Contentsmath.xmu.edu.cn/group/ga/dg_files/surface_notes.pdf · example, Chapter 1 to 3 in [2] discuss the classical theory of curves and surfaces in the context of Riemannian geometry

NOTES ON DIFFERENTIAL GEOMETRY

BO YANG

Abstract. From Fall 2018 the author taught an undergraduate course on differential geom-etry, with Klingenberg [9] as the textbook. Originally these expository notes are written to

complement [9], and now they can be read more or less independently without [9].

During the preparation of this notes, we find [9], [4], [5], [3], [2], and [10] helpful. Forexample, Chapter 1 to 3 in [2] discuss the classical theory of curves and surfaces in the

context of Riemannian geometry. The role of integration in the study of curvature problem

of surfaces is emphasized in [10]. There are interesting pictures of curves and surfaces in [1],[6], [7], and [8]. We also include more references in various footnotes.

These notes are still work in progress. I would like to thank students in class for catching

typos and suggesting corrections.

Contents

1. Introduction 31.1. The organization of these notes 31.2. Notations 31.3. A review on implicit function theorems in Euclidean spaces 41.4. A review on basic results on ODEs 52. Curves in Rn: local theory 82.1. Frenet Equations 82.2. Plane curves 82.3. Space curves 92.4. Fundamental theorem on curves with prescribed curvature functions 103. Plane and space curves: some global results 123.1. From parametrized curves to global definitions of curves 123.2. Rotation number and the Umlaufsatz 133.3. Isoperimetric inequalities in the plane 173.4. Euler’s elastica 264. Surfaces in R3: local theory 294.1. The first and second fundamental forms 294.2. Fundamental theorem on surfaces with prescribed fundamental forms 314.3. Examples of surfaces 375. Surfaces in R3: from local to global 485.1. Smooth 2-manifolds 485.2. The geodesic curvature of a curve on a surface 535.3. The Gauss-Bonnet Theorem 575.4. Riemannian 2-manifolds 636. Global theory of surfaces: selected topics 796.1. Ovaloids and their noncompact analogues 796.2. Existence of surfaces with prescribed curvature 796.3. Isometric embeddings into R3 and the sign of Gauss curvature 826.4. Prescribing Gauss curvature 836.5. Conformal deformation to constant curvature 836.6. More on minimal surfaces and CMC surfaces 83

Date: Version of 05/25/2020.

1

2 BO YANG

7. Global theory of curves continued 867.1. The curve shortening flow: a short introduction 867.2. Further reading 88References 89

NOTES ON DIFFERENTIAL GEOMETRY 3

1. Introduction

1.1. The organization of these notes.Materials we do not cover and might be added in the future include

(i) Proof of Brunn-Minkowski inequality when n = 2.(ii) The construction of various coordinates systems, include geodesic orthogonal (Fermi)

geodesic normal, geodesic polar, and isothermal coordinates.(iii) More on compact Riemannian 2-manifolds, including divergence theorem, Laplacian

operators, and basic spectral geometry.(iv) More on complete Riemannian 2-manifolds, including Cohn-Vossen inequality and har-

monic functions, Li-Tam on finite total curvature.(v) Second variation of length and area.

Materials which we deliberately avoid (however they provide useful alternative viewpoints)include.

(i) Differential forms, and structural equations which express Gauss and Codazzi equationsin differential forms, and the solvability of Pfaff system in terms of differential forms.

(ii) Degree of a mapping, index of a vector field, and its connection to Gauss-Bonnettheorem.

1.2. Notations. We set up some notations which will be used throughout these notes.

(i) We use column vectors unless otherwise stated, let u, v be two vectors in Rn, assumeu = (u1, · · · , un)T , the Euclidean inner product will be u · v = uT v. The norm of uwill be |u| = √u · u.

(ii) On Euclidean space Rn, there is a natural distance function d(x, y) = |x− y|, we saya map F : Rn → Rn is an isometry of Rn if d(F (x), F (y)) = d(x, y) for any x, y ∈ Rn.As an important fact, any isometry F of Rn may be written of the form

F (x) = Ax+ b,

whereA is an orthogonal matrix and b is some vector in Rn. We say F is an orientation-preserving isometry if detA = 1, it is orientation-reversing if detA = −1.1

(iii) Let TpRn be the tangent space of Rn at p, it is a n-dim vector space whose elementsare (p, x) ∈ p×Rn, it can be identified as Rn. A natural choice of as a basis of TpRn isei = (0, · · · 0, 1, 0, · · · , 0)T 1≤i≤n, here 1 is the i-th component. For the convenience

of notations we also identify ∂∂xi

= ei.

(iv) With the Euclidean metric topology induced by d(x, y), let U be an open subset of Rn,we define a map F : U → Rm by F = (F1, · · · , Fm) is continuous or differentiable atx0 ∈ U . In particular, let dFx0

be the derivative of F at x0, it can be viewed as linearmapping from Rn to Rm, whose matrix representation under the standard basis is thefollowing Jacobian matrix:

∂F1

∂x1· · · ∂F1

∂xi· · · ∂F1

∂xn...

......

......

∂Fm∂x1

· · · ∂Fm∂xi

· · · ∂Fm∂xn

.Using the identification of tangent space Tx0

Rn and Rn, we call dFx0: Tx0

Rn →TF (x0)Rm the tangent map of F at x0. Let L(Rn,Rm) be the set of all linearmaps from Rn to Rn, then we will define the derivative (differential) of F asdF : U → L(Rn,Rm) = Rmn since dF (x0) ∈ L(Rn,Rm) for any x0 ∈ U . From this wemay define F : U → Rm to be a C1 map if it is differentiable, and dF is continuous on

1The crucial step to verify this fact is to prove any isometry on Rn is linear modulo a translation (i.e. affine).Though we do not need it, the stronger Mazur-Ulam theorem (1932) states that every bijective isometry betweenreal (infinite-dimensional) normed spaces is also affine.

4 BO YANG

U . Similarly, we may define higher order derivatives of F . and we call F : U → Rm isa Ck map if it is k-times differentiable, and the k-th derivative, dkF , viewed as a mapfrom U to Rc(k) for some c(k) (which depends on k,m, and n) is continuous on U .

(v) The Kroneckers symbol δij or δji are defined to be 1 if i = j, and 0 otherwise.(vi) The Einstein convention is used for convenience, for example, Aibi :=

∑iA

ibi. The

inverse matrix A−1 of A = aij (or A = aji) is denoted by A−1 = aij or (or

A−1 = aji), hence we have aijajk = δik or ajiakj = δki .

1.3. A review on implicit function theorems in Euclidean spaces. In this subsection wereview the implicit function theorem in Euclidean spaces, and explain some details on Section0.5 of [9].

Theorem 1.3.1 (Implicit function theorem). Let G : Rm+n → Rn be a C1 map

G(x, y) = (g1(x, y), · · · , gn(x, y)) x ∈ Rm and y ∈ Rn.

Pick some c = (c1, · · · , cn) ∈ Rn, and consider (x0, y0) on the level surface G−1(c). Assume thematrix

∂G

∂y= [

∂gi∂yj

]n×n

is invertible at (x0, y0), then there exists some δ > 0 such that all points in G−1(c)∩Bδ((x0, y0))can be expressed as (x, F (x)) = (x, f1(x), · · · , fn(x)) for some function F : U → Rn where U issome open set in Rm obtained by projecting G−1(c)∩Bδ((x0, y0)) to the x-direction. Moreover,[∂G

∂x+∂G

∂y

∂F

∂x

]ij

=∂gi∂xj

+

n∑p=1

∂gi∂yp

∂fp∂xj

= 0, 1 ≤ i ≤ n, 1 ≤ j ≤ m.

Theorem 1.3.2 (Inverse function theorem). Let F : U → Rn be a C1 map where U ⊂ Rnis open. Assume

∂F

∂x= [

∂fi∂xj

]n×n

is invertible at some point x0 ∈ Rn, then there exists some δ such that F |Bδ(x0)Bδ(x0) →F (Bδ(x0)) has an inverse map F−1. Moreover, F−1 is also C1, and [dF−1

F (x)] = [dFx]−1.

Note that Theorem 1.3.2 is a special case of Theorem 1.3.1 if we set G = F (x)− y and c = 0.Now we present some variants of Theorem 1.3.1 and Theorem 1.3.2, which appear as Theorem

0.5.2 on p.6 from [9].

Corollary 1.3.3. Let U be an open set containing 0 ∈ Rn. Suppose F : U → Rm is a C1 mapwith F (0) = 0.

(i) If dF0 : Rn → Rm is injective, then there exists a C1 diffeomorphism (i.e. it is bijectiveonto its image, C1 and admits a C1 inverse) G : V → G(V ) ⊂ Rm where V is on anopen set containing 0 ∈ Rm such that

(1) G F (x1, · · · , xn) = (x1, · · · , xn, 0, · · · , 0).

holds on an open set containing 0 ∈ Rn.(ii) If dF0 : Rn → Rm is surjective, then there exists a C1 diffeomorphism G : V → G(V ) ⊂

Rn where V is on an open set containing 0 ∈ Rn such that

(2) F G(y1, · · · , ym, ym+1, · · · , yn) = (y1, · · · , ym).

holds on an open set containing 0 ∈ Rm.

Proof of Corollary 1.3.3. Let us write F = (f1(x1, · · · , xn), · · · , fm(x1, · · · , xn)).

(i) Since dF0 : Rn → Rm is injective, we have n ≤ m and the rank of the matrix [ ∂fi∂xj]n×m

is n. Therefore we may assume a n× n matrix [ ∂fi∂xj] where 1 ≤ i, j ≤ n is invertible, it follows


from Theorem 1.3.2 that we may find a C1 inverse map H : V1 → Rn where 0 ∈ V1 ⊂ Rn isopen. Let us write H(y1, · · · , yn) = (h1(y), · · · , hn(y)) and

H (f1, · · · , fn)(x1, · · · , xn) = (x1, · · · , xn).

Now we define a new map G : V1 × Rm−n → Rm where 0 ∈ V ⊂ Rm is open.

g1(y1, · · · , ym) = h1(y1, · · · , yn)...

gn(y1, · · · , ym) = hn(y1, · · · , yn)

gn+1(y1, · · · , ym) = yn+1 − fn+1(H(y1, · · · , yn))...

gm(y1, · · · , ym) = ym − fm(H(y1, · · · , yn))

(3)

One may check that the Jacobian matrix of dG0 is of m × m and invertible. It follows fromTheorem 1.3.2 again that G is a C1 diffeomorphism on some open set 0 ∈ V ⊂ V1 × Rn−m. Itis direct to see that (1) is true.

(ii) If dF0 : Rn → Rm is surjective, we have m ≤ n and the rank of the matrix [ ∂fi∂xj]n×m is

m. Therefore we may assume a m×m matrix [ ∂fi∂xj] where 1 ≤ i, j ≤ m is invertible.

Let G(x, y) = F (x) − y where x ∈ Rn, y ∈ Rm and c = 0 in the notation of Theorem 1.3.1.We may conclude that there exists a C1 map H : V1 → Rm where V1 is an open set in Rn withxi = hi(y1, · · · , ym, xm+1, · · · , xn) with 1 ≤ i ≤ m.

Now we may define the desired function G = (g1, · · · , gn) as

g1(y1, · · · , ym, xm+1, · · · , xn) = h1(y1, · · · , ym, xm+1, · · · , xn)...

gm(y1, · · · , ym, xm+1, · · · , xn) = hm(y1, · · · , ym, xm+1, · · · , xn)

gm+1(y1, · · · , ym, xm+1, · · · , xn) = xm+1

...

gn(y1, · · · , ym, xm+1, · · · , xn) = xn

(4)

Similarly, one may check that the Jacobian matrix of dG0 is of n×n and invertible. Theorem1.3.2 implies that G is a C1 diffeomorphism on some open set in Rn containing 0. It is alsodirect to see that (2) is true.

Remark 1.3.4. Roughly speaking, Corollary 1.3.3 states that ‘modulo a local diffeomorphism’a C1 map with an injective or surjective tangent map at some point is locally injective (asan inclusion map) or surjective (as a projection map). Corollary 1.3.3 can be generalized todifferentiable manifolds and has important consequences, see for example Lemma 5.1.3.

1.4. A review on basic results on ODEs. We recall some baisc existence, uniqueness, andextension theorems for solutions to ordinary differential equations (ODEs).2

We consider the following initial value problem (IVP):dx(t)dt = F (t, x(t));

x(t0) = x0.(5)

The basic assumption is that F : E → Rn is continuous, where E is an set in Rn+1 containing(t0, x0). If E is not open, we mean F is continuous on an open set containing E. We call adifferentiable function x(t) : I → Rn where I an interval containing t0 is a solution to (5) if

2Results in this subsection are from Hartman’s book “Ordinary differential equations”. Classics in AppliedMathematics, 38. Society for Industrial and Applied Mathematics (SIAM), 2002.

6 BO YANG

x(t0) = x0, (t, x(t)) ⊂ E, and x′(t) = f(t, x(t)) for any t ∈ I. We will make E more explicit inthe following theorems.

Theorem 1.4.1 (Existence: Peano). Let E = [t0− a, t0 + a]×B(x0, b) and F is continuous onE. Assume

M = sup(t,x)∈E

|F (t, x)|, ε = min(a,b

M).

Then IVP (5) has at least one solution on [t0 − ε, t0 + ε].

Theorem 1.4.2 (Existence and uniqueness: Picard-Lindelof). Let E = [t0−a, t0+a]×B(x0, b),assume F (t, x) is continuous on E and uniformly Lipschitz continuous with respect to x. Assume

M = sup(t,x)∈E

|F (t, x)|, ε = min(a,b

M),

then IVP (5) has a unique solution on [t0 − ε, t0 + ε].

Remark 1.4.3. A typical example of nonuniqueness is as follows, Consider (t, x) ∈ R2 and

F (t, x) = |x| 12 , and the initial condition x(0) = 0, then besides the trivial zero solution, we canfind a family of solutions

x(t) =

− 1

4 (t− α)2 t ≤ α,0 α ≤ t ≤ β,14 (t− β)2, t ≥ β.

(6)

Next we are interested in the extension of solutions constructed in Theorem 1.4.2 and 1.4.1.Recall that even in the case E = R × R, it is not always possible that (5) has a solution witht defined on the whole of R. For example, u′(t) = [u(t)]2 with u(0) = c, where the solutionu(t) = c

c−t where t ∈ (−∞, c). We now state a theorem to characterize ‘maximal interval ofexistence’ of solutions.

Definition 1.4.4 (The maximal interval of existence). Assume F (t, x) is continuous on E andx(t) solves the IVP (5) on some interval I containing t0, then we call (ω−, ω+) the maximalinterval of existence of x(t) if the following conditions hold:

(i) x(t) can be extended to a solution to (5) on (ω−, ω+).(ii) There exists no interval J which properly contains (ω−, ω+) so that x(t) can be extended

to a solution to (5) on J .

Theorem 1.4.5 (Extension of solutions). Assume F (t, x) is continuous on E and consider asolution x(t) of the IVP (5) defined on some interval I containing t0. Then x(t) can be extendedto a maximal interval of existence (ω−, ω+). Moreover, for any compact K ⊂ E, we have(t, x(t)) ∈ E \K as t→ ω− or ω+.

In most applications, we choose E = I × U , Theorem 1.4.5 implies we can always extend asolution x(t) defined on (a, b) beyond the endpoints if

lim inft→a+or b−

1

dist(x(t), ∂U)+ |x(t)| < +∞.

Next we study continuous dependence on initial conditions and parameters of ODEs. Weconsider the following IVP with a parameter:

dxdt = F (t, x, u);

x(t0) = x0.(7)

The basic assumption is that F : E1 → Rn is continuous, where E1 is an set in R1×Rn+1×Rkcontaining (t0, x0, u) where u ∈ Rk is the set of parameters.


Theorem 1.4.6 (Continuous dependence on initial conditions and parameters). Assume E1 ⊂R1 × Rn × Rk is open, F is continuous on E1, and in addition, for any fixed (t0, x0, λ) ∈E, the IVP (7) has a unique solution x(t, t0, x0, λ), whose maximal interval of existence is(ω−(t0, x0, λ), ω+(t0, x0, λ)).

Then ω+(t0, x0, λ) (or ω+(t0, x0, λ)) is a lower (upper) semi-continuous function of (t0, x0, λ) ∈E, and x(t, t0, x0, λ) is continuous on (t, t0, x0, λ) ∈ (ω−, ω+)× E. Here we allow ω+(t0, x0, λ)(or ω−(t0, x0, λ)) takes values as +∞ (or −∞).

Remark 1.4.7. In general, ω+(t0, x0, λ) need not to be continuous on E. Here is a simpleexample, consider E = R2 \ 0 × R1 and F (t, x, λ) = 0, then we have

ω+(t0, x0, λ) =

undefined if x0 = 0, t0 = 0,

0 if x0 = 0, t0 < 0,

+∞ otherwise.

In particular, in the proof of the fundamental theorem of surfaces, we are interested in thecontinuous dependence on parameters for IVP with fixed initial condition. Let us state a specialcase of Theorem 1.4.6.

Corollary 1.4.8. Assume E1 = [0, 1] × Rn × B(λ0, c), F (t, x, λ) is continuous on E1, anduniformly Lipschitz continuous with respect to x. Then for any initial data and a parameter(t0, x0, λ) ∈ E1 the IVP (7) has a unique solution x(t, x0, λ) for all t ∈ [0, 1]. Moreover, for anyfixed time t ∈ [0, 1] and initial data (t0, x0) ∈ [0, 1]× Rn, x(t, t0, x0, λ) is a continuous functionof λ ∈ B(λ0, c).

8 BO YANG

2. Curves in Rn: local theory

This section is a summary of main results in Chapter 1 of [9].

2.1. Frenet Equations.

Definition 2.1.1 (Parametrized curve). Let I be an interval, a C∞ (smooth) mappingc : I → Rn is called a parametrized curve, it is regular if c′(t) 6= 0 for any t ∈ I. If I has anendpoint, we define a parametrized curve by its open extension, i.e. there exists an open intervalI1 containing I and c : I1 → Rn satisfying c|I = c.

Definition 2.1.2 (Frenet frames). A Frenet frame for a parametrized curve c : I → Rn is acollection of vector fields along c, denoted by e1(t), · · · , en(t) such that it is orthonormal atevery t ∈ I, and for all 1 ≤ k ≤ n, c(k)(t) lies in the span of e1(t), · · · , ek(t).

Proposition 2.1.3 (Existence of Frenet frames). If a parametrized curve c : I → Rn satisfiesthe property that for any t ∈ I c′(t), c′′(t), · · · , c(n−1)(t) are linearly independent. Then thereexists a unique Frenet frame e1(t), · · · , en(t) such that

(i) For 1 ≤ k ≤ n−1 and any t ∈ I, the transformation matrix between c′(t), c′′(t), · · · , c(k)(t)and e1(t), · · · , ek(t) has positive determinant.

(ii) e1(t), · · · , en(t) is positively oriented, i.e. det[e1(t), · · · , en(t)] = 1.

Next we state the moving equation for Frenet frame for a parametrized curve satisfying theassumption in Proposition 2.1.3.

d

dt

e1(t)T

e2(t)T

...en−1(t)T

en(t)T

=

0 ω12 · · · 0 0−ω12 0 · · · 0 0

......

......

...0 0 · · · 0 ωn−1,n

0 0 · · · −ωn−1,n 0

e1(t)T

e2(t)T

...en−1(t)T

en(t)T

(8)

The only nonzero entries in the matrix appearing on the right hand side (8) are used to definethe i-th curvature function of c

ki(t) =ωi i+1

|c′(t)| .

Proposition 2.1.4 (Properties of curvature functions of a parametrized curve). Let c : I → Rnsatisfying the assumption in Proposition 2.1.3.

(i) Assume B is an orientation-preserving isometry of Rn (i.e. its orthogonal part has

determinant +1), and c = B c : I → Rn has the corresponding curvature functions ki,

then we have ki(t) = ki(t) for any 1 ≤ i ≤ n− 1.(ii) Assume c : J → Rn is an orientation preserving reparametrization of c in the sense

c(u) = c φ(u) with φ : J → I satisfying φ′(u) > 0. Then ki(u) = ki(φ(u)).(iii) Asssume n ≥ 3, ki(t) > 0 for all 1 ≤ i ≤ n− 2.

The proof of Proposition 2.1.4 can be found on pp.12-14 in [9].

2.2. Plane curves. Any parametrized curve c : I → R2 which is regular satisfies the assumptionin Proposition 2.1.3, therefore we could find a Frenet frame. However, we could work in anexplicit way.

Consider a plane curve c : I → R2 which is parametrized by its arc length. let J denote acounter-clockwise rotation of a vector by 90 degrees, then we could choose

e1 = c′(s), e2 = Je1(9)

as the Frenet frame. Sometimes we also denote e1 = T and e2 = N .Assume e′1(s) · e2 = k(s), is not hard to find the Frenet equation:


d

ds

[e1(s)T

e2(s)T

]=

[0 k(s)

−k(s) 0

] [e1(s)T

e2(s)T

](10)

The function k(s) = c′′(s) · e2 is called the curvature of the plane curvature c. Note that

k(s) = Jc′(s) · c′′(s) and J has the matrix expression

[0 −11 0

], we conclude

k(s) = det(c′(s), c′′(s)).

Using (134), we may also calculate the curvature of any regular plane curve c which is notnecessarily parametrized by its arc length.

Proposition 2.2.1. Any parametrized regular plane curve c : I → R2 has its (signed) curvature

k(t) =det(c′(t), c′′(t))|c′(t)|3 .(11)

In particular, if c is parametrized by its arc length

k(s) = det(c′(s), c′′(s)).(12)

Proposition 2.2.2 (Plane curves with constant curvature). Given any plane curve c : I → R2,the following two conditions are equivalent.

(i) k(t) is constant.(ii) c is a piece of circular arc or straight line.

Proof of Proposition 2.2.2. We only prove one direction. Assume k = 0, then (134) impliese′1(s) = 0, after integration we get the equation of a straight line. If k 6= 0, then (134) impliesd2

ds2 e1(s) + k2e1(s) = 0. the general solution of this ODE is

e1(s) = (C1 cos(ks+ C2), C3 sin(ks+ C4))T .

In order that |e1(s)| = 1 for all s, the only possible solution is

e1(s) = (± cos(ks+ C1),± sin(ks+ C1)).

After an integration we get

c(s) = (±1

ksin(ks+ C1) + C2,±

−1

kcos(ks+ C1) + C3).

This is a parametrization of a circle equation (x−C2)2 +(y−C3)2 = 1k2 . There is an alternative

proof on p.17 [9].

2.3. Space curves. Consider a regular parametrized space curve c : I → R3 which satisfies theassumption in Proposition 2.1.3. For example, if c is parametrized by its arc length and c′′ 6= 0,then c′(s) and c′′(s) are linearly independent. In this case we could also work out a Frenet frameexplicitly:

e1 = c′(s), e2 =c′′(s)|c′′(s)| , e3 = e1 × e2.

The corresponding Frenet equation is:

d

ds

e1(s)T

e2(s)T

e3(s)T

=

0 k(s) 0−k(s) 0 τ(s)

0 −τ(s) 0

e1(s)T

e2(s)T

e3(s)T

(13)

In particular, the first and the second curvatures, k(s) and τ(s), are called the curvature andtorsion of a space curve parametrized by arc length. Similarly, working out (13) in terms of ageneral parametrization, we may also calculate the curvature and torsion of any parametrizedspace curve.

10 BO YANG

Proposition 2.3.1. Any parametrized regular space curve c : I → R3 which satisfies the as-sumption in Proposition 2.1.3 has

k(t) =|c′(t)× c′′(t)||c′(t)|3 , τ(t) =

det(c′(t), c′′(t), c(3)(t))

|c′(t)× c′′(t)|2 .(14)

In particular, any such a curve parametrized by arc length has

k(s) = |c′′(s)|, τ(s) =det(c′(s), c′′(s), c(3)(s))

|c′′(s)|2 .(15)

Proposition 2.3.2 (Taylor expansion of a space curve, on p.18 in [9]). Suppose c : I → R3 bea space curve prametrized by arc length, then

c(s)− c(s0) =[(s− s0)− (s− s0)3

6k2(s0)

]e1(s0)(16)

+[ (s− s0)2

2k(s0) +

(s− s0)3

6k′(s0)

]e2(s0)

+[ (s− s0)3

6k(s0)τ(s0)

]e3(s0) + o(|s− s0|3).

2.4. Fundamental theorem on curves with prescribed curvature functions.

Theorem 2.4.1 (Existence and uniqueness for curves with prescribed curvatures, p.14 in [9]).Let k1(s), · · · , kn−1(s) be smooth functions defined on an open interval I containing 0, andki(s) > 0 for any 1 ≤ i ≤ n − 2 (if n = 2, no need to assume). There exists a smooth curvec : I → R3 with the following properties:

(i) |c′(s)| = 1 for any s ∈ I.(ii) It satisfies the assumption in Proposition 2.1.3.(iii) The i-th curvature functions is exactly ki(s) for 1 ≤ i ≤ n− 1.

Moreover, any such two curves c : I → Rn and c : I → Rn are related by an isometry B of Rn,i.e. c = B c where B is some orientation-preserving isometry of Rn.

Remark 2.4.2. We give several remarks before the proof.

(i) The crucial step is to apply Theorem 1.4.2 and Theorem 1.4.5 on systems of first orderlinear ODEs, those theorems ensure that we can solve the desired curve c in a smallersubinterval and extend the solution uniquely on the whole open interval I.

(ii) The isometry in Theorem 2.4.1 could always be arranged as orientation preserving.Here is a simple example, consider c1(t) = (cos t, sin t, 0) and c2(t) = (cos t,− sin t, 0)where t ∈ [0, 2π]. We make check they have the same k(t) = 1 and τ(t) = 0. Now c1and c2 differ by a rotation about x−axis with the angle π, i.e. x→ x, y → −y, z → −z.

(iii) In Theorem 2.4.1 we consider smooth curves with smooth curvature and torsion func-tions. The smoothness assumption can be weakened. For example choose n = 3, for k1

is positive and C1, and k2 is C0. we may construct a curve c(t) which is C3 with thesame conclusion as in Theorem 2.4.1.

(iv) It is necessary to assume ki is positive in Theorem 2.4.1 for 1 ≤ i ≤ n − 2, otherwisethis situation might be complicated. Take n = 3 for example, we may check the followingtwo curves in R3:

c1(t) =

(t, e−

1t , 0) t > 0

(0, 0, 0) t = 0

(t, 0, e−1t ) t < 0

, c2(t) =

(t, e−

1t , 0) t > 0

(0, 0, 0) t = 0

(t, e−1t , 0) t < 0

Both curves have the same k(t) : R → R and k(0) = 0, and the same τ(t) = 0 for allt 6= 0, and c′′1(0) = c′′2(0) = 0. So τ(0) is undefined. Let us view these two curves areboth with the same k and τ with k(0) = 0 and τ ≡ 0 by defining τ(0) = 0. Now, canwe expect c1 and c2 are related by an isometry in R3? The answer is NO! The reasonis that c1 is in xy−plane and xz−plane, and c2 is a xy−plane curve.


Proof of Theorem 2.4.1. The proof is given on p.14 in [9] for the proof, we give a summary.Assume Y (s) =

[Y1 · · · Yn

]a n × n matrix-valued function with respect to s ∈ I, and

each Yi is a column vector.Recall the matrix appearing in (8):

A(s) =

0 k1(s) · · · 0 0

−k1(s) 0 · · · 0 0...

......

......

0 0 · · · 0 kn−1(s)0 0 · · · −kn−1(s) 0

Consider the following initial value problem (IVP) of ODE:

ddsY (s) = Y (s)A(s)

T, s ∈ I,

Y (0) = Y0, Y0 is any fixed orthogonal matrix with det = 1.(17)

Note that the RHS of the ODE in (17) is linear with respect to Y (s), hence Lipschitz contin-uous. We may apply Theorem 1.4.2 to solve Y (s) on some sub-interval [−δ, δ] of I, then applyTheorem 1.4.5 we may extend this solution uniquely on the whole open interval I.

Since A(s) is skew-symmetric for any s ∈ I, we may check

d

ds(Y (s)Y (s)T ) = [

d

dsY (s)]Y (s)T + Y (s)[

d

dsY (s)]T = Y ATY T + Y AY T = 0.(18)

Therefore Y (s)Y (s)T = Y (0)Y (0)T = Id. Moreover since detY (s) is continuous function anddetY (s) = 1, we conclude that Y (s) is a an orthogonal matrix with detY (s) = 1.

Now define c(s) =∫ s

0Y1(u)du+ c(0), one may check the desired properties listed in Theorem

2.4.1 indeed holds.Finally, if any two curves c : I → Rn and c : I → Rn satisfies the properties listed, then their

Frenet frames, denoted by Y = [e1, · · · , en] and Y = [e1, · · · , en], both satisfy the same ODEin (17). Now pick an orthogonal matrix R with detR = 1 such that the Frenet frames of two

curves at s = 0 differ by B, i.e. Y (0) = RY (0), we may apply the uniqueness of the solution to

IVP for ODE (Theorem 1.4.2) to conclude Y (s) = RY (s) for any s ∈ I. In particular, we havee1(s) = Re1(s) Recall c(s) =

∫ s0e1(u)du + c(0) and c(s) =

∫ s0e1(u)du + c(0), we conclude that

c(s) = Rc(s) + b for some constant b ∈ Rn, and B(x) = R(x) + b is the desired isometry.

12 BO YANG

3. Plane and space curves: some global results

This section is a selection of main results in Chapter 2 of [9]. We also adopt the definition of‘simple curve’ on p.138 in [10].

To discuss global aspects of curves, we make it clear the meaning of ‘curve’ we want to discuss,i.e. the parametrization map I → Rn or its image as a subset in Rn.

3.1. From parametrized curves to global definitions of curves.

Definition 3.1.1 (Closed curves). A parametrized curve c : [a, b]→ R3 is closed if there existsanother parametrized c : R→ R3 such that

c|[a,b] = c, and c(t+ b− a) = c(t) ∀ t ∈ R.

Here b− a is called a period of c. Note that we can always find a minimal positive period if c isregular.

Definition 3.1.2 (Simple curve in R2 or R3, p.138 in [10]). A subset C ⊂ R2 is a simplecurve if

(i) C is connected with respect to the induced topology from R2.(ii) for any point p ∈ C there exists an open set Vp of R2 containing p, such that we can

find a regular parametrized curve c : I → R2 which is a homeomorphism from I ontoc(I) = Vp ∩ C.

An interesting example of a non-simple curve is a part of the folium of Descartes defined by

c(t) = (3t

1 + t3,

3t2

1 + t3), where t ∈ (−1,+∞).

While c(t) is a smooth regular paramatrized curve, the intersection of any open set containingc(0) = (0, 0) with the curve c(t) is not homeomorphic to an interval.

Figure 1. A non-simple smooth curve

The following important result clarifies how far a simple curve is to a parametrized curve.

Proposition 3.1.3 (Classification of simple curves, Theorem 9.10 on p.318 in [10]). Any simplecurve in R2 or R3 must be either of the following

(i) It is diffeomorphic to the real line R, in this case it can be covered by a single parametriza-tion.

(ii) It is diffeomorphic to a closed circle, in this case it can be realized as the image of aclosed parametrized curve.


Remark 3.1.4. According to Definition 3.1.2, if a regular prametrized curve is a homeomor-phism onto it its image (with the induced topology from R3), then its image is a simple curve.In other words, a simple curve is a union of regular prametrized curves assembled in a ‘smooth’way. For later purposes, we point out that simple curves are exactly smooth embedded 1-manifoldin R2 or R3.

A parametric curve c : [a, b] → R2 is defined to be ‘simple closed’ on p.21 in [9] if it isclosed in the sense of Definition 3.1.1 and c is one-to-one on [a, b). By Proposition 3.1.3 this isthe same as ‘compact simple curve’ Definition 3.1.2 in R2.

From now on we focus on plane curves, as explained in Remark 3.1.4, we will use thefollowing equivalent notations for the same object without distinction:

(i) closed embedded curve;(ii) compact simple curve;(iii) simple closed curve.

Many results for smooth parametrized curves can be generalized to piecewise smooth curves,so we introduce:

Definition 3.1.5 (piecewise smooth (parametrized) curve). A piecewise smooth (parametrized)curve is a continuous map c : [a, b]→ Rn together with partition of [a, b]

a = a0 < a1 < · · · < ak−1 < ak = b

such that cj.= c|[aj ,aj+1] is smooth for 0 ≤ j ≤ k − 1. The points c(ai)(0 ≤ i ≤ k) are called

corner points of c. We call c regular if each cj is regular, and c is closed if c(a) = a(b), c issimple closed if it is closed and c|[a,b) is one-to-one.

3.2. Rotation number and the Umlaufsatz. If we a consider a closed parametrized curvec(t) : I → R2, we want to define a quantity measuring how much the curve winds when theparameter exhausts a complete period. For any t ∈ I, we may find a small interval I ′ of theparameter t, we may introduce a differentiable function θ to measure the angle between the unittangent vector and a fixed unit vector, say the positive x-axis. Therefore

e1(s) = (cos θ(s), sin θ(s)), e2(s) = (− sin θ(s), cos θ(s)).(19)

It follows from the Frenet equation (134) that θ′(s) = k(s) on I ′. If we normalize 0 ≤ θ < 2π,then θ is uniquely determined. However, this θ is not continuous if the variation of c′(t) passesthrough the positive x-axis. Nevertheless, by allowing θ to be out of [0, 2π) we can construct aglobally continuous function θ along c(t).

Proposition 3.2.1. Let c : [a, b]→ R2 be a regular smooth parametrized curve, then there existsa differentiable function θ : [a, b]→ R such that

e1 =c′(t)|c′(t)| = (cos θ(t), sin θ(t)),

dθ

dt= k(t)|c′(t)|.(20)

Moreover, the difference θ(b)− θ(a) is independent of the choice of θ.

Proof of Proposition 3.2.1. The proof is on p.22 of [9]. The idea goes as follows, we give apartition of [a, b], denoted by a0 = a < a1 < · · · < ak−1 < ak = b so that we can define θ whichis differentiable on [a0, a1] then we continuously extend θ onto [a1, a2] by requiring (20). Hereθ(s) ∈ (−∞,+∞). And repeat this process to get θ defined on [a, b] and satisfying (20).

Any such two θ function must differ by 2πφ(t) where φ(t) is integer-valued. Since φ(t) iscontinuous, hence constant. We may conclude θ(b)− θ(a) is independent of the choice of θ.

The following Figure 2 is an example of θ when the curve is c(t) = (cos(2t), sin(2t)) wheret ∈ [0, 2π)

After we do the piecewise continuation outlined in the proof of Proposition 3.2.1, we choosea continuous θ = 2t+ π

2 for t ∈ [0, 2π).

Definition 3.2.2 (Rotation number). We consider smooth and piecewise smooth curves sep-arately.

14 BO YANG

Figure 2. θ is not continuous if we restricted to [0, 2π)

(i) Let c : [a, b] → R2 be a closed regular smooth parametrized curve, then its rotationnumber is defined as

nc =1

2π(θ(b)− θ(a)),(21)

where θ is defined by Proposition 3.2.1. Equivalently,

nc =1

2π

∫[a,b]

k(t)c′(t)dt =1

2π

∫C

k(s)ds.

(ii) Let c : [a, b] → R2 be a closed regular piecewise smooth curve defined by a partition of[a, b] where a = a0 < a1 < · · · < ak−1 < ak = b, then its rotation number is

nc =1

2π

k−1∑j=0

(θj(aj+1)− θj(aj)) +1

2π

k−1∑j=0

αj ,(22)

where θi is defined for each cj = c|[aj ,aj+1] defined by Proposition 3.2.1, and −π < αj <π is the oriented angle from c′(aj−) to c′(aj+).

Remark 3.2.3. We need to show both (21) and (22) are necessarily integers. In the smoothcase, it directly follows from Proposition 3.2.1. Since the angle θ(b) and θ(a) represents theangle of the same tangent vector e1(a) with the positive x−axis. In the piecewise smooth case,we may rewrite

nc =1

2π

k−1∑j=0

(θj−1(aj)− θj(aj) + αj), here we set θ−1(a0) := θk−1(ak)(23)

Then nc is an integer since each summand in (23) must be an integer multiple of 2π.

Theorem 3.2.4 (Umlaufsatz for a smooth curve). Let c : [a, b]→ R2 be a simple closed curve,then its rotation number is ±1.


Figure 3. Some examples of rotation numbers for non-simple curves

Theorem 3.2.5 (Umlaufsatz for a piecewise smooth curve). Let c : [a, b] → R2 be a regularpiecewise smooth curve, assume it is closed (i.e. c(a) = c(b) and c is injective on [a, b)), thennc = ±1.

Figure 4. A spiralFigure 5. A piecewise smoothsimple closed curve

To see why Theorem 3.2.4 and 3.2.5 are nontrivial. Just check examples in Figure 4 and 5.Can you solve their rotation numbers?

There are two approaches (essentially equivalent though), due to H. Hopf (1935). In Method1, we define a continuous angle function θ on a half diagonal region. A proof of Theorem 3.2.4can be found in [4], and Theorem 3.2.5 was proved on pp.24-26 of [9]. See Figure 6. In Method2, we define the rotation number as a mapping degree and use homotopy. This is done in Section5-7 of [5] and Section 9.4 of [10].

Next we state an application of Umlaufsatz.

Definition 3.2.6 (Convex parametrized curves). A regular plane curve c : I → R2 isconvex if for any t ∈ I, the curve lies entirely on one side of the tangent line at c(t). Moroever,it is called strictly convex if for any t ∈ I the curve touches the tangent line at c(t) exactly asingle point.

Theorem 3.2.7 (Convexity in terms of curvature). A closed simple curve is convex if and onlyif it can be parametrized so that k(t) ≥ 0 for all t ∈ I or k(t) ≤ 0 for all t ∈ I.

Proof of Theorem 3.2.7. This proof is copied from p.93 in [4], which is the same as the proof ofTheorem 2.3.2 in [9].

16 BO YANG

Figure 6. Copied from p.25 in [9], which in turn is from p.396 in [5]

(1) Assume C is a simple closed parametrized curve with k(t) ≥ 0 for all t ∈ [a, b], we willprove C is convex.

Since θ defined in Proposition 3.2.1 satisfies θ′(s) = k(s), we know θ is monotone. By theUmlaufsatz (Theorem 3.2.4), θ(b)−θ(a) = 2π. Therefore, if there exists two points A, B on thecurve C with the same tangent direction, then θ(A) = θ(B).3 It follows from the monotonicitythat θ is constant along the part of C connecting A and B. In other words, the segment AB iscontained as a part of C. Assume C is not convex, there must be a point D on C such that thereare points of C at both sides of the tangent line l at D. Consider the oriented distance from anypoint on C to the line l, since this function is continuous we can find a maximum and minimumat the points M and N on C. Note that both the tangents at M and N are parallel to l, theremust be at least one of them, say M , with the same direction of l. The above observation showsa segment MD is contained as part of C, which is a contradiction since M is away from l.

Figure 7. Proof of Theorem 3.2.7 explained

(2) Assume C is a closed simple curve which is convex, we will prove θ(t) is monotone.

3The application of this Umlaufsatz is essential here. It is easy to find examples of non-simple closed curveswith k(s) ≥ 0 but not convex.


To show θ : [a, b] → R is monotone, it suffices to show whenever θ(t1) = θ(t2) for somea ≤ t1 < t2 ≤ b, then θ is constant on [t1, t2].

If there exist two points A = c(t2) and B = c(t1) on C with θ(t1) = θ(t2), then A and B havethe same tangent direction. Since C is convex, these these two tangent line must coincide, call itl. We will prove the segment AB must be contained as a part of C. Otherwise, consider a pointD on AB which is not on C. Draw through D a line perpendicular to l, which intersects C withat least two points. This is because the closed curve C is completely on one side of l. Let Fthe nearest intersection point from l and G the farthest one. Since F lies in the interior of thetriangle ABG, we find that C has points at both side the tangent at F . This is a contradictionto the convexity of C. Therefore the segment AB has to be a part of C, and θ(t) must beconstant on [t1, t2].

3.3. Isoperimetric inequalities in the plane. The isoperimetric problem in the plane has along history. It might first appear in a legend of Queen Dido (around 850 B.C.), the legendaryfounder of Carthage. The Queen wanted to purchase a piece of land along the coastline of NorthAfrica. According to the lander owner, Dido could only have as much land as she could encloseby the hide of an ox. Dido have the ox-hide cut, made a long thin strip, and used it to to enclosea semi-circle along with the coastline, the area is where the city of Carthage (Tunisia) was built.

The mathematical problem behind can be rephrased as follows: Fix a positive number L > 0,consider a simple plane curve which lies above x-axis and has its total length L, assume two ofits endpoints (however not fixed) are symmetric on x-axis, then what is the shape of curve sothat the area enclosed by the curve and the x-axis is maximized?

The answer is the semi-circle with radius Lπ ! As a warm-up exercise, you might consider a

special case of the above problem. Namely among all circular arcs with length L, the semi-circleencloses the maximal area.

If we disregard the interval on the x-axis, and consider the area enclosed by a closed simplecurve instead, then this is the well-known isoperimetric problem. It is concerned with maximiz-ing the area enclosed by plane curves with a fixed length, or minimizing the length among allplane curves with a fixed enclosed area. The answer is the following isoperimetric inequality.

Theorem 3.3.1 (Isoperimetric inequality on R2). Let C be a simple closed curve of lengthL(C) in R2 and Ω its inner domain. Then

4π[Area(Ω)] ≤ [L(C)]2.(24)

with the equality holds if and only if C is a circle.

Remark 3.3.2. Here the assumption of C being a simple closed curve is essential. the Jordancurve theorem states any such curve divides R2 into two disjoint regions with common bound-ary C, one of which is bounded. We call this region the inner domain Ω. We also mention ageneralization due to Banchoff-Pohl to closed non-simple curves in Theorem 3.3.12.

Proof of Theorem 3.3.14. We offer four ways to approach this problem. The first and secondmethod introduce an important tool in the study of differential geometry, but the proofs areincomplete. Both the third and the fourth proof are complete and based on other inequalitieswhich implies isoperimetric inequalities.

Method 1: Calculus of variations with constraints.

To prove (24), it suffices to show the following:

Claim 3.3.3. the maximal area of a region enclosed any simple curve with fixed length L isnecessarily realized by the circle of its perimeter L.

Assume we know the maximal area is indeed realized by some simple curve (i.e. the optimalcurve is not a sequence of simple curves with bad regularity), it is reasonable to expect thatsome compactness property is needed to guarantee this assumption holds. Here we are onlyconcerned with showing that the optimal curve must be a circle under this assumption.

18 BO YANG

Consider any simple curve C with one point fixed at the origin, assume it is parametrized byc(t) = (x(t), y(t)) where 0 ≤ t ≤ t0 and c(0) = c(t0) = 0, it follows from the Green’s formulathat the area of Ω which is the bounded region enclosed by C can be solved. It suffices to solvethe following extremal problem with constraint:

Maximize: A(x, y)

.= 1

2

∫ t00

(−ydx+ xdy)

subject to: (x(t), y(t)) are smooth functions on [0, t0], x(0) = y(0) = x(t0) = y(t0) = 0,

and L(x, y) :=∫ t0

0

√(x′(t))2 + (y′(t))2dx− L = 0.

(25)

Assume the optimal solution is parametrized by (y∗(t), y∗(t)) where 0 ≤ t ≤ t0, then itseems natural to consider a small perturbation x(t) = x∗(t) + ε1η1(t) and y(t) = y∗(t) + ε1δ1(t)where η1(t), δ1(t) are arbitrary smooth functions satisfying η1(0) = η1(t0) = δ1(0) = δ1(t0) = 0.However there is no way to guarantee (x(t), y(t)) satisfies the constraint in (25) if η1(t), δ1(t) arechosen arbitrarily. In other words, there might be cases that for some η1(t), δ1(t) chosen, theonly ε1 satisfying L(x, y) = 0 is ε1 = 0. To overcome this difficulty, we consider a two-parameterperturbation

x(t) = x∗(t) + ε1η1(t) + ε2η2(t), y(t) = y∗(t) + ε1δ1(t) + ε2δ2(t).(26)

Here η1(t), δ1(t) are two arbitrary smooth functions vanishing at endpoints, and η2(t), δ2(t) aretwo suitable smooth functions vanishing at endpoints. Note that η2(t), δ2(t) serve as ‘correction’terms so that (x(t), y(t)) satisfy the last constraint in (25).

After we fix any arbitrary η1(t), δ1(t) and suitable η2(t), δ2(t) satisfying conditions in (26),we may view (x∗(t), y∗(t) (when ε1 = ε2 = 0) is the critical point of the following problem,

Maximize: A(ε1, ε2) = 12

∫ t00

(−ydx+ xdy)

subject to: L(ε1, ε2) =∫ t0

0

√[x′(t)]2 + [y′(t)]2dx− L = 0, and x(t), y(t) satisfy (26).

(27)

It follows from the method of Lagrange multipliers that

∇(A(ε1, ε2)− λ∇L(ε1, ε2))|ε1=ε2=0 = 0(28)

for some constant λ to be determined.Now we introduce F (ε1, ε2) and f(ε1, ε2) such that

F (ε1, ε2) := A(ε1, ε2)− λL(ε1, ε2) =

∫ t0

0

f(ε1, ε2)dt.

where f(ε1, ε2) = 12 (xy′(t)− yx′(t))− λ(

√(x′(t))2 + (y′(t))2 − L

t0). It follows from (28) that

∂F

∂ε1|ε1=ε2=0 =

∫ c

−c

[ ∂f∂ε1|ε1=ε2=0

]dx = 0.(29)

By integration by parts, we get∫ t0

0

η1(t)[y′∗(t)− λ

d

dt(

x′∗(t)√[x′∗(t)]2 + [y′∗(t)]2

)]dt

+

∫ t0

0

δ1(t)[− x′∗(t)− λ

d

dt(

y′∗(t)√[x′∗(t)]2 + [y′∗(t)]2

)]dt = 0(30)

Lemma 3.3.4 (The fundamental lemma of calculus of variations). Let h : [a, b] → R be asmooth function, assume for any smooth function η : [a, b] → R with compact support in (a, b)we have ∫ b

a

h(x)η(x)dx = 0.

Then h(x) = 0 on [a, b].


In view of Lemma 3.3.4, we end up with the following Euler-Lagrange equation satisfied by(x∗(t), x∗(t)): −x

′∗(t)− λ d

dt (y′∗(t)√

[x′∗(t)]2+[y′∗(t)]

2) = 0,

y′∗(t)− λ ddt (

x′∗(t)√[x′∗(t)]

2+[y′∗(t)]2) = 0.

(31)

We may check that either of (31) leads to

x′∗(t)y′′∗ (t)− x′∗(t)y′′∗ (t)

([x′∗(t)]2 + [y′∗(t)]2)32

=1

λ.(32)

Since the right hand side represents the curvature of (x∗(t), y∗(t)), by Proposition 2.2.2, weconclude (x∗(t), y∗(t)) is a circle.

Method 2: Steiner symmetrization.

As another way to study (24), we begin with

Claim 3.3.5. the minimal perimeter of a region enclosed any simple curve with fixed area A,we will show that the optimal perimeter is only realized by the circle of its area A.

There is a simple argument showing Claim 3.3.5 is equivalent to Claim 3.3.3 studied in Method1. Indeed, pick any noncircular plane region Ω and a circular disk D such that L(∂Ω) = L(∂D),then we need to show A(Ω) < A(D). If not, we may choose a circle concentric to ∂D so that theresulting enclosed disk D′ so that A(Ω) = A(D′). It follows from Claim 3.3.3 L(∂Ω) > L(∂D′).Since L(∂Ω) > L(∂D′) ≥ L(∂D), we have a contradiction.

We will use the Steiner symmetrization to study Claim 3.3.5. It is a beautiful idea in thestudy of rigidity properties of geometric shapes. For the plane region it is illustrated by thefollowing picture, which is the symmetrization of a region Ω with respect to a line l.

Figure 8. Steiner symmetrization copied from p. 28 in [2]

20 BO YANG

Lemma 3.3.6. Assume Ω is enclose by two parallel lines, and the line l is the line perpendicularto these two lines. Let Ω1 is the symmetrization of Ω with respect to a line l, then the perimeterof Ω1 is no larger than and that of Ω, while A(Ω1) = A(Ω1).

Proof of Lemma 3.3.6. Assume the line l is given by y = 0, if we parametrize Ω in Figure 8 by(x(t), y2(t)) for the part above y = 0 and (x(t), y1(t)) for the part below, here a ≤ t ≤ b. ThenΩ1 in Figure 8 can be parametrized by (x(t), 1

2 [y2(t)− y1(t)]) and (x(t),− 12 [y2(t)− y1(t)]). Note

that both Ω1 and Ω1 are enclosed by two parallel lines x = x(a) and x = x(b), and y1(a) = y2(a)and y1(b) = y2(b).

It suffices to show the following is true

2

∫ b

a

√[x′(t)]2 + [

y′2(t)− y′1(t)

2]2dt ≤

∫ b

a

(√[x′(t)]2 + [y′1(t)]2 +

√[x′(t)]2 + [y′2(t)]2

)dt.

Moreover, when the equality holds y′1(t) = −y′2(t) on [a, b].What can we say if the perimeters of Ω1 and Ω are the same? In this case, we see that

y2(t) + y1(t) = 2y1(a) = 2y1(b) for any t ∈ [a, b], therefore Ω must be symmetric aroundy = y1(a) = y2(a).

Using Steiner symmetrization, we will prove

Proposition 3.3.7. Assume we know the minimal perimeter in Claim 3.3.5 is indeed realizedby some simple curve, then that optimal curve must be a circle.

Proof of Proposition 3.3.7. Assume C is the optimal curve and Ω the region enclosed by C. Sincewe can always enclose Ω by two parallel lines with any given slope, it follows from Lemma 3.3.6that Ω has a line of symmetry along any direction in R2. In the following we prove a strongerresult: any C1 curve which encloses a region with lines of symmetry along all directions mustbe a circle.

Let C be the curve and Ω denote the region, pick two lines of symmetry which are perpen-dicular to each other, call them x-axis and y-axis. and O their intersection point. Since thecomposition of reflection around x-axis followed by reflection around y-axis is exactly a rotationby 180 degrees (a half rotation) about O, then Ω is symmetric with respect to O by a halfrotation.

Note that the line of symmetry must be unique after its direction fixed, this is obvious sincesuch aline must divide the area of Ω into equal halves. Now we see that the line connectingevery point of of C and O must be a line of symmetry. Note that such a line already dividesthe area of Ω into equal halves since Ω is symmetric with respect to O, therefore it must be aline of symmetry. In other words, all line of symmetry of Ω intersects at O.

Now fix a point P ∈ C, for any point Q ∈ C, pick the unique point which divide the arc ofC connecting P and Q by equal halves. Since the line l through this point and O is the line ofsymmetry, it is easy to see that Q and P must be related by a reflection around l. Therefore weconclude that |OP | = |OQ| for any Q ∈ C. This implies all points of C has a equal distance toO, hence C must be a circle.

Let us end the discussion of Method 2 by the following observation:

Question 3.3.8. Consider a C1 curve C which encloses a region with lines of symmetry along aninfinite number of directions, and assume at least two of them are perpendicular, then is C mustbe a circle? If the answer is affirmative, then we could get a sequence of Steiner symmetrizationalong a sequence of chosen directions. Does this sequence converge to a circle?

Though I have not checked details, this problem might be answered in Almut Burchard’slecture notes “Rearrangement inequalities”.

Method 3: Writinger’s inequality and the method of Hurwitz.


We follow the exposition on p.99 in [4]. Since compact simple curve admits a closed parametriza-tion which is a periodic function. It is not quite surprising we could use Fourier analysis. Infact, one could prove the following result by Parseval’s formula.

Lemma 3.3.9 (Writinger’s inequality). Let f : R1 → R be a C1 function of period 2π, assume∫ 2π

0f(t)dt = 0, then ∫ 2π

0

f ′(t)dt ≥∫ 2π

0

[f(t)]2dt.

Moreover, the equality holds if and only if f(t) = a cos t+ b sin t for some constants a and b.

Assume Lemma 3.3.9, we will prove (24) for any compact simple curve C with a parametriza-tion c(s) = (x(s), y(s)) where 0 ≤ s ≤ L. Here c is parametrized by arc length and it encloses Ω.By a scaling it suffices to prove (24) in the case of L = 2π, i.e. A(Ω) ≤ π. By Green’s formulawe have

A(Ω) =

∫ 2π

0

xy′(s)ds.

Note that [x′(s)]2 + [y′(s)]2 = 1, we conclude

2(π −A(Ω)) =

∫ 2π

0

([x′(s)]2 + [y′(s)]2 − 2xy′(s)

)ds

=

∫ 2π

0

([x′(s)]2 − [x(s)]2

)ds+

∫ 2π

0

(x(s)− y′(s)

)2

ds.

After a suitable translation of C we may assume∫ 2π

0x(s)ds = 0, then it follows from Lemma

3.3.9 that A(Ω) ≤ π. Moreover, when the equality holds,

x(s) = a cos s+ b sin s, y′(s) = x(s).

This will imply x = a cos s + b sin s and y − C = a sin s − b cos s for some constant a, b and C,hence

x2 − (y − C)2 = a2 + b2.

In the general case, ( 12π

∫ 2π

0x(s)ds, 1

2π

∫ 2π

0y(s)ds) represents the center of mass of the curve

c(s) = (x(s), y(s)). We can always make a translation so that its center of mass of the newcurve sits on y-axis and the previous argument works. Note that all quantities in (24) remainsunchanged under a translation.

Method 4: The Brunn-Minkowski inequality.

This proof is adapted from p.195 in [10] where they treated isoperimetric inequality in R3.

Lemma 3.3.10 (Brunn-Minkowski inequality on R2, for the proof check4). Let A and B be twobounded open sets of R2, and define

A+B = a+ b | a ∈ A, b ∈ B.Then

[Area(A)]12 + [Area(B)]

12 ≤ [Area(A+B)]

12 .

Consider a compact simple curve C with a parametrization by its arc length in the counter-clockwise orientation. Here c(s) = (x(s), y(s)) where 0 ≤ s ≤ L

.= L(C) and C encloses the

region Ω. Now we consider Cε which is a family of deformation of C along e2(s) (the innernormal direction) defined by cε(s) = c(s) + εe2(s). There exists some ε0 > 0 such that for any0 < ε < ε0, Cε is also a simple curve and enclose a region denoted by Ωε.

4The proof of this inequality can be found in Simon’s book convexity an analytic viewpoint.

22 BO YANG

Observe Ωε+Dε = Ω where Dε is the disk with radius ε centered at the origin. Apply Lemma3.3.10, we have

A(Ω0) ≥A(Ωε) +A(Dε) + 2√A(Ωε) ·

√A(Dε),

A(Ω0)−A(Ωε)

ε≥2√π ·√A(Ωε).(33)

To compute the LHS of (33), we consider a bijection F : [0, L]× (0, ε]→ Ω \ Ωε by F (s, ε) =c(s) + εe2(s). Then we have

∂F

∂s= e1 + ε(−k)e1 = (1− kε)e1,

∂F

∂ε= e2.

Therefore

A(Ω)−A(Ωε) =

∫ L

0

∫ ε

0

det(JacF (s, τ))dτds

=

∫ L

0

∫ ε

0

(1− k(s)ε)dτds

=εL− ε2

2

∫ L

0

k(s)ds.(34)

Here we implicitly assume F (s, τ) is a bijection which parametrizes a family of simple closedcurves. This is true for ε small enough.

Combining (33) and (34) and letting ε→ 0+, we conclude

[L(C)]2

A(Ω)≥ 4π =

[L(Sr)]2

A(Dr).(35)

Here Sr represents a circle with radius r.It remains to show the circle is the unique minimizer of the LHS of (36). Assume there is a

compact simple curve C such that (36) holds as an equality, we will prove it must be a circle.This is exactly the situation of Method 1. Here we intend to give another argument basedon calculus of variations whose calculation is simpler and more intrinsic compared to that inMethod 1.

Consider any smooth function η defined on the curve C, and consider Cε(η), a deformation ofC by cε(s) = c(s)+ εη(s)e2(s). Assume Cε(η) encloses Ωε(η) and consider the following functiondefined on [−ε0, ε0] for some small ε0:

h(ε) = [L(Cε(η))]2 − 4πA(Ωε(η)).

By (36) we have h(0) = 0 and h(ε) ≥ 0 for any ε ∈ [−ε0, ε0], then h′(0) = 0.The above calculation of (34) can be generalized to Ωε(η) and the case of ε < 0.

A(Ω)−A(Ωε) = ε

∫ L

0

η(s)ds− ε2

2

∫ L

0

η2(s)k(s)ds..(36)

We may also calculate

L(Cε(η)) =

∫ L

0

|(1− εη(s)k(s))e1(s) + εη′(s)e2(s)|ds

=

∫ L

0

√[1− εη(s)k(s)]2 + [εη′(s)]2ds.

∂L(Cε(η))

∂ε|ε=0 =− ε

∫ L

0

η(s)k(s)ds..(37)

It follows from (36), (37), and h′(0) = 0 that∫ L

0

η(s)[−2L(C)k(s) + 4π]ds = 0


Since η(s) is arbitrary smooth function along C, it is a periodic smooth function on [0, L], bya similar version of Lemma 3.3.4 we conclude k(s) = 2π

L(C) . By Proposition 2.2.2, we conclude

C is a circle.

Remark 3.3.11. We give some remarks on the four approaches we described in the proof ofTheorem 3.3.14:

(i) Our exposition of Method 1 assumes the existence of optimal curve with suitable regu-larity, hence not complete. The argument of calculus of variations is used to find theEuler-Lagrange equation satisfied by the critical points, which in turns concludes it mustbe a circle. We only use the optimal curve must be a critical point, instead of being amaximizer.

(ii) Our exposition of Method 2 also assumes the existence of optimal curve with suitableregularity, hence the proof is not complete either. Schmidt (1939) gave a complete proofof isoperimetric inequalities with a method similar to Steiner symmetrization [4].

(iii) It is unclear if Method 3 can be generalized to higher dimensions, for example one mayuse spherical harmonics to replace Fourier series (p.27 in [2]).

(iv) The Brunn-Minkowski Inequality holds on Rn, and Method 4 can be used to proveisoperimetric inequality on Rn.

It is worth pointing out that isoperimetric inequalities can be generalized to non-simple closedcurves. The idea is to use the area weighted with the square of the winding number.

Theorem 3.3.12 (Isoperimetric inequality for non-simple curves, Banchoff-Pohl (1971)5).For any closed parametrized curve C on R2,

[L(C)]2 ≥ 4π

∫p∈R2

[n(C, p)]2dA,(38)

where n(C, p) is the winding number of C with respect to p for any fixed p ∈ R2.6 Moreover, theequality (38) holds if and if C is any of the following

(i) a circle,(ii) a circle winded in the same direction any number of times,(iii) several coincident circles each winded in the same direction any number of times.

For example, c(t) = (cos(2t), sin(2t)) for t ∈ [0, 2π], the LHS of (38) is (4π)2, the RHS is4π × (2)2 × π, and the equality holds and c(t) is a circle winded in the counter-clockwise twice.An example of case (iii) is c(t) where t ∈ [0, 4π], with c(t) = (cos(t), sin(t)) for t ∈ [0, 2π] andc(t) = (cos(t), sin(t)) for t ∈ [2π, 4π]. Here we view c(t) as two coincident circles in the samedirection.

3.3.1. References on isoperimetric problems.

Remark 3.3.13. Several further remarks on isoperimetric inequalities are:

(i) The isoperimetric inequality is true for C1 curves or curves with less regularity, if wecan make sense of the length of the curve and the enclosed area. For example, we coulddefine length for any rectifiable curve. Such a generalization is beyond this course andone may check Federer’s book “Geometric measure theory”.

(ii) The Steiner symmetrization is a powerful method in differential geometry and partialdifferential equations (PDEs). It is useful to study geometric equalities, and relatedto the method of moving planes/spheres in the study of rigidity of solutions to PDEs.

5Banchoff, T. F.; Pohl, W. F. A generalization of the isoperimetric inequality. J. Differential Geometry 6(1971-72), 175-192. https://projecteuclid.org/euclid.jdg/1214430403.

6It is not surprising that winding number appears here. In fact, the rotation number for a closed smooth andregular parametrized curve defined in (21) is exactly the winding number of the tangent mapping of c : [a, b]→ R2

with respect to the origin, here the tangent mapping is defined as e1(t) =c′(t)|c′(t)| .

https://projecteuclid.org/euclid.jdg/1214430403

24 BO YANG

Check Sec 9.5.2 in Evans’s book “Partial differential equations”, 2nd edition, AMS,2010.

(iii) Besides the Steiner symmetrization, Steiner also applied some beautiful arguments inattempting to prove isoperimetric problem, one of results states that for any closedsimple curve which is not a circle there is there is another simple curve which is of thesame perimeter and encloses a larger area.

(iv) Gromov has an interesting proof of the isoperimetric inequality which works for Rn forany n ≥ 2, the proof makes use of the Stokes theorem (see Section 6.6.9 in [3]).

(v) Recently, optimal transportation can be used to prove functional inequalities with geo-metric background, For example, both isoperimetric and Brunn-Minkowski inequalitiescan be proved by optimal transportation. See for example Sec 6.1 in Villani’s book“Topics in optimal tranportation”, AMS, 2003) or pp.103-106 in Figalli’s book (“TheMonge-Ampere equation and its applications”, EMS, 2017).

In the end, let us mention some useful survey articles on isoperimetric inequalities.

(i) Bandle, C.. Dido’s Problem and Its Impact on Modern Mathematics. Notices Amer.Math. Soc. 64 (2017), no. 9, 980-984. http://dx.doi.org/10.1090/noti1576.

(ii) Osserman, R.. The isoperimetric inequality. Bull. Amer. Math. Soc. 84 (1978), no.6, 1182-1238.

(iii) Talenti, G.. The standard isoperimetric theorem. Handbook of convex geometry, Vol.A,73-123, North-Holland, 1993.

(iv) Chavel, I.. Topics in isoperimetric inequalities. Geometry, spectral theory, groups, anddynamics, 89-110, Contemp. Math., 387, Amer. Math. Soc., Providence, RI, 2005.

(v) Ros, A.. The isoperimetric problem. Global theory of minimal surfaces, 175-209, ClayMath. Proc., 2, Amer. Math. Soc., 2005.

3.3.2. An isodiametric inequality. To be done.

Theorem 3.3.14 (Isodiametric inequality on R2). Let C be a simple closed curve of lengthL(C) in R2 and Ω its inner domain. Let the diameter of Ω to be

d(Ω) = supp,q∈Ω

|p− q|.

Then

Area(Ω) ≤ π(d(Ω)

2

)2.(39)

with the equality holds if and only if C is a circle.

a standard proof via Steiner symmetrization is in Evans-Gariepy. or by Brunn-Minkowski.

Area(Ω) ≤ Area(1

2(Ω− Ω)) ≤ Area(B(O,

d

2)).

If we assume convex, then isoperimetric implies isodiametric once we use the Rosenthal-Szaszstheorem and its equality case follows from the fact

For a convex curve in the plane, width is defined by the minimum of all possible distanceof two parallel tangent lines, while diameter d, originall defined as the sup of distance of twopoints in its interior region. It turns out that the diameter is the same as the the maximum ofall possible distance of two parallel tangent lines along the curve.

Rosenthal-Szaszs theorem says p ≤ πd for any convex curves. The equality holds for convexcurves of constant width d. Examples.

the role of support functions. p.56 of Bonnessen-Fenchel book or Fudan exercise book.

3.3.3. A digression on discrete isoperimetric problems. The discrete analogue of Theorem 3.3.14states

http://dx.doi.org/10.1090/noti1576


Theorem 3.3.15. Of all simple n−gons (n ≥ 3) (a simple polygon does not intersect itself andhas no holes.) with the same perimeter, the regular n−gon has the greatest area. In particular,

4n tan(π

n)[Area(P )] ≤ [L(∂P )]2.(40)

holds for any simple n-gon P . Note 4n tan(πn ) 4π as n→ +∞.

In below we outline steps in the proof of Theorem 3.3.15:

(i) There is an elementary proof of Theorem 3.3.15 when n = 3 and 4. The triangle casefollows from the Heron’s formula, and there is a similar formula in the quadrilateralcase.

(ii) Consider all simple n-gon with given ordered side lengths, there is one which attainsthe greatest area, and it is inscribed a circle.

(iii) Consider all simple n-gon with a given perimeter, there is one which attains the greatestarea, and it is the regular n-gon.

Next we give an account of an interesting problem first studied by Erdos (1935) and Nagy(1939). Given a simple polygon P0 in R2, a pocket is a maximal connected region interior tothe convex hull and exterior to the polygon. A (pocket) flip is the reflection of a pocket alongthe bounding edge of the convex hull C(P0). After one flip, we recalculate the convex hull ofthe new polygon P 1 and do a flip again. Repeat this process, we get a sequence of polygonsP, P 1, P 2, · · · , Pn, · · · , . Does this sequence converge?

This problem has been studied by many people, we refer to the book of Demaine-O’Rourke’sbook “Geometric Folding Algorithms Linkages, Origami, Polyhedra” for more details.7 Here aretwo examples of pocket flips, in particular, the second example in Figure 10 can be modified toshow the flip sequence can be arbitrarily long. Precisely, given any positive integer k, there areexamples of quadrilaterals whose flip sequence has at least k terms.

P1: SBT0521857570c05 CUNY758/Demaine 0 521 81095 7 February 25, 2007 7:7

5.3. Reconfiguration Without Self-Crossing 75

Figure 5.22. Flipping several pockets simultaneously can lead to crossings (de Sz. Nagy 1939).

(a)

(b)

Figure 5.23. (a) Flipping a polygon until it is convex. Pockets are shaded. (b) The first flip shownin 3D.

Figure 5.24. Quadrangles can require arbitrarily many flips to convexify.

pocket flipping for arbitrary polygons.4 This beautiful result has been rediscovered andreproved several times, as uncovered by Grunbaum and Toussaint, and detailed in theirhistories of the problem (Grunbaum 1995; Toussaint 2005).

Before we come to Nagy’s finiteness proof, it is algorithmically important to mentionthat the number of required flips can be arbitrarily large in terms of the number of ver-tices, even for a quadrangle, a fact first proved by Joss and Shannon in 1973. Figure 5.24

4 http://www.cs.mcgill.ca/∼cs507/projects/1998/mas/.

Figure 9. Copied from p.75 in Demaine-O’Rourke’s book

P1: SBT0521857570c05 CUNY758/Demaine 0 521 81095 7 February 25, 2007 7:7

5.3. Reconfiguration Without Self-Crossing 75

Figure 5.22. Flipping several pockets simultaneously can lead to crossings (de Sz. Nagy 1939).

(a)

(b)

Figure 5.23. (a) Flipping a polygon until it is convex. Pockets are shaded. (b) The first flip shownin 3D.

Figure 5.24. Quadrangles can require arbitrarily many flips to convexify.

pocket flipping for arbitrary polygons.4 This beautiful result has been rediscovered andreproved several times, as uncovered by Grunbaum and Toussaint, and detailed in theirhistories of the problem (Grunbaum 1995; Toussaint 2005).

Before we come to Nagy’s finiteness proof, it is algorithmically important to mentionthat the number of required flips can be arbitrarily large in terms of the number of ver-tices, even for a quadrangle, a fact first proved by Joss and Shannon in 1973. Figure 5.24

4 http://www.cs.mcgill.ca/∼cs507/projects/1998/mas/.

Figure 10. Copied from p.75 in Demaine-O’Rourke’s book

The main result states:

7Demaine, E. D. and O’Rourke, J.. “Geometric folding algorithms. Linkages, origami, polyhedra.” CambridgeUniversity Press, 2007.

26 BO YANG

Theorem 3.3.16 (Due to many people). Any simple n-polygon P admits a finite sequence ofpocket flips, with last member being a convex polygon, of k ≤ n sides.

The proof we present is from a paper by Demaine-Gassend-O’Rourke-Tousaint “Polygonsflip finitely: flaws and a fix”. Let P = P0, assume there is an infinite sequence of polygonsP k = vk0 , · · · , vkn, where each P k derived from the previous P k−1 by exactly one pocket flip.We will find a contradiction by showing

(i) The distance from each vertex vki to any fixed point x ∈ P 0 is nondecreasing withrespect to k.

(ii) Each corresponding vertices of P k converges, hence P k converges a limit polygon P ∗

(iii) The directed angle θki at vertex vki converges, and the limit limk→∞ θki ∈ (0, 2π).(iv) Any vertex vki that moves infinitely many times converges to a flat vertex of P ∗.(v) The infinite sequence P k cannot exist, it must end after a finite number of terms.

According to Wegert,8 Theorem 3.3.16 is the motivation for him to propose the famouspentagon problem in IMO 1886. It remains an interesting question to bound the number ofterms of the flip sequence on Theorem 3.3.16 in terms of some natural quantities of the initialpolygon P .

3.4. Euler’s elastica. In this subsection, we discuss Euler’s elastica from calculation of varia-tions. The question is

Question 3.4.1. Minimize∫Ck2ds among all smooth curve C parametrized by c(t) : [0, t0]→ R2

with both end points c(0) and c(t0) fixed, and the total length fixed.

The problem was proposed by Bernoulli in 1691 and solved by Euler in 1744. The optimalcurve have been called “elastica” ever since then. It was also rediscovered by many authors invarious works. In particular Euler’s elastica has found several applications including:

(i) Elasticity, computer graphics, and computer vision.9

(ii) Modelling of the shape of a river bend. This “energy-minimizing” feature of the Euler’selastica is useful for the river-bed shape hypothesis. We refer to the book by Varlamov-Aslamazov,10and copy the following picture.

February 15, 2012 16:54 WorldScientific/ws-b9x6-0 Main

How meanders are formed 11

periodic structures. In Figure 1.6 you can see a real river-bed and the Euler

curve (the dashed line) that approximates its shape best of all.

Fig. 1.6: a --- Modeling meanders in laboratory.

Channel-bed develops periodic Euler bends (the

dashed line). b --- A real river and the nearest

Euler curve (the dashed line).

By the way, the word “meander” itself is of ancient origin. It comes

from the Meander, a river in Turkey famous for its twists and turns (now

called the Menderes). Periodic deflections of ocean currents and of brooks

that form on surfaces of glaciers are also called meanders. In each of these

cases, random processes in a homogeneous medium give rise to periodic

structures; and though the reasons that bring meanders about may differ,

the shape of the resulting periodic curves is always the same.

z zz

JJ

JJ?

Show that the surface of a uniformly (rigidly) re-

volving liquid takes a parabolic form.

Figure 11. copied from p.11 in Varlamov-Aslamazov

8Wegert, E. and Reiher, C.. “Relaxation procedures on graphs”. Discrete Appl. Math. 157 (2009), no. 9,2207-2216.

9Horn, B. K. P.. The curve of least energy. ACM Trans. Math. Software 9 (1983), no. 4, 441-460. andMumford, David. Elastica and computer vision. Algebraic geometry and its applications, 491-506, Springer,

New York, 1994.10 See p.10, Chapter 1 ‘Meandering down to the sea’ in Varlamov, A.; Aslamazov, L.. The wonders of physics,

3rd Edition, World Scienfitic, 2012. Available at https://www.worldscientific.com/doi/suppl/10.1142/8300/

suppl_file/8300_chap01.pdf

https://www.worldscientific.com/doi/suppl/10.1142/8300/suppl_file/8300_chap01.pdf

https://www.worldscientific.com/doi/suppl/10.1142/8300/suppl_file/8300_chap01.pdf


Proposition 3.4.2 (ODE satisfied by elastica). Let c(s) (s being its arc length) denote theoptimal solution of Question 3.4.1, then

2d2k

ds2+ k3 = λk(41)

for some constant λ.

Proof of Proposition 3.4.2. We follow p.500 in Mumford’s paper “Elastica and computer vision”mentioned in the above footnote.

Since c(s) is parametrized by is its arc length, we consider a deformation of c(s) with its endpoints fixed, and the total length L fixed. Given any smooth function δ on [0, L] vanishing atend points. let cε(s) be a deformation of c(s) along its normal direction

cε(s) = c(s) + εδ(s)e2(s)(42)

for any real number ε. Recall e1(s), e2(s) is the Frenet frame of c(s).

Let s, e1, e2, and k denote the corresponding quantity for the new curve cε(s). we compute

dcε(s)

ds= (1− εkδ(s))e1(s) + εδ′(s)e2(s);(43)

ds

ds=√

((1− εkδ(s)))2 + ε2(δ′(s))2 = (1− εkδ(s))√

1 +O(ε2)

= 1− εkδ(s) +O(ε2), here we used√

1 + x = 1 +x

2+O(x2);(44)

e1 =1

|dsds |dcε(s)

ds=

(1− εkδ(s))e1 + εδ′(s)e2

1− εkδ(s) +O(ε2)

= e1 + εδ′(s)e2 +O(ε2).(45)

e2 = −εδ′(s)e1 + e2 +O(ε2).(46)

de1

ds=

1

1− εδk(s) +O(ε2)

[d(e1 + εδ′(s)e2 +O(ε2))

ds

]= (1 + εδk(s) +O(ε2))

[(k + εδ′′(s))e2 − εktδ′(s)

]= (k + εδk2 + εδ′′(s))e2 − (εkδ′(s))e1 +O(ε2).(47)

k =de1

ds· e2 = k + εδk2 + εδ′′(s) +O(ε2).(48)

Next we compute the derivatives of∫ds and

∫k2ds with respect to ε at ε = 0.

d

dε

∫ds∣∣∣ε=0

=d

dε

∫(1− εδk +O(ε2))ds

∣∣∣ε=0

= −∫δk(s)ds;(49)

d

dε

∫k2ds

∣∣∣ε=0

=d

dε

∫ [k + εδk2 + εδ′′(s) +O(ε2)

]2ds∣∣∣ε=0

=

∫2kδ′′(s) + k3δds

=

∫(2k′′ + k3)δ(s)ds, where we use integration by parts.(50)

So far, we have proved that among all smooth curves with end points fixed, the minimizer ofthe

∫(k2 + λ)ds, if exists, must satisfy the equation

2d2k

ds2+ k3 = λk(51)

If we consider a two-parameter deformation as c(s) + (ε1δ1 + ε2δ2)e2(s), then we may applythe method of Lagrange multipliers as in Method 1 in proof of Theorem 3.3.14 to show:

28 BO YANG

Among all smooth curves with end points fixed and fixed total length L, the minimizer ofthe

∫k2ds, if exists, must satisfy the equation

2d2k

ds2+ k3 = λk(52)

for some constant λ.

We have not answered whether the minimizer of Question 3.4.1 necessarily exists. Moreover,once we derived (41), it is natural to know whether we can solve it. It turns out its solutionrelies on elliptic integrals and can be solved in terms of theta functions. We refer to the followingtwo papers on further mathematical aspects of elastica.

(i) Bryant, Robert; Griffiths, Phillip. Reduction for constrained variational problems and∫12k

2 ds. Amer. J. Math. 108 (1986), no.3, 525-570.(ii) Mumford, David. Elastica and computer vision. Algebraic geometry and its applica-

tions, 491-506, Springer, New York, 1994.


4. Surfaces in R3: local theory

Definition 4.0.1 (a (parametrized) surface patch). Let U be an open set in R2, a smoothmap f : U → R3 such that for any u ∈ U , dfu (the derivative of f at u ∈ U) is an injectivelinear map from TuU and R3. Here TuU is the tangent space of U at u, it can be identified as

R2 = spane1 = (1, 0)T , e1 = (0, 1)T = span ∂

∂u1,∂

∂u2.

In other words, if write f(u1, u2) = (x(u1, u2), y(u1, u2), z(u1, u2)), the Jacobian matrix

duf =

∂x∂u1

∂x∂u2

∂y∂u1

∂y∂u2

∂z∂u1

∂z∂u2

has rank 2 for any (u1, u2) ∈ U .

We call Tuf.= dfu(TuU) = spandfu2

, dfu2 ⊂ R3 the tangent space of f at u ∈ U , where

we simply write dfu(e1) = fu1and dfu(e2) = fu2

.

This section is a summary of main results in Chapter 3 in [9]. First we introduce the first andsecond fundamental forms for a (parametrized) surface patch, then we define mean curvaturesand Gauss curvatures, and finally state Theorem Egregium due to Gauss and the basic existencetheorem of surfaces with prescribed the first and second fundamental forms.

4.1. The first and second fundamental forms. Recall a fact on bilinear forms, let L : S → Tbe a linear map between two vector spaces, let β be a bilinear form on T , it induces anotherbilinear form on S by α(X,Y ) = β(LX,LY ). Then β being positive definite implies so being αif L is injective.

Definition 4.1.1 (The first fundamental form). For any u ∈ U , Iu is a bilinear formdefined on TuU = R2, induced from the standard Euclidean inner product via dfu : TuU → R3.

Iu(X,Y ) = dfu(X) · dfu(Y ).

Note that Iu is positive definite and one may write the corresponding matrix representation as

gij.= Iu(ei, ej) = Iu(

∂

∂ui,∂

∂uj).

Definition 4.1.2 (The Gauss map). For any u ∈ U , define the unit normal vector of f by

nu =fu1 × fu2

|fu1× fu2

| .

Then Gauss map n : U → S2 ⊂ R3 defined by n(u) = nu.

Note that the unit normal vector might be differ by a minus sign if we choose a differentparametrization. For example, if we consider S2

+ \ N where N is the north pole by f1(u, v) =

(u, v,√

1− u2 − v2) where U1 = 0 < u2 + v2 < 1. then n = f1. However if we considerf2(u, v) = (cosu cos v, cosu sin v, sinu) where U2 = (0, π2 )× R, then n = −f2.

Definition 4.1.3 (The second fundamental form). For any u ∈ U , IIu is a bilinear formdefined on TuU = R2.

IIu(X,Y ) = −dnu(X) · dfu(Y ).

Since n · n = 1, we have n · dnu = 0, we conclude dnu(TuU) ∈ Tuf . Since dfu is injective, onemay also view IIu as a bilinear form defined on Tuf . 11 Let us write the correspondingmatrix representation as hij

.= IIu(ei, ej).

11We adopt this point of view in Definition 4.1.5 of the Weingarten map.

30 BO YANG

Note that

h12 = −dnu(e1) · dfu(e2) = −∂(nu · dfu(e2))

∂u1+ nu ·

∂2f

∂u1∂u2,

h21 = −dnu(e2) · dfu(e1) = −∂(nu · dfu(e1))

∂u2+ nu ·

∂2f

∂u2∂u1.

We have the following observation.

Proposition 4.1.4. IIu is a symmetric bilinear form for any u ∈ U .

Let us recall another fact from linear algebra: any symmetric bilinear form α on the Euclideanspace V can be associated with a self-adjoint linear transformation A : V → V by α(X,Y ) =AX · Y .

Definition 4.1.5 (The Weingarten map). For any u ∈ U , define the Weingarten map to bethe linear transformation L : Tuf → Tuf associated to IIu. In other words,

IIu((df−1u )(Z), (df−1

u )(W )) = (−dnu df−1u )(Z) ·W = L(Z) ·W, where Z,W ∈ Tuf.

Since all eigenvaluess of any self-adjoint linear transformation on a vector space must be real,this allows us to define

Definition 4.1.6 (Principal curvatures, Gauss curvatures, and mean curvatures). The eigenval-ues k1(u) and k2(u) of the Weingarten map L : Tuf → Tuf are called principal curvatures,and the corresponding eigenvectors are called principal directions.

The Gauss curvature and the mean curvature are defined by

K(u) = k1(u)k2(u), H(u) =1

2(k1(u) + k2(u)).

Proposition 4.1.7. Assume under the basis Tuf = spanfu1, fu2, the Weingarten map has

the follow expression:

−[nu1 nu2

]= L

[fu1 fu2

]=[fu1 fu2

]·[a1

1 a12

a21 a2

2

].

Then we conclude12

− ni =

2∑k=1

fkaki , hij =

2∑k=1

gikakj ;(53)

aki =

2∑j=1

gkjhji, here gij denote the inverse matrix of gij ;(54)

K(u) =det(hij)

det(gij), 2H(u) =

2∑i=1

2∑j=1

gijhji.(55)

4.1.1. The geometric meaning of the second fundamental form.

Definition 4.1.8 (an umbilic point and a planar point). Let f : U → R3 be a surface patch. Apoint u0 ∈ U is called umbilic if k1(u0) = k2(u0), and an umbilic point with k1(u0) = k2(u0) = 0is called planar.

Proposition 4.1.9 (surfaces with entirely umbilic points). A surface patch with all umbilicpoints must be a piece of plane or a piece of sphere.

Recall one could prove invariant properties of the first and second fundamental forms fromtheir definitions.

12Caution: The matrix aki is not symmetric in general, unless fu1 and fu2 form an orthonormal basis.


Proposition 4.1.10 (Invariance of the first and second fundamental forms). (1) Let R be an

isometry of R3, then consider two surface patches f = R f and f both defined on U , we have

Iu(X,Y ) = Iu(X,Y ), X, Y ∈ TuU ;

IIu(X,Y ) = ±IIu(X,Y ) X,Y ∈ TuU,where ± depends on the sign of the determinant of the orthogonal part of R.

(2) Let φ be a diffeomorphism V → U , then consider two surface patches f = f φ and fwith the same image in R3, we have

Iv(X,Y ) = Iφ(v)(dφ(X), dφ(Y )), X, Y ∈ TvV ;

IIv(X,Y ) = ±IIφ(v)(dφ(X), dφ(Y )), X, Y ∈ TvV,where ± depends on the sign of the Jacobian determinant of φ.

We call such a φ a change of coordinates for the surface patch f . In particular, these implythe Gauss curvature is invariant under change of coordinates and isometries of R3, andthe mean curvature is invariant under orientation-preserving change of coordinatesand isometries of R3.

Let we study the Taylor expansion of a surface on a neighborhood of u0 ∈ U under theframe fu1(u0), fu2(u0), n(u0). The following result can be viewed as a surface analogue ofProposition 2.3.2 for space curves.

Proposition 4.1.11. Given any surface patch f : U → R3 and u0 ∈ U . Then there existsU0 ⊂ U an open neighborhood of u0 such that for any u = (u1, u2) ∈ U0,(

f(u)− f(0))· n(0) =

1

2

∑i,j

huiuj (u0)uiuj + o(|u|2).(56)

In fact, a stronger conclusion holds if you allow a change of variables. Namely, there is adiffeomorphism φ : V → U1 ⊂ U where U1 is some open neighborhood of u0 with φ(0) = u0 such

that f.= f φ satisfies

f(u)− f(0) = v1fv1 + v2fv2 +(1

2

∑i,j

hvivj (0)vivj

)n(0) + o(|u|2).(57)

Moreover, vi = ui − u0i + o(|u− u0|) and hvivj (0) = huiuj (u0).

Proposition 4.1.11 is proved on p.50 in [9], It states that under a suitable change of variablesthe second fundamental form at a given point measures the second order deviation of the surfacefrom its tangent plane at that point.

4.2. Fundamental theorem on surfaces with prescribed fundamental forms.Let f : U → R3 be a surface patch and u ∈ U , we call the frame fu1

, fu2, n(u) the Gauss

frame of f . Just as the role of Frenet equations in the study of curves, we are interested in theequation describing the movement of such a frame as u varies on U .

Theorem 4.2.1. Let 1 ≤ i, j ≤ 2, fi.= fui , fij

.= fuiuj = ∂

∂uj( ∂f∂ui ), and ni

.= nui . If we define

Γlik.=

2∑j=1

1

2glj(

∂gij∂uk

+∂gkj∂ui

− ∂gik∂uj

),(58)

and

Γikp.=

2∑l=1

Γlikglp =1

2(∂gip∂uk

+∂gkp∂ui

− ∂gik∂up

).(59)

Here Γikp and Γlik and are called Christoffel symbols of the first and the second kindrespectively.

32 BO YANG

Then

fik =∑l

Γlikfl + hikn, ni = −∑l,k

hilglkfk.(60)

See p.62 of [9] for a proof of the above. We just remark that the crucial observation is

∂gij∂uk

=∑p

Γpikgpj +∑p

gipΓpjk.(61)

Theorem 4.2.2. (1) (Gauss equations) Let us introduce

Rmikj :=∂Γmik∂uj

−∂Γmij∂uk

+∑l

(ΓlikΓmlj − ΓlijΓmlk).(62)

Then we may compare the fi-components of fijk = fikj and show

Rmikj =∑l

((hikhjl − hijhkl)glm

).(63)

(2) (Codazzi-Mainardi equations) If we compare the n-compnents of fijk = fikj or equiv-alently, compare the fi and n components of nij = nji, we may show:∑

l

(Γlikhlj − Γlijhlk) +∂hik∂uj

− ∂hij∂uk

= 0.(64)

Proof of Theorem 4.2.2. We use the Einstein notation where ApBp :=∑pA

pBp, then:

fikj = (Γpikfp + hikn)j

=∂Γpik∂uj

fp + Γpik(Γqpjfq + hpjn) +∂hik∂uj

n− hikhjpgpqfq

= (∂Γpik∂uj

+ ΓqikΓpqj − hikhjqgpq)fp +(

Γpikhpj +∂hik∂uj

)n.

Definition 4.2.3 (Curvature tensor of f). Pick any four tangent vectors from Tuf = spanf1, f2:

X =∑i

Xifi, Y =∑i

Y ifi, Z =∑i

Zifi, W =∑i

W ifi.(65)

Then we introduce introduce a multi-linear map R : Tuf × Tuf × Tuf × Tuf → R.

R(X,Y, Z,W ) =

2∑i,l,k,j=1

RilkjXiY lZkW j .(66)

where

Rilkj =∑p

Rpikjgpl.(67)

According to our notation, R(fi, fl, fk, fj) = Rilkj.

Immediately, we may rewrite (63) as

Rilkj = hikhjl − hijhkl.(68)

Caution: Here our notation on Rmikj and Rijkl in (62) and (67) is the same as

in Def 3.8.4 on p.63 in [9]. There is an alternative choice of notations which aremore consistent with those in textbooks on Riemannian geometry. For example, indo Carmo’s book’s on Riemannian geometry, the right hand side of (62) is definedas Rmkji and the corresponding (67) should be Rkjil = Rpkjigpl. In either choice of

notations, the Gauss equation (68) is the same.


It is direct to deduce either from (62) and (67), or from (68) that Rikjl = −Rkijl = −Riklj =Rjlik, and

R(X,Y, Z,W ) = −R(Y,X,Z,W ) = −R(X,Y,W,Z) = R(Z,W,X, Y ).(69)

Since all the indices are either 1 or 2, the only non-vanishing terms are R1212 = −R2112 =−R1221 = R2121.

Similarly, we find the original 16 equations in (63) and 8 equations in (64) reduces to a singleequation in (63) and 2 equations in (64). Let us summarize in below:

R1212 = h11h22 − h12h21,∑2k=1(Γk11hk2 − Γk12hk1) + ∂h11

∂u2− ∂h12

∂u1= 0,∑2

k=1(Γk21hk2 − Γk22hk1) + ∂h12

∂u2− ∂h22

∂u1= 0.

(70)

Theorem 4.2.4 (Theorema Egregium). The Gauss curvature can be computed from the firstfundamental form and its first and second order partial derivatives.

K(u) =R1212

det(gij).(71)

Here R1212 is defined in (67) and (62).

Example 4.2.5 (Isothermal coordinates). Given any surface patch f : U → R3 and u0 ∈ U ,there exists a change of coordinates φ : V → U0 where U0 is some open neighborhood of u0 such

that the first fundamental form for f(v).= f φ(v) is

gij = Eδij(72)

where E is a positive function and δij is the Kronicker symbol. Coordinates under which the firstfundamental form satisfies (72) are called isothermal coordinates. It is a nontrivial resultdue to Korn and Lichtenstein 13 that we can always find isothermal coordinates near any pointon surface patches.

It is direct to check that under isothermal coordinates, Gauss curvature has a nice formula

K(v1, v2) = − 1

2E(∂2 lnE

∂v21

+∂2 lnE

∂v22

) = − 1

2E∆ lnE.(73)

where ∆ is the usual Laplace operator on R2.

As a slightly more general case, we consider orthogonal coordinates on a surface patchf : U → R3, i.e. assume its first fundamental form satisfies g11 = E, g22 = G, and g12 = g21 = 0.a lengthy calculation following Theorem 4.2.4 leads to:

K(v) = − 1

2√EG

((Eu2√EG

)u2+ (

Gu1√EG

)u1).(74)

Next we turn to the existence problem of a surface patch with prescribed the first and secondfundamental forms. Note that the first and second fundamental forms of a surface patch is nottotally independent, it must satisfy the Gauss and Codazzi-Mainardi equations (63) and (64) inTheorem 4.2.2. Are (63) and (64) the only obstruction to solve the desired surface patch? Sincewe use fijk = fikj and nij = nji to derive (63) and (64), presumably we may come up with newequations if we take derivatives further, such as in fijkl = fijkl and nijk = nikj? The followingresult is a surface analogue of Theorem 2.4.1 which is on space curves.

Theorem 4.2.6 (Fundamental theorem on surfaces with prescribed first and second fundamen-tal forms). Let U be a simply-connected open set in R2. Suppose Iu and IIu are two bilinearforms defined on TuU for any u ∈ U , their coefficients gij and hij are C2 and C1 functions of

13More precisely they proved this result under the assumption that components gij are Holder continuous

near u0 ∈ U . Moreover, each entry of the Jacoian matrix ∂v∂u

is Holder continuous. In our case if we assume

f : U → is C4, it follows from their result that E is C2 under isothermal coordinates v. Therefore, (73) makes

sense.

34 BO YANG

u respectively. If Iu is positive definite for any u, and Iu and IIu satisfy Gauss and Codazzi-Mainardi equations (63) (or (68) equivalently) and (64) (or a more explicit form in (70)). Thenthere exists a surface patch f : U → R3 where f is a C3 map such that the first and secondfundamental forms of f are Iu and IIu.

Moreover, any two surfaces f : U → R3 and f : U → R3 which have the same Iu and IIusatisfy f(u) = R f(u) for some isometry R of R3.

An ‘almost complete’ proof of Theorem 4.2.6. The proof of Theorem 4.2.6 is on p.65 in [9]. Itis our purpose here to provide some details on some crucial points.

Step 1: Solve a first order linear PDE system.The crucial ingredient is an existence and uniqueness result of certain system of first order

linear partial differential equations (PDEs). It is our purpose here to provide more backgroundon this PDE system.14

Let U an open set in RN , F (U) = (F1, · · · , Fn), and each Fi : U → Rn, as a column vectorin Rn. Assume Γi where 1 ≤ i ≤ N is a n × n matrix-valued function on U . A Pfaff systemis the following system of linear PDEs.

∂∂ui

[F1(u), · · · , Fn(u)] = [F1(u), · · · , Fn(u)] Γi(u), u ∈ U, 1 ≤ i ≤ N.F (u0) = F 0, u0 ∈ U, and F 0 is a n× n constant matrix.

(75)

If we assume F (u) is a nonsingular n× n matrix for every u ∈ U , a direct calculation shows

∂2F

∂uk∂ui=

∂2F

∂uk∂ui⇔ ∂Γk

∂ui− ∂Γi∂uk

+ ΓiΓk − ΓkΓi = 0. 1 ≤ i, k ≤ N.(76)

The main result on the solvability of Pfaff system is summarized in the following:

Lemma 4.2.7. Let U be a simply-connected open set RN , and each Γi(1 ≤ i ≤ N) be a C1

n × n matrix-valued map on U which satisfies (76). Then the Pfaff system (75) has a uniquesolution F which is a C2 matrix-valued map on U for any initial condition F (u0) = F 0.

Proof of Lemma 4.2.7. The proof on the existence part of Lemma 4.2.7 lies in the idea ofreduction to a ODE system. Namely, we consider d

dtF (ξ(t)) where 0 ≤ t ≤ 1 a C1 pathξ(t) = (ξ1(t), , · · · , ξn(t)) from u0 to any point u1:

dFi(ξ(t))dt =

∑nj,k=1 Fj(ξ(t))Γ

jki(ξ(t))

dξk(t)dt ,

F (ξ(0)) = F 0.(77)

This is a first order linear ODE system, and a solution always exists by Theorem 1.4.2 andTheorem 1.4.5. Then one must show F (u1) = F (ξ(1)) is independent of choice of the path ξ(t),this is where simply-connectedness of U comes into play. Rougly speaking, consider ξ1(t) andξ2(t) two C1 paths connecting u0 and u1, one need to consider a C2 homotopy map15

χ(t, s) = (χ1(t, s), χ2(t, s)) : [0, 1]× [0, 1]→ U

where

χ(0, s) = u0, χ(0, s) = u1, χ(t, 0) = ξ1(t), χ(t, 1) = ξ2(t).

It suffices to show ∂F (χ(1,s))∂s = 0 to conclude F (ξ1(1)) = F (ξ2(1)). Follow Sec 6.20 in Ciarlet’s

book, we define a new quantity

Pi(t, s).=∂Fi(χ(t, s))

∂s−

n∑j,k=1

ΓkijFk(χ(t, s))∂χj(t, s)

∂s.(78)

14A nice reference: Sec 6.20 of the book by Ciarlet, P. G. “Linear and nonlinear functional analysis with

applications”. Society for Industrial and Applied Mathematics, 2013. In fact, we follow this approach in the

proof Lemma 4.2.7.15We assume χ is C2 since we need ∂2χ

∂s∂t= ∂2χ

∂t∂s.


Note that we have (76) and the symmetry Γkij = Γkji, a beautiful calculation in Sec 6.20in Ciarlet’s book leads to the following identity

∂Pi(t, s)

∂t=

n∑j,k=1

Γkij∂χj(t, s))

∂tPk(t, s).(79)

For any fixed s ∈ [0, 1], we consider the following initial value problem for an ODE.dPi(t,s)dt =

∑nj,k=1 Γkij

dχj(t,s))dt Pk(t, s),

Pi(0, s) = 0.(80)

From the uniqueness of solutions to ODE systems (Theorem 1.4.2), we conclude Pi(t, s) = 0for any t, s ∈ [0, 1], In particular, we have

Pi(1, s) =∂Fi(χ(1, s))

∂s−

n∑j,k=1

ΓkijFk(u1)∂χj(1, s)

∂s=∂Fi(χ(1, s))

∂s= 0.(81)

This establishes the desired result F (ξ1(1)) = F (ξ2(1)), therefore we may well define F (u1) =F (ξ(1)) for any C1 curve ξ(t) connecting u0 to u1. It remains to show F constructed above isindeed a C2 map on U and solves (75) to finish the existence part of Lemma 4.2.7. To that end,we argue by three steps.

Step A: We show F is continuous on U .For any u1 ∈ U , consider a small neighborhood B(u1, δ) such that ξ(u, t) where 0 ≤ t ≤ 1 is

a C1 curve connecting u0 and u. We further assume dξ(u,t)dt is continuous on u for any t ∈ [0, 1].

Consider the following IVP with a parameter u ∈ B(u1, δ):dFi(ξ(u,t))

dt =∑nj,k=1 Fj(ξ(u, t))Γ

jki(ξ(u, t))

dξk(u,t)dt ,

F (ξ(0)) = F 0.(82)

Then it follows from the continuous dependence of solutions to ODEs (Corollary 1.4.8) thatFi(ξ(u, t)) is a continuous function on u for t = 1. Hence F (u) = Fi(ξ(u, 1)) is continuous on u.

Step B: We show F is C1 on U .Given any point u ∈ U and fix 1 ≤ p ≤ n and ep the unit vector whose p-th component is

1 and zero elsewhere, we can always construct a C1 path from ξ(t) where 0 ≤ t ≤ 1 such thatξ(0) = u0, ξ(τ) = u for some τ ∈ (0, 1). Moreover, there exists a sub-interval Iδ = (τ − δ, τ + δ)such that the restriction of ξ(t)|Iδ = u + (t − τ)ep. From (77) we do a Taylor expansion neart = τ .

Fi(ξ(t)) = Fi(ξ(τ)) + (t− τ)dFi(ξ(τ))

dt|t=τ + o(t− τ)

= Fi(ξ(τ)) + (t− τ)

n∑j,k=1

Fj(ξ(τ))Γjki(ξ(τ))dξk(t)

dt|t=τ + o(t− τ).(83)

Therefore,

Fi(u+ (t− τ)ep) = Fi(u) + (t− τ)

n∑j=1

Fj(u)Γjpi(u) + o(t− τ),(84)

which implies

∂Fi(u)

∂up=

n∑j=1

Fj(u)Γjpi(u).(85)

Since Γjpi(u) is C1, it follows from Step A that all the first partial derivatives of F are

continuous, hence F is C1 on U .Step C: F is C2 on U .Since F is C1, (85) implies F is C2 on U , again note that Γjpi(u) is C1.

36 BO YANG

The uniqueness part of Lemma 4.2.7 is easier and can be directly proved by the uniquenessof the ODE system satisfied by F (ξ(t)), we omit here as the same argument will appear in Step3 in below.

Now that Lemma 4.2.7 is true, in our case we choose N = 2, n = 3, F 0 = [X1, X2, N ] where

Xi ·Xj = gij(u0), Xi ·N = 0, N ·N = 1.(86)

Recall any positive definite matrix [gij ]2×2 can be written as ATA for some non-singular matrixof order 2. Let us pick column vectors of A and add to each of them a third component as zero,call it X1 and X2, then X1 and X2 are vectors in R3 and we further set N = X1×X2

|X1×X2| . Moreover,

Γi =

Γ11i Γ1

2i −∑2p=1 hipg

p1

Γ21i Γ1

2i −∑2p=1 hipg

p2

h1i h2i 0

i = 1, 2.(87)

It can be checked (as in the proof of Theorem 4.2.2) that (76) is equivalent to (63) and(64). From this point, we may conclude from Lemma 4.2.7 that (75) has a unique solution[F1(u), F2(u), F3(u)] with our choice of Γi and F (u0) = F 0.

Step 2: Construct the surface patch map.Define f : U → R3 by

f(u) =

∫C

F1(u)du1 + F2(u)du2(88)

where C is any curve connecting u0 to u. Since two equations from the system (75) reads∂F1

∂u2= ∂F2

∂u1, here we make use of the symmetry of Christoffel symbols Γkij = Γkji, the integral

(88) is independent of choice of the path because U is simply-connected.Since f1 = F1 and f1 = F2 by definition (88). To conclude f is a surface patch, it remains to

check F1(u) and F2(u) are linearly independent for any u ∈ U , so far we only know this is trueat u0 ∈ U .

Step 3: Verify f has the desired first and second fundamental forms.On p.65 in [9] Klingenberg derived a new linear PDE system satisfied by G : U → R6 where

G(u).= [G1, G2, G3, G4, G5, G6]

.= [F1 · F1, F1 · F2, F2 · F2, F1 · F3, F2 · F3, F3 · F3].(89)

Indeed, applying (75) and the definition of Γi in (87), we can show

∂G

∂uk= G

2Γ11i Γ1

2i 0 −∑2p=1 hipg

p1 0 0

2Γ21i Γ1

1i + Γ22i 2Γ2

1i −∑2p=1 hipg

p2 −∑2p=1 hipg

p1 0

0 Γ22i 2Γ2

2i 0 −∑2p=1 hipg

p2 0

2h1i h2i 0 Γ1i1 Γ1

i2 −2∑2p=1 hipg

p1

2h1i h1i 2h2i Γ2i1 Γ2

i2 −2∑2p=1 hipg

p2

0 0 0 hi1 hi2 0

.

(90)

Here i = 1, 2, and G satisfies the initial condition G(u0) = [g11(u0), g12(u0), g22(u0), 0, 0, 1].Now we conclude the solution (if any) of this new system (90) is unique. Otherwise, there

must be two solutions G and G with G(u1) 6= G(u1) for some u1 ∈ U . Now pick any curve ξ(t)

where 0 ≤ t ≤ 1 connecting u0 and u1, G(ξ(t)) and G(ξ(t)) solve the same ODE with the same

initial value, we may apply Theorem 1.4.2 to conclude G(ξ(1)) = G(ξ(1)) which is contradiction.Next back to our case, since we already knew [g11(u), g12(u), g22(u), 0, 0, 1] is a solution of

this new system, it follows from the uniqueness property we just explained that

[F1 · F1, F1 · F2, F2 · F2, F1 · F3, F1 · F3, F3 · F3] = [g11(u), g12(u), g22(u), 0, 0, 1].

Thus far, we proved the map f constructed in Step 2 satisfies fi · fj = gij on U . Since [gij ] ispositive definite, this also implies f1 and f2 are linearly independent on U , hence f is indeed a


well-defined surface patch. Moreover, F3 is the normal vector of f , since f1, f2, F3 is postivelyoriented at u0, we conclude f1, f2, F3 is positively oriented everywhere on U .

Finally, we see the second fundamental form of f satisfies

−ni · fi = n · fij = F3 ·∂Fj∂ui

= hij ,

where the last equality follows from (75) and the definition of Γi.Step 4: Uniqueness of the surface patch up to isometry of R3.This again follows from the uniqueness result of the PDE system (75), which can be proved

by a similar argument as in Step 3. We refer to p.66 in [9] for the rest of proof.

Remark 4.2.8. Two further remarks are in order:

(i) Lemma 4.2.7 is related to some integrability-type results in terms of exterior differentialforms due to Frobenius. One form of such results can be found in Flanders’s book.16

(ii) The regularity assumption made on Theorem 4.2.6 can be weakened. For example, anexistence theorem still holds if I is C1 and II is continuous as a matrix-valued map onU , the resulting f : U → R3 is C2, in this case, conditions (63) and (64) are assumedto hold in the sense of distributions. There are also generalizations to Sobolev spaceswhere f ∈W 2,p. See Sec 8.16 (in particular p.652) in Ciarlet’s book for more details.

4.3. Examples of surfaces. We are mainly interested in following examples of surface patches.

(i) Surfaces of constant (Gauss) curvature.(ii) Surfaces of H = 0, they are also called minimal surfaces.(iii) Surfaces with constant mean curvature (CMC surfaces for short).(iv) Weingarten surfaces.17 Given a surface patch f : U → R3, assume two principal

curvatures k1(u) ≥ k2(u). (i.e. k1 = H +√H2 −K, k2 = H +

√H2 −K.) Let

Ω = (k1(u), k2(u)) ∈ R2 | u ∈ U.Ω is called the curvature diagram of f . We assume that there exists a differentiablefunction ϕ defined on Ω such that ∇ϕ 6= 0 and ϕ(k1(u), k2(u)) = 0 for all u ∈ U .Obviously Weingarten surfaces include the previous three examples.

Note that all the above classes of surfaces is defined for a (local) surface patch. In particular,we are interested in looking for examples among surfaces of revolution, ruled surfaces, andgraphs of functions.

4.3.1. Surfaces of revolution. If we revolve the curve (h(u), 0, k(u)) with (h′(u))2 + (k′(u))2 6= 0and h ≥ 0 about z-axis, The surface of revolution obtained has the following parametrization.

f(u, v) = (h(u) cos v, h(u) sin v, k(u)), v ∈ [0, 2π) or v ∈ R.

16On p.97 of Flanders “Differential forms with applications to the physical sciences”. 2nd edition. Dover

Publications, 1989. It states the following: Assume n = r+s, let ω1, · · · , ωr be 1-forms on Rn linearly independentnear 0, Suppose there are 1-forms θij satisfying

dωi =

r∑j=1

θij ∧ ωj , 1 ≤ i ≤ r.

Then there exist functions f ij and gj at a neighborhood of 0 such that

ωi =r∑j=1

f ijdgj , 1 ≤ i ≤ r.

This result implies the local solvability of certain first order PDE system. See p.104 in this book for an examplewhere a nonlinear Pfaff system is solved.

17For more background on Weingarten surfaces, see p.127 of Hopf’s book. “Differential geometry in thelarge”. Lecture notes in Mathematics, No. 1000, Springer, 1983.

38 BO YANG

We list some general facts:

I =

[(h′)2 + (k′)2 0

0 h2

], II =

1√(h′)2 + (k′)2

[−k′h′′ + h′k′′ 0

0 hk′

];(91)

λ1 =−k′h′′ + h′k′′

((h′)2 + (k′)2)32

, λ2 =k′

h((h′)2 + (k′)2)12

.(92)

There are two special cases where H or K has a nice form:

(i) If we assume the curve is parametrized with its arc length, then (h′(u))2 + (k′(u))2 = 1for any u. Then

K = −h′′

h, H =

(hk′)′

(h2)′.(93)

(ii) If we assume the curve is of the form (h, k) = (h(u), u), then H = 0 is equivalent to

h′′h = 1 + (h′)2.(94)

A careful analysis of the ODE h′′(u) + Ch(u) = 0 leads to a classification of surfaces ofrevolution with constant Gauss curvature.

Proposition 4.3.1 (See p.66 in [9]). There is a complete classification of surfaces of rev-olution with constant Gauss curvature. In particular, there is a model of a surface patchwith K = −1 which is called a pseudosphere. It is generated by the curve (h(u), k(u)) =

(e−u,∫ u

0

√1− e−2tdt) with u ∈ (0,∞). Such a curve is called the tractrix.

Besides the above form, there are at least two forms of the tractrix in various literature,neither parametrized by arc length.

(i) If we set e−u = sin t where t ∈ (π2 , π), then we get

(h(u), k(u)) = (sin t, cos t+ ln[tan(t

2)]).

(ii) If we further set sechu = sin t where u ∈ (0,+∞), then cos t = − tanhu, we get

(h(u), k(u)) = (sechu, u− tanhu).

There is an intuitive description of the tractrix.18 Consider a person standing in the xz-planeat the origin. The person holds a string attached to a rock at (1, 0), and the person proceed towalk upward along positive z-axis dragging the rock behind. The velocity of the rock is alwaysstraight towards the person, and its distance to the person is always 1 (if we assume the stringdoes not stretch). The curve traced out by the rock is a tractrix. One may check that the

tractrix is the solution to the differential equation −√

1− x2

x=dz

dx.

There is a standard way to solve the minimal surface equation H = 0 explicitly from either(93) or (94), one may prove:

Proposition 4.3.2 (Bonnet (1860)). Any surface of revolution which is a minimal surface isa piece of either

(i) a plane.(ii) or a catenoid with h(u) = 1

C1cosh(C1k(u) +C2) where C1 > 0 and C2 are constants,

cosh(x).= ex+x−x

2 is the hyperbolic cosine function. The plane curve (t, 1C1

cosh(C1t))is called a catenary.

18I first learned from Danny Calgary’s notes on hyperbolic geometry https://lamington.wordpress.com/

2010/04/08/hyperbolic-geometry-157b-notes-1/. We may find an animation for this process on this wiki pagehttps://en.wikipedia.org/wiki/Tractrix.

https://lamington.wordpress.com/2010/04/08/hyperbolic-geometry-157b-notes-1/

https://lamington.wordpress.com/2010/04/08/hyperbolic-geometry-157b-notes-1/

https://en.wikipedia.org/wiki/Tractrix


Figure 12. a pseudosphere,copied from the wiki page

1. PARAMETRIZED CURVES 7

along the positive y-axis. COMMENT: See the Wikipedia “tractrix” page fora nice animation of this.

Figure 1.5. The tractrix from Exercise 1.4. This curvehas acoustic advantages in the design of horn speakers. Theterm “tractrix horn” features prominently in the marketingmaterial of Klipsch speakers

Exercise 1.5. Prove Proposition 1.7.

Exercise 1.6. If all three component functions of a space curve γ arequadratic functions, prove that the image of γ is contained in a plane.

Exercise 1.7. Find a plane curve parametrized by arc length that tra-verses the unit circle clockwise starting at (0,−1).

Exercise 1.8. Compute the arc length of γ(t) =(2t, 3t2), t ∈ [0, 1].

Exercise 1.9. Let a, b > 0. Find the maximum and minimum speed ofthe ellipse γ(t) = (a cos t, b sin t).

Exercise 1.10. Prove that the arc length, L, of the graph of the polarcoordinate function r(θ), θ ∈ [a, b], is

L =

∫ b

a

√r(θ)2 + r′(θ)2 dθ.

Exercise 1.11. Figure 1.6 shows a polygonal approximation of theregular curve γ : [a, b] → R

n. This polygonal approximation is determinedby a partition, a = t0 < t1 < t2 < · · · < tk = b. The sum of the lengths of the

line segments equals L =∑k−1

i=0 |γ(ti+1) − γ(ti)|. The mesh of the partitionis defined as δ = maxti+1 − ti. Prove that L converges to the arc length ofγ for every sequence of partitions for which δ → 0.

Figure 1.6. A polygonal approxima-tion of γ

HINT: For notational simplicity,suppose first that n = 2, and de-note the component functions byγ(t) = (x(t), y(t)). For eachi ∈ 0, 1, . . ., k−1, the mean valuetheorem guarantees a sample pointt∗i ∈ [ti, ti+1] such that x′(t∗i ) =

Figure 13. a tractrix, copiedfrom Tapp’s book p.7

358 M Ito and T Sato

Figure 1. A catenary curve with radius r(z) and minimum radius a labelled (left panel). A catenoidis made by rotating the catenary curve around the z-axis (centre panel). A three-dimensionalschematic of a catenoid (right panel, created by using Mathematica 5.1).

In the proposed experiment, we observe the dynamic behaviour of a catenoid, whichis the shape a soap film takes between two coaxial rings. The catenoid can be explainedusing dynamic mechanics and differential geometry. We propose that in situ observation andmathematical analysis of a catenoid soap film are suitable for an educational curriculum inexperimental physics.

This paper is organized as follows. In section 2, we describe the mathematical propertiesof the minimal surface and analyse the shape of the catenoid. In section 3, we explain howto observe the catenoid and present some results of in situ observation of a catenoid. Insection 4, we interpret the behaviour of the catenoid using dynamical mechanics. Finally, wegive a brief summary and conclude in section 5.

2. Mathematics of the catenoid

In this section, we explain the mathematics of a minimal surface. For a given boundary,the minimal surface is that which has an extremal area [2, 3, 5]. If a minimal surfaceis parametrized in a three-dimensional orthogonal coordinate system (x, y, f (x, y)), theEuler–Lagrange equation leads to(

1 + f 2y

)fxx +

(1 + f 2

x

)fyy − 2fxfyfxy = 0, (1)

where fi denotes the partial derivative with respect to index i. Because the partial differentialequation (1) is rather complicated, it is difficult to solve. However it can be easily solvedby imposing some assumptions (e.g., symmetry or boundary condition). Some well-knownminimal surfaces include the catenoid, Helicoid, Scherk’s surface, Enneper’s surface, etc[2, 3, 8–10]. In this study, we focus on the catenoid, which is the minimal surface betweentwo coaxial circular rings.

If the minimal surface described by equation (1) is restricted to a surface of revolutionwith axial symmetry, we obtain the equation of a catenoid [2, 3, 8–10]. As shown in figure 1,a catenoid is a surface of revolution generated from a catenary curve with radius

r(z) = a coshz

a, (2)

Figure 14. Copied from Ito-Sato’s paper in Eur. J. Phys. 31 (2010) 357-365

Proposition 4.3.3 (Delaunay19). Any surface of revolution which is a CMC surface must be apiece of any of the following:

(i) a plane.(ii) a circular cylinder.(iii) a sphere.(iv) a catenoid.(v) an unduloid, see Figure 15.

19For a detailed exposition of Delaunay’s theorem, we recommend the paper of Eells, J. “The surfaces of

Delaunay”, Math. Intelligencer 9 (1987), 53-57, and Sec 3.2 of the book by Kenmotsu, K. “Surfaces withconstant mean curvature”. Translations of Mathematical Monographs, 221. American Mathematical Society,2003.

40 BO YANG

(vi) a nodoid, see Figure 16.

Figure 15. An undulary to generate an unduloid (from Kenmotsu’s book)

Figure 16. An nodary to generate an nodoid (from Kenmotsu’s book)

4.3.2. A short introduction on ruled surfaces.

Definition 4.3.4 (a ruled surface or a developable surface). Let f : U → R3 be asurface patch if for each u0 ∈ U there exists a neighborhood on U0 and a change of variablesφ : (s, t) ∈ V → U0 such that

f(s, t) = f φ(s, t) = sX(t) + c(t).

Here X(t) is a vector field along the curve c(t). Any t =const curve is a line having X(t) as itsdirection vector, it is called a generator or a ruling. Any s =const curve is called a directrixof f . In some literature, only the curve c(t) which corresponds to s = 0 is called the directrix.

In other words, a surface in R3 is ruled if through every point there is a straight line whichlies on the surface.

If in addition, the unit normal vector n(s, t) is constant along each generator. We call f isdevelopable.

NOTES ON DIFFERENTIAL GEOMETRY 413C Surfaces of rotation and ruled surfaces 87

Figure 3.14. Helicoidal ruled surface and a one-sheeted hyperboloid of

revolution as ruled surface

−JX ′. For constant F and λ it follows that c′′ = FX ′ + λX ×X ′′ =

(F − λJ)X ′. Hence c′ is a constant multiple of X plus an additive

constant Y0, where Y0 is perpendicular to the plane which is spanned

by the circle X (i.e., the X ′, X ′′-plane). To see this one just has

to calculate 〈Y0, X′〉 = 〈Y0, X

′′〉 = 0. Therefore c′ coincides with

the tangent to a helix, and one further integration determines c as a

helix. This in turn determines a screw-motion, and the surface arises

as the trajectory of a line under the one-parameter group of all of

these screw-motions. Conversely, for a helicoidal ruled surface the

three determining quantities λ, F, J must be constant, since they are

invariant under the one-parameter group of Euclidean motions. The

case of FJ + λ = 0 reduces to that of pure surfaces of rotation, in

which the screw-motion degenerates into a rotation because in this

case we have Det(c′, c′′, c′′′) = 0.

For the proof of (ii), the fact that J = 0 implies that X is a great

circle with constant X ×X ′, and F = 0 implies c′ = λX ×X ′. Hence

c is a line in the direction of X × X ′. If we choose X × X ′ as the

vector (0, 0, 1) in space, then we get the above parametrization of the

right helicoid.

Part (iii) is an easy exercise: If the rotating line meets the axis of

rotation, then we get either a plane or a double cone. If it is parallel

to the axis of rotation, we get a cylinder, and if they have different

Figure 17. A one-sheeted hyperboloidwith rotational sym-metry, copied fromp.87 of Kuhnel’s book

Figure 18. A helicoid, from“Minimal Surface GraphicsProject” by Matthias Weber

A developable surface without planar points is much understood, it must have zero Gausscurvature and it is almost a cylinder, a cone or a tangential developable. see Example 3.7.6,Proposition 3.7.9, and 3.7.10 on pp.57-60 in Klingenberg’s book [9].

The following results require more detailed local analysis on ruled surfaces, their proofs canbe found elsewhere.

Proposition 4.3.5 (Ruled surfaces of revolution). Any surface of revolution which is also aruled surface must satisfy either of the following

(i) it has K = 0, hence by Proposition 4.3.1, it is contained in a circular cone, a circularcylinder, or a plane.

(ii) it is a piece of the one-sheeted hyperboloid with the equation x2 + y2 − z2

a2 = c2. SeeFigure 17.

In fact, one-sheeted hyperboloids and hyperbolic paraboloids are both doubly ruled (i.e.through every one of its points there are two distinct lines that lie on the surface), we refer tohttps://en.wikipedia.org/wiki/Ruled_surface for the explanation.

Theorem 4.3.6 (Catalan (1842)20). Any ruled surface with H = 0 and nonflat (its Gausscurvature is nonzero somewhere) must be contained in a helicoid (Figure 18), which is definedby f : R2 → R3

f(u, v) = u(cos v, sin v, 0) + (0, 0, v).(95)

4.3.3. Graphs of functions.By the implicit function theorem (Theorem 1.3.1), given any surface patch f : U → R3 and

a point u0 ∈ U , there exist two open subsets V and U0 3 u0 of R2, and a local diffeomorphismφ : V → U0 where U0 contains u0, such that f φ|V = (v1, v2, φ(v1, v2)).

In particular, any C2 function ϕ : U → R defines f : U → R:

f(u, v) = (u, v, ϕ(u, v)).

We call f a graphical surface patch or a surface in non-parametric form.

20A proof can be founded on p. 207 in [5]

https://en.wikipedia.org/wiki/Ruled_surface

42 BO YANG

A direct calculation shows such a graph has its mean curvature

2H(u, v) =(1 + ϕ2

v)ϕuu + (1 + ϕ2u)ϕvv − 2ϕuvϕuϕv

(1 + |∇ϕ|2)32

= div(∇ϕ√

1 + |∇ϕ|2),

and its curvature curvature

K(u, v) =ϕuuϕvv − ϕ2

uv

(1 + |∇ϕ|2)2.

On the other hand, note that the area of f(U) ⊂ R3 can be calculated by

Area(f(U)) =

∫U

|fu × fv|dudv =

∫U

√1 + |∇ϕ|2.

Now consider the area functional applied on a variation of ϕ defined by ϕt = ϕ+ tη where η isany compactly supported smooth function on U , we have

d

dtArea(ϕt(U))|t=0 =

∫U

∇ϕ · ∇η√1 + |∇ϕ|2

dudv,

= −∫U

η div(∇ϕ√

1 + |∇ϕ|2)dudv,

In the last step we use the divergence theorem on R2.

Proposition 4.3.7. A graphical surface with H = 0 is the critical point of the area functionalamong variations with compact support.

In fact, a stronger result is true.

Proposition 4.3.8 (Area-minimizing property21). A graphical surface with H = 0 is area-minimizing among surfaces in the cylinder U × R. i.e. Let fϕ(u, v) = (u, v, ϕ(u, v)) satisfiesH = 0, and Σ ⊂ U × R is any other surface with ∂Σ = ∂U , then

Area(fϕ(U)) ≤ Area(Σ).

Let us call a C2 function ϕ : U → R solves the minimal surface equation if

div(∇ϕ√

1 + |∇ϕ|2) = 0.(96)

Now we are ready to state a beautiful result in the study of graphical minimal surfaces.

Theorem 4.3.9 (Bernstein). If ϕ : R2 → R is a C2 solution to (96) for any (u, v) ∈ R2, wecall ϕ an entire solution of (96). Then ϕ(u, v) = au+ bv + c for constants a, b, c ∈ R.

The surprising feature of Theorem 4.3.9 is the linearity of ϕ follows from the only fact thatϕ is defined on the whole R2. Compared with the Liouville theorem for harmonic functionsand solutions of more general linear PDEs, usually we need the additional assumption on theboundedness or growth conditions of a solution to conclude it is constant or else. There are manyproofs of Theorem 4.3.9, for example we refer to Colding-Minicozzi’s book for a proof which isrelated to other intrinsic geometric properties of minimal surfaces.22 The proof in below followsSec 3.8 in [4] and it is due to Nitsche.23 We choose to present Nitsche’s proof because of itsbeautiful idea to connect complex analysis and PDEs.

21Lemma 1.1 on p.3 in Colding, T. H.; Minicozzi, W. P., II “A course in minimal surfaces”, Graduate Studiesin Mathematics, 121. American Mathematical Society, 2011.

22In Sec 5 of Chapter 1 in Colding, T. H.; Minicozzi, W. P., II “A course in minimal surfaces”. The proof

they wrote is due to L. Simon (1997).23Nitsche, J. C. C. , Elementary proof of Bernstein’s theorem on minimal surfaces. Ann. of Math. (2) 66

(1957), 543-544.


A sketchy proof of Theorem 4.3.9. Let ϕ(u, v) is an entire solution of (96), set

p = ϕu, q = ϕv, W =√

1 + p2 + q2.

Then the minimal surface equation reads

(1 + q2)ϕuu + (1 + p2)ϕvv − 2ϕuvpq = 0.(97)

By a direct calculation

(1 + p2

W)v − (

pq

W)u = − q

W 3

[(1 + q2)ϕuu + (1 + p2)ϕvv − 2ϕuvpq

]= 0,(98)

(pq

W)v − (

1 + q2

W)u =

p

W 3

[(1 + q2)ϕuu + (1 + p2)ϕvv − 2ϕuvpq

]= 0.(99)

Since both (98) and (98) holds on R2, we conclude there exists a C2 function φ such that[φuu φuvφvu φvv

]=

[1+p2

WpqW

pqW

1+q2

W

].(100)

Now we define a transformation which was introduced by H. Lewy.

(ξ, η) = (u, v) + (φu, φv).

The curical step of the proof is to show: The map (ξ, η) := F (u, v) is a C1-differmorphismon R2.

Intuitively speaking, the Jacobian determinant of F is 1W +W +2, F is always a local one-to-

one by the inverse function theorem. In fact F is distance-increasing, i.e. for F (u1, v1) = (ξ1, η1)and F (u2, v2) = (ξ2, η2), we have√

(u2 − u1)2 + (v2 − v1)2 ≤√

(ξ2 − ξ1)2 + (η2 − η1)2.(101)

To see this, recall that φu = ξ − u, φv = η − v, we define h : R→ Rh(t) = φ(u1 + t(u2 − u1), v1 + t(v2 − v1)),

It follows from the definition of φ in (100) that h′′(t) > 0. Therefore h′(1) > h′(0), i.e.

(ξ2 − u2)(u2 − u1) + (η2 − v2)(v2 − v1) > (ξ1 − u1)(u2 − u1) + (η1 − v1)(v2 − v1).

After rearranging we have

(ξ2 − ξ1)(u2 − u1) + (η2 − η1)(v2 − v1) > (u2 − u1)(u2 − u1) + (v2 − v1)(v2 − v1).

It follows from the arithmetic mean inequality that

(u2 − u1)2 + (v2 − v1)2 <1

2

[(ξ2 − ξ1)2 + (u2 − u1)2 + (η2 − η1)2 + (v2 − v1)2

],

which leads to (101).Therefore F is injective, it remains to show F (u, v) = (ξ, η) is surjective on R2. Assume there

is a point p0 not in the image of F , then there are two possibilities:(1) There exists a sequence pn = F (un) such that pn → p0, then the distance-dilation property

concludes un → u0, F (un)→ F (u0), and p0 = F (u0). Contradiction!(2) Assume dist(p0, F (R2)) = δ > 0. then there exists a sequence un such that

dist(F (un), p0) ≤ δ +1

nfor any positive integer n.

Again by the distance-dilation property of F , un must be bounded and hence subconvergesto some u0 with dist(F (u0), p0) = δ. Since F is a local diffeomorphism (hence locally anopen mapping) at u0, we can always find some F (u∗) in the neighborhood of F (u0) withdist(F (u∗), p0) < δ, which contradicts to the assumption δ is the minimal distance of p0 tothe image of F .

Now we view (ξ, η) = F (u, v) as a change of coordinates, the original graphical surface f is

reparametrized by f(ξ, η) = (u, v, ϕ(u, v)). It turns out that f satisfies following remarkableproperties:

44 BO YANG

(i) The new coordinate (ξ, η) are isothermal coordinates, its first fundamental form hascomponents gij = ( W

W+1 )2δij .

(ii) The gradient of ϕ with respect to (ξ, η) is

(ϕξ, ϕη) = (p

W + 1,

q

W + 1).(102)

(iii) ϕ is a harmonic function with respect to (ξ, η) ∈ R2, i.e. ϕξξ + ϕηη = 0. This followsfrom the minimal surface equation (97) by a direct calculation.24

As an immediate consequence the above properties, ϕξ −√−1ϕη is a entire holomorphic

function with respect to the complex coordinates z = ξ +√−1η. In the meantime,

|ϕξ −√−1ϕη| = |

p−√−1q

W + 1| ≤ 1.

Hence ϕξ −√−1ϕη must be constant by the Liouville theorem. Since p

W+1 and qW+1 are both

constants, we have p = a and q = b for some constants a, b. Finally, we have ϕ(u, v) = au+bv+cfor constants a, b, c ∈ R.

Remark 4.3.10. We have several remarks on the above proof:

(i) It has been known that Bernstein’s theorem is a consequence of Jorgens’s theoremon some real Monge-Ampere equation on R2. Jorgens’ theorem states if a C2 functionf is a solution to

fuufvv − f2uv = 1, fuu > 0.(103)

for all (u, v) ∈ R2, then f(u, v) must be a quadratic polynomial.Indeed, a slight modification of the above proof of Theorem 4.3.9 above proves Jorgens’s

theorem at the same time. To see the connection, just note that (100) implies φ satisfies(103). It is possible to construct a bounded entire holomorphic function without aid ofminimal surface equation (97), We refer to p.121 in Sec 3.8 of [4] or Nitsche’s originalpaper for more details.

(ii) The function defined in (102) has the following geometric meaning. Recall the unitnormal vector of the graph (u, v, ϕ(u, v)) is given by (−pW , −qW , 1

W ) and the stereographic

projection St with respect to the south pole S of S2 sends (x, y, z) ∈ S2 \ S to( x

1+z ,y

1+z ) on R2. Therefore, the image of unit normal (Gauss map) after stereographic

projection back to R2 is exactly

St n(u, v) = (−p

W + 1,−q

W + 1).(104)

It is a general fact on minimal surfaces that Stn is conjugate-holomorphic with respectto isothermal coordinates (ξ, η). That means p

W+1−√−1 q

W+1 is a holomorphic function

of z = ξ +√−1η. This explains the argument in the end of the above proof.

(iii) In the case of Bernstein’s theorem, the entire solution forces the image of Gauss map tobe constant! In the global theory of minimal surfaces, we study minimal surfaces whichare complete.25 It is not hard to show, a entire graphical surface is indeed complete,note that the image of the Gauss map of any graphical surface lies in the open northernhemisphere. There are further works by R. Osserman (1959), F. Xavier (1981, 1982),and H. Fujimoto (1988) where they study the density of the image of Gauss map of acomplete minimal surfaces. The final result states that if a complete minimal surfacein R3 is not a plane, then the image of its Gauss map of can not miss more than fourpoints on S2 ⊂ R3.

24It is an exercise to show: for any minimal surface f : U → R3 by f(u, v) = (x, y, z) with (u, v) are isothermalcoordinates, each coordinate function x, y, z is a harmonic function.

25Any surface can be endowed with a metric space topology where the distance function defined by its firstfundamental form. By ‘complete’ we mean this metric space is complete, i.e. any Cauchy sequence converges. Itis the notion of a complete Riemannian manifold. Check Theorem 5.4.17 for more discussions.


4.3.4. More on minimal surfaces.We plan to give a quick introduction on some classical aspects of minimal surfaces, the

emphasis will be placed on its connection to complex analysis. Topics include

(i) The Enneper-Weierstrass Representation,(ii) The Schwartz reflection principle and the Riemann-Schwartz example.26

Consider a minimal surface surface f : U → R3 with isothermal coordinates (u, v)

f(u, v) = (x(u, v), y(u, v), z(u, v)),(105)

Then x, y, z are all harmonic functions of (u, v). In fact, this is an equivalent condition withH = 0, and we may use it to define minimal surfaces.

Now we may define a conjugate (adjoint) minimal surface f∗ : U → R3 by

∂f

∂u=∂f∗

∂v,∂f

∂v= −∂f

∗

∂u.(106)

This notion is similar to a conjugate harmonic function, as long as U is simply connected onR2, we can always find a conjugate minimal surface.

Definition 4.3.11 (Isotropic holomorphic curve). Let U be an connected open set on R2 = C1

and set z = u+√−1v, we call F = (F1, F2, F3) : U → C3 be a holomorphic curve if each Fi is

holomorphic on U . In addition, F is isotropic if

∂F

∂z· ∂F∂z

:= (∂F1

∂z)2 + (

∂F2

∂z)2 + (

∂F3

∂z)2 = 0.(107)

Recall by Cauchy-Riemann condition, we have ∂Fi∂z = ∂ ReFi

∂u +√−1∂ ImFi

∂u .

Proposition 4.3.12 (Minimal surfaces via isotropic holomorphic curves).

(i) Assume U is a simply connected open set on R2, if f : U → R3 is a minimal surfacewith isothermal coordinates (u, v) and f∗ : U → R3 its conjugate minimal surface, thenF := f +

√−1f∗ : U → C3 is a nonconstant isotropic holomorphic curve.

(ii) Conversely, if U is an open set on R2 and F : U → C3 a nonconstant isotropicholomorphic curve, assume we have

∂F

∂z· ∂F∂z

:=

3∑i=1

|∂Fi∂z|2 > 0(108)

for any z ∈ U . Then ReF (u, v) : U → R3 and ImF (u, v) : U → R3 defines twoconjugate minimal surfaces with isothermal coordinates. Here z = u+

√−1v.

In view of (107) and Proposition 4.3.12, we need to find three holomorphic functions (Φ1,Φ2,Φ3)

satisfying∑3i=1 Φ2

i = 0 to generate a isotropic holomorphic curve. The solution is given in below.

Proposition 4.3.13 (The Enneper-Weierstrass Representation). Given a simply connecteddomain U ⊂ R2, if f : U → R3 is a minimal surface which is not contained in a plane and(u, v) ∈ U are the isothermal coordinates of f , now we consider functions of u +

√−1 ∈ U ⊂

C1, then we can find a holomorphic function µ, a meromorphic function ν such that µν2 isholomorphic on U ,27 Moreover, µ and µν2 have no common zeros. If we define

(Φ1,Φ2,Φ3) := (1

2µ(1− ν2),

√−1

2µ(1 + ν2), µν

),

f(u, v) = f(z0) + Re

∫ z

z0

(Φ1,Φ2,Φ3)dξ, z = u+√−1v ∈ U(109)

where f(z0) ∈ R3 is constant, and ξ is any curve in U which connects z0 and z.Conversely, (109) with µ and ν satisfying the above property defines a minimal surface if U

is simply connected.

26This construction gives a complete embedded triply periodic minimal surface with infinite genus, and theimage of its Gauss map is the entire S2. https://en.wikipedia.org/wiki/Triply_periodic_minimal_surface.

27This means at every point where ν has a pole of order m, µ must have a zero of order at least 2m.

https://en.wikipedia.org/wiki/Triply_periodic_minimal_surface

46 BO YANG

Next we solve the Gauss mapping of the minmal surface f defined by (109), a direct calculationshows

n =(2 Re ν, 2 Im ν, |ν|2 − 1)

|ν|2 + 1.(110)

This implies the stereographic projection of the meromorphic function ν is the Gauss map of f !We may also check the corresponding first fundamental form (with isothermal coordinates) is

E =1

4|µ|2(1 + |ν|2)2 > 0(111)

And its Gauss curvature of f is

K = −( 4|∂ν∂z |2|µ|(1 + |v|2)2

)2

.(112)

Example 4.3.14 (Catenoid revisited). Although Proposition 4.3.13 assumes that U is simplyconnected. In some cases, (109) also works for some non-simply connected domains. Thepoint is that the integral (109) still defines a minimal surface on U as long as the integral∫γ(Φ1,Φ2,Φ3)dξ is purely imaginary for any closed curve γ in U .

Now pick U = C \ 0 and set µ = 1z2 and ν = z, we get

(Φ1,Φ2,Φ3) =(1

2(

1

z2− 1),

√−1

2(

1

z2+ 1),

1

z

).

We may check that (109) represents the catenoid, indeed

f(u, v) = Re(−1

2(z +

1

z),

√−1

2(z − 1

z), ln z) + const, (u, v) ∈ R2 \ 0.

If we introduce the polar coordinates z = u+√−1v = reiθ, we get

f(r, θ) = (−1

2(r +

1

r) cos θ,−1

2(r +

1

r) cos θ, ln r) + const, r ∈ [0,+∞), θ ∈ [0, 2π).

After a further change of coordinates u = ln r and v = θ + π, we get the standard catenoid

f(u, v) = (cosh u cos v, cosh u sin v, u) + const, u ∈ R, v ∈ [0, 2π).(113)

We may also use (112) to show the total Gauss curvature of a catenoid is finite∫f(U)

KdA = −4π,

where dA is the surface area element of the catenoid f : U → R3.

Example 4.3.15 (Helicoid revisited). Now pick U = C and set ν = eiz and µ = e−iz, we get

(Φ1,Φ2,Φ3) =(1

2(e−iz − eiz),

√−1

2(e−iz + eiz), 1

).

Use (109) we solve

f(u, v) = Re(−1

2√−1

(eiz + e−iz),1

2(eiz − e−iz), z) + const

= (sinh(v) sinu,− sinh(v) cosu, u) + const.

After a change of coordinates by u = sinh(v) and u = v+ π2 , we get the standard parametrization

of a helicoid (95)

f(u, v) = (u cos v, u sin v, v) + const, (u, v) ∈ R2.

Example 4.3.16 (A family of minimal surfaces). If we pick U = C and keep ν = eiz and setµ = eiθe−iz where θ ∈ [0, 2π), then we get a family of minimal surfaces f(θ) parametrized by θ.For example θ = 0 corresponding to a helicoid, and θ = π

2 to a catenoid which we now explain.28

28There is a video demonstration at https://www.youtube.com/watch?v=E6JtYMVayeI.

https://www.youtube.com/watch?v=E6JtYMVayeI


Indeed, when θ = π2 we get

f(u, v) = (− cosh v cosu,− cosh v sinu,−v) + const, (u, v) ∈ R2.

which can be easily identified with the standard catenoid after we set u = u+ π and v = −v.

f(u, v) = (cosh u cos v, cosh u sin v, u) + const, u, v ∈ R.(114)

Strictly speaking, it is better to distinguish the catenoid (113) and (114) since the range of vparameter is different. We may consider (114) is an infinite number of copies of (113), we maysay that (114) is the universal cover of (113).

For further reading, we recommend:

(i) Osserman, R.. ‘A survey of minimal surfaces’. Dover Publications, 1986.(ii) “Minimal Surfaces” by Matthias Weber for pictures on various examples of minimal

surfaces.29

(iii) Colding, T. H.; Minicozzi, W. P., II “A course in minimal surfaces”, Graduate Studiesin Mathematics, 121. American Mathematical Society, 2011.

(iv) Xin, Y.. “Minimal submanifolds and related topics”. 2nd edition. Nankai Tracts inMathematics, 16. World Scientific, 2019.

29Check http://www.indiana.edu/~minimal/toc.html

http://www.indiana.edu/~minimal/toc.html

48 BO YANG

5. Surfaces in R3: from local to global

5.1. Smooth 2-manifolds. First we clarify the objects to discuss in this section.

Definition 5.1.1 (An embedded surface in R3). A connected set M ⊂ R3 is an embeddedsurface in R3 if when M is endowed in the induced topology from R3, for every point p ∈M ,there exists an open set Uα ⊂ R2, an open set Mα ⊂M containing p, and a C∞ map fα : Uα →Mα with the following property.

(i) fα is smooth and is a homeomorphism from Uα to Mα.(ii) dfα(u) is injective for any u ∈ Uα, i.e. fα is a parametrized surface patch.

Definition 5.1.2 (A smooth (Ck, or real analytic) n-dimensional manifold). Let M be a con-nected Haursdorff topological space with a countable topological basis such that there exists anopen cover Mαα∈Λ and a family of homeomorphisms uα : Mα → Uα ⊂ Rn where Uα are opensets in Rn. Moreover, for any p ∈Mα∩Mβ the transition map (or change of coordinates)

φαβ.= uα u−1

β : uβ(Mα ∩Mβ) → uβ(Mα ∩Mβ) is C∞ as a map between open sets in Rn.

We call M a smooth n-manifold and (Mα, uα)α∈Λ a smooth atlas of M . Similarly, wemay define a Ck manifold or real analytic manifold if we change the ‘smoothness’ of thetransition maps.

Lemma 5.1.3 (Lemma 6.1.2 on p.124 in [9]). Let M be an embedded surface in R3, then theinverse of the local parametrized patch fα can be used to define a smooth atlas on M , hence Mis a smooth 2-manifold.

To prove the above, it suffices to show f−1β fα is a smooth map. Apriori we do not know if

f−1β is smooth since it is defined on Mα which is an closed subset in R3, it follows from part (1)

of Corollary 1.3.3 that we can use smooth diffeomorphisms to modify fβ and rewrite f−1β fα

so that it is defined on an open set in some plane contained in R3. We refer p.124 in [9] for thedetails.

Definition 5.1.4 (Smooth maps, immersions, and embeddings). Let M and N be two smoothmanifolds with dimension m and n respectively, and ψ : M → N a continuous map. Assumem ≤ n.

(i) Let (M,uα) and (N, vi) be their corresponding smooth atlases, ψ is called a smoothmap if any vi ψ u−1

α is a C∞ map from open sets in Rm to Rn. It is easy to seethis definition is independent of the choice of smooth atlases on both M and N .

(ii) Let m ≤ n, a smooth map ψ : M → N is called an immersion if for every p ∈M theJacobian of vi ψ u−1

α , as a nonsingular m×n matrix, is of rank m. Note that ψ doesnot need to be one-to-one.

(iii) Let m ≤ n, a smooth map ψ : M → N is called an embedding if it is an immersion,one-to-one, and M is homeomorphic to ψ(M) when ψ(M) endowed with the subsettopology induced from N .

By an immersed (embedded) surface in R3 we mean an immersion (embedding)from a smooth 2-manifold to R3. In view of Lemma 5.1.3, we conclude this definition of anembedded surface in R3 is the same as Definition 5.1.1. Indeed, in Definition 5.1.1, we just needto pick the inclusion map M → R3 and it is an embedding in the sense of Definition 5.1.4.

In this section we are interested in both extrinsic and intrinsic geometry of embedded surfacesin R3. Here by intrinsic we mean geometric quantities which only depends on the first funda-mental forms of surfaces, why extrinsic quantities are related to second fundamental forms, forexample, the mean curvature. Most of time, embedded surfaces (Definition 5.1.1) are sufficientfor our purposes, though occasionally we mention results on immersed surfaces in R3.

5.1.1. How to work on the whole manifold?The first question is: given an embedded surface M in R3, can we define a global Gauss map

n : M → S2 ⊂ R3?


Any point p ∈M , choose a local patch fα : Uα →Mα or uα := f−1α : Mα → Uα, define

n =(fα)uα × (fα)vα|(fα)uα × (fα)vα |

.

Now if p ∈Mα ∩Mβ , and we try another patch fβ : Uβ →Mβ , then we should get

n =(fβ)uβ × (fβ)vβ|(fβ)uβ × (fβ)vβ |

.

We leave as an exercise to check

(fβ)uβ × (fβ)vβ = det[Jac(uα u−1β )](fα)uα × (fα)vα .

Therefore in order to make sure these two definitions coincide on the overlapping region, weneed to assume the transition map f−1

α fβ = uα u−1β have a positive Jacobian determinant.

Of course this is an assumption on the transition functions, we can define this condition for anysmooth manifold.

Definition 5.1.5 (Orientable manifolds). Let M be a smooth manifold, we say M is orientableif it admits a smooth atlas such that each of the transition maps has a positive Jacobian deter-minant.

Compact embedded surface is orientable, in fact Proposition 5.4.24 says any embedded surfacewhich is a closed subset of R3 is orientable. Such a manifold has two choices of its unit normalvector in R3. In fact, orientability is closed related to Jordan-Brouwer separation property,any embedded surface which is a closed subset of R3 divides R3 into two connected components.If the surface is compact, then one component is bounded. We refer to p.116 of [10] for theproof.

Assume M is a compact embedded surface in R3, then M encloses a bounded region Ω, wewould like to understand how Area(M) and V ol(Ω) changes if we vary the surface M . To makeit precise, we will prove

Theorem 5.1.6 (First variation of area and volume). Assume f : M → R3 is the originalembedding of M into R3, by definition we could set f = id, we may also choose n : M → S2 asits Gauss map so that it points to the inner normal of M .

Consider a family of deformations f(·, t) : M → R3 such that

(i) f(·, 0) = f ,

(ii) ∂f(·,t)∂t |t=0 = X, therefore X is a vector field along M .

(iii) When τ ∈ (−t, t), the image of f(·, τ) is a disjoint family of compact embedded surface.Each of them, denoted by f(τ)(M), encloses a bounded region Ω(τ).

Then we have

d V ol(Ω(t))

dt|t=0 = −

∫M

X · ndA,(115)

dArea(f(t)(M))

dt|t=0 = −

∫M

2HX · ndA.(116)

Remark 5.1.7. As an important case, if we consider f(·, t) = f + t η n where η is a smoothfunction defined on M . It can be proved that for t sufficiently small, such a deformation indeedsatisfies the above assumption. It is called a tubular neighborhood of M , we refer to p.120of [10] for a detailed discussion. Theorem 5.1.6 implies

d V ol(Ω(t))

dt|t=0 = −

∫M

ηdA,(117)

dArea(f(t)(M))

dt|t=0 = −

∫M

2HηdA.(118)

To understand the sign in the above equation, take M as the unit sphere S2, since the Gaussmap is the inner normal, then both principal curvatures are +1, so H = +1. If η > 0, f(t)(M)is shrinking, therefore Area(f(t)(M)) and V ol(Ω(t)) are decreasing.

50 BO YANG

Corollary 5.1.8 (Geometric meaning of constant mean curvature). A compact embedded sur-face in R3 with constant mean curvature is the critical point of the area functional under theconstraint that its enclosed region has a fixed volume.

Proof of Theorem 5.1.6. We argue by four steps.Step 1: Consider the special case f(·, t) = f + t η n, we first solve the area element of

Area(f(t)(M)), choose a local patch fα(u1, u2) : Uα → Mα for M , for notational convenience,

we still denote f and let L(fi) = Ajifj its Weingarten map.

f(t)1 :=∂f(·, t)∂u1

= (1− tηA11)f1 − tA2

1ηf2 + tη1n,(119)

f(t)2 :=∂f(·, t)∂u2

= −tA12ηf1 + (1− tηA2

2)f2 + tη2n.(120)

The new area element

|f(t)1 × f(t)2| = |(1− 2tηH +O(t2))f1 × f2 + t(η1n× f2 + η2f1 × n) +O(t2)|.(121)

Note that, both n× f1 and n× f2 are along the tangent directions, we conclude

d|f(t)1 × f(t)2|dt

|t=0

=d

dt|t=0

√(1− 2tηH)2|f1 × f2|2 + t2|η1n× f2 + η2f1 × n|2 +O(t3)

=− 2ηH|f1 × f2|,

Though we do not need it, we could solve the term η1n× f2 + η2f1 × n more explicitly:

n× f1 =−g22√

g11g22 − g212

f1 +g12√

g11g22 − g212

f2,

n× f2 =−g12√

g11g22 − g212

f1 +g11√

g11g22 − g212

f2.

η1n× f2 + η2f1 × n = −[(g1jηj)f1 + (g2jηj)f2

].

The last term in the bracket is called the gradient vector field of η, and we will discuss it inlater sections.

Now we are ready to to integration on the whole f(t)(M), first we may check that on eachf(t)(Uα), the area element |f(t)1 × f(t)2|du1du2 is independent of change of local parametriza-tion, we may cover f(t)(M) by a finite number of disjoint subsets Vi such that each of them liesin some single local parametrization. It follows that

dArea(f(t)(M))

dt|t=0 =

∑i

∫Vi

d|f(t)1 × f(t)2|dt

|t=0du1du2

=∑i

∫Vi

−2Hη|f1 × f2|du1du2

=

∫M

−2HηdA.


Step 2: We solve the change of V ol(Ω(t)) in the special case f(·, t) = f + t η n. Since n isthe inner normal, we get

V ol(Ω(0))− V ol(Ω(t))

=

∫ t

0

∫M

|Jac(f(τ))|dAdτ

=

∫ t

0

∫M

|det(f(τ)1, f(τ)2, f(τ)τ )|du1du2dτ

=

∫ t

0

∫M

(1− 2tηH +O(t2))η|f1 × f2|du1du2dτ

=

∫ t

0

∫M

ηdA+O(t2).

Therefore

dV ol(Ω(t))

dt|t=0 = −

∫M

ηdA.

Step 3: The general case, where f(t) satisfies dfdt |t=0 = X and f(0) = f , We need to

differentiate an area element |f(t)1 × f(t)2|du1du2 again. Here the idea is since this expressionis independent of choice of coordinates of M , we compute in an isothermal coordinate (u1, u2)so that fi · fj = Eδij . Keep in mind now we have |f1 × f2| = E, and

2H =1

E(h11 + h22) =

1

E(f11 + f22) · n.(122)

Moreover, we have

(f11 + f22) · f1 =1

2(|f1|2)1 + (f1 · f2)2 − f12 · f2(123)

=1

2E1 −

1

2(|f2|2)1 = 0,

(f11 + f22) · f2 = +(f1 · f2)1 − f12 · f1 +1

2(|f2|2)2(124)

= −1

2(|f1|2)2 +

1

2E2 = 0.

It follows from (122), (123), and (124) that

f11 + f22 = 2Hn.(125)

Now we calculate under an isothermal coordinate

d|f(t)1 × f(t)2|dt

|t=0

=1

E

[(f1 ·X1)E2 + (f2 ·X2)E2

]=(f1 ·X1 + f2 ·X2)|f1 × f2|.

A similar argument as in Step 1. we have

dArea(f(t)(M))

dt|t=0 =

∫M

d|f(t)1 × f(t)2|dt

|t=0du1du2

=

∫M

(f1 ·X1 + f2 ·X2)|f1 × f2|du1du2

=

∫M

−(f11 + f22) ·XdA+

∫M

[(f1 ·X)1 + (f2 ·X)2]dA

=

∫M

−2Hn ·XdA+

∫M

[(f1 ·X)1 + (f2 ·X)2]dA.

52 BO YANG

Note that dA = Edu1 du2, and we use (125) in the last step. Recall in Step 1, we choose X asthe normal direction, the second bracket term in the above equation is automatically zero, nowwe have to use Green’s formula on a local patch.∫

M

[(f1 ·X)1 + (f2 ·X)2]dA

=∑α

∫Uα

[(f1 ·X)1 + (f2 ·X)2]du1du2

=∑α

∫∂Uα

(f1 ·X, f2 ·X) · µαds

=∑α

∫∂Mα

X · µαds.

Here ∂Uα is the boundary of Uα, µα its outer normal and ds its arc length. On the other hand,∂Mα is the boundary of Mα = f(Uα), µα is its outer normal of ∂Mα, which is tangent alongMα, and ds the arc length of ∂Mα. Since we work on isothermal coordinates, if we expressX = af1 + bf2 and µα = (c, d), then µα = 1

E (cf1 + df2), and ds = Eds. These relations explainthe last equality in the above.

Since outer normals µα take opposite directions along the common boundary, all these bound-ary terms cancel. we conclude∫

M

[(f1 ·X)1 + (f2 ·X)2]|f1 × f2|du1du2 = 0.

Therefore we prove

dArea(f(t)(M))

dt|t=0 =

∫M

−2Hn ·XdA.

Step 4: The general case, the change of V ol(Ω(t)) can be solved similarly as in step 2. Weleave it as an exercise.

5.1.2. Digression: a non-orientable 2-manifold.

Example 5.1.9 (The open Mobius strip as an embedded surface). Topologically, we glue to-gether a rectangle [−1, 1] × [0, 2π] along two parallel shorter edges with opposite orientations,the resulting object is called a called a Mobius strip. If we do the same for the rectangle(−1, 1)× [0, 2π], we end up with an open Mobius strip.

Now we realize an open Mobius strip as an embedded surface in R3; in the case of a Mobiusstrip we hope to make it an embedded surface with boundary. To this end, consider D = [−1, 1]×[0, 2π]→ R3 and

f(u, v) = ((1 +u

2cos

v

2) cos v, (1 +

u

2cos

v

2) sin v,

u

2sin

v

2).

Note that f(1, 2π) = f(−1, 0), the boundary of f(D) is connected and topologically a circleS1, in fact it is nontrivial Z2-bundle over the unit circle S1. Moreover, f(u, 0) = f(−u, 2π), itfollows that f(D) is homeomorphic to a Mobius strip.

If we pick D1 = (−1, 1)× [0, 2π]→ R3, then f(D1) is an open Mobius strip.

Next we show the open Mobius strip f(D1) is an embedded surface in R3 in the sense ofDefinition 5.1.1. It suffices to consider

f1 : U1 = (−1, 1)× (0, 2π), f1(u, v) = f(u, v).

f2 : U2 = (−1, 1)× (−π, π), f2(ξ, η) = ((1 +ξ

2cos

η

2) cos η, (1 +

ξ

2cos

η

2) sin η,

ξ

2sin

η

2).


It is direct to check each of fi is a surface patch, and it is injective and homeomorphism ontoits image. Moreover, the overlapping region f1(U1)∩ f2(U2) has two connected components, oneach we have

on f1((−1, 1)× (0, π)), f−12 f1(u, v) = (u, v).

on f1((−1, 1)× (π, 2π)), f−12 f1(u, v) = (−u, v − 2π).

each of them is a smooth map.Recall in Definition 5.1.5 a manifold is orientable if it admits a smooth atlas so that all the

transition maps have positive Jacobian determinants. Obviously, f(D) is not orientable for theabove choice of f1 and f2. In fact, a Mobius strip is not orientable for any choice of its smoothatlas. Alternatively, if we pick the point f(0, 3π

2 ) on the center circle f(0 × [0, 2π]), then its

normal vector is (0, 1√2,− 1√

2) using the local parametrization f1, and n2 = −(0, 1√

2,− 1√

2) in

terms of f2. However, we do have the same normal in terms of f1 and f2 for the point f(0, π2 ).To sum up there is no consistent way to define the normal vector for points on f(D).

If M is an embedded surface in R3, it is equivalent to say that M is non-orientable if thereexists a closed curve in M such that the normal to M , when transported around the curve in acontinuous fashion, comes back with the opposite direction. In the Mobius strip, it suffices tochoose the loop f(0, v) and check the normal at v = 0 and v = 2π, see Figure 20.

Figure 19. The parametriza-tion f does not have zero Gausscurvature

Figure 20. No consistent nor-mal vector along a Mobius strip

We should point out that the Gauss curvature of f(D) defined above is not identically zero,and the image curve f(u× [0, 2π]) has length longer than 2π unless u = 0. On the other hand,the topological procedure to construct a Mobius strip from a long thin rectangle involves onlybending and gluing, but no stretching. It is desirable to find an embedded model of a Mobiusstrip which inherits the flat structure of a rectangle, thus has zero Gauss curvature. We willreturn to this topic in Subsection 5.4.8.

5.2. The geodesic curvature of a curve on a surface.We consider a surface patch f : U → R3, assume c(t) = f(u(t)) : I → R3 is a curve on f , X

is a tangent vector field along c(t), we define the covariant derivative of X with respect tot by

∇X(t)

dt= (

dX(t)

dt)T , the tangential componnet of

dX(t)

dt.

54 BO YANG

Let c(t) = f(u(t)) = f(u1(t), u2(t)), X =∑k ξ

k(u(t))fk(u(t)) where we take fk = ∂f∂uk

as usual.

dX(t)

dt=

2∑k,p

(∂ξk∂up

+∑q

ξqΓkqp

)dupdt

fk +

2∑q,p

ξqhqpdup

dtn,(126)

∇X(t)

dt=

2∑k,p

(∂ξk∂up

+∑q

ξqΓkqp

)dupdt

fk(127)

Both dX(t)dt and ∇X(t)

dt is independent of change of coordinates in the following sense: consider

a diffeomorphism φ : V → U , and a reparmetrization of the curve c(t) = f(v(t)) := f φ v(t),

both dX(t)dt and ∇X(t)

dt remain the same.

It follows from (127) that ∇X(t)dt |t=0 only depends on u(0) and u′(0), and it is independent

on any other information of c(t). Consider another tangent vector field Y and use it to generatea curve c(t) on f , motivated by (127) we may define the covariant derivative of a tangentvector field X with respect to another tangent vector field Y by

∇YX =

2∑k,p

(∂ξk∂up

+∑q

ξqΓkqp

)ηpfk, where X =

∑k

ξk(u)fk, Y =∑p

ηp(u)fp.(128)

Lemma 5.2.1. (127) and (128) are equivalent.

Proof of Lemma 5.2.1. Firstly, for any curve c(t) = f(u1(t), u2(t)) on the surface f(U), we maycheck

∇c′(t)X =∇X(t)

dt.(129)

On the other hand, given any tangent vector field Y along the surface, write Y = η1fu1 +η2fu2

where η1, η2 are functions of u1, u2, by (128) we get the value ∇YX at any p0 = (u1, u2) ∈ U .Next we show there is a unique curve c(t) = f(u1(t), u2(t)) satisfying (u1(0), u2(0)) = p0 ∈ U

and c′(t) = Y (c(t)), Indeed, from Theorem 1.4.2 and 1.4.5, we may solve the following ODEsystem

du1(t)dt = η1(u1(t), u2(t)),

du2(t)dt = η2(u1(t), u2(t)).

(130)

combined with some initial condition (u1(0), u2(0)) = p0 ∈ U . Comparing with both (127) and(128), we see

∇YX = ∇c′(t)X =∇X(t)

dt.

Next we list some elementary properties of (127) and (128), for example (128) implies∇fifj =∑l Γ

ljifl. And ∇YX is linear in both X and Y , i.e. given any constants a, b and tangent vector

fields X,Y, Z, the following holds.

∇Y (aX + bZ) = a∇YX + b∇Y Z,∇(aY+bZ)X = a∇YX + b∇Y Z.

Next we given any smooth function λ defined along the surface, in fact we should writeλ(f(u1, u2)), for simplicity, we also write λ(u1, u2) = λ(f(u1, u2)). From Y , we may solve (130)to get the curve c(t) with initial value (u1(0), u2(0)) = p0 ∈ U and c′(t) = Y (c(t)). Now wedefine the directional derivative of λ along Y at p0

Y (λ)(p0) :=dλ(c(t))

dt|t=0.(131)

As an exercise, if we write Y =∑i ηifui , then Y (λ) =

∑i ηi

∂λ∂ui .


Now we list two additional properties of (128), which are quite useful in calculations.

∇λYX = λ∇YX,(132)

∇Y (λX) = Y (λ)X + λ∇YX.(133)

Consider a curve c(t) = f(u(t)), then we could choose an orthornormal frame along c(t) by

e1 =c′(t)|c′(t)| , e2 = n× e1

as the Frenet frame tangent to the surface. Assume s the arc length of c(t), then e1 = c′(s).We have the following ‘curved’ analogue of Frenet equation for planar curves.

∇ds

[e1(s)e2(s)

]=

[0 kg−kg 0

] [e1(s)e2(s)

](134)

The function kg = ∇e1ds · e2 is called the geodesic curvature of c.

Proposition 5.2.2 (Geodesic curvature is invariant under change of coordinates). Given anycurve c(t) = f(u1(t), u2(t)) on the surface f : U → R3, the geodesic curvature kg is independent

of choice of coordinates. Indeed, consider a diffeomorphism φ : V → U and c(t) = f(v1(t), v2(t))

where f := f φ : V → R3 as the new surface patch, then the geodesic curvature of c(t) remainsthe same for any t.

By Proposition 5.2.2, we may calculate the geodesic curvature of a curve on a surface after anychoice of your favorite coordinates on the surface. In particular, we will derive the correspondingformulas under geodesic orthogonal coordinates, isothermal, and orthogonal coordinates.(a). kg in terms of geodesic orthogonal coordinates.

It follows from Lemma 4.3.6 on p.80 in [9] that we can always find geodesic orthogonalcoordinates f(u, v) : U → R3 such that the first fundamental form is I = du2 + Edv2. A directcalculation shows

Γ111 = Γ2

11 = Γ112 = 0, Γ2

12 =1

2EEu, Γ1

22 = −1

2Eu, Γ2

22 =1

2EEv.

From Theorema Egregium we have K(g) = R1212

det(gij),

K =R1212

E=

Γ112v − Γ212u +∑p Γp21Γ12p −

∑p Γp11Γ22p

E= − (

√E)uu√E

.(135)

There is a natural choice of orthonormal frame E1 = fu, E2 = 1√Efv. Assume the tangent

vector of curve c(s) forms the angle θ between E1, then we could write e1 = c′(s) = cos(θ)E1 +sin(θ)E2, θ is locally well-defined smooth function with respect to s. Next we will find a formulaof kg in terms of θ and E. It is easy to see e2 = n× e1 = − sin(θ)E1 + cos(θ)E2. Applying (132)and (133), we get

56 BO YANG

kg =∇e1

ds· e2,

=(∇ds

(cos θE1 + sin θE2))·(− sin θE1 + cos θE2

),

=(d cos θ

dsE1 + cos θ

∇E1

ds+d sin θ

dsE2 + sin θ

∇E2

ds

)·(− sin θE1 + cos θE2

),

= sin2 θdθ

ds+ cos2 θ

∇E1

ds· E2 − sin2 θ

∇E2

ds· E1 + cos2 θ

dθ

ds,

=dθ

ds+∇E1

ds· E2, here we apply

∇Eids· Ej = −∇Ej

ds· Ei,

=dθ

ds+∇(cos θE1+sin θE2)E1 · E2,

=dθ

ds+ cos θ∇E1E1 · E2 + sin θ∇E2E1 · E2.

Next we compute

∇E1E1 · E2 = ∇fufu ·

1√Efv =

1√E

Γ211fv · fv = 0,

∇E2E1 · E2 =1√E∇fvfu ·

1√Efv = T 2

21 =Eu2E

.

To sum up, we conclude under geodesic coordinates,

kg =dθ

ds+ sin θ

Eu2E

=dθ

ds+ (√E)u

dv

ds,(136)

where we use dvds = sin θ√

E.

(b). kg in terms of isothermal coordinates.

Recall in Example 4.2.5 we state that there exist isothermal coordinates f(u, v) : U → R3

such that the first fundamental form is I = E(du2 + dv2). In this case we have

Γ111 = Γ2

12 =Eu2E

, Γ112 = Γ2

22 =Ev2E

, Γ211 = −Ev

2E, Γ1

22 = −Eu2E

.

Instead the natural choice of orthonormal frame E1 = 1√Efu, E2 = 1√

Efv. Similarly, if we

may assume the unit tangent vector of curve c(s) is e1 = c′(s) = cos(θ)E1 + sin(θ)E2, ande2 = n× e1 = − sin(θ)E1 + cos(θ)E2.

A similar calculation as in geodesic coordinates shows

kg =∇e1

ds· e2

=dθ

ds+ cos θ∇E1E1 · E2 + sin θ∇E2E1 · E2

=dθ

ds+ cos θ

1√E∇fu(

1√Efu) · 1√

Efv + sin θ

1√E∇fv (

1√Efu) · 1√

Efv

=dθ

ds+

cos θ√E

Γ211 +

sin θ√E

Γ212

=dθ

ds− cos θ

2√E

(lnE)v +sin θ

2√E

(lnE)u.

To sum up, we conclude under isothermal coordinates, here we use duds = cos θ√

E, dvds = sin θ√

E.

kg =dθ

ds− cos θ

2√E

(lnE)v +sin θ

2√E

(lnE)u =dθ

ds− 1

2

((lnE)v

du

ds− (lnE)u

dv

ds

).(137)

(c). kg in terms of orthogonal coordinates.


Assume there exists local coordinates (u, v) in U such that the first fundamental form off(u, v) : U → R3 is I = Edu2 +Gdv2. Then its Gauss curvature is

K(v) = − 1

2√EG

((Ev√EG

)v + (Gu√EG

)u).(138)

And Liouville’s formula solves the geodesic curvature of a curve c(s) with s being its arclength:

kg =dθ

ds− cos θ

2√G

(lnE)v +sin θ

2√E

(lnG)u.(139)

5.3. The Gauss-Bonnet Theorem.

Theorem 5.3.1 (Gauss-Bonnet, local and global versions). Let f : M → R3 be an embeddedsurface which is orientable.

(1) Assume C a smooth simple closed curve on f(M), and C is the boundary of a set Ω whichis the diffeomorphism image from some closed disk in R2, then∫

C

kgds = 2π −∫

Ω

KdAf(M).(140)

(2) Assume C is a closed simple curve on f(M) which is piecewise smooth, If C is theboundary of a set Ω which is the diffeomorphism iamge from some closed polygon in R2, then∫

C

kgds = 2π −∑j

aj −∫

Ω

KdAf(M).(141)

Here −π ≤ αj ≤ π denotes the oriented exterior angles at the vertices of Ω.(3) Let χ(f(M)) denote the Euler characteristic number of f(M).∫

f(M)

KdAf(M) = 2πχ(f(M)).(142)

Remark 5.3.2. As we shall see in the proof, the local version of Gauss-Bonnet can be generalizedto a general compact domain Ω ⊂ f(M) with piecewise smooth boundary. The correspondingconclusion is ∫

∂Ω

kgds = 2πχ(Ω)−∑j

aj −∫

Ω

KdAf(M).(143)

Here aj denotes the oriented exterior angles at the vertices of Ω and χ(Ω) is the Euler character-istic number of Ω. Recall that the Euler characteristic can be defined for both compact surfaceswith or without boundary. If a compact surface Ω (possibly with boundary) can be divided into fopen discs by drawing e edges which connect v vertexes, then we define χ(Ω) = f−e+v. A factin topology says that any compact orientable surface Σg,n with g holes and n boundary circleshas its Euler characteristic as 2− 2g−n, and it is independent of your choice of decomposition.The latter also follows from the Gauss-Bonnet formula.

It is important to have a simpler decomposition to calculate the Euler number. For example,try to count the Euler number of a torus with two holes by Figure 21.

Proof of Theorem 5.3.1.We present three approaches. The first two replies on a special choice of coordinates, namely

geodesic orthogonal coordinates and isothermal coordinates. The proofs are given in Theorem6.3.2 on p.139 in [9] and Theorem 3.6.1 on p.111 in [4] respectively. Both proofs relies on theformula of Gauss curvature and geodesic curvature in the special coordinates. Therefore, onehas to first work in a small region which lies in a single coordinate chart, then move to largerregions by cutting into small pieces. We learned the third proof from Donaldson’s lectures notes.It uses polar coordinates in a larger region and is independent of Theorema Egregium.30

Method 1 and 2: Use geodesic coordinates and isothermal coordinates.

30See http://wwwf.imperial.ac.uk/~skdona/lecturenotes/GAUSS.PDF.

http://wwwf.imperial.ac.uk/~skdona/lecturenotes/GAUSS.PDF

58 BO YANG

Figure 7: Example of a surface with genus 2 glued from triangles.

Figure 8: Flattening a triangular subdivision of the sphere.

That means curvature is only an interesting property if the sides of ourtriangles (of minimizing paths) we use to measure it enclose at least one vertex.Let v1, . . . , vV denote the set of vertices in X. Consider a small trianglearound every vertex vi in X and denote by αi the sum of the angles. As wehave seen above αi < π would correspond to negative curvature, αi = π wouldmean that at vi we have glued triangles in a plane, while αi > π corresponds topositive curvature.

The Gauß–Bonnet theorem tells us that we can detect the Euler characteristicby examining the curvature; more precisely it states

V∑

i=1(αi − π) = 2πχ(X).

For example, if we have no curvature at all, then X must have Euler charac-teristic zero. On the other hand if we know that the Euler charasteristic is zero,as we do for the torus – the surface of a doughnut, which has precisely one hole– then the sum on the left hand side has to vanish as well, that is, there have tobe points, or areas, with positive curvature that compensate points, or areas,

9

Figure 21. The above figure is copied from ‘The Interaction of Curvature andTopology’ by Kordaß on https://publications.mfo.de/handle/mfo/3692

Step 1: We prove the theorem after assuming Ω lies in a single coordinate system f : U →Mα ⊂M as either of two choices in below.

In our previous calculation on geodesic curvature, we use the fact that the angle function θformed by e1 = c′(s) and E1 is locally well-defined. In fact, a similar result as Proposition 3.2.1holds, we can always find a function θ which is continuous, and piecewisely smooth if the curvec : [a, b] → M is smooth or piecewisely smooth. Moreover, θ(b) − θ(b) is independent of thechoice θ.

(1a):Use geodesic orthogonal coordinates.

Now that Ω lies in one geodesic coordinate system f : U →Mα ⊂M . Since f is a homeomor-phism, we may simply assume the restriction of f on some K ⊂ U realizes the diffeomorphismfrom P to Ω. Therefore the boundary of P , denoted by ∂P is the preimage of C ∈M . Since Cis a closed simple curve, so is ∂P = f−1(C).

From (135), we have∫Ω

KdAf(M) =

∫P

−(√E)11√E

√Edudv

=

∫P

div((−√E)1, 0)dudv

=

∫∂P

(−(√E)u, 0) · (dv,−du) we use the divergence theorem on R2

=

∫∂P

−(√E)udv

=

∫∂P

−(√E)u

dv

dsds caution: we are using s as the arc length of C

=

∫∂P

(dθ

ds− kg)ds here we use (136)

=

∫∂P

dθ −∫∂P

kgds

=

∫C

dθ −∫C

kgds.

In the last step we change the domain of integral since C and ∂P are diffeomorphic.We are done with the proof by the following lemma.

https://publications.mfo.de/handle/mfo/3692


Lemma 5.3.3 (A ‘curved’ analogue of Umlaufsatz). In the above notation, if C is simple closedsmooth (or piecewise smooth) curve, we have∫

C

dθ = 2π (or 2π −∑j

aj),(144)

where aj denotes the oriented exterior angles at the vertices of Ω ⊂M .

Proof of Lemma 5.3.3. The starting point of the proof is that there exists some integer nc sothat ∫

C

dθ = 2πnc (or 2πnc −∑j

aj),(145)

This can be seen in the proof of Proposition 3.2.1, the function θ which is continuous, andpiecewise smooth along the closed curve c : [a, b] → M . Moreover, θ(b) − θ(a) is a multiple of2π.

Lemma 5.3.3 was proved by Klingenberg on p.140 in [9] via a continuous deformation of thefirst fundamental forms g(t) = du2 + (t + (1 − t)E)dv2, then apply the Umlaufsatz for planarcurves. (See Theorem 3.2.5.)

(1b): Use isothermal coordinates.

Since Ω lies in one isothermal coordinate system f : U → Mα ⊂ M . similarly as in the caseof geodesic coordinate system, we may assume the restriction of f on some K ⊂ U realizes thediffeomorphism from P to Ω. Therefore ∂P is diffeomorphic to C ⊂M . From (73), we have∫

Ω

KdAf(M) =

∫P

− 1

2Ediv((lnE)1, (lnE)2)E dudv

=

∫∂P

−1

2((lnE)1, (lnE)2) · (dv,−du)

= −1

2

∫∂P

(lnE)udv − (lnE)vdu

=

∫∂P

(dθ

ds− kg)ds, here we use (137)

=

∫∂P

dθ −∫∂P

kgds

Now let us consider the curve C ⊂M parametrized by c(s) = f(u(s), v(s)), since s is the arclength parameter of c and θ is the angle between c′(s) and E1 = 1√

Efu, we conclude

(du

ds)2 + (

dv

ds)2 =

1

E, cos θ = c′(s) · E1 =

√Edu

ds.(146)

Next we turn to the plane curve ∂P ⊂ U parametrized by (u(s), v(s)), the unit tangent vector

of ∂P is given by T =√E(u′(s), v′(s)) and T · (1, 0) =

√Eu′(s) = cos θ. This means we view θ

function, originally defined as angle between c′(s) and E1, is also the angle between T and (1, 0)on the plane. Moreover, as another benefit of working under isothermal coordinates we see thatthe oriented exterior angles at the vertices of Ω ⊂ M is the same as those of ∂P ⊂ U ⊂ R2.Apply Theorem 3.2.5 the Umlaufsatz for the plane curve, we conclude∫

Ω

KdAM =

∫∂P

dθ −∫∂P

kgds = 2π −∑j

aj −∫C

kgds.(147)

Step 2: The general case when Ω no longer lies in a single coordinate system f : U →Mα ⊂M . We omit this part and refer to the proof on p.140 in [9].

Step 3: A similar argument as in Step 2 leads to the global version, This part is proved onp.142 in [9].

60 BO YANG

Method 3: Use polar coordinates.

Step 1: When the boundary curve C is smooth.

Lemma 5.3.4. Let P ∈ R3 be plane through the origin, and N its unit normal vector. For anythree vectors X,Y, Z ∈ P , we have

det(X,Y,N)Z + det(Z,X,N)Y + det(Y, Z,N)X = 0.(148)

Proof of Lemma 5.3.4. Assume aX + bY + cZ = 0, then taking cross product with X,Y, Zrespectively, then inner product with N , we end up with

b det(Y,X,N)Y + cdet(Z,X,N)Z = 0,

adet(X,Y,N)X + cdet(Z, Y,N)Z = 0,

adet(X,Z,N)X + bdet(Y,Z,N)Y = 0.

Let us assume Y and Z is not linear independent, then we pick a = det(Y, Z,N), then we maysolve b = det(Z,X,N) and c = det(Y,Z,N).

Alternatively we may argue by plane geometry, first note that (148) is linear, it suffices toconsider X,Y, Z which are all unit vectors. Consider the unit circle in P centered at the origin,assume their angles are α, β, γ respectively, we only need to verify

sin(α)Z + sin(γ)Y + sin(β)X = 0.

Note that sin(α) = sin(2π − β − γ) = − sinβ cos γ − cosβ sin γ. we have

sin(α)Z + sin(γ)Y + sin(β)X = (sinβ)(X − cos γZ) + (sin γ)(X − cosβZ)

= sinβ cos(γ − π

2)− sin γ cos(β − π

2)

= 0.

Yet there is an alternative way via the plane geometry, let X,Y, Z be unit vectors and considerthe triangle in which the unit circle determined by X,Y, Z is inscribed, Let l1, l2, l3 denote thecorresponding edges, then divergence theorem implies l1X + l2Y + l3Z = 0. Comparing areasof some triangles we could see

l1 : l2 : l3 = det(Y, Z,N) : det(Z,X,N) : det(X,Y,N).

By assumption Ω ⊂ f(M) is the image of diffeomorphism φ from some disk D ∈ R2, note thatf : M → R3 is an embedding and injective, we consider f−1 φ : D →M which naturally servesa piece of coordinate chart of M . With a slight abuse of notation, let f : D → Ω ⊂ f(M)(with its boundary f : ∂D → C) denote the parametrized surface we will work on.

Assume D = Dr ∈ R2 is a disk of radius r centred at the origin, in terms of polar co-ordinates we may parametrize the boundary circle ∂D = f−1(C) by c(θ) = f(u(θ), v(θ)) =f(r cos θ, r sin θ). Now we have the Frenet frame along f−1(C):

e1 =c′(θ)|c′(θ)| , e2 = n× e1;

kg = ∇e1e1 · e2 = (de1

ds)T · (n× e1) = −(

de1

ds× e1) · n, s is the arc length of c(θ).

We drop the sign of tangential component, since the normal component of de1ds cancels with n

automatically in the expression.

Lemma 5.3.5. Note that the frame is well defined for any circle Dr ∈ D as 0 < r < r0, andwe set

S.=

∂

∂r[(∂e1

dθ× e1) · n]− ∂

∂θ[(∂e1

∂r× e1) · n].(149)

Then S = K (fr × fθ) · n = K |fr × fθ|.


Proof of Lemma 5.3.5. Let T = e1, then

S = [(Tθ × T ) · n]r − [(Tr × T ) · n]θ

= 2[(Tθ × Tr) · n] + [(Tθ × T ) · nr]− [(Tr × T ) · nθ]= [(T × nr) · Tθ]− [(T × nθ) · Tr] here we note Tθ, Tr, n are on the same plane

= [(T × nr) · n](Tθ · n)− [(T × nθ) · n](Tr · n) we use T × nr is parallel to n

=(− [(T × nr) · n]nθ + [(T × nθ) · n]nr

)· T

=(

det(nr, T, n)nθ + det(T, nθ, n)nr

)· T

= −det(nθ, nr, n)T · T here we use Lemma 5.3.4

= det(nr, nθ, n).

Recall the Weingarten map L(fr) = −nr, L(fθ) = −nθ, we have

S = det(nr, nθ, n) = K det(fr, fθ, n) = K |fr × fθ|.

It follows from Lemma 5.3.5 that

d

dr

(∫f(∂Dr)

kgds)

= − d

dr

(∫ 2π

0

(∂e1

ds× e1) · nds

dθdθ)

= − d

dr

(∫ 2π

0

(∂e1

dθ× e1) · ndθ

)= −

∫ 2π

0

(S +

∂

∂θ[(∂e1

∂r× e1) · n]

)dθ

= −∫ 2π

0

K |fr × fθ|dθ.(150)

On the other hand,

d

dr

(∫f(Dr)

KdAf(M)

)=

d

dr

(∫ r

0

∫ 2π

0

K|fρ × fθ|dρdθ)

=

∫ 2π

0

K|fr × fθ|dθ.(151)

Combining (150) and (151), we conclude for any 0 < r < r0∫f(∂Dr)

kgds+

∫f(Dr)

KdAf(M) = const(152)

The smooth boundary case of Gauss-Bonnet will be done once we show the following

Lemma 5.3.6.

limr→0

(∫f(∂Dr)

kgds)

= 2π(153)

Proof of Lemma 5.3.6. Intuitively speaking, when r goes to 0, the curve f(∂Dr) shrinks to thecenter f(O), and the first fundamental form nearby goes closer to gij(f(O)). So the geometryof the surface approximates the plane in smaller regions, and

∫f(∂Dr)

kgds gets more closer to

the curvature integral of a planar circle.Therefore, when r is small enough, we may use geodesic normal coordinates (u, v) near f(O),

such that

gij(f(O)) = δij ,(154)

∂gij∂u

(f(O)) =∂gij∂v

(f(O)) = 0, for any 1 ≤ i, j ≤ 2.(155)

62 BO YANG

On one hand, for the curve c(θ) = f(u(θ), v(θ)) = f(r cos θ, r sin θ) with s its arc length, wehave

e1 =c′(θ)|c′(θ)| =

− sin θfu + cos θfv√g11 sin2 θ + g22 cos2 θ − 2g12 sin θ cos θ

,(156) ∫f(∂Dr)

kgds =

∫ 2π

0

de1

dθ· (n× e1)dθ = −

∫ 2π

0

|de1

dθ× e1|dθ.(157)

On the other hand, we consider the curve c(θ) = (u(θ), v(θ)) ∈ R2 where R2 is endowed withthe standard Euclidean inner product gij(O) = δij . For this curve c(θ), we have

e1 =c′(θ)|c′(θ)| =

(− sin θ, cos θ)√g11(O) sin2 θ + g22(O) cos2 θ − 2g12(O) sin θ cos θ

= (− sin θ, cos θ),(158)

∫ 2π

0

de1

dθ· (n(O)× e1)dθ = −

∫ 2π

0

|de1

dθ× e1|dθ =

∫ 2π

0

dθ = 2π.(159)

It follows from (155) that all the Christoffel symbols at f(O) vanishes. Moreover, using both(154) and (155) it is direct to check

e1 → − sin θfu + cos θfv,de1

dθ→ −fu cos θ − fv sin θ, as r → 0.(160)

Finally, comparing both right hand sides of (157) and (159), we conclude

limr→0

(∫f(∂Dr)

kgds− 2π)

= 0.

In fact, we cheat a bit in the above as we use geodesic normal coordinates satisfying (157)and (159). It involves a change of coordinates and the circle under consideration will become asmooth simple closed curve. It is not difficult to rewrite a rigorous proof with aid of Theorem3.2.4 (Umlaufsatz for the smooth planar closed curve).

Step 2: When the boundary curve C is piecewise smooth.

Once we prove the case when the boundary curve C (the region Ω ⊂ f(M)) being a dif-feomorphic image of a circle (a disk), By Proposition 3.1.3, any smooth simple closed curveis diffeomorphic to a circle, we conclude (140) holds when the boundary is any smooth simpleclosed curve. Next we turn to the case when the boundary curve C (the region Ω) being a diffeo-morphic image of a piecewise smooth curve (a polygon), again we simply call this diffeomorphismf .

For simplicity, we consider the curve C, parametrized by its arc length, c(s) = f(u(s), v(s)) :[0, L] → f(M), has one corner point c(s∗) = P∗. Therefore the corresponding preimage c(s) =(u(s), v(s)) has a corner at f−1(P∗) we may defined a sequence of smooth curves ci(t) : [0, L]→f(M) approximating c(s) by ci, see Figure 22 in below.

ci(t) =

c(t) 0 ≤ t ≤ s∗ − 1

i

hi(t), a arc of small size which converges to the corner from the outside

c(t) s∗ + 1i ≤ t ≤ L

On one hand, assume Ωi ⊂ Ω is the region bounded by Ci.= f(ci).

limi→∞

∫Ci

kg(Ci))dsi = 2π − limi→∞

∫Ωi

KdAf(M) = 2π −∫

Ω

KdAf(M).(161)


Figure 22. An smooth arc hi(t) approximating the corner from outside

On the other hand,

limi→∞

∫Ci

kg(Ci)dsi

= limi→∞

(∫ s∗− 1i

0

kg(Ci)dt+

∫ L

s∗+1i

kg(Ci)dt+

∫ s∗− 1i

s∗− 1i

kg(Ci)dt)

=

∫C

kg(C)ds+ limi→∞

∫ s∗+1i

s∗− 1i

kg(Ci)dt.(162)

As i→∞, the curve Ci shrinks to a corner point P∗, the first fundamental form gij aroundCi becomes constant g(P∗). We may apply a similar argument as in the proof of Lemma 5.3.6 toreduce to the case of curves on the Euclidean plane. In this case, the arc part hi are parametrizedby (ri cos θ, ri cos θ) with ri → 0 as i→ +∞. Hence

limi→∞

∫ s∗+1i

s∗− 1i

kg(Ci)dt = limi→∞

∫ s∗+1i

s∗− 1i

dθ(hi) = α.(163)

Here α is the oriented exterior angle formed by c′(s∗−) and c′(s∗+). Alternatively, we may show(163) holds by making use of (136).

Therefore, the polygon version of Gauss-Bonnet follows from (161), (162) and (163).

Step 3: To the global version.

Suppose any compact embedded surface in R3 can be triangulated, then apply the triangleversion of Gauss-Bonnet proved in Step 2 and summing. This concludes the global version ofGauss-Bonnet. As a corollary we see that the Euler characteristic number is independent of thechoice of the triangulation.

5.4. Riemannian 2-manifolds.

5.4.1. Isometry and conformal equivalence.

Definition 5.4.1 (Isometry and conformally equivalence between surface patches).

Let f : U → R3 and f : V → R3 be two surface patches with first fundamental forms Iu and

Iv respectively.

64 BO YANG

f (or f(U)) and f (or f(V )) are conformally equivalent if there exists a smooth diffeo-morphism φ : V → U and a positive smooth function ρ on V such that

Iφ(v)(dφ(X), dφ(Y )) = ρ2(v)Iv(X,Y ), ∀v ∈ V, and X,Y ∈ TvR2.(164)

We call φ a conformal diffeomorphism. In particular, if ρ = 1 is the constant function, we

call φ an isometry, and f (or f(U)) and f (or f(V )) are isometric. Note that (164) has thematrix form

(dφ)T Iφ(v)dφ = ρ2(v)Iv, ∀v ∈ V.(165)

In the following we list several standard examples on conformal diffeomorphisms.

(i) The most standard example of an isometry is change of coordinates. One may checkthat any diffeomorphism φ : V → U must be an isometry between f : U → R3 and

f : V → R3, where f = f φ.

(ii) In case both f and f is the trivial parametrization of z = 0 = R2, with the standardEuclidean metric as their first fundamental forms, in terms of complex coordinateson C, it follows from Cauchy-Riemann equation that any holomorphic (or conjugateholomorphic) injective maps φ from V ⊂ C to φ(V ) ⊂ C is a conformal diffeomorphism.

(iii) The above example can be generalized to any two surfaces f and f as long as theyare both with isothermal coordinates. let φ : V → U be the diffeomorphism such thatif v1 +

√−1v2 = φ(v1 +

√−1v2) is holomorphic or (conjugate holomorphic). Using

Cauchy-Riemann equation we find that (164) reduces to

Iφ(v)|∂φ

∂z|2 = ρ2(v)Iv, ∀v ∈ V.(166)

if φ is holomorphic, and

Iφ(v)|∂φ

∂z|2 = ρ2(v)Iv, ∀v ∈ V.(167)

if φ is conjugate holomorphic.(iv) Another example of isometry comes from the Enneper-Weierstrass representation, in

Example 4.3.16 we have a family of minimal surfaces f(θ) parmetrized by θ ∈ [0, 2π).each of them is isometric to each other.

As an immediate consequence of Theorema Egregium (Theorem 4.2.4), we know that theGauss curvature is invariant under isometries.

Proposition 5.4.2. Let f : U → R3 and f : V → R3 two surface patches, K and K their

corresponding Gauss curvatures. Assume φ : V → U is an isometry, then K(f(φ(v))) = K(f(v))for any v ∈ V .

Proof of Proposition 5.4.2. From Proposition 4.1.10 we already knew Gauss curvature is invari-ant under change of coordinates. Therefore to compute Gauss curvature of f : U → R3, we maycompute the Gauss curvature of f φ : V → R3 instead. Now compare two surfaces patches

f φ : V → R3 and f : V → R3, their first fundamental forms are exactly the same since φ isan isometry. By Theorem 4.2.4, their corresponding Gauss curvature must be the same.

Let f : M → R3 be a smooth embedded surface in R3, we may define I as the first fundamentalform of M by a collection of first fundamental forms on each of its surface patches. Namely, interms of atlas uα : Mα → Uα ⊂ R2, we may write local components of I on each Mα (or Uα) as

g(α)ij =∂(f u−1

α )

∂ui· ∂(f u−1

α )

∂uj.(168)

Note that g(α)ij are smooth functions on Uα and uβ u−1α : uα(Mα ∩Mβ) → uβ(Mα ∩Mβ) is

an isometry with respect to gα and gβ in the sense of Definition 5.4.1. To sum up, any smoothembedded surface admits a natural global first fundamental form after gluing isometric surfacepatches together.


This motivates us to introduce a smooth 2-manifold M endowed with a ‘global’ first fun-damental form. We define a smooth Riemannian manifold of dimension 2 as a smooth2-manifold M with an assignment g which sends every p ∈ M to a positive definite innerproduct gp on TpM in a smooth way. Here TpM is the ‘abstract’ tangent space of M . Thismeans on each of its atlas uα : Mα → Uαα∈Λ, there is a positive definite inner product gαdefined on TuUα which varies smoothly with respect to u ∈ Uα. Moreover, for any α, β ∈ Λ, thetransition map (also called change of coordinates)

φαβ := uα u−1β : uβ(Mα ∩Mβ)→ uβ(Mα ∩Mβ)

is an isometry between two ‘abstract’ surface patches u−1α : Uα → Mα and u−1

β : Uβ → Mα ontheir overlapping region Mα ∩Mβ , in the following sense:

(dφαβ)T gα|φαβ(v)dφαβ = gβ |v, ∀v ∈ uβ(Mα ∩Mβ).(169)

We call such an assignment g a Riemannian metric on M , Usually, we also introduce

ds2 =

2∑i,j

gijduiduj =

2∑i,j

g(∂

∂ui,∂

∂ui)duiduj .(170)

From now on, we call (M, g) (or (M,ds2)) a Riemannian manifold of dimension 2, and itis direct to generalize the notion of isometry (conformal diffeomorphism) in Definition 5.4.1 toan isometry (conformal diffeomorphism) between two Riemannian manifolds.

Definition 5.4.3. Let φ be a diffeormophism between two Riemannian manifolds (M, g) and(N, g), we say φ is an isometry if

gφ(p)(dφ(X), dφ(Y )) = gp(X,Y ), ∀p ∈M, X, Y ∈ TpM.(171)

Let us state without proof some important facts on conformal diffeomorphisms and isometries.

(i) Let Iso(M, g) be the set of all isometries of a Riemannian manifold, then it has a groupstructure by composition of maps. We call Iso(M, g) the isometry group of (M, g).For example, the Iso(Rn, gE) contains a subgroup O(n) which fixes the origin, and anormal subgroup Rn which consists of all translations.

(ii) Let (Sn, ground) be the standard round metric on the unit sphere and p : Sn\N → Rnthe standard stereographic projection with respect to the north pole, then for anyconformal diffeormophism φ of (Sn, ground), p φ p−1 is a Mobius transformationon Rn ∪ ∞. Recall a Mobius transformation is defined by a composition oftranslations, dilations, orthogonal transformations and inversions. A typical exampleis of the form

φ = b+ αA(x− a)

|x− a|2 , or φ = b+ αA(x− a)(172)

where b means a translation, α a dilation, A a rotation in R3, and a the center of theinversion or rotation. Conversely any Mobius transformations on Rn ∪ ∞ can belifted by stereographic projections to a conformal diffeormophism of (Sn, ground). Inparticular, translations, rotations and inversions lift to isometries of (Sn, ground).

(iii) If we consider domains in R2, by the Riemann Mapping Theorem conformal maps canbe found between any proper simply-connected domains and the unit disk. However,conformal maps are more rigid in higher dimensions. A theorem of Liouville statesany conformal mapping on an open domain Ω ⊂ Rn where n ≥ 3 must be a Mobiustransformation.31

31A proof of Liouville’s theorem, due to R.Nevanlinna, is given in Blair’s book “Inversion theory and conformal

mapping.” Student Mathematical Library, 9. American Mathematical Society, 2000. See also p.138. of Book 1of Dubrovin-Fomenko-Novikov, it is further shown that the group of conformal diffeomorphisms of (Sn, ground)is isomorphic to O(1, n+ 1) when n ≥ 3.

66 BO YANG

Many previous results on first fundamental forms of embedded surfaces in R3 carries over togeneral Riemannian 2-manifolds. For example, area, length, angles, and Christoffel symbols canbe defined similarly, In particular, motivated by Theorema Egregium for surfaces in R3, we maydefine the Gauss curvature (also called Riemannian sectional curvature in dimesion2) of (M, g) by

K =R1212

det(gij)=

∑i

(∂Γi11∂u2− ∂Γi12

∂u1+∑p Γp11Γip2 −

∑p Γp12Γip1

)gi2

det(gij).(173)

In fact the above (173) is defined on a local atlas u−1α : Uα →Mα. The following result shows

the Gauss curvature on Riemannian manifolds is invariant under change of coordinates, hencethe same on any overlapping region Mα ∩Mβ .

Proposition 5.4.4. The Gauss curvature (173) of a Riemannian manifold (M, g) is invariantunder change of local coordinates, hence it is globally defined on (M, g). Moreover, if there is

an isometry φ between two Riemannian manifolds (M, g) and (N, g), then K(p) = K(φ(p)) forany p ∈M .

Proof of Proposition 5.4.4. If we check the proof in Proposition 5.4.2, we conclude that thefollowing expression ∑

i

(∂Γi11∂u2− ∂Γi12

∂u1+∑p Γp11Γip2 −

∑p Γp12Γip1

)gi2

det(gij)(174)

is invariant under any isometry between two surface patches.By definition of a Riemannian metric, we know the transition map φαβ := uα u−1

β : uβ(Mα∩Mβ) → uβ(Mα ∩Mβ) is an isometry between two ‘abstract’ surface patches u−1

α : Uα → Mα

and u−1β : Uβ → Mα on their overlapping region Mα ∩Mβ . It is exactly (174) that we use to

define Gauss curvature on Riemannian manifolds, hence Gauss curvature on (M, g) is the sameon Mα ∩Mβ .

Remark 5.4.5. In view of Proposition 5.4.4, one may wonder if any two Riemannian mani-folds with a same Gauss curvature function are necessarily connected by a isometry (or a localisometry). The answer is, except for the case of constant Gauss curvature, we can always findexamples of two Riemannian manifolds with same Gauss curvature function, but not isomet-ric. We refer to two mathoverflow posts 32for more information. In below, we include such anexample for surface patches.

Example 5.4.6. Consider a surface of revolution and the helicoid

f(u1, u2) = (u1 cosu2, u1 sinu2, lnu1), (u1, u2) ∈ (0,+∞)× (0,+∞),(175)

f(v1, v2) = (v1 cos v2, v1 sin v2, v2), (u1, u2) ∈ (0,+∞)× (0,+∞).(176)

A simple calculation shows f has

I =

[u21+1

u21

0

0 u21

], H =

1

2u1(1 + u21)2

, K = − 1

(1 + u21)2

.(177)

While f has

I =

[1 00 u2

1 + 1

], H = 0, K = − 1

(1 + v21)2

.(178)

If there is a map (u1, u2) = φ(v1, v2) such that K(φ(v)) = K(v) then we have

− 1

(1 + u21)2

= − 1

(1 + v21)2

.

32See https://mathoverflow.net/questions/100281/does-the-curvature-determine-the-metric andhttps://mathoverflow.net/questions/181502/examples-of-non-isometric-surfaces-having-the-same-curvature-function

https://mathoverflow.net/questions/100281/does-the-curvature-determine-the-metric

https://mathoverflow.net/questions/181502/examples-of-non-isometric-surfaces-having-the-same-curvature-function


Therefore

(u1, u2) = φ(v1, v2) = (v1, ϕ(v1, v2)) for some function ϕ.

However, we will show the above φ can never be an isometry, indeed from the isometry condition

(165) we have (dφ)T I|φ(v)dφ = I|v and conclude1 =

1+u21

u21

+ u21( ∂ϕ∂v1 )2,

0 = u21( ∂ϕ∂v1 )( ∂ϕ∂v2 ),

1 + u21 = u2

1( ∂ϕ∂v2 )2.

(179)

Obviously, such a system has no solution. Therefore, there is no isometry between f and f .

5.4.2. Laplace operators on Riemannian 2-manifolds.Assume a surface patch f : U → R3 by f(u, v) = (x(u, v), y(u, v), z(u, v)). Given any smooth

function ψ defined on U , we introduce an operator:

∆fψ.=∑i,j

gij(∂2ψ

∂ui∂uj−∑k

Γkij∂ψ

∂uk)

where u1 = u, u2 = v, and all the summation involved are from 1 to 2. This is called theLaplace operator on the surface f .

There is an alternative way to define ∆f , first we define gradψ which is called the gradient of

any smooth function ψ defined on U such that for any tangent vector field ξ =∑i ξi ∂∂ui ∈ TuU ,

ξ(ψ) = I(ξ, gradψ),

where ξ(ψ) is the usual directional derivative of ψ along ξ. It is direct to check

gradψ =∑p,j

gpj∂ψ

∂uj∂

∂up.(180)

Since the first fundamental form can be defined either on TuU or Tuf , from now on we willnot distinguish gradψ on TuU or Tuf , and use the following interchangeably.

gradψ =∑p,j

gpj∂ψ

∂uj∂

∂up, or gradψ =

∑p,j

gpj∂ψ

∂ujfup .

Next we define the divergence of any tangent vector field by

trace∇X =∑ij

gij(∇ ∂

∂uiX) · fuj ,(181)

divX.= trace∇X.(182)

Recall the covariant derivative is defined by (128), it follows from (181) and (182) that

divX =∑i

∂ξi

∂ui+∑i,q

ξqΓiqi =1√

det(g)

∑k

∂

∂uk(√

det(g) ξk).(183)

∆fψ = div(gradψ) =1√

det(g)

∑j,k

∂

∂uk(√

det(g) gkj∂ψ

∂uj).(184)

The operators div, grad, and ∆f are all independent of the change of coordinates.Indeed all the formula (180), (181), (182), (183), and (184) remain invariant under change ofcoordinates. For example, for any smooth function ψ defined on U , assume φ : V → U adiffeomorphism, then consider f φ : V → R3 as the new surface patch, we have

(∆fφ(ψ φ))v = (∆fψ)φ(v), ∀v ∈ V.This invariant property allows us to define div, grad, and Laplace operators on any Riemannian2-manifold (M, g).

68 BO YANG

In the proof of Gauss-Bonnet for compact embedded surface M in R3, we used divergencetheorem in Euclidean space. Now with the divergence operator defined on M , we could get adivergence theorem directly on M . It states

Theorem 5.4.7 (Divergence theorem for surfaces in R3, p.117 in [9]). Let M be a compactembedded surface in R3, D a bounded region on M with piecewise smooth boundary, and X is atangent vector field on M . ∫

D

divXdA =

∫∂D

(X · ν)ds.(185)

Here ν is the unit outer normal of the boundary ∂D and ν is tangent along M . In particular,∫M

divXdA = 0.(186)

Further plan: TBD, state coarea formula. Example of spherical caps, and else.As an application of the Laplace operator, we present an alternative proof of the Bernstein

theorem (Theorem 4.3.9) due to S.-S. Chern.

Another proof of Theorem 4.3.9.Step 1: We solve its Gauss curvature

K = ∆f ln(1 +1

W), where W =

√1 + (ϕu)2 + (ϕv)2.

Here ∆f is the Laplacian operator with respect to the surface f . The quick way to see this toconsider the standard construction of isothermal coordinates (ξ, η) = φ(u, v). Note that (ξ, η)is also defined on the whole R2.

Step 2: We need a crucial lemma.

Lemma 5.4.8 (Liouville theorem for subharmonic functions). Let u be a C2 subharmonicfunction on the complex plane except the origin, assume

lim infr→0

max|z|=r u(r)

|ln r| = lim infr→∞

max|z|=r u(r)

|ln r| = 0,

then u is constant.In particular, any C2 subharmonic function on the complex plane with

lim infr→∞

max|z|=r u(r)

|ln r| = 0,

is constant.

For simplicity, we have assumed that u is C2, then the subharmonic condition on R2 means∆u ≥ 0 for the usual Laplacian on Euclidean space C1 = R2.

A quick proof of Theorem 5.4.8 makes use of Hadamard’s three-circle theorem. Another proofdepends on Phragmen-Lindelof principle.33

Step 3: Consider the Laplacian ∆f which only differs the standard Euclidean Laplacian(ξ, η)

∆(ξ,η) =∂2

∂ξ2+

∂2

∂η2,(187)

by a positive multiple. Therefore, we may use Theorem 5.4.8 for ∆(ξ,η) to conclude W is

constant. Since both H = K = 0, we conclude that any point on the surface f : R2 → R3 is

umbilic, which implies the unit normal (−ϕu,−ϕu,1)W is constant. Therefore ϕ is a linear function.

Remark 5.4.9. It would be desirable to prove an analogue of Theorem 5.4.8 directly on ∆f .We hope to come back to this topic later.

33For example it can be found on p.31 in the book by Ransford, “Potential theory in the complex plane”.London Mathematical Society Student Texts, 28. Cambridge University Press, 1995.


5.4.3. Gauss-Bonnet for Riemannian 2-manifolds. The Gauss-Bonnet theorem holds for anycompact oriented Riemannian 2-manifold, which includes a closed embedded surfaces in R3 asa special case.

Theorem 5.4.10 (Gauss-Bonnet for Riemannian 2-manifold). Assume (M, g) is a compactoriented Riemannian manifold of dimension 2. Let χ(M) = 2 − 2g(M) denote its Euler char-acteristic number where g(M) is the genus of M .∫

M

KdAg = 2πχ(M).(188)

There are many approaches to Theorem 5.4.8, and its higher dimensional generalization isan important topic in the early development of global Riemannian geometry. We will sketcha proof from Phong’s lecture notes at Columbia.34 This proof utilizes variational aspects ofRiemannian metrics. In general dimensions, the calculation of first variation of the Einstein-Hilbert functional is fundamental in Riemannian geometry.

A sketchy proof of Theorem 5.4.8.Set up: Assume M is a closed oriented smooth 2-manifold, according to the classification

of 2-manifolds, M can be obtained by a 2-sphere attached by g handles, and g is defined as thegenus of M . Any such 2-manifold M admits infinitely many Riemannian metrics on M , Indeed,the space of all Riemannian metrics on a fixed smooth 2-manifold is itself an infinite dimensionalobject. Here we do not bother its structure, and content with defining the following functionalon it.

(M, g) 7→ 1

2π

∫M

K(g)dAg,(189)

where Ag is the area element of (M, g) whose local expression is√

det(gij) dx2 dx2.

Recall by definition K(g) = R1212

det(gij). In the context of Riemannian geometry it is more

convenient to rewrite

K = R1212 det(gij) = R1212(g11g22 − g12g21) =1

2

2∑i,j,k,l

gikgjlRijkl.(190)

Step 1: The crucial observation is the following claim.

Claim 5.4.11. Assume (M, g) is a compact oriented Riemannian 2-manifold, Consider a path

g(t) in the space of Riemannian metrics, where g(0) = g and ∂g(t)∂t |t=0 = η. Here η is any

arbitrary bilinear form and represents is a direction of variation of g.

d

dt

( 1

2π

∫M

K(g(t))dAg(t)

)t=0

= 0.(191)

The proof of Claim 5.4.11 depends on some basic calculation in Riemannian geometry andthe divergence theorem, i.e. a Riemannian geometry version of Theorem 5.4.7.

Let us only list some important formulas involved

d

dtlog(det(gij))|t=0 = gijηij ,

d

dtΓkij |t=0 =

1

2gkl(∇jηil +∇iηjl −∇lηij),

d

dtK(g(t))|t=0 = −1

2

(− ηikgjlRijkl + gpjgqi∇p∇qηij −∇lηij −∆(gijηij)

).

And the divergence theorem on compact Riemannian manifolds with boundary states∫M

divXdAg =

∫∂M

g(X,µ)ds∂M(192)

34See Phong, D.. Course notes on complex analysis and Riemann surfaces at Columbia prepared by You Qi,Fall 2008.

70 BO YANG

where µ is the unit outer normal of the boundary ∂M . When M is compact without boundary,then the right hand side is 0.

As an immediate consequence of Claim 5.4.11, we conclude 12π

∫MK(g)dAg is independent

of the choice of Riemannian metric g. We may always choose our favorite metric to calculatethis quantity!

Step 2: We do some surgery on M , reduce the calculation of 12π

∫MK(g)dAg into combina-

tions of corresponding terms on spheres and hemispheres.Firstly if the genus of M is 0, M is topologically 2-sphere, we may choose our favorite metric

as the induced metric g0 of unit sphere S2 ⊂ R3. In this case, K(g0) = 1,

1

2π

∫M

K(g)dAg =1

2πA(S2) =

4π

2π= 2.

Secondly if the genus of M is 1, M is topologically 2-torus, the sum of two handles. On eachhandle, we could add a hemisphere on each of its end, make it a topological sphere again!, Andthese two hemisphere add together into another sphere. In the end,

1

2π

∫M

K(g)dAg = 2× 1

2π

∫one handle

K(g)dAg

= 2×( 1

2π

∫one sphere

K(g)dAg −1

2π

∫one sphere

K(g)dAg

)= 0.

Thirdly if the genus of M is 2, M is topologically a body with the sum of two handles. Notethat the body becomes a sphere with each of its four boundary circles added a hemisphere. Asimilar calculation as above shows

1

2π

∫M

K(g)dAg =1

2π

∫a body

K(g)dAg + 2× 1

2π

∫one handle

K(g)dAg

=1

2π

∫one sphere

K(g)dAg −4

2

1

2π

∫one sphere

K(g)dAg

= −2.

Finally if the genus of M is g, a similar argument allows us to conclude 12π

∫MK(g)dAg =

2− 2g.

5.4.4. Geodesics (on an ellipsoid). First we define the geodesic on a surface patch f : U → R3,and then calculate the first variation of length of curves. We will also discuss these notions ona 2-dim Riemannian manifold.

Let c(t) = f(u1(t), u2(t)) be a curve on a surface patch f : U → R3, we call c(t) is a geodesic

if ∇c′(t)dt = 0. An immediate consequence is that |c′(t)| is constant along t. Conversely, any

regular curve with its geodesic curvature kg = 0, reparametrized in terms of arc length, is ageodesic. It follows from the local existence theorem for ODE (Theorem 1.4.2) that given anyu0 ∈ U , and X ∈ Tu0f , there exists a unique geodesic c(t) = f(u(t)) on t ∈ (−ε,+ε) for someε > 0 with the initial condition c(0) = f(u0), c′(0) = X.

Proposition 5.4.12 (First variation of the arc length). Consider a family of curvesγ(s, t) = f(u1(s, t), u2(s, t)) where (s, t) ∈ [0, L] × (−δ, δ) on f : U → R3, assume both end-points γ(0, t) and γ(L, t) are fixed, and c(s)

.= γ(s, 0) is parametrized by its arc length.

dL(γ(s, t))

dt|t=0 = −

∫ L

0

∇c′(s)ds

· ∂γ(s, t)

∂t|t=0ds.(193)

Proposition 5.4.12 also holds true for a Riemannian manifold (M, g). As a corollary, any

smooth curve is a geodesic if for every variation field ∂γ(s,t)∂t |t=0 along γ(s, 0), dL(γ(s,t))

dt |t=0 = 0.In this sense, we say a geodesic is a critical point of the variation of the length functional.

Next we give examples of geodesics on embedded surfaces, in particular spheres and ellipsoids.A basic lemma is the following


Lemma 5.4.13 (Geodesics via isometry).(1) Let c(s) be a geodesic on an embedded surface M ⊂ R3, then for any isometry R of R3,

R(c(s)) is a geodesic on R(M) ⊂ R3.(2) Let φ be an isometry between (M, g) and (N, g), i.e. φ : M → N is a diffeomorphism and

gφ(p)(dφ(X), dφ(X)) = gp(X,Y ), p ∈M, X, Y ∈ TpM.(194)

Then φ sends any geodesic on M to a geodesic on N .

First we consider the unit sphere S2 ⊂ R3, by Lemma 5.4.13 we can show any great circle cutby a plane through the origin will be a geodesic. We may see it by first checking the equator isa geodesic and then apply a rotation in O(3) to get any great circle.

Next we consider an ellipsoid E ⊂ R3 defined by

E = x2

a2+y2

b2+z2

c2= 1. a > b > c > 0.(195)

and a local parametrization

f(u, v) = (a cosu cos v, b cosu sin v, c sinu),(196)

where U = (u, v) ∈ R2 | − π2 < u < π

2 ,−∞ < v < +∞. Note that f parametrizes all theellipsoid but the north and south poles N and S.

There are four umbilic points on the ellipsoid.

(a

√a2 − b2a2 − c2 , 0, c

√b2 − c2a2 − c2 ), (a

√a2 − b2a2 − c2 , 0,−c

√b2 − c2a2 − c2 ),

(−a√a2 − b2a2 − c2 , 0, c

√b2 − c2a2 − c2 ), (−a

√a2 − b2a2 − c2 , 0,−c

√b2 − c2a2 − c2 ).

Let us begin with a simple oberservation:

Proposition 5.4.14. Let c(s) = (x(s), y(s), z(s)) be a geodesic on E parametrized by its arclength, then there exists some function λ(s) such that

x′′(s) + λ(s)x(s)a2 = 0, y′′(s) + λ(s)y(s)

b2 = 0, z′′(s) + λ(s) z(s)c2 = 0,

(x′(s))2 + (y′(s))2 + (z′(s))2 = 1.(197)

In particular, the intersection of the plane P = αx+ βy + γz = 0 with E is a geodesic if andonly if P is one of the coordinate planes x = 0, y = 0, and z = 0.

Proof of Proposition 5.4.14. Note that any point (x, y, z) ∈ E has its normal parallel to ( xa2 ,yb2 ,

zc2 ),

The first formula in (197) is simply equivalent to

∇c′(s)ds

= (dc′(s)ds

)T = 0.(198)

Next we consider the curve c which is the intersection of z = 0 and E, it is direct to check (via196) that fu and fv are orthogonal along c, and fu is parallel to (0, 0, 1). It follows that c, as aplane curve in z = 0, must have c′′(s) orthogonal to both fu and fv, hence it is parallel to n.A similar argument shows the intersection of y = 0 with E, and that of z = 0 with E are bothgeodesics. These correspond to the cases when only one of (α, β, γ) is nonzero.

How about the general case? If a general intersection curve c(s) is a geodesic, we see from(198) that (α, β, γ), the normal vector of the plane, must be orthogonal to the normal of E.After plugging into (196), we have

αcosu cos v

a+ β

cosu sin v

a+ γ

sinu

c= 0,(199)

αa cosu cos v + β b cosu sin v + γ c sinu = 0.(200)

72 BO YANG

Some simple cancelation leads to

βa2 − b2

bcosu sin v + γ

a2 − c2c

sinu = 0,(201)

αc2 − a2

acosu cos v + β

c2 − b2b

cosu sin v = 0,(202)

αb2 − a2

acosu cos v + γ

b2 − c2c

sinu = 0.(203)

Assume α = 0, β 6= 0 and γ 6= 0, it follows from (202) and (203) that sinu = sin v = 0, henceit is isolated point, not counted as a geodesic. Similarly, we exclude all the cases when only oneof (α, β, γ) is zero.

Finally if none of (α, β, γ) is zero. from (202) we conclude α c2−a2a cos v + β c

2−b2b sin v = 0,

hence v is constant, however from (201) u is also constant, we arrive at isolated points again.

The geodesics obtained by Proposition 5.4.14 are three simple and closed ones. There arealso other closed geodesics (but much longer and complicated) on the ellipsoid. It is related tothe construction of second order confocal surfaces, we refer to Example 3.7.3 on p.55 in [9], p.38in [2], pp.264-266 in Arnold’s book.35, and in particular a comprehensive study in Sec 3.5 ofKlingenberg’s book.36 We just mention that Jacobi firstly studied geodesics on an ellipsoid, thistopics falls into the subject of integrable systems. Among others, an interesting phenomenon isthat every geodesic starting at one umbilic point on E again converges to the antipodal umbilicpoint. Moreover all have the same length between two umbilic points, with only the geodesicformed by the intersection of the plane y = 0 and E is closed.

Let us mention some important progress on closed geodesics on an ellipsoid.

Theorem 5.4.15 (Morse 1934, and Klingenberg?).(1) The ellipsoid E defined by (195) with a, b, c sufficiently close to each other admits exactly

only three closed geodesics which are simple (without self-intersection).(2) There are infinitely many distinct closed geodesics on the ellipsoid defined by (195). In

general they have self-intersections.

Since the ellipsoid is diffeomorphic to the unit sphere S2, the ellipsoid with the induced metricfrom R3 can be viewed as a Riemannian metric on S2 which is not round. Let us mention twoimportant results on closed geodesics on S2. The first one is due to Lusternik-Schnirelmann(1929), it states that any Riemannian metric on S2 admits three closed simple geodesics, notethat the ellipsoid in Theorem 5.4.15 shows this result is optimal. The second one is due to Franks(1992) and Bangert (1993), they proved that any Riemannian metric on S2 admits infinitelymany closed geodesics.

5.4.5. Complete Riemannian 2-manifolds. Any Riemannian manifold (M, g) is a metric space.

Definition 5.4.16 (Non-continuable). Let (M, g) be a Riemannian manifold, we say M iscontinuable if there exists an isometry f to another Riemannian manifold (M1, g1) where theimage F (M) is a proper subset of M1. Otherwise, we say (M, g) is non-continuable.

Theorem 5.4.17 (Hopf-Rinow (1931)). Let (M, g) be a Riemannian manifold, then M has anatural metric space structure induced by g, this metric topology is the same as the originaltoplogical structure on M . The following are equivalent:

(i) Every divergence path has an infinite length. By a divergence path, we mean a C1 mapc : [0,+∞)→M such that for every compact subset K ⊂M , there exists some t0 withthe property that c(t) 6∈ K for t > t0.

(ii) M is a complete metric space.i.e. every Cauchy sequence converges.

35Arnol’d, V. I. “Mathematical methods of classical mechanics.” Corrected reprint of the 2nd (1989) edition.Springer-Verlag.

36Klingenberg, W.. “Riemannian geometry.” 2nd edition. De Gruyter Studies in Mathematics, 1., 1995.


(iii) Every bounded closed set of M is compact.

A Riemannian manifold which satisfies any of the above property is called complete.

Closed manifolds are obviously non-continuable and complete. However, there are noncom-pact examples of Riemannian manifolds of dim 2 which are non-continuable but not complete.

5.4.6. Models of surfaces of with constant curvature. In this subsection we will define an ‘ab-stract’ surface of revolution and study its geodesics. In the end, we drive sine laws and cosinelaws in non-Euclidean geometry.37

Let us begin with a surface of revolution in R3

f(u, v) = (h(u) cos v, h(u) sin v, k(u)).

Such a surface is generated by revolving a regular curve (h(u)k(u)) in the plane y = 0 aboutz-axis. Now we further assume the generating curve passes through the origin and the resultingsurface is smooth at the origin, therefore, we conclude that the tangent of curve satisfies k′ = 0at the origin. Now if we apply the change of variables (r, θ) such that

dr =√

(k′(u))2 + (h′(u))2du, v = θ, ϕ(r) = h(u),

Then the first fundamental form of f becomes dr2 +ϕ2(r)dθ2. Note that we have ϕ(0) = 0 andϕ′(0) = 1. This motivates us to define

Definition 5.4.18 (An ‘abstract’ surface of revolution). Given any smooth function ϕsatisfying ϕ(0) = 0 and ϕ′(0) = 1, we call the 2-dim Riemannian manifold

(M(ϕ), ds2) = (r, θ) ∈ R2 | ds2 = dr2 + ϕ2(r)dθ2(204)

an ‘abstract’ surface of revolution once we specify the following rules for (r, θ)

(i) If ϕ(r) > 0 for any r > 0, then the domain of r is (0,∞),(ii) If b = infr0 > 0 | ϕ(r0) = 0 is finite, then the domain of r is [0, b].(iii) (r, θ) and (r, θ+2π) represents the same point on M(ϕ), and (0, θ) represents the same

point. If (ii) happens, then (b, θ) also represents the same point on M(ϕ).

On an abstract surface of surface of revolution, the Riemannian metric ds2 has the form thesame as in geodesic orthogonal coordinates. It follows from (135) and (136) that

K = −ϕrrϕ, kg =

dα

ds+ ϕ′(r)

dθ

ds.(205)

Here ∂∂r and 1

ϕ∂∂θ give an orthonormal frame, and α denotes the angle between a curve c′(s)

and ∂∂r . Therefore we have

dr

ds= cosα, ϕ(r)

dθ

ds= sinα.(206)

Example 5.4.19 (Models of constant curvature). We may solve K = −ϕrrϕ = k which is

constant and ϕ(0) = 0 and ϕ′(0) = 1, there are three cases, we call these model spaces M(k).(1) If k = 0, ϕ(r) = r on [0,∞). This is the Euclidean plane R2 with polar coordinates.

(2) If k > 0, ϕ(r) = 1√k

sin(√kr) on [0, π√

k]. This represents the round sphere S2( 1√

k) ⊂ R3.

(1) If k < 0, ϕ(r) = 1√−k sinh(

√−kr) on [0,∞). This is the hyperbolic space with curvature

k.

As an immediate consequence of (205) and (206), we have a characterization of geodesics onan abstract surface of revolution.

Proposition 5.4.20. Any geodesic c(s) = (r(s), θ(s)) on (M(ϕ), ds2) satisfies

ϕ(r(s)) sinα(s) = constant.(207)

37Most materials in this subsection is from Chapter 6 of the book by Wuyi Hsiang, Shenhuai Wang, andYanglian Pan. “Classical geometry”, Higher Education Press, 2014.

74 BO YANG

Consider the hyperbolic space with Gauss curvature −1, we have ds2 = dr2 + sinh2(r)dθ2.Introduce a change of coordinates

x = tanh(r

2) cos θ, y = tanh(

r

2) sin θ.(208)

We have

D2 = x2 + y2 < 1, ds2 =4

(1− x2 − y2)2(dx2 + dy2).(209)

(209) is called the Poincare disk model of the hyperbolic plane. A remarkable propertyis that the Mobius transformation ψ : (D2, ds2)→ (D2, ds2) acts as an (orientation-preserving)isometry. Here we set z = x+

√−1y, a ∈ D2, θ ∈ R fixed, then

ψ(z) = e√−1θ z − a

1− az .(210)

In general, we call a Riemannian manifold is homogeneous if given any two points P andQ there is an isometry which maps P to Q. From the previous discussion we know all modelsof constant curvature in Example 5.4.19 are homogeneous. Conversely as in Proposition ??, weknow that any homogeneous 2-dim Riemannian manifold must have constant Gauss curvature.Hence all homogeneous abstract surfaces of revolution are exactly model spaces determined byExample 5.4.19. Now we state the sine and cosine laws on these model spaces.

Theorem 5.4.21 (Bolyai’s absolute sine laws). Consider (M(k), ds2) an abstract surfaceof revolution with curvature k, then any geodesic triangle 4ABC on M(k) satisfies

sinA

ϕ(a)=

sinB

ϕ(b)=

sinC

ϕ(c)(211)

where ϕ(r) is of the form in Example 5.4.19.

Theorem 5.4.22 (Cosine laws). Consider (M(k), ds2) an abstract surface of revolution withcurvature k and 4ABC any geodesic triangle, then

c2 = a2 + b2 − 2ab cosC (k = 0)(212)

cos(c√k) = cos(a

√k) cos(b

√k) + sin(a

√k) sin(b

√k) cosC (k > 0)(213)

cosh(c√−k) = cosh(a

√−k) cosh(b

√−k)− sinh(a

√−k) sinh(b

√−k) cosC (k < 0).(214)

Making use of the fact that these model spaces are homogeneous, both Theorem 5.4.21 andTheorem 5.4.22 can be derived from Proposition 5.4.20.

It is worth noting that there is no notion of similar triangles in neither spherical nor hy-perbolic geometry. If two triangles have the same angles then they are congruent. In hyperboliccase, they only differ by a Mobius tranformation, possibly with a reflection as z → z. To see it,we may assume two triangles with equal interior angles and both have the origin as a vertex.Frin Theorem 5.4.21 and Theorem 5.4.22, all the ordered side lengths of a triangle with threeinterior angles given can be uniquely determined. Then we may apply the Schwartz lemma incomplex analysis to show that any such two triangles are related by rotation around the origin,possibly plus a reflection about the x-axis as z → z. We may make this angle-determine-sideprocess more explicit by stating

Theorem 5.4.23 (Additional cosine laws). Under the same assumption as in Theorem5.4.22, we have

cos(c√k) =

cosA cosB + cosC

sinA sinB(k > 0)(215)

cosh(c√−k) =

cosA cosB + cosC

sinA sinB(k < 0).(216)


As another application of Proposition 5.4.20, we study geodesics in the disk model of thehyperbolic plane (209). Consider a curve c(s) = (r(s), θ(s)), by the definition of α(s) and (208)we have

c′(s) = cosα∂

∂r+

sinα

sinh(r)

∂

∂θ

= cosαdt

dr

∂

∂t+

sinα

sinh(r)

∂

∂θ

= cosαdt

dr(cos θ

∂

∂x+ sin θ

∂

∂y) +

sinα

sinh(r)(−y ∂

∂x+ x

∂

∂y)

= x′(s)∂

∂x+ y′(s)

∂

∂y.

Therefore, we have

x′(s) = cosα1− t2

2tx− sinα

1− t22t

y

y′(s) = cosα1− t2

2ty + sinα

1− t22t

x.

From these two equation, we solve

(sinα)(x2 + y2)1− t2

2t= xy′(s)− yx′(s).

Now we apply Proposition 5.4.20 to conclude any geodesic on the Klein model must satisfy

xy′(s)− yx′(s)(1− x2 − y2)2

= constant,(217)

4[(x′(s))2 + (y′(s))2]

(1− x2 − y2)2= 1. since |c′(s)| = 1.(218)

As an exercise, let us verify that a piece of a circle which intersects x2 +y2 = 1 orthogonally isa geodesic. Any such circle could be determined by two parameters µ which is the angle formedthe segment connecting its center to the origin, and β which is 1

2 of the angle facing the arcbetween two intersection points. We may parametrize such a circle via an argument in planegeometry:

x(η) = 1cos β cosµ+ tanβ cos η,

y(η) = 1cos β sinµ+ tanβ sin η.

(219)

Note that in this case dsdη = 2 tan β

(1−x2−y2)2 , it is a straightforward calculation that (219) satisfies

(217) and (218), hence it is a geodesic of (209). It is much easier to see any segment of diameteris also a geodesic.

However, it is a mystery how one can solve (217) and (218) directly to conclude any geodesicof the Poincare disk model must be a piece of arc or a segment of diameter which intersectsx2 + y2 = 1 orthogonally. Indeed, from (217) it is direct to see any geodesic must intersect theunit circle orthogonally. However, to conclude it is a piece of a circle, one must show (x(s), y(s))has constant curvature as a plane curve.

x′(s)y′′(s)− y′(s)x′′(s)[(x′(s))2 + (y′(s))2]

32

= constant,(220)

It is unclear how to derive (220) from (217) and (218).A traditional way to bypass this difficulty is to use Lemma 5.4.13. First we conclude any

geodesic through the origin in (209) must be a straight segment forming the diameter. Thiscan be see from (217) and (218), and the uniqueness of geodesic at a given point with a givendirection. Next since (209) is a homogeneous Riemannian manifold, it follows from Lemma5.4.13 that any geodesic through a given point p ∈ D2 must the image of a straight segment

76 BO YANG

through the origin under a Mobius transformation in the form of (210). Such a image must bea piece of circular arc which intersects x2 + y2 = 1 orthogonally.

Alternatively, we may first determine all geodesics of the Poincare upper half planemodel of hyperbolic geometry (p.95 in [9]) and apply an isometry (a Mobius tranformation)between these two models.

H = (x, y) ∈ R2 | y > 0, ds2 =dx2 + dy2

y2,(221)

In the rest of this subsection we would like to give a quick introduction on 2-dimensionalhyperbolic space and its isometries. The 2-dimmensional hyperbolic space (H2, g) has threemodels:

(i) The (Poincare) unit disk model,(ii) The the upper half plane model,(iii) The hyperboloid model.

There is also a synthetic (axiomatic) definition of 2-dimensional hyperbolic space as how wedefine Euclidean geometry on R2.

5.4.7. Examples of spectral geometry. In this subsection we discuss spectral geometry on S2 andH2. Some useful references are: Taylor, Michael E. Partial differential equations II. Qualitativestudies of linear equations. 2nd edition. Springer, New York, 2011. and Shubin’s book. For H2,there is also Iwaniec’s Spectral Methods of Automorphic Forms.

There is a good example on explicit calculation in a paper by Bates-Gibson, “a geometrywhere everything is better than nice”.

5.4.8. Examples of (non-orientable) Riemannian 2-manifolds embedded in R3. First of all, allcompact (without boundary) smooth 2-manifolds (orientable and non-orientable) has been clas-sified in topology. The orientable ones include, sphere S2, the torus T 2 (a S2 attached with onehandle), the surface Σg obtained from the sphere with g handles (g ≥ 2). We have used thisfact in the proof of Gauss-Bonnet Theorem. The non-orientable ones include the real projectiveplane RP2 which is obtained from attaching a closed disk D2 along the boundary of a closedMobius strip, and the Klein bottle, and the surface obtained from attaching RP2 with g smallopen disks removed, with g Mobius strips (g > 2). As a special case of Whitney’s embeddingtheorem, any compact orientable surfaces admit smooth embeddings in R3, while those compactnon-orientable ones admit smooth embeddings in R4. Instead of explaining Whitney’s result,we content ourselves with the following result.

Proposition 5.4.24. 38 Any embedded surface which is a closed subset in R3 must be orientable.

If we consider complete Riemannian 2-manifolds with zero Gauss curvature, those are calledflat and are classified: those orientable ones include the plane, the cylinder, the 2-dimensionaltorus; those non-orientable ones are the complete unbounded Mobius strip39and the Klein bot-tle.40

38See Hatcher, Chapter 3, p.256 for a proof via Alexander duality. For an alternative proof on smoothmanifolds in Samelson, Hans, “Orientability of hypersurfaces in Rn”. Proc. Amer. Math. Soc. 22 (1969),

301-302.39A complete unbounded Mobius strip is the quotient of the infinite cylinder x2 + y2 = 1,−∞ ≤ z ≤ +∞

by the antipodal map (x, y, z) → (−x,−y,−z). If we consider the open cylinder x2 + y2 = 1,−1 ≤ z ≤ 1, the

quotient is an open Mobius strip. Since we only consider complete Riemannian manifolds here, we disregardopen Mobius strip.

40Two equivalent ways to describe a Klein bottle are useful here: (1) The quotient of R2 by the groupgenerated by (x, y) → (x + 1, y) and (x, y) → (1 − x, y + 1). (2) First we define a torus T 2 as the quotient ofR2 by the group generated by translations (x, y) → (x + 2, y) and (x, y) → (x, y + 2), then a Klein bottle is a

quotient of T 2 by the map (x, y) → (−x, y + 1). It is interesting to note that this quotient can also be viewedas the quotient by the antipodal map (x1, y1, z1) → (−x1,−y1,−z1) after we view the torus T 2 as a standardsurface of revolution in R3.


Definition 5.4.25 (Isometric embedding (immersion)). We say a Riemannian 2-manifold(M, g) admits an isometric embedding (immersion) in the Euclidean space (RN , gE) if thereexists an embedding (immersion) map f : M → RN as in Definition 5.1.4 such that

gE(df(X), df(Y )) = df(X) · df(Y ) = gp(X,Y ), ∀p ∈M, X, Y ∈ TpM.(222)

One of the central questions in Riemannian geometry is to find an isometric realizationof a given Riemannian manifold in an ambient Euclidean space. In general, one may considervarious settings of this type of questions by

(i) weakening the smoothness assumption of the map in Definition 5.4.25,(ii) looking for the lowest dimension N of the ambient Euclidean space,(iii) considering immersion instead of embedding,(iv) making more assumptions on the curvature of the given Riemannian manifold.

These questions of isometric embedding was first studied by John Nash, Nicolaas Kuiper, MikhailGromov, Matthias Gunther, and others. 41 Here our plan is moderate and only to give someremarks on some specific Riemannian 2-manifolds.

A flat 2-torus can be isometrically embedded in R3 by a C1 map (by the Nash-Kuiper embed-ding theorem) but not by a C2 map (see the proposition in below). While it is rather counter-intuitive to imagine a C1 embedding of a flat 2-torus in R3, there is some recent progress on amore explicit construction of such a map. 42

On the other hand, the Clifford torus provides a smooth isometric embedding of a flat torusin R4. The Clifford torus in defined in S3 ⊂ R4 as:

S1(1√2

)× S1(1√2

) = (x1, x2, x3, x4) ∈ R4 | x21 + x2

2 = x23 + x2

4 =1

2.

It is a simple exercise to check f(u, v) = ( 1√2

cosu, 1√2

sinu, 1√2

cos v, 1√2

sin v) where u, v ∈ [0, 2π]

is an isometric smooth (in fact real analytic) embedding in R4 in the sense of Definition 5.1.4,and the induced metric on the Clifford torus is flat (a constant multiple of Euclidean metric).

Proposition 5.4.26 (see Theorem 6.1.8 on p.129 of [9]). On a compact surface M embeddedin R3, there must be a point p with positive Gauss curvature.

Now we consider closed and open Mobius strips defined in Example 5.1.9, we have pointed outthat the parametrization there does not induce a flat Riemannian metric on the Mobius strip.However, a closed Mobius strip do admits a flat smooth embedding in R3, found by Sadowskyin 1930.43 For complete unbounded flat Mobius strip, there exists a smooth embedding into R4,we refer to p.39 of Han-Hong’s book listed in the footnote above and the paper by Sabitov.44

Next we turn to real projective plane. Since RP2 is non-orientable, by Proposition 5.4.24we can not expect an smooth embedded model in R3, instead there are interests in looking forimmersions (possibly not smooth, with singular points) in R3. For example, there are Romansurfaces and Boy surfaces as a model of RP2 in R3.45 If we raise the dimension of the ambientspace, then RP2 admits a real analytic isometric embedding in R5, this is the classical Veroneseembedding.

At last, we consider the Klein bottle, by the Nash-Kuiper theorem, there exists a C1 isometricembedding of the flat Klein bottle into R4. To the best of our knowledge, such an embedding

41We refer to the book of Qing Han and Jia-Xing Hong, “Isometric embedding of Riemannian manifolds in

Euclidean spaces”, American mathematical society, 2006.42Borrelli, V.; Jabrane, S.; Lazarus, F.; Thibert, B.. “Flat tori in three-dimensional space and convex

integration”. Proc. Natl. Acad. Sci. USA 109 (2012), no. 19, 7218-7223 and also a website by those authorshttp://hevea-project.fr/ENPageToreDossierDePresse.html.

43We refer to the interesting article by Gideon E. Schwarz, “The dark side of the Mobius strip” The American

Mathematical Monthly, Vol. 97, No. 10 (1990), pp. 890-897 for real analytic models, even algebraic models ofMobius strip in R3.

44Sabitov, I. Kh.; Isometric immersions and embeddings of a flat Mobius strip into Euclidean spaces. (Russian)

Izv. Ross. Akad. Nauk Ser. Mat. 71 (2007), no. 5, 197-224; translation in Izv. Math. 71 (2007), no. 5, 1049-1078.

45We recommend Urich Pinkall’s article “Models of the real projective plane” in [6].

http://hevea-project.fr/ENPageToreDossierDePresse.html

78 BO YANG

has not been constructed explicitly.46 Instead, if we only require isometric immersion, then asmooth isometric immersion of the flat Klein bottle into R4 was discovered by Tompkins (Bull.Amer. Math. Soc. 1941), it is defined by

f(u, v) = (cosu cos v, cosu sin v, 2 sinu cos(v

2), 2 sinu sin(

v

2)).

Since f(u + 2π, v) = f(u, v) and f(2π − u, v + 2π) = f(u, v), consider the restriction of f on[0, 2π]× [0, 2π], it is direct to see the first fundamental form is

(1 + 3 cos2 u)du2 + dv2.

Obviously it has zero Gauss curvature, hence f gives an isometric smooth immersion of flatKlein bottle into R4, however, it is not an embedding since f fails to be one-to-one, for examplef(0, 0) = f(π, π).

As an exercise, one may check if we modify

f(u, v) = ((3 + cosu) cos v, (3 + cosu) sin v, sinu cos(v

2), sinu sin(

v

2)).

where (u, v) ∈ [0, 2π]× [0, 2π] is a smooth isometric embedding of a topological Klein bottle inR4, however the induced metric is not flat.

Let us end this subsection by the following open question:

Question 5.4.27 (Yau47). Prove every smooth Riemannian 2-manifold can be (smoothly) iso-metrically embedded in R4.

46A smooth isomeric embedding of a flat Klein bottle in R4 was proved in a paper by Bushmelev, A. V.“Isometric embeddings of an infinite flat Mobius strip and a flat Klein bottle in R4. (Russian) Vestnik Moskov.Univ. Ser. I Mat. Mekh. 1988, no. 3, 38-41; translation in Moscow Univ. Math. Bull. 43 (1988), no. 3, 30-33.

Unfortunately, we do not have access to check the details.47Problem 22 in “Open problems in geometry”. Differential geometry: partial differential equations on

manifolds, 1-28, Proc. Sympos. Pure Math., 54, Part 1, Amer. Math. Soc., 1993.


6. Global theory of surfaces: selected topics

In this section we list some classical results on global geometry of surfaces, most of them arepicked from Nirenberg’s notes “Seminar on differential geometry in the large”.48 These topicswill be good for a second course on differential geometry on surfaces.

6.1. Ovaloids and their noncompact analogues.A closed embedded surfaces in R3 with K > 0 is called an ovaloid whose literally meaning

is egg-surfaces.

Theorem 6.1.1 (Hadamard (1987)). Let f : M → R3 be an ovaloid, then

(i) f(M) is orientable. Fix an orientation, the Guass map n : M → S2 is a diffeomor-phism.

(ii) M is strictly convex, i.e., for every f(p) ∈ f(M), M lies entirely on one side of thetangent plane at f(p), with the only intersection point being f(p).

(iii) f(M) is the boundary of a convex body set Ω ⊂ Rn. i.e. any segment between twopoints in Ω lies in Ω.

If M is a complete non-compact embedded surface R3, a similar result is due to Stoker (1936).

6.2. Existence of surfaces with prescribed curvature.We give a quick overview of Minkowski’s problem and Christoffel’s problem.Given any function ϕ : S2 → R, we would like to introduce the following condition:

∫S2

ϕ(x)xi dAS2 = 0, 1 ≤ i ≤ 3.(223)

Here x = (x1, x2, x3) ∈ S2 and dAS2 the standard surface measure on S2.Let us consider a closed strictly convex surface f : M → R3. It follows from Theorem

6.1.1 we know such a surface f(M) must enclose a convex region D in R3 and its Gauss mapis a diffeomorphism. Therefore, after composed with the inverse Gauss map, we may view1λ1· 1λ2

= 1K and 1

λ1+ 1

λ2= 2H

K as smooth functions defined on S2. Here λ1 and λ2 are two

principal curvatures of f at n−1(ξ) which is the preimage of ξ under the Gauss map. We willshow both of these two functions satisfy Condition (223).

The first one is straightforward, for example we take i = 1, recall a formula on change of thesurface area element

dAS2 = det(Jac(dn df−1))dAM = KdAM

Then by the divergence theorem.

∫S2

1

K(x)x1dAS2 =

∫f(M)

(1, 0, 0) · ndAf(M) =

∫D

div((1, 0, 0))dVD = 0.

48See the book: Nirenberg, L.. “Lectures on differential equations and differential geometry”, Classical Topicsin Mathematics 7, Higher Education Press, 2018.

80 BO YANG

To show the second one, we consider a local coordinate (u, v) ∈ M , recall the Weingarten

map −[nu, nv] = [fu, fv]

[a1

1 a21

a12 a2

2

]we have

(n× fu)v − (n× fv)u=nv × fu − nu × fv=− (a2

1fu + a22fv)× fu + (a1

1fu + a12fv)× fv

=(a11 + a2

2)fu × fv

=2H

KK√

det(I)n

=(1

λ1+

1

λ2) |nu × nv|n.

Let us divide M by disjoint open sets Ui so that each of them is sufficiently small and containedin a single local chart, it follows from the divergence theorem that∫

S2

(1

λ1+

1

λ2) (ξ1, ξ2, ξ3) dAS2

=∑i

∫Ui

(1

λ1+

1

λ2)n |nu × nv|dudv,

=∑i

∫Ui

((n× fu)v − (n× fv)u

)dudv,

=∑i

∫∂Ui

(−n× fv, n× fu) · ν ds.

=0

The last equality holds since the outer normal ν takes the opposite sign along the shared bound-ary of two adjacent Ui’s and all terms cancel. Therefore (223) is true.

Given a smooth positive function ϕ : S2 → R satisfying Condition (223), Minkowski’sproblem asks whether one can construct a closed embedded strictly convex surface f : M → R3

such that its Gauss cuvature satisfies 1K = ϕ n, here n : M → S2 ⊂ R3 is the Gauss map of f .

H.Lewy (1938) proved that existence and uniqueness of such a surface after assuming ϕ satisfies(223) and is an real analytic function on S2. Minkowski’s problem was completely solved byNirenberg (1953) and Pogorelov (1952).

Remark 6.2.1. As expected, the curve analogue of Minkowski’s problem is much easier. Let λbe a 2π-periodic positive function, then λ is the curvature of a smooth simple curve if and onlyif the following is satisfied ∫ 2π

0

sin θ

λ(θ)dθ =

∫ 2π

0

sin θ

λ(θ)dθ = 0.(224)

We may think Condition (224) is a curve analogue of Condition (223).

Given any smooth positive function ϕ : S2 → R satisfying (223), Christoffel’s problemasks if we find a strictly convex surfaces f : M → R3 whose 1

λ1+ 1

λ2= 2H

K (n−1(x)) = ϕ(x) for

any x ∈ S2. The uniqueness of solution to Christoffel’s problem is known. Christoffel (1910)and Hurwitz (1902) proved if two closed strictly convex surfaces fi : Mi → R3 with i = 1, 2share the same 1

λ1+ 1

λ2, then f1 and f2 differ by a translation in R3. For the existence part of

Christoffel’s problem, Alexandrov (1937) pointed out that (223) was not sufficient to guaranteethe existence, and Firey (1967) found the necessary and sufficient conditions on ϕ to solveChristoffel’s problem.

Solutions to both Minkowski’s problem and Christoffel’s problem rely on the solvability ofcertain PDEs, in below we will explain how to set up the corresponding PDE problem.


Assume a closed strictly convex surfaces f : M → R3, by Theorem 6.1.1 implies that itsGauss map n : M → S2 is a diffeomorphism and f(M) encloses a convex body Ω, We choose anorientation on f(M) so that n is the outer normal of f(M), and both λ1 and λ2 are positive.Now we introduce the Minkowski support function u : S2 → R of a convex body Ω:

u(x) = maxy∈Ω

y · x = f n−1(x) · x, x ∈ S2.(225)

Let us point out a duality relation between x and f n−1(x) = (y1(x), y2(x), y3(x)) via thesupport function u.

If we may extend u as a homogeneous function of degree one in R3 by.

u(x1, x2, x3) = |x|∇u(x

|x| ) = f n−1(x

|x| ) · x = y(x

|x| ) · x.

Then at any point x ∈ S2 and take i = 1 as example, we have

∂

∂x1u(x)| = ∂

∂x1(f n−1(

x

|x| ) · x),

=∂f n−1( x

|x| )

∂x1· x+ y1(x)

= y1(x) +(

limt→0

y( x+tε1|x+tε1| )− y( x

|x| )

t

)· x, where ε1 = (1, 0, 0)

= y1(x)

In the last two steps we explained that∂fn−1( x

|x| )

∂xiis along the tangent direction at f n−1(x).

The following result shows a ‘magic’ relation between the support function on S2 and principalcurvatures of f .

Proposition 6.2.2. Let e1, e2 be two orthonormal frame on S2, consider a matrix-valuedfunction W on S2:

Wij.= uij + u δij , 1 ≤ i, j ≤ 2.

Here uij indicates the second order covariant derivatives of u with respect to e1, e2. Then theeigenvalues of W are 1

λ1and 1

λ2.

Proof of Proposition 6.2.2. It suffices to prove it for any fixed point on S2, first we will chooselocal coordinates which allows us to calculate W more directly.

Consider any point x ∈ S2, since the tangent space of S2 at x can be identified as the tangentspace of f(M) at f n−1(x). Assume (u1, u2) are local coordinates around n−1(x) ∈ M , wemay assume ni

.= dn( ∂ui ) = ei at n−1(x) ∈ M , moreover, fi

.= df( ∂ui ) are principical directions

at f n−1(x). Therefore, we have the expression of Weingarten map L = dn df−1

−ni = L(fi) = λifi.

82 BO YANG

Note that e1, e2 are orthonormal, which implies dn(ei) · dn(ej) = δij . Therefore, we concludefi × fj = 1

λiδij .

ei(u) = ei(f n−1(x) · x) = f · ni;ej ei(u) = (f · ni)j

= fj · ni + f · ∂2

∂uj∂uin;

∇ejei(u) = (∇ ∂∂uj

∂

∂ui)(f · n)

= (∇ ∂∂uj

∂

∂ui)f · n+ f · (∇ ∂

∂uj

∂

∂ui)n;

uij.= ej ei(u)−∇ejei(u)

= fj · ni + f · ∂2

∂uj∂uin− f · (∇ ∂

∂uj

∂

∂ui)n

= fj · ni + II(∂

∂ui,∂

∂uj)f · n(n), here II is the second fundamental form of S2,

= fj · ni − δiju

=1

λiδij − δiju.

Therefore, we have proved that uij + uδij has the eigenvalue 1λi

at x ∈ S2.

Therefore to solve Minkowski’s problem it is crucial to solve a smooth function u : S2 → Rsuch that

det(uij + uδij) = 1ϕ(x)

(uij + uδij) is positive definite(226)

for any given positive smooth function ϕ : S2 → R.In the case of Christoffel’s problem one need to solve a smooth function u : S2 → R such that

∆S2u+ 2u = u11 + u22 + 2u = ϕ(x), ∆S2 is the spherical Laplace operator on S2

(uij + uδij) is positive definite

(227)

for any given positive smooth function ϕ : S2 → R.Once we solve the support function u defined on S2, we may extend it as a homogeneous

function of degree one in R3, still called u. Then we may set

f n−1(x) = ∇u(x), where ∇u(x) means the gradient in R3

for any (x1, x2, x3) ∈ S2. By this we may recover the surface F (M), uij + uδij being positivedefinite allows us to conclude it is strictly convex.

The solvability of these equations (227) and (226) and their higher-dimensional generalizationsis a central problem in the study of geometric PDEs.49

6.3. Isometric embeddings into R3 and the sign of Gauss curvature.

(i) Hilbert (1901) proved there are complete embedded surfaces with Gauss curvature −1in R3, and another proof is given by Holmgren (1902). A much more general non-existence result was proved by Efimov (1966).

(ii) Schilt (1937) studied continuous isometric deformation of surfaces with K ≤ 0, andconstructed examples of surface patches which are isometric, but can not be deformedinto each other.

49We refer to Pengfei Guan’s lecture notes on http://www.math.mcgill.ca/guan/notes.html

http://www.math.mcgill.ca/guan/notes.html


(iii) Blaschke (1912, 1921) and Weyl (1917), any closed embedded surface in R3 with K > 0is infinitesimally rigid, i.e. if there is a family of isometric embeddings ft : (M, g)→ R3

where t is a parameter, such that f , Dft, and D2ft are all continuous with respect tot, then the result of varying t is varying isometries in R3.

(iv) Liebmann (1899) and Hilbert (1909) proved that the only closed embedded surfacewhose K is a constant c must be isometric to a round sphere. Chern (1935) generalizedthis result and showed that an embedded Weingarten surface whose principal curvaturesare both positive is also isometric to sphere.

(v) Cohn-Vossen (1927), and a modification due to Zhitomirsky (1938), any closed embed-ded surface in R3 with K > 0 is rigid, i.e. any isometric embedding into R3 differsby an isometry in R3. Cohn-Vossen in fact proved a stronger result which holds forK ≥ 0 with an additional assumption on the zero locus of K. This rigidty result showsthe uniqueness part of the so-called Weyl’s problem. Weyl’s problem asks if thereis an isometrical embedding of a compact two-dimensional Riemannian manifold withK > 0 into R3. It was solved by Lewy (1938) after assuming the Riemnanian metric isreal analytic, and in full generality by Nirenberg (1953) and Pogorelov (1952).

6.4. Prescribing Gauss curvature. Kazdan-Warner identity, Nirenberg’s problem.

6.5. Conformal deformation to constant curvature. Negative case, try method of sup andsub-solutions.

PDE approach

6.6. More on minimal surfaces and CMC surfaces.To be done:positive Gauss cuvature on surfaces to two closed geodesic intersects.generalized toFrankel property positive Ricci curvature.stable sphere theorem minimal surfaces do CarmoThere are many important recent works on minimal surfaces and CMC surfaces in R3, let us

only mention a few.

Theorem 6.6.1 (Some results on existence and uniqueness of CMC surfaces in R3).

(i) (Hopf (1951)). A closed immersed CMC surface of genus 0 in R3 must be isometric toa round sphere.

(ii) (Alexandrov (1958)). A closed embedded CMC (in general, Weingarten) surface in R3

must be isometric to a round sphere.(iii) (Wente (1984)). There exist countably many immersed CMC surfaces of genus 1 in

R3.(iv) (Kapouleas (1990) and others, more recent works). There exist more examples of com-

pact or complete noncompact immersed CMC surfaces in R3.

The set of points with negative curvature on a CMC surface form an extremal domain (wherethe second variation has λ1 = 0.). Describe this set for Delaunay surfaces (nodary) and Wentetori.

Boundary value problems on CMC graphs, capillary surfaces. Given Ω ⊂ R2 a bounded openset, and ν the outer unit normal along ∂Ω.

div(∇ϕ√

1 + |∇ϕ|2) = ku, Ω

∇ϕ√1 + |∇ϕ|2

· ν = cos γ, ∂Ω.(228)

There are many theories proposed on how water moves upwards through the roots to thetop of a plant, among of which are capillary action and transpirational pull. Capillary actionis the attraction between water molecules and the walls of the xylem vessels of the plant. TheConcus-Finn theorem explains this phenomenon in a mathematical way.

84 BO YANG

Definition 6.6.2 (Willmore energy). Let f : Σ → R3 be any (immersed or embedded) closedsurface and H its mean curvature. the Willmore energy is defined as

W (Σ) =

∫Σ

H2dAΣ,(229)

where dAΣ is the area element induced by the first fundamental form of f .

Proposition 6.6.3. W (Σ) is invariant under a conformal transformation of R3.

Remark 6.6.4. According to Liouville’s theorem we only need to consider Mobius transforma-tions on R3, and the main point is to check W (Σ) is invariant under inversions, here we needto assume the center of inversion is away from f(Σ). The proof in below is independent ofLiouville’s theorem.

Proof of Proposition 6.6.3. Given an immersion f : Σ → R3 and a conformal equivalence φ :

R3 → R3, now there are two immersions we need to compare: f and f = φ f : Σ→ R3 → R3.Let g = df · df and h = −dn · df be the first and second fundamental forms of f . and g and g

be those of f .

The crucial observation is that f can be viewed as the isometric immersion

f : Σ→ (R3, φ∗(gE)ij = ρ2δij).

Here ∇ is the Levi-Civita connection for (R3, φ∗(gE)) and Γkij its Christoffel symbols.

Γrpq =1

2ρ2δrs[∂(ρ2δps)

∂xq+∂(ρ2δqs)

∂xp− ∂(ρ2δpq)

∂xs

]=

1

ρ(ρqδpr + ρqδpr − ρrδpq).

Then the second fundamental form of f is

hij = (∇df( ∂∂xj

)df(∂

∂xi)) · 1

ρn

= (fij +∂fp

∂ui

∂fq

∂ujΓkpq

∂

∂xk) · 1

ρn

=1

ρhij +

1

ρ2

(∇ρ · ∂f

∂ui

∂f

∂uj· n+∇ρ · ∂f

∂uj

∂f

∂ui· n− (∇ρ · n)

∂f

∂ui· ∂f∂uj

)=

1

ρhij −

1

ρ2(∇ρ · n)gij . where ∇ρ is the Euclidean gradient.

Therefore the Weingarten matrix L of f is

Lki = hij gkj = hij

1

ρ2gkj

=1

ρhijg

kj − 1

ρ2δki (∇ρ · n).

Therefore the mean curvature and Gauss curvature of f and f satisfy

H2 − K =(λ1 − λ2)2

4=

1

ρ2(H2 −K).

It follows that ∫f(Σ)

(H2 − K)dAf(Σ) =

∫f(Σ)

(H2 −K)dAf(Σ).

The proof is done once we invoke the Gauss-Bonnet theorem.

Willmore proved that W (Σ) ≥ 4π, with equality if and only if Σ is an embedded roundsphere, then he continued the study of his energy and tried to find the minimizing shape amongthe class of tori (the genus-1 case).


Theorem 6.6.5 (Conjectured by Willmore (1965), proved by Marques-Neves (2014)).

(i) Let Σ be any (immersed or embedded) closed surface of genus 1 in R3, then

W (Σ) ≥ 2π2.(230)

(ii) If the equality (230) holds, then up to an isometry in R3, Σ must be the torus ofrevolution:

Σ√2,1 = ((√

2 + cosu) cos v, (√

2 + cosu) sin v, sinu) ∈ R3 | −∞ < u, v < +∞.The connection between minimal surfaces and Theorem 6.6.5 is manifested by the fact that

Σ√2,1 is conformally equivalent to the Clifford torus in S3:

S1(1√2

)× S1(1√2

) = (x1, x2, x3, x4) ∈ R4 | x21 + x2

2 = x23 + x2

4 =1

2.

Indeed, recall the the stereographic map φ : R3 → S3 \ N has the formula:

φ(u1, u2, u3) = (2u1

u21 + u2

2 + u23 + 1

,2u2

u21 + u2

2 + u23 + 1

,2u3

u21 + u2

2 + u23 + 1

,u2

1 + u22 + u2

3 − 1

u21 + u2

2 + u23 + 1

).

It is direct to check

φ(Σ√2,1) = (cos v√

2,

sin v√2,

sinu

2 +√

2 cosu,

1 +√

2 cosu

2 +√

2 cosu).

Note that the domain of sinu√2+cosu

is the whole [−1, 1], if we reparametrize by sinu2+√

2 cosu= cosw√

2,

then

φ(Σ√2,1) = (cos v√

2,

sin v√2,

cosw√2,

sinw√2

) = S1(1√2

)× S1(1√2

).

Since in this course we are only interested in minimal surfaces in R3, we just remark thatthe Cliford torus is a minimal surface in S3 and it plays an important role in various works onminimal surfaces in S3.

86 BO YANG

7. Global theory of curves continued

7.1. The curve shortening flow: a short introduction.Consider the heat equation on the unit circle, it is equivalent to solve the following periodic

boundary value problem on [0, 2π]× [0,∞):∂u∂t (x, t) = ∂2u

∂x2 (x, t),

u(x, 0) = f(x),∂(k)u∂x(k) (0, t) = ∂(k)u

∂x(k) (2π, t) for any k ∈ N.(231)

It is a standard exercise on separation of variables that the formal series solution of (231)

u(x, t) =

∞∑i=1

[(An cos(nx) +Bn sin(nx))e−n

2t]

+A0,(232)

A0 =1

2π

∫ 2π

0

f(x)dx,An =1

π

∫ 2π

0

f(x) cos(nx)dx,Bn =1

π

∫ 2π

0

f(x) sin(nx)dx.(233)

Once we have more estimates on the growth of An and Bn in (233), it is reasonable to expectthe infinite series in (232) converges and it solves (231). Immediately we will find

limt→∞

u(x, t) =1

2π

∫ 2π

0

f(x)dx.

Therefore the heat equation has the homogenization effect: the initial value at t = 0 convergesits average as time tends to infinity. This property motivates many studies on heat equationsin various geometric problems.

In this subsection, we give a short introduction to a nonlinear heat equation in the study ofglobal curve theory. We say X(u, t) ((u, t) ∈ [a, b]×[0, T )), a one-parameter family of immersionsof a smooth curve into R2, solves the curve shortening flow (CSF) if

∂X

∂t= kN(234)

where T and N are unit tangent and normal direction of X(u, t) for any fixed t. In our notationN refers to the inner normal if X(∗, t) parametrizes a closed curve (9). Let ds = |Xu|du isthe arc length of X(u, t), we may rewrite (234) as

Xt = Xss.(235)

However since s depends on t, (235) is not the linear heat equation in (231).

Lemma 7.1.1 (CSF for graphs).If a family of graphs of y = f(x, t) defined by X(x, t) = (x, f(x, t)) satisfies

(∂X

∂t) ·N = k, i.e. ft =

fxx1 + (fx)2

(236)

then after a suitable reparametrization,

X(u, t) = (x(u, t), y(u, t)) with x(u, t) = x, y(u, t) = f(x, t).(237)

X(u, t) solves the CSF equation (234).

Proof of Lemma 7.1.1. We will determine (237) in the course of proof. Note that the term kNis independent of parametrization, and

Xt = (xt, yt) = (xt, fxxt + ft) = xt(1, fx) + (0, ft) = xt(1, fx) +Xt.

It suffices to make sure Xt = kN . Since (1, fx) goes along the tangential direction of Xt, we get

(0, ft)T + xt(1, fx) = 0,(238)

(0, ft)N = kN.(239)


Here (0, ft)T and (0, ft)

N are tangential and normal components of Xt respectively. Note that(239) is exactly (236), and (238) leads to

xt = − ftfx1 + (fx)2

.(240)

In conclusion, once f(t, x) solves (236), we can always find x = x(u, t) satisfying (240), hence

X(u, t) in (237) solves (234). As an example we may pick x = x(u, t) as the solution to thefollowing IVP for ODE.

dx

dt= − ftfx

1 + (fx)2, x(0) = u.(241)

Example 7.1.2 (Grim Reaper).Let f : (−π2 , π2 ) × (−∞,+∞) → R1 defined by f(x, t) = t + ln secx. Then it solves the

graphical CSF equation (236). This is called the Grim Reaper solution to the CSF. From (241),if we set

tanx = e−t tanu, y = t+ ln secx,

then (x(u, t), y(u, t)) where u ∈ (−π2 , π2 ) solves the original CSF (234).

Next we list some formulas on evolution of geometric quantities associated to CSF. Let L bethe arc length of the curve X(∗, t), recall if we set θ such that T = (cos θ, sin θ), then θs = k.

∂(ds)

∂t=∂(|Xu|du)

∂t= −k2 |Xu|du = −k2 ds, hence

dL

dt= −

∫k2ds,(242)

fst = fts + k2fs, where f is any C2 function,(243)

Formula (243) is fundamental in all the calculations in below:

Nt = −ksT, θt = ks by (243),(244)

kt = kss + k3 use θs = k and (243).(245)

When X(∗, t) is simple closed, let Ω the region enclosed by X(∗, t) and A(Ω) its area.

A(Ω) =1

2

∫X · (−N)ds by Green’s formula,(246)

dA(Ω)

dt= −

∫kds = −2π, by (244) and Theorem 3.2.4 in the second identity.(247)

Formulas from (242) to (247) play an important role in understanding behaviors of CSF. Oneof the most important results on CSF is the following convergence result.

Theorem 7.1.3 (Gage-Hamilton, Grayson (1987)).Let X0 : [a, b] → R2 be a closed simple smooth curve (compact embedded curve), and let Ω0

the enclosed region by X0.Then the solution to CSF (234) with initial data X(∗, 0) = X0 exists on [0, T ) for

T =A(Ω0)

2π.

For t sufficiently close to T , X(∗, t) gives an embedding of a strictly convex (k > 0) curve intoR2, we may pick pt as the center of the circle of the largest radius which is enclosed by X(∗, t).Now we define the rescaled embeddings

X(∗, t) =X(∗, t)− pt√

2(T − t).(248)

Then X(∗, t) converges smoothly to the unit circle centered at the origin.50

50See https://www.youtube.com/channel/UCuKrtmZ6GtPKIIwTox9O-KQ for some video demonstrations of CSF.

https://www.youtube.com/channel/UCuKrtmZ6GtPKIIwTox9O-KQ

88 BO YANG

For closed simple curves, Grayson also proposed to study a variant of CSF (234), the so-calledarea-preserving CSF. This flow has the advantage of keeping the enclosed area fixed whiledecreasing the length.

∂X

∂t= (k − 2π

L)N(249)

where L its arc length of X(∗, t) in R2.To see why (249) preserves the enclosed area, we need to work out the formulas from (242)

to (247) in a general setting. Assume the velocity of the evolution is Xt = φN , then a similarcalculation as from (242) to (247) leads to

∂(ds)

∂t= −φkds, dL

dt= −

∫Ct

φkds(250)

fst = fts + φkfs, f is any C2 function,(251)

Nt = −φsT(252)

dA

dt= −

∫φds.(253)

In the case of area-preserving CSF, φ = k− 2πL , we have dA

dt = 0 again by Theorem 3.2.4. In the

meantime, since (∫k2ds)L ≥ (2π)2 by Holder’s inequality, we know

dL

dt= −

∫k2ds+

(2π)2

L≤ 0,

hence the length is non-increasing.It seems harder to study the convergence of area-preserving CSF, there is some progress when

the initial curve is convex, i.e. of nonnegative curvature.

Theorem 7.1.4 (Grayson (1986)). Let X0 be a closed simple convex curve (k ≥ 0), then thearea-preserving CSF (249) with initial data X(∗, 0) = X0 has a smooth solution on [0,∞),Moreover, as t→ +∞, X(∗, t) converges smoothly to a round circle in R2.

Besides Grayson’s convergence Theorem 7.1.3 for smooth simple closed curves, there hasbeen other progress on the study of CSF in the past years. For example, the study of CSF withambient space being R3 or a general Riemannian manifold, CSF with rough initial data, andCSF for noncompact curves. We refer to a forthcoming book by Andrews, Chow, and others51

for more introduction on this topic.

7.2. Further reading. The four vertex theorem,The converse of the four vertex theorem, Gluck.Total curvature of a space curveEstimates the total curve of a closed curve in R3. Results of Fenchel, Fary, and Milnor.

51Ben Andrews, Bennett Chow, Christine Guenther, Mat Langford; “Extrinsic Geometric Flows” GraduateStudies in Mathematics, vol.206, AMS, 2020


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http://www.msri.org/publications/sgp/jim/geom/index.html

http://virtualmathmuseum.org/

Documents

Contentsmath.xmu.edu.cn/group/ga/dg_files/surface_notes.pdf · example, Chapter 1 to 3 in [2] discuss the classical theory of curves and surfaces in the context of Riemannian geometry