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The plane can be tiled without gaps by congruent two-dimensional regular simplices (i.e., equilateral triangles). The three-dimensional Euclidean space cannot be tiled by congruent three-dimensional simplices (i.e., equilateral tetrahedra). Can this be done for higher dimensions? I have a vague recollection that I saw a proof that for some dimensions the tiling exists and that the existence is somehow related to the divisibility of the dimension by four, but this may be false. Can anybody please formulate the correct statement and provide either a proof or a reference? Thanks!

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Don't look it up. This a good exercise! Calculate the relevant angles using inner products. –  Tom Goodwillie Nov 22 '11 at 1:27
    
Perhaps the result you are recalling is related to regular simplices whose vertices are in $\mathbb{Z}^n$, which is OEIS sequence A096315? –  ARupinski Nov 22 '11 at 4:48

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up vote 9 down vote accepted

Senechal's survey paper has a beautiful historical overview and some standard references. Recently, we proved in this paper that one cannot tile $\Bbb R^d$, $d\ge 4$, with congruent copies of an acute simplex (this answers your question as well). Interestingly, existence of an acute simplex in $\Bbb R^3$ which tiles the space is open since Sommerville's classification does not allow reflections (see e.g. this paper). Curiously, any polytope which tiles $\Bbb R^3$ or $\Bbb R^4$ must be scissors congruent to a cube, even if the tiling is aperiodic (see here).

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Thanks a lot. So my hope that for some dimensions this can be done failed... which is good. My main problem is not purely mathematical, it is more about applications, and no dimensions is better than some dimensions... –  garikz Nov 22 '11 at 23:40

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