# Differentiable Manifolds/Maximal atlases, second-countable spaces and partitions of unity

## Maximal atlases[edit | edit source]

**Definition 3.1**:

Let be a -dimensional manifold of class and let be it's atlas. We call the set

the **maximal atlas of **.

**Lemma 3.2**: We have .

**Proof**: This is because if , then by definition of an atlas it is compatible with all the elements of and hence, by definition of , contained in .

**Theorem 3.3**: The maximal atlas really is an atlas; i. e. for every point there exists such that , and every two charts in it are compatible.

**Proof**:

1.

We first show that for every point there exists such that :

From lemma 3.2 we know that the atlas of is contained in .

Let now . Due to the definition of an atlas, we find an such that . Since , we obtain .

2.

We prove that every two charts such that , are compatible.

So let such that be 'arbitrary' (of course we still require ).

If we have , this directly implies compatibility (recall that we defined compatibility so that if for two charts , then the two are by definition automatically compatible).

So in this case, we are finished. Now we shall prove the other case, which namely is .

Due to the definition of compatibility of class , we have to prove that the function

is contained in and

is contained in .

Let . Since is the atlas of , we find a chart such that . Due to the definition of , and are compatible and and are compatible. Hence, the functions

and

are -times differentiable (or, if , continuous), in particular at , respectively. Since was arbitrary, since

and

(which you can show by direct calculation!) and since are bijective, this shows the theorem.

**Theorem 3.4**:

Let be a -dimensional manifold with atlas , and let be it's maximal atlas. There does not exist an atlas such that (this notation shall mean that is contained in , but is 'strictly larger than ' (by this obscure saying we shall mean that there exists at least one element in which is not contained in )).

This is, in fact, the reason why the word *maximal atlas* for does not completely miss the point.

**Proof**: We show that there does not exist an atlas of such that .

Assume by contradiction that there exists such an atlas. Then we find an element . But since is an atlas, is compatible to all other charts for which . This means, due to lemma 3.2, that it is compatible to every . Hence, due to the definition of , . This is a contradiction!

## Second-countable spaces[edit | edit source]

**Definition 3.5**:

Let be a topological space and let be a set of open sets. We call a **basis** of the topology of iff every open set can be written as the union of elements of , i. e.

where .

**Definition 3.6**:

Let be a topological space. We call **second-countable** iff 's topology has a countable basis.

## Locally finite refinements and partitions of unity[edit | edit source]

**Definition 3.7**:

Let be a topological space. An **open cover of ** is a set of open subsets of such that

**Example 3.8**:

The set is an open cover of the real numbers.

**Definition 3.9**:

Let be a manifold of class . We say that ** admits partitions of unity** iff for every open cover there exist functions such that:

- For all , there exists a such that ,
- for all and , AND
- for all , .

**Definition 3.10**:

Let be a topological space and let be an open cover of . A **locally finite refinement of ** is defined to be another open cover of , say , such that:

- for each , there exists a such that , AND
- for each , the set is finite.

We will now prove a few lemmas, which will help us to prove that every manifold whose topology has a countable basis admits partition of unity. Then, we will prove that every manifold whose topology has a countable basis admits partition of unity :-)

**Lemma 3.11**:

Let be a manifold with a countable basis. Then has a countable basis such that for each , is compact.

**Proof**:

Let be a countable basis of . For each , we choose a chart such that . Then we choose . Since in , sets are compact if and only if bounded and closed, is compact. There is a theorem from topology, which states that the image of a compact set under a homeomorphism is again compact. Hence, is a compact subset of .

Further, if is an cover of by open subsets of , then the set is a cover of by open subsets of . Since is compact in , we may pick out of the latter a finite subcover . Then, since

, the set is a finite subcover of . Thus, is also a compact subset of .

As is a homeomorphism, is open in , and from , it follows . Thus, also

since the closure of is, by definition (with the definition of *some* lectures), equal to

Further, another theorem from topology states that closed subsets of compact sets are compact. Hence, is compact.

Since was a basis, each of the can be written as the union of elements of . We choose now our new basis as consisting of the union over of the elements of with smallest index , such that and . Now the closures of the are compact: From follows that , and since , is compact as the closed subset of a compact set.

Since our new basis is a subset of a countable set, it is itself countable (we include finite sets in the category 'countable' here). Thus, we have obtained a countable basis the elements of which have compact closure.

**Lemma 3.12**:

Let be a manifold with a countable base (i. e. a second-countable manifold). Then for every cover of there is a locally finite refinement.

**Proof**:

Let be a cover of . Due to lemma 3.11, we may choose a countable basis of such that each is compact. We now define a sequence of compact sets inductively as follows: We set . Once we defined , we define

, where is smallest such that we have:

This is compact, since a theorem from topology states that the finite union of compact sets is compact. Since, as mentioned before, there is a theorem from topology stating that closed subsets of compact sets are compact, the sets defined by and

for (intuitively the closed annulus) are compact. Further, the sets , defined by , and

for (intuitively the next bigger open annulus) are open, and we have for all :

Now since is covered by , so is each of the sets . Now we compose our locally finite refinement as follows: We include all the sets, which are the intersection of and the (by compactness existing) sets of the finite subcovers of out of . This is a locally finite refinement.

**Lemma 3.13**:

Let be a -dimensional manifold of class with atlas , let , let be open in (with respect to the subspace topology and let and let be such that . If we define

, and

- ,

then we have .

**Proof**:

Let . Then we have for :

This function is times differentiable (or continuous if ) as the composition of times differentiable (or continuous if ) functions.

**Theorem 3.14**:

Let be a manifold of class with a countable basis, i. e. a second-countable manifold. Then admits partitions of unity.

**Proof**:

Let be an open cover of .

We choose for each point an atlas such that . Further, we choose an arbitrary in the open cover such that . By definition of the subspace topology we have that is open in . Therefore, due to lemma 3.13, we may choose such that . Since is continuous, all the are open; this is because they are preimages of the open set . Further, since there is a for every , and always , the form a cover of . Due to lemma 3.12 we may choose a locally finite refinement. This open cover, this set of open sets we shall denote by .

We now define the function

This function is of class as a finite sum (because for each there are only finitely many such that , because was a locally finite subcover) of functions (that finite sums of functions are again follows from theorem 2.22 and induction) and does not vanish anywhere (since for every there is a such that is in it; remember that a finite refinement is an open cover), and therefore follows from theorem 2.26, that all the functions are contained in . It is not difficult to show that these functions are non-negative and that they sum up to at every point. Further due to the construction, each of their supports is contained in one . Thus they form the desired partition of unity.

## Sources[edit | edit source]

- Lang, Serge (2002).
*Introduction to Differentiable Manifolds*. New York: Springer. ISBN 0-387-95477-5.