# Exercises: Chapter 5, Section 1

1. If is a -dimensional manifold with boundary, prove that is a -dimensional manifold and is a =dimensional manifold.

The boundary of is the set of points which satisfy condition . Let be as in condition ; then the same works for every point in such that . In particular, each such is in . Further, also is map which shows that condition is satisfied for each such . So is a manifold of dimension , and because those points which don't satisfy must satisfy , it follows that is a manifold of dimension .

2. Find a counter-example to Theorem 5-2 if condition (3) is omitted.

Following the hint, consider defined by

Let , , , . Then condition holds except for part (3) since

1. Let be an open set such that boundary is an -dimensional manifold. Show that is an -dimensional manifold with boundary. (It is well to bear in mind the following example: if , then is a manifold with boundary, but .)

Since is open, each of its points satisfies condition with . Let . Then satisfies with , say with the function . Let be one of the half-planes or Suppose there is a sequence of points of such that the all lie in and converge to . If there is no open neighborhood of such that , then there is a sequence of points of such that the sequence converges to . But then the line segments from to must contain a point o the boundary of , which is absurd since the points of U in the boundary of all map to points with last coordinate 0. It follows that h restricted to an appropriately small open subset of either satisfies condition or condition . This proves the assertion.

2. Prove a similar assertion for an open subset of an n-dimensional manifold.

The generalization to manifolds is proved in the same way, except you need to restrict attention to a coordinate system around . By working in the set of condition , one gets back into the case where one is contained within , and the same argument applies.

3. Prove a partial converse to Theorem 5-1: If is a -dimensional manifold and , then there is an open set containing and a differentiable function such that and has rank when .

Let be as in condition applied to , , and be defined by . Then the function satisfies all the desired conditions.

4. Prove that a -dimensional (vector) subspace of is a -dimensional manifold.

Let be a basis for the subspace, and choose so that all the together form a basis for . Define a map by . One can verify that satisfies the condition .

5. If , the graph of is . Show that the graph of is an -dimensional manifold if and only if is differentiable.

If is differentiable, the map defined by is easily verified to be a coordinate system around all points of the graph of f; so the graph is a manifold of dimension n.

Conversely, suppose is as in condition for some point in the graph. Let be the projection on the last coordinates. Then apply the Implicit function theorem to . The differentiable function obtained from this theorem must be none other than since the graph is the set of points which map to zero by .

6. Let . If is a -dimensional manifold and is obtained by revolving around the axis , show that is a -dimensional manifold. Example: the torus (Figure 5-4).

Consider the case where n = 3. If is defined in some open set by , then is defined by . The Jacobian is Since either or is non-zero, it is easy to see that the Jacobian has the proper rank.

In the case where n > 3, it is not obvious what one means by ``rotate".

1. If is a -dimensional manifold in and , show that has measure 0.

For each , one has condition holding for some function . Let be the domain of one of these functions, where we can choose to be a ball with center at rational coordinates and rational radius. Then is a countable cover of . Now each maps points of in to points with the last coordinates 0. Take a thin plate including the image of ; its inverse image has volume which can be bounded by where is the volume of the plate (by the change of variables formula). By choosing the thickness of the plate sufficiently small, we can guarantee that this value is no more than for the element of the cover. This shows the result.

2. If is a closed -dimensional manifold with boundary in , show that the boundary of is . Give a counter-example if is not closed.

Clearly, every element of is in the boundary of by the condition . If is in the boundary of , then since is closed. So if , it must satisfy condition . But then is in the interior of because the dimension of is n.

The open unit interval in is a counter-example if we do not require to be closed.

3. If is a compact -dimensional manifold with boundary in , show that is Jordan-measurable.

By part (b), the boundary of is . By Problem 5-1, is an -dimensional manifold contained in . By part (a), it follows that is of measure 0. Finally, since is bounded, the definition of Jordan measurable is satisfied.