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.
