Consider a tame quiver $Q$ whose underlying undirected graph is connected. So that undirected graph is one of the extended, simply laced Dynkin graphs; it's either $\tilde A_n$ for some $n\ge 1$ or $\tilde D_n$ for some $n\ge 4$ or $\tilde E_6$ or $\tilde E_7$ or $\tilde E_8$. And consider some vector $d$ of nonnegative integers. The general theory of Kac and others tells us that $Q$ has a unique isomorphism class of indecomposable representations of dimension $d$ just when $d$ is a real root of the associated root system. Furthermore, when $d$ is such a real root, we always have $q(d) = 1$, where $q$ is the Tits quadratic form associated with the quiver $Q$.
But what about the reverse implication? Let $d$ be any vector of nonnegative integers that satisfies $q(d) = 1$. Is it always the case that $d$ is a real root? That is, does $d$ always lie in the Weyl orbit of some simple root?
If the undirected graph underlying $Q$ is the extended Dynkin graph $\tilde A_n$ for some $n\ge 1$, then the reverse implication does hold. In these cases, the only nonnegative integral vectors $d$ with $q(d) = 1$ are the cyclic shifts of $(m, m, \dots,m,m+1,m+1,\dots,m+1)$, for some $m\ge 0$; that is, $d$ assigns dimension $m$ to a contiguous substring of the linear spaces, going around the cycle, and dimension $m+1$ to the rest. In their text Representations of Finite-Dimensional Algebras, Gabriel and Roiter explicitly construct indecomposable representations for every dimension vector of this form, and it follows that all such vectors are real roots.
But does the reverse implication hold also for the extended Dynkin graphs of types D and E? That is my question.
Note that the assumption of connectedness is critical. If the undirected graph underlying the quiver $Q$ consisted of, say, a 2-cycle and an isolated vertex, then the dimension vector $d = (1, 1, 1)$ would satisfy $q(d) = 1$, but would not be a real root.