This is a cross post from Math SE that no one seemed able to solve.

Here is a problem that came up in a conversation with a professor after I made a false assumption about the geometry of $\mathbb{Z}^n$. I do not know if he knew the answer (and told me none of it) and has since passed so I can no longer ask him about it.

Let $C$ be a lattice cube in $\mathbb{R}^n$. Characterize all possible volumes for $C$. A cube is called a lattice cube if and only if every vertex has integer coordinates.

I broke this proof into three cases, the last of which I am having trouble with in one direction. We will let $V(n)$ be the set of all numbers $V$ for which there exists a lattice cube of volume $V$ in dimension $n$. We will break into three cases based on the value mod 4.

\begin{align*} V(2k+1)&=\{a^n:a\in\mathbb{N}\} \\ V(4k)&=\{a^\frac{n}{2}:a\in\mathbb{N}\} \\ V(4k+2)&\supseteq\{(a^2+b^2)^\frac{n}{2}:a,b\in\mathbb{N}\} \end{align*}

These statements I have proven (none of them are hard), and conjecture that the last one is an equality. I've been trying to use a collapsing dimension argument to show if I can make a cube of side length $s$ in $\mathbb{R}^{4k+2}$ then I can in $\mathbb{R}^{4k-2}$, at which point the theorem follows since I have proven the special case of $n=2$, unfortunately this does not seem to be fruitful. After trying general collapsing arguments, I got into technical arguements about the matrices whose rows are the vectors that define the cubes, but again, no avail.


1 Answer 1


[Edited to add the variant with $n-2$ instead of $n+2$]

The conjecture is true; in fact for $n=4k+2$ one cannot even form an $n$-cube of side $D^{1/2}$ in ${\bf R}^n$ with rational coordinates unless $D$ is the sum of two squares. We prove this using Witt's cancellation theorem for quadratic forms.

Let $D$ be a positive rational number. Translating one vertex of the cube to the origin, we see that an $n$-cube in ${\bf Q}^n$ of side $D^{1/2}$ is equivalent to an orthogonal basis for ${\bf Q}^n$ consisting of vectors each of length $D^{1/2}$, which in turn is tantamount to an isomorphism between the quadratic form $D \sum_{m=1}^n X_m^2$ and the standard inner product $\sum_{m=1}^n X_m^2$. But we already know that such a cube exists in dimension $n+2 = 4(k+1)$, that is, that the quadratic form $\sum_{m=1}^{n+2} X_m^2$ is isomorphic with the form $D \sum_{m=1}^{n+2} X_m^2$. By Witt it follows that the forms $X_{n+1}^2 + X_{n+2}^2$ and $D(X_{n+1}^2 + X_{n+2}^2)$ are isomorphic. In particular, $D = D(1^2 + 0^2)$ is the sum of two squares, Q.E.D.

Alternatively we could have used the fact that $n-2 = 4k$ is a multiple of $4$: it follows that $\sum_{m=1}^{n-2} X_m^2$ is isomorphic with the form $D \sum_{m=1}^{n-2} X_m^2$, and then we could cancel a form of rank $n-2$ to deduce again an isomorphism between the diagonal forms $X_{n+1}^2 + X_{n+2}^2$ and $D(X_{n+1}^2 + X_{n+2}^2)$.

[To obtain the isomorphism in dimension $n \pm 2$: it is enough to do it in dimension $4$ and then use direct sums to get all multiples of $4$. Write $D$ as a sum of $4$ squares, which is to say as the norm of a quaternion $q$. Then the quaternions $q$, ${\bf i}q$, ${\bf j}q$, ${\bf k}q$ give the desired orthogonal basis.

Likewise if $D = a^2+b^2$ then $(a,b)$ and $(-b,a)$ [corresponding to complex numbers $a+ib$ and $i(a+ib)$] are sides of a square of side $D^{1/2}$, and by taking direct sums we obtain a cube of any even dimension. When $n$ is odd we can use determinants to confirm that the side of an integral (or rational) $n$-cube must itself be an integer or rational number.]


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