XU\_0\rightarrow X$. So pulling back $f\_0$ gives us isomorphisms $$p\_1^\*Y'\rightarrow Y\times\_X(U\_0\times\_XU\_0)\rightarrow p\_2^\*Y'.$$


This is your $\alpha\_{0,0}$; if it satisfies the co-cycle condition--this amounts to the required compatibility between the pull-backs of $\alpha\_{0,0}$ to 
$U\_0\times\_{X}U\_0\times\_{X}U\_0$ under the *three* different projections to $U\_0\times\_XU\_0$--then it gives you descent data for $Y'$. In the Galois setting, if you use $f\_0$ to identify $Y'$ with $Y\times\_XU\_0$, then $\alpha\_{0,0}$ is giving you a `twisted' action of $G$ on $Y\times\_XU\_0$.


This data is not always effective. I would highly recommend the chapter on descent in Bosch-Lutkebohmmert-Raynaud's `Neron Models' for an explanation of all these things.


