Chernov–Maguire conjecture on Khovanov detection of Legendrian simple knots

For every smooth knot type KK that is Legendrian simple—meaning that any two Legendrian representatives L1,L2L_1,L_2 of KK with tb⁡(L1)=tb⁡(L2)\operatorname{tb}(L_1)=\operatorname{tb}(L_2) and r(L1)=r(L2)r(L_1)=r(L_2) are Legendrian isotopic—and every knot type K′K', if Kh⁡(K′)≅Kh⁡(K)\operatorname{Kh}(K')\cong\operatorname{Kh}(K), then K′K' is equivalent to KK. In other words, Khovanov homology detects every Legendrian-simple knot type.

References

Primary source

arXiv

Progress summary

Refreshed
Claimed solved

A new preprint claims the conjecture is false by giving a first counterexample, but no independent verification has appeared.

Chernov and Maguire proposed that Khovanov homology distinguishes every Legendrian-simple knot: matching homology should force equivalence of the knots. Their 2022 work presented this as a conjecture supported by computation, not a theorem.

Known results

  • Computations through 2020 crossings found no matching Khovanov polynomials for the tested torus, twist, or conjecturally Legendrian-simple knots (Chernov–Maguire, 2022).
  • Known detection cases include the unknot, trefoils, figure-eight knot, and, over Z/2Z\mathbb{Z}/2\mathbb{Z}, the cinquefoil T(5,2)T(5,2) (Chernov–Maguire, 2022).

August 25, 2026 stated counterexample

The preprint Legendrian simple knots not detected by Khovanov homology claims the first counterexample, indicating that Khovanov homology does not detect all Legendrian-simple behavior asserted by the conjecture. This claim is currently supported only by the preprint and remains unverified.

Current status (as of August 2026): The conjecture has a claimed counterexample, but its refutation is unverified; the earlier computational evidence remains the only established support recorded here.

Sources

Solutions 0

No solutions have been posted yet.