DIFFERENTIAL GEOMETRY 1
PROBLEM SET 2 SOLUTIONS
Lee: 2-1c, 2-6, 2-8, 2-10
CFB: 1.4.2:1-5
Remark 0.1. Note that problem 1.4.2.2 was corrected. The errata for the text “Cartan For
Beginners” can be found at
http://www.math.tamu.edu/~jml/cfbcorrectionsMay2006.pdf
CFB 1.4.2.1 V a vector space with nondegenerate inner product h·, ·i. Let v (t) be a curve
in V such that F (t) := hv (t) , v (t)i constant. Show that v (t)0 ⊥ v (t) for all t. Show that the
converse is true.
Proof:
First note that
F 0 (t) = hv 0 (t) , v (t)i + hv (t) , v 0 (t)i = 2hv 0 (t) , v (t)i.
Then F is constant if and only if 0 = F 0 (t) for all t.
Thus F is constant if and only if 0 = 2hv 0 (t) , v (t)i for all t.
Assuming the field doesn’t have characteristic 2 we have that 0 = 2hv 0 (t) , v (t)i if and
only if 0 = hv 0 (t) , v (t)i.
But this inner product vanishes for all t if an only if v 0 (t) ⊥ v (t) for all t.
Therefore F (t) is constant if and only if v 0 (t) ⊥ v (t) for all t.
Date: February 7, 2012.
1
2
Rt
CFB 1.4.2.2 Suppose c is regular. Let s (t) = 0 |c0 (τ )| dτ and consider c parameterized
by s instead of t. Since s gives the length of the image c : [0, s] → E2 , s is called an
1
arclength parameter. Show that in this preferred parameterization, κ (s) = he2 , de
i.
ds
Proof:
Recall that c is regular if and only if c0 (t) 6= 0. Thus v (t) = |c0 (t)| =
6 0 and hence we may
divide by v (t) without issue. With this in mind we calculate
de1
de1 ds
=
by the chain rule
dt
ds dt
Z
de1 d t 0
=
|c (t)| dt
ds dt 0
de1 0
|c (t)| by the fundamental theorem of calculus.
=
ds
de1
v (t)
=
ds
However, dedt1 = λ (t) e2 . So we have
de1
λ (t)
=
e2 = κ (t) e2
ds
v (t)
1
, e2 i = κ (t).
Thus h de
ds
CFB 1.4.2.3 Show that κ (t) is constant if and only if the curve is an open subset of a line
(if κ = 0) or a circle of radius κ1 .
Proof:
We treat the line and the circle cases separately. First the line.
Claim 0.2. c is an open subset of a line if and only if κ = 0.
Proof:
(⇒). If c is an open subset of a line then we may parameterize c by c (t) = (a1 t + b1 , a2 t + b2 ).
0
Differentiating we find
p c (t) = (a1 , a2 ).
c(t)
But v (t) = |c0 (t)| = a21 + a22 and hence e1 = v(t)
= √ 21 2 (a1 , a2 ).
a1 +a2
We immediately see that e1 is constant and hence dedt1 = 0.
Since e2 6= 0 and dedt1 = λ (t) e2 we can conclude e2 = 0.
Moreover, c is an open subset of a line and hence a1 , a2 cannot both vanish and so
v (t) 6= 0.
Therefore κ (t) = λ(t)
= 0.
v(t)
(⇐) Suppose κ (t) = 0.
3
Given the definition of κ (t) we immediately conclude that λ (t) = 0 and hence dedt1 = 0.
Applying the definition of e1 we can conclude that c0 (t) is constant and thus is expressible as (a1 , a2 ) ∈ R2 .
Integrating gives c (t) = (a1 t + b2 , a2 t + b2 ).
Claim 0.3. c is an open subset of a circle of radius
nonzero.
1
κ
if and only if κ is constant and
Proof:
(⇒) Supposing c is an open subset of a circle of radius κ1 . Then since we’re after computing κ which is a differential invariant we may translate and rotate the circle without
affecting κ. Thus c may be parameterized as c (t) = κ1 (cos (at) , sin (at)). Of course κ
is parameterization invariant and so we may assume, without loss of generality, that
a = 1.
Differentiating we find c0 (t) = κ1 (− sin (t) , cos (t)) and v (t) = |c0 (t)| = κ1 .
0
Thus e1 (t) = |cc0 (t)
= (− sin (t) , cos (t)) from which it follows that e2 = (− cos (t) , − sin (t)).
(t)|
Differentiating e1 we find
de1
= (− cos (t) , − sin (t)) = e2 (t)
dt
1
Therefore λ (t) = 1 and so κ (t) = λ(t)
= 1/κ
= κ.
v(t)
(⇐) Suppose that κ is a nonzero constant.
Then taking a unit speed parameterization we have dedt1 = κe2 .
2
Differentiating again we find ddte21 = κ dedt2 since κ is constant.
However, 0 = dtd 1 = dtd he2 , e2 i = 2h dedt2 , e2 i and 0 = dtd 0 = dtd he1 , e2 i = h dedt1 , e2 i +
h dedt2 , e1 i = κ + h dedt2 , e1 i. Therefore dedt2 = −κe1 . This result should be familiar to those
that have experience with Frenet frames.
2
Therefore, ddte21 = −κ2 e1 .
This is a familiar equation and integrating it reveals e1 = Aei(κt+ϕ0 ) . Since e1 has
magnitude 1 we may take A = 1.
Integrating again to find c (t) gives c (t) = κ1 ei(κt+ϕ0 ) .
4
CFB 1.4.2.4 Let c (t) = (x (t) , y (t)) be given in coordinates. Calculate κ (t) in terms of
x (t), y (t), and their derivatives.
Proof:
Note: Below the t dependence will be suppressed to avoid notational
clutter.
q
2
First we compute the speed of this curve and find v = |c0 | = (x0 ) + (y 0 )2 .
0 0
This immediately gives e1 = xv , yv . Since e1 and e2 constitute an orthonormal frame in
0 0
2
E we may take e2 = −y
, xv .
v
Next we compute v 0 and find
x0 x00 + y 0 y 00
v0 =
v
Applying this to compute
de1
dt
de1
=
dt
gives
x00 x0 v 0 y 00 y 0 v 0
− 2 ,
− 2
v
v
v
v
x00 (y 0 )2 − x0 y 0 y 00 y 00 (x0 )2 − y 0 x0 x00
=
,
v3
v3
0 00
x y − y 0 x00
=
e2
v2
Thus λ =
x0 y 00 −y 0 x00
v2
and so κ =
λ
v
=
x0 y 00 −y 0 x00
3/2
((x0 )2 +(y0 )2 )
!
.
CFB 1.4.2.5 Calculate the function κ (t) for an ellipse. Characterize the points on the ellipse
where the maximum and minimum values of κ (t) occur.
Proof:
Since κ is a differential invariant we may assume, without loss of generality, that c (t) may
be parameterized as c (t) = (x, y) = (c1 cos (t) , c2 sin (t)) with c1 ≥ c2 > 0.
Now if c1 = c2 then c parameterizes a circle and so by CFB 1.4.2.3 we know that the
curvature is constant. Thus we will henceforth take c1 > c2 .
To make use of the results of CFB 1.4.2.4 we compute the following derivatives
x0 = −c1 sin (t)
x00 = −c1 cos (t)
y 0 = c2 cos (t)
y 00 = −c2 sin (t)
5
Applying the previous problem gives
κ (t) =
c1 c2
3/2
c21 cos2 (t) + c22 sin2 (t)
Differentiating κ to find the extrema gives
0 = κ0 =
−3c1 c2 0 00
(x x + y 0 y 00 )
2v 7/2
q
where v = (x0 )2 + (y 0 )2 .
1 c2
is defined everywhere and is nonzero. Thus extremal
Examination of v reveals that −3c
2v 7/2
0 00
0 00
curvatures occur only for x x = −y y .
Applying our above derivatives this condition reads
0 = c21 − c22 sin (t) cos (t)
Since c1 > c2 we must have sin (t) = 0 or cos (t) = 0 and so t ∈ 0, π2 , π, 3π
.
2
An application of the second derivative test reveals that 0 and π are maxima and π2 and
3π
are minima. This makes intutive sense, since at t = 0 and t = π we are along the
2
semi-major axis and the curve is bending the most while at t = π2 and t = 3π
we are along
2
the semi-minor axis where the ellipse is flattest.
Lee 2-1c Compute the coordinate representation of F : S3 → S2 given by F (z, w) =
(zw + wz, iwz − izw, zz − ww) in stereographic coordinates.
Use this result to prove
that2 F
2
2
3
2
is smooth. Note that here we think of S as the subset (w, z) ∈ C | |z| + |w| = 1 ⊂ C .
Proof:
Identifying C2 with R4 via (x1 + ix2 , x3 + ix4 ) ↔ (x1 , x2 , x3 , x4 ) we have
1 3
1
2
3
4
1
2
3
4
2 4
2 3
1 4
1 2
2 2
3 2
4 2
f x , x , x , x = F x + ix , x + ix = 2x x + 2x x , 2x x − 2x x , x + x − x − x
Since R4 and C2 are diffeomorphic it suffices to show that f is smooth.
Next recall that stereographic projection was given by
1
1
3
x
,
.
.
.
,
x
defined on S3 \ {(0, 0, 0, 1)}
4
1−x
1
1
2
3
1 2
2 2
3 2
x1 , x2 , x3 =
2x
,
2x
,
2x
,
x
+
x
+
x
−
1
(x1 )2 + (x2 )2 + (x3 )2 + 1
σ 3 x1 , . . . , x 4 =
σ3−1
and σ̃3 : S3 \ (0, 0, 0, −1) was defined by σ̃3 (x) = −σ3 (−x) and σ̃3−1 = −σ3−1 (−x).
We have similar functions on S2 which will be denoted by σ2 and σ̃2 .
6
With these functions direct calculation reveals
2x1
2
−1
1
2
3
,x
σ2 ◦ f ◦ σ3 x , x , x =
(x1 )2 + (x2 − 1)2 + (x3 )2
2x1
x2
−1
1
2
3
σ2 ◦ f ◦ σ̃3 x , x , x =
,
(x1 )2 + (x2 − 1)2 + (x3 )2 (x1 )2 + (x2 )2 + (x3 )2
2x1
2
−1
1
2
3
,x
σ̃2 ◦ f ◦ σ̃3 x , x , x =
(x1 )2 + (x2 + 1)2 + (x3 )2
2x1
x2
−1
1
2
3
σ̃2 ◦ f ◦ σ3 x , x , x =
,
(x1 )2 + (x2 + 1)2 + (x3 )2 (x1 )2 + (x2 )2 + (x3 )2
These are all rational functions and hence are smooth on their domains of definition.
Therefore, f is smooth.
But f is merely F under the C2 ∼
= R4 diffeomorphism and hence F is smooth.
Lee 2-6 For any topological space M, let C ∞ (M) denote the algebra of continuous functions
f : M → R. If F : M → N is a continuous map, define F ∗ : C ∞ (N ) → C ∞ (M) by
F ∗ (f ) = f ◦ F .
(a) Show that F ∗ is a linear map.
(b) If M and N are smooth manifolds, show that F is smooth if and only if F ∗ (C ∞ (N )) ⊂
C ∞ (M).
(c) If F : M → N is a homeomorphism between smooth manifolds show that it is a
diffeomorphism if and only if F ∗ restricts to an isomorphism from C ∞ (N ) to C ∞ (M).
Proof:
Claim 0.4. F ∗ is a linear map.
Proof:
Let f, g ∈ C ∞ (N ) and a, b ∈ R then
F ∗ (af + bg) = (af + bg) ◦ F = (af ) ◦ F + (bg) ◦ F = a (f ◦ F ) + b (g ◦ F ) = aF ∗ f + bF ∗ g
Since f, g, a, b were arbitrary we can conclude that F ∗ is a linear map.
Claim 0.5. If M and N are smooth manifolds, show that F is smooth if and only if
F ∗ (C ∞ (N )) ⊂ C ∞ (M).
Proof:
7
(⇒) Suppose F is smooth and f ∈ C ∞ (N ).
Then F ∗ f = f ◦ F : M → R is a composition of smooth maps and hence is smooth by
Lee’s Lemma 2.4.
Thus F ∗ f ∈ C ∞ (N ) for any f ∈ C ∞ (N ). In other words F ∗ (C ∞ (N )) ⊂ C ∞ (M).
(⇐). Suppose that F ∗ (C ∞ (N )) ⊂ C ∞ (M).
Fixing m ∈ M and f ∈ C ∞ (N ).
Since F ∗ (C ∞ (N )) ⊂ C ∞ (M) we know that g := F ∗ f is smooth at m. Furthermore,
f is smooth at F (m).
Then by the inverse function theorem we know that there is a chart (U, ϕ) on N containing F (m) on which f locally has a smooth inverse.
So restricting to U we know that g = F ∗ f = f ◦ F and so F = f −1 ◦ g.
However, on U , f −1 ◦ g is smooth and so F is smooth on U and in particular at m.
Since this holds for any m ∈ M we can conclude that F is smooth.
Claim 0.6. If F : M → N is a homeomorphism between smooth manifolds show that it is
a diffeomorphism if and only if F ∗ restricts to an isomorphism from C ∞ (N ) to C ∞ (M).
Proof:
(⇒) Since F is a diffeomorphism we know that F −1 : N → M exists and is smooth.
∗
So by part (b) we can conclude that (F −1 ) C ∞ (M) ⊂ C ∞ (N ).
Furthermore, by (a) we know that F ∗ and (F ∗ )−1 are linear.
∗
∗
So it suffices to show that (F −1 ) = (F ∗ )−1 and that F ∗ (rep. (F −1 ) ) respects the
multiplicative structure on C ∞ (N ) (resp.C ∞ (M)). These results are achieved by direct
calculation, let f, g ∈ C ∞ (N ) then
∗
F −1 ◦ F ∗ f = (f ◦ F ) ◦ F −1 = f ◦ IdN = f
F ∗ (f · g) = (f · g) ◦ F = (f ◦ F ) · (g ◦ F ) = (F ∗ f ) · (F ∗ g)
∗
Thus (F −1 ) ◦ F ∗ = IdC ∞ (M) and F ∗ respects the product (point-wise multiplication)
∗
in C ∞ (N ). The calculations for (F −1 ) are completely analogous.
∗
Thus (F −1 ) = (F ∗ )−1 and F ∗ are algebra homomorphisms and hence C ∞ (N ) ∼
=
C ∞ (M).
(⇐). Suppose that C ∞ (N ) ∼
= C ∞ (M) and that F is a homeomorphism.
−1
Then F : N → M exists and is continuous. Furthermore, examining the (⇒) proof
∗
we see that (F −1 ) = (F ∗ )−1 did not rely on the smoothness of F or F −1 and so we
∗
can still conclude that (F −1 ) = (F ∗ )−1 .
∗
But (F −1 ) = (F ∗ )−1 : C ∞ (M) → C ∞ (N ) and F ∗ : C ∞ (N ) → C ∞ (M) are isomor∗
phisms by hypothesis and so (F −1 ) C ∞ (M) ⊂ C ∞ (N ) and F ∗ : C ∞ N ⊂ C ∞ M.
So we can apply (b) to conclude that F and F −1 are smooth whence F is a diffeomorphism.
8
Lee 2-8 Show that n : Rn → Tn defined in Example 2.8(d) is a smooth covering map.
Proof:
We first need to show that n is smooth. If the book is to be believed then n is a Lie
algebra homomorphism and hence is smooth. However, the book asserts this without proof
and so we will show smoothness for completeness.
To do this note that n (t1 , . . . , tn ) = (1 (t1 ) , . . . , 1 (tn )) and so it suffices to show that
1 : R → S1 is smooth.
To do this we recall that stereographic projection is given by σ : S1 \ {(0, 1)} → R and
σ̃ : S1 \ {(0, −1)} → R where
x1
1 − x2
2x x2 − 1
−1
,
σ (x) =
x2 + 1 x2 + 1
σ̃ x1 , x2 = −σ −x1 , −x2
σ x1 , x2 =
σ̃ −1 (x) = −σ −1 (−x)
Then
cos (2πt)
1 − sin (2πt)
cos (2πt)
σ̃ ◦ 1 (t) =
1 + sin (2πt)
σ ◦ 1 (t) =
which are smooth. Thus 1 is smooth from which we can conclude that n is smooth.
So it remains to show that n is a covering map. To do this it again suffices to show that
1 is a covering map since coordinate-wise n is given by the 1 .
So consider e2πit ∈ S1 and set δ = 1/4. Then
a
−1 e2πiτ | t − δ < τ < t + δ =
(t − δ + n, t + δ + n)
n∈Z
Restricting 1 to one component of this disjoint union (t − δ + n, t + δ + n) we see that maps it onto {e2πiτ | t − δ < τ < t + δ} since e2πit = e2πit+2nπi . Furthermore, ker 1 = Z
and so the restriction of 1 to (t − δ + n, t + δ + n) is injective.
Thus 1 : R → S1 is a smooth covering map and hence n : Rn → Tn is a smooth covering
map.
9
Lee 2-10Let CPn denote n-dimensional complex projective space, as defined in Problem 1-7.
(a) Show that the quotient map π : Cn+1 \ 0 → CPn is smooth.
(b) Show that CP1 is diffeomorphic to S2 .
Proof:
Claim 0.7. π : Cn+1 \ 0 → CPn is smooth
Proof:
Let f : R2n+2 → Cn+1 be the diffeomorphism defined by (x1 , y 1 , . . . , xn+1 , y n+1 ) 7→
(x1 + iy 1 , . . . , xn+1 + iy n+1 ). Furthermore, let (Ui , ϕi ) be the charts for CPn defined in
problem 1-7.
Then we have
where z j = xj + iy j
ϕi ◦ π ◦ f x1 , y 1 , . . . , xn+1 , y n+1 = ϕi ◦ π z 1 , . . . , z n
= ϕi z 1 , . . . , z n
1
z
z i−1 z i+1
zn
=
,··· , i , 1 ,..., i
zi
z
z
z
But each component of this result is a rational function and hence is smooth on its
domain of definition.
Since this holds for all i we can conclude that π is smooth.
Claim 0.8. CP1 is diffeomorphic to S2
Proof:
Let (U1 , ϕ1 ) and (U2 , ϕ2 ) be the charts on CP1 given by U1 = CP1 \ {[0 : 1]} and
U2 = CP1 \ {[1 : 0]} with ϕ1 ([z1 : z2 ]) = zz12 and ϕ2 ([z1 : z2 ]) = zz12 . These are of course
the charts from problem 1-7 and so we know that they constitute an atlas.
Next let V1 = S2 \(0, 0, 1) and V2 = S2 \(0, 0, −1). Then (V1 , ψ1 ) and (V2 , ψ2 ) constitute
an atlas were
a
b
+i
ψ1 (a, b, c) =
1−c
1−c
a
b
ψ2 (a, b, c) =
−i
1+c
1+c
This is sterographic projection in complex notation and so there is no need to check
that it is a smooth atlas (this was done in last weeks homework).
−1
We now define f : CP1 → S2 by f ◦ ϕ−1
i (z) = ψi (z).
To see that this function is well defined we note that ψ1−1 (z1 ) = ψ2−1 (z2 ) if and only
if z2 = ψ2 ◦ ψ1−1 (z1 ) = z1−1 = ϕ2 ◦ ϕ−1
1 (z1 ) if and only if ϕ1 (z1 ) = ϕ1 (z2 ). While
10
−1
ψi−1 (z1 ) = ψi−1 (z2 ) if and only if z1 = z2 if and only if ϕ−1
i (z1 ) = ϕi (z2 ) for all i.
Since f is well defined on all of CP−1 we need to show that f is smooth and has a
smooth inverse. To do this we note
ψi ◦ f ◦ ϕ−1
i (z) = z
1
ψj ◦ f ◦ ϕi−1 (z) =
for i 6= j and z 6= 0.
z
Furthermore, f −1 : S2 → CP1 is clearly given by f −1 ◦ ψi−1 (z) = ϕ−1
i (z).
−1
We know that f is well defined by arguments completely symmetric to those given
for f . Moreover, f ◦ f −1 = Id and f −1 ◦ f = Id by construction.
Finally, we see that f −1 is smooth since
ϕi ◦ f −1 ◦ ψi−1 (z) = z
1
for i 6= j and z 6= 0
ϕi ◦ f −1 ◦ ψj−1 (z) =
z
Thus f : CP1 → S2 is smooth with smooth inverse and hence these two manifolds are
diffeomorphic.