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Let $X$ be a connected and compact $d$-dimensional smooth manifold; where $d$ is a positive integer. Does (or rather, when does) there exist a metric $\rho$ on $X$ generating $X$'s topology and a countable number of sets $\{X_n\}_{n}$ such that

  • $\bigcup_n\,X_n = X$,
  • Each $X_n$ is $\rho$-geodesically convex,
  • Each $X_n$ is contractable,
  • $\mu(X_n\cap X_m)=0$ whenever $n\neq m$ (almost disjoint),
  • There is a $d$-dimensional Ahfors regular measure on $(X,\rho)$ such that $$ n\neq m \Rightarrow \rho(X_n\cap X_m)=0. $$

Comment: Ideally, I would like $\{X_n\}_n$ to be a finite set; but I'm open to countably infinite also..

Definition:

  • A measure $\mu$ is Ahfors regular if there are $q,c,C>0$ such that $$ c\,r^q \le \mu(Ball_{\rho}(x,r)) \le C\,r^q. $$
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  • $\begingroup$ Could you clarify the question a bit? As it stands, couldn't you just Whitney embed $M \to \mathbf{R}^{2d}$, take the induced Riemannian metric for $\rho$, the volume measure for $\mu$, and take $\{ M \}$ as a partition? $\endgroup$
    – Leo Moos
    Commented Sep 21, 2022 at 15:06
  • $\begingroup$ @LeoMoos I modified the question to simplify the required condition greatly. $\endgroup$
    – ABIM
    Commented Sep 21, 2022 at 16:18
  • $\begingroup$ Did you mean $\mu(X_n\cap X_m)=0$? $\endgroup$
    – Anton Petrunin
    Commented Sep 22, 2022 at 13:18
  • $\begingroup$ @PierrePC Do you have a reference for this? Also there must be some limitation (ie cases where this doesn't work) by Anton's comment below. $\endgroup$
    – ABIM
    Commented Sep 22, 2022 at 13:31
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    $\begingroup$ @PierrePC no, it is not true, see the second part of my answer. $\endgroup$
    – Anton Petrunin
    Commented Sep 22, 2022 at 14:13

1 Answer 1

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Choose a triangulation of $X$. Let us equip $X$ with a length metric such that each simplex is standard. We may think that $X$ subcomplex of a standard simplex $S$ of large dimension.

Since each face of $S$ is convex; it follows that each simplex is a convex set in $X$. Therefore the covering of $X$ by the simplexes of maximal dimension meets your requirement.

Let me also mention that from our result with Alexander Lytchak it follows that there is no finite covering for generic Riemannian metric on $X$. Namely, for any covering $\{X_i\}$ of $X$, we have that $X_i\cap X_j$ has nonempty interior for some $i\ne j$; hence $\mu(X_i\cap X_j)>0$.

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    $\begingroup$ Would you mind expanding on how your result implies that a generic metric on $X$ cannot satisfy the requirements? I am struggling to find where the extremal points come into play. Also, are your generic Riemannian metrics smooth? (I am not sure if the $\mathcal C^2$ regularity refers only to the topology.) $\endgroup$
    – Pierre PC
    Commented Sep 22, 2022 at 16:57
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    $\begingroup$ @AntonPetrunin Thanks a lot, let me ask, why do you use a spherical simplex? This I can't see (why not a typical Euclidean simplex?) $\endgroup$
    – ABIM
    Commented Sep 23, 2022 at 13:15
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    $\begingroup$ @AIM The statement easily follows from Gromov's flag test for CAT(1) spaces. The original paper of Berestovskii is in Russian: В. Н. Берестовский. «Задача Борсука о метризации полиэдра». Докл. АН СССР 268.2 (1983), 273—277. $\endgroup$
    – Anton Petrunin
    Commented Sep 23, 2022 at 16:17
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    $\begingroup$ @AIM yes, μ might be the d-dimensional Hausdorff measure or any Ahfors regular measure. $\endgroup$
    – Anton Petrunin
    Commented Sep 25, 2022 at 13:57
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    $\begingroup$ @AIM Otherwise proving that each simplex is convex will be harder. I do not see a proof now, but I am sure it is true. $\endgroup$
    – Anton Petrunin
    Commented Sep 25, 2022 at 22:05

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