Anybody can ELI5?

Are you at least a little familiar with linear algebra?

If so, you've probably heard of the determinant. It's a certain way of "summarizing" a matrix with one value.

The determinant in this case is of the Jacobian, which is a matrix you can construct from a multi-variable function. Each term is the partial derivative with respect to each variable (x, y, z, etc.), with one line per output variable (vector element).

The Jacobian of a polynomial function is, in general, going to be a matrix where every term is some polynomial expression. And the determinant of that will also be a complicated expression. But in some cases all the variable terms cancel out and you're left with a single constant (0 or some other value).

The conjecture says that if the Jacobian determinant is constant (i.e., all the terms cancel out), then there must be a polynomial inverse. And the key condition for an inverse is that there must not be two input points that evaluate to the same output. It's just like y=x^2. It's not invertible, because both +2 and -2 square to +4.

So if you can find a function where the Jacobian determinant is constant and also find two or more points that evaluate to the same output, then you've found a counterexample to the conjecture. And that's what's been done. And remarkably, the counterexample is pretty simple. It would be tedious but a bright high school student could verify it.

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krackers a day ago | parent | next [-]

I think another way to understand it is the generalization of the inverse function theorem. The inverse function theorem gives you local invertibility, but even being "locally invertible" everywhere does not imply global invertibility (you don't need too pathological an example to see this, a periodic function serves iirc).

The Jacobian conjecture roughly asks what whether local invertibility gives you global invertibility when you restrict only to polynomials (which we might hope "behave nicely"). Apparently for polynomials over reals this was disproved a while back, but up until now the general case of polynomials over complex numbers was open.

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tptacek a day ago | parent | prev | next [-]

Nice!

(It's gotta be a nonzero constant, right, a nonsingular matrix).

	▲	SyzygyRhythm a day ago \| parent [-]
		Yeah, the only possible counterexamples would have a constant but non-zero determinant (in this case it was -2).

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spuz a day ago | parent | prev [-]

Thanks that's a very nice summary.

	▲	SyzygyRhythm a day ago \| parent [-]
		Thanks, though I realize (thanks to tptacek) that I didn't properly emphasize that the determinant must be constant but non-zero. If the determinant were zero, then (for basically the same reason as why you can't divide by zero) the matrix wouldn't be invertible. You wouldn't expect the polynomial to be invertible in that case, either. If the determinant had free variables left over, then there is some combination of variables that will make it zero (e.g., if it came out to x^2-1, then it would be 0 for x=1 or -1). So maybe inversion won't fail everywhere, but it will fail in those spots, which would also lead one to think the polynomial is not invertible. But if it's constant and non-zero, then there is no place where you can't invert the Jacobian. So it seems very plausible that this would apply to the polynomial as well. That's essentially what motivated the conjecture in the first place (leaving out some details, of course).