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Let $Q$ and $R$ be two acyclic quivers which differ only in the directions of their arrows (i. e., the underlying undirected graphs are the same).

1. Does there exist an isomorphism of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?

At the moment, I am not even 100% sure about the weaker statement that there exists an equivalence of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$. (My intuition for this is pretty much zero.)

In before reflection functors - they are not isomorphisms. (But yes, it's a nice exercise to see that we can obtain $R$ from $Q$ by a sequence of admissible reflections, where an "admissible reflection" means choosing some sink in $Q$ and switching all the arrows to $Q$. Unfortunately, the corresponding reflection functors may turn some nonzero representations of $Q$ to zero.)

2. Are the representation rings of $Q$ and $R$ isomorphic? The isomorphy of their additive groups follows from Gabriel's theorem, but it is not clear to me what this does to tensor products.

1+2. Does there exist an isomorphism of tensor categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?

3. If some of these answers are No, what if we restrict ourselves to Dynkin quivers?

I am new to quivers, so I'm sorry if this has been already talked over a dozen of times here.

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2. Representations of a quiver do not come with a tensor product. –  Bruce Westbury Mar 12 '10 at 13:02
    
Uhm, maybe this is not standard notation; what I meant was tensoring corresponding vector spaces and tensoring the maps between them. –  darij grinberg Mar 12 '10 at 13:16
    
But yes, this is not the same kind of tensor product as for groups. –  darij grinberg Mar 12 '10 at 13:16
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3 Answers

up vote 11 down vote accepted

The categories are not equivalent. In fact, an acyclic quiver is determined (up to non-unique isomorphism) by the equivalence class of its category of representations. The construction is simple: Find the isomorphism classes of simple objects. These are in bijection with the vertices. For any two simple objects $S$ and $S'$, the number of arrows from vertex $S$ to vertex $S'$ is the dimension of $\mathrm{Ext}^1 (S, S')$. The nonuniqueness is because we need to choose a basis of $\mathrm{Ext}^1 (S, S')$.


There is a lot more to say on this subject, but it doesn't go in the direction your questions are pointing. When the underlying graphs of $Q$ and $R$ are trees, then it is true that the bounded derived categories of $\mathrm{Rep} \ Q$ and $\mathrm{Rep} \ R$ are isomorphic. The basic point here is that, although the reflection functors are not equivalences of categories, the derived reflection functors are equivalences of derived categories. Of course, every Dynkin diagram is a tree, so in particular this is true for Dynkin diagrams.

There is a version of this for non-trees, but I don't know a reference for it nor the exact statement and I don't want to hunt one down until I know whether derived categories are something that you love or that you fear. The proof, of course, is completely different.

Here is a result of Kac you might be happier with. Let our ground field be a finite field $\mathbb{F}_q$ and fix a dimension vector $d$. Then the number of isomorphism classes of representations of $Q$ and $R$, of dimension $d$, over $\mathbb{F}_q$, is the same. (Infinite root systems, representations of graphs and invariant theory, Theorem 1.)

Morally, one wants to work over an arbitrary field. The statement then is a statement about the stacks of $Q$ and $R$ representations of dimension $d$. But, again, to formulate this you need to know about a lot of machinery: stacks, perverse sheaves, derived categories again.


In short, the categories are not equivalent. There are very close relations between them, but the best formulations of those relations use sophisticated category theory. You can often see shadows of these relations by counting points over $\mathbb{F}_q$.

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That's a lot of stuff I want to understand one day. As for derived category, all I know is the definition. Will try to prove that the derived reflection functors are isomorphisms. –  darij grinberg Mar 12 '10 at 14:10
    
Hmm, could you please be more precise about what "the derived reflection functors are equivalences of derived categories" means? I am understanding this as: "if A -> B -> C is a complex of representations of a quiver Q, and i is a sink of Q, then Ker (F_i^+ B -> F_i^+ C) / Im (F_i^+ A -> F_i^+ B) is isomorphic to Ker (B -> C) / Im (A -> B)". But I can't confirm this. I'd be glad to know whether I'm trying to prove the right thing at all. –  darij grinberg Mar 12 '10 at 21:10
    
I think you may need to take some time to learn more about derived categories if you want to understand this statement. The derived reflection functor is a functor between two derived categories, and is an equivalence in the usual sense: it has an inverse up to isomorphism of functors. –  Ben Webster Mar 12 '10 at 22:27
    
And, as you might guess, that inverse is the opposite derived reflection functor. A big hint here is to understand that Rep Q is hereditary. A consequence is that every object in the dervied category is a direct sum of shifts of objects from Rep Q. –  David Speyer Mar 13 '10 at 0:30
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As you now know, there is no equivalence of the categories. However, the (positive) reflection functors that you mentioned get you about as close as you'll ever come to finding one. The reflection functor $F_x^+$ associated with a particular sink, x, takes the category of representations of the original quiver to the category of representations of the reoriented quiver, where the sink has become a source. The reflection functor $F_x^+$ kills precisely one nonzero representation, the simple one associated with the vertex x.

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No, there cannot be an equivalence of categories, not even for Dynkins.

Take .-->.-->.-->. versus .-->.<--.-->.

the simples corresponding to the vertices have to be mapped to themselves because of the two outer arrows, but then if there was an equivalence the 2-dml extension between the two inner simples on the left has to be mapped to an extension between the very same simples on the right, so the direction of the arrow cannot change! In fact one can even suffice with

.-->.-->. versus .-->.<--.

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Thanks, but I don't quite see why simples must go to themselves. I only see why they go to simples. –  darij grinberg Mar 12 '10 at 13:55
    
Nevermind. I see what you want to say: if we make a directed graph where the vertices are the simple representations and they're connected by arrows whenever their Ext is nonzero, then this directed graph is simply our quiver. So if the representation categories are equivalent, then the quivers must be isomorphic (as directed graphs). –  darij grinberg Mar 12 '10 at 14:02
    
that's it. or use that source-vertices must map to sources and sinks to sinks. –  lieven lebruyn Mar 12 '10 at 14:07
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