The anti-canonical polar cylinder criterion for K-polystability of Fano varieties

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Let XX be a Fano variety with at most Kawamata log terminal singularities. An anti-canonical polar cylinder means a cylinder polarized by −KX-K_X. The anti-canonical cylinder conjecture. If XX does not contain (−KX)(-K_X)-polar cylinders, then XX is KK-polystable. This conjecture proposes that the absence of anti-canonical polar cylinders gives a sufficient condition for KK-polystability, complementing the role of cylinders as potential sources of destabilizing test configurations. Its general status is unresolved.

References

Primary source

Adrien Dubouloz, In-Kyun Kim, Takashi Kishimoto and Joonyeong Won, “Cylinders in weighted Fano varieties”, arXiv:2603.11490 (2026).

Additional references

3 papers in this index state this conjecture (2020–2026). The statement above is taken from the most recent of them; the others are arXiv:2311.13192, arXiv:2007.14207.

Progress summary

Refreshed
Claimed solved

A March 2026 paper claims counterexamples to the conjecture, but the alleged refutation has not been independently verified.

The conjecture asserts that a Fano variety with at most Kawamata log terminal singularities and no anti-canonical polar cylinder is KK-polystable. No proposer or original date is identified in the retrieved sources.

Known results

  • For log-canonical Fano varieties, α(X)≥1\alpha(X)\geq 1 implies absence of anti-canonical polar cylinders.
  • δ(X)≥1\delta(X)\geq 1 implies KK-semistability, while δ(X)>1\delta(X)>1 implies KK-stability; neither implication proves the cylinder criterion.

March 13, 2026 claimed counterexamples

Adrien Dubouloz, In-Kyun Kim, Takashi Kishimoto, and Joonyeong Won claim that del Pezzo hypersurfaces in families 11 through 2727, for n≥4n\geq 4, are KK-unstable despite having no anti-canonical polar cylinders. Family 44 is given explicitly; if correct, this refutes even the version with KK-semistability. The claim is unverified.

Current status (as of September 2026): The conjecture has a claimed counterexample, but its correctness is unverified, so the general problem remains unresolved.

Sources

Solutions 0

No solutions have been posted yet.