A Geometric Higher Dimensional Phenomenon?

Geometry Level 3

In 3D \text{3D} it's possible to make a regular tetrahedron with integer coordinates that all lie on the vertices of a cube.

Does this phenomenon occur in any other dimensional space?

Or specifically, for how many n N n \in \mathbb{N} is it possible to construct a regular n n -dimensional simplex with integer co-ordinates that lie on an n n -dimensional hypercube in Z n \mathbb{Z}^n ?

Details and Assumptions :

  • A regular n n -dimensional simplex in Z n \mathbb{Z}^n has n + 1 n+1 vertices that are all an equal distance apart. (It's like an n n -dimensional version of an equilateral triangle!)

  • Here is the Wikipedia article on hypercubes. (It's like an n n -dimensional version of a square!)

  • Note that the analogous phenomena is not possible in 2 2 dimensions. That is, we cannot create an equilateral triangle with integer co-ordinates that all lie on a square. Check out why, here .

Infinitely many! It occurs in finitely many dimensions greater than 1! In 3D only!

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1 solution

Michael Mendrin
Mar 30, 2016

Let a n n cube have vertex coordinates in binary form for all k = 0 k=0 to 2 n 1 {2}^{n}-1 , such as for example for a square, { ( 0 , 0 ) , ( 0 , 1 ) , ( 1 , 0 ) , ( 1 , 1 ) } \{(0,0),(0,1),(1,0),(1,1)\} . A square is a 2 2 -hypercube, where n = 2 n=2 .

Let H p {H}_{p} be a square matrix such that

H 1 = ( 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 ) {H}_{1}=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{pmatrix}

H p + 1 = ( H p H p H p I p H p ) {H}_{p+1}=\begin{pmatrix} { H }_{ p } & { H }_{ p } \\ { H }_{ p } & { I }_{ p }-{ H }_{ p } \end{pmatrix}

where I p {I}_{p} is a square matrix of the same size as the H p {H}_{p} but with all elements 1 1 .

Then, each H p {H}_{p} matrix contains the coordinates of the n + 1 = 2 p + 1 n+1={2}^{p+1} vertices of a n n simplex sharing the same vertices as the n n hypercube, which are the n + 1 n+1 rows with the first column left off.

The Pythagorean distance between any two vertices of the n n simplex as described by the H p {H}_{p} matrix are all the same, here being 2 p \sqrt{{2}^{p}} , which is a property of regular simplexes.

Since this will generate an infinite sequence of such H p {H}_{p} square matrices, there’s an infinity of hypercubes with this property. The H p {H}_{p} square matrices belong to a special class called Hadamard Matrices.

This is a simplified proof. A fuller proof takes longer. Below contains the coordinates for the 7 7 -Simplex.

H 2 = ( 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 1 ) {H}_{2}=\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}

@Roberto Nicolaides You should mention (the well known fact) that it is not possible in 2-D.

Calvin Lin Staff - 5 years, 2 months ago

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That's true. It's a non-trivial matter figuring out which n n doesn't permit such n-simplexes embedded in n-hypercubes. I have no idea how to systemically determine that.

Michael Mendrin - 5 years, 2 months ago

Good idea, I'll add this now :)

Roberto Nicolaides - 5 years, 2 months ago

I'm blown away! This is much nicer than I had in mind. Thanks for introducing me Hadamard Matrices.

Roberto Nicolaides - 5 years, 2 months ago

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Maybe I should have added that there's many more n-hypercubes with this property than those with simplexes "constructed" with this method. This merely proves that there's an infinity of them. Likewise, the class of Hadamard matrices is far more numerous than those "constructed" here.

Michael Mendrin - 5 years, 2 months ago

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This is a question that I've been thinking about for a while now :

For exactly which n n does this property exist? I think I've found a way of showing it to be true in a very similar way to your solution for all 2 k 1 2^k-1 . Is it possible to show that these aren't the only cases?

Roberto Nicolaides - 5 years, 2 months ago

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@Roberto Nicolaides These are hardly the only cases, but because of the property that regular simplexes have edges of all the same lengths, we need to look at Hadamard matrices, because they have this property. The two are closely related. Hadamard matrices are much studied and have wide application because of their special properties.

Michael Mendrin - 5 years, 2 months ago

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@Michael Mendrin Interesting, I'll spend some more time later reading the wikipedia article. I wonder if this is an open question.

Roberto Nicolaides - 5 years, 2 months ago

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@Roberto Nicolaides Feel free to start the Brilliant wiki on hypercubes as well!

Andrew Ellinor - 5 years, 1 month ago

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@Andrew Ellinor I think the subject of simplexes is a real hair-puller, in spite of its deceptively simple sound name.

Michael Mendrin - 5 years, 1 month ago

@Michael Mendrin I've had a look at the Wikipedia page on Hadamard matrices but found it quite difficult to grasp. For clarity, do you think its possible to find an n-simplex with the described properties if and only if n = 2 k 1 n = 2^k -1 for some k? I want to try prove this but to be honest, I'm not even sure if its true.

EDIT : For anyone that's interested, I think I managed to prove that this is the case. Maybe I'll write a note on it if anyone asks.

EDIT : Turns out I made a mistake, so have not proved it!

Roberto Nicolaides - 5 years, 2 months ago

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