5
$\begingroup$

Given a topological space X, i'd like to find Der X - the derived category of sheaves of abelian groups on X - to be a closed monoidal category. Hom should be cohomological and the internal-hom should be triangulated.

  1. Is this possible in full generality? (Unbounded complexes, no restrictions on X)
  2. Consider a sheaf of rings R or equivalently a ring of sheaves. This gives us two things: An abelian category of left R modules that we can derive; let's call this one Der R; A monoid R in Der X whose category of modules we denote dMod R. Is Der R = dMod R'? Ff not: how do they relate?
  3. Given monoids R,S,T in Der X do we get the usual adjunctions in two variables between their categories of bimodules?
  4. Given rings R,S,T in Sh X do we get the usual adjunctions in two variables between their derived categories of bimodules?

Now for the question: What is the right setting to do this? As i understand it, there's no suitable model structure that gives 4 in full generality.

$\endgroup$

2 Answers 2

7
$\begingroup$

I think the natural context to answer this and related questions is that of symmetric monoidal stable $(\infty,1)$-categories. In that context Lurie's DAG I-III give all the necessary foundations to straightforwardly generalize all of the usual abelian story you refer to to the derived category. There is such an object for sheaves on X (unbounded complexes thereof), and we can talk about associative algebra objects therein, their stable module categories, bimodules etc. and everything works just as you'd expect. In particular the homotopy categories of all these will be what you expect. (If you're looking for a worked-out example of playing these games in a similar context, I can selfishly suggest this.)

$\endgroup$
6
$\begingroup$

The most direct answer to question (2) is no, even in the case of sheaves on a point. In this case we simply have a ring R and we are asking whether the derived category of R-modules is equivalent to the category of R-module objects in the derived category of abelian groups. For example, let $R = Z/p^2$. Then in the derived category of R-modules, we have nonzero Ext groups $$Ext^n_{Z/p^2}(Z/p,Z/p) = Hom_{D(Z/p^2)}(Z/p,Z/p[n])$$ for all positive degrees $n$. However, because no object in the derived category of Z-modules has any nontrivial Ext-groups in high degrees, we have for $n$ large $$Hom_{dMod(Z/p^2)}(Z/p,Z/p[n]) = 0.$$ The problem is that dMod(R) doesn't really remember that R is an actual ring, but instead only remembers that it can be lifted to a chain complex with a chain-homotopy associative multiplication map. These kinds of ring objects are too weak to do really serious homological algebra and define a proper derived category.

The answer to question 3 is correspondingly no, because we can't really construct proper derived functors of the tensor product given only a monoid in Der(X).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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