Math 147, Homework 6 Solutions
Due: May 22, 2012
1. Let T = S 1 × S 1 be the torus. Is it possible to find a finite set S = {P1 , . . . , Pn }
of points in T and an embedding of the complement T \ S into R2 ? [Hint: You
may find the Jordan-Brouwer separation theorem helpful.]
Solution.
No. Suppose for a contradiction that there were an embedding f : T \ S →
R2 . Since S is finite, we can find a point x ∈ S 1 such that {x} × S 1 ⊂ T \ S.
Note that T \(S ∪S 1 ) is connected. By the Jordan-Brouwer separation theorem,
the complement of f (S 1 ) in R2 consists of two disjoint open sets, D0 and D1 .
To argue that f (T \ (S ∪ S 1 )) intersects both D0 and D1 , note that f is a local
diffeomorphism about any point y in {x}×S 1 and hence maps any neighborhood
about y to both D0 and D1 . This is a contradiction: the image of the connected
set T \ (S ∪ S 1 ) under a continuous map f cannot intersect two disjoint open
sets D0 and D1 . Hence no such embedding f : T \ S → R2 can exist.
2. Suppose f : M → N is a diffeomorphism of connected oriented manifolds with
boundary. Show that if dfx preserves orientation at one point x, then it preserves
orientation at every point.
Solution.
Since f is a diffeomorphism, every point y ∈ N is a regular value. Also, for
any point y ∈ N the set f −1 (y) is a single point. Hence deg(f ; y) = sign dff −1 (y) .
Let
S+ = {y ∈ N | deg(f ; y) = +1}
and
S− = {y ∈ N | deg(f ; y) = −1}.
We know that deg(f ; y) is locally constant by page 27 of Milnor. Hence both
S+ and S− are open. Note that N is the disjoint union N = S+ ∪ S− , and
that S+ is nonempty because f (x) ∈ S+ . Since N is connected, S+ = N . So
f −1 (S+ ) = M , and dfx preserves orientation at every point x ∈ M .
3. Let M ⊂ Rm+1 be a smooth manifold. The Gauss map on M is the smooth
map:
fM : M → S m ,
where fM (x) is the unit length, outward pointing pointing vector in T Mx⊥ .
(a) Compute the degree of the Gauss map for S 2 ⊂ R3 .
Solution.
Note that the Gauss map for S 2 is the identity map from S 2 to itself, and
hence it has degree one.
(b) Compute the degree of the Gauss map for the surface of revolution Y ⊂ R3
diffeomorphic to S 1 × S 1 from Homework 1.
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Solution.
Every point on S 2 except for the north pole (0, 0, 1) and the south pole
(0, 0, −1) is a regular value for the Gauss map fY . Note that (1, 0, 0) is a
regular value for fY and fY−1 (1, 0, 0) = {(3, 0, 0), (−1, 0, 0)}. Map fY preserves
orientation at each point in {(x, y, z) ∈ Y | x2 +y 2 > 2} and reverses orientation
at each point in {(x, y, z) ∈ Y | x2 + y 2 < 2}. Hence fY preserves orientation
at (3, 0, 0) and reverses orientation at (−1, 0, 0). We have
X
degf =
sign dfx
x∈fY−1 (1,0,0)
= sign df(3,0,0) + sign df(−1,0,0)
= 1 + (−1)
= 0.
4. Can you find a function f : R3 → R1 with a regular value c so that the interior
of f −1 (c) is diffeomorphic to the Möbius band? Explain why or why not.
Solution.
No. Let f : R3 → R be a smooth function with regular value c. The
orientations for R3 and for R produce an orientation for f −1 (c), as follows.
Given x in f −1 (c), let (v1 , v2 , v3 ) be a positively oriented basis for T R3x with
v1 and v2 tangent to f −1 (c). Then (v1 , v2 ) determines the required orientation
for T (f −1 (c))x if and only if dfx carries v3 into a positively oriented basis for
T Rf (x) . Hence f −1 (c) is an oriented manifold. So the interior of f −1 (c) is an
oriented manifold and can’t be diffeomorphic to the Möbius band, which is not
orientable.
5. Suppose f and g are smooth maps from M to S n that satisfy |f (x) − g(x)| < 2
for each x ∈ M . Show that f and g are smoothly homotopic.
Solution.
Let x ∈ M . Since |f (x) − g(x)| < 2, points f (x) and g(x) are not antipodal.
Hence there is a unique great arc on S 2 between f (x) and g(x). Let px : I → S 2
be the path of constant speed along this great arc, with px (0) = f (x) and
px (1) = g(x). Define H : M → I by
H(x, t) = px (t)
for all x ∈ M and t ∈ I. Note that H is a smooth homotopy between f and g.
6. There are two ways we can orient S 2 : as the boundary of the disk D3 and as
the level set f −1 (1) of the smooth map f : R3 → R1 given by f (x) = |x|2 . Are
these the same orientations?
Solution.
2
Let e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1). Note e1 ∈ S 2 and that
(e2 , e3 ) is a basis for T Se21 .
Since e1 is an outward vector in T De31 and (e1 , e2 , e3 ) is a positively oriented
basis for T De31 = R3 , basis (e2 , e3 ) for T Se21 is positively oriented under the
orientation of S 2 as the boundary of the disk D3 .
Since (e2 , e3 , e1 ) is a positively oriented basis for T R3e1 and since dfe1 carries e1
into a a positively oriented basis for T Rf (e1 ) , basis (e2 , e3 ) for T Se21 is positively
oriented under the orientation of S 2 as the level set f −1 (1).
Hence these are the same orientations.
7. Let p : R2 → S 1 × S 1 be the smooth function defined by
p(t1 , t2 ) = (cos(2πt1 ), sin(2πt1 )) × (cos(2πt2 ), sin(2πt2 ))
and let LA : R2 → R2 be a linear map associated to a 2 × 2-matrix A with
integer coefficients.
(a) Show that there is a smooth function fA : S 1 × S 1 → S 1 × S 1 with the
property that p ◦ LA = fA ◦ p.
Solution.
Define fA : S 1 × S 1 → S 1 → S 1 by setting fA (x) equal to p ◦ LA (y), where
y is any point in p−1 (x). One can show that this definition does not depend on
the choice of y ∈ p−1 (x).
(b) Compute the degree of fA .
Solution.
We claim that deg(fA ) = det(A).
If det(A) = 0, then the image of LA is either a point or a line in R2 , and so
p ◦ LA is not surjective. Since p ◦ LA = fA ◦ p, it follows that fA is not surjective.
So deg(fA ) = 0.
Now suppose det(A) 6= 0. Then LA is a diffeomorphism, and is orientationpreserving if and only if det(A) > 0. Since p is orientation preserving and
p ◦ LA = fA ◦ p, it follows that fA is orientation preserving (or reversing) at each
point in S 1 × S 1 if and only if det(A) > 0 (or det(A) < 0). Note that every
point of S 1 × S 1 is a regular value of fA , and in particular x = ((1, 0) × (1, 0))
is a regular value. So
deg(fA ) = sign det(A) · #fA−1 (x).
Since p : R2 → S 1 × S 1 restricts to a bijection on [0, 1) × [0, 1), we have
n
o
−1
−1
−1
#fA (x) = # p (fA (x)) ∩ [0, 1) × [0, 1)
n
o
−1
= # L−1
(p
(x))
∩
[0,
1)
×
[0,
1)
A
n
o
−1
= # p (x) ∩ LA [0, 1) × [0, 1) ,
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which is the number of lattice points (or points
with integer coordiantes) in the
half-open parallelogram LA [0, 1) × [0, 1) .
Let’s count the number of lattice points in the half-open parallelogram
LA [0, 1) × [0, 1) . Note the area of the closed parallelogram LA [0, 1] × [0, 1] is
| det(A)|. By Pick’s Theorem (http://en.wikipedia.org/wiki/Pick’s theorem),
this area | det(A)| is equal to i + b/2 − 1, where i is the number of interior lattice points in the closed parallelogram LA [0, 1] × [0, 1] , and b is the number
of boundary lattice points on this closed parallelogram. Of the b points on the
boundary of the closed parallelogram, b − 4 of them are not vertices of the parallelogram, and (b − 4)/2 of them are also non-vertex boundary lattice points
in the half-open parallelogram. We also have one vertex (0, 0) in the half-open
parallelogram, giving (b − 4)/2 + 1 = b/2 − 1 lattice points on the boundary
of the half-open parallelogram. We have i lattice points on the interior of the
half-open parallelogram. Hence in total there are i + b/2 − 1 lattice points in
the half-open parallelogram LA [0, 1) × [0, 1) . To summarize, we have shown
| det(A)| = area of closed parallelogram LA [0, 1] × [0, 1]
= i + b/2 − 1
= i + (b − 4)/2 + 1
= number of lattice points in the half-open parallelogram LA [0, 1) × [0, 1)
= #fA−1 (x).
Putting everything together, we have
deg(fA ) = sign det(A) · #fA−1 (x)
= sign det(A) · | det(A)|
= det(A).
Solution.
Let p(z) = an z n + . . . + a1 z + a0 with ai ∈ C and an 6= 0. Then p is a
polynomial of degree n. Let q(z) = an z n . Let fq : S 2 → S 2 be the induced map
on the sphere. Note that deg(fq ) = n. One way to see this is to note that an is
a regular value of fq and that
X
deg(fq ; an ) =
sign d(fq )x
x∈fq−1 (an )
X
=
sign d(fq )x
x∈{ζ1 ,...,ζn }
X
=
x∈{ζ1 ,...,ζn }
= n,
where {ζ1 , . . . , ζn } are the n-th roots of unity.
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Define H : C × I → C by
H(z, t) = an z n + (1 − t)(an−1 z n−1 + . . . + a1 z + a0 ).
So H is a homotopy from H(z, 0) = p(z) to H(z, 1) = q(z). Since limz→∞ (z, t) =
∞ for all t ∈ I, we get an induced homotopy fH : S 2 × I → S 2 with fH (x, 0) =
fp (x) and fH (x, 1) = fq (x) for all x ∈ S 2 . By Theorem B on page 28 of Milnor,
deg(fp ) = deg(fq ) = n.
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