Math-2011 · mathematics · Graph theory

Math-2011

Some Questions — Question 12

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Fix a prime $p$. Consider finite undirected graphs $G_j$ of uniformly bounded degree, with an integer label on each edge and, if needed, an integer diagonal label at each vertex. Let $A_j$ be the corresponding symmetric integer adjacency matrix: an off-diagonal entry is the edge label, or zero for a missing edge, and a diagonal entry is the vertex label.

Root $G_j$ at a uniformly selected vertex. Local convergence means that for every radius $r$, the distribution of the rooted, labeled radius-$r$ neighborhood converges to a probability distribution on such rooted neighborhoods, consistently as $r$ varies. Labels are retained in these distributions, not discarded.

Must the normalized ranks after reducing all entries modulo $p$ converge? \[ \lim_{j\to\infty}\frac{\operatorname{rank}_{\mathbb F_p}(A_j)}{|V(G_j)|} \quad\hbox{exists?} \] Equivalently, ask whether the normalized dimensions of the kernels over the finite field $\mathbb F_p$ converge, since rank plus nullity equals the matrix size.

The matrices are sparse because degrees are bounded, but the dependence among their rows is a global algebraic feature. Real eigenvalue convergence therefore does not by itself prove the asserted finite-field rank convergence. The question includes the ordinary adjacency matrices with every edge label equal to one. The prime is fixed throughout a sequence; varying it with the graph would be a different problem.

Short English statement

Does local convergence of bounded-degree integer-labeled graphs force convergence of adjacency-matrix rank over a fixed finite field?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.