I am currently reading some notes where an operator, originally defined on functions over Euclidean Space is now transferred to the setting of a compact, smooth Riemannian manifold.

There is a statement in the notes which says the following:

"Let $M$ be a smooth compact Riemannian manifold without boundary. Cover $M$ by a finite number of coordinate charts $U_i$ with diffeomorphisms $h_i : O_i \to U_i$ where the $O_i$ are open subsets of $\mathbb{R}^m$ with compact closure. We assume the coordinate charts $U_i$ are chosen so that the union $U_i \cup U_j$ is also contained in a larger coordinate chart for all $(i,j)$."

This last statement confuses me a little, because I cannot see this as being trivial. How can I justify this assumption ?

That is, given I have a smooth atlas $\{ U_i \}$ for $M$ (which we can take to consist of finitely many charts since $M$ is compact), how do I adjust this chart so that the last sentence above holds ?

Many thanks for your help!

  • $\begingroup$ Are you sure you are citing this correctly? In this generality (for each pair $(i,j)$) I doubt that this can be achieved. Is there something missing? E.g. for each pair $(i,j)$ such that $U_i\cap U_j$ is not empty? $\endgroup$
    – user20266
    Apr 25, 2012 at 12:47
  • $\begingroup$ @Thomas thanks for your comment ! I have copied the whole definition now into the post above in order to reduce the possibility that I have missed something. $\endgroup$
    – harlekin
    Apr 25, 2012 at 12:55
  • $\begingroup$ This is interesting. You would need to heavily restrict. For example, you could not take the two charts on $S^n$ given by stereographic projection. $\endgroup$
    – Neal
    Apr 25, 2012 at 13:59
  • 1
    $\begingroup$ I'd start checking what happens with the cover and what is really needed. $\endgroup$
    – user20266
    Apr 25, 2012 at 14:38
  • $\begingroup$ Hm ... ok it sounds like I have to search for a result that says something like "for each compact manifold there exists such a cover", instead of a statement of the form "any given cover can be adjusted so that the statement holds." $\endgroup$
    – harlekin
    Apr 25, 2012 at 15:04

1 Answer 1


I'm interpreting the statement as follows:

There is a finite collection of distinguished coordinate charts $U_i$ covering $M$ where each $U_i\cup U_j$ is contained in another (perhaps nondistinguished) coordinate chart $W_{ij}$.

I am not necessarily claiming that $W_{ij}\cup W_{kl}$ is contained in a chart $Z_{ijkl}$ and each $Z_{ijkl}\cup Z_{mnop}$ is contained in a chart ...., just that the union of the $U$s are contained in charts.

Here's a proof that such things always exist, though it may use tools you don't have access to.

Equip $M$ with any Riemannian metric $g$. This gives rise to several functions. First, for each point, there is a map called the exponential map, $\operatorname{exp}_p :T_p M\rightarrow M$ with the property that $\operatorname{exp}_p$ is a diffeomorphism onto its image when restricted to a small enough ball around the origin in $T_p M$. It is a fact that when $\operatorname{exp}_p$ is restricted to a ball of radius $r$ in $T_p M$, the image consists of all points in $M$ a distance $r$ away from $p$.

Another important function coming from a choice of metric is the function $\operatorname{inj}:M\rightarrow\mathbb{R}$. It is called the injectivity radius and defined as the largest radius of a ball around the origin in $T_p M$ such that $\operatorname{exp}_p:B(r)\rightarrow M$ is a diffeomorphism onto its image. It is known that $\operatorname{inj}$ is a continuous function and, on a compact manifold, bounded away from $0$.

Let $\rho$ denote the minimum value of $\operatorname{inj}$ on $M$. Then, by definition, at every point $p$, if we restrict $\operatorname{exp}_p$ to the ball of radius $\rho$, it's a diffeomorphism onto its image. In particular, we can use $\operatorname{exp}_p(B(\rho))$ as a chart on $M$.

Let $U_p$ denote $\operatorname{exp}_p(B(\frac{\rho}{2}))$. I claim that the collection of $\{U_p\}$ has the property that the union of any 2 are contained in a coordinate chart and that they cover $M$. Once we establish all this, use compactness of $M$ to extract a finite subcollection of the $U_p$, giving the desired collection of charts.

First, $p\in U_p$ since $\operatorname{exp}_p(0) = p$ always. Thus, these do cover.

Now, why do they have the property that the union is contained in a coordinate chart? We'll break into 2 cases depending on whether or not $U_p\cap U_q$ is empty or not.

If $U_p\cap U_q =\emptyset$, then $U_p\cup U_q$ is a (disconnected) chart. The diffeomorphism between it and an open subset of $\mathbb{R}^n$ is given by $$\begin{cases} \operatorname{exp}_p^{-1}(r) & r\in U_p \\ \operatorname{exp}_q^{-1}(r) + v & r\in U_q\end{cases}, $$ where $v$ is some vector of very large (compared to $\rho$) length. The point of $v$ is to shift the ball $\operatorname{exp}_q^{-1}(U_q)$ in $T_q M\cong \mathbb{R}^n$ far enough away from the origin so that it wont intersect the image of $\operatorname{exp}_p^{-1}(U_p)$ in $T_p M \cong\mathbb{R}^n$

On the other hand, if $U_p\cap U_q \neq \emptyset$, choose $r\in U_p\cap U_q$. I claim that $U_p\cup U_q \subseteq \operatorname{exp}_r(B(\rho))$. To see this, we'll use the triangle inequality. Using $d$ to denote distance, we have for any $s\in U_p$, that $d(s,r)\leq d(s,p) + d(p,r) < \frac{\rho}{2} + \frac{\rho}{2} = \rho$. This same argument works for $U_q$, so we have $U_p\cup U_q\subseteq \operatorname{exp}_r(B(\rho))$.

  • 2
    $\begingroup$ Incidentally, if this is applied to $S^2$, the unit sphere in $\mathbb{R}^3$ with usual metric, then one finds that $\operatorname{exp}_p$ is nothing but stereographic projection from the antipodal point. Hence, $U_p$ is the open hemisphere containing $p$ in the center and it's easy to see these really do work. $\endgroup$ May 16, 2012 at 14:59

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