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Let $M$ be an object in an $k$-linear abelian category with enough projectives. Then one can construct an $A_\infty$-structure on the Ext algebra $$Ext^\bullet(M,M)$$ as follows: One chooses projective resolution $P\rightarrow M$ and forms the Hom complex $$Hom^\bullet (P,P)$$ Now the cohomology of this complex is the Ext algebra and in the case where $k$ is a field, one can choose a "homotopy retraction" $Ext^\bullet(M,M)\rightarrow Hom^\bullet (P,P)$ and transfer the dg-algebra structure on $Hom^\bullet (P,P)$ along it.

My questions are:

  1. Why does this construction (up to $A_\infty$-isomorphism) not depend on a choice of projective resolution?
  2. One could try the same thing with an injective resolution, why is the result still the same?
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In the construction of the Ext-algebra, don't you form the Hom complex $Hom^\bullet (P,M)$ and then take homology? It seems like $Hom^\bullet (P,P)$ would be a double complex... –  MTS Mar 12 '12 at 17:04
    
I guess one can do both. I mean the total complex of the double complex $Hom(P,P)$. On $Hom(P,M)$ the dg-algebra structure is not so obvious to me, thats why I have choosen $Hom(P,P)$. –  Jan Weidner Mar 12 '12 at 17:24
    
Ah, I see. Thanks for the clarification. I don't know the answer, but it's an interesting question for sure. –  MTS Mar 12 '12 at 18:04
7  
There's a very basic lemma on homological algebra which says that projective resolutions are essentially unique, in the sense that two of them are homotopy equivalent, therefore their endomorphism DG-algebras become homotopy equivalent too (through a zig-zag of equivalences, though). The same holds for injective resolutions by the same reason that $Ext$ can be computed both kinds of resolutions. –  Fernando Muro Mar 12 '12 at 18:18

1 Answer 1

up vote 6 down vote accepted

Let $P\to M$ be a projective resolution of $M$ and $M\to J$ be an injective resolution. Consider the composition of the morphisms of complexes $P\to M\to J$ and set $C$ to be the cone of the morphism $P\to J$. Then the complex $C$ has a subcomplex isomorphic to $J$ with the quotient complex isomorphic to $P[1]$. Moreover, the short exact sequence of complexes $J\to C\to P[1]$ splits as a short exact sequence of graded objects in your abelian category (i.e., after the differentials are forgotten).

Consider the subcomplex $D$ in the complex $Hom^\bullet(C,C)$ formed by all the homogeneous morphisms of graded objects $C\to C$ taking $J\subset C$ to $J\subset C$. This condition on morphisms $C\to C$ is preserved by the composition, so $D$ is a DG-algebra over $k$. The restriction of morphisms $C\to C$ to the subcomplex $J$ defines a DG-algebra morphism $D\to Hom^\bullet(J,J)$; and the passage to the induced morphism of the quotient complexes defines a DG-algebra morphism $D\to Hom^\bullet(P[1],P[1])\simeq Hom^\bullet(P,P)$.

It is claimed that both these DG-algebra morphisms are quasi-isomorphisms. E.g., the morphism $D\to Hom^\bullet(J,J)$ is surjective and its kernel is the complex $Hom^\bullet(P[1],C)$, which is acyclic as $P$ is a complex of projectives (bounded from above) and $C$ is acyclic. The argument for the second DG-algebra morphism is similar.

Now, being quasi-isomorphic DG-algebras, $Hom^\bullet(P,P)$ and $Hom^\bullet(J,J)$ have $A_\infty$-isomorphic minimal $A_\infty$-models. The proof of the independence of these from the choice of the resolution $P$ and/or $J$ is analogous.

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Thank you for your explanation! Do you know, if one could take more generally instead of a projective/injective resolution an end-acyclic resolution in order to compute the $A_\infty$ structure? Here a complex X is called end-acyclic, if its shifted endomorphisms in the homotopy category and in the derived category coincide: Hot(X,X[n])=D(X,X[n]) for all n –  Jan Weidner Mar 13 '12 at 18:23
    
Yes, I think one can: if X is an end-acyclic complex, P is a bounded above complex of projectives, and P -> X is a quasi-isomorphism, setting C to be the cone of the latter and arguing as above, one only needs the complexes Hom(P,C) and Hom(C,X) to be acyclic for the argument to hold. As H^*(Hom(P,P)) = H*(Hom(P,X)) = D(X,X[]) = Hot(X,X[]) = H^*(Hom(X,X)), this seems to be true. –  Leonid Positselski Mar 13 '12 at 19:17

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