# Exercises: Chapter 5, Section 2

1. Show that consists of the tangent vectors at of curves in with .

Let be a coordinate system around in ; by replace with a subset, one can assume that is a rectangle centered at . For and , let be the curve . Then ranges through out as and vary.

Conversely, suppose that is a curve in with . Then let be as in condition for the point . We know by the proof of Theorem 5-2, that is a coordinate system about where . Since , it follows that the tangent vector of is in .

2. Suppose is a collection of coordinate systems for such that (1) For each there is which is a coordinate system around ; (2) if , then . Show that there is a unique orientation of such that is orientation-preserving for all .

Define the orientation to be the for every , , and with . In order for this to be well defined, we must show that we get the same orientation if we use use and . But analogous to the author's observation of p. 119, we know that implies that

where . Let be such that . Then we have

i.e. as desired.

Clearly, the definition makes orientation preserving for all , and this is only orientation which could satisfy this condition.

3. If is an -dimensional manifold-with-boundary in , define as the usual orientation of (the orientation so defined is the usual orientation of . If , show that the two definitions of given above agree.

Let and be a coordinate system about with and . Let where , and is perpendicular to . Note that is the usual orientation of , and so, by definition, is the induced orientation on . But then is the unit normal in the second sense.

1. If is a differentiable vector field on , show that there is an open set and a differentiable vector field on with for .

Let be the projection on the first coordinates, where is the dimension of . For every , there is a diffeomorphism satisfying condition . For , define where . Then is a differentiable vector field on which extends the restriction of to .

Let and be a partition of unity subordinate to . For , choose a with non-zero only for elements of . Define

Finally, let . Then is a differentiable extension of to .

2. If is closed, show that we can choose .

In the construction of part (a), one can assume that the are open rectangles with sides at most 1. Let . Since is closed, is compact, and so we can choose a finite subcover of . We can then replace with the union of all these finite subcovers for all . This assures that there are at most finitely many which intersect any given bounded set. But now we see that the resulting is a differentiable extension of to all of . In fact, we have now assured that in a neighborhood of any point, is a sum of finitely many differentiable vector fields .

Note that the condition that was needed as points on the boundary of the set of part (a) could have infinitely many intersecting every open neighborhood of . For example, one might have a vector field defined on by . This is a vector field of outward pointing unit vectors, and clearly it cannot be extended to the point in a differentiable manner.

4. Let be as in Theorem 5-1.
1. If , let be the essentially unique diffeomorphism such that and . Define by . Show that is 1-1 so that the vectors are linearly independent.

The notation will be changed. Let , and be defined, as in the proof of the implicit function theorem, by ; let and . Then and so . Also, . Let be defined by . We have changed the order of the arguments to correct an apparent typographical error in the problem statement.

Now

which is 1-1 because is a diffeomorphism. Since it is 1-1, it maps its domain onto a space of dimension and so the vectors, being a basis, must map to linearly independent vectors.

2. Show that the orientations can be defined consistently, so that is orientable.

ince is a coordinate system about every point of , this follows from Problem 5-10 with .

3. If , show that the components of the outward normal at are some multiple of .

We have and so by considering the components, we get This shows that is perpendicular to as desired.

5. If is an orientable -dimensional manifold, show that there is an open set and a differentiable so that and has rank 1 for .

Choose an orientation for . As the hint says, Problem 5-4 does the problem locally. Further, using Problem 5-13, we can assume locally that the orientation imposed by is the given orientation . By replacing with its square, we can assume that takes on non-negative values. So for each , we have a defined in an open neighborhood of . Let , , and be a partion of unity subordinate to . Each is non-zero only inside some , and we can assume by replacing the with sums of the , that the are distinct for distinct . Let be defined by . Then satisfies the desired conditions.

6. Let be an -dimensional manifold in . Let be the set of end-points of normal vectors (in both directions) of length and suppose is small enough so that is also an -dimensional manifold. Show that is orientable (even if is not). What is if is the M"{o}bius strip?

Let , and be as in Problem 5-4 in a neighborhood of . Let be as in Problem 5-13. Then we have a coordinate systems of the form and of the form . Choose an orientation on each piece so that adding (respectively ) gives the usual orientation on . This is an orientation for .

In the case of the M"{o}bius strip, the is equivalent to a single ring .

7. Let be as in Theorem 5-1. If is differentiable and the maximum (or minimum) of on occurs at , show that there are , such that

The maximum on on is sometimes called the maximum of subject to the constraints . One can attempt to find by solving the system of equations. In particular, if , we must solve equations

in unknowns , which is often very simple if we leave the equation for last. This is Lagrange's method, and the useful but irrelevant is called a Lagrangian multiplier. The following problem gives a nice theoretical use for Lagrangian multipliers.

Let be a coordinate system in a neighborhood of the extremum at . Then and so . Now the image of is just the tangent space , and so the row of is perpendicular to the tangent space . But we also have for all near , and so . In particular, this is true at , and so the rows of are also perpendicular to . But, is of rank and is of dimension , and so the rows of generate the entire subspace of vectors perpendicular to . In particular, is in the subspace generated by the , which is precisely the condition to be proved.

1. Let be self-adjoint with matrix , so that . If , show that . By considering the maximum of on show that there is and with .

One has

Apply Problem 5-16 with , so that the manifold is . In this case, is a Lagrangian multiplier precisely when . Since is compact, takes on a maximum on , and so the maximum has a for which the Lagrangian multiplier equations are true. This shows the result.

2. If , show that and is self-adjoint.

Suppose . Then

and so . This shows that . Since as a map of is self-adjoint and , it is clear that as a map of is also self-adjoint (cf p. 89 for the definition).

3. Show that has a basis of eigenvectors.

Proceed by induction on ; the case has already been shown. Suppose It is true for dimension . Then apply part (a) to find the eigenvector with eigenvalue . Now, is of dimension . So, has a basis of eigenvectors with eigenvalues respectively. All the together is the basis of eigenvectors for .