# When can a regular simplex be inscribed in a regular hypercube?

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?

• @hardmath Yes, sorry if I didn't make that clear enough. Apr 27 '20 at 5:22
• 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. Apr 27 '20 at 6:26
• @quasi Doesn’t this contradict the $n=7$ example? I’ve checked it and it certainly seems to work out. Apr 27 '20 at 6:49
• This question is treated in different articles. See for example Theorem 4.5 in this reference Apr 27 '20 at 8:16

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.
• 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. Apr 27 '20 at 6:46
• @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. Apr 27 '20 at 7:01