Post #3950678

2026-07-20 04:37 UTC

The Jacobian conjecture is false! The Jacobian conjecture proposed that if you have a map F:C^n→C^n with polynomials as the components, and the determinant of the matrix of partial derivatives of F is a non-zero constant, then F will have a polynomial inverse. https://en.wikipedia.org/wiki/Jacobian_conjecture But someone has now found a counterexample. This is easy to check, so there’s no doubt that it’s correct. https://x.com/__alpoge__/status/2079028340955197566 𝐹(𝑥,𝑦,𝑧) = (𝑦²(3𝑥𝑦+4)(𝑥𝑦+1)+𝑧(𝑥𝑦+1)³, 3𝑥𝑦²(3𝑥𝑦+4)+3𝑥𝑧(𝑥𝑦+1)²+𝑦, 2𝑥−𝑥³𝑧−3𝑥²𝑦) F has a constant Jacobian determinant of -2, but it cannot have an inverse because it is not one-to-one: F(0, 0, -1/4) = F(1, -3/2, 13/2) = F(-1, 3/2, 13/2) = (-1/4, 0, 0)

Replies (6)

@johncarlosbaez@mathstodon.xyz 2026-07-20 07:42

@gregeganSF@mathstodon.xyz - Wow, that's the most amazing thing I've heard for a while! I've never heard any good reason to think this conjecture is true, but I assumed people had looked for counterexamples so hard that any counterexample would have to be much more complicated than this one. "The Jacobian conjecture proposed that if you have a map F:C^n→C^n with polynomials as the components, and the determinant of the matrix of partial derivatives of F is a non-zero constant, then F will have a polynomial inverse." I find that this statement, on Wikipedia, sort of weakens the blow. Instead of saying "is a nonzero constant" I would say "everywhere nonzero", since that makes the conjecture sound more general. It's equivalent, but it sounds more impressive. Talking to myself, I'd actually say "if F: ℂⁿ→ℂⁿ is a polynomial map whose Jacobian is invertible everywhere, F has a polynomial inverse." That clarifies the charm of the conjecture: it's an insane relative of the inverse function theorem, which says that a smooth function whose linearization at a point is invertible must have a smooth inverse in some neighborhood. We change "smooth" to "polynomial", "at a point" to "at every point", and "in some neighborhood" to "everywhere". Puzzle: Find a smooth F: ℝⁿ→ℝⁿ whose Jacobian is invertible at each point but where F does not have a smooth inverse.

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@oantolin@mathstodon.xyz 2026-07-20 04:51

@gregeganSF@mathstodon.xyz Surprisingly low degree, right? Wow.

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@abdullahkhalids@anarchism.space 2026-07-20 05:32

@gregeganSF@mathstodon.xyz I wonder if more counter-examples can be found now that we know they exist.

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@ProfKinyon@mathstodon.xyz 2026-07-20 05:43

@gregeganSF@mathstodon.xyz I worked on it as a grad student and would have put money on it being true.

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@aeva@mastodon.gamedev.place 2026-07-20 06:30

@gregeganSF@mathstodon.xyz eat it jacob!

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@Quantensalat@scicomm.xyz 2026-07-20 07:16

@gregeganSF@mathstodon.xyz Is it maybe still true with some additional constraints which were implicit in peoples' intuition all along?

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