Special chart of a compact connected manifold.

Question

A friend of mine asked me if the following question is true:

Question: Let $M$ be a compact connected smooth manifold without boundary, then if $a_1,$ $a_2$ $\in$ $M$, there is a chart $(\varphi,U)$ of $M$ such that $a_1,$ $a_2$ $\in$ $U$.

Despite this seems like a lie, I wasn't able to find a counterexample or prove this statement.

Anyone know how to demonstrate this or know a counterexample for this question?

Answer 1

Here's a general fact:

Theorem: For any $x,y \in M$, any path $\gamma : [0,1] \to M$ such that $\gamma(0)=x$ and $\gamma(1)=y$, and any open subset $V \subset M$ containing $\text{image}(\gamma)$, there exists a diffeomorphism $f : M \to M$ such that $f(x)=y$ and such that $f(z)=z$ for all $z \in M-V$.

You can apply this theorem to your problem by starting with any chart $(\phi',U')$ containing $a_1$. If $U'$ contains $a_2$ then you're done. Otherwise, pick a point $a'_2 \ne a_1 \in U'$. Let $\gamma : [0,1] \to M$ be a path with $\gamma(0)=a_2$ and $\gamma(1)=a'_2$, and perturb $\gamma$ so that $a_1 \not\in\text{image}(\gamma)$. Now take $V$ to be any open set that contains $\text{image}(\gamma)$ and does not contain $a_1$. Then apply the above theorem to get a diffeomorphism $f$, and we obtain the desired chart $(\phi,U)$ where $U = f^{-1}(U')$ and $\phi = \phi' \circ f$.

The theorem above is not hard to prove. One starts with any covering of $\text{image}(\gamma)$ by coordinate charts of $M$ that are contained in $V$. Then one applies the Lebesgue number lemma to subdivide $\gamma$ into finitely many subpaths each of which is contained in a single coordinate chart that is contained in $V$: $$\gamma = \gamma \mid [t_0,t_1] * ... * \gamma \mid [t_{K-1},t_K] \qquad 0=t_0 < t_1 < .... < t_{K-1} < t_K=1 $$ where $\gamma \mid [t_{k-1},t_k]$ is contained in a coordinate chart $(\phi_k,U_k)$ such that $U_k \subset V$. For each $k$ one can easily produce a diffeomorphism $f_k : M \to M$ which is the identity on $M-U_k$ and such that $f_k(t_{k-1})=t_k$. Then one defines $f = f_K \circ ... \circ f_1$.