0

$(\mathbb{R}, +)$ is a group. But it’s not immediately clear where the symmetries are.

I can answer this problem in two steps.

First I can find a substitute group for $(\mathbb{R}, +)$, that will have the exact same behaviour, but will be easier to understand intuitively, because it will be of the type (symmetries, composition). Specifically in our case this will be the set of translations of $\mathbb{R}$, obtained by applying Cayley's theorem to our original group.

Second, I need to determine from which object these symmetries come from. This will allow me to visualize what’s going on. I don't fully understand this step.

My own research on this question has led be to believe it has something to do with Frucht's theorem making it so any group can be thought of as the automorphism group of a graph (which through some procedure, can be turned into an object in some d-dimensional space).

This is a follow up on my previous question. I'd like to note that a similar question already exists, but I don't find the answers satisfactory because they focus on finite group (which is why I have $(\mathbb{R}, +)$ as the main example in mind).

Davide Radaelli
  • 199
  • 2
    What about the answer provided in the first linked question don't you understand? In particular, that answer gives a geometric object, namely the real line, on which $\mathbb{R}$ acts as the group of symmetries. – Noah Solomon Mar 26 '22 at 12:39
  • I would like to find the object from which these symmetries (the translations) are derived. In the same way that the set {1,2,3,4} is the object from which the symmetries $S_4$ arise. – Davide Radaelli Mar 26 '22 at 13:34
  • The object is the set of real numbers and the symmetries in question are translations. Perhaps something you are finding confusing is that symmetries in question are not the full set of permutations on $\mathbb{R}$, but rather just the translations. – Noah Solomon Mar 26 '22 at 16:53
  • So what structure must $\mathbb{R}$ be equipped with to guarantee that its symmetries are exactly the translations? – Davide Radaelli Mar 26 '22 at 22:22
  • I'm not sure exactly what you mean. The word symmetries is in general not precisely defined so we need to specify what we mean when we use it. In this context, by symmetries of $\mathbb{R}$ I mean linear functions on $\mathbb{R}$ which restrict to bijections on $\mathbb{Z}$. The fancy term for (the more general version of) this is is isometry, and I recommend reading the wikipedia page for that. – Noah Solomon Mar 27 '22 at 22:24

0 Answers0