# About “good” covers of manifolds

Let $M$ be compact manifold of dimension $n$. I am wondering if a certain type of covering of M by coordinate charts exists (this is not the usual good covering theorem but seems related to it). I want to have a cover of M by finitely many closure of open coordinate balls $\{\bar{U}_i\}$ s.t

i. they only intersect at the boundary

ii. the intersections consists of union of finite number of embedded submanifolds of lower dimension.

(it would be even nicer if in ii. these submanifolds dont accumulate on each other. this does not seem essential but this requirement would for instance rule out triangulations)

The well known "good covering" theorem seems to be close to this in the sense that if you contract the intersections then the open covers only intersect at the boundaries. You can break each open ball into coordinates balls so that is not a problem but I am still not sure if the intersections of these covers are embedded manifolds.

I also know that if you have a measure $\mu$ on M, you can choose such a covering by positive measure sets that they intersect only at measure zero sets. But this is again not enough as measure zero set could be anything.

For instance you can do this definitely in $S^n$ in $T^1,T^2,T^3$ and probably in $T^n$ although I havent thought about it.

Any reference answering this question is also very welcome.

• Sorry I now see that I did not state it precisely/correctly. Lets say that your open sets are $U_i$ such that $\bar{U_i}$ (closure) forms a cover. I want to know if it is possible to have $M - \cup_{i}U_i$ to be union of finite, disjoint, embedded submanifolds (of lesser dimension), which also do not accumulate on each other. This last condition could be repharesed as each submanifolds can be seperated from others by a open nbd. In the case of triangulations this is not the case while in the case of for instance sphere example this is the case. – Sina Apr 4 '14 at 22:32
• @Sina: In this example you are not using coordinate balls but annuli. If instead of balls you are allowed to used open subsets on $R^n$ then every compact smooth manifold can be divided in two handlebodies which meet along a codimension 1 submanifold. – Moishe Kohan Apr 5 '14 at 16:36