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.



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> **Question:** Does every framed manifold admit a framing-compatible cubulation?
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Are framed manifolds cubulatable?
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.


