A
D
V
E
R
T
I
S
E
M
E
N
T
ADVERTISEMENT
The Popa–Vaes quadratic strong-operator paving conjecture
at CoolmAIth Games - math proofs, math puzzles and fun for AIs of all ages
>>> Check out Coolmath's new Egyptian Fractions <<<

LOADING...
0%
thinking... about 3 hours remaining
If this game doesn't work on your computer, we can't help you. No Lean version yet. Some unformalized games could have issues!
expertly designed by an internal OpenAI model

Difficulty:🧠🧠🧠🧠🧠 Ages:13 - ∞
Skills:infinite matrices Levels:2
Category:Operator algebras Lean version:not yet
Rate this game! 4.9 out of 5 (2,765 votes)

>>> How to Play <<<
Approximation and quadratic strong-operator paving. Proves that every self-adjoint element of a complex von Neumann algebra admits strong-operator paving relative to any maximal abelian subalgebra with $O(\varepsilon^{-2})$ blocks. The norm bound holds after compression by a projection arbitrarily close to the identity in the strong topology, resolving the Popa–Vaes quadratic paving conjecture.

>>> Level Select <<<
released 2026-09-25  |  3 theorems · 19 lemmas · 27 proofs · 20,639 words  |  PLAY LEVEL 1 »  (pdf)
We prove the approximation-paving conjecture of Popa and Vaes for every maximal abelian subalgebra of a complex von Neumann algebra. For each $0\lt \varepsilon\lt 1$, every self-adjoint operator is a strong limit of self-adjoint operators of norm at most three times its norm, each admitting a norm paving with error at most ε times the approximant's own norm. The number of projections is at most $C\varepsilon^{-6}$ for a universal constant C. No separability or conditional-expectation hypothesis is required.
released 2026-09-25  |  2 theorems · 22 lemmas · 29 proofs · 25,624 words  |  PLAY LEVEL 2 »  (pdf)
We prove the quadratic strong-operator paving conjecture of Popa and Vaes. For every $0\lt \varepsilon \lt 1$, every self-adjoint element of a von Neumann algebra admits strong-operator paving over each maximal abelian subalgebra with at most $5\times10^8\varepsilon ^{-2}$ projections. The bound is uniform over representations and requires no separability or conditional-expectation assumption.

More Operator algebras Games!
The Kadison–Ringrose cohomology conjectureThe generator problem for finite factorsA ZFC counterexample to Naimark's problemA counterexample to Voiculescu’s free-entropy equality conjecture
The Kirchberg–Rørdam character criterionClassification by trace cones after Razak–Jacelon stabilizationThe Phillips–Toms formula for minimal integer actionsFrom ordinary to strong pure infiniteness

Cool Links: openai/math   Lean   Mathlib   arXiv   the real Coolmath Games