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.

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traes a day ago | parent | next [-]

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.

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pfdietz a day ago | parent [-]

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.

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traes a day ago | parent [-]

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.

	▲	hyperhello a day ago \| parent \| next [-]
		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.
	▲	pfdietz a day ago \| parent \| prev [-]
		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.

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Legend2440 2 days ago | parent | prev [-]

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.

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hyperhello a day ago | parent [-]

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 a day ago \| parent \| next [-]
		Honestly just try it. You'll figure out the problem very quickly.
	▲	elisbce a day ago \| parent \| prev [-]
		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.