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
$$
c\,r^q \le \mu(Ball\_{\rho}(x,r)) \le C\,r^q.
$$



Partitioning a smooth manifold into geodesically convex sets
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](https://arxiv.org/abs/2103.15189) 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 $\mu(X\_i\cap X\_j)>0$.


