What is a topological group on $ \mathbb{R}^n\ $

Question

While working on another problem, I have to use the idea of the topological group on $ \mathbb{R}^n\ $. The on-line definitions don’t help much because they just say something like “$ \mathbb{R}^n\ $is a topological group under addition,” without further explanation.

Addition of what? What are the elements of the group? Will it be clear how they add?

Just to make it interesting, I have the same question concerning the topological group on $\ S^1 $.

Please keep it simple. I am no expert.

Answer 1

Addition of what? What are the elements of the group? Will it be clear how they add?

If $\mathbb R^n$ is the group, then elements of $\mathbb R^n$ are the elements of the group. If you have experience adding vectors then you can figure out how to add the elements of this group.

$S^1$ is a multiplicative group of complex numbers with absolute value 1. If you are familiar with the multiplication of complex numbers, particularly in exponential form, then this is also simple.

If you don't know what a topological space is then you will probably have a tough time understanding the "topological" part of it.

Answer 2

I won't go through checking everything very carefully, but intuitively, the operation $+:\Bbb{R}^n\times \Bbb{R}^n\to \Bbb{R}^n$ is given in the usual way. Given a pair of vectors $v=(x_1,\ldots, x_n)$ and $(y_1,\ldots, y_n)=w$, $$ +(v,w)=v+w.$$ We know (from algebra) that this defines an Abelian group structure on $\Bbb{R}^n$, but more is true. Indeed, it is easy to see that $+$ is continuous as a map $\Bbb{R}^n\times \Bbb{R}^n\to \Bbb{R}^n$. This is because $+$ is given componentwise by $(x_i,y_i)\mapsto (x_i+y_i)$, which is continuous (in fact smooth). We also need to check that the inversion map (sending a vector to its additive inverse) is continuous, but this is nothing but the antipodal map $$ (x_1,\ldots, x_n)\mapsto (-x_1,\ldots,-x_n),$$ which is continuous.

As for the circle, we can view $S^1\subseteq \Bbb{C}^\times$ as a subgroup under multiplication. If we represent elements of $S^1$ as $e^{i\theta}$ for $\theta \in \Bbb{R}$ it becomes clear that the group operation is given by $(e^{i\theta},e^{i\psi}) \mapsto e^{i\theta}e^{i\psi}=e^{i(\theta+\psi)}$. You can check that this is again continuous. The inversion map can be viewed as the restriction of complex conjugation to $S^1$. I.e. $z\mapsto \overline{z}$. This map is a linear transformation of $\Bbb{C}$ (regarded as an $\Bbb{R}-$vector space) and hence continuous. It restricts to a continuous map $S^1\to S^1$ given in angular representations by $e^{i\theta}\mapsto e^{-i\theta}$.