Saxl’s staircase tensor-square conjecture
Follow the latest progress by email.
For the staircase partition (m,m−1,…,1), does the tensor square of its Specht representation of S_{m(m+1)/2} contain every irreducible complex representation?
Why it matters and what counts as progress
Why it matters: The square conjecture is stronger than the known tensor-cube result and constituent-family or asymptotic coverage results.
What counts as progress: A proof including every constituent for every m≥1, a larger proved constituent family, or a rigorously missing constituent.
How the problem developed
- 2012 · Conjecture [1]
Saxl Conjectured that the square of the staircase Specht representation contains every irreducible representation of the corresponding symmetric group.
- 2023 · Partial result [1]
Chris Harman and Mark Ryba Proved that the tensor cube of the staircase representation contains every irreducible, while the tensor-square conjecture remained open.
- 24 Sept 2026 · Proof claim [2]
OpenAI Released a manuscript claiming that the staircase Specht representation tensor-squared contains every irreducible complex representation for every staircase size.
AI contribution
A Cyclic Polytabloid Proof of Saxl's Conjecture
For every m≥1, the tensor square of the staircase Specht representation of the symmetric group S_{m(m+1)/2} contains every irreducible complex representation.
Sources
Website timeline
Publication dates and editorial changes.