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I'm wondering how to find indecomposable representations of a given quiver explicitely. In particular, I'm interested in the maximal indecomposable representation of $\mathbb{E}_8$(I'm working over $\mathbb{C}$).

Since this branch of mathematics is quite popular, I think representations of at least Dynkin quivers were already calculated but I didn't manage to find these results.

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  • $\begingroup$ This is indeed popular; Dynkin diagrams and only those have been classified. Have you tried Google to begin with? What about Quiver (mathematics) $\endgroup$
    – Alex Degtyarev
    Commented Jun 3, 2014 at 18:08
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    $\begingroup$ Or indeed en.wikipedia.org/wiki/Gabriel%27s_theorem $\endgroup$
    – David Roberts ♦
    Commented Jun 3, 2014 at 22:15
  • $\begingroup$ @AlexDegtyarev: There is also a complete classification of indecomposable representations of affine quivers. $\endgroup$
    – Hugh Thomas
    Commented Jun 24, 2014 at 2:53

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Neither of the references linked in the comments seem to solve the OP's question. Gabriel's theorem says that the indecomposables correspond to positive roots. The way this correspondence works is as follows: you take the positive root you are interested in, and express it as a sum of simple roots (with multiplicities). The multiplicity of the simple root corresponding to node $i$ tells you the dimension of the vector space at node $i$.

However, this still doesn't explicitly construct a representation! All it tells you is the dimension of the underlying vector spaces, without telling you about the linear maps between them. A crucial piece of additional information is that in Dynkin type, if there is an indecomposable representation with dimension vector $\vec d$, then a generic representation of that dimension vector will be indecomposable. So, one answer is to say "take a generic representation of that dimension vector".

For more detail, I suggest the book of Gabriel and Roiter. Its section on hereditary algebras (several chapters in the middle) can be read without reference to what comes before or after, and goes into a lot of detail.

Edited to add: oh, another thing: for Dynkin quivers, there is a way to construct any indecomposable by applying a sequence of "reflection functors" starting from a simple indecomposable, but the sequence of reflections one would have to apply to get to the highest root of $E_8$ is long enough that it probably wouldn't be especially useful to work this out.

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  • $\begingroup$ A quick addition. As Hugh mentions, if $d$ is a positive root for a Dynkin quiver $Q$, then the general representation of dimension vector $d$ is indecomposable. So a reasonably efficient way of finding explicit matrices would be to make some suitably generic choices and then computing the endomorphism ring. If you get that the endomorphism ring is the base field, then you chose wisely. $\endgroup$
    – Andrew Hubery
    Commented Aug 6 at 17:19
  • $\begingroup$ A more exact solution would be take the projective (or injective) for the central vertex, and then repeatedly compute the (inverse) Auslander-Reiten translate. This construction is detailed in many places. $\endgroup$
    – Andrew Hubery
    Commented Aug 6 at 17:23
  • $\begingroup$ @AndrewHubery How do you know the representation corresponding to the highest root will be in the AR translation orbit of the projective at the central vertex? $\endgroup$
    – Hugh Thomas
    Commented Aug 7 at 5:48
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    $\begingroup$ For $E_8$ you can argue as follows. Let $\theta$ be the largest root. This is fixed under all but one reflection, say $s$ corresponding to the leaf on the longest arm, and $s(\theta)=\theta-e$. If $c$ is the Coxeter transformation, then one of $c^\pm(\theta)$ equals $\theta-e$. Choosing one, say $c$, we can then compute $\dim\mathrm{Hom}(\tau X,X)$ using the Euler form, where $X$ is indecomposable of dimension vector $\theta$. This has dimension 2, so the Auslander-Reiten sequence has three middle terms. $\endgroup$
    – Andrew Hubery
    Commented Aug 7 at 6:30
  • $\begingroup$ @AndrewHubery very cool, thanks! $\endgroup$
    – Hugh Thomas
    Commented Aug 7 at 9:03

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