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It seems to me that a smooth manifold should admit the structure of a cubical complex by Morse theory, since handle attachments seem to be perfectly cubical maps.

Is this cubical structure "essentially unique," whatever that means cubically? Or is it non-unique as in the simplicial case (PL structures are unique for smooth manifolds -- the Hauptvermutung fails only for more general spaces, such as topological manifolds!)?

EDIT: In the interest of concreteness, let's say that a cubical complex consists of:

  • a (normal [1]) cubical set $X$

  • such that each nondegenerate $n$-cell $\sigma \in X_n$ is uniquely determined (up to automorphism [2]) by its 0-skeleton.

One subtlety is that there are presheaf categories which may be called "categories of cubical sets" -- one may or may not have one or both connections, and one may or may not have symmetries (aka "exchanges" aka "extensions") which allow one to permute the axes of a cube, and may or may not have reversals, which allow one to flip the direction of an edge or higher cube. See Grandis and Mauri for a discussion. For the purposes of this definition, let's assume we've fixed one of these categories of cubical sets -- and why not just plain cubical sets (the version described at the link).

Define as usual the star of a nondegenerate cell $\sigma \in X_n$ to be the set of cells $\tau \in X_m$ containing $\sigma$, the closure of a set $S$ of cells of $X$ to be the smallest cubical subcomplex containing $S$, and the link of a nondegenerate cell $\sigma \in X_n$ to be the closure of the star of $\sigma$ minus the star of the closure of $\sigma$. Say that $X$ is PL if

  • the link of any $n$-cell is topologically a sphere.

Here we're using the geometric realization functor from cubical sets to topological spaces, which sends the $n$-cube to $[0,1]^n$.

I'm not quite sure when to say that two cubical complexes are "equivalent", though.

[1] The word "normal" just means that the automorphism group of the $n$-cube acts freely on the set of nondegenerate $n$-cells of $X$. This automorphism group is trivial unless we have symmetries and / or reversals, so the "normal" condition is vaccuous if we're considering e.g. plain cubical sets.

[2] Again, this only matters if we have symmetries and /or reversals.

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  • $\begingroup$ What about the counterexamples to the classical Hauptvermutung? Are thes triangulations "cubulable"? $\endgroup$
    – ThiKu
    Commented Mar 11, 2019 at 23:48
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    $\begingroup$ Do you have a definition of cubical complex? $\endgroup$
    – Ryan Budney
    Commented Mar 12, 2019 at 1:38
  • $\begingroup$ @RyanBudney I've added some discussion, but there are still some details that are hazy. $\endgroup$
    – Tim Campion
    Commented Mar 12, 2019 at 15:10
  • $\begingroup$ One thing to keep in mind: the second barycentric subdivision of a cubulation (whatever that might be) should be a triangulation in the ordinary sense of the word. $\endgroup$
    – Lee Mosher
    Commented Mar 12, 2019 at 16:26

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