Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 3 (Representation Theory and Quivers)

Write a finite quiver as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A representation of a quiver assigns a vector space $X_i$ to each vertex and a linear map $f_{\rho }:X_{s(\rho )}\to X_{t(\rho )}$ to each arrow. A quiver representation morphism $h:X\to Y$ is a family satisfying $h_{t(\rho )}{f}_{\rho }=g_{\rho }{h}_{s(\rho )}$.

The path algebra $A=kQ$ has every directed path, including each length-zero path $e_i$, as a basis. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The orthogonal idempotents satisfy $1={\sum }_i{e}_i$.

The path-algebra module equivalence is explicit. From a representation, form $M={⨁}_i{X}_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-module gives $X_i=e_{i}M$ and $f_{\rho }(x)=\rho x$. An $A$-module homomorphism restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.

$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle. For a finite acyclic quiver, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.

Choose only the orientation $1\to 2\to 3$. The interval representations of an equioriented three-vertex quiver $I[a,b]$ have $k$ at vertices $a,\dots ,b$, zero elsewhere, and identity arrows within that interval. The complete list isHere is an elementary proof, without the Gabriel theorem. For $X_1\stackrel{f}{\to }{X}_2\stackrel{g}{\to }{X}_3$, set $K=\mathrm{im}f\cap \ker g$. Choose $F$ complementing $K$ in $\mathrm{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\mathrm{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has endomorphism ring $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.

For a finite acyclic quiver, the arrow ideal of a path algebra $R$ is nilpotent, and $A/R\cong {\prod }_{i\in Q_0}k$. If $S$ is a simple module, its submodule $RS$ is either zero or $S$. The latter would imply $R^{d}S=S$ for every $d$, contradicting nilpotence. Thus $RS=0$, and a simple module over the product of fields is supported at one coordinate. Therefore the simples are exactly $S(i)$, with $k$ at $i$, zero elsewhere, and zero arrows; the vertex $i$ is unique.

A finite-dimensional semisimple module is consequently ${⨁}_{i}S(i{)}^{\oplus n_i}$, where $n_i=\dim X_i$. Its dimension vector of a quiver representation determines its isomorphism class.

For an arbitrary finite quiver, cycles allowed, the vertex projective module of a path algebra is $P(i)=A{e}_i$. Its space at vertex $j$ has basis all paths from $i$ to $j$, and an arrow acts by adjoining that arrow at the end of the path. Its endomorphism ring iswhere $e_{i}A{e}_i$ is spanned by the closed paths based at $i$. The opposite multiplication appears because endomorphisms act by right multiplication.

The evaluation isomorphism for a vertex projective isFor $x\in X_i$, its inverse sends a path $p$ starting at $i$ to $px$. This proves both injectivity and surjectivity, and is natural in $X$. Vertex evaluation is exact, so $P(i)$ is a projective module; alternatively it is a direct summand of the free module $A$.

The closed-path corner of a path algebra is a domain: in a product of two nonzero linear combinations, choose their longest path lengths. Concatenation in that top degree has a unique cut at those lengths, so a product of two nonzero top coefficients cannot cancel. Hence its only idempotents are zero and one. The same is true of the opposite ring, proving $P(i)$ is an indecomposable module, even when cycles make it infinite-dimensional.

For $1\to 2\leftarrow 3$, the paths starting at vertex $1$ are $e_1$ and the arrow $1\to 2$. Thus the displayed $k\stackrel{1}{\to }k\leftarrow 0$ is $P(1)$ and is projective.

The extension group ${\mathrm{Ext}}_Q^1(V,W)$ consists of equivalence classes of short exact sequences $0\to W\to E\to V\to 0$, with the zero class represented by a split sequence and addition given by the Baer sum. The extension complex of quiver representations givesThe printed map has $\gamma (u{)}_{\rho }=u_j{f}_{\rho }-g_{\rho }{u}_i$, so its kernel is ${\mathrm{Hom}}_Q(V,W)$ and its cokernel is ${\mathrm{Ext}}_Q^1(V,W)$. Reversing the overall differential sign changes neither identification.

For dimension vectors $\mathbf{v},\mathbf{w}$, the Ringel form isThe first expression makes its dependence only on the dimension vectors explicit.

For the one-loop representation $V=k$ with loop scalar $\lambda$, $\gamma (u)=u\lambda -\lambda u=0$ on $k$. Both cochain spaces have dimension one, so $\dim {\mathrm{Ext}}_Q^1(V,V)=1$, for every $\lambda$. Concretely, a self-extension has loop matrix $\left(\begin{array}{cc}\lambda & c\\ 0& \lambda \end{array}\right)$, with $c$ the extension parameter.

The four-subspace quiver has four one-dimensional sources and a two-dimensional sink. Its Tits form of a quiver at this dimension vector is $q=4\cdot 1^2+2^2-4(1\cdot 2)=0$.

An endomorphism comprises source scalars $a_1,\dots ,a_4$ and a sink matrix $T$. The first two columns $e_1,e_2$ force $T=\mathrm{diag}(a_1,a_2)$. The third column $e_1+e_2$ then forces $a_1=a_2=a_3$, making $T$ scalar. Since the fourth column $(\lambda ,1{)}^T$ is always nonzero, its scalar is also the same, for every $\lambda$.

Thus ${\mathrm{End}}_Q(V)=k$, so the representation is a brick module. The Ringel form givesThis conclusion also covers $\lambda =0,1$, where the fourth line repeats one of the earlier lines; the first three lines already force scalar endomorphisms.

A unipotent algebraic group admits a faithful linear representation in which every group element is a unipotent matrix. For $E={\mathrm{End}}_Q(X)$, invertibility is the nonvanishing condition ${\prod }_i\det h_i\ne 0$. Therefore ${\mathrm{Aut}}_Q(X)=E^{\times }$ is a nonempty Zariski-open subset of the vector space $E$.

Take a Krull-Schmidt decomposition $X\cong {⨁}_{a=1}^r{M}_a^{\oplus m_a}$, with pairwise nonisomorphic indecomposable modules. The Fitting lemma makes each ${\mathrm{End}}_Q(M_a)$ a local endomorphism ring. Its residue division algebra is $k$: over an algebraically closed field, every element of a finite-dimensional division algebra has an eigenvalue and hence must be scalar. The semisimple quotient of a module endomorphism algebra consequently gives, for the Jacobson radical $J=J(E)$,is surjective with kernel $1+J$. The nilpotence of $J$ makes $(1+j{)}^{-1}=1-j+j^2-\cdots$ finite and each $1+j$ unipotent. The kernel is closed and normal. Acting on the multiplicity spaces embeds the product of general linear groups back into $E^{\times }$ and splits this quotient. This proves the Levi decomposition of a quiver automorphism groupSince $1+J$ is the unipotent radical, a nonzero $X$ is indecomposable exactly when ${\mathrm{Aut}}_Q(X)/R_u({\mathrm{Aut}}_Q(X))\cong {\mathbb{G}}_m$: the product has a single factor of size one.

For the base change action on quiver representations, the orbit map is $g\mapsto (g_j{x}_{\rho }{g}_i^{-1}{)}_{\rho :i\to j}$. Substituting $g_i=I+ϵu_i$, with ${ϵ}^2=0$, shows its differential isIts kernel is ${\mathrm{End}}_Q(X)$. The stabilizer is smooth because it is open in that vector space. Hence the differential has rank $\dim \mathrm{GL}(\mathbf{n})-\dim {\mathrm{Aut}}_Q(X)=\dim {\mathcal{O}}_X$, and its image is the Zariski tangent space $T_x{\mathcal{O}}_X$. The normal space to a quiver orbit is thereforeThe ambient quiver representation space is an irreducible affine space, and orbits are locally closed. An orbit is open exactly when its dimension equals that ambient dimension, equivalently when ${\mathrm{Ext}}_Q^1(X,X)=0$. This proves that rigid quiver representations have open orbits. Such an orbit is dense and unique, since two nonempty open subsets of an irreducible space intersect.

For a finitely generated associative algebra $A$, the representation variety of an associative algebra consists of generator matrices satisfying all defining polynomial relations. With fixed orthogonal idempotents $1={\sum }_{i=1}^m{e}_i$, require a representation on ${⨁}_i{k}^{n_i}$ to send $e_i$ to the standard vertex projector. This is the meaning of ${\mathrm{Rep}}_A(\mathbf{n})$; without specified idempotents, use the usual single dimension $r$ and ${\mathrm{Rep}}_A(r)={\mathrm{Hom}}_{k\text{-alg}}(A,M_r(k))$. Polynomial relations cut out a closed affine variety in the space of generator matrices.

The group $G={\prod }_i{\mathrm{GL}}_{n_i}(k)$ acts by conjugation, preserving the prescribed projectors. Its stabilizer is ${\mathrm{Aut}}_A(X)$, a nonempty open subset of ${\mathrm{End}}_A(X)$. The orbit dimension formula consequently givesA degeneration of a module $X$ to $Y$ means that ${\mathcal{O}}_Y\subseteq \overline{{\mathcal{O}}_X}$, equivalently that one representative of $Y$ lies in this closure.

For $0\to X^{\prime }\to X\to X^{\prime \prime }\to 0$, choose a vector-space splitting, so every generator has block matrix $x_g=\left(\begin{array}{cc}{x}_g^{\prime }& b_g\\ 0& x_g^{\prime \prime }\end{array}\right)$. Conjugation by $\mathrm{diag}(tI,I)$, $t\ne 0$, givesThis polynomial family extends to $t=0$, retains all algebra relations, and at zero represents $X^{\prime }\oplus X^{\prime \prime }$. Thus splitting an extension gives a module degeneration. Iterating along a composition series gives $X⇝\mathrm{gr}X={⨁}_j{X}_j/X_{j-1}$. Degenerations are transitive because an orbit closure is closed and invariant under base change.

For the one-arrow quiver with dimension vector $(n_1,n_2)$,The action is $A\mapsto g_{2}A{g}_1^{-1}$. Its rank orbits of a matrix under left-right multiplication are indexed by $r=0,\dots ,\min (n_1,n_2)$, with a representative containing an $r\times r$ identity block and zeros elsewhere. Their dimensions are $r(n_1+n_2-r)$; their closures contain exactly matrices of rank at most $r$.

Write the two arrow matrices as $(A,B)\in M_2(k{)}^2$. The base change action on quiver representations sends them to $(g_{2}A{g}_1^{-1},g_{3}B{g}_2^{-1})$. Starting from $(I,I)$ yields precisely the pairs with both matrices invertible: given such a pair, choose $g_1=I$, $g_2=A$, $g_3=BA$. HenceThis is a nonempty Zariski-open subset of the irreducible affine space $M_2(k{)}^2$, so its closure is the entire representation space. Its boundary in that closure is $\{\det A\det B=0\}$.

The rank classification of a two-step linear map says an orbit is determined by $(r,s,t)=(\mathrm{rank}A,\mathrm{rank}B,\mathrm{rank}(BA))$. Indeed, the six interval multiplicities from the elementary decomposition areThey are nonnegative exactly when $0\le r,s\le 2$ and $\max (0,r+s-2)\le t\le \min (r,s)$. Apart from the open orbit $(2,2,2)$, there are nine boundary orbits. Put $D=\mathrm{diag}(1,0)$ and $E=\mathrm{diag}(0,1)$; representatives areThe two rank-one/rank-one cases differ by whether $\mathrm{im}A\subseteq \ker B$; the individual arrow ranks alone do not distinguish them.

A nonzero finite-dimensional representation is a brick module when its endomorphism ring is a division algebra. Over the algebraically closed field $k$, this means ${\mathrm{End}}_Q(X)=k$: for any endomorphism $f$, an eigenvalue $\lambda$ makes $f-\lambda I$ noninvertible, hence zero in a division algebra.

For a counterexample to the converse of “brick implies indecomposable”, take the one-loop representation $k^2$ with nilpotent Jordan block $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Its endomorphism ring is $k[N]\cong k[t]/(t^2)$, a local endomorphism ring of dimension two. It has no nontrivial idempotents, so the module is indecomposable, but it is not a brick.

For the one-arrow quiver, splitting the kernel, image and target complement decomposes any representation into copies of $k\to 0$, $0\to k$ and $k\stackrel{1}{\to }k$. Each has endomorphism ring $k$. Therefore every indecomposable of the one-arrow quiver is a brick.

For the Kronecker quiver representation with arrows $I$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, an endomorphism $(P,Q)$ satisfies $Q=P$ and $PN=NP$. Solving the second equation givesThus this representation is indecomposable but is not a brick: its endomorphism ring is a local endomorphism ring, while the nonzero endomorphism $N$ is nilpotent.

For a general indecomposable non-brick, the proof of Ringel lemma on bricks finds a proper indecomposable submodule with nonzero self-extensions. Repetition in strictly decreasing dimension reaches a brick module $Y\subset X$ with ${\mathrm{Ext}}_Q^1(Y,Y)\ne 0$. The linked proof supplies the minimal-rank, retraction and hereditary-extension steps.

Now assume the Tits form of a quiver is positive definite. If an indecomposable $X$ were not a brick, this $Y$ would give the contradictionHence $X$ is a brick. For its nonzero dimension vector $\mathbf{n}$, positivity and integrality then implyThus every indecomposable in this case is a rigid brick. This deduction uses the Ringel lemma on bricks and the Ringel form, without requiring the full Gabriel theorem.