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Suppose you have a manifold $M$ and a finite set $S = \{x_1, \ldots, x_n\}$ of points in the same connected component of $M$. I am looking for criteria that ensures $S$ is contained in a connected chart.

Partial progress:

  • $n = 2$: by the discussion in comments, there exists an embedded arc from $x_1$ to $x_2$. A tubular neighborhood will be the desired chart.
  • $M$ compact, $x_i$'s very close: Take the lebesgue number $r$ of an atlas of $M$, with respect to the distance induced by some riemannian metric on $M$. If $d(x_1, x_i) < r$ for $i=2, \ldots, n$, then $S \subset B_r(x_1)$ which is in a chart by the lebesgue number property.
Andrea Marino
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  • Charts are fixed, they are part of the definition of manifold. Are you asking about neighbourhood containing all points homeomorphic to Euclidean space? Or an equivalent atlas on $M$ with a chart containing all points? – freakish Aug 01 '23 at 09:07
  • Btw your $n=2$ case is incorrect. The path connecting $x_1$ to $x_2$ may be a space filling curve, i.e. entire $M$. – freakish Aug 01 '23 at 09:09
  • @freakish: paths are assumed to be continous – Andrea Marino Aug 01 '23 at 09:17
  • And yes, a manifold can have several atlases. "A" chart means an open set that is diffeomorphic to the euclidean space. – Andrea Marino Aug 01 '23 at 09:18
  • What makes you think a space filling curve is not continuous? Any compact connected manifold $M$ can be covered by a continuous curve $[0,1]\to M$. Read here: https://en.wikipedia.org/wiki/Space-filling_curve As a consequence, for any two points $x_1,x_2$ on a compact, connected manifold $M$ there is a continuous curve $\lambda:[0,1]\to M$ such that $\lambda(0)=x_1$, $\lambda(1)=x_2$ and $im(\lambda)=M$. – freakish Aug 01 '23 at 09:22
  • You are right, continuity is not enough. However, Ryan Budney observed here (https://math.stackexchange.com/questions/35780/given-two-points-in-a-manifold-can-one-always-find-a-topological-ball-that-cont) that one can always put a complete riemannian metric on $M$, and a geodesic between the two points will be an embedding. – Andrea Marino Aug 01 '23 at 09:29
  • Yes, that is correct. But as you can see it requires a stronger argument. – freakish Aug 01 '23 at 09:32
  • Is the chart required to be connected? – Deane Aug 01 '23 at 13:02
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    Here's an idea. Take a path $\lambda:[0,1]\to M$ that connects all points. Now foreach $\lambda(t)$ there is a an open ball $B_t$ containing $\lambda(t)$ homeomorphic to the Euclidean space. Since $\lambda^{-1}(B_t)$ cover $[0,1]$, then only finitely many cover $[0,1]$. And so we have a finite sequence of balls containing our points and such that consecutive have nonempty intersection. If $dim\geq 2$ then I think from this we can construct injective smooth curve containing all points (by piecewise glueing lines and careful avoidance of intersections) and then tabular neighbourhood works. – freakish Aug 01 '23 at 15:49
  • If $\lambda$ is made of geodesics, I think you could be right, by generalizing the above argument. – Andrea Marino Aug 02 '23 at 15:31
  • The key point is that a connected smooth manifold with dimension greater than 2 remains connected if you remove a compact smoothly embedded curve segment. Therefore there is a smooth closed curve that starts at $x_1$, ends at $x_n$, and contains all of the points. A tubular neighborhood of the curve is diffeomorphic to a ball and therefore a connected coordinate chart. – Deane Aug 03 '23 at 00:48

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As it is asked, the answer to your question is trivially "yes". Just take charts $(U_i,\phi_i : U_i \to U'_i \subset \mathbb R^n)$ such that

  1. $x_i \in U_i$.
  2. $U_i \cap U_j = \emptyset$ for $i \ne j$.
  3. $U'_i \cap U'_j = \emptyset$ for $i \ne j$.

Then $U = \bigcup U_i$ contains $S$ and the $\phi_i$ define a chart $(U, \phi : U \to U' = \bigcup U'_i)$ such that $\phi \mid_{U_i} = \phi_i$.

The set $U$ is not connected. If that is an additional requirement, more subtle arguments are needed.

Paul Frost
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  • Hi Paul, we probably have different definitions of chart. For me, a chart is a neighborhood diffeomorphic to an euclidean ball. However, in case of variants like yours, I added the connection assumption. What is your definition, by the way? – Andrea Marino Aug 02 '23 at 15:30
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    @AndreaMarino Most authors define a chart as a homeomorphism from an open subset of $M$ (no further restrictions!) to an open subset of $\mathbb R^n$. A smooth atlas is a collection of charts such that all transition functions are smooth. – Paul Frost Aug 02 '23 at 20:32