Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 5

A representation of the one-arrow quiver is an $n_2\times n_1$ matrix $F$. If $r=\mathrm{rank}F$, changes of basis put it in the form with an $r\times r$ identity block and all other entries zero. Since multiplication by invertible matrices preserves matrix rank,This describes the orbit for every $0\le r\le \min (n_1,n_2)$, including the zero orbit. Its dimension isIndeed, choosing its $r$-dimensional image contributes $r(n_2-r)$ parameters in a Grassmannian, and choosing a surjective map from $k^{n_1}$ onto that image contributes $r{n}_1$. This is a rank orbit of a matrix under left-right multiplication.

If $y=g\cdot x$, the component maps $g_i$ are invertible and satisfy $g_{t(\rho )}{f}_{\rho }=f_{\rho }^{\prime }{g}_{s(\rho )}$. Thus they define a quiver representation isomorphism $X_x\to X_y$. Conversely, any such isomorphism consists of invertible maps $g_i$, and its commuting squares rearrange to $f_{\rho }^{\prime }=g_{t(\rho )}{f}_{\rho }{g}_{s(\rho )}^{-1}$. ThereforeThe base change action on quiver representations consequently identifies its orbits exactly with isomorphism classes at fixed dimension vector of a quiver representation.

Split the sequence as vector spaces at each vertex. In these coordinates, the middle arrow maps have block formFor $t\in k^{\times }$, change basis by $\mathrm{diag}(t{I}_{X_i},I_{Z_i})$ at every vertex. It changes the off-diagonal block to $t{\xi }_{\rho }$. These arrow matrices depend polynomially on $t$ and at $t=0$ give $X\oplus Z$. Thus the split extension as a degeneration showsWe must check that the two orbits are different. If $Y\cong X\oplus Z$, apply ${\mathrm{Hom}}_Q(Z,-)$ to the original short exact sequence. Exactness makes the kernel of $\mathrm{Hom}(Z,Y)\to \mathrm{End}(Z)$ equal to $\mathrm{Hom}(Z,X)$. The assumed isomorphism of the middle representation gives $\dim \mathrm{Hom}(Z,Y)=\dim \mathrm{Hom}(Z,X)+\dim \mathrm{End}(Z)$, so this map is surjective. In particular the identity of $Z$ lifts to $Z\to Y$, splitting the sequence, a contradiction. Hence the orbits are indeed distinct.

The group $\mathrm{GL}(\mathbf{n})$ is irreducible, so $\overline{{\mathcal{O}}_Y}$ is irreducible. An algebraic-group orbit is locally closed, hence open in its closure; its boundary is a proper closed subset of smaller dimension. The distinct orbit ${\mathcal{O}}_{X\oplus Z}$ lies in that boundary. Therefore

An acyclic finite quiver has integer vertex weights $w_i$ such that $w_j>w_i$ for every arrow $i\to j$; for example, let $w_i$ be the maximum length of a path ending at $i$. For $t\ne 0$, act by $g_i(t)=t^{w_i}{I}_{n_i}$. ThenEvery exponent is positive. The arrow matrices therefore extend polynomially to $t=0$ with all arrows zero, which is the origin of the quiver representation space. The values for $t\ne 0$ lie in ${\mathcal{O}}_X$, soThis proves that acyclic quiver representations degenerate to zero, also when some vertex spaces are zero.