My question is a little bit vague. I want to know if an arbitrary compact Riemannian manifold (M^d,g) admits partitions that are uniform in some sense. To be more precise, I need for every eps > 0 a partition of M into finitely many measurable subsets P\_1,...,P\_k such that 


(i) The diameter of each P\_i is smaller than eps.


(ii) The number k goes like 1/(eps)^d w.r.t. eps (asymptotically).


(iii) The boundary of each P\_i is pieced together from finitely many pieces of smooth submanifolds.


(iv) I have some control over the (d-1)-dimensional volumes of the boundaries and also over their curvatures, measured by the second fundamental form. The ideal case would be, when the boundaries are pieced together by totally geodesic submanifolds such that the second fundamental form vanishes on each piece and such that the volume of the boundary of each P\_i asymptotically behaves like the volume of a (d-1)-dimensional Euclidean sphere.


Any hint or idea would be helpful.



Uniform partitions of a compact Riemannian manifold
Assume first that your manifold $M$ is embedded in an Euclidean space $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ and the metric is the induced metric. (Nash's embedding theorem shows that this is always possible.)


For each $\newcommand{\ve}{\varepsilon}$ $\ve>0$ denote by $\newcommand{\eL}{\mathscr{L}}$ $\eL\_{\ve}$ the "cubulation" of $\bR^n$ defined by the infinite collection of hyperplanes $\newcommand{\bZ}{\mathbb{Z}}$


$$\frac{x\_i}{\ve}\in \bZ,\;\;i=1,\dotsc n. $$


The vertices of this cubulation fill the lattice $\bigl(\,\ve^{-1}\bZ\bigr)^n$.


Consider the intersection of $\eL\_\ve$ with $M$. If $M$ is in general position (meaning you may have to rotate and translate $M$ a little bit), then this intersection determines a polygonal decomposition of $M$ with the properties you desire. For details see Whitney's gem "*Geometric Integration Theory*", Chap. IV, Sec. B.


