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

Set $A=kQ$, $X_i=e_{i}X$, and let $f_{\rho }$ be the action of the arrow $\rho$ on $X_{s(\rho )}$. Write $P_i=A{e}_i$, a left projective module. The standard projective resolution of a quiver representation isThe algebra acts on the first tensor factor. The augmentation is $\epsilon (p\otimes x)=px$. On the summand for $\rho :i\to j$, the differential sends $p\otimes x$ to $p\rho \otimes x$ in the $i$-summand minus $p\otimes f_{\rho }(x)$ in the $j$-summand. Here $p\in A{e}_j$, so $p\rho \in A{e}_i$. These formulas specify every term and map, and $\epsilon d=0$.

Exactness is the standard path resolution fact: the relations identify a path acting on a vector with successively applying its arrows; uniqueness of the first traversed arrow supplies injectivity of the relation map. Each $A{e}_i$ is a direct summand of $A$ because $e_i$ is an idempotent, and tensoring with a $k$-vector space gives a direct sum of copies of $A{e}_i$. Both terms preceding $X$ are consequently projective modules. Thus the displayed sequence is a projective resolution of length at most one.

The same path resolution works for arbitrary left modules, with possibly infinite-dimensional $X_i$ and arbitrary direct sums of projective modules. Hence every left $A$-module has projective dimension at most one. Equivalently, $A$ is a left hereditary ring: if $N\subseteq P$ with $P$ projective, dimension shifting gives ${\mathrm{Ext}}_A^1(N,M)\cong {\mathrm{Ext}}_A^2(P/N,M)=0$ for every $M$, so $N$ is projective. This argument also covers quivers with oriented cycles.

For dimension vectors $\mathbf{n},\mathbf{m}\in {\mathbb{R}}^{Q_0}$, define the Ringel formApply ${\mathrm{Hom}}_A(-,Y)$ to the projective resolution. Evaluation at $e_i$ identifies ${\mathrm{Hom}}_A(P_i{\otimes }_{k}V,Y)$ with ${\mathrm{Hom}}_k(V,Y_i)$. Consequently there is an exact sequencewhereThis is the extension complex of quiver representations. Taking its alternating dimension sum yields

Let $K=\ker f_{\rho }$ and $C=Y_j/\mathrm{im}{g}_{\rho }$. Consider the map from the arrow term of the extension complex of quiver representations to ${\mathrm{Hom}}_k(K,C)$ that restricts its $\rho$-component to $K$ and then takes the quotient in $Y_j$. This map is surjective: a map $K\to C$ can be lifted to $Y_j$ and extended from $K$ to $X_i$.

Every coboundary is killed by this map, since for $x\in K$,It therefore induces a surjectionBoth vector spaces in the final Hom functor are nonzero, so its dimension is positive. Hence ${\mathrm{Ext}}_Q^1(X,Y)\ne 0$. This is the kernel-cokernel obstruction to splitting a quiver extension and remains valid for loops and repeated arrows elsewhere in the quiver.

If an arrow map failed to have maximal matrix rank, it would have both a nonzero kernel of a linear map and a nonzero cokernel. Apply the preceding obstruction with $Y=X$. It would give ${\mathrm{Ext}}_Q^1(X,X)\ne 0$, contrary to the assumption. Therefore every arrow map is injective or surjective, equivalently its rank is the minimum of the two vertex dimensions. This is a necessary property of a rigid quiver representation.

We first record why rigid quiver representations have open orbits. Let $\mathbf{n}=\dim X$, $G={\prod }_i\mathrm{GL}(X_i)$ and $R={\mathrm{Rep}}_Q(\mathbf{n})$. The stabilizer is ${\mathrm{Aut}}_Q(X)$, the nonempty open set of units in ${\mathrm{End}}_Q(X)$, so it has dimension $\dim {\mathrm{End}}_Q(X)$. The dimension formula for an algebraic group homomorphism, or the same constant-fiber argument for the orbit map, givesBy the Ringel form identity and ${\mathrm{Ext}}_Q^1(X,X)=0$, this equals ${\sum }_{\sigma }{n}_{s(\sigma )}{n}_{t(\sigma )}=\dim R$. An algebraic-group orbit is locally closed; since $R$ is an irreducible affine space, this full-dimensional orbit is open and dense.

Write $q=\rho \pi$ for the composed path. Each matrix entry of $q$ is a polynomial function on $R$. Since $qX=0$, change of basis makes it zero on all of ${\mathcal{O}}_X$, hence on all of $R$ by density. Suppose instead that $X_{t(\rho )}\ne 0$. The condition $\pi X\ne 0$ ensures that every vertex space visited by $\pi$ is nonzero. Choose a vector $v_a\ne 0$ and a functional ${\ell }_a$ with ${\ell }_a(v_a)=1$ at every vertex visited by $q$. Assign to every arrow $\sigma$ occurring in $q$ the map $v_{t(\sigma )}{\ell }_{s(\sigma )}$, and assign arbitrary maps, say zero, to the other arrows. At this representation, the path $q$ carries its initial chosen vector to its final chosen vector, so $q$ is nonzero, a contradiction.

The construction uses a single assigned map per arrow, so it still works if an arrow or vertex occurs repeatedly in the path. ThereforeThis proves the path identities in a rigid quiver representation claim for an arbitrary quiver. Maximal rank of the individual arrows alone would not justify the conclusion about their composition.