Smooth curves on a path connected smooth manifold

Suppose that $M$ is a path connected smooth manifold, so any two points $p,q\in M$ can be joined with a continuous curve on $M$. Is it true that any two points can be joined with a smooth (I mean $C^{\infty}$) curve on $M$?

Solution 1:

Since the accepted answer is rather incomplete, i'll post an answer here. As @Elchanan Solomon did, for any two points $p$ and $q$ on a connected manifold $M$, we can obtain a piecewise smooth curve $\gamma : [0,1] \to M$ such that $\gamma(0)=p$ and $\gamma(1)=q$. That is there are finite partition $0=t_0<t_1<\dots<t_{n-1}<t_n=1$ such that $\gamma|_{[t_i,t_{i+1}]} : [t_i,t_{i+1}] \to M$ are smooth. To smoothing the curve $\gamma$ we can use smooth bump function (as the answer here) or we can use Whitney Approximation Theorem for Function.

For any point of corner $\gamma(t_i)$ where $\gamma$ not smooth, we choose a smooth chart $(U_i,\varphi)$ containing it. Now we pick a portion of $\gamma$ that still contain in $U_i$. That is, consider the restriction of $\gamma$ to a closed interval $[a,b]$ where $a<t_i<b$ and $\gamma([a,b]) \subset U_i$. By composing with chart map $\varphi : U \to \mathbb{R}^n$, we obtain a piecewise smooth curve in $\mathbb{R}^n$, $$ \alpha := \varphi \circ \gamma|_{[a,b]} : [a,b] \to \mathbb{R}^n. $$ For a small number $\epsilon>0$, the subset $A=[a,t_i-\epsilon] \cup [t_i+\epsilon,b]$ is closed in $[a,b]$ and $\alpha$ is smooth on $A$. By Whitney Approximation theorem, there are smooth map $\tilde{\alpha} : [a,b] \to \mathbb{R}^n$ such that $\tilde{\alpha}$ is agree with $\alpha$ on $A$. Mapping back to $M$ and replacing the map $\varphi^{-1} \circ \tilde{\alpha} : [a,b] \to M$ with $\gamma|_{[a,b]}$ before. Doing this for all corners, we obtain the desired smooth map joining $p$ and $q$.

Solution 2:

No need for a partition of unity. Let $\gamma$ be a path from $p$ to $q$, not necessarily smooth. Take finitely many local charts that cover the compact image of $\gamma([0,1])$. Call these local charts $U_1, \cdots, U_n$. Let us have ordered these charts so that the overlap between $U_{i}$ and $U_{i+1}$ is nonempty, but rather some point $r_i$. Pragmatically, you can do this by pulling back the $U_i$ to a cover of $[0,1]$ by the $\gamma^{-1}(U_i)$, and then ordering them by left-endpoints. Now, in each of these local charts the manifold looks like $\mathbb{R}^n$, so we can find a smooth path from $p$ to $r_1$ in $U_1$, then a smooth path from $r_1$ to $r_2$ in $U_2$, etc. Connecting all these smooth paths gives us a smooth path from $p$ to $q$.

Solution 3:

Consider the following equivalence relation on the points of $M$. Two points $x,y\in M$ are equivalent if they are connected by a smooth path.

Let us prove the equivalence classes of this relation are open. This means every point $x$ has a neighborhood of points smoothly connected to it. Any chart about $x$ provides such a neighborhood, since any two points in Euclidean space may be smoothly connected (as Euclidean space is locally convex).

Consequently $M$ is partitioned into open subsets given by the equivalence classes. By connectedness this partition has a single equivalence class, proving every two points are connected by a smooth curve as desired.