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Claude Fable produced a counterexample to the Jacobian Conjecture

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Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#24

Post by rowbin »

hliyan wrote: post #17

I suspect the LLM was able to synthesize a counterexample because of the availability of a lot of prior work: > The Jacobian conjecture is notorious for the large number of published and unpublished proofs that turned out to contain subtle errors. https://en.wikipedia.org/wiki/Jacobian_conjecture#cite_note-...

Maybe, maybe not. The "proofs" may not have helped at all with finding a counterexample. Either way, it doesn't matter. A counterexample was found, no one found one before even though clearly a lot of people have tried who also had access to the prior "proofs".

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#25

Post by hyperhello »

loubbrad wrote: post #2

Context: https://en.wikipedia.org/wiki/Jacobian_conjecture

> Jacobian conjecture [...] states that if a polynomial function from an n-dimensional space to itself has a Jacobian determinant which is a non-zero constant, then the function has a polynomial inverse.

> ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)

But 1 != -1 and -3/2 != 3/2 . So it's not its own inverse. Is the conjecture that it is its own inverse or that is has an inverse?

Edit: it was worded a bit strangely, but it is saying that [ (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) ].map(F) all produce (-1/4, 0, 0). Thus it has no inverse and indeed disproves the Jacobian conjecture.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#27

Post by nullsanity »

krackers wrote: post #13
luciana1u wrote: post #10

the conjecture held for 85 years and the counterexample was announced in a format that expires after seven days

The surprising thing is that the counterexample seems relatively "simple" in that it's low degree, with coefficients that aren't too large. Does anyone more familiar with this know why this _wasn't_ found earlier, when it seems like you could brute-force through some low-order polynomials?

Same awnser as much of the LLM Proofs - people cared about other things. There isn't a lot of money in academic math, and the ones that love it don't look for low value findings. Proofs like these are, funnily enough, usually the domain of hobbyists - but over the last few years, the "Monetize everything" mentality and struggling first world economy has pushed people away from interesting academic pursuits on their free time.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#28

Post by cyphar »

aurareturn wrote: post #12
scarmig wrote: post #11

"Any idiot could have done this, it's just high school calculus and just a counterexample anyway. Stochastic parrot, spicy autocomplete, AI psychosis. Wake me up when an AI does something real."

The goal posts have moved. People generally stopped saying this stuff now. Even if you go to the ultimate anti-AI subreddit r/betteroffline, they've changed from "AI is useless" to "AI is good but the AI bubble will collapse soon" over the last 6 months.

I don't think the (fairly factual) description of these systems as stochastic parrots means that they will never do useful work, just that they are not intelligent in the way we believe animals to be (to "push back" on your anecdata, I've also heard fewer people claiming that LLMs are actually conscious in the past year -- maybe we're reaching the happy medium?). That was the point the stochastic parrots paper and Chinese room thought experiments were making -- nobody claimed that the man in the Chinese room would be unable to accurately translate Chinese text.

Fuzzers are another kind of stochastic generator but nobody would claim they don't do useful work in a way that is hard to replicate through deterministic methods. (I still find the code these models produce kind of awful, but advancements in harnesses do mean that they can finally produce code that works most of the time.)

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#29

Post by meowface »

bigyabai wrote: post #26
pyuser583 wrote: post #21

I asked Fable to verify and it absolutely freaked out! I have no idea what any of this stuff even means, but my AI thinks I’m a legend level mathematician!

https://en.wikipedia.org/wiki/Sycophancy_(artificial_intelli...

The interesting thing about using Claude Fable 5 is it's nearly as irritatingly sycophantic as past Claudes while genuinely being smarter than the previous models. So you get a kind of yo-yoing of it glazing you as a creative genius and disappointedly revealing to you that your ideas are bad and dumb.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#30

Post by krackers »

hyperhello wrote: post #25
loubbrad wrote: post #2

Context: https://en.wikipedia.org/wiki/Jacobian_conjecture

> Jacobian conjecture [...] states that if a polynomial function from an n-dimensional space to itself has a Jacobian determinant which is a non-zero constant, then the function has a polynomial inverse. > ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0…

I think the map sends (1, -3/2, 13/2), -> (-1/4, 0, 0) and also (-1, 3/2, 13/2) -> (-1/4, 0, 0) so it's not invertable which disprove the jacobian conjecture that polynomial maps over complex numbers with a jacobian that's non-zero are globally invertible.

(Just as a note for myself, I had to think of why the fact that such jacobians are constant is a byproduct, I guess it's because of lioville's theorem implying that any polynomial over C that never hits 0 must be a constant [because the reciprocal is bounded and thus must also be a constant])

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