The Steenrod-span conjecture for motivic cohomology operations
Let be a field. Write for the algebra of motivic Steenrod operations and for the corresponding bigraded operations on motivic mod- cohomology. Steenrod-span conjecture. For all fields , the injection
is an equality. This asks whether all motivic mod- cohomology operations are generated by the motivic Steenrod operations, and is part of the paper's motivation for studying phenomena arising from -adic coefficients.
References
Primary source
Toni Annala and Elden Elmanto, “Motivic Steenrod operations at the characteristic via infinite ramification”, arXiv:2506.05585 (2026).
Progress summary
The conjecture is proved away from the field characteristic, but the equal-characteristic case remains open and no complete proof has appeared.
The conjecture asks whether every bistable motivic mod- cohomology operation is generated by motivic Steenrod operations for every field . The all-fields equality remains unproved; existing results separate the case from the difficult equal-characteristic case.
Known results
- For , admissible monomials in Bocksteins and power operations form a basis of all bistable operations (2013).
- In characteristic , the relevant map is a split monomorphism, not known to be an equivalence (2017).
- Characteristic- Steenrod power operations satisfy instability, Adem, and Cartan relations, but generation is not established (2019).
- Hoyois proved the related Hopkins–Morel statement after inverting the exponential characteristic; unpublished work of Hopkins and Morel covers characteristic zero.
June 2025 development
The paper “Motivic Steenrod operations at the characteristic via infinite…” constructs characteristic- power operations and gives a basis for the image of the Steenrod algebra, but still states the all-fields equality as Conjecture 1.7; it reports no proof or counterexample.
Current status (as of September 2026): The conjecture is settled for and in characteristic zero only in the cited qualified senses, while the full equal-characteristic equality remains open.
Solutions 0
No solutions have been posted yet.