Djoković’s Conjecture A for PU(n,1)

For every integer n≥2n\ge 2, the involution length of PU(n,1)\mathrm{PU}(n,1) is exactly 44: every g∈PU(n,1)g\in\mathrm{PU}(n,1) can be written as a product of at most four nontrivial involutions, and some element cannot be written as a product of three involutions. Moreover, every g∈PU(n,1)g\in\mathrm{PU}(n,1) is a single commutator, i.e. there exist a,b∈PU(n,1)a,b\in\mathrm{PU}(n,1) such that g=aba−1b−1g=aba^{-1}b^{-1}.

References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A September 2026 preprint claims to settle the conjecture in every dimension, but the result has not yet been refereed.

Djoković’s Conjecture A concerns the exact involution length and single-commutator property of PU(n,1)\mathrm{PU}(n,1) for all nn. The latest preprint claims both statements for the full family.

Known results

  • PU(2,1)\mathrm{PU}(2,1) has involution length 44 and commutator length 11; for PU(n,1)\mathrm{PU}(n,1) with n≥3n\geq 3, only the bound ≤8\leq 8 was previously known (Gongopadhyay and Thomas, 2016).

September 2026 preprint

Zhongqi Wang and Shihai Yang claim that the higher-dimensional involution-length bound improves to the optimal value 44 and that every element of PU(n,1)\mathrm{PU}(n,1) is a single commutator. This would settle Conjecture A for all nn, but the preprint is unrefereed.

Current status (as of September 2026): A preprint claims the conjecture is solved for all PU(n,1)\mathrm{PU}(n,1), extending the verified two-dimensional results, but its full proof remains unverified.

Sources

Solutions 0

No solutions have been posted yet.