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According to arXiv:math/0503542 by Suter (Fact 5.1) the sizes of the conjugacy classes of $G$ other than the trivial one are 1, 30, 20, 20, 12, 12, 12, 12. These numbers (plus 1 for the trivial) sum …

Always. If the quiver has a cycle, then the root system is not of finite type (since the quivers corresponding to finite type root systems are trees), and any root system not of finite type has an in …

You may not like this answer -- I am just using (some!) Nakayama algebras. Let $Q$ be the quiver with vertices $1$ to $n$, and arrows from $i$ to $\pi(i)$. Then kill all relations of length 2. No …

Part of what makes this question more interesting outside type $A$ is that the highest root can't be projective or injective. A test case to consider is $D_4$, say with all three arrows pointing to t …

This will certainly work fine in finite type. Folding $Q$ to $Q'$ corresponds to an inclusion of $W'$ into $W$, where the reflections of $W'$ are mapped to products of commuting reflections in $W$. …

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 …

First of all, you have to assume that $A$ is non-semi-simple. For a semi-simple Artin algebra, the representation dimension is defined to be 1. For a non-semi-simple algebra, the representation di …

I am going to give a negative answer for the first question, under a stronger notion of canonicity. The approach I want to take is to consider the poset of tilting modules. They are ordered by in …

It turns out the answer is "no". 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. …

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 a …

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 …

There is another way to relate representations of $Q$ to representations of $Q'$: reflection functors. These are quite easy to describe combinatorially. One downside is that the way they work is by …

I don't think (*) is correctly copied from the paper. The corresponding claim in the paper is that every morphism from an indecomposable summand of $M$ except for the identity morphism from $T$ to $T …

This is an answer to Alexander's combinatorial reformulation of the question in comments to Bruce's answer. dim $V_\lambda$/$n$! is the chance that you will get a standard Young tableau if you assi …

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 …