DIFFERENTIALGEOMTRIE Aufgabenblatt 1 › ~pako › EDG2019 › WS2019-EDG-Solutions.pdf ·...

Panagiotis Konstantis Universitat zu KolnAlexander Caviedes Castro Wintersemester 2019

ELEMENTARE DIFFERENTIALGEOMTRIE

Aufgabenblatt 1Abgabe: Diese Ubungen mussen bis spatestens 11:55 Uhr Montag 21.10.2019 in den Briefkasten im

studentischen Arbeitsraum des MI (3 Stock) abgegeben werden. Schreiben Sie Ihren Namen, IhreMatrikelnummer und Ihre Ubungsgruppe auf Ihre Abgabe und tackern Sie alles zusammen.

Hausaufgaben (50 Punkte)

A1.1 Wir bezeichnen mit 〈·, ·〉 das standard euklidische Skalarprodukt auf Rn . Seien α, β : I → Rn

zwei parametrisierten Kurven. Beweisen Sie, dass

〈α(t), β(t)〉· = 〈α(t), β(t)〉+ 〈α(t), β(t)〉

gilt. (10)

Proof. Let us write

α(t) = (α1(t), . . . , αn(t))

β(t) = (β1(t), . . . , βn(t)) .

Then,

〈α(t), β(t)〉· =n∑

i=1

(αi(t)βi(t))· =

n∑i=1

(αi(t)βi(t) + αi(t)βi(t)

)= 〈α(t), β(t)〉+ 〈α(t), β(t)〉

and we are done.

A1.2 Die folgende regulare parametrisierte Kurve wird als Traktrix oder auch Schleppkurve bezeichnet:

c : (0, π/2)→ R2,

t 7→(

sin tcos t+ ln tan t/2

).

Zeigen Sie, dass die Schleppkurve die folgende Eigenschaft hat, die auch ihren Namen erklart: Furjeden Kurvenpunkt hat die Strecke auf der Tangente vom Kurvenpunkt zur y-Achse die Lange 1. (10)

Proof. The derivative of c at t0 ∈ (0, π/2) is given by

c(t0) =

(cos(t0)

cos(t0) cot(t0)

).

The equation of the line tangent to the curve at the point c(t0) is

cot(t0)(x− sin(t0))− (y − (cos(t0) + ln(tan(t0/2)))) = 0 .

The line intersects the y-axis at the point(0

ln(tan(t0/2))

),

whose distance to c(t0) equals one.

A1.3 Wir definieren eine Relation auf der Menge der parametrisierten Kurven im Rn, fur die genaudann c : I → Rn in Relation zu c′ : J → Rn steht, wenn c eine Umparametrisierung von c′ ist.Zeigen Sie, dass dies eine Aquivalenzrelation ist. (10)

Proof. We recall the definition of reparametrization: a curve c : I → Rn is a reparametrization ofc′ : J → Rn if there exists a smooth bijective map ϕ : J → I with smooth inverse such that

c′ = c ◦ ϕ : J → Rn .

The relation is of equivalence as it is reflexive (consider the identity map), symmetric (considerthe inverse of ϕ) and transitive (consider the composition of the corresponding bijective mapsdefining the reparametrization).

A1.4 Sei c : I → Rn eine parametrisierte Kurve, und sei F ∈ E(n) eine euklidische Bewegung.Zeigen Sie: Ist c nach (bzw. proportional zur) Bogenlange parametrisiert, so ist auch F ◦ c nach(bzw. proportional zur) Bogenlange parametrisiert. (10)

Proof. Let A ∈ O(n) and b ∈ Rn such that F (x) = Ax+ b . The chain rule implies that

(F ◦ c)(t)· = A ◦ c(t) .

As A is a linear isometry and |c(t)| equals to a constant, |(F ◦ c)(t)·| equals to the same constantand we are done.

A1.5 Sei c eine ebene nach Bogenlange parametrisierte Kurve. Sei F ∈ E(2) eine orientierungserhal-tende euklidische Bewegung, F (x) = Ax+ b mit A ∈ SO(2) und b ∈ R2. Zeigen Sie, dass F ◦ cdieselbe Krummung hat wie c .

Was geschieht bei einer orienterungsumkehrenden euklidischen Bewegung F ?

Proof. By assumption, for every t|c(t)| = 1 .

Let κ(t) be the curvature of the curve so

c(t) = κ(t) · n(t),

where n(t) is the vector normal to the curve, which by definition it is equal to(0 −11 0

)· c(t) .

By the previous exercise, we know that F ◦ c is parameterized by arc length. It is not hard to seethat

(F ◦ c)··(t) = A ◦ c(t) .

As the matrix A belongs to SO(2) it commutes with the matrix(0 −11 0

)and

A ◦ c(t) = A ◦ κ(t)n(t) = κ(t)A

(0 −11 0

)c(t)

= κ(t)

(0 −11 0

)(F ◦ c)·(t) ,

2

and we are done. If the determinant of A is minus one instead of one, the curvature changes its

sign, as in that case the matrix A anti commutes with(0 −11 0

).

(10)

3


ELEMENTARE DIFFERENTIALGEOMETRIE



Hausaufgaben

A2.1 Die logarithmische Spirale ist gegeben durch

c : R→ R2

t 7→(et/10 · cos tet/10 · sin t

).

Finden Sie eine Umparametrisierung c′ : R → R2 dieser Kurve, sodass c′ nach Bogenlangeparametrisiert ist.

Proof. We compute first |c(t)| :

|c(t)| =√( 1

10et/10 cos t− et/10 sin t

)2+( 1

10et/10 sin t+ et/10 cos t

)2=

1

10et/10

√(cos t− 10 sin t

)2+(sin t+ 10 cos t

)2=

√101

10et/10

Let us consider the function defined by the integral

ψ(s) :=

∫ s

−∞

√101

10et/10dt =

√101es/10

and its inverse

ϕ := ψ−1 :(0,∞)→ R

t 7→ 10 ln( t√

101

)(Here there is a mistake at the statement of the exercise, the reparamaterization is not necessarilydefined on the whole real numbers. Also in principle we should consider the integral

∫ st0|c(t)|dt

where t0 is any real number, but we actually just care about the fact that ψ(t) = |c(t)|).The reparameterization that we are looking for is given by the composition

c ◦ ϕ(t) =

t√101· cos

(10 ln

( t√101

))t√101· sin

(10 ln

( t√101

)) .

A2.2 Sei c = (c1, c2) : [a, b] → R2 eine ebene nach Bogenlange parametrisierte Kurve. Nehmen Siean, dass es t0 ∈ (a, b) gibt, sodass

c([a, t0]) ⊂ {(x, y) ∈ S1 : x > 0}c([t0, b]) ⊂ {(x, y) ∈ S1 : y > 0}

gilt. Beweisen Sie, dass

θ : [a, b]→ R

t 7→

arctan

( c2(t)c1(t)

), falls t ∈ [a, t0]

π/2− arctan( c1(t)c2(t)

), falls t ∈ [t0, b]

eine glatte Funktion darstellt.

Proof. Let U1 = {(x, y) ∈ R2 : x > 0} and U2 = {(x, y) ∈ R2 : y > 0}. We define the function

F : U1 ∪ U2 ⊂ R2 → R

(x, y) 7→

arctan(yx

), if x > 0

π/2− arctan(xy

), if y > 0

The function F is smooth as arctan(yx

)and π/2 − arctan

(xy

)are smooth on the open sets U1

and U2 and they coincide at the intersection U1 ∩ U2 as

arctan(yx

)= π/2− arctan

(xy

)for x, y > 0 .

The function θ is smooth as it is equal to the composition of functions F ◦ c .

A2.3 Sei c eine regulare ebene (nicht notwendigerweise nach Bogenlange) parametrisierte Kurve. Zei-gen Sie, dass die Krummund der Kurve durch die Formel

κ(t) =det(c(t), c(t))

‖c(t)‖3

gegeben ist. Genauer heißt das: Ist c = c ◦ϕ eine orienterungshaltende Umparametrisierung nachder Bogenlange mit Krummung κ, so gilt

κ = κ ◦ ϕ .

Proof. Note first that ϕ(t) =1

|c(ϕ(t))|which follows from the facts that 〈 ˙c(t), ˙c(t)〉 = 1 and the

reparameterization preserves the orientation.

Now the exercise follows from the following calculation where n denotes the normal vector of c:

κ = 〈¨c(t), n(t)〉 = det(˙c(t), ¨c(t)

)= ϕ(t) · det(c(ϕ(t)), ¨c(t))

= ϕ(t) · det(c(ϕ(t)) , ϕ(t)2c(ϕ(t)) + ϕ(t)c(ϕ(t))

)= ϕ(t)3 · det

(c(ϕ(t)), c(ϕ(t))

)=

det(c(ϕ(t)), c(ϕ(t))

)|c(ϕ(t))|3

2

A2.4 Sei c : R→ R2 die parametrisierten Kurven gegeben durch

c(t) =(sin(t),

sin(2t)

2

).

(a) Skizzieren Sie die Spuren der Kurven c und ermitteln Sie die Periode von c .

(b) Ist die Kurve c einfach geschlossen? Begrunden Sie Ihre Antwort.

(c) Berechnen Sie die Krummung der Kurven c .

Proof. The curve is the Lemniscate. As sin(t) and sin(2t) have period 2π and π, respectively, theperiod of the whole function is 2π. The curve is not simply closed as it passes two times throughthe point (0, 0) at t = 0, π ∈ [0, 2π].

In order to compute the curvature, we use the previous exercise:

κ(t) :=

det

(cos(t) − sin(t)cos(2t) −2 sin(2t)

)|(cos(t), cos(2t))|3

=− sin(t)(1 + 2 cos2(t))√4 cos4(t)− 3 cos2(t) + 1

.

(The curvature seems not to have further simplification that would be worth to write).

A2.5 Gegeben sei eine geschlossene, ebene Kurve c : I → R2 mit Lange L. Zeigen Sie es existiert einPunkt p ∈ R2 mit

Bild(c) ⊂ BL/4(p) := {x ∈ R2 : ‖x− p‖ ≤ L/4} .

Proof. Following Artem’s solution:

Find two points c(t0), c(t1) such that the length of the two Curves arcs equals to L/2.

Let p be the midpoint between c(t0) and c(t1). After an isometry, wee can assume that p is theorigin and c(t0) = (m, 0), c(t1) = (−m, 0) for a m ≥ 0.

Suppose there is a point c(t) = (x, y) that

|c(t)− p| > L/4

We have that|c(t0)− c(t)|+ |c(t)− c(t1)| ≤ L/2

3

as |c(t0)− c(t)|+ |c(t)− c(t1)| is less or equal to the arc length between c(t0) and c(t1) .

Use the assumptions (c(t0) = (m, 0), c(t1) = (−m, 0), c(t) = (x, y)) and write for t ≤ 0, thefunction

f(t) :=√(x− t)2 + y2 +

√(x+ t)2 + y2

Now by assumption f(0) > L/2 and f(t) is monotone growing (calculation). Here we achieve acontradiction as f(m) ≤ L/2 .

4





Hausaufgaben

A3.1 Zeigen Sie, dass sich das Vorzeichen der Krummung unter der Parametertransformation t 7→ −tandert.

Proof. Let c : (a, b)→ R2 be a regular curve and p = c(t0) be a point of the curve. The curvatureof the curve at the point p equals

κc(t0) =det(c(t), c(t))

|c(t)|3∣∣∣t=t0

.

Now consider the reparameterization of the curve c given by

d : (−b,−a)→ R2

t 7→ c(−t) .

The point p lies on the graph of the curve d when t = −t0 . The curvature of d at p equals

κd(−t0) =det(d(t), d(t))

|d(t)|3∣∣∣t=−t0

=det(−c(−t), c(−t))

|c(−t)|3∣∣∣t=−t0

= −κc(t0) ,

and we are done.

Remark: There is of course some ambiguity if the curve passes several times through the samepoint, but the curvature will change its sign accordingly.

A3.2 Sei c : R→ R2 eine einfach geschlossene nach Bogenlange parametrisierte Kurve mit Periode Lund sei X die Menge gegeben durch

X := {(t1, t2) ∈ R2 : 0 ≤ t1 ≤ t2 ≤ L} .

Betrachten Sie die Abbildung

f : X → S1

(t1, t2) 7→

c(t2)− c(t1)|c(t2)− c(t1)|

, falls t2 > t1 und (t1, t2) 6= (0, L)

c(t), t2 = t1 = t

−c(0), (t1, t2) = (0, L) .

Beweisen Sie, dass f stetig auf X ist.

Proof. Consider the following function defined by

F : R2 → R2

(x, y) 7→

c(y)− c(x)y − x

, it x 6= y

c(t), x = y = t .

The mean value theorem implies that for every x 6= y, there exist t1, t2 ∈ [min(x, y),max(x, y)]such that

(c1(y)− c1(x), c2(y)− c2(x)) = (c(t1)(y − x), c(t2)(y − x)) .

As c is continuous, if (x, y)→ (t, t), then (c1(t1), c2(t2))→ (c1(t), c1(t)), which implies that Fis continuous. Also, the following function is continuous

G =F

|F |: {(x, y) ∈ R2 : y − x /∈ Z · L\{0}} → S1

(x, y) 7→

c(y)− c(x)|c(y)− c(x)|

, if y > x

− c(y)− c(x)|c(y)− c(x)|

, if x > y

c(t), x = y = t .

Notice that if 0 < x ≤ y < L and (x, y) → (0, L), then for the variable y′ = y − L it holds thaty′ < 0 < x and (x, y′)→ (0, 0) . We now show that f is continuous at (x, y) = (0, L) on X :

lim(x,y)→(0,L)

c(y)− c(x)|c(y)− c(x)|

= lim(x,y′)→(0,0)

c(y′ + L)− c(x)|c(y′ + L)− c(x)|

= lim(x,y′)→(0,0)

c(y′)− c(x)|c(y′)− c(x)|

= −c(0) .

A3.3 Sei c : R → R2 eine regulare periodische Kurve mit Periode L und mit Krummung κ (sieheAufgabe A2.3). Wir definieren die Windungszahl der Kurve durch die Formel

W c :=1

2π

∫ L

0κ(t) · |c(t)| dt .

Sei c ◦ϕ eine orientierungshaltende Umparametrisierung nach der Bogenlange mit WindungszahlWc◦ϕ. Zeigen Sie, dass

W c =Wc◦ϕ

gilt.

Proof. We already know that

ϕ(t) =1

|c(ϕ(t))|.

Let κ and κ be the curvatures of c and c ◦ ϕ, respectively. From a previous HWA, we know that

κ = κ ◦ ϕ .

Let us assume that the period of c ◦ ϕ is L′ . The exercise follows from the Integration by substi-tution theorem:

Wc◦ϕ :=1

2π

∫ L′

0κ(s)ds =

1

2π

∫ L′

0κ(ϕ(s))ds

=1

2π

∫ L

0κ(t)

1

ϕ(s)dt =

1

2π

∫ L

0κ(t) · |c(t)| dt =W c .

2

A3.4 Sei c : R→ R2 die parametrisierte Kurve

c(t) =(sin(t),

sin(2t)

2

).

Zeigen Sie, dassW c = 0

gilt.

Proof. The curve has period 2π, and we have to compute the integral

W c =1

2π

∫ π

−πf(t)dt,

where

f(t) = κ(t) · |c(t)| =det

(cos(t) − sin(t)cos(2t) −2 sin(2t)

)|(cos(t), cos(2t))|2

The exercise follow from the fact that f is odd, i.e, f(−t) = −f(t) .

3



Aufgabenblatt 4Abgabe: Diese Ubungen mussen bis spatestens 11:55 Uhr Dienstag 12.11.2019 in den Briefkasten im


Hausaufgaben

A4.1 Seien p0, v ∈ R2 mit |v| = 1 . Sei G die Spur von der parametrisierten Geraden γ : R → R2

gegeben durchγ(t) = p0 + t · v .

Geben Sie eine euklidische Bewegung Φ ∈ E(R2) an, sodass

Φ(G) = x-Achse

gilt.

Proof. As |v| = 1, there exists θ such that v = (cos(θ), sin(θ)) . The transformation defined by

Φ : R2 → R2(xy

)7→(

cos(θ) sin(θ)− sin(θ) cos(θ)

)((xy

)− p0

)does the job.

A4.2 Sei c = (c1, c2) : I → R2 eine regulare parametrisierte, ebene, konvexe Kurve. Beweisen Sie,dass es fur alle t0 ∈ I eine euklidische Bewegung Φ ∈ E(R2) gibt, sodass die Tangente inΦ(c(t0)) der Kurve Φ ◦ c : I → R2 die x-Achse ist, und die abgeschlossene obere Halbebene dieSpur von Φ ◦ c enthalt.

Proof. Let us recall the definition of convex curve: assume first that c is parametrized by arc lengthand n denotes its normal vector. The curve c is convex if for all t, t0 ∈ I,

〈c(t)− c(t0), n(t0)〉 ≥ 0

or〈c(t)− c(t0), n(t0)〉 ≤ 0

The previous exercise allows us to assume that for t0 ∈ I we have that c(t0) = 0, the tangent ofthe curve at c(t0) is the x-axis and n(t0) = (0, 1). As we can move the tangent at c(t0) through atranslation and then an orthogonal transformation, the above inequalities are preserved and underour assumptions we have for every t ∈ I the following inequality holds

〈c(t), (0, 1)〉 = c2(t) ≥ 0,

and we are done.

A4.3 Geben Sie eine nach Bogenlange parametrisierte Kurve c : I → R2 an mit Krummung

κ(t) =1

1 + t2.

Proof. Since the curve is parametrized by arc length, we can write the curve as

c(t) =

(c1(t)c2(t)

)=

(cos(θ(t))sin(θ(t))

),

for some smooth function θ : I → R2 . We want to solve the differential equation(c1(t)c2(t)

)=

1

t2 + 1

(−c2(t)c1(t)

),

or equivalently

c1(t) = cos(θ(t))· = − sin(θ(t)) ˙θ(t) =−1

t2 + 1sin(θ(t)) =

1

1 + t2c2(t) .

A solution of the differential equation

θ(t) =1

t2 + 1

is θ(t) = arctan(t) , and we can take

(c1(t)c2(t)

)=

(cos(arctan(t))sin(arctan(t))

)

=

1√t2 + 1t√

t2 + 1

,

so (c1(t)c2(t)

)=

∫ dt√

t2 + 1= sinh−1(t) + C∫ t dt√

t2 + 1=√t2 + 1 + C ′ .

,

A4.4 Sei c : (a, b)→ R2 eine einfach geschlossene regulare Kurve. Nehmen Sie an, dass die Spur von cin der abgeschlossenen oberen Halbebene enthalten ist und die x-Achse nur in zwei verschiedenenPunkten schneidet. Zeigen Sie, dass es ein kleines ε > 0 gibt, sodass die Gerade gegeben durch

y = ε

mindestens vier Punkte mit der Spur von c gemeinsam hat.

Proof. Let us write c(t) = (c1(t), c2(t)) and let t0, t1 ∈ (a, b) be the only real numbers such thatc2(t0) = c2(t1) = 0. By assumption c2(t) ≥ 0 for every t ∈ (a, b) which implies that zero is alocal minimum of c2 which is achieved only at t0 and t1 . As a consequence c2(t1) = c2(t2) = 0.But as c is parametrized by its arc length, then c1(t1) = c1(t2) = ±1 . The inverse functiontheorem implies that there exist δt1 , δt2 > 0 such that

[t0 − δt0 , t0 + δt0 ], [t1 − δt1 , t1 + δt1 ] ⊂ I,

and c1 has a smooth inverse c−11 on these intervals. We can write the curve c on these intervals asthe graph of a function

y = f(x)

2

where f(x) := c2(c−11 (x)) . Take any

0 < ε < f(t0 ± δt0), f(t1 ± δt1)

and the statement follows from the intermediate value theorem.

Remark: Recall that the one variable inverse function theorem is equivalent to the well knownstatement in Calculus that says that the derivative of a C1-function is positive if and only if thefunction is increasing, or, the derivative of a C1-function is negative if and only if the function isdecreasing.

3





Hausaufgaben

A5.1 Sei c : I → R3 eine nach Bogenlange parametrisierte Kurve, mit nicht-verschwindender Krummung,und sei (e1(t), e2(t), e3(t)) das begleitende Dreibein von c, mit t ∈ I. Wir bezeichnen mit κ undτ die Krummung und Torsion von c .

(a) Beweisen Sie, dass e1(s)e2(s)e3(s)

=

0 κ(s) 0−κ(s) 0 τ(s)

0 −τ(s) 0

e1(s)e2(s)e3(s)

gilt.

Proof. Let us recall that

e1(s) = c(s),

e2(s) =c(s)

|c(s)|,

e3(s) = e1(s)× e2(s),κ(s) = |c(s)| , τ(s) = 〈e2(s), e3(s)〉 .

In particulare1(s) = c(s) = κ(s)e2(s)

Note that for every i ∈ {1, 2, 3}

〈ei(s), ei(s)〉 = 0

〈ei(s), ei(s)〉+ 〈ei(s), ei(s)〉 = 0,

In particular〈e1(s), e1(s)〉 = −κ2(s) .

As e1(s), e2(s), e3(s) is an orthonormal basis,

e2(s) = 〈e1(s), e2(s)〉e1(s) + 〈e2(s), e2(s)〉e2(s) + 〈e3(s), e2(s)〉e3(s)

By definition τ(s) = 〈e3(s), e2(s), on the other hand

〈e1(s), e2(s)〉 = 〈e1(s),( e1(s)κ(s)

)·〉 = 〈e1(s),

( e1(s)κ(s)− e1(s)κ(s)κ2(s)

)〉 = −κ(s),

and we conclude thate2(s) = −κ(s)e1(s) + τ(s)e3(s)

As the cross product satisfies the Leibniz rule (why?), we end with the following calculation

e3(s) = e1(s)× e2(s) + e1(s)× e2(s)= κ(s)e2(s)× e2(s) + e1(s)× (−κ(s)e1(s) + τ(s)e3(s))

= τ(s)e1(s)× e3(s) = −τ(s)e2(s) .

(b) Wir wissen, dass e3(t) = e1(t)× e2(t) gilt. Zeigen Sie, dass

e1(t) = e2(t)× e3(t), und e2(t) = e3(t)× e1(t) .

Proof. The cross product and the scalar product are related through the following equality

〈u, v × w〉 = det(u, v, w) .

Also|v × w|2 = |v|2|w|2 − 2〈v, w〉2

This equalities allow us to conclude for instance that e1, e2, e3 are an orthonormal basis ase1, e2 are orthogonal unit vectors.

On the other hand, note that

〈e1, e2 × e3〉 = det(e1, e2, e3) = det(e3, e1, e2) = 〈e3, e1 × e2〉 = |e3|2 = 1 .

We conclude that e1 = e2 × e3 and likewise e2 = e3 × e1 .Remark: Alternatively, one can use the Grassmann identities:

(i) a× (b× c) = 〈a, c〉b− 〈a, b〉c(ii) (a× b)× c = 〈a, c〉b− 〈b, c〉a

For instance,

e2 × e3 = e2 × (e1 × e2) = 〈e2, e2〉e1 − 〈e2, e1〉e2 = e1

Likewise, e3 × e1 = e2 .

(c) Sei ϕ : J → I ein orientierungserhaltender Diffeomorphismus und sei c = c ◦ ϕ eine Um-parametrisierung der Kurve. Das begleitende Dreibein (e1(s), e2(s), e3(s)) von c im Punkts ist definiert als das begleitende Dreibein (e1(t), e2(t), e3(t)) von c im Punkt t = ϕ(s).Zeigen Sie, dass

e1(s) =˙c(s)

| ˙c(s)|, e2(s) =

˙e1(s)

| ˙e1(s)|,

e3(s) =˙c(s)× ¨c(s)

| ˙c(s)× ¨c(s)|.

Proof. Similarly as we did in previous HWAs, it is not hard to conclude that

ϕ(s) = | ˙c(s)| .

By definition

e1(s) = e1(ϕ(s)), e2(s) = e2(ϕ(s)), e3(s) = e3(ϕ(s)) .

Now we prove the formulas for every ei without giving too much explanations:

(i) e1(s) =˙c(s)

| ˙c(s)|:

e1(s) = c(ϕ(s)) = c(ϕ(s))ϕ(s)1

ϕ(s)=

˙c(s)

| ˙c(s)|

2

(ii) e2(s) =˙e1(s)

| ˙e1(s)|:

Note that˙e1(s) = c(ϕ(s))ϕ(s) .

Hence,

e2(s) = e2(ϕ(s)) =c(ϕ(s))

|c(ϕ(s))|

=c(ϕ(s))ϕ(s)

|c(ϕ(s))|ϕ(s)=

˙e1(s)

| ˙e1(s)|

(iii) e3(s) =˙c(s)× ¨c(s)

| ˙c(s)× ¨c(s)|:

˙c(s)× ¨c(s) = c(ϕ(s))ϕ(s)×(c(ϕ(s))ϕ(s)2 + c(ϕ(s))ϕ(s)

)= ϕ(s)3c(ϕ(s))× c(ϕ(s))= | ˙c(s)|3κ(s)e1(s)× e2(s) = | ˙c(s)|3κ(s)e3(s)

and hence

˙c(s)× ¨c(s)

| ˙c(s)× ¨c(s)|= e3(s) .

(d) Zeigen Sie, dass die Frenet-Gelichungen als ˙e1(s)˙e2(s)˙e3(s)

= | ˙c(s)|

0 κ(s) 0−κ(s) 0 τ(s)

0 −τ(s) 0

e1(s)e2(s)e3(s)

geschrieben werden konnen, wobei κ(s) := κ(t) und τ(s) := τ(t).

Proof. The exercise follow from part a) and the fact that for i ∈ {1, 2, 3}

˙ei(s) = ei(ϕ(s)) · | ˙c(s)| .

(e) Zeigen Sie, dass

κ(s) =| ˙c(s)× ¨c(s)|| ˙c(s)|3

.

Proof. From part c-iii), we got the following equality

˙c(s)× ¨c(s) = | ˙c(s)|3κ(s)e3(s) .

And the exercise follows from the fact that |e3(s)| = 1 .

3

A5.2 Sei c : R→ R3 die Kurve parametrisiert durch

c(s) =

s

12s

2

13s

3

(a) Kann man die Formeln aus Aufgabe A5.1 benutzen?

Proof. Yes, because the curve is regular and any regular curve can be parametrized by arclength. (Here in this exercise, do not be picky about what students answered, as it is kind ofopen to interpretation).

(b) Berechnen Sie das begleitende Dreibein (e1(s), e2(s), e3(s)) der Kurve c .

Proof. (i) e1(s) :

e1(s) =˙c(s)

| ˙c(s)|= (1 + s2 + s4)−1/2

1ss2

(ii) e2(s)

˙e1(s) =− (s+ 2s3)(1 + s2 + s4)−3/2

1ss2

+ (1 + s2 + s4)−1/2

012s

= (1 + s2 + s4)−3/2

(−(s+ 2s3)

1ss2

+ (1 + s2 + s4)

012s

)

= (1 + s2 + s4)−3/2

−2s3 − s−s4 + 1s3 + 2s

and

e2(s) =˙e1(s)

| ˙e1(s)|= (s8 + 5s6 + 6s4 + 5s2 + 1)−1/2

−2s3 − s−s4 + 1s3 + 2s

(iii) e3(s) :

˙c(s)× ¨c(s) =

1ss2

× 0

12s

=

s2

−2s1

and

e3(s) =˙c(s)× ¨c(s)

| ˙c(s)× ¨c(s)|= (s4 + 4s2 + 1)−1/2

s2

−2s1

4

(c) Berechnen Sie die Krummung κ der Kurve c .

Proof.

κ(s) =| ˙c(s)× ¨c(s)|| ˙c(s)|3

=(s4 + 4s2 + 1)1/2

(s4 + s2 + 1)3/2

A5.3 Sei c : I → Rn eine Frenet-Kurve im Rn mit begleitendem Frenet-n-Bein e1, . . . , en . Sei κi diei-te Frenet Krummung. Beweisen Sie, dass das Vorzeichen von κi das gleich ist wie das von

〈c(i+1), ei+1〉 > 0 .

Proof. First we note the following claim from linear algebra:

Claim: Let {v1, . . . , vn} be a basis of a vector space V with inner product 〈, 〉. Let {e1, . . . , en}be the corresponding orthonormal basis constructed from GS process. Then for every i

〈vi, ei〉 > 0 .

Proof of the claim: Let

fi = vi − 〈e1, vi〉e1 − . . .− 〈e1, vi〉ei−1 .

GS states thatei =

fi|fi|

We take the inner product of both sides of the equality with ei to obtain

1 =〈vi, ei〉|fi|

,

and the claim is proven.

From the Linear algebra claim we infer that

〈c(i+1), ei+1〉 > 0 .

On the other hand, let us write

ci =

i∑k=1

〈ci, ek〉ek

If we differentiate the equality, we obtain

ci+1 =i∑

k=1

(. . .)ek +i∑

k=1

〈ci, ek〉ek

and〈ci+1, ei+1〉 = 〈ci, ei〉〈ei, ei+1〉 = κi〈ci, ei〉 .

as ei+1⊥ span{e1, . . . , ei−1} ⊂ span{e1, . . . , ei} .The statement follows from mathematical induction.

5





Hausaufgaben

A6.1 Seien r, a positive reelle Zahlen mit a > r . Seien S der Kreis in der yz-Ebene mit Mittelpunkt(0, a, 0) und Radius r und T der Torus die Flache, die erzeugt wird, indem man den Kreis S umdie z-Achse rotiert. Zeigen Sie, dass T eine regulare Flache ist.

Proof. The equation of the circle in the yz-plane is

(y − a)2 + z2 = r2,

and the points of the torus T are obtained by rotating this circle around the z-axis and they satisfythe equation

z2 = r2 − (√x2 + y2 − a)2 .

Thus, T is the inverse image of r2 by the function

f(x, y, z) = z2 + (√x2 + y2 − a)2 .

This function is differentiable for (x, y) 6= (0, 0), and

∂f

∂x=

2x(√x2 + y2 − a)√x2 + y2

,∂f

∂y=

2y(√x2 + y2 − a)√x2 + y2

,∂f

∂z= 2z ,

r2 is a regular value of f. Then the torus T is a regular surface.

A6.2 Sei S ⊂ R3 eine regulare Flache, W ⊂ R3 offen. Zeigen Sie, dass W ∩ S auch eine regulareFlache ist.

Proof. Let us review first the definition of regular surface: a subset S ⊂ R3 is a regular surfaceif for every point p ∈ S, there exist an open neighborhood V of p in R3, such that there exist anopen set U ⊂ R2 and a smooth function F : U → R3, such that

(a) F (U) = S ∩ V and F : U → S ∩ V is a homeomorphism.

(b) The Jacobian matrix DuF at every point u ∈ U has rank 2.

If p ∈ W ∩ S, the result follows easily because (W ∩ S) ∩ V is an open set of S ∩ V, and as Fis a homeomorphism, there exists an open set U ′ ⊂ U ⊂ R2 such that F (U ′) = (W ∩ S) ∩ V.The restriction F |U ′ is a homeomorphism onto the image, and the first condition of the definitionis satisfied. The second condition of the definition holds for every u ∈ U ′ as U ′ ⊂ U .

A6.3 Seien S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1} die Sphare und N = (0, 0, 1) der Nordpol.

(a) Sei L(x,y) die Gerade durch (x, y, 0) und N . Zeigen Sie, dass L(x,y) ∩ (S2\N) genau einenPunkt enhalt.

Wir bezeichnen diesen Punkt mit F (x, y). Teil (a) zeigt, dass F eine wohldefinierte FunktionF : R2 → R3 ist.

(b) Geben Sie eine Formel fur F an.

(c) Zeigen Sie, dass F (R2) = R2\{N} und dass F : R2 → S2\{N} eine Bijektion ist.

(d) Geben Sie eine Formel fur F−1 : S2\{N} → R2 an.

(e) Argumentieren Sie, dass dasselbe fur S2\{S} gilt, wobei S = (0, 0,−1) der Sudpol ist, unddass S2 eine regulare Flache ist.

Proof. A parameterization of the line is given by

L(x,y)(t) = (tx, ty, 1− t) .

The line intersects the sphere at points where

t2(x2 + y2) + (1− t)2 = 1

ort(t(x2 + y2 + 1)− 2) = 0 .

The equations is solved when

t = 0, t =2

x2 + y2 + 1

Note that L(x,y)(0) = (0, 0, 1) and the line intersects the sphere at just another point when t =2

x2 + y2 + 1. The function F is then given by

F (x, y) =( 2x

x2 + y2 + 1,

2y

x2 + y2 + 1,x2 + y2 − 1

x2 + y2 + 1

).

In order to find the inverse of F, we take (x, y, z) ∈ S2\{(0, 0, 1)} and consider the line withparamaterization

t(x, y, z) + (1− t)(0, 0, 1)

passing through (x, y, z) and (0, 0, 1). The line intersects the plane R2×{0} when t =1

1− z, so

the inverse of F is given by

F−1 : S2\{(0, 0, 1)} → R2

(x, y, z)→( x

1− z,

y

1− z

).

The function is well defined, and as it is an inverse of F, so the function F is bijective, and inparticular F (R2) = S2\{(0, 0, 1)} .Likewise for the south pole, a parameterization of the line L(x,y) passing through (x, y, 0) andS = (0, 0,−1)

L(x,y)(t) = (tx, ty, t− 1) .

The line intersects the sphere at t = 0 and t =2

x2 + y2 + 1. Just as before we define a function

G : R2 → S2

(x, y) 7→( 2x

x2 + y2 + 1,

2y

x2 + y2 + 1,1− x2 − y2

x2 + y2 + 1

)and its inverse is

G−1 : S2\{(0, 0,−1)} → R2

(x, y, z)→( x

1 + z,

y

1 + z

).

2

A6.4 Sei F : (0, π)× (0, 2π)→ R3 definiert durch

F (θ, ϕ) =

sin(θ) cos(ϕ)sin(θ) sin(ϕ)

cos(θ)

(a) Skizzieren Sie fur (x, y, z) im Bild von F die Koordinaten θ und ϕ.

(b) Bestimmen Sie das Bild von F .

(c) Finden Sie vier lineare Isometrien g1, g2, g3, g4 : R3 → R3, sodass die Bilder von

g1 ◦ F, g2 ◦ F, g3 ◦ F, g4 ◦ F : (0, π)× (0, 2π)→ R3

die Sphare S2 uberdecken. Skizzieren und Bestimmen Sie die Bilder von g1 ◦F, g2 ◦F, g3 ◦F, g4 ◦ F .

Proof. The image of F is the set

S2\{(x, y, z) ∈ R3 : x2 + z2 = 1, y = 0, x ≥ 0} .

The spherical coordinates are obtained by rotating every point of the unit circle in the xy-planewith polar coordinates (cos(ϕ), sin(ϕ), 0) an angle θ between the z-axis and the line segmentfrom the origin to the point (cos(ϕ), sin(ϕ), 0).

Let us consider the following orthogonal matrices

A1 =

1 0 00 1 00 0 1

, A2 =

−1 0 00 1 00 0 1

,

A3 =

1 0 00 0 −10 1 0

, A4 =

−1 0 00 0 −10 1 0

.

We denote by {gi : R3 → R3}4i=1 the linear transformation determined by the matrices Ai . Themap g1 is the identity, the map g2 is the reflection with respect to the zy-plane, the map g3 is acounterclockwise rotation of 90◦ around the x-axis, and the map g4 is the composition of g2 andg3 . The images of g1 ◦ F, g2 ◦ F cover the sphere with the exception of the north and south pole.The images of g3◦F, g4◦F cover the sphere except the points (0, 1, 0) and (0,−1, 0). The imagesof the four maps cover the full sphere.

A6.5 Betrachten Sie der Doppelkegel

S = {(x, y, z) ∈ R3 : x2 + y2 = z2}.

Beweisen Sie, dass S bei (0, 0, 0) keine regulare Flache ist.

Proof. We show that S is not locally homeomorphic to an open set of R2 around (0, 0, 0).Let us suppose that S is a regular surface around p = (0, 0, 0) . So there exist open setsU ⊂ R2, V ⊂ R3,and a smooth function F : U → V, so F (U) = S ∩ V and F : U → S ∩ V is a homeomorphism.Let u0 = F−1(0, 0, 0) ∈ U. Then U is an open neighborhood of u0 and we can assume without loss ofgenerality that is an open disk with centre u0. Clearly the image of F contains points p1 = (x1, y1, z1)with z1 > 0 and p2 = (x2, y2, z2) with z2 < 0 .Let ui := F−1(pi), i = 1, 2 . In U we connect the points u1 and u2 with a continuous path c that doesnot pass through u0 . The path F ◦ c connects p1 with p2 without passing through F (u0) = (0, 0, 0) .

3

That is a contradiction due to the intermediate value theorem applied to the third component of F ◦ c ..Remark: Somehow the idea is to show that if around the origin, the set S were locally homeomorphicto a ball, the corresponding set obtained by removing the origin will have at least two path connectedcomponents which is a contradiction.

4





Hausaufgaben

A7.1 Seien a, b, c positiven reellen Zahlen. Gegeben ist die regulare Flache

S = {(x, y, z) ∈ R3 :x2

a2+y2

b2+z2

c2= 1} .

Skizzieren Sie diese Flache und Berechnen Sie die Tangentialebene TpS von S im jeden Punktp ∈ S.

Proof. At a point (x, y, z) ∈ S, a vector normal to the tangent space at (x, y, z) is the gradient

∇(x2a2

+y2

b2+z2

c2

)=

2x/a2

2y/b2

2z/c2

The tangent space at (x, y, z) equals

{(u, v, w) ∈ R3 :xu

a2+yv

b2+zw

c2= 0} .

A7.2 Sei S die Menge der Punkte (x, y, z) ∈ R3, die die Gleichung x2 + y2 = 1 erfullen.

(a) Zeigen Sie, dass S eine regulare Flache ist.

Proof. Let G(x, y, z) = x2 + y2 . It is easy to check that 1 is a regular value of G : at(x, y, z) ∈ R3, the differential of G

dG(x,y,z)(a, b, c) = (2x, 2y, 0) · (a, b, c)

has rank equal to one as far as (x, y, z) 6= (0, 0, z), which is the case as x2 + y2 = 1 .

(b) Zeigen Sie, dass die folgende Abbildung

f : S → R(x, y, z) 7→ xyz

glatt ist, und berechnen Sie dpf fur alle p ∈ S .

Proof. The function is smooth as it is the restriction of a smooth function defined on R3 .Let p = (x, y, z). The differential dpf is the restriction of the differential

TpR3 → R(u, v, w) 7→ (yz, xz, xy) · (u, v, w)

to TpS = {(u, v, w) ∈ R3 : (u, v, w) · (x, y, 0) = 0} .

Another approach to compute the differential is to consider a smooth curve c = (c1, c2, c3) :I → S such that c(0) = (x, y, z) and c′(0) = (u, v, w). We know that

dpf(u, v, w) =d

dt(f ◦ c)(0) =

d

dt(c1(t)c2(t)c3(t))

∣∣∣t=0

= c′1(0)c2(0)c3(0) + c1(0)c′2(0)c3(0) + c1(0)c2(0)c′3(0)

= uyz + xvz + xyw

= 〈(u, v, w), (yz, xz, xy)〉

A7.3 Seien S eine regulare Flache, f : S → R eine glatte Abbildung und (U,F, V ) eine zentrierteKarte um einen Punkt p ∈ S. Beweisen Sie, dass fur jedes u ∈ U, v ∈ TF (u)S und w ∈ R2 mitDuF (w) = v

Du(f ◦ F )(w) = dF (u)f(v)

gilt.

Proof. Let U ′ = {x ∈ R2 : (x+ u) ∈ U} and

L : U ′ → U

x 7→ x+ u

The triple (U ′, F ◦L, V ) is a chart centered around F (u), . We apply the definition of differentialto this chart to obtain the differential dF (u)f(v) : Note first that

d0L : T0U′ ∼= R2 → TuU ∼= R2

is just the identity. Let w′ = d0L−1(w). Then

dF (u)f(v) = D0(f ◦ F ◦ L)(w′) = Du(f ◦ F )(w) .

A7.4 Es sei f : S → R gegeben durchf(p) = |p− p0|2,

wobei p ∈ S und p0 ein fester Punkt des R3 ist. Zeigen Sie, dass

dfp(w) = 2〈w, p− p0〉, w ∈ Tp(S),

ist.

Proof. Let us consider the map

F : R3 → Rp 7→ |p− p0|2 .

The differential of F equals

dFp : TpR3 → Rw 7→ 2〈w, p− p0〉 .

The differential dfp is just the restriction of the differential dFp to TpS .

2

Just in exercise A7.2, let us consider a smooth curve c = (c1, c2, c3) : I → S such that c(0) =(x1, x2, x3) = p and c′(0) = w = (w1, w2, w3). Let us write p0 = (p01, p02, p03). Then

dpf(w) =d

dt(f ◦ c)(0) =

d

dt((c1(t)− p01)2 + (c2(t)− p02)2 + (c3(t)− p03)2)

∣∣∣t=0

= 2(c1(0)− p01)c′1(0) + 2(c2(0)− p02)c′2(0) + 2(c3(0)− p03)c′3(0)

= 2(x1 − p01)w1 + 2(x2 − p02)w2 + 2(x3 − p03)w3

= 2〈w, p− p0〉

A7.5 SeiP (z) ein nicht konstantes Polynom mit komplexen Koeffizienten. Betrachten Sie die 2-dimensionaleSphare

S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1}

und die stereografische Projektion

h : S2\{(0, 0, 1)} → R2

(siehe Aufgabe A6.3). Wir identifizieren R2 mit der komplexen Zahlenebene und wir betrachtenP (z) als eine Abbildung der komplexen Zahlenebene auf sich selbst. Sei f : S2 → S2 gegebendurch

f(x) =

{h−1Ph(x) x 6= (0, 0, 1)

(0, 0, 1) x = (0, 0, 1).

Zeigen Sie, dass f eine glatte Abbildung ist.

Proof. Let k be the stereographic projection from the south pole (0, 0,−1) and set

Q(z) = kfk−1(z) .

Note thathk−1(z) = z/|z|2 = 1/z

Now if P (z) = a0zn + a1z

n−1 + . . .+ an, with a0 6= 0, then a short computation shows

Q(z) = zn/(a0 + a1z + . . .+ anzn) .

Thus Q is smooth in a neighborhood of 0, and it follows that f = k−1Qk is smooth in a neigh-borhood of (0, 0, 1) .

A7.6 Sei S eine regulare Flache. Beweisen Sie, dass fur alle p ∈ S

TpS = {v ∈ R3 : es gibt eine glatte parametrisierte Kurve

c : I → S mit c(0) = p und c(0) = v}

gilt.

Proof. Let c : I → S be a curve with c(0) = p and c′(0) = v. Let F : U → R3 be a parametriza-tion of S. We define a curve c : I → U by

c(t) = F−1 ◦ c(t)

3

(here we restrict the interval I if necessary). Let

c′(0) = v1e1 + v2e2 .

Then

v = DF−1(p)F (c′(0)) = v1DF−1(p)F (e1) + v2DF−1(p)F (e2) ∈

span{ ∂

∂u1(p) = DF−1(p)F (e1),

∂

∂u2(p) = DF−1(p)F (e2)} = TpS

This shows the inclusion

{v ∈ R3 : es gibt eine glatte parametrisierte Kurve

c : I → S mit c(0) = p und c(0) = v} ⊂ TpS .

The other inclusion follows from taking a parametrization F : U → R3 centered around p ∈ Sand considering the curves defined by

γ1(t) = F (te1) , γ2(t) = F (te2) .

4





Hausaufgaben

A8.1 Sei S ⊂ R3 eine regulare Flache und sei Wp : TpS → TpS die Weingarten Abbildung in p ∈ S .Die dritte Fundamental form III von S wird definiert als

IIIp(v, w) := Ip(W2p v, w) = Ip(Wpv,Wpw)

fur alle v, w ∈ TpS .Zeigen Sie, dass zwischen den drei Fundamentalformen I, II, III die folgende Gleichung be-steht:

III − Spur(W )(II) + Det(W )I = 0 .

Proof. Let us recall the Cayley-Hamilton theorem from linear algebra:

Every square matrix over the complex numbers satisfies its own characteristic equation.

The characteristic equation of Wp : TpS → TpS is

W 2p − Trace(Wp)Wp +Det(Wp)I = 0

and in particular for every v, w ∈ TpS

〈W 2p v − Trace(Wp)Wpv +Det(W0)v, w〉 = IIIp(v, w)− Trace(Wp)IIp(v, w) + Det Ip(v, w) = 0 .

A8.2 Eine Flache heißt Drehflache oder Rotationsflache, wenn sie durch Drehung einer regularen, ebe-nen Kurve (der sogenannten Meridiankurve)

t 7→ (r(t), h(t))

um die z-Achse im R3 entsteht, also wenn sie das Bild der folgenden Abbildung zulaßt:

f : I × [0, 2π]→ R3

(t, θ) 7→

r(t) cos(θ)r(t) sin(θ)h(t)

.

(a) Zeigen Sie, dass eine Drehflache eine regulare Flache ist.

(b) Berechnen Sie die erste und zweite Fundamentalformen von einer regularen Flache mit Me-ridiankurve

t 7→ (r(t), h(t)) .

Proof. We compute the partial derivatives of f :

∂f

∂t= (r cos(θ), r sin(θ), h)

∂f

∂θ= (−r sin(θ), r cos(θ), 0)

and thus the first fundamental form in these coordinates is

I(t,θ) =

(r2 + h2 0

0 r2

)(Note that r2 + h2 6= 0 because the curve is regular. When r 6= 0, the function f is animmersion, but the surface may have selfintersections, so it is not a regular surface in thesense defined in class, but it is for instance in Kuhnel...In order to assure that the solid ofrevolution is a regular surface we need to assume for instance that r > 0 and the curve issimply closed).

In order to compute the second fundamental form, we compute the Normal vector and thesecond partial derivatives of f with respect to the coordinates (t, θ) (recall that IIp(∂i, ∂j) =

〈N, ∂2f

∂uj∂ui〉):

N(t,θ) =

∂f

∂t× ∂f

∂θ∣∣∣∂f∂t× ∂f

∂θ

∣∣∣ =1√

r2 + h2

(−h cos(θ),−h sin(θ), r

)∂2f

∂t2= (r cos(θ), r sin(θ), h)

∂2f

∂t∂θ= (−r sin(θ), r cos(θ), 0)

∂2f

∂θ2= (−r cos(θ),−r sin(θ), 0)

and

II(t,θ) =1√

r2 + h2

(−rh+ rh 0

0 rh

).

(c) Berechnen Sie die erste und zweite Fundamentalformen des Torus (siehe Aufgabe A6.1).

Proof. In the case of the torus,

r(t) = a+ r cos(t) , h(t) = r sin t .

So in this case,

I(t,θ) =

(r2 00 (a+ r cos(t))2

)II(t,θ) =

(r 00 (a+ r cos(t)) cos(t)

).

A8.3 Es sei ϕ : S1 → S2 ein Diffeomorphismus zwischen zwei regulare Flachen. Zeigen Sie, dass S1genau dann orientierbar ist, wenn S2 orientierbar ist.

Proof. Suppose that S1 is orientable. This means that there exist a covering (Ui, Fi, Vi)i∈I of localparametrizations such that for all i, j ∈ I such that Vi ∩ Vj 6= 0, the differential of the transitionmatrix

F−1j ◦ Fi : F−1i (Vi ∩ Vj)→ F−1j (Vi ∩ Vj)

2

has positive determinant. The triples (Ui, ϕ ◦Fi, ϕ(Vi))i∈I is a covering of local parametrizationsof S2 and for i, j ∈ I such that ϕ(Vi) ∩ ϕ(Vj) 6= 0, the differential of the transition matrices

(ϕ ◦ Fj)−1 ◦ (ϕ ◦ Fi) = F−1j ◦ Fi

has positive determinant, and S2 is also orientable.

A8.4 Betrachten Sie den Zylinder

S = {(x, y, z) ∈ R3 : x2 + y2 = 1} .

und einen Punkt p = (x, y, z) ∈ S .

(a) Zeigen Sie, dass die Vektoren

b1 :=

001

b2 :=

−yx0

eine Basis von TpS bilden.

(b) Berechnen Sie die Matrixdarstellung der Weingarten Abbildung Wp in der basis {b1, b2} .

Proof. The normal vector of S at p = (x, y, z) is

N(x, y, z) = (x, y, 0) ,

The tangent space isTpS = {(v1, v2, v3) ∈ R3 : xv1 + yv2 = 0}.

It is not hard to see that the two vectors b1 and b2 are linearly independent, and they belong to thetangent space TpS .

The Weingarten map is just

Wp : TpS → TpS

(v1, v2, v3) 7→ (−v1,−v2, 0)

andWp(b1) = 0 , Wp(b2) = −b2

and the matrix presentation of Wp with respect to the basis {b1, b2} is(0 00 −1

)

3





Hausaufgaben

A9.1 Seien S eine regulare Flache und p ein Punkt in S. Zeigen Sie, dass die Hauptkrummungen κ1, κ2von S im punkt p die Extrema der Abbildung

Q : {v ∈ TpS : |v| = 1} → Rv 7→ IIp(v, v)

sind.

Proof. As Wp is a self adjoint operator on TpS, there exists an orthonormal basis {e1, e2} of TpSsuch that

Wpe1 = κ1e1 , Wpe2 = κ2e2

where κ1, κ2 are the eigenvalues of Wp (or in other words, the main curvatures of S at p). Let usassume from now on that κ1 ≥ κ2.Note that

IIp(e1, e2) = 0 .

We need to show that e1, e2 are a maximum and minimum point of Q on {v ∈ TpS : |v| = 1}.Indeed, for any v = xe1 + ye2 in V with x2 + y2 = 1 we have

IIp(v, v) = IIp(e1, e1)x2 + IIp(e2, e2)y

2 = κ1x2 + κ2y

2 ≥ κ2(x2 + y2) = κ2 .

Likewise,IIp(v, v) ≤ κ1

and we are done.

A9.2 Fur jede reelle Zahl λ ∈ [0, 1] geben man eine nach Bogenlange parametrisierte Kurve auf demZylinder

cλ : (−ε, ε)→ {(x, y, z) ∈ R3 : x2 + y2 = 1}

mit cλ(0) = (1, 0, 0) an, welche

II(1,0,0)(cλ(0), cλ(0)) = λ

gilt.

Proof. We know from previous HWAs, that

TpS = {(v1, v2, v3) ∈ R3 : xv1 + yv2 = 0}.

and

Wp : TpS → TpS

(v1, v2, v3) 7→ (v1, v2, 0)

We are looking for a curve cλ that satisfies all the following equations:cλ1(t)

2 + cλ2(t)2 = 1

cλ1(t)2 + cλ2(t)

2 + cλ3(t)2 = 1

cλ1(0) = 1, cλ2(0) = 0, cλ3(0) = 0

cλ1(0)2 + cλ2(0)

2 = λ

The curvecλ(t) = (cos(

√λt), sin(

√λt),√1− λt)

does the job.

Remark: Here we are doing calculations with respect to the orientation defined by the normalvector N = (−x,−y, 0).

A9.3 Rechen Sie die Gauß-Krummung und die mittlere Krummung des Torus aus (siehe Aufgabe A6.1).

Proof. We use the same convention of Aufgabe 8.2. The matrix that represents the Weingartenmap W(t,θ) is

II(t,θ)I−1(t,θ) =

(r 00 (a+ r cos(t)) cos(t)

)(r2 00 (a+ r cos(t))2

)−1

=

1/r 0

0cos(t)

a+ r cos(t)

The mean curvature and the Gauss-curvature are

a+ 2r cos(t)

2r(a+ r cos(t)),

cos(t)

r(a+ r cos(t)),

respectively.

A9.4 Sei I ⊂ R ein offenes Intervall, und sei c : I → R3 eine parametrisierte Raumkurve. Sei v :I → R3 eine glatte Abbildung mit v(t) 6= (0, 0, 0) fur alle t ∈ I. Sei J ⊂ R ein weiteres offenesIntervall. Wir setzen

F : I × J → R3

(t, s) 7→ c(t) + sv(t) .

Eine regulare Flache S ⊂ R3, die durch Parametrisierungen der Form F : I × J → R3 uberdecktwerden kann, heißt Regelflache .

(a) Zeigen Sie, dass das Rotations-hyperboloid

S = {(x, y, z) ∈ R3 : 1 + z2 = x2 + y2}

eine Regelflache bildet.

Proof. We start by taking a curve passing through the surface, for instance

c(t) = (cos(t), sin(t), 0) .

2

We would like to find the equations of lines that pass through c(t) and are totally containedin the hyperboloid. The equation of the line is of the form

l(s) = sv(t) + c(t),

for some v(t) ∈ R3 that depends on t .

The following equation should be satisfied

(cos(t) + sv1(t))2 + (sin(t) + sv2(t))

2 = 1 + (sv3(t))2

or equivalently

s2(v21(t) + v22(t)− v23(t)) + 2s(v1 cos(t) + v2 sin(t)) = 0 .

As the last equation must hold for every s, we must have

v21(t) + v22(t)− v23(t) = 0 , v1 cos(t) + v2 sin(t) = 0

We can take for instancev(t) = (− sin(t), cos(t), 1) .

If we define

F : [0, 2π)× R→ R3

(t, s) 7→ (cos(t)− s sin(t), sin(t) + s cos(t), s),

it is not hard to see that∂F

∂s,∂F

∂tare linearly independent (see calculations below), the image

of F is the full cylinder, and F is a homeomorphism onto the image with inverse given by

(x, y, z) 7→ (t, z)

with t given by the polar coordinate (or argument) of the complex numberx+ iy

1 + iz.

(b) Berechnen Sie die Gauß-Krummung des Rotationshyperboloid .

Proof. We compute the first and second fundamental form with respect to the parametriza-tion given by

F (t, s) = (cos(t)− s sin(t), sin(t) + s cos(t), s).

We compute the first and second fundamental forms and do the usual computations:

∂

∂t=

− sin(t)− s cos(t)cos(t)− s sin(t)

0

,∂

∂s=

− sin(t)cos(t)1

.

I(t,s) =

(1 + s2 1

1 2

)

N =∂t × ∂s|∂t × ∂s|

=1√

1 + 2s2

cos(t)− s sin(t)sin(t) + s cos(t)

−s

∂2

(∂t)2=

− cos(t) + s sin(t)− sin(t)− s cos(t)

0

,∂2

∂t∂s=

∂2

∂s∂t=

− cos(t)− sin(t)

0

,∂2

(∂s)2=

000

.

3

II(t,s) =1√

1 + 2s2

(−(1 + s2) −1−1 0

)The Gauss curvature is just equal to

det(II)

det(I)=

−1(1 + 2s2)2

.

(c) Zeigen Sie: Falls F (t, s) = c(t) + sv(t) eine regulare Parametrisierung einer Regelflacheist, dann gilt:

K(F (s, t)) < 0

genau dann wenn, c(t), v(t) und v(t) linear unabhangig sind.

Proof. If we use the formula

K(F (s, t)) =det(II(s,t))

det(I(s,t)),

we conclude that K(F (s, t)) < 0 if and only if

〈 ∂2F

∂s∂t,∂F

∂s× ∂F

∂t〉 = 〈(c+ sv)× v, v〉,

is nonzero as∂2F

(∂s)2= 0 and det(I(s,t)) > 0 .

But the above expression equals to

det(c+ sv, v, v) = det(c, v, v),

which is non zero if and only if c, v, v are linearly independent.

A9.5 (a) Beweisen Sie den Satz von Rodriguez:

Sei S ⊂ R3 eine orientierbare regulare Flache, sei N : S → S2 die Gauß-Abbildung. Seic : I → S eine nach Bogenlange parametrisierte Kurve. Dann ist c Krummungslinie auf Sgenau, wenn es eine Funktion λ : I → R gibt mit

d

dtN(c(t)) = λ(t) · c(t), t ∈ I .

In diesem Fall ist −λ(t) die entsprechende Hauptkrummung.

Proof. The statement follows from the chain rule and the definition of line of curvature(Krummungslinie):

d

dtN(c(t)) =

dN

dt

∣∣∣c(t)

(c(t))

= −Wc(t)(c(t)) = −λ(t) · c(t) .

4

(b) Beweisen Sie, dass auf der Sattelflache

S = {(x, y, z) ∈ R3 : z = y2 − x2},

eine nach Bogenlange parametrisierte Kurve c = (c1, c2, c3) : I → S mit c(0) = (0, 0, 0)und c1 ≡ 0 (oder c2 ≡ 0) eine Krummungslinie ist.

Proof. Let us consider the parametrization

F (u1, u2) = (u1, u2, u22 − u21) .

The tangent space at the point F (u1, u2) is spanned by

∂F

∂u1=

10−2u1

,∂F

∂u2=

01

2u2

.

We recall that

IIp

( ∂F∂ui

,∂F

∂uj

)=⟨ ∂2F

∂ui∂uj, N(F (u1, u2))

⟩In these coordinates, we have that

∂2F

(∂u1)2=

00−2

,∂2F

(∂u2)2=

002

,∂2F

∂u1∂u2=

∂2F

∂u2∂u1=

000

and

N =∂u1 × ∂u2|∂u1 × ∂u2 |

=1√

4u21 + 4u22 + 1

2u1−2u21

With respect to the basis ∂u1 , ∂u2 , the first and second fundamental form are represented bythe matrices (

1 + 4u21 −4u1u2−4u1u2 1 + 4u22

)and

1√4u21 + 4u22 + 1

(−2 00 2

),

respectively. Note that the matrix that represents W with respect to this basis is

1√4u21 + 4u22 + 1

(−2 00 2

)(1 + 4u21 −4u1u2−4u1u2 1 + 4u22

)−1

.

When u1 = 0, this matrix is just

1

(1 + 4u22)3/2

(−2(1 + 4u22) 0

0 2

).

For a curve c = F (c1, c2) : I → S with c1 ≡ 0, we have that

d

dtN(c(t)) = −Wc(t)(c(t)) = −Wc(t)

( ∂

∂u2

∣∣∣c(t)· c2(t)

)= − 2

(1 + 4c22(t))3/2

0c2(t)

2c2(t)c2(t)

,

and we are done as the curve satisfies the Rodriguez equation. Similarly, we obtain the resultwhen c2 ≡ 0 .

5





Hausaufgaben

A10.1 (a) Sei U ⊂ R eine offene Menge und f : U → R eine glatte Abbildung. Rechnen Sie dieGauß-Krummung und die mittlere Krummung des Funktionsgraphen der Abbildung f aus.

Proof. Let us consider the following map

F : R2 → R3

(u1, u2) 7→ (u1, u2, f(u1, u2)) .

We compute the first and second fundamental forms and do the usual computations:

∂F

∂u1=

10∂f

∂u1

,∂F

∂u2=

01∂f

∂u2

I(u1,u2) =

1 +( ∂f∂u1

)2 ∂f

∂u1

∂f

∂u2∂f

∂u1

∂f

∂u21 +

( ∂f∂u2

)2

N =∂Fu1 × ∂Fu2|∂Fu1 × ∂Fu2 |

=1√

1 +( ∂f∂u1

)2+( ∂f∂u2

)2− ∂f

∂u1

− ∂f

∂u21

∂2F

(∂u1)2=

00∂2f

(∂u1)2

,∂2F

∂u1∂u2=

∂2F

∂u2∂u1=

00∂2f

∂u1∂u2

,∂2F

(∂u2)2=

00∂2f

(∂u2)2

II(u1,u2) =1√

1 +( ∂f∂u1

)2+( ∂f∂u2

)2

∂2f

(∂u1)2∂2f

∂u1∂u2∂2f

∂u1∂u2

∂2f

(∂u2)2

The Gauss curvature is equal to

det(II)

det(I)=fu1u1fu2u2 − f2u1u2(1 + f2u1 + f2u2)2

.

and the mean curvature is equal to

(1 + f2u1)fu2u2 + (1 + f2u2)fu1u1 − 2fu1fu2fu1u22(1 + f2u1 + f2u2)3/2

.

(b) Berechnen Sie die Gauß-Krummung des Funtktionsgraphen der folgenden Abbildungenf, g : R2 → R

f(u1, u2) = u31 − 3u22u1, g(u1, u2) = (u21 + u22)2 .

Proof. For the first function

fu1 = 3u21 − 3u22 , fu2 = −6u1u2 ,

fu1u1 = 6u1 , fu1u2 = −6u2 , fu2u2 = −6u1 .

The corresponding Gauss curvature equals

6u1(−6u1)− (−6u2)2

(1 + (3u21 − 3u22)2 + (−6u1u2)2)2

=−36(u21 + u22)

(1 + (3u1 + 3u2)2)2.

For the second function

gu1 = 4(u31 + u22u1) , gu2 = 4(u32 + u21u2) ,

gu1u1 = 4(3u21 + u22) , gu1u2 = 8u1u2 , gu2u2 = 4(3u22 + u21) .

The corresponding Gauss curvature equals

4(3u21 + u22)4(3u22 + u21)− (8u1u2)2

(1 + (4(u31 + u22u1))2 + (4(u32 + u21u2)

2))2=

48(u21 + u22)2

(16(u21 + u22)3 + 1)2

.

(20)

A10.2 Berechnen Sie das Integral ∫ ∞−∞

∫ ∞−∞

1

(1 + (u21 + u22))2du1 du2 .

Hinweis: Betrachten Sie den Diffeomorphismus

Φ : (0,∞)× (0, 2π)→ R2\{x > 0}(r, θ) 7→ (r cos(θ), r sin(θ)) .

Proof. From the transformation theorem, the integral equals∫ ∞0

∫ 2π

0

r

(1 + r2)2dθ dr = 2π

∫ ∞0

r

(1 + r2)2dr

= −π 1

1 + r2

∣∣∣∞0

= π .

(10)

A10.3 Berechnen Sie den Flacheninhalt des Ellipsoids

{(x, y, z) ∈ R3 :x2

a2+ y2 + z2 = 1} , a > 0 .

Proof. We consider the following parametrization of the Ellipsoid that covers it up to a set ofmeasure zero:

F (θ, ϕ) = (a cos(θ), sin(θ) cos(ϕ), sin(θ) sin(ϕ))

where0 ≤ θ ≤ π , 0 ≤ ϕ ≤ 2π .

2

We compute the partial derivatives and then the first fundamental form:

∂F

∂θ=

−a sin(θ)cos(θ) cos(ϕ)cos(θ) sin(ϕ)

,∂F

∂ϕ=

0− sin(θ) sin(ϕ)sin(θ) cos(ϕ)

I(θ,ϕ) =

(a2 sin2(θ) + cos2(θ) 0

0 sin2(θ)

)and √

det(Iθ,ϕ) = | sin(θ)|√a2 sin2(θ) + cos2(θ) .

The surface area equals∫ π

0

∫ 2π

0

√det(Iθ,ϕ)dϕ dθ =

∫ π

0

∫ 2π

0| sin(θ)|

√a2 sin2(θ) + cos2(θ)dϕ dθ

= 2π

∫ π

0sin(θ)

√a2 sin2(θ) + cos2(θ)dθ

= 2π

∫ 1

−1

√a2 + (1− a2)u2du

= 2πa

∫ 1

−1

√1 + eu2du

where e =1

a2−1 . This integral has different values depending on whether e > 0, e < 0 or e = 0.

In the first case, the integral equals

2πa(√

e+ 1 +sinh−1(

√e)√

e

)= 2πa

(1

a+

sinh−1(√ 1

a2− 1)

√1

a2− 1

).

In the second case, the integral equals

2πa(√

e+ 1 +sin−1(

√−e)√

−e

)= 2πa

(1

a+

sin−1(√

1− 1

a2

)√

1− 1

a2

)

When e = 0, the integal equals to 4π that correponds to the surface area of a sphere of radius1. (10)

A10.4 Sei S eine regulare Fache und sei f : S → R eine Funktion. Wir sagen, dass f integrierbar ist,falls folgendes erfullt ist: Es existiert eine Uberdeckung von S durch Karten {(Ui, Fi, Vi)}ki=1,sodass die Funktionen

f1 : = χV1 · f,f2 : = χV2\V1 · f,

...

fk : = χVk\(V1∪...∪Vk−1) · f

3

integrierbar sind. Ist dies der Fall, so ist das Integral von f uber S definiert als∫SfdA :=

k∑i=1

∫SfidA .

Beweisen Sie, dass diese Definition nicht von der Wahl der Uberdeckung anhangt.

Proof. Let (U,F, V ) be a local parametrization of S and f a function such that f |S\V ≡ 0 . Inthis Exercise we denote∫

SfdA =

∫Uf ◦ F (u1, u2)

√det(gij(u1, u2))du1 du2

by∫U F

∗(fdA) . This is a standard convention but not used yet on the lectures...

Let {(Ui, Fi, Vi)}ki=1 and {(Uj , Fj , Vj)}lj=1 be two different parametrizations of S.

We definehi,j := χ(Vi\(V1∪...∪Vi−1))∩(Vj\(V1∪...∪Vj−1))

· f

Note that for every 1 ≤ i ≤ k, 1 ≤ j ≤ l

fi =l∑

j=1

hi,j , fj =k∑i=1

hi,j .

Proposition 3.2 of the Skript implies that∫Ui

F ∗i (hijdA) =

∫Uj

F ∗j (hijdA)

Now with all these identities in mind, we check that the definition of the integral is independentof the covering:

k∑i=1

∫SfidA =

k∑i=1

∫Ui

F ∗i (fidA) =k∑i=1

∫Ui

F ∗i (l∑

j=1

hi,jdA)

=k∑i=1

l∑j=1

∫Ui

F ∗i (hi,jdA) =l∑

j=1

k∑i=1

∫Ui

F ∗i (hi,jdA)

=l∑

j=1

k∑i=1

∫Uj

F ∗j (hi,jdA) =l∑

j=1

∫Uj

F ∗j (fjdA)

=l∑

j=1

∫SfjdA .

(10)

4



Aufgabenblatt 11Abgabe: Diese Ubungen mussen bis spatestens 11:55 Uhr Dienstag 14.01.2019 in den Briefkasten im


Hausaufgaben

A11.1 Sei U ⊂ R2 eine offene Menge und f : U → R eine glatte Abbildung. Rechnen Sie den Flachen-inhalt des Funktionsgraphen der Abbildung f aus.

Proof. Let

F : U → R3

(u, v) 7→

uv

f(u, v)

From the previous HWA, the first fundamental form in terms of the coordinates (u, v) equals

I(u,v) =

(1 + f2u fufvfufv 1 + f2v

).

The surface area in terms of the coordinates (u, v) equals∫U

√det(I(u,v))dudv =

∫U

√1 + f2u + f2v dudv =

∫U

√1 +∇f2

(10)

A11.2 Es seien S1, S2 und S3 regulare Flachen. Beweise:

(a) Ist ϕ : S1 → S2 eine Isometrie, so ist ϕ−1 : S2 → S1 auch eine Isometrie.

(b) Sind ϕ : S1 → S2 und ψ : S2 → S3 eine Isometrie, so ist ψ ◦ ϕ : S1 → S3.

Proof. We recall the definition of isometry: a diffeomorphism Φ : (S, g)→ (S, g) is an isometryif for all p ∈ S and every v1, v2 ∈ TpS

gp(v1, v2) = gϕ(p)(dϕp(v1), dϕp(v2)) =: (ϕ∗g)p(v1, v2)

When ϕ : (S, g)→ (S, g) is the identity diffeomorphism, we clearly have that

ϕ∗(g) = g .

On the other hand, it follows from the chain rule that for two isometries ϕ : (S1, g1) → (S2, g2)and ψ : (S2, g2)→ (S3, g3), the identity

(ψ ◦ ϕ)∗(g3) = ψ∗ ◦ ϕ∗(g3) = ψ∗g2 = g1

holds, and the second part of the exercise is done.

The chain rule also implies that when ϕ = ψ−1,

g2 = (IdS2)∗g2 = (ϕ−1 ◦ ϕ)∗g2

= (ϕ−1)∗(ϕ∗g2) = (ϕ−1)∗(g1),

and the first part of the exercise is done.

(5)

A11.3 Gegeben sei die Menge

K := {(x, y, z) ∈ R3 : x2 + y2 =1

3z2 , z > 0} .

Betrachten Sie weiterhin die differenzierbare Abbildung R2\{0} → K gegeben durch

f : R2\{0} → K(xy

)7→ 1

2√x2 + y2

x2 − y22xy√

3(x2 + y2)

zwischen den regularen Flachen R2\{0} und K (muss nicht bewiesen werden). Zeigen Sie: DieAbbildung f ist eine lokale Isometrie.

Proof. In standard polar coordinates (r, θ), the first fundamental form equals

I(r,θ) =

(1 00 r2

).

In terms of these coordiantes, the map f is written as

f(r, θ) =1

2

r cos(2θ)r sin(2θ)√

3r

The exercise follows from the identity between matrices

I(r,θ) =

〈∂f∂r , ∂f∂r 〉〈∂f∂r , ∂f∂θ 〉〈∂f∂r,∂f

∂θ〉〈∂f

∂θ,∂f

∂θ〉

,

which can be seen easily from the explicit calculation of partial derivatives:

∂f

∂r=

1

2

cos(2θ)sin(2θ)√

3

,∂f

∂θ=

1

2

−2r sin(2θ)2r cos(2θ)

0

(10)

A11.4 Sei S der ZylinderS = {(x, y, z) ∈ R3 : x2 + y2 = 1} .

Fur (x, y, z) ∈ S betrachte

Φ : S → Sxyz

7→cos(θ0) − sin(θ0) 0

sin(θ0) cos(θ0) 00 0 1

xyz

.

Beweisen Sie, dass die Abbildung Φ eine Isometrie ist.

2

Proof. Recall that for a point p = (x, y, z) ∈ S,

TpS = {(v1, v2, v3) ∈ R3 | xv1 + yv2 = 0},

and

dΦp : TpS → TpSv1v2v3

7→cos(θ0) − sin(θ0) 0

sin(θ0) cos(θ0) 00 0 1

v1v2v3

.

The exercise follows as it is not hard to check that

〈dΦp(v1, v2, v3), dΦp(w1, w2, w3)〉 = 〈(v1, v2, v3), (w1, w2, w3)〉 .

Remark: Indeed, the map Φ extends to a isometry on the whole R3 .

(10)

A11.5 Es sei

t 7→ (r(t), h(t))

eine nach Bogenlange parametrisierte Kurve mit r(t) > 0 fur alle t ∈ I und

f : I × [0, 2π]→ R3

(t, θ) 7→

r(t) cos(θ)h(t)

r(t) sin(θ)

.

Nehmen Sie an, dass das Bild von f eine regulare Rotationsflache definiert. Zeigen Sie, dass dieChristoffelsymbole der Rotationsflache gegeben sind durch Γ1

11 = Γ211 = Γ1

12 = Γ222 = 0 und

Γ122 = −r(t) · r′(t) , Γ2

12 =r′(t)

r(t).

Proof. The Christoffel symbols are defined by the equation

∂2f

∂ui∂uj= Γ1

ij

∂f

∂u1+ Γ2

ij

∂f

∂u2+ hijN

(for us u1 = t, u2 = θ). As ft, fθ, N are all orthogonal, we have the formulae

Γkij =

〈 ∂2f

∂ui∂uj,∂f

∂uk〉

〈 ∂f∂uk

,∂f

∂uk〉

We compute now the partial derivatives

∂f

∂t=

r′(t) cos(θ)h′(t)

r′(t) sin(θ)

,∂f

∂θ=

−r(t) sin(θ)0

r(t) cos(θ)

∂2f

∂t2=

r′′(t) cos(θ)h′′(t)

r′′(t) sin(θ)

,∂2f

∂θ∂t=

∂2f

∂t∂θ=

−r′(t) sin(θ)0

r′(t) cos(θ)

,∂2f

∂θ2=

−r(t) cos(θ)0

−r(t) sin(θ)

3

We obtain

Γ111 =

d

dt

1

2(r′(t)2 + h′(t)2)

r′(t)2 + h′(t)2= 0 , Γ1

12 = Γ121 = 0 , Γ1

22 =−r(t) · r′(t)r′(t)2 + h′(t)2

= −r(t) · r′(t) ,

Γ211 = 0 , Γ2

12 = Γ221 =

−r(t) · r′(t)r(t)2

= −r′(t)

r(t), Γ2

22 = 0 .

(15)

4





Die Matrixdarstellung der ersten Fundamentalform, der zweiten Fundamentalform und der Weingar-ten Abbildung bezuglich einer regulare Parametrisierung (U,F, V ) werden mit (gij), (hij) und (wj

i )bezeichnen.

Hausaufgaben

A12.1 Zeigen Sie, dass fur die Gauß-Krummung

K =h12w

21 − h11w

22

g11

gilt.

Proof. The matrices (gij), (hij) and (wji ) are related through the equality(

w11 w2

1

w12 w2

2

)(g11 g12g21 g22

)=

(h11 h12h21 h22

)If we multiply on the left by the adjoint matrix of (wj

i ), we obtain the equality

K

(g11 g12g21 g22

)=

(w22 −w2

1

−w12 w1

1

)(h11 h12h21 h22

)Comparing term by term, we get the equality

K =h11w

22 − h12w

21

g11

Remark: The signs are reversed at the statement of the exercise!

A12.2 Sei (U,F, V ) eine orthogonale Parametrisierung mit Koordinaten (u, v), d.h. g12 = g21 = 0 .Zeigen Sie, dass fur die Gauß-Krummung

K = − 1

2√g11 · g22

( ∂

∂v

( ∂vg11√g11 · g22

)+

∂

∂u

( ∂ug22√g11 · g22

))gilt.

Proof. We write

(gij) =

(E FF G

)In class the following equality was proven:

K =1

E((Γ2

11)v − (Γ212)u + Γ1

11Γ212 − Γ1

12Γ211 + Γ2

11Γ222 − (Γ2

12)2)

As F = g12 = g21 = 0, we use the following identities of Christoffel symbols:

Γ111 =

Eu

2E, Γ2

11 =−Ev

2G

Γ112 =

Ev

2E, Γ2

12 =Gu

2G

Γ122 =

−Gu

2E, Γ2

22 =Gv

2G

We compute now the right hand side of the equation appearing at the statement of the exercise:

− 1

2√EG

( ∂

∂v

( Ev√EG

)+

∂

∂u

( Gu√EG

))=

− Evv

2EG+

Ev(EvG− EGv)

4E2G2− Guu

2EG+

Gu(EuG− EGu)

4E2G2

=EuGu

4E2G− (GEvv − EvGv)

2EG2− EvGv

4EG2+

EvEv

4E2G− (GGuu −GuGu)

2EG2− GuGu

4EG2

=EuGu

4E2G− 1

E

(Ev

2G

)v− EvGv

4EG2+

EvEv

4E2G− 1

E

(Gu

2G

)u− GuGu

4EG2

=1

E

(Γ111Γ

212 + (Γ2

11)v + Γ211Γ

222 − Γ1

12Γ211 − (Γ2

12)u − (Γ212)

2),

and we are done!

A12.3 Prufe nach, dass die Flachen

F (u, v) = (u cos(v), u sin(v), log(u)),

G(u, v) = (u cos(v), u sin(v), v)

dieselbe Gaußsche Krummung in den Punkten F (u, v) and G(u, v) haben, die Abbildung G◦F−1aber keine Isometrie ist. Das zeigt, daßdie ”Umkehrung “des Theorem Egregium nicht richtig ist.

Proof. First we compute the first fundamental forms of both parametrizations:

IF =

(1 +

1

u20

0 u2

)

IG =

(1 00 u2 + 1

)If we apply the previous exercise, we see that both parametrization have Gauss curvature equal to

− 1

2√u2 + 1

∂

∂u

( 2u√u2 + 1

).

Let (S, g1) and (T, g2) be the surfaces defined by F and G, respectively. Here g1 and g2 are justthe restrictions of the standard metric of R3 on S and T, respectively.

If G ◦ F−1 : (S, g1)→ (T, g2) were an isometry, then

(G ◦ F−1)∗(g2) = (F−1)∗ ◦G∗(g2) = g1

orG∗(g2) = F ∗(g1)

2

as metrics in R2, but they are not equal as the first fundamental forms IF and IG are not the same,so G ◦ F−1 is not an isometry. 1

A12.4 Sei c : [a, b]→ R2 eine parametrisierte Kurve. Beweisen Sie, dass fur die Lange von c

L[c] ≥ |c(b)− c(a)|

gilt.

Proof. We certainly can use the triangular inequality of integrals (which follows from Cauchy-Schwarz and the standard triangular inequality applied to double integrals):2

|c(b)−c(a)| = |(c1(b)−c1(a), c2(b)−c1(a))| =∣∣∣∫ b

a(c′1(t), c

′2(t))dt

∣∣∣ ≤ ∫ b

a|(c′1(t), c′2(t))|dt = L[c] .

1For a parametrization (U,F, V ) of a surface S with coordinates (u1, u2),

F ∗((dx1 ⊗ dx1 + dx2 ⊗ dx2 + dx3 ⊗ dx3)|S) = dF1 ⊗ dF1 + dF2 ⊗ dF2 + dF3 ⊗ dF3∑k,i,j

∂Fk

∂ui

∂Fk

∂ujdui ⊗ duj =

∑i,j

〈 ∂F∂ui

,∂F

∂uj〉dui ⊗ duj =

∑i,j

gijdui ⊗ duj

where gij = 〈 ∂F∂ui

,∂F

∂uj〉 are the matrix components of the first fundamental form IF of F .

2 ∣∣∣∫ b

a

(f1(t), f2(t))dt∣∣∣2 =

⟨∫ b

a

(f1(t), f2(t))dt,

∫ b

a

(f1(t), f2(t))dt⟩=

∫ b

a

∫ b

a

〈(f1(t), f2(t)), (f1(s), f2(s))〉dt ds

≤∫ b

a

∫ b

a

|(f1(t), f2(t))| · |(f1(s), f2(s))|dt ds =(∫ b

a

|(f1(t), f2(t))|dt)2

3

DIFFERENTIALGEOMTRIE Aufgabenblatt 1 › ~pako › EDG2019 › WS2019-EDG-Solutions.pdf ·...

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