Intersecting $B^d_\infty$ with the hyperplane $H = \{x\in \Bbb R^d: x_1 + \ldots + x_d = 0\}$

Question

The following problem appears in Section $2.3$ (Exercise $13$) of these notes on Convex and Discrete Geometry (see Pg. $16$).

Consider the following vertices of $B_\infty^4 := \{x\in \Bbb R^4: \|x\|_\infty \le 1\}$: $$a_1 = (-1,1,1,-1), \quad a_2 = (1,-1,1,-1), \quad a_3 = (1,1,-1,-1),$$ $$b_1 = (1,-1,-1,1), \quad b_2 = (-1,1,-1,1), \quad b_3 = (-1,-1,1,1).$$

1. Show that $K := \operatorname{conv}\{a_1,a_2,a_3,b_1,b_2,b_3\}$ is a three-dimensional regular octahedron.
2. Prove that $B_\infty^4 \cap H = K$, where $H = \{x\in \Bbb R^4: x_1 + \ldots + x_4 = 0\}$.
3. What is the analogous result in $\Bbb R^3$, i.e., what do we obtain by intersecting the cube $[-1, 1]^3$ with the hyperplane $H = \{x \in \Bbb R^3: x_1 + x_2 + x_3 = 0\}$?

My thoughts.

1. Every $a_i$ and $b_i$ (for $i = 1,2,3$) belongs to the hyperplane $H$, which is a three-dimensional subspace of $\Bbb R^4$. As a result, $\operatorname{conv}\{a_1,a_2,a_3,b_1,b_2,b_3\} \subset H$, and it's clear that $\dim K \le 3$. How does one show that $K$ is a regular octahedron, though?

2. Using the definitions, one can easily check $B_\infty^4 \cap H \supset K$. Why is $B_\infty^4 \cap H \subset K$?

3. Seems hard to visualize. What's going on?

Hints and solutions are welcome. Thanks!

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• $\operatorname{conv} S$ is the convex hull of $S\subset \Bbb R^n$.

Answer 1

Question 1. To prove that $K$ is a regular octahedron, we'll use an isometric coordinates change so as to express its vertices as those of the following octahedron in $\mathbb{R}^3$ (cf. cartesian coordinates): $(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)$.

In the new basis, the first three basis vectors are colinear to $a_1, a_2, a_3$, and the fourth one is orthogonal to them all: $(1,1,1,1)$. Using unit vectors, this gives the following change-of-basis orthogonal matrix:

$$ \begin{pmatrix} -\frac 1 2 & \frac 1 2 & \frac 1 2 & -\frac 1 2 \\ \frac 1 2 & -\frac 1 2 & \frac 1 2 & -\frac 1 2 \\ \frac 1 2 & \frac 1 2 & -\frac 1 2 & -\frac 1 2 \\ \frac 1 2 & \frac 1 2 & \frac 1 2 & \frac 1 2 \end{pmatrix} $$

Multiplying this matrix by $a_1, a_2, a_3, b_1, b_2, b_3$, we get: $(2,0,0,0), (0,2,0,0), (0,0,2,0), (-2,0,0,0), (0,-2,0,0), (0,0,-2,0)$
which is the octahedron quoted above, scaled by an homothety of $2$.

Question 2. $B_{\infty}^4 \cap H$ is the intersection of a full hypercube with an hyperplane. It is the convex hull of the intersection of $H$ with the edges of the hypercube $[-1,1]^4$.

Those edges are obtained by having three coordinates with a value of $1$ or $-1$, and the fourth in $[-1,1]$. There are $32$ edges in total.
As all coordinates are integers except one, the equation $x_1+x_2+x_3+x_4=0$ imposes that all four coordinate be integers. So intersections are vertices of the hypercube.

This leaves us with verifying which vertices of the hypercube are in $H$. This is those quadruplets of $1$ and $-1$ whose sum is $0$. This imposes to have two $1$ and two $-1$, so there are ${4 \choose 2} = 6$ of them, which are the $a_i$ and $b_i$.

Question 3. Using the same process in dimension 3, we want the intersections of the cube $[-1,1]^3$ with the plane $x_1+x_2+x_3=0$. The edges are $(1,1,[-1,1]), (1,-1,[-1,1]), \dots, (1, [-1,1], 1), \dots, ([-1,1],-1,-1)$.
In order for the sum of coordinates to be null, we need one $1$, one $-1$, and one $0$ in $[-1,1]$.
This leaves us with the following vertices:
$(1,-1,0), (-1,1,0), (1,0,-1), (-1,0,1), (0,1,-1), (0,-1,1)$.

As in question 1, we change coordinates, with two vectors inside the plane $x_1+x_2+x_3=0$, and the third one orthogonal to this plane. Ths gives for example the following orthogonal matrix:

$$ \begin{pmatrix} \frac 1 {\sqrt 6} & \frac 1 {\sqrt 6} & -\frac 2 {\sqrt 6} \\ \frac 1 {\sqrt 2} & -\frac 1 {\sqrt 2} & 0 \\ \frac 1 {\sqrt 3} & \frac 1 {\sqrt 3} & \frac 1 {\sqrt 3} \end{pmatrix} $$

Applying this matrix to the vertices above, this gives:
$(\sqrt 2, 0, 0), (-\sqrt 2, 0, 0), (\frac {\sqrt 3} {\sqrt 2}, \frac 1 {\sqrt 2}, 0), (-\frac {\sqrt 3} {\sqrt 2}, -\frac 1 {\sqrt 2}, 0), (\frac {\sqrt 3} {\sqrt 2}, -\frac 1 {\sqrt 2}, 0), (-\frac {\sqrt 3} {\sqrt 2}, \frac 1 {\sqrt 2}, 0)$

Dividing by $\sqrt 2$, we have:
$(1, 0, 0), (-1, 0, 0), (\frac {\sqrt 3} 2, \frac 1 2, 0), (-\frac {\sqrt 3} 2, -\frac 1 2, 0), (\frac {\sqrt 3} 2, -\frac 1 2, 0), (-\frac {\sqrt 3} 2, \frac 1 2, 0)$
where we recognize the regular hexagon, $\frac {\sqrt 3} 2$ being $\sin(\frac {\pi} 3)$ and $\frac 1 2$ being $\cos (\frac {\pi} 3)$.