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
.