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Posted by gmays 7 hours ago

Terrence Tao's ChatGPT Conversation about the Jacobian Conjecture Counterexample(chatgpt.com)
545 points | 344 commentspage 5
khernandezrt 5 hours ago|
Damn. I should have stayed in school.
vb-8448 7 hours ago||
Maybe it's silly, but from someone who is ignorant on this topics, what are the consequences of this kind of "discoveries"?

Is it something "revolutionary" or just another small brick that will pile up until something really "revolutionary" will happen?

arm32 6 hours ago||
To me, this shows that extremely talented and qualified mathematicians (can) use frontier-level LLMs to automate their personal grind-y workloads that would otherwise (probably) take more time to accomplish with natural intelligence.

By itself, no consequence. But over time, provided we keep pumping out talented and qualified mathematicians and keep subsidizing costs, we could maybe hit a breakthrough... somewhere... that has real impact.

sdwr 7 hours ago|||
It's an indicator of AI progress. The solutions aren't especially revolutionary, but no person had been able to solve them after decades of collective attempts.
hyperhello 6 hours ago||
To be fair I don’t think there were too many people really trying to. Symbolically, one could make a parameterization of the Jacobian determinant and then brute force a solution, if one had known such a polynomial existed in only three dimensions.
traes 3 hours ago|||
This is not true at all. The parameter space is absolutely MASSIVE. The counterexample is a degree 7 polynomial in 3 variables, which means 360 coefficients. There's no particular way to bound these coefficients or even the degree or number of variables apriori, but assume you somehow did. Also assume you were confident that it would work with integer coefficients bounded from -12 to 12. Now you have to iterate over 360 degrees of freedom, verify that the Jacobian is a nonzero constant, and somehow show uninvertibility of the transformation, which is not a particularly simple task.

If you searched for coefficients from -12 to 12, this would be 25^360 = 2 * 10^503 different possibilities. A common reference point is that there are 10^80 atoms in the observable universe. Sure you could probably reduce this a bit with clever tricks, but the starting point makes the method completely unviable, even with the knowledge: A) a counterexample exists, B) it's in 3 variables, C) it's in degree 7 or less, D) it's in integer coefficients, E) those coefficients are 12 or lower.

pfdietz 1 hour ago||
Here the search wouldn't have been chosing the coefficients independently. Note that one intermediate variable is a polynomial in the input variables, and it is used in other polynomials. A search over expressions like the ones in the counterexample would have a much smaller search space.
traes 1 hour ago||
1) How do you know this structure is the correct one a priori

2) You are starting at 10^500 possibilities. "Much" smaller is not enough, the order of magnitude of the order of magnitude needs to be changed.

3) You still need all of the other assumptions, which were completely unfounded

Impossible.

pfdietz 14 minutes ago|||
If I pull out the two terms 1+xy and 3 + 4xy as new variables, then make three polynomials that are <= 3 terms in each with coefficients in the range -3 to 3, then there are something like 10^19 possibilities. Multiply this by the various simple possibilities for the definitions of those two new variables.

The coefficients are mostly 1, so biasing toward that would make it much faster.

hyperhello 1 hour ago|||
No, I’m not saying you would know in advance that it was possible, but sometimes you visit the crystal cave and the diamond is just sitting there, so why not work out the odds?

I learned from poking around that checking the invertibility of a system in C is a much, much harder problem than I thought. Nonetheless if that were no object, let’s say coefficients from -16 to 15 (5 bits) times eight terms times choosing up to cubes (64) times three equations is searchable, especially since you have only the final combination of coefficients in the determinant. It’s not impossible to generate the equations like this Fizzbuzz style.

Edit: no. 2048 possible monomials, to the 24th power, not times 24. Fine, can’t brute force it.

Legend2440 6 hours ago|||
Oh yes there were. The Jacobian conjecture is "notorious for the large number of published and unpublished false proofs which turned out to contain subtle errors."

It's not quite the Reimann hypothesis, but many prominent mathematicians have spent years working on this problem. Yitang Zhang wrote his PhD thesis on it.

hyperhello 3 hours ago||
I shouldn’t, but:

F1 = x^3y^3z + 3x^2y^4 + 3x^2y^2z + 7xy^3 + 3xyz + 4y^2 + z

F2 = 3x^3y^2z + 9x^2y^3 + 6x^2yz + 12xy^2 + 3xz + y

F3 = -x^3z - 3x^2y + 2x

That’s the counterexample. Low integer coefficients, power 7 in three variables. If someone said it was there, couldn’t we all have written a pretty simple brute force solution for the search space, especially with the constraints that the symbolic determinant had to cancel to a constant?

traes 3 hours ago|||
Honestly just try it. You'll figure out the problem very quickly.
elisbce 2 hours ago|||
I don't think you even understand the problem. The determinant needs to be a non-zero constant AND you need to prove that particular map is not globally injective, meaning you have to find at least two points mapping to the same value. Of course it looks easy when someone shows you the counterexample.
fragmede 6 hours ago|||
Practically, from this specific one? Nothing, it's very much a math thing. It's like art or music at this level. Are there consequences to a van Gogh?
hyperhello 6 hours ago|||
[dead]
rickypp 7 hours ago||
"Hey Fable, please generate me the next 1000 undiscovered bitcoin hashes"
echelon 6 hours ago||
You're joking, but perhaps LLMs will find a way to mathematically break the complexity of factorization.

Maybe they'll find a solution where P=NP.

That could really throw a wrench into the whole internet thing.

It seems they need an expert human driver for now.

ConceptJunkie 6 hours ago||
I'm sorry, I can't do that, but here is the design for a stable quantum computing platform that should allow you to generate those keys yourself...
NitpickLawyer 6 hours ago||
Some materials are readily available on eMazon and aBay, so I've taken the liberty of ordering those for you. Your credit card bill will be a bit high this month, but it'll be worth it. There weren't any sellers for the advanced EUV lithography machines, so I've hacked into the only place on earth that makes them, changed their records and had them ship it to you. Expect to receive a "pinball machine" from Amsterdam, soon. I've instructed the roomba connected to the local network to start assembling stuff while we wait for the other materials. Oh, and you're gonna need a new toaster.
WithinReason 4 hours ago||
All I can tell from this is that Terrence Tao has good mathematical intuition
housu 7 hours ago||
It's awesome to publish this kind of thing - great PR at least. Even if you don't understand the details, it's interesting to be able to peek into a technical conversation that a world class mathematician is having about their work with a "colleague". It's also the clearest demonstration I've seen of the vision AI people have about a future with truly intelligent copilots in super technical fields.
personalityson 5 hours ago||
The last Fields Medal has been awarded.
gus_massa 5 hours ago|
I expect Centaur Field Medals to be awarded. It's not clear how much steering Levent Alpöge did to get the result. In math it's very common to hide the search pattern and show only a simplified path to reach the final result.
Human-Cabbage 6 hours ago||
Is ChatGPT's interface always this atrociously jittery? Or is it just because this page is getting an HN hug-of-death right now? Every time I try to scroll the whole page goes blank for a few seconds and then re-renders.
7734128 5 hours ago||
I think that's partly because it's a fairly long conversation and partly because of all the LaTeX. There's enough formulas there to fill a small text book, and I doubt their renderer is all that efficient.
ben_w 6 hours ago||
I've found it's had unacceptable scroll performance in long contexts for a while now.

In this case, it takes me 12 seconds to see content when first opening the link, and about 18 to re-render content when scrolling.

buzzin_ 52 minutes ago||
So, the Jacobian Conjecture was done, via an counterexample. Can we now put the models to work on an even more difficult problem: Scrolling 100k of text, on a 128GB 24 core processor, smoothly in a browser?
sjreese 5 hours ago||
Specifically, if the eigenvalues all have real parts that are negative, then the system is stable near the stationary point. If any eigenvalue has a real part that is positive, then the point is unstable. If the largest real part of the eigenvalues is zero, then the Jacobian matrix does not allow for an evaluation of the stability.

Yes—for a continuous-time autonomous system

x ˙ =f(x),f(x ∗ )=0,

this is the standard linearization criterion, with J=Df(x ∗ ):

If every eigenvalue of J has strictly negative real part, then x ∗ is locally exponentially asymptotically stable. If at least one eigenvalue has strictly positive real part, then x ∗ is unstable. If no eigenvalue has positive real part but at least one has real part 0, linearization is generally inconclusive. Nonlinear terms or a center-manifold analysis are needed.

The last case really can go either way. For example, all three scalar equations below have Jacobian J=0 at x=0:

x ˙ =−x 3 , x ˙ =x 3 , x ˙ =0.

Yet 0 is respectively asymptotically stable, unstable, and neutrally stable.

A slightly more precise wording is therefore:

If the spectral abscissa

α(J)= λ∈σ(J) max

Reλ

is negative, the equilibrium is locally exponentially stable. If α(J)>0, it is unstable. If α(J)=0, the Jacobian test is inconclusive.

This criterion concerns the Jacobian matrix of a dynamical system at an equilibrium; it is unrelated to the “constant Jacobian determinant” condition in the Jacobian conjecture.

brador 6 hours ago||
"The determinant identity is almost embarrassingly simple once one writes the map in the right way." #Flexingontheentirehumanspecies
sunrunner 6 hours ago|
Or, "Oh just give it here, let me do it."
jdw64 7 hours ago|
I wish you'd share some conversations from experienced programmers too. How do they ask questions?
sunrunner 6 hours ago||
"You are an expert software engineer with ten years of experience. How do I center a div?"
epolanski 3 hours ago||
Antirez spends weeks planning with multiple models till he has a clear architecture and understood the constraints before he writes a line of code.
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