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 .
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.
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.
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 .
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.
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.
ince is a coordinate system about every point of , this follows from Problem 5-10 with .
We have and so by considering the components, we get This shows that is perpendicular to as desired.
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.
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 .
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.
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.
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).
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 .