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An abstract polytope is a certain kind of partially-ordered set. Its elements or "faces" are ranked by "dimension" and also partially ordered via a pairwise "incidence" relation between elements of adjacent ranks.

For some abstract polytope $P$, of which $F$ and $G$ are faces such that $G \le F$, the set of faces $H$, such that $G \le H \le F$, is a section of $P$ and is written $F/G$.

I define a sub-polytope of $P$ as a subset of $P$ which is also an abstract polytope.

What is the relationship between sections and sub-polytopes?

  • Are all sections of $P$ necessarily (sub-)polytopes in their own right?
  • Are all sub-polytopes of $P$ necessarily sections?
Guy Inchbald
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  • I suppose this depends on exactly how one axiomatises the notion of abstract polytope. – Angina Seng Jul 05 '20 at 19:18
  • As far as I know all current definitions are equivalent. See for example the one on Wikipedia: https://en.wikipedia.org/wiki/Abstract_polytope – Guy Inchbald Jul 05 '20 at 20:10
  • "[a] sub-polytope of P [is] any abstract polytope which sits within the structure of P". I don't understand this definition. Can you elaborate? What does it mean to "sit in the structure of something"? – M. Winter Jul 21 '20 at 18:27
  • Sorry. A sub-polytope of $P$ is any subset of $P$ which is also an abstract polytope. I'll make the change. – Guy Inchbald Jul 21 '20 at 20:31

2 Answers2

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Any polytope $P$ has a single unique nullity $Ø≤P$.

Any subpolytope or subelement $F$ then necessarily requires $Ø≤F≤P$ and yes, both the polytope itself as also that subpolytope can be identified with the according sections: $$P\cong P/Ø,\ F\cong F/Ø$$ therefore all subpolytopes indeed are sections.

Whereas when $Ø<G≤F≤P$ then $F/G$ certanly is a section, but not a subpolytope (subelement) of $P\cong P/Ø$.

None the less, $P/G$ happens to be the $G$-figure of $P$. Thence any $F/G$ can be considered a polytope on its own right for sure. In fact it just represents the subdiagram of the Hasse diagram of $P$ (ie spanned between $Ø$ and $P$) which is spanned between $G$ and $F$. And that one for sure is a Hasse diagram itself.

--- rk

Dr. Richard Klitzing
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  • Johnson pointed out some years ago that identifying $F/Ø$ with $F$ is not rigorous and is no more than a popular convenience; $Ø≤P$ is not a valid relation and we should write $Ø≤F_n$ (where $n$ is the dimension of $P$ and $F_n$ is the maximal face). Now, $Ø≤F≤F_n$ is trivially true for all $F$. It does not make $F$ a polytope. The set of all faces $F$ which meet this condition, i.e. the section $F_n/Ø$, is just $P$. So I do not see how this argument can take us forward. – Guy Inchbald Jul 06 '20 at 08:29
  • the elements of P are simply all the downward complete Hasse diagram substructures, just as the figures are the upwards complete Hasse diagram substructures. All these are Hasse diagrams for their own sake. But clearly for $Ø<G≤F<P$ neither G is the lowest nor F is the upmost end of the Hasse diagram of P. – Dr. Richard Klitzing Jul 06 '20 at 12:13
  • Yes, but what is the relevance to my question? A section is still not an element and its null element is not necessarily the null element of P. Don't be confused by the convention of identifying the null element with the empty set, it is only that, a convention and as false as identifying a face with its span. – Guy Inchbald Jul 06 '20 at 12:47
  • If you resist from the identification of P and P/Ø, then sections are fully separate things. If you'd allow for that identification, then all subpolytopal elements (faces, down-based) as well as the dual (top-based) figures can be considered (identified with) sections. But even then not all sections are subpolytopes - at least not of P. – Dr. Richard Klitzing Jul 06 '20 at 17:30
  • Can you give an example of a section which is not a subpolytope and add it to your answer? Please bear in mind that allowing the identification of a face with its span (where the span of $F$ is just the section $F/Ø$), as you suggest, is an optional interpretation. Thanks. – Guy Inchbald Jul 06 '20 at 18:53
  • Eg. the edge figure of a facet is $F_{n-1}/F_1$. It neither contains the topmost element $P=F_n$ nor the bottom one $Ø=F_{-1}$. So it is neither a figure nor an element of $P$. - But, as I said already, it clearly is a polytope on its own right. Simply not directly correlated to $P$. – Dr. Richard Klitzing Jul 07 '20 at 06:02
  • It is an abstract polytope and it is a subset of $P$. Surely that defines it as a sub-polytope of $P$. Is there a more formal definition of a "sub-polytope" than the one I give in my original question? – Guy Inchbald Jul 07 '20 at 08:31
  • Ah, I see the issue here. You have defined a sub-polytope as an element of $P$, i.e. a face. This is not at all how my question defines a sub-polytope. For example a vertex section meets my definition of a sub-polytope but it is not a face. – Guy Inchbald Jul 07 '20 at 19:10
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Every section of a polytope is a polytope. See for example:

As the question defines a sub-polytope, a section of $P$ is therefore indeed a sub-polytope of $P$.

I can find no direct statement going the other way, however it seems to follow easily enough:

Every sub-polytope has a maximal and a minimal element. Every connected max/min pair $G \le F$ has a section $F/G$ which is a polytope.

It is more or less by definition that no sub-polytope of the same rank (i.e. with the same max/min pair) as the original can exist. Therefore the section $F/G$ must be the unique polytope with max/min pair $G \le F$.

Sections exist for every possible pair $G \le F$. They are the only sub-polytopes for each pair. No other sub-polytopes can therefore exist. The set of sections of a polytope is therefore also the full set of sub-polytopes.

So I conclude that the sections of $P$ are just its sub-polytopes.

Guy Inchbald
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  • For each pair $G\lt F$, the set ${G,F}$ is a sub-polytope that is not a section unless the ranks of $G$ and $F$ differ by exactly 1. I think the only counterexamples are of this form, though. – Karl Feb 14 '23 at 22:19