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John Baez (Jul 20 2026 at 08:33):The Jacobian conjecture is a crazy relative of the inverse function theorem.
The inverse function says that if is a smooth function and its matrix of partial deriatives is invertible at some point, has a smooth inverse in some neighborhood of that point.
The Jacobi Conjecture says that if is a polynomial function and its matrix of partial deriatives is invertible at every point, has a polynomial inverse.
This seems inherently implausible, but it was open since 1939, resisting attempts to find a counterexample... until yesterday.
John Baez (Jul 20 2026 at 08:35):The mathematician Levent Alpöge, who works for Anthropic, found a counterexample to the Jacobi conjecture using Claude Fable. It's easy enough to check:
F(𝑥,𝑦,𝑧) = (𝑦²(3𝑥𝑦+4)(𝑥𝑦+1)+𝑧(𝑥𝑦+1)³, 3𝑥𝑦²(3𝑥𝑦+4)+3𝑥𝑧(𝑥𝑦+1)²+𝑦, 2𝑥−𝑥³𝑧−3𝑥²𝑦)
has Jacobian determinant equal to -2 everywhere, but it's not invertible since
F(0, 0, -1/4) = F(1, -3/2, 13/2) = F(-1, 3/2, 13/2) = (-1/4, 0, 0)
John Baez (Jul 20 2026 at 08:37):Omar Antolin passed on some tweets where Claude Fable kept rechecking its work, raising various questions, some quite hilarious.
"Therefore the overwhelmingly likely resolution: THE MAP IS NOT ACTUALLY WHAT IT SEEMS - maybe it's not a map ℂ³→ℂ³?? It is."
Claude Fable, freaking out:
"... but come ON. If this simple example were valid, [the Jacobi Conjecture] would have been settled long ago: this is clearly designed as a puzzle for AI/humans."
But no: sometimes famous open conjectures have fairly simple solutions!
Jens Hemelaer (Jul 20 2026 at 08:47):Any idea already what the consequences are for the closely related Dixmier conjecture? There are some subtleties regarding determinant 1 vs. nonzero determinant and n vs. 2n.
Morgan Rogers (he/him) (Jul 20 2026 at 08:49):Ah yes, how could we have not spotted this trio of polynomials in 3 variables featuring at least one term of degree 7? Very careless in retrospect.
Martti Karvonen (Jul 20 2026 at 08:53):Jens Hemelaer said:
Any idea already what the consequences are for the closely related Dixmier conjecture? There are some subtleties regarding determinant 1 vs. nonzero determinant and n vs. 2n.
Here's a claimed negative resolution built on this.
fosco (Jul 20 2026 at 09:13):John Baez said:
Claude Fable kept rechecking its work, raising various questions, some quite hilarious.
"Therefore the overwhelmingly likely resolution: THE MAP IS NOT ACTUALLY WHAT IT SEEMS - maybe it's not a map ℂ³→ℂ³?? It is."
hmm hmm hmm... I wonder why the training set of Claude is skewed towards these affectations and not others.
fosco (Jul 20 2026 at 09:15):If anything, this shows the sheer power of AI of getting rid of all the factoids that are a few brute-force attempts away from being answered. In a perfect world this would skew the power dynamics of mathematics towards structural thinking over computation... in our world, I don't know.
fosco (Jul 20 2026 at 09:16):One thing: "this seems inherently implausible"
does it?!
John Baez (Jul 20 2026 at 11:24):I don't see anything saying "this seems inherently implausible". But I Fable is correct to worry
This example is suspiciously compact. If a 3-to-1 Keller map on this simple existed, it would have been found decades ago by computer searches.
Kevin Carlson (Jul 20 2026 at 11:25):you said “this seems inherently implausible” about the conjecture itself, John! I do wonder why Fable is wrong that this should have been found long ago by old-fashioned computer search.
John Baez (Jul 20 2026 at 11:30):Oh, okay. I thought Fosco was talking about Fable's solution. Yeah, the conjecture itself seems inherently implausible in a naive prima facie way, not based on any deep knowledge of algebraic geometry.
Why should local invertibility imply global invertibility? For example, everyone learns in school an analytic map such that the determinant of its Jacobian is invertible everywhere, so is locally one-to-one, but fails to be one-to-one. But the Jacobian conjecture says this can't happen for polynomials. So somehow the magic would need to come from replacing "analytic" by "polynomial".
Of course, a lot of magic comes from replacing "analytic" by "polynomial". So maybe if I actually understood algebraic geometry, I'd know something that makes the Jacobian conjecture seem plausible. But I don't, so it always seemed amazing-if-true.
Oisín Flynn-Connolly (Jul 20 2026 at 14:24):Has anyone tried trying to resolve the Hadamard conjecture this way? I guess it's a slightly different type of conjecture but is something that strikes me as something likely to be in the convex hull of human knowledge (and, as of yesterday I would have imagined it to be infinitely easier than the Jacobian conjecture).
John Baez (Jul 20 2026 at 14:31):I hadn't even known about the Hadamard conjecture:
For every integer there's a matrix whose entries are either +1 or −1 and whose rows are mutually orthogonal.
Wikipedia says that in 2014, there were 12 multiples of 4 less than 2000 for which no Hadamard matrix of that order was known.[9] They are: 668, 716, 892, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948, and 1964.
So, the easiest thing for a LLM to try might be to find a Hadamard matrix of order 668. Proving that none exists could be hard.
Unlike the Jacobian conjecture, where the counterexample was very easy to check, a counterexample to the Hadamard conjecture is a statement that at least matrices don't have rows that are mutually orthogonal!
Oisín Flynn-Connolly (Jul 20 2026 at 14:39):In principle, sure, but if it's false, there's likely some kind of combinatorial obstruction that may already exist in some form the literature (or a general construction, but my guess is that it is probably false, else one would already have been found).
John Baez (Jul 20 2026 at 15:00):By the way: an "old question of Grothendieck about whether every finite free group scheme of order n was killed by n" was resolved in the negative on July 11th by ChatGPT Sol.
Morgan Rogers (he/him) (Jul 20 2026 at 15:36):I find concerning Buzzard's attitude is that grad students should be paying out of pocket for these things if he deems that they're necessary for the job:
A few days earlier I had got an email from a professor in the maths department here at Imperial, expressing surprise that some of our graduate students were paying $200 per month to access models such as Sol and Fable. He said that he thought that these people were crazy. I did not immediately respond. But after meeting with Andrew I emailed the professor back and told him that in my opinion, any PhD student who was not paying $200 per month to access these tools was crazy.
I also think it would be a pretty devastating blow to the future of funding of mathematics research (or at least of mathematics researchers) if his suggested use of AI to find a counterexample to the Hodge conjecture were to be successful. Eek.
Kevin Carlson (Jul 20 2026 at 15:40):Yes, that's likely to be more than a tenth of a graduate student's gross income, utterly tasteless of Buzzard. Unionized grad student populations may well start demanding funding for this kind of subscription, though it may be quite self-destructive.
John Baez (Jul 20 2026 at 15:41):I guess he thinks if you're poor you're crazy.
fosco (Jul 20 2026 at 16:46):is he rich tho?
John Baez (Jul 20 2026 at 16:48):He's probably doing quite well, being the face of mathematical formalization in Lean.