Max formula for cyclic covering dimensions of invariant direct sums

Let $q$ be a prime power, let $\sigma$ be an operator with $\sigma^n=\operatorname{id}$ , and let $W_1$ and $W_2$ be $\sigma$ -invariant subspaces satisfying

\mathbb{F}_{q^n}=W_1\oplus W_2.

Write $h_{\sigma}(V)$ for the corresponding cyclic covering parameter of a $\sigma$ -invariant subspace $V$ . The max-formula conjecture. One has

h_q(n)=h_{\sigma}(\mathbb{F}_{q^n})=\max\{h_{\sigma}(W_1),h_{\sigma}(W_2)\}.

The preceding discussion establishes the inequality with the maximum as a lower bound; the conjecture asserts equality for every such invariant direct-sum decomposition. No resolution is given in the source.

References

Primary source

Yangcheng Li, Pingzhi Yuan, Shuang Li and Yuanpeng Zeng, “On cyclically covering subspaces of F^n_q”, arXiv:2602.04558 (2026).

finite-fields linear-algebra

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written construction claims a counterexample in dimension $15$ , but no independent verification has appeared.

Li, Yuan, Li, and Zeng posed the max-formula conjecture in a February 2026 preprint. It asks whether the cyclic covering parameter of every invariant direct sum equals the larger parameter of its two summands.

Known results

The general lower bound $h_{\sigma}(\mathbb{F}_{q^n})\ge\max\{h_{\sigma}(W_1),h_{\sigma}(W_2)\}$ is proved.
Equality holds if one summand has parameter $0$ .
Equality holds when $h_q(n)=1$ .
Equality holds for the known values $h_q(q^d-1)=d-1$ .

Posted attempt

A reader-written construction claims a complete counterexample with $q=2$ , $n=15$ , and invariant summands having parameters $1$ and $2$ , while the whole space has parameter $3$ . Thus it claims $h_{\sigma}(\mathbb{F}_{2^{15}})>\max\{h_{\sigma}(W_1),h_{\sigma}(W_2)\}$ . The attempt has not been independently verified.

Current status (as of August 2026): The lower bound and listed special cases are settled, while the universal max-formula conjecture is disputed by an unverified claimed counterexample.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture is false, even when its first equality holds. Take $q=2$ and $n=15$ , and identify the $\mathbb F_2$ -vector space $\mathbb F_{2^{15}}$ with

V=\mathbb F_4\oplus\mathbb F_2^9\oplus\mathbb F_{16}.

Choose $\alpha\in\mathbb F_4^\times$ of order $3$ and $\beta\in\mathbb F_{16}^\times$ of order $5$ . Define the $\mathbb F_2$ -linear automorphism

\sigma(a,t,b)=(\alpha a,t,\beta b).

Then $\sigma$ has order $15$ . Set $W_1=\mathbb F_4\oplus\mathbb F_2^9$ and $W_2=\mathbb F_{16}$ ; these are $\sigma$ -invariant and $V=W_1\oplus W_2$ .

For $W_1$ , the codimension-one subspace $U_1=\mathbb F_2\oplus\mathbb F_2^9$ is $\sigma$ -covering because $\alpha$ acts transitively on $\mathbb F_4^\times$ . Conversely, every $\sigma$ -covering subspace must contain the nine-dimensional fixed summand $\mathbb F_2^9$ : each fixed vector appearing in any $\sigma$ -translate already belongs to the original subspace. Its image in the $\mathbb F_4$ quotient must also be nonzero. Therefore

h_\sigma(W_1)=1.

For $W_2$ , regard $\mathbb F_4$ as the unique subfield of $\mathbb F_{16}$ with four elements. Since $|\langle\beta\rangle|=5$ and $|\mathbb F_4^\times|=3$ ,

\mathbb F_{16}^\times=\langle\beta\rangle\mathbb F_4^\times.

Consequently $\bigcup_{j=0}^4\beta^j\mathbb F_4=\mathbb F_{16}$ , so $h_\sigma(W_2)\ge2$ . A one-dimensional subspace has only one nonzero vector, and its five translates contain at most five nonzero vectors, whereas $\mathbb F_{16}$ has fifteen. Hence

h_\sigma(W_2)=2.

Now $U=U_1\oplus\mathbb F_4$ has codimension $3$ in $V$ . For every $(a,t,b)\in V$ , choose residues $j_3\bmod3$ and $j_5\bmod5$ such that $\alpha^{-j_3}a\in\mathbb F_2$ and $\beta^{-j_5}b\in\mathbb F_4$ . The Chinese remainder theorem supplies $j\bmod15$ satisfying both conditions. Thus $\sigma^{-j}(a,t,b)\in U$ , proving that $U$ is $\sigma$ -covering. Conversely, a codimension-four subspace has at most $2^{11}$ elements, so its fifteen translates contain at most $15\cdot2^{11}<2^{15}$ elements. Therefore

h_\sigma(V)=3.

Finally, the known identity $h_q(q^d-1)=d-1$ gives $h_2(15)=3$ . Hence

h_2(15)=h_\sigma(V)=3> 2=\max\{h_\sigma(W_1),h_\sigma(W_2)\},

contradicting the proposed formula.