16
$\begingroup$

The Cayley-Menger determinant gives the squared volume of a simplex in $\mathbb{R}^n$ as a function of its $n(n+1)/2$ edge lengths:

$$v_n^2 = \frac{(-1)^{n+1}}{(n!)^2 2^n} \begin{vmatrix} 0&d_{01}^2&d_{02}^2&\dots&d_{0n}^2&1\\ d_{01}^2&0&d_{12}^2&\dots&d_{1n}^2&1\\ d_{02}^2&d_{12}^2&0&\dots&d_{2n}^2&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ d_{0n}^2&d_{1n}^2&d_{2n}^2&\dots&0&1\\ 1&1&1&\dots&1&0 \end{vmatrix}$$ where $d_{i j}$ is the distance from vertex $i$ to vertex $j$ of the simplex.

For the regular simplex with all edge lengths 1, i.e. $d_{i j}=1$ for all $i,j$, we have:

$$(n!)^2 2^n v_n^2 = (-1)^{n+1} \begin{vmatrix} 0&1&1&\dots&1&1\\ 1&0&1&\dots&1&1\\ 1&1&0&\dots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&1&1&\dots&0&1\\ 1&1&1&\dots&1&0 \end{vmatrix} = n+1$$

This follows from the fact that the $(n+2)\times(n+2)$ matrix here has eigenvector $(1,1,1,\dots,1)$ with eigenvalue $n+1$, along with $n+1$ eigenvectors of the form $(1,0,0,\dots,-1,0,0,\dots)$ with eigenvalue $-1$.

Any odd integer $2k+1$ has a square that is equal to 1 modulo 8, since:

$$(2k+1)^2 = 4k(k+1)+1$$

and either $k$ or $k+1$ must be even. It follows that, for any simplex whose edge lengths are all odd integers, the quantity $(n!)^2 2^n v_n^2$ must equal $n+1$ modulo 8.

When $n+1$ is not a multiple of 8, this means a simplex in $\mathbb{R}^n$ whose edge lengths are all odd integers cannot be degenerate, i.e. it cannot have zero volume.

However, this does not settle the same question when $n=8k-1$.

Are there known examples of degenerate simplices in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths? Or is there a proof for their nonexistence that completes the proof that applies in other dimensions?

Acknowledgement: This question arose from a discussion on Twitter between Thien An and Ian Agol.

Edited to add: For all cases where $n = 7$ mod 16, it is possible to rule out a degenerate simplex by working modulo 16, where any squared odd integer must equal either 1 or 9. Computing the determinant when $x$ is added to any single squared edge length gives a quadratic in $x$ that has even coefficients for $x$ and $x^2$ (given that all the original entries are integers), from which it follows that adding 8 to any squared edge length preserves the determinant modulo 16. Since the determinant when all squared edge lengths are equal to 1 is $n+1$, changing any number of the squared edge lengths from 1 to 9 can never yield a determinant divisible by 16.

$\endgroup$
1
  • 3
    $\begingroup$ By the same argument, there are no degenerate simplices where no edge length is a multiple of three except possibly in $\mathbb{R}^{3k-1}$ for some $k$. Maybe solving this related problem for e.g. $\mathbb{R}^2$ or $\mathbb{R}^5$ could give some intuition as to what happens in the odd length case. $\endgroup$
    – pregunton
    Commented Aug 28, 2021 at 10:08

1 Answer 1

12
$\begingroup$

This is answered in a paper by R. L. Graham, B. L. Rothschild & E. G. Straus "Are there $n+2$ Points in $E_n$ with Odd Integral Distances?". Such simplexes exist iff $n+2 \equiv 0 \pmod {16}$. They also consider the related problem of integral distances relatively prime to 3 and 6.

$\endgroup$
1
  • 3
    $\begingroup$ Thanks! To be clear, their $n$ is one less than I’ve used in the question here; it’s the dimension of the degenerate subspace. $\endgroup$
    – Greg Egan
    Commented Aug 29, 2021 at 4:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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