I am studying a category of representations of an infinite-dimensional Lie algebra, which is an abelian category. Unfortunately, the category does not have enough injectives/projectives. I would like to find some references to see how one computes the Ext-groups in this kind of situations.

More precisely, let $R$ be a unital ring (even a unital associative $\mathbb{K}$-algebra over a field $\mathbb{K}$). For an abelian subcategory $\mathcal{A}$ of the category of unitary left $R$-modules, possibly without enough injectives/projectives, how do I compute $\text{Ext}_\mathcal{A}^n\left(M,N\right)$ for each $M,N\in\mathcal{A}$ and $n\in\mathbb{N}$ (it is especially important to consider the case where $M$ and $N$ are simple $\mathcal{A}$-modules)? Does it help if $\mathcal{A}$ is assumed to be a full subcategory? Does it help if there exists duality in $\mathcal{A}$, i.e., an exact contravariant functor $(\_)^\vee:\mathcal{A}\to\mathcal{A}$ such that $\left(M^\vee\right)^\vee$ is naturally isomorphic to $M$ for all $M\in\mathcal{A}$? Does it help if $\mathcal{A}$ is the universal enveloping algebra of a Lie algebra (potentially infinite-dimensional, but countable-dimensional simple Lie algebras are of particular interest), especially if the base field is algebraically closed and of characteristic $0$?

I would like to establish the blocks of this category by studying extensions between two simple objects. I can find some full abelian subcategories of this category which have enough injectives/projectives, but not all simple objects are in such subcategories. So, at this moment, I think I really need to understand how Ext-groups are computed without the assumption that the (abelian) category has enough injectives/projectives.

For those who are curious how one defines Ext-groups without the use of injectives/projectives, please take a look at http://stacks.math.columbia.edu/tag/06XP. Even better, you can read "An Introduction to Homological Algebra" by Charles A. Weibel.

  • 4
    $\begingroup$ How are you defining Ext, if not in terms of projective/injective resolutions? If I remember correctly, once one has defined Ext one can usually compute it in terms of acyclic resolutions, but this presupposes that it has a definition to begin with... $\endgroup$ – Yemon Choi Mar 30 '17 at 15:14
  • 6
    $\begingroup$ @YemonChoi Ext-groups can be defined more generally in terms of Yoneda equivalence (of exact sequences), and it coincides with the definition we regularly use when there are enough injectives/projectives. See stacks.math.columbia.edu/tag/06XP, for example. I have a hard time using this definition to make any meaningful computation. $\endgroup$ – Batominovski Mar 30 '17 at 16:21
  • 1
    $\begingroup$ @YemonChoi Unfortunately Ext-acyclics are projectives/injectives (depending of which variable you are deriving) so acyclic resolutions are not very useful in this context. $\endgroup$ – Denis Nardin Mar 30 '17 at 16:50
  • 1
    $\begingroup$ To the OP: ah, you are using what I have sometimes seen denoted by Yext. (I thought this might be the case but wanted to check.) While I don't know the answer, perhaps some of the people on the category theory mailing list might have suggestions? This kind of foundational issue seems like it should have been considered in the 1960s and 1970s $\endgroup$ – Yemon Choi Mar 30 '17 at 17:29
  • 1
    $\begingroup$ It $A$ is an abelian category, you can turn it into a dg category by taking the dg category of complexes in $A$. If $A$ is a dg category, then $D(A)$ is the homotopy category of a certain dg category (the idempotent completion of the essential image of the Yoneda embedding of $A$ into $Fun(A^{op}, Ch)$, dg-localized with respect to quasi-isomorphisms). In this localized category of modules I think you can compute Exts using semi-free resolutions, but I don't know a reference. Not sure if this makes sense or is overkill? $\endgroup$ – Harrison Chen Mar 30 '17 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.