Interval representations of an equioriented three-vertex quiver ($I[a,b],1\le a\le b\le 3$)

For $1\to 2\to 3$, $I[a,b]$ has $k$ on the interval of vertices $a,\dots ,b$, zero elsewhere and identity maps along its internal arrows. These six representations are exactly the indecomposables. To decompose $V_1\stackrel{f}{\to }{V}_2\stackrel{g}{\to }{V}_3$, split $\mathrm{im}f\cap \ker g$, its complements inside each of $\mathrm{im}f$ and $\ker g$, and a complement to their sum. Lift the image basis through $f$ and transport the complementary nonkernel basis through $g$. This yields interval blocks directly, without the Gabriel theorem.

For fixed dimensions $(d_1,d_2,d_3)$, a pair $V_1\stackrel{f}{\to }{V}_2\stackrel{g}{\to }{V}_3$ is classified up to vertexwise change of basis by its three ranks $r,s,t$. The multiplicities of its interval representations of an equioriented three-vertex quiver are $m_{13}=t$, $m_{12}=r-t$, $m_{23}=s-t$, $m_{11}=d_1-r$, $m_{22}=d_2-r-s+t$ and $m_{33}=d_3-s$. Nonnegativity is precisely the existence condition. For dimensions $(2,2,2)$ there are ten orbits, of which $(2,2,2)$ is open; its boundary has nine orbits. The two cases $(1,1,0)$ and $(1,1,1)$ show why the individual arrow ranks are insufficient.