Download Solutions to homework problems

Solutions to homework problems
November 25, 2015
Contents
1 Homework Assignment # 1
1
2 Homework assignment #2
6
3 Homework Assignment # 3
9
4 Homework Assignment # 4
14
5 Homework Assignment # 5
20
6 Homework Assignment # 6
25
7 Homework Assignment # 7
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8 Homework Assignment # 8
37
9 Homework Assignment #9
41
10 Homework Assignment # 10
46
1
Homework Assignment # 1
1. Show that the -δ definition of continuity of a map f : X → Y between metric spaces is
equivalent to the definition of continuity in terms of open subsets.
Proof. Suppose f is -δ-continuous (i.e., continuous in the sense of the -δ definition of
continuity) and that U is an open subset of Y . We need to show that f −1 (U ) is an open
subset of X. The definition of an open subset U of a metric space implies that U can be
written as a union of open balls Bi , i ∈ I. Then
[
[
f −1 (U ) = f −1 ( Bi ) =
f −1 (Bi ),
i∈I
i∈I
and hence it suffices to show that the inverse image of a ball B in Y is an open subset of
X. So let B = B (y0 ) = {y ∈ Y | d(y, y0 ) < } and let x0 be an element of f −1 (B). Then
1
the continuity of f in the -δ sense implies that there is some δ > 0 such that Bδ (x0 ) ⊂
f −1 (B (y0 )). This shows that f is continuous.
Conversely, assume that f is continuous (as defined via open sets). To show that f is
-δ-continuous let x ∈ X and > 0. Then f −1 (B (f (x)) is an open subset of X containing
x. Hence there is some δ > 0 such that the ball Bδ (x) is contained in f −1 (B (f (x))). In
other words, f (Bδ (x)) ⊂ B (f (x)), which implies
∀y ∈ X
d(y, x) < δ ⇒ d(f (y), f (x)) < 2. a) Suppose T, T 0 are topologies on a set X which are generated by bases B and B0 ,
respectively. Show that the topologies agree if and only if the bases B, B0 are equivalent in
the sense that every B ∈ B is the unions of subsets belonging to B0 and every B 0 ∈ B0 is the
union of subsets belonging to B.
b) Show that the metric topology on Rn determined by the
P Euclidean metric is equal to
the metric topology determined by the `1 -metric d1 (x, y) = ni=1 |xi − yi |.
c) Show that the product topology on Rm × Rn (with each factor equipped with the
metric topology) agrees with the metric topology on Rm+n = Rm × Rn . Hint: Use the
freedom provided by part (b) to choose whether to work with the Euclidean metric or the
`1 -metric to make this easier.
Proof. Part (a). Suppose U ∈ T, i.e., U is an open subset of X w.r.t. the topology T. Then
U is the union of subsets Bi belonging to B. By assumption, each Bi is the union of subsets
belonging to B0 , and hence U is the union of subsets belonging to B0 and consequently
U ∈ T 0 . Analogously, U ∈ T 0 implies U ∈ T, and hence the topologies T and T 0 coincide.
Part (b). We recall that the metric topology on a set X equipped with a metric d(x, y) is
generated by the collection of subsets of X consisting of the open balls B (x) = {y ∈ X |
d(y, x) < } of any radius > 0 around any point x ∈ X. Specializing to X = Rn , let us
denote by B (x) (resp. B0 (x)) the open ball of radius around x ∈ Rn for the Euclidean
metric d (resp. the `1 -metric d1 ), and by B (resp. B0 ) the collection of all balls B (x) (resp.
B0 (x)). By part (a) it suffices to show that
1. every ball B (x) is a union of balls Bδ0 (y) ⊂ B (x), and
2. every ball B0 (x) is a union of balls Bδ (y) ⊂ B0 (x).
Comparing balls defined w.r.t. the Euclidean metric d with those w.r.t. the metric d1 will
require inequalities comparing the distances d(x, y) and d1 (x, y) for x, y ∈ Rn . We recall
that d(x, y) = ||x − y||, and d1 (x, y) = ||x − y||1 , where
q
||v|| := v12 + · · · + vn2
and
||v||1 := |v1 | + · · · + |vn |
is the standard norm (resp. the 1-norm) of a vector v ∈ Rn . These norms are related by the
inequalities
||v|| ≤ ||v||1
||v||1 ≤ n||v||.
2
The first inequality follows from
||v||2 = |v1 |2 + · · · + |vn |2 ≤ (|v1 | + · · · + |vn |)(|v1 | + · · · + |vn |) = ||v||21 .
The second from
||v||1 = |v1 | + · · · + |vn | ≤ n max |vi | = n
1≤i≤n
q
p
max |vi |2 ≤ n |v1 |2 + · · · + |vn |2 = n||v||.
1≤i≤n
We note that these inequalities imply
B0 (x) ⊆ B (x)
and
B/n (x) ⊆ B0 (x).
In particular, for y ∈ B (x) and δ := − d(y, x) > 0, we have
Bδ0 (y) ⊆ Bδ (y) ⊆ B (x).
(1.1)
Here the second inclusion is a consequence of the triangle inequality for the metric d, since
if z ∈ Bδ (y), then
d(z, x) ≤ d(z, y) + d(y, x) < δ + d(y, x) = ,
and hence z ∈ B (x). This proves (1). Statement (2) is proved similarly: assume y ∈ B0 (x)
and set δ := − d(y, x) > 0. Then
Bδ/n (y) ⊆ Bδ0 (y) ⊆ B0 (x),
where the second inclusion is a consequence of the triangle inequality for the metric d1 .
Part (c). For (x, y) ∈ Rm × Rn = Rm+n , the relationship between ||x||1 , ||y||1 and ||(x, y)||1
is given by
||(x, y)||1 = |x1 | + · · · + |xm | + |y1 | + · · · + |yn | = ||(x, y)||1 ,
which is simpler than the corresponding relationship between the Euclidean norms of these
vectors. This equality implies in particular
B0 1 (x) × B0 2 (y) ⊆ B0 1 +2 ((x, y))
and
B0 ((x, y)) ⊆ B0 (x) × B0 (y).
(1.2)
This motivates us to compare the d1 metric topology on Rm+n with the product topology of
Rm and Rn , with each factor equipped with the d1 metric topology. A basis for the metric
topology for Rm+n is given by the balls B0 ((x, y)) ⊂ Rm+n . A basis for the product topology
is given by U × V , where U ⊂ Rm , V ⊂ Rn are subsets which are unions of open balls in
Rm and Rn , respectively. In particular, any such product U × V is a union of subsets of the
form B0 1 (x) × B0 2 (y) which shows that these products of balls form a basis for the product
topology. As in part (b), the inclusions (1.2) imply that every open ball B0 ((x, y)) is a union
of product of balls, and that every product of balls is a union of balls. By part (a) this
implies that the topologies they generate agree.
3. Let X, Y1 , Y2 be topological spaces. Then the projection maps pi : Y1 × Y2 → Yi are
continuous and a map f : X → Y1 × Y2 is continuous if and only if the component maps
fi = pi ◦ f are continuous for i = 1, 2.
3
Proof. If U1 is an open subset of Y1 then p−1
1 (U1 ) = U1 ×Y2 is an open subset of the Cartesian
product Y1 × Y2 and hence p1 is continuous; similarly for p2 .
Now assume that f : X → Y1 ×Y2 is continuous. Then the i-th component map fi = pi ◦f
is continuous as a composition of the continuous maps pi and f .
Conversely, let us assume that the component maps f1 and f2 are continuous. To show
that f is continuous, it suffices to show that f −1 (B) is open for an open subset B ⊂ Y1 × Y2
belonging to the basis B for the product topology consisting S
of products B = U1 × U2 of
open sets S
Ui ⊂ Yi (since any open set is of the form U = Bi for Bi ∈ B and hence
f −1 (U ) = f −1 (Bi ) which is an open subset of X, provided the subsets f −1 (Bi ) are open).
Now
f −1 (U × V ) = f1−1 (U ) ∩ f2−1 (V )
is the intersection of open subsets of X (since fi is continuous and Ui is an open subset of
Yi ) and hence open.
2
4. Show that the map f : GLn (R) → GLn (R), A 7→ A−1 is continuous (here GLn (R) ⊂ Rn
is equipped with the metric topology).
Proof. We recall from a lemma in class that f is continuous if and only if the composition
i
f
GLn (R) −→ GLn (R) ,→ Rn
2
is continuous, where i is the inclusion map. Moreover, this composition is continuous if and
2
only if all component maps pij ◦ i ◦ f are continuous (thinking of Rn as the space of all
2
n × n-matrices, pij for 1 ≤ i, j ≤ n is the projection map that takes a matrix M ∈ Rn to
its component Mij ∈ R). We conclude that it suffices to show that for 1 ≤ i, j ≤ n the map
GLn (R) → R, A 7→ (A−1 )ij is continuous.
We recall from linear algebra that the inverse of A can be calculated by the formula
A−1 =
Ct
,
det(A)
where det(A) is the determinant of A, and C t is the transpose of the n × n matrix C whose
entry Cij is (−1)i+j times the (i, j)-minor of A (the determinant of the (n − 1) × (n − 1)
matrix that results from deleting row i and column j of A).
, where p(A) and q(A) are
This shows that each matrix entry (A−1 )ij is of the form p(A)
q(A)
polynomial functions of the matrix entries of A. In particular the functions p, q : GLn (R) → R
are continuous. Since q(A) = det(A) in non-zero for all A ∈ GLn (R), we conclude with a
lemma from class that the function p(A)
is continous.
q(A)
5. Consider the following topological spaces
• The subspace T1 := {v ∈ R3 | d(v, S) = r} ⊂ R3 equipped with the subspace topology,
where S = {(x, y, 0) ∈ R2 | x2 + y 2 = 1} and 0 < r < 1.
• The product space T2 := S 1 × S 1 equipped with the product topology.
• The quotient space T3 := ([−1, 1] × [−1, 1])/ ∼ equipped with the quotient topology,
where the equivalence relation is generated by (s, −1) ∼ (s, 1) and (−1, t) ∼ (1, t).
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These three spaces are homeomorphic, but we don’t yet have the tools to prove this in an
efficient way. For this homework problem construct two continuous bijections between pairs
of these spaces. Hint: An important part of this problem is to figure out which of these
three spaces should be the (co)domains of the two maps you are looking for. Of course,
any bijection has an inverse, but determining continuity is a different matter. Think about
which type of topological space (subspace, product space, quotient space) you would prefer
to have as domain (resp. codomain) for a map whose continuity you want to prove.
Proof. We recall from class that checking continuity of a map f : X → Y is easy if the domain
X is a quotient space and the range Y is a subspace or a product space. This suggests to
construct continuous bijections
f : T3 → T1
g : T3 → T2 .
and
We first discuss the simpler map g given by the formula
g([s, t]) := ((cos πs, sin πs), (cos πt, sin πt)) ∈ S 1 × S 1 ⊂ R2 × R2 .
(1.3)
To argue that this is a well-defined bijection, we note that the map [−1, 1] → S 1 , s 7→
(cos πs, sin πs) is surjective, and its only failure to be injective is due to the fact that it maps
s = −1 and s = +1 to the same point in S 1 . The map above when precomposed with the
projection map [−1, 1] × [−1, 1] → T3 is the product of two copies of this map. Hence it is
surjective, and its only failure of injectivity comes from points (s, t) with s = ±1 or t = ±1.
This shows that g is in fact a bijection. To prove continuity of g it suffices to prove continuity
of the composition
[−1, 1] × [−1, 1]
pr
/
g
[−1, 1] × [−1, 1]/{±1}
/
i
S1 × S1
/
R2 × R2 = R4 ,
where pr is the projection map, the i is the inclusion map. This map is continuous, since all
its component functions, given explicitely in equation (1.3) are continuous.
Now we define the map f , first describing it geometrically. From this point of view, it
will be obvious that f is bijective. Then we will derive an explicit formula for f , which will
make it easy to argue that f is continuous.
We note that there is a surjective map p : T1 → S which sends a point v ∈ R3 to the
unique point (x, y, 0) ∈ S which is closest to v. Similarly, there is a surjective map
q : T3 = ([−1, 1] × [−1, 1])/ ∼ −→ [−1, 1]/{±1}
given by
[s, t] 7→ [s].
We want to construct the map f such that it fits into a commutative diagram
T3 = ([−1, 1] × [−1, 1])/ ∼
f
/ T1
q
,
p
f0
[−1, 1]/{±1}
/
S
where f0 is the homeomorphism we discussed in class that sends [s] to (cos πs, sin πs, 0) ∈
S ⊂ R3 . We note that the fiber q −1 ([s]) for any fixed [s] ∈ [−1, 1]/{±1} is a circle. Similarly,
for any point (x, y, 0) of the circle S ⊂ R3 , the fiber p−1 (x, y, 0) ⊂ T1 is a circle of radius r
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in the plane spanned by the unit vectors (x, y, 0) and (0, 0, 1) with center (x, y, 0). We note
that the map
[−1, 1]/{±1} −→R3
[t] 7→(x, y, 0) + r cos πt(x, y, 0) + r sin πt(0, 0, 1)
= ((1 + r cos πt)x, (1 + r cos πt)y, r sin πt)
is a bijection onto the circle p−1 (x, y, 0). In particular, defining f on each fiber q −1 ([s]) to
be this map, we obtain a bijection between T3 and T1 .
To argue that f is continuous, we write down f : T3 → T1 explicitly: f0 sends a point
[s] ∈ [−1, 1]/{±1} to (x, y, 0) = (cos πs, sin πs, 0) and hence
f ([s, t]) =((1 + r cos πt)x, (1 + r cos πt)y, r sin πt)
=((1 + r cos πt) cos πs, (1 + r cos πt) sin πs, r sin πt).
(1.4)
Then f is continuous if and only if the composition
[−1, 1] × [−1, 1]
pr
/
[−1, 1] × [−1, 1]/{±1}
f
/ T1
i
/
R3
is continuous. This is continuous since its component functions are, which can be read off
from equation (1.4).
2
Homework assignment #2
1. Show that the quotient space Dn /S n−1 is homeomorphic to the sphere S n .
Proof. Generalizing the case n = 1 which we did in class, we define a map
f : Dn −→ R ⊕ Rn = Rn+1
by
f (v) := (cos π||v||, v
sin π||v||
)
||v||
We recall from Calculus that t → sintπt is a continuous function on R \ {0} which extends to
a continuous function on all of R (by defining its value at t = 0 to be π). It follows that f
is a continuous function since all its components are continuous.
It’s easy to check that f (v) belongs to S n ⊂ Rn+1 : f (0) = (0, 1) ∈ S n , and for v 6= 0 we
have
||f (v)||2 = cos2 π||v|| + sin2 π||v||2 = 1
We conclude that the map Dn → S n , v 7→ f (v) is continuous w.r.t. the subspace topology
on S n since its composition with the inclusion map i : S n → Rn+1 is the map f . Abusing
notation, we write again f : Dn → S n .
We note that f maps any unit vector v ∈ Dn to the point (−1, 0) ∈ Rn ⊕ R. Hence f
induces a well-defined map
f¯: Dn /S n−1 −→ S n
[v] 7→ f (v)
This map is continuous since its composition with the projection map p : Dn → Dn /S n−1 is
the map f which is continuous. We won’t present the details that f¯ is bijective (if you can
come up with a map which in fact is a bijection, I believe that you could prove that it is).
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So f¯: Dn /S n−1 → S n is a continuous bijection, and hence it is a homeomorphism, since
its range is Hausdorff (since S n is a metric space), and its domain is compact (Dn ⊂ Rn is
closed and bouneded and hence compact by the Heine-Borel Theorem; it follows that the
quotient space Dn /S n−1 is compact).
2. Use the Heine-Borel Theorem to decide which of the topological groups GLn (R), O(n),
SO(n) are compact. Provide proofs for your statements. Hint: A strategy often useful
for proving that a subset C of Rn is closed is to show that C is of the form f −1 (C 0 ) for
some closed subset C 0 ⊂ Rk (often C 0 consists of just one point) and some continuous map
f : Rn → Rk .
2
Proof. The general linear group GLn (R) ⊂ Mn×n (R) = Rn is not compact, since it is not
bounded. To see this, consider the sequence of diagonal matrices


i
 1



Ai = 
.
.
.


.
1
2
Considered as a vector in Rn , the norm squared of Ai is given by
||Ai ||2 = i2 + 12 + · · · + 12 = i2 + n − 1
2
which shows that GLn (R) is an unbounded subset of Rn .
We claim that O(n) and SO(n) are compact. We begin by showing that they are bounded
2
subspaces of Rn . So let A ∈ O(n) be a matrix with column vectors v1 , . . . , vn ∈ Rn . The
assumption that A belongs to O(n) means that the vi ’s form an orthogonal basis of Rn .
Hence
||A||2 = ||v1 ||2 + · · · + ||vn ||2 = n,
2
which shows that O(n) and hence also SO(n) ⊂ O(n) are bounded subsets of Rn .
2
To show that O(n) is a closed subset of Rn we note that a matrix A with column vectors
vi ∈ Rn belongs to O(n) if and only if the vi ’s are orthogonal, i.e., if
hvi , vj i = δi,j
for all 1 ≤ i, j ≤ n,
where hv, wi denotes the scalar product of vectors v, w, and the delta symbol δi,j is defined
by declaring δi,i = 1, and δi,j = 0 for i 6= j. More elegantly, this can be rephrased by saying
A ∈ O(n) if and only if At A = I, where At is the transpose of A, and I is the identity
matrix. In particular, if we define
f : Mn×n −→ Mn×n
by
A 7→ At A,
then O(n) = f −1 (I). Since f is continuous (since each component is), and the one-element
2
2
subset {I} ⊂ Mn×n = Rn is closed, it follows that O(n) is a closed subset of Rn .
To show that SO(n) is a closed subset, we note that SO(n) = O(n) ∩ SLn (R), where
SLn (R) is the special linear group consisting of all n × n matrices with determinant 1. In
other words, SLn (R) = det−1 (1), where det : Mn×n → R is the map that sends a matrix to
its determinant. Since the determinant is a polynomial function of the entries of the matrix,
2
this is a continuous function, and hence SLn (R) = det−1 (1) is a closed subset of Rn . This
implies that the intersection SO(n) = O(n) ∩ SLn (R) is closed as well.
7
4. Let X be a topological space which is the union of two subspaces X1 and X2 . Let f : X →
Y be a (not necessarily continuous) map whose restriction to X1 and X2 is continuous.
(a) Show f is continuous if X1 and X2 are open subsets of X.
(b) Show f is continuous if X1 and X2 are closed subsets of X.
(c) Give an example showing that in general f is not continuous.
Remark. This result is needed to verify that various constructions (e.g., concatenations of
paths) in fact lead to continuous maps. In a typical situation, we have continuous maps
f1 : X1 → Y and f2 : X2 → Y which agree on X1 ∩ X2 and hence there is a well-defined map
(
f1 (x) x ∈ X1
f : X −→ Y
given by
f (x) =
f2 (x) x ∈ X2
The above result then helps to show that this map is continuous.
Proof. To prove part (a) let U ⊂ Y be open. Then Ui := fi−1 (U ) is an open subset of Xi ,
i.e., it is of the form Ui = Vi ∩ Xi , where Vi ⊂ X is open. If follows that Ui is an open subset
of X, and hence f −1 (U ) = f1−1 (U ) ∪ f2−1 (U ) = U1 ∪ U2 is an open subset of X, proving the
continuity of f .
To prove part (b) we note that f : X → Y is continuous if and only if f −1 (U ) is a closed
subset of X for every closed subset U ⊂ Y . Then repeating the previous sentences with open
replaced by closed provides a proof of part (b).
For part (c) consider the map f : X = R → Y = R given by f (t) = 0 if t ∈ (−∞, 0), and
f (t) = 1 if t ∈ [0, ∞). The restrictions of f to (−∞, 0) resp. [0, ∞) are constant and hence
continuous, but f is not.
5. Which of the topological groups GLn (R), O(n), SO(n) are connected? Hint: Use without
proof the fact that every element in SO(n) is represented by a matrix of block diagonal form
whose diagonal blocks are the 1 × 1 matrix with entry +1 and/or 2 × 2 rotational matrices
cos φ − sin φ
R=
.
sin φ cos φ
Here “block diagonal” means that all other entries are zero.
Proof. The determinant function det : Mn×n → R is a continuous function. Since the determinant for matrices in GLn (R) or O(n) is non-zero, its restriction to G = GLn (R), O(n)
provides a continuous map
f : G −→ R \ {0}.
It follows that G = f −1 ((−∞, 0)) ∪ f −1 ((0, ∞)) is a decomposition of G into a disjoint union
of open subsets. Both of these are not empty, since the determinant of the identity matrix
is 1, and the determinant of the diagonal matrix with diagonal entries (−1, 1, . . . , 1) is −1.
This shows that the spaces GLn (R) and O(n) are both not connected.
To prove that SO(n) is connected it suffices to show that SO(n) is path connected,
i.e., any two elements A, B ∈ SO(n) can be connected by a path. We will show this by
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constructing for every A ∈ SO(n) a path γA : [0, 1] → SO(n) with γA (0) = I (the identity
matrix) and γA (1) = A. This implies that any two points A, B ∈ SO(n) can be connected by
a path γAB , obtained by first taking a path from A to I, by running the path γA backwards,
and then taking the path γB from I to B. In formulas, the path γAB is given by
(
γA (1 − 2t) for t ∈ [0, 21 ]
γAB (t) =
γB (2t − 1) for t ∈ [ 12 , 1]
This is in fact a path (i.e., a continuous map) by the previous problem.
Let us first construct γA for n = 2, using the fact that every A ∈ SO(2) is a rotation,
given by a matrix of the form
cos θ − sin θ
A=
.
sin θ cos θ
We define
γA : [0, 1] −→ SO(2)
cos tθ − sin tθ
γA (t) :=
sin tθ cos tθ
by
which is continuous since it components are continuous functions of t, and has the desired
property γA (0) = I and γA (1) = A.
For a general dimension n, we use the fact that for any A ∈ SO(n) there is a choice
of basis for Rn such that the matrix representing the isometry A is of block diagonal form
as described in the hint. Replacing for each block Ri the rotation angle θi (which may be
different for the different blocks) by tθi we obtain a path γA : [0, 1] → SO(n) from I to A,
generalizing our construction in the n = 2 case.
3
Homework Assignment # 3
1. Show that the connected sum RP2 #RP2 of two copies of the real projective plane is
homeomorphic to the Klein bottle K.
Proof. Consider the topological spaces given by identifying egdes with the same label for the
disjoint union of polygons given by the following pictures.
b
b
a
a
b
≈
a
c
c
c
a
≈
b
c
a
b
c
a
b
≈
c
b
b
The left picture is our standard construction of the Klein bottle K. The second and third
picture shows the quotient obtained from two disjoint triangles by identifying edges with the
same label. The second space is homeomorphic to the first since identifying the two edges
labeled “c” in the second picture, we obtain the first. The third space is homeomorphic to
the second, since the top right triangle of picture 2 is just redrawn as the left triangle of
9
picture 3 (flipped over such that the a-edge of that triangle lines up with the a-edge of the
second triangle). The fourth space is homeomorphic to the third since it is obtained from
the third space by identifying the two edges labeled “a” in the third picture.
Reading off the word bbcc by starting at the top left corner the last picture and going
around counterclockwise, the topological space given by that picture is Σ(bbcc). From class
we know the homeomorphisms
Σ(bbcc) ≈ Σ(bb)#Σ(cc) ≈ RP2 #RP2 .
Composing all the homeomorphisms mentioned above we obtain the desired homeomorphism
between the Klein bottle K and the connected sum RP2 #RP2 .
2. a) Show that the surface of genus g
Σg = T # . . . #T
| {z }
g
and the connected sum
RP2 # . . . #RP2
{z
}
|
k
are both homeomorphic to the topological space Σ(W ) associated to a word W . What is
the word W in these two cases?
b) Calculate the Euler characteristic of these manifolds.
Proof. Part (a) From class we know that
• T ≈ Σ(aba−1 b−1 ), RP2 ≈ Σ(cc), and
• Σ(W )#Σ(W 0 ) if W, W 0 are words from disjoint alphabets.
This implies
−1
−1 −1
Σg = T # . . . #T ≈ Σ(a1 b1 a−1
1 b1 )# . . . #Σ(ag bg ag bg )
−1
−1 −1
≈ Σ(a1 b1 a−1
1 b1 . . . ag bg ag bg )
(3.1)
and
RP2 # . . . #RP2 ≈ Σ(a1 a1 )# . . . #Σ(ak ak ) ≈ Σ(a1 a1 . . . ak ak ).
(3.2)
Part (b). The homeomorphism 3.1 implies that Σg is the quotient space of a polygon with
−1
−1 −1
4g edges (the length of the word a1 b1 a−1
1 b1 . . . ag bg ag bg ) and edge identification encoded
by this word. After gluing, this polygon will provide a “pattern of polygons” on Σg with
one face and 2g edges (of the original 4g edges of the polygon, the pairs of edges with the
same label are glued, resulting in 2g edges after gluing). Each of the original 4g vertices is
indentified with any other by gluing, resulting in one vertex in the quotient space (this is
easy to check for the first four vertices corresponding to the first torus, and then arguing
inductively). It follows that
χ(Σg ) = #V − #E + #F = 1 − 2g + 1 = 2 − 2g.
10
The argument for RP2 # . . . #RP2 is completely analogous. The homeomorphism 3.2 implies
that this surface is the quotient space of a 2k-gon with edge identifications encoded in the
word a1 a1 . . . ak ak . Using the resulting “pattern of polygons” on RP2 # . . . #RP2 to calculate
its Euler characteristic, we obtain
χ(RP2 # . . . #RP2 ) = #V − #E + #F = 1 − k + 1 = 2 − k.
{z
}
|
k
3. Important examples of quotient spaces are orbit spaces of a group G acting on a topological
space X. We recall that a (left) of a group G on a set X is given by a map
G × X −→ X
typically written as
(g, x) 7→ gx,
such that g1 (g2 x) = (g1 g2 )x for g1 , g2 ∈ G, x ∈ X (associativity) and ex = x for e the
identity element element of G, x ∈ X (identity element property). Given a G-action on a
topological space X the orbit space denoted X/G is the quotient space X/ ∼ where two
elements x, y ∈ X are declared equivalent if and only if there is some g ∈ G with gx = y. In
particular, the equivalence class of x is the subset {gx | g ∈ G}, which is called the orbit of
x, and hence X/G is the space of orbits. Although we haven’t used this terminology, we’ve
already encountered orbit spaces, namely
RPn = S n /{±1}
and
CPn = S 2n+1 /S 1 .
and
S 1 × S 2n+1 → S 2n+1
,
(z, (z0 , . . . , zn )) 7→ (zz0 , . . . , zzn )
Here the actions are given by
{±1} × S n → S n
(t, (v0 , . . . , vn )) 7→ (tv0 , . . . , tvn )
where z ∈ S 1 ⊂ C and (z0 , . . . , zn ) ∈ S 2n+1 ⊂ Cn+1 .
(a) Consider the action Z2 × R2 → R2 , (m, n), (x, y) 7→ (x + m, y + n). Show that the
quotient space R2 /Z2 is homeomorphic to the torus, described as the quotient space
b
a
a
b
Use without proof the fact that this orbit space is Hausdorff (this will come up later this
semester).
(b) Consider the action G × R2 → R2 where G is the subgroup of the group of isometries of
the metric space R2 generated by the isometries g, h : R2 → R2 defined by
g(x, y) = (x + 1, y)
and
11
h(x, y) = (−x, y + 1)
Show that the quotient space R2 /G is homeomorphic to the Klein bottle, described as
the quotient space of the square [0, 1] × [0, 1] with edge identifications
b
a
a
b
Again, use without proof the fact that this orbit space is Hausdorff. Hint: Show that
every orbit can be represented by a point (x, y) ∈ [0, 1] × [0, 1]. To do this, it might be
helpful to argue that the composition ghgh−1 is the identity and to use this to show that
every element of G can be uniquely written in the form g m hn with m, n ∈ Z.
Proof. Part (a). We recall that the torus T is the quotient (I × I)/ ∼ where I = [0, 1] and
the equivalence relation is generated by (x, 0) ∼ (x, 1) and (0, y) ∼ (1, y). Consider the map
f : T = (I × I)/ ∼−→ R2 /Z2
[(x, y)] 7→ Z2 (x, y),
where we write Z2 (x, y) for the orbit of the Z2 -action through (x, y).
Next we will argue that f is well-defined, surjective, injective and continuous. This
implies that f is a homeomorphism since its domain is compact (as the image of the closed
bounded subset I × I ⊂ R2 under the projection map I × I → (I × I)/ ∼, and we use the
fact that R2 /Z2 is Hausdorff.
• The map f is well-defined since the points (x, 0), (x, 1) ∈ I ×I belong to the same orbit
of the Z2 -action on R2 , since (0, 1)(x, 0) = (x, 1). Similarly, (0, y) and (1, y) belong to
the same Z2 -orbit.
• For every (x, y) ∈ R2 the element (x − [x], y − [y]) belongs to I × I, where [x] ∈ Z is the
largest integer smaller or equal to x ∈ R. Since ([x], [y]) ∈ Z2 applied to (x−[x], y −[y])
gives (x, y), this shows that (x − [x], y − [y]) and (x, y) belong to the same orbit and
hence the map f is surjective.
• Two elements (x, y), (x0 , y 0 ) ∈ R2 belong to the same Z2 -orbit if and only if x − x0 ∈ Z
and y − y 0 ∈ Z. Hence the only points (x, y) ∈ I × I mapping to the same orbit are of
the form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I × I/ ∼, the
map f is injective.
• To prove continuity of f consider the commutative diagram
I ×I
p
(I × I)/ ∼
/
i
f
/
R2
.
q
R2 /Z2
The inclusion map i and the projection maps p, q are continuous, and so the composition q ◦ i is continuous. By commutativity of the diagram it follows that f ◦ p = q ◦ i
is continuous, and hence f is continuous.
12
Part (b). The proof is analogous to that of part (a). Explicitly the Klein bottle K is the
quotient (I × I)/ ∼ where the equivalence relation is generated by (x, 0) ∼ (1 − x, 1) and
(0, y) ∼ (1, y). Consider the map
f : T = (I × I)/ ∼−→ R2 /Z2
[(x, y)] 7→ G(x, y),
where we write G(x, y) for the orbit of the G-action through (x, y). We will prove this in
the same steps as in (a), but first we prove the following result.
Lemma 3.3. The G-orbit G(x, y) of a point (x, y) ∈ R2 is given explicitly by
G(x, y) = {(m + (−1)n x, n + y) | m, n ∈ Z}.
To prove this result we first verify that the relation ghgh−1 = 1 holds in G:
h−1
g
g
h
(x, y) 7→ (−x, −1 + y) 7→ (1 − x, −1 + y) 7→ (−(1 − x), 1 + (−1 + y)) = (−1 + x, y) 7→ (x, y)
A general element of G is a word made from the letters g, g −1 , h, h−1 . The relation just
proved implies hgh−1 = g −1 which implies
h±1 g ±1 = g ∓1 h±1 .
In particular, we can move all factors g ±1 to the left of the factors h±1 , which shows that
every element of G can be written in the form g m hn for m, n ∈ Z. Since
g m hn (x, y) = g m ((−1)n x, n + y) = (m + (−1)n x, n + y),
this proves the lemma.
• The map f is well-defined since by the lemma the points (x, 0), (x, 1) ∈ I × I (resp.
the points (0, y) and (1, y)) belong to the same G-orbit.
• The map f is surjective since by the lemma for the orbit through any point (x, y) ∈ R2 ,
there is a unique point (x0 , y 0 ) ∈ [0, 1) × [0, 1) in the same orbit as (x, y).
For every (x, y) ∈ R2 the element (x − [x], y − [y]) belongs to I × I, where [x] ∈ Z is the
largest integer smaller or equal to x ∈ R. Since ([x], [y]) ∈ Z2 applied to (x−[x], y −[y])
gives (x, y), this shows that (x − [x], y − [y]) and (x, y) belong to the same orbit and
hence the map f is surjective.
• Two elements (x, y), (x0 , y 0 ) ∈ R2 belong to the same Z2 -orbit if and only if x − x0 ∈ Z
and y − y 0 ∈ Z. Hence the only points (x, y) ∈ I × I mapping to the same orbit are of
the form (0, y), (1, y) or (x, 0), (x, 1). Since these points are identified in I × I/ ∼, the
map f is injective.
• To prove continuity of f consider the commutative diagram
I ×I
p
(I × I)/ ∼
/ R2
i
f
/
.
q
R2 /Z2
The inclusion map i and the projection maps p, q are continuous, and so the composition q ◦ i is continuous. By commutativity of the diagram it follows that f ◦ p = q ◦ i
is continuous, and hence f is continuous.
13
4. a) Show that a subspace of a Hausdorff space is Hausdorff and that a product of Hausdorff
spaces is Hausdorff.
b) Show that a subspace of a second countable space is second countable and that a product
of second countable spaces is second countable.
Proof. Part (a). Suppose A ⊂ X is a subspace of a Hausdorff space X. To show that
A is Hausdorff, let x, y ∈ A be distinct points. Since X is Hausdorff, there are disjoint
open subsets U, V ⊂ X with x ∈ U and y ∈ V . Then U ∩ A and V ∩ A are disjoint open
neighborhoods of x resp. y in A, proving that A is Hausdorff.
Suppose X, Y are Hausdorff spaces. To show that X ×Y is Hausdorff, let (x, y), (x0 , y 0 ) ∈
X × Y be distinct points. Without loss of generality we may assume x 6= x0 . This allows
us to find disjoint open neighborhoods U, V ⊂ X of x resp. x0 . Then U × Y and V × Y are
disjoint open neighborhoods of (x, y) resp. (x0 , y 0 ), which proves that X × Y is Hausdorff.
Part (b). Let A be a subspace of a second countable space X. Let {Ui }i∈I be a countable
basis for X. We claim that the countable collection {Ui ∩ A}i∈I is a basis for A, thus proving
that A is second countable. To prove the claim, we recall that every open subset of A is of
the form U ∩ A for some open subset U . Now U can be expressed as a union of Ui ’s for i
belonging to some subset J ⊂ I. It follows that
[
[
U ∩ A = ( Ui ) ∩ A = (Ui ∩ A),
i∈J
i∈J
proving that {Ui ∩ A}i∈I is a basis for the topology of A.
Let X, Y be second countable spaces with countable bases {Ui }i∈I and {Vj }j∈J , respectively. We claim that the countable collection {Ui × Vj }(i,j)∈I×J is a basis for X × Y , thus
showing that X × Y is second countable. To prove the claim, we need to show that every
open neighborhood W of (x, y) ∈ X ×Y contains a neighborhood of (x, y) of the form Ui ×Vj .
By definition of the product topology, W contains a neighborhood of the form U × V , where
U is an open neighborhood of x ∈ X and V is an open neighborhood of y ∈ Y . Since the
Ui ’s form a basis for the topology of X, this implies that there is some Ui with x ∈ Ui and
Ui ⊂ U . Similarly, we find some Vj with y ∈ Vj ⊂ V . Then
(x, y) ∈ Ui × Vj ⊂ U × V ⊂ W
which finishes the proof.
4
Homework Assignment # 4
1. Let ωn : (I, ∂I) → (S 1 , 1) be the based loop defined by ωn (s) = e2πins . Show that the map
Φ : Z → π1 (S 1 , 1) given by n 7→ [ωn ] is a group homomorphism.
Proof. We note that
Φ(m + n) = [ωm+n ],
and
Φ(m)Φ(n) = [ωm ][ωn ] = [ωm ∗ ωn ]
14
and so we need to construct a homotopy relative endpoints between ωm+n and the concatenation ωm ∗ ωn (note that they are not the same path unless m = n: both wind around the
circle m + n times, but ωm+n goes around the circle at a constant speed, while ωm ∗ ωn winds
around the circle n times on the interval [0, 1/2], while its winds around m times on [1/2, 1]).
To produce the desired homotopy we note that ωm+n = p ◦ α1 and ωm ∗ ωn = p ◦ α2 , where
p : R → S 1 is defined by p(s) = e2πis , and the paths αi : [0, 1] → R are given by
(
2ns
0 ≤ s ≤ 1/2
α1 (s) = (m + n)s
α2 (s) =
2ms − m + n 1/2 ≤ s ≤ 1
We note that paths α1 and α2 have the same start/end point and hence they are homotopic
via the linear homotopy Ht connecting them. Then p ◦ Ht is the desired homotopy between
the paths ωm+n = p ◦ α1 and ωm ∗ ωn = p ◦ α2 .
2. Let γ : I → X be a path in a topological space X, and let γ̄ : I → X be the path γ run
backwards, that is, γ̄(s) = γ(1 − s). Then there are the following homotopies
γ̄ ∗ γ ' cγ(0)
γ ∗ γ̄ ' cγ(1)
γ ∗ cγ(0) ' γ
cγ(1) ∗ γ ' γ,
(1)
where cx for x ∈ X denotes the constant path at x.
(a) Prove the first and the third homotopy.
(b) Use the homotopies (1) to finish our proof that π1 (X, x0 ) is a group. In other words,
show that the constant map cx0 gives an identity element for π1 (X, x0 ), and that [γ̄] is
the inverse for [γ] ∈ π1 (X, x0 ).
(c) Let β be a path from x0 to x1 . Show that the map
Φβ : π1 (X, x0 ) −→ π1 (X, x1 )
[γ] 7→ [β ∗ γ ∗ β̄]
is an isomorphism of groups.
Proof. Part (a). The strategy of constructing these homotopies is analoguous to the strategy
used in problem 1. We note that
γ̄ ∗ γ = γ ◦ α1
and
cγ(0) = γ ◦ α2 ,
where α1 , α2 : [0, 1] → [0, 1] are the paths in [0, 1] given by
(
2s
0 ≤ s ≤ 21
and
α1 (s) =
2(1 − s) 21 ≤ s ≤ 1
α2 (s) = 0.
Let Ht be the linear homotopy between α1 and α2 . This is a homotopy of paths since the
initial and terminal points of the paths α1 and α2 agree. This implies that γ ◦ Ht is the
desired homotopy between γ ◦ α1 = γ̄ ∗ γ and γ ◦ α1 = cγ(0) .
Similarly,
γ ∗ cγ(0) = γ ◦ β1
and
γ ∗ cγ(0) = γ ◦ β2 ,
15
where β1 , β2 : [0, 1] → [0, 1] are the paths given by
(
0
0 ≤ s ≤ 12
β1 (s) =
2(s − 1) 12 ≤ s ≤ 1
and β2 (s) = 0. As before, the linear homotopy Gt between these maps is a path homotopy
since β1 (0) = β2 (0) and β1 (1) = β2 (1). It follows that γ ◦ Gt is a homotopy between γ ∗ cγ(0)
and γ as desired.
Part (b). Let γ be a loop in X, based at x0 ∈ X. Then the third and fourth homotopy in
part (a) implies the equalities in π1 (X, x0 )
[γ][cx0 ] = [γ]
[cx0 ][γ] = [γ].
This shows that [cx0 ] is the identity element for the multiplication in π1 (X, x0 ).
Similarly, the first and second homotopy from part (a) imply the equalities
[γ̄][γ] = [cx0 ]
[γ][γ̄] = [cx0 ]
in π1 (X, x0 ). These equalities imply that [γ̄] is the inverse of [γ]. In particular, π1 (X, x0 ) is
a group.
Part (c). First we show that Φβ : π1 (X, x0 ) −→ π1 (X, x1 ) is a homomorphism. So let γ, δ
be loops in X based at x0 . Then
Φβ ([γ][δ]) = Φβ ([γ ∗ δ]) = [β ∗ γ ∗ δ ∗ β̄] = [β ∗ γ ∗ β ∗ β̄ ∗ δ ∗ β̄]
= [β ∗ γ ∗ β̄][β ∗ δ ∗ β̄] = Φβ ([γ])Φβ ([δ])
Here we avoid using parantheses for the concatenation operation; this is OK by the fact
that the paths resulting from different ways of putting in parantheses leads to homotopic
paths. The first and second equality is just the definition of multiplication in π1 (X, x0 )
resp. the definition of Φβ . The third equation holds due to the homotopy β ∗ β̄ ' cx0 (an
application of the second homotopy of part (a)), and the fact that we can insert at any point
an appropriate constant path into an iterated concatenation without changing the homotopy
class of the path (follows from either the third or the fourth homotopy of part (a)). The last
two equations again hold by construction of multiplication in the fundamental group resp.
the map Φβ .
It remains to show that Φβ is an isomorphism. We claim that its inverse is given by
Φβ̄ : π1 (X, x1 ) → π1 (X, x0 ).
Φβ̄ Φβ (γ) = Φβ̄ ([β ∗ γ ∗ β̄]) = [β̄ ∗ β ∗ γ ∗ β̄ ∗ β] = [cx1 ∗ γ ∗ cx0 ] = [γ]
Here the first two equations is just by construction of Φβ resp. Φβ̄ , the third equation holds
thanks to the first two homotopies of part (a), and the last equation holds due to the last
two homotopies of part (a).
3. A subspace A ⊂ X of a topological space X is called a retract of X if there is a continuous
map r : X → A whose restriction to A is the identity.
16
(a) Show that S 1 is not a retract of D2 . Hint: Show that the assumption that there is a
continuous map r : D2 → S 1 which restricts to the identity on S 1 leads to a contradiction
by contemplating the induced map r∗ of fundamental groups.
(b) Brouwer’s Fixed Point Theorem states that every continuous map f : Dn → Dn has a
fixed point, i.e., a point x with f (x) = x. Prove this for n = 2. Hint: show that if f has
no fixed point, then a retraction map r : D2 → S 1 can be constructed out of f .
Proof. We prove part (a) be contradiction. Suppose that r : D2 → S 1 is a retraction of D2
onto S 1 . This means that r makes the following diagram commutative:
i /
D2
BB
BB
r
B
id BB! S1B
S1
Choosing a basepoint x0 ∈ S 1 and applying the fundamental group functor gives the commutative diagram
i∗ /
π1 (D2 , x0 )
NNN
NNN
NN
r∗
id∗ =id NNN'
π1 (S 1 , x0 )
π1 (S 1 , x0 )
This gives the desired contradiction since according to the diagram the identity map on
π1 (S 1 , x0 ) ∼
= Z factors through the trivial group π1 (D2 , x0 ).
To prove part (b) we assume that f : Dn → Dn does not have a fixed point, i.e., that for
all x ∈ Dn , f (x) 6= x. Then the line through the points f (x) and x intersects the sphere
S n−1 in two points. Let r(x) ∈ S n−1 be the intersection point closer to x. It is clear from the
construction that r : Dn → S n−1 has the property r(x) = x for x ∈ S n−1 , and so it remains
to show the continuity of r to derive the desired contradiction from part (a).
To show that r is continuous, we note that r(x) can be written in the form
r(x) = x + α(x)(x − f (x)),
where α(x) ∈ R is the unique non-negative solution of the quadratic equation
||x + α(x)(x − f (x))||2 = 1
(4.1)
(the solutions of this equation correspond to the two intersections of the line through x
and f (x) with S n−1 ; hence it is clear geometrically, that there are two solution for every
x, and that exactly one solution is non-negative). Writing out the quadratic equation (4.1)
explicitly as
||x − f (x)||2 α2 (x) + 2hx, x − f (x)iα(x) + ||x||2 − 1 = 0
shows that its coefficients are continuous functions of x, and hence the quadratic formula
shows that its non-negative solution α(x) is continuous function of x. We conclude that r(x)
is a continuous function of x since its components are linear combinations of products of
continuos functions.
17
4. We recall that the torus T and the Klein bottle K can both be described as quotient
spaces of the square by identifying edges with the same label as shown in the following
pictures.
b
T =
a
b
a
a
K=
b
a
b
Each edge of the square can be interpreted as a path in the square, which projects to a based
loop in the quotient space T resp. K. Here the base point x0 of the quotient space is the
projection of each of the four vertices of the square. Abusing language, let us denote by a
resp. b the elements in the fundamental group of T (resp. K) represented by these loops.
Prove the following identities:
aba−1 b−1 = 1 ∈ π1 (T, x0 )
abab−1 = 1 ∈ π1 (K, x0 ).
Proof. We begin with the torus T = ([0, 1] × [0, 1])/ ∼ with equivalence relation generated
by (0, y) ∼ (1, y) and (x, 0) ∼ (x, 1). For i = 1, 2, let αi , βi : I → [0, 1] × [0, 1] be the paths
in the square [0, 1] × [0, 1] given by its edges, with the direction provided by the arrow on
each edge, as shown in the picture
x
e0
[0, 1] × [0, 1] =
β1
α1
α2
β2
It should be emphasized that this is a precise description by adopting the convention that
all of these paths are the linear paths connecting their endpoints. Going counterclockwise
around the boundary of the square, we obtain the loop
γ := α1 ∗ β1 ∗ ᾱ2 ∗ β̄2 ,
based at the top left corner point x̃0 (strictly speaking we should put parentheses here, since
we are talking about paths, but we don’t want to burden the notation). Let p : I × I → T be
the projection map. We note that the projection p ◦ αi of the paths αi is a loop in T based
at the point x0 := p(x̃0 ). In fact, p ◦ α1 and p ◦ α2 are the same loop in T which we denote
by α, and [α] ∈ π1 (T, x0 ) is the generator called “a”. Similarly, projecting the paths βi we
obtain a loop β in T based at x0 with [β] = b ∈ π1 (T, x0 ). Hence
aba−1 b−1 = [α][β][ᾱ][β̄] = [α ∗ β ∗ ᾱ ∗ β̄] = [p ◦ (α1 ∗ β1 ∗ ᾱ2 ∗ β̄2 )] = [p ◦ γ].
Since γ is a loop based at x
e0 in the convex subset [0, 1]2 ⊂ R2 , the linear homotopy Ht
between this concatenation and the constant loop cxe0 is a homotopy relative endpoints (we
18
want to emphasize that the fact that γ is a loop is essential here; by contrast, the linear
homotopy between the path α1 and cxe0 is not a homotopy relative endpoints). It follows
that p ◦ Ht : I → T is a homotopy between p ◦ γ and the constant loop cx0 , which by (??)
implies the relation aba−1 b−1 = 1 in the fundamental group of the torus.
The argument in the case of the Klein bottle K is completely analogous. As before we
have paths αi , βi for i = 1, 2 in [0, 1]2 which project to loops α, β based at x0 = p(e
x0 ) ∈ K
under the projection map p : [0, 1]2 → ([0, 1]2 / ∼) = K. Here is the picture:
x
e0
[0, 1] × [0, 1] =
β1
α1
α2
β2
The only difference to the previous situation is that we have the loop
δ := α1 ∗ β1 ∗ α2 ∗ β̄2
in [0, 1]2 , which is homotopic to the constant loop cxe0 . The image of the loop δ under p is
the loop α ∗ β ∗ α ∗ β̄ which represents the element abab−1 in the fundemental group of the
Klein bottle. This proves the relation abab−1 = 1.
5. Let Z2 act on R2 by translation (as in problem # 3 of the last set). Show that the
projection p : R2 → R2 /Z2 is a covering map.
Proof. We note that for (x, y) ∈ R2 the subset p−1 (p(x, y)) ⊂ R2 is the orbit through (x, y)
(this is true in general for the projection map p : X → X/G to the orbit space for a G-action
on a topological space X). More explicitly:
p−1 (p(x, y)) = {(x + m, y + n) ∈ R2 | m, n ∈ Z}.
In particular, if U ⊂ R2 is an open subset, then
[
p−1 (p(U )) =
(m + n) + U
(4.2)
(m,n)∈Z2
is an open subset of R2 , since it is the union of the open subsets (m, n) + U given by the
image of U under the homeomorphism R2 → R2 , (x, y) 7→ (x + m, y + n). It follows that
p(U ) is an open subset of R2 /Z2 . In other words, p is an open map, which means that p
takes open subsets of its domain to open subsets of its range.
We claim that the open subset p(U ) ⊂ R2 /Z2 is evenly covered, provided the intersection
((m, n) + U ) ∩ U is empty for (m, n) 6= (0, 0). To prove the claim, assume that ((m, n) + U ) ∩
U = ∅ for (m, n) 6= (0, 0). This implies that the the sets (m, n) + U in the decomposition
(4.2) of p−1 (p(U )) are pairwise disjoint, since if (x, y) ∈ ((m, n) + U ) ∩ ((m0 , n0 ) + U ) for
(m, n) 6= (m0 , n0 ), then (x − m0 , y − n0 ) ∈ ((m − m0 , n − n0 ) + U ) ∩ U , contradicting our
assumption that this intersection is empty. This implies that the restriction of p to (m, n)+U
19
is a continuous bijection between (m, n) + U and p(U ). Moreover, this map is open (as the
restriction of the open map p), and hence it is a homeomorphism. This finishes the proof of
the claim above.
To finish the proof we note that for any (x, y) ∈ R2 the open ball U = B1/2 (x, y) of radius
1/2 around (x, y) does not intersect any translated ball (m, n) + U . In particular p(U ) is an
evenly covered open subset. Since every point of R2 /Z2 is contained in an open subset of
this form, this proves that p : R2 → R2 /Z2 is a covering map.
5
Homework Assignment # 5
1. Let f : S 1 → S 1 be defined by f (z) = z n for some n ∈ Z. Calculate the induced
homomorphism
f∗ : π1 (S 1 , 1) −→ π1 (S 1 , 1).
Clarification: We’ve proved that the fundamental group π1 (S 1 , 1) is isomorphic to Z. In
particular, any endomorphism of π1 (S 1 , 1) is given by multiplication by some integer k ∈ Z.
“Calculating f∗ ” means determining that integer k for the endomorphism f∗ ∈ End(π1 (S 1 , 1)).
Proof. We recall that the homotopy class [ω1 ] ∈ π1 (S 1 , 1) of the based loop ω1 : I → S 1 ,
defined by ω1 (s) = e2πis , is a generator of π1 (S 1 , 1). So it suffices to calculate f∗ [ω1 ] =
[f ◦ ω1 ] ∈ π1 (S 1 , 1):
f ◦ ω1 (s) = f (e2πis ) = (e2πis )n = (e2πins ) = ωn (s).
Hence f∗ ([ω1 ]) = [ωn ] = n[ω1 ] ∈ π1 (S 1 , 1), where the second equation was proved in problem
#1 of assignment # 4. This shows that the endomorphism f∗ is given by multiplication by
n ∈ Z.
e −→ X be a covering space such that for each x ∈ X the fiber p−1 (x) is countable.
2. Let p : X
e Hint: to show that X
e has
(a) Show that if X is a manifold of dimension n, then so is X.
a countable basis, first show that X has a countable basis consisting of evenly covered
open subsets.
(b) If X is compact manifold of dimension 2 and p is a d-fold covering map, what is the
e in terms of the Euler characteristic of X? Here a d-fold covering
Euler characteristic of X
means that each fiber p−1 (x) consists of d ∈ N points.
e −→ X is a d-fold covering, d ∈ N where X is a surface of genus g
(c) Suppose that p : X
e is a surface of genus ge. Give a formula expressing ge in terms of g and d.
and X
e let x = p(e
Proof. Part(a). Given a point x
e ∈ X,
x) ∈ X, and let U ⊂ X be an open
≈
neighborhood of x and φ : U −→ V be a homeomorphisms between U and an open subset
V ⊂ Rn . Replacing U by a smaller neighborhood of x if necessary, we may assume that U is
20
evenly covered. Then p−1 (U ) is the disjoint union of open subsets Uα such that p|Uα : Uα → U
is a homeomorphism. Then x
e is contained in some Uα , and the composition
Uα
p|Uα
/
φ
U
/
V ⊂ Rn
e is locally homeomorphic
provides a homeomorphism between Uα and V . This proves that X
to Rn .
e is Hausdorff, let x
e with x
To show that X
e, ye ∈ X
e 6= ye. If p(e
x) 6= p(e
y ), then we can find
disjoint open neighborhoods Ux , Uy ⊂ X of x := p(e
x) resp. y := p(e
y ), and then p−1 (Ux ),
p−1 (Uy ) are disjoint open neighborhoods
e resp. ye. If x = y, let U be an evenly covered
` of x
−1
neighborhood of x, and let p (U ) = α∈A Uα as above. If x
e belongs to Uα , then ye belongs
to Uβ for β 6= α, and hence Uα , Uβ are disjoint open neighborhoods of x and y.
e is second countable, let B0 be a countable basis for the topology of X,
To show that X
and let B be the subcollection of those open sets U ⊂ X belonging to B0 which are evenly
covered. We claim that this is a basis for the topology on X. To prove this, let x ∈ X and
let V be an open neighborhood of x. Intersecting with an evenly covered open set containing
x, we may assume that V is evenly covered. Since B0 is a basis, there is some U ∈ B0 with
x ∈ U and U ⊆ V . Since V is evenly covered, so is U , and hence U belongs to B. This
proves that B is a basis for the topology on X; as a subcollection of the countable collection
B0 it is countable.
e of X
e as follows. For each U ∈ B, pick a point
Now we construct a countable basis B
U
x ∈ U . By construction U is evenly covered, and hence p−1 (U ) is a disjoint union of open
e such that the restriction of p to each of these subsets is a homeomorphism. In
subsets of X,
particular, each of these subsets contains exactly one of the points x
e ∈ p−1 (xU ), allowing us
exe for the subset containing
to index these sets by the elements of p−1 (xU ): we will write U
−1 U
the point x
e ∈ p (x ). With this notation, we have
a
exe,
p−1 (U ) =
U
x
e∈p−1 (xU )
e be the following collection of subsets of X:
exe −→ U is a homeomorphism. Let B
e
and p|Uexe : U
e := {U
exe | U ∈ B, x
B
e ∈ p−1 (xU )}.
This is a countable collection, since B is countable, and each fiber p−1 (xU ) is countable.
e is a basis for X.
e To show this, it suffices to prove that for every ye ∈ X
e
We claim that B
e with ye ∈ U
e there is some U
exe ∈ B
exe and U
exe ⊆ Ve . Let
and open neighborhood Ve ⊆ X
e be the subset of p−1 (U ) that
y := p(e
y ), let U ∈ B an open set containing y, and let Uxe ∈ B
contains the point ye. The open neighborhood Uxe of ye might not yet be what we are looking
for; it might be too big in the sense that it might not be a subset of Ve . In order to fix this,
consider Uxe ∩ Ve , a smaller open neighborhood of ye that via p maps homeomorphically to an
open neighborhood of y. Then there is smaller open neighborhood U 0 of y belonging to the
0
exe0 is an open subset belonging
basis B, and hence for some x
e0 ∈ p−1 (xU ) the open subset U
e which contains the point ye and is contained in Ve . This shows that B
e is indeed a basis
to B,
e
for the topology of X.
21
Part (b). Let P be a pattern of polygons on X. By passing to a refinement of P is necessary
we can assume that each face is contained in an evenly covered open subset of X. Then the
e Since
preimages of vertices, edges and faces of P will form a pattern of polygons Pe on X.
the preimage of each vertex (resp. edge resp. face) of P consists of d vertices (resp. edges
resp. faces) of Pe. In particular, if we denote by V (resp. E resp. F ) the number of vertices
e resp. Fe) the corresponding number for Pe,
(resp. edges resp. faces) of P and by Ve (resp. E
we have
e = Ve − E
e + Fe = dV − dE + dF = dχ(Σ).
χ(X)
Part (c). We recall that the Euler characteristic of a surface Σg of genus g is given by
χ(Σg ) = 2 − 2g. This implies
1
g = (2 − χ(Σg )).
2
It follows that
1
e = 1 (2 − dχ(X)) = 1 (2 − d(2 − 2g)) = g − d + 1.
ge = (2 − χ(X))
2
2
2
3. Show that assigning to a pointed space (X, x0 ) its fundamental group π1 (X, x0 ), and
to each basepoint preserving map f : (X, x0 ) → (Y, y0 ) the induced map f∗ : π1 (X, x0 ) →
π1 (Y, y0 ) defines a functor
π1 : Top∗ −→ Gp
from the category Top∗ of pointed topological spaces to the category Gp of groups.
Proof. To prove that π∗ is a functor, we need to show that it is compatible with compositions
and that it maps identity morphisms in Top∗ to identity morphisms of Gp. To show that it
is compatible with compositions, suppose
f : (X, x0 ) → (Y, y0 )
and
g : (Y, y0 ) → (Z, z0 )
are morphisms in Top∗ . We need to show that
(g ◦ f )∗ = g∗ ◦ f∗
as group homomorphisms from π1 (X, x0 ) to π1 (Z, z0 ). So let γ : I → X be a based loop in
(X, x0 ) representing the element [γ] ∈ π1 (X, x0 ). Then
(g∗ ◦ f∗ )([γ]) = g∗ (f∗ ([γ])) = g∗ ([f ◦ γ]) = [g ◦ (f ◦ γ)]
(g ◦ f )∗ ([γ]) = [(g ◦ f ) ◦ γ].
The associativity of maps implies that the based loops g ◦ (f ◦ γ) and (g ◦ f ) ◦ γ in (Z, z0 )
agree and proves compatibility with composition.
To show that identity morphisms are preserved, let idX be the identity morphism of the
pointed space (X, x0 ). Then
(idX )∗ ([γ] = [idX ◦γ] = [γ],
which shows that the induced homomorphism (idX )∗ : π1 (X, x0 ) → π1 (X, x0 ) is the identity
as required.
22
4. Given a set S, let F [S] be the free group generated by S. The elements of F [S] are equivalence classes of words whose letters consist of the symbols s and s−1 for s ∈ S (including the
empty word). If a word contains letters s and s−1 adjacent to each other, the word obtained
by deleting those two letters is declared equivalent to the original word, and the equivalence
relation is generated by these relations. The multiplication is given by concatenation of
words, and the identity element is provided by the empty word.
(a) Show that there is a functor F : Set → Gp which on objects sends a set S to the free
group F [S].
(b) Show that for any group G the map
Gp(F [S], G) −→ Set(S, G)
given by
f 7→ f|S
is a bijection. Here we identify S with a subset of F [S] by regarding s ∈ S as a one
letter word.
Proof. Part (a). We need to construct the functor F on morphisms, and show that our
construction is compatible with composition and identity morphisms.
To construct F on morphisms, we need to associate to each map of sets
f: S →T
a group homomorphism
F (f ) : F (S) → F (T ).
Every element of F (S) is given by a word s11 . . . skk with si ∈ S and i ∈ {±1}. We define
F (f ) by
F (f )(s11 . . . skk ) := f (s1 )1 . . . f (sk )k .
(5.1)
We note that this is well defined. If si+1 = si and i+1 = −i , the word s11 . . . skk is equivalent
i+1
from it. Their images under
to the shorter word obtained by removing the segment si i si+1
F (f ) are again equivalent, since the shorter one is obtained by removing the segment
f (si )i f (si+1 )i+1
from the longer word.
Next we need to show that our map F (f ) : F (S) → F (T ) is indeed a group homomorphism. It is clear that it maps the identity element of F (S), given by the empty word, to
the identity element of F (T ). It is also clear that it maps a product a · b in F (S) to the
product F (f )(a) · F (f )(b), since the product is given by concatenation of words.
The formula (5.1) shows that if f : S → S is the identity map of S, then F (f ) is the
identity map of F (S).
Finally we need to show that F is compatible with compositions. So let f : S → T and
g : T → U be maps between sets. Then
(F (g) ◦ F (f ))(s11 . . . skk ) = F (g)(F (f )(s11 . . . skk ))
= F (g)(f (s1 )1 . . . f (sk )k )
= g(f (s1 ))1 . . . g(f (sk ))k
= ((g ◦ f )(s1 ))1 . . . ((g ◦ f )(sk ))k
= F (g ◦ f )(s11 . . . skk )
23
Part (b). To show that the map Φ : Gp(F [S], G) −→ Set(S, G) is a bijection, we consider
the map
Ψ : Set(S, G) −→ Gp(F [S], G)
that sends a map g : S → G to the group homomorphism Ψ(g) : F [S] → G defined by
Ψ(g)(s11 . . . skk ) := g(s1 )1 . . . g(sk )k .
By the same arguments as in part (a) this map is well-defined, and a group homomorphism.
Applying the homomorphism Ψ(g) : F [S] → G to one the one-letter word s ∈ S ⊂ F [S],
we have Ψ(g)(s) = g(s), which shows Φ ◦ Ψ = id. Conversely, given a group homomorphism
f : F [S] → G, then (Ψ ◦ Φ)(f ) is another group homomorphism F [S] → G which agrees with
f on the generators s ∈ S ⊂ F [S] of the group F [S], and hence (Ψ ◦ Φ)(f ) = f .
5. Let Vα , α ∈ A be a collection of vector spaces over a fixed field k. Let
(
)
Y
X
Vα :=
vα α | vα ∈ Vα
α∈A
α∈A
be the set of all formal linear combinations of elements in A such that the coefficient vα of
α ∈ A belongs to Vα . This is a k-vector space with addition (resp. scalar multiplication)
given by adding coefficients (resp. multiplying all coefficients by a fixed element of k).
Q
(a) Show that α∈A Vα is a product of the objects Vα , α ∈ A in the category Vectk .
(b) Let
M
Vα ⊂
Y
Vα
α∈A
α∈A
P
be the subspace consisting of those linear combinations
α∈A vα α such that only finitely
L
many coefficients vα are non-zero. Show that α∈A Vα is a coproduct of the objects Vα
in Vectk .
Q
Proof. Part (a). To show that α∈A Vα is a product of the Vα ’s, define linear map
Y
X
pβ :
Vα −→ Vβ
by
vα α 7→ vβ .
α∈A
α∈A
We need to show that for any vector space W the map
Y
Y
Vect(W,
Vα ) −→
Vect(W, Vα )
α∈A
f 7→ (pα ◦ f )α∈A
(5.2)
α∈A
is a bijection. Let us write fα := pα ◦ f ∈ Vect(W, Vα ). We note that for w ∈ W we have
X
Y
f (w) =
fα (w)α ∈
Vα .
(5.3)
α∈A
α∈A
In particular, f (w) is determined by the collection fα (w) ∈ Vα , which shows that the map
(5.2) is injective. Conversely, given a collection of linear maps fα ∈ Vect(W, Vα ), the formula
24
Q
(5.3) defines a homomorphism f : Vect(W, α∈A Vα ), thus showing that the map (5.2) is
surjective.
L
Part (b). To show that
α∈A Vα is a coproduct of the Vα ’s, define for β ∈ A the linear
map
(
M
X
v α=β
iβ : Vβ →
Vα
by
iβ (v) :=
vα α, vα =
0 α 6= β
α∈A
α∈A
We claim that for any vector space W the map
M
Y
Φ : Vect(
Vα , W ) −→
Vect(Vα , W )
α∈A
f 7→ (f ◦ iα )α∈A
α∈A
L
is a bijection, which implies that α∈A Vα is a coproduct of the Vα ’s. To prove this claim
we consider the map
Y
M
Ψ:
Vect(Vα , W ) −→ Vect(
Vα , W )
α∈A
α∈A
defined by
(fα )α∈A 7→ (
X
vα α 7→
α∈A
X
fα (vα )).
α∈A
P
We note that the sum α∈A fα (vα ) gives a well defined vector in W since it is a finite sum,
since all but finitely many of the L
vα ’s vanish. It is straightforward to check that the formula
above defines a linear map from
Vα to W , and that Ψ is the inverse of Φ.
6
Homework Assignment # 6
1. Show that the following five topological spaces are all homotopy equivalent:
1. the circle S 1 ,
2. the open cylinder S 1 × R,
3. the annulus A = {(x, y) | 1 ≤ x2 + y 2 ≤ 2},
4. the solid torus S 1 × D2 ,
5. the Möbius strip
Hint: show that each of the spaces (2)-(5) contains a subspace S homeomorphic to the circle
S 1 which is a deformation retract of the bigger space it is contained in.
Proof. As suggested by the hint, for each of the spaces X in cases (2)-(5) we construct a
continuous injection i : S 1 → X. This gives a continuous bijection from S 1 to the subspace
S := i(S 1 ) ⊂ X. Since S 1 is compact and the spaces X under consideration are all Hausdorff,
25
the subspace S is homeomorphic to S 1 . To show that S is a deformation retract of X we
need to construct a continuous map
H : X × I −→ X
with the properties
(a) H(x, 1) = x for all x ∈ X;
(b) H(x, 0) ∈ S for all x ∈ X;
(c) H(x, t) = x for x ∈ S.
This map is the homotopy between the identity of X and the retraction r : X → S ⊂ X
given by r(x) := H(x, 0), showing that S is a deformation retract.
Here are the maps i : S 1 → X and H : X × I → X for each of the cases (2)-(5). It
is straightforward to show that the maps i, H are continuous, since we describe them by
explicit formulas, to prove the injectivity of i, and to show that H has properties (a), (b)
and (c).
For the cylinder S 1 × R we define
i : S 1 −→ X = S 1 × R
z 7→ (z, 0)
H : S 1 × R × I −→ S 1 × R
(z, s, t) 7→ (z, st)
For the annulus A = {(x, y) ∈ R2 | 1 ≤ x2 + y 2 ≤ 2} = {z ∈ C | 1 ≤ |z|2 ≤ 2} we define
i : S 1 −→ A
H : A × I −→ A
z 7→ z
(z, t) 7→ (1 − t)
z
+ tz
|z|
For the solid torus S 1 × D2 we define
i : S 1 −→ S 1 × D2
z 7→ (z, 0)
H : S 1 × D2 × I −→ S 1 × D2
(z, w, t) 7→ (z, tw)
For the Möbius strip M = (I × [−1, 1])/ ∼ with the equivalence relation (0, s) ∼ (1, −s)
it will be convenient to replace S 1 by the homeomorphic space R/Z. We define
i : R/Z −→ M
[r] 7→ [r, 0]
H : M × I −→ M
([r, s], t) 7→ [r, ts]
2. Let H, G1 , G2 be groups, and f1 : H → G1 , f2 : H → G2 be homomorphisms. Show that
f1
f2
/ G2 , that is, show that
the amalgamated product G1 ∗H G2 is the pushout of G1 o
H
there are group homomorphisms gi : Gi → G1 ∗H G2 such that the commutative diagram
f2
H
f1
G1
g1
/
/ G2
g2
G1 ∗H G2
is a pushout diagram.
26
(6.1)
Proof. Let us recall the definition of G1 ∗ G2 and G1 ∗H G2 . The elements of the free product
G1 ∗ G2 are equivalence classes of words a1 . . . am whose letters ai are elements of G1 or G2 .
The amalgamented product G1 ∗H G2 is the quotient of G1 ∗ G2 by the normal subgroup N
generated by the words of the form f1 (h)(f2 (h))−1 .
Let Gi → G1 ∗ G2 for i = 1, 2 be the homomorphism that sends an element g ∈ Gi to the
element of G1 ∗ G2 given by the one letter word g ∈ G1 ∗ G2 , and let gi : Gi → G1 ∗H G2 be
the composition of that homomorphism and the projection map G1 ∗ G2 → G1 ∗H G2 . This
makes the diagram (6.1) commutative, since g1 ◦ f1 sends an element h to [f1 (h)] ∈ G1 ∗H G2
(the element of the quotient G1 ∗H G2 represented by the one letter word f1 (h)), while g2 ◦ f2
sends an element h to [f2 (h)]. These are different elements in G1 ∗ G2 , but the quotient
G1 ∗H G2 is taylor made for these elements to agree in.
To show that the diagram (6.1) is a pushout diagram we need to check the universal
property of the pushout, that is, we need to show that for any group K the map
Φ : Gp(G1 ∗H G2 , K) −→ Gp(G1 ) ×Gp(H,K) Gp(G2 , K)
k 7→ (k ◦ g1 , k ◦ g2 )
is a bijection. To show this, we define a map
Ψ : Gp(G1 ) ×Gp(H,K) Gp(G2 , K) −→ Gp(G1 ∗H G2 , K)
that is an inverse to Φ. Given (k1 , k2 ) ∈ Gp(G1 ) ×Gp(H,K) Gp(G2 , K), we define Ψ(k1 , k2 ) ∈
Gp(G1 ∗H G2 , K) by
(
k1 (a) if a ∈ G1
Ψ(k1 , k2 )([a1 . . . am ]) := k(a1 ) . . . k(am )
where
k(a) :=
.
k2 (a) if a ∈ G2
It is clear that Ψ(k1 , k2 ) constructed as above is a well-defined homomorphism from G1 ∗ G2
to K, but we need to check that Ψ(k1 , k2 ) sends the generators f1 (h)f2 (h−1 ) ∈ G1 ∗ G2 of
the normal subgroup N to the identity element of K:
Ψ(k1 , k2 )(f1 (h)f2 (h−1 )) = k(f1 (h))k(f2 (h−1 )) = k1 (f1 (h))k2 (f2 (h−1 ))
= k2 (f2 (h))k2 (f2 (h−1 )) = 1 ∈ K.
Here the third equality holds since (k1 , k2 ) is an element of
Gp(G1 ) ×Gp(H,K) Gp(G2 , K) = {(k1 , k2 ) ∈ Gp(G1 ) × Gp(G2 , K) | k1 ◦ f1 = k2 ◦ f2 }.
Checking that Ψ is an inverse to Φ is straightforward from the construction of Ψ.
3. Use the Van Kampen Theorem to show π1 (S n , x0 ) = {1} for n > 1.
Proof. Let U, V be the open subsets of S n obtained by removing the northpole N = (0, . . . , 0, 1)
resp. the south pole S = (0, . . . , 0, −1). Via the stereographic projection U and V are homeomorphic to Rn , and hence their fundamental group is trivial. We claim that U ∩ V is path
connected, provided n > 1. This allows us to apply the Van Kampen Theorem which implies
π1 (S n ) ∼
= π1 (U ) ∗π1 (U ∩V ) π1 (V ) = {1} ∗π1 (U ∩V ) {1} = {1}.
27
To prove that U ∩ V = S n \ {N, S} is pathconnected, suppose that x, y ∈ U ∩ V . We try
to construct a path γ connecting x and y by taking the linear path δ : I → Rn+1 between x
and y, given by δ(s) = (1 − s)x + sy, and then dividing by the norm to obtain a path in S n .
In other words,
δ(s)
.
γ: I → U ∩ V
is given by
γ(s) =
|δ(s)|
We note that there are two potential issues with this construction: the denominator could
become 0 for some s ∈ I and γ(s) could be equal to N or S for some s ∈ I. The first
problem doesn’t occur if δ doesn’t pass through 0 ∈ Rn+1 , and the second problem doesn’t
occur if δ(s) doesn’t belong to the subspace spanned by N for any s ∈ I. In particular, if y
doesn’t belong to the 2-dimensional subspace of Rn+1 spanned by x and N , then the curve
γ constructed above is a path in U ∩ V connecting x and y.
If n > 1 and x, y ∈ U ∩ V we can choose some z ∈ S n such that z neither belongs to
the plane spanned by x and N nor the plane spanned by y and N . Then the construction
above gives paths connecting x and z resp. y and z, and hence their concatenation gives a
path between x and y.
4. We recall that if G × X → X is the action of a group G on a set X, then the subgroup
Gx := {g ∈ G | gx = x} ⊆ G is the isotropy subgroup of the point x ∈ X. The action is
called free if the isotropy subgroup Gx is the trivial group for all x ∈ X.
(a) Show that if G is a finite group which acts freely and continuously on a Hausdorff space
X, then the projection map p : X → X/G to the orbit space X/G is a covering map.
Hint: Use the assumptions that the action is free and X is Hausdorff to show that for
every x ∈ X there is an open neighborhood U such that the subsets gU ⊂ X for g ∈ G
are mutually disjoint.
(b) Show that if X is a manifold of dimension n, then also the orbit space X/G is a manifold
of dimension n (don’t forget to argue that X/G is Hausdorff and second countable).
(c) Show that the map
Z/k × S 2n−1 −→ S 2n−1
given by
(m, v) 7→ e2πim/k v
is a continuous free action of the cyclic group Z/k on the sphere S 2n−1 ⊂ Cn . By part
(b) the orbit space S 2n−1 /Z/k is then a manifold of dimension 2n − 1, known as a lens
space.
Proof. Part (a). For any point x ∈ X the points gx for g ∈ G are all distinct, since the action
is free (if gx = hx for g 6= h, then h−1 gx = x and hence h−1 g ∈ Gx , contradicting the freeness
assumption). This implies that there are open neighborhoods Ugx of gx ∈ X which are
mutually disjoint, since X is Hausdorff. We note that g −1 Ugx is another open neighborhood
of x (it is open, since it is the image of the open set Ugx under the homeomorphism given
by the action of the group element g −1 ). Hence the finite intersection
\
U :=
g −1 Ugx
g∈G
28
is again an open neighborhood of x.
We claim that the subsets gU for g ∈ G are mutually disjoint. To prove this, assume
that y ∈ gU ∩ hU for g 6= h. Then in particular y belongs to g(g −1 Ugx ) = Ugx and to
h(h−1 Uhx ) = Uhx contradicting Ugx ∩ Uhx = ∅.
Given the point [x] ∈ X/G
S the subset V = p(U ) ⊂ X/G is an open neighborhood of [x];
−1
it is open since p (V ) = g∈G gU is open. This open set is evenly covered, since for any
g ∈ G the restriction
p|gU : gU −→ V
is a homeomorphism, since it is a continuous bijection and an open map. To argue that p
is open, suppose
S that U ⊂ X is an open subset. Then p(U ) is an open subset of X/G since
p−1 (p(U )) = g∈G gU is an open subset of X.
Part (b). To show that X/G is locally homeomorphic to Rn , let y ∈ X/G and let V ⊂ X/G
be an evenly covered open neighborhood of y. Then p−1 (V ) is the disjoint of open subsets
Vα ⊂ X such that p|Vα : Vα → V is a homeomorphism. Let x be the unique point in Vα
with p(x) = y. Since X is a manifold of dimension n a possibly smaller neighborhood of
x is homeomorphic to an open subset of Rn . Via the restriction of p|Vα to this smaller
neighborhood we conclude that a neighborhood of y is homeomorphic to an open subset of
Rn .
To show that X/G is second countable, let B be a basis for X. We claim that the
countable collection of subsets of X/G given by p(U ) for U ∈ B is a basis for X/G. To see
this, let y ∈ X/G, and let V ⊂ X/G be an open neighborhood of y. Since B is a basis for the
topology of X for x ∈ p−1 (y) ⊂ p−1 (V ), there is some U ∈ B which is an open neighborhood
of x contained in p−1 (V ). It follows that p(U ) ⊂ V is an open neighborhood of y, proving
that {p(U ) | U ∈ B} is in fact a basis for the topology of X/G.
To show that X/G is Hausdorff, let y, y 0 ∈ X/G with y 6= y 0 , and pick points x ∈ p−1 (y),
x0 ∈ p−1 (y 0 ). Proceeding like in part (a) the points gx, g 0 x0 for g, g 0 ∈ G are distinct points
of X. Hence there are mutually disjoint open neighborhoods Ugx , Ug0 x0 of these points. It
follows that
\
\
U :=
g −1 Ugx
U 0 :=
(g 0 )−1 Ug0 x0
g 0 ∈G
g∈G
are open neighborhoods of x resp. x0 . Moreover their image under p are disjoint open
neighborhoods of y resp. y 0 since p−1 (p(U ) and p−1 (p(U 0 )) are disjoint. This proves that
X/G is Hausdorff.
Part (c). To show that the Z/k action on S 2n−1 is free, assume [m] ∈ Z/k belongs to the
isotropy subgroup of some v = (v1 , . . . , vn ) ∈ S 2n−1 , that is, e2πim/k v = v. Since v is a unit
vector, it has a non-zero component vi ∈ C. Then e2πim/k vi = vi which implies e2πim/k = 1,
which in turn implies that m ∈ Z must be a multiple of k. In particular, [m] is the identity
element in Z/k, proving that the action is free.
e → X be a covering map. Assume that X
e is path connected, let x
e and let
5. Let p : X
e0 ∈ X
x0 = p(e
x0 ) ∈ X.
e x
(a) Show that the kernel of the induced homomorphism p∗ : π1 (X,
e0 ) → π1 (X, x0 ) is trivial.
Hint: use the lifting property for families of paths.
29
(b) Show that the map
Φ : π1 (X, x0 ) −→ p−1 (x0 )
defined by
[γ] 7→ γ
e(1)
e is the lift of γ with γ
is a well-defined surjection. Here γ
e: I → X
e(0) = x
e0 .
(c) Show that Φ(g1 ) = Φ(g2 ) for g1 , g2 ∈ G := π1 (X, x0 ) if and only if Hg1 = Hg2 , where
e x
Hgi ⊂ G are right cosets for the subgroup H = p∗ (π1 (X,
e0 )) ⊆ G.
We note that putting parts (b) and (c) together, we obtain a bijection between the set of
right cosets H\G and the fiber p−1 (x0 ).
e be a based loop in X
e representing an element in the kernel
Proof. Part (a). Let γ
e: I → X
of p∗ . This means that the loop p ◦ γ
e is homotopic to the constant loop cx0 via a homotopy
H : I × I → X. Writing this out more explicitly we have
H(s, 0) = p(γ(s))
H(s, 1) = x0 = H(0, t) = H(1, t)
In particular, the first condition implies that the diagram
I × {0}
γ
e
/
e
X
p
H
I ×I
/
X
e → X to obtain
commutes, and hence we can use the lifting property for the covering space X
e as shown in the diagram
a map H
γ
e
e
/X
;
xx
x
x
p
xx
xx
x
x
H
/X
×I
I × {0}
e
H
I
e t) ∈ p−1 (x0 ), since H
e is a lift of H and H has the property H(0, t) = x0 .
We note that for H(0,
Since the map
e
e t)
I → p−1 (x0 ) ⊂ X
given by
t→
7 H(0,
is a continuous map from a connected space to a discrete space, it must be constant. Since
e 0) = γ
e t) = x
H(0,
e(0) = x
e0 we must in fact have H(0,
e0 for all t ∈ I. Similarly, we can show
e
that H(1, t) = x
e0 for all t ∈ I. Applying the same argument to the map
e
I → p−1 (x0 ) ⊂ X
given by
e 1)
s 7→ H(s,
e 1) = x
e show that H
e is a homotopy between
we conclude that H(s,
e0 . These properties of H
the loop γ
e and the constant loop cxe0 , which implies that [e
γ ] is the identity element in
e
π1 (X, x
e0 ).
30
Part (b). To show that Φ is well-defined, let γ, γ 0 : I → X be loops in X based at x0
representing the same element in π1 (X, x0 ). Then we can interpret a homotopy H : I ×I → X
between these loops as a family of loops parametrized by the point s ∈ I, that is
H(0, t) = γ(t)
H(1, t) = γ 0 (t)
H(s, 0) = x0 = H(s, 1).
e be the unique lift of this homotopy determined by the commutative diagram
Let H
I
cxe0
e
/X
w;
w
e ww
H
w
p
ww
w
w
w
H /
×I
X,
I × {0}
e Restricting to s = 0, 1 we obtain paths
where cxe0 is the constant map to the point x
e0 ∈ X.
e t) and γ
e t) which are the unique lifts of the loops γ, γ 0 to paths in
γ
e(t) := H(0,
e0 (t) := H(1,
e starting at x
X
e0 .
e given by s 7→ H(s,
e 1). Since H(s, 1) = x0 for all s, this is
Now consider the path I → X
−1
a path in the discrete space p (x0 ), and hence as in part (a) we conclude that
e 1) = H(1,
e 1) = γ
γ
e(1) = H(0,
e0 (1)
which shows that the map Φ is well-defined.
e be a path with
Surjectivity of Φ is easy: given any point x
e ∈ p−1 (x0 ) let γ
e: I → X
γ
e(0) = x
e0 and γ
e(1) = x
e. Then γ := p ◦ γ
e is a loop in X based at x0 , and γ
e is the unique lift
of γ with starting point x
e0 . This shows that Φ maps [γ] ∈ π1 (X, x0 ) to γ
e(1) = x
e.
Part (c). Suppose that Φ(g1 ) = Φ(g2 ) for elements gi ∈ G = π1 (X, x0 ). Representing gi by
e is the unique lift
loops γi in X based at x0 , this means that γ
e1 (1) = γ
e2 (1), where γ
ei : I → X
e2 is defined, and is a loop in X
of γi with γ
ei = x
e0 . In particular, the concatenation γ
e1 ∗ γ
based at x
e0 . Then
p∗ ([e
γ1 ∗ γ
e2 ]) = [(p ◦ γ
e1 ) ∗ (p ◦ γ
e2 )] = [γ1 ∗ γ̄2 ] = g1 g2−1
e x
is an element of H = p∗ (π1 (X,
e0 )) which implies Hg1 = Hg2 .
Conversely, assume Hg1 = Hg2 . Then g2 = hg1 for some element h ∈ H. Let g1 ∈
π1 (X, x0 ) be represented by the based loop γ1 , and let γ
e1 be the unique lift of this loop to a
e starting at x
e x
path in X
e0 . Let e
h ∈ π1 (X,
e0 ) be the unique element mapping to h ∈ π1 (X, x0 )
e
e representing e
via p∗ , and let δ be a based loop in X
h. Then δ := p ◦ δe is a based loop in
X representing h ∈ π1 (X, x0 ). We note that γ2 := δ ∗ γ1 is a loop representing g2 , and that
γ
e2 = δe ∗ γ
e1 is the unique lift of this loop (since δe is a loop). By part (b) this implies
Φ(g2 ) = γ
e2 (1) = (δe ∗ γ
e1 )(1) = γ
e1 (1) = Φ(g1 ).
31
7
Homework Assignment # 7
1. Let p : (E, e0 ) → (B, b0 ) be a covering space, and let f : (X, x0 ) → (B, b0 ) be a map with
X path-connected and locally path-connected. Show:
(a) A lift fe: (X, x0 ) → (E, e0 ) of f exists if and only if f∗ (π1 (X, x0 )) ⊆ p∗ (π1 (E, e0 )). Hint:
use the path lifting property for the covering space E → B.
(b) There is at most one such lift.
Proof. Part (a). Suppose that a lift fe: (X, x0 ) → (E, e0 ) of f exists. This means that we
have the following commutative diagram of pointed spaces and basepoint preserving maps:
(E, e0 )
9
t
fe ttt
t
tt
tt
(X, x0 )
/
f
p
(B, b0 )
Applying the fundamental group functor results in the following commutative diagram in
the category of groups:
π81 (E, e0 )
q
fe∗ qqqq
qqq
qqq
π1 (X, x0 )
f∗
p∗
/ π1 (B, b0 )
This shows that f∗ (π1 (X, x0 )) is contained in p∗ (π1 (E, e0 )).
Conversely, let us assume f∗ (π1 (X, x0 )) ⊆ p∗ (π1 (E, e0 )). To construct the desired lift
e
f : (X, x0 ) → (E, e0 ) we choose for a given point x ∈ X a path γ : I → X with γ(0) = x0
and γ(1) = x; such a path exists due to the assumption that X is path-connected. Applying
the path lifting property for the covering space E → B to the path δ := f ◦ γ : I → B, there
e = e0 . More explicitly, we apply the homotopy lifting
is a unique lift δe: I → E of δ with δ(0)
property to the diagram
e0
/
{0}
n7 E
I
δen
n
n n
p
n
n nγ /
X
f
6/ B
δ
e ∈ E. It remains to show:
Then we define fe(x) := δ(1)
(i) the map fe: X → E is well-defined and basepoint preserving.
(ii) fe is continuous.
For (i), let us assume that γ 0 is another path from the basepoint x0 to the given point x ∈ X,
let δ 0 := f ◦ γ 0 , and let δe0 : I → E be a lift of δ 0 with δe0 (0) = e0 . Then the concatenation
γ̄ ∗ γ 0 is a based loop in (X, x0 ) and its image
f ◦ (γ̄ ∗ γ 0 ) = (f ◦ γ̄) ∗ (f ◦ γ 0 ) = δ̄ ∗ δ 0
32
is a based loop in (B, b0 ). By assumption, the element f∗ ([γ̄ ∗ γ 0 ]) = [δ̄ ∗ δ 0 ] ∈ π1 (B, b0 ) is in
the image of p∗ : π1 (E, e0 ) → π1 (B, b0 ).
According to a proposition in class a based loop α : I → B represents an element in the
image of p∗ : π1 (E, e0 ) → π1 (B, b0 ) if and only if the unique lift α
e : I → E with α
e(0) = e0 is
0
0
e
e
a loop. Applying this to the loop α = δ̄ ∗ δ shows that the lifts δ and δ of δ, δ 0 have the
same endpoint: restricting the loop α
e to the two subintervals [0, 1/2] and [1/2, 1] we write α
e
as the concatenation of two paths, which will be lifts of the two paths δ̄, δ 0 that α = δ̄ ∗ δ 0 is
built from. Then uniqueness of pathlifting implies that these paths must be b̄
δ and δb0 . The
e
e
fact that α
e is a loop then implies that the two endpoints of the paths δ and δ 0 must agree.
The map fe is basepoint preserving by construction (for x = x0 we choose γ to be the
constant path at x0 , and then δ, δe will be the constant paths at b0 and e0 , respectively.
To prove continuity of fe, we note that continuity of a map is a local property, i.e., it
suffices to prove continuity of fe restricted to a sufficiently small open neighborhood V of
every point x ∈ X. Given x ∈ X we construct V as follows: the point f (x) ∈ B has a
neighborhood U such that p−1 (U ) is a union of disjoint open subsets Ui ⊂ E, i ∈ I, such
that the restriction p|Ui : Ui → U is a homeomorphism for each i ∈ I (by the definition of
a covering map). By continuity of f the preimage f −1 (U ) is then an open neighborhood of
x ∈ X. The assumption that X is locally path-connected implies that there is a possibly
smaller open neighborhood V ⊂ f −1 of x which is path connected.
We claim that if fe(x) belongs to Ui , then fe(y) ∈ Ui for every y ∈ V . To see this, suppose
as above that γ is a path in X from x0 to x, and δe is a path in E starting at e0 which is a
e
lift of f ◦ γ. Then fe(x) = δ(1).
To determine fe(x0 ) for a point x0 ∈ V we choose γ 0 = ∗ γ,
where is a path in V starting at x and ending in x0 . Then by construction fe(x0 ) is the
endpoint of the lift of δ 0 := (f ◦ ) ∗ (f ◦ γ) = (f ◦ ) ∗ δ. This is the concatenation of δe
(which goes from e0 to fe(x)) and of a lift β of the path (f ◦ ) with β(0) = fe(x). The
≈
homeomorphism p|Ui : Ui −→ U restricts to a homeomorphisms between Vi := Ui ∩ p−1 (V )
and V , which implies that Vi is path-connected, since V is. In particular, since β(0) = fe(x)
belongs to Vi , so does the endpoint β(1) = fe(x0 ). This proves the claim above.
We note that the composition
X
fe
/E
p
/B
is the map f which is contininuous by assumption. This does not imply that the factor
f is continuous (think of a counterexample in the case where X is the unit interval I and
fe: I → E is a path in E). However, upon restriction of fe to V the factorization takes the
following form:
V
fe|V
/
Ui
p
/7 U
.
f|V
Since p : Ui → U is a homeomorphism, the continuity of f|V does then indeed imply the
continuity of fe|V .
Part (b). If fe, : (X, x0 ) → (E, e0 ) is a lift of f , the value fe(x) ∈ E can be described as in
b
part (a) as fe(x) = δ(1),
where we chose a path γ from x0 to x and a lift δe of the image path
f ◦ δ. The uniqueness of path-lifting then implies the uniqueness of the lift fe.
33
2. Show that if a path-connected, locally path-connected space X has finite fundamental
group, then every map X → S 1 is homotopic to a constant map. Hint: Use the covering
space R → S 1 and the previous homework problem.
Proof. Let f∗ : π1 (X, x0 ) → π1 (S 1 , y0 ), y0 := f (x0 ) be the homomorphism of fundamental
groups induced by the map f : X → S 1 . We note that every element a ∈ π1 (X, x0 ) has
finite order and hence f∗ (a) ∈ π1 (S 1 ) ∼
= Z has finite order which implies that f∗ (a) = 0.
In particular f∗ (π1 (X, x0 )) is contained in the image of p∗ : π1 (R) → π1 (S 1 ) induced by the
covering map p : R → S 1 . By a result proved in class, this implies that f factors in the form
f = p ◦ fe for some map fe: X → R. The map fe is homotopic to the constant map sending
every point x ∈ X to ye0 := fe(x0 ) via the linear homotopy
e : X × I −→ R
H
given by
e t) = tfe(x) + (1 − t)e
H(x,
y0
e : X × I → S 1 is a homotopy between f and the constant map at y0 .
Then H := p ◦ H
3. Let U± := S n \ {(±1, 0, . . . , 0)}. We recall that the stereographic projection is the map
1
P± : U± → Rn given by P± (x0 , x1 , . . . , xn ) = 1∓x
(x1 , . . . , xn ).
0
(a) Show that (U+ , P+ ) and (U− , P− ) are charts for S n .
(b) Show that {(U+ , P+ ), (U− , P− )} is a smooth atlas for S n .
(c) Show that the atlas above is smoothly compatible with the smooth atlas we’ve dis±
n
cussed in class, consisting of the charts (Ui± , φ±
i ), where Ui ⊂ S consists of the vectors
±
n
(x0 , . . . , xn ) ∈ S such that xi > 0 resp. xi < 0, and φi (x0 , . . . , xn ) = (x0 , . . . , xbi , . . . , xn ).
Proof. Part (a). Let ∈ {±1}. We need to show that the map P : U → Rn is a homeomorphism. To do this, we calculate the inverse of P explicitly (we need the explicit form of
the inverse for later parts anyway). So, given y = (y1 , . . . , yn ) ∈ Rn we want to show that
there is a unique x = (x0 , x1 , . . . , xn ) ∈ U = S n \ (−, 0, . . . , 0) satisfying the equation
(y1 , . . . , yn ) = P (x0 , . . . , xn ) =
1
(x1 , . . . , xn ).
1 − x0
Taking the norm squared of both sides we obtain
||y||2 =
x21 + · · · + x2n
||x||2 − x20
1 − x20
=
=
.
(1 − x0 )2
(1 − x0 )2
(1 − x0 )2
This leads to the quadratic equation
||y||2 (1 − x0 )2 = 1 − x20 ,
for x0 . In standard form this equation becomes
(||y||2 + 1)x20 − 2||y||2 x0 + (||y||2 − 1) = 0
34
(7.1)
and hence we have solutions
p
2||y||2 ± 4||y||4 − 4(||y||2 + 1)(||y||2 − 1)
2||y||2 ± 2
||y||2 ± 1
x0 =
=
=
.
2(||y||2 + 1)
2(||y||2 + 1)
||y||2 + 1
2
+
= , since that implies xi = 0 for
We observe that we don’t want the solution x0 = ||y||
||y||2 +1
1 ≤ i ≤ n, and hence forces x = (, 0, . . . , 0) which does not belong to U . This forces
x0 =
(||y||2 − 1)
.
||y||2 + 1
Using equation (7.1) this implies
xi = yi (1 − x0 ) = yi (1 − ||y||2 − 1
2
(||y||2 − 1)
)
=
y
(1
−
) = yi
i
2
2
||y|| + 1
||y|| + 1
||y||2 + 1
These considerations imply that the inverse map P−1 : Rn −→ U is given by
P−1 (y1 , . . . , yn ) =
1
(||y||2 − 1), 2y1 , . . . , 2yn .
2
||y|| + 1
Since all components of P−1 are continuous functions, this is a continuous map and hence
P is a homeomorphism.
Part (b). We need to show that the transition map
P−1
P (U ∩ U− )
/
U ∩ U−
P−
/
P− (U ∩ U− )
is a smooth map for = ±1. We note that U ∩ U− = S n \ {±1, 0, . . . , 0} and hence
P± (U ∩ U− ) = Rn \ {0}.
Explicitly, for y = (y1 , . . . , yn ) ∈ Rn \ {0} we have
1
2
(||y||
−
1),
2y
,
.
.
.
,
2y
)
1
n
||y||2 + 1
1
2y1
2yn
=
,...,
2
||y||2 + 1
1 + (||y||2 −1) ||y||2 + 1
P− (P−1 (y)) = P− (
||y|| +1
=
=
||y||2
1
(2y1 , . . . , 2yn )
+ 1 + (||y||2 − 1)
1
(y1 , . . . , yn )
||y||2
This map is smooth, since all its components are smooth functions (the function 1/||y||2 is
smooth on Rn \ {0}, the domain of P− ◦ P−1 ).
Part (c). For δ = ±1 the map φδi : Uiδ → D̊n is given by forgetting the i-th coordinate.
Hence the composition φδi ◦P−1 is a smooth function, since all components of P−1 are smooth.
35
To check smoothness of the composition P ◦ (φδi )−1 we recall that (φδi )−1 : D̊n −→ Uiδ is
given by
p
(φδi )−1 (y1 , . . . , yn ) = (y1 , . . . , yi−1 , δ 1 − ||y||2 , yi , . . . , yn ).
It follows that
P ◦
(φδi )−1 (y)
=


√1
(y1 , . . . , yn )
1−δ 1−||y||2
p
 1 (y2 , . . . , yi−1 , δ 1 −
1−y1
i=0
||y||2 , yi , . . . , yn ) i 6= 0
We note that there are two potential problems with the smoothness of these maps:
• The square root function is not differentiable at 0; so there is a potential issue when
1 − ||y||2 , the argument of the square root, becomes zero. This doesn’t happen since
y ∈ D̊2 , the interior of the 2-disk.
• The function is undefined if the denominator becomes 0. We could discuss for which y
this happens in either case (i = 0 or i 6= 0) and show that those y are not in the domain
of the transition map. It is easier to say that by construction the transition map is
well-defined on its domain, and hence we don’t need to worry about those points.
4. Let Ui := {[z0 , z1 , . . . , zn ] ∈ CPn | zi 6= 0} for i = 0, . . . , n and let
z0
zbi
zn
n
φi : Ui −→ C
be defined by
φi ([z0 , z1 , . . . , zn ]) =
,..., ,...,
.
zi
zi
zi
(a) Show that (Ui , φi ) is a chart for CPn .
(b) Show that {(Ui , φi ) | i = 0, . . . , n} is a smooth atlas for CPn .
Proof. Part (a). It is easy to check that the inverse of φi : Ui −→ Cn is given by
φ−1
i (w1 , . . . , wn ) = [w1 , . . . , wi , 1, wi+1 , . . . , wn ] ∈ Ui
for
(w1 , . . . , wn ) ∈ Cn .
This map is the composition of the map
Cn → Cn+1 \ {0}
(w1 , . . . , wn ) 7→ (w1 , . . . , wi , 1, wi+1 , . . . , wn )
given by
composed with the projection map p : Cn+1 \{0} → ((Cn+1 \ {0})/ ∼) = CPn . The first map
is continuous since its components are, projection maps onto quotient spaces are continuous
by the construction of the quotient topology and hence the composition φ−1
is continuous.
i
This shows that φi : Ui → Cn is a homeomorphism. In other words, (Ui , φi ) is a chart for
CPn .
Part (b). The Ui ’s form an open cover of CPn and so the collection (Ui , φi ) is an atlas for
CPn . It remains to show that it is a smooth atlas, i.e., that the transition map
Cn ⊃ φi (Ui ∩ Uj )
φ−1
i
/
Ui ∩ Uj
36
φj
/ φj (Ui
∩ Uj ) ⊂ Cn
is smooth for all i, j = 0, . . . , n. For w = (w1 , . . . , wn ) ∈ φi (Ui ∩ Uj ), i 6= j we compute
φj (φ−1
i (w) = φj ([w1 , . . . , wi , 1, wi+1 , . . . , wn ]

wi
j+1
1
wn
 w1 , . . . , wd
,
.
.
.
,
,
,
.
.
.
,
0≤j<i
w
wj+1
wj+1 wj+1
wj+1
= wj+1
wj
 1 , . . . , wi , 1 , wi+1 , . . . , c
, . . . , wwnj
i<j≤n
wj
wj wj
wj
wj
As in the previous homework problem this shows that the transition map is smooth since its
components are all smooth functions (since the transition function is well-defined those w’s
for which the denominator is 0 can’t be in the domain of the transition function).
5. Show that the Cartesian product of M × N of smooth manifolds of dimension m resp. n
is again a smooth manifold of dimension m + n.
Proof. Let (Ui , φi )i∈I be a smooth atlas for M , and (Vj , ψj )j∈J for N . We claim that the
collection of charts of M × N given by
M × N ⊃ Ui × Vj
open
φi ×ψj
/ φi (Ui ) × ψj (Vj )
⊂ Rm × Rn = Rm+n
open
for (i, j) ∈ I × J is a smooth atlas for M × N . In particular, M × N is a smooth manifold
of dimension m + n.
−1
It is clear that φi × ψj is a homeomorphism with inverse given by φ−1
i × ψj . Also it is
clear that the open subsets Ui × Vj cover M × N . Hence this collection is an atlas, and it
only remains to show that this is a smooth atlas.
The transition map between the charts φi × ψj and φi0 × ψj 0 is given by (after restricting
the domains in the obvious way) by
−1
−1
−1
(φi0 × ψj 0 ) ◦ (φi × ψj )−1 = (φi0 × ψj 0 ) ◦ (φ−1
i × ψj ) = (φi0 ◦ φi ) × (ψj 0 ◦ ψj )
The map φi0 ◦ φ−1
is a transition map for a smooth atlas for M and hence a smooth map
i
between open subsets of Rm . Similarly the map ψj 0 ◦ ψj−1 is a smooth map between open
subsets of Rn . It follows that the Cartesian products of these maps is a smooth map (between
open subsets of Rm ×Rn = Rm+n ) and hence the atlas constructed above is in fact smooth.
8
Homework Assignment # 8
1.
(a) Let M be a smooth manifold and let f : N → M be a homeomorphism. Show that there
is a unique smooth structure on N such that f is a diffeomorphism.
(b) Give the space {x ∈ Rn+1 | max{|xi |} = 1} (the boundary of the (n + 1)-cube) the
structure of a smooth manifold.
(c) Applying part (a) to N = {(x, y) ∈ R2 | y 2 = x3 } and the homeomorphism f : N →
M = R given by f (x, y) = y gives N the structure of a smooth manifold. Is the inclusion
map i : N → R2 a smooth map?
37
Proof. Part (a). Let (hα : Uα → Uα0 )α∈A be the maximal atlas giving the smooth structure
on M . For α ∈ A, let Vα := f −1 (Uα ) (which is an open subset of N ), and let gα be the
composition
Vα
f
≈
/ Uα hα / U 0
≈
α
We claim that
1. (gα : Vα → Uα0 )α∈A is a smooth atlas for N .
2. This atlas is maximal.
3. With this smooth structure on N , the homeomorphism f is a diffeomorphism.
4. If (g : V → U 0 ) is a chart belonging to a maximal atlas of N such that f is a diffeomorphism, then g = gα for some α ∈ A.
Claims 1, 2 and 3 show that there exists a smooth structure on N such that f is smooth.
Claim 4 shows that any smooth structure on N such that f is a diffeomorphism is equal to
the smooth structure constructed above; in particular, there is a unique smooth structure
on N such that f is a diffeomorphism, proving part (a).
To prove that (gα : Vα → Uα0 )α∈A is a smooth atlas we consider the transition map gβ ◦gα−1
relating the charts gα and gβ for α, β ∈ A. The following commutative diagram shows that
this transition map is equal to the transition map hβ ◦ h−1
α and hence is smooth:
∩ Vβ VVV
VVVV g
VVVVβ
VVVV
VVVV
≈
V+
f ≈
hβ (Uα ∩ Uβ ) = gβ (Vα ∩ Vβ )
h Vα
hhhh
h
h
h
hhhh
hhhh ≈
s hhh
h
gα
gα (Vα ∩ Vβ ) = hα (Uα ∩ Uβ )
kVVVV
VVVV ≈
VVVV
VVVV
VVVV
hα
hhh4
≈hhhhhh
Uα ∩ Uβ
hhh
hhhh hβ
hhhh
To prove that the smooth atlas (gα )α∈A is maximal we need to show that if g0 : V0 → V00
is a chart for N such that all transition maps gα ◦ g0−1 for charts (gα )α∈A are smooth, then
g0 is equal to the chart gα for some α ∈ A. To show this, let h0 be the composition
U0 := f (V0 )
f −1
≈
/ V0 g0 / U 0
≈
0
−1
Then the diagram above for β = 0 shows that the transition maps h0 ◦ h−1
α and hα ◦ h0 for
any α ∈ A are smooth. It follows that the atlas (hα )α∈Aq{0} is a smooth atlas; by maximality,
the chart h0 must be equal to hα0 for some α0 ∈ A. It follows that the chart g0 is equal to
gα0 and hence the atlas (gα )α∈A is a maximal smooth atlas.
To prove claim 4 we note that the assumptions imply that hα ◦ f ◦ g −1 = gα ◦ g −1 and
g ◦ f −1 ◦ hα = g ◦ gα−1 are smooth. In other words, the chart (V, g) is smoothly compatible
with any of the charts (Vα , gα ), α ∈ A. The maximality of the atlas (gα )α∈A then implies
g = gβ for some β ∈ A.
38
Part (b). To prove part (b) we note that the map
f
X = {x ∈ Rn+1 | max{|xi |} = 1} −→ S n
defined by
f (x) =
x
||x||
is continuous. It is a homeomorphism since its inverse can be described explicitly as f −1 (x) =
x
where ||x||max = max{|xi |}, which shows in particular that f −1 is continuous (the map
||x||max
Rn → R, x 7→ ||x||max is continuous thanks to the estimate ||x||max ≤ ||x||). Then the results
of part (a) allow us to use this homeomorphism to equip X with the structure of a smooth
manifold.
Part (c). In general for a smooth n-manifold N a map h : N → Rk is smooth if and only if
the composition
ϕ−1
U0
/
U ⊂N
h
/
Rk
ϕ
is smooth for every chart N ⊃ U → U 0 ⊂ Rn in the smooth atlas defining the smooth
structure on N . For the manifold N at hand, the smooth structure on N constructed by the
f
method of part (a) is given by the atlas consisting of the single chart N = U → U 0 = R. So
i is smooth if and only if the composition
i ◦ f −1 : R
/
R2
given by
y 7→ (y 2/3 , y)
is smooth. This shows that the inclusion map i is not smooth, since the first component of
the map i ◦ f −1 is not smooth (the derivative of y 2/3 is −y −1/3 for y 6= 0, but does not exist
for y = 0).
2. Show that S n ⊂ Rn+1 is a smooth manifold by proving that 1 is a regular value of the
smooth function f : Rn+1 → R, x 7→ ||x||2 .
Proof. We need to show that for every x ∈ S n = f −1 (1) the differential
(Df )x : Rn+1 −→ R
is surjective. As in class we calculate (Df )x via (Df )x (y) =
∂f (x+sy)
(0).
∂s
f (x + sy) = ||x + sy||2 = hx + sy, x + syi = hx, xi + 2shx, yi + s2 hy, yi
and hence
∂f (x + sy)
(0) = 2hx, yi
∂s
Taking y = x we see that (Df )x is a surjective linear map for x ∈ f −1 (1).
(Df )x (y) =
3. Let Mn×k (R) be the vector space of n × k-matrices. For A ∈ Mn×k (R) let At ∈ Mk×n (R)
be the transpose of A, and let Sym(Rk ) = {B ∈ Mk×k (R) | B t = B} be the vector space of
symmetric k × k-matrices.
(a) Show that the map Φ : Mn×k (R) → Sym(Rk ), A 7→ At A is smooth.
(b) Show that the identity matrix is a regular value of the map Φ.
39
(c) Let Vk (Rn ) ⊂ Mn×k (R) be the subset consisting of those n × k matrices whose column
vectors are unit length and pairwise perpendicular. Show that Vk (Rn ) is a smooth
manifold of dimension kn − k(k + 1)/2. Hint: what is the relationship between Vk (Rn )
and the map Φ?
We remark that identifying Mn×k (R) in the usual way with the vector space Hom(Rk , Rn )
of linear maps f : Rk → Rn , a matrix belongs to Vk (Rn ) if and only if the corresponding
linear map f is an isometry, that is, if f preserves the length of vectors in the sense that
||f (v)|| = ||v||, or equivalently, if f preserves the scalar product in the sense that
hf (v), f (w)i = hv, wi
for all v, w ∈ Rk .
The manifold Vk (Rn ) is called the Stiefel manifold. We observe that Vn (Rn ) is the orthogonal
group O(n) of isometries Rn → Rn . In particular, this problem shows that O(n) is a manifold
of dimension n2 − n(n + 1)/2 = n(n − 1)/2.
Proof. The entries of the matrix At A are quadratic polynomials in the entries of A and hence
the map Φ is smooth.
To prove part (b) we first calculate the differential (DΦ)A (C), C ∈ Mn×k (R) as in the
previous problem as the derivative of the function Φ(A + sC) at s = 0.
Φ(A + sC) = (A + sC)t (A + sC) = At A + s(C t A + At C) + s2 C t C
and hence
∂Φ(A + sC)
(0) = C t A + At C
∂s
−1
To show that for A ∈ Φ (I) (I ∈ Mk×k (R) is the identity matrix) the differential
(DΦ)A : Mn×k (R) → Sym(Rk )
is surjective, let B ∈ Sym(Rk ) and let C = AB ∈ Mn×k (R). Then
(DΦ)A (C) = C t A + At C = (AB)t A + At (AB) = B t At A + At AB = B t + B = 2B,
which proves surjectivity.
According to a theorem proved in class, if f : V → W is a smooth map between finite
dimensional real vector spaces, then the preimage f −1 (w) of a regular value w ∈ W is a
manifold of dimension dim V −dim W . Parts (a) and (b) show that this theorem is applicable
in our situation and hence Vk (Rn ) is a manifold of dimension
dim Mn×k (R) − dim Sym(Rk )
It remains to show that
dim Mn×k (R) = kn
dim Sym(Rk ) = k(k + 1)/2
and
The first statement follows from the fact that a basis of Mn×k (R) is provided by the matrices
{A(i, j)}, 1 ≤ i ≤ n, 1 ≤ j ≤ k, whose entries are all zero except the ij-th entry which is
1. There are nk matrices A(i, j) and hence dim Mn×n (R) = nk. Similarly, a basis for the
40
symmetric k × k matrices is given by the matrices {B(i, j)}, 1 ≤ i ≤ j ≤ k; here B(i, j) is
the matrix whose entries are zero except the ij-th entry and the ji-th entry which are both 1
(obviously, these are the same entries for i = j). For a fixed j between 1 and n, the number
of matrices B(i, j) with 1 ≤ k ≤ j is j. It follows that
dim Sym(Rk ) = 1 + 2 + 3 + · · · + k = k(k + 1)/2
4. Let M , N be non-empty smooth manifolds of dimension m resp. n. Show that if M is
diffeomorphic to N , then m = n.
Proof. Let f : M → N be a diffeomorphism and x ∈ M . Let φ : U → U 0 ⊂ Rm be a smooth
chart for M with x ∈ U , and let ψ : V → V 0 ⊂ Rn be a smooth chart for N with f (x) ∈ V .
Replacing U , V by smaller open subsets, we can assume that f restricts to a diffeomorphism
from U to V . Then the composition
Rm ⊃ U 0
φ−1
open
/
U
f|U
/V
ψ
/
V 0 ⊂ Rn
open
is a diffeomorphism g between U 0 ⊂ Rm and V 0 ⊂ Rn . The differential of g at any point
y ∈ U 0 is then an isomorphism from Rm to Rn (whose inverse is the differential of g −1 at the
point g(y)). This implies m = n.
9
Homework Assignment #9
1. Let M1 , M2 be smooth manifolds, and let πj : M1 × M2 → Mj be the projection onto the
j-th factor for j = 1, 2. Show that for any (p1 , p2 ) ∈ M1 × M2 the map
α : T(p1 ,p2 ) (M1 × M2 ) −→ Tp1 M1 ⊕ Tp2 M2
defined by
α(v) = (Tp1 π1 (v), Tp2 π2 (v))
is an isomorphism. Remark: Using this isomorphism, we will routinely identify Tpj Mj with
a subspace of T(p1 ,p2 ) (M1 × M2 ).
Proof. For ` = 1, 2 let i` : M` → M1 × M2 be the inclusion map given by i1 (x1 ) = (x1 , p2 ),
i2 (x2 y) = (p1 , x2 ). Then for `, j = 1, 2 the composition
πj
i
`
M` −→
M1 × M2 −→ Mj
is the identity map for ` = j, and the constant map x` 7→ pj otherwise. Differentiating it, it
follows that the composition
(πj )∗
(i` )∗
Tp` M` −→ T(p1 ,p2 ) (M1 × M2 ) −→ Tpj Mj
41
is the identity map for ` = j and the trivial map otherwise. It follows that the composition
(i` )∗
α
Tp` M` −→ T(p1 ,p2 ) (M1 × M2 ) −→ Tp1 M1 ⊕ Tp2 M2
sends a vector X` ∈ Tp` M` to (X, 0) for ` = 1 and (0, X) for ` = 2. Since every pair (X1 , X2 )
can be written as a sum of pairs of this type, the map α is surjective. This implies that α is
an isomorphism, since the dimension of domain and range is both dim M1 + dim M2 .
2. Let M , W be smooth manifolds of dimension dim M = n, dim W = k, and let f : M → W
be a smooth map. We recall that w ∈ W is a regular value of f if every x ∈ f −1 (w) is a
regular point of f , i.e., the differential Tx f : Tx M → Tw W is surjective.
(a) Show that N = f −1 (w) is a smooth submanifold of W of dimension n − k. Hint: reduce
to the situation proved in class where M , W are open subsets of Euclidean space.
(b) Show that for x ∈ N = f −1 (w) the tangent space Tx N ⊂ Tx M is the kernel of the
differential f∗ : Tx M → Tw W .
ϕ
Proof. Part (a). Given a point x ∈ M := f −1 (p), we need to find a chart N ⊃ U −→ U 0 ⊂
χ
Rn such that ϕ(U ∩ M ) = U 0 ∩ Rn−k . Let P ⊃ W → W 0 ⊂ Rk be a chart with p ∈ W , and
ψ
let N ⊃ V → V 0 ⊂ Rn be a chart with x ∈ V . Choosing a smaller open neighborhood V of
x if necessary, we can assume that the smooth map f restricts to a smooth map f : V → W .
Identifying V (resp. W ) with the open subsets V 0 (resp. W 0 ) of Euclidean space via the
diffeomorphisms ψ (resp. χ), we can apply our theorem from class according to which the
level set f −1 (p) ⊂ V of the regular value p ∈ W is a submanifold of dimension n − k, giving
us in particular the desired submanifold chart (U, ϕ) at the point x ∈ V .
Part (b). To show that i∗ is injective consider the following commutative diagram of smooth
maps
U ∩M
i
ϕ|U ∩M
/ U0
ϕ
U
∩ Rn−k ⊂ Rn−k
/
j
U 0 ⊂ Rn
Here i0 : U ∩ M → U ⊂ N is the restriction of the inclusion map i : M ,→ N to the open
subset U ∩ M ; since i and i0 agree in a neighborhood of x ∈ M , we have the equality i∗ = i0∗
for their differentials. The map j is the inclusion of Rn−k in Rn given by j(v1 , . . . , vn−k ) =
(v1 , . . . , vn−k , 0, . . . , 0). The horizontal maps are charts for M resp. N containing the point x.
In particular, these a local diffeomorphisms, whose differentials are therefore isomorphisms
of tangent spaces. Functoriality of the derivative then leads to the following commutative
diagram of linear maps
Tx M
i∗
Tx N
(ϕU ∩M )∗
∼
=
ϕ∗
∼
=
/ Tϕ(x) Rn−k
j∗
/ Tϕ(x) Rn
42
= Rn−k
= Rn
This shows that the map i∗ is injective, since it corresponds via the horizontal isomorphisms
of the above diagram to the injective map j∗ = j (the derivative of the linear map j is again
j).
To show that the image of i∗ : Tx M → Tx N is equal to the kernel of f∗ : Tx N → Tp P , we
note that the composition
M
i
/
N
f
/P
is the constant map that sends every point of M = f −1 (p) to p. This implies that the
derivative of f ◦ i is zero. Since (f ◦ i)∗ = f∗ ◦ i∗ , this implies that the image of i∗ is
contained in the kernel of f∗ . To prove im i∗ = ker f∗ , we will argue that these vector
spaces have the same dimension, and hence im i∗ ⊂ ker f∗ implies their equality. Since i∗ is
injective, dim im i∗ = dim Tx M = dim M = n − k. Since f∗ : Tx N → Tp P is surjective by the
assumption that p is a regular value of f , we have
dim ker f∗ = dim Tx N − dim im f∗ = dim Tx N − dim Tp P = n − k.
3. Let O(n) be the orthogonal group (which is the submanifold of Mn×n (R) consisting of
those n × n matrices A with At A = e; here e denotes the n × n identity matrix which is
the identity element of the group O(n)). Identify the tangent space Te O(n) as a subspace of
Te Mn×n (R) = Mn×n (R). Hint: Use part (b) of the previous problem.
Proof. Let Φ : Mn×n (R) → Mn×n (R) be the map given by A 7→ At A. Then by definition the
orthogonal group O(n) is the level set Φ−1 (11) where 11 ∈ Mn×n (R) is the identity matrix. We
remark that 11 can’t possibly be a regular value of the smooth map f : Mn×n (R) → Mn×n (R),
since otherwise O(n) = f −1 (11) would be a submanifold of Mn×n (R) whose dimension is the
difference of the dimensions of the vector spaces that are the domain resp. range of f , leading
to the obviously wrong conclusion that O(n) is a manifold of dimension 0!
To fix this problem, we note that At A is always a symmetric matrix, since (At A)t =
t
A (At )t = At A. This allows us to think of Φ as a smooth map
Φ : Mn×n (R) −→ Sym(Rn )
to the vectorspace Sym(Rn ) of symmetric n × n-matrices.
In problem 3 of the last homework set we computed the differential of Φ (actually, the
map Φ of that problem is more general, but specializing to k = n it agrees with the map
considered here). We showed that the differential
(DΦ)A : TA Mn×n (R) −→ TΦ(A) Sym(Rn )
is given by
(DΦ)A (C) = C t A + At C,
and went on to argue that (DΦ)A is surjective for A ∈ Φ−1 (11) = O(n) which shows that
O(n) is a submanifold of Mn×n (R).
By part (b) of problem 2 of this set, the tangent space of a level set Φ−1 (λ) of a regular
value λ at a point A ∈ Φ−1 (λ) is equal to the kernel of the differential (DΦ)A . In the case
at hand, this means that the tangent space TI O(n) at the identity element e = I ∈ O(n) is
equal to the kernel of the differential
(Dφ)I : TI Mn×n (R) = Mn×n (R) −→ TI Sym(Rn ) = Sym(Rn )
43
which we computed to be (DΦ)I (C) = C t + C. This shows that TI O(n) = ker(DΦ)I consists
of the vector space of skew-symmetric matrices, that is, matrices C satisfying C t = −C.
4. Show that the projection map p : S 2n+1 → CPn is a submersion, i.e., the differential
p∗ : Tz S 2n+1 → Tp(z) CPn is surjective for every point z ∈ S 2n+1 . (In particular, by 2(a) each
fiber p−1 (L), L ∈ CPn is a submanifold of dimension 1). Hint: The projection map p extends
to a projection map P : Cn+1 \ {0} → CPn . Argue first that P is a submersion, using the
obvious identification of tangent spaces of the domain with Cn+1 , and using the differential
of our standard charts for CPn to identify the tangent spaces of the range with Cn .
Proof. Let z = (z0 , . . . , zn ) ∈ S 2n+1 ⊂ Cn+1 , and let p(z) = [z0 , . . . , zn ] ∈ CPn be the image
of z under the projection map. We recall from class that the tangent space Tz S 2n+1 can be
identified with the vectors w = (w0 , . . . , wn ) ∈ Cn which are perpendicular to z, i.e., those
w for which the Hermitian inner product
hz, wi = z̄0 w0 + z̄1 w1 + · · · + z̄n wn
vanishes. This suggests to extend the map p to all of Cn \ {0} by defining
P : Cn \ {0} −→ CPn
by
P (z0 , . . . , zn ) := [z0 , . . . , zn ].
Then p∗ is the restriction of P∗ to Tx S 2n+1 ⊂ Tx (Cn \ {0}) = Cn+1 .
To calculate P∗ we use the standard charts
ϕi
CPn ⊃ Ui = {[z0 , . . . , zn ] ∈ CPn } −→ Ui0 = Cn
zbi
zn
z0
[z0 , . . . , zn ] 7→ ( , . . . , , . . . , )
zi
zi
zi
to identify the tangent space Tp(z) CPn at a point p(z) ∈ Ui with Tϕi (p(z)) Ui0 = Cn via the
differential (ϕi )∗ of the local diffeomorphism ϕi .
To calculate (ϕi )∗ ◦ P∗ = (ϕi ◦ P )∗ we first compute
ϕi ◦ P (z0 , . . . , zn ) = ϕi ([z0 , . . . , zn ]) = (
z0
zbi
zn
, . . . , , . . . , ).
zi
zi
zi
(9.1)
Then the differential of ϕi ◦ P at z = (z0 , . . . , zn ) ∈ Cn+1 with zi 6= 0 can be calculated as
D(ϕi ◦ P )z (w) =
∂ϕ ◦ P (z + sw)
.
∂s
|s=0
By equation (??) the k-th component of (ϕi ◦ P )(z) ∈ Cn has the form zj /zi for j = k − 1
for k < i and j = k for k ≥ i. Hence the k-th component of D(ϕi ◦ P )z (w) is given by
∂
zj + swj
wj zi − zj wi
=
.
∂s |s=0 zi + swi
zi2
(9.2)
We claim that the differential D(ϕi ◦ P )z : Tz (Cn+1 \ {0}) → TP (z) Cn = Cn is surjective. To
show that a vector v = (v1 , . . . , vn ) ∈ Cn is in the image of D(ϕi ◦ P )z , we need to find a
vector w = (w0 , . . . , wn ) ∈ Cn+1 such that for each k = 1, . . . , n we have
wj zi − zj wi
= vk ,
zi2
44
(9.3)
where as above j = k − 1 for k < i and j = k for k ≥ i. This is easy to solve: choose wi = 0;
since zi 6= 0 by assumption, we can solve equation (9.3) uniquely for wj . This shows that the
differential of P : Cn \ {0} → CPn is surjective. To show that the differential of p (which is
the restriction of P to S 2n+1 ⊂ Cn \ {0}) is surjective, we need to argue that we can choose
the vector w ∈ Cn+1 to be perpendicular to z ∈ Cn+1 . We observe that if w = λz for some
λ ∈ C, then the calculation (9.2) shows that w is in the kernel of DPz . In particular, DPz (w)
doesn’t change if we replace w by its projection to z ⊥ .
5. Let h : RPn → R be the function defined by
h([x0 , . . . , xn ]) = (
n
X
`x2` )(x20 + x21 + · · · + x2n )−1 .
`=0
(a) Show that h is a smooth function.
(b) Determine the critical points of h, i.e., the points of RPn which are not a regular points
for h.
ϕi
Hint: Use the smooth atlas consisting of the charts RPn ⊃ Ui −→ D̊n with
p
Ui = {[x0 , . . . , xn ] ∈ RPn | xi 6= 0} and ϕ−1 (v1 , . . . , vn ) = [v1 , . . . , vi , 1 − ||v||2 , vi+1 , . . . , vn ].
Proof. Part (a). To show that h is smooth it suffices to show that the composition
n
h ◦ ϕ−1
i : D̊ → R
is smooth for 0 ≤ i ≤ n. We compute:
(h ◦ ϕ−1
i )(v1 , . . . , vn ) = h([v1 , . . . , vi ,
=
i−1
X
2
`v`+1
p
1 − ||v||2 , vi+1 , . . . , vn ])
2
+ i(1 − ||v|| ) +
`v`2
`=i+1
`=0
=
n
X
i−1
X
n
X
`=0
`=i+1
2
(` − i)v`+1
+i+
(9.4)
(` − i)v`2
This shows that h ◦ ϕ−1
i is a quadratic polynomial and hence in particular a smooth function.
Part (b). A point [x0 , . . . , xn ] ∈ Ui is a critical points of h if and only if (v1 , . . . , vn ) =
ϕ([x0 , . . . , xn ]) ∈ D̊n is a critical point of f := h ◦ ϕ−1
i . Moreover, v = (v1 , . . . , vn ) is a
critical point of f if and only if the gradient of f vanishes at v. We compute:
grad f = (
∂f
∂f
,...,
) = (2(−i)v1 , 2(1−i)v2 , . . . , 2(i−1−i)vi , 2(i+1−i)vi+1 , . . . , 2(n−1)vn ).
∂v1
∂vn
Since the coefficient in front of vk is non-zero for each 1 ≤ k ≤ n, this shows that the gradients
of f only vanishes at (v1 , . . . , vn ) = (0, . . . , 0). This implies that each open subset Ui ⊂ RPn ,
0 ≤ i ≤ n contains exactly one critical point, namely φ−1
i (0, . . . , 0) = [0, . . . , 0, 1, 0, . . . , 0]
(the non-zero component is in the i-th slot).
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10
Homework Assignment # 10
1. Show that the bracket of vector fields satisfies the Jacobi identity, that is, if U, V, W are
vector fields on a smooth manifold M , then
[[U, V ], W ] + [[W, U ], V ] + [[V, W ], U ] = 0.
In particular, the bracket of vector fields gives the vector space X(M ) of vector fields on M
the structure of a Lie algebra (the other two requirements for the bracket, its linearity in
each slot and its skew symmetry are evident).
Proof. We recall that [U, V ] = U V − V U , where U V resp. V U are the compositions as
endomorphisms of the vector space C ∞ (M ). We compute:
[[U, V ], W ] = [U V −V U, W ] = (U V −V U )W −W (U V −V U ) = U V W −V U W −W U V +W V U.
Cyclically permuting the symbols U , V , W then leads to the following 12 term formula:
[[U, V ], W ] + [[W, U ], V ] + [[V, W ], U ] =U V W − V U W − W U V + W V U
+V W U − W V U − U V W + U W V
+W U V − U W V − V W U + V U W
Inspection shows that each of the six possible permutations of U , V , W occurs twice with
opposite sign, thus proving the Jacobi identity for the Lie bracket of vector fields.
2. Let V x , V y , V z be the following vector fields in R3 which are the infinitesimal generators
of rotation around the x-axis (resp. y-axis resp. z-axis).
Vx =y
∂
∂
−z
∂z
∂y
Vy =z
∂
∂
−x
∂x
∂z
Vz =x
∂
∂
−y .
∂y
∂x
Show that
[V x , V y ] = −V z
[V y , V z ] = −V x
[V z , V x ] = −V y .
In particular, the span of these vector fields in the space X(R3 ) of vector fields on R3 is a
Lie algebra of dimension 3.
Proof. For f ∈ C ∞ (R3 ) we have
[Vx Vy ]f = Vx Vy f − Vy Vx f.
Let us first calculate Vx Vy f :
∂
∂
∂f
∂f
Vx Vy f = y
−z
z
−x
∂z
∂y
∂x
∂z
2
2
∂f
∂ f
∂ 2f
∂ 2f
2 ∂ f
=y
+ yz
−z
− yx 2 + zx
∂x
∂z∂x
∂y∂x
∂z
∂y∂z
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∂
We note that there are five terms, since evaluating the vector field y ∂z
on the product
∂f
function z ∂x necessitates the use of the product rule. The corresponding expression for
Vy Vx f is obtained by simply permuting the symbols x, y on the right hand side of the
previous equation. We see that for their difference Vx Vy f − Vy Vx f most terms cancel, leaving
us only with
∂f
∂f
−x
= −Vz .
Vx Vy f − Vy Vx f = y
∂x
∂y
The other two equations follow by cyclically permuting the symbols x, y, z.
3. Let F : M → N be a smooth map, and let F ∗ : C ∞ (N ) → C ∞ (M ) be the induced map
of functions given by f 7→ f ◦ F .
(a) Let x ∈ M and let v ∈ Tx M = Derx (C ∞ (M ), R) be a tangent vector. Show that the
composition
∗
C ∞ (N ) F / C ∞ (M ) v / R
is an element of TF (x) N = DerF (x) (C ∞ (N ), R). We remark that this shows that we can
define the differential F∗ : Tx M → Ty N by v 7→ v ◦ F ∗ .
(b) Let V ∈ X(M ) = Der(C ∞ (M )) be a vector field. For x ∈ M define Vx to be the
composition
evx /
V /
C ∞ (M )
R
C ∞ (M )
where evx is the evaluation map which sends f ∈ C ∞ (M ) to f (x) ∈ R. Show that Vx is
an element of Derx (C ∞ (M ), R). In other words, Vx is a tangent vector at the point x.
(c) Let V ∈ X(M ), W ∈ X(N ) be vector fields on M resp. N . Show that the following two
conditions are equivalent.
(i) F∗ (Vx ) = WF (x) ∈ TF (x) N for every point x ∈ M .
(ii) V ◦ F ∗ = F ∗ ◦ W : C ∞ (N ) → C ∞ (M ).
If these conditions are met, then the vector fields V, W are called F -related.
(d) Show that the Lie bracket is natural in the following sense. Let Vi ∈ X(M ) be vector
fields that are F -related to vector fields Wi ∈ X(N ) for i = 1, 2. Show that [V1 , V2 ] is
F -related to [W1 , W2 ].
Proof. Part (a). We observe that F ∗ : C ∞ (N ) → C ∞ (M ) is an algebra homomomorphism,
since for f, g ∈ C ∞ (N ), x ∈ M we have
(F ∗ (f g))(x) = (f g)(F (x)) = f (F (x))g(F (x)) = ((F ∗ f )(x))((F ∗ g)(x)) = ((F ∗ f )(F ∗ g))(x).
It follows that v ◦ F ∗ is a derivation at F (x) since
(v ◦ F ∗ )(f g) =v(F ∗ (f g)) = v(F ∗ (f )F ∗ (g))
=v(F ∗ (f ))(F ∗ (g)(x)) + (F ∗ (f )(x))v(F ∗ (g))
=(v ◦ F ∗ )(f )g(F (x)) + f (F (x))(v ◦ F ∗ )(g).
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Part (b). For f, g ∈ C ∞ (M ) we have
Vx (f g) = evx (V (f g)) = evx (V (f )g + f V (g))
=(V (f )(x))g(x) + f (x)(V (g)(x)
=(evx (V (f )))g(x) + f (x)(evx (V (g)))
which proves that evx ◦V is a derivation at x ∈ M .
Part (c). Evaluating the tangent vector of both sides of the equation in part (i) on a
function f ∈ C ∞ (N ) we obtain:
(F∗ (Vx ))(f ) = Vx (F ∗ f ) = evx (V (F ∗ f )) = ((V ◦ F ∗ )(f ))(x)
WF (x) (f ) = evF (x) (W (f )) = (W (f ))(F (x)) = (F ∗ (W (f )))(x) = ((F ∗ ◦ W )(f ))(x)
This shows that evaluating the left (resp. right) hand side of equation (i) on a function
f ∈ C ∞ (N ) is equal to the left (resp. right) hand side of equation (ii) evaluated on f (giving
elements of C ∞ (M )), and then evaluated at the point x ∈ M of part (i), giving real numbers.
In particular, the conditions (i) and (ii) are equivalent.
Part (d).
[V1 , V2 ]F ∗ = V1 V2 F ∗ − V2 V1 F ∗ = V1 F ∗ W2 − V2 F ∗ W1 = F ∗ W1 W2 − F ∗ W2 W1 = F ∗ [W1 , W2 ]
Here the second and third equation hold since Vi F ∗ = F ∗ Wi , the assumption that the vector
field Vi is F -related to Wi . This shows that [V1 , V2 ] is F -related to [W1 , W2 ].
4. Let G be a Lie group. For g ∈ G let Lg : G → G, h 7→ gh be the smooth map given by
left multiplication by g.
(i) Show that the following conditions are equivalent for a vector field V ∈ X(G):
(a) (Lg )∗ (Vh ) = Vgh ∈ Tgh G for all g, h ∈ G;
(b) L∗g ◦ V = V ◦ L∗g for all g ∈ G.
A vector field on G satisfying these conditions is called left invariant.
(ii) Show that if V , W are left invariant vector fields on G, then so is their bracket [V, W ].
In particular, the vector space of left-invariant vector fields on G is a Lie algebra, called
the Lie algebra of G.
Proof. Part (a). These conditions are equivalent by part (c) of the previous problem.
Part (b). Comparing the definition of left-invariance with the definition of F -related vector
fields of part (c) of the previous problem, we see that a vector field V on G is left invariant
if and only if V is Lg -related to itself for all g ∈ G. Part (d) of the previous problem shows
that if V and W are Lg -related to V (resp. W ), then [V, W ] is Lg -related to [V, W ] which
proves the statement.
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