Math 147, Homework 2 Solutions
Due: April 17, 2012
1. Show that 0 is the only critical value of the map f : R3 → R1 defined by:
f (x, y, z) = x2 + y 2 − z 2 .
Prove that if a and b are either both positive or both negative, then f −1 (a) and
f −1 (b) are diffeomorphic and sketch what happens at f −1 (0).
Solution.
The derivative of f is
df(x,y,z) = 2x 2y −2z .
Note that df(x,y,z) has rank less than 1 only if (x, y, z) = (0, 0, 0). Hence (0, 0, 0)
is the only critical point of f and f (0, 0, 0) = 0 is the only critical value of f .
Suppose that a and b are either both positive or both negative (we will
handle both cases at once). Let G : R3 → R3 be defined by
p
p
p
G(x, y, z) = (x b/a, y b/a, z b/a).
p
That is, G is scalar multiplication by b/a. Then G is clearly a diffeomorphism
from R3 to R3 , and the restriction of G to f −1 (a) is a diffeomorphism from
f −1 (a) to f −1 (b). Hence f −1 (a) and f −1 (b) are diffeomorphic.
The set f −1 (0) is a cone, as drawn in Figure 1-2 in Guillemin and Pollack.
2. Show that 1 is a regular value for the function:
p
f (x, y, z) = z 2 + ( x2 + y 2 − 2)2 .
√
What is the tangent space to the level set f −1 (1) at (1, 3, 1)?
Solution.
The derivative of f is
2
2
2y 1 − √ 2 2
2z .
df(x,y,z) = 2x 1 − √x2 +y2
x +y
So if (x, y, z) is a critical point of f , then all entries of this matrix are zero,
meaning z = 0 and either x = y = 0 or x2 + y 2 = 4. Hence f (x, y, z) = 4 or 0.
Thus 1 is a regular value of f .
√
The tangent space to the level set f −1 (1) at (1, 3, 1) is the set of all points
orthogonal to the gradient vector
√
2
2
p
p
, 2y 1 −
, 2z |(1,√3,1)
∇f (1, 3, 1) = 2x 1 −
x2 + y 2
x2 + y 2
= (0, 0, 2).
1
Hence
T f −1 (1) (1,√3,1) = {(x, y, z) ∈ R3 | 0x + 0y + 2z = 0}
= {(x, y, z) ∈ R3 | z = 0}.
3. Let M ⊂ Rk be a smooth manifold, x be a point in M and v be a vector in
T Mx . Show that there is a smooth map from an interval in R1 to M :
f : (−t, t) → M,
with f (0) = x and df0 (1) = v.
Solution.
Let M be a manifold of dimension m. Choose a diffeomorphism g : U →
W ∩ M where U is an open subset of Rm containing ~0, where W is an open
set in Rk containing x, and where g(~0) = x. Since T Mx = dg(~0) by definition,
there exists a vector u ∈ Rm satisfying dg~0 (u) = v.
Since U is an open subset of Rm containing ~0, we can pick t > 0 such that
{su | s ∈ (−t, t)} ⊂ U . Define the function h : (−t, t) → U by h(s) = su. Note
that h(0) = ~0 and dh0 (1) = u. Define f : (−t, t) → M by f = g ◦ h. Function f
is smooth because it is the composition of the smooth functions g and h. Note
f (0) = g(h(0)) = g(~0) = x.
and
df0 (1) = d(g ◦ h)0 (1)
= dgh(0) (dh0 (1)) by the chain rule
= dg~0 (u)
= v by definition of u.
Hence we have constructed a smooth map satisfying the necessary properties.
4. The tangent bundle of a smooth m-manifold M in Rk is the subset of M × Rk :
T M = {(x, v) : x ∈ M, v ∈ T Mx } .
Show that T M is a smooth manifold of dimension 2m.
Solution.
This is a Proposition on the top of page 51 in Chapter 1 §8 of Guillemin
and Pollack. See pages 50 and 51 for a complete proof.
5. Find the critical values of the smooth function:
det : Mn (R) → R1 .
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sending an n × n matrix to its determinant. Show that the special linear group
SLn (R) consisting of n × n matrices of determinant 1 is a smooth manifold.
Compute the tangent space of SLn (R) at the identity matrix.
Solution.
2
Let A ∈ Mn (R). We identify Mn (R) with Rn , and hence T (Mn (R))A =
Mn (R). Also T (R1 )det(A) = R1 . So
d(det)A : Mn (R) → R1 .
First we show that 0 is a critical value of det. Let 0 ∈ Mn (R) be the matrix
of all zeros. Let A be any matrix in Mn (R). We have
det(0 + tA) − det(0)
t→0
t
det(tA)
= lim
t→0
t
tn det(A)
= lim
t→0
t
= 0.
d(det)0 (A) = lim
Since A ∈ Mn (R) was arbitrary, this means that d(det)0 is singular, that 0 is a
critical point of det, and that 0 = det(0) is a critical value of det.
Next we show that det has no other critical values. Suppose det(A) 6= 0.
We’ll show that d(det)A is nonsingular by considering d(det)A (A). We have
det(A + tA) − det(A)
t→0
t
det((1 + t)A) − det(A)
= lim
t→0
t
(1 + t)n det(A) − det(A)
= lim
t→0
t
(1 + t)n − 1
= det(A) lim
t→0
t
= det(A) n
6= 0,
d(det)A (A) = lim
since the limit in the third to the last line is the derivative of (1 + x)n evaluated
at x = 0, which is equal to n(1 + x)n−1 evaluated at x = 0, which is n. Since
d(det)A (A) 6= 0, this means that d(det)A : Mn (R) → R1 is nonsingular. Hence
det(A) 6= 0 implies that A is a regular value of det, and so the only critical
value of det is 0.
Since 1 is a regular value of det and since SLn (R) = det−1 (1), Lemma 1 on
page 11 in Chapter 1 of Milnor implies that SLn (R) is a smooth manifold of
dimension n2 − 1.
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Finally, we compute the tangent space of SLn (R) at the identity matrix
I. By the Proposition on page 24 in Chapter 1 §4 of Guillemin and Pollack,
we have T (SLn (R))I = ker(d(det)I ). To find ker(d(det)I ), first note that for
A ∈ Mn (R) we have
det(I + tA) = 1 + t Tr(A) + O(t2 )
where Tr(A) is the trace of A and O(t2 ) is a polynomial in t with all terms of
degree at least two. Hence
det(I + tA) − det(I)
t→0
t
1 + t Tr(A) + O(t2 ) − 1
= lim
t→0
t
= Tr(A).
d(det)I (A) = lim
So T (SLn (R))I = ker(d(det)I ) is the set of all matrices with trace equal to zero.
6. (a) Show that the function f : R1 → R1 defined by:
(
2
e−1/x x > 0
f (x) =
0
x≤0
is smooth.
Solution.
We need to show that all derivatives f (i) of f exist and are continuous. For
x < 0, all derivatives of f are identically zero. Hence we can restrict attention
to x > 0. We need to show all derivatives f (k) exist on x > 0 and satisfy
limx→0+ f (i) = 0 on x > 0.
Note f 0 = 2x−3 f . In fact, every derivative f (i) is of the form f (i) = p(x−1 )f
where p is a polynomial and hence p(x−1 ) is a polynomial in x−1 . We prove this
by induction. The base case is above. The inductive step is
0
f (i+1) = f (i) = [p(x−1 )f ]0 = −x−2 p0 (x−1 )f + 2x−3 pf = (−x−2 p0 (x−1 ) + 2x−3 p)f.
Hence each f (k) is a linear combination of terms of the form x−k f for k ≥ 0.
It remains to show that limx→0+ x−k f = 0. The logarithm of x−k f is −k log x −
1/x2 , and we have
h
h 1 k log x
i
1i
lim+ −k log x − 2 = lim+ − 2
+
1
.
x→0
x→0
x
x
x2
The first term on the right hand side goes to −∞ and the second term goes to
1 by L’Hopital’s rule. Hence the limit of the logarithm is −∞, and therefore
the limit of its expoential is limx→0+ x−k f = 0, as desired.
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(b) Show that, for a < b, the function g(x) = f (x − a)f (b − x) is smooth,
positive on (a, b) and zero elsewhere.
Solution.
Function g is smooth because it is the product of two smooth functions. It
is positive on (a, b) because both f (x − a) and f (b − x) are positive on (a, b). It
is zero on x < a because f (x − a) is zero here, and it is zero on x > b because
f (b − x) is zero here.
Rx
R∞
(c) Show that the function h(x) = −∞ g(t)dt/ −∞ g(t)dt is smooth and satisfies
h(x) = 0 for x < a, h(x) = 1 for x > b and 0 < h(x) < 1 for x in (a, b).
Solution.
R∞
Function h is smooth because its first derivative is g/ −∞ g(t)dt, and we
showed
R x in part (b) that g is smooth. If x < a then g(t) = 0 for all t < x,
so −∞ g(t)dt = 0, so h(x) = 0. If x > b then g(t) = 0 for all t > x, so
Rx
R∞
Rx
R∞
g(t)dt = −∞ g(t)dt, so h(x) = 1. Since 0 < −∞ g(t)dt < −∞ g(t)dt for x
−∞
in (a, b), we have 0 < h(x) < 1 for x in (a, b).
(d) Construct a smooth function on Rk that equals 1 on the ball of radius a,
zero outside a ball of radius b, and is strictly between 0 and 1 at intermediate
points. (A function of this form is sometimes called a bump function.)
Solution.
Define g : Rk → R by g(x) = 1 − h(kxk). Note that g equals 1 on the ball
of radius a, zero outside a ball of radius b, and is strictly between 0 and 1 at
intermediate points. Since the function x → kxk is smooth on Rk \{~0}, function
g is clearly smooth on Rk \ {~0}. But since g is constant on the ball of radius a,
function g is also smooth at the origin, and hence is smooth on all of Rk .
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