Download Manifolds and Topology MAT3024 2011/2012 Prof. H. Bruin

Manifolds and Topology
MAT3024
2011/2012
Prof. H. Bruin
Department of Mathematics
University of Surrey
1
Basic notions in topology
Study advice:
• Proofs in topology almost always start with definitions. Hence know your definitions.
In working out proofs, write down the definitions to check what exactly is needed.
• Learn definitions always with an example in mind, and a counter-example.
• It also helps to draw a picture (although the picture may be more abstract than in
the Euclidean plane).
• Focus on examples to understand the basic concepts. You should know the properties of Rn , Tn , Sn and the disk D. Are they Hausdorff, connected, simply connected,
compact? What is their dimension? How to parametrise them, and are they pairwise
homeomorphic? Why (not)?
1.1
Topological Spaces
So far, the mathematical spaces with the least structure you may have seen are metric
spaces. A metric d is a general version of distance. We call (X, d) a metric space if the
following four properties hold:
1. For all x, y ∈ X, 0 ≤ d(x, y) < ∞.
2. d(x, y) = 0 if and only if x = y.
3. Symmetry: For all x, y ∈ X, d(x, y) = d(y, x).
4. Triangle inequality: For all x, y, z ∈ X, d(x, z) ≤ d(x, y) + d(y, z).
pPn
2
Examples: (i) Euclidean space: X = Rn and d2 (x, y) =
i=1 (xi − yi ) .
(ii) Discrete space. d(x, y) = 0 if x = y and d(x, y) = 1 otherwise.
(iii) Sup-metric for a space of real-valued functions: d∞ (f, g) = supx |f (x) − g(x)|.
P
(iv) Sum-metric for Rn : d1 (x, y) = ni=1 |xi − yi |.
Once given a metric space (X, d), you can define an open ball of radius r, centered at x
by
B(x; r) = {y ∈ X : d(x, y) < r}.
Definition 1 Given a set A ⊂ X, a point a ∈ X is
• interior point of A if there is r > 0 such that B(a; r) ⊂ A.
• exterior point of A if there is r > 0 such that B(a; r) ⊂ Ac , where Ac = X \ A is
the complement of A.
• boundary point of A if for every r > 0 both B(a; r)∩A 6= ∅ and B(a; r)∩Ac 6= ∅.
1
The set A is open if every a ∈ A is an interior point; it is closed if Ac is open.
Some sets are neither open nor closed, for example the half-open interval [0, 1) in onedimension Euclidean space R. Sets can be both open and closed at the same time - these
are called clopen - for example R and ∅. In fact, the following four properties hold for
every metric space (X, d):
1. X is open.
2. ∅ is open.
3. The union of any collection of open sets is open.
4. The intersection of any two open sets is open.
Corollary 2 The intersection of finitely many open sets is open.
The idea of topology is that properties 1.-4. are enough to prove all kind of things
about space, even if a metric is not available. In other words:
Definition 3 A collection O of subsets of X is a topology if
1. X ∈ O.
2. ∅ ∈ O.
3. The union of any collection of elements of O belongs to O.
4. The intersection of any two elements of O belongs to O.
We say that O is the collection of the open sets of X. Any open set containing x is called
a neighbourhood of x. Every set of the form X \ U for some U ∈ O is called closed.
Equipped with topology O, X is called a topological space.
Five Examples: (i) Every metric space is a topological space when we call a set U open
if every u ∈ U is interior point of U .
(ii) Discrete topology: every subset of X is open.
(iii) Trivial topology: O = {X, ∅}.
(iv) Co-finite topology: O = {U ⊂ X : U c is finite} ∪ ∅.
(v) X = R and U is open if it is the union of sets of the form [a, b). The line with this
topology is called the Sorgenfrey1 line.
Remark: If A is a subset of a topological space, then A is a topological space on its
own right, with topology inherited from X if we say that a subset U ⊂ A is open if and
only if there is an open set V ⊂ X such that U = A ∩ V . This is called the relative
topology. For example, if X = R with Euclidean topology, and A = [0, 2), then the
interval U = [0, 1), although not open in X, is open in the relative topology on A. Indeed
V = (−1, 1) is open in X and U = V ∩ A.
1
Robert Henry Sorgenfrey (1915–1996) was an American mathematician and professor at the University of California, Los Angeles.
2
Figure 1: Surjective, injective and bijective functions.
Definition 4 In a topological space X, the closure A of a set A is the smallest closed
set in X containing A. The interior A◦ of A is the largest open set contained in A. A
set A ⊂ X is dense in X if A = X. The boundary ∂A of A is the set of points x such
that every neighbourhood U of x intersects both A and its complement Ac .
1.2
Cardinality
Let us recall some terminology regarding functions first. Usually, we denote functions by
letters f, g, h, but this is not obligatory. More formally, when we write
f : A → B,
it means that f assigns to every element x in the domain A, a value f (x) in the codomain B. So formally, f (x) is not the function, it is the value f takes at x. As an
example, the function f : [−5, 10) → R defined by f (x) = x2 , has the interval [−5, 10) as
domain and R as co-domain (because we wrote so), but not every value in R is actually
assumed. The values that are assumed form the interval [0, 100) (check with the domain!),
and this is called the range of f . Let us also recall the following, see Figure 1:
Definition 5 A function f : X → Y is called surjective (or onto) if ∀ y ∈ Y ∃x ∈ X
such that f (x) = y. (In this case, the range of g is the whole co-domain Y .)
We call g : X → Y injective (or one-to-one) if ∀ x, x0 ∈ X, x 6= x0 , also g(x) 6= g(x0 ).
Finally, if h : X → Y is both injective and surjective, then h is bijective.
Cardinality refers to the number of elements in a set. We use the notation2 #(A), so for
example #({2, 4, 6}) = 3. However, when dealing with set with infinitely many elements,
just saying “#(A) is infinite” is not as precise as can be. Instead, we say:
2
Other books might use the notation card(A) and |A| is used.
3
Definition 6 Two sets A and B have the same cardinality if there is a bijection f :
A → B between A and B. If f : A → B is a surjection, then we say #(A) ≥ #(B). If
#(A) ≥ #(B) but #(A) 6= #(B), then we have strict inequality #(A) > #(B).
Thus the legs of a single lobster and the fingers on two hands have the same cardinality, which is less than the cardinality of the set of ants in the world. It becomes more
interesting with infinite sets. We know that
N⊂Z⊂Q⊂R
#(N) ≤ #(Z) ≤ #(Q) ≤ #(R).
and therefore
(1)
For example, we can make a surjection f : R → Q be setting f (x) = x if x ∈ Q and
f (x) = 0 if x ∈
/ Q. But the inclusions in (1) are all strict, so does that mean that
#(N) < #(Z) < #(Q) < #(R)?
The 19th century mathematician Georg Cantor developped the theory of cardinalities
and infinite set theory. In his view, N and Z are equally big:
Theorem 7 N and Z have the same cardinality.
Proof. The function f : N → Z defined by
( n
f (n) =
if n is even,
2
1−n
2
if n is odd,
is a bijection. (Check!). Therefore #(N) = #(Z).
Definition 8 A set C is countable (or more precisely countably infinite) if it has
the same cardinality as N, and the notation for this is #(C) = ℵ0 . (Read: Aleph zero,
for the Hebrew letter ℵ.) In this case, there is a bijection f : N → C, so we can write
C = {cn }n∈N where cn = f (n). We call {cn }n∈N a denumeration of C.
Theorem 9 N and Q have the same cardinality.
Proof. Figure 2 gives a way to find a denumeration {pn }n∈N of the positive rationals. So
p1 = 1, p2 = 2, p3 = 21 , p3 = 13 , p4 = 3 (we skip 22 because 22 = 1 which we already had in
the list), p4 = 4, p5 = 32 , etc. By skipping all rationals that appeared earlier in the list,
we make sure that pi 6= pj for all i 6= j. (The denumeration becomes injective.) Convince
yourself that the denumeration is also surjective: every positive rational will eventually
appear in the list.
This shows that N has the same cardinality as the positive rationals. Finally, set

0 if n = 1,

pn/2 if n is even,
qn =

−p(n−1)/2 if n > 1 is odd.
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Figure 2: Finding a denumerable list of all positive rationals
Then {qn }n∈N is a denumeration of Q (so also the non-positive rationals), and therefore
n 7→ qn gives a bijection between N and Q.
However, there are infinite sets that are uncountable i.e., sets that have cardinality
strictly larger than ℵ0 . The main example are the real numbers, and the proof is based
on Cantor’s diagonal argument.
Theorem 10 The cardinality #(R) is strictly larger than #(N ).
The notation that is sometimes used for the cardinality of the real numbers is #(R) =
c. (Here c stands for continuum.)
Proof. Since [0, 1] ⊂ R, it suffices to show that #([0, 1]) > ℵ0 . We will prove this by
contradiction.. Assume that there is a denumeration (xn )n∈N of [0, 1]. We write down the
decimal expansions of each xn in a countable list.
x1 = 0.x11 x12 .x13 x14 x15 x16 .x17 x18 . . .
x2 = 0.x21 x22 .x23 x24 x25 x26 .x27 x28 . . .
x1 = 0.x31 x32 .x33 x34 x35 x36 .x37 x38 . . .
x2 = 0.x41 x42 .x43 x44 x45 x46 .x47 x48 . . .
..
..
..
.
.
.
where xij ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is the j-th digit of the i-th number on the list. Now
form the diagonal number
5
if xjj 6= 5
d = 0.d1 d2 d3 d4 d5 d6 . . . where dj =
.
6
if xjj = 5
5
Hence the j-th digit of d is always different from the j-th digit of the j-th number on
the list. Therefore d 6= xj for each j ∈ N. In other words, d is not included in the
denumeration (xn )n∈N of [0, 1], which contradicts that (xn )n∈N is a good denumeration (it
is not surjective). From this contradiction, we conclude that [0, 1] is not countable.
However, there is a surjection f : [0, 1] → N. For example,
(
n if x = n1 for some n ∈ N,
f (x) =
1 if x ∈ [0, 1] is not of that form
is such a surjection. Therefore #([0, 1]) ≥ #(N). We already know that #([0, 1]) 6= #(N),
and therefore #(R) ≥ #([0, 1]) > #(N) is the only possibility.
The following theorem (or rather its proof) is beyond the scope of these notes, but
very relevant here:
Theorem 11 (Schröder-Bernstein) If there are both a surjection f : X → Y and an
injection g : X → Y , then #(X) = #(Y ).
In other words, if then #(X) ≥ #(Y ) and #(X) ≤ #(Y ), then #(X) = #(Y ). This may
seem trivial, but if you look closely at how ≤ and ≥ are defined for cardinalities, it is not
simple at all.
1.3
Separable and second countable spaces
Definition 12 A basis B of a topology O is a collection of open sets (so B ⊂ O), such
that every U ∈ O is a union of elements of B. If X has a countable basis, we call X
a second countable topological space.3 A topological space X is called separable if it
contains a countable dense set.
Example: The Euclidean space (R, d) is second countable, because we can take B =
{(a, b) : a, b ∈ Q}. It is also separable, because Q is a countable dense subset of R.
Proposition 13 A separable metric space is second countable.
Proof. Let A be a countable dense subset of the separable metric space X. Write
C(a, n) = B(a; 1/n). We claim that
B := {C(a, n) : a ∈ A, n ∈ N}
is a countable basis of the topology. Since the Cartesion product of countable sets (here
A and N) is countable, B is a countable collection of open sets. Now let U ⊂ X be an
arbitrary open set. Each u ∈ U is interior, so there is ε > 0 such that B(u; ε) ⊂ U . Let
A space X is called first countable if for every x ∈ X, the collection of neighbourhoods of x is
generated by a countable basis of neighbourhoods of x. This is a stronger property than second countable,
and we will not use it in these notes.
3
6
nu be an even integer such that
by the triangle inequality,
1
nu
<
ε
3
and find au ∈ A such that an ∈ B(u; 1/nu ). Then
u ∈ C(au , nu /2) = B(au ; 2/nu ) ⊂ B(u; ε) ⊂ U.
It follows that U = ∪u∈U C(au , nu ) can be written as union of elements of B.
1.4
Continuous functions and homeomorphisms
Definition 14 Let f : X → Y be a map from topological spaces X to Y . Then f is
continuous if for every open set U ⊂ Y , f −1 (U ) is open in X.
Remark: This definition shows that continuity is a topological notion: it depends
only on the topology and requires no metric. This may be a bit surprising, because the
definition that you were familiar with (see (2) below) was expressed in terms of a metric.
Example: Let the map f : R → R (where both spaces R are Euclidean) be defined as
f (x) = bxc. This is not a continuous map. For example, U = (− 21 , 21 ) is an open subset
of the co-domain, but f −1 (U ) = [0, 1) is not an open subset of the domain.
However, if we give R the topology of the Sorgenfrey line (both in domain and codomain), then f becomes continuous.
Proposition 15 Let f : R → R be a map from the Euclidean line to itself. Then f is
continuous in the topological sense if and only if
∀x ∈ R ∀ε > 0∃δ > 0 ∀y ∈ R d(x, y) < δ ⇒ d(f (x), f (y)) < ε.
(2)
Proof. ⇒ Take x ∈ R and ε > 0 arbitrary, and let U = B(f (x); ε). This is an open set,
so f −1 (U ) is open by assumption. Therefore x is an interior point of f −1 (U ), so there is
δ > 0 such that B(x; δ) ⊂ f −1 (U ). But this means that if d(x, y) < δ (so y ∈ B(x; δ)),
then f (y) ∈ f (B(x; δ)) ⊂ f ◦ f −1 (U ) ⊂ U = B(f (x); ε) and hence d(f (x), f (y)) < ε. This
proves formula (2).
⇐ Assume the formula (2) is true. Take U open and x ∈ f −1 (U ). Then f (x) is interior to
U , so there is ε > 0 such that B(f (x); ε) ⊂ U . Take δ as in formula (2), and V = B(x; δ).
Then f (V ) ⊂ U , so V ⊂ f −1 (U ). This shows that x is an interior point of f −1 (U ), and
because x ∈ f −1 (U ) is arbitrary, f −1 (U ) is open.
Example: Continuity is a topological notion; uniform continuity is not. Indeed, X2 =
(− π2 , π2 ) with Euclidean metric d2 (x, y) = |x − y| and then again with metric Xt = (− π2 , π2 )
with metric ρ(x, y) = | tan(x) − tan(y)|. Then the identity map f : X2 → X2 given by
f (x) = x is a homeomorphism, and uniformly continuous. However, f : X2 → Xt is also
a homeomorphism, but not uniformly continuous.
7
Definition 16 A sequence (an )n∈N in a topological space X converges to α if for every
neighbourhood U of α, an ∈ U for all but finitely many indices n. (That is, ∀U ∈
O with α ∈ U, ∃N ∈ N ∀n ≥ N an ∈ U )
Example: A convergent sequence is a topological notion, but a Cauchy sequence is
not. Indeed, convergence has a completely topological definition: A sequence (an )n∈N is
convergent if there is a limit α such that for every neighbourhood U 3 α, there is N ∈ N
such that an ∈ U for all n ≥ N .
To explain why Cauchy sequence is not a topological notion, let us take X = (− π2 , π2 )
first with Euclidean metric d2 (x, y) = |x − y| and then with metric ρ(x, y) = | tan(x) −
tan(y)|. Both metrics give the same topology. However, the sequence (an )n∈N given by
an = π/2 − 1/n is Cauchy in d but not in ρ.
Definition 17 A map h : X → Y between topological spaces is called a homeomorphism if
(i) h is a bijection,
(ii) h is continuous, and
(iii) the inverse h−1 is continuous.
If there is such a homeomorphism, X and Y are called homeomorphic, and we write
X 'Y.
Examples: (i) The letters S and L are homeomorphic. So are Q and P. But H and F are
not homeomorphic.
(ii) (0, 1) and R (both with Euclidean topology) are homeomorphic, for example via the
2x−1
.
homeomorphism h(x) = x(1−x)
(iii) The interval [0, 1) and the circle S1 = {z ∈ C : |z| = 1} are not homeomorphic. There
is a bijection h : [0, 1) → S1 given by h(x) = e2πix , and h is also continuous, but h−1 is
not continuous. This, however, does not show that [0, 1) and S1 are non-homeomorphic,
i.e., that there is no homeomorphism between [0, 1) and S1 whatsoever. An easy way of
doing that is probably by noting that S1 is compact, but [0, 1) is not.
Under a homeomorphism, topological properties are preserved.
This means that compactness, (the degree of) connectedness, dimension, cardinality, separability, and all other topological properties of sets discussed later on are not changed
by taking the image under a homeomorphism.
1.5
The Hausdorff Property and Quotient Spaces
Definition 18 A topological space X is a Hausdorff space if for every distinct x, y ∈ X,
there are neighbourhoods U 3 x and V 3 y such that U ∩ V = ∅.
8
Proposition 19 In a Hausdorff space, every convergent sequence has exactly one limit.
Proof. Assume that an → α, and let β 6= α. By the Hausdorff property, there are open
sets U 3 α and V 3 β such that U ∩ V = ∅. Since an → α there is N ∈ N such that
an ∈ U for all n ≥ N . Therefore an ∈
/ V for such n and β cannot be a limit of (an )n∈N .
Therefore there can be no other limit than α.
An equivalence relation ∼ on a space X is a relation on a set which is
1. Reflexive: x ∼ x.
2. Symmetric: x ∼ y if and only if y ∼ x.
3. Transitive: x ∼ y and y ∼ z imply x ∼ z.
The set {y ∈ X : y ∼ x} is the equivalence class of x; it is denoted as [x].
Definition 20 Given an equivalence relation ∼ on a topological space, the quotient
space Y = X/ ∼ is the space in which the points are the equivalence classes of ∼, i.e.,
y ∈ Y equals [x] for some x ∈ X. A subset U ⊂ Y is open if the union of all points in X
that belong to an equivalence class [x] ∈ A form an open set in X.
Examples: Quotient spaces emerge from glueing parts of a space to other parts of itself:
(i) Let X = [0, 1] with Euclidean topology, and 0 ∼ 1. Then X/ ∼ is a circle S1 .
(ii) Let X = {z ∈ C : |z| ≤ 1} with Euclidean topology be the closed unit disk, and
eiφ ∼ eiψ for all angles φ, ψ ∈ [0, 2π). That is, we identify all boundary points of the disk.
Then X/ ∼ is homeomorphic to a two-dimensional sphere S2 . (The n-dimensional sphere
Sn = {x ∈ Rn+1 : kxk2 = 1}.)
(iii) Let X = [0, 1] × [0, 1] be the unit square with Euclidean topology, and take the
equivalence relation (0, t) ∼ (1, t) and (t, 0) ∼ (t, 1) for all t ∈ [0, 1]. Then X/ ∼ is a
torus. In this case it is the two-dimensional torus T2 . The same procedure in dimension
n yields the n-dimensional torus Tn . Note that T1 ' S1 , but Tn 6' Sn for n ≥ 2.
(iv) Let X = [0, 1] × [0, 1] be the unit square with Euclidean topology, and take the
equivalence relation (0, t) ∼ (1, t) and (t, 0) ∼ (1 − t, 1) for all t ∈ [0, 1]. This means that
the left boundary and the right boundary of the unit square are identified “with a flip”.
The resulting quotient space K = X/ ∼ is called the Klein bottle. It is a non-orientable
surface, and not homeomorphic to T2 .
(v) Let X = [0, 1] × [0, 1] be the unit square with Euclidean topology, and take the
equivalence relation (0, t) ∼ (1, 1 − t) and (t, 0) ∼ (1 − t, 1) for all t ∈ [0, 1]. This means
that now the left boundary and the right boundary of the unit square are identified “with a
flip”, and also the top boundary and the bottom boundary of the unit square are identified
“with a flip”. The resulting quotient space X/ ∼ is called the projective plane. It is
again a non-orientable surface, and not homeomorphic to K or to T2 , see Figure 3.
Proposition 21 If ∼ has an equivalence class [x] that is not closed in X, then Y = X/ ∼
is not Hausdorff.
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Figure 3: The torus, Klein bottle and projective plane
Proof. Let us call the projection from X to its quotient space π : X → Y ; that is,
π(x) = [x]. It is important to realise here that [x] means two things: (i) an equivalence
class (and hence subset) of X, and (ii) a point in the quotient space Y . Thus π(x) is a
single point of Y .
Suppose that [x] is not closed in X, and y ∈ ∂[x] is a boundary point of [x] not belonging to [x]. We claim that in Y = X/ ∼, there are no open neighbourhoods U of [x]
and V of [y] that are disjoint. Indeed, if V ∈ [y] is open in Y , this means by definition
that π −1 (V ) is open, but π −1 (V ) 3 y, and there is no open neighbourhood of y that is
disjoint from the equivalence class [x]. Therefore V 3 [x] and Y cannot be Hausdorff. Example: All equivalence classes being closed is not sufficient to guarantee that Y =
X/ ∼ is Hausdorff. For example, let X = R with Euclidean topology, and set 1 − n1 ∼
−1 + n1 for every n ∈ N. Then [1] and [−1] are still distinct points in Y , but there are no
disjoint open neighbourhoods U 3 [1] and V 3 [−1] in Y .
1.6
Compactness
Definition 22 A subset A of a topological space X is coverings-compact if every open
cover of A contains a finite subcover. (Here a cover C = {Ui }i of A is a collection of open
sets such that ∪i Ui ⊃ A. A subcover is a subcollection of a cover which is itself covers A.)
A subset A of a topological space X is sequentially compact if every sequence (an )n∈N
contains a subsequence which converges in A (hence the limit belongs to A).
Remark: In a second countable Hausdorff space, coverings-compact and sequentially
compact are equivalent notions; hence we simply speak of compact. There are nonHausdorff spaces and also non-second countable spaces in which coverings-compact and
sequentially compact are really different notions.
Lemma 23 In a Hausdorff space X, every compact set is closed.
Proof. Let A ⊂ X be compact, and assume by contradiction that it is not closed. Then
there is a boundary point a ∈ ∂A \ A. By the Hausdorff property, for each x ∈ A, there
are disjoint neighbourhoods Ux 3 x and Vx 3 a. The collection {Ux }x∈A is an open cover
of A, and compactness of A implies that there is a finite set {x1 , . . . , xN } ⊂ A such that
10
N
{Uxi }N
i=1 is a finite subcover of A. Furthermore ∩i=1 Vxi is a neighbourhood of a that is
disjoint from ∪N
i=1 Uxi and hence disjoint from A. But this contradicts that a is a boundary
point of A.
Lemma 24 In a metric space, every compact set is bounded.
Proof. Let A be a compact subset of metric space (X, d). Then {B(x; 1)}x∈A is an open
cover of A, and by compactness, there are x1 , . . . , xN ∈ A such that {B(xi ; 1)}N
i=1 is a
finite subcover. Let K = max{d(xi , xj ) : 1 ≤ i, j ≤ N }. Since the maximum is taken
over a finite set, it exists and is finite. Let x, y ∈ A be arbitrary, and find i, j such that
x ∈ B(xi ; 1) and y ∈ B(xj ; 1). Then by the triangle inequality
d(x, y) ≤ d(x, xi ) + d(xi , xj ) + d(xj , y) < K + 2.
Hence A is bounded; it has diameter sup{d(x, y) : x, y ∈ A} ≤ K + 2.
Theorem 25 (Heine-Borel) In Euclidean space Rn , a set is compact if and only if it
is closed and bounded.
Proposition 26 Let f : X → Y be a continuous map between topological spaces. If A is
compact in X, then f (A) is compact in Y .
Proof. Let {Vi }i be an open cover of f (A), and let Ui = f −1 (V ). By continuity of f ,
Ui is open, so {Ui }i is an open cover. Since A is compact, we can take a finite subcover
N
{Ui }N
i=1 . Then {Vi }i=1 is a cover of f (A). Since {Vi }i was arbitrary, this means that f (A)
must be compact.
Proposition 27 A closed subset A of a compact topological space X is compact.
Proof. Let U be an open cover of A. Since A is closed, X \ A is open, and together
with U, it forms an open cover of X. But X is compact. Therefore we can select a finite
subcover of (X \ U ) ∪ U, which is then also a finite subcover of A.
1.7
Connected Sets
Definition 28 Topological space X is called a disconnected if it is the union of two
non-empty disjoint open sets. If there are no such sets, then X is connected.
Note that the two sets, say U and V , are both open and closed, because U = X \ V
and V = X \ U are the complements of open sets. Recall that a set which is both open
and closed is called clopen. The only clopen subsets of R (with Euclidean topology) are
R itself and the empty set. Therefore R is connected. Similarly, Rn is connected for each
n.
11
Definition 29 A topological space X is arc-wise connected if for every two points
x, y ∈ X there is a continuous map φ : [0, 1] → X such that φ(0) = x and φ(1) = y. Such
a map φ is called a path from x to y. For this reason, arc-wise connected is also called
path-connected.
Example: Not every connected space is arc-wise connected. A standard example here is
the sin x1 -continuum. This subset of R2 is the union of an arc {0} × [−1, 1] with the graph
of the function sin x1 , i.e., {(x, sin x1 ) : x ∈ (0, 1]}. In this case there is no continuous path
from (0, 0) to (0, 1/π), but the sin x1 -continuum is connected.
Definition 30 Given a point x in a topological space X, the connected component of
x is the largest connected subset of X that contains x. A topological space X is totally
disconnected if each of its connected components is a singleton.
Definition 31 A point x of a topological space X is isolated if it has a neighbourhood
that is disjoint from the rest of X. Any point that is not isolated is called a limit point or
accumulation point of X. The space X is called discrete if all its points are isolated.
Example: The middle third Cantor set C is a subset of [0, 1] obtained as follows. First
remove the open interval of length 31 from the middle of I0 = [0, 1] (i.e., remove the middle
third). The remainder is the union I1 = [0, 31 ] ∪ [ 32 , 1]. Repeat the process of removing
the middle third of these two intervals. Then you have I2 = [0, 91 ] ∪ [ 29 , 13 ] ∪ [ 23 , 79 ] ∪ [ 89 , 1].
Continue this way. At step, the set In consists of 2n intervals of length 3−n . The middle
third Cantor set C = ∩n≥0 In .
I0 = [0, 1]
I1 = [0, 13 ] ∪ [ 23 , 1]
I2 = [0, 19 ] ∪ [ 29 , 13 ]
∪[ 32 , 79 ] ∪ [ 98 , 1]
Figure 4: Steps towards creating the middle third Cantor set C = ∩n In .
(a) It is non-empty; for example each endpoint of In for each n also belongs to C.
(b) On the other hand, C contains no interval, because if it did, say (a, b) ⊂ C, then this
interval has length b − a > 0. Therefore there is n such that 3−n < b − a, and this means
that (a, b) 6⊂ In .
(c) The same argument shows that every connected component of C is a singleton.
(d) C is a closed subset of [0, 1] since it is the intersection of the closed sets In . By the
Heine-Borel this means that C is compact.
(e) C has no isolated points. Indeed, if x ∈ C were isolated, then there is ε > 0 such that
B(x; ε) contains no other points of C than x. Now take n so large that 3−n < ε/2. Since
x ∈ C, x ∈ In as well. But that means that the endpoints a, b of the subinterval [a, b]
of In that contains x are also contained in C. However, d(x, a), d(x, b) < ε, so B(x; ε)
contains at least one other point of C than x itself.
Hence the middle third Cantor set satisfies the following definition of a general Cantor
set.
12
Definition 32 A non-empty set X is called a Cantor set if it is compact, totally disconnected and has no isolated points. (Any Cantor set is homeomorphic to the middle
third Cantor set C.)
1.8
Exercises of Section 1
Exercise 1 For X = R2 and p ≥ 1, define the p-metric
p
dp (x, y) = p |x1 − y1 |p + |x2 − y2 |p and d∞ (x, y) = max(|x1 − y1 | , |x2 − y2 |).
Sketch the unit balls Bp ((0, 0); 1) for p = 1, 32 , 2, 3, ∞.
Exercise 2 Prove the following assertions: An open set doesn’t contain any of its boundary points. A closed set contains all its boundary points.
Exercise 3 Prove that in a metric space, the four defining properties of a topology hold.
Exercise 4 Why can’t we say that the intersection of infinitely many open sets is open?
Give a counter-example.
Exercise 5 You can equip Rn not only with Euclidean metric d2 , but just as well with the
P
sum metric d1 (x, y) = ni=1 |xi − yi | or with the sup metric d∞ (x, y) = maxni=1 |xi − yi |.
Show that for all x, y ∈ Rn ,
d∞ (x, y) ≤ d2 (x, y) ≤ d1 (x, y) ≤ nd∞ (x, y).
Hence show that these three metrics produce the same topology (i.e., the same collection
of open sets).
Exercise 6 Derive the properties of closed sets:
1. X is closed.
2. ∅ is closed.
3. The intersection of any collection of closed sets is closed.
4. The union of any two closed sets is closed.
Exercise 7 Which of the following sets are open in the Sorgenfrey line?
{0}
(0, 1)
(−∞, 1)
(−∞, 1]
Exercise 8 Are the following statements true or false (in topological space and in metric
space):
(a) A ∩ Ac = ∂A.
(b) A ∪ ∂A = A.
(c) A = A \ ∂A.
13
Exercise 9 Which of the above five examples on page 2 is metrisable? That is, for
which of the example is there a metric d on X such that O consists of exactly those sets
for which every point is interior w.r.t. metric d?
Exercise 10 Show the following statements.
a) If A ⊃ B then #(A) ≥ #(B).
b) If g : A → B is an injection, then #(A) ≤ #(B). (Hint: Given g, find a surjection
f : B → A.)
Exercise 11 Let A and B be countable. Show that also the union A∪B and the Cartesian
product A × B are countable.
Exercise 12 Let A be a set of at least two elements. Show that the infinite product AN
is uncountable. (This infinite product consists of all sequences (an )n∈N with an ∈ A.
Exercise 13 Which of the above five examples on page 2 is separable and/or second
countable? Conclude that the Sorgenfrey line is a different topological space from the
Euclidean line.
Exercise 14 Let the map f : R → R (where both spaces R are Euclidean) be defined as
f (0) = 0 and f (x) = sin x1 if x =
6 0. Show that f is not continuous.
Exercise 15 Show that if X has discrete topology, then every map f : X → Y is continuous. If X is trivial topology, then f : X → Y is continuous if and only if f is
constant.
Exercise 16 Show that the map f : R \ {0} → R \ {0} (domain and co-domain with
Euclidean topology) given by f (x) = 1/x is continuous.
Exercise 17 Show the following statements about the five examples on page 2. (a) If
X has discrete topology, then the only convergent sequences are those that are eventually
constant.
(b) If X has trivial topology, then every sequence is convergent, and every x ∈ X is a
limit of every sequence.
(c) If R is the Sorgenfrey line, and (an )n∈N is a bounded and strictly monotone sequence.
Then (an )n∈N is convergent if and only if it is decreasing.
Exercise 18 Prove that a metric space is Hausdorff. Which of the five examples of
topological spaces on page 2 are not Hausdorff ?
Exercise 19 How many topologies can you find on a space {a, b, c} of three points? Which
of these are Hausdorff ?
Exercise 20 Prove that in a Hausdorff space every singleton set {x} is closed.
14
Exercise 21 The definition of an equivalence relation is standard. Yet, it seems that 1.
follows from 2. and 3. Can you explain why 1. is not superfluous?
Exercise 22 Show that the projection π : X → Y in the above proof is continuous.
Exercise 23 Find a metric space in which some closed and bounded set is not compact.
Exercise 24 Show that every arc-wise connected space is connected.
Exercise 25 Show that Q is a totally disconnected subset of R. Show that Q2 is a totally
disconnected subset of R2 .
Exercise 26 Show that if X has isolated point x of a Hausdorff space X, then X is not
connected. (We can argue about whether x is an isolated point of X if X = {x}.)
Exercise 27 Show that the middle third Cantor set C contains points that are not endpoints of any In . For example, 41 is such a point. Show that C is uncountable.
Exercise 28 Let f : X → Y be a continuous map between topological spaces. If A is
connected in X, show that f (A) is connected in Y . If A is arc-wise connected in X, show
that f (A) is arc-wise connected in Y .
1.9
Solutions to Exercises
Exercise 1: The unit balls Bp ((0, 0); 1) billow out from a diamond-shape for p = 1, a
round ball for p = 2 to a square for p = ∞.
Exercise 2: If a ∈ ∂A, then for every r > 0, B(a; r) ∩ Ac 6= ∅, hence a is not an
interior point of A. So an open set cannot contain any of its boundary points. By the
same token, if B is closed, then B c is open, and therefore contains no point of ∂(B c ). But
the definition of boundary is symmetric, so ∂(B c ) = ∂B, and hence ∂B ⊂ B.
Exercise 3: 1. If x ∈ X is arbitrary, then B(x; r) ⊂ X for arbitrary r > 0. Hence x
is interior to X, so X is open.
2. Every x ∈ ∅ is interior to ∅, because there are no exceptions.
3. If x ∈ ∪α Uα , then x ∈ Uα0 for some α0 . All Uα are open, so x is an interior point of
Uα0 and there is r > 0 such that B(x; r) ⊂ Uα0 . But then B(x; r) ⊂ ∪α Uα as well.
4. If x ∈ U ∩ V for open sets U, V , then there are radii rU > 0 and rV > 0 such that
B(x; rU ) ⊂ U and B(x; rV ) ⊂ V . Take r = min{rU , rV }. Then B(x; r) ⊂ U ∩ V . This
shows that every x ∈ U ∩ V is interior to U ∩ V .
Exercise 4: Let Un = (− n1 , n1 ). These are all open intervals, but ∩n Un = {0} is not
open.
15
Exercise 5:
d∞ (x, y) =
v
u n
q
uX
2
max |xi − yi | =
max |xi − yi | ≤ t
|xi − yi |2
i=1,...,n
= d2 (x, y) ≤
= d1 (x, y) ≤
i=1,...,n
n
X
i=1
n
X
i=1
|xi − yi |
i=1
by the Cauchy-Schwarz inequality
max |xj − yj | = n max |xj − yj | = nd∞ (x, y).
j=1,...,n
j=1,...,n
This gives
r
B∞ (x; r) ⊃ B1 (x, y) ⊃ B2 (x, y) ⊃ B∞ (x; )
n
so if x is interior to A in dt , then there is r such that Bt (x; r) ⊂ A, and by replacing
r by r/n if necessary, we have Bu (x; r) ⊂ A as well. This works for every choice of
t, u ∈ {1, 2, ∞},
Exercise 6: 1. X is closed because X c = ∅ is open.
2. ∅ is closed because ∅c = X is open.
3. If Dα are all closed, then ∩α Dα = (∪α Dαc )c is the complement of a union of open sets,
and hence closed.
4. If D and E are all closed, then D ∪E = (Dc ∩E c )c is the complement of the intersection
of two open sets, and hence closed.
Exercise 7: On the Sorgenfrey line: {0} is not open, (0, 1) = ∪n∈N [ n1 , 1) is open,
(−∞, 1) = ∪n∈N [−n, 1) is open, but (−∞, 1] is not open.
Exercise 8: (a) is true.
(b) is true.
(c) is false. For example, take any closed set with non-empty interior, or take A = Q in
R with Euclidean topology.
Exercise 9: (i) Metric space is metrisable by definition.
(ii) Discrete topology is generated by discrete metric, and hence metrisable.
(iii) Trivial topology on any space consisting of more than one point is not Hausdorff and
therefore not metrisable.
(iv) Co-finite topology on any infinite space is not Hausdorff and therefore not metrisable.
(v) The Sorgenfrey line is separable (because Q is a countable and dense subset) but not
second countable (because not all uncountable many set [α, ∞), α ∈ R, can be generated
by any countable collection of open sets). Thus, in view of Proposition 13, the Sorgenfrey
line is not metrisable.
Exercise 13: (i) There are both separable and non-separable metric spaces, but if
the space is separable, then it is also second countable by Proposition 13.
(ii) Separable and second countable if and only if the space is countable.
(iii) Always separable and second countable.
(iv) Separable and second countable if and only if the space is countable.
(v) See Exercise 9.
16
Exercise 14: Take U = (− 12 , 21 ), then 0 ∈ f −1 (U ), but 0 is not an interior point of
f −1 (U ), because arbitrarily close to 0 there are points 1/((k + 12 )π), k ∈ N, that map
outside U . Therefore f −1 (U ) is not open and f is not continuous.
Exercise 15: (a) In discrete topology, every set is open, so also every set of the form
−1
f (U ). Hence f is continuous.
(b) Without knowing the topology of Y you cannot answer this, but assuming that Y
is Hausdorff and f is not a constant function, then there are x, y ∈ Y and disjoint open
sets Ux 3 x, Uy 3 y such that f −1 (Ux ) 6= ∅ and f −1 (Uy ) 6= ∅. In trivial topology on X,
these two sets are not open, so f is not continuous. If, on the other hand, f is constant x,
then f −1 (U ) = X or ∅ depending on whether the (open) set U contains x or not. Thus
a constant function f is (always) continuous.
Exercise 16: The intervals (x, y) with either 0 < x < y or x < y < 0 generate the
topology of R \ {0}. For each such interval U , f −1 (U ) is again an open interval, so f −1 (V )
is an open for every open V ⊂ R \ {0}.
Exercise 17: (a) If there is N such that an = α for all n ≥ N , then B(α, r) contains
an for all n ≥ N . regardless what r > 0 is. Therefore an → α. If an 6= α infinitely often,
then for r = 21 , and in discrete topology, {α} = B(α; r) 63 an infinitely often, so an 6→ α.
(b) The only non-empty neighbourhood of any α ∈ X is X itself, so every sequence
converges to α, and every α is a limit of every sequence.
(c) If (an )n∈N is decreasing and α = inf n an , then for any r > 0, there is N ∈ N such
that an ∈ [α, α + r) for all n ≥ N . Hence α = limn an . Conversely, if (an )n∈N is strictly
increasing and α = supn an , then for any r > 0 and β ≥ α, [β, β + c) is disjoint from
(an )n∈N , so β 6= limn an . However, if β < α, and r = (α − β)/2, then an ∈ [β, β + r) for
at most finitely many n, so also β 6= limn an . In conclusion: (an )n∈N diverges.
Exercise 18: Take x 6= y in a metric space (X, d), and let r = d(x, y). Then
0 < r < ∞, and B(x; r/3) and B(y; r/3) are disjoint. Indeed, if z ∈ B(x; r/3) ∩ B(y; r/3),
then by the triangle inequality, r = d(x, y) ≤ d(x, z) + d(z, y) < r/3 + r/3 = 2r/3, which
is a contradiction. Hence X is Hausdorff. Of the five examples, the co-finite topology (on
an infinite space) and the trivial topology are not Hausdorff.
Exercise 19: Disregarding all symmetries, there are eight different topologies on
{a, b, c}, but only the discrete topology is Hausdorff.
Exercise 20: For every y 6= x, let Uy 3 x and Vy 3 y be disjoint open sets. Then
∪y6=x Vy is open and {x} = (∪y6=x Vy )c is closed.
Exercise 21: Without the first property, if x is not equivalent to any other y, there
is no way of concluding that x ∼ x.
Exercise 22: The set U ⊂ Y = X/ ∼ is open by definition if ∪u∈U [u] is open in X.
But ∪u∈U [u] = ∪u∈U π −1 (u) = π −1 (U ), so π is continuous.
Exercise 23: The closed unit ball in the infinite dimensional space `∞ is B(0; 1) =
{(xk )k∈N : |xk | ≤ 1 for all k ∈ N}. If ek = (0, 0, . . . , 0, 1, 0, 0)˙ (with the 1 at position k),
then (ek )k∈N ⊂ B(0; 1) is a sequence without convergent subsequence. Indeed, d(ek , el ) = 1
for all k 6= l. So B(0; 1) is closed and bounded, but not compact.
Another example is N itself with discrete topology. Then it is bounded (with diameter
17
1), and closed (every subset is open, so also closed), but it is not compact.
Exercise 24: Suppose that X is arc-connected. Assume by contradiction that X is
not connected, so there are two disjoint non-empty clopen subsets U, V ⊂ X such that
X = U ∪ V . Take u ∈ U , v ∈ V and let γ : [0, 1] → X be a continuous path connecting
u = γ(0) and v = γ(1). Let U 0 = γ −1 (U ) and V 0 = γ −1 (V ). Then U 0 and V 0 are open
by continuity of γ, 0 ∈ U 0 , 1 ∈ V 0 and U 0 ∪ V 0 = [0, 1] and also U 0 ∩ V 0 = ∅ because
U and V are disjoint. Hence U 0 and V 0 are two disjoint non-empty clopen subsets such
that [0, 1] = U 0 ∪ V 0 . But this shows that [0, 1] with Euclidean topology is disconnected,
a contradiction.
Exercise 25: If C ⊂ Q is a connected component containing two different point x < y,
and we can take an irrational point z ∈ (x, y) and Cx = C ∩ (−∞, z) and Cy = C ∩ (z, ∞).
It follows that Cx and Cy are both open in the relative topology on C, they are disjoint,
non-empty and their union is C. This contradicts that C is connected.
For Q2 we use the same argument. If a connected component contains two different
point x, y ∈ Q2 , then there is at least one coordinate, say the first, such that x1 6= y1 .
Without loss of generality, we assume that x1 < x2 . Take an irrational point z ∈ (x1 , y1 )
and Cx = {a ∈ C : a1 < z} and Cy = {a ∈ C : a1 > z}. Then Cx and Cy are both open,
disjoint, non-empty and their union is C. This contradicts that C is connected.
Exercise 26: Let x be isolated, so there is a neighbourhood U 3 x such that X ∩ U =
{x}, i.e., U = {x} is itself open. Since X is Hausdorff, {x} is closed as well. Take
V = X \ {x}. If X contains more than one point, then U , V are now clopen non-empty,
disjoint sets such that X = U ∪ V . Hence X is disconnected.
Exercise 27: The middle third Cantor set can be written as those points x ∈ [0, 1]
P
−k
whose ternary expansion x = ∞
only requires digits xk = 0 or 2. Since there are
k=1 xk 3
uncountably many such sequence (xk )k∈N , and different sequences correspond to different
points in C, C is uncountable.
For endpoints of In , the sequence (xk )k∈N ends with 0, 0, 0, 0, . . . or with 2, 2, 2, 2, . . . .
There are only countably many sequences of this type. Naturally, there are many other seP
P
quences, for example the sequence 0, 2, 0, 2, 0, 2, 0, 2, . . . results in k∈N xk 3−k = j∈N 2 ·
3−2j = 41 .
Exercise 28: (a) Suppose by contradiction that f (A) is disconnected. so there are
disjoint non-empty clopen sets U, V ⊂ f (A) such that U ∪ V = f (A). Let U 0 = f −1 (U )
and V 0 = f −1 (V ). Then U 0 and V 0 are non-empty, disjoint and open (by continuity of f )
and U 0 ∪ V 0 = A. This shows that A is disconnected, a contradiction.
(b) Let â = f (a) and b̂ = f (b) be two points in f (A). Since A is arc-connected, there is
a continuous path γ : [0, 1] → A connecting a and b. But then f ◦ γ : [0, 1] → f (A) is
a continuous path connecting â and b̂, so f (A) is arc-connected. (Here we used that the
composition f ◦ γ of two continuous mappings is continuous again. Indeed, if U is any
open set, then f −1 (U ) is open, and therefore γ(f −1 (U )) = (f ◦ γ)−1 (U ) is open as well.)
18
2
Manifolds
2.1
What are Manifolds?
Examples of manifolds are the sphere and the torus. Most of the time, we think of these
as two-dimensional objects (i.e., surfaces), but you can construct spheres and tori in any
dimension. The term manifold M is used for the n-dimensional generalisation of surface.
It is an object that locally looks like Rn , but globally can be more complicated. Three
thousand years ago, we thought that the earth was flat, because on small scale that’s what
it looks like. About 2500 years ago, scientists changed their mind about it and decided it
was probably a sphere (Eratosthenes gave a fairly accurate estimate of its circumference).
Similarly, until the theories of Einstein and 20-th century cosmology, it was thought that
space was R3 , but now we know that space is curved (and its dimension is 3 or 4 or 10,
depending which type of relativity theory or string theory you believe in).
Definition 33 An n-dimensional manifold M is a compact, second countable (i.e., having a countable basis), Hausdorff topological space, in which every point has a neighbourhood which is homeomorphic to (an open ball in) Rn .
This means that every p ∈ M has a neighbourhood V ⊂ M and a homeomorphism
φ:U →V
U an open ball in Rn
giving local coordinates near p. Such φ is called a local chart or patch of M .
Since M is compact, the whole manifold can be covered by finitely many charts, say
φi : Ui → Vi
with M = ∪si=1 Vi .
The collection of all charts is called an atlas.
We call M smooth if, whenever Vi ∩ Vj 6= ∅,
−1
−1
ψij := φ−1
j ◦ φi : φi (Vi ∩ Vj ) → φj (Vi ∩ Vj )
is smooth.4
Example (Parametrising the unit circle): The unit circle S1 = {(x, y) ∈ R2 :
x2 + y 2 = 1} is a smooth one-dimensional manifold. It allows many different atlases
(parametrisations):
√
φ1 : (−1, 1) → S1
t 7→ (t, 1 − t2 ),
√
φ2 : (−1, 1) → S1
t 7→ (t, − 1 − t2 ),
together cover S1 , except for (1, 0) and (−1, 0). So we can take two more charts:
√
φ3 : (−1, 1) → S1
t 7→ ( 1 − t2 , t),
√
φ4 : (−1, 1) → S1
t 7→ (− 1 − t2 , t).
4
Unfortunately, smooth is a bit vague; it can mean differentiable, or infinitely differentiable, or just
as often differentiable as needed.
19
It is more standard to use polar coordinates:
φ1 : (0, 2π) → S1
t 7→ (cos t, sin t)
covers everything, except (1, 0), so we need to add
φ2 : (−π, π) → S1
t 7→ (cos t, sin t).
Thus φ−1
2 ◦ φ1 (t) = t for all t = (0, π) ∪ (π, 2π) is indeed smooth.
A third type of parametrisation is called the stereographic projection. Let N =
(0, 1) and (0, −1) = S be the north and south pole of the circle. To obtain the stereographic projection from the north pole, we draw the tangent line to S1 at the south
pole, and draw the cord from N to some point p ∈ S1 . Extending this cord, we intersect
the tangent line at some point (ξ, −1), see Figure 5. The coordinate ξ parametrises S1 ,
except for the north pole. A little computation shows that if p = (sin ψ, cos ψ), then
ξ = 2 cotan(ψ/2). Expressing p in terms of ξ is possible too, but a bit messier. To obtain
Ns ζ = 2 tan(ψ/2)
'$
aa
s
ψ apa= (sin ψ, cos ψ)
s
aa
aa
aa
s
&%
S
ξ = 2 cotan(ψ/2)
Figure 5: The stereographic projection of the circle from the north and south pole.
the stereographic projection from the south pole, we draw the tangent line to S1 at the
north pole, and draw the cord from S to p ∈ S1 . Extending this cord, we intersect the
tangent line at some point (ζ, 1), see Figure 5. The coordinate ζ parametrises S1 , except
for the south pole, via ζ = 2 tan(ψ/2). So both stereographic projections together form
an atlas of the circle.
Since ξ·ζ = 4, it is easy to express ζ = 4/ξ, and this is a smooth map for all ξ ∈ R\{0},
that is, for all ξ that map to S1 \ {N, S}.
The same idea of stereographic projection works for the sphere Sn in any dimension
n ≥ 1.
Example: The closed disk D = {(x, y) ∈ R2 : x2 + y 2 ≤ 1} is two-dimensional, but not
quite a manifold, because the points on the edge {(x, y) ∈ R2 : x2 + y 2 = 1} don’t have
neighbourhoods that are homeomorphic to R2 . We call this a manifold with boundary.5
Removing the edge (i.e., the boundary) doesn’t make it into a manifold, because then it
is no longer compact.
5
Unfortunately, boundary used here does not have precisely the same meaning as ∂D in Section 1,
because it is independent of the ambient space. But when D is embedded ni R2 , so R2 is the ambient
space, then indeed ∂D = {(x, y) ∈ R2 : x2 + y 2 = 1}.
20
Example: A circle with diagonal
(x, y) ∈ R2 : x2 + y 2 = 1
(x ∈ [−1, 1], y = 0)
or
is one-dimensional, but not a manifold. The neighbourhoods of the points (1, 0) and
(−1, 0) are more complicated than an open ball in R.
A circle with a separate sphere
(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1
or
((x − 2)2 + y 2 = 1, z = 0)
is not a manifold, because some part is 2-dimensional, whereas another part is 1-dimensional.
Note also that this example is not connected.
You should be able to find parametrisations of many types of manifolds, so we give
some more examples, without bothering too much if these parametrisations form a complete atlas.
Parametrising the n-sphere using spherical coordinates:
S1 = {(x, y) ∈ R2 : x2 + y 2 = 1},
dim(S1 ) = 1
φ ∈ (0, 2π),
x = cos φ
y = sin φ.
S2 = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1},
dim(S2 ) = 2
x = sin θ cos φ
φ ∈ (0, 2π),
y = sin θ sin φ
θ ∈ (0, π),
z = cos θ.
dim(S3 ) = 3
S3 = {(x, y, z, u) ∈ R4 : x2 + y 2 + z 2 + u2 = 1},
x = sin ψ sin θ cos φ
φ ∈ (0, 2π),
y = sin ψ sin θ sin φ
θ ∈ (0, π),
z = sin ψ cos θ
ψ ∈ (0, π),
u = cos ψ.
Et cetera.
The 2-torus can be embedded in R3 . Using radii R > r > 0. we get
p
dim(T2 ) = 2
T2 = {(x, y, z) ∈ R3 : ( x2 + y 2 − R)2 + z 2 = r2 },
x = (R + r cos φ) cos θ
φ ∈ (0, 2π),
y = (R + r cos φ) sin θ
θ ∈ (0, π),
z = r sin φ.
21
Figure 6: A surface of revolution.
Example: Surface of Revolution. Let {y = f (x)} be the graph of some function.
When we revolve this graph in R3 around the x-axis, see Figure 6, we obtain a surface
of revolution, that can be parametrised as
t∈R
x = t,
y = f (t) cos φ,
φ ∈ (0, 2π)
z = f (t) sin φ.
Many surfaces (e.g. spheres, tori, cones, cylinders) can be obtained this way.
Although an n-dimensional manifold is an abstract object, we usually embed it in Euclidean space Rk for some k ≥ n. Otherwise it becomes hard to imagine it (although
imagining a 28-dimensional sphere remains a challenge). Fortunately, every manifold,
whether orientable or not, can indeed be embedded in Euclidean space.
Theorem 34 (Whitney’s Embedding Theorem) Every n-dimensional manifold can
be embedded in Rk for every k ≥ 2n + 1.
For example, you need R4 to embed the 2-dimensional Klein bottle. You need R3 to
embed the 1-dimensional trefoil not, that can be parametrised as
t 7→ ((2 + cos t) cos 3t, (2 + cos t) sin 3t, cos t) .
When embedded in Rk , is it often possible to express an n-dimensional manifold as the
level set of some function f : Rk → Rk−n . For example, the unit 2-sphere in R3 is the
level 1 set of f (x, y, z) = x2 + y 2 + z 2 .
2.2
Tangent Spaces and Tangent Bundles
Tangent spaces are the n-dimensional generalisation of tangent lines (to curves) and tangent planes (to surfaces). A smooth manifold M has a well-defined tangent space Tp M
at each point p ∈ M . To define them properly, we will first discuss curves on manifolds.
22
Figure 7: The tangent space to a point X on a sphere (left) and m on a torus (right).
Let p ∈ M be some point on a n-dimensional manifold. A curve α : (−1, 1) → M goes
through p if α(t) = p for some t ∈ (−1, 1). Without loss of generality, we can assume that
α(0) = p. It would be logical to denote the velocity at which this curves runs through p
by dα(t)
. The problem, however, is that M is an abstract manifold, and doesn’t allow us
dt
to define dα(0)
properly as the limit of a differential quotient. Therefore, we have to make
dt
use of local coordinate in a neighbourhood V of p. Suppose φ : U → V is a chart near p.
Then we can decompose
α = φ ◦ β or β = φ−1 ◦ α,
where β : (−1, 1) → U ⊂ Rn is a curve in Euclidean space. So now, assuming this curve
is smooth, dβ(0)
= limε→0 β(ε)−β(0)
can be properly defined.
dt
ε
Two curves α and α̃ through p can of course have the same velocity at p, even if they
= dβ̃(0)
. The tangent space Tp M of M
are different. It just depends on whether dβ(0)
dt
dt
at p is the space of all of these velocity vectors of smooth curves through p; it is defined
as the quotient space
Tp M = {α : (−1, 1) → M : α is smooth, α(0) = p} / ∼
quotiented out over the equivalence relation α ∼ α̃ if
dβ(0)
dt
=
dβ̃(0)
.
dt
When M is embedded in Rk for some k ≥ n, then we can picture Tp M as an n-dimensional
space glued tangentially to M at p. If M near p is a level set of some function f : Rk →
Rk−n , say {x ∈ Rk : f (x) = 0}, then Tp M is the orthogonal complement of the gradient
∇f (p) at p:
Tp M ' x ∈ Rk : x − p ⊥ ∇f (p) .
Example: Let M be the 2-spere of radius R embedded in R3 with center C = (Cx , Cy , Cz ).
Then it is the level R2 set of the function f (x, y, z) = (x − Cx )2 + (y − Cy )2 + (z − Cz )2 .
23
In p = (px , py , pz ), the gradient is


2(px − Cx )
∇f (p) = 2(py − Cy ) = 2(p − C).
2(pz − Cz )
The tangent plane Tp M is given as
{v ∈ R3 : v − p ⊥ 2(p − C)} = {v ∈ R3 : hv − p, 2(p − C)i = 0}.
The tangent bundle of an n-dimensional manifold M is the union of all tangent spaces:
[
TM =
Tp M,
p∈M
with coordinates (p, v), p ∈ M , v ∈ Tp M . The tangent bundle is a 2n-dimensional
manifold itself.
Remark: The tangent bundle might look like the Cartesian product of M and Rn , but
this is in general not true. If we have two different paths α and α̃ connecting distinct
points p and q ∈ M , then due to the curvature of the manifold M , it means that if a fixed
vector v ∈ Tp M is transported via α and α̃ to Tq M , the resulting vectors can be different.
If f : M1 → M2 is a smooth map from one manifold to another, then it is the derivative
Dfp that maps the tangent spaces Tp M1 to Tq M2 for q = f (p). This derivative Dfp is a
linear transformation that automatically takes care that the image of tangent space Tp M1
becomes tangent to M2 at q = f (p).
Example: Let M1 = {(x, y) ∈ R2 : x2 + y 2 = 1} be the unit sphere and M2 = {(x, y) ∈
R2 : x2 + 4y 2 = 4} be an ellips,
both
embedded in R2 . Then the map f (x, y) = (2x, y)
√
√
maps M1 to M2 . Take p = ( 21 2, 12 2) ∈ M1 . The tangent space Tp M1 is subtended by
the vector v = (−1, 1)T , or in other words, the tangent line to M1 through p has slope
√
−1. You can√
also verify this by noting that near p, M1 is the graph of s1 (x) = 1 − x2 ,
−x
√
and at x = 12 2, the derivative is s01 (x)|x= 1 √2 = √1−x
2 |x= 1 2 = −1.
2
2
√ 1√
2 0
. Therefore, the tangent
Under f , p is mapped to q = ( 2, 2 2), and Dfp =
0 1
space Tq M2 is subtended by
2 0
−1
−2
w := Dfp (v) =
=
,
0 1
1
1
so this tangent space (line) has slope − 21 . You can √verify this by noting that near q,
√
M2 is the graph of s2 (x) = 12 4 − x2 , and at x = 2, the derivative is s02 (x)|x=√2 =
√−x | √ = − 1 .
2
2 4−x2 x= 2
24
2.3
Topological Dimension
You probably first encountered the dimension of space in linear algebra, where it denotes
the number of vectors in a basis of a linear space. Hence dim(Rn ) = n. An n-dimensional
manifold is a manifold that is locally homeomorphic to an open ball in Rn . A manifold
can not always be embedded in Rn , for example, the 2-sphere doesn’t fit in R2 , and the
Klein bottle doesn’t even fit in R3 . But Whitney’s Embedding Theorem states that a
smooth n-dimensional manifold can always be embedded in R2n+1 .
There are in fact many definitions of dimension in geometry and topology. Some are
adapted to baroque shapes, assigning fractal dimension to them (Hausdorff dimension,
upper and lower box dimension, etc). For example, the middle third Cantor set C (see
below) has Hausdorff dimension dimH (C) = log 2/ log 3.
There are purely topological definitions of dimension as well. We give the classical one
(which is surely not the only one). Recall that any collection of open sets {Ui }i such that
A ⊂ ∪i Ui is called a cover of A. A cover {Vj }i is a refinement of {Ui }i if it is a cover
of A on its own right, and for each j there is i such that Vj ⊂ Ui .
Definition 35 Given a topological space A, dim(A) ≤ n if for every cover {Ui }i of A
there is a subcover {Vj }i such that any intersection of n + 1 distinct elements of {Vj }i is
empty. The least n for which this works is dim(A).
Example: Take Ui = (i − 23 , i + 23 ), then {Ui }i is a cover of R (with Euclidean topology),
and at most 2 distinct elements can have a non-empty intersection. Therefore dim(R) ≤ 1.
However, it is not possible to find a finer cover such that all its elements are pairwise
disjoint. So dim(R) > 0 as well. So we arrive at the well-known fact that dim(R) = 1.
Topological dimension is invariant under homeomorphisms. This gives an easy argument
why for example R and R2 are not homeomorphic. There are, however, continuous maps
h : R → R2 that are surjective. These are called space-filling or Peano curves, see
Figure ??. They don’t have a continuous inverse, though.
Proposition 36 If A 6= ∅ and dim(A) = 0, then A is totally disconnected.
Proof. Assume by contradiction that x 6= y ∈ A belong to the same connected component C. Let {Ui }i be a cover of A so that x and y belong to different elements of this
cover. Since dim(A) = 0, such a refinement {Vi }i can be found with Vi ∩ Vj = ∅ for i 6= j.
Renumber the indices such that x ∈ V1 . Then y ∈
/ V1 , and V1 and ∪i6=1 Vi are disjoint open
sets. Then also C ∩ V1 and C ∩ (∪i6=1 Vi ) are mutually disjoint open sets (in the relative
topology on C). This contradicts that C is connected.
Theorem 37 The dimension of the Cartesian product is equal to the sum of the dimensions of the factors: dim(X × Y ) = dim(X) + dim(Y ).
25
2.4
Exercises of Section 2
Exercise 29 Parametrise the cylinder C ⊂ R3 with radius 1 around the z-axis. Is it a
manifold? Can it be expressed as the level set of some function?
Exercise 30 Let C be the cone in R3 with top at the origin, symmetric around the z-axis
and with top-angle of 90◦ . Parametrise C. Is it a manifold? Express C as the level set of
some function?
Exercise 31 Parametrise a Möbius band.
Exercise 32 Find parametrisations of spiral curves on the cylinder and cone of Exercises 29 and 30.
Exercise 33 What is the topological dimension of the Sorgenfrey line?
2.5
Solutions to Exercises
Exercise 29: The cylinder is parametrised by (z, t) → (cos t, sin t, z). It is not a manifold,
because it is not compact. It is the level set at level 1 of the function f (x, y, z) = x2 + y 2 .
Exercise 30: The cone is parametrised by (z, t) → (z cos t, z sin t, z). It is not a manifold, because it is not compact, and because (0, 0, 0) has no neighbourhood homeomorphic
to R2 . It is the level set at level 0 of the function f (x, y, z) = x2 + y 2 − z 2 .
Exercise 31: Take R > 1, ψ ∈ (0, 2π) and t ∈ (−1, 1). Then
φ(ψ, t) = ((R + t cos ψ/2) cos ψ, (R + t cos ψ/2) sin ψ, sin ψ/2)
parametrises a Möbius band.
Exercise 32: α(t) = (cos t, sin t, t) for t ∈ R parametrises a spiral on the cylinder.
α(t) = (t cos t, t sin t, t) for t ∈ R parametrises a spiral on the cone.
Exercise 33: The Sorgenfrey line is zero-dimensional. Every cover can be refined by
a cover of the type {[nε, (n + 1)ε) : n ∈ Z} for ε > 0 sufficiently small, and the elements
of this open cover are all disjoint.
3
3.1
Classification of Topological Spaces
Euler Characteristics
For a polyhedron, let us denote the number of faces, edges and vertices by F , E and V
respectively. For example, the five Platonic solids have counts as follows:
polyhedron
tetrahedron
cube
octahedron
dodecahedron
icosahedron
F
4
6
8
12
20
26
E
6
12
12
30
30
V
4
8
6
20
12
Figure 8: The five Platonic solids
Remark: There is a duality relation between these polyhedra. If you take the midpoint
of each face of a Platonic solid and connect it with straight arcs to the midpoints of
neighbouring faces, you get a smaller new Platonic solid with as many vertices as the
previous one had faces. Taking one more step, you get a smaller copy of the original
polyhedron. In summary:
tetrahedron
↔
cube
↔
dodecahedron ↔
tetrahedron
octahedron
icosahedron
Euler6 proved that the following relation is valid for all polyhedra:
χ := F − E + V = 2
(3)
The number χ is called the Euler characteristic of the polyhedron.
Definition 38 Let ∆ stand for a solid closed triangle in the plane. Topologically, a
triangle is the same as a closed disk, but we try to emphasise with this choice that ∂∆
consists of three well-defined arcs. A triangulation of a surface S is a finite collection
of maps fi : ∆ → S, say i ∈ {1, . . . , N } such that
• each fi is a homeomorphism from ∆ to its image ∆i = fi (∆),
• S = ∪N
i=1 ∆i , and
• for distinct i, j, the images ∆i and ∆j intersect in a whole common edge, or at a
single common vertex, or not at all.
By drawing in diagonals in faces of a polyhedron, you can always create a triangulation.
This doesn’t change the Euler characteristic, because every extra diagonal increases the
6
Leonhard Euler (1707 - 1783) was the most prolific and arguably the best mathematician of the 18-th
century
27
number of edges and the number of faces by exactly one, and doesn’t change the number
of vertices.
By triangulation, you can assign an Euler characteristic to every surface. For example:
Surface
sphere
disk
Möbius strip
torus
Klein bottle
projective plane
“pretzel”
χ
2
1
0
0
0
1
−2
orientable?
yes
yes
no
yes
no
no
yes
Every orientable surface S without boundary can be obtained by gluing handles to a
sphere. One handle gives a torus, two handles a “pretzel”, etc. The number of handles g
is called the genus of the surface, and we have the relation
χ(S) = 2 − 2g.
Proof of Euler’s formula (3). First complete the triangulation of the polyhedron by
drawing in diagonals in faces until no non-triangular face remains. Then remove one of
the faces (decreasing the Euler characteristic by one) and flatten out the remaining hull
to a graph in the plane consisting of F − 1 triangles. Next remove edges and vertices from
the graph according to the following rules.
B
@
B@
B @
B
B
B
HH
HH B
B
HH
B
s
s
s
A
A
A
(a) AA
(a) (b)
s A
s
s
@A
@A
@
@A
@
@A
s
@As s
As s
@s
@
s
s
s
(a)
(b)
s
s
s
s
s
Figure 9: Illustration of the proof of Euler’s formula for the tetrahedron.
(a) If removing an edge from the graph doesn’t disconnect the graph, remove it. Since
this either merges two faces, or merges a face with the exterior of the graph, both the
number of faces and the number of edges decreases by one. The Euler characteristic
remains unchanged.
(b) If removing an edge from the graph separates a single vertex from the rest of the
graph, remove both edge and vertex. Both the number of edges and the number of
edges decreases by one. The Euler characteristic remains unchanged.
28
You stop when you have one edge connecting two vertices left. This configuration has
Euler characteristic 0 − 1 + 2 = 1. Therefore the original polyhedron, with the face that
was initially removed restored, has Euler characteristic 2.
3.2
Homotopy and the Fundamental Group
Topology is about the continuous movement of shapes into other shapes. A way to express
this is by means of homotopy. Given a topological space X, a path is a continuous map
f : [0, 1] → X. It connects the points a = f (0) with b = f (1). If f (0) = f (1), then we
call f a loop.
Definition 39 Two paths f and g are called homotopic if there is a family of paths
ht : [0, 1] × [0, 1] → X continuous in both variables x and t, such that ht (0) = a and
ht (1) = b for all t ∈ [0, 1], and h0 = f and h1 = g. The family ht is called the homotopy.
Sometimes we stipulate a little more. If f and g have the same starting point a =
f (0) = g(0) and end point b = f (1) = g(1), then they are homotopic relative to a and
b if there is a homotopy ht as above with ht (0) = a and ht (1) = b for all t ∈ [0, 1]. In this
case, the points a and b are fixed throughout the homotopic motion.
Example: Let X = R2 \ {(0, 0)} with Euclidean topology. We consider three loops
e(x) = (cos x, sin x), f (x) = (2 cos x, 2 sin x) and g(x) = (3 + cos x, sin x). Are these loops
homotopic?
NB: Since we are speaking of loops here, we stipulate that ht (0) = ht (1) for all
t ∈ [0, 1], so the homotopy is not allowed to open up loops.
Clearly e and f are homotopic, because we can choose ht (x) = ((t + 1) cos x, (t +
1) sin x). This homotopy expands the radius of the circle continuously from 1 to 2. It looks
that also e and g are homotopic, via homotopy h̃t (x) = (3t + cos x, sin x), which moves the
center of the circle sideways from (0, 0) to (3, 0). However, for t = 31 , h̃t ([0, 1]) 3 (0, 0), so
h̃t ([0, 1]) is no longer a subset of X. In fact, there is no way of unlinking the loop e from
the hole (0, 0), so e and g are not homotopic. (They would be homotopic in R2 , though.)
Definition 40 A topological space is simply connected if it is connected, and all loops
in X are homotopic to each other.
If we look at the punctured plane X = R2 \ {(0, 0)} once more, and at all loops with
starting and end point b, say b = (1, 0), then we can ask: How many different loops at
base point b are there if we consider homotopic loops to be the same.
In other words, what is
π1 (X, b) := {f : [0, 1] → X continuous : f (0) = f (1) = b}/ ∼,
where loops f ∼ g if they are homotopic relative to b?
29
(4)
The homotopy class of f depends on its winding number around the hole (0, 0),
i.e., how many times f loops around the hole in counter-clockwise direction. A counterclockwise loop is not homotopic to a clockwise loop, so the winding number can be negative. Hence, the set of integers Z is a short way of expressing (4).
We can piece loops together as well. Given f and g with the same base point b, define
f (2x)
if x ∈ [0, 21 ];
f ∗ g(x) =
g(2x − 1) if x ∈ [ 21 , 1].
Since f (1) = g(0) = b, this is well-defined: f ∗ g( 12 ) = b.
Concatenation of loops have the following properties (see Exercise 36):
1. f −1 (x) = f (1 − x) is a loop with base point b.
2. f −1 ∗ f ∼ id, where id stands for the constant loop id(x) ≡ b.
3. f ∗ id ∼ id ∗ f ∼ f .
4. (f ∗ g) ∗ h ∼ f ∗ (g ∗ h).
Properties 1.-4. above show that π1 (X, b) with operation ∗ has a group structure. It
is called the fundamental group of X.
For the punctured plane X = R2 \ {(0, 0)}, the operation ∗ acts on π1 (X) exactly
as the addition on Z: winding numbers add up. For Y = R2 \ {(0, 0), (0, 1)}, the plane
with two holes, you have loops around the different holes that don’t cancel; they don’t
even commute! If loops f and g with common base point b = (1, 0) go counter-clockwise
around (0, 0) and (0, 1) respectively, then f ∗ g 6∼ g ∗ f . The fundamental group π1 (Y ) is
not commutative. In fact, there are no cancellations of loops other than the very obvious
ones: f −1 ∗f = id and g −1 ∗g = id. Therefore π1 (Y ) is the free group with two generators
hf, gi.
Example: Let X = T2 be the two-dimensional torus. There are two loops, say f going
around the torus horizontally and g going around the torus vertically. Every loop however
complicated can be expressed as concatenation of loops f , g, f −1 and g −1 . Furthermore,
f ∗ g = g ∗ f , but there are no further cancelling relation. Therefore the fundamental
group is a commutative group with two generators: π1 (T2 ) = Z2 .
3.3
Exercises of Section 3
Exercise 34 Show that being homotopic is an equivalence relation.
Exercise 35 Which of the following spaces (see Figure 10) are simply connected: S2 , T2 ,
T3 , the Klein bottle, the Möbius7 strip?
7
August Möbius (1790 - 1868) was a German professor of mathematics. He is best known for his work
in topology, but his famous strip was in fact used before him by Listing. Later on, we will see Möbius’
name attached to a class of transformations.
30
Figure 10: A flattened-out torus, pretzel, Klein bottle and projective plane and their
“triangulations”.
Exercise 36 Show the group properties of concatenation of loops, i.e., :
1. f −1 (x) = f (1 − x) is a loop with base point b.
2. f −1 ∗ f ∼ id, where id stands for the constant loop id(x) ≡ b.
3. f ∗ id ∼ id ∗ f ∼ f .
4. (f ∗ g) ∗ h ∼ f ∗ (g ∗ h).
Exercise 37 Show that in an arc-connected space X, π1 (X, b) and π1 (X, a) have exactly
the same group structure for any a, b ∈ X. Hence we can simply write π1 (X) in this case.
Exercise 38 What are the fundamental groups of the following spaces: S2 , T3 , the Klein
bottle, the Möbius strip?
Exercise 39 A painting is to be hung from two nails in the wall using a single string
attached to the two upper corners of the painting. Can you hang it in such a way that
whenever you remove a nail, the painting falls to the ground? Can you extend your strategy
to any number of nails?
3.4
Solutions to Exercises
Exercise 36:
31
1. f −1 (x) = f (1 − x) satisfies f −1 (1) = f (0) = b and f −1 (0) = f (1) = b, so f −1 is also
a loop with base point b.
2. Use the homotopy
ht (x) =
f −1 (2tx)
f (2t(1 − x))
if 0 ≤ x ≤ 21 ,
if 12 ≤ x ≤ 1.
Then h0 (x) = id(x) ≡ b and h1 = f −1 ∗ f .
3. Use the homotopy
ht (x) =
f (2tx)
f (2(1 − t)x + 2t − 1)
if 0 ≤ x ≤ 21 ,
if 12 ≤ x ≤ 1.
Then h0 = f ∗ id and h1 = id ∗ f , and note also that h1/2 = f .
4. We have
and


 f (4x)
g(4x − 1)
F (x) = (f ∗ g) ∗ h(x) =

 h(2x − 1)
if 0 ≤ x ≤ 41 ,
if 14 ≤ x ≤ 21 ,
if 12 ≤ x ≤ 1.


 f (2x)
g(4x − 2)
G(x) = f ∗ (g ∗ h(x)) =

 h(4x − 3)
if 0 ≤ x ≤ 21 ,
if 12 ≤ x ≤ 43 ,
if 34 ≤ x ≤ 1.
So both paths trace the same image, but with different speeds. Take

if 0 ≤ x ≤ 41 ,

 (1 + t)x
x + t/4
if 41 ≤ x ≤ 12 ,
ψt (x) =

 (1 + t/2)x − t/2 if 3 ≤ x ≤ 1.
4
The F ◦ ψt (x) is the required homotopy.
Exercise 37: Let p be a path connecting a to b. Then for every f ∈ π1 (X, b) the
path T (f ) = p ∗ f ∗ p−1 belongs to π1 (X, a), and the operation T gives a concatenationpreserving one-to-one correspondence between π1 (X, a) and π1 (X, b).
Exercise 38: π1 (S2 ) = {0} because all loops are null-homotopic.. π1 (T3 ) = Z3 ; same
argument as for T2 . π1 (K) = ha, bi/abab−1 . π1 (Möbius strip) = Z.
Exercise 39: Let a be a clock-wise loop around nail 1 and b be a clock-wise loop
around nail 2. Then wrapping the chord around the nails as indicated by aba−1 b−1 will
do the trick. For three nails we use an extra generator c, and b−1 a−1 caba−1 b−1 c−1 ba will
do the trick.
32
Figure 11: The tractrix (left) and the tractrix plus catenary curve (right).
4
4.1
Curves in Rn
Parametrised curves
A curve or path is the image of a map α : (a, b) → Rn , where α is called the parametrisation of the curve. Sometimes it is useful to include the endpoints in the domain. For
example, the straight line connecting P = (1, 0) and Q = (0, 3) in R2 has parametrisation
α : [0, 1] → R2
α(t) = (1 − t, 3t).
(5)
A loop or closed curve is obtained if α(a) = α(b). For example, the unit circle in R2 is
usually parametrised as
β : [0, 1] → R2
β(t) = (cos 2πt, sin 2πt).
(6)
Definition 41 A curve in Rn is regular if it has a differentiable parametrisation α :
(a, b) → Rn such that α0 (t) 6= 0 for all t ∈ (a, b). (If the curve is a closed curve, then we
also stipulate that α0 (a) = α0 (b).
Example: A stick of unit length lies in the sand. At one end, it has a spike pointing
into the sand. You draw the stick at the other end along a line perpendicular to the
stick’s initial position. The curve that the spike draws in the sand is called a tractrix
(“drag-line” in latin).
Let y = y(x) be the graph of this curve. It has two smooth parts, meeting in a cusp at
(0, 1). The horizontal axis is the line along which you draw the stick. Then the tangent
line to y at x intersects the horizontal axis at x1 such that the points (x, y(x)) and (x1 , 0)
have Euclidean distance one. Therefore
y(x) = y 0 (x)(x − x1 ),
y(x)2 + (x − x1 )2 = 1,
and we can rewrite the differential equation:
(
y(0) = 1,
y0 = ∓ √ y
1−y 2
,
where the − is for x > 0 and the + for x < 0. This can be solved as
!
p
1 + 1 − y2 p
x = ± ln
+ 1 − y2
y
33
(7)
or in parametric form
t
t 7→ (ln(tan ) + cos t, sin t),
2
t ∈ (0, π).
(8)
If you draw a line through (x, y(x)) perpendicular to the tractrix, then the envelope
of these line form another curve, see Figure 11. This catenary curve is given by
ex + e−x
.
2
This is also the shape that you get if you suspend a chain from its endpoints, hence the
name (catena = chain in latin).
c(x) = cosh(x) =
The derivative α0 (t) gives the tangent vector to the curve at α(t). For example, α0 (t) =
(−1, 3) for all t in the straight arc of (5), and β 0 (t) = (−2π sin 2πt, 2π cos 2πt) for the
circle as in (6). The tangent vector not only depends on the curve, but also on the
parametrisation. For example, the unit circle can also be parametrised as
γ : [0, 2π] → R2
γ(t) = (cos t, sin t),
(9)
and then γ 0 (t) = (− sin t, cos t). This is a scalar multiple of β 0 (t/2π), i.e., tangent vectors
of different parametrisations at the same point of the curve, are scalar multiples of each
other. The direction of the tangent vector depends on the curve, the size of the tangent
vector of the parametrisation.
If parameter t is thought of as time, then it is customary to use Newton’s notation for
...
derivatives: γ̇(t) = γ 0 (t). Second and third derivative are γ̈(t) and γ (t). If kα̇(t)k = 1
for all t, then we say that the curve has a unit speed parametrisation, or that it is
parametrised by arc-length. The latter is motivated by the formula for the length of
a curve γ : (a, b) → Rn , namely
Z b
q
(10)
l(γ) =
kγ̇(t)k dt
where kxk = x21 + · · · + x2n .
a
Hence, if γ were parametrised by arc-length, then l(γ) = b − a.
Proposition 42 A curve can be parametrised by arc-length if and only if it is regular.
Proof. ⇒ If kγ 0 (t)k = 1 for all t, then γ is differentiable by default, and γ 0 (t) 6= 0. So γ
is a regular parametrisation.
⇐ Assume that γ : (a, b) → Rn is a regular parametrisation, and define
Z t
s(t) =
kγ 0 (u)k du.
a
0
= kγ (t)k > 0, and s(a) = 0. By the Inverse Function Theorem, t is a difThen
dt
ferentiable function of s, and the chain rule gives ds
= ( ds
)−1 = kγ 0 (t)k−1 . Now define
dt
γ̃(s) = γ ◦ t(s). Then γ̃(0) = γ ◦ t(0) = γ(a), and the derivative is
ds
dt
dγ̃
dγ ◦ t
dγ dt
dγ
(s) =
(s) =
· (s) =
· kγ 0 (t)k−1 ,
ds
ds
dt ds
dt
34
whence
dγ
(t)
dγ̃
k (s)k = k dt0
k = 1.
ds
kγ (t)k
In short, the recipe for parametrising a curve γ : [0, 1] → R by arc-length is
Rt
1. Compute s(t) = 0 kγ̇(u)k du.
3
2. Invert s(t) to t(s).
3. Now the curve γ̃(s) = γ(t(s)) is parametrised by arc-length.
In practice, reparametrising a curve by arc-length boils down to enormous calculations,
involving most unfriendly integrals, so there are only a few examples where you can
compute the arc-length parametrisation explicitly.
Lemma 43 If α is a curve in Rn such that kα(t)k ≡ 1, then hα̇(t), α(t)i = 0 for all t.
Remark: This lemma applies to α̇ just as well, so if α is a unit-speed curve, then
α̇(t) ⊥ α̈(t) for all t.
Proof. Because α has unit length, 0 =
d
kαk2
dt
=
d
hα, αi
dt
= hα̇, αi + hα, α̇i = 2hα̇, αi. Definition 44 For a planar unit speed curve γ : (a, b) → R2 , the curvature at point
P = γ(s) is defined as κ(s) = kγ̈(s)k
Example: If γ traces a circle of radius R, then we can take

s
s

 γ(s) = (R cos R , R sin R )
γ̇(s) = (− sin Rs , cos Rs )

 γ̈(s) = (− 1 cos s , − 1 sin s )
R
R
R
R
and so the curvature κ(s) ≡ R1 , the inverse radius of the circle. For planar curves, you
should think of curvature as the inverse radius of the circle that best fits the curve at
γ(s). For straight lines, this circle would be infinitely big, and indeed, the curvature of a
straight line is zero. For non-planar curves, this interpretation makes little sense.
4.2
Frenet frames in R3
For curves in R3 , we have the extra tool of the cross product ×. This enables us to define
a canonical Frenet frame of vectors that traces the bends of the curve through space.
It is spanned by {t, n, b}, which stand for tangent vector, normal vector and binormal
vector, and it has a right-handed orientation, which means that if you consider (t, n, b)
35
as the columns of a 3 × 3-matrix, then the determinant is positive. In fact, for unit speed
curves, they are three perpendicular unit vectors, so det(t, n, b) = 1, and
t×n=b
n×b=t
b × t = n.
(11)
Recall that the curvature of a unit speed curve γ with tangent vector t(s) = γ̇(s) is
defined as κ(s) = kγ̈(s)k = kṫ(s)k. If curvature κ = κ(s) 6= 0, define n by
ṫ = κn.
It is a unit vector by definition, and by Lemma 43 perpendicular to t. Hence the name
normal vector. Furthermore b = t × n is a unit vector perpendicular to both n and t;
it is called the binormal vector, and standard cross product identities imply that (11)
holds. Now
ḃ = ṫ × n + t × ṅ = κn × n + t × ṅ = t × ṅ,
so ḃ is perpendicular to t, and by Lemma 43 also to b. Hence ḃ must be parallel to n
and there is τ = τ (s) such that
ḃ = −τ n.
The quantity τ is called the torsion, and the minus sign is purely conventional.
Lemma 45 For a unit speed curve γ, the torsion is τ =
1
κ2
...
hγ̇ × γ̈, γ i.
Proof. For unit speed curves, we have
τ = −hn, ḃi = −hn, ṫ × n + t × ṅi = −hn, t × ṅi
Since n = κ1 ṫ = κ1 γ̈, this gives
1
d 1
τ = − hγ̈, γ̇ × ( γ̈)i
κ
ds κ
1
1 ... κ̇
= − hγ̈, γ̇ × ( γ − 2 γ̈)i
κ
κ
κ
...
1
= − 2 hγ̈, γ̇ × γ i
κ
1 ...
= 2 h γ , γ̇ × γ̈i.
κ
...
Remark: For curves at non-unit speed, we have the formula τ =
hγ̇×γ̈, γ i
.
kγ̇×γ̈k2
Theorem 46 If γ : (a, b) → R3 is a space curve with non-vanishing curvature, then γ is
contained in a plane if and only if the torsion is constant zero.
36
Proof of Theorem 46. Assume that unit speed curve γ is planar, so there is a vector
β ∈ R3 and d ∈ R such that8 hγ(s), βi = d for all s. Taking derivatives gives us
hγ̇, βi = 0
and
hγ̈, βi = 0.
Therefore t = γ̇ and n = κγ̈ are always perpendicular to β. It follows that b = t × n is
a unit vector parallel to β, and hence 0 = ḃ = τ n, and τ (s) = 0 for all s.
Conversely, if τ (s) ≡ 0, then ḃ = τ n = 0 (which holds when curvature κ 6= 0 because
otherwise n is not well-defined). Hence b is a constant vector. Let d := hγ(s0 ), bi, then
d
hγ, bi = hγ̇, bi = ht, bi = 0
ds
and so hγ(s), bi = d for all s.
Proposition 47 There is an skew-symmetric matrix such that
  
 
t
0
κ 0
t
d   


n = −κ 0 τ
n
ds
b
0 −τ 0
b
Proof. The matrix equation comprises three linear equations. The first and third are
just the definition of curvature and torsion. For the second, the relation n = b × t gives
ṅ = ḃ × t + b × ṫ = −τ n × t + κb × n = −κt + τ b
4.3
Exercises of Section 4
Exercise 40 Solve (7) to verify that the given solutions are correct. Show that the tractrix
has an infinitely sharp cusp at (0, 1) (where t = π/2). Hence, there is no tangent line
at (0, 1). How can you reconcile this with the parametric solution, which looks entirely
smooth. Is the tractrix a regular curve?
Exercise 41 The curve γ(t) = (cos t − 12 cos 2t, sin t − 1s sin 2t) is called a cardioid. Sketch
the image of γ. Is the cardioid a regular curve?
Exercise 42 Reparametrise the helix
h : R → R3 ,
h(t) = (cos t, sin t, ct)
(12)
by arc-length.
If kβk = 1, then d is the signed distance of this plane to the origin. Indeed, we can rewrite hγ(s), βi = d
to hγ(s) − βd, βi = 0
8
37
Exercise 43 Reparametrise the curve γ : (0, ∞) → R2 defined as γ(t) = (t2 , 32 t3 ), by
arc-length.
Exercise 44 Compute the arc-length of a piece of parabola: α : [0, 1] → R2 , α(t) = (t, t2 ).
Exercise 45 Starting at the point (0, 1) for t = π2 in (8), show that the arc-length is
s(t) = − ln | sin t|. Use this to show that the parametrisation of the right half of the
tractrix by arc-length is
√
√
(13)
γ(s) = s + 1 − e−2s + ln(1 − 1 − e−2s ) , e−s .
Exercise 46 Show that the helix of (12) has curvature
1
1+c2
and torsion
c
.
1+c2
Exercise 47 Show that the assumption that the curvature is non-vanishing is essential.
That is, give an example of a curve with vanishing torsion, but which is not confined to a
plane.
4.4
Solutions to Exercises
√
Rtp
Exercise 42: We have s(t) = 0 (− sin u)2 + (cos u)2 + c2 du = 1 + c2 t. Therefore
√
t = s/ 1 + c2 and the arc-length parametrisation becomes
√
√
√
h(s) = (cos(s/ 1 + c2 ), sin(s/ 1 + c2 ), cs/ 1 + c2 ).
√
Exercise 43: We have γ̇(t) = (2t, 2t2 ), so kγ̇(t)k = 2t 1 + t2 . This gives the arclength:
t
Z t √
3
3
2
2
2
s(t) =
2u 1 + u2 du = (1 + u ) 2 = ((1 + t2 ) 2 − 1).
3
3
0
0
q
2
This inverts to s(t) = ( 23 s + 1) 3 − 1, so the reparametrisation by arc-length becomes:
2
2
3
3
3
γ̃(s) = ( s + 1) 3 − 1 , (( s + 1) 3 − 1) 2 .
2
2
p
Exercise 44: Since α̇(t) = 1 + (2t)2 , the arc-length between (0, 0) and (t, t2 ) is
Z t√
Z sinh−1 (2t)
1
s(t) =
1 + 4u2 du =
cosh2 (x)dx
2
0
0
Z sinh−1 (2t)
1
1
1
sinh−1 (2t)
=
(1 + cosh 2x) dx = [x + sinh(2x)]0
4 0
4
2
−1
1
sinh (2t)
=
[x + sinh(x) cosh(x)]0
4
q
1
1
1 √
sinh−1 (2t)
=
[x + sinh(x) 1 + sinh2 (x)]0
= sinh−1 (2t) + t 1 + 4t2 .
4
4
2
Exercise 45 : Using the parametrisation for the branch of positive x-values
t
γ(t) = (ln(tan ) + cos t, sin t),
2
38
π
t ∈ ( , π),
2
we get (using the angle doubling formula sin t = 2 cos 2t sin 2t )
1
1
1
γ̇(t) =
−
sin
tt,
cos
t
=
(
− sin tt, cos t).
t
t
sin t
tan 2 2 cos2 2
Hence
r
kγ̇(t)k =
1
− 2 + sin2 t + cos2 t =
sin2 t
r
s
1
−1=
sin2 t
1 − sin2 t
cos t
.
=−
2
sin t
sin t
Note that for positive x-values, t ∈ ( π2 , π), and so we need the minus sign for the square
root.. The arc-length from the cusp is
Z t
Z t
cos u
s(t) =
kγ̇(u)k du =
−
du = − ln | sin(u)|tπ/2 = − ln | sin t|.
sin u
π/2
π/2
This inverts as t(s) = arcsin e−s . Insert in the initial parametrisation to give:
arcsin e−s
−s
−s
γ̃(s) = γ(t(s)) =
ln(tan
) + cos arcsin e , sin arcsin e
2
√
arcsin e−s
−s
−2s
=
ln(tan
) + 1 − e ,e
.
2
Using angle doubling formulas
sin 2t
2 sin 2t cos 2t
sin t
t
=
tan =
t =
t
2
2
1 + cos t
cos 2
cos 2 − 1 + 1
we can simplify the above formula to
√
e−s
−s
−2s
√
γ̃(s) =
ln(
) + 1 − e ,e
1 + 1 − e−2s
√
√
e−s (1 − 1 − e−2s )
−s
=
ln(
) + 1 − e−2s , e
1 − (1 − e−2s )
√
√
= s + ln(1 − 1 − e−2s ) + 1 − e−2s , e−s
as stated.
Exercise 46: Continuing with the arc-length parametrisation
√
√
√
h(s) = (cos(s/ 1 + c2 ), sin(s/ 1 + c2 ), cs/ 1 + c2 ).
of Exercise 42, we find
√


− sin(s/√ 1 + c2 )
1
 cos(s/ 1 + c2 ) 
ḣ(s) = √
1 + c2
c
√


− cos(s/√ 1 + c2 )
1 
ḧ(s) =
− sin(s/ 1 + c2 )  = κn.
1 + c2
0
39
Therefore the curvature is κ =
1
1+c2
for all t. Also
√

sin(s/ √
1 + c2 )
c
− cos(s/ 1 + c2 )
b=t×n= √
1 + c2
0

and
√


cos(s/√ 1 + c2 )
c 
c
ḃ(t) =
n
sin(s/ 1 + c2 )  = −
2
1+c
1 + c2
0
c
Therefore the torsion is τ = 1+c
2 for all t.
Exercise 47: Take a curve γ : [0, 1] → R3 that is confined to the xy-plane for t ∈ [0, 13 ],
belongs to the y-axis for t ∈ [ 13 , 23 ] and is confined to the yz-plane for t ∈ [ 32 , 1]. Then γ is
not planar, but the torsion is constant zero. The linear part for t ∈ [ 31 , 23 ] is necessary to
move from the horizontal to the vertical plane whilst keeping torsion zero. In this linear
part, however, curvature vanishes.
40
5
5.1
Surfaces in R3
The first fundamental form
Let S be a surface in R3 and σ : U ⊂ R2 → S be a patch. Then a curve γ : (a, b) → S
through this patch can be obtained as γ(t) = σ ◦ α(t), where α : (a, b) → U with
α(t) = (u(t), v(t)) is a path through U . Using the chain rule, we find γ̇ = σu u̇ + σv v̇ and
kγ̇k2 = hγ̇, γ̇i = hσu , σu iu̇2 + 2hσu , σv iu̇v̇ + hσv , σv iv̇ 2 .
Definition 48 The first fundamental form is
ds2 = E du2 + 2F du dv + G dv 2
where E = hσu , σu i, F = hσu , σv i and G = hσv , σv i.
It is a 2-form using the first derivatives of the parametrisation of the surface. In this
terminology, the arc-length of a curve is
Z √
Z √
Z
2
2
2
ds =
Edu + 2F dudv + Gdv = kγ̇(t)k dt.
Example: The torus with “outer radius” R (the distance between the center of the torus
and the center circle through the tube of the torus) and “inner radius” r (the distance
between circle through the tube and the edge of the torus) can be parametrised by


(R + r cos u) cos v
u ∈ [0, 2π), v ∈ [0, 2π)
σ(u, v) =  (R + r cos u) sin v  ,
r sin u
Taking derivatives gives


−r sin u cos v
σu =  −r sin u sin v 
r cos u
Therefore, E = r2 ,


−(R + r cos u) sin v
σv =  (R + r cos u) cos v  .
0
and
F = hσu , σv i = 0,
G = hσv , σv i = (R + r cos u)2 .
Now let γ be a “straight” curve that wraps around the torus once, both horizontally and
vertically. What is its length? Take α : [0, 2π) → U = [0, 2π) × [0, 2π), α(t) = (t, t), and
γ(t) = σ ◦ α. Then we find
Z 2π p
l(γ) =
(R + r cos t)2 + r2 dt,
0
which is, unfortunately, a most unfriendly integral.
The normal vector N to a surface is given by
N=
σu × σv
.
kσu × σv k
41
(14)
Figure 12: Archimedes’s untimely end and a sphere with circumscribed cylinder adorning
his grave.
It is obviously perpendicular to (the tangent plane of) the surface. Also, the triple of
infinitesimal vectors σu du, σv dv and σu × σv form a small parallelepiped with volume
kσu × σv kdudv equal to the area of its base (spanned by σu du and σv dv). Another way of
computing this base is by recalling that the area of a parallelogram is the product of the
sides and the sine of the included angle. Note that σu and σv span Tp S, so this included
angle φ 6= 0 and φ 6= π.
p
0 < kσu kkσv k sin φ = kσu kkσv k 1 − cos2 φ
p
kσu k2 kσv k2 − kσu k2 kσv k2 cos2 φ
=
p
=
hσu , σu ihσv , σv i − hσu , σv i2
√
=
EG − F 2 .
(15)
This leads to the following proposition.
Proposition 49 If σ : R → S is a region on a surface, then its area is
ZZ
ZZ √
Area(S) =
kσu × σv k du dv =
EG − F 2 du dv
R
R
Corollary 50 A smooth map f : S1 → S2 preserves area if and only if their fundamental
forms satisfy E1 G1 − F12 = E2 G2 − F22 . (Or more precisely (E1 G1 − F12 )(x) = (E2 G2 −
F22 )(f (x)) for every x ∈ S1 .)
Example: You can circumscribe a sphere S (centered at the origin and with radius
R) with a cylinder C (radius R, height 2R, centered around the vertical axis, without
bottom and top). Define the map f : S \ {north pole, south pole} → C by horizontal
projection outwards from the vertical axis. It turns out that this map preserves area,
as was discovered by Archimedes well over 2000 years ago. Using this, he was the first
to discover that the area of a sphere is 4πR2 . Archimedes considered this as his main
achievement of his career; at least a picture of a sphere with circumscribed cylinder as
engraved on his tomb, see Figure 12.
42
5.2
Geodesics
Let γ(t) = σ(u(t), v(t)) be a unit speed curve on surface S. Then kγ̇k ≡ 1 and γ̇ is
a tangent vector to S, and therefore γ̇ ⊥ N. Therefore γ̇, N and N × γ̇ are mutually
perpendicular, and γ̇ ⊥ γ̈ because kγ̇k is constant (see Lemma 43). Hence γ̈ is a linear
combination of N and N × γ̇:
γ̈ =
κn N + κg N × γ̇
|{z}
|{z}
normal
(16)
geodesic
curvature
curvature
and since N and N × γ̇ are perpendicular unit vectors, we have κ = kγ̈k = κ2n + κ2g by
Pythagoras’ Theorem. The normal and geodesic curvature indicate how much a curve
bends “towards the normal vector” and “within the surface” respectively. If κg ≡ 0, then
we call γ a geodesic. In Newtonian mechanics, if a particle of mass m travels along γ,
we have the exerted force equals F = mγ̈. This means that geodesic motion only feels a
force perpendicular to the surface S it is confined to, but none within S. If the particle
is unaware of what is outside S, it thinks it is going along a straight line.
Example: Let S be the unit sphere. Parametrised by spherical coordinates, we find






− cos u sin v
− sin u cos v
cos u cos v
σ(u, v) =  sin u cos v  , σu =  cos u cos v  , σv =  − sin u sin v  .
cos v
0
sin v
Therefore

 

 
2
− sin u cos v
− cos u sin v cos u cos2 v  cos u cos v  ×  − sin u sin v  =  sin u cos v  = cos v.
cos v sin v cos v
0
and N(u, v) = σ(u, v). Now take γ(t) = σ(u(t), v(t)) to be one of the meridians, so
u(t) = u0 and v(t) = t for t ∈ (−π, π). Then






− cos u0 cos t
cos u0 cos t
− cos u0 sin t
γ(t) = σ(u, v) =  sin u0 cos t  , γ̇(t) =  sin u0 sin t  , γ̈(t) =  − sin u0 cos t  .
cos t
− sin t
sin t
Therefore γ̈(t) = −N and so the normal curvature is κn = −1 (γ curves away from
N) and geodesic curvature κg = 0. Thus, meridians (and in fact all great circles) are
geodesics. When sailing along a “straight line” on the ocean surface, you actually go
along a geodesic, and this the straightest you can go on a sphere.
Lemma 51 If γ is a geodesic, then kγ̇k is constant.
Proof. Since the geodesic curvature κg = 0, γ̈ is parallel to N, and therefore perpendicular to S. It follows that dtd kγ̇k2 = dtd hγ̇, γ̇i = 2hγ̈, γ̇i = 0.
43
Lemma 52 If a constant speed curve γ satisfies γ̈ ⊥ S, then it is a geodesic.
Proof. Since γ̈ = κn N + κg (N × γ̇) and γ̈ is parallel to N, it follows that κg = 0.
Example: In a plane H, geodesics are precisely the straight lines (traversed at constant
speed). Indeed, if γ(t) = at + b ∈ H, for some fixed vectors a, b ∈ R3 , then γ̇ = a is
constant and γ̈ = 0.
Conversely, since the unit normal vector N for H is constant, we find
d
hγ̇, Ni = hγ̈, Ni + hγ̇, Ṅi = hγ̈, Ni.
dt
For a geodesic, γ̈ = κn N, so the normal curvature κn = hγ̈, Ni = 0 and γ̈ = 0. Integrating
twice, we find γ(t) = at + b.
0≡
Definition 53 A normal section of a surface S is the intersection of a plane H with
S such that H ⊥ S at every point of intersection.
Corollary 54 Suppose γ : (a, b) → S is a constant speed curve and its image is contained
in a normal section of S. Then γ is a geodesic.
Proof. At every point γ(t) ∈ S ∩ H, the plane H is spanned by γ̇(t) and N. Also
γ̈(t) ⊥ γ̇(t) because γ is a constant speed curve. Since the curve is contained in H (H
is the ‘plane of movement”), γ̈ is parallel to H, and hence γ̈ ⊥ S. By Lemma 52, γ is a
geodesic.
Theorem 55 If γ is a curve through a patch σ : U → S with first fundamental form
Edu2 + 2F dudv + Gdv 2 , then γ is a geodesic if and only if the geodesic equations
( d
E u̇ + F u̇ = 12 Eu u̇2 + 2Fu u̇v̇ + Gu v̇ 2
dt
(17)
d
1
2
2
F
v̇
+
G
v̇
=
E
u̇
+
2F
u̇
v̇
+
G
v̇
v
v
v
dt
2
We will not prove this theorem, but remark on its applications. An explicit solution
of (17) is almost always too hard to get. However, on a smooth surface S, the functions
E, F and G are smooth and we can rewrite (17) to a quadruple of first order ordinary
differential equations

u̇ = x



 ẋ = H (u, x, v, y)
1
(18)

v̇ = y



ẏ = H2 (u, x, v, y)
where H1 and H2 are smooth functions (obtained by the Implicit Function Theorem). The
theory of ODEs states that given an initial condition (u(0), x(0), v(0), y(0)) = (u0 , x0 , v0 , y0 ),
there is a unique solution to (18) with this initial condition. In other words, if we specify
a starting point P0 ∈ S and velocity t0 ∈ TP0 S, there is always a unique geodesic with
γ(0) = P0 and γ̇(0) = t0 .
44
Theorem 56 A geodesic between P and Q is locally the shortest path connecting P and
Q.
Remark: Here “locally” means that we cannot make the length of the path γ smaller by
a small change. It doesn’t mean that γ is the shortest path. For instance Quito (Ecuador)
and Kampala (Uganda) both lie more or less on the equator. You can travel from Quito
to Kampala going along the equator in either direction; in both ways we go along the
geodesic, but only one way gives the shortest distance globally.
Sketch of Proof. The proof is by method of Calculus of Variations. Assume that γ is
the curve of locally shortest distance between P and Q. Let γ τ , τ ∈ [0, 1] be a homotopy
of curves connecting P and Q so that γ 0 = γ and γ τ (a(τ )) = P and γ τ (b(τ )) = Q for all
τ ∈ [0, 1]. We want to show that the length form
Z b(τ )
L(τ ) =
kγ τ (s)kds
a(τ )
is indeed minimised at γ, regardless of the choice of homotopy. This means that dτd L(τ ) =
0 at γ regardless of the choice of homotopy. This turns out to hold exactly when γ satisfies
the geodesic equations (17).
Remark: Although introduced for surfaces, geodesics exist for smooth manifolds of any
dimension, and they are the curves of locally shortest distance, as well as curves for which
γ̈ is always perpendicular to the manifold.
5.3
The second fundamental form and the principal curvatures
If γ is a unit speed curve, and ∆t a small time unit, then we have from Taylor’s Theorem
that
1
γ(t + ∆t) − γ(t) = γ̇(t)∆t + γ̈(t)(∆t)2 + h.o.t.
2
where “h.o.t.” stands for higher order terms in ∆t, and these are negligible compared to
(∆t)2 . The infinitesimal displacement of the curve from its tangent line can be obtained
by taking the inner product with the unit normal vector n, so this is
1
hγ(t + ∆t) − γ(t), ni = hγ̇(t), ni∆t + hγ̈(t), ni(∆t)2 + h.o.t.
2
1
κ(∆t)2 + h.o.t.
=
2
if we recall that γ̇ ⊥ n and the curvature κ satisfies γ̈ = κn.
We have an analogue for a surface S embedded in R3 . Let σ : U → S be chart for the
surface with unit normal vector N (see (14)). Then Taylor’s Theorem in dimension two
states
σ(u + ∆u, v + ∆v) − σ(u, v) = σu ∆u + σv ∆v
1
+
σuu (∆u)2 + 2σuv ∆u∆v + σvv (∆v)2 + h.o.t.
2
45
Taking the inner product with normal vector N gives the infinitesimal displacement of
the surface from its tangent plane.
hσ(u + ∆u, v + ∆v) − σ(u, v), Ni = hσu , Ni∆u + hσv , Ni∆v +
1
h σuu (∆u)2 + 2σuv ∆u∆v + σvv (∆v)2 , Ni + h.o.t.
2
1
= h σuu (∆u)2 + 2σuv ∆u∆v + σvv (∆v)2 , Ni + h.o.t.
2
1
L(∆u)2 + 2M ∆u∆v + N (∆v)2 + h.o.t.
=
2
where L = hσuu , Ni, M = hσuv , Ni and N = hσvv , Ni. The quantity
Ldu2 + 2M dudv + N dv 2
is called the second fundamental form; it is a 2-form involving second derivatives of
σ.
We can capture the contents of the first and second fundamental forms in symmetric
matrices
E F
L M
FI =
and FII =
.
(19)
F G
M N
Recall from (15) that det(FI ) > 0, so we can compute FI−1 . The matrix product
W = FI−1 · FII
(20)
is used frequently; it is known as the Weingarten matrix in coordinates σu and σv .
Definition 57 The principal curvatures κ1 and κ2 at p in a surface S are the two
solutions of the quadratic equation
det(FII − κFI ) = 0
For each solution κi (i = 1, 2), there is a tangent vector wi ∈ Tp S such that wi =
ξi σu + ηi σv and
ξ
(FII − κi FI ) i = 0.
(21)
ηi
These tangent vectors wi are called the principal vectors.
If we multiply the matrices in (22) by FI−1 , then we get a true eigenvalue equation:
ξ
(22)
(W − κi I) i = 0.
ηi
Proposition 58 The principal curvatures κ1 and κ2 are always real. If κ1 = κ2 , then
every vector is a principal vector. If κ1 6= κ2 , then the principal vectors w1 and w2 are
perpendicular.
46
Proof. We have the parametrisation σ : U → S, with U ⊂ R2 and S embedded in R3 .
Assume that σ(0, 0) = p. To be very formal, FI , FII and W are linear maps on the
tangent space T(0,0) U .
We know from linear algebra, that symmetric matrices have real eigenvalues and perpendicular eigenvectors. Although FI and FII and also FI−1 are symmetric,
that
doesn’t
ξ
ξ2
1
−1
mean that W = FI FII is symmetric. So we cannot conclude that
and
are
η1
η2
perpendicular. But that is not what the
proposition
claims.
ξ1
ξ2
Instead, it says that if you map
and
∈ T(0,0) U (via the derivative map
η1
η2
Dσ) to w1 and w2 ∈ Tp S, then they become perpendicular. The matrix FI can be used
the express the inner product h , i on Tp S as an inner product h , iI on T(0,0) U . Namely,
if x and y are tangent vectors, say x = x1 σu + x2 σv and y = y1 σu + y2 σv , then the inner
product
hx, yi = hx1 σu + x2 σv , y1 σu + y2 σv i = Ex1 y1 + F (x1 y2 + x2 y1 ) + Gx2 y2
x1
y1
x1
y
=
, FI
=:
, 1
.
x2
y2
x2
y2
I
The defining properties of a complex-valued inner product on a linear space X are:
hx, xi ≥ 0 ∀x ∈ X
hx, yi = hy, xi ∀x, y ∈ X where the bar
means complex conjugate (23)
hαx, yi = αhx, yi ∀x, y ∈ X, ∀α ∈ C
hx + x̂, yi = hx, yi + hx̂, yi ∀x, x̂, y ∈ X,
You can check that the dot product as well as the inner product h , i above satisfy these
four properties.
Linearity in the first coordinate and skew-symmetry of (23) give hx, αyi = αhx, yi. If
h , i is applied to real vectors and real α, you won’t notice the complex conjugate, but it
is still useful to prove the following claim:
If linear operator W : R2 → R2 is Hermitian, i.e., hWx, yi = hx, Wyi for
all x, y ∈ R2 , then W has real eigenvalues with perpendicular eigenvectors.
To prove this, assume that x is an eigenvector with eigenvector κ, then
κhx, xi = hκx, xi = hWx, xi = hx, Wxi = hx, κxi = κhx, xi.
Hence κ = κ and κ is real. Now if κ1 and κ2 are different eigenvalues, with eigenvectors
x1 and x2 , then
κ1 hx1 , x2 i = hκ1 x1 , x2 i = hWx1 , x2 i = hx1 , Wx2 i = hx1 , κ2 x2 i = κ2 hx1 , x2 i.
Now κ2 = κ2 , because eigenvalue κ2 is real. But as κ1 6= κ2 , we have hx1 , x2 i = 0, so
x1 ⊥ x2 . Finally, if κ1 = κ2 = κ, i.e., W has adouble eigenvalue, then we take x1 to be
an eigenvalue, and take x2 ⊥ x1 . Then
hx1 , Wx2 i = hWx1 , x2 i = κhx1 , x2 i = 0,
47
Figure 13: Two paraboloids, left an elliptic one with positive curvature at the origin O,
right a hyperbolic paraboloid with negative curvature at O.
so Wx2 ⊥ x1 . Since the dimension of the space is only two, this means that Wx2 is a
scalar multiple of x2 , so x2 is an eigenvector. But κ is the only eigenvalue, so Wx2 = κx2
(in fact, Wx = κx for every x ∈ R2 ), and the claim is proved.
To finish the proof of the proposition, we show that the Weingarten matrix W is
Hermitian w.r.t. inner product h , iI :
hWx, yiI = hWx, FI yi
= hFI−1 FII x, FI yi
(FI−1 is symmetric)
= hFII x, FI−1 FI yi
(FII is symmetric)
= hx, FII yi = hx, FI (FI−1 FII y)i = hx, FI (Wy)i = hx, WyiI ,
and this finishes the proof.
The principal vector indicate the direction in which the surface curves “the most” and
“the least”. Here, “the least” may just mean that κ2 < 0 < κ1 , and in the direction of
w2 the surface curves away from itself in the direction of w1 .
Example: Let the paraboloid S be given by the equation z = ax2 +by 2 , see Figure 13.
This can be parametrised by σ : R2 → S, σ(u, v) = (u, v, au2 + bv 2 ). This gives




 
−2au
1
0
1
 −2bv  ,
σu =  0  , σv =  1  , N = √
2
2 + 4b2 v 2
1
+
4a
u
1
2au
2bv

and
σuu
This gives
FI =

 
 
0
0
0





= 0 , σuv = 0 , σvv = 0  .
2a
0
2b
1 + 4a2 u2
4abuv
4abuv
1 + 4b2 v 2
FII
and
1
=√
2
1 + 4a u2 + 4b2 v 2
If we evaluate this at the origin (u, v) = (0, 0), it simplifies to
2a 0
1 0
FII − κFI =
−κ
.
0 2b
0 1
48
2a 0
.
0 2b
Therefore the principal curvatures are κ1 = 2a with principal vector w1 = (1, 0, 0)T and
κ2 = 2b with principal vector w1 = (0, 1, 0)T . If γi , i = 1, 2 are two unit speed curves
through the origin O such that γ̇i = wi , then the normal curvature of γi at O is equal to
the principal curvature κi . Any other unit speed curve has normal curvature somewhere
between κ1 and κ2 . We call O

elliptic
if κ1 · κ2 > 0,






parabolic
if κ1 · κ2 = 0,


hyperbolic if κ1 · κ2 < 0,




flat
if κ1 = 0 and κ2 = 0,




umbilic
if κ1 = κ2 .
For example, if a = −1, b = 1, then the principal vectors have opposite curvature (a
hyperbolic paraboloid), and the point O is hyperbolic, and if we take the unit speed
curve γ through γ(0) = O with γ̇(0) = √12 (w1 + w2 ), then the normal curvature of γ is 0
(i.e., halfway between κ1 and κ2 ). For umbilic points, the curvature of a unit speed curve
γ is the same, no matter what direction it goes.
Lemma 59 Let γ̇ = σu u̇ + σv v̇ be the tangent vector of a curve. The normal curvature
of γ can be written as
u̇
κn = (u̇, v̇) FII
= Lu̇2 + 2M u̇v̇ + N v̇ 2 .
v̇
Proof. By direct calculation:
d
(u̇σu + v̇σv ), Ni
dt
= hüσu + u̇2 σuu + u̇v̇σuv + v̈σv + v̇ u̇σvu + v̇ 2 σvv , Ni
κn = hγ̈, Ni = h
= Lu̇2 + 2M u̇v̇ + N v̇ 2
Writing out the matrix product gives the same result.
The fact that the normal curvature of a curve through a point P is always a convex
combination of the principal curvatures at P is made precise in a theorem due to Euler.
Theorem 60 (Euler’s Theorem) Let γ be a curve within surface S through a point P ,
and let θ be the angle between γ̇ and the principal direction w1 . Then the normal curvature
of γ at P is
κn = κ1 cos2 θ + κ2 sin2 θ.
(24)
Proof. First, if κ1 = κ2 , then the normal curvature of every curve through P is equal to
κ1 = κ2 , and so (24) holds. Let us assume that κ1 6= κ2 , so the principal vectors w1 , w2
at P = γ(0) are perpendicular. Without loss of generality, we can assume that w1 and
w2 are unit vectors.
49
Write γ̇(0) = cos θ w1 + sin θ w2 . In local coordinates σu and σv of the tangent plane
at P , this becomes γ̇(0) = ξσu + ησv for wi = ξi σu + ηi σv , i = 1, 2. Therefore
ξ
ξ
ξ1
= cos θ
+ sin θ 2 .
η
η1
η2
By Lemma 59 the normal curvature is
ξ
κn = (ξ, η)FII
η
ξ
ξ1
+ sin θ 2
= [cos θ · (ξ1 , η1 ) + sin θ · (ξ2 , η2 )] FII cos θ
η2
η1
ξ1
ξ
2
2
= cos θ · (ξ1 , η1 )FII
+ sin θ · (ξ2 , η2 )FII 2
η1
η2
ξ
ξ
+ sin θ cos θ (ξ2 , η2 )FII 1 + (ξ1 , η1 )FII 2
.
η1
η2
Now using that wi satisfies (22), and then the the inner product from (23) this is equal
to
ξ1
ξ
2
2
κn = cos θ · (ξ1 , η1 )κ1 FI
+ sin θ · (ξ2 , η2 )κ2 FI 2
η1
η2
ξ
ξ
+ sin θ cos θ (ξ2 , η2 )κ1 FI 1 + (ξ1 , η1 )κ2 FI 2
η1
η2
= cos2 θκ1 hw1 , w1 i + sin2 θκ2 hw2 , w2 i + sin θ cos θ (κ1 hw2 , w1 i + κ2 hw1 , w2 i) .
But w1 ⊥ w2 , and kw1 k = kw2 k = 1 respectively, so the above simplifies to κn =
κ1 cos2 θ + κ2 sin2 θ, finishing the proof.
5.4
The Gaussian and mean curvature of a surface
Definition 61 If κ1 and κ2 are the principal curvatures of a surface S at a point p ∈ S,
then
1
κM (p) = (κ1 + κ2 ) and κG (p) = κ1 κ2
2
are called the mean curvature and Gaussian9 curvature of S at p.
9
Carl Friedrich Gauss, besides being the strongest mathematician in his time, was one of the founders
of non-Euclidean geometry.
50
As a recap:
κ =
κn =
κg =
κ1 , κ2 =
κM =
κG =

curvature of a (planar) curve


normal curvature
for a curve


geodesic curvature

principal curvatures 

mean curvature
for a surface


Gaussian curvature
Proposition 62 The Gaussian and mean curvature can be expressed in terms of FI and
FII as follows:
LN − M 2
det FII
= det(W) =
κG =
det FI
EG − F 2
and
LG − 2M F + N E
κM =
.
2(EG − F 2 )
Proof. A quadratic equation with roots κ1 , κ2 can be written as
0 = (κ − κ1 )(κ − κ2 ) = κ2 − (κ1 + κ2 )κ + κ1 κ2 .
Writing out the equation det(FII − κFI ) in full gives
(EG − F 2 )κ2 − (LG − 2M F + N E)κ + (LN − M 2 ) = 0.
Divide by (EG − F 2 ), and we find Gaussian curvature κG = κ1 κ2 =
F +N E
curvature κM = 12 (κ1 + κ2 ) = 12 LG−2M
as required.
EG−F 2
LN −M 2
EG−F 2
and mean
Example: Let α(t) = (f (t), 0, g(t)) be a curve in the x-z-plane parametrised by arclength, so f˙2 + ġ 2 ≡ 1. The surface of revolution that emerges by revolving this curve
around the z-axis can be parametrised by
σ(u, v) = f (u) cos v, f (u) sin v, g(u) .
This gives






−ġ(u) cos v
f˙(u) cos v
−f (u) sin v
σu =  f˙(u) sin v  , σv =  f (u) cos v  , N =  −ġ(u) sin v  ,
0
f˙(u)
ġ(u)
and
σuu






f¨(u) cos v
−f˙(u) sin v
−f (u) cos v
=  f¨(u) sin v  , σuv =  f˙(u) cos v  , σvv =  −f (u) sin v  .
0
g̈(u)
0
This gives
FI =
1
0
0 f 2 (u)
and
FII =
51
f˙(u)g̈(u) − ġ(u)f¨(u)
0
.
0
f (u)ġ(u)
Figure 14: The tractrix and its surface of revolution, the pseudo-sphere.
Using Proposition 62 we find
κG =
det FII
f ġ(f˙g̈ − ġ f¨)
ġ(f˙g̈ − ġ f¨)
=
=
.
det FI
f2
f
But f˙2 + ġ 2 ≡ 1, and by differentiating 2(f˙f¨+ ġg̈) = 0. Hence ġ(f˙g̈ − ġ f¨) = −f˙2 f¨− ġ 2 f¨ =
−f¨, and the Gaussian curvature of the surface of revolution is
f¨
κG = − .
f
(25)
Example: If we take the tractrix (say only for x ≥ 0) and revolve it around the x-axis, we
obtain a surface called the pseudo-sphere Ψ, see Figure 14. By (13), we can parametrise
the right half of the tractrix by arc-length as
√
√
γ(s) = e−s , s + 1 − e−2s + ln(1 − 1 − e−2s ) = f (s), g(s)
Therefore, the surface of revolution becomes σ(u, v) = f (u) cos v, f (u) sin v, g(u) , and
by (25) the curvature is
f¨
e−s
κG = − = − −s = −1.
f
e
Hence the pseudo-sphere has constant curvature −1. However, it has a boundary. In
the chapter on hyperbolic geometry, we will see that there are compact surfaces without
boundary, but with constant curvature −1. However, they cannot be embedded in Euclidean 3-space R3 . For R3 , David Hilbert10 proved that you cannot do better than the
pseudo-sphere: every surface S ⊂ R3 of constant curvature −1 is isometric to a patch of
the pseudo-sphere, and Ψ is the largest surface in R3 with constant curvature −1.
Definition 63 The Gauss map N : S → S2 of a surface S is the map that assigns to
every P ∈ S its unit normal vector N(P ) ∈ S2 .
Let P0 = σ(u0 , v0 ) be a point in S and Dr (P ) = σ(Br (u0 , v0 )) the image of a small
disk in local coordinates. We can find the Gaussian curvature by comparing the area of
Dr with the area that the normal vector P 7→ N(P ) sweeps out in the unit 2-sphere,
when P varies over Dr .
10
David Hilbert (1862 - 1943) was one of the leading mathematicians around 1900. His list of 23
problems for future mathematics, presented in a lecture in 1900 in Paris, is still immensely influential.
52
Proposition 64 The Gaussian curvature
Area(N(Dr ))
.
r→0
Area(Dr )
κG (P0 ) = lim
(26)
Examples: If S = S2 is the unit 2-sphere, then the Gauss map is the identity. Therefore
Area(N(Dr )) = Area(Dr ), and the quotient in (26) is obviously one, which is indeed
the Gaussian curvature of the unit sphere. If, on the other hand, S is a plane, then
N(P ) ≡ N0 is a constant map. Therefore Area(G(Dr )) = Area(N0 ) = 0, and naturally,
the Gaussian curvature of plane is zero.
Proof of Proposition 64. The normal vector N is a function of the local coordinates
(u, v), but it has constant length, so (similar to Lemma 43) the partial derivatives Nu ⊥ N
and Nv ⊥ N. So there must be coefficients a, b, c, d ∈ R such that
Nu = a · σu + b · σv
We claim that the matrix
a c
b d
and
Nv = c · σu + d · σv
= −FI−1 FII = −W.
To prove this, note that σu ⊥ N, so 0 =
d
hσu , Ni
du
(27)
= hσuu , Ni + hσu , Nu i, and hence
−L = −hσuu , Ni = hσu , Nu i.
Similarly −M = hσv , Nu i = hσu , Nv i and −N = hσv , Nv i. Using the coefficients a, b, c, d ∈
R from above, we find
−L = hσu , Nu i = hσu , a · σu + b · σv i = aE + bF,
−M = hσv , Nu i = hσv , a · σu + b · σv i = aF + bG,
−M = hσu , Nv i = hσu , c · σu + d · σv i = cE + dF,
−N = hσv , Nv i = hσv , c · σu + d · σv i = cF + dG.
In matrix form, this becomes
L M
E F
a c
−
=
·
.
M N
F G
b d
Multiplying both sides with FI−1 proves the claim (27).
We can now write the cross product
Nu × Nv = (a · σu + b · σv ) × (c · σu + d · σv )
= ad · σu × σv + bc · σv × σu
= (ad − bc) σu × σv
a c
= det
σ × σv
b d u
= det(−FI−1 FII ) σu × σv
=
by formula (27)
det(FII )
σu × σv
det(FI )
= κG (σ(u, v)) · σu × σv
53
by Proposition 62.
On a small region Dr , Gaussian curvature is almost constant. In fact, by Taylor’s Theorem,
κG (σ(u, v)) = κG (P0 ) + ψ(u, v) · r
for (u, v) ∈ Br (u0 , v0 ),
where ψ is some bounded function depending on the parametrisation σ and its first and
second order derivatives. Therefore
RR
kNu × Nv k dudv
Area(N(Dr ))
B (u .v )
= RR r 0 0
Area(Dr )
kσu × σv k dudv
B (u .v )
RR r 0 0
κ (σ(u, v)) · kσu × σv k dudv
Br (u0 .v0 ) G
RR
=
kσu × σv k dudv
Br (u0 .v0 )
RR
(κG (P0 ) + ψ(u, v)r) · kσu × σv k dudv
Br (u0 .v0 )
RR
=
kσu × σv k dudv
B (u .v )
RR r 0 0
ψ(u, v) · kσu × σv k dudv
Br (u0 .v0 )
RR
= κG (P0 ) + r
.
kσu × σv k dudv
Br (u0 .v0 )
As r → 0, this tends to κG (P0 ).
Without proof we give one further interpretation of Gaussian curvature.
Proposition 65 Let LP (r) be the length of the boundary curve ∂B(P ; r). This is the
loop of points that are at distance r from P . Then the Gaussian curvature at P is
κG (P ) = lim 6
r→0
2πr − LP (r)
.
2πr3
(28)
Example: On the plane, LP (r) = 2πr, because this is just the circumference of a round
disk of radius r. Inserting this in (28), we get κG (P ) ≡ 0. At an elliptic point P of a
surface, the circumference of the round disk is a little bit less, and this produces positive
Gaussian curvature. Indeed, if we take the sphere centered at O with radius R and the
north pole P , then the “diameter through the sphere” of a circle C of “radius over the
surface” is 2πR sin ϕ, where ϕ is the angle made by OP and an arc connecting O with
any point of C. Therefore
r
Pr
ϕ= R
'$
R
LP (r) = 2πR sin ϕ
O
&%
κG (P ) = lim 6
r→0
= 2πR( Rr − 16 ( Rr )3 + h.o.t.)
2πr − 2πR( Rr − 16 ( Rr )3 + h.o.t.)
2πr − LP (r)
1
=
lim
6
= 2.
3
3
r→0
2πr
2πr
R
54
5.5
The Gauss-Bonnet Theorem
Definition 66 Let γ : [0, 1] → M and δ : [0, 1] → M be two smooth curves such that
p = γ(1) = δ(0); then γ̇(1) and δ̇(0) span a (two-dimensional) plane H in Tp M . The
external angle θ between them is the angle (in clock-wise direction within H) by which
you have to turn γ̇(1) to obtain δ̇(0). The internal angle α = π − θ.
(If γ̇(1) and δ̇(0) have the same direction, then θ = 0; if γ̇(1) and δ̇(0) have opposite
direction, then θ = ±π according to the convention in Figure ??.)
The Gauss-Bonnet Theorem unites curvatures of lines, surfaces and the Euler characteristic of a surface. In its most general form, it reads as follows.
Theorem 67 (Gauss-Bonnet) Let U be a surface, such that ∂U = ∪N
i=1 γi consists of
N curves parametrised by arc-length, and the external angle between γi and γi+1 is θi
(and θN the external angle between γN and γ1 ). Also, for any v ∈ U , let pan(v) be the
‘panorama” of v, that is, the total number of radians of a small circular circle around v.
If pan(v) 6= 2π, then we call v a singularity. Then
ZZ
κG dΩ +
U
N Z
X
i=1
κg (s) ds +
γi
N
X
i=1
X
θi +
singularities
(2π − pan(v)) = 2πχ(U ),
(29)
v
where κG is Gaussian curvature, dΩ is an infinitesimal area element, κg is geodesic curvature, ds an infinitesimal length element, and χ(U ) the Euler characteristic of U .
Examples: 1: The torus T can be considered as a flat square [0, 1]2 (so Gaussian curvature κG = 0) with opposite sides identified. If U = T, then ∂U = ∅, so there is no
RR
contribution from γi s or θi s. Therefore (37) reduces to 0 = T 0 = 2πχ(T), and χ(T) = 0
is indeed the Euler characteristic of the torus.
If, on the other hand, we embed the torus in R3 , then the curvature is not constant
RR
zero. Yet T κG dΩ = 0, so the torus, embedded this way, has points of positive curvature
(elliptic points) as well as points of negative curvature (hyperbolic points).
2: The “pretzel” P (two-holed torus) can be seen as a flat octagon with opposite sides
identified. The same reasoning as in the previous example would suggest that χ(P ) = 0
as well, but this is not true. The problem is that under the identification of opposite
sides, the eight vertices of the octagon are one and the same point x ∈ P , but the total
panorama of x is not 2π, but 6π radians. Check this by drawing a little “circle” around
x. So x doesn’t have a flat neighbourhood; we call it a singularity.
3: A geodesic n-gon is a region U that is homeomorphic with a two-dimensional disk,
and for which ∂U consists of n geodesic arcs. Therefore, U has n vertices, with internal
angles αi = π − θi . Since χ(U ) = 1, Gauss-Bonnet gives:
ZZ
κG dΩ + (n − 2)π =
U
n
X
i=1
55
αi .
Figure 15: A spherical, hyperbolic and flat triangle, with angle sum > π, < π and = π
respectively.
For a flat triangle (κG = 0, n = 3), we retrieve the standard fact that the angles of a
triangle add to π radians. If U is a triangle on the unit sphere, κG ≡ 1, and we find that
the sum of the angles minus π is the area of the triangle, so in particular, the sum of the
angles of a triangle is more than π. Similarly, on the pseudo-sphere, the area of a geodesic
triangle is π minus the sum of the angles, and the sum of the angles of a triangle is less
than π, see Figure 15.
5.6
Isometries and Theorema Egregium
An isometry is a map that preserves distances.
Theorem 68 A smooth map f : S1 → S2 is an isometry if and only if their fundamental
forms coincide: E1 (x) = E2 ◦ f (x), F1 (x) = F2 ◦ f (x) and G1 (x) = G2 ◦ f (x) for every
x ∈ S1 .
Corollary 69 Isometries preserve geodesics. In other words, if γ1 : (a, b) → S1 is a
geodesic and f : S1 → S2 is an isometry, then γ2 := f ◦ γ1 : (a, b) → S2 is also a geodesic.
Proof. An isometry preserves the first fundamental form, whilst γ1 satisfies the geodesic
equations (17). That means that also the curve γ2 satisfies the geodesic equations; after
all, (once transformed to the coordinates of S2 ) the coefficients of the geodesic equations
are the same. Hence γ2 is a geodesic.
A map is called conformal if it preserves angles.
Theorem 70 A smooth map f : S1 → S2 is conformal if and only if their fundamental
forms are proportional, i.e., there is λ 6= 0 such that E1 du21 + 2F1 du1 dv1 + G1 dv12 =
λ(E2 du22 + 2F2 du2 dv2 + G2 dv22 ).
It follows immediately that an isometry is conformal.
Theorem 71 (Theorema Egregium of Gauss) Gaussian curvature is preserved under isometries.
56
5.7
Exercises of Section 5
Exercise 48 You can circumscribe a sphere S (centered at the origin and with radius R)
with a cylinder C (radius R, height 2R, centered around the vertical axis, without bottom
and top). The map f : S \ {north pole, south pole} → C can be defined by horizontal
projections outwards from the vertical axis. Show that f preserves area.
Exercise 49 Show that every normal section is a geodesic.
Exercise 50 Assume that S is a surface of revolution, obtained by revolving the curve
y = f (x) around the x-axis. Show that if x0 is such that f 0 (x0 ) = 0, then S ∩ {x = x0 } is
a geodesic.
Exercise 51 If a surface S is the graph of a smooth function f = f (x, y), show that at
stationary points, the Weingarten matrix coincides with the Hessian
!
2
2
H=
∂ f
∂x2
∂2f
∂x∂y
∂ f
∂x∂y
∂2f
∂y 2
.
Exercise 52 Let C be a lower half-cylinder of radius R and central axis (t, 0, R) for t ∈ R.
You can roll this cylinder to a flat rectangle R in the horizontal plane R × [−πR/2, πR/2].
Show that the corresponding map f : C → R is an isometry.
Exercise 53 a) Give a parametrisation of the torus embedded in R3 .
b) Compute the first and second fundamental form; hence compute Gaussian curvature.
c) Which points on the torus are ellptic, hyperbolic and parabolic?
d) Compute the Weingarten matrix and principal curvature. Are there any umbilic points
on the torus?
5.8
Solutions to Exercises
Exercise 48: We have parametrisations, derivatives and first fundamental forms




 
cos u
− sin u
0





cos u
σ(u, v) = sin u
σu =
σv = 0  ,
v
0
1
so E = 1, F = 0 and G = 1 for the cylinder, and

 √

√
2 cos u
2 sin u
1
−
v
−
1
−
v
√
√
τu =  1 − v 2 cos u 
τ (u, v) =  1 − v 2 sin u 
v
0

τv =
√ −v
1−v 2
 √ −v 2
1−v

cos u
sin u  ,
1
so E = 1 − v 2 , F = 0 and G = 1/(1 − v 2 ) for the sphere. In both cases EG − F 2 = 1, so
the parallel projection (which preserves vertical coordinate z = v and polar angle φ = u)
is area preserving.
57
Exercise 53: For radii R > r > 0, we have parametrisation, derivatives, normal
vectors and fundamental forms






(R + r cos u) cos v
−r sin u cos v
−(R + r cos u) sin v
σ(u, v) =  (R + r cos u) sin v  σu =  −r sin u sin v  σv =  (R + r cos u) cos v  ,
r sin u
r cos u
0
so E = r2 , F = 0 and G = (R + r cos u)2 for the torus, and


−(R + r cos u)r cos u cos v
σu × σv =  −(R + r cos u)r cos u sin v  , so normalised
−(R + r cos u)r sin u
Second derivatives:


−r cos u cos v
σuu =  −r cos u sin v 
−r sin u

σuv

r cos u sin v
= −r sin u cos v 
0
σvv


− cos u cos v
N =  − cos u sin v 
− sin u


−(R + r cos u) cos v
=  −(R + r cos u) sin v  ,
0
so L = r, M = 0 and (R + r cos u) cos u for the second fundamental form. Therefore the
Gaussian curvature is
cos u
det(FII )
=
.
κG =
det (FI )
r(R + r cos u)
This is positive (elliptic points) for the outer half of the torus u ∈ (− π2 , π2 ), negative
), and zero (parabolic points) for
(hyperbolic points) for inner half of the torus u ∈ ( π2 , 3π
2
π π
upper and lower circle of the torus u ∈ {− 2 , 2 }. The Weingarten matrix is
1
0
−1
r
W = FI · FII =
0 (R+rcoscosu u)3
with eigenvalues (principal curvatures) κ1 = 1r and κ2 = (R+rcoscosu u)3 . Equating these gives
r cos u = (R + r cos u)3 , which has no solution since R > r. Therefore there are no umbilic
points.
6
6.1
Hyperbolic Geometry
The hyperbolic upper half plane H
There are several different hyperbolic surfaces (i.e., surfaces of negative curvature), but
1
many of them are derived from the upper half plane H with hyperbolic metric gH (z) = Im(z)
(this gH is closely related to the first fundamental form), so we start with this one:
(
H = {z ∈ C : Im(z) > 0},
1
gH (z) = Im(z)
.
This means that the inner product on the tangent space Tz H at z is
1
(u1 v1 + u2 v2 )
hu, viz =
Im(z)2
kukz =
and
58
p
1
hu, uiz =
Im(z)
q
u21 + u22 .
The cosine of the angle θ between to vectors u, v ∈ Tz is given by the usual formula
cos(θ) =
hu, viz
u 1 v 1 + u2 v 2
=p 2
kukz kvkz
(u1 + u22 ) · (v12 + v22 )
(30)
because the factors 1/Im(z) cancel from the equation. Hence angles in H are the same as
the usual Euclidean angles.
A path γ : [0, 1] → H has length
Z 1
1
lH (γ) =
|γ 0 (t)| dt,
Im(γ(t))
0
where instead of just taking |γ 0 (t)| as in Euclidean geometry, we multiply the integrand
1
with gH (γ(t)) = Im(γ(t))
. For example, the points w = ai and z = bi (say with 0 < a < b)
can be connected with a straight arc γ : t 7→ ai + (b − a)it, and its length is (using the
change of coordinates s = a + (b − a)t)
Z 1
Z b
1
1
lH (γ) =
(b − a) dt =
ds = log b − log a.
0 a + (b − a)t
a s
Definition 72 A path connecting w and z is a geodesic if it is a path of locally minimal
length.
Remark: Here “locally” means that we cannot make the length of the path smaller
by a small change, and this is important for various surfaces, such as a torus. Indeed, a
path connecting w to w that wraps around a torus horizontally once is a geodesic, because
you cannot make it shorter by a small change. The shortest path connecting w to w is
just the path γ(t) ≡ w, and it has length 0. On H, however, geodesics are entirely unique.
6.2
Möbius transformations
The question we are going to answer in this section is: What is the geodesic connecting
general w and z in H (so not necessarily vertically above one another). For this, we need
to study the class of Möbius transformations Möb.
ψ ∈ Möb if there are a, b, c, d ∈ C, ad − bc 6= 0 such that ψ(z) =
az + b
.
cz + d
We want to extend ψ to the Riemann sphere Ĉ = C ∪ {∞} by setting
d
a
ψ −
ψ(∞) =
= ∞.
c
c
We can always scale ψ by a number such that ad − bc = 1. Indeed, if κ = (ad − bc)−1/2 ,
then
aκz + bκx
az + b
=
and (aκ)(dκ) − (bκ)(cκ) = 1.
ψ(z) =
cz + d
cκz + dκx
So let us from now on always assume that ad − bc = 1.
59
Every Möbius transformation is invertible. Indeed,
ψ −1 (w) =
dw − b
.
−cw + a
(31)
From Exercise 57 it follows that Möb is a group under composition. It is generated
by the following types of transformations:
Translations
Dilatations
Inversions
Tb : z 7→ z + b
Dk : z 7→ kz
Inv : z 7→ − z1
a=√
d = 1, c = 0√
a = k, d = 1/ k, b = c = 0
a = d = 0, b = c = 1
For the dilatations, we assume that k 6= 0. There is no requirement that k > 1, or even
k ∈ R. For example, if k = eiφ , then Dk is a rotation over angle φ.
Lemma 73 Every Möbius transformation ψ can be written as a composition Tα ◦ Dβ ◦
Inv ◦ Tγ .
Proof. If c = 0, then ψ(z) is a simple linear transformation, so the lemma is immediate.
If c 6= 0, we write
a(z + dc ) + b −
az + b
=
cz + d
c(z + dc
ad
c
a b − ad
1
c
=
+
c
c z + dc
a ad − bc −1
=
+
c
c2 z + dc
= Tα ◦ Dβ ◦ Inv ◦ Tγ
for α = ac , β =
ad−bc
c2
and γ = + dc .
A circle with center z0 and radius ρ has the equation |z − z0 |2 = ρ2 . We can rewrite
this to
zz − z0 z − z0 z + z0 z0 − ρ2 = 0.
So if we write β = −z0 and γ = z0 z0 − ρ2 ∈ R, this becomes
zz + βz + βz + γ = 0.
A straight line of the form y = ax + c can be written in coordinates x =
as
−i − a
i−a
z
+z
− c = 0.
2
2
Therefore, if β = 12 (−a − i) and γ = −c ∈ R, then we find
βz + βz + γ = 0.
(32)
z+z
2
and y =
z−z
2i
(33)
Therefore, lines and circles have very much the same equation, it is just a matter if the
coefficient of zz is 0 or 1.
60
Proposition 74 Every Möbius transformation ψ transforms circles and straight lines
into circles and straight lines. (More precisely, ψ can map a circle into a straight line or
vice versa, but the set of all circles and straight lines is mapped by ψ into the set of all
circles and straight lines.)
Proof. Clearly, translations preserve straight lines and circles. For the inversion, we
argue as follows. Let
αzz + βz + βz + γ = 0.
(34)
be the equation of a circle (if α = 1) or line (if α = 0). Note also that this circle or line
passes through the origin if and only if γ = 0.
is its own inverse: z = ψ −1 (w) = −1
by (31). Substituting
The inversion ψ(z) = −1
z
w
into (34) gives
γww − βw − βw + α = 0
If γ = 0, then we have a line. If γ 6= 0, then we can divide the equation by γ, and obtain
a circle. Hence the inversion turns lines and circles into lines and circles.
If we do the same for the dilatation Dk (z) = kz with inverse D1/k (w) = wk , and
multiply with kk afterwards, we get
αww + (βk) w + (βk) w + γkk = 0,
so this is a cricle or line again.
The proposition follows now from Lemma 73.
6.3
Möbius transformations on H
We say that ψ ∈ Möb(H) if there are a, b, c, d ∈ R, ad − bc > 0, such that ψ(z) =
Since a, b, c, d are real, ψ(R ∪ {∞}) = R ∪ {∞}, and since ad − bc > 0,
ψ(i) =
az+b
.
cz+d
ai + b
(ai + b) · (ci + d)
(ac + bd) + (ad − bc)i
=
=
ci + d
|ci + d|2
(ci + d) · (ci + d)
has positive imaginary part. Therefore ψ maps H homeomorphically into itself, and this
explains the name Möb(H). As before, we can scale ψ such that ad − bc = 1. Similarly
to Lemma 73, every ψ ∈ Möb(H) can be written as Tα ◦ Dβ ◦ Inv ◦ Tγ for
Translations
Dilatations
Inversions
Tb : z 7→ z + b
Dk : z 7→ kz
Inv : z 7→ − z1
b∈R
k>0
a=√
d = 1, c = 0√
a = k, d = 1/ k, b = c = 0
a = d = 0, b = c = 1
Here we restricted the translations and dilatations to those that preserve H and the circle
at infinity R ∪ {∞}.
Theorem 75 Every ψ ∈ Möb(H) preserves hyperbolic length and angles.
61
Remark: Angle preserving maps are called conformal. The proof below shows that
every complex differentiable map f is conformal wherever f 0 (z) 6= 0.
Proof. For ψ(z) =
az+b
,
cz+d
ψ 0 (z) =
it is easy to compute
ad − bc
(cz + d)2
and
Im(ψ(z)) =
ad − bc
· Im(z).
|cz + d|2
Therefore, if γ : [0, 1] → H is a path, and we write z = γ(t), then the length of image
path is (recall that ad − bc = 1 ∈ R by convention.)
Z 1
1
lH (ψ ◦ γ) =
|(ψ ◦ γ)0 (t)| dt
Im(ψ(z))
0
Z 1
ad − bc
|cz + d|2
0
dt
·
γ
(t)
=
2
0 (ad − bc)Im(z) (cz + d)
Z 1
1
=
|γ 0 (t)| dt = lH (γ).
Im(z)
0
Now for the angles, the argument works for every complex analytic function. Indeed, if
f (z) = f (x + iy) = u(x, y) + iv(x, y) is complex differentiable, then the Cauchy-Riemann
equations hold:
∂u
∂v
∂u
∂v
=
and
=− .
(35)
∂x
∂y
∂y
∂x
The derivative Df at a point z acts as a linear transformation on the tangent space Tz ,
so it is given by the matrix
ux uy
Df (z) =
,
vx vy
where ux =
∂u
,
∂x
etc. denote the partial derivatives. The inner product behaves under f as
hDf (ξ), Df (η)if (z) = g 2 (z)hξ, (Df )T Df (η)if (z)
for ξ, η ∈ Tz and Riemann metric g. Using the Cauchy-Riemann equations (35), we can
compute that
ux v x
ux uy
T
(Df ) Df =
u y vy
vx vy
2
2
ux + v x
ux uy + v x v y
1 0
2
2
= (ux + uy )
=
uy ux + v y v x
u2y + vy2
0 1
is a scalar multiple of the identity. This scalar multiple as well as g(z) cancel from the
quotient when computing the angle between Df (ξ) and Df (η), see (30):
hDf (ξ), Df (η)if (z)
hξ, ηiz
=
kDf (ξ)kf (z) kDf (η)kf (z)
kξkz kηkz
Therefore, f preserves angles. Since every ψ ∈ Möb(H) is analytic, and ψ 0 (z) 6= 0 for all
z ∈ H, this applies to ψ.
62
If we consider the equations (32) and (33) for β, γ ∈ R, then we have either
p a line
Re(z) = −γ/(2β) perpendicular to R, or a circle with center β ∈ R and radius β 2 − γ ∈
R. (Here we need β 2 > γ!) In either case, wherever the curve intersects R, the intersection
is perpendicular.
Theorem 76 A geodesic in H is contained in either a vertical line, or a semi-circle
perpendicular to R.
Proof. Suppose that γ is the geodesic between w and z ∈ H. Find ψ ∈ Möb(H) such
that ψ(w) = ai and ψ(z) = bi for some 0 < a < b. From Exercise 54 we know that the
geodesic between ai and bi is the straight arc [ai, bi] between them. Since ψ preserves
hyperbolic length, ψ(γ) = [ai, bi]. Extend this arc to [0, bi], and apply ψ −1 to it. Since
ψ −1 maps H to H, and maps the right angle between [0, bi] and R to a right angle between
ψ −1 ([0, bi]) and R, γ must be contained in a circle or line perpendicular to R.
6.4
Curvature of H
Let LP (r) be the length of the boundary curve ∂BH (p; r). This is the loop of points that
are at hyperbolic distance r from P .
Theorem 77 The Gaussian curvature at the point P is given by the formula
2πr − LP (r)
.
(36)
r→0
2πr3
We want to use formula (36) to compute the curvature of the hyperbolic plane H.
Since there is an isometry between every two points in H, the curvature will be the same
at every point. So let us fix p = i. The next thing to find is the loop of points at
hyperbolic distance r away from i. By Exercise 54, e−r i and er i are two points on this
loop, but to guess the entire loop is a bit trickier. However, since Möbius transformations
are isometries, we can look at the Möbius transformation ψt that fixes i and moves the
point 0 to t ∈ R. The geodesic arc (0, i] will be moved by ψt to a geodesic through t and
i, and ψt (e−r i) is a point on this geodesic that is still at hyperbolic distance r away from
i. We find (leaving the convention that ad − bc = 1)
κG (P ) = lim 6
ψt (z) =
z+t
−tz + 1
ψt (e−r i) =
and
e−r i + t
.
−te−r i + 1
e−r i+t
is a parametrisation of
−te−r i+1
e−r (1+t2 )
1−e−2r
and γ 0 (t) = (1−te
−r )2 . Therefore
|1−te−r |2
In other words, γ : R ∪ {∞} → H given by γ(t) =
loop ∂BH (i, r). We can compute Im(γ(t)) =
Z ∞
1
Lp (r) =
|γ 0 (t)| dt
Im(γ(t))
−∞
Z ∞
1 − e−2r
=
dt
−r
2
−∞ e (1 + t )
Z ∞
1
r
−r
= (e − e )
dt = 2π sinh(r).
2
−∞ 1 + t
63
the
Recall the McLaurin series sinh(r) = r + 16 r3 + · · · +
1
r2n−1
(2n−1)!
+ . . . . This gives
2π(r − (r + 16 r3 + h.o.t.))
2πr − Lp (r)
= lim 6
= −1.
κG (i) = lim 6
r→0
r→0
2πr3
2πr3
Thus we have proved that:
Corollary 78 The Gaussian curvature is κG (p) = −1 at every point p ∈ H.
6.5
Geodesic triangles and polygons
Definition 79 A region P bounded by n subarcs of geodesics is called a geodesic n-gon.
Figure 16: Two geodesic triangles and a triangular tiling of H.
The hyperbolic area AreaH (P ) of a region P ⊂ H is computed using the formula
ZZ
ZZ
1
2
AreaH (P ) =
κG dσ =
dz 2 .
2
P
P (Im(z))
Theorem 80 (Gauss-Bonnet) If a curvilinear polygon P is made of n curves si , making exterior angles θi , then
ZZ
n Z
n
X
X
kg (s)ds +
κG dσ +
θi = 2π
(37)
i=1
si
P
i=1
where kg denotes the geodesic curvature and κG the Gaussian curvature.
Corollary 81 A geodesic n-gon in H satisfies
−AreaH (P ) +
n
X
θi = 2π.
(38)
i=1
Proof. A geodesic polygon has geodesic boundary curves by definition, and their geodesic
curvature is zero. The Gaussian curvature of H is constant −1. So (38) follows directly
from (37)
In particular, a geodesic triangle P in the hyperbolic plane has sum of interior angles
φi = π − θi equal to
3
X
0≤
φi = π − AreaH (P ) ≤ π.
i=1
64
The internal angle sum tends to the “Euclidean” π as AreaH (P ) → 0. The smallest
internal angle sum is obtained when all interior angles are zero; this happens “in the
limit”, when the vertices of P lie on the circle at infinity. In this case, AreaH (P ) = π,
and this is the largest area that a geodesic triangle in H can have, cf. Exercise 60. At the
same time, such triangles have infinite diameter!
6.6
The Poincaré disk D
The Möbius transformation χ : z 7→
−z+i
iz−1
maps H to the open unit disk D. In particular,

0



1
χ:
∞



i
7→
7
→
7
→
7→
−i
1
i
0
2
Equipped with the metric form gD (z) = 1−|z|
2 , we call D the Poincaré disk. Let Möb(D)
be the class of Möbius transformations that map D homeomorphically to itself.
Lemma 82 ψ ∈ Möb(D) if ψ(z) =
αz+β
αz+β
where α, β ∈ C satisfy |α|2 = |β|2 .
Proof. Möbius transformation ψ preserves D precise then when it can be written as
ψ = χ ◦ φ ◦ χ−1 for some χ ∈ Möb(H). Using the matrix representation as in (39), we find
α β
a b
1 −i
a b
−1 i
−1
Mψ = Mχ ·
· Mχ =
=
c d
i −1
c d
−i 1
α β
for α = (d−a)+(c−b)i and β = (b+c)+(a−d)i, and therefore |α|2 = (d−a)2 +(c−b)2 =
|β|2 .
Proposition 83 The Möbius transformation χ : H → D above is an isometry from
(H, gH ) to (D, gD ).
Proof. Let γ : [0, 1] → D be a path in D, so χ−1 ◦ γ is a path in H. Direct computation
1−|z|2
−2
−1
(z)) = |−iz+1|
gives (χ−1 )0 (z) = (−iz+1)
2 and Im(χ
2 . Therefore
lH (χ
−1
Z
1
◦ γ) =
0
Z
1
=
0
Z
=
1
−1 0
1
(χ ) (γ(t)) · γ 0 (t) dt
Im(χ−1 (γ(t)))
2
|γ 0 (t)| dt
1 − |γ(t)|2
gH (z) |γ 0 (t)| dt = lD (γ)
0
as required.
65
Figure 17: The tractrix and its surface of revolution, the pseudo-sphere. On the right a
pentagonal tiling of the Poincaré disk.
Corollary 84 The Poincaré disk has constant curvature −1.
Proof. Curvature is preserved by isometries (according to Gauss’ Theorema Egregium),
in particular by the isometry χ.
Corollary 85 Geodesics on the Poincaré disk are diameters (i.e., straight lines through
0) or circular arcs perpendicular to ∂D.
Proof. Since χ : H → D is an isometry, χ maps geodesics γ in H (i.e., lines and semicircles perpendicular to R ∪ {∞}) to geodesics in D. By the proof of Theorem 75, χ is
conformal, so χ(γ) is a straight or circular arc perpendicular to ∂D.
6.7
Fuchsian groups and modular surfaces
Let Γ ⊂ Möb(H) be a subgroup of Möbius transformation. That is, Γ contains the
identity, is closed under composition, and contains ψ −1 whenever ψ ∈ Γ. If z ∈ H, then
the collection
orbΓ (z) = {ψ(z) : ψ ∈ Γ}
is called the orbit of z. For example, if Γ is the collection of translations Tb : z 7→ z + b,
for b ∈ R, then orb(z) is a horizontal line at height Im(z). If Γ is the discrete collection
of translations Tb : z 7→ z + b, for b ∈ Z, then orbΓ (z) = {z + n : n ∈ Z}.
Remark: Recall that a set S is discrete if every x ∈ S in it is an isolated point of S.
This is not the same as totally disconnected. Every discrete set is totally disconnected,
but the Cantor set is totally disconnected, but contains no isolated points.
Definition 86 A group is discrete is orb(z) is a discrete set for each z. Any discrete
subgroup of Möb(H) is called a Fuchsian11 group.
11
Lazarus Fuchs (1833 - 1902) was German professor working mostly on analysis and differential equations.
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6.8
Exercises of Section 6
Exercise 54 Argue that the straight arc between ai and bi is a geodesic. Conclude that
the real axis is infinitely removed from any point in H. For this reason, R ∪ {∞} is called
the circle at infinity for H ∪ {∞}.
Exercise 55 Let us say that {z ∈ C : |z| > R} ∪ {∞} are neighbourhoods of ∞ for any
R ∈ R. Show that Ĉ is homeomorpic to the standard 2-sphere S2 .
Similarly, R̂ = R ∪ {∞} is homeomorpic to the circle S1 , if we say that {x ∈ R : |x| >
R} ∪ {∞} are neighbourhoods of ∞.
Exercise 56 If ad − bc = 0, show that ψ(z) =
a
c
for all z ∈ Ĉ.
Exercise 57 To each Möbius transformation ψ(z) =
Mψ =
az+b
,
cz+d
we can assign a matrix
a b
.
c d
Show that inverse and composition of Möbius transformations work in the same way as
matrix multiplication and inverse, i.e.,
Mψ · Mχ = Mψ◦χ
and
Mψ−1 = Mψ−1 .
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Exercise 58 For every two triples (z1 , z2 , z3 ) and (w1 , w2 , w3 ) (where all zi are distinct,
and all wi are distinct), show that there is a Möbius transformation ψ such that ψ(zi ) = wi
for i = 1, 2, 3. This ψ is unique up to a factor ±1 in the four parameters a, b, c, d.
Exercise 59 For every two pairs z1 6= z2 ∈ H and w1 6= w2 ∈ H, show that there is a
unique ψ ∈ Möb(H) such that ψ(zi ) = wi for i = 1, 2.
Exercise 60 Let P ⊂ H be the geodesic triangle {z ∈ H : |z| > 1, −1 < Re(z) < 1}, see
Figure 16, far left. Show that the hyperbolic area AreaH (P ) = π.
1+x
), so ∂D is infinitely far
Exercise 61 Let x ∈ (0, 1). Compute that dD (0, x) = log( 1−x
from every z ∈ D. Also show that the arc [0, x] is the unique geodesic connecting 0 and x.
Exercise 62 Let Γ = hInv, T1 i be the group generated by the inversion z 7→ −1/z and
translation z 7→ z + 1. Describe orbΓ (i). Is Γ discrete?
6.9
Solutions to Exercises
67
References
[1] J. Anderson, Hyperbolic geometry, London, Springer, (1999).
[2] V. I. Arnol’d, Mathematical methods of classical mechanics, Springer-Verlag,
New York, Heidelberg, Berlin (1984).
[3] M. do Carmo, Differential geometry of curves and surfaces, Prentice Hall, New
Jersey, 1976.
[4] M. Crossley, Essental topology, Springer-Verlag, London 2005.
[5] J. Hocking, G. Young, Topology, Addison Wesley, Reading Massachussetts,
1961.
[6] James R. Munkres, Topology, Prentice Hall, Upper Saddle River, New Jersey,
2000.
[7] A. Pressley, Elementary differential geometry, Springer-Verlag, London 2007.
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