Think 3-D Is Cool? Wait Till You Try 4-D!

Geometry Level 4

A lattice hypercube is a cube in 4-dimensional space, whose vertices are lattice points. Describe the set of possible side lengths of a lattice hypercube.

Note : A lattice point is a point whose coordinates are all integers. $(0,0,1)$ is a lattice point, but $\left(\frac{1}{2}, \sqrt{2}, -0.2 \right)$ is not a lattice point.

If you are interested in Lattice cubes, try the 3-D version .

For similar problems, you can read my note on Construction .

\{ n^2 \mid n \text{ is a non-negative integer.} \}

\{ \sqrt{n} \mid n \text{ is a non-negative integer.} \}

\{ n \mid n

is a non-negative integer not divisible by a prime of the form

4k+3. \}

\{ n \mid n \text{ is a non-negative integer.} \}

1 solution

Calvin Lin Staff
Mar 7, 2014

The answer is $\{ \sqrt{n} \mid n$ is a non-negative integer $\}$ . As this is a construction problem, we need to show that the description is both necessary, and sufficient.

First, we establish a necessary condition. The distance between any 2 lattice points is of the form $\sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2 + (w_1 - w_2)^2 }$ , and so must be of the form $\sqrt{n}$ where $n$ is a non-negative integer.

Second, we have to establish existence. (This is not guaranteed, see for example, the 3-D version). We use the Lagrange 4 square theorem, which states that for any non-negative integer $n$ , it can be written as the sum of 4 squares, namely $n = a^2 + b^2 + c^2 + d^2$ .

Consider the 4 column vectors in the following matrix: $\begin{pmatrix} a & b & c & d \\ b & -a & -d & c \\ c & d & -a & -b \\ d & -c & b & -a \end{pmatrix}$ It is clear that they each have side length $\sqrt{n}$ , and we can check that they are mutually orthogonal. Hence, the cube that is spanned by these 4 vectors, is a cube of side length $\sqrt{n}$ . This establishes the sufficient condition.

Is this true for N dimensional cube?

Maharnab Mitra - 7 years, 3 months ago

That is a good question.

You can try the 3-D case .

The 1-D case is pretty obvious, where the answer is just non-negative integers $n$ . The 2-D case requires slight Number Theory, but there is a very simple classification involving Fermat's Two Square theorem.

You can explore larger values of $N$ . There is a classification, which is easy to obtain for cases of $N \equiv 0, 1, 3, \pmod{4}$ .

Calvin Lin Staff - 7 years, 3 months ago

Think 3-D Is Cool? Wait Till You Try 4-D!

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