Per mentalizzare, capire il senso della congettura, ricordiamo:
occorre considerare tutte le derivate parziali possibili, che sono
organizzate in una matrice, detta jacobiana
Teo: se il determinante jacobiano e' continuo e diverso da zero in un intorno,
allora e' invertibile nell'intorno.
NdR forse sarebbe meglio dire "congettura dello jacobiano".
Levent Alpöge at Harvard University wrote on X 2:19 AM · Jul 20, 2026
hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final
((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)
cmt:
Per fortuna c'e' Terence, anche lui ha bisogno di "digerire" un boccone cosi' servito
https://terrytao.wordpress.com/2026/07/21/a-digestion-of-the-jacobian-conjecture-counterexample/
Mon, Jul 20 at 7:17 AM
prompt: The following polynomial from C^3 to C^3 was just announced as a counterexample to the jacobian conjecture! Det[D[{(1+x y)^3 z + y^2 (1 + x y) (4 + 3x y), y + 3x (1 + x y)^2 z + 3x y^2 (4 + 3x y), 2x - 3x^2 y - x^3 z}, {{x,y,z}}]] . It is remarkable that the jacobian is constant, that is an exceptional amount of cancellation. Does this polynomial map have any symmetry or other structure that makes this cancelation less miraculous?
visualizer of the counterexample map
#comment-693677 e i 3 seguenti