# Do smooth manifolds admit unique cubical structures?

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.

• What about the counterexamples to the classical Hauptvermutung? Are thes triangulations "cubulable"? – ThiKu Mar 11 at 23:48
• Do you have a definition of cubical complex? – Ryan Budney Mar 12 at 1:38
• @RyanBudney I've added some discussion, but there are still some details that are hazy. – Tim Campion Mar 12 at 15:10
• 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. – Lee Mosher Mar 12 at 16:26