# Recovering an abelian category from the Ext of its simple objects

Let $$C$$ be an abelian category, assume for simplicity that $$C$$ is enriched over $$Vect_k$$ (vector spaces over $$k$$) for some fixed field $$k$$.

Suppose also that $$C$$ is both Artinian and Noetherian, so that for any object $$X$$ there is a sequence of objects $$0=X_0 \hookrightarrow \ldots \hookrightarrow X_n = X$$ with $$X_i/X_{i-1}$$ simple. Finally suppose that $$C$$ has enough injective/projective objects so that $$\operatorname{Ext}_C$$ can be defined.

Given $$C$$, we build a new category $$S$$, enriched over graded $$k$$-vector spaces, in the following way:

• The objects of $$S$$ are the simple objects of $$C$$
• If $$X,Y\in Ob(S)$$ then $$\operatorname {Hom}_S(X,Y) = \bigoplus_{n\geq0}\operatorname{Ext}^n_C(X,Y)$$
• Compositions of morphisms are defined using the natural maps $$\operatorname{Ext}^n_C(X,Y)\otimes \operatorname{Ext}^m_C(Y,Z)\to\operatorname{Ext}^{n+m}_C(X,Z)$$

My question is: Can we recover $$C$$ from $$S$$ (say up to equivalence)?

Assuming the answer is "yes", I guess that there is an analogue for when $$C$$ is only enriched over $$Ab$$, maybe if we redefine $$S$$ so that $$\operatorname{Hom}_S(X,Y)=\operatorname{Hom}_{D(C)}(X,Y)$$ or something

Here's a counterexample that appears in nature.

Fix a prime $$p$$ and a field $$k$$ of characteristic $$p$$, and let $$G=C_{p^{n}}$$ be a cyclic group of order $$p^{n}$$ (where $$n\geq1$$ if $$p$$ is odd, and $$n\geq2$$ if $$p=2$$).

Then the category $$\operatorname{mod}kG$$ of finitely generated $$kG$$-modules has only one simple module: the trivial module $$k$$.

As graded $$k$$-algebras $$\operatorname{Ext}^{\ast}_{kG}(k,k)=H^{\ast}(G,k)\cong k[s,t]/(s^{2}),$$ with $$s$$ and $$t$$ in degree $$1$$ and $$2$$ respectively, independent of $$n$$.

But $$\operatorname{mod}kG$$ determines $$n$$, as $$p^n$$ is the length of an indecomposable projective module.

So, for example, $$\operatorname{mod}\mathbb{F}_{2}C_{4}$$ and $$\operatorname{mod}\mathbb{F}_{2}C_{8}$$ are not equivalent, but have the same $$\operatorname{Ext}$$-algebra.

Alternatively, this generalizes to looking at $$\operatorname{mod}k[x]/(x^{m})$$ for any field $$k$$ and varying values of $$m\geq3$$.

So a slightly smaller counterexample is $$\operatorname{mod}k[x]/(x^{3})$$ and $$\operatorname{mod}k[x]/(x^{4})$$.

• I presume one can try to use some higher structure in Ext to distinguish them (and even recover the derived categories?) in the cases you mentioned? As the other answer shows, at least in the last two examples, the culprit is indeed the existence of non trivial higher products. I know this works in char zero but I don't know what happens in the modular case. – Pedro Tamaroff May 1 at 15:48
• @PedroTamaroff Yes, if I remember correctly, if $B$ is a finite dimensional algebra, you can recover $\operatorname{mod}B$ from the $\operatorname{Ext}$-algebra of the direct sum of the simple $B$-modules with its natural $A_\infty$-structure. I don't think the characteristic of the field is relevant. – Jeremy Rickard May 1 at 19:31
• @PedroTamaroff since the operad of associative algebras is non-symmetric, assumptions on the ground field are not necessary. Some finite-dimensionality conditions might be useful just to ensure that things do not explode. – Vladimir Dotsenko May 2 at 6:14

This will only be possible when the abelian category $$C$$ is "Koszul" or formal in some sense.

What will always be true is that the bounded derived category $$D^{b}(C)$$ (with its $$dg$$ or stable infinity enhancement) can be recovered from the full $$dg$$-subcateory $$\mathcal S$$ spanned by the simple objects concentrated in degree zero (and $${\rm RHom}$$ between them). In fact $$D^{b}(C) \simeq {\rm Fun}(\mathcal S^{\rm op}, D^{b}(k)),$$ where one direction of the equivalence is by taking $$x \in C \mapsto {\rm RHom}(-,x)|_{\mathcal S}$$.

To recover the abelian cateory $$C$$ from $$S$$, you will need $$\mathcal S$$ to be equivalent to $$S$$-- a formality property. Further, you will need to identify the heart of $$D^{b}(C)$$ inside of $$D^{b}({\rm Fun}(S,k))$$.

A common situation where this is possible is when $$C$$ is the category of modules over a finite Koszul ring.