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The AR quiver of the regular representations of an affine quiver consists of infinitely many "tubes". A tube of rank $r$ has $r$ modules on what you call the border. Let me number them $B_1, B_2, \dot …

The quiver given in the question has five simple modules, six which correspond to a single arrow, and the remaining representations have support as follows: 123, 124, 125, 235, 345, 1235, 12235 (note …

For other tame quivers with no relations over an algebraically closed ground field, the situation is slightly worse: the natural indexing set for the representations whose dimension vector is the null …

Let $Q$ be a quiver, and let $d=(d_i)$ be a dimension vector. We can consider Rep($Q,d$), the affine space consisting of representations of $Q$ with dimension vector $d$. The general linear $GL(d)= \p …

There is an example in section 3.4 of Riedtmann's paper "Degenerations for representations of quivers with relations", Ann. sci. Éc. Norm. Sup. v. 19 (1986), 275-301. …

This answer perhaps says things that are all obvious to the OP. $\textrm{Ext}^i(A,B)$ is a vector space over the ground field, so its cardinality is $q^d$ where $d$ is the dimension of the vector sp …

Let $Q$ be a quiver without loops (cycles of length 1). Kac proved that if $K$ is algebraically closed, the dimension vectors of indecomposable representations of $Q$ over $K$ are exactly the positiv …

This shouldn't be surprising, though: path algebras of quivers have global dimension one, so you shouldn't expect their derived categories to agree with derived categories of sheaves on higher-dimensional …

Reflection functors take you between categories of representations of different orientations of the same quiver and preserve indecomposability (up to the fact that a reflection functor destroys a sing …

As a module over $kQ$, a finite-type preprojective algebra is a direct sum of each of the indecomposable $kQ$-modules once. Thus, the total dimension is the sum over all positive roots of the height …

Edited to be more of an answer. But, unfortunately, to be quite wrong. I apologize for having been borne away by my enthusiasm. The picture of the free modular lattice above shows thirty elements. …

Dag has already answered the case where the quiver is finite and acyclic, and given a conjecture in the case that cycles are allowed. I will prove his conjecture. Suppose we have an element $x$ of …

Two references, neither of which exactly addresses your question, are as follows: Ringel, Claus Michael Exceptional modules are tree modules. Linear Algebra Appl. 275/276 (1998), 471–493. In this p …

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 …

Let me first point out, as it confused me initially, that one expects a Dynkin diagram's worth of these embeddings, and the quiver involved is affine. So we can't have a bijection between vertices of …