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Let $A = KQ$, where $Q$ is the quiver $$\begin{array}{ccc} & \alpha & \\ 1 & \rightleftarrows & 2 \\ & \beta& \end{array}$$ are there simple right $A$-modules with dimension $\geq3$?

In generally, how to find all simple modules for the given path algebra, especially the infinitely dimensional case?

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As xymatrix is not working here, you should probably write down that this is the quiver with two vertices 1 and 2 and two arrows, one going $1\to 2$ and one going $2\to 1$. –  Julian Kuelshammer Nov 16 '12 at 14:08
    
Do you assume the field to be algebraically closed? –  Julian Kuelshammer Nov 16 '12 at 15:02
    
Hint: In your example you can use Gauss normal formal to get the identity matrix as one of the matrices and Jordan normal form to get the other matrix to Jordan normal form. From that you easy see "canonical" subrepresentations. –  Julian Kuelshammer Nov 16 '12 at 15:20
    
As you can see from Julian's answer, and also his answer to your previous question math.stackexchange.com/questions/227909/… , it is in general difficult to determine simples of path algbera which are infinite dimension. For finite dimensional ones, standard result says they correspond to putting 1-dim vector space on one vertex...But I guess this is not you want to know? –  Aaron Nov 16 '12 at 19:05
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I forgot to mention this related mathoverflow question: mathoverflow.net/questions/73989/… –  Julian Kuelshammer Nov 18 '12 at 2:21

1 Answer 1

Let me give a partial answer:

1) Your example (in the case that $K$ is assumed to be algebraically closed): Recall that a representation of a quiver is an assignment of a vector space to each vertex and a linear map to each arrow. So in your case $(V,W,f:V\to W, g:W\to V)$. As a first step note that the two linear maps have to be injective, otherwise e.g. for $f$ injective, $(\ker(f),0)$ would be a subrepresentation. Hence by dimension formula you know that $V$ and $W$ are equidimensional and by choosing basis you can assume that they are $k^n$ (assuming you are talking about finite-dimensional representations). Now by choosing bases for $V$ and $W$ respectively you can assume that $f$ is the identity matrix (this is what Gaussian elimination tells you). Now you know that $g$ is similar to a matrix in Jordan normal form. Meaning you can conjugate $g$ (or the matrix of $g$ if you want to be more precise) to a matrix $J$ in Jordan normal form, that has to have one block since otherwise your representation would decompose. But if you conjugate the identity matrix you will only get the identity matrix. This way every indecomposable representation with $f$ and $g$ invertible is isomorphic to $(k^n,k^n,I_n,J_n(\lambda))$, where $I_n$ is the identity matrix and $J_n(\lambda)$ is a Jordan block of size $n$ and eigenvalue $\lambda$. Now this certainly has $(\langle e_1\rangle, \langle e_1\rangle)$ as a subrepresentation, where $e_1$ is just the first standard basis vector. Therefore the only case left is $n=1$ and you are left with $(k,k,1,\lambda)$.

2) The above procedure will give you a simple module for every oriented cycle. So there are always infinitely many simple representations if your field is infinite.

3) You can have more than one parameter, e.g. take the polynomial ring $R$. Then one version of Hilbert Nullstellensatz tells you that the maximal ideals are of the form $\mathfrak{m}=(X_1-\lambda_1,\dots,X_n-\lambda_n)$ and each of them will give you a non-isomorphic simple module $R/\mathfrak{m}$. However in contrast to the situation of classification of the indecomposable representations this won't tell you that the simple modules are not "classifiable" in a certain sense, i.e. wild, as you can see from this example. I'm not sure what I should expect in general.

4) I found a paper from Derksen and Weyman, The combinatorics of quiver representations that gives a possibility to give at least the dimension vectors of the simple representations, see Corollary 6.8 and in particular Example 6.11 which gives simple representations of a certain quiver of arbitrary dimension.

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Thanks, Julian Kuelshammer. I learned sonething new from your answer. If the field is algebraically closed, does the path algebra $A=KQ$ above and the polynomial ring $R$ have infinitely dimensional simple representation? –  Aimin Xu Nov 18 '12 at 0:39
    
I don't know much about infinite-dimensional representations: Here www1.maths.leeds.ac.uk/~ahubery/RHAlgs.pdf it is left as an exercise to the reader that the 2-loop quiver admits an infinite-dimensional simple representation. In contrast there is a theorem by Amitsur and Small stating that for a (commutative) polynomial ring every simple representation is finite-dimensional, see e.g. in this link: books.google.de/… , Theorem 17.6 –  Julian Kuelshammer Nov 18 '12 at 2:25

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