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 Are any examples known? Or if not, is there a lower bound on the number of vertices required?
Gro-Tsen (temp account)@mjd @gro-tsen.bsky.social There seems to be an explicit construction with 33 vertices (theorem 9.2.1 in Richter-Gebert's book), but I admit I've only glanced at it.
0xDE@gro_tsen_test @mjd @gro-tsen.bsky.social Moritz Firsching has enumerated 274148 combinatorial types of 9-vertex 4-polytopes and found rational coordinates for all of them, so any irrational 4-polytope needs at least 10 vertices: https://arxiv.org/abs/1803.05205 . There is still a big gap between 10 and 33, though.
The complete enumeration of 4-polytopes and 3-spheres with nine vertices
We describe an algorithm to enumerate polytopes. This algorithm is then implemented to give a complete classification of combinatorial spheres of dimension 3 with 9 vertices and decide polytopality of those spheres. In particular, we completely enumerate all combinatorial types of 4-dimensional polytopes with 9 vertices. It is shown that all of those combinatorial types are rational: They can be realized with rational coordinates. We find 316014 combinatorial spheres on 9 vertices. Of those, 274148 can be realized as the boundary complex of a four-dimensional polytope and the remaining 41866 are non-polytopal.