Keevash–Mubayi simplex-cluster conjecture

Let k>d≥2k>d\ge 2 and let n≥k(d+1)/dn\ge k(d+1)/d. A dd-simplex-cluster is a collection of d+1d+1 distinct members A1,…,Ad+1A_1,\dots,A_{d+1} of a family F⊆([n]k)\mathcal F\subseteq\binom{[n]}{k} such that ⋂i=1d+1Ai=∅\bigcap_{i=1}^{d+1}A_i=\varnothing, ⋂i≠jAi≠∅\bigcap_{i\ne j}A_i\ne\varnothing for every j∈[d+1]j\in[d+1], and ∣⋃i=1d+1Ai∣≤2k\left|\bigcup_{i=1}^{d+1}A_i\right|\le 2k. The conjecture states that if F\mathcal F contains no dd-simplex-cluster, then ∣F∣≤(n−1k−1)|\mathcal F|\le\binom{n-1}{k-1}. Moreover, equality holds if and only if there exists x∈[n]x\in[n] such that F={A∈([n]k):x∈A}\mathcal F=\{A\in\binom{[n]}{k}:x\in A\}, namely, F\mathcal F is a full star.

References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A new unrefereed paper claims the conjecture is completely proved, but the claim has not been independently checked.

Keevash and Mubayi’s conjecture predicts the star bound for every k>d≥2k>d\ge2 and n≥k(d+1)/dn\ge k(d+1)/d, with equality only for a full star. It is a strengthening of the Erdős–Chvátal simplex and Mubayi cluster conjectures.

Known results

  • Keevash–Mubayi: for sufficiently large nn in a specified linear range of kk, the star bound and equality case hold.
  • A 2018 paper: the bound holds for d<k≤dd+1nd<k\le \frac{d}{d+1}n, with equality only for a star, plus an asymptotic version.
  • Currier, 2020: the conjecture is resolved for 4≤d+1≤k4\le d+1\le k and n≥2k−d+2n\ge2k-d+2.

October 2026 claimed proof

Wu and Feng’s preprint claims that every cluster-free family satisfies ∣F∣≤(n−1k−1)|\mathcal F|\le\binom{n-1}{k-1} for all k>d≥2k>d\ge2 and n≥k(d+1)/dn\ge k(d+1)/d, with equality only for a full star. This covers the entire conjectured range, but the retrieved record contains no independent verification or referee assessment.

Current status (as of October 2026): Earlier parameter ranges are proved, while the full conjecture remains open as a verified theorem because the claimed complete proof is unverified.

Sources

Solutions 0

No solutions have been posted yet.