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Let a quiver be $Q=(Q_0, Q_1, s, t)$, where $Q_0$ is $\{1, 2\}$. The quiver has only one arrow: $\alpha: 2 \to 2$. What are all representations of $Q$? Thank you very much.

In my original question,the quiver has only one arrow from 2 to 2, not 1 to 2. But it was edited. I changed the question to the original one.

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So your quiver is the union of two quivers, one with one vertex and no arrows and one with a loop? –  Julian Kuelshammer Apr 24 '13 at 16:02
    
I have edited my answer to reflect the rollback of the edit (I think I might even have posted it before the edit, but saw a quiver with two vertices and one arrow and assumed it was $\mathsf{A}_2$). –  Matt Pressland Apr 24 '13 at 16:45
    
(Oops, it was actually my edit! Sorry about that.) –  Matt Pressland Apr 24 '13 at 21:33
    
Oops! I should delete my answer instead. –  Aaron Apr 24 '13 at 22:11
    
@JulianKuelshammer, yes, the quiver is the union of two quivers. –  LJR Apr 25 '13 at 1:29
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up vote 3 down vote accepted

Presumably you mean representations up to isomorphism, else the question is almost vacuous. You can think about representations of the two components separately; the representations of a single vertex with no arrows are just vector spaces up to isomorphism. For the loop on one vertex, you should (if the field is algebraically closed) think about Jordan normal forms; the classification of endomorphisms of vector spaces is just the classification of square matrices up to similarity (where $A$ and $B$ are similar if there exists an invertible $P$ with $B=P^{-1}AP$). (This is still true if the field is not algebraically closed, but the classification isn't given by Jordan normal form).

Then representations of the quiver are just direct sums of indecomposable representations of each of the two components.

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One should note that for the classification to be in terms of Jordan normal form you assume an algebraically closed ground field. –  Julian Kuelshammer Apr 24 '13 at 21:24
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I do indeed (I'm allergic to non-algebraically closed fields). Added to body of answer... –  Matt Pressland Apr 24 '13 at 21:31
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Another point probably worth to add is that the path algebra $kQ \cong k\oplus k[X]$, and the module (representation) category naturally decomposes into direct sums of module category of $k$ (i.e. the category of vector spaces) and that of polynomial algebra $k[X]$. –  Aaron Apr 24 '13 at 22:16
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