This is a question posed by Adam Chalcraft. I am posting it here because I think it deserves wider circulation, and because maybe someone already knows the answer.


A *polyomino* is usually defined to be a finite set of unit squares, glued together edge-to-edge. Here I generalize it to mean a finite set of unit hypercubes, glued together facet-to-facet.


Given a polyomino $P$ in $\mathbb{R}^m$, I can *lift* it to a polyomino in a higher-dimensional Euclidean space $\mathbb{R}^{m+n}$ by crossing it with a unit $n$-cube: the lifted polyomino is just $P\times [0,1]^n$.


Obviously, not all polyominos tile space.



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> Is it true that given *any* polyomino $P$ in $\mathbb{R}^m$, there exists some $n$ such that the lifted polyomino $P\times [0,1]^n$ tiles $\mathbb{R}^{m+n}$?
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Many people's first instinct is that multiply-connected polyominos (those with "holes" in them) can't possibly tile, but you can get inside holes if you lift to a high enough dimension.



Does every polyomino tile R^n for some n?
A positive answer to this question has just appeared in the arXiv:
Tiling with arbitrary tiles;
Vytautas Gruslys, Imre Leader, Ta Sheng Tan;
<http://arxiv.org/abs/1505.03697>


