Gro-TsenHere's a mathematical fact which I find amazingly counter-intuitive:
There exists a polytope (=the convex hull of a finite set) in ℝ⁴ which is not combinatorially equivalent to one with rational vertex coordinates (=the convex hull of a finite subset of ℚ⁴)!
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Mark Dominus@gro-tsen.bsky.social What prevents one from perturbing the vertices to points with rational coordinates by sufficiently small amounts that the topology is unaffected?
Oscar Cunningham@mjd @gro-tsen.bsky.social If you perturb the vertices of a quadrilateral, how do you guarantee that they remain coplanar? In fact I'm not surprised that this doesn't hold in 4 dimensions so much as I'm surprised that it does hold in 3 dimensions.
Simon Tatham@OscarCunningham @mjd @gro-tsen.bsky.social ... I think I begin to see. If you have a face of degree > 3, then perturbing its vertices might leave it nonplanar. My knee-jerk answer is to instead perturb the face normals instead of the vertices. But that's only doing the same thing in duality, so you have the flip-side problem that a _vertex_ of degree > 3 might separate into two vertices.
So if you had a solid in which all faces had degree 4 or more, and so did all vertices, you might manage to arrange that either kind of perturbation destroyed one or the other.
Except that that doesn't work in 3 dimensions, because those constraints are incompatible with Euler's formula.
But in four dimensions, Euler's formula isn't quite the same. So presumably the answer is that it _does_ permit the analogous lower bounds to coexist?
@simontatham @mjd @gro-tsen.bsky.social What if a polyhedron has some vertices of degree 3 and some of degree > 3, and likewise for faces? I don't see why we could find a perturbation that works in that case.