Math 147, Homework 7 Solutions
Due: May 29, 2012
1. Let M ⊂ Rk be a smooth manifold without boundary and let N M be the
normal bundle on M :
N M = (x, v) : x ∈ M and v ∈ T Mx⊥ .
(a) Show that N M is a smooth manifold of dimension k.
Solution.
See the Proposition on page 71 of Guillemin and Pollack.
(b) For > 0, show that N M = {(x, v) ∈ N M : |v| ≤ } is a smooth k-manifold
with boundary.
Solution.
Consider the map f : N M → R given by f (x, v) = |v|2 . Note that this is a
smooth function and that df(x,v) is not surjective if and only if v = 0. Hence 2
is a regular value of f for > 0. By Lemma 3 on page 12 of Milnor, the set of
points (x, v) in N M with f (x, v) ≤ 2 is a smooth k-manifold with boundary.
This set is equal to
N M = {(x, v) ∈ N M : |v| ≤ } .
(c) Now suppose M is compact. Show that the smooth map φ : (x, v) →
7
x
+
v
maps
N
M
diffeomorphically
onto
the
closed
-neighborhood
M
=
k
y ∈ R : |x − y| ≤ for some x ∈ M .
Solution.
This is only true for sufficiently small. We adapt the proof of the Product
Neighborhood Theorem on page 46 of Milnor.
Consider the mapping Φ : N M → Rk given by Φ(x, v) = x + v. Consider
(x, 0) ∈ N M . Since dΦ(x,0) is nonsingular, Φ is a diffeomorphism on some
neighborhood about (x, 0).
We will show that the restriction of Φ to N M for some sufficiently small
is is a diffeomorphism onto its image. If not, then there would exist pairs
(x, v) 6= (x0 , v 0 ) ∈ N M with kvk and kv 0 k arbitrarily small and with Φ(x, v) =
Φ(x0 , v 0 ). Since M is compact, we could choose a sequence of such pairs (xi , vi ) 6=
(x0i , vi0 ) ∈ N M with xi converging to x0 , with x0i converging to x00 , and with vi
and vi0 converging to 0. Then necessarily x0 = x00 , and so this sequence of pairs
contradicts the fact that Φ is a diffeomorphism on some neighborhood about
(x0 , 0) = (x00 , 0). Hence the restriction of Φ to N M for some sufficiently small
is a diffeomorphism. Call this restiction φ : N M → Rk .
It remains to show that φ(N M ) = M . Since kφ(x, v) − xk = kvk ≤ , we
have φ(N M ) ⊂ M . Conversely, suppose w ∈ M . Since M is compact, let y
1
be a point in M closest to w, so kw − yk ≤ . (A priori there may be multiple
such points y, but we will show in problem 2 that there is a unique such point
y. We do not need this uniqueness here.) If c(t) is a curve on M with c(0) = y,
then the smooth function kw − c(t)k2 has a minimum at 0. Note
d
kw − c(t)k2 = 2c0 (t) · (w − (c(t))
dt
and plugging in t = 0 we get
0 = 2c0 (0) · (w − y).
Since this is true for all such curves c, this shows that w − y ∈ T My⊥ . Since
kw − yk ≤ , we have (y, w − y) ∈ N M with φ(y, w − y) = w. Hence φ(N M ) =
M .
In summary, for some sufficiently small the smooth map φ : N M → M is a diffeomorphism.
2. Let φ : N M → M be the diffeomorphism from Problem 1. Show that the
closest point on M to φ(x, v) is φ(x, 0). Also show that φ(x, v) − φ(x, 0) is
orthogonal to T Mφ(x,0) .
Solution.
We did many of the following steps in problem 1(c) as well.
Since M is compact, let y be a point in M closest to φ(x, v) (a priori there
may be multiple such points y). If c(t) is a curve on M with c(0) = y, then the
smooth function kφ(x, v) − c(t)k2 has a minimum at 0. Note
d
kφ(x, v) − c(t)k2 = 2c0 (t) · (φ(x, v) − (c(t))
dt
and plugging in t = 0 we get
0 = 2c0 (0) · (φ(x, v) − y).
Since this is true for all such curves c, this shows that φ(x, v) − y ∈ T My⊥ . Since
φ is injective, it must be that y = x = φ(x, 0). Hence we have shown that the
closest point on M to φ(x, v) is y = x = φ(x, 0), and that φ(x, v) − φ(x, 0) ∈
⊥
T Mφ(x,0)
.
3. Suppose M ⊂ Rm+1 is a compact m-manifold without boundary. Show that
N M is diffeomorphic to M × R1 .
Solution.
Since M ⊂ Rm+1 , as in problem 3 of homework 6 we can define the Gauss
map g : M → S m . For x ∈ M , point g(x) is the unit length, outward pointing
vector in T Mx⊥ .
Define f : M × R1 → N M by f (x, t) = (x, tg(x)). Note that f is a smooth
bijective map, with smooth inverse given by f −1 (x, v) = (x, v · g(x)). Hence f
is a diffeomorphism, so N M is diffeomorphic to M × R1 .
2
4. Let M ⊂ Rk be a compact smooth manifold without boundary. For v ∈ Rk , let
πv : M → Rk be the vector field whose value at x is the orthogonal projection
of v onto T Mx :
πv (x) = projT Mx (v).
Show that, for a full-measure set of v, the vector field πv has isolated zeros.
Solution.
Define f : N M → Rk by f (x, v) = v. Suppose that vector field πv has a
non-isolated zero. We will show in the paragraph below that f −1 (v) is not a
smooth manifold of dimension dim(Rk ) − dim(N M ) = 0. Hence by Lemma 1
on page 11 of Milnor, v ∈ Rk is a critical value. By Sard’s Theorem, the set
of all such v has measure zero. So the vector field πv has isolated zeros for a
full-measure set of v in Rk .
If vector field πv has a non-isolated zero, then there is a sequence of points
{xi } converging to x ∈ M with πv (xi ) = 0 for all i, and hence πv (x) = 0.
We have v ∈ T Mx⊥i for all i and v ∈ T Mx⊥ . So (xi , v) ∈ f −1 (v) for all i,
(x, v) ∈ f −1 (v), and {(xi , v)} converges to (x, v) in f −1 (v). This shows that no
neighborhood about (x, v) in f −1 (v) is diffeomorphic to R0 . Hence f −1 (v) is
not a smooth manifold of dimension dim(Rk ) − dim(N M ) = 0. We are done by
the paragraph above.
5. (a) Let Y ⊂ R3 be the surface of revolution diffeomorphic to S 1 × S 1 and
v = (1, 0, 0). Show that the vector field πv (defined in Problem 4) has isolated
zeros and compute the index of πv at each of its zeros.
Solution.
The vector field πv has only four zeros in Y : at (3, 0, 0), (−3, 0, 0), (1, 0, 0),
and (−1, 0, 0). Hence these zeros are isolated.
At (3, 0, 0) vector field πv is a sink. See Figure 3-18(b) on page 133 of
Milnor. Hence the index is +1 (see the second to last paragraph on page 133).
At (−3, 0, 0) the vector field is a source, as shown in Figure 3-18(c), so the index
is also +1. At points (1, 0, 0) and (−1, 0, 0) the vector field is a saddle, as shown
in Figure 3-18(d), so the index is −1.
(b) Write down a nowhere vanishing vector field on Y .
Solution.
Define vector field u : Y → R3 by u(x, y, z) = (−y, x, 0). Fix a point
(x0 , y0 , z0 ) ∈ Y . To see that u(x0 , y0 , z0 ) ∈ T Y(x0 ,y0 ,z0 ) , let circle C be the
intersection of Y with the plane z = z0 . It is easy to see that u(x0 , y0 , z0 ) ∈
T C(x0 ,y0 ,z0 ) , and since T C(x0 ,y0 ,z0 ) ⊂ T Y(x0 ,y0 ,z0 ) , we have u(x0 , y0 , z0 ) ∈ T Y(x0 ,y0 ,z0 ) .
Hence u is a vector field. To see that u is non-vanishing, note that
Y ∩ {(x, y, z) | x = y = 0} = ∅.
3
6. Can you construct a smooth vector field v : S 2 → R3 that only vanishes at
(0, 0, 1)?
Solution.
Yes. First, consider the vector field w : R2 → R2 given by w(y) = (1, 0) for
all y ∈ R2 . Let f : R2 → S 2 \ {(0, 0, 1)} be the stereographic diffeomorphism
given by
2v
−1 + u2 + v 2 2u
.
,
,
f (u, v) =
1 + u2 + v 2 1 + u2 + v 2 1 + u2 + v 2
We have


2
2
df(u,v) =
2(1−u +v )
(1+u2 +v 2 )2
 −4uv
 (1+u2 +v2 )2
4u
(1+u2 +v 2 )2
−4uv
(1+u2 +v 2 )2
2(1+u2 −v 2 ) 
.
(1+u2 +v 2 )2
4v
(1+u2 +v 2 )2
Note that as (u, v) goes to infinity, df(u,v) (1, 0) goes to (0, 0, 0). Hence we can
define a smooth vector field v : S 2 → R3 by
(
f∗ w(x) for x 6= (0, 0, 1)
v(x) =
(0, 0, 0) for x = (0, 0, 1).
Note that df(u,v) (1, 0) 6= (0, 0, 0) for all (u, v) ∈ R2 , and hence the vector field v
vanishes only at (0, 0, 1).
7. For each n > 0, find a vector field on S n with isolated zeros and compute its
index.
Solution.
Let e1 ∈ Rn+1 be the first standard basis vector. Let πe1 : S n → Rn+1 be
the vector field defined as in problem 4. Note that πe1 (x) = 0 if and only if
x = ±e1 , hence this vector field has isolated zeros.
Near the point e1 ∈ S n the vector field radiates outwards. Hence the vector
field acts like the identity map on a small (n − 1)-sphere about e1 . The identity
map on the (n − 1)-sphere has degree 1, and hence the index of πe1 at e1 is 1.
Near the point −e1 ∈ S n the vector field radiates inwards. Hence the vector
field acts like the antipodal map on a small (n − 1)-sphere about −e1 . The
antipodal map on the (n − 1)-sphere has degree (−1)n , by page 30 of Milnor.
Hence the index of πe1 at −e1 is (−1)n .
So the index of vector field πe1 on S n is 1 + (−1)n , which is 2 for n even and
0 for n odd. See also page 39 of Milnor for a description of this problem.
8. Suppose Rn has a division algebra structure m : Rn × Rn → Rn . Show that
T S n−1 is diffeomorphic to S n−1 × Rn−1 .
Solution.
Suppose Rn is a division algebra. As a first case, we’ll also assume that
this division algebra has an identity element e ∈ S n−1 satisfying m(e, x) =
4
m(x, e) = x for all x ∈ Rn . Now, for each point x ∈ S n−1 , define the map
Lx : S n−1 → S n−1 by
m(x, y)
Lx (y) =
km(x, y)k
for y ∈ S n−1 . Since Rn is a division algebra, for each element y 0 ∈ S n−1
there exists exactly one element y ∈ S n−1 with Lx (y) = y 0 . Hence Lx is a
diffeomorphism. Note also Lx (e) = x. Thus d(Lx )e : T Sen−1 → T Sxn−1 is a linear
isomorphism. We define a diffeomorphism f : S n−1 × T Sen−1 → T S n−1 by
f (x, v) = (x, d(Lx )e (v)).
So S n−1 × T Sen−1 = S n−1 × Rn−1 is diffeomorphic to T S n−1 .
Though the division algebra may not have such an identity element e, we
show how to modify the multiplication map m to produce an identity. Choose
a vector e ∈ S n−1 . After composing the multiplication with an invertible linear
map Rn → Rn taking m(e, e) to e, we may assume that m(e, e) = e. Let α
be the map x → m(x, e) and let β be the map x → m(e, x). Define the new
multiplication map m0 to be
m0 (x, y) = m(α−1 (x), β −1 (x)).
Note that
m0 (x, e) = m(α−1 (x), β −1 (e)) = m(α−1 (x), e) = x
and similarly
m0 (e, x) = x.
Hence we have produced a new multiplication map m0 : Rn → Rn with an
identity element e ∈ S n−1 . So the paragraph above finishes the problem.
9. Let G ⊂ Mn (R) be a Lie group. Show that G admits a nowhere vanishing
vector field.
Solution.
Let I ∈ G be the identity matrix. Pick some nonzero matrix h ∈ T GI ,
and consider the map v : G → Mn (R) given by v(g) = hg. Since h ∈ T GI ,
that means that there is some smooth path α : (−, ) → G with α(0) = I
and α0 (0) = h. Hence gα : (−, ) → G is a smooth path with gα(0) = g and
gα0 (0) = gh. This shows that v(g) = hg ∈ T Gg for all g ∈ G, and hence our
map v is a vector field. Since each g ∈ G is invertible and h is not the zero
matrix, this shows that each v(g) = gh is not the zero matrix, and hence vector
field v is non-vanishing.
5