Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This can be seen as a follow up my question here:

Is there a notion of "fibered category with boxproducts"?

Given a monoidal fibration $f:E\rightarrow B$ (i.e. a strict monoidal functor between monoidal categories which is a fibration of ordianary categories) where the base is a cartesian monoidal category endowed with a grothendieck topology. What are the right conditions for such a fibration be called a stack?

I guess it is not enough to ask that $E(X)\rightarrow Desc(X,U)$ is an equivalence of ordinary categories. Insted one should need some further condition that ensures the following:

"if $(\phi_i)$ can be glued to $\phi$ and $(\psi_j)$ can be glued to $\psi$ than $(\phi_i\boxtimes \psi_j)$ can be glued to $\phi\boxtimes \psi$"

Does this notion exist yet? What would be the right condition?

Examples I have in mind are

$B$=geometric objects for example smooth varieties and

$E$=sheaves for example $\mathcal{D}_X$-modules

share|improve this question
3  
It looks like what you are asking for is that $E(X) \to Desc(X,U)$ is a monoidal equivalence. –  Jeffrey Giansiracusa Jul 9 '10 at 11:48
    
Yes, I also thought about this, however I was confused and saw a problem which was no problem. Do you know whether there is a criterion when s monoidal functor is an monoidal equivalence? Is full, faithful and essential surjective enough? –  Jan Weidner Jul 9 '10 at 19:14
    
@ Jan. Yes. If you have a monoidal functor, which is an equivalence of categories after forgetting the monoidal part, then it is in fact a monoidal equivalence. Hence fully-faithful and essentially surjective is enough. –  Chris Schommer-Pries Jul 10 '10 at 14:29
    
Thanks, so, I am a bit confused. The functor $E(X)\rightarrow Desc(X,U)$ is monoidal, so no extra condition is needed, since it is automatically an monoidal equivalence whenever it is an equivalence? –  Jan Weidner Jul 10 '10 at 18:05
add comment

1 Answer

up vote 3 down vote accepted

I would take the view point that a monoidal category is a bicategory with one object. (Then a category fibered in monoidal categories should be the same thing as a weak functor into bicategories that "only hits monoidal categories".) In other words, what you should have is that this fibration is a 2-stack when viewed as a fibration in bicategories. The descent condition should then be that the canonical map $E|_X \to Desc(X,U)$ be an equivalence of bicategories, where each of these monoidal categories is viewed as a bicategory, which is equivalent to Jeff's comment; this is just saying that $E|_X \to Desc(X,U)$ is a monoidal equivalence.

share|improve this answer
    
I am a bit worried about what $Desc(X,U)$ should mean here, since where are going to 2-stacks. You should define it to be $Hom(S_U,E)$, where the $Hom$ is maps of fibrations of bicategories, where $S_U$ is the sieve generated by the covering family $U$ –  David Carchedi Jul 9 '10 at 15:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.