Closures of orbits in the space of representations of a quiver

Question

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 $GL(d)= \prod_i GL(d_i)$ acts naturally on Rep($Q,d$) in such a way that its orbits are the isomorphism classes of representations of $Q$ with dimension vector $d$.

Let $X$ be such a representation, and consider the Zariski closure of the orbit corresponding to the representations isomorphic to $X$.

There is a simple argument, given in Kirillov's book "Quiver representations and quiver varieties" which shows that if we have a filtration of $X$ as $$X=X_r > X_{r-1} > \dots > X_0 = 0$$ then the orbit corresponding to $\bigoplus X_i/X_{i-1}$ is contained in the Zariski closure of the orbit of $X$.

Kirillov also gives an argument, citing Kempf, that if the orbit corresponding to $Z$ is closed and is contained in the closure of the orbit corresponding to $X$, then the converse holds, i.e., $Z$ is the direct sum of the subquotients of a filtration of $X$.

My question is: what if we have some orbit corresponding to $Z$ contained in the closure of the orbit corresponding to $X$, but the orbit corresponding to $Z$ is not closed. Is it necessarily isomorphic to the direct sum of the subquotients of some filtration of $X$?

Answer 1

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.

The quiver has vertices 1 and 2, with an arrow from 1 to 2 and a loop at 2. The dimension vector is $d=(1,2)$. In $X$, the linear transformation associated to the loop is a single nilpotent Jordan block $N$, and the linear transformation associated to the arrow sends the generator at 1 to an element which is not in the kernel of $N$.

In $Z$, the linear transformation associated to the loop is the same, but the arrow sends the generator at 1 to the kernel of $N$.

It isn't hard to show that $Z$ is in the closure of the orbit of $X$.

$Z$ isn't a direct sum in the way that I asked for because in fact it is indecomposable.