Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 3 / 5

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.