# 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 $M$?

• Use finitely many local charts and approximate the given curve with a smooth one locally. Commented Nov 23, 2012 at 17:36
• Do I need partition of unity? Commented Nov 23, 2012 at 17:44

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 $$0=t_0 such that $$\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 as demonstrated below:

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 $$a and $$\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$$.

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 $q$.

• if the domain of the curve is $(0,1)$ then the image is not compact, but the proof works. Is it right? Commented Nov 23, 2012 at 20:59
• You generally need endpoints so that one goes to $p$ and the other goes to $q$. That's the definition of a continuous path, no? Commented Nov 23, 2012 at 21:01
• You also might specify that within a neighborhood of the intersection of two paths the restriction of the path to these two segments can also be made smooth by this neighborhood also being diffeomorphic to $R^n$ Commented Apr 24, 2013 at 14:39
• How do we know that when we patch these paths together the result will remain smooth at the $r_i$? Commented Aug 29, 2015 at 6:54
• I believe that you have only constructed a piecewise-smooth curve. Commented Oct 27, 2016 at 21:16

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.

• You need to prove that this relation is transitive, which amounts to smoothing the corners of a piecewise-smooth curve; so your answer has the same gap as Elchanan's. Commented Mar 29, 2018 at 15:18