Let $\cal A$ be an abelian category, say linear over a field, with enough injectives and $\cal P$ be the heart of a t-structure on the bounded derived category $D^b(\cal A )$. Assume that $\cal P$ also has enough injectives. Suppose that the realization functor

$$real:D^b(\cal P)\rightarrow D^b(\cal A)$$

is an equivalence. Now given two objects corresponding through this equivalence, are their Ext-algebras $A_\infty$-quasi-isomorphic?

Edit: If I understand it correctly the realization functor is constructed in BBD as follows:

  1. They take the homotopy category of injective complexes of objects in $\cal A$, equipped with descending filtration, whose filtration steps lie in $\cal P$. This category is called $DF_{bete}$.

  2. Using the boundary map of the triangles $gr^{i+1}F\rightarrow F^{i-1}/F^{i+1} \rightarrow gr^{i} F$ they construct a functor to the category of complexes $DF_{bete}\rightarrow C^b(\cal P)$ and show that it is an equivalence.

  3. Forgetting of the filtration on $DF_{bete}$ translates to a functor $C^b(\cal P)\rightarrow D^b(\cal A)$. The derived functor of this is the realization functor.

  • $\begingroup$ What's the realization functor? $\endgroup$ Mar 22 '12 at 20:13
  • $\begingroup$ Well, in the above situation there is a canonical functor real, extending the inclusion of C in D^b(A).This is not obvious and wrong for an arbitrary triangulated category instead of D^b(A). You can read about it for example in Beilinson, Bernstein, Deligne "Faisceaux pervers" Asterisque 100. $\endgroup$ Mar 23 '12 at 8:05
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    $\begingroup$ @Jan I see. I had guessed your answer. There's no 'canonical' functor extending the inclusion of the heart unless you enrich $\mathcal{D}^{b}(\mathcal{A})$ with extra structure. The problem is canonicity. Even more, the Ext-algebra of an object in a triangulated category is just a plain algebra. If you want to equip it with a possibly non-trivial A-infinity structure you need more structure again. The triangulated structure cannot remember more than a little bit of the triple product $m_3$. Once you put extra structure so that everything is well defined, the answer may be 'yes'. $\endgroup$ Mar 23 '12 at 8:52
  • $\begingroup$ I am aware that the definitions of (an isomorphism class of) A infinity structure on the Ext algebras and the realization functor rely on the fact that we don't have arbitrary triangulated categories here but derived categories. Yet I would say everything can be constructed "without choices" from the data of the abelian category A and the t-structure, so my question does make sense. Am I wrong? $\endgroup$ Mar 23 '12 at 11:28
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    $\begingroup$ 'The' realization functor is actually a choice. If you wish to choose it invoking tilting theory, then it's born as a derived functor, compare Rickard, Jeremy "Derived equivalences as derived functors" J. London Math. Soc. (2) 43 (1991), no. 1, 37–48, "Morita theory for derived categories" J. London Math. Soc. (2) 39 (1989), no. 3, 436–456, and Dugger, Daniel; Shipley, Brooke "K-theory and derived equivalences" Duke Math. J. 124 (2004), no. 3, 587–617. $\endgroup$ Mar 23 '12 at 14:56

OK, I still doubt that if you divide out chain homotopies in (1) in the category of filtered complexes, you can get the category of complexes (and not the homotopy category) as a target in (2). I also doubt that you get an equivalence in this way unless you invert something in source and target. Sorry that I'm too lazy to look now at BBD.

Suppose that for any bounded complex $X$ in $\mathcal{P}$ you can construct a filtered complex of injectives $F$ in $\mathcal{A}$ such that each $F^{n}/F^{n+1}$ is an injective object in $\mathcal{P}$ and $gr^{\ast}(F)=F^{\ast}/F^{\ast+1}$ equipped with the differential $\delta$ obtained as in (2) is a bounded below complex quasi-isomorphic to $X$ in $C(\mathcal{P})$. You can probably check in BBD that all this is possible (at leas under reasonable assumptions).

Consider the following zig-zag of DG-algebra morphisms:

$\operatorname{End}_{C(\mathcal{P})}(gr^{\ast}(F),\delta)\leftarrow \operatorname{End}_{\mbox{filtered}}(F)\rightarrow \operatorname{End}_{C(\mathcal{A})}(F)$

By definition, the $\operatorname{Ext}$ algebra of $X$ is the cohomology of the DG-algebra on the left. According to (3), the $\operatorname{Ext}$ algebra of $real(X)$ is the cohomology of the DG-algebra on the right. Moreover, the $A$-infinity structures on these $\operatorname{Ext}$ algebras are obtained from these DG-algebras by the standard transfer procedure, choosing bases of cohomology vector spaces and representing cocycles for the elements in these bases. By definition, the transfer maps are $A$-infinity quasi-isomorphisms between the $\operatorname{Ext}$ $A$-infinity algebras and these DG-algebras.

Your claim in (2) should instead say, I think, that the functor you sketch induces an equivalence after inverting quasi-isomorphisms on the right, and something appropriate on the left (filtered quasi-isos?). Therefore $\leftarrow$ above is a quasi-isomorphism. Moreover, since you assume that $real$ is fully faithful, $\rightarrow$ above is also a quasi-isomorphism. Now you can get your desired $A$-infinity quasi-isomorphism between the $\operatorname{Ext}$ $A$-fininity algebras by inverting and composing.

  • $\begingroup$ Thank you for your answer and sorry that it took me so long to reply. I don't however understand how to construct the map $\leftarrow$. Given a filtered endomorphism which is not closed or not of degree $0$ I don't see how to construct something on the left hand side. $\endgroup$ May 7 '12 at 11:53

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