lOMoARcPSD|17273780
Do Carmo Differential Form and Application Solution
Differential Geometry (中国科学院大学)
Studocu is not sponsored or endorsed by any college or university
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
olutions for Do Carmo Differential
Form and Application
S
C
hapter. 1
1)
ufficiency:
S
if
is alternate, then for all
ecessity:
N
if for
meets, then for arbitrary
which means
:
is alternate.
2)
int tells the solution
H
3)
emark1: if
re
a
R
forms, then the exterior product
1-
a
grees with the
-
form
previously defined by
P
roof:
we define a
form as
-
by definition, for an 1-form
we have
recall that
so for the
form above we have
-
where
or according to whether permutation
is odd or even. Notices that element
is not in , so that matrix
can be used to construct a new matrix
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
then we have
by using algebraic cofactor.
nd is there
A
a
matrix
he answer is yes and we have
? T
then
4)
let denotes a certain permutation indicates by
then
nd so does
,
a
we have
can be rewritten as
5)
it is easy to compute
6)
by definition we have
nd
a
notice that number of
(which is actually
there must be duplicates. then we have
is less than
so
)
so in the permutation
,
7)
exterior product of n copies of
can be denoted as
, a
nd
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
,
whose length is
,
lOMoARcPSD|17273780
for alternativity, we have
then
8)
so
9)
a)
since multilinearity, we have
in which
is the set of all possible permutations of number
nd
, a
where
if the permutation is odd and 2 if even. Due to this we can construct a matrix
in place at the -th row and -th column, then we have
by putting
b)
firstly
then if there exists and
form
-
that meets the properties and is not equal to
,
we have
that means
by rearranging this equation we get
for set
all
is the basis, which means they are linear independent, so the equation holds if and only if
. From the relationship between
and
we have
so
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
that means
.
odge star operation)
10) (H
a)
it is easy to verify the odevity of permutation and compute
.
b)
just compute it
c)
firstly, we know that in
, the absolute value of each term is the same as
(along the basis). Then for
a certain permutation in
denoted as
, the corresponding permutation in
(denoted
as
) is certain.
oreover,
in
M
moving
,
the
permutation
we can deal it just like what author did in proof in Prop.2.b):
so we have
11) (
the divergence)
a)
it is easy to verify that
where
is dual of
. A
nd from the perspective of multivariable function, we have Jacobian
whose trace at is
nd
a
is equal to the norm of projection along
(
which is
),
by omitting we have
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
to
lOMoARcPSD|17273780
b)
for function
,
we have
then we have isomorphism for some vector
where
where
is the canonical isomorphism of
12) (
so we have
denotes the permutation
nd the derivative of
a
.
is
-
form that
the gradient)
a)
from definition we have (in basis
)
so
for canonical isomorphism
we have
b)
for any vector on the "level surface", we have the vector "in" it, denoted as
which can be seen as "sea level", and for
we have
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
where
denoted a vector whose -th component is
a
nd else are
.
c)
not a good proof )
(
that is soluve this optimize problum:
it is natural that the inner product reaches maximum when is a unit vector with the same direction as
that is
,
13) (
the Laplacian)
a)
from definition of gradient and divergence
b)
first
then
c)
we have
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
nd from exercise (11).(b) we have
a
combine these two equations we have
14) (
the rotational)
a)
b)
b1)
there is a canonical isomorphism that:
b2)
15) (
geometrical definition of Hodge star)
a)
firstly we have
because of
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
so the non-zero values only contains items just have no duplicate , which is the same choose all
each
, with the total number of possible items is
a.k.a.
.
where and
or even. and
denotes all possible permutations of
,
or
a
ccording to whether
for
is odd
therefore
b)
(
just follow the hint)
the basis of
can be any n linearly independent (i.e.
form can be rewritten as
nd
,
so
otice that for any
,
. So for the wedge product between
a
.
forms:
)1-
,
,
N
a
,
that
means
nd for any
,
a
nd the basis
e write down again that
W
so it mean
nd we have
, a
let us proof the determinant part. We have
that implies that
then every 1-
.
c)
is decomposable. that means we have:
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
nd the induced vectors
meets
a
lso we have another set of induced vectors
a
it is natural that sets
can be written as
a
nd
a
re both linearly independent, and the subspace generated by them
now what to prove can be written as
that means the above equation is equivalent to
nd vice versa. Here we just prove one part of it, i.e.
a
otice that from above exercises
N
a) a
nd b) we know that
then we have
so
n similar way we can prove that
I
so
d)
we can write
so the
in linear combination of canonical basis
volume of
is
-
lso we have the
a
of
volume of
-
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
otice that
N
so we have
nd for multi-linearity we have
a
so we have
ow we proved the independence.
N
e)
we have the subspace of
defined by
the
orthonormal
denoted as
basis
of
it
then we have the orthogonal complement
,
is
.
And
the vectors in basis of
share the same orientation, which means it is a "positive" basis.
hen from canonical isomorphism we have
, now we can define that
T
for each
,
. F
urther set the volume of
to be
for property i):
it is natural that the subspace of
is
,
so it is perpendicular to
.
for property ii):
due to the volume of orthonormal basis
,
so
for property iii):
for the positive basis
,
we have
f)
he subspace of
plane containing
dimension space
T
A
in
is the
space spanned by
nd . In the same way we know that
with basis
a
, denoted as
. Geometrically, is a
is the norm vector of , which is a 1-
nd the vector product
for the volume of
we have
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
t is not so intuitive. But geometrically,
further written as
is the projection length of
I
where is the angle between
area) of the parallelogram of
a
and
nd
. S
o
on
,
so the above can be
is the 2-dimension volume (or more normally, the
.
o the *vector product *
is an orthogonal vector with the norm equal to the volume of
or we can say, it is the oriented volume of
.
,
S
g)
n fact
I
I
do not really understand the description of
D
oes
mean
a
unit
form?
1-
nyway I will try to prove it under my understanding.
A
irstly let us go on without extra assumptions of
Hodge dual.
,
F
nd the set
a
spans a k-dimension subspace
then we try to do something on its
.
ollow what we did in e), we have
F
which meets the 3 properties above, and we have described its geometrical meaning.
for convenience, we suppose that
isomorphic vector
o
.S
,
is 1, and 0 otherwise.
whose element at position �㗘
is the basic
1
ow we need to prove is
N
i.e.
nd the volume
a
is a
-
order square matrix arranged by
with such elements:
�㗘
whose determinant is exactly
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
1-
form,
a
nd its
lOMoARcPSD|17273780
m not good at multivariable calculus so the following solutions will be post later.
I a
oincaré's Lemma)
16) (P
irstly we write
F
down
that means
nd for
a
we have
reformulate it we get
17)
a)
for linear independency we know
iff
nd we know that
. A
b)
c)
18)
a)
b)
c)
1. I
know it is ugly,but let it be! ↩
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)
lOMoARcPSD|17273780
Downloaded by david segura (davidstivenseguragonzalez@gmail.com)