Elementary Di erential Geometry: Lecture Notes · [2] L. P. Eisenhart An introduction to di...

Elementary Differential Geometry:

Lecture Notes

Gilbert Weinstein

Contents

Preface 5

Chapter 1. Curves 71. Preliminaries 72. Local Theory for Curves in R3 83. Plane Curves 104. Fenchel’s Theorem 15Exercises 16

Chapter 2. Local Surface Theory 191. Surfaces 192. The First Fundamental Form 213. The Second Fundamental Form 234. Examples 255. Lines of Curvature 286. More Examples 307. Surface Area 338. Bernstein’s Theorem 379. Theorema Egregium 39Exercises 41

Chapter 3. Local Intrinsic Geometry of Surfaces 451. Riemannian Surfaces 452. Lie Derivative 473. Covariant Differentiation 484. Geodesics 505. The Riemann Curvature Tensor 536. The Second Variation of Arclength 56Exercises 59

Chapter 4. Global Theory of Surfaces 611. Differentiable Manifolds 612. Some Multi-linear Algebra 643. Integration 65Exercises 65

Index 67

3

Preface

These notes are for a beginning graduate level course in differential geometry.It is assumed that this is the students’ first course in the subject. Thus the choiceof subjects and presentation has been made to facilitate as much as possible aconcrete picture. For those interested in a deeper study, a second course would takea more abstract point of view, and in particular, could go further into Riemanniangeometry.

Much of the material is borrowed from the following sources, but has beenadapted according to my own taste:

[1] M. P. Do Carmo, Differential geometry of curves and surfaces, Prentice-Hall.

[2] L. P. Eisenhart An introduction to differential geometry with use of the ten-sor calculus, Princeton University Press.

[3] W. Klingenberg, A course in differential geometry, Springer-Verlag.

[4] B. O’Neill Elementary differential geometry, Academic Press.

[5] M. Spivak, A comprehensive introduction to Differential Geometry, Publishor Perish.

[6] J. J. Stoker, Differential Geometry, Wiley & Sons.

The prerequisites for this course are: linear algebra, preferably with some ex-posure to multilinear algebra; calculus up to and including the inverse and implicitfunction theorem; the fundamental theorem of ordinary differential equations con-cerning existence of solutions, uniqueness, and continuous dependence on parame-ters, and some knowledge of linear systems of ordinary differential equations; linearfirst order partial differential equations; complex analysis including Liouville’s the-orem; and some elementary topology.

It is highly recommended for the students to complete all the exercises includedin these notes.

Gilbert WeinsteinBirmingham, Alabama

April 2000

5

CHAPTER 1

Curves

1. Preliminaries

Definition 1.1. A parametrized curve is a smooth (C∞) function γ : I → Rn.A curve is regular if γ′ 6= 0.

When the interval I is closed, we say that γ is C∞ on I if there is an intervalJ and a C∞ function β on J which agrees with γ on I.

Definition 1.2. Let γ : I → Rn be a parametrized curve, and let β : J → Rnbe another parametrized curve. We say that β is a reparametrization (orientation-preserving reparametrization) of γ if there is a smooth map τ : J → I with τ ′ > 0such that β = γ τ .

Note that the relation β is a reparametrization of γ is an equivalence relation.A curve is an equivalence class of parametrized curves. Furthermore, if γ is regularthen every reparametrization of γ is also regular, so we may speak of regular curves.

Definition 1.3. Let γ : I → Rn be a regular curve. For any compact interval[a, b] ⊂ I, the arclength of γ over [a, b] is given by:

Lγ([a, b]) =∫ b

a

|γ′| dt.

Note that if β is a reparametrization of γ then γ and β have the same length.More specifically, if β = γ τ , then

Lγ([τ(c), τ(d)

])= Lβ([c, d]).

Definition 1.4. Let γ be a regular curve. We say that γ is parametrized byarclength if |γ′| = 1

Note that this is equivalent to the condition that for all t ∈ I = [a, b] we have:

Lγ([a, t]) = t− a.

Furthermore, any regular curve can be parametrized by arclength. Indeed, if γ is aregular curve, then the function

s(t) =∫ t

a

|γ′| ,

is strictly monotone increasing. Thus, s(t) has an inverse function τ(s) function,satisfying:

dτ

ds=

1|γ′|

.

It is now straightforward to check that β = γ τ is parametrized by arclength.

7

8 1. CURVES

2. Local Theory for Curves in R3

We will assume throughout this section that γ : I → R3 is a regular curve inR3 parametrized by arclength and that γ′′ 6= 0. Note that γ′ · γ′′ = 0.

Definition 1.5. Let γ : I → R3 be a curve in R3. The unit vector T = γ′

is called the unit tangent of γ. The curvature κ is the scalar κ = |γ”|. The unitvector N = κ−1T ′ is called the principal normal . The binormal is the unit vectorB = T × N . The positively oriented orthonormal frame (T,N,B) is called theFrenet frame of γ.

It is not difficult to see that N ′ + κT is perpendicular to both T and N , hencewe can define the torsion τ of γ by: N ′ + κT = τB. Note that the torsion, unlikethe curvature, is signed. Finally, it is easy to check that B′ = −τN . Let X denotethe 3×3 matrix whose columns are (T,N,B). We will call X also the Frenet frameof γ. Define the rotation matrix of γ:

(1.1) ω :=

0 κ 0−κ 0 τ0 −τ 0

Proposition 1.1 (Frenet frame equations). The Frenet frame X = (T,N,B)

of a curve in R3 satisfies:

(1.2) X ′ = Xω.

The Frenet frame equations, Equation (1.2), form a system of nine linear ordi-nary differential equations.

Definition 1.6. A rigid motion of R3 is a function of the form R(x) = x0 +Qxwhere Q is orthonormal with detQ = 1.

Note that if X is the Frenet frame of γ and R(x) = x0 +Qx is a rigid motionof R3, then QX is the Frenet frame of R γ. This follows easily from the fact thatQ is preserves the inner product and orientation of R3.

Theorem 1.2 (Fundamental Theorem). Let κ > 0 and τ be smooth scalarfunctions on the interval [0, L]. Then there is a regular curve γ parametrized byarclength, unique up to a rigid motion of R3, whose curvature is κ and torsion isτ .

Proof. Let ω be given by (1.1). The initial value problem

X ′ = Xω,

X(0) = I

can be solved uniquely on [0, L]. The solution X is an orthogonal matrix withdetX = 1 on [0, L]. Indeed, since ω is anti-symmetric, the matrix A = XXt isconstant. Indeed,

A′ = XωXt +XωtXt = X(ω + ωt)Xt = 0,

and since A(0) = I, we conclude that A ≡ I, and X is orthogonal. Furthermore,detX is continuous, and detX(0) = 1, so detX = 1 on [0, L]. Let (T,N,B) bethe columns of X, and let γ =

∫T , then (T,N,B) is orthonormal and positively

oriented on [0, L]. Thus, γ is parametrized by arclength, γ′ = T , and N = κ−1T ′ is

2. LOCAL THEORY FOR CURVES IN R3 9

the principal normal of γ. Similarly B is the binormal, and consequently, κ is thecurvature of γ and τ its torsion.

Now suppose that γ is another curve with curvature κ and torsion τ , and let Xbe its Frenet frame. Then there is a rigid motion R(x) = Qx+ x0 of R3 such thatRγ(0) = γ(0), and QX(0) = X(0). By the remark preceding the theorem, QX isthe Frenet frame of the curve R γ, and thus both QX and X satisfy the initialvalue problem:

Y ′ = Y ω,

Y (0) = QX(0).

By the uniqueness of solutions of the initial value problem, it follows that QX = X.In particular, (R γ)′ = γ′, and since R γ(0) = γ(0) we conclude R γ ≡ γ.

Assuming γ(0) = 0, the Taylor expansion of γ of order 3 at s = 0 is:

γ(s) = γ′(0)s+12γ′′(0)s2 +

16γ′′′(0)s3 +O(s4).

Denote T0 = T (0), N0 = N(0), B0 = B(0), κ0 = κ(0), and τ0 = τ(0). We haveγ′(0) = T0, γ′′(0) = κ0N0, and γ′′′(0) = κ′(0)N0 + κ0(−κ0T0 + τ0B0). Substitutingthese into the equation above, decomposing into T , N , and B components, andretaining only the leading order terms, we get:

γ(s) =(s+O(s3)

)T +

(κ2s2 +O(s3)

)N +

(τ6s3 +O(s4)

)B

The planes spanned by pairs of vectors in the Frenet frame are given specialnames:

(1) T and N — the osculating plane;(2) N and B — the normal plane;(3) T and B — the rectifying plane.

We see that to second order the curve stays within its osculating plane, where ittraces a parabola y = (κ/2) s2. The projection onto the normal plane is a cuspto third order: x =

((3τ/2) y

)2/3. The projection onto the rectifying plane is tosecond order a line, whence its name.

Here are a few simple applications of the Frenet frame.

Theorem 1.3. Let γ be a regular curve with κ ≡ 0. Then γ is a straight line.

Proof. Since |T ′| = κ = 0, it follows that T is constant and γ is linear.

Theorem 1.4. Let γ be a regular curve with κ > 0, and τ = 0. Then γ isplanar.

Proof. Since B′ = 0, B is constant. Thus the function ξ = (γ − γ(0)) · Bvanishes identically:

ξ(0) = 0, ξ′ = T ·B = 0.

It follows that γ remains in the plane through γ(0) perpendicular to B.

Theorem 1.5. Let γ be a regular curve with κ constant and τ = 0. Then γ isa circle.

10 1. CURVES

Proof. Let β = γ + κ−1N . Then

β′ = T +1κ

(−κT + τB) = 0.

Thus β is constant, and |γ − β| = κ−1. It follows that γ lies in the intersectionbetween a plane and a sphere, thus γ is a circle.

3. Plane Curves

3.1. Local Theory. Let γ : [a, b]→ R2 be a regular plane curve parametrizedby arclength, and let κ be its curvature. Note that κ is signed, and in fact changessign (but not magnitude) when the orientation of γ is reversed. The Frenet frameequations are:

e′1 = κ e2, e′2 = −κ e1Proposition 1.6. Let γ : [a, b] → R2 be a regular curve with |γ′| = 1. Then

there exists a differentiable function θ : [a, b]→ R such that

(1.3) e1 = (cos θ, sin θ).

Moreover, θ is unique up to a constant integer multiple of 2π, and in particularθ(b)− θ(a) is independent of the choice of θ. The derivative of θ is the curvature:θ′ = κ.

Proof. Let a = t0 < t1 < · · · < tn = b be a partition of [a, b] so that thediameter of e1([ti−1, ti]) is less than 2, i.e., e1 restricted to each subinterval mapsinto a semi-circle. Such a partition exists since e1 is uniformly continuous on [a, b].Choose θ(a) so that (1.3) holds at a, and proceed by induction on i: if θ is definedat ti then there is a unique continuous extension so that (1.3) holds. If ψ is anyother continuous function satisfying (1.3), then k = (1/2π)(θ − ψ) is a continuousinteger-valued function, hence is constant. Finally, e2 = (− sin θ, cos θ) hence

e′1 = κ e2 = θ′(− sin θ, cos θ),

and we obtain θ′ = κ.

3.2. Global Theory.

Definition 1.7. A curve γ : [a, b]→ Rn is closed if γ(k)(a) = γ(k)(b). A closedcurve γ : [a, b] → Rn is simple if γ|(a,b) is one-to-one. The rotation number of asmooth closed curve is:

(1.4) nγ =1

2π(θ(a)− θ(b)

),

where θ is the function defined in Proposition 1.6.

We note that the rotation number is always an integer. For reference, we alsonote that the rotation number of a curve is the winding number of the map e1.Finally, in view of the last statement in Proposition 1.6, we have:

nγ =1

2π

∫[0,L]

κ ds.

Theorem 1.7 (Rotation Theorem). Let γ : [0, L] → R2 be a smooth, regular,simple, closed curve. Then nγ = ±1. In particular

12π

∫[0,L]

κ ds = ±1.

3. PLANE CURVES 11

For the proof we will need the following technical lemma. We say that a setΩ ⊂ Rn is star-shaped with respect to x0 ∈ Ω if for every y ∈ Ω the line segmentx0y lies in Ω.

Lemma 1.8. Let Ω ⊂ Rn be star-shaped with respect to x0 ∈ Ω, and let e : Ω→S1 be a continuous function. Then there exists a continuous function θ : Ω → Rsuch that:

(1.5) e = (cos θ, sin θ).

Moreover, if ψ is another continuous function satisfying (1.5), then θ − ψ = 2πkwhere k is a constant integer.

In fact, it is sufficient to assume that Ω is simply connected, but we will notprove this more general result here.

Proof. Define θ(x0) so that (1.5) holds at x0. For each x ∈ Ω define θcontinuously along the line segment x0x as in the proof of Proposition 1.6. SinceΩ is star-shaped with respect to x0, this defines θ everywhere in Ω. It remainsto show that θ is continuous. Let y0 ∈ Ω. Since x0y0 is compact, it is possibleto choose δ small enough that the following holds: y′ ∈ x0y0 and |y − y′| < δimplies |e(y)− e(y′)| < 2 or equivalently e(y) and e(y′) are not antipodal. Let0 < ε < π. Then there exists a neighborhood U ⊂ Bδ(y0) of y0 such that y ∈ Uimplies θ(y) − θ(y0) = 2πk(y) + ε′(y) where |ε′(y)| < ε and k(y) is integer-valued.It remain to prove that k ≡ 0. Let y ∈ U and consider the continuous function:

φ(s) = θ(x0 + s(y − x0)

)− θ(x0 + s(y0 − x0)

), 0 6 s 6 1.

Since ∣∣(x0 + s(y − x0))−(x0 + s(y0 − x0)

)∣∣ = |s(y − y0)| < δ,

it follows from our choice of δ that e(x0 + s(y−x0)

)and e

(x0 + s(y0−x0)

)are not

antipodal. Thus, φ(s) 6= π for all 0 6 s 6 1, and since φ(0) = 0 we conclude that|φ| < π. In particular

|2πk(y) + ε′(y)| = |θ(y)− θ(y0)| = |φ(1)| < π,

and it follows that

|2πk(y)| 6 |2πk(y) + ε′(y)|+ |ε′(y)| < 2π.

Since k(y) is integer-valued this implies k(y) = 0.

Proof of the Rotation Theorem. Pick a line which intersects the curveγ and pick a last point p on this line, i.e., a point with the property that one rayof the line from p has no other intersection points with γ. Let h be the unit vectorpointing in the direction of that ray. We assume without loss of generality that γis parametrized by arclength, γ(0) = γ(L) = p. Now, let Ω = (t1, t2) ∈ R2 : 0 6t1 6 t2 6 L, and note that Ω is star-shaped. Define the S1-valued function:

e(t1, t2) =

γ′(t1) if t1 = t2;−γ′(0) if (t1, t2) = (0, L);γ(t2)− γ(t1)|γ(t2)− γ(t1)|

otherwise.

12 1. CURVES

We claim that e is continuous on Ω. Indeed, let ∆ = (t1, t2) : t1 = t2 denote thediagonal in Ω, let Ω′ = Ω \∆ ∪ (0, L) and suppose that (t1, t2) ∈ Ω′, t0 ∈ [0, L].Then we can write:

γ(t2)− γ(t1) = (t2 − t1) (γ′(t0) +Q(t1, t2)) ,

where

Q(t1, t2) =1

t2 − t1

∫ t2

t1

(γ′(t)− γ′(t0)

)dt.

Thus, if we can show that Q(t1, t2)→ 0 as (t1, t2) ∈ Ω′ tends to (t0, t0), we get:

γ(t2)− γ(t1)|γ(t2)− γ(t1)|

=γ′(t0) +Q(t1, t2)|γ′(t0) +Q(t1, t2)|

→ γ′(t0).

and we will have shown continuity at (t0, t0) on diagonal ∆. We now show thatQ(t1, t2)→ 0 as (t1, t2)→ (t0, t0). Let ε > 0, then since γ′ is continuous at t0 thereδ such that if |t− t0| < δ, then |γ′(t)− γ′(t0)| < ε. Now, if the distance between(t1, t2) and (t0, t0) is less than δ, then |t1 − t0| , |t2 − t0| < δ, hence |t− t0| < δ forall t ∈ [t1, t2], and thus it follows that |γ′(t)− γ′(t0)| < ε. Consequently, we find:

|Q(t1, t2)| =∣∣∣∣ 1t2 − t1

∫ t2

t1

(γ′(t)− γ′(t0)

)dt

∣∣∣∣ 6 1t2 − t1

∫ t2

t1

∣∣γ′(t)− γ′(t0)∣∣ dt 6 ε.

A similar argument, using in addition the fact that γ is closed, can be used to showcontinuity of e at (0, L).

Now by the Lemma, there is a continuous function θ : Ω → R such that e =(cos θ, sin θ). We claim that θ(L,L) − θ(0, 0) = ±2π which proves the theorem,since θ(t, t) is a continuous function satisfying (1.3) in Proposition 1.6, and thuscan be used on the right-hand side of (1.4) to compute the rotation number.

To prove this claim, note that, for any 0 < t < L, the unit vector

e(0, t) =γ(t)− γ(0)|γ(t)− γ(0)|

is never equal to h. Hence, there is some value α such that θ(0, t)−θ(0, 0) 6= α+2πkfor any integer k. Thus, |θ(0, t)− θ(0, 0)| < 2π, and since e(0, L) = −e(0, 0) itfollows that θ(0, L)− θ(0, 0) = ±π.

Since the curves e(0, t) and e(t, L) are related via a rigid motion, i.e., e(t, L) =Re(0, t) where R is rotation by π, it follows that ψ(t) =

(θ(t, L) − θ(0, L)

)−(

θ(0, t) − θ(0, 0))

is a constant. Since clearly ψ(0) = 0, we get θ(0, L) − θ(0, 0) =θ(L,L)− θ(0, L), and we conclude:

θ(L,L)− θ(0, 0) =(θ(t, L)− θ(0, L)

)+(θ(0, t)− θ(0, 0)

)= ±2π.

Definition 1.8. A piecewise smooth curve is a continuous function γ : [a, b]→Rn such that there is a partition of [a, b]:

a = a0 < a1 < · · · < bn = b

such that for each 1 6 j 6 n the curve segment γj = γ|[aj−1,aj ] is smooth.The points γ(aj) are called the corners of γ. The directed angle −π < ψj 6 πfrom γ′(aj−) to γ′(aj+) is called the exterior angle at the j-th corner. Define

3. PLANE CURVES 13

θj : [aj−1, aj ] → R as in Proposition 1.6, i.e., so that γ′j = (cos θj , sin θj). Therotation number of γ is given by:

nγ =1

2π

n∑j=1

(θj(aj)− θj(aj−1)

)+

12π

n∑j=1

ψj .

Again, nγ is an integer, and we have:

nγ =1

2π

∫[a,b]

κ ds+1

2π

n∑j=1

ψj .

The Rotation Theorem can be generalized to piecewise smooth curves providedcorners are taken into account.

Theorem 1.9. Let γ : [0, L]→ R2 be a piecewise smooth, regular, simple, closedcurve, and assume that none of the exterior angles are equal to π. Then nγ = ±1.

3.3. Convexity.

Definition 1.9. Let γ : [0, L] → R2 be a regular closed plane curve. We saythat γ is convex if for each t0 ∈ [0, L] the curve lies on one side only of its tangentat t0, i.e., if one of the following inequality holds:(

γ − γ(t0))· e2 6 0,(

γ − γ(t0))· e2 > 0.

Theorem 1.10. Let γ : [0, L]→ R2 be a regular simple closed plane curve, andlet κ be its curvature. Then γ is convex if and only if either κ > 0 or κ 6 0.

We note that an orientation reversing reparametrization of γ changes κ > 0into κ 6 0 and vice versa. Thus, ignoring orientation, those two conditions areequivalent. We also note that the theorem fails if γ is not assumed simple.

Proof. We may assume without loss of generality that |γ′| = 1. Let θ : [0, L]→R be the continuous function given in Proposition 1.6 satisfying:

e1 = (cos θ, sin θ),

and θ′ = κ.Suppose that γ is convex. We will show that θ is weakly monotone, i.e., if

t1 < t2 and θ(t1) = θ(t2) then θ is constant on [t1, t2]. First, we note that since γ issimple, we have nγ = ±1 by the Rotation Theorem, and it follows that e1 is ontoS1, see Exercise 1.5. Thus, there is t3 ∈ [0, L] such that

e1(t3) = −e1(t1) = −e1(t2).

By convexity, the three parallel tangents at t1, t2, and t3 cannot be distinct, henceat least two must coincide. Let p1 = γ(s1) and p2 = γ(s2), s1 < s2 denote thesetwo points, then the line p1p2 is contained in γ. Otherwise, if q is a point on p1p2

not on γ, then the line through q perpendicular to p1p2 intersects γ in at leasttwo points r and s, which by convexity must lie on one side of p1p2. Withoutloss of generality, assume that r is the closer of the two to p1p2. Then r lies inthe interior of the triangle p1p2s. Regardless of the inclination of the tangent atr, the three points p1, p2 and s, all belonging to γ, cannot all lie on one side ofthe tangent, in contradiction to convexity. If p1p2 6= γ(s) : s1 6 s 6 s2, then

14 1. CURVES

p1p2 = γ(s) : s2 6 s 6 L ∩ γ(s) : 0 6 s 6 s1. However, in that case, we wouldhave θ(s2)− θ(s1) = θ(L)− θ(0) = 2π, a contradiction. Thus, we have

p1p2 = γ(s) : s1 6 s 6 s2 = γ(t) : t1 6 t 6 t2.In particular θ(t) = θ(t1) = θ(t2).

Conversely, suppose that γ is not convex. Then, there is t0 ∈ [0, L] such thatthe function φ =

(γ − γ(t0)

)· e2 changes sign. We will show that θ′ also changes

sign. Let t+, t− ∈ [0, L] be such that

min[0,L]

φ = φ(t−) < 0 = φ(t0) = φ(t+) = max[0,L]

φ.

Note that the three tangents at t−, t+ and t0 are parallel but distinct. Sinceφ′(t−) = φ′(t+) = 0, we have that e1(t−) and e1(t+) are both equal to ±e1(t0).Thus, at least two of these vectors are equal. We may assume, after reparametriza-tion, that there exists 0 < s < L such that e1(0) = e1(s). This implies that

θ(s)− θ(0) = 2πk, θ(L)− θ(s) = 2πk′

with k, k′ ∈ Z. By the Rotation Theorem, nγ = k + k′ = ±1. Since γ(0) and γ(s)do not lie on a line parallel to e1(t0), it follows that θ is not constant on either[0, s] or [0, L]. If k = 0 then θ′ changes sign on [0, s], and similarly if k′ = 0 then θ′

changes sign on [s, L]. If kk′ 6= 0, then since k + k′ = ±1, it follows that kk′ < 0and θ′ changes sign on [0, L].

Definition 1.10. Let γ : [0, L] → R2 be a regular plane curve. A vertex of γis a critical point of the curvature κ.

Theorem 1.11 (The Four Vertex Theorem). A regular simple convex closedcurve has at least four vertices.

Proof. Clearly, κ has a maximum and minimum on [0, L], hence γ has at leasttwo vertices. We will assume, without loss of generality, that γ is parametrized byarclength, has its minimum at t = 0, its maximum at t = t0 where 0 < t0 < L,that γ(0) and γ(t0) lie on the x-axis, and that γ enters the upper-half plane inthe interval [0, t0]. All these properties can be achieved by reparametrizing androtating γ.

We now claim that p = γ(0) and q = γ(t0) are the only points of γ on thex-axis. Indeed, suppose that there is another point r = γ(t1) on the x-axis, thenone of these points lies between the other two, and the tangent at that point must,by convexity, contain the other two. Thus, by the argument used in the proof ofTheorem 1.10 the segment between the outer two is contained in γ, and in particularpq is contained in γ. If follows that κ = 0 at p and q where κ has its minimumand maximum, hence κ ≡ 0, a contradiction since then γ is a line and cannot beclosed. We conclude that γ remains in the upper half-plane in the interval [0, t0]and remains in the lower half-plane in the interval [t0, L].

Suppose now by contradiction that γ(0) and γ(t0) are the only vertices of γ.Then it follows that:

κ′ > 0 on [0, t0], κ′ 6 0 on [t0, L].

Thus, if we write γ = (x, y), then we have κ′y > 0 on [0, L], and x′′ = −κy′, hence:

0 =∫ L

0

x′′ ds = −∫ L

0

−κy′ ds =∫ L

0

κ′y ds.

4. FENCHEL’S THEOREM 15

Since the integrand in the last integral is non-negative, we conclude that κ′y ≡ 0,hence y ≡ 0, again a contradiction.

It follows that κ has another point where κ′ changes sign, i.e., an extremum.Since extrema come in pairs, κ has at least four extrema.

4. Fenchel’s Theorem

We will use without proof the fact that the shortest path between two pointson a sphere is always an arc of a great circle. We also use the notation γ1 + γ2 todenote the curve γ1 followed by the curve γ2.

Definition 1.11. Let γ : [0, L] → Rn be a regular curve parametrized by ar-clength. The spherical image of γ is the curve γ′ : [0, L]→ Sn−1. The total curvatureof γ : [0, L]→ Rn is:

Kγ =∫I

|γ′′| ds.

We note that the total curvature is simply the length of the spherical image.

Theorem 1.12. Let γ be a regular simple closed curve in Rn parametrized byarclength. Then the total curvature of γ is at least 2π:

Kγ > 2π,

with equality if and only if γ is planar and convex.

The proof will follow from two lemmata which are interesting in their own right.

Lemma 1.13. Let γ : [0, L] → Rn be a regular closed curve parametrized byarclength. Then the spherical image of γ cannot map into an open hemisphere. Ifγ′ maps into a closed hemisphere, then γ maps into an equator.

Proof. Suppose, by contradiction, that there is v ∈ Sn−1 such that γ′ · v > 0.Then

0 = γ · v |L − γ · v |0 =∫ L

0

γ′ · v ds > 0.

If γ′ ·v > 0, then the same inequality shows that γ′ ·v ≡ 0, hence γ lies in the planeperpendicular to v through γ(0).

Lemma 1.14. Let n > 3, and let γ : [0, L]→ Sn−1 be a regular closed curve onthe unit sphere parametrized by arclength.

(1) If the arclength of γ is less than 2π then γ is contained in an open hemi-sphere.

(2) If the arclength of γ is equal to 2π then γ is contained in a closed hemi-sphere.

Proof. (1) First observe that no piecewise smooth curve of arclength lessthan 2π contains two antipodal points. Otherwise the two segments of of the curvebetween p and q would each have length at least π, and hence the length of thecurve would have to be at least 2π. Now pick a point p on γ and let q on γ bechosen so that the two segments γ1 and γ2 from p to q along γ have equal length.Note that p and q cannot be antipodal. Let v be the midpoint along the shorter ofthe two segments of the great circle between p and q. Suppose that γ1 intersectsthe equator, the great circle v · x = 0. Let γ1 be the reflection of γ with respectto v, then the length of γ1 + γ1 is the same as the length of γ hence is less than

16 1. CURVES

2π. But γ1 + γ1 contains two antipodal points, a contradiction. Thus, γ1 cannotintersect the equator. Similarly, γ2 cannot intersect the equator, and we concludeγ stays in the open hemisphere v · x > 0.

(2) If the arclength of γ is 2π, we refine the above argument. If p and q areantipodal, then both γ1 and γ2 are great semi-circle, thus, γ stays in a closedhemisphere.1 So we can assume that p and q are not antipodal and proceed asbefore, defining v to be the midpoint on the shorter arc of the great circle betweenp and q. Now, if γ1 crosses the equator, then γ1 + γ1 contains two antipodal pointson the equator, and the two segments joining these points enter both hemispheres.Thus, these segments are not semi-circle, and consequently both have arclengthstrictly greater than π. Thus the arclength of γ1 + γ1 is strictly larger than 2πa contradiction. Similarly, γ2 does not cross the equator, and we conclude that γstays in the closed hemisphere v · x > 0.

Proof of Fenchel’s Theorem. Note that the total curvature is simply thearclength of the spherical image of γ. By Lemma 1.13 γ′ is not contained in anopen hemisphere, so by Lemma 1.14

Kγ =∫I

|γ′′| ds > 2π.

If the arclength of γ′ is 2π, then by Lemma 1.14, γ′ is contained in a closed hemi-sphere, and by Lemma 1.13, γ maps into an equator. If n > 3, we may proceedby induction until we obtain that γ is planar. Once we have that γ is planar, theRotation Theorem gives nγ = ±1. Without loss of generality,2 we may assume thatnγ = 1. Hence

0 6∫I

(|κ| − κ

)ds = Kγ − 2π = 0,

and it follows that κ = |κ| > 0, which by Theorem 1.10 implies that γ is convex.

Exercises

Exercise 1.1. A regular space curve γ : [a, b]→ R3 is a helix if there is a fixedunit vector u ∈ R3 such that e1 · u is constant. Let κ and τ be the curvature andtorsion of a regular space curve γ, and suppose that κ 6= 0. Prove that γ is a helixif and only if τ = cκ for some constant c.

Exercise 1.2. Let γ : I → R4 be a smooth curve parameterized by arclengthsuch that γ′, γ′′, γ′′′ are linearly independent. Prove the existence of a Frenet frame,i.e., a positively oriented orthonormal frame X = (e1, e2, e3, e4) satisfying e1 = γ′,and X ′ = Xω, where ω is anti-symmetric, tri-diagonal, and ωi,i+1 > 0 for i 6 n−2.The curvatures of γ are the three functions κi = ωi,i+1, i = 1, 2, 3. Note thatκ1, κ2 > 0, but κ3 has a sign.

Exercise 1.3. Prove the Fundamental Theorem for curves in R4: Given func-tions κ1, κ2, κ3 on I with κ1, κ2 > 0, there is a smooth curve γ parameterized byarclength on I such that κ1, κ2, κ3 are the curvatures of γ. Furthermore, γ is uniqueup to a rigid motion of R4.

1In fact, since γ is smooth, γ1 and γ1 are contained in the same great circle, and hence γ is

itself a great circle.2Reversing the orientation of γ if necessary.

EXERCISES 17

Exercise 1.4. Let γ : [a, b] → R2 be a regular plane curve with non-zero cur-vature κ 6= 0, and let β = γ + κ−1N be the locus of the centers of curvature ofγ.

(1) Prove that β is regular provided that κ′ 6= 0.(2) Prove that each tangent ` of β intersects γ at a right angle.

A curve satisfying (1) and (2) is called an evolute of γ.(3) Prove that each regular plane curve γ : [a, b]→ R2 has at most one evolute.

Exercise 1.5. A convex plane curve γ : [a, b]→ R2 is strictly convex if κ 6= 0.Prove that if γ : [a, b]→ R2 is a strictly convex simple closed curve, then for everyv ∈ S1, there is a unique t ∈ [a, b] such that e1(t) = v.

Exercise 1.6. Let γ : [0, L]→ R2 be a strictly convex simple closed curve. Thewidth w(t) of γ at t ∈ [0, L] is the distance between the tangent line at γ(t) and thetangent line at the unique point γ(t′) satisfying e1(t′) = −e1(t) (see Exercise 1.5).A curve has constant width if w is independent of t. Prove that if γ has constantwidth then:

(1) The line between γ(t) and γ(t′) is perpendicular to the tangent lines atthose points.

(2) The curve γ has length L = πw.

Exercise 1.7. Let γ : [0, L] → R2 be a simple closed curve. By the JordanCurve Theorem, the complement of γ has two connected components, one of whichis bounded. The area enclosed by γ is the area of this component, and accordingto Green’s Theorem, it is given by:

A =∫γ

x dy =∫γ

xy′ dt,

where the orientation is chosen so that the normal e2 points into the boundedcomponent. Let L be the length of γ, and let β be a circle of width 2r equal tosome width of γ. Prove:

(1) A = 12

∫γ(xy′ − yx′) dt.

(2) A+ πr2 6 Lr.(3) The isoperimetric inequality : 4πA 6 L2.(4) If equality holds in (3) then γ is a circle.

Exercise 1.8. Prove that if a convex simple closed curve has four vertices,then it cannot meet any circle in more than four points.

CHAPTER 2

Local Surface Theory

1. Surfaces

Definition 2.1. A parametric surface patch is a smooth mapping:

X : U → R3,

where U ⊂ R2 is open, and the Jacobian dX is non-singular.

Write X = (x1, x2, x3), and each xi = xi(u1, u2), then the Jacobian has thematrix representation:

dX =

x11 x1

2

x21 x2

2

x31 x3

2

where we have used the notation fi = fui = ∂f/∂ui. According to the definition,we are requiring that this matrix has rank 2, or equivalently that the vectors X1 =(x1

1, x21, x

31) and X2 = (x1

2, x22, x

32) are linearly independent. Another equivalent

requirement is that dX : R2 → R3 is injective.

Example 2.1. Let U ⊂ R2 be open, and suppose that f : U → R is smooth.Define the graph of f as the parametric surface X(u1, u2) = (u1, u2, f(u1, u2)). Toverify that X is indeed a parametric surface, note that:

dX =

1 00 1f1 f2

so that clearly X is non-singular.

A diffeomorphism between open sets U, V ⊂ R2 is a map φ : U → V which issmooth, one-to-one, and whose inverse is also smooth. If det(dφ) > 0, then we saythat φ is an orientation-preserving diffeomorphism.

Definition 2.2. Let X : U → R3, and X : U → R3 be parametric surfaces.We say that X is reparametrization of X if X = X φ, where φ : U → U is adiffeomorphism. If φ is an orientation-preserving diffeomorphism, then X is anorientation-preserving reparametrization.

Clearly, the inverse of a diffeomorphism is a diffeomorphism. Thus, if X is areparametrization of X, then X is a reparametrization of X.

Definition 2.3. The tangent space TuX of the parametric surface X : U → R3

at u ∈ U is the 2-dimensional linear subspace of R3 spanned by the two vectors X1

and X2.1

1Note that the tangent plane to the surface X(U) at u is actually the affine subspace X(u) +TuX. However, it will be very convenient to have the tangent space as a linear subspace of R3.

19

20 2. LOCAL SURFACE THEORY

If Y ∈ TuX, then it can be expressed as a linear combination in X1 and X2:

Y = y1X1 + y2X2 =2∑i=1

yiXi,

where yi ∈ R are the components of the vector Y in the basis X1, X2 of TuX.We will use the Einstein Summation Convention: every index which appears twicein any product, once as a subscript (covariant) and once as a superscript (con-travariant), is summed over its range. For example, the above equation will bewritten Y = yiXi. The next proposition show that the tangent space is invariantunder reparametrization, and gives the law of transformation for the componentsof a tangent vector. Note that covariant and contravariant indices have differenttransformation laws, cf. (2.1) and (2.2).

Proposition 2.1. Let X : U → R3 be a parametric surface, and let X = X φbe a reparametrization of X. Then Tφ(u)X = TuX. Furthermore, if Z ∈ Tφ(u)X,and Z = ziXi = zjXj, then:

(2.1) zi = zj∂ui

∂uj,

where dφ = (∂ui/∂uj).

Proof. By the chain rule, we have:

(2.2) Xj =∂ui

∂ujXi.

Thus TuX ⊂ Tφ(u)X, and since we can interchange the roles of X and X, weconclude that TuX = Tφ(u)X. Substituting (2.2) in zjXj , we find:

ziXi = zj∂ui

∂ujXi,

and (2.1) follows.

Definition 2.4. A vector field along a parametric surface X : U → R3, is asmooth mapping Y : U → R32. A vector field Y is tangent to X if Y (u) ∈ TuX forall u ∈ U . A vector field Y is normal to X if Y (u) ⊥ TuX for all u ∈ U .

Example 2.2. The vector fields X1 and X2 are tangent to the surface. Thevector field X1 ×X2 is normal to the surface.

We call the unit vector field

N =X1 ×X2

|X1 ×X2|the unit normal . Note that the triple (X1, X2, N), although not necessarily or-thonormal, is positively oriented. In particular, we can see that the choice of anorientation on X, e.g., X1 → X2, fixes a unit normal, and vice-versa, the choice ofa unit normal fixes the orientation. Here we chose to use the orientation inheritedfrom the orientation u1 → u2 on U .

Definition 2.5. We call the map N : U → S2 the Gauss map.

2We often visualize Y (u) as being attached at X(u), i.e. belonging to the tangent space ofR3 at X(u); cf. see footnote 1.

2. THE FIRST FUNDAMENTAL FORM 21

The Gauss map is invariant under orientation-preserving reparametrization.

Proposition 2.2. Let X : U → R3 be a parametric surface, and let N : U → S2

be its Gauss map. Let X = X φ be an orientation-preserving reparametrization ofX. Then the Gauss map of X is N φ.

Proof. Let v ∈ V . The unit normal N(v) of X at v is perpendicular toTvX. By Proposition 2.1, we have Tφ(v)X = TvX. Thus, N(v) is perpendicularto Tφ(v)X, as is N(φ(v)). It follows that the two vectors are co-linear, and henceN(v) = ±N(φ(v)). But since φ is orientation preserving, the two pairs (X1, X2)and (X1, X2) have the same orientation in the plane TvX. Since also, the two triples(X1(φ(v)), X2(φ(v)), N(φ(v))) and (X1(v), X2(v), N(v)) have the same orientationin R3, it follows that N(φ(v)) = N(v).

2. The First Fundamental Form

Definition 2.6. A symmetric bilinear form on a vector space V is functionB : V × V → R satisfying:

(1) B(aX + bY, Z) = aB(X,Z) + bB(Y,Z), for all X,Y ∈ V and a, b ∈ R.(2) B(X,Y ) = B(Y,X), for all X,Y ∈ V .

The symmetric bilinear form B is positive definite if B(X,X) > 0, with equality ifand only if X = 0.

With any symmetric bilinear form B on a vector space, there is associated aquadratic formQ(X) = B(X,X). Let V andW be vector spaces and let T : V →Wbe a linear map. If B is a symmetric bilinear form on W , we can define a symmetricbilinear form T ∗Q on V by T ∗Q(X,Y ) = Q(TX, TY ). We call T ∗Q the pull-back ofQ by T . The map T is then an isometry between the inner-product spaces (V, T ∗Q)and (W,Q).

Example 2.3. Let V = R3 and define B(X,Y ) = X · Y , then B is a positivedefinite symmetric bilinear form. The associated quadratic form is Q(X) = |X|2.

Example 2.4. Let A be a symmetric 2× 2 matrix, and let B(X,Y ) = AX ·Y ,then B is a symmetric bilinear form which is positive definite if and only if theeigenvalues of A are both positive.

Definition 2.7. Let X : U → R3 be a parametric surface. The first funda-mental form is the symmetric bilinear form g defined on each tangent space TuXby:

g(Y, Z) = Y · Z, ∀Y,Z ∈ TuX.

Thus, g is simply the restriction of the Euclidean inner product in Example 2.3to each tangent space of X. We say that g is induced by the Euclidean innerproduct.

Let gij = g(Xi, Xj), and let Y = yiXi and Z = ziXi be two vectors in TuX,then

(2.3) g(Y, Z) = gijyizj .

Thus, the so-called coordinate representation of g is at each point u0 ∈ U aninstance of Example 2.4. In fact, if A =

(gij), and B(ξ, η) = ξ ·Aη for ξ, η ∈ R2 as

in Example 2.4, then B is the pull-back by dXu : R2 → TuX of the restriction ofthe Euclidean inner product on TuX.


The classical (Gauss) notation for the first fundamental form is g11 = E, g12 =g21 = F , and G = g22, i.e., (

gij)

=(E FF G

)Clearly, F 2 < EG, and another condition equivalent to the condition that X1 andX2 are linearly independent is that det

(gij)

= EG−F 2 > 0. The first fundamentalform is also sometimes written:

ds2 = gij dui duj = E (du1)2 + 2F du1 du2 +G (du2)2.

Note that the gij ’s are functions of u. The reason for the notation ds2 is that thesquare root of the first fundamental form can be used to compute length of curveson X. Indeed, if γ : [a, b]→ R3 is a curve on X, then γ = X β, where β is a curvein U . Let β(t) =

(β1(t), β2(t)

), and denote time derivatives by a dot, then

Lγ([a, b]) =∫ b

a

|γ| dt∫ b

a

√gij βiβj dt.

Accordingly, ds is also called the line element of the surface X.Note that g contains all the intrinsic geometric information about the surface

X. The distance between any two points on the surface is given by:

d(p, q) = infLγ : γ is a curve on X between p and q.

Also the angle θ between two vectors Y,Z ∈ TxX is given by:

cos θ =g(Y, Z)√

g(Y, Y ) g(Z,Z),

and the angle between two curves β and γ on X is the angle between their tangentsβ and γ. Intrinsic geometry is all the information which can be obtained from thethree functions gij and their derivatives.

Clearly, the first fundamental form is invariant under reparametrization. Thenext proposition shows how the gij ’s change under reparametrization.

Proposition 2.3. Let X : U → R3 be a parametric surface, and let X = X φbe a reparametrization of X. Let gij be the coordinate representation of the firstfundamental form of X, and let gij be the coordinate representation of the firstfundamental form of X. Then, we have:

(2.4) gij = gkl∂uk

∂ui∂ul

∂uj,

where dφ = (∂ui/∂uj).

Proof. In view of (2.2), we have:

gij = g(Xi, Xj) = g

(∂uk

∂uiXk,

∂ul

∂ujXl

)=∂uk

∂ui∂ul

∂ujg(Xk, Xl) = gkl

∂uk

∂ui∂ul

∂uj.

3. THE SECOND FUNDAMENTAL FORM 23

3. The Second Fundamental Form

We now turn to the second fundamental form. First, we need to prove atechnical proposition. Let Y and Z be vector fields along X, and suppose thatY = yiXi is tangential. We define the directional derivative of Z along Y by:

∂Y Z = yiZi = yi∂Z

∂ui.

Note that the value of ∂Y Z at u depends only on the value of Y at u, but dependson the values of Z in a neighborhood of u. In addition, ∂Y Z is reparametrizationinvariant, but even if Z is tangent, it is not necessarily tangent. Indeed, if we writeY = yiXi, then we see that:

yi ∂iZ = yi∂Z

∂ui= yj

∂ui

∂uj∂Z

∂uk∂uk

∂ui= yj∂jZ.

The commutator of Y and Z can now be defined as the vector field:

[Y,Z] = ∂Y Z − ∂ZY.

Proposition 2.4. Let X : U → R3 be a surface, and let N be its unit normal.(1) If Y and Z are tangential vector fields then [Y, Z] ∈ TuX.(2) If Y, Z ∈ TuX then ∂YN · Z = ∂ZN · Y .

Proof. Note first that since X is smooth, we have Xij = Xji, where we haveused the notation Xij = ∂2X/∂ui∂uj . Now, write Y = yiXi and Z = zjXj , andcompute:

∂Y Z − ∂ZY = yizjXji + yi∂izjXj − yizjXij − zj∂jyiXi

=(yi∂iz

j − zi∂iyj)Xj .

To prove (2), extend Y and Z to be vector fields in a neighborhood of u, and use (1):

∂YN · Z − ∂ZN · Y = −N ·(∂Y Z − ∂ZY

)= 0.

Note that while proving the proposition, we have established the followingformula for the commutator:

(2.5) [Y,Z] =(yi∂iz

j − zi∂iyj)Xj

Definition 2.8. Let X : U → R3 be a surface, and let N : U → S2 be itsunit normal. The second fundamental form of X is the symmetric bilinear form kdefined on each tangent space TuX by:

(2.6) k(Y,Z) = −∂YN · Z.

We remark that since N ·N = 1, we have ∂YN ·N = 0, hence ∂YN is tangen-tial. Thus, according to (2.6), the second fundamental form is minus the tangentialdirectional derivative of the unit normal, and hence measures the turning of the tan-gent plane as one moves about on the surface. Note that part (2) of the propositionguarantees that k is indeed a symmetric bilinear form. Note that it is not neces-sarily positive definite. Furthermore, if we set kij = k(Xi, Xj) to be the coordinaterepresentation of the second fundamental form, then we have:

(2.7) kij = Xij ·N.


This equation leads to another representation. Consider the Taylor expansion of Xat a point, say 0 ∈ U :

X(u) = X(0) +Xi(0)ui +12∂ijX(0)uiuj +O(|u|3)

Thus, the elevation of X above its tangent plane at u is given up to second-orderterms by: (

X(u)−X(0)−Xi(0)ui)·N =

12kij(0)uiuj +O(|u|3).

The paraboloid on the right-hand side of the equation above is called the osculatingparaboloid . A point u of the surface is called elliptic, hyperbolic, parabolic, or planar ,depending on whether this paraboloid is elliptic, hyperbolic, cylindrical, or a plane.

In classical notation the second fundamental form is:(kij)

=(L MM N

).

Clearly, the second fundamental form is invariant under orientation-preservingreparametrizations. Furthermore, the kij ’s, the coordinate representation of k,changes like the first fundamental form under orientation-preserving reparametriza-tion:

kij = k(Xu, Xj) = kml∂um

∂ui∂ul

∂uj,

Yet another interpretation of the second fundamental form is obtained by con-sidering curves on the surface. The following theorem is essentially due to Euler.

Theorem 2.5. Let γ = X β : [a, b] → R3 be a curve on a parametric surfaceX : U → R3, where β : [a, b] → U . Let κ be the curvature of γ, and let θ be theangle between the unit normal N of X, and the principal normal e2 of γ. Then:

(2.8) κ cos θ = k(γ, γ).

Proof. We may assume that γ is parametrized by arclength. We have:

γ = βiXi,

andκe2 = γ = βiXi + βiβj Xij .

The theorem now follows by taking inner product with N , and taking (2.7) intoaccount.

The quantity κ cos θ is called the normal curvature of γ. It is particularly inter-esting to consider normal sections, i.e., curves γ on X which lie on the intersectionof the surface with a normal plane. We may always orient such a plane so that thenormal e2 to γ in the plane coincide with the unit normal N of the surface. In thatcase, we obtain the simpler result:

κ = k(γ, γ).

Thus, the second fundamental form measures the signed curvature of normal sec-tions in the normal plane equipped with the appropriate orientation.

4. EXAMPLES 25

Definition 2.9. Let X : U → R3 be a parametric surface, and let k be itssecond fundamental form. Denote the unit circle in the tangent space at u bySuX = Y ∈ TuX : |Y | = 1. For u ∈ U , define the principal curvatures of X atu by:

k1 = minY ∈SuX

k(Y, Y ), k2 = maxY ∈SuX

k(Y, Y ).

The unit vectors Y ∈ SuX along which the principal curvatures are achieved arecalled the principal directions. The mean curvature H and the Gauss curvature Kof X at u are given by:

H =12

(k1 + k2), K = k1k2.

If we consider the tangent space TuX with the inner product g and the uniquelinear transformation ` : TuX → TuX satisfying:

(2.9) g(`(Y ), Z) = k(Y,Z), ∀Z ∈ TuX,

then k1 6 k2 are the eigenvalues of ` and the principal directions are the eigenvectorsof `. If k1 = k2 then k = λg and every direction is a principal direction. A pointwhere this holds is called an umbilical point. Otherwise, the principal directionsare perpendicular. We have that H is the trace and K the determinant of `. Let(gij) be the inverse of the 2× 2 matrix (gij):

gimgmj = δij .

Set `(Xi) = `jiXj , then since kij = g(`(Xi), Xj) = `mi gmj , we find:

`ji = kimgmj .

It is customary to say that g raises the index of k and to write the new objectkij = kimg

mj . Here since kij is symmetric, it is not necessary to keep track of theposition of the indices, and hence we write: `ji = kji . In particular, we have:

(2.10) H =12kii, K =

det(kij)

det(gij) .

Now, kij = gimgjlklm, and we have

|k|2 = kijkij = tr `2 = k2

1 + k22 = 4H2 − 2K.

Hence, we conclude

(2.11) K = 2H2 − 12|k|2

4. Examples

In this section, we use u1 = u, and u2 = v in order to simplify the notation.

4.1. Planes. Let U ⊂ R2 be open, and let X : U → R3 be a linear function:

X(u, v) = Au+Bv,

with A,B ∈ R3 linearly independent. Then X is a plane. After reparametrization,we may assume that A and B are orthonormal. In that case, the first fundamentalform is:

ds2 = du2 + dv2.


Furthermore, |A×B| = 1, and N = A×B is constant, hence k = 0. In particular,all the points of X are planar, and we have for the mean and Gauss curvatures:H = K = 0.

It is of interest to note that if all the points of a parametric surface are planar,then X(U) is contained in a plane. We will later prove a stronger result: X has areparametrization which is linear.

Proposition 2.6. Let X : U → R3 be a parametric surface, and suppose thatits second fundamental form k = 0. Then, there is a fixed vector A and a constantb such that X ·A = b, i.e., X is contained in a plane.

Proof. Let A be the unit normal N of X. Let 1 6 i 6 2, and note that Niis tangential. Indeed, N · N = 1, and differentiating along ui, we get N · Ni = 0.However, since k = 0 it follows from (2.6) that Ni ·Xj = −kij = 0. Thus, Ni = 0 fori = 1, 2, and we conclude that N is constant. Consequently, (X ·N)i = Xi ·N = 0,and X ·N is also constant, which proves the proposition.

4.2. Spheres. Let U = (0, π)× (0, 2π) ⊂ R2, and let X : U → R2 be given by:

X(u, v) = (sinu cos v, sinu sin v, cosu).

The surface X is a parametric representation of the unit sphere. A straightforwardcalculation shows that the first fundamental form is:

ds2 = du2 + sin2 u dv2,

and the unit normal is N = X. Thus, Ni = Xi, and consequently kij = −Ni ·Xj =−Xi · Xj = −gij , i.e., k = −g. In particular, the principal curvatures are bothequal to −1 and all the points are umbilical. We have for the mean and Gausscurvatures:

H = −1, K = 1

Proposition 2.7. Let X : U → R3 be a parametric surface and suppose thatall the points of X are umbilical. Then, X(U) is either contained in a plane or asphere.

Proof. By hypothesis, we have

(2.12) Ni = λXi.

We first show that λ is a constant. Differentiating (2.12), we get Nij = λjXi+λXij .Interchanging i and j, subtracting these two equations, and taking into accountNij −Nji = Xij −Xji = 0, we obtain λiXj − λjXi = 0, e.g.,

λ1X2 − λ2X1 = 0.

Since X1 and X2 are linearly independent, we conclude that λ1 = λ2 = 0 andit follows that λ is constant. Now, if λ = 0 then all points are planar, and byProposition 2.6, X is contained in a plane. Otherwise, let A = X − λ−1N , then Ais constant:

Ai = Xi − λ−1Ni = 0,

and |X −A| = |λ|−1 is also constant, hence X is contained in a sphere.

4. EXAMPLES 27

4.3. Ruled Surfaces. A ruled surface is a parametric surface of the form:

X(u, v) = γ(u) + vY (u)

for a curve γ : [a, b] → R3, and a vector field Y : [a, b] → R3 along γ. The curve γis the directrix , and the lines γ(u) + tY (u) for u fixed are the generators of X. Wemay assume that Y is a unit vector field. Provided Y 6= 0. We will also assume thatY 6= 0. In this case, it is possible to arrange by reparametrization that γ · Y = 0,in which case γ is said to be a line of striction. Indeed, if this is not the case, thenwe can set φ =

(γ · Y

)/|Y |2, and note that the curve

α = γ + φY

lies on the surface X, and satisfies α · Y = 0. Consequently, the surface:

X(s, t) = α(s) + tY (s)

is a reparametrization of X. Furthermore, there is only one line of striction on X.Indeed, if β and γ are two lines of striction, then since both β is a curve on X wemay write β = γ + φY for some function φ and consequently:

β = γ + φY + φY .

Taking inner product with Y and using the fact that Y is a unit vector, we obtainφ∣∣Y ∣∣2 = 0 which implies that φ = 0 and thus, β = γ.

We have Xu = γ + vY , Xv = Y , and Xvv = 0. Thus, the first fundamental is:(gij)

=

(1 + v2|Y |2 γ · Yγ · Y 1

)and

det(gij)

= 1 + v2|Y |2 +(γ · Y

)2> v2|Y |2.

Hence, dX is non-singular except possibly on the line of striction. Furthermore,kvv = N · Xvv = 0, hence det

(kij)

= −k2uv and if det

(kij)

= 0 then Nv · Xu =Nv · Xv = 0, i.e., N is constant along generators. We have proved the followingproposition.

Proposition 2.8. Let X be a ruled surface. Then X has non-positive Gausscurvature K 6 0, and K(u) = 0 if and only if N is constant along the generatorthrough u.

4.3.1. Cylinders. Let γ : [a, b]→ R3 be a planar curve, and A be a unit normalto the plane which contains γ. Define X : [a, b]× R→ R3 by:

X(u, v) = γ(u) + vA.

The surface X is a cylinder . The first fundamental form is:

ds2 = du2 + dv2,

and we see that for a cylinder dX is always non-singular. After possibly reversingthe orientation of A, the unit normal is N = e2. Clearly, Nv = 0, and Nu = −κe1.Thus, the second fundamental form is:

κ du2

The principal curvatures are 0 and κ. We have for the mean and Gauss curvatures:

H = −12κ, K = 0.


A surface on which K = 0 is called developable.4.3.2. Tangent Surfaces. Let γ : [a, b]→ R3 be a curve with nonzero curvature

κ 6= 0. Its tangent surface is the ruled surface:

X(u, v) = γ(u) + vγ(u).

Since γ · γ = 0, the curve γ is the line of striction of its tangent surface. We haveXu = e1 + vκ e2 and Xv = e1, hence the first fundamental form is:

(gij)

=

(1 + v2κ2 1

1 1

).

The unit normal is N = −e3, and clearly Nv = 0. Thus,4.3.3. Hyperboloid. Let γ : (0, 2π) → R3 be the unit circle in the x1x2-plane:

γ(t) =(cos(t), sin(t), 0

). Define a ruled surface X : (0, 2π)× R→ R3 by:

X(u, v) = γ(u) + v(γ(u) + e3

)=(cos(u)− v sin(u), sin(u) + v cos(u), v

).

Note that (x1)2 + (x2)2 − (x3)3 = 1 so that X(U) is a hyperboloid of one sheet. Astraightforward calculation gives:

N =1√

1 + 2v2((cos(u)− v sin(u), sin(u) + v cos(u),−v

),

and

|Nv|2 =2

1 + 4v2 + 4v4.

It follows from Proposition 2.8 that X has Gauss curvature K < 0.

5. Lines of Curvature

Definition 2.10. A curve γ on a parametric surface X is called a line ofcurvature if γ is a principal direction.

The following proposition, due to Rodriguez, characterizes lines of curvatureas those curves whose tangents are parallel to the tangent of their spherical imageunder the Gauss map.

Proposition 2.9. Let γ be a curve on a parametric surface X with unit normalN , and let β = N γ be its spherical image under the Gauss map. Then γ is a lineof curvature if and only if

(2.13) β + λγ = 0.

Proof. Suppose that (2.13) holds, then we have:

∂γN + λγ = 0.

Let ` be the linear transformation on TuX associated with k as defined by (2.9).Then, we have for every Y ∈ TuX:

g(`(γ), Y ) = k

(γ, Y

)= −∂γN · Y = λg

(λγ, Y

).

Thus, `(γ)

= λγ, and γ is a principal direction. The proof of the converse issimilar.

5. LINES OF CURVATURE 29

It is clear from the proof that λ in (2.13) is the associated principal curvature.The coordinate curves of a parametric surface X are the two family of curves

γc(t) = X(t, c) and βc(t) = X(c, t). A surface is parametrized by lines of curvatureif the coordinate curves of X are lines of curvature. We will now show that any non-umbilical point has a neighborhood in which the surface can be reparametrized bylines of curvature. We first prove the following lemma which is also of independentinterest.

Lemma 2.10. Let X : U → R3 be a parametric surface, and let Y1 and Y2 belinearly independent vector fields. The following statements are equivalent:

(1) Any point u0 ∈ U has a neighborhood U0 and a reparametrization φ : V0 →U0 such that if X = X φ then Xi = Yi φ.

(2) [Y1, Y2] = 0.

Proof. Suppose that (1) holds. Then Equation (2.5) shows that [X1, X2] = 0.However, since the commutator is invariant under reparametrization, it follows that[Y1, Y2] = 0.

Conversely, suppose that [Y1, Y2] = 0. Express Xi = ajiYj and Yi = bjiXj , andnote that

(bji)

is the inverse of(aji). We now calculate:

0 = [Xi, Xj ]

=[aki Yk, a

ljYl]

=(ali∂Yl

akj − alj∂Ylaki)Yk + aki a

lj [Yk, Yl]

=(alib

ml ∂ma

kj − aljbml ∂maki

)Yk

=(∂ia

kj − ∂jaki

)Yk.

Since Y1 and Y2 are linearly independent, we conclude that:

(2.14) ∂iakj − ∂jaki = 0.

Now, fix 1 6 k 6 2, and consider the over-determined system:

∂uk

∂ui= aki , i = 1, 2.

The integrability condition for this system is exactly (2.14), hence there is a solutionin a neighborhood of u0. Furthermore, since the Jacobian of the map ψ(u1, u2) =(u1, u2) is dψ =

(aki), and det

(aki)6= 0, it follows from the inverse function theorem,

that perhaps on yet a smaller neighborhood, ψ is a diffeomorphism. Let φ = ψ−1,then φ is a diffeomorphism on a neighborhood V0 of ψ(u0), and if we set X = X φ,then:

Xi = Xj∂uj

∂ui= Xjb

ji = Yi.

Proposition 2.11. Let X : U → R3 be a parametric surface, and let Y1 andY2 be linearly independent vector fields. Then for any point u0 ∈ U there is aneighborhood of u0 and a reparametrization X = X φ such that Xi = fi Yi φ forsome functions fi.


Proof. By Lemma 2.10 is suffices to show that there are function fi such thatf1 Y1 and f2 Y2 commute. Write [Y1, Y2] = a1Y1 − a2Y2, and compute:

[f1Y1, f2Y2] = f1f2(a1Y1 − a2Y2

)+ f1

(∂Y1f2

)Y2 − f2

(∂Y2f1

)Y1.

Thus, the commutator [f1Y1, f2Y2] vanishes if and only if the following two equa-tions are satisfied:

∂Y2f1 − a1f1 = 0∂Y1f2 − a2f2 = 0.

We can rewrite those as:

∂Y2 log f1 = a1

∂Y1 log f2 = a2.

Each of those equation is a linear first-order partial differential equation, and canbe solved for a positive solution in a neighborhood of u0.

In a neighborhood of a non-umbilical point, the principal directions define twoorthogonal unit vector fields. Thus, we obtain the following Theorem as a corollaryto the above proposition.

Theorem 2.12. Let X : U → R3 be a parametric surface, and let u0 be anon-umbilical point. Then there is neighborhood U0 of u0 and a diffeomorphismφ : U0 → U0 such that X = X φ is parametrized by lines of curvature.

If X is parametrized by lines of curvature, then the second fundamental formhas the coordinate representation:(

kij)

=

(k1g11 0

0 k2g22

)Definition 2.11. A curve γ on a parametric surface X is called an asymptotic

line if it has zero normal curvature, i.e., k(γ, γ) = 0.

The term asymptotic stems from the fact that those curve have their tangent γalong the asymptotes of the Dupin indicatrix , the conic section kijξ

iξj = 1 in thetangent space. Since the Dupin indicatrix has no asymptotes when K > 0, we seethat the Gauss curvature must be non-positive along any asymptotic line.

The following Theorem can be proved by the same method as used above toobtain Theorem 2.12.

Theorem 2.13. Let X : U → R3 be a parametric surface, and let u0 be a hyper-bolic point. Then there is neighborhood U0 of u0 and a diffeomorphism φ : U0 → U0

such that X = X φ is parametrized by asymptotic lines.

6. More Examples

A surface of revolution is a parametric surface of the form:

X(u, v) =(f(u) cos(v), f(u) sin(v), g(u)

),

where(f(t), g(t)

)is a regular curve, called the generator , which satisfies f(t) 6= 0

. Without loss of generality, we may assume that f(t) > 0. The curves

γv(t) =(f(t) cos(v), f(t) sin(v), g(t)

), v fixed.

6. MORE EXAMPLES 31

are called meridians and the curves

βu(t) =(f(u) cos(t), f(u) sin(t), g(u)

), u fixed.

are called parallels. Note that every meridian is a planar curve congruent to thegenerator and is furthermore also a normal section, and every parallel is a circleof radius f(u). It is not difficult to see that parallels and meridians are lines ofcurvature. Indeed, let γv be a meridian, then choosing as in the paragraph followingTheorem 2.5 the correct orientation in the plane of γv, its spherical image under theGauss map is σv = N γv = e2, and by the Frenet equations, σv = −κe1 = −κγv.Thus, using Proposition 2.9 and the comment immediately following it, we see thatγv is a line of curvature with associated principal curvature κ. Since the parallelsβu are perpendicular to the meridians γv, it follows immediately that they are alsolines of curvature. We derive this also follows from Proposition 2.9 and furthermoreobtain the associated principal curvature. A straightforward computation gives thatthe spherical image of βu under the Gauss map is:

τu = N βu = cβu +B

where B ∈ R3 and c ∈ R are constants. Thus, τu = cβu and βu is a line of curvaturewith associated principal curvature c.

The plane, the sphere, the cylinder, and the hyperboloid are all surfaces ofrevolution. We discuss one more example.

The catenoid is the parametric surface of revolution obtained from the gener-ating curve

(cosh(t), t

):

X(u, v) =(cosh(u) cos(v), cosh(u) sin(v), u

).

The normal N is easily calculated:

N(u, v) =(− cos(v)cosh(u)

,− sin(v)cosh(u)

,sinh(u)cosh(u)

)If γv(t) is a meridian, then σv(t) = N(t, v) is its spherical image under the Gaussmap, and differentiating with respect to t, we get the principal curvature associatedwith meridians: κ(u, v) = −1/ cosh(u). Similarly, the principal curvature associatedwith parallels is: 1/ cosh(u). Thus, we conclude that

H = 0, K = − 1cosh(u)2

.

Definition 2.12. A parametric surface X is minimal if it has vanishing meancurvature H = 0.

For example, the catenoid is a minimal surface. The justification for the ter-minology will be given in the next section. The following proposition is immediatefrom (2.11).

Proposition 2.14. Let X be a minimal surface. Then X has non-positiveGauss curvature K 6 0, and K(u) = 0 if and only if u is a planar point.

We will set out to construct a large class of minimal surfaces. We will use theWeierstrass Representation.

Definition 2.13. A parametric surface X is conformal if the first fundamentalform satisfies g11 = g22 and g12 = 0. A parametric surface X is harmonic if∆X = X11 +X22 = 0.


Proposition 2.15. Let X : U → R3 be a parametric surface which is bothconformal and harmonic. Then X is a minimal surface.

Proof. We can write the first fundamental form(gij), its inverse

(gij), and

the second fundamental form(kij)

as:

(gij)

=

(λ 0

0 λ

),(gij)

=

(λ−1 0

0 λ−1

),(kij)

=

(X11 ·N X12 ·NX12 ·N X22 ·N

).

Thus, the mean curvature vanishes:

H = gijkij = λ−1(X11 +X22

)·N = 0.

In order to construct parametric surfaces which are both conformal and har-monic, we will use complex analysis in the domain U . Let ζ = u+iv where i denotes√−1, and let f(ζ) and h(ζ) be two complex analytic functions on U . Define

F1 = f2 − h2, F2 = i(f2 + h2

), F3 = 2fh.

We have: (F1

)2 +(F2

)2 +(F3

)2 = 0.

If we write Fj = ξj + iηj , then this can be written as:

3∑j=1

[(ξj)2 − (ηj)2]2 + 2i

3∑j=1

ξjηj = 0.

Now, in any simply connected subset of U , we can always find analytic functionsGj = xj + iyj satisfying

(Gj)ζ

= Fj . We let X = (x1, x2, x3). Then X is conformaland harmonic. Indeed, xj being the real parts of complex analytic functions, areharmonic, and hence X is harmonic. Furthermore, we have

(xj)u

= ξj , and by theCauchy-Riemann equations

(xj)v

= −(yj)u

= −ηj . Thus, we see that

Xu ·Xu −Xv ·Xv =3∑j=1

[(ξj)2 − (ηj)2]2 = 0,

and

Xu ·Xv = −3∑j=1

ξjηj = 0,

and hence, X is conformal.3 Since X is real analytic, the zeroes of det(Xi ·Xj

)are

isolated. Removing the set Z of those zeroes from U , we get that X : U \ Z → R3

is a harmonic and conformal parametric surface, hence X is a minimal surface4.If we carry out this procedure starting with the complex analytic functions

f(ζ) = 1 and h(ζ) = 1/ζ, then X is another parametrization of the catenoid,cf. 2.6.

3Of course, Y = (y1, y2, y3) is also conformal, cf. 2.5.4X is also said to be a branched minimal surface on U . The zeroes of det

`gij

´are called

branched points.

7. SURFACE AREA 33

7. Surface Area

In this section we will give interpretations of the Gauss curvature and the meancurvature. Both of these involve the concept of surface area. Before introducingthe definition, we first prove a proposition which will show that the definition isreparametrization invariant.

Proposition 2.16. Let X : U → R3 be a parametric surface with first funda-mental form

(gij), and V ⊂ U . Let X : U → R3 be a reparametrization of X, let

V = φ−1(V ), and let(gij)

be the coordinate representation of the first fundamentalform of X. Then, we have:

(2.15)∫V

√det(gij)du1 du2 =

∫V

√det(gij)du1 du2.

Proof. By (2.4) we have√det(gij)

=√

det(gij) ∣∣det

(φij)∣∣

where φij = ∂ui/∂uj . Thus, for any open subset V ⊂ U , and V = φ−1(V ), we have:∫V


∫V

√det(gij) ∣∣det

(φij)∣∣ du1 du2 =

∫V

√det(gij)du1 du2

Thus, the integral on the right-hand side of (2.15) is reparametrization invari-ant. This justifies the following definition.

Definition 2.14. Let X : U → R3 be a parametric surface and let(gij)

be itsfirst fundamental form. The surface area element of X is:

dA =√

det(gij)du1 du2.

If V ⊂ U is open then the surface area of X over V is:

(2.16) AX(V ) =∫V

dA =∫V

√det(gij)du1 du2

By Proposition 2.16, the surface area of X over V is reparametrization invari-ant, and we can thus speak of the surface area of X(V ).

Definition 2.15. Let X : U → R3 be a parametric surface, and let V ⊂ U beopen. The total curvature of X over V is:

KX(V ) =∫V

K dA.

It is easy to show, as in the proof of Proposition 2.16 that the total curvatureof X over V is invariant under reparametrization. We now introduce the signedsurface area, a variant of Definition 2.14 which allows for smooth maps Y into asurface X, with Jacobian dY not necessarily everywhere non-singular, and whichalso accounts for multiplicity.

Definition 2.16. Let X : U → R3 be a parametric surface, and let Y : U →X(U) be a smooth map. Define σ(u) to be 1, −1, or 0, according to whether thepair Y1(u), Y2(u) has the same orientation as the pair X1(u), X2(u), the opposite


orientation, or is linearly dependent, and let hij = Yi · Yj . If V ⊂ U is open thenthe signed surface area of Y over V is:

AY (V ) =∫V

σ√

det(hij)du1 du2

For a regular parametric surface, this definition reduces to Definition 2.14.Next, we prove that the total curvature of a surface X over an open set U is thearea of the image of U under the Gauss map counted with multiplicity.

Theorem 2.17. Let X : U → R3 be a parametric surface, and let V ⊂ U beopen. Let N : U → S2 be the Gauss map of X, then:

KX(V ) = AN (V ).

Proof. We first derive a formula which is of independent interest:

(2.17) Ni = −kjiXj

To verify this formula, it suffices to check that the inner product of both sides withthe three linearly independent vectors X1, X2, N are equal. Since N · N = 1, wehave N ·Ni = 0 = −kjiXj ·N = 0, and −kjiXj ·Xl = −kji gjl = −kij = −Ni ·Xk.In particular, if hij = Ni ·Nj , then we find:

hij =(kmi Xm

)·(knjXn

)= kmi k

nj gmn = kimkjng

mn.

In particular,

det(hij)

=

(det(kij))2

det(gij)

Note also that Equation (2.17) implies that the pair N1, N2 has the same orientationas X1, X2 if and only if det

(kij)> 0. Furthermore, since N(u) is also the outward

normal to the unit sphere at N(u), and since X1, X2, N is positively oriented inR3, it follows that X1(u), X2(u) also gives the positive orientation on the tangentspace to the S2 at N(u). Thus, we deduce that sign det

(kij)

= σ. Consequently, inview of Equation (2.10), we obtain:

σ√

det(hij)

=sign det

(kij) ∣∣det

(hij)∣∣√

det(gij) = K

√det(gij)

The proposition follows by integrating over V .

We now turn to an interpretation of the mean curvature. Let X : U → be aparametric surface. A variation of X is a smooth family F (u; t) : U × (−ε, ε)→ R3

such that F (u; 0) = X. Note that since dF (u; 0) is non-singular, the same is trueof dF (u; t0) for any fixed u0, perhaps after shrinking the interval (−ε, ε). Thus, allthe maps F (u; t0) for t0 close enough to 0 are parametric surfaces. The generatorof the variation is the vector field dF/dt(u; 0). The variation is compactly supportedif F (u; t) = X(u) outside a compact subset of U . The smallest such compact set iscalled the support of the variation F . Clearly, if a variation is compactly supported,then the support of its generator is compact in U . We say that a variation istangential if the generator is tangential; we say it is normal if the generator isnormal. Suppose now that the closure V is compact in U . We consider the areaAF (V ) of F (u; t) as a function of t. The next proposition shows that the derivativeof this function depends only on the generator, and in fact is a linear functional inthe generator.

7. SURFACE AREA 35

Proposition 2.18. Let X : U → R3 be a parametric surface, and let F (u; t) bea variation with generator Y . Then:

(2.18)dAF (V )

dt

∣∣∣∣t=0

=∫V

gij Xi · Yj dA

We first need the following lemma from linear algebra. We denote by Sn×n

the space of n× n symmetric matrices, and by Sn×n+ the subset of those which arepositive definite.

Lemma 2.19. Let B : (a, b) → Sn×n+ be continuously differentiable. Then wehave:

(2.19)(log detB

)′ = tr(B−1B′

).

Proof. First note that (2.19) follows directly if we assume that B is diagonal.Next, suppose that B is symmetric with distinct eigenvalues. Then there is acontinuously differentiable orthogonal matrix Q such that B = Q−1DQ, where Dis diagonal. Note that dQ−1/dt = −Q−1(dQ/dt)Q, hence:

B−1B′ = −Q−1D−1Q′Q−1DQ+Q−1D−1D′Q+Q−1Q′,

and in view of tr(AB) = tr(BA), we obtain:

tr(B−1B′

)= tr

(D−1D′

).

We also have that detB = detD. Thus taking into the account that (2.19) holdsfor for D: (

log detB)′ =

(log detD

)′ = tr(D−1D′

)= tr

(B−1B′

).

In order to prove the general case, it is more convenient to look at the equivalentidentity:

(2.20)(detB

)′ = tr((detB)B−1B′

).

Note that by Kramer’s rule, the matrix (detB)B−1 is the matrix of co-factorsof B, hence its components being determinants of minors of B, are multivariatepolynomials in the components of B. Thus, both sides of the identity (2.20) arelinear polynomials

p(B′;B) =n∑

i,j=1

pij(B)b′ij , q(B′;B) =n∑

i,j=1

qij(B)b′ij ,

in the components b′ij of B′, whose coefficients pij(B) and qij(B) are themselvesmultivariate polynomials in the components bij of B. Since the set of matriceswith distinct eigenvalues is an open set U ⊂ Sn×n+ , we have already proved thatp(B′;B) = q(B′;B) holds for all values of B′, and all B ∈ U . For each suchB ∈ U the equality p(B′;B) = q(B′;B) for all B′ implies that pij(B) = qij(B) fori, j = 1, . . . , n. Since this holds for all B in an open set, we conclude that pij = qij ,and hence p = q.

We remark that the more general identity (2.20) in fact holds, as easily shown,for all square matrices B. An immediate consequence of the proposition is that:

(2.21)(√

detB)′

=12

tr(B−1B′

)√detB,

for any continuously differentiable family of symmetric positive definite matricesB. We are now ready to prove the proposition.


Proof of Proposition 2.18. Differentiating the area (2.16) under the inte-gral sign, and using (2.21), we get:

dAF (V )dt

=12

∫V

gijdgijdt


12

∫V

gijdgijdt

dA.

Since Y is smooth, we have at t = 0 that dFi/dt = (dF/dt)i = Yi, and thus

gijdgijdt

= gij(Yi ·Xj +Xi · Yj

)= 2gijXi · Yj .

This completes the proof of the proposition.

Since the variation of the area dAF (V )/dt is a linear functional in the generatordF/dt of the variation, it is possible to decompose any variation into tangential andnormal components. We begin by showing that the area doesn’t change under atangential variation. This is simply the infinitesimal version of Proposition (2.16).

Proposition 2.20. Let X : U → R3 be a parametric surface, and let F (u; t) bea compactly supported tangential variation. If V ⊂ U is open with V compact inU , and the support of F contained in V , then dAF (V )/dt = 0.

Proof. Let Y be the generator of F (u; t). We will show that there is a smoothfamily of diffeomorphisms φ : U × (−δ, δ)→ U such that Y is also the generator ofthe variation G = X φ. This proves the proposition since Proposition 2.16 givesthat AG(U) is constant. Since Y is tangential, we can write Y = yiXi. Considerthe initial value problem:

dvi

dt= yi(v), vi(0) = ui.

Since the yi’s are compactly supported, a solution v = v(u; t) exists for all t.Defining φ(u; t) = v(u; t), then an application of the inverse function theorem showsthat φ(u; t) is a diffeomorphism for t in some small interval (−δ, δ). Finally, we seethat:

dX φ

dt= Xi

dvi

dt= Xiy

i = Y.

Our next theorem gives an interpretation of the mean curvature as a measureof surface area variation under normal perturbations.

Theorem 2.21. Let X : U → R3 be a parametric surface, and let F (u; t) be acompactly supported variation with generator Y . If V ⊂ U is open with V compactin U , and the support of F contained in V , then

(2.22)dAF (V )

dt= −2

∫V

(Y ·N)H dA.

Proof. By Propositions 2.18 and 2.20, it suffices to consider normal variationswith generator Y = fN . In that case, we find that Yj = fjN + fNj , so thatgij Xi · Yj = fgij Xi · Nj = −fkii = −2fH. The theorem follows by substitutinginto (2.18).

Definition 2.17. A parametric surface X is area minimizing if AX(U) 6AX(U) for any parametric surface X such that X = X on the boundary of U .A parametric surface X : U → R3 is locally area minimizing if for any compactlysupported variation F (u; t), the area AF (U) has a local minimum at t = 0.

8. BERNSTEIN’S THEOREM 37

Clearly, an area-minimizing surface is locally area-minimizing. The followingtheorem is an immediate corollary of Theorem 2.21.

Theorem 2.22. A locally area minimizing surface is a minimal surface.

Note that in general a minimal surface is only a stationary point of the areafunctional.

8. Bernstein’s Theorem

In this section, we prove Bernstein’s Theorem: A minimal surface which is agraph over an entire plane must itself be a plane. We say that a surface X is agraph over a plane Y : R2 → R3, where Y is linear, if there is a function f : R2 → Rsuch that X = Y + fN where N is the unit normal of Y .

Theorem 2.23 (Bernstein’s Theorem). Let X be a minimal surface which is agraph over an entire plane. Then X is a plane.

We may without loss of generality assume that X is a graph over the planeY (u, v) = (u, v, 0), i.e. X(u, v) =

(u, v, f(u, v)

)as in example 2.1. It is then

straightforward to check that X is a minimal surface if and only if f satisfies thenon-parametric minimal surface equation:

(2.23) (1 + q2)pu − 2pqpv + (1 + p2)qv = 0,

where we have used the classical notation: p = fu, q = fv. We say that a solutionof a partial differential equation defined on the whole (u, v)-plane is entire. Thus,to prove Bernstein’s Theorem, it suffices to prove that any entire solution of (2.23)is linear.

Proposition 2.24. Let f be an entire solution of (2.23). Then f is a linearfunction.

By Exercise 2.7, if f satisfies (2.23), then p and q satisfy the following equations:

∂

∂u

(1 + q2√

1 + p2 + q2

)=

∂

∂v

(pq√

1 + p2 + q2

),(2.24)

∂

∂u

(pq√

1 + p2 + q2

)=

∂

∂v

(1 + p2√

1 + p2 + q2

).(2.25)

Since the entire plane is simply connected, Equation (2.25) implies that there existsa function ξ satisfying:

ξu =1 + p2√

1 + p2 + q2, ξv =

pq√1 + p2 + q2

,

and Equation (2.24) implies that there exists a function η satisfying:

ηu =pq√

1 + p2 + q2, ηv =

1 + q2√1 + p2 + q2

.

Furthermore, ξv = ηu, hence there is a function h so that hu = ξ, hv = η. TheHessian of the function h is:(

hij)

=

(huu huv

hvu hvv

)=

(ξu ξv

ηu ηv

),


hence h satisfies the Monge-Ampere equation:

(2.26) det(hij)

= 1.

In addition, h11 > 0, thus(hij)

is positive definite, and we say that h is convex .Proposition 2.24 now follows from the following result due to Nitsche.

Proposition 2.25. Let h ∈ C2(R2) be an entire convex solution of the Monge-Ampere Equation (2.26). Then h is a quadratic function.

Proof. The proof uses the following transformation introduced by H. Lewy:

ϕ : (u, v) 7→ (ξ, η) = (u+ p, v + q)

where p = hu, and q = hv. Clearly, ϕ is continuously differentiable, and its Jacobianis:

dϕ =

(1 + r s

s 1 + t

),

where r = huu, s = huv, and t = hvv. Since det(dϕ)

= 2 + r + t > 0, it followsfrom the inverse function theorem that ϕ is a local diffeomorphism, i.e., each pointhas a neighborhood on which ϕ is a diffeomorphism. In particular, ϕ is open.

In view of the convexity of the function h, we have, according to Exercise 2.8:(u2 − u1

)(ξ2 − ξ1

)+(v2 − v1

)(η2 − η1

)=(u2 − u1

)2 +(v2 − v1

)2 +(u2 − u1

)(p2 − p1

)+(v2 − v1

)(q2 − q1

)>(u2 − u1

)2 +(v2 − v1

)2,

and therefore: (u2 − u1

)2 +(v2 − v1

)26(ξ2 − ξ1

)2 +(η2 − η1

)2,

i.e., ϕ is an expanding map. This implies immediately that ϕ is one-to-one. Ac-cording to Exercise 2.9, ϕ is also onto. Thus, ϕ has an inverse (u, v) = ϕ−1(ξ, η)which is also a diffeomorphism. Consider now the function

f(ξ + iη) = u− p− i(v − q) = 2u− ξ + i(−2v + η),

where i =√−1. In view of

dϕ−1 =

(uξ uη

vξ vη

)=

12 + r + t

(1 + t −s−s 1 + r

),

it is straightforward to check that f satisfies the Cauchy-Riemann equations, andconsequently f is analytic. In fact, f is an entire functions and so is f ′. Further-more,

f ′(σ) =(t− r) + 2is

2 + r + t, |f ′(σ)|2 = 1− 4

2 + r + t< 1,

and Liouville’s Theorem gives that f ′ is constant. Finally, the relations:

r =|1− f ′|2

1− |f ′|2, s =

−i(f ′ − f ′

)1− |f ′|2

, t =|1 + f ′|2

1− |f ′|2,

show that r, s, t are constants.

9. THEOREMA EGREGIUM 39

9. Theorema Egregium

In this section, we prove that the Gauss curvature can be computed in termsof the first fundamental form and its derivatives. We then prove the FundamentalTheorem for surfaces in R3, analogous to Theorem 1.2 for curves, which states that aparametric surface is uniquely determined by its first and second fundamental form.Partial derivatives with respect to ui will be denoted by a subscript i following acomma, unless there is no ambiguity in which case the comma may be omitted.

Proposition 2.26. Let X : U → R3 be a parametric surface. Then the follow-ing equations hold:

Xij = ΓmijXm + kijN,(2.27)

where,

Γmij =12gmn

(gni,j + gnj,i − gij,n

),(2.28)

and(gij)

and(kij)

are the coordinate representations of its first and second fun-damental form.

Proof. Clearly, Xij can be expanded in the basis X1, X2, N of R3. We al-ready saw in Equation (2.7), that the component of Xij along N is kij , henceEquation (2.27) holds with the coefficients Γmij given by

Xij ·Xm = Γnijgmn.

In order to derive (2.28), we differentiate gij = Xi · Xj , and substitute the aboveequation to obtain:

(2.29) gij,m = Γnimgnj + Γnjmgni.

Now, permute cyclically the indices i, j,m, add the first two equations and subtractthe last one:

gij,m + gmi,j − gjm,i = 2Γnjmgni.

Multiplying by gil and dividing by 2 yields (2.28).

The coefficients Γmij are called the Christoffel symbols of the second kind .5 Itis important to note that the Christofell symbols can be computed from the firstfundamental form and its first derivatives. Furthermore, they are not invariantunder reparametrization.

Theorem 2.27. Let X : U → R3 be a parametric surface. Then the followingequations hold:

Γmij,l − Γmil,j + ΓnijΓmnl − ΓnilΓ

mnj = gmn

(kijkln − kilkjn

),(2.30)

kij,l − kil,j + Γmijklm − Γmil kjm = 0.(2.31)

Proof. If we differentiate (2.27), we get:

Xijl =(ΓmijXm

)l+(kijN

)l

= Γmij,lXm + ΓmijXml + kij,lN + kijNl.

5The Christoffel symbols of the first kind are: Γijm =1

2

`gim,j + gjm,i − gij,m

´.


Substituting Xml from (2.27) and Nl from (2.17), and decomposing into tangentialand normal components, we obtain:

Xijl = AmijlXm +BijlN,

where:

Amijl = Γmij,l + ΓnijΓmnl − gmnkijkln,

Bijl = kij,l + Γmijklm.

Taking note of the fact that Xijl = Xilj , we now interchange j and l and subtractto obtain (2.30) and (2.31).

Equation (2.30) is called the Gauss Equation, and Equation (2.31) is calledthe Codazzi Equation. The Gauss Equation has the following corollary which hasbeen coined Theorema Egregium. It’s discovery marked the beginning of intrinsicgeometry, the geometry of the first fundamental form.

Corollary 2.28. Let X : U → R3 be a parametric surface. Then the Gausscurvature K of X can be computed in terms of only its first fundamental form

(gij)

and its derivatives up to second order:

K =12gij(Γmij,m − Γmim,j + ΓnijΓ

mnm − ΓnimΓmnj

),

where Γmij are the Christoffel symbols of the first kind.

Proof. Combine (2.30) and (2.11).

We now show, in a manner quite analogous to Theorem 1.2, that providedthey satisfy the Gauss-Codazzi Equations, the first and second fundamental formuniquely determine the parametric surface up to rigid motion.

Theorem 2.29 (Fundamental Theorem). Let U ⊂ R2 be open and simply-connected, let

(gij)

: U → S2×2+ and

(kij)

: U → S2×2 be smooth, and suppose thatthey satisfy the Gauss-Codazzi Equations (2.30)–(2.31). Then there is a parametricsurface X : U → R3 such that

(gij)

and(kij)

are its first and second fundamentalforms. Furthermore, X is unique up to rigid motion: if X is another parametricsurface with the same first and second fundamental forms, then there is a rigidmotion R of R3 such that X = R X.

Proof. We consider the following over-determined system of partial differen-tial equations for X1, X2, N :6

Xi,j = ΓmijXm + kijN,(2.32)

Ni = −kijgjmXm,(2.33)

where Γmij is defined in terms of(gij)

by (2.28). The integrability conditions forthis system are: (

ΓmijXm + kijN)l

=(ΓmilXm + kilN

)j

(2.34) (kijg

jmXm

)l

=(kljg

jmXm

)i.(2.35)

6Here Xi is not to be understood as the derivative of X with respect to ui until later in theproof.

EXERCISES 41

The proof of Theorem 2.27 also shows that the Gauss-Codazzi Equations (2.30)–(2.31) imply (2.34) if Xi and N satisfy (2.32) and (2.33). We now check that (2.31)also implies (2.35). First note that since Γmij is defined by (2.28), we have

Γmij gmn =12(gni,j + gnj,i − gij,n

).

Interchanging n and i and adding, we get (2.29). Now, differentiate (2.33), andtaking into account that gij,l = −giagab,lgbj , substitute (2.29) to get:

Ni,l = −kij,lgjmXm + kijgja(Γnalgnb + Γnblgna

)gbmXm

− kijgjm(ΓamlXa + kmlgjmN

)=(−kij,l + kinΓnjl

)gjmXm + kijkmlg

jmN.

Note that the last term is symmetric in i and l so that interchanging i and l, andsubtracting, we get:

Ni,l −Nl,i =(−kij,l + kil,j − Γnijkln + Γnilkjn

)gjmXm

which vanishes by (2.31). Thus, it follows that (2.35) is satisfied. We conclude thatgiven values for X1, X2, N at a point u0 ∈ U there is a unique solution of (2.32)–(2.33) in U . We can choose the initial values to that Xi ·Xj = gij , N ·Xi = 0, andN · N = 1 at u0. Using (2.32) and (2.33), it is straightforward to check that thefunctions hij = Xi ·Xj , pi = N ·Xi and q = N ·N , satisfy the differential equations:

hij,l = Γnilhnj + Γnjlhni + kilpj + kjlpi,

pi,j = −kjlglmhmi + Γmijpm + kijq,

qi = −2kijgjmpm.

However, the functions hij = gij , pi = 0 and q = 1 also satisfy these equations, aswell as the same initial conditions as hij = Xi ·Xj , pi = N ·Xi and q = N ·N at u0.Thus, by the uniqueness statement mentioned above, it follows that Xi ·Xj = gij ,N ·Xi = 0, and N · N = 1. Clearly, in view of (2.32) we have Xi,j = Xj,i, hencethere is a function X : U → R3 whose partial derivatives are Xi, cf. foonote 6. Since(gij)

is positive definite we have that X1, X2 are linearly independent, hence X is aparametric surface with first fundamental form

(gij). Furthermore, it is easy to see

that the unit normal of X is N , and Ni ·Xj = −N ·Xij = −kij , hence the secondfundamental form of X is kij . This completes the proof of the existence statement.

Assume now that X is another surface with the same first and second fun-damental forms. Since X and X have the same first fundamental form, it fol-lows that there is a rigid motion R(x) = Qx + y with Q ∈ SO(n; R) such thatR(X(u0)

)= X(u0), QXi(u0) = Xi(u0), QN(u0) = N(u0). Let X = R X. Since

the two triples (X1, X2, N) and (X1, X2, N) both satisfy the same partial differen-tial equations (2.32) and (2.33), it follows follows that they are equal everywhere,and consequently X = X = R X.

Exercises

Exercise 2.1. Let X : U → R3 and X : U → R3 be two parametric surfaces.The angle θ between them is the angle between their unit normals: cos θ = N · N .Let γ be a regular curve which lies on both X and X, and suppose that the angle


between X and X is constant along γ. Show that γ is a line of curvature of X ifand only if it is a line of curvature of X.

Exercise 2.2. Let X : U → R3 be a parametric surface, and let γ be anasymptotic line with curvature κ 6= 0, and torsion τ . Show that |τ | =

√−K

Exercise 2.3. Denote by SO(n) the set of orthogonal n × n matrices, andby D(n) the set of n × n diagonal matrices. Let A : (a, b) → Sn×n be a Ck func-tion, and suppose that A maps into the set of matrices with distinct eigenvalues.Show that there exist Ck functions Q : (a, b) → SO(n) and Λ: (a, b) → D(n) suchthat Q−1AQ = Λ. Conclude the matrix function A has Ck eigenvector fieldse1, . . . , en : (a, b)→ Rn, Aej = λjej . Give a counter-example to show that this lastconclusion can fail the eigenvalues of A are allowed to coincide.

Exercise 2.4. Let Mn×n be the space of all n×n matrices, and let B : (a, b)→Mn×n be continuously differentiable. Prove that:(

detB)′ = tr(B∗B′),

where B∗ is the matrix of co-factors of B.

Exercise 2.5. Two harmonic surfaces X,Y : U → R3 are called conjugate, ifthey satisfy the Cauchy-Riemann Equations:

Xu = Yv, Xv = −Yu,where (u, v) denote the coordinates in U . Prove that if X is conformal then Y isalso conformal. Let X and Y be conformal conjugate minimal surfaces. Prove thatfor any t:

Z = X cos t+ Y sin tis also a minimal surface. Show that all the surfaces Z above have the same firstfundamental form.

Exercise 2.6. Prove that setting f(ζ) = 1, g(ζ) = 1/ζ in the Weierstrassrepresentation, we get the catenoid. Find the conjugate harmonic surface of thecatenoid.

Exercise 2.7. Let U ⊂ R2, let f : U → R be a smooth function, and letX : U → R3 be given by (u, v, f(u, v)), where (u, v) denote the variables in U .Show that X is a minimal surface if and only if it satisfies the non-parametricminimal surface equation:

(1 + q2)pu − 2pqpv + (1 + p2)qv = 0,

where we have used the classical notation: p = fu, q = fv. Show that if f satisfiesthe equation above then the following equations are also satisfied:

∂

∂u

(1 + q2√

1 + p2 + q2

)=

∂

∂v

(pq√

1 + p2 + q2

),

∂

∂u

(pq√

1 + p2 + q2

)=

∂

∂v

(1 + p2√

1 + p2 + q2

).

EXERCISES 43

Exercise 2.8. Let f ∈ C2(U) be a convex function defined on a convex openset U , and let ∇f = (p, q) : U → R2 denote the gradient of f . Prove that for anyu1, u2 ∈ U the following inequality holds:(

u2 − u1

)·(∇f(u2)−∇f(u1)

)> 0.

Exercise 2.9. Let U ⊂ Rn be open. A map ϕ : U → Rn is expanding if|x− y| 6 |ϕ(x)− ϕ(y)| for all x, y ∈ U . Let ϕ : U → Rn be an open expanding map.Show that the image of the ball BR(x0) of radius R centered at x0 ∈ U containsthe disk BR

(ϕ(x0)

)of radius R centered at ϕ(x0). Conclude that if U = Rn, then

ϕ is onto Rn.

CHAPTER 3

Local Intrinsic Geometry of Surfaces

In this chapter, we change our point of view, and study intrinsic geometry, inwhich the starting point is the first fundamental form. Thus, given a parametricsurface, we will ignore all information which cannot be recovered from the firstfundamental form and its derivatives only. In particular, we will ignore the Gaussmap and the second fundamental form. Thanks to Gauss’ Theorema Egregium, wewill still be able to take the Gauss curvature into account.

1. Riemannian Surfaces

Definition 3.1. Let U ⊂ R2 be open. A Riemannian metric on U is a smoothfunction g : U → S2×2

+ . A Riemannian surface patch is an open set U equippedwith a Riemannian metric.

The tangent space of U at u ∈ U is R2. The Riemannian metric g defines aninner-product on each tangent space by:

g(Y, Z) = gijyizj ,

where yi and zj are the components of Y and Z with respect to the standardbasis of R2. We will write |Y |2g = g(Y, Y ), and omit the subscript g when it is notambiguous.

Two Riemannian surface patches (U, g) and (U , g) are isometric if there is adiffeomorphism φ : U → U such that

(3.1) gij = glmφliφmj ,

where φli = ∂ul/∂ui. In fact, Equation (3.1) reads:

dφ∗g = g,

where dφ∗g is the pull-back of g by the Jacobian of φ at u. We then say that φ isan isometry between (U, g) and (U , g). As before, we denote by gij the inverse ofthe matrix gij .

As in Chapter 2, we also denote the Riemannian metric:

ds2 = gij dui duj ,

and at times refer to it as a line element. The arclength of a curve γ : [a, b]→ U isthen given by:

Lγ =∫ b

a

√gij γi γj dt.

Note that the arclength is simply the integral of√g(γ, γ).

45

46 3. LOCAL INTRINSIC GEOMETRY OF SURFACES

Example 3.1. Let U ⊂ R2 be open, and let(δij)

be the identity matrix,then (U, δ) is a Riemannian surface. The Riemannian metric δ will be called theEuclidean metric.

Example 3.2. Let X : U → R3 be a parametric surface, and let g be thecoordinate representation of its first fundamental form, then (U, g) is a Riemanniansurface patch. We say that the metric g is induced by the parametric surface X. IfX = X φ : U → R3 is a reparametrization ofX and g the coordinate representationof its first fundamental form, then (U , g) is isometric to (U, g).

Example 3.3 (The Poincare Disk). Let D = (u, v) : u2 + v2 < 1 be the unitdisk in R2, and let

gij =4

(1− r2)2δij

where r =√u2 + v2 is the Euclidean distance to the origin. We can write this line

element also as

(3.2) ds2 = 4du2 + dv2

(1− u2 − v2)2.

The Riemannian surface (D, g) is called the Poincare Disk. Let U = (x, y) : y > 0be the upper half-plane, and let

hij =1y2δij .

Then it is not difficult to see that(D, gij

)and

(U, hij

)are isometric with the

isometry given by:

φ : (u, v) 7→ (x, y) =(

2v(1 + u)2 + v2

,1− u2 − v2

(1 + u2) + v2

).

In fact, a good bookkeeping technique to check this type of identity is to computethe differentials :

dx = −4v(1 + u)(

(1 + u)2 + v2)2 du+ 2

(1 + u)2 − v2((1 + u)2 + v2

)2 dvdy = −2

(1 + u)2 − v2((1 + u)2 + v2

)2 du+ 4v(1 + u)(

(1 + u)2 + v2)2 dv,

substitute intodx2 + dy2

y2,

and then simplify using du dv = dv du to obtain (3.2). It is not difficult to see thatthis is equivalent to checking (3.1).

Definition 3.2. Let (U, g) be a Riemannian surface. The Christoffel symbolsof the second kind of g are defined by:

(3.3) Γmij =12gmn

(gni,j + gnj,i − gij,n

).

The Gauss curvature of g is defined by:

(3.4) K =12gij(Γmij,m − Γmim,j + ΓnijΓ

mnm − ΓnimΓmnj

).

2. LIE DERIVATIVE 47

If (U, g) is induced by the parametric surface X : U → R3, then these definitionsagree with those of Section 9.

2. Lie Derivative

In this section, we study the Lie derivative. We denote the standard basis onR2 by ∂1, ∂2. Let f be a smooth function on U , and let Y = yi∂i ∈ TuU be a vectorat u ∈ U . The directional derivative of f along Y is:

(3.5) ∂Y f = yi∂if = yifi.

Since yi = ∂Y ui where (u1, u2) are the coordinates on U , we see that Y = Z

follows from ∂Y = ∂Z as operators. The next proposition shows that the directionalderivative of a function is reparametrization invariant.

Proposition 3.1. Let φ : U → U be a diffeomorphism, and let Y be a vectorat u ∈ U . Then for any smooth function f on U , we have:(

∂dφ(Y )f)φ = ∂Y (f φ).

Proof. Denoting the coordinates on U by uj and the coordinates on U by ui,we let φji = ∂uj/∂ui, and we find, by the chain rule:

∂Y (f φ) = yi∂i(f φ) = yi(∂jf)φji =(∂dφ(Y )f

)φ.

We define the commutator of two tangent vector fields Y = yi∂i and Z = zi∂i,as in Section (3), Equation (2.5):

(3.6) [Y,Z] =(yi∂iz

j − zi∂iyj)∂j .

Note that

(3.7) ∂[Y,Z]f = ∂Y ∂Zf − ∂Z∂Y f.This observation together with Proposition 3.1 are now used to show that thecommutator is reparametrization invariant.

Proposition 3.2. Let Y and Z be vector fields on U , and let φ : U → U be adiffeomorphism, then

dφ([Y , Z]

)=[dφ(Y ), dφ(Z)

].

Proof. For any smooth function f on U , we have:

(3.8) ∂dφ([Y ,Z]

)f = ∂[Y ,Z](f φ) = ∂Y ∂Z(f φ)− ∂Z∂Y (f φ)

= ∂Y(∂dφ(Z)f

)φ− ∂Z

(∂dφ(Y )f

)φ = ∂dφ(Y )∂dφ(Z)f − ∂dφ(Z)∂dφ(Y )f

= ∂[dφ(Y ),dφ(Z)

]f,and the proposition follows.

We note for future reference that in the proofs of propositions 3.1 and 3.2, onlythe smoothness of the map φ is used, and not the fact that it is a diffeomorphism.

The operator Z 7→ LY Z = [Y,Z], also called the Lie derivative, is a differentialoperator, in the sense that it is linear and satisfies a Leibniz identity: LY (fZ) =(∂Y f)Z + fLY Z. However, LY Z depends on the values of Y in a neighborhood ofa point as can be seen from the fact that it is not linear over functions in Y , but


rather satisfies LfY Z = fLY Z − (∂Zf)Y . Hence the Lie derivative cannot be usedas an intrinsic directional derivative of a vector field Z, which should only dependon the direction vector Y at a single point1.

3. Covariant Differentiation

Definition 3.3. Let (U, g) be a Riemannian metric, and let Z be a vector fieldon U . The covariant derivative of Z along ∂i is:

(3.9) ∇iZ =(∂iz

j + Γjikzk)∂j .

Let Y ∈ TuU , the covariant derivative of Z along Y is:

∇Y Z = yiZ;i.

We write the components of ∇iZ as:

(3.10) zj ;i = zj ,i + Γjikzk,

so that ∇Y Z = yizj ;i∂j . Furthermore, note that

(3.11) ∇i∂j = Γkij∂k.

Our first task is to show that covariant differentiation is reparametrizationinvariant. However, since the metric g was used in the definition of the covari-ant derivative, it stands to reason that it would be invariant only under thosereparametrization which preserve the metric, i.e., under isometries.

Proposition 3.3. Let φ : (U , g) → (U, g) be an isometry. Let Y ∈ TuU , andlet Z be a vector field on U . Then

(3.12) dφ(∇Y Z) = ∇dφ(Y )dφ(Z).

Proof. This proof, although tedious, is quite straightforward, and is relegatedto the exercises.

Note that on the left hand-side of (3.12), the covariant derivative ∇ is thatobtained from the metric g.

Our next observation, which follows almost immediately from (2.27), givesan interpretation of the covariant derivative when the metric g is induced by aparametric surface X.

Proposition 3.4. Let the Riemannian metric g be induced by the parametricsurface X. Then the image under dX of the covariant derivative dX(∇iZ) is theprojection of ∂iZ onto the tangent space.

Proof. Note that dX(∂i) = Xi. Thus, if Z = zj∂j then we find:

dX(∇iZ) = zj ;iXj = zj ,iXj + ΓjikzkXj = ∂i

(zjXj

)− kijzjN,

which proves the proposition.

We now show that covariant differentiation is in addition well-adapted to themetric g.

1Indeed ∂Y Z as defined in Chapter 2 does depend only on the value of Y at a single pointand satisfies ∂fY Z = f∂Y Z.

3. COVARIANT DIFFERENTIATION 49

Proposition 3.5. Let (U, g) be a Riemannian surface, and let Y and Z bevector fields on U . Then, we have

(3.13) ∂ig(Y, Z) = g(∇iY,Z) + g(∇iY,Z).

Proof. We first note that, as in the proof of Theorem 2.29, the definition ofthe Christoffel symbols (3.3) implies (2.29):

(3.14) gij,l = Γkilgkj + Γkjlgki.

Now, setting Y = yi∂i and Z = zi∂i, we compute:

∂ig(Y, Z) = ∂igjkyjzk = Γmjigkmy

jzk + Γmkigmjyjzk + gjky

j,iz

k + gjkyjzk,i

= gjk(yj ,i + Γjmiym)zk + gjky

j(zk,i + Γkmizm) = g(Y;i, Z) + g(Y, Z;i).

This completes the proof of (3.13) and of the proposition.

Definition 3.4. Let Y = yi∂i be a vector field on the Riemannian surface(U, g). Its divergence is the function:

div Y = ∇iyi = ∂iyi + Γiijy

j .

Note that:

Γiij =12gim(gmi,j + gmj,i − gij,m) =

12gimgim,j = ∂j log

√det g.

Thus, we see that:

(3.15) div Y =1√

det g∂i

(√det g yi

)Observe that this implies∫

U

div Y dA =∫U

∂i

(√det g yi

)du1 du2.

Thus, Green’s Theorem in the plane implies the following proposition.

Proposition 3.6. Let Y be a compactly supported vector field on the Riemann-ian surface (U, g). Then, we have:∫

U

div Y dA = 0.

Definition 3.5. If f : U → R is a smooth function on the Riemannian surface(U, g), its gradient ∇f is the unique vector field which satisfies g(∇f, Y ) = ∂Y f .The Laplacian of f if the divergence of the gradient of f :

∆f = div∇f.

It is easy to see that ∇f = gijfj∂j , hence

(3.16) ∆f =1√

det g∂i

(gij√

det g fj).

Thus, in view of Proposition 3.6, if f is compactly supported, we have:∫U

∆f dA = 0.


4. Geodesics

Definition 3.6. Let (U, g) be a Riemannian surface, and let γ : I → U bea curve. A vector field along γ is a smooth function Y : I → R2. The covariantderivative of Y = yi∂i along γ is the vector field:

∇γY =(yi + Γijky

j γk)∂i.

Note that if Z is any extension of Y , i.e., a any vector field defined on aneighborhood V of the image γ(I) of γ in U , then we have:

∇γY = ∇γZ = γiZ;i.

Thus, any result proved concerning the usual covariant differentiation, in particularProposition 3.5 holds also for the covariant differentiation along a curve.

Definition 3.7. A vector field Y along a curve γ is said to be parallel alongγ if ∇γY = 0.

Note that if Y and Z are parallel along γ, then g(Y, Z) is constant. This followsfrom Proposition 3.5:

∂γg(Y, Z) = g(∇γY,Z) + g(Y,∇γZ) = 0.

Proposition 3.7. Let γ : [a, b] → U be a curve into the Riemannian surface(U, g), let u0 ∈ U , and let Y0 ∈ Tu0U . Then there is a unique vector field Y alongγ which is parallel along γ and satisfies Y (a) = Y0.

Proof. The condition that Y is parallel along γ is a pair of linear first-orderordinary differential equations:

yi = −Γijk(γ)γjyj .

Given initial conditions yi(a) = yi0, the existence and uniqueness of a solution on[a, b] follows from the theory of ordinary differential equations.

The proposition together with the comment preceding it shows that paralleltranslation along a curve γ is an isometry between inner-product spaces Pγ : TaU →TbU .

Definition 3.8. A curve γ is a geodesic if its tangent γ is parallel along γ:

∇γ γ = 0.

If γ is a geodesic, then |γ| is constant and hence, every geodesic is parametrizedproportionally to arclength. In particular, if β = γ φ is a reparametrization of γ,then β is not a geodesic unless φ is a linear map.

Proposition 3.8. Let (U, g) be a Riemannian surface, let u0 ∈ U and let0 6= Y0 ∈ Tu0U . Then there is and ε > 0, and a unique geodesic γ : (−ε, ε) → U ,such that γ(0) = u0, and γ(0) = Y0.

Proof. We have:∇γ γ =

(γi + Γijkγ

j γk)∂i.

Thus, the condition that γ is a geodesic can written as a pair of non-linear second-order ordinary differential equations:

γi = −Γijk(γ(t))γj γk.

Given initial conditions γi(0) = ui0, γi(0) = yi0, there is a unique solution on definedon a small enough interval (−ε, ε).

4. GEODESICS 51

Definition 3.9. Let γ : [a, b] → U be a curve. We say that γ is length-minimizing, or L-minimizing, if:

Lγ 6 Lβfor all curves β in U such that β(a) = γ(a) and β(b) = γ(b).

Let γ : [a, b] → U be a curve. A variation of γ is a smooth family of curvesσ(t; s) : [a, b]×(−ε, ε)→ I such that σ(t; 0) = γ(t) for all t ∈ [a, b]. For convenience,we will denote derivatives with respect to t as usual by a dot, and derivativeswith respect to s by a prime. The generator of a variation σ is the vector fieldY (t) = σ′(t; 0) along γ. We say that σ is a fixed-endpoint variation, if σ(a; s) = γ(a),and σ(b; s) = γ(b) for all s ∈ (−ε, ε). Note that the generator of a fixed-endpointvariation vanishes at the end points. We say that a variation σ is normal if itsgenerator Y is perpendicular to γ: g(γ, Y ) = 0. A curve γ is locally L-minimizingif

Lσ(s) =∫ b

a

√g(σ, σ) dt

has a local minimum at s = 0 for all fixed-endpoint variations σ. Clearly, anL-minimizing curve is locally L-minimizing.

If γ is locally L-minimizing, then any reparametrization β = γ φ of γ is alsolocally L-minimizing. Indeed, if σ is any fixed-endpoint variation of β, then τ(t; s) =σ(φ−1(t); s) is a fixed-endpoint variation of γ, and since reparametrization leavesarclength invariant, we see that Lτ (s) = Lσ(s) which implies that Lσ also hasa local minimum at at s = 0. Thus, local minimizers of the functional L arenot necessarily parametrized proportionally to arclength. This helps clarify thefollowing comment: a locally length-minimizing curve is not necessarily a geodesic,but according to the next theorem that is only because it may not be parametrizedproportionally to arclength.

Theorem 3.9. A locally length-minimizing curve has a geodesic reparametriza-tion.

To prove this theorem, we introduce the energy functional:

Eγ =12

∫ b

a

g(γ, γ) dt

We may now speak of energy-minimizing and locally energy-minimizing curves.Our first lemma shows the advantage of using the energy rather than the arclengthfunctional: minimizers of E are parametrized proportionally to arclength.

Lemma 3.10. A locally energy-minimizing curve is a geodesic.

Proof. Suppose that γ is a locally energy-minimizing curve. We first notethat if Y is any vector field along γ which vanishes at the endpoints, then settingσ(t; s) = γ(t) + sY (t), we see that there is a fixed-endpoint variation of γ whosegenerator is Y . Since γ is locally energy-minimizing, we have:

E′σ(0) =∫ b

a

12(g(σ, σ)

)′∣∣s=0

dt = 0.

We now observe that:(σj)′∣∣

s=0=( ddtσj)′∣∣

s=0=

d

dt

(σj)′∣∣

s=0= yj .


where Y = yi∂i is the generator of the fixed-endpoint variation σ, and:(gij)′∣∣

s=0= gij,k

(σk)′∣∣

s=0= gij,k y

k.

Thus, we have:

12(g(σ, σ)

)′∣∣s=0

=12(gij σ

iσj)′∣∣

s=0=

12(gij)′∣∣

s=0σiσj + gij σ

i(σj)′∣∣

s=0

=12gij,k y

kγiγj + gij γiyj .

Since Y vanishes at the endpoints, we can substitute into E′σ(0), and integrate byparts the second term to get:

E′σ(0) = −∫ b

a

[d

dt

(gij γ

i)− 1

2gik,j γ

iγk]yj .

Since:d

dt

(gij γ

i)

= gij γi + gij,k γ

iγk = gij γi +

12(gij,k + gkj,i

)γiγk,

We now see that:

E′σ(0) = −∫ b

a

[gij γ

i +12(gmj,k + gkj,m − gmk,j

)γmγk

]yj dt = −

∫ b

a

g(∇γ γ, Y ) dt.

Since E′σ(0) = 0 for all vector fields Y along γ which vanish at the endpoints, weconclude that ∇γ γ = 0, and γ is a geodesic.

The Schwartz inequality implies the following inequality between the lengthand energy functional for a curve γ.

Lemma 3.11. For any curve γ, we have

L2γ 6 2Eγ (b− a),

with equality if and only if γ is parametrized proportionally to arclength.

Finally, the last lemma we state to prove Theorem 3.9, exhibits the relationshipbetween the L and E functionals.

Lemma 3.12. A locally energy-minimizing curve is locally length-minimizing.Furthermore, if γ is locally length-minimizing and β is a reparametrization of γ byarclength, then β is locally energy-minimizing.

Proof. Suppose that γ is locally energy-minimizing, and let σ be a fixed-endpoint variation of γ. For each s, let βs(t) : [a, b] → U be a reparametrizationof the curve t 7→ σ(t; s) proportionally to arclength. Let τ(t; s) = βs(t), then it isnot difficult to see, using say the theorem on continuous dependence on parametersfor ordinary differential equations, that τ is also smooth. By Lemma 3.10, γ is ageodesic, hence by Lemma 3.11, L2

γ = 2Eγ(b− a). It follows that:

L2σ(0) = L2

γ = 2Eγ(b− a) = 2Eτ (0)(b− a) 6 2Eτ (s)(b− a) = L2τ (s) = L2

σ(s).

Thus, γ is locally length-minimizing proving the first statement in the lemma.Now suppose that γ is locally length-minimizing, and let β be a reparametriza-

tion of γ by arclength. Then β is also locally length-minimizing, hence for anyfixed-endpoint variation σ of β, we have:

Eσ(0) = Eβ =L2β

2(b− a)6

L2σ(s)

2(b− a)6 Eσ(s).

5. THE RIEMANN CURVATURE TENSOR 53

Thus, β is locally energy-minimizing.

We note that the same lemma holds if we replace locally energy-minimizing byenergy-minimizing. The proof of Theorem 3.9 can now be easily completed withthe help of Lemmas 3.10 and 3.12.

Proof of Theorem 3.9. Let β be a reparametrization of γ by arclength. ByLemma 3.12, β is locally energy-minimizing. By Lemma 3.10, β is a geodesic.

5. The Riemann Curvature Tensor

Definition 3.10. Let X,Y, Z,W be vector fields on a Riemannian surface(U, g). The Riemann curvature tensor is given by:

R(W,Z,X, Y ) = g([∇X ,∇Y

]Z −∇[X,Y ]Z,W

).

We first prove that R is indeed a tensor , i.e., it is linear over functions. Clearly,R is linear in W , additive in each of the other three variables, and anti-symmetricin X and Y . Thus, it suffices to prove the following lemma.

Lemma 3.13. Let X,Y, Z,W be vector fields on a Riemannian surface (U, g).Then we have:

R(W,Z, fX, Y ) = R(W, fZ,X, Y ) = fR(W,Z,X, Y ).

Proof. We have:

∇fX∇Y Z−∇Y∇fXZ−∇[fX,Y ]Z = f∇X∇YX−∇Y(f∇XZ

)−∇f [X,Y ]−(∂Y f)XZ

= f∇X∇Y Z −(∂Y f

)∇XZ − f∇Y∇XZ − f∇[X,Y ]Z +

(∂Y f

)∇XZ

= f(∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z

).

The first identity follows by taking inner product with W . In order to prove thesecond identity, note that:

∇X∇Y fZ = ∇X((∂Y f

)Z)

+∇X(f∇Y Z

)=(∂X∂Y f

)Z +

(∂Y f

)(∇XZ

)+(∂Xf

)(∇Y Z

)+ f∇X∇Y Z.

Interchanging X and Y and subtracting we get:[∇X ,∇Y

]fZ =

(∂[X,Y ]f

)Z + f

[∇X ,∇Y

]Z.

On the other hand, we have also:

∇[X,Y ]fZ =(∂[X,Y ]f

)Z + f∇[X,Y ]Z.

Thus, we conclude:[∇X ,∇Y

]fZ −∇[X,Y ]fZ = f

([∇X ,∇Y

]Z −∇[X,Y ]Z

).

The second identity now follows by taking inner product with W .

LetRijkl = R(∂i, ∂j , ∂k, ∂l),

be the components of the Riemann tensor. The previous proposition shows that ifX = xi∂i, Y = yi∂i, Z = zi∂i, W = wi∂i, then

R(W,Z,X, Y ) = wizjxkylRijkl,

that is, the value of R(W,Z,X, Y ) at a point u depends only on the values of W ,Z, X, and Y at u.


Proposition 3.14. The components Rijkl of the Riemann curvature tensor ofany metric g satisfy the following identities:

Rijkl = −Rijlk = −Rjikl = Rklij(3.17)

Rijkl +Riljk +Riklj = 0.(3.18)

Proof. We first prove (3.18). Since [∂k, ∂l] = 0, it suffices to prove

(3.19)[∇k,∇l

]∂j +

[∇j ,∇k

]∂l +

[∇l,∇j

]∂k = 0.

Note that (3.11) together with the symmetry Γmlj = Γmjl imply that ∇l∂j = ∇j∂l.Thus, we can write: [

∇k,∇l]∂j = ∇k∇j∂l −∇l∇k∂j .

Permuting the indices cyclically, and adding, we get (3.19). The first identityin (3.23) is obvious from Definition 3.10. We now prove the identity:

Rijkl = −Rjikl.Using Proposition 3.5 repeatedly, we observe that:

g(∇k∇l∂j , ∂i) = ∂kg(∇l∂j , ∂i)− g(∇l∂j ,∇k∂i)= ∂k

(∂lgji − g(∂j ,∇l∂i)

)− ∂lg(∂j ,∇k∂i) + g(∂j ,∇l∇k∂i)

= ∂k∂lgji − ∂kg(∂j ,∇l∂i)− ∂lg(∂j ,∇k∂i) + g(∂j ,∇l∇k∂i).It is easy to see that the first term, and the next two taken together, are symmetricin k and l. Thus, interchanging k and l, and subtracting, we get:

Rijkl = g([∇k,∇l

]∂j , ∂i

)= g

(∂j ,[∇l,∇k

]∂i)

= −g([∇k,∇l

]∂i, ∂j

)= −Rjikl.

The last identity in (3.17) now follows from the first two and (3.18). We prove thatBijkl = Rijkl−Rklij = 0. Note that Bijkl satisfies (3.17) as well as Bijkl = −Bklij .Now, in view of the identities already established, we see that:

Rijkl = −Riljk −Riklj = −Rlikj −Riklj = Rljik +Rlkji −Riklj = Bljik +Rklij ,

hence Bijkl = Bljik. Using the symmetries of Bijkl, we can rewrite this identity as:

Bijkl +Biklj = 0.(3.20)

We now permute the first three indices cyclically:

Bkijl +Bkjli = 0,(3.21)

Bjkil +Bjilk = 0,(3.22)

add (3.20) to (3.21) and subtract (3.22) to get, using the symmetries of Bijkl:

Bijkl +Biklj +Biklj +Bkjli −Bkjli −Bijkl = 2Biklj = 0.

This completes the proof of the proposition.

It follows, that all the non-zero components of the Riemann tensor are deter-mined by R1212:

R1212 = −R2112 = R2121 = −R1221,

and all other components are zero. The proposition also implies that for any vectorsX,Y, Z,W , the following identities hold:

R(W,Z,X, Y ) = −R(W,Z, Y, Z) = −R(Z,W,X, Y ) = R(X,Y,W,Z),(3.23)

R(W,Z,X, Y ) +R(W,Y,Z,X) +R(W,X, Y, Z) = 0.(3.24)

5. THE RIEMANN CURVATURE TENSOR 55

Proposition 3.15. The components Rijkl of the Riemann curvature tensor ofany metric g satisfy:

(3.25) gmjRimkl = Γjik,l − Γjil,k + ΓnikΓjnl − ΓnilΓjnk.

Furthermore, we have:

(3.26) K =R1212

det(g),

where K is the Gauss curvature of g.

Proof. Denote the right-hand side of (3.25) by Sjikl. We have:

∇l∇k∂i = ∇k(Γjik∂j

)=(Γjik,l + ΓnikΓjnl

)∂j ,

or equivalently:

Γjik,l + ΓnikΓjnl = gjmg(∇l∇k∂i, ∂m).

Interchanging k and l and subtracting we get:

Sjikl = gjmg([∇l,∇k

]∂i, ∂m

)= gjmRmilk = gjmRimkl.

According to 3.4 and (3.25), we have:

K =12gikSjikj =

12gikgjlRijkl.

In view of the comment following Proposition 3.14, the only non-zero terms in thissum are:

K =12(g11g22R1212 + g12g21R1221 + g21g12R2112 + g22g11R2121

)= det

(g−1

)R1212,

which implies (3.26)

Corollary 3.16. The Riemann curvature tensor of any metric g on a surfaceis given by:

(3.27) Rijkl = K(gikgjl − gilgjk

).

Proof. Denote the right-hand side of (3.27) by Sijkl, and note that it satis-fies (3.17). Thus, the same comment which follows Proposition 3.14 applies andthe only non-zero components of Sijkl are determined by S1212:

S1212 = −S2112 = S2121 = −S1221.

In view of (3.27), we have R1212 = S1212, thus it follows that Rijkl = Sijkl

In particular, we conclude that:

(3.28) R(Z,W,X, Y ) = K(g(W,X) g(Z, Y )− g(W,Y ) g(Z,X)

).


6. The Second Variation of Arclength

In this section, we study the additional condition E′′σ(0) > 0 necessary for aminimum. This leads to the notion of Jacobi fields and conjugate points.

Proposition 3.17. Let γ : [a, b] → U be a geodesic parametrized by arclengthon the Riemannian surface (U, g), and let σ be a fixed-endpoint variation of γ withgenerator Y . Then, we have:

(3.29) E′′σ(0) =∫ b

a

(∣∣∇γY ∣∣2 −K γ (|Y |2 − g(γ, Y )2))

dt,

where K is the Gauss curvature of g.

Before we prove this proposition, we offer a second proof of the first variationformula:

(3.30) E′σ(0) = −∫ b

a

g(∇γ γ, Y ) dt,

which is more in spirit with our derivation of the second variation formula. Firstnote that if σ is a fixed-endpoint variation of γ with generator σ′ = Y , and withσ = X, then [X,Y ] = 0. Here Y denotes the vector field σ′ along σ rather thanjust along γ. Indeed, since X = dσ(d/dt) and Y = dσ(d/ds), it follows, as inPropositions 3.1 and 3.2, that for any smooth function f on U , we have

∂[X,Y ]f =[d

dt,d

ds

]f σ = 0.

In view of the symmetry Γijk = Γikj , this implies:

∇YX −∇XY = [X,Y ] = 0.

We can now calculate:

E′σ(s) =12

∫∂Y g(X,X) dt =

∫ b

a

g(∇YX,X) dt =∫ b

a

g(∇XY,X) dt

=∫ b

a

d

dtg(Y,X) dt−

∫ b

a

g(Y,∇XX) dt = g(Y,X)|ba −∫ b

a

g(Y,∇XX) dt

Setting s = 0, (3.30) follows.

Proof of Proposition 3.17. We compute:

E′′σ =12

∫ b

a

∂Y ∂Y g(X,X) dt =∫ b

a

∂Y g(∇YX,X) dt =∫ b

a

∂Y g(∇XY,X) dt

=∫ b

a

(g(∇Y∇XY,X) + g(∇XY,∇YX)

)dt

=∫ b

a

(g(∇X∇Y Y,X) + g

([∇Y ,∇X

]Y,X

)+ g(∇XY,∇XY )

)dt

=∫ b

a

(d

dtg(∇Y Y,X)− g(∇Y Y,∇XX) +R(X,Y, Y,X) + g(∇XY,∇XY )

)dt,

6. THE SECOND VARIATION OF ARCLENGTH 57

where as aboveX = σ, and Y = σ′. Now, the first term integrates to g(∇Y Y,X)|ba =0, and when we set s = 0, the second term also vanishes since ∇XX = ∇γ γ = 0.Furthermore, the last term becomes g(∇γY,∇γY ). Hence, we conclude:

(3.31) E′′σ(0) =∫ b

a

(∣∣∇γY ∣∣2 −R(X,Y,X, Y ))dt.

The proposition now follows from (3.28).

Thus, E′′σ(0) can be viewed as a quadratic form in the generator Y . Thecorresponding symmetric bilinear form is called the index form of γ:

I(Y, Z) =∫ b

a

(g(∇γY,∇γZ)−K γ

(g(Y, Z)− g(γ, Y ) g(γ, Z)

))dt.

It is the Hessian of the functional E, and if E has a local minimum, I is positivesemi-definite. We will also write I(Y ) = I(Y, Y ).

Definition 3.11. Let γ be a geodesic parametrized by arclength on the Rie-mannian surface (U, g). A vector field Y along γ is called a Jacobi field , if it satisfiesthe following differential equation:

∇γ∇γY +K (Y − g(γ, Y )γ) = 0.

Two points γ(a) and γ(b) along a geodesic γ are called conjugate along γ if thereis a non-zero Jacobi field along γ which vanishes at those two points.

The Jacobi field equation is a linear system of second-order differential equa-tions. Hence given initial data specifying the initial value and initial derivative ofY , a unique solution exists along the entire geodesic γ.

Proposition 3.18. Let γ be a geodesic on the Riemannian surface (U, g). Thengiven two vectors Z1, Z2 ∈ Tγ(a)U , there is a unique Jacobi field Y along γ suchthat Y (a) = Z1, and ∇γY (a) = Z2.

In particular, any Jacobi field which is tangent to γ is a linear combination ofγ and tγ. The significance of Jacobi fields is seen in the following two propositions.We say that σ is a variation of γ through geodesics if the curves t 7→ σ(t; s) aregeodesics for all s.

Proposition 3.19. Let γ be a geodesic, and let σ be a variation of γ throughgeodesics. Then the generator Y = σ′ of σ is a Jacobi field.

Proof. As before, denote X = σ and Y = σ′. We first prove the followingidentity: [

∇Y ,∇X]X = −K

(Y − g(X,Y )X

).

Indeed, in the proof of Lemma 3.13, it was seen that the left-hand side above isa tensor, i.e., is linear over functions, and hence depends only on the values ofthe vector fields X and Y at one point. Fix that point. If X and Y are linearlydependent, then both sides of the equation above are zero. Otherwise, X and Yare linearly independent, and it suffices to check the inner product of the identityagainst X and Y . Taking inner product with X, both sides are zero, and equa-tion (3.27) implies that the inner products with Y are equal. Since ∇XX = 0, weget:

0 = ∇Y∇XX = ∇X∇YX +[∇Y ,∇X

]X = ∇X∇XY −K

(Y − g(X,Y )X

).

Thus, Y is a Jacobi field.


We see that Jacobi fields are infinitesimal generators of variations throughgeodesics. If there is a non-trivial fixed endpoint variation of γ through geodesics,then the endpoints of γ are conjugate along γ. Unfortunately, the converse is nottrue but nevertheless, a non-zero Jacobi field which vanishes at the endpoints canbe perceived as a non-trivial infinitesimal fixed-endpoint variation of γ throughgeodesics. This makes the next proposition all the more important.

Proposition 3.20. Let γ be a geodesic, and let Y be a Jacobi field. Then, forany vector field Z along γ, we have:

(3.32) I(Y, Z) = g(∇γY,Z)|ba.In particular, if either Y or Z vanishes at the endpoints, then I(Y,Z) = 0.

Proof. Multiplying the Jacobi equation by Z and integrating, we obtain:

0 =∫ b

a

(g(∇γ∇γY, Z)−K

(g(Y,Z)− g(γ, Y ) g(γ, Z)

))dt

=∫ b

a

(d

dtg(∇γY,Z)− g(∇γY,∇γZ)−K

(g(Y,Z)− g(γ, Y ) g(γ, Z)

))dt

= g(∇γY, Z)|ba − I(Y, Z).

Thus, a Jacobi field which vanishes at the endpoints lies in the null space ofthe index form I acting on vector fields which vanish at the endpoints.

Theorem 3.21. Let γ : [a, b]→ (U, g) be a geodesic parametrized by arclength,and suppose that there is a point γ(c) with a < c < b which is conjugate to γ(a).Then there is a vector field Z along γ such that I(Z) < 0. Consequently, γ is notlocally-length minimizing.

Proof. Define:

V =

Y a 6 t 6 c

0 c 6 t 6 b

and let W be a vector field supported in a small neighborhood of c which satisfiesW (c) = −∇γY (c) 6= 0. We denote the index form of γ on [a, c] by I1, and the indexform on [c, b] by I2. Since V is piecewise smooth, we have, in view of (3.32):

I(V,W ) = I1(V,W ) + I2(V,W ) = I1(Y,W ) = −|∇γY (c)|2 < 0

It follows that:

I(V + εW, V + εW ) = I(V ) + 2εI(V,W ) + ε2I(W ) = 2εI(V,W ) + ε2I(W )

is negative if ε > 0 is small enough. Although V + εW is not smooth, there is forany δ > 0 a smooth vector field Zδ, satisfying |Y |2 + |∇γZδ|2 6 C uniformly inδ > 0, which differs from V + εW only on (c − δ, c + δ). Since the contributionof this interval to both I(V + εW, V + εW ) and I(Zδ, Zδ) tends to zero with δ,it follows that also I(Zδ, Zδ) < 0 for δ > 0 small enough. Thus, γ is not locallyenergy-minimizing. Since it is parametrized by arclength, if it was locally length-minimizing, it would by Lemma 3.12 also be locally energy-minimizing. Thus, γcannot be locally length-minimizing.

A partial converse is also true: the absence of conjugate points along γ guar-antees that the index form is positive definite.

EXERCISES 59

Theorem 3.22. Let γ : [a, b]→ (U, g) be a geodesic parametrized by arclength,and suppose that no point γ(t), a < t 6 b, is conjugate to γ(a) along γ. Then theindex form I is positive definite.

Proof. Let X = σ, and let Y be a Jacobi field which is perpendicular to X,and vanishes at t = a. Note that the space of such Jacobi fields is 1-dimensional,hence Y is determined up to sign if we also require that |Y (a)| = 1. Since Y isperpendicular to X, it satisfies the equation:

∇X∇XY +KY = 0.

Furthermore, since Y never vanishes along γ, the vectors X and Y span Tγ(t)Ufor all t ∈ (a, b]. Thus, if Z is any vector field along γ which vanishes at theendpoints, then we can write Z = fX + hY for some functions f and h. Note thatf(a) = f(b) = h(b) = 0 and hY (a) = 0. We then have:

I(Z,Z) = I(fX, fX) + 2I(fX, hY ) + I(hY, hY ).

Since R(X, fX,X, fX) = 0 and ∇XfX = fX, it follows from (3.31) that:

I(fX, fX) =∫ b

a

g(fX, fX) dt =∫ b

a

f2 dt.

Furthermore,

I(fX, hY ) =∫ b

a

g(fX,∇XhY ) dt

= g(fX, hY )|ba −∫ b

a

g(∇X fX, hY ) dt = −∫ b

a

g(fX, hY ) dt = 0.

Finally, since |∇XhY |2 = g(∇XY,∇Xh2Y ) + h2 |Y |2, it follows from Proposi-tion 3.20 that:

I(hY, hY ) =∫ b

a

h2 |Y |2 dt+ I(Y, hY ) =∫ b

a

h2 |Y |2 dt.

Thus, we conclude that:

I(Z,Z) =∫ b

a

(f2 + h2 |Y |2

)dt > 0.

If I(Z,Z) = 0, then f = 0 and hY = 0 on [a, b]. Since Y 6= 0 on (a, b], we concludethat h = 0 on (a, b], and in view of h(b) = f(b) = 0, we get that Z = 0. Thus, I ispositive definite.

Exercises

Exercise 3.1. Two Riemannian metrics g and g on an open set U ⊂ R2 areconformal if g = e2λg for some smooth function λ.

(1) Prove that a parametric surface X : U → R3 is conformal if and only if itsfirst fundamental form g is conformal to the Euclidean metric δ on U .

(2) Let g = e2λg be conformal metrics on U , and let Γkij and Γkij be theirChristoffel symbols. Prove that:

Γkij = Γkij + δki λj + δkj λi + gij gkm λm


(3) Let g and g be two conformal metrics on U , g = e2λg, and let K and Kbe their Gauss curvatures. Prove that:

K = e−2λ(K −∆λ).

CHAPTER 4

Global Theory of Surfaces

In this chapter, we begin our study of the global theory of differentiable surfaces.We present a five global results:

• The Hopf-Rinow Theorem establishes the equivalence of several propertiessuch as: the ability to extend geodesics to all values of the parameters,the ability to connect any two points by a minimizing geodesic, and thecompleteness of the space as a metric space.

• The Hadamard Theorem states that if then Gauss curvature is non-positiveand the space is complete, then the space is homeomorphic to the plane.

• The Bonnet-Myers Theorem states that if the Gauss curvature is boundedbelow by a positive constant, then the space is compact.

• Hadamard’s Theorem on convex surfaces states that if a compact surfacein R3 has positive Gauss curvature, then it bounds a convex body.

• The Gauss-Bonnet Theoremrelates the integral of the Gauss curvature toa topological invariant.

With the exception of the Hopf-Rinow Theorem, these establish a relation be-tween local geometric properties and global mostly topological ones. The HadamardTheorem and the Bonnet-Myers Theorem are in some sense dual, both using re-sults on geodesics. Hadamard’s Theorem on convex surfaces is the 2-dimensionalanalog of Theorem 1.10. All these results, with the exception of the Gauss-BonnetTheorem, have straightforward generalizations to higher dimension.

1. Differentiable Manifolds

Before, we can begin our study, we need to define the concept of a differentiablemanifold and a differentiable surface.

Definition 4.1. Let M be a Hausdorff topological space with a countablebasis. A differentiable atlas on M is a collection of pairs (Uα, ϕα), α ∈ A whereUα ⊂M is an open set, and ϕα : Uα → Rn is a homeomorphism of Uα onto an opensubset Vα = ϕα(Uα) of Rn, satisfying the following conditions:

(i) Uα, α ∈ A is a covering of M .(ii) If Uα ∩ Uβ 6= ∅, then the mapping ϕαβ = ϕβ ϕ

−1α : Vα → Vβ is a diffeo-

morphism.A n-dimensional differentiable manifold is a Hausdorff topological space with acountable basis equipped with a differentiable atlas. The set Uα is called a coordi-nate patch, and the functions ϕα = (u1

α, u2α) are called local coordinates on Uα. The

pair (Uα, ϕα) is called a chart. The functions ϕαβ are called transition functions.A differentiable surface is a 2-dimensional differentiable manifold.

Remark 4.1.

61

62 4. GLOBAL THEORY OF SURFACES

(a) For short, we will use manifold to mean differentiable manifold and surfaceto mean differentiable surface.

(b) One may replace C∞ by Ck to get the definition of a Ck-manifold.(c) Clearly, if M is a manifold and U ⊂ M is an open set, then U is also a

manifold.(d) It is a straightforward exercise in topology to check that any differentiable

manifold admits a countable atlas.

A given space may have more than one atlas. Two atlases are equivalent iftheir union is again an atlas. It is easy to see that this is indeed an equivalencerelation. A differentiable structure on M is an equivalence class of atlases.

Definition 4.2. Let f : M → R be a continuous function defined on a manifoldM . We say that f is Ck if for each chart (Uα, ϕα), the function f ϕ−1

α : ϕα(Uα)→R is Ck. A curve γ : I → M is Ck if for each chart (Uα, ϕα) and any subintervalJ ⊂ I such that γ(J) ⊂ Uα, the curve ϕα γ : J → ϕα(Uα) is Ck. Let F : M → Nbe a continuous map between two manifold. We say that F is Ck if for charts(Uα, ϕα) on M and (Wβ , ψβ) on N , and for V = ϕα

(Uα ∩ F−1(Wβ)

)the function

ψβ F ϕ−1α : V → Wβ is Ck. A diffeomorphism between two manifolds M and

N is a smooth (C∞) map F : M → N which is 1-1, onto, and whose inverseF−1 : N →M is also smooth.

Note that if for some α ∈ A, the function f ϕ−1α : Uα → R is smooth, then

since so is the function f ϕ−1β : Uα ∩ Uβ → R since f ϕ−1

β = f ϕ−1α ϕβα. It

follows that the smoothness of f is independent of the particular atlas within adifferentiable structure. The same is true of curves, and maps between manifolds.

Example 4.1. The unit sphere S in R3 is a differentiable surface, see Exer-cise 4.1. More generally, if f1, . . . , fn−2 are smooth real functions on an open subsetΩ ⊂ Rn such that their gradients ∇f1, . . . ,∇fn−2 are linearly independent in Ω,then the zero set M = x ∈ Ω: f1(x) = · · · = fn(x) = 0 is a differentiable sur-face. Indeed, suppose that p ∈ M , then there are two vectors e1, e2 ∈ Rn suchthat ∇f1(p), . . . ,∇fn−2(p), e1, e2 is a basis of Rn. Define a map Fp : Ω → Rn byFp(x) =

(f1(x), . . . , fn−2(x), e1·(x−p), e2·(x−p)

). Clearly, Fp(p) = 0, Fp is smooth,

and in fact dFp = (∇f1, . . . ,∇f2, e1, e2). Thus, dFp(p) is non-singular, hence bythe inverse function theorem, there is a neighborhood Wp ⊂ Rn of 0 with compactclosure in Ω such that Fp has a smooth inverse Gp on a neighborhood of W p. LetUp = G(Wp) ∩M , and define ϕp : Up → R2 by ϕp(x) =

(e1 · (x − p), e2 · (x − p)

).

Then Up is open, and ϕp is continuous. In fact, if π is the projection on thelast two coordinates in Rn, then ϕp is simply the restriction of π Fa to Up. LetVp = ϕp(Up), then Vp = π(Wp), and Vp is open. The inverse of ϕp on Vp isϕ−1p (u, v) = Gp(0, . . . , 0, u, v). It now follows easily that ϕp is a homeomorphism of

Up onto Vp. We claim that the collection (Up, ϕp), p ∈M is an atlas. It remainsto check that the transition functions ϕpq are diffeomorphisms. So, suppose thatUp ∩ Uq 6= ∅, then Wpq = Wp ∩Wq 6= ∅, and clearly the map Fq Gp is smooth onFp(Wpq). Since ϕpq is the restriction of Fq Gp to π

(Fp(Wpq)

), it is also smooth.

Similarly ϕ−1pq = ϕqp is smooth, and we conclude that ϕpq is a diffeomorphism.

Let M be a surface, let p ∈M , and let us denote by Fp(M) the real algebra ofsmooth functions defined on a neighborhood of point p ∈M . If f, g ∈ Fp(M), thenf + g and fg are defined on the intersection of the domains of f and g.

1. DIFFERENTIABLE MANIFOLDS 63

Definition 4.3. Let M be a surface. A tangent vector at p ∈ M is a linearoperator X : Fp(M)→ R satisfying:

(4.1) X(fg) = X(f)g(a) + f(a)X(g).

The set of tangent vectors at p is clearly a vector space over the real numbers. Itis called the tangent space of M at p, and is denoted by TpM .

We will write Xf for X(f) when no ambiguity arises. First, note that if X is atangent vector at p then Xf only depends on the values of f in an arbitrarily smallneighborhood of p. Indeed, suppose f = 0 in a neighborhood of p, and let g be asmooth cut-off function supported in the set x ∈M : f(x) = 0, see Exercise 4.3,then fg = 0 hence X(fg) = 0 by linearity, and we get:

0 = X(fg) = (Xf)g(p) + f(p)Xg = Xf

It follows by linearity that if f = g in a neighborhood of p, then Xf = Xg.Similarly, if f = c is constant in a neighborhood of p, then Xf = 0, since:

Xf = cX(1) = cX(1 · 1) = c(X(1) · 1 + 1 ·X(1)

)= 2cX(1) = 2Xf.

Now, let (u1, u2) be local coordinates in a neighborhood U of p, and denote by∂i the partial derivative at p with respect to u1:

∂if =∂f ϕ−1

α

∂uiϕα((p)).

Then, it is easy to check that ∂i is a tangent vector at p. Furthermore, since∂ix

j = δji , it follows that ∂1, ∂2 are linearly independent, see the comment followingEquation (3.5). We will now show that these two vectors also span TpM . We willneed the following elementary lemma which is a straightforward consequence ofTaylor’s Expansion.

Lemma 4.1. Let f be a smooth function defined on a neighborhood V of 0 ∈R2, then there are smooth functions f1, f2 defined on V such that f(u) = f(0) +u1f1(u) + u2f2(u).

Note that fi(0) = ∂f/∂ui(0). We can now prove the following theorem:

Theorem 4.2. Let M be a surface, and let a ∈ M . The vectors ∂1, ∂2 are abasis of TaM . Consequently, TaM is 2-dimensional.

Proof. It suffices to show that ∂1, ∂2 span TaM . Let X be a tangent vectorat a, and let ϕ = (u1, u2) be local coordinates defined on a neighborhood U of a.Without loss of generality we may assume that ϕ(a) = 0. Let ξi = Xui, and letY = ξi∂i. We claim that X = Y . Indeed, if f ∈ F1

a(M), we may assume that f isdefined on U , and by Lemma 4.1, we can write f ϕ−1(u) = f(0) + uifi(u), wherefi(0) = ∂if . Thus, we have:

Xf = X(f(0) + uifi(u)

)= Xuifi(0) + 0 ·Xfi = ξi∂if(0) = Y f.

Example 4.2. Let 0 be an interior point of the interval I, and let γ : I → Mbe a C1 curve into M with γ(0) = a. If f ∈ Fa(M), then there is an interval J ⊂ Isuch that 0 ∈ J and f γ : J → R is C1. Indeed, if (U, u) is a chart with a ∈ Uthen f γ =

(f ϕ−1

)(ϕ γ

). It is now easy to see that Xf = (f γ)′(0) is a

tangent vector. We call this vector the tangent of γ at a and denote it γ(a). If


ϕ = (u1, u2) are the local coordinates in the given chart, and we write γi = ui γthen ϕ γ = (γ1, γ2), and X = γi∂i, where the dot represent differentiation withrespect to the parameter in J . Note that if X is a tangent vector at a, then is acurve γ through a whose tangent at a is X.

We call the coefficients ξi = Xui, the components of X in the local coordinates(u1, u2). If (v1, v2) is another system of local coordinates in a neighborhood of a,yi = Xvi, and ∂′i is the partial derivative with respect to vi, then we have:

X = xi∂i = yi∂′i,

where yi = dϕij xj and dϕij = ∂ui/∂vj is the Jacobian of the transition function

ϕ : u 7→ v. Similarly, ∂i = dϕji ∂′j . Clearly, it now follows from the expansion

X = xi∂i that X can be extended as an operator to F1a(M), the algebra of C1

functions defined on a neighborhood of a. We will denote the dual of the vectorspace TaM by T ∗aM . Elements of T ∗aM are called co-vectors.

Definition 4.4. If f ∈ F1a(M) we define its differential is the element of T ∗a (M)

given by df(X) = Xf .

Thus for example, dui(X) are the components of X with respect to (u1, u2),hence dui(∂j) = δij . It follows that du1, du2 is the basis of T ∗aM dual to thebasis ∂1, ∂2 of TaM . Any θ ∈ T ∗aM can be expressed as θ = θidu

i. If (v1, v2) isanother system of local coordinates, we have dvi = dϕij u

j , hence θ = ωidvi, where

θi = dϕij ωj .For any open subset U ⊂ M let TU = ∪p∈UTpM . If (Uα, ϕα) is a chart and

a ∈ Uα, we define the map dϕα : TaM → R2 by dϕα(X) =(du1(X), du2(X)

), and

the map Φα : TUα → Uα × R2 by Φα : X 7→(ϕα(a), dϕα(Y )

)if X ∈ TaM . By

requiring that Φα be a homeomorphism, the set TUα acquires a topology, and infact (TUα,Φα) is a chart. If (Uβ , ϕβ) is another chart with Uα ∩ Uβ 6= ∅, then thetopologies induced by Φα and Φβ on T (Uα ∩ Uβ) are equal and Φαβ = Φβ Φ−1

α =(ϕαβ , dϕαβ

): TUα → TUβ is clearly smooth since dϕαβ is a linear map on R2. It

follows that if (Uα, ϕα) : α ∈ A is an atlas on M , then (TUα,Φα) : α ∈ A isan atlas on TM . We conclude that TM is a 4-dimensional manifold. We call TMthe tangent bundle of M . The natural projection π : TM → M is the map whichsends each tangent vector X ∈ TpM to p. Similarly, the 4-dimensional manifoldT ∗M = ∪p∈MT ∗pM is called the co-tangent bundle and it has its natural projectionπ : T ∗M →M .

Definition 4.5. A vector field is a smooth map X : M → TM such thatπ X = id. A one-form is a smooth map θ : M → T ∗M such that π θ = id.

Thus a vector field is a map X on M which assigns to each p ∈ M a tangentvector X(p) ∈ TpM , and such that for each chart (Uα, ϕα), the map dϕα(X) : Uα →R2 is smooth. If X is a vector field on M , and θ is a one-form, then θ(X) is a smoothfunction on M . Thus, the space of one-forms is the dual of the space of vector fields,when viewed as a module over the ring of smooth functions on M .

2. Some Multi-linear Algebra

Recall that if V is a finite dimensional vector space over the field of real num-bers, and if V ∗ is its dual, we can define the tensor product V ∗ ⊗ V ∗ as thespace of bi-linear maps on V × V . If θ, ω ∈ V ∗, we define θ ⊗ ω ∈ V ∗ ⊗ V ∗

EXERCISES 65

by θ ⊗ ω(v, w) = θ(v)ω(w). It is easy to see that the set θ ⊗ ω : θ, ω ∈ V ∗ spansV ∗ ⊗ V ∗. Similarly V ⊗ V is defined as the space of bi-linear maps on V ∗ × V ∗,v⊗ x ∈ V ⊗W is defined by v⊗w(θ, ω) = θ(v)ω(w), and v⊗w : v, w ∈ V spansV ⊗V . We can similarly define higher order tensor products, and even mixed tensorproducts, e.g., V ∗⊗V is defined as the space of multi-linear maps on V ×V ∗. Notethat V ∗⊗V is naturally identified with L(V, V ), the space of linear transformationof V onto itself, via the unique linear transformation Φ: V ∗ ⊗ V → L(V, V ) whichsatisfies Φ θ ⊗ v(w) = θ(w)v. Clearly Φ−1T (v, θ) = θ

(T (v)

).

Let e1, . . . , en be a basis of V , then we write v = viei and we call vi thecomponents of v with respect to this basis. Let ωi be the dual basis of V ∗, definedby ωi(ej) = δij , then vi = ωi(v). Similarly, if θ ∈ V ∗ then θ = θiω

i, and thecomponents are given by θi = θ(ei). A basis for any tensor product can now beformed by these two bases, e.g., a basis for V ∗ ⊗ V is ωi ⊗ ej where i, j = 1, . . . , n.If α ∈ V ∗ ⊗ V ∗ then α = αji ω

i ⊗ ej , and αji = α(ei, ωj).A tensor α ∈ V ∗ ⊗ V ∗ is symmetric if α(v, w) = α(w, v) for all v, w ∈ V .

Equivalently, its component satisfy αij = αji. The symmetrization of a tensor αis the symmetric tensor β given by β(v, w) = 1

2

(α(v, w) + α(w, v)

). A basis for the

space of symmetric tensors is ωiωj with 1 6 i 6 j 6 n. Consequently, the spaceof symmetric tensors is n(n + 1)/2. One can also define higher order symmetrictensors by averaging over all permutations of the arguments.

A tensor α ∈ V ∗ ⊗ V ∗ is antisymmetric if α(v, w) = −α(w, v) for v, w ∈ V .The antisymmetrization of a tensor α is the antisymmetric tensor β given byβ(v, w) = − 1

2

(α(v, w)−α(w, v)

). We denote by θ∧ω 2 times the antisymmetrization

of θ⊗ω. The space of antisymmetric tensors in V ∗⊗V ∗ is denoted∧2

V ∗. A basisfor∧2

V ∗ is ωi ∧ωj for 0 6 i < j 6 n. The space of antisymmetric tensors of orderk is

∧kV ∗, the wedge product of k elements is:

θ1 ∧ · · · ∧ θk =∑σ∈Sk

sign(σ)θσ(1) ∧ · · · ∧ θσ(k),

where Sk is the k-th symmetric group, and signσ is the sign of a permutationσ ∈ Sk:

sign(σ) =

1 if σ is even−1 if σ is odd.

A basis for∧k

V ∗ is ωi1∧· · ·∧ωik where 0 6 i1 < · · · < ik. In particular,∧k

V ∗ hasdimension

(nk

). By convention

∧0V ∗ = R . Note that

∧kV ∗ are trivial when k > n,∧n

V ∗ is one dimensional. For example in dimension n = 2, we have∧0

V ∗ = R,∧1V ∗ = V ∗, and

∧2V ∗ ∼= R. However, this last isomorphism is not canonical.

Let M be an n-dimensional manifold. For each tensor product of TpM andT ∗pM we can repeat the construction before Definition 4.5

3. Integration

Exercises

Exercise 4.1. Prove that the unit sphere in R3:

S2 = x ∈ R3 : |x| = 1,

is a differentiable surface by constructing explicitly a finite differentiable atlas.


Exercise 4.2. Prove Lemma 4.1.

Exercise 4.3. Let 0 < r < R <∞. Show that there is a smooth function φ onRn such that 0 6 φ 6 1, φ = 1 inside Br(0), and φ = 0 outside BR(0). Concludethat if U is an open set in a manifold M and a ∈ U , there is a function cut-offfunction g such that 0 6 g 6 1, g = 1 in a neighborhood of a, and g = 0 outside U .

Index

angle, 22

between surfaces, 41

exterior, 12

arclength, 7, 45

area minimizing, 36

asymptotic line, 30, 42

atlas, 61

Bernstein’s Theorem, 37

binormal, 8

Bonnet-Myers, 61

bundle

co-tangent, 64

tangent, 64

catenoid, 31, 42

Cauchy-Riemann equations, 32, 38, 42

chart, 61

Christoffel Symbols, 39

Christoffel symbols, 46

co-tangent bundle, 64

co-vectors, 64

Codazzi Equation, 40

commutator, 23, 47

conformal, 31, 59

conjugate, 42

conjugate points, 56, 57

convex, 12, 38

strictly, 16

coordinate

patch, 61

coordinates

local, 61

corner, 12

curvature

center of, 16

Gauss, 25

line of, 29, 42

mean, 25

normal, 24

of a curve, 8

principal, see also principal, curvature

Riemann tensor, 53

total, see also total curvature

curvature, line of, 28

curve

closed, 10

parametrized, 7

piecewise smooth, 12

regular, 7

simple, 10

cut-off function, 63

cylinder, 27

derivative

covariant, 48

directional, 23, 47

Lie, 47

developable, 28

diffeomorphism, 19, 62

differentiable

manifold, 61

structure, 62

differential, 64

differentials, 46

directrix, 27

distance, 22

divergence, 49

Dupin indicatrix, 30

Einstein summation convention, 20

entire, 37

Euclidean metric, 45

Euler, 24

evolute, 16

expanding map, 38, 43

Fenchel’s Theorem, 14

form

first fundamental, 21

quadratic, 21

second fundamental, 23

symmetric bilinear, 21

Four Vertex Theorem, 13

Frenet frame, 8

Frenet frame equation, 8

Fundamental Theorem

for curves in R3, 8

67

68 INDEX

for surfaces, 39, 40

Gauss curvature, 46

Gauss Equation, 40

Gauss map, 20

generator, 27

geodesic, 50

geometry

intrinsic, 22

gradient, 43, 49

graph, 19, 37

Hadamard, 61

harmonic, 31

helix, 16

Hessian, 37

Hopf-Rinow, 61

hyperboloid, 28

index

contravariant, 20

covariant, 20

raise, 25

index form, 57

induced metric, 46

intrinsic geometry, 45

isometry, 21, 45

isoperimetric inequality, 17

Jacobi field, 57

Jacobi fields, 56

Laplacian, 49

Leibniz, 47

length-minimizing, 51

locally, 51

line element, 22, 45

manifold, 62

meridian, 31

minimal surface, 31

non-parametric, 37, 42

Monge Ampere equation, 38

natural projection, 64

Nitsche, 38

normal section, 24

one-form, 64

orientation, 7, 19, 20

osculating paraboloid, 24

parallel, 31

plane, 25

normal, 9

osculating, 9

rectifying, 9

Poincare Disk, 46

point

ellitpic, 24

parabolic, 24

planar, 24, 26

umbilical, 25, 26

point,hyperbolic, 24

principal

curvature, 25

direction, 25

principal normal, 8

pull-back, 21

reparametrization

of curves, 7

of surfaces, 19

representation

coordinate, 21

Riemannian metric, 45

Riemannian surface patch, 45

Rodriguez, 28

rotation number, 10, 12

Rotation Theorem, 10, 12

sphere, 26

spherical image

of a curve, 14

under Gauss map, 28

star-shaped, 10

striction, line of, 27

surface, 62

differentiable, 61

minimal, see also minimal surface

of revolution, 30

generator, 30

parametric, 19

ruled, 27

tangent, 28

surface area, 33

element, 33

signed, 34

tangent bundle, 64

tangent plane, 19

tangent space, 19, 63

tangent vector, 63

tensor, 53

Theorema Egregium, 39, 40

torsion, 8

total curvature

of a curve, 14

of a surface, 33

transition functions, 61

unit normal, 20

unit tangent, 8

upper half-plane, 46

variation, 34, 51

fixed-endpoint, 51

vector field, 20, 64

vertex, 13

Weierstrass representation, 31, 42

INDEX 69

width, 16

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