The converse-type conjecture for sparse Haar-random matrices

About 2 years old · traced to

For each nn, let Σn=(σn,1,…,σn,n)\Sigma_n=(\sigma_{n,1},\ldots,\sigma_{n,n}), where each σn,j\sigma_{n,j} is a subset of [n]={1,2,…,n}[n]=\{1,2,\ldots,n\}. Let Xn∈M⁡n(Zp)X_n\in\operatorname{M}_n(\mathbb{Z}_p) be Haar-random on this support: (Xn)i,j=0(X_n)_{i,j}=0 for i∉σn,ji\notin\sigma_{n,j}, while the entries with i∈σn,ji\in\sigma_{n,j} are independent Haar-random elements of Zp\mathbb{Z}_p. Define

∣Σn∣:=∑i=1n∣σn,i∣,|\Sigma_n|:=\sum_{i=1}^n|\sigma_{n,i}|,

and write cok⁡(Xn)\operatorname{cok}(X_n) converges to CL when its distribution converges to the Cohen–Lenstra distribution as n→∞n\to\infty. Converse-type conjecture. For every sequence (an)n≥1(a_n)_{n\geq 1} satisfying n≤an≤n2n\leq a_n\leq n^2 and

lim⁡n→∞(ann−log⁡pn)=∞,\lim_{n\to\infty}\left(\frac{a_n}{n}-\log_p n\right)=\infty,

there is a sequence (Σn)n≥1(\Sigma_n)_{n\geq1} such that cok⁡(Xn)\operatorname{cok}(X_n) converges to CL and ∣Σn∣=an|\Sigma_n|=a_n for all nn.

The conjecture is presented as a converse-type result to the necessary lower bound for convergence to the Cohen–Lenstra distribution, and is described as best possible with respect to that bound. No resolution evidence is supplied in the source.

References

Primary source

Dong Yeap Kang, Jungin Lee and Myungjun Yu, “Random p-adic matrices with fixed zero entries and the Cohen–Lenstra distribution”, arXiv:2409.01226 (2026).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.