I am student in mechanical engineering.
I am trying to study nonlinear dynamical systems, confused with the dimension of manifold and the space whose coordinate function maps to. I am using this web page to learn the concept of manifold. http://lavalle.pl/planning/node385.html

In example 8.11, a 2 dimensional circle $(x^{2} + y^{2} =1)$ is considered. I understand that this circle is a manifold of dimension 1, denoted by $n=1$, embedded in $\mathbb{R}^{2}$ space. I also understand that a coordinate function $\phi = \tan(y/x)$ maps a point on the manifold to $\mathbb{R}$ which can be seen as $\mathbb{R}^{n}$ with $n=1$.

In example 8.12, a manifold $M=SO(3)$ is considered. I understand that this a manifold of dimension $n=2$ as given in the web page. However, it says that "this means that any coordinate neighborhood must map a point in $SO(3)$ to a point in $\mathbb{R}^{3}$ ". Following the same reasoning as in the case of the circle in example 8.11, shouldn’t this be in $\mathbb{R}^{2}$, because the dimension of the manifold is $n=2$?


1 Answer 1


Everything in the text of Example 8.12 refers to the (correct) fact that $SO(3)$ is a manifold of dimension $3$, except for that one (incorrect) equation $n=2$. It's just an error, so feel free to change that $2$ to a $3$.

  • $\begingroup$ Thank you. So, $SO(3)$ is a 3 dimensional manifold embedded in $\mathbb{R}^{3}$ ? $\endgroup$ Jul 26 at 21:23
  • $\begingroup$ Yes, $SO(3)$ is a 3-dimensional manifold. No, $SO(3)$ is not embedded in $\mathbb R^3$. $\endgroup$
    – Lee Mosher
    Jul 26 at 21:26
  • $\begingroup$ Thank you. I see, it should be $\mathbb{R}^{9}$. $\endgroup$ Jul 26 at 21:33

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