Finding null-homologous curves via the matrix equation $AB^iC^jx=0$

Question

Motivation: I have a family of curves obtained from a single curve by repeatedly applying two automorphisms of the surface (Dehn twists to be specific). I am interested in the images of these curves under a covering map. In particular, I want to know which curves are null-homologous. The automorphisms give rise to automorphisms $B,C$ of the homology group, while the covering gives rise to a (non-injective) homomorphism $A$ from one homology group to the other. I represent all of these as matrices and let $x$ be the homology class of the starting curve.

Let $A$ be a fixed $m\times n$ matrix, $B,C$ fixed invertible $n\times n$ matrices and $x$ a fixed vector. Is there any way to determine all integers $i,j$ such that $AB^iC^jx=0$? If $B,C$ were diagonalizeable then it would at least be possible to place bounds on $i,j$ by converting the matrix exponentiation to exponentiation on the diagonal. Also if we had some $m\times m$ matrix $B'$ such that $B'A=AB$, or similarly for $C$, this would simplify matters. Sadly neither matrix is diagonal or has such a $B'$ or $C'$. Is there a way to solve this problem in general?

Answer 1

I am pretty sure your general question is hopeless. However, your motivating question is not. Notice that a Dehn twist gives rise not to some random invertible matrix but to a transvection, which is to say $B(v) = v + a <v, u> u$ for some fixed scalar $a$ and vector $u$ (in your case the scalar is $1$ but that is not relevant; $< , >$ denotes the symplectic pairing). This acts by identity on $u^\perp$ and generally adds a scalar multiple of $u$ to any vector. You can easily compute its powers in closed form (you will just be multiplying $a$ by the exponent), and similarly with your $C.$ You are then solving a linear system in $i, j.$