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Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $\mathbb R^n$. For instance, the face $[0,1]^{k-1} \times \{1\} \times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} \times \{0\} \times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(\mathbb R/\mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] \to \mathbb R/\mathbb Z$); the Klein bottle is not cubulated in my sense.

One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.

It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM \cong \mathbb R^n \times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.

Question: Does every framed manifold admit a framing-compatible cubulation?

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    $\begingroup$ Does the 3-sphere admit a cubulation in your sense? $\endgroup$ – Chris Schommer-Pries Feb 14 '19 at 20:38
  • $\begingroup$ @ChrisSchommer-Pries A hypercube provides one, I believe. $\endgroup$ – მამუკა ჯიბლაძე Feb 14 '19 at 20:58
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    $\begingroup$ Your restrictions on the cubulation seem to imply that $M$ has a flat metric and that the action of $\pi_1 M$ on the universal covering space $\mathbb R^n$ is an action by translation, so $\pi_1 M$ is isomorphic to $\mathbb Z^n$. $\endgroup$ – Lee Mosher Feb 14 '19 at 22:25
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If you glue together the Riemannian metrics on the various cubes you obtain a flat metric on your cubulated manifold. So e.g. $S^3\cong SU(2)$ is framed but not cubulated.

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