## Twisted forms and $\check{H}^1$

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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 $Y$.

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## 2 Answers

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.

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Dear Keerthi, thank you for you answer. I will look at the chapter in "Neron models". Is it easy to answer from what general theorem does it follow that in the case of vector bundles or Severi-Brauer varieties the descent data is always effective? In the chapter of Milne's book I have mentioned, SGA1.I.VIII.7.8 is referenced, but so far I couldn't extract the precise statement that would imply the bijection between isomorphism types of twisted forms and cocycle classes. – Dima Sustretov Apr 21 2011 at 19:19
Dear Dmitry--Yes, I believe BLR has a resume of effectivity results. For vector bundles, it's the fact that descent data for quasi-coherent sheaves are always effective. For Severi-Brauer varieties, it's the fact that varieties with an ample line bundle always have effective descent data. I don't have BLR at hand, so I can't give you references for the precise statements there. Once you have effectivity, the bijection you need follows formally. Again, I don't know of a good reference where this is laid out cleanly, though I can say that SGA1 is quite clearly written in general. – Keerthi Madapusi Pera Apr 21 2011 at 20:43

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If your $Y$ is an object over $X$ of a stack in the étale topology, then you can use the cocycles as descent data to get a twisted form (because by definition of a stack all descent data are effective).

Some examples other than the ones you mentioned are quasi-coherent sheaves and affine morphisms of schemes (i.e. quasi-coherent sheaves of algebras).

This is far from a necessary condition on $Y$, I don't know if there is a sharper characterization.

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