# Is there a degenerate simplex in $\mathbb{R}^{8 k-1}$ with odd integer edge lengths?

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.

• 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. Commented Aug 28, 2021 at 10:08

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.
• 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. Commented Aug 29, 2021 at 4:56