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Timeline for Homotopy of quivers

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Jun 15, 2020 at 7:27 history edited CommunityBot
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Dec 25, 2013 at 21:33 comment added Qiaochu Yuan @Johannes: I think you need to equip the category algebra with extra structure to get something like that. For example, because we start with a groupoid above, the category algebra is equipped with an involution extending taking inverses. I don't know what the right notion of Morita equivalence for rings with involution is though.
Dec 25, 2013 at 21:05 comment added Johannes Hahn Oh wait, that's not true in general what I just wrote. But there is a better-than-morita-equivalence somewhere in here. Your theorem about category algebras of groupoids being isomorphic matrix rings over group rings is stronger: The morita equivalence comes from the equivalence to (a disjoint union of) one-object categories but there are a lot of rings morita equivalent to a group ring but not isomorphic to a group ring (modular representation theory is full of examples). I can single out the isomorphism class by hand but is there a general nonsense argument for that?
Dec 25, 2013 at 20:41 comment added Johannes Hahn Thank you very much for your answer. It took me a while to get behind the idea of an "upgrade" to 2-categories. Follow-up question: A general morita equivalence between rings $R\simeq S$ means that there is an isomorphism $R\cong eS^{d\times d}e$ for some $d\in\mathbb{N}$ and some idempotent $e\in S^{d\times d}$. Does it follow from general nonsense that there is an isomorphism $R\cong S^{d'\times d'}$ or $S\cong R^{d''\times d''}$ if $R$ and $S$ are the category algebras of two equivalent $k$-linear categories (with finitely many objects of course) or does this need a separate proof?
Dec 25, 2013 at 19:05 vote accept Johannes Hahn
Nov 21, 2013 at 19:19 comment added Qiaochu Yuan One lesson here is that even if all you care about is Morita equivalence for rings it's convenient to pass through Morita equivalence for linear categories. For example, to show that $R$ and $M_n(R)$ are Morita equivalent it's convenient to pass through the linear category consisting of $n$ objects, all of which are isomorphic, and all of which have endomorphism ring $R$.
Nov 19, 2013 at 21:50 history edited Qiaochu Yuan CC BY-SA 3.0
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Nov 19, 2013 at 21:45 history edited Qiaochu Yuan CC BY-SA 3.0
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Nov 19, 2013 at 21:36 history edited Qiaochu Yuan CC BY-SA 3.0
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Nov 19, 2013 at 21:15 history answered Qiaochu Yuan CC BY-SA 3.0