Manifolds

  OVERVIEW: Manifolds constitute an important class of objects to study in many branches of mathematics. We also discuss, very briefly, the notion of a smooth manifold.

Manifolds form a collection of familiar of geometric objects encountered in the study of calculus such as, curves, surfaces, 3-dimensional regions, etc. Roughly speaking, a manifold is a subset which locally looks like Euclidian space, or Euclidian half-space.

Definition 8020

Let $X \subseteq {\bf R^n}$, X is a manifold with boundary, of dimension k, if for every point $x \in X$,there is an open subset, U, of X, 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 ${\bf R}^{k-1}$.   

Remark 8026

    

If X is an empty set then, in a vacuous way, it is an n-dimensional manifold for any n. As peculiar at this might first seem, it is traditional to allow this usage. $\diamondsuit$

Propositions 1.04 and 1.05, below, are important basic results. Complete proofs are beyond the scope of this text. We sketch, the proofs using Proposition 1.03.

Proposition 8032

   

(Invariance of domain) ${ \bf R^n}$ and ${\bf R^m}$ are not homeomorphic if $n \neq m$.   

Proof: Assume€$n \neq m$, and suppose there is a homeomorphism $h:{ \bf R^n} \to { \bf R^m}$. Let $\vec{x_0} = h(\vec{0})$, and let $X= { \bf R^n} - \{\vec{0}\}$ and $Y= { \bf R^m} - \{\vec{x_0}\}$. Then $h\vert _X :X \to Y$ is a homeomorphism, and thus X and Y have the same homotopy type.

But X has the homotopy type of S^n-1 as one can see using the deformation retraction given by:

$$H_t(\vec{x} ) = (1-t) \vec{x} + t \frac{\vec{x} }{\vert\vec{x}\vert}, \mbox{ where } 0 \leq t \leq 1.$$

Similarly, Y has the homotopy type of S^m-1. This is not possible since $H_{n-1} (S^{n-1}) \neq 0$ and H_n-1 (S^m-1) = 0. width4pt height6pt depth1.5pt

We can use use Proposition to show that a 1-dimensional manifold cannot be homeomorphic to a two-dimensional manifold,. In general, Proposition 1.04, the dimension of a manifold is a topological invariant:

Proposition 8059

   

If $X \subseteq { \bf R^k}$ is a n-dimensional manifold and $Y \subseteq { \bf R^{k^\prime}}$ is an m-dimensional manifold, if X and Y are not empty and $n \neq m$ then X and Y are not homeomorphic.  

Proposition 8064

Conditions a) and b) in the above definition are mutually exclusive.  

We can use the above result to make the following two definitions.

Definition 8074

If $X \subseteq { \bf R^k}$ is a manifold with boundary, then the interior of X, Int(X), is defined by $Int(X) = \{x \in X\colon$ there is an open subset, U, of X, with U homeomorphic to ${\bf R^k}$ }.    

Definition 8086

If $X \subseteq { \bf R^k}$ is a manifold with boundary, then the boundary of X , $\partial X$ , is defined by $\partial X = \{x \in X\colon$ there is an open subset , U, of X, with U homeomorphic to R^k₊ and x corresponds to a point of ${\bf R}^{k-1}\}.$   

Remark 8093

        

We warn the reader not to confuse boundary of a set, Definition 0.34, and boundary of a manifold, Definition 1.07. Similarly, we have interior of a set, Definition 0.31, and interior of a manifold, Definition 1.06. We use a capital I in ``Int'' when referring to interior of a manifold and the lower case i in int for interior of a subset.

This is a potential source of confusion, however these are all standard definitions. Some topologists use the term frontier of a subset to denote the boundary of the subset.   $\diamondsuit$

Example 8117

           

Consider the standard unit ball in the plane : $D^2 = \{(x,y) \in { \bf R^2} \mid x^2 + y^2 \leq 1 \}$. The boundary of the set D² and the boundary of the manifold are the same, namely the unit circle in the plane. Also the interior of the set D² and the interior of the manifold D² are the same, namely $\{(x,y) \in { \bf R^2} \colon x^2 + y^2 < 1 \}$. So, $Bd(D) = \partial D$ and int(D) = Int(D) This example, explains why mathematicians would use the same word for two distinct concepts.

On the other hand, there are times when the two uses of boundary and interior do not agree. The standard inclusion of ${ \bf R^3}$ into ${ \bf R^3}$ sends D² to $D = \{(x,y,z) \in { \bf R^3} \colon x^2 + y^2 \leq 1 \mbox{ and } z=0 \}$. The boundary of the manifold D is the circle $\{(x,y,z) \in { \bf R^3} \mid x^2 + y^2 = 1 \mbox{ and } z=0 \}$ and the interior of the manifold D is $\{(x,y,z) \in { \bf R^3} \mid x^2 + y^2 < 1 \mbox{ and } z=0\}$.The interior of the set D is the empty set and the boundary of the set D is all of D, so in this case $Bd(D) \neq \partial D$ and $int(D) \neq Int(D)$. $\diamondsuit$

 

Example 8120

   

Any interval in ${ \bf R^1}$ (open, closed, half-open) is a 1-dimensional manifold.  

Example 8130

   

For each k, S^k is a k-dimensional manifold. Also ${\bf R^k}$ is a k-dimensional manifold, as well as any open subset of ${\bf R^k}$ All these examples have empty boundary. $\diamondsuit$ 

Example 8138

   

For each k, D^k is a k-dimensional manifold and $\partial D^k = S^{k-1}$. $\diamondsuit$ 

Example 8152

     

Use polar coordinates, and let $A = \{(\rho$, $\theta ) \in {\bf R^2} \mid 1 \leq \rho \leq 2 \}$. This 2-dimensional manifold is an annulus, see Definition . Here $\partial A$ consists of two disjoint circles: $\rho =1$, and $\rho = 2$.

Using spherical coordinates, we could let $M = \{ (\rho, \theta, \phi \} \in { \bf R^3} \mid 1 \leq \rho \leq 2$. This is a 3-dimensional manifold and $\partial M$ is the disjoint union of two concentric spheres: $\rho =1$, and $\rho = 2$. $\diamondsuit$ 

Using Proposition 1.05 one can show:

Proposition 8160

     

If X is an n-dimensional manifold, then int(X) is an n-dimensional manifold, and $\partial X$ is an (n-1)-dimensional manifold (possibly empty). width4pt height6pt depth1.5pt  

Consider that $\partial D^2$ is a circle, and the circle has empty boundary. Similarly, $\partial D^3$ is a sphere, and the boundary of this sphere is the empty set. In general we have the following.

Proposition 8166

   

If X is a manifold, then $\partial (\partial X) = \emptyset$. width4pt height6pt depth1.5pt  

Proposition 8170

   

If $X \subseteq { \bf R^k}$ is an n-manifold, it is locally connected.  

Remark 8174

    

Proposition 1.16 implies that the topologist's $\sin(1/x)$ curve is not a manifold.) $\diamondsuit$

Here is a basic question about manifolds:

Question 8178

a given natural number, n, how many n-dimensional manifolds are there? By this we mean--how many homeomorphism classes of such manifolds are there? Are there uncountably many homeomorphism classes of manifolds ?

 

Remark 8184

        

One-dimensional manifolds are familiar subsets.

An example of a compact, disconnected one dimensional manifold is a union of two disjoint circles. By considering unions of more and more circles, one sees that there are infinitely many one-dimensional manifolds, no two of which are homeomorphic.

An example of a connected one-dimensional manifold which is not compact is ${ \bf R^1}$. An example of a compact, connected one-dimensional manifold with non-empty boundary is I. $\diamondsuit$

Proposition 8189

       

If X is a connected, compact, one-dimensional manifold with empty boundary, then X is homeomorphic to a circle.  

One of the goals in topology is to capture the nature of certain subsets terms of topological properties.

Proposition 1.20 is a statement of this sort. But we cannot extend this result to the two dimensional sphere easily since there are compact, connected, two-dimensional manifolds which are not homeomorphic to S². One example of this is the torus, see Definition 1.22.

 

Figure 1-1: The standard torus in ${ \bf R^3}$, see Definition 1.22 Also shown are a meridian and longitude circles, see Example 1.24. The meridian ``goes through the hole'' and the longitude ``goes around the hole''.  

 

Figure 1-2: The inverse image, via P of a point on the torus, see Example 1.23.  

Example 8235

     

The torus is an important 2-dimensional manifold, see Figure 1-1. We describe it as a subset of ${ \bf R^3}$.Informally, we an speak of the torus as the ``surface of a donut'', D<sub>(a,b)</sub>, where the ``thickness of the donut" is 2b and ``size of the hole'' is 2(a - b). Of course we need b<a. We may describe a torus as a surface obtained ``by spinning a circle about about the z-axis''. Let C be the circle in the x-y-plane with center at the origin and radius a. The circle C will correspond to the ``core circle or the donut''. Using cylindrical coordinates, we describe C as all points $(r, \theta, z)$ with r=a, z=0.

For any angle, $\alpha$ consider the page $H^2_\alpha$ (recall Definition 0.13). In cylindrical coordinates, let $m(\alpha)$ be the circle in $H^2_\alpha$ with center $(a,\alpha, 0)$ and radius b, and let $D(\alpha)$ be the closed disk in $H^2_\alpha$ with the same center and radius. Let $T_{(a,b)} = \bigcup_{0 \leq \alpha < 2 \pi} m(\alpha)$, and $D_{(a,b)} = \bigcup_{0 \leq \alpha < 2 \pi} D(\alpha)$.

You may think of the letter ``D'' of D<sub>(a,b)</sub>, here as referring to ``donut''; the circles $m(\alpha)$, will later be referred to as ``meridians''.

The torus is the image of a continuous map that will play an important role in later Chapters. Let a and b be numbers with 0 < a < b . Define a function, $P \colon {\bf R^2} \rightarrow {\bf R^3}$ , (using Cartesian coordinates, this time) by

$$P(x,y) = \Big( (b + a \cos( x)) \cos( y) , (b + a \cos( x)) \sin( y), a \sin( x) \Big).$$

The map, P, is an immersion; the image of this map is T_(a,b).

No particular choice of a and b is mathematically special, but for the sake of being specific we chose to focus on T_(1,2). $\diamondsuit$

 

Definition 8246

The standard torus T², in ${ \bf R^3}$ is defined to be T_(1,2). The standard covering map $P \colon {\bf R^2} \rightarrow T^2$ is the map described above with a = 2 and b = 1.

The set D_(2,1) is called the standard solid torus in ${ \bf R^3}$ . The circle C is called the core of the standard solid torus.          

Later, we will examine the torus as a subset of ${ \bf R^4}$, Definition .

 

Figure 1-3: The line l_(2,3,0) see Example 1.24.  

 

Figure 1-4: The circle L_(2,3,0) see Example 1.24.  

 

Figure 1-5: The line l_(3,5,0) see Example 1.24.  

 

Figure 1-6: The circle L_(3,5,0) see Example 1.24.  

 

Figure 1-7: A line with irrational slope, see Example 1.24.  

Example 8271

us examine, in detail, the map P.

Certainly P is not a one-to-one function.

In fact if p and q are integers then $P(p (2 \pi),q (2 \pi)) = (3, 0, 0)$. More generally, for any real numbers x and y and integers p and q, we have $P(x +p (2 \pi), y + q (2 \pi)) = P(x, y)$. Thus every point of T_(a,b) is the image of infinitely many points, see Figure 1-2. Note however that P is one-to-one on the subset $[0,2 \pi) \times [0,2 \pi)$. $\diamondsuit$

 

Example 8295

   

We can use the map P to describe interesting subsets of the torus. Suppose that l_(A,B,C) is the line in the plane with equation A x + B y = C (with A and B not both zero). Then we can define L_(A,B,C) = P(l_(A,B,C)). The sets L_(1,0,0) and L_(0,1,0) are circles on the torus. Generally, any of the circles L_(1,0,C) are called a longitude of the torus, a circle L_(0,1,C) is call a meridian of the torus, see Figure 1-1  

In our figures, we can visually verify the type of curve using a grid for the plane. When we transfer this the image of such a grid to the torus, using P we get a curvilinear grid. For example the line and grid shown in Figure 1-3 transfers to the circle and circular grid as shown in the bottom figure of Figure 1-4. A similar example is shown in the pair Figure 1-5 and Figure 1-4

For any two rational numbers, r and q, L_(r,q,0) is homeomorphic to a circle. Such a set can be described as ``the image, under P, of a line in the plane through the origin with rational slope''. In Figure 1-4 we see the curve l_(3,-2,0). Here, of course L_(3,-2,0) is a line with slope 3/2 thus traverses 3 squares in the y-direction for every two squares in the x-direction , see Figure 1-3. On the torus we can visually verify the type of curve by noting that it traverses the curvilinear grid three units in a meridian direction for every two in the longitude direction, Figure 1-4. Similarly one has the curve l_(5,-3,0), see Figures 1-5 and 1-6. $\diamondsuit$

 

 

Figure 1-8: The standard Möbius band in ${ \bf R^3}$, see Example 1.25.  

 

Figure 1-9: Above is the standard Möbius band M, below is mirror image, $M^\prime$, see Example 1.25.  

 

Figure 1-10: Two other embeddings of the Möbius band into ${ \bf R^3}$. These are mirror images of each other, but are not equivalent subsets to those shown in Figure 1-9, see Example 1.25.  

Example 8351

   

We describe an interesting subset of ${ \bf R^3}$, the Möbius band. A rough description of this is: a band with a half twist. The Möbius band in ${ \bf R^3}$ will be described as a union of line segments whose endpoints lie on a circle.

Consider the circle L(2,-1,0), in ${ \bf R^3}$. Clearly each page, $H_\alpha$, of ${ \bf R^3}$ intersects L(2,-1,0) in exactly two points. For any $0 \leq \alpha \leq 2 \pi$, let $I_\alpha$ be the line segment in $H_\alpha$ with those two points as endpoints. (We note that $I_0 = I_{2 \pi}$.). Define: $M = \bigcup\limits_{0 \leq \alpha \leq 2 \pi} I_\alpha$, see Figure 1-8.

We will call M the standard Möbius band in ${ \bf R^3}$. Generally, any subset homeomorphic to the standard Möbius band will be called a Möbius band. We will call M the standard Möbius band in ${ \bf R^3}$. It is clear from the Figure 1-8 that M is a 2-dimensional manifold and $\partial M = L(2,-1,0)$.

We could similarly consider the circle L(2,1,0). This also meets each $H_\alpha$ in two points. For any $0 \leq \alpha \leq 2 \pi$, let $J_\alpha$ be the line segment in $H_\alpha$ between these points. We let $M^\prime$ be the union of these line segments: $M^\prime = \bigcup\limits_{0 \leq \alpha \leq 2 \pi} J_\alpha$, see Figure 1-9.

Clearly, M and $M^\prime$ are homeomorphic, in fact, by a homeomorphism that takes $I_\alpha$ to $J_\alpha$. We can describe $M^\prime$ also as a band with a half twist, but this half twist is in the opposite direction as if seen in a mirror. Consider, $m \colon { \bf R^3} \to { \bf R^3}$ defined by m(x,y,z) = (x, y -z). We can see m is a homeomorphism, in fact, a rigid motion, of ${ \bf R^3}$ which takes M to $M^\prime$. Sometimes m is described as a mirror reflection in the x-y-pane. We call $M^\prime$ the mirror image of M.

Of course one can get other Möbius bands with other curves. In Figure 1-10 we show Möbius bands constructed using the circles L(3,-1,0) and L(3,1,0). The fact that the Möbius bands of Figure 1-9 and Figure 1-10 look so different is a reflection of the fact (which we do not prove) that these are not equivalent subsets of ${ \bf R^3}$.

$\diamondsuit$

 

Example 8392

   

One gets interesting subsets of the torus by taking lines with irrational slopes. Consider a line l = m x, with irrational slope; let L = P(l). Claim: that L is dense in T², see Definition 0.06 for definition of dense.

As an example, we focus at one point, Z = (3,0,0) of T², and leave as an exercise proof for an arbitrary point of T². Let U be an open subset of T² containing Z, we want to find a point of L in U.

Using P, We transfer this problem to a problem in the plane. Let $\tilde Z = P^{-1} (Z)$; $\tilde Z = \{(p 2 \pi, q 2 \pi \}_{p,q \in Z}$.In Figure 1-7, we show the line L; $\tilde Z$ corresponds to the grid points. We need to show we can find points of l arbitrarily close to a grid point.

For any integer, N, the line l intersects the vertical line $x = N 2 \pi$ at $(N 2 \pi, m N 2 \pi)$. Since m is irrational, $m N 2 \pi$ is an irrational multiple of $2 \pi$. So we need to show that, given $\epsilon \gt 0$, we can find an N so that, for some Q, the distance from $(N 2 \pi, m N 2 \pi)$ and $(N 2 \pi, Q 2 \pi)$ is less than $\epsilon$.This follows from the proof of Example 0.41. There we showed we can find multiples of angle $m N 2 \pi$ arbitrarily close to a multiple of $2 \pi$.

Not only is L a dense subset of T², it has some other interesting properties. Further analysis shows that L fails to be locally connected at every point! In particular this shows that L is not a manifold, even though it is the image of a manifold and P|_l is an immersion. $\diamondsuit$ 

Example 8422

      We give an alternate description of the torus and solid torus and some related manifolds.

The standard solid torus, D_(2,1), in ${ \bf R^3}$ is a 3-dimensional manifold whose boundary is the standard torus, T² see Definition 1.22.

We can describe the torus using the idea of distance from a point to a set. Let C the circle as in Example 1.21--centered at the origin, in the x-y-coordinate plane, of radius 2. Then T² consists of those points in ${ \bf R^3}$ whose distance to C is 1; D_(2,1) consists of those points whose distance to C is greater than or equal to 1. More formally, consider the map $f \colon { \bf R^3} \to { R^1_+}$ defined by $f(\vec{x}) = \delta( \vec{x},C)$. Then f^-1 (1) = T².

Let $U = f^{-1} \Big((1, \infty)\Big)$, $V= f^{-1} \Big( [0,1) \Big)$. It is not hard to show that both U and V are connected. In fact, they are path connected. It follows then that ${\bf R^3} - T^2$ has two components: V, which is bounded and U, which is unbounded.

Also ${ \bf R^3} - Int(D_{(2,1)}) = U \cup T^2$ is an interesting non-compact 3-dimensional manifold whose boundary is a torus.

Our final example is, W, consists of a ball of radius 4 with the interior of the standard solid torus removed. If we let $W = \overline{N}_4 (\vec{0}, { \bf R^3}) - U$, then W is a compact 3-dimensional manifold whose boundary has two components--one of which is homeomorphic to a sphere and the other to a torus. $\diamondsuit$ 

 

Figure 1-11: On the left a solid 2-holed torus, on the right a solid 3-holed torus, see Example 1.28.  

 

Figure 1-12: A smoother version of a 3-holed torus, see Example 1.28 

Example 8440

  We say that the torus has a hole (thinking of the ``hole in the donut"). Some other examples of manifolds of dimensions 2 and 3 are variations of the torus and solid torus ``with several holes''.

Picture two- and three- holed ``bricks'' as in Figure 1-11. Considered topologically, these are called the solid 2-holed torus and the solid 3-holed torus, respectively. The surfaces of these bricks are called the 2-holed torus and 3-holed torus respectively. A more traditional, smoother version of a 3-holed torus is shown in Figure 1-12

More generally we can have an n-holed torus. In contrast to the torus, it is not easy (nor profitable for us) to describe these examples with equations. $\diamondsuit$       

Question 8444

is a mathematical definition of ``hole" as used above? That is, is there a topological property ``the number of holes of X'' so that the torus has one hole, etc? $\diamondsuit$ 

The answer to the above question is the beginning of the subject of algebraic topology and is beyond the scope of this book.

At this point it is handy to introduce a definition:

Definition 8448

If $M \subseteq { \bf R^n}$ is a manifold which is compact with empty boundary, then we say that the manifold M is closed.

   

Remark 8460

          

Once again we see that words are used for manifolds which do not have the same meaning as for subsets.

If M is a closed manifold and $M \subseteq X \subseteq { \bf R^n}$, then M is also a closed subset of X, since all compact subsets are closed subsets. The closed n-ball Dⁿ, 0 < n, is a compact subset of ${ \bf R^n}$ which is not a closed manifold, since $\partial D^n = S^{n-1}, \neq \emptyset$. $\diamondsuit$

There are other similarities between the notion of boundary of a set and boundary of a manifold. Compare the result below to similar statements about the boundary of a set: Proposition 0.37 c), also consult Remark 0.39 and Figure 0-7.

Proposition 8474

     

Suppose that M is an m-manifold and N is an n-manifold. Then $Int(M \times N) = Int(M) \times Int(N)$.

Let $A = M \times \partial N$ and $B = \partial M \times N$. Then $M \times N$ is an (n+m)-manifold, and $\partial (M \times N)$ is the union of A and B . Furthermore, $A \cap B = \partial M \times \partial N$. width4pt height6pt depth1.5pt  

The proof of the above is not too difficult if either $\partial M = \emptyset$or $\partial N = \emptyset$.

We can obtain bountiful supply of manifolds (generally non-compact) by using the next proposition.

Proposition 8484

    If X is an n-dimensional manifold and C is a closed subset of int(X), then X - C is an n-dimensional manifold. width4pt height6pt depth1.5pt  

Remark 8496

    

In particular, any open subset of ${ \bf R^n}$ is an n-dimensional manifold. Also, if we remove a finite number of points from an n-holed torus, we get a two-dimensional manifold.

Suppose K is a subset of ${ \bf R^3}$ such that K is homeomorphic to S¹. Examples where K is ``knotted" give rise to interesting three-dimensional manifolds by considering ${\bf R^3} - K$. This will be discussed in more detail in Chapter . $\diamondsuit$

Proposition 8501

     

Suppose $M \subseteq { \bf R^n}$ is a connected k-dimensional manifold, then M is path connected.  

Proof: Let $p \in M$. Let U be the set of points, q, of M, such that there is a path, in M, from p to q. Let V be the set of points, q, of M that there is no path in M from p to q. Our goal is to show that $V = \emptyset$.

Certainly, $U \neq \emptyset$ since $p \in U$. We next show that U is an open subset of M. Suppose $x \in U$; then there is a path $\alpha$ from p to x. Since M is a manifold, there is an open subset, W, of M, with $x \in W$ such that W is either homeomorphic to ${\bf R^k}$ or R^k₊; call this homeomorphism h. Let $y \in W$. Since ${\bf R^k}$ and R^k₊ are path connected, we can find a path $\beta$ from h(x) to h(y). Let $\alpha^\prime =h^{-1} \circ \beta$, then $\alpha^\prime$ is a path in W from x to y. Finally $\alpha * \alpha^\prime$ is a path from p to y, where * denotes concatenation of paths, recall Definition 0.40. Thus U is an open subset of M since, for every point of U, there is an open subset, W, of M, which is contained in U.

A variation of the above argument shows that V is open. Suppose $x \in V$; we can find a W, open in M, homeomorphic to ${\bf R^k}$ or R^k₊. Since any point of W can be joined to x be a path in W, it follows that no point of W can be joined by a path in M to p. This means that V is an open subset of M.

We conclude that V must be an empty subset since, otherwise, U and V would form a disconnecting partition of M contradicting the assumption that M is connected. width4pt height6pt depth1.5pt

Remark 8580

      

Proposition 1.05 basically says that, topologically, boundary points do not look like interior points. In particular, this means that a manifold with non-empty boundary cannot be homogeneous. However the next Proposition states that a connected manifold with empty boundary is homogeneous.

One needs the connectedness hypothesis if the manifold has components which are not homeomorphic. We will later prove that a sphere and a torus are not homeomorphic. If X is a subset with two components, S and T with S homeomorphic to the sphere and T homeomorphic to a torus, then X cannot be homogeneous since a homeomorphism must take a component to a component. $\diamondsuit$

Proposition 8587

    Suppose $M \subseteq { \bf R^n}$ and M is a connected manifold with $\partial M = \emptyset$, then M is homogeneous.

 

Remark 8637

  The proof of Proposition 1.37 is difficult and depends on basic results too technical to prove here. However we can give a sketch of a proof. Suppose x and y are points of M.

The idea is to find an open subset, U, of M homeomorphic to ${\bf R^k}$ which contains x and y, this is the ``hard part'' of the proof. Suppose we had such a homeomorphism, h. From this point it is not difficult to finish. The idea is to use homogeneity of ${ \bf R^n}$ move x to y inside U, while keeping things fixed outside of U.

Using h we obtain another homeomorphism, $h^\prime \colon U \to { \bf R^k}$ so that $h^\prime (x)$ and $h^\prime (y)$ are contained in the standard open unit ball in ${\bf R^k}$.We know that ${\bf R^k}$ is homogeneous. Moreover, one can find a homeomorphism, $f \colon { \bf R^k} \to { \bf R^k}$ such that $f(h^\prime (x)) = h^\prime (y)$ and $f(\vec{z}) = \vec{z} \mbox{ if } 2 \leq \vert z\vert$.Define $F \colon M \to M$ by

$$F(p) = \left\{ \begin{array} {ll} p & \mbox{ if } p \notin ... ...circ f \circ h^\prime & \mbox{ if } p \in U \end{array}\right.$$

Continuity of F follows from the gluing lemma. F is a homeomorphism of M with F(x) = y. Details of this paragraph is left as an exercise for the reader.

Here is the idea of how to find U. By Proposition 1.35 we know that there is a path, $\alpha$ from x to y, $\alpha \colon I \to M$. With a non-trivial amount of work we can improve this to show that, in fact there is a path from $\alpha$ from x to y such that $\alpha$ is an embedding. We then obtain U as a ``thin'' open subset containing $\alpha$--we chose a small $\epsilon$ and then U is the set of points if M a distance less than $\epsilon$ from a point of $\alpha$, The next step is to show that for a small enough choice of $\epsilon$, this subset is homeomorphic to ${ \bf R^n}$. $\diamondsuit$

 

Figure 1-13: Example of an ambient isotopy of the square, see proof of Proposition 1.40. At the left is shown a regular grid and two points $P = (.2, .6) \mbox{ and } Q = (.7, .9)$. On the left we show image of the grid under an isotopy (as defined in the proof) which takes P to Q. Note, in the left figure,that P is the intersection of the third vertical line (beginning on left) and the 7-th horizontal line (beginning at bottom) and that, in the right figure, Q is the image of the third vertical line and the 7-th horizontal line.  

 

Figure 1-14: Graphs of the functions $f(x) \mbox{ and } g(x)$ for the points of Figure 1-13, see proof of Proposition 1.40. The graph of f passes through the points $(0,0), (0.2, 0.7) \mbox{ and } (1,1)$ while the graph of g passes through the points $(0,0), (0.6, 0.9), \mbox{ and } (1,1)$.  

 

Figure 1-15: Changing from g to the identity map, see proof of Proposition 1.40. We see a sequence of functions that begin with the identity, (whose graph is the diagonal), and, at the end, is g(x), as shown in Figure 1-14.  

 

Figure 1-16: The map $G^\prime$, see see proof of Proposition 1.40. Shown are images of regular grid under $G^\prime$.  

As noted in Remark 1.36, a manifold M with non-empty boundary is not homogeneous, but we can obtain a version of homogeneity by treating $\partial M$ specially.

In Proposition 1.40, we prove a special case of Proposition 1.39, below. We will use the particular special case in a discussion in in Chapter .

Proposition 8685

     

Suppose $M \subseteq { \bf R^n}$ and M is a compact connected manifold with $\partial M \neq \emptyset$, and if p and q are points of Int(M), then there is a homeomorphism, h, of M onto M such that h(p) = q and $h\vert _{\partial M} = Id\vert _{\partial M}$.

 

Proposition 8696

     

Suppose D² and be standard unit ball. If p and q are points of Int(D²), then there is a homeomorphism, h, of D² such that h(p) = q and $h\vert _{\partial D^2} = Id\vert _{\partial D^2}$.

 

Proof: First we investigate a ``lower dimensional'' version of our problem: given two points of Int(D¹) is there a homeomorphism fixed on $\partial D^1$ that takes one point to the other? Since D¹ is a closed interval [-1,1], and $\partial D^1$ is the two point set $\{-1,1\}$, this is not hard to do. We then apply this result to obtain a map of a square to itself, considering one coordinate at a time.

This map is not quite we want. It sends the boundary of the square to itself, but is not the identity map there. The next step is to improve this by changing the map near the boundary of the square. Finally, since the square is homeomorphic to D², we ``transfer'' this to a map to the square to itself to a map of D² to itself.

Now for the details.

Let h be a homeomorphism: $h \colon D^2 \to I \times I$, and let $P = h(p) \mbox{ and } Q = h(q)$. Write coordinates of these points as: $P =(x_P, y_P) \mbox{ and } Q =(x_Q, y_Q)$. Define $f \colon I \to I$ to be the piecewise linear function such that $f(0) = 0, f(x_P) = x_Q, \mbox{ and } f(1) =1$. Also define $g \colon I \to I$ to be the piecewise linear function such that $g(0) = 0, g(y_P) = y_Q, \mbox{ and } g(1) =1$, see Figure 1-14. (Note: f and g are solutions of a corresponding 1-dimensional problem.) Define a map $G \colon I \times I \to I \times I$ by G(x,y) = (f(x), g(y)). Since f and g are homeomorphisms, G is a homeomorphism with, G(P) = Q.

We can check that $G(\partial (I \times I)) = \partial (I \times I)$, but, as can be seen in Figure 1-15, $G\vert _{\partial (I \times I)} \neq Id\vert _{\partial (I \times I)}$.

We find a sub-square, X, of $(I \times I)$ which contains P and Q. Our idea is to use the map we have on this sub-square, and define a different map on $I \times I - X$. That is, we find a homeomorphism $G^\prime \colon I \times I \to I \times I$, such that $G^\prime\vert _X = G\vert _X$, and $G^\prime\vert _{\partial (I \times I)} = Id\vert _{\partial (I \times I)}$. Figure 1-16 gives the idea of how this is to be done.

Further details shed little further light on the matter. But, as a concrete example of how one constructs such a homeomorphism, here are some of those details.

Let $\epsilon = \min (x_P,y_P, x_Q, y_Q, 1-x_P, 1-y_P, 1-x_Q, 1-y_Q)$, and let $X = [\epsilon, 1-\epsilon] \times [\epsilon, 1-\epsilon]$. We have P and Q points of X.

For any $x \in I$, $G\vert \{x\} \times I$ maps the interval $\{x\} \times I$ to itself by moving the point (x,y) to the point (x,g(y)) but on $\{0\} \times I$ we want (0,y) to go to (0,y). In other words, looking at vertical line segments, we want a map ``changes from g to the identity map'' as x gets close to zero. Looking at Figure 1-15 we get an idea of what we want to do. A standard way to get a family of functions that change from the identity to g is to consider (1-s) x + s g(x); If s=0 then this function is x, if s=1 it is g(x). We want to arrange it so that for $s = \epsilon$, the map we get is g(x). For $0 \leq s \leq \epsilon$, let $g_s (x) = \big(1 - \frac{s}{\epsilon} \big) x + \big( \frac{s}{\epsilon}\big) g(x)$. Then g₀ (x) = x and $g_{\epsilon} (x)= g(x)$.

Consider the other three sides of ${\partial (I \times I)}$. Completing the above analysis, we end up with the following definitions:

$$g_s (x) = \left\{ \begin{array} {ll} \big(1 - \frac{s}{\eps... ...(x) & \mbox{ if } 1- \epsilon \leq s \leq 1 \end{array}\right.$$

and

$$f_s (x) = \left\{ \begin{array} {ll} \big(1 - \frac{s}{\eps... ...x) & \mbox{ if } 1- \epsilon \leq s \leq 1 \end{array}\right. .$$

Now define $G^\prime (x,y) = (f_y (x), g_x (y) )$. For any y, f_y is a homeomorphism (it is a monotone increasing function, since it is a linear combination of monotone increasing functions). From this we conclude that $G^\prime$ is 1-to-1 and onto. Since each of the coordinate functions is continuous, it follows that $G^\prime$ is continuous. Since $I \times I$ is compact, we conclude that $G^\prime$ is a homeomorphism. Since we have explicit formulas, we can check that $G^\prime\vert _X = G\vert _X$, and $G^\prime\vert _{\partial (I \times I)} = Id\vert _{\partial (I \times I)}$.

As a final step we transfer this function to a homeomorphism of D² by conjugation by h. Let $H = h^{-1} \circ G^\prime \circ h$, then H is a homeomorphism of D², fixed on $\partial D^2$ such that H(p) = q.

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Historically, the interest in the topic of manifold, arose from analysis. There focus is on subsets of ${ \bf R^n}$ for which we can apply the basic concepts of calculus such as: tangent lines to curves, and tangent planes for surfaces. A differentiable function, is continuous, but there are continuous functions such as f(x) = |x| which are continuous but not differentiable.

A smooth function is a differentiable function, in fact generally in topology one assumes that the functions involved are (infinitely) differentiable. For most purposes, the hypothesis that functions have continuous second derivatives is all that is required. So assuming higher derivatives is more convenience than necessity. Most definitions of derivative require an open domain so the definition below is in two parts, one for open subsets, a second for arbitrary subsets.

Definition 8795

  Suppose U is an open subset of ${ \bf R^n}$ and $f \colon U \to { \bf R^m}$ is a function written as

$$(x_1, \ldots, x_n) = \big( f_1(x_1, \ldots, x_n), \ldots, f_m(x_1, \ldots, x_n) \big) .$$

We say f is smooth if each of the functions, f_i, have derivatives of all orders.

If $X \subseteq {\bf R^n}$ and $f \colon X \to { \bf R^m}$, we say f is smooth if there is an open subset U of ${ \bf R^n}$, and a smooth map $F \colon U \to { \bf R^m}$, such that F|_X = f and F is smooth.  

Definition 8804

Suppose $X \subseteq {\bf R^n}$, $Y \subseteq { \bf R^m}$ and $f \colon X \to Y$ is a homeomorphism. If f is also smooth then we say F is a diffeomorphism. If a diffeomorphism exists, then we say that X and Y are diffeomorphic.      

Definition 8820

Let $X \subseteq {\bf R^n}$, X is a smooth manifold with boundary, of dimension k, if for every point $x \in X$,there is an open subset, U, of X such that either

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

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

   

Example 8835

${ \bf R^n}$ is and Rⁿ₊ are smooth manifolds, for any n. Similarly any open subset of ${ \bf R^n}$ or Rⁿ₊ is a smooth manifold.

To show that Sⁿ is a smooth manifold is not hard. We can use stereographic projection, see Proposition 0.20, to get a smooth map from $S^n -\{P\}$ where $P = (0, 0, \ldots , 0, -1)$ is the ``south pole'',. Then we use a similar map for $S^n -\{Q\}$ where $Q = (0, 0, \ldots , 0, 1)$ is the ``north pole'', . $\diamondsuit$

 

The following idea is is a familiar one from calculus. A regular value is a value which is not the image of any critical point.

Definition 8844

  Suppose $f \colon { \bf R^n} \to { \bf R^1}$ is a smooth function is $y \in { \bf R^1}$. We say that y is a regular value of f if for all points, $x \in f^{-1} (y)$, some partial derivative, $\frac{\partial f}{\partial x_j} (x) \neq 0$. 

With the next Proposition, we can find many smooth manifolds.

Proposition 8852

Suppose $f \colon { \bf R^n} \to { \bf R^1}$ is a smooth function is $y \in { \bf R^1}$ and y is a regular value for f. Then f^-1 (y) is a smooth (n-1)-dimensional manifold with empty boundary.  

Remark 8857

      

Manifolds obtained by the Proposition 1.46, are familiar constructions. For maps of ${ \bf R^2}$, these are the level curves; for ${ \bf R^3}$ these are level surfaces. $\diamondsuit$

A smooth manifolds is sometimes called a differentiable manifold  . The study of smooth manifolds is called differential topology.    

To go much further into this topic would entail a whole book (such as [GP,HR2]), but at least we can state one basic problem.

Question 8861

X and Y are homeomorphic smooth manifolds are they diffeomorphic?  

Remark 8872

 The answer to this question is ``no''! This is a long and interesting story, we mention only a few of the highlights. Basically, in low dimensions, there is not much problem. But there are smooth 7-dimensional manifolds that are homeomorphic to S⁷ but not diffeomorphic to S⁷. As a matter of fact, there are exactly 27 different (mutually non-diffeomorphic) such examples.

As a rule, in topology, a situation is simpler in the compact case compared to the non-compact case. The case Question of 1.48 for, ${ \bf R^n}$, is a notable exception to this rule: if X is homeomorphic to ${ \bf R^n}$, then it is diffeomorphic to ${ \bf R^n}$, except for ${ \bf R^4}$, in which case there are infinitely many non-diffeomorphic examples. $\diamondsuit$

PROBLEMS FOR CHAPTER

Problem 8875

that an n-dimensional manifold has dimension n, according to Definition .  

Problem 8880

that if X is a two-dimensional manifold and if Y is a one dimensional manifold that X and Y are not homeomorphic.  

Problem 8885

that D² is a 2-dimensional manifold and $\partial D^2 = S^{1}$. ( Hint: for points of S¹ you might consider the complex exponential map. )  

Problem 8888

Proposition 1.32 in the case that if either $\partial M = \emptyset$or $\partial N = \emptyset$. 

Problem 8889

Proposition 1.16.  

Problem 8894

M be a k-dimensional manifold. Show that each component of M is a k-dimensional manifold.

Problem 8902

$M \subseteq { \bf R^n}$ and M is a k-dimensional manifold. Show that $int(M) \subseteq Int(M)$. (Note: be careful to distinguish int and Int.) Also show that $\partial M \subseteq Bd(M)$.

Problem 8908

that if M is a connected k-manifold with 1 <k and $x \in M$ then $M - \{x\}$ is a connected set.

Problem 8909

that the standard torus does not have the fixed point property.

Problem 8913

the statement of Remark 1.24 that for any two rational numbers, r and q that L_(r,q,0) is a circle.  

Problem 8915

the proof in Exercise 1.26 for an arbitrary point of T².  

Problem 8919

that T² - L and L fail to be locally connected at every point, where L is as in Example 1.26.  

Problem 8922

that the subsets U and V of Example 1.27are path connected.  

Problem 8926

that for the subsets defined in Example 1.27, that $V \cup T^2 = D_{(2,1)}$ and also that V = int(D_(2,1)) = Int(D_(2,1) and $Cl(V,{ \bf R^3})= D_{(2,1)}$.

Problem 8927

the details of the claims of the middle paragraph of Remark 1.38.  

Problem 8930

formulas for the piecewise functions f and g as described in the proof of Proposition 1.40.  

Problem 8933

, directly, that for any n, Sⁿ is homogeneous.

Problem 8937

, directly, that the torus , T² is homogeneous. (Hint: use the map $P \colon { \bf R^2} \to T^2$ and the homogeneity of ${ \bf R^2}$).

Problem 8939

that the circle, S² is a smooth manifold, see Example 1.44.