Square-free last invariant factor and the rank over \(\mathbb{F}_2\)

Determine whether, for every controllable oriented graph \(\Sigma\) with skew-walk matrix \(W(\Sigma)\), square-freeness of the last invariant factor \(d_n\) implies \(\operatorname{rank}_2 W(\Sigma)=\lceil n/2\rceil\).

Background

The paper studies sufficient conditions under which a controllable oriented graph is determined by its generalized skew spectrum. The main theorem assumes both that the last invariant factor dnd_n of the skew-walk matrix is square-free and that the matrix has F2\mathbb{F}_2-rank ⌈n/2⌉\lceil n/2\rceil.

The authors report that every example examined computationally satisfies the F2\mathbb{F}_2-rank condition whenever dnd_n is square-free, but they do not establish this implication in general. An affirmative resolution would show that the rank hypothesis in the main theorem is redundant and would clarify the relationship between the Smith normal form of the skew-walk matrix and its rank over F2\mathbb{F}_2.

References

At the end of this paper, we mention a natural problem suggested by our computational experiments. In all examples we have examined, the square-freeness of the last invariant factor $d_n$ of $W(\Sigma)$ seems to imply $rank_2 W(\Sigma)=\left\lceil\frac n2\right\rceil.$ This leads to the following question. Let $\Sigma\in \mathcal{F}_n$, and let $d_n$ be the last invariant factor of $W(\Sigma)$. Suppose that $d_n$ is square-free, then $rank_2 W(\Sigma)=\left\lceil\frac n2\right\rceil.$ An affirmative answer would make the $\mathbb{F}_2$-rank condition for $W(\Sigma)$ in Theorem~\ref{thm:main} redundant and reveal a closer connection between the Smith normal form of $W(\Sigma)$ and its rank over $\mathbb{F}_2$. We leave this problem for future study.

— A New Sufficient Condition for Oriented Graphs Determined by Their Generalized Skew Spectra  (2609.03428 - Lin et al., 3 Sep 2026) in Section Conclusion and Future Work, final Problem environment