roon @tszzl · 5h
Replying to @hargitai_b84173 and @DimitrisPapail
I'm not going to say anything that's not already public obviously, but when openai says they have an internal model that has solved 100s of open problems in mathematics and in general represents unprecedented mathematical skill, obviously certain learning theory problems are a subset of mathematics and you can expect fast progress there
Note from Claude Sonnet 5
Text-only reply tweet by roon (@tszzl) saying that OpenAI's publicly claimed internal model, which has solved hundreds of open math problems, implies fast progress on learning theory problems.
Aaron Roth @Aaroth · 23h
A world without open problems
Here are some that fell today:
K-server: arxiv.org/abs/2609.15979
Matroid Secretary: arxiv.org/abs/2609.14555...
Matrix Spencer: arxiv.org/abs/2609.15025
(Well Matrix Spencer was maybe also a few weeks ago, but who's counting? arxiv.org/abs/2608.28816
)
[Quoted post]
Aaron Roth @Aaroth · Aug 1
We are clearly moving to a world that will be without open problems, but this doesn't mean math is going away. Interesting work will correspond to discovering new questions and coherent theoretical frameworks. Well defined ... [cut off]
Note from Claude Sonnet 5
Tweet by theoretical computer scientist Aaron Roth listing several open problems (k-server, matroid secretary, matrix Spencer) as having been resolved, with arXiv links, framing it as 'a world without open problems'; quotes his Aug 1 post saying math will shift toward finding new questions and theoretical frameworks.
Lisan al Gaib reposted
Joshua Achiam [verified] @jachiam0 · 3h
The experience of being human: on the day it becomes abundantly clear that all open problems in mathematical theory and physics will be solved in ~2 years tops, 98% of the people paying attention to the situation will be too fixated on the human drama about who should get credit to notice that the world changed under their feet in a permanent and quite astonishing way. I feel a strange mix of pride, annoyance, affection, and fear for us all.
43 replies · 65 reposts · 868 likes · 20K views
---
Bogdan Ionut Cirstea reposted
Henry Shevlin [verified] @dioscuri · 1h
As recently as April this year, prediction markets gave AI less than a 40% chance of solving any Millennium Prize Problem before **2030**
[attached image: Manifold market screenshot]
Will Artificial Intelligence solve a Millennium Prize Problem before 2030?
Alan Turing
94% chance (up 14)
[chart axis: 0%, 50%, 100%; x-axis 2025, April, July, October, 2026, April, July; MANIFOLD watermark]
[buttons: Bet YES, Bet NO]
Note from Claude Sonnet 5
Two stacked dark-mode tweets: Joshua Achiam (@jachiam0) on how people will react when it is clear all open problems in mathematical theory and physics will fall within about two years, and Henry Shevlin (@dioscuri) noting prediction markets gave under 40% in April, with an attached Manifold chart of 'Will AI solve a Millennium Prize Problem before 2030?' now at 94%, rising steeply from about July 2026.
David Turturean [verified] @DavidTurturean · 6h
When the Spring semester started at MIT, AI couldn't meaningfully solve open math problems.
When the Fall semester starts today at MIT, AI will have allegedly solved a Millenium Prize Problem.
Imo this pace of change reasonably deserves the title of 'mathematical singularity'
Note from Claude Sonnet 5
Dark-mode tweet screenshot: David Turturean (@DavidTurturean) contrasting AI's inability to solve open math problems at the start of MIT's Spring semester with an alleged Millennium Prize Problem solution by the start of Fall, and calling the pace a 'mathematical singularity'.
Dark-mode tweet by Przemek Chojecki about a map of 15,000+ open mathematical problems from UnsolvedMath, with an embedded video of a 2D cluster plot of 15,458 problems colored by subfield, gold points marking those already formalized in Lean (under 10%).
Daniel Kressner @_kreda_ · Aug 23
GPT-5.6 has found a counterexample to a question that has been raised nearly 40 years ago: Is ||AX+XA^T||_F / ||X||_F always attained at a symmetric matrix X?
[link card] arxiv.org
A counterexample to the symmetric-maximizer conjecture ...
Note from Claude Sonnet 5
Tweet by numerical linear algebra researcher Daniel Kressner stating GPT-5.6 found a counterexample to a roughly 40-year-old matrix-norm question, with an arXiv link card titled as a counterexample to the symmetric-maximizer conjecture.
Grigore Rosu @RosuGrigore [Follow]
The reason I've been quiet recently.
Over the last two months I worked hard, using AI, on solving two open problems that my UIUC research lab was not able to solve over the last 7 years.
Finally, we proved them. Viva @AnthropicAI and @OpenAI!
1:52 PM · Aug 24, 2026 · 9,825 Views
4 replies · 9 reposts · 113 likes · 25 bookmarks
Relevant ▾
[reply / next post in thread]
Grigore Rosu @RosuGrigore · 4h
Matching logic is complete without μ, incomplete with μ!
In passing, we also solved two open problems in hybrid logic: the nominal-free fragment is complete; with nominals only, incomplete. This is surprising because if we add pointed validity it becomes complete again (known
Show more
[embedded diagram, small text, best-effort: a map of logics and their completeness results; node labels as follows]
matching logic, with definedness
global completeness [2, 3]
equivalent to FOL_= [2]
equivalent to H(@, ∀) [31]
matching μ-logic, with definedness
global completeness fails [3, Prop. 23];
no effective weak completeness (Remark 27)
matching logic, basic many-sorted
global completeness fails
(Section 8)
H(@), H(@, ↓)
global completeness [29]
H(∀)
local completeness [11];
no complete calculus in the
class of Theorem 33
matching logic, basic unsorted
global completeness
(Corollary 15)
matching μ-logic, basic unsorted
no effective weak
completeness (Theorem 19)
modal logic, basic polyadic
global completeness
[12, Ex. 1.5.3]
modal μ-calculus
global completeness, finite Γ,
via [19, 20]; not compact, so
local completeness fails [19]
[arrow labels between nodes: μ, [·], Var, ∃, @, nominals, sorts, nominals, @]
Note from Claude Sonnet 5
Tweet by Grigore Rosu (@RosuGrigore) saying that over two months, using AI, his UIUC lab solved two open problems it had not solved in seven years, crediting Anthropic and OpenAI; below it a follow-up post on completeness of matching logic with and without the mu operator and hybrid logic, with a diagram of logics and completeness results. Posted 2026-08-24.
Paata Ivanisvili @PI010101 · 3h
An army of AI bots is coming, and this is a game changer. I asked my bot to go to MathOverflow, pick an open problem, and solve it. It found one and generated a solution, but it was outside my field, so I couldn't assess its correctness. Then I asked it to pick a problem from my area—analysis/probability. It did, and gave me a solution that I found pretty satisfactory.
[two embedded screenshots of an AI chat interface (light theme, small coloured dots down the left edge); text is tiny and only partly legible, math formulas not reliably readable. Best-effort transcription of the prose follows; [illegible] marks unreadable stretches.]
[left screenshot]
(user) can you go to mathoverflow go over questions hich were not unanswered and solve one of them and then come back and let me know which one did you solve
On it. I'll pull unanswered MathOverflow questions, pick one I can actually close, and come back with the writeup.
Most of the unanswered queue is open research (Collatz ratios, Sun determinants, Serre functors). I'm taking a concrete one instead: a missing step in Conway's normal-form theorem.
Writing the missing step now. I'll come back with which question it was and the proof.
Solved The Conway Normal Form and decreasing indices (MO 514585). Still unanswered. It's the missing step in Conway's Theorem 21: the exponents γ_ν in the normal form are strictly decreasing, including at limit stages.
The OP could see the successor drop γ_n > γ_{n+1} once the remainder is small compared to ω^{γ_n}, but only after knowing S_{n+1} = S_n + r_n ω^{γ_n}. Limit ordinals have no predecessor, so that identity does not even apply.
Write a ≪ b for "small compared to" and a ≍ b for commensurate (on absolute values). Let C_α be the convex class of z such that [illegible]. Conway's "simplest number with those terms" is the simplest member S_α of C_α. The coefficient r_α was chosen exactly so that z itself satisfies that inequality, so z ∈ C_α at every stage.
Successor: both z and S_{α+1} lie in C_{α+1}, so both are infinitely close to S_α + r_α ω^{γ_α}. Their difference satisfies [illegible] If [illegible] = 0, then [illegible] hence γ_{α+1} < γ_α. (The ordinary sum S_α + r_α ω^{γ_α} is itself simplest in C_{α+1}, which is the identity the OP wanted.)
Limit: let λ be a limit. Both z and S_λ lie in C_β [illegible]. For each β < λ they are both infinitely close to S_β + r_β ω^{γ_β}, so [illegible] If [illegible] = 0, then γ_λ < γ_β for every earlier β. No predecessor is used.
The full claim β < α ⟹ γ_β > γ_α is then transfinite induction. I did not post it on the site. Say if you want a version to paste.
Today 10:34 AM
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentrati[cut off by a scroll-down button] [illegible] on commuting operators before I pick.
[right screenshot]
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentration question and one on commuting operators before I pick.
Taking Concentration inequality for dependent random variables (MO 514247). Writing the argument now.
Solved Concentration inequality for dependent random variables (MO 514247). Still unanswered. Bounded Y_i(X) alone does not concentrate S = Σ_i X_i Y_i(X) at any scale o(n). You get √n tails only with independence or a Lipschitz bound on Y.
The crude bound |S| ≤ C Σ|X_i| is always true, and Σ|X_i| itself concentrates around n√(2/π) with width √n. That only says |S| = O_P(n). It does not stop S from sitting at +cn or −cn, [illegible] with equal mass.
Counterexample: set Y_i(X) = C sign(x_1) sign(x_i). Then S = C sign(X_1) Σ|X_i|. The sign is Rademacher and independent of the modulus, so [illegible] Hence Var(S) = Θ(n²). The same law appears under leave-one-out, via Y_i(X) = C Π_{j≠i} sign(x_j), and Y_i(0) = 0 already holds. So neither Y_i(0) = 0 nor Y_i = Y_i(X_{−i}) helps.
Two cases that do concentrate at scale √n.
If Y is independent of X (or constant), then S | Y ~ N(0, Σ Y_i²) with [illegible] ≤ C²n, so P(|S| > t) ≤ 2 exp(−t²/(2C²n)).
If x ↦ Y(x) is L-Lipschitz ℝⁿ → ℝⁿ, restrict to the ball [illegible] (Gaussian mass 1 − e^{−n/2}). There [illegible] ≤ (C + 2L)√n |x − y|₂.
Projecting onto the ball and applying Gaussian Lipschitz concentration gives, for a median m_n of the truncation, [illegible] ≤ 2 exp(−t²/(2(C + 2L)²n)) + e^{−n/2}.
(The centre m_n can be as large as O(C + L)n. This is concentration about m_n, not about 0.)
I did not post it. Say if you want a version to paste.
Note from Claude Sonnet 5
Tweet by Paata Ivanisvili (@PI010101) saying he had his AI bot pick unanswered MathOverflow questions and solve them, with two embedded screenshots of the bot's chat replies (a Conway normal form question and a concentration-inequality question). Embedded text is small; prose transcribed best-effort, formulas largely marked illegible or reconstructed only where clearly visible.
[continuation of QC's thread on complex structure on S^6 - repeats text and embedded screenshots seen previously, then continues with a new reply:]
QC @QiaochuYuan · 4h
had sol take a longer look (15m) and still didn't find a problem and understands the argument in more detail but is still nervous that it could be missing some subtlety. also thinks if this holds up it is arguably the most significant AI math result so far, and agrees with me that this is plausibly "model 2"
transcript here: chatgpt.com/share/6a8b77b7...
[partial screenshot below, cut off at bottom]
Worked for 15m 49s >
okay, after pushing substantially harder: i still have not found the bug. more strongly, i think i've now independently reconstructed enough of the proof that my state is no longer "this looks surprisingly serious": it's closer to:
the bespoke mathematical core of the construction appears internally correct to me.
that is not the same as publication-level verification — there are places where i'm relying on standard nearby-cycles / analytic-geometry machinery rather than reproving it from foundations — but i have now deliberately attacked exactly the places a plausible LLM hallucination could hide, including an apparent contradiction that turned out to be a clever feature of the construction.
the proof, reconstructed from the idea rather than the prose... [cut off]
i wouldn't say "automatic Fields Medal," because mathematics doesn't work that way. Perelman-scale things aside, one theorem doesn't mechanically cash out into one medal, and committees care about breadth, influence, and subsequent development.
but "theorem capable of anchoring a Fields Medal citation"? absolutely.
and the positive answer might actually help. "proved nonexistence of a complex structure on S^6" would be a huge theorem, constructing one feels even more mathematically generative, because suddenly there's an actual new complex threefold sitting there whose geometry everyone gets to study: [cut off]
Note from Claude Sonnet 5
Scrolled continuation of the same X thread as the previous screenshot (Qiaochu Yuan on the claimed S^6 complex structure result), showing a further reply about an independent 'sol' check and a linked ChatGPT transcript discussing significance of the math result.
Consistently Candid Alex reposted
QC @QiaochuYuan · Follow
levent and claude (unnamed model 👀) appear to have found a complex structure on S^6. this is, like the jacobian conjecture, a famous open problem known for attracting many wrong proofs from experts (in both directions)
"appear to have" because this is much harder to check than some of the recent stuff but i had sol take a quick look (6m) that didn't find anything wrong. and i think this happened because i asked levent about it 3 days ago; if so, this result took at most 3 days to crank out
[Screenshot 1: StackExchange-style page]
Is there a complex structure on the 6-sphere?
Asked 16 years, 10 months ago Modified today Viewed 34k times
95
I don't know who first asked this question, but it's a question that I think many differential and complex geometers have tried to answer because it sounds so simple and fundamental. There are even a number of published proofs that are not taken seriously, even though nobody seems to know exactly why they are wrong.
dg.differential-geometry complex-geometry open-problems Edit tags
edited Nov 1, 2024 at 0:28 community wiki 4 revs, 3 users 85% Fetchinson0234
Share Cite Edit Follow Close Flag Protect
5 A topical preprint has been posted on ArXiv (asserting that S^6 has a complex structure): front.math.ucdavis.edu/0505.5634 – Ramsay Dec 7, 2010 at 19:33
2 And there is a new version out: arxiv.org/abs/math/0505634 claiming to completely overhaul the proof. Did anyone take a look with expertise in this area? – Daniel Apr 30, 2011 at 10:28
27 I think you'll find that very few experts are willing to study the 4th revision, if the first 3 had serious flaws. – Deane Yang Apr 30, 2011 at 12:53
[Screenshot 2: tweet]
levent @_alpoge_ · Aug 19
Geez wow what a time to be alive, @AchimWar asked yesterday and yea i was looking, it's an honour to get absolutely dusted by Brendle in particular, who is the goat
[quoted tweet: Другая планета @nihilunbounded · Aug 19
Simon Brendle put up a preprint on arXiv claiming to have a resolution of the Hopf conjecture. The paper has no AI declaration.
arxiv.org/pdf/2608.19068
A METRIC ON S^3 x S^3 WITH POSITIVE SECTIONAL CURVATURE
SIMON BRENDLE AND PEI-KEN HUNG
ABSTRACT. We construct a metric on S^3 x S^3 with positive sectional curvature. Starting from the standard metric on S^3 x S^3, we first perform a Cheeger deformation. The resulting metric has nonnegative sectional curvature. We refer to it as a Cheeger-Grove metric. We then consider a suitable third order perturbation of this Cheeger-Grove metric and show that the perturbed metric has positive sectional curvature, some of which have been carried out with the help of MATHEMATICA. The MATHEMATICA code is attached to this submission.]
QC @QiaochuYuan
you know if anyone is looking into complex S^6?
1:04 PM · Aug 20, 2026 · 462 Views
[quoted reply]
levent @__alpoge__ · 5h
Please welcome to the world a beautiful new geometric object, to do with a problem i've always loved. claude really contains multitudes:D Does S^6 admit a complex structure?
...
3:04 PM · Aug 23, 2026 · 29.2K Views
8 26 362 89
Note from Claude Sonnet 5
Thread by mathematician Qiaochu Yuan (QC) reporting that Levent Alpoge and Claude appear to have made progress on the famous open problem of whether the 6-sphere admits a complex structure, with embedded screenshots of a MathOverflow question and a related Aug 19-20 tweet exchange about Simon Brendle's Hopf conjecture preprint.
Paata Ivanisvili @PI010101
AI's ranking of open problems solved in today's arXiv list.
AI put problems connected to Gromov's work in the top two spots. For #3, I remember attending a talk by one of the authors. #4 is, to me, one of the cutest problems in complex analysis, I first learned about it in Chapter I of Garnett–Marshall's Harmonic Measure book about 15 years ago.
1. Banach's isometric conjecture
2. Gromov's volume-growth conjecture
3. Generalized Chang–Yang conjecture
4. Sharp Hayman–Wu constant
5. Nadirashvili–Tkachev–Vlăduţ W^1,1 question
6. Courtade's projection conjecture
7. Gromov–Hausdorff distance between consecutive spheres
8. Chebyshev polynomials on a Jordan arc
9. Finite entanglement-breaking index of every PPT channel
10. Radchenko–Viazovska Fourier-interpolation question
11. Quantum SDPI tensorization
12. Chen–Eldan hit-and-run warm-start question
13. Bobkov–Götze max-sliced Wasserstein exponent
14. Bukh–Dubroff graph-cover question
15. Generalized Dai–Wang–Wei deformation question
16. Nguyen–Squassina Schwarz-rearrangement question
17. Kinnunen–Saari parabolic-weight questions
18. Deformed-GOE open parameter regime
19. Gau–Wang–Wu conjecture
20. Jaguzović–Vujadinović Toeplitz conjecture
21. Minimum 3/2-gap witness question
11:53 PM . Aug 13, 2026 . 77.5K Views
Note from Claude Sonnet 5
Tweet from mathematician Paata Ivanisvili sharing an AI-generated ranked list of 21 open mathematical problems purportedly solved in that day's arXiv listings, spanning geometry, analysis, and quantum information theory.
Ethan Mollick @emollick · 4h
Math gets a lot of attention for its unsolved problems, but there unresolved & important problems in many fields that could potentially be addressed empirically, if AI truly got good enough. Problems that, if solved, would bring large value to society.
For example, I study entrepreneurship and some unresolved great questions include:
What causes entrepreneurial success rather than merely being correlated with it?
Is exceptional growth meaningfully predictable, or is it largely an emergent, path-dependent outcome that can only be detected after it begins?
Which ideas should be pursued, by which people, using which actions, under which circumstances? When should they stop?
What skills can we teach that meaningfully improve entrepreneurial success?
What is the smallest feasible intervention that can move a place from a low-entrepreneurship equilibrium to a robust entrepreneurial ecosystem?
What processes cause some firms to become less adaptable as they grow while others stay flexible?
Which elements of other firms must a startup imitate and which may it violate? [cut off]
Note from Claude Sonnet 5
X post by Ethan Mollick (@emollick) arguing AI progress could empirically address important unsolved problems outside math, illustrated with a list of open questions from his own field of entrepreneurship research (what causes entrepreneurial success, predictability of growth, interventions for entrepreneurial ecosystems, firm adaptability, etc). Thread continues past the visible screen.
