Math 147, Homework 3 Solutions
Due: April 24, 2012
1. Let Y ⊂ R3 be the surface of revolution diffeomorphic to S 1 × S 1 you studied
in Homework 1 and let f and g be the smooth functions:
f, g : Y → R1
defined by f (x, y, z) = x and g(x, y, z) = z. What are the critical values of
f ? What are the critical values of g? Sketch the level sets of f and g at their
regular values.
Solution.
Recall Y is a two-dimensional manifold. Since df(x,y,z) = [1, 0, 0], the map
df(x,y,z) : T Y(x,y,z) → R1 is surjective unless T Y(x,y,z) = {(x, y, z) | x = 0}. One
can check that T Y(x,y,z) = {(x, y, z) | x = 0} happens if and only if (x, y, z) =
(−3, 0, 0), (−1, 0, 0), (1, 0, 0), or (3, 0, 0). These are the only critical points of f
on Y , and so the only critical values of f are −3, −1, 1, and 3.
Since dg(x,y,z) = [0, 0, 1], the map dg(x,y,z) : T Y(x,y,z) → R1 is surjective unless
T Y(x,y,z) = {(x, y, z) | z = 0}. One can check that T Y(x,y,z) = {(x, y, z) | z = 0}
happens if and only if x2 + y 2 = 2 and z = ±1. These are the only critical
points of g on Y , and so the only critical values of g are −1 and 1.
A level set of f at a regular value is either a single loop or two adjacent
loops. A level set of g at a regular value is a union of two circles, one inside the
other.
2. Construct a smooth function f : R1 → R1 whose critical values are dense in R1 .
Solution.
We will construct a smooth map f : R → R whose set of critical values is
the rational numbers Q, which is dense in R.
Since the rationals are countable, we can write all rational numbers in a
sequence r0 , r1 , r2 , . . .. Using Question 6(d) on Homework 2, for each integer
0 ≤ i we can construct a smooth bump function gi : [i, i + 1] → R1 that is equal
to 0 on [i, i+1/5] and on [i+4/5, i+1] and that is equal to ri on [i+2/5, i+3/5].
Function gi has ri as a critical value.
Define f : R → R by
(
0
if x < 0
f (x) =
gi (x) if i ≤ x < i + 1.
Note that f is smooth and has each ri as a critical value. Hence the critical
values of f are dense in R1 .
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3. The unit tangent bundle of a smooth m-manifold M ⊂ Rk is the set of unit
vectors in T M :
T 1 M = (x, v) : x ∈ M, v ∈ T Mx and |v|2 = 1 .
Show that T 1 M is a smooth manifold of dimension 2m − 1. Also show that
T 1 M is compact whenever M is compact.
Solution.
Define f : T M → R by f (x, v) = kvk2 . Then for any (u, w) ∈ T M(x,v) we
have
df(x,v) (u, w) = 2v · w.
So df(x,v) is surjective if v 6= ~0, and so in particular 1 is a regular value of
f . Therefore f −1 (1) = T 1 M is a smooth manifold of dimension dim(T M ) −
dim(R) = 2m − 1 by Lemma 1 in §2 of Milnor, page 11.
Suppose M is compact. Then M is bounded, so there exists some c ∈ R1
with kxk ≤ c for all x ∈ M . Hence for all (x, v) ∈ T 1 M , we have
k(x, v)k ≤ kxk + kvk ≤ c + 1.
So T 1 M is bounded.
Next we show that T 1 M is closed. Suppose (xi , vi ) is a sequence of points in
T 1 M converging to some point (x, v) in R2k . We must show that (x, v) ∈ T 1 M .
Since M is compact hence closed, we have x ∈ M . Next we’ll show that (x, v) ∈
T M . Consider a local parameterization φ : U → W where W is an open set in X
containing x, and where U is an open set in Rk . From page 50 of Guillemin and
Pollack, we know dφ : T (U ) → T (W ) is a diffeomorphism. Here T (U ) = U × Rk
and dφ acts on a point (u, z) ∈ T (U ) by dφ(u, z) = (φ(u), dφu (z)). Since xi
converges to x, for i sufficiently large each xi is in U . And since each vi is in
T Mxi , for i sufficiently large each (xi , vi ) is in T (W ). One can show that their
preimages (dφ)−1 (xi , vi ) = (ui , zi ) for i sufficiently large lie in some compact
subset of T (U ), hence have a subsequence converging to a point (u, z) in T (U ).
By continuity of dφ we get a subsequence of (xi , vi ) converging to dφ(u, z). By
uniqueness of limit points, we must have
(x, v) = dφ(u, z) ∈ T (W ) ⊂ T M.
Finally, since k · k is a continuous function and kvi k = 1 for all i, we have
kvk = 1, and so (x, v) is also in T 1 M . So T 1 M is closed, bounded, and hence
compact.
4. Show that the set SO(3) of 3 × 3 orthogonal matrices of determinant one is
diffeomorphic to T 1 S 2 .
Solution.
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Recall
T 1 S 2 = {(x, v) : x ∈ S 2 , v ∈ T Sx2 and kvk2 = 1}
= {(x, v) ∈ R3 × R3 : kxk = kvk = 1 and x · v = 0}.
Define f : T 1 S 2 → SO(3) by letting f (x, v) be the matrix with first column x,
second column v, and third column x × v. Map f lands in SO(3) because the
resulting matrix has orthonormal columns and has determinant one. Map f is
clearly injective.
Let’s see why f is surjective. An orthogonal matrix M satisfies M t M = I,
and it follows that the first two columns of any orthogonal matrix are orthogonal
unit vectors, call them x and v. The third column must be a unit vector
orthogonal to the first two, and hence must equal either x × v or −x × v. But
for M to also have determinant one, the third column must be x × v. Hence f
is surjective.
The inverse f −1 is given by sending the first and second columns of a matrix
in SO(3) to x and v respectively, forming (x, v) ∈ T 1 S 2 . Since f can be extended
to a smooth map F : R3 × R3 → M3 (R) given by the same formula, and since
f −1 can be extended to a smooth map G : M3 (R) → R3 × R3 given by the same
formula, it follows that f and f −1 are smooth. Hence f : T 1 S 2 → SO(3) is a
diffeomorphism.
5. (a) In class we showed that T O(n)I is the set of anti-symmetric matrices in
Mn (R). Show that AB − BA is in T O(n)I whenever A and B are in T O(n)I .
Solution.
Suppose A and B are in T O(n)I and hence anti-symmetric. Then −AT = A
and −B T = B. Note that
−(AB − BA)T = −B T AT + AT B T
= −BA + AB
= AB − BA.
So AB − BA is also anti-symmetric and hence also in T O(n)I .
(b) Let G ⊂ Mn (R) be a collection of matrices that is both a smooth manifold
and a group. Show that if A and B are in T GI , then so is AB − BA.
Solution.
Since A and B are in T GI , from Question 3 on Homework 2 we have smooth
maps α, β : (−t, t) → G with α(0) = β(0) = I, dα0 = A, and dβ0 = B.
Let g ∈ G. We show that if B ∈ T GI , then the conjugate gBg −1 is also in
T GI . Define curve γ : (−t, t) → G by γ(t) = gβ(t)g −1 , which lands in G because
G is a group. Since γ(0) = I and γ 0 (0) = gBg −1 , this shows that gBg −1 is in
T GI for any g ∈ G.
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Define curve δ : (−t, t) → T GI by δ(t) = α(t)Bα(t)−1 − B. This curve lands
in T GI because α(t)Bα(t)−1 is in T GI for each t by the paragraph above, and
since T GI is a linear subspace.
Since T GI is a linear subspace, the derivative of curve δ also lies in T GI .
Note we have
δ 0 (0) = α0 (0)Bα(0)−1 + α(0)B[α(t)−1 ]0t=0
= ABI + IB[α(t)−1 ]0t=0
= AB + B[α(t)−1 ]0t=0 .
Since I = α(t)α(t)−1 , we calculate
0 = [I]0t=0
= [α(t)α(t)−1 ]0t=0
= α0 (0)α(0)−1 + α(0)[α(t)−1 ]0t=0
= AI + I[α(t)−1 ]0t=0
= A + [α(t)−1 ]0t=0 .
Hence [α(t)−1 ]0t=0 = −A and so
δ 0 (0) = AB + B[α(t)−1 ]0t=0
= AB − BA.
Hence AB − BA is in T GI .
6. Suppose f : M → N is a smooth map between manifolds, x is in M and dfx is
one-to-one. Show that there is a neighborhood of U of x in M such that the
restriction of f to U is a diffeomorphism onto its image f (U ).
Solution.
Let m and n be the dimensions of M and N . Since dfx : T Mx → T Nf (x)
is one-to-one, we know that m ≤ n. Moreover, there exists a linear map
L : Rn−m → T Nf (x) whose image in T Nf (x) is orthogonal to dfx (T Mx ). Hence
(f, L) : M × Rn−m → N
is a smooth map whose derivative
(dfx , L) : T Mx × Rn−m → T Nf (x)
at the point (x, 0) is an isomorphism. By the Inverse Function Theorem on page
13 of Guillemin and Pollack, (f, L) is a local diffeomorphism at (x, 0). That is,
there is a neighborhood (U, V ) of (x, 0) in M × Rn−m such that the restriction
of (f, L) to (U, V ) is a diffeomorphism onto its image (f (U ), L(V )). Hence the
restriction of f to the neighborhood U of x is a diffeomorphism onto its image
f (U ).
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7. Suppose that f : M → N is a smooth map between m-manifolds and that dfx
is non-singular at each x in M . Show that f is an open mapping, i.e. f (U ) is
open whenever U ⊂ M is open.
Solution.
Let U ⊂ M be open, and let y be any point in f (U ). Let f (x) = y.
Since M and N have the same dimension and dfx is non-singular, dfx is also
an isomorphism. By the Inverse Function Theorem on page 13 of Guillemin
and Pollack, f is a local diffeomorphism at x. So the restriction of f to some
neighborhood U 0 ⊂ U ⊂ M about x is a diffeomorphism onto its image f (U 0 ).
Since U 0 is open in M and f is a diffeomorphism here, f (U 0 ) is an open set in
N that contains y and that is contained in f (U ). Since y was an arbitrary, we
have shown that f (U ) is open in N .
8. (a) Let U be an open set in Rn and f : U → Rn be a smooth map. Show that
if A ⊂ U has measure zero, then f (A) has measure zero. [Hint: First show that
you can cover A by countably many sets A1 , . . . , An , . . . whose closures Āi are
compact and contained in U . Then show that f distorts volumes of cubes by a
bounded amount near Āi .]
Solution.
This is proven in a Theorem in Appendix 1 of Guillemin and Pollack, on
the bottom of page 204. How do we write A as a countable union of subsets
A1 , A2 , A3 , . . . whose closures are compact and contained in U ? One choice is
to let Ai = A ∩ {x ∈ Rn : kxk ≤ i}.
(b) Use part (a) to prove mini-Sard’s: If f : U → Rn is a smooth map, with U
open in Rm and m < n, then f (U ) has measure zero in Rn .
Solution.
Define F : U × Rn−m → Rn by F (u, x) = f (u). Map F is smooth because f
is smooth. We know that Rm × {0} has measure zero in Rn because m < n, and
hence its subset U ×{0} is also a set of measure zero in U ×Rn−m . We apply part
(a) to see that F (U × {0}) has measure zero in Rn . Since F (U × {0}) = f (U ),
we are done.
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