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Definition. A convex polytope is a compact finite intersection of hyperplanes in $\mathbb{R}^n$

Definition. The polycomplex is the following data set:

  1. a set of convex polytopes, closed under transitions to a face
  2. a set of affine isomorphisms between them, closed under composition and not containing non-trivial automorphisms ("we can affinely identify different polytopes, but we cannot glue a polytope with itself")

Comment. I prefer the definition: a polycomplex is a functor from a small thin groupoid to the category of convex polytopes, but the above will (probably) attract more people.

Comment.

  1. Δ-complex is a special case of a polycomplex, when all polytope are simplices.
  2. A polyhedron is a special case of a polycomplex, when each polytope "includes" no more than once in each polyhedron of higher dimension.
  3. A simplicial complex is both a polyhedron and a Δ-complex.

Definition. $P_n$ is the set of all n-dimensional oriented polytopes of the polycomplex $P$ (for each n-dimensional polytope $P_n$ contains two elements: $X$ with one orientation and the other). Denote the natural involution on $P_n$ by $\neg$.

Definition. The polyhedral chain complex $P$ is the following chain complex:​

  • $C_n(P) = FreeAb(P_n)/\sim$, where $\neg X \sim - X$ and $X \sim f(X)$ for each affine isomorphism $f$ from $P$ (an affine isomorphism transfers the orientation c $X$ on $f(X)$)
  • $\partial \colon C_n(P) \to C_{n-1}(P)$ is defined on the basis as follows: $\partial X$ is the sum of its faces with induced orientations (for definiteness, the last frame vector enters the face, the rest set the orientation).

It is obvious from the orientation properties that $C(P)$ is a chain complex. The homology of $C(P)$ is called the polyhedral homology of $P$.

Comment. It is easy to see that the polyhedral homology of the Δ-complex coincides with its simplicial ones, since the resulting chain complexes are simply isomorphic.

I am interested in the following question: how to prove that the polyhedral homology of $P$ coincides with the polyhedral homology of any of its subdivisions $P'$? (i.e. $P'$ is a polycomplex obtained from $P$ only by a subdivision of the polytopes $P$ ). I'm sure it's true and probably very simple, but I can't think of a proof.

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    $\begingroup$ Did you try the method of acyclic models? $\endgroup$
    – John Wiltshire-Gordon
    Commented Feb 23, 2022 at 20:38
  • $\begingroup$ @JohnWiltshire-Gordon Thank you very much for reminding me of this theorem. Taking as models all (pairwise distinct) convex polytopes on the one hand and the standard $Δ^n$ simplices on the other, I can obtain two oppositely directed natural transformations between polyhedral and singular homology functors. I can prove that one of the compositions is equal to id by taking the standard simplices as models (the singular homology functor is both acyclic and free with respect to them). $\endgroup$
    – Arshak Aivazian
    Commented Mar 7, 2022 at 5:33
  • $\begingroup$ @JohnWiltshire-Gordon To prove that another composition is equal to id, we need to find models with respect to which the polyhedral homology functor is acyclic and free. Obviously, for a functor to be free, it is necessary to take exactly all (pairwise distinct) convex polytopes. Now we need to somehow prove that the polyhedral homology of a polytope is trivial. $\endgroup$
    – Arshak Aivazian
    Commented Mar 7, 2022 at 5:34
  • $\begingroup$ To do that, maybe you can find a chain homotopy contraction. Perhaps something like choosing a basepoint for each polytope and then forming the cone with the basepoint as apex? $\endgroup$
    – John Wiltshire-Gordon
    Commented Mar 8, 2022 at 0:34
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    $\begingroup$ Ah, I thought that a basis for $C_2(P)$ was all affine linear maps from all other polytopes. So a basis is just given by the usual cells? Then $C(P)$ should be isomorphic to the cw-chains on $P$? I confess that I did not read the details of your construction. $\endgroup$
    – John Wiltshire-Gordon
    Commented Mar 8, 2022 at 20:59

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