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These days I have been reading about Keller's cube tyling conjecture, which asks if in any covering of $\mathbb{R}^n$ by translates of $[0,1]^n$ with disjoint interiors there are two cubes sharing one $n-1$-dimensional face (the problem was completely solved recently in The Resolution of Keller’s Conjecture , by Brakensiek, Heule, Mackey and Narváez).

I have not found any information on the same problem but without assuming that the cubes are translates of each other, that is, for any covering of $\mathbb{R}^n$ by n-dimensional hypercubes of side one (sets isometric to $[0,1]^n$) with disjoint interiors, can we always find two hypercubes sharing one face? The interesting case is $n\leq7$, for $n\geq8$ there are counterexamples to the original conjecture.

There are coverings of $\mathbb{R}^3$ by disjoint cubes such that they are not all translates of each other, so at least this is not a vacuous question.

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  • $\begingroup$ What would 'sharing a face' mean in the context of distinctly sized cubes? The faces being identical, or one being a subset of the other? $\endgroup$ Commented Nov 7, 2022 at 17:52
  • $\begingroup$ @StevenStadnicki In my question the cubes are all of the same size, because they are all isometric to $[0,1]^n$. Maybe an interesting question could be posed with cubes of different sizes $\endgroup$
    – Saúl RM
    Commented Nov 7, 2022 at 18:46
  • $\begingroup$ Ahhh! Somehow I missed that and thought you were talking about arbitrary sized cubes as well as orientations; mea culpa. Do you have a pointer to the coverings of R^3 that you mention? I'm trying to mentally grasp the mechanism there. $\endgroup$ Commented Nov 8, 2022 at 3:05
  • $\begingroup$ @StevenStadnicki You can take a tiling of $\mathbb{R}^2$, multiply it by $[0,1]$ and stack it infinitely many times. Then you can rotate each layer so that they are not parallel $\endgroup$
    – Saúl RM
    Commented Nov 8, 2022 at 3:17

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