Munkres Analysis on Manifolds, Study Guides, Projects, Research of Mathematics

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Solution to selected problems of Munkres

Analysis on Manifolds Book

Herman Jaramillo

May 10, 2016

Chapter 4: Change of Variables

Section 16: Partitions of Unity

Problem 1.

Prove that the function f of Lemma 16.1 is of class C

∞ as follows: Given

any integer n ≥ 0, define fn : R → R by the equation

fn(x) =

(e

− 1 /x /x

n ) for x > 0 ,

0 for x ≤ 0.

(a) Show that fn is continuous at 0. [Hint: Show that a < e

a for all a. Then

set a = t/ 2 n to conclude that

t

n

e

t

(2n)

n

e

t/ 2

Set t = 1/x and let x approach 0 through positive values.]

sln. Let us first show that a < e

a

. The function f (a) = e

a − a, has

derivative f

′ (a) = e

a − 1. That is, f

′ (a) > 0 , ∀a > 0. That is the func-

tion is monotonically increasing in the positive a axis and monotonically

decreasing in the negative a axis. That is,

a < e

a ∀a > 0

If a = 0, from 0 < 1, and if a < 0, from a < 0 < e

a , ∀a < 0, we see that

in general a < e

a

. In particular, if a = t/ 2 n we get

t

2 n

< e

t/ 2 n

(d) Show that fn is of class C

∞

. The recursion formula 4.0.2 shows that

the derivative of fn can be written in terms of functions fn+1 and fn+2,

weighted with coefficients 1 and −n respectively. To the derivative exists

and its continuous for all n (an induction argument).

Problem 2.

Show that the functions defined in Example 1 form a partition of unity on

R. [Hint: Let

fm(x) = f (x − mπ), for all integers m. (4.0.3)

Show that

f 2 m(x) = (1 + cos x)/ 2. (4.0.4)

Then find

f 2 m+1(x)].

sln. I believe the hint make the problem more complicated. Let us perform

a direct evaluation of

φi(x).

Let us consider the case of x ∈ [−π, π] where f has a support. here are

two cases

• Even indices:

φ 2 m(x) = f (x + mπ), m ≥ 1.

If −π ≤ x ≤ π and −π ≤ x + mπ ≤ π then

Here the only possible value is m = 1, since a higher value will shift the

argument from 0 by 2π or greater and the function f would evaluate

to zero there, and it is necessary that x ≤ 0. Hence, for −π ≤ x ≤ 0,

φ 2 m(x) = f (x + π) =

1 + cos(x + π)

1 − cos x

• Odd indices:

φ 2 m+1(x) = f (x − mπ), m ≥ 0.

If −π ≤ x ≤ π and −π ≤ x − mπ ≤ π then we have two possibilities

m = 0, 1. That is,

φ 1 (x) = f (x) = (1 + cos x)/ 2 , −π ≤ x ≤ π

φ 3 (x) = f (x − π) = (1 + cos(x − π))/2 = (1 − cos x)/ 2 , 0 ≤ x ≤ π

so

φ 2 m+1(x) =

(1 + cos x)/ 2 −π ≤ x ≤ 0 ,

1 0 ≤ x ≤ π,

We now add all parts, from equations 4.0.5 and 4.0.

φi(x) =

(φ 2 m(x) + φ 2 m+1(x)) =

1 −π ≤ x ≤ 0

1 0 ≤ x ≤ π

That is

φi(x) = 1.

in the interval [−π, π].

Note that for the negative branches we added (1 − cos x)/2 + (1 +

cos x)/2 = 1, and for the positive branches we added (1 + cos x)/2 +

(1 − cos x)/2 = 1.

Section 19: Proof of Change of Variables The-

orem

Problem *6.

Let B

n (a) denote the closed ball of radius a in R

n , centered at 0,

(a) Show that

v(B

n (a)) = λna

n

for some constant λn. Then λn = v(B

n (1)).

The Jacobian of this transformation is given by:

Dβ =

r 0 · · · 0 u 1

0 r 0 · · · 0 u 2

0 0 · · · r un− 1

ru 1 un

ru 2 un

run− 1 un

un

Since un = ±

∑n− 1

i=1 u

2 i , that is,^ un^ is multivalued and we need to

consider two patches. Each patch (due to symmetry) has the same area

and volume. We then only evaluate one patch and duplicate our result.

Let us evaluate det Dβ(u, r). For this we perform Gaussian elimination

to put zeroes in the last raw, except for the last entry of that raw. We

find

det Dβ(u, r) =

un

u 1

det

ru 1 un

u^21

un

0 r 0 · · · 0 u 2

0 0 · · · r un− 1

ru 1 un

ru 2 un

run− 1 un

un

un

u 1

det

ru 1 un

u^21

un

0 r 0 · · · 0 u 2

0 0 · · · r un− 1

ru 2 un

run− 1 un

u^21 +u^2 n un

un

u 1

un

u 2

det

ru 1 un

u^21

un

ru 2 un

u^22 un

0 0 · · · r un− 1

run− 1 un

u^21 +u^22 +u^2 n un

sln. We found the formula

v(B

n (a)) = 2

B +n(a)

det Dβ(u, r)

a

0

r

n− 1 dr

B

n− 1

• (1)

∑n− 1

i=1 u

2 i

du 1 · · · dun− 1.

2 a

n

n

∫

B

n− 1

• (1)

∑n− 1

i=1 u

2 i

du 1 · · · dun− 1.

Let us test this formula with a few simple cases

• If n = 1 (in the one–dimensional space, then

v(B

0 (a)) =

2 a

= 2a,

• If n = 2 (in the 2D space)

v(B

1 (a)) =

2 a

2

− 1

1 − u

2 1

du 1 = πa

2 .

• If n = 3 (in the 3D space)

v(B

2 (a)) =

2 a

3

− 1

1 − u

2 1 −^ u

2 2

du 1 du 2

2 a

3

− 1

du 1

− 1

du 2

b

2 − u

2 2

where b

2 = 1 − u

2 1.^ The substitution^ u^2 =^ b^ sin^ θ,^ du^2 =^ b^ cos^ θ,

where θ ∈ [−π/ 2 , π/2] provides

v(B

2 (a)) =

2 a

3

− 1

du 1

∫ (^) π/ 2

−π/ 2

b cos θdθ

b cos θ

2 a

3

− 1

du 1 π

4 πa

3

which corresponds to the three dimensional volume of a sphere.

In the last derivation we assumed that the variables u 1 and u 3 can be

integrated iteratively. That is they are independent variables along or-

thogonal directions. This is not always true and we were lucky in ob-

taining the right answer with a weak assumption. In general we could

have trouble.

In general, let us assume that we are considering the n–dimensional

space. We will assume that the variable of integration r is decoupled

from the rest of the variables (and this is right) there is not dependence

between the radius and any of the polar/azimuthal directions. So we can

write

v(B

n (a)) =

2 a

n

n

∫

B

n− 1

• (1)

∑n− 1

i=1 u

2 i

du 1 · · · dun− 1.

To solve this integral we assume that the denominator is bounded away

from zero, so we can apply Fubini’s rule, and then after we apply the

rule we can take the limit as the denominator (un) goes to zero. Recall

that the domain of integration B

n− 1

• (a) is the manifold of^ n^ −^ 1–tuples

(u 1 , · · · , un− 1 ) under the mapping

u

2 1 +^ · · ·^ +^ u

2 n− 1 = 1^ −^ u

2 n,

and since 0 < un ≤ 1 then 1 −

n− 1 i=

u

2 i >^ 0.

Let us take the last coordinate un− 1 and let it be in the interval un− 1 ∈

[−1 + , 1 − ], 0 <  << 1 and write

v(B

n (a)) =

2 a

n

n

∫ 1

− 1

dun− 1

∫

B

n− 1

• (1)

∑n− 1

i=

u

2 i

du 1 · · · dun− 2.

with B

n− 1

• =^ B

n− 2

• ×[−^1 ,^ 1].^ B

n− 2

• is the manifold defined by the (n−2)–

tuples (u 1 , · · · un− 2 ) such that

u

2 1 +^ · · ·^ u

2 n− 2 ≤^1.

Problem *7.

(a) Find the centroid of the upper half–ball

B

n +(a) =^ {x|x^ ∈^ B

n (a) and xn ≥ 0 }

in terms of λn and λn− 1 and a, where λn = v(B

n (1)).

sln. From the definition in exercise 3, we have

ck(B

n +(a)) =^

v(B

n +(a))

Bn +(a)

πk =

I

v(B

n +(a)

with

I =

Bn +(a)

πk.

We know that half the sphere has half the volume (isn’t this obvious?,

since the density is constant :)) So

v(B

n +(a)) =^

a

n λn

, and so ck(B

n +(a)) =^

2 I

anλn

Next we evaluate the integral I. For this evaluation we use the change

of variables defined by the function β in the mapping 4.0.7, for which we

already know (see equation 4.0.8 ) that

V (Dβ) =

r

n− 1

un

We should evaluate the integral

I =

Bn +(a)

πk.

and since πk = xk = ruk, then

I =

B +n(a)

ruk det D(β(u, r))

∫ (^) a

0

r

n dr

Bn +(1)

uk

un

du 1 · · · dun− 1

a

n+

n + 1

Bn +(1)

uk

un

du 1 · · · dun− 1 (4.0.15)

Let us will consider two cases

(a) k < n. From equation 4.0.

I =

a

n+

n

Bn +(1)

du 1 · · · duk− 1 duk+1 · · · dun− 1

∫ (^) d

c

duk

uk

un

where the integration bounds c and d are to be defined. The ques-

tion of Fubini’s rule always should be addressed. I will assume that

the k–integration can be moved to the end as I did here (and I

could, but will not prove it).

From

i

u

2 i = 1,^ i^ = 1,^ · · ·^ , n^ we have

un =

n− 1 ∑

i=

u

2 i

n− 1 ∑

i=1,i 6 =k

u

2 i −^ u

2 k

b^2 − u

2 k

where

b

2 = 1 −

n− 1 ∑

i=1,i 6 =k

u

2 i.

This defines the bounds of integration as |uk| ≤ b. That is c = −b

and d = b. Since, uk is odd (recall uk = xk/a )

I =

a

n+

n + 1

∫ (^) d

c

duk

uk

un

a

n+

n + 1

∫ (^) b

−b

duk

uk √ b^2 − u

2 k

(b) k = n. Initially, from 4.0.11 we have the recursion

λn+1 =

2 π

n + 1

λn− 1. (4.0.17)

sln. We will only compare the non–zero components.

c(B

n +(a))

c(B

n− 2

• (a))^

aλn+

πλn

πλn− 2

aλn− 1

λn+1λn− 2

λnλn− 1

and by using equation 4.0.

c(B

n +(a))

c(B

n− 2

• (a))^

π

(n+1)/ 2

n+ 2

π

(n−2)/ 2

n− 2 2

n 2

π

n/ 2

n− 1 2

π

(n−1)/ 2

(n + 2)/ 2

(n + 3)/ 2

n + 2

n + 3

So the relationship looked for is

c(B

n +(a)) =

n + 2

n + 3

c(B

n− 2

• (a)).