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It is well-known that there are finitely many indecomposable module over the preprojective algebra associated to a quiver $Q$ if and only if $Q=A_2,A_3,A_4$ and tame type for $A_5$ and wild for others. Now let say $Q$ is a ADE Dynkin diagram, and $V$ be an indecomposable preprojective algebra. What can we say about the dimension vector $|V|$ about $V$, does $|V| \in \mathbb{N}_0^I$ have an explicit bound?

I think for $Q=A_5$, for any indecomposable module the dimension vector of which is bounded by $(1,2,2,2,1)$. What about other types?

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    $\begingroup$ Your question seems to have typos which mean that I'm not 100% sure what you want to ask, but for every finite dimensional algebra that does not have finite representation type, there is no bound on the dimension of indecomposable modules. This is the first Brauer-Thrall conjecture. $\endgroup$ – Jeremy Rickard May 25 '18 at 8:41
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    $\begingroup$ Your first sentence is incorrect. The preprojective algebra of type $D_4$ is also tame. See GLS for a description of the root system in type $A_5$. $\endgroup$ – Jabby May 25 '18 at 8:58
  • $\begingroup$ The errors reflect the wrong belief I held before asking this question. Thank you for correcting me. $\endgroup$ – Ben May 25 '18 at 12:57
  • $\begingroup$ @Jeremy I think your comment answered my question. Can u post it as an answer? I will accept it. $\endgroup$ – Ben May 25 '18 at 12:57
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For every finite dimensional algebra that does not have finite representation type, there is no bound on the dimension of indecomposable modules. This is known as the first Brauer-Thrall conjecture (now a theorem).

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