This is slightly vague as I've not yet come to terms with what I'm actually looking for.

On $S^2$ we may choose charts (stereographic projection) such that the image of a disk (i.e. all points within distance $r$ from a fixed point $p\in S^2$) completely contained inside the domain of the chart , maps onto a disk in $\mathbb{R}^2$.

Let's say $M$ is a connected Riemanninan manifold and $p\in M$. Can I always find such a disk-preserving chart around $p$? Note that for any other $q$ inside the chart, the image of a disk around $q$ contained in the domain should also map to a disk.

edit: I used $S^2$ and $\mathbb{R}^2$ for simplicity. Note that I am asking for $n$-disks on $n$-dimensional manifolds.

  • $\begingroup$ A four-set Venn diagram (i.e. one that shows all 16 possible unions and intersections) is not realizable with disks in $\mathbb R^2$. If it is realizable with disks on a different Riemannian manifold, that would answer your question in the negative. $\endgroup$ – Rahul Mar 21 '16 at 16:55
  • $\begingroup$ Oh, I see that I have been sloppy. Let me try an edit. Sorry about that. $\endgroup$ – M.B. Mar 21 '16 at 16:57
  • $\begingroup$ The stereographic projection only preserves disks centred on the pole - others are distorted. If you really want all disks to map to disks then I'm almost certain the chart has to be a composition of a homothety and an isometry; i.e. it's only possible when $M$ is flat. $\endgroup$ – Anthony Carapetis Mar 22 '16 at 9:57
  • $\begingroup$ @AnthonyCarapetis: I don't believe that is true. As long as your circle doesn't intersect the north pole, then the image is a circle in the plane. $\endgroup$ – M.B. Mar 22 '16 at 14:44
  • $\begingroup$ @M.B.: ah, you're right of course; I assumed the stronger condition that the central points of the discs are preserved. $\endgroup$ – Anthony Carapetis Mar 27 '16 at 11:30

Note that stereographic projection maps $B_R(p)$ onto $B_r(0)$ for some $R,\ r$ where $0$ is a origin of ${\bf R}^2$.

In fact any Riemannian manifold $M$ has a chart : $\exp_p : B_r(0)\mapsto B_r(p)$ where $B_r(0)$ is $r$-ball in tangent space and $B_r(p)$ is $r$-ball in $M$

  • $\begingroup$ Hmm.. But I want this to be true for all points inside the same chart. I.e, for (small enough) radii $r_1$ and $r_2$ I want the image of $B_{r_1}(p)$ and $B_{r_2}(q)$ to be disks. $\endgroup$ – M.B. Mar 21 '16 at 16:53

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