= Solution
First split off the kernel of each source-to-sink arrow. Each kernel contributes copies of the simple representation supported at that source. The remaining arrows are injective, so identify the three source spaces with subspaces $U_1,U_2,U_3$ of the sink space $V$. We now give an elementary <three-subspace decomposition>.
Split off the common intersection $U_1\cap U_2\cap U_3$: any projection onto it preserves all three subspaces. Next split off each pairwise intersection. For example, once the triple intersection has gone, $W=U_1\cap U_2$ is disjoint from $U_3$; a projection onto $W$ that kills $U_3$ preserves $U_1,U_2,U_3$. This extracts blocks whose membership is exactly $\{1,2\}$. Repeating for the other pairs leaves pairwise disjoint subspaces.
For each $i$, choose a complement $W_i$ to $U_i\cap(U_j+U_\ell)$ in $U_i$. A projection onto $W_i$ killing $U_j+U_\ell$ preserves the triple, so these complements split off as blocks of membership exactly $\{i\}$. In the remaining triple, every $U_i$ lies in the sum of the other two and all pairwise intersections are zero. Thus the sum is $U_1\oplus U_2$, and both projections of $U_3$ onto $U_1,U_2$ are isomorphisms. Hence $U_3$ is the graph of an isomorphism $T:U_1\to U_2$. Choose a basis $u_a$ of $U_1$ and the basis $Tu_a$ of $U_2$; this graph decomposes into two-dimensional blocks with three distinct lines. A complement to $U_1+U_2+U_3$ in $V$ contributes sink-only simple blocks.
The complete list of indecomposable blocks is therefore:
* \b[Three source simples], with zero sink space.
* \b[Eight blocks with one-dimensional sink], one for each subset $S\subseteq\{1,2,3\}$: the source space is $k$ for $i\in S$ and zero otherwise, and every nonzero arrow is the identity.
* \b[One exceptional block], with all three source spaces $k$, sink $k^2$, and arrows $1\mapsto e_1$, $1\mapsto e_2$, $1\mapsto e_1+e_2$.
Each of the first eleven blocks has <endomorphism ring> $k$. For the <exceptional indecomposable of the three-subspace quiver>, an endomorphism of $k^2$ preserving the first two lines is diagonal; preserving the third makes its two diagonal entries equal. Its <endomorphism ring> is also $k$, so all twelve are indecomposable. Their <dimension vectors of quiver representations> distinguish them. The decomposition argument proves completeness, and hence \b[there are exactly $12$ isomorphism classes]. No general classification theorem is needed.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-3-three-subspace-exception.png]
{title=Exceptional three-subspace representation with a two-dimensional sink and three distinct image lines}
Back to article page
