Challenge in 4 Dimensions

Using Lagrange Four Square Theorem, classify the set of side lengths of a lattice hypercube in $ \mathbb{Z}^4 $.

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Comments

Alright. I'm going to take a crack at this. Thank you for your patience.

Theorem: The set of side lengths of hypercubes in $\mathbb{R}^4$ with vertices in $\mathbb{Z}^4$ is precisely the image of the positive integers under the principal square root.

Proof: Let $\sqrt{\mathbb{Z}_+}$ denote the image of the positive integers under the principal square root. I wish to note that, if we wanted to consider the case of a degenerate hypercube as well, we would instead be considering the image of the nonnegative integers under the principal square root.

For the sake of comfort, I will provide an equivalent and facile definition of a hypercube. We define a hypercube as a square matrix composed of mutually orthogonal column vectors of the same magnitude. This is equivalent to defining, in our case, 4 vectors in $\mathbb{R}^4$ that are orthogonal and of the same magnitude, which then form the edges of the hypercube.

We wish to prove that a side length being an element of $\sqrt{\mathbb{Z}_+}$ is a necessary and sufficient condition for there to exist a hypercube of that side length with vertices in $\mathbb{Z}^4$ .

The necessity is satisfied because of the Lagrange Four Square Theorem. If we consider a vector $\vec{x}\in\mathbb{R}^4$ with integer components, then its magnitude will be of the form $\|\vec{x}\|=\sqrt{m^2+a^2+t^2+h^2}$ , with $m,a,t,h\in\mathbb{Z}$ . Every positive integer can be written in the form $m^2+a^2+t^2+h^2$ , and every $m^2+a^2+t^2+h^2$ is a positive integer, so the side lengths we consider will necessarily be in $\sqrt{\mathbb{Z}_+}$ .

To show that the condition is sufficient, we construct a hypercube for a given side length. Suppose $\|\vec{x}\|^2=m^2+a^2+t^2+h^2$ . Then, we consider the hypercube $\begin{bmatrix}m & -a & h & -t \\ a & m & t & h \\ t & -h & -a & m \\ h & t & -m & -a\end{bmatrix}$ . Clearly, every column $\vec{x}$ is orthogonal to the others and satisfies $\|\vec{x}\|^2=m^2+a^2+t^2+h^2$ . Thus, there must exist a desired hypercube if the side length is in $\sqrt{\mathbb{Z}_+}$ . This completes the proof. $\boxed{}$

Jacob Erickson - 7 years, 5 months ago

Thumbs up to the choice of variables.

Bob Krueger - 7 years, 5 months ago

I'm not exactly sure how this uses the Four Square Theorem, but here's my guess:

Since the side lengths are defined as the distance between two points, and all side lengths are the same, take any two points of the tesseract and calculate that distance by $d = \sqrt{(a_1-a_2)^2+(b_1-b_2)^2+(c_1-c_2)^2+(d_1-d_2)^2}$ where each point is defined by the ordered quadruple $(a_i,b_i,c_i,d_i)$ . Then, since every number can be expressed as the sum of at most four squares, the set of side lengths of a lattice hypercube is the set of all positive square roots of positive integers.

Bob Krueger - 7 years, 6 months ago

In $\mathbb{Z}^3$ , is there a lattice cube of side length $\sqrt{1+1+1} = \sqrt{3}$ ?

Where does your above logic break down?

Calvin Lin Staff - 7 years, 6 months ago

Clearly, there is no such lattice cube in $\mathbb{Z}^3$ , since we cannot have integers $a,b,c$ such that $a+b+c=0$ and $a^2+b^2+c^2=3$ .

I'm going to see if I can finish a solution. I think I have a partial answer and a large enough margin. However, I have a history project that I need to finish by tomorrow, so I'll have to get back to this on Friday.

Jacob Erickson - 7 years, 5 months ago

@Jacob Erickson – Can you explain your thinking step by step? Why do we need $a + b + c = 0$ ?

Calvin Lin Staff - 7 years, 5 months ago

@Calvin Lin – We have the vector $\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ with magnitude $\sqrt{3}$ . We wish to find a vector orthogonal to it under the given inner product. Thus, we want to find $\begin{bmatrix}a \\ b \\ c\end{bmatrix}$ satisfying $a+b+c=0$ from the "dot" product and $a^2+b^2+c^2=3$ for the magnitude.

Jacob Erickson - 7 years, 5 months ago

@Jacob Erickson – Thanks for the explanation. The important aspect here is that "There is no orthogonal vector", which is why no such 3-dimensional cube exists. Thus, going back to the 4-D case, we need to find 4 orthogonal vectors, or show that they do not exist.

For clarity, It seems like you made an assumption that one of the vectors must be $(1,1,1,)$ . Of course, this need not be the case, as we start with vectors of the form $(\pm 1, \pm 1, \pm1)$ , and it is best to add WLOG to the start.

Calvin Lin Staff - 7 years, 5 months ago

@Calvin Lin – My apologies. Indeed, I should note that I performed the above argument under the assumption that we all understood there was no loss of generality in assuming the starting vector was $\begin{bmatrix}1 & 1 & 1\end{bmatrix}^T$ , since $a$ , $b$ , and $c$ can be positive or negative (or zero).

Jacob Erickson - 7 years, 5 months ago

How are you defining a hypercube in this case? Are we assuming that $\mathbb{Z}^4$ inherits the standard Euclidean metric?

Jacob Erickson - 7 years, 5 months ago

It would be the 4-D analogue of a cube. You would have to define $2^4$ points. Alternatively, you could define $4$ vectors which form the edges of the cube from 1 vertex, where the angles are all $90^\circ$ .

Yes, standard Euclidean metric where distance between the points is $\sqrt{ (x_1-y_1)^2 + (x_2-y_2)^2 + (x_3-y_3)^2 + (x_4-y_4)^2 }$ .

Calvin Lin Staff - 7 years, 5 months ago

m only read lagrange mean value theorem :(

Manoj Kumar - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$