Are framed manifolds cubulatable?

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?

• Does the 3-sphere admit a cubulation in your sense? – Chris Schommer-Pries Feb 14 '19 at 20:38
• @ChrisSchommer-Pries A hypercube provides one, I believe. – მამუკა ჯიბლაძე Feb 14 '19 at 20:58
• 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 $\mathbb Z^n$. – Lee Mosher Feb 14 '19 at 22:25

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.