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Here is a problem that came up in a conversation with a professor. 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, 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 $n=2$ (which is quite trivial - there is no way to write down a square whose volume is not of the specified form in $2$D.

The above assertions (sans my conjecture) are proven here

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What is a lattice cube? – Will Orrick Mar 14 '14 at 20:35
    
A lattice cube is a cube with vertices that lie in $\mathbb{Z}^n$ – Stella Biderman Mar 14 '14 at 22:19
    
You define $V(k)$ to be "the set of possible volumes". Can you be more precise? Does this mean the set of volumes that satisfy some necessary condition, or the set of volumes that satisfy both necessary and sufficient conditions, i.e. the set of volumes that actually exist? – Will Orrick Mar 14 '14 at 23:06
    
I ask because the latter seems very hard. Wouldn't the question subsume the Hadamard conjecture as a special case? – Will Orrick Mar 14 '14 at 23:08
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This should have been more obvious to me than it was, but to keep anyone else from slipping up the same way: these cubes are not in 1:1 correspondence with the set of integer-coordinate vectors for any $n\gt 2$; even in 3 dimensions, for instance, one can't produce two mutually orthogonal vectors with integer coordinates that are orthogonal to $(1, 1, 0)$ and also of length $2$. – Steven Stadnicki Mar 16 '14 at 17:43
up vote 2 down vote accepted

An excellent, but theoretically advanced proof can be found here at my cross-post to MO, thanks to Noam D. Elkies.

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