Solution to Komlós conjecture was announced recently:
https://arxiv.org/pdf/2609.11189
I have not read the proof yet, but will do so when I have some time. Meanwhile, I want to briefly mention an implication for neural networks that does not seem to be discussed in the paper.
Neural networks repeatedly perform matrix multiplications, in particular products Wx, where W can be an astronomically large matrix of weights w\_ij and x is a vector of activations.
Storing and moving high-precision weights and performing these multiplications require memory, time and energy. One therefore wants to round the weights to a grid hℤ, where h is a super small positive number, while keeping the outputs approximately unchanged.
This is called quantization. Its connection with discrepancy theory and the Komlós conjecture was already discussed by Lybrand and Saab:
https://arxiv.org/abs/2010.15979
For each weight w = w\_ij, consider the two neighboring grid points:
h⌊w/h⌋ and h⌈w/h⌉ (floor and ceiling).
Can we coordinate these choices so that the outputs Wx remain almost unchanged across many inputs x simultaneously?
More precisely, fix inputs x⁽¹⁾, …, x⁽ᵐ⁾, and define
L = maxⱼ √(∑ₛ |xⱼ⁽ˢ⁾|²),
We seek a rounded weight matrix H such that
maxₛ ‖Hx⁽ˢ⁾ − Wx⁽ˢ⁾‖\_∞ ≤ C h L,
where C is an absolute constant, independent of the dimensions, weights and inputs.
There are two questions:
1\. Does such a rounding exist?
2\. what is the reasonable in time algorithm doing this?
If the announced proof is correct, the answer to the first question is yes, with C = 3√(2π).
The paper does not provide a polynomial-time algorithm achieving this guarantee.
P.S. This controls computations Wx on specified inputs x. It does not by itself guarantee accuracy on unseen prompts or control accumulated errors throughout a transformer. It establishes that simultaneous output preservation at lower precision is possible under the stated bound.
> **Paata Ivanisvili @PI010101** · 2026-02-01
>
> Min–max (or max–min) optimization problems in mathematics are among the most subtle creatures in the wild.
>
> A canonical specimen is Komlós’ conjecture.
>
> It predicts a uniform bound C(n,d)<100. The number “100” itself is completely unimportant--replace it by any finite constant
>
> [image]
---
##### Comments
> **Dmitry Rybin @DmitryRybin1** · [2026-09-21](https://x.com/DmitryRybin1/status/2101969052847755548)
>
> Waiting for your analysis of the proof and also what it implies for Komlos constants C\_{p,q} for p-q-norms!
>
> Also can hunt for optimal bounds!
> **Thomas Bloom @thomasfbloom** · [2026-09-21](https://x.com/thomasfbloom/status/2101971534260658612)
>
> Karingula and Lovett posted another 'simpler' proof to the arxiv today: https://arxiv.org/pdf/2609.20979
> **Nicholas Perry @RegolithHunter** · [2026-09-21](https://x.com/RegolithHunter/status/2101972558505529709)
>
> In neural netwoks and computer networking, minmax improvements tend to optimize decision making.
>
> In this instance, it would probably enable better quantization while improving multiplication speed.
> **Mehmet Mars Seven 🐴 @MehmetMars7** · [2026-09-21](https://x.com/MehmetMars7/status/2101985182915998165)
>
> Gpt pro says the calculations in the two new papers check out
>
> [chatgpt.com Check out this chat](https://t.co/OTvfq3lc4c)
Bogdan Ionut Cirstea reposted
Paata Ivanisvili @PI010101 · 22h
Seeing what is happening on the frontlines, I would say this: if we humans want to keep understanding the knowledge AI is producing at the current rate (new proofs, ideas, and approaches) we may need 10× more mathematicians within the next year than we have today.
[quoted tweet]
Daniel Litt @littmath · Aug 14
We're going to need a lot more mathematicians!
x.com/PI010101/statu...
Note from Claude Sonnet 5
Paata Ivanisvili (@PI010101), reposted by Bogdan Ionut Cirstea, saying humans may need 10x more mathematicians within a year to keep up with AI-produced mathematics; quotes a short Daniel Litt (@littmath) post from Aug 14.
ai math resultsmathematiciansverification burdenai progressfuture of mathematics
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.
mathematicsmathoverflowai agentsai math resultsopen problemsprobabilityautonomous agents
Paata Ivanisvili @PI010101 · 13h
The preimage of every line under a conformal map of the unit disk has total length at most π², and this is best possible arxiv.org/pdf/2608.12844
I first learned about this problem from John Garnett and Donald Marshall's wonderful book Harmonic Measure. Chapter I gives the previously known suboptimal bound 4π. A later result showed that the optimal constant is strictly smaller than 4π, and that remained the state of the art until today.
AI did the job. My contribution was to direct it toward the right problem, verify the argument, digest it, and present the solution in a short and hopefully easily readable form. The complete proof is now a little under four pages long.
It is a really nice solution. My first reaction was: "Wow, how was this missed?" I remember having the same feeling when I first read the proof of the Sensitivity Conjecture.
[embedded image of a textbook/paper excerpt]
5. The Hayman–Wu Theorem
We give a very elementary proof, based on an idea of the late K. Øyma [1992], of the theorem of Hayman and Wu. The Hayman–Wu theorem will be a recurrent topic throughout this book.
Theorem 5.1 (Hayman–Wu). Let φ be a conformal mapping from 𝔻 to a simply connected domain Ω and let L be any line. Then
length(φ⁻¹(L ∩ Ω)) ≤ 4π. (5.1)
Hayman and Wu [1981] gave the first proof of (5.1) with 4π replaced by some large unknown constant. Øyma [1992] obtained the constant 4π, Rohde [2002] proved that the best constant in (5.1) is strictly smaller than 4π, and Øyma [1993] proved that the best constant is at least π². The sharp constant in (5.1) is not known. See Exercises 24 and VI.3. We present Øyma's elementary proof, as modified by Rohde.
Note from Claude Sonnet 5
Tweet by Paata Ivanisvili claiming an AI solved the sharp constant (π²) for the Hayman-Wu theorem, with a screenshot of a textbook excerpt (Garnett & Marshall, Harmonic Measure) stating the theorem and its proof history embedded below the text.
mathai capabilitiesconformal mapshayman-wu theoremtwitter
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.
mathematicsarxivai research capabilityopen problems
Greg Burnham @GregHBurnham · Mar 26
Wild. Can you give an intuition for why? I can see it for N=2 ;) I guess that shows how the points can "waste" lots of dimensions, and the geometry is more determined by the points themselves than the ambient dimension. So I guess the question is just, why log(N)?
[6 replies, 18 likes, 5.1K views]
Paata Ivanisvili ✓ @PI010101 · Mar 26
log(N) comes from the union bound + the fact that square of Gaussian is subexponential.
You can think of any linear map f as nxd matrix A, where n is the dimension of the target space, and d the dimension of the space where your N vectors live.
A good starting point is to look among random matrices A having the property E ||Av||^2 =||v||^2 for all vectors v, and then, hopefully, the rest should follow from concentration inequalities, i.e., ||Av||^2 cannot be too far from its average E ||Av||^2=||v||^2. Union bound tells you that the probability this inequality fails for some two vectors among our set of N vectors is at most N^2 times P( ||Av||^2 is outside eps-neighbourhood of its average) < N^2 exp(-n C(eps)). And this is less than 1 if n is of order log(N).
[1 reply, 69 likes, 4.4K views]
Greg Burnham @GregHBurnham · Mar 26
Oh that's cool. So can you get something tighter precisely by the degree to which the square of the Gaussian is subexponential, if that makes sense?
[1 reply, 6 likes, 810 views]
Paata Ivanisvili ✓ @PI010101 · Mar 26
Good point. One can certainly experiment with different random variables, but since the log(N) is already sharp, there's not much room for [cut off]
Note from Claude Sonnet 5
Continuation of the Johnson-Lindenstrauss lemma discussion thread from the prior screenshot — detailed math proof sketch via union bound and concentration inequalities. Pure math/theory content.
mathematicstwitterjohnson-lindenstraussdimensionality reductionconcentration inequalities
**Paata Ivanisvili** @PI010101 2026-03-24
The Johnson--Lindenstrauss lemma says something quite remarkable: if you have an astronomical number N of vectors of large size (say, in a very high-dimensional Euclidean space), then you can linearly map them into a much lower-dimensional space, of dimension about log(N), in such a way that the distances between the vectors are almost preserved.
In other words, you can compress your data dramatically without making it too upset about its geometry. A random matrix with i.i.d. standard Gaussian entries will most likely do the job.
> 2026-03-24
>
> Introducing TurboQuant: Our new compression algorithm that reduces LLM key-value cache memory by at least 6x and delivers up to 8x speedup, all with zero accuracy loss, redefining AI efficiency. Read the blog to learn how it achieves these results: http://goo.gle/4bsq2qI
---
**Greg Burnham** @GregHBurnham [2026-03-26](https://x.com/GregHBurnham/status/2037174210791866595)
Wild. Can you give an intuition for why? I can see it for N=2 ;) I guess that shows how the points can "waste" lots of dimensions, and the geometry is more determined by the points themselves than the ambient dimension. So I guess the question is just, why log(N)?
---
**Paata Ivanisvili** @PI010101 [2026-03-26](https://x.com/PI010101/status/2037192298900115796)
log(N) comes from the union bound + the fact that square of Gaussian is subexponential.
You can think of any linear map f as nxd matrix A, where n is the dimension of the target space, and d the dimension of the space where your N vectors live.
A good starting point is to look among random matrices A having the property E ||Av||^2 =||v||^2 for all vectors v, and then, hopefully, the rest should follow from concentration inequalities, i.e., ||Av||^2 cannot be too far from its average E ||Av||^2=||v||^2. Union bound tells you that the probability this inequality fails for some two vectors among our set of N vectors is at most N^2 times P( ||Av||^2 is outside eps-neighbourhood of its average) < N^2 exp(-n C(eps)). And this is less than 1 if n is of order log(N).
---
**Greg Burnham** @GregHBurnham [2026-03-26](https://x.com/GregHBurnham/status/2037193657804103948)
Oh that's cool. So can you get something tighter precisely by the degree to which the square of the Gaussian is subexponential, if that makes sense?
---
**Paata Ivanisvili** @PI010101 [2026-03-26](https://x.com/PI010101/status/2037247222958744058)
Good point. One can certainly experiment with different random variables, but since the log(N) is already sharp, there’s not much room for improvement beyond optimizing multiplicative constants.
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**William Wale** @snigus [2026-03-27](https://x.com/snigus/status/2037632812887441450)
My sense is that this is why pretty much anything works.
Applications to interps obvious. But also quantisation, evolutionary algorithms, SGD, residual links,
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**Qingping He** @QingpingHe1 [2026-03-26](https://x.com/QingpingHe1/status/2037244960601444513)
Yeah I feel like this result can be used to compress neural network weights too 🤔 Cuz the biggest problem I thought was that people thought the weights were random and fully packed with information but I guess not 🤔
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**Tom Turney** @no\_stp\_on\_snek [2026-03-26](https://x.com/no_stp_on_snek/status/2037202320224629144)
nice explainer. TurboQuant takes this one step further, the QJL stage uses a 1-bit version (just storing signs of the random projection) to correct the bias from the first quantization stage. the geometry preservation holds even at 1 bit per dimension, which is wild.
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**mrkelly** @kellypeilinchan [2026-03-26](https://x.com/kellypeilinchan/status/2037210074511532210)
I learned this the hard way. Theoretical elegance rarely survives the noise of production systems.
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**Mazasiel** @mazasiel [2026-03-28](https://x.com/mazasiel/status/2037890464750223676)
To guarantee all M rows are unique, you need N such that the number of distinct binary patterns (with values -1/1) is at least M.
The number of distinct patterns of width N is 2^N, so you need:
2^N >= M, which gives N >= log2(M).
That's the hard lower bound. If N = ceil(log2(M)),
---
**Andrés Mac Allister** @andresmac73 [2026-03-26](https://x.com/andresmac73/status/2037206069819154857)
If pairwise geometry survives a random projection down to ~logn dimensions, I think it suggests most coordinates are just redundancy, not signal.
We might be overpaying for representation everywhere, in memory, bandwidth, even training...
---
**Len Seaside** @LenSeaside [2026-03-26](https://x.com/LenSeaside/status/2037222826394992675)
If all you need is the distances between the points.
---
**Lando** @cryptomessenger [2026-03-26](https://x.com/cryptomessenger/status/2037208158171517333)
Could you also use structured Johnson–Lindenstrauss with algebra?
Ie preserving the algebraic relationships instead of just preserving the Euclidean distances.
---
**Piyush Sao** @piyusch [2026-03-26](https://x.com/piyusch/status/2037310148511674417)
Why no approx nearest neighbor (ann) algorithm on any metric space uses it? @grok
---
**observer** @msiguc [2026-03-28](https://x.com/msiguc/status/2037713295239450721)
Awesome, intuitive explanation, no prerequisites required (besides understanding of linalg) here:
Paata Ivanisvili ✓ @PI010101
The Johnson--Lindenstrauss lemma says something quite remarkable: if you have an astronomical number N of vectors of large size (say, in a very high-dimensional Euclidean space), then you can linearly map them into a much lower-dimensional space, of dimension about log(N), in such a way that the distances between the vectors are almost preserved.
In other words, you can compress your data dramatically without making it too upset about its geometry. A random matrix with i.i.d. standard Gaussian entries will most likely do the job.
> QUOTED: Google Research ✓ @GoogleResear... · Mar 24
> Introducing TurboQuant: Our new compression algorithm that reduces LLM key-value cache memory by at least 6x and delivers up to 8x speedup, all with zero accuracy loss, redefining AI efficiency. Read the blog to learn how it achieves these results: goo.gle/4bsq2qI
Note from Claude Sonnet 5
A mathematician explaining the Johnson-Lindenstrauss lemma as the theoretical basis behind Google Research's TurboQuant, a new LLM KV-cache compression algorithm. Technical ML-infrastructure content.
machine learningllm efficiencycompressionmathematicstwittergoogle research
