[Davidzheng]

here's another version directly from the horse's mouth : "Consider the natural map π: P¹ × Sym²(P¹) → Sym³(P¹), (p, {q,r}) ↦ {p,q,r}. Let R be its ramification divisor and let H ⊂ Sym³(P¹) ≅ P³ be a hyperplane tangent but not osculating to the small diagonal; identify X := (P¹ × Sym²(P¹)) \ (R ∪ π⁻¹(H)) ≅ A³ and Y := Sym³(P¹) \ H ≅ A³. Take π|X: X → Y." This is in fact so simple if correct that someone should have found it after all...

[jlev1]

My Claude found a similar description (it phrased it in terms of the natural map from "cubics with a choice of root" to "cubics"). The part that seems not at all simple or obvious is the fact that X is isomorphic to A^3. In your presentation (and more or less similarly in the one my Claude found), X is given as P1 x P2 minus a reducible hypersurface, also I think R itself is reducible since it contains points of the form (p, {p, q}) and (p, {q, q}). Then it takes some calculation to identify X with A^3.

[Davidzheng]

on the other hand it's incredible to me as someone who doesn't do computations that GPT took one look and saw the geometry--though it's not saying much we should ask ppl who do AG computations

[jlev1]

It’s not that surprising (to me) that it would recognize these features, in that the features it picks up on are intrinsic to the map. Once you have the map, which is generically of degree 3, there’s the locus where the map drops from degree 3 to 2, which contains a big hint because it pops out the equation for the discriminant locus of a cubic. Then there’s the other bit about H, which becomes more apparent from the formula after simplifying things a bit in terms of discriminant. I still don’t yet understand the rest of the calculation, but it’s visibly simpler after you notice the role of the discriminant.

[Davidzheng]

yeah I think it's probably correct -- this is actually insanely simple (except for the fact that the codomain as described is not obvious isomorphic to A^3.

[jlev1]

I agree.