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TECHNISCHE UNIVERSITÄT MÜNCHEN
Zentrum Mathematik
Prof. Dr. M. Wolf
A. Müller-Hermes
Differential Topology
http://www-m5.ma.tum.de/Allgemeines/MA5122 2014S
Sommers. 2014
Lösungsblatt 1
(09.04.2014)
Tutoraufgaben
1. Homeomorphism between the projective line and the circle
Show that RP 1 ' S 1 , where RP 1 := R2 \ {0} / ∼ for the equivalence relation x ∼ y ⇔
∃λ ∈ R : λx = y on R2 \ {0}, denotes the real projective line.
Lösung:
1 / ∼ , where S 1 = {x ∈
The proof goes in two steps. First we show, that RP 1 ' S+
1
+
1
S |x2 ≥ 0} denotes the upper half-circle. Here x ∼1 y ⇔ (x = y) ∨ {x, y} = {−1, 1}
1 . Take the function
denotes the equivalence relation identifying the points −1 and 1 on S+
1 / ∼ defined via
g : R2 \ {0} → S+
1

x

, x2 > 0

 kxk
g(x) =
x
− kxk


{−1, 1}
, x2 < 0
, else
This map is surjective and has the properties that g (x) = g (y) iff x ∼ y and also V ⊂
1 / ∼ is open (wrt quotient topology on S 1 / ∼ ) iff g −1 (V ) is open in R2 \ {0}. Now
S+
1
1
+
1 / ∼ → RP 1 via h(x) = q g −1 ({x}) , where q : R2 \{0} →
define a homeomorphism h : S+
1
RP 1 maps representatives of an equivalence class, to this equivalence class in RP 1 (see
definition in lecture notes). Using the properties of g it is easy to show, that h is indeed
1 / ∼ is open iff g −1 (V ) = q −1 (h (V )) is open iff h (V )
bijective and it holds, that V ∈ S+
1
is open wrt the quotient topology of RP 1 showing continuity in both directions.
1 / ∼ ' S 1 . For this construct a homeomorphism via the
Finally we have to show, that S+
1
1
1
map f : S+ / ∼1 → S defined via
(
(1, 0)
, φ ∈ {0, π}
f ((cos(φ), sin(φ))) =
(cos(2φ), sin(2φ)) , else
using polar coordinates. It is easy to see, that this is a homeomorphism, which finishes
the proof.
2. Connectedness
A topological space X is called path-connected if for all x, y ∈ X there is a continuous
function γ : [0, 1] → X, a path, such that γ (0) = x and γ (1) = y.
(a) Show that a topological manifold is connected iff it is path-connected.
(b) Give an example of a topological space that is connected, but not path-connected.
Lösung:
(a) Let M be the topological manifold. By H1.1 we know, that if M is path-connected it
also is connected. So we only have to show the other direction. Therefore consider the
set
Cx := {y ∈ X|∃γ : [0, 1] → X path , γ(0) = x, γ(1) = y}.
If this set is open and closed in M we are finished, because we can consider the partition
M = Cx ∪ (M \ Cx ) into open and disjoint sets. When M is connected one of the two
sets has to be empty, which shows that Cx = M as x ∈ Cx 6= ∅. It remains to show that
Cx is open and closed. Therefore take an arbitrary point y ∈ Cx . As M is a topological
manifold, there is a neighborhood Uy ⊆ M and a homeomorphism h : Uy → Rn for
some n ∈ N. Consider an open ball B (h (y)) ⊂ h (Uy ) ⊂ Rn around h(y). As this
ball is open and connected, there is a path for every h (z) ∈ B (h (y)) connecting it
to h (y) ∈ B (h (y)). Because h is a homeomorphism every such path leads to a path
from z to y in Uy , which shows, that there is an open set Ũy := h−1 (BS(h (y))) ⊂ Cx
that contains y. As this is true for every y ∈ Cx , we have Cx = y∈Cx Ũy and
therefore Cx open. To show that Cx is closed consider a limit point y ∈ M of a
sequence (yk )k∈N ∈ CxN . Consider now an open neighborhood Uy ⊆ M of y ∈ M
such that there is a homeomorphism h : Uy → h(Uy ) ⊆ Rn . Again there is an open
ball B (h (y)) ⊂ h (Uy ) ⊂ Rn around y. By continuity there is a K ∈ N such that
h (yk ) ∈ B (h (y)) for all k ≥ K and B (h (y)) is an connected subset of Rn , there is a
path from h(yk ) to h(y) for an arbitrary k ≥ K. The homeomorphism maps this path
to a path in M, which connects yk and y. Concatenating this path with a path from
x to yk ∈ Cx gives a path from x to y and shows y ∈ Cx . Therefore Cx is open and
closed which finishes the proof.
(b) An example is given by the set S = ({0} × [−1, 1]) ∪ graph sin x1 ⊂ R2 , which is
called the ’topologist’s sine curve’. It is clear, that S is not path-connected, as there
cannot be a path connecting (e, sin(1/e)) and (0, 0). To see that S is connected, we
show S = S+ for S+ := {(x, y) ∈ S : x > 0}. It is easy to see, that S+ is connected
(even path-connected). Now consider a separation of S = S+ into two closed (wrt
subset topology on S), disjoint subsets U1 , U2 ⊂ R2 . These lead to a separation of S+
into two closed (wrt subset topology on S+ ), disjoint subsets U1 ∩ S+ and U2 ∩ S+ . As
S+ is connected we have wlog U1 ∩ S+ = S+ . As U1 is closed wrt the subset topology
on S, there is a closed (now wrt to the R2 topology) set C ⊂ R2 , with U1 = C ∩ S.
But this gives S+ ⊂ U1 ⊂ C for a closed subset C, which shows that S+ ⊂ C which
gives U1 = S+ . This finishes the proof.
Note: We just proved that if a set is connected so is its closure. The example shows,
that unlike connectedness, path-connectedess is not preserved under the operation of
taking the closure of a set.
Hausaufgaben
1.1. Connected vs. Path-Connected
A topological space X is called path-connected if for all x, y ∈ X there is a continuous
function γ : [0, 1] → X, i.e. a path, such that γ (0) = x and γ (1) = y.
(a) Show that every path-connected topological space is also connected.
(b) Is connectedness preserved under homeomorphisms?
(c) Is R2 homeomorphic to R1 ?
Lösung:
(a) We will first show that for every path γ : [0, 1] → X the image γ ([0, 1]) is connected.
Therefore take two open (wrt subspace topology on γ ([0, 1])) non-empty sets Z1 , Z2
and assume γ ([0, 1]) = Z1 ∪ Z2 . As [0, 1] is connected, we get from [0, 1] = γ −1 (Z1 ) ∪
γ −1 (Z2 ), that γ −1 (Z1 )∩γ −1 (Z2 ) 6= ∅. Here we used that γ −1 (Zi ) is open (continuity)
and also non-empty for i = 1, 2. But this means that Z1 ∩ Z2 6= ∅, which finishes the
argument. Now consider a partition of the space X = X1 ∪ X2 for X1 , X2 open and
non-empty. As for every path γ : [0, 1] → X connecting x ∈ X1 to y ∈ X2 , we have a
partition γ ([0, 1]) = (γ ([0, 1]) ∩ X1 ) ∪ (γ ([0, 1]) ∩ X2 ) into non-empty and open sets,
we are finished as X1 ∩ X2 ⊇ (γ ([0, 1]) ∩ X1 ) ∩ (γ ([0, 1]) ∩ X2 ) 6= ∅.
(b) Yes, as a partition X = X1 ∪ X2 into two non-empty, open sets yields a partition
h (X) = h (X1 ) ∪ h (X2 ) into two non-empty, open sets and X1 ∩ X2 6= ∅ iff h (X1 ) ∩
(X2 ) 6= ∅.
(c) No, as can be seen by considering R2 \ {0}, which is still connected. Removing any
point from R leaves it disconnected and as connectedness is preserved under homeomorphisms we would get a contradiction.
Remark: This is a special case of the “Topological invariance of dimension Theorem”
that Rn and Rm are not homeomorphic if n 6= m. This theorem is astounishingly hard
to proof and there is currently no elementary proof known. On the other hand it is
very easy to prove the corresponding smooth version of the statement, that Rn and
Rm are not diffeomorphic if n 6= m. This illustrates the differences between continuous
and differential topology.
1.2. Some topology
Let X denote a topological space and Y ⊆ X a subspace equipped with the subset
topology. A topological space X is called second countable if there exists a countable
basis for the topology of X.
(a) Show that Rn is a second countable Hausdorff space.
(b) Show Y is a topological Hausdorff space if X is a topological Hausdorff space.
(c) Show Y is second-countable if X is second countable.
Note: This implies that any open subset of a topological manifold is also a topological
manifold.
Lösung:
a) Consider the basis given by {Bq (p) ⊆ Rn |q ∈ Q, p ∈ Qn }, i.e. balls around rational
points and with rational radii. This is clearly a countable set, and as Qk is dense in
Rk for all k ∈ N, this is also a basis.
b) Consider Y ⊆ X with the subset topology. Take two distinct points y1 , y2 ∈ Y . As
X is a Hausdorff space, there are disjoint open neighborhoods Uk 3 yk for k = 1, 2.
But this means that U1 ∩ Y, U2 ∩ Y ⊆ Y are disjoint open (wrt subset topology)
neighborhoods of y1 , y2 in Y, which finishes the proof.
c) For a countable basis {Uk }k∈N of the topology of X, we consider the set {Uk ∩ Y }k∈N .
This is a basis of the subset topology on Y and countable by construction.
1.3. Real Projective spaces are topological manifolds
Show that RP n equipped with the quotient topology is a topological manifold. Therefore
show that
(a) RP n is a Hausdorff space.
(b) RP n is second-countable.
(c) Now construct charts φi : Ui → Rn , where Ui ⊂ RP n denote open sets covering RP n .
(Hint: A possible choice is Ui = q (Ui0 ) for the open sets Ui0 = {x ∈ Rn+1 \{0}|xi 6= 0},
where q : Rn+1 \ {0} → RP n denotes the quotient map used to define RP n . Why are
the Ui open?)
Lösung:
(a) Take two points x, y ∈ RP n such that x 6= y. By definition there are sx , sy ∈ S n which
are representatives of x, y. Now we use that S n is a Hausdorff space as can be verified
easily, to construct open neighborhoods Ux , Uy ⊂ S n of sx and sy with the following
properties:
Ux ∩ Uy = ∅
−Ux ∩ Uy = ∅
Ux ∩ −Uy = ∅
−Ux ∩ −Uy = ∅
Here −Ux := {s ∈ S n : −s ∈ Ux }. This construction is possible as one can use the
Hausdorff property of S n to construct first Ux such that sy ∈
/ Ux and −sy ∈
/ Ux . Then
construct a candidate for Uy in each of the 4 set equations above. An intersection of
these candidates yields Uy . It is now easy to see, that q (Ux ) ⊂ RP n and q (Uy ) ⊂ RP n
are disjoint, as z ∈ p (Ux ) ∩ p (Uy ) would imply the existence of s ∈ Ux and r ∈ Uy
and λ, µ ∈ {±1} such that λs = z = µr. But depending on λ, µ this would contradict
one of the 4 set equations above. It is clear that q (Ux ) and q (Uy ) contain open sets,
which finishes the proof.
Another way: For x ∈ RP n consider the map αx : RP n → R defined by αx (y) =
1 − | hỹ, x̃i |2 for representatives ỹ, x̃ ∈ S n of x, y. Here h•, •i denotes the standard
scalar product on Rn+1 . This is a well-defined map and by realizing that αx ◦ q is
continuous, we see that αx is continuous too (see (c)). Now consider x, y ∈ RP n with
x 6= y and note that | hỹ, x̃i | = 1 implies that
|x̃ − ỹ hỹ, x̃i |2 = hx̃, x̃i + hỹ, ỹi | hỹ, x̃i |2 − | hỹ, x̃i |2 − | hỹ, x̃i |2 = 0
for representatives x̃, ỹ ∈ S n of x and y. This shows that x 6= y implies that 0 =
αx (x) 6= αx (y) and as R is hausdorff there are two open disjoint neighborhoods
Ux , Uy ⊂ R of αx (x) and αx (y). By continuity the disjoint sets αx−1 (Ux ) and αx−1 (Ux )
are open in RP n , which finishes the proof.
Third way: With Lemma from the lecture it isenough to show, that Γ = {(x, y) :
x ∼ y} is closed in Rn+1 \ {0} × Rn+1 \ {0} . Therefore consider the functions
fij (x, y) = xi yj − yi xj . As these functions are continuous, every set fij−1 ({0}) is
closed. As intersections of closed sets are closed it is also true, that the set where all
fij vanish is closed. But this set coincides with Γ, so the proof is finished.
Important Note: The above proofs use the properties of RP n and that the equivalence classes, which define this space, are sufficiently nice. There are however examples
of quotient spaces of Hausdorff-spaces that are not Hausdorff themselves. An example
is given by [0, 1] / ∼ for x ∼ y iff |x − y| ∈ Q, as its only open sets wrt the quotient
topology are ∅ and [0, 1] / ∼.
n+1 \ {0} is open
(b) Consider an open set U ⊂ Rn+1 \ {0}.
S As λU := {λx : x ∈ U } ⊂ R
−1
for λ 6= 0, we see that q (q (U )) = λ∈R\{0} λU is also an open set. This shows that
q (U ) ⊂ RP n is open for an open set U ⊂ Rn+1 \ {0}.
Now we take a countable basis {Uk }k∈N for the topology of Rn+1 . By definition for
any open neighborhood V ⊂ RP n of x ∈ RP n with representative x̃ ∈ Rn+1 \ {0}, the
set q −1 (V ) is open and contains x̃. Therefore there is an element Ul ⊂ q −1 (V ) from
the countable basis, which is mapped to the open set q (Ul ) ⊂ V . But this means that
{q (Uk )}k∈N is a countable basis for the topology on RP n .
(c) In order to show, that RP n is a topological manifold, we will now construct local
charts mapping a neighborhood of a point x ∈ RP n to Rn . As in the hint above,
we consider Ui = q (Ui0 ) for the open sets Ui0 = {x ∈ Rn+1 \ {0}|xi 6= 0}. The
n
sets Ui are open
by the argument given in (b). Now we define maps φi : Ui → R
x̃
x̃
x̃
by φi (x) = x̃x̃1i , . . . , x̃i−1
, x̃i+1
, . . . , n+1
for a representative x̃ of x. These maps
x̃i
i
i
are well-defined on Ui as a scaling by λ 6= 0 on the rhs does not matter. They are
0
n for an open set
continuous as can be seen by considering V = φ−1
i (V ) ⊂ RP
0
n
−1
n+1
V ∈ φi (Ui ) ⊂ R . By definition V is open if q (V ) ⊂ R
\ {0} is open. This
n+1
0
condition is thus equivalent to saying that φi ◦ q : R
\ {0} ⊃ Ui → Rn is continuous
(this is the general definition
of continuity in the quotient topology). It is now easy to
x̃1
, x̃x̃i+1
, . . . , x̃n+1
is a continuous map. The inverses
see that φi ◦ q (x̃) = x̃i , . . . , x̃x̃i−1
x̃i
i
i
φ−1
: φi (Ui ) → Ui of the charts are given by φ−1
i
i (y) = q (y1 , . . . , yi−1 , 1, yi , . . . , yn ),
which are also continuous. This shows that the φi are homeomorphisms and realizing
that the Ui are covering RP n finishes the proof.
1.4. Properties of matrix groups
Let B : Rn × Rn → R denote a bilinear form on Rn . We define the invariance group of
B via
On (B) := {T ∈ Rn×n |∀x, y ∈ Rn : B (T x, T y) = B (x, y)}
(a) Show that On (B) is closed in Rn×n for every bilinear form B : Rn × Rn → R.
(b) The orthogonal group is O (n) := On (h•, •i), where h•, •i : Rn × Rn → R denotes the
usual scalar product on Rn . Show that O (n) is compact.
(c) Show that GL (n) := {X ∈ Rn×n |det (X) 6= 0} is open and dense in Rn×n . Fun Fact:
When B is non-degenerate, On (B) ⊂ GL (n)
(d) On R4 , we define a bilinear form

hx, yiM
−1

0
:= xT 
0
0
0
1
0
0
0
0
1
0

0
0
y
0
1
Let O (3, 1) denote the invariance group of this bilinear form called the Lorentzgroup. Show that the special Lorentz-group SO (3, 1) := {L ∈ O (3, 1) |det (L) =
1} is not connected.
Lösung:
(a) Let (Tk )k∈N ∈ On (B)N be a convergent sequence with limit point T ∈ Rn×n . Then we
have
0 = lim B (T x − Tk x, T y − Tk y)
k→∞
= B (T x, T y) − lim B (Tk x, T y) − lim B (T x, Tk y) + B (x, y)
k→∞
k→∞
= −B (T x, T y) + B (x, y) ,
which implies T ∈ On (B). Here we used continuity of B and the definition of On (B).
(b) According to (a) On is closed. Note that T ∈ On iff kT xk = kxk for all x ∈ Rn by
definition. This shows, that kT k = 1 for all T ∈ On and therefore On is bounded in
operator norm.
(c) The continuity of det implies that GL (n) = det−1 (R \ {0}) is open. Let A ∈ Rn×n be
arbitrary. Choose the sequence Q
Xk = A + ka I where a < min{|λ| : λ ∈ spec (A) \ {0}}.
Then Xk → A and det (Xk ) = nl=1 λl + ka 6= 0 for all k ∈ N.
(d) Note that A ∈ O (3, 1) iff AT M A = M , where

−1 0 0
0 1 0
M := 
0 0 1
0 0 0

0
0
 .
0
1
By looking at the matrix components of the above identity, we obtain
A200 = 1 +
3
X
(Ak0 )2 ≥ 1 .
k=1
From this we get the two components SO+ (3, 1) = {A ∈ SO (3, 1) |A00 ≥ 1} and
SO− (3, 1) = {A ∈ SO (3, 1) |A00 ≤ −1}. It is easy to see, that SO (3, 1) = SO+ (3, 1) ∪
SO− (3, 1) and that the two components are closed and disjoint, which shows that
SO (3, 1) is not connected. The component SO+ (3, 1) is also called the proper orthochronous Lorentz group. It contains all Lorentz transformations, which preserve the direction of time. Likewise SO− (3, 1) contains all Lorentz transformations,
which reverse the direction of time. Together with the choice det (A) = ±1 this shows,
that the Lorentz-Group O (3, 1) has 4 connected components.
1.5. Bonus: A topological riddle
It is easy to see, that one can draw uncountably many circles into the 2-dimensional
plane, such that no two of them intersect (just draw a circle for every radius r ∈ (0, 1]
with common center). We call a ’topological Eight’ a dot with two infinitely thin nonintersecting handles attached to it (see Figure). These handles can be deformed, shrinked,
etc. but they cannot be made into a line or made vanish.
(a) Is it possible to draw uncountably many topological eights in the 2-dimensional plane,
such that no two of them intersect in a point? (See Figure for a drawing of finitely
many topological eights)
(b) How about topological Y’s, defined as a point with three arms coming out of it?(See
Figure)
Lösung:
No solution to riddles. This would spoil all the fun.
Further Information:
Lecture: Mi, 10:15 - 11:45, MI 00.07.011
Excercise Class: Mo (every second week), 14:15 - 15:45, MI 03.08.011
First Excercise Class: 14.04.2014