I am reading a text book and it says the following:
If $G$ is an $r$-dimensional Lie group acting on a manifold $X$. If $G$ acts freely, then its orbits have the same dimension as $G$ itself.
I am very confused about this statement. Can anyone please give me an example to see this?
For example, let $X$ be a torus, in its usual embedding in $\mathbb R^3$. Then $SO(2)$ acts freely on it, by rotation around the $z$-axis. The orbits of points on the torus are horizontal circles, which have dimension $1$ just like $SO(2)$ does.
On the other hand, suppose $X$ is $\mathbb R^2$, with the natural action of $SO(2)$. This action is not free -- the stabilizer of $(0,0)$ is all of $SO(2)$ -- and actually there is an orbit whose dimension is different from $1$, namely $\{(0,0)\}$ which has dimension $0$.