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A cube's vertices can be trivially divided into two sets, each forming the vertices of a regular tetrahedron. I was wondering if such a construction could be generalized to higher dimensions.

I've found here a set of coordinates for the $7$-simplex, which imply that it can be inscribed in a $7$-hypercube. (There’s a pretty cool connection to projective geometry there). So, I'm only aware of the cases $n=3,7$ at the moment.

For a particular dimension $n$, we could check the cases for all possible edge lengths of the $n$-simplex: if we choose a vertex and an edge length, we can check whether our remaining points allow us to build the $(n-1)$-simplex of correct edge length we need to build our $n$-simplex. I've checked $n=4$, doesn't seem to work. But this approach doesn't seem generalizable.

We could also try to construct examples by playing around with the coordinates of the hypercube $(\pm1,\pm1,\ldots,\pm1)$, but so far I haven't found anything.

So, for which dimensions can a regular $n$-simplex be built from the vertices of a regular $n$-hypercube?

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  • $\begingroup$ @hardmath Yes, sorry if I didn't make that clear enough. $\endgroup$
    – ViHdzP
    Apr 27, 2020 at 5:22
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    $\begingroup$ Solving via Maple for the case $n=4$, there does not exist a regular $4$-simplex whose $5$ vertices are vertices of a $4$-cube. $\endgroup$
    – quasi
    Apr 27, 2020 at 6:26
  • $\begingroup$ @quasi Doesn’t this contradict the $n=7$ example? I’ve checked it and it certainly seems to work out. $\endgroup$
    – ViHdzP
    Apr 27, 2020 at 6:49
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    $\begingroup$ This question is treated in different articles. See for example Theorem 4.5 in this reference $\endgroup$
    – Jean Marie
    Apr 27, 2020 at 8:16

1 Answer 1

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Rows of a Hadamard matrix form a regular simplex.
Hadamard matrices are known for many multiples of 4 including all multiples of 4 up to 664. $n$ is one less than the multiple of 4.

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  • $\begingroup$ A regular $(n-1)$-simplex perhaps, but the OP's question was whether the $n+1$ vertices of a regular $n$-simplex can be taken from the vertices of an $n$-cube. $\endgroup$
    – quasi
    Apr 27, 2020 at 6:46
  • $\begingroup$ @quasi Given a Hadamard matrix of size $n$, each row may be kept or inverted, such that the first entry of every row becomes equal. The new matrix remains a Hadamard matrix. Reading each row as a point then creates an $(n-1)$-simplex embedded in an $(n-1)$-hypercube. $\endgroup$
    – ViHdzP
    Apr 27, 2020 at 7:01
  • $\begingroup$ It would be cool to know whether this is an if and only if condition. $\endgroup$
    – ViHdzP
    Apr 27, 2020 at 7:03
  • $\begingroup$ Yep, it is. Jean Marie's reference on my post confirms this via theorem 5.3. $\endgroup$
    – ViHdzP
    Apr 27, 2020 at 12:49

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