Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 2

The path algebra $kQ$ is the $k$-vector space with basis all directed paths, including the length-zero path $e_i$ at each vertex. Extend path composition bilinearly, declaring the product zero when the paths cannot be composed. We use the left-module convention: $pq$ means first traverse $q$, then $p$; for an arrow $\rho :i\to j$, $e_j\rho e_i=\rho$.

The constant paths are mutually orthogonal idempotents. For any path $p:i\to j$,Consequently, for the finite vertex set used here,It acts as an identity on every basis path, hence on the entire path algebra. Finiteness of the vertex set is what makes this a genuine element of the algebra; an infinite-vertex path algebra instead has finite vertex sums as local units.

There is one constant path and $r$ loop generators; every other path is a word in those loops. Distinct words are distinct basis paths, so there are no relations among the generators. Thereforethe free associative algebra on $r$ generators, with each loop sent to its corresponding generator. For $r=1$ this is $k[x_1]$; for $r\ge 2$ it is noncommutative. The quiver with one loop is the $r=1$ case of the quiver with r loops.

Number the source vertices $1,\dots ,r$ and the sink $c=r+1$, and write $a_i:i\to c$ for the arrows. The only basis paths are $e_1,\dots ,e_r,e_c,a_1,\dots ,a_r$. Using standard matrix units, defineThe relations $e_c{a}_i=a_i=a_i{e}_i$, orthogonality of the vertex idempotents, and $a_i{a}_j=0$ agree exactly with multiplication of these matrix units. The map is therefore an algebra homomorphism, and its images form a basis of the algebraThis is an explicit isomorphism of the path algebra of the subspace quiver with a $(2r+1)$-dimensional triangular matrix algebra.

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 $T{u}_a$ 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:

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 there are exactly $12$ isomorphism classes. No general classification theorem is needed.