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Francisco Santos's user avatar
Francisco Santos's user avatar
Francisco Santos
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A rational polytope that is not a 01-polytope?
Embedding in higher dimension does not give you additional 0/1 polytopes. If you have a $d$-polytope $P$ embedded in dimension $n$ for some $n>d$ there are always $n-d$ coordinates that you can forget, to obtain a projection of $P$ in $\mathbb R^d$ with the same combinatorial type. If $P$ was 0/1, the projection still is.
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A rational polytope that is not a 01-polytope?
Changed dimension to $n$, as in the original post
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Simplicial polytope with regular cones
As the OP says, the condition implies that "the vertices of [each of] the facets and the origin form a unimodular simplex". This implies that the facet can be written as $ax \le 1$ for an integer vector $a$. Thus, the polar polytope has integer vertices, namely the $a$'s coming from the facets. Together with the fact that $P_0$ has integer vertices this says that $P_0$ is reflexive.
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Product of vertex degrees of an edge in a planar graph
Another easy way to construct triangulations with minimum degree five: For every even 𝑛≥12: start with an antiprism with n-2 vertices (that is, take two (n-2)/2 -gons and join them with a zig-zag of edges) then cap the anti-prism with two (n-2)/2-pyramids. You get all vertices of degree five except for the two apices of pyramids, of degree (n-2)/2. For odd 𝑛≥15: do the same (getting n-1 vertices, with the two apices of degree k≥6) and at the end split one of the apices into two vertices of degrees i and j with i+j=k+4, which allows you to take both i and j≥5 (e.g., i=5 and j=k-1).
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Triangulations of polyhedra
As Pak and O'Rourke point below, the asymptotics of the number of triangulations of cyclic polytopes is more or less known, and covered in my book on Triangulations with de Loera and Rambau. But the original question is about the exact number of triangulations of the cyclic 3-polytope with n vertices. I just want to point out that a nice closed formula for that could exist, although we do not know it. Related to this, I have a paper where we compute the exact number of triangulations of the cyclic d-polytope with d+4 vertices (doi.org/10.1007/s00454-001-0050-y)
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Every triangulation of an oriented matroid is partitionable
Since I am an author of the two "excellent sources" I should give my opinion: no, as far as I know this is not published. When I was writing the "Triangulations of Oriented Matroids" paper I thought about this and the way I recall it (this was 20 years ago) is that at some point I thought I had a proof but then was not convinced by it and left the question out of the paper.
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Not especially famous, long-open problems which anyone can understand
"It has been proved that infinitely many numbers appear twice; similarly three times, four times". I guess most of this is easy: every prime number $p$ appears exactly twice, every number of the form $\binom{p}{2}$ ($p\ge 5$) appears exactly four times, and every number of the form $\binom{2k}{k}$ appears an odd and at least three number of times.
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Triangulation of S^2xS^2
This other paper of Lutz is more relevant to your question: arxiv.org/pdf/math/0506372.pdf. In particular, Theorem 40 says that the vertex-minimal triangulations of $S^2\times S^2$ have 11 vertices. It is not unique but Lutz conjectures that the $f$-vector $(11, 55 , 150, 170, 68)$ is component-wise minimal. (I am assuming you want a simplicial triangulation, as opposed to Ryan Budney's answer),
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Lattice question
Requiring the fundamental unit not to have lattice points other than its vertices excludes two-dimensional counter-examples (since the only lattice polygons with that property are a unit parallelogram and half of it) and it also excludes counter-examples constructed using a Minkowski sum of segments as a fundamental unit. The latter was convenient for counter-examples since Minkowski sums have few different edge vectors. But I would expect the statement to still fail in this restrict version.
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Lattice question
Excluding interior points in the "fundamental unit" is certainly not enough: a counter-example with interior points can be converted into one without by Cartesian product (of both $P$ and the fundamental unit) with a segment in one extra dimension. I guess you mean that the fundamental unit does not have any lattice point apart of its vertices.