# Dimension of a manifold and coordinate function.

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 $$n=2$$?

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