Suppose I have a faithfully flat cover of schemes $\phi:X\to Y$, and a sheaf $F$ on $Y$. I might be interested in so-called ``twisted forms for $F$." That is, sheaves $F'$ on $Y$ such that $\phi^\ast(F)\cong \phi^\ast(F')$. In the case that $\phi$ is a cover by some collection $\{U_i\}$, this is often stated as saying that $F$ and $F'$ are locally isomorphic on this cover. Okay, great. You'll have to trust me that this is something I might be interested in.

It is ``known" that if we compute $\check{H}^1(\phi,Aut(F))$, we can determine all twisted forms for $F$ (up to isomorphism class) along $\phi$. Or at least, that is what I believe people to be saying.

1) Is this the correct statement?

2) If so, can someone at least give me a rough explanation of why this is true, if not a complete proof?

3) Instead of starting out with a sheaf on $Y$, can I start off with an effective descent datum for $\phi$ and determine all twisted forms of that descent datum? In other words, can I start with a sheaf $F''$ on $X$ with effective descent data (which tells us that it comes from something over $Y$) and then ask for sheaves on $Y$ which pull back to (something isomorphic to) $F''$? How is their canonical descent datum (coming from pulling back) related to the descent datum on $F''$?

4) What can I do/say if $\phi$ is not faithfully flat? Can I still make these kinds of statements? What if I am interested in only finding twisted forms for one particular descent datum?

I know this is a ton of questions. Sorry if it's way too much.



  • $\begingroup$ It might be a good idea to make your question more precise. For example, what do you mean by a sheaf? Is it quasi-coherent, what is the topology, ... Same for the Cech cohomology. $\endgroup$ – naf Dec 24 '13 at 6:36
  • $\begingroup$ Oh my, perhaps this question shouldn't even exist, given the following: mathoverflow.net/questions/62528/twisted-forms-and-checkh1?rq=1 $\endgroup$ – Jonathan Beardsley Dec 26 '13 at 4:23

In general the statement is slightly different. What you typically have is a presheaf $F$ of groupoids (i.e. for every element $X$ you can have $QCoh(X)$ the groupoid of quasicoherent sheaves and isomorphisms, but maybe you are interested in the groupoid of algebras, Azumaya algebras, Hopf algebras etc..) and an object $A$ of $F(Y)$.

Then a twisted form of $A$ is an object $B$ of $F(Y)$ which becomes isomorphic to $A$ when they are pulled back to $X$. In this context you have is a map of pointed sets from the set of isomorphism classes of twisted forms of $A$ to $\check{H}^1(X,Aut(A))$, given by $B\mapsto Iso(A,B)$ where $Iso(A,B)$ is seen as an $Aut(A)$-torsor via precomposition. It is clear that if this torsor is trivial (that is it has a section on $Y$) then $B$ is isomorphic to $A$. Your question can be rephrased to asking conditions under which this map is an isomorphism.

The only case in which I know of a general theorem of this type is when the presheaf of groupoids is obtained by $QCoh$ via some sort of algebraic structure. At example call a signature $S$ a collection of pairs $(n_i,m_i)_{i\in I}$. Then the ''algebraic structures of signature $S$'' are the quasicoherent sheaves $F$ with a collection of maps $\Phi_i:F^{\otimes n_i} \to F^{\otimes m_i}$. Then for any signature $S$ the groupoid of ''algebraic structures of signature $S$'' and isomorphisms is such that the map of the previous paragraph is an isomorphism.

You can find a complete discussion when $X\to Y$ is a field extension in the book by Gille and Szamuely Central simple algebras and Galois cohomology. From that discussion and the general descent theorem is not hard to get the general case.

Sorry I pressed the submit button by mistake, hence the incomplete answer.

  • $\begingroup$ During Christmas lunch I realized that the proof I know works for every sheaf of groupoids $F$ (a.k.a. stack) on your site, so if you want you can ignore the description of ''algebraic structures'' (apart from being nice examples of stacks. Oh and merry christmas everyone! $\endgroup$ – Denis Nardin Dec 25 '13 at 14:48
  • $\begingroup$ Thanks Denis!! Are you assuming the cover is faithfully flat here? (And Merry Christmas to you too!) $\endgroup$ – Jonathan Beardsley Dec 26 '13 at 3:44
  • $\begingroup$ And I might add, in case anyone doesn't know, that cohomology group rightfully computes torsors for $Aut(A)$ over $X$, hence Denis only explaining how $B$ gets taken to an $Aut(A)$-torsor. $\endgroup$ – Jonathan Beardsley Dec 26 '13 at 3:56
  • $\begingroup$ Well I am imagining $F$ to be a sheaf of groupoids on some site, which could be the fpqc if you want (if you want I can write down the proof, it's actually easier than I remembered at first) $\endgroup$ – Denis Nardin Dec 26 '13 at 8:57
  • $\begingroup$ Oh and just to be precise: I never require $Aut(X)$ or any torsor to be representable, so $Aut(X)$ is a sheaf of groups in our site and a torsor $T$ is a sheaf of sets with an action $T\times Aut(X)\to T$ which is simply transitive (i.e. $T\times Aut(X)\to T\times T$ is an isomorphism). These are the objects classified by the first Cech cohomology group. $\endgroup$ – Denis Nardin Dec 26 '13 at 9:04

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