Lyla Carson

2022-09-23

What does mean $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{{e}_{1},-{e}_{l},\dots ,{e}_{d},-{e}_{d}\}$ and $\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{\{+1,-1{\}}^{d}\}.$

I could not understand, need a simple explanation.

I could not understand, need a simple explanation.

Cremolinoer

Beginner2022-09-24Added 11 answers

Let us consider the case d=3.

The set $\{\{+1,-1{\}}^{d}\}$ is the set of all triples (there are ${2}^{3}=8$ of them)

$\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),\cdots (1,1,1)\}$

i.e., the vertices of a cube.

More generally, $\{+1,-1{\}}^{d}$ is the set of vertices of a hypercube in ${\mathbb{R}}^{\mathbb{d}}$

As a consequence:

$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{\{+1,-1{\}}^{d}\}$

is the "solid hypercube" in ${\mathbb{R}}^{d}$

The set $\{\{+1,-1{\}}^{d}\}$ is the set of all triples (there are ${2}^{3}=8$ of them)

$\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),\cdots (1,1,1)\}$

i.e., the vertices of a cube.

More generally, $\{+1,-1{\}}^{d}$ is the set of vertices of a hypercube in ${\mathbb{R}}^{\mathbb{d}}$

As a consequence:

$\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{v}\{\{+1,-1{\}}^{d}\}$

is the "solid hypercube" in ${\mathbb{R}}^{d}$

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