I would like to check my understanding of the following.
Let M be a smooth embedded submanifold of $R^n$, and let $q_1,q_2$ be points in M.
Define "a path on M between $q_1, q_2$" as a $C^2$ map $p: \mathbb{R} \supseteq [a,b] \rightarrow M$, for some $a,b$, such that $p(a)=q_1, p(b)=q_2$.
My question is:
- is it correctly understood that if $M$ is not connected, such a path might not exist between arbitrary points $q_1, q_2$?
- Context: I'm studying mechanics and calculus of variations on manifolds (Lie groups in particular), reading Marsdens Introduction to mechanics and symmetry and Darryl Holms Geometric mechanics and symmetry. There it seems to be assumed that such paths (and "variations" of these) always exist, possibly by partitioning $[a,b]$ into subpaths contained in different coordinate domains. But this "patching" of subpaths does not create a continuous path, does it?
My background: I studied some differential geometry through Lee's Introduction to smooth manifolds.
Edit:
Construction of a path between arbitrary points, $q_1, q_2$:
Let $q_1$ be contained in the chart $(U \subset M, \varphi: U \rightarrow \mathbb{R}^n)$, and $q_2$ in $(V, \phi)$, where $U \cap V \neq \emptyset$ (this is without loss of generality).
Pick a point $q \in U \cap V$. Then we can find a path $p_1$ from $q_1$ to $q$ in $U$, $p_1 = \varphi^{-1} \circ h : \mathbb{R} \rightarrow U$, where $h : \mathbb{R} \rightarrow \mathbb{R}^n$ is some continuous parameterized curve in $\mathbb{R}^n$. I.e. it is continuous and fulfills $p_1(a)=q_1$ and $p_1(k) = q$ for some $a,k \in \mathbb{R}$. Likewise, we can find a path $p_2$ in $V$ such that $p_2(k)=q$, $p_2(b) = q_2$.
I would think that, by some homeomorphic transformation of the $\mathbb{R}^n$-coordinates of each path, we can choose $a < k < b$, and construct the paths such that $p_1(k)=p_2(k)$ (as assumed above).
Now
$$ p(t)= \begin{cases} p_1(t) ,& \text{if } t\in [a,k)\\ p_2(t) ,& \text{if } t\in [k,b] \end{cases} $$
is a continuous path between $q_1, q_2$.
If the above construction is correct, doesn't this show that M is connected? I.e. a contradiction.
Thank you!