Let $\Gamma$ be a finite subgroup of $SL_2({\mathbb C})$, and $Q$ the set of its irreducible representations. McKay makes $Q$ into a directed graph by having $V \to W$ if $W \leq V \otimes {\mathbb C}^2$, where the latter comes from the natural action of $\Gamma$ on ${\mathbb C}^2$. (But since ${\mathbb C}^2 \otimes {\mathbb C}^2$ has an $SL_2$-invariant hence $\Gamma$-invariant vector, the directed graph is actually undirected: each edge comes with its reverse.) In this way we get a graph with a vector space at each edge.

McKay observes that the graphs so arising are exactly the simply-laced affine Dynkin diagrams, with the trivial rep as the affine node. (In particular, the extra symmetry of the affine diagram comes here from $(\Gamma / \Gamma')^*$, which is therefore identifiable with $Z(G)$, for $G$ the corresponding simply-connected Lie group. I wonder if there's some larger correspondence there... but that's not my question.)

If we toss that node, and orient the edges (i.e. throw out half), we can look at the "roots", or indecomposable representations, of the resulting quiver. McKay observes further (in effect) that the largest such quiver representation has the same-dimensional vector spaces as in the first construction. But now, since it's a quiver representation, there are maps between the spaces. So my question:

In McKay's construction, the vertices of a Dynkin diagram are labeled by nontrivial irreps $\{V\}$ of $\Gamma$. Given an orientation on the diagram and an edge $V \to W$, is there a natural linear map $V \to W$, such that the result is the largest indecomposable quiver representation?

Obviously these maps aren't $\Gamma$-equivariant. The natural map is $V \otimes {\mathbb C}^2 \to W$, so maybe these other maps correspond to choosing a vector, or a list of vectors, in ${\mathbb C}^2$. So a more specific version of the question:

If $\vec x$ is a generic vector in ${\mathbb C}^2$, e.g. with no $\Gamma$-stabilizer, do the resulting composite maps $V \cong V \otimes \vec x \hookrightarrow V \otimes {\mathbb C}^2 \twoheadrightarrow W$ give the largest indecomposable? If so, what if $\vec x$ isn't generic?

Feel free to add tags; I couldn't think of anything other than rt.representation-theory.