From c0a3e71283511c976b757b490e2b9fba1cdb494a Mon Sep 17 00:00:00 2001 From: Glen Whitney Date: Sat, 10 Feb 2024 01:09:07 +0000 Subject: [PATCH] Update Engine prototype --- Engine-prototype.md | 1 + 1 file changed, 1 insertion(+) diff --git a/Engine-prototype.md b/Engine-prototype.md index 9c8dd9a..5a91bb7 100644 --- a/Engine-prototype.md +++ b/Engine-prototype.md @@ -5,6 +5,7 @@ One strategy for exploring a positive-dimensional solution variety is to enumera * [Sage](https://doc.sagemath.org/html/en/reference/schemes/sage/schemes/affine/affine_rational_point.html) and [Macaulay2](https://macaulay2.com/doc/Macaulay2-1.22/share/doc/Macaulay2/RationalPoints2/html/_rational__Points.html) can enumerate rational points over various fields. For $\mathbb{Q}$, however, they turn out to use mostly brute force. * Let's try the classic strategy of taking [rational-slope slices](https://www.quantamagazine.org/how-simple-math-reveals-rational-points-on-curves-20210722/) through a known point. In our case, the known point is the totally degenerate solution where all the points coincide and all the spheres coincide. * The OP of [this question](https://mathoverflow.net/q/255567) suggests this strategy for finding rational points on a sphere. +* We discussed in person birational isomorphisms of a single quadratic surface, by projecting from a known point, and the possible limitations of that in attacking the simultaneous solutions of multiple quadratics. But when it happens that all the quadratics share a solution, as in our current toy case of incidence-and-tangency only constraints, is there any chance it would work to parametrize the solutions to one of the quadratics, and then project all the other conditions into the parameter space using the selected isomorphism? Would they remain quadratic in the parameter space and could we then repeat? Does this question I am asking even make sense? ## Basis optimization