For which $n$ is there a regular $n$-simplex with vertices in $\mathbb{Z}^n$ (or equivalently, in $\mathbb{Q}^n$)?

Some easy observations:

  • Such an $n$-simplex exists with vertices in $\mathbb{Z}^{n+1}$: just take the $n+1$ points $(0, \ldots, 0, 1, 0, \ldots, 0)$.

  • If $n$ is even and $n+1$ is not a perfect square, then no regular $n$-simplex with vertices in $\mathbb{Z}^n$ exists: on the one hand, the side length $x$ is the square root of an integer, and thus the volume, namely $\frac{\sqrt{n+1}}{n! 2^{n/2}} x^n$, is irrational; on the other hand the volume must be rational by the formula using determinants.

  • For $n=3$ a regular tetrahedron with vertices in $\mathbb{Z}^3$ does exist: $\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$.

  • If there exists a Hadamard matrix of order $n+1$, we can normalize it so the first column is all ones and then the remaining $n$ columns give the coordinates of the $n+1$ vertices of a regular $n$-simplex with vertices in $\{1,-1\}^n$.


As shown in the accepted answer to this question, the Hadamard matrix condition is necessary and sufficent, so you have answered your own question...

EDIT As pointed out by Noam, I misread the linked-to question, and to atone: Robin Chapman gives the answer to the actual OP question, and here is his 1998 answer (slightly reformatted):

This is a good problem. It reduces to a question in the theory of rational quadratic forms. Let's ask for which $n$ an $n$-simplex can be embedded $n$-space with integral coordinates. The answer is if and only if

  • $n + 1$ is an odd square, or

  • $n + 1$ is a sum of $2$ odd squares, or

  • $n + 1 \equiv 0 \mod 4.$

It's equivalent to consider rational coordinates, as we can scale, and we can also translate to put one of the vertices at the origin. Let $v_1,\dots, v_n$ be the other vertices. Then for some rational number m we have $v_i\cdot v_i = 2m,$ and $v_i\cdot v_j = m$ for $i \neq j.$ This means that the quadratic forms $Q_1 = x_1^2 + x_2^2 + ... + x_n^2$ and $2m Q_2$ where $ Q_2 = x_1^2 + x_1 x_2 + x_1 x_3 + \dots + x_1 x_n + x_2^2 + x_2 x_3 + \dots + x_n^2$ are equivalent over the rationals. Indeed this is a necessary and sufficient condition. One can use the Hasse-Minkowski theory of rational quadratic forms to determine when an m exists so that this is the case, and doing so leads, after some effort, to the stated condition.

  • $\begingroup$ Thanks, I somehow failed to find that question! But the paper by Adams, Zvengrowski and Laird mentioned there proves that there is a regular $n$-simplex in $\{0,1\}^n$ if and only if there exists a Hadamard matrix of order $n+1$. Is it easy to see that if there is any regular $n$-simplex in $\mathbb{Z}^n$ there must also be one in $\{0,1\}^n$? $\endgroup$ – Omar Antolín-Camarena Aug 15 '17 at 3:43
  • 5
    $\begingroup$ The accepted answer answers the question for simplices inscribed in The accepted answer answers the question for simplices inscribed in a cube of the same dimension. Robin Chapman's answer claims a criterion for simplices with arbitrary integer coordinates: "The answer is yes iff $n+1$ is the sum of one, two, four or eight odd squares.". So for example $n=8,9$ are possible even though there's no Hadamard matrix of order $9$ or $10$ (any Hadamard matrix has order $1$, $2$, or a multiple of $4$). $\endgroup$ – Noam D. Elkies Aug 15 '17 at 4:39
  • 1
    $\begingroup$ @NoamD.Elkies Oops, you are right! $\endgroup$ – Igor Rivin Aug 15 '17 at 4:43
  • 1
    $\begingroup$ @OmarAntolín-Camarena see the edit, after Noam's sage commentary. $\endgroup$ – Igor Rivin Aug 15 '17 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.