Comment by Davidzheng
2 months ago
I'm going to paraphrase what GPT told me: Consider the canonical degree 3 (subvariety of the trivial P1 bundle consisting of zeros) cover of the projectivization of homogenous polynomials of degree 3 in 2 variables (so it's a 3fold cover of P^3). The top space is P1 x P2 and if you take a standard affine open of the base and look at the cover over that restricted to a subset where the zero of the cubic is simple you get the map for some choice of coordinates...
I honestly have no idea if it's correct lol I didn't check it (I should given I actually work in AG) but it doesn't look impossible at first sight
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...
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.
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
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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.
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