Terence Tao's ChatGPT conversation about the Jacobian Conjecture counterexample

Math isn’t necessarily hard, but it’s incredibly dense

A simple statement like let f(x) be a continuous function can carry a lot of definitions

In that statement, if you missed the day in class where they covered continuous functions it might not even register that it’s a well defined term

And that’s the most over simplistic example I could think of

As a math major, I remember that being one of the first lessons I learned, that every single word could be carrying a lot of weight so to look things up in detail if I was ever struggling on a problem. One of the oldest entries in my memory.md file

Some areas are hard in different ways. I could never quite wrap my head around the way logicians have to think. A clever combinatorial bijection is a work of art you probably can explain to a undergrad class easily but good luck coming up with it. And number theorists will throw the kitchen sink at their problems: no area of math is safe from getting used by them.

People who do this have spent years of their life thinking in this language and studying it, so it is going to be hard. We're also not good at communicating the intuition which for algebraic geometry often comes from other fields.

Yes—for a continuous-time autonomous system

x ˙ =f(x),f(x ∗ )=0,

this is the standard linearization criterion, with J=Df(x ∗ ):

If every eigenvalue of J has strictly negative real part, then x ∗ is locally exponentially asymptotically stable. If at least one eigenvalue has strictly positive real part, then x ∗ is unstable. If no eigenvalue has positive real part but at least one has real part 0, linearization is generally inconclusive. Nonlinear terms or a center-manifold analysis are needed.

The last case really can go either way. For example, all three scalar equations below have Jacobian J=0 at x=0:

x ˙ =−x 3 , x ˙ =x 3 , x ˙ =0.

Yet 0 is respectively asymptotically stable, unstable, and neutrally stable.

A slightly more precise wording is therefore:

If the spectral abscissa

α(J)= λ∈σ(J) max

Reλ

is negative, the equilibrium is locally exponentially stable. If α(J)>0, it is unstable. If α(J)=0, the Jacobian test is inconclusive.

This criterion concerns the Jacobian matrix of a dynamical system at an equilibrium; it is unrelated to the “constant Jacobian determinant” condition in the Jacobian conjecture.