Recovering an abelian category from the Ext of its simple objects

Question

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

Answer 1

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 $\operatorname{mod}k[x]/(x^{4})$.

Answer 2

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.