Let $1\to \cdots\to n$ be a linear quiver of length $n$. Let $\mathbf{d}=(d_1,\dots,d_n)$ be a dimension vector. It's well known (for example, by Gabriel's theorem, but also by basic linear algebra) that the orbits of $G=\prod GL(\mathbb{C}^{d_i})$ on the representation space $E=\oplus \mathrm{Hom}(\mathbb{C}^{d_i},\mathbb{C}^{d_{i+1}})$ are classified by ways of writing $\mathbf{d}$ as a sum of segments, that is, vectors of the form $(0,\dots, 0,1,\dots, 1,0,\dots, 0)$.

It's well known that each one of these orbits has a resolution of singularities given by choosing sequences $(i_1,\dots, i_d)\in [1,n]$ and $(a_1,\dots, a_d)\in \mathbb{Z}_{> 0}^d$, and looking at pairs of points in $E$, and flags $\cdots \subset V^j_{k-1}\subset V^j_k\subset\cdots \subset \mathbb{C}^{d_j}$ such that $\dim V^j_k$ is the sum $\sum_{p\leq k,i_p=j}a_p$. That is, this flag only jumps in one of the spaces at each step, $i_p$ tells us which one, and the $a_p$ by how much.

Sometimes these maps are isomorphisms to the orbit closure, sometimes they are small resolutions, sometimes they are semi-small, but in general none of these properties hold.

What I would like to do is find good examples (or at least a good method for finding examples) where none of these resolutions are small.

Now, I do know of one way of finding such examples: if a resolution is small, then the intersection cohomology of the orbit closure is the same as the cohomology of the resolution. Since the letter is easily seen to be torsion-free, we can prove that an orbit can't have a small resolution of this form if it has torsion in its intersection cohomology. Such examples do exist, but they are big, not easy to find, and I strongly suspect that this lack of small resolutions is much more general. I would like to have some other tools in my pocket for figuring this out.

**EDIT**: Let me make this more concrete. For $n=1$, the question is trivial (every orbit is a point). For $n=2$, a quiver rep is just a map $\mathbb{C}^{d_1}\to \mathbb{C}^{d_2}$, and the orbits are classified by rank $r$. There is always a small resolution: if $d_1\leq d_2$, it corresponds to the sequence $(2,1,2)$ for $(d_1-r,d_2,r)$ and if $d_2\leq d_1$, to $(1,2,1)$ for $(r,d_1,d_2-r)$.

I suspect very strongly that there is an orbit without small resolution for $n=3$, but don't know how to find it or prove it doesn't exist.