Manifolds as topological space--the intrisic point of view

 

OVERVIEW: We discuss a definition of manifold from the point of view of general topology, and relate these two viewpoints. We also discuss the notion of smooth manifold.

The point of studying manifolds is that they are a simple, but rich, collection of objects, which relate to important topics in a number of areas of mathematics.

We have looked at a circle as being the standard circle in the plane, the boundary of a square in the plane or even a non self-intersecting closed curve in space. These subsets are all homeomorphic, or putting it in other words they are images of different embeddings

The point of using an abstract viewpoint is focus on properties of the manifold itself, and not properties of the particular embedding.

We might propose the following as the ``obvious generalization'' of an n-dimensional manifold : a topological space $\{ X, \mathcal T \}$ so that for each $x \in X$ there is an open subset homeomorphic to ${ \bf R^n}$ ( or Rⁿ₊ ). But this is not quite what we want, as we see from this next example

Example 8969

   

Let write the x-axis in the plane as $X = \{ (x,y) \in { \bf R^2} \mid x= 0 \}$, let x₀ be the point (0,1) and let $M = \{x_0\} \cup X$. Let $\mathcal S$ denote the open subsets of X derived from the standard topology of ${ \bf R^2}$.

Define a topology $\mathcal T$, for M, as follows. The open subsets of $\mathcal T$, which do not contain x₀, are the open $\mathcal S$. The open subsets of $\mathcal T$, which do contain x₀, consist sets of the form $(S - (0,0)) \cup \{x_0\}$ where $S \in \mathcal S$.

It can be verified that $\{ M, \mathcal T \}$ is a non Hausdorff topological space, and that every point of M is contained in an open subset homeomorphic to ${ \bf R^1}$. (The Hausdorff property fails for the two points (0, 0) and (0,1).) $\diamondsuit$

 

Example 2.01 shows that our proposed definition was not entirely successful. What we really have in mind, is to define manifold, abstractly, as a topological space $\{ M, \mathcal T \}$, so that any such M will have an embedding into ${ \bf R^n}$ for some n. Example 2.01 cannot be embedding in any metric space, let alone ${\bf R^k}$.

There are some other problems as well.

Example 8989

$\mathcal T$ be the standard topology for ${ \bf R^1}$, let $\mathcal D$ be the discrete topology for ${ \bf R^1}$, and let X be the topological space $\{ { \bf R^1} , \mathcal T \} \times \{ { \bf R^1} , \mathcal D \}$. One can, roughly, describe X as: uncountably many disjoint copies of ${ \bf R^1}$. Clearly, X is a 1-dimensional manifold according to our proposed definition. Now the problem is that X is not second countable, but any subset of ${ \bf R^n}$ must be second countable. $\diamondsuit$

 

These are the only problems. Leading us to the following definition:

Definition 9004

  A topological k-manifold is a topological space, $\{ M, \mathcal T \}$, which is Hausdorff, second countable and such that for each $x \in M$ there is an open subset, U, in $\mathcal T$such that either

a) U is homeomorphic to ${\bf R^k}$,

b) U is homeomorphic to R^k₊ and x corresponds to a point of ${\bf R}^{k-1}$.

 

The next theorem, not an easy one, shows that this definition satisfies our expectations:

Proposition 9012

Let $\{ M, \mathcal T \}$ be a topological k-manifold. Then there is an embedding of M into ${ \bf R^n}$, for some n.  

Furthermore, in a high enough dimension, all these embeddings are equivalent:

Proposition 9022

Let $\{ M, \mathcal T \}$ be a topological k-manifold. and let $f \colon M \to { \bf R^n}$ and $g \colon M \to { \bf R^m}$, be two embeddings Then f and g are stably equivalent. n.  

If M is a topological k-manifold, it is possible to define a concept of smooth manifold without the assumption that $M \subseteq { \bf R^n}$, for some n. To simplify the discussion, we will assume that $\partial M = \emptyset$; the case of non-empty boundary is not difficult, only verbose.

We will not be working with these definitions in this text. The point is to show how one can go about transferring ideas, which seem to be firmly rooted in ${ \bf R^n}$, to the more abstract world of general topology.

Definition 9044

Suppose M is a topological k-manifold and U and V open subsets of M, as in Definition 2.03 a), and assume that $U \cap V \neq \emptyset$. So we have homeomorphisms $\phi \colon U \to { \bf R^k}$ and $\psi \colon V \to { \bf R^k}$. Such maps will be called local coordinate maps.  .

Let $W = \phi (U \cap V)$. Since $\phi$ is a homeomorphism, W is an open subset of ${\bf R^k}$. Now consider $\psi \circ \phi^{-1} \colon W \to { \bf R^k}$, such a homeomorphism is called a transition function.    

It is easy to see that $\phi$ is an embedding. For a manifold to be smooth, we require that these be smooth embeddings:

Definition 9056

  Suppose $M \subseteq { \bf R^n}$ is a topological k-manifold written as a union of open subsets, $M = \cup_\alpha U_\alpha$, such that each $U_\alpha$ is homeomorphic to ${\bf R^k}$ by coordinate function $f_\alpha \colon U_\alpha \to { \bf R^k}$. If we can find choices for $U_\alpha$ and $f_\alpha$ so that all transition functions are smooth, then we say that M is a smooth manifold.

 

We can similarly transfer the idea of smooth function:

Definition 9070

Suppose M is a smooth k-manifold, and $M^\prime$ a smooth $k^\prime$-manifold, in the sense of Definition ref 2.07. Suppose $f \colon M \to M^\prime$ a continuous function. We say that f is smooth if, for all coordinate functions of M, $\phi: U_\alpha \to { \bf R^k}$ and for all coordinate functions of $M^\prime$,$\phi^\prime: U^\prime_{\alpha^\prime} \to { \bf R^{k^\prime}}$, with $Im(\phi) \cap U^\prime_{\alpha^\prime} = W \neq \emptyset$, the map $\phi^\prime \circ f \circ \phi^{-1}$ is smooth.    

Note in the above definition the domain of definition of $\phi^\prime \circ f \circ \phi^{-1}$ is the set ,Now we seem to have conflicting definitions for smooth manifold, Definitions 1.43 and 2.07.

The following definition, resolves the issue:

Proposition 9079

If $X \subseteq {\bf R^n}$ is a smooth manifold in the sense of Definition 1.43, it is a smooth manifold in the sense of Definition 2.07.

If X is a topological space which is a smooth manifold in the sense of Definition 2.07, there exists a smooth embedding, in the sense of Definition 2.08, of X into some ${ \bf R^n}$ so that its image is a smooth a smooth manifold in the sense of Definition 1.43.