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

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.