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(Crossposted from math.SE.)

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). Does it then follow that $C, D$ are 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, Mariano's answer on math.SE shows that the result is true if $C, D$ are either both taken to be free categories on graphs or both taken to be posets by basic results in the representation theory of quivers. (This and idle curiosity are the main motivation for the question.)

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Sorry for the stupid question, but why does $\mathrm{Fun}(G,\mathrm{Vect})$, which is the category of representations of $G$ over $k$, only depend on the number of conjugacy classes of $G$? If this was true, representation theory would not be such a big subject? –  Martin Brandenburg Jan 20 '12 at 7:34
    
Side remark: Sefi Ladkani has studied this type of (derived) Morita equivalence for posets; you can find all his papers on his homepage. –  Martin Brandenburg Jan 20 '12 at 7:38
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@Martin: he is only looking at the abelian structure, not the tensor structure. –  S. Carnahan Jan 20 '12 at 8:38
    
Ok. Does the following work? $\mathrm{Fun}(G,\mathrm{Vect})$ is the Ind-category of $\mathrm{Fun}(G,\mathrm{f.d.Vect})$. If $V_1,...,V_n$ are the irreducible representations of $G$, then $\mathrm{Fun}(G,\mathrm{f.d.Vect}) \to \mathrm{f.d.Vect}^n, V \mapsto (\mathrm{Hom}(V_i,V))_i$ is an equivalence of categories. Thus the whole thing only depends on $n$. –  Martin Brandenburg Jan 20 '12 at 13:48
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To each finite category C you can associate a category algebra kC with basis the arrows C and tbe product extending that of C by making undefined compositions 0. In the paper of Mitchell mentioned by Tom it is proved the category of f.d. kC-modules is equivalent to the category Func(C,Vect). Now if C is acyclic, then the algebra kC is basic (its semisimple quotient is a product |Obj(C)| copies of k). So by classical theory it follows kC and kD are Morita equivalent iff they are iso. So your question really is does $kC\cong kD$ is $C\cong D$. Mariano writes this in the quiver language. –  Benjamin Steinberg Jan 20 '12 at 18:04
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1 Answer

up vote 2 down vote accepted

The answer to your question is "no". A counterexample is given in this paper by Leroux in Example 1.6.

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Nb, Leroux's incidence algebra for a finite category is what I called the category algebra in my comment above. Leroux uses Mitchell's terminology (attributed to Rota) and calls an acyclic category a Delta. –  Benjamin Steinberg Jan 20 '12 at 18:47
    
PS. I am not on MSE so maybe link to this answer there. –  Benjamin Steinberg Jan 20 '12 at 18:48
    
Leroux proves the free category and poset case in his paper without explicitly using quivers. –  Benjamin Steinberg Jan 20 '12 at 18:50
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