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Call a category acyclic if only the identity morphisms are invertible and the endomorphism monoid of every object is trivial. Let $C, D$ be two finite acyclic categories. Suppose that they are Morita equivalent in the sense that the abelian categories $\text{Fun}(C, \text{Vect})$ and $\text{Fun}(D, \text{Vect})$ are equivalent (where $\text{Vect}$ is the category of vector spaces over a field $k$, say algebraically closed of characteristic zero). Are $C, D$ equivalent? (If so, can we drop the finiteness condition?)

Without the acyclic condition this is false; for example, if $G$ is a finite group regarded as a one-object category, $\text{Fun}(G, \text{Vect})$ is completely determined by the number of conjugacy classes of $G$, and it is easy to write down pairs of nonisomorphic finite groups with the same number of conjugacy classes (take, for example, any nonabelian group $G$ with $n < |G|$ conjugacy classes and $\mathbb{Z}/n\mathbb{Z}$).

On the other hand, I believe this result is known to be true if $C, D$ are free categories on finite graphs by basic results in the representation theory of quivers, and I believe it's also known to be true if $C, D$ are finite posets.

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2 Answers

Suppose $C$ is finite. Then acyclicity implies that there is a well-determined subgraph $Q_C$ in $C$ which generates $C$ as a category and such that its arrows are irreducible in $C$, that is, they cannot be factorized non-trivially. The category of functors $C\to\mathrm{Vect}$ is the category of modules on a certain quotient $A_C$ of the path algebra on the quiver $Q_C$, and the ideal involved (which is generated by all binomials expressing all relations between the arrows of $Q_C$ in $C$) is admissible in the sense of Gabriel's theorem.

If $C$ and $D$ are two acyclic finite categories with equivalent categories of functors to $\mathrm{Vect}$, then $A_C$ and $A_D$ are Morita equivalent, and by the above observations, they are actually isomorphic.

In particular, the quivers $Q_C$ and $Q_D$ are isomorphic. This immediately gives the case of free categories (and if one of the two is free and they are equivalent, so is the other: freeness is the same thing as global dimension at most $1$, and this is a Morita invariant)

If $C$ and $D$ are posets, then the ideals which present the algebras $A_C$ and $A_D$ are uniquely determined, so in that case, the categories $C$ and $D$ are isomorphic.

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up vote 2 down vote accepted

On MO, Benjamin Steinberg links to a paper of Leroux with a counterexample.

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