Solution to exercises , manifolds

An Introduction to Manifolds, by TU

page 1

Exercise Exercise 1.5

 B (0, 1) the Let  0  0  = (0, 0) be 0) be the origin and  B 1) the open unit disk in  R2 . To find a diffeomorphism  between  B (0, 1) and  R2 , we identify  R2 with the  xy -plane in  R2 and introduce the lower open  hemisphere  S  : x2 + y 2 + (z   1)2 = 1, z <  1

−

in  R3 as an intermediate space. First note that the map f  :  B (0, 1)



f (a, b) = a,b, 1

is a bijection.

  − − 1

→ S  given by 



 (a2 + b2 )

2 (a) The  stereographic The  stereographic projection g : S   from  (0, 0, 1)  is the map that send a point  R  from  S  to the intersection of the line through  (0, 0, 1) and  (a,b,c)   with the  xy (a,b,c) plane. Show that it is given by 

→

 ∈

(a,b,c)

→   (u, v) =



a

1

,

b

− c 1 − c



, c  = 1

−

√ 

1

2

2

− a − b ,

with inverse  (u, v )

 →  √ 

u

1 + u + v 2

2

,

v

√ 1 + u

2

+ v

2

,1

1

− √ 1 + u

2

+ v

2



.

(b) Composing Composing the two maps  f  and  g  gives the map (g  f )(a, b) =  h (a, b) =

◦

 √ 

a

b

√  1 − a − b 1 − a − b 2

2

,

2

2



.

Find a formula for  h 1 (u, v )  and conclude that  h  is a diffeomorphism of the open disk  B (0, 1) with  1)  with  R2 . −

(c) Gener Generalize (b) to

n R .

Solution.

We first wish to observe that



f (a, b) = a,b, 1

  −



(a + b ) 2

2

   − −

is a bije biject ctio ion. n. It’s It’s easy easy to see that that it is inje inject ctiv ive, e, for for if ( x1 , y1 , 1 1  (x21 +  y12 )) = (x2 , y2 , 1 1  (x22 +  y22 )), then x1 = x2 and y1 = y2 , hence (x1 , y1 ) = (x2 , y2 ), as desired. The fact that it is surjective follows from observing that we can describe the set S  as

  − −

S  =



(x , y , z ) 

3

  − −



∈ R : z  = 1 1  (x + y ) . Now it is easy to see that any ( x , y , z )  ∈ S , (x, y ) ∈  B (0, 1) maps to (x , y , z ). )  . Thus, the map is, indeed, a bijection as claimed.

2

2

page 2

(a) To see that the stereographic projection is given by

 →

(a,b,c)

a

1

b

− c , 1 − c



,

we consider the line that passes through (0 , 0, 1) and (a,b,c) given by the equation s(t) = (0, 0, 1) +  t(a,b, (c

− 1)) = (at, bt, 1 + (c − 1)t).

Since we’ve identified R2 with the xy -plane, we need only solve for when 1+ (c 1)t  = 0. Easy algebra yields 1 t  = , 1  c which means that the claim for the stereographic projection defined above is correct.

−

−

To verify the inverse, we apply a similar idea. The equation of the line between ( u,v, 0) and (0, 0, 1) is given by (u,v, 0) +  t( u, v, 1) = (u(1

− −

− t), v(1 − t), t).

Using the distance formula, we find that

 

(u(1

precisely when

− t) − 0)

2

+ (v (1

t  = 1

− t) − 0)

− √ 1 + u1

2

+ (t  1)2 = 1

−

.

+ v 2 Thus, the inverse image of the stereographic projection is exactly as it was claimed, since when t  = 1  (1 +  u2 + v 2 ) 1/2 , we have

−

(u, v )

2

−

 →  √ 

u

1 +  u + v 2

2

,

v

√ 1 + u

2

+ v

2

,1

1

− 1 + u

2

+ v 2



.

(b) Again, basic knowledge of the composition of functions and algebra show that h(a, b) = (g  f )(a, b) =  h (a, b) =

◦



a

  − 1

 (a2 + b2 )

,

b

  − 1

 (a2 + b2 )



.

To find the inverse, all we need to do is compose the inverse of  g and f , i.e., compute (g  f ) 1 (a, b) = f  1  g 1 (a, b). The inverse of  f  is clearly just the projection of the first two coordinates, and since we already have the inverse of  g , we have

◦

−

−

−1

h

◦

−

−1

(a, b) = (f 

◦ g

−1

)(a, b) =

 √ 

a

1 +  a + b 2

2

,

b

√ 1 + a

2

+ b

2



.

Since each component is clearly C  , we have that h(a, b) and h 1 (a, b) are each C  . ∞

−

∞

page 3

(c) It’s easy to see we can generalize this to h (  x) =



x1

1

xn

−  x , ··· , 1 −  x



,

with the obvious analog for h 1 (  x). −

Q.E.D.