22

Philosophical questions deserve philosophical answers, so I am afraid no amount of references and specific results will probably satisfy you. Let me try to explain it in a somewhat generic way.
Think about it this way - why care about sequences like $\{n!\}$, Fibonacci or Catalan numbers? The honest answer is "because they come up all the time". Now, ...

21

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler
In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty ...

answered Oct 2 '20 at 12:36

Moritz Firsching

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17

The origins of associahedra go back to the thesis work in homotopy theory of Jim Stasheff in the early 1960's.
He did graduate work at Oxford, working with Ioan James, who in the mid 1950's had proved lots of beautiful theorems about the free associative topological monoid $JX$ generated by a based topological space $X$; in particular, showing that, when $...

13

Pick's theorem says these two convex lattice polygons have area
$$i+\frac{b}{2}-1 = 4 + 10/2 -1 = 8 \;,$$
and they both have perimeter $8 + 2 \sqrt{2}$.
You can see I've "bumped out" two corners of an underlying octagon.
(I am interpreting the OP's phrase "the same area and perimeter"
as "the same area and the same perimeter" as ...

answered Mar 13 '21 at 20:56

Joseph O'Rourke

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12

In my opinion there are two answers to this question.
The first is that these particular classes of polytopes have fascinating combinatorial properties and structure. Presumably you're aware of the work of Postnikov and others in this direction. In my view, and the view of many others, these properties make the polytopes worth studying in their own right.
...

answered Jan 9 '20 at 16:33

Timothy Chow

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There are remarkable combinatorial formulas for the face numbers and the volumes (of certain geometric realizations of) of these polytopes and a more general family ("generalized permutohedra" a.k.a. "polymatroids") to which they belong. These numbers include classical sequences like the Eulerian numbers, Catalan numbers, $(n+1)^{n-1}$, etc. This is a major ...

9

I'm not an expert, but perhaps this could help:
Hohlweg, Christophe. "Permutahedra and associahedra: Generalized associahedra from the geometry of finite reflection groups." arXiv:1112.3255 (2011):
"Permutahedra are a class of convex polytopes arising naturally from the study of finite reflection groups, while generalized associahedra are a class of ...

answered Jan 9 '20 at 14:27

Joseph O'Rourke

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9

The answer is No, there are no other such polytopes. The proof is quite laborious in parts, and I wrote it down in this article.
Theorem. In dimension $d\ge 4$, an edge-transitive polytope is vertex-transitive.
The idea is as follows: first, show that every edge-transitive polytope $P$ that is not vertex-transitive has the following three properties:
all ...

8

There are other polytopes. To construct one let's do the following. Remember first that in the hyperbolic $4$-space there exists a regular compact right-angled 120-cell. Here, right-angled means that any two adjacent faces intersect under angle $\frac{\pi}{2}$. Regular means, that all the faces are isomeric, and the polytope has the same group of self-...

8

This is true in all dimensions, and can be proved by induction (on $d$) applied to the following (slightly stronger) hypothesis:
Theorem: If $P$ is a convex $d$-polytope with $k$-in-spheres for all $k \in [0, d-1]$, then:
$P$ is regular.
$P$ is determined (up to an element of the orthogonal group $O(d)$) by that $d$-tuple $(r_0, r_1, \dots, r_{d-1})$ of $k$...

6

It is not clear what do you mean by tightest bound --- in which sense tightest?
Also you did not say from above or from below. Anyway, let me say something, hope it will help.
Let
$$Q = \{x \in \mathbb{R}^n : Ax \leq \mathbb{1}\},$$
where $\mathbb{1}=(1,\dots,1)$
Note that
$$P'=P+\varepsilon\cdot Q.$$
A. You can get lower bounds from Brunn–Minkowski ...

answered Jun 6 '14 at 23:58

Anton Petrunin

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6

If you're asking about the combinatorics of the polytope, then there is an easy answer. If you sample with a continuous distribution, then you will get a combinatorial permutohedron with probability 1. The only way to get something else is to take a point on one or more of the hyperplanes x_i=x_j. Then your orbit (i.e. set of vertices of the polytope) ...

6

You are looking for the edgewise subdivision:
Edelsbrunner, H.; Grayson, D. R., Edgewise subdivision of a simplex, Discrete Comput. Geom. 24, No. 4, 707-719 (2000). ZBL0968.51016.
The basic idea is to slice your simplex k times along each coordinate hyperplane. Then you get some small uniform shapes, which are not simplices except at the corners (or in ...

5

Using the unimodular scaling of the Leech lattice, the length of each minimal vector is $\sqrt{4}$. Fixing a particular minimal vector $u$, the remaining minimal vectors $v$ are:
1 vector $v$ with $\langle u, v \rangle = 4$ (namely $v = u$);
4600 vectors $v$ with $\langle u, v \rangle = 2$;
47104 vectors $v$ with $\langle u, v \rangle = 1$;
93150 vectors $v$...

answered Oct 24 '20 at 22:05

Adam P. Goucher

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5

Edited to add: I now think the answer below is completely wrong. The three-dimensional cyclohedron has 12 facets, while the three-dimensional polytope the OP is looking for should have 14. This is the number of facets of the permutohedron, and none of them collapse. (In fact, if I have correctly understood things, the facets of the polytope the OP wants ...

5

Newton polytopes and the polynomials they support
We will use the standard notion $\mathbf{x}^{\mathbf{a}} := \prod_{i=1}^{n} x_i^{a_i}$ to represent monomials in a multivariate (Laurent) polynomial ring.
The Newton polytope of a polynomial is the convex hull of its exponent vectors. We say that a polytope $P$ supports a polynomial $p$ if $P$ is the Newton ...

5

It is possible to have an abstract polytope where the automorphism group acts transitively on the faces of each rank (fully transitive) but does not act transitively on flags (not regular).
For any flag $\Phi$ and any $j$ let $\Phi_j$ denote the face of rank $j$ in $\Phi$ and let $\Phi^j$ denote the flag adjacent to $\Phi$ which differs only in the rank $j$ ...

5

This recent paper
Sartipizadeh, Hossein, and Tyrone L. Vincent. "Computing the Approximate Convex Hull in High Dimensions." arXiv:1603.04422 (2016).
includes a summary of previous work on approximate convex hulls. The time complexity of their algorithm is independent of the dimension,
and quadratic in the number of points.
"The proposed algorithm uses a ...

answered Oct 25 '16 at 9:51

Joseph O'Rourke

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5

This is Theorem 1 (actually, Satz 1) of Roswitha Blind, Konvexe Polytope mit kongruenten regulären $(n- 1)$-Seiten im $\Bbb{R}^n$ $(n \ge 4)$, Comment. Math. Helvetici 54 (1979) 304--308. The short proof follows from two short lemmas, one of which cites Coxeter's Regular Polytopes and an article by Shephard.
You might also be interested in a related ...

5

Note that $x+\varepsilon y\in K$ for small $\varepsilon$ if and only if $y$ satisfies the inequalities $y_i\leqslant y_{i+1}$ if $x_i=x_{i+1}$; $y_1\geqslant 0$ if $x_1=0$; $y_n\leqslant 0$ if $x_n=0$. We may think that $y_i$ are i.i.d. Gaussian, then the probability of a chain of an inequalities like $0\leqslant y_1\leqslant y_2\leqslant \ldots \leqslant ...

answered Oct 3 '20 at 18:40

Fedor Petrov

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5

This answers question 2 as well. But I think both questions are way more suitable for math.stackexchange.
I add another example because, unlike the first, it has the property that multiplied by $I^{n-2}$ it gives further examples in $\mathbb{R}^n$ as well:

4

In fact, there are many to be found on Wikipedia under isogonal figures, even in three dimensions.
Examples in dimension four are obtained as dual polytope of runcinated 4-simplex or runcinated 24-cell. This works, because these both runcinations are
edge-transitive, and
each edge is contained in exactly four facets.
Since in the dual edges become 2-...

4

If you consider a tiling of 3-space to be a 4-dimensional polytope, then the Rhombic dodecahedral honeycomb would work.
Other possibilities are limited by the potential 3-faces.
Because every edge has one endpoint in each of two vertex orbits,
the 2-faces must all have evenly many sides.
If the edge-transitivity descends to the 3-faces, then the 3-faces ...

4

Observe that your set of inequalities is $S_n$-invariant, hence your polytope is, hence your set of vertices is. So it's enough to understand the case $x_1 \leq \ldots \leq x_n$.
Now you don't need to think about general $S$; for each $|S|$ the tightest inequality comes from $S = [n-|S|+1,n]$ a terminal interval. The $|S|=n$ inequality is actually your ...

4

Now that I think of it, my answer was wrong. I was just reproducing examples mentioned by the question poser (Bing's, as seen in Ziegler's book). These examples cannot be reduced keeping the boundary to be a 2-sphere, but the question is whether there are examples that cannot be reduced keeping the boundary a 2-manifold.

4

The answer is yes. In other words, if the order of the orbit $x_n=T^n(x)$ of each point $x\in K$ is finite then $T^n$ is identity map for some $n$.
Choose a simplex $\triangle$ of maximal dimension $m$.
Set
$$E_n=\{\,x\in\triangle\mid T^n(x)=x\,\}.$$
By Baire theorem, $E_n$ has nonempty interior for some $n$;
fix such a value $n_1$.
Note that $E_{n_1}$ ...

answered Dec 12 '14 at 4:50

Anton Petrunin

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4

Maybe Lerman-Tolman's paper Hamiltonian torus action on symplectic orbifolds and toric varieties http://www.ams.org/journals/tran/1997-349-10/S0002-9947-97-01821-7/S0002-9947-97-01821-7.pdf is useful for your question. In that paper, a "labeled polytope" is defined to be a convex rational simple polytope, plus a positive integer attached to each open facet, ...

4

In $R^3$, since the spheres are concentric, not only all faces are regular, but also all edges are of the same length, and all faces are inscribed in circles of the same radius, hence are congruent. Also, all dihedral angles between faces with a common edge are equal, which implies that all vertices are of the same valence. This makes the polytope regular. ...

4

First, a simple remark: If a polytope with congruent facets is inscribed in a sphere, then it is circumscribed about a sphere as well, and the two spheres are concentric.
Next, there is a series of examples described and pictured in my old question
Can the sphere be partitioned into small congruent cells? . Each of these examples is what you want in $R^3$. ...

4

Multiple polytopes can have the same data, as pictured below.
Take a pyramidal frustum, and twist it slightly clockwise or slightly counterclockwise. Make one polytope by gluing two identical versions, and make another polytope by gluing two opposite versions.
These will have the same combinatorial type, the same edge lengths, and the same distances from the ...

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