Notes on Math 321

Notes on Math 321

September 2, 2007

ii

c©2006 Guowu Meng. All rights reserved.

Contents

Preface v

1 Curves 11.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Frenet Trihedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Local canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Regular Surfaces 72.1 Digression: Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.1.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.1.2 Inverse Function Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2.1 Proof of theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.2 Proof of theorem 2.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Existence of tangent planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4 Local Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Gauss Map 153.1 Smooth maps between regular surface . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Orientation, Gauss map and curvatures . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Digression: Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3.1 One-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3.2 Two-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.3.3 The area form on an oriented Euclidean plane . . . . . . . . . . . . . . . . . . 19

3.4 Fundamental Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.4.1 The total differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.4.2 The first fundamental two-form . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4.3 The second fundamental two-form . . . . . . . . . . . . . . . . . . . . . . . . 223.4.4 Area Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

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iv CONTENTS

4 Intrinsic Theory 254.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.1.1 Lie derivatives of smooth functions . . . . . . . . . . . . . . . . . . . . . . . . 264.1.2 Gradient vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.1.3 Lie derivatives of Rm-valued smooth functions . . . . . . . . . . . . . . . . . 27

4.2 Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.2.1 Directional derivatives of Rm-valued smooth functions . . . . . . . . . . . . . 274.2.2 Covariant derivatives of vector fields . . . . . . . . . . . . . . . . . . . . . . . 284.2.3 The covariant differentiation is non-commutative in general . . . . . . . . . . 29

4.3 Riemann Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.3.1 Digression: Lie bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3.2 Curvarure Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.4 The covariant differentiation is intrinsic . . . . . . . . . . . . . . . . . . . . . . . . . 324.5 Riemann curvature and Gauss curvature are all intrinsic . . . . . . . . . . . . . . . . 334.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Selected topics 355.1 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.1.1 Digression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Preface

Rough speaking, geometry is the study of interesting shapes and relations among them. (Moregenerally, the core of mathematics is the study of interesting objects and relations among them.The same is true for physics.) Geometry is intuitive, one can quickly get many nice facts by justlooking at the geometric shapes — a picture is worth a thousand words. For example, two straightlines on a plane either don’t intersect or intersect exactly at one point. Surely, everybody agreeson this simple fact, but most people cannot give a rigorous proof. To give a proof, one needsalgebra, and that lead us to analytic geometry. In a sense algebra is a tool which makes ourgeometric intuition rigorous. Moreover, algebra extends our understanding of geometry beyond ourintuition; for example, the fact that one cannot trisect an angle by using a compass and a ruler isnot intuitively clear, but can be proved by using algebra.

For the study of more sophisticated shapes, algebra is not enough, we need a great tool, namelycalculus, and that leads us to differential geometry. For example, what is the volume of a ball withradius r? Algebra alone cannot give you the answer, but calculus surely can. Roughly speaking,differential geometry is the study of interesting shapes and relations among them by using calculus.But I prefer to view differential geometry as college calculus III — calculus on curved spaces. Tome, there is a huge psychological benefit to adopt this view: differential geometry is just a furtherextension of our regular calculus, so we are just repeating many of the same old calculus ideas mostof the time during the learning process; in other words, we are fearless while wandering because weare not far away from our “mother” (i.e., the regular calculus).

In this course, we only study curves and surfaces inside R3; these are nice geometric shapesof dimension one and two respectively. Something new such as curvatures and torsion appearnaturally, they tell us how the the curves and surfaces sit inside R3 up to the rigid motions ofR3 (i.e., the transformations of R3 which preserve both lengths and angles.) These facts arequite intuitive. While the curves are intrinsically flat (i.e., can be straightened locally withoutstretching); for surface, much less intuitive facts were discovered by Gauss: 1) the Gauss curvatureK is intrinsic, i.e., stays the same no matter how you bend the surface without stretching, 2) thetotal Gauss curvature of an (oriented) closed surface is topological, i.e., stays the same no matterhow you stretch the surface.

It is worth to mention that the Gauss curvature K also has a meaning in calculus! Recallthat, in regular calculus of two variables, for smooth vector field F , (∂x∂y − ∂y∂x)F is zero. If weextend the calculus to a surface, then (∂x∂y − ∂y∂x)F = KF , so the Gauss curvature measures thethe failure of commutativity for mixed partial derivatives. This observation is quite natural andfruitful; as a matter of fact, if we emphasize the view that differential geometry is the calculus on

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vi PREFACE

curved spaces, we could rediscover pretty much all mathematics that we need for writing down thefundamental laws of physics: from Riemannian geometry to geometry of vector bundles.

Chapter 1

Curves

Prerequisite: Calculus I, i.e., calculus for vector-valued functions of one variable — the kind ofcalculus needed for the study of the Newtonian mechanics.

Curves are one dimensional nice geometric objects, and it is very useful to view it as thetrajectory of a moving particle in R3 or R2.

1.1 Basic Definitions

Some standard notations:

• I — an open interval, say (−1, 2) or (−∞,∞) or (2,∞), etc.;

• t0 — a number in I;

• f — a function on I taking values in R or R2 or R3;

• α — a parameterized smooth curve;

• C — a regular curve.

Some standard terminologies:

• f is smooth — each component function of f is a smooth (i.e., infinitely differentiable)real-valued function;

Definition 1.1.1. A parameterized smooth curve in R3 is just a smooth map α: I // R3.The image of α (viewed as a map) is called the trace of α or a curve. We say α is regular if α′

is nonzero everywhere on I. A regular curve in R3 is just the trace of a regular parameterizedsmooth curve in R3.

In other word, a regular curve in R3 is just a subset of R3 which can be parameterized bya smooth map α with nowhere vanishing first derivative. By the way, curves in R2, i.e., planecurves, can be similarly defined.

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2 CHAPTER 1. CURVES

Example 1.1.1. α(t) = (t, t) is a regular parameterized smooth curve in R2, and its trace ( astraight line) is a regular plane curve.

Example 1.1.2. α(t) = (t3, t3) is a parameterized smooth curve in R2, but not regular at t = 0.However, its trace ( a straight line) is a regular plane curve.

So a regular curve can have a non-regular parametrization.

Example 1.1.3. α(t) = (t3, t2) is a parameterized smooth curve in R2, but it is not a regularparameterized smooth curve. One can show that its trace is not a regular curve1.

So a non-regular curve can have a smooth parametrization.

Example 1.1.4. α(t) = (t, |t|) is not a parameterized smooth curve in R2 because it is not smoothat t = 0. One can show that its trace is not a regular curve either2.

Example 1.1.5. Let f be a real valued smooth function on I. Then the graph of f :

Γ(f) := {(t, f(t) | t ∈ I}

is a regular plane curve.

Theorem 1.1.1 (Arclength Parametrization). Let C be a regular curve, then C admits an arclengthparametrization, i.e., C is the trace of a regular parameterized smooth curve α such that |α′| = 1everywhere.

Proof. Since C is a regular curve, it must be the trace of a parameterized smooth curve β: J //R3

such that |β′| 6= 0 everywhere. Let s = g(t) =∫ tt0|β′(x)| dx. Since3 g′ = |β′| > 0 on J , g is smooth

and has a smooth inverse t = f(s) with

f ′(s) =1

g′(f(s))=

1|β′(f(s))| .

Let α(s) = β(f(s)) for s ∈ I (where I is the image of f). Then |α′(s)| = |β′(f(s))| · f ′(s) = 1.

Example 1.1.6. The unit circle centered at (0, 0) is a regular plane curve. α(s) = (cos t, sin t) isone of its arclength parameterizations.

Definition 1.1.2. Suppose that α: I // R3 is a regular parameterized smooth curve and t0 ∈ I.The trace of the linear approximation of α around t0 is called the tangent line of α at t0.

Note that the tangent line of α at t0 is the trace of this parameterized curve:

t 7→ α(t0) + α′(t0)(t− t0).

1The proof is given in example 1.3.1.2I leave its proof to you.3I will prove this statement in class.

1.2. FRENET TRIHEDRON 3

1.2 Frenet Trihedron

Suppose that α: I // R3 is a regular parameterized smooth curve. By a re-parametrization ifnecessary, we can assume that |α′| = 1 on I. Furthermore, we assume that |α′′| 6= 0 on I.

Let ~t = α′ and we call ~t(s) the tangent vector at s. Since ~t · ~t = 1, ~t · ~t′ = 0 after takingderivatives. In view of the fact that ~t′ = α′′ 6= ~0, we can introduce unit vector

~n =~t′

|~t′| =α′′

|α′′| .

We shall call ~n(s) the normal vector at s, that is because ~t(s) · ~n(s) = 0. The plane spanned by~t(s) and ~n(s) is called osculating plane at s. Let ~b = ~t× ~n, then ~b(s) is a unit normal vector ofthe osculating plane at s and is called the binormal vector at s. Note that (~t(s), ~n(s),~b(s)) is anorthornormal frame for R3 and is called the Frenet trihedron at s.

Proposition 1.2.1. There are smooth functions k > 0 and τ on I such that

~t′ = k~n

~n′ = −k~t −τ~b~b′ = τ~n

Proof. The proof is just a bunch of easy computations.

Remark 1.2.1. k(s) is called the curvature at s and τ(s) is called the torsion of s. Roughlyspeaking, k(s) measures the extent of the bending of tangent line and τ(s) measures the extent ofthe bending of osculating plane.

We say a map T : R3 // R3 is a rigid motion of R3 if T preserves both distance and angle.

Exercise 1.2.1. Assume T is a rigid motion of R3 such that T (~0) = ~0. Show that1) T (~x) · T (~y) = x · y.2) T (~x + ~y) = T (~x) + T (~y).3) T is continuous, hence T is a linear transformation.4) A rigid motion of R3 must be an affine linear transformation.

Theorem 1.2.1 (Fundamental Theorem of the Local Theory of Curves). Given smooth functionsk(s) > 0 and τ(s), s ∈ I, there exists a regular parameterized smooth curve α: I //R3 such thats is the arc length, k(s) is the curvature and τ(s) is the torsion of α. Moreover, α is unique up toan orientation preserving rigid motion of R3.

Proof. It is a consequence of existence and uniqueness theorem for ordinary differential equations.

4 CHAPTER 1. CURVES

1.3 Local canonical form

Let α: I // R3 be an arc length parameterized smooth curve, 0 ∈ I. Then

α(s) = α(0) + α′(0)s +12α′′(0)s2

+16α′′′(0)s3 + · · ·

= α(0) + ~t(0)s +12k(0)~n(0)s2

+16α′′′(0)s3 + · · · (1.1)

Since α′′′ = k′~n + k~n′ = k′~n + k(−k~t − τ~b) = −k2~t + k′~n − kτ~b. Up to a rigid motion, we mayassume α(0) = ~0 and (~t(0), ~n(0),~b(0)) = (~i,~j,~k), then

α(s) = ~is +12k(0)~js2

+16(−k(0)2~i + k′(0)~j − k(0)τ(0)~k)s3 + · · ·

=

s− k(0)2

6 s3

k(0)2 s2 + k′(0)

6 s3

−k(0)τ(0)6 s3

+ o(s3) (1.2)

For simplicity we write k(0) as k, k′(0) as k′, and τ(0) as τ . It is not hard to see that s =x + k2

6 x3 + o(x3), it is then clear that the piece of curve near s = 0 is just the graph of map F :I // R2 where

F (x) = (k

2x2 +

k′

6x3,−kτ

6x3) + o(x3). (1.3)

Exercise 1.3.1. Find the curvature and torsion of the graph of F (x) = (cos x, (sinx)3) at x = 0.

Example 1.3.1. The trace of α(t) = (t3, t2) is not a regular curve. Otherwise, there were aregular parameterized curve β(t) = (x(t), y(t)) whose trace is the same as that of α, so we have

y3(t) = x2(t). (1.4)

We may assume that β(0) = (0, 0). Since the tangent line at (0, 0) would be the y-axis by thecontinuity argument, so x′(0) = 0, then y′(0) = a 6= 0 because β is regular at t = 0. Therefore,we have x(t) = bt2 + o(t2) for some number b and y(t) = at + o(t). In view of eq. (1.4), we have(at + o(t))3 = (bt2 + o(t2))2 = o(t3), i.e., a3t3 = o(t3) as t // 0; then we must have a3 = 0,contradicting to the fact that a 6= 0.

Definition 1.3.1. A subset of R3 is called an one-manifold inside R3 if, up to a rigid motion ofR3, locally it is the graph of a smooth function of the form (y, z) = F (x).

1.4. EXERCISES 5

Here, “locally” means the part of subset inside an open set (could be very small) of R3. Soa non-simple (i.e., self-crossing) regular curve is not an one-manifold. Since figure eight is notsimple, it is not an one-manifold. A simple regular curve is called a “nice” curve if, for any pointp in the curve, the part of curve inside a sufficiently small open ball of R3 centered at p consists ofa single connected piece4. A “nice” curve is an one-manifold.

An one-manifold may not be a “nice” curve because it might be the disjoint union of two “nice”curves.

Example 1.3.2. A line is an one-manifold. The graph of a smooth real-valued function of one-variable is an one-manifold. All conics are one-manifolds, but which one of which is not a curve?

In the forthcoming notes, I will talk about the two-manifolds, i.e., the regular (non self-crossing)surfaces defined in chapter two of the textbook. The above definition of one-manifold can be easilygeneralized as follows:

Definition 1.3.2. A subset of R3 is called a two-manifold inside R3 if , up to a rigid motion ofR3, locally it is the graph of a smooth function of the form z = F (x, y).

Please note that, a regular surface defined in the textbook is nothing but a two-manifold definedas above, so it must be non self-crossing; however, a regular curve defined in the textbook can beself-crossing.

Please also note that, a curve must be connected, and a regular surface may not be connected,just like the fact that an one-manifold may not be connected.

In summary, we have the following relationships among the terminologies:

• regular surface = 2-manifold in R3;

• “nice” curve = connected 1-manifold in R2.

1.4 Exercises

Exercise 1.4.1. This exercise tests your understanding of a few basic concepts.1) Find a non-smooth function whose graph is a regular curve.2) Find a parameterized smooth curve whose trace is not a regular curve.3) Find a parameterized smooth curve which is not regular, but its trace is a regular curve.4) Find a regular curve which is not a 1-manifold.5) Find a 1-manifold which is not a regular curve.

Exercise 1.4.2. Let α(t) = (cos t, sin t, t) (where −∞ < t < ∞) be a parameterized smooth curve,C be the trace of α.

1) Show that C is a simple regular curve.2) Show that C is a 1-manifold.3) Find an arc length parametrization for C.4) Find the Frenet trihedron at t = 0.5) Find the torsion and curvature at any point of the curve.6) Find the local canonical form for C at t = 0.

4See examples in class for explanations

6 CHAPTER 1. CURVES

Chapter 2

Regular Surfaces

Prerequisite: Calculus II, i.e., multi-variable calculus — the kind of calculus needed for the studyof electromagnetism.

Some standard notations:

• U — an open set of Rn;

• ~r0 — a point in U ;

• f — a smooth map from U or Rm;

• Df(~r0) — the differential of f at ~r0, it is a linear map from Rn to Rm;

• Jf(~r0) — the standard matrix for Df(~r0), called the Jacobian matrix of f at ~r0;

• Σ — a regular surface;

• (Σ, p) — a pair such that p ∈ Σ;

• φ: (R2, 0)loc

// (Σ, p) is a local diffeomorphism, called the local parametrization of Σ aroundp.

Some standard terminologies:

• An open set of Rn — a subset of Rn such that, for any point of this subset, there is a ballcentered at this point which is entirely inside this subset.

• An open set of A ⊂ Rn — the intersection of A with an open set of Rn.

• f : U //Rm is smooth — each component function of f is a smooth (i.e., infinitely differ-entiable) real valued function;

• f : A // B is smooth where A ⊂ Rn and B ⊂ Rm are arbitrary subsets — f has a smoothextension to an open neighborhood of A, i.e., there is an open set U of Rn containing A anda smooth map f : U // Rm (not into B in general) such that f = f on A.

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8 CHAPTER 2. REGULAR SURFACES

• f : A // B is a diffeomorphism — f is one-to-one and onto, both f and its inverse f−1:B // A are smooth;

• f : (A, a)loc

// (B, b) is a local diffeomorphism — there is an open set A′ of A and an

open set B′ of B such that, a ∈ A′, b ∈ B′, and f is a map from A′ to B′ which maps a to b,moreover, f is a diffeomorphism.

2.1 Digression: Calculus

Let U be an open set of Rn, f : U // Rm a smooth (i.e., infinitely differentiable) map, ~r0 ∈ U .Then

f(~r) = f(~r0) + Df(~r0)(~r − ~r0) + o(|~r − ~r0|)where o(|~r − ~r0|) is a term that goes to zero faster than |~r − ~r0| does as ~r // ~r0. Note that thei-th component of Df(~r0)(~r − ~r0) is ∇f i(~r0) · (~r − ~r0), i.e.,

3∑

k=1

∂f i

∂xk(~r0)(xk − xk

0).

Remark 2.1.1. Df(~r0): Rn //Rm is a linear map whose (coordinate) standard matrix is calledthe jacobian matrix, denoted by Jf(~r0) whose (i, j)-entry is ∂f i

∂xj (~r0).

2.1.1 Chain Rule

Theorem 2.1.1 (Chain Rule). Let f : Rn // Rm and g: Rm // Rl be smooth maps. Then

D(g ◦ f)(x) = Dg(f(x)) ◦Df(x), i.e.,∂(g ◦ f)i

∂xj(x) =

∑

k

∂gi

∂yk(f(x))

∂fk

∂xj(x)

for all x ∈ Rn.

(Hint for the proof: Try to linearize g ◦ f around x.) In terms of Jacobian matrices, we haveJ(g ◦ f)(x) = Jg(f(x)) · Jf(x).

2.1.2 Inverse Function Theorems

Theorem 2.1.2 (Inverse Function Theorem). Let f be a smooth map from Rn to Rn. Then f isa local diffeomorphism around 0 if and only if Df(0) is an isomorphism. (we write this statementas follows: f : (Rn, 0) ∼=loc (Rn, f(0)) ⇐⇒ Df(0) is an isomorphism.)

(I.e., f is locally invertible as a smooth map if and only if it is infinitesimally invertible as alinear map. )

Corollary 2.1.3 (Inverse Function Theorem-Injective Form). Let f be a smooth map from Rn

to Rn+k. Suppose Df(0) is injective, then there is a local diffeomorphism φ: (Rn+k, f(0)) ∼=loc

(Rn+k, f(0)) such that for all x closer enough to 0.

φ ◦ f(x) = (x, 0).

2.2. BASIC DEFINITIONS 9

(I.e., f is infinitesimally injective if and only if locally it is left equivalent to a standard injection.)

Proof. We may assume the first n columns of Jf(0) are linearly independent. Then, if we let F :Rn+k → Rn+k be a smooth map defined as

F (x, y) = f(x) + (0, y),

then DF (0, 0) must be invertible. So there is an open set W containing 0 such that F |W is adiffeomorphism onto F (W ) ∈ Rn+k. Let φ = F−1|F (W ), then φ ◦ f(x) = φ ◦ F (x, 0) = (x, 0) for(x, 0) in W . The rest is clear.

Corollary 2.1.4 (Inverse Function Theorem-Surjective Form). Let f be a smooth map from Rn+k

to Rn with f(0) = 0. Suppose Df(0) is surjective, then there is a local diffeomorphism ψ: :(Rn+k, 0) ∼=loc (Rn+k, 0) such that for all (x, y) closer enough to 0,

f ◦ ψ(x, y) = x.

(I.e., f is infinitesimally surjective if and only if it is right equivalent to a standard projection.)

Proof. We may assume the first n rows of Jf(0) are linearly independent. Then, if we let F :Rn+k → Rn+k be a smooth map defined as

F (x, y) = (f(x, y), y),

then DF (0, 0) must be invertible. So there is an open set W containing (0, 0) such that F |W is adiffeomorphism onto F (W ) ∈ Rn+k. Let ψ = F−1|F (W ), then f ◦ ψ(x, y) = x for (x, y) in W .

Corollary 2.1.5 (Implicit Function Theorem). Let f : Rn+k //Rn be a smooth map with f(0, 0) =0. Suppose that Df(0, 0) maps Rn×0 isomorphically onto Rn. Then there is an open neighborhoodof 0 ∈ Rk on which there defines a unique smooth map g into Rn such that g(0) = 0 and

f(g(y), y) = 0.

Proof. Let F (x, y) = (f(x, y), y), then F (0, 0) = (0, 0) and DF (0, 0) is invertible. Then F is a localdiffeomorphism around (0, 0). Let ψ be the inverse of F in a product neighborhood of (0, 0), sayU × V ⊂ Rn+k. Define g on V by letting g(y) = p1 ◦ ψ(0, y) for any y ∈ V . Then g(0) = 0 and

0 = p1 ◦ Fψ(0, y) = p1 ◦ F (g(y), y) = f(g(y), y).

The uniqueness comes from the uniqueness of the inverse of F .

2.2 Basic Definitions

Definition 2.2.1. A subset of R3 is called a regular surface if locally it is the graph of a smoothfunction of two variables taking value in an one dimensional Euclidean space; i.e., up to a rigidmotion of R3, locally it is the graph of a smooth function of the form z = f(x, y).


Remark 2.2.1. In other word, for Σ ⊂ R3 to be qualified as a regular surface, for any point p ∈ Σ,we must be able to find a 2-dimensional vector subspace V of R3, an open set U of V and a smoothmap f : U // V ⊥ (the orthogonal complement of V ) such that Σ ∩W = Γf where W is an openset of R3 containing p. Of course, up to a rigid motion (which is smooth), we can assume that Vis the xy-coordinate plane, so V ⊥ is the z-coordinate axis and then the function is z = f(x, y) for(x, y) in an open set of the xy-coordinate plane.

Remark 2.2.2. Let φ: R3 // R3 be a rigid motion of R3 and Σ ⊂ R3. Then, Σ is a regularsurface if and only if φ(Σ) is a regular surface.

Example 2.2.1. Let U be an open set of R2 and f : U // R a smooth function. Then the graphof f is a regular surface.

Example 2.2.2. S2 := {~r ∈ R3 | |~r| = 1} is a regular surface. Let ~r0 = (x0, y0, z0) ∈ S2. Up to arotation, we may assume that z0 > 0. Then locally around point ~r0, S2 is the graph of z = f(x, y) =√

1− x2 − y2. In fact, if we take W = {(x, y, z) ∈ R3|z > 0} and U = {(x, y) ∈ R2|x2 + y2 < 1},then S2 ∩W = Γf .

Let Σ, W , V , U and f be as in the remark. Let π: R3 // V be the orthogonal projection ontoV , then π|Γf

: Γf//U is the inverse of ι: U //Γf which maps (x, y) to (x, y, f(x, y)). Note that

ι is smooth and its inverse π|Γfis smooth in the sense that it has a smooth extension to an open

set in R3 (e.g., W ), such an ι is called a diffeomorphism of U onto Σ∩W ( = Γf ). Therefore, wehave shown that, locally, a regular surface must be diffeomorphic to an open set of R2. In fact, thefollowing is true.

Theorem 2.2.1. A subset Σ of R3 is a regular surface if and only if locally it is diffeomorphicto an open set of R2; i.e., (Σ, p) ∼=loc (R2, 0) for any p ∈ Σ.

Corollary 2.2.2. Let Σ ⊂ R3 be a regular surface, φ: R3 // R3 be a diffeomorphism. Then theimage of Σ under φ is also a regular surface. In particular, the image of a regular surface under arigid motion of R3 is still a regular surface.

Let f : U //R be a smooth map. We say that y ∈ R is a regular value of f if either f−1(y)is empty or for any x ∈ f−1(y), the rank Df(x) is 1 (=the dimension of R).

Theorem 2.2.3. Let U be an open set of R3, f : U //R be a smooth map. Suppose that z0 is aregular value of f . Then f−1(z0) is a regular surface provided that it is non-empty.

Example 2.2.3. All but one of the six standard non-degenerate quadrics are regular surfaces.

2.2.1 Proof of theorem 2.2.1

Proof. Let φ: (R2, 0) ∼=loc (Σ, p). I.e., 1) φ: U // Σ is a one-to-one map (where U is an open setof R2 containing 0), smooth when viewed as a map into R3 ; 2) φ−1, when viewed as a map intoR2, has a smooth extension to Z — an open set of R3 containing the image of φ; 3) φ(0) = p.

Since Dφ(0) has rank 2, there exists a coordinate plane V in R3 such that if π: R3 // V is theorthogonal projection onto V , then D(π ◦φ)(0) = Dπ(p)◦Dφ(0) = π ◦Dφ(0) is an isomorphism, so

2.3. EXISTENCE OF TANGENT PLANES 11

π ◦ φ is a local diffeomorphism by the inverse function theorem. From the following commutativetriangle

(V, p) (Σ, p)ooπ

(R2, 0)

(V, p)

π◦φ²²

(R2, 0)

(Σ, p)

φ

ÂÂ???

????

????

where φ and π ◦ φ are local diffeomorphisms, we conclude that π in the triangle is a local diffeo-morphism, too.

Since R3 = V ⊕ V ⊥, viewed as a map into R3, π−1 has the following form:

π−1(u, v) = (u, v, f(u, v)).

provided that V is the uv-coordinate plane. Here f is a locally defined smooth function. Then(Σ, p) =loc (Γf , p).

2.2.2 Proof of theorem 2.2.3

Proof. Let p ∈ f−1(z0). We may assume that z0 = 0. Since 0 is a regular value of f , by the inversefunction theorem, there is a local diffeomorphism φ: (R3, 0)

loc// (R3, p) such that f ◦φ(x, y, z) = z.

Then,

(f−1(0), p)φ−1

∼= loc((f ◦ φ)−1(0), 0) =loc (R2, 0).

In view of theorem 2.2.1, we are done.

2.3 Existence of tangent planes

Let Σ ⊂ R3 be a regular surface and p ∈ Σ. Then there is a local diffeomorphism φ: (R2, 0) ∼=loc

(Σ, p). Such a φ is called a local parametrization and its inverse is called a local coordinatemap. It is customary to write (u, v) for the local parameters, so φ is a R3-valued function of (u, v).

The tangent plane of Σ at p, denoted by TpΣ, is defined to be the linear span of

{∂uφ(0), ∂vφ(0)}with the understanding that φ is a R3-valued function of (u, v).

One can show that this definition of tangent plane is actually independent of the choice of thelocal parameterizations: suppose that ψ: (R2, 0) ∼=loc (Σ, p) is another local parametrization, thenφ = ψ ◦ λ where λ: (R2, 0) ∼=loc (R2, 0) is a local diffeomorphism. By chain rule, we have

(∂uφ(0), ∂vφ(0)) = (∂uψ(0), ∂vψ(0)) · Jλ(0).

Since Jλ(0) is nonsingular, we have

span{∂uφ(0), ∂vφ(0)} = span{∂uψ(0), ∂vψ(0)}.


Remark 2.3.1. Actually, the tangent plane of Σ at p is the parallel translate of span{∂uφ(0), ∂vφ(0)}by p, i.e.,

TpΣ = p + span{∂uφ(0), ∂vφ(0)}.Since the identification of TpΣ with span{∂uφ(0), ∂vφ(0)} is quite natural, we simply write

TpΣ = span{∂uφ(0), ∂vφ(0)}in the above.

Example 2.3.1. The solution set of equation x2 + y2 = z2 is not a regular surface because itstangent plane at (0, 0, 0) does not exist.

Let Σ be a regular surface, α: I // R3 be a parameterized smooth curve. We say α is insideΣ if the trace of α is inside Σ. Assume that p ∈ Σ and α(0) = p, then we claim that α′(0) ∈ TpΣ.To show that, we choose a local parametrization φ: (R2, 0) ∼=loc (Σ, p). Locally around t = 0, wehave α(t) = φ(φ−1 ◦ α(t)) = φ(β(t)) where β = φ−1 ◦ α ≡ (β1, β2); then

α′(0) = Dφ(β(0))(β′(0))= Dφ(0)(β′(0))= β′1(0) ∂uφ(0) + β′2(0) ∂vφ(0)∈ TpΣ. (2.1)

In fact, more can be proved:

Exercise 2.3.1. Let Σ be a regular surface and p ∈ Σ. Show that

TpΣ ={α′(0)|α(t) is a parameterized smooth curve inside Σ with α(0) = p

}.

2.4 Local Canonical Form

Let Σ be a regular surface and p ∈ Σ, φ: (R2, 0) ∼=loc (Σ, p) be a local parametrization. Up to arigid motion of R3, we may assume that p is the origin of R3, TpΣ is the xy-coordinate plane. Letπ be the orthogonal projection of R3 onto the xy-coordinate plane, then D(π ◦ φ)(0) = π ◦Dφ(0)is an isomorphism, so π ◦ φ: (R2, 0) ∼=loc (Σ, p) is a local diffeomorphism. From the followingcommutative triangle

(R2, 0) (Σ, p)ooπ

(R2, 0)

(R2, 0)

π◦φ²²

(R2, 0)

(Σ, p)

φ

ÂÂ???

????

????

where φ and π ◦ φ are local diffeomorphisms, we conclude that π in the triangle is a local diffeo-morphism with π−1(u, v) = (u, v, f(u, v)) being a local parametrization. Since p = 0 and TpΣ isthe xy-coordinate plane, we have the Taylor expansion of f around (0, 0) as follows:

f(u, v) = Q(u, v) + o(u2 + v2)

2.5. EXERCISES 13

where Q is a quadratic form. We may further assume that Q is in diagonal form. Therefore, wehave proved that, up to a rigid motion of R3, locally, a regular surface is the graph of a functionof the following form:

z = f(x, y) =12

(κ1 x2 + κ2 y2

)+ o(x2 + y2) (2.2)

where κ1 and κ2 are some real numbers.

2.5 Exercises

Exercise 2.5.1. Show that quadric ax2+by2+cz2 = 1 is diffeomorphic to quadric x2+y2+z2 = 1when a, b and c are all positive.

Exercise 2.5.2. Show that the six standard non-degenerate quadrics listed in my notes on quadricsare mutually non-diffeomorphic except for a pair.

Exercise 2.5.3. Let A ⊂ R2 be the unit circle centered at the origin, B ⊂ R3 be the intersectionof plane x + y + z = 1 with the unit sphere centered at the origin. Show that A is diffeomorphic toB.

Exercise 2.5.4. Pick up a quadric, and a point on that quadric.1) Find the tangent plane of the chosen quadric at the chosen point.2) Find the local canonical form of the chosen quadric at the chosen point.


Chapter 3

Gauss Map

3.1 Smooth maps between regular surface

Let Σ1 and Σ2 be two regular surfaces and f : Σ1// Σ2 be a map. We say that f is smooth, if

as a map into R3, it has a smooth extension to an open set of R3 containing Σ1.Assume that p1 ∈ Σ1, p2 ∈ Σ2 and f(p1) = p2. We claim that f induces a linear map from Tp1Σ1

to Tp2Σ2. To show this claim, we let α be a parameterized smooth curve inside Σ1. Then f ◦ α isa parameterized smooth curve inside Σ2; here the smoothness can be seen this way: f ◦ α = f ◦ αwhere f is a smooth extension of f to W — an open set of R3 containing Σ1, i.e., f : W //R3 issmooth and f = f on Σ1. Now α′(0) ∈ Tp1Σ1 and

Tp2Σ2 3 (f ◦ α)′(0)= (f ◦ α)′(0)= Df(p1)(α′(0)) (3.1)

In view of exercise 2.3.1, we have shown that f induces a linear map

Tfp1 : Tp1Σ1// Tp2Σ2. (3.2)

Note that Tfp1(~w) = Df(p1)(~w), however, in view of the first line of the Eq. (3.1), Tfp1 is actuallyindependent of the choice of f .

Remark 3.1.1. Actually, what Tfp1 really does is to map p1 + ~w to p2 + Df(p1)(~w).

Remark 3.1.2. Tfp1 is well-defined for local map f : (Σ1, p1)loc

// (Σ2, p2) provided that f is

smooth.

3.2 Orientation, Gauss map and curvatures

Let Σ be a regular surface and p ∈ Σ. We say that Σ is oriented at p if a unit normal vector np

to TpΣ is chosen. We say that Σ is oriented if it is smoothly oriented everywhere, i.e., 1) Σ isoriented at each of its point; 2) for any p ∈ Σ, there is a local parametrization φ: (R2, 0) ∼=loc (Σ, p)

15

16 CHAPTER 3. GAUSS MAP

such that (∂uφ× ∂uφ)|φ−1(q) is parallel to nq for all q in the domain of φ−1. Let us call such a φ anorientation-preserving local parametrization.

Let Σ be an oriented regular surface and p ∈ Σ. Then the Gauss map

n : Σ // S2

which maps p to np is a smooth map. To show that, let φ: (R2, 0) ∼=loc (Σ, p) be an orientation-preserving local parametrization, then

n(q) =∂uφ× ∂uφ

|∂uφ× ∂uφ|(φ−1(q)) =

∂uφ× ∂uφ

|∂uφ× ∂uφ|(φ−1(q))

for all q in the domain of φ−1, hence n is smooth.The previous section says that we then have a linear map Tnp: TpΣ // Tn(p)S2. Note that,

as a unit vector in R3, n(p) is perpendicular to both TpΣ and Tn(p)S2, so we have the naturalidentification: Tn(p)S2 = TpΣ. Therefore, we have shown that Tnp is an endomorphism of TpΣ. Infact, we have

Theorem 3.2.1. Tnp is a self-adjoint operator on TpΣ.

There is a proof based on direct computation. However, there is more interesting way to provethis theorem. The basic idea is that one just needs to prove this theorem in the simpler case whenΣ is represented locally in a canonical form. (that is what canonical form is good for). Note that,to get the canonical form, we may need a rigid transformation R(x) = T (x)+ a where a is a vectorin R3 and T is an orthogonal transformation of R3. You are invited to do the following exercise.

Exercise 3.2.1. Under the rigid transformation R(x) = T (x) + a, Tnp is turned into

T ◦ Tnp ◦ T−1.

So Tnp is self adjoint if and only if T ◦ Tnp ◦ T−1 is.

Now, up to a rigid motion, we have a local a local parametrization φ: (R2, 0) ∼=loc (Σ, p) of theform

φ(x, y) = (x, y,12

(κ1 x2 + κ2 y2

)+ o(x2 + y2)),

then for (x, y, z) ∈ Σ, we have

n(x, y, z) =∂uφ× ∂uφ

|∂uφ× ∂uφ|= (−κ1x,−κ2y, 1) + o(

√x2 + y2). (3.3)

Let n(x, y, z) be equal to the righthand side of the last line in the above equation for any (x, y, z)in a sufficiently small open neighborhood U of 0 ∈ R3, then n is a smooth extension of n|U . Then,for any ~w = (w1, w2) ∈ TpΣ = R2 = R2 × 0 ⊂ R3, we have

Tnp(~w) = Dn(0)(~w) = (−κ1w1,−κ2w2).

3.3. DIGRESSION: LINEAR ALGEBRA 17

So, with the identification: TpΣ = R2, the standard matrix for Tnp is(−κ1 0

0 −κ2

).

So Tnp is self adjoint. In view of exercise 3.2.1, we know that {κ1, κ2} is uniquely determined by(Σ, p) because it is the set of eigenvalues of −Tnp, and is called the set of principal curvatures1

of Σ at p.

Definition 3.2.1. The mean curvature of Σ at p, denoted by Hp, is defined to be −12trTnp. The

Gauss curvature of Σ at p, denoted by Kp, is defined to be det Tnp

In terms of the principal curvatures κ1, κ2 at p, it is clear that Hp = κ1+κ22 and Kp = κ1κ2.

3.3 Digression: Linear Algebra

Let V be a real vector space of dimension n.

3.3.1 One-form

An 1-form on V is a linear map T : V //R. The set of all 1-forms on V is called the dual spaceof V , denoted by V ∗. Let v = (v1, · · · , vn) be a basis for V ,

φ : V // Rn

be the coordinate map with respect to v,

πi : Rn // R

be the projection to the i-th component of Rn, and

vi = πi ◦ φ.

Then v = (v1, · · · , vn) is a basis for V ∗ — the dual basis.

3.3.2 Two-form

A 2-form on V is a bilinear map T : V ×V //R. Here “bilinear” means that T is a linear in eachvariable, i.e., T (x, y) is linear in x for any fixed y and linear in y for any fixed x. We say that the2-form is symmetric if T (x, y) = T (y, x) for any x, y; we say that the 2-form is antisymmetricif T (x, y) = −T (y, x) for any x, y. An inner product 〈, 〉 on V is nothing but a symmetric andpositive-definite 2-form on V , here positive-definite means that 〈x, x〉 ≥ 0 for any x and 〈x, x〉 = 0if and only if x = 0.

Remark 3.3.1. Let λ be a two-form on V , then λ defines a linear map λ∗: V // V ∗ in a naturalway: λ∗(x)(y) = λ(x, y).

1their geometric meaning will be explained in class.


The set of 2-forms, the set of symmetric 2-forms and the set of antisymmetric 2-forms are allvector spaces. They are denoted by T 2V ∗, S2V ∗ and ∧2V ∗ respectively. It is clear that

T 2V ∗ = S2V ∗ ⊕ ∧2V ∗.

Let α and β be two 1-forms on V . For any x, y in V , we define

α⊗ β : V × V // R

as follows: α ⊗ β(x, y) = α(x)β(y). It is clear that α ⊗ β ( called the tensor product of α with β)is a 2-form on V , i.e., α⊗ β ∈ T 2V ∗.

Let

αβ = 12 (α⊗ β + β ⊗ α) ,

α ∧ β = 12 (α⊗ β − β ⊗ α) .

(3.4)

Then αβ is a symmetric 2-form on V and α ∧ β is an antisymmetric 2-form on V .

Exercise 3.3.1. Let V be a real vector space of dimension n, θ = (θ1, · · · , θn) be a basis for V ∗.Show that

1) {θi ⊗ θj | 1 ≤ i, j ≤ n} is a minimal spanning set for T 2V ∗.2) {θiθj | 1 ≤ i ≤ j ≤ n} is a minimal spanning set for S2V ∗.3) {θi ∧ θj | 1 ≤ i < j ≤ n} is a minimal spanning set for ∧2V ∗.

4) dim T 2V ∗ = n2, dimS2V ∗ = n(n+1)2 and dim∧2V ∗ = n(n−1)

2 .

Exercise 3.3.2. Let λ be a symmetric 2-form on V . Let θ = (θ1, . . . , θn) be a basis for V ∗ andv = (v1, . . . , vn) be the corresponding dual basis of V . Then λ =

∑λijθ

iθj for some (unique) realsymmetric matrix [λij ]. Show that [λij ] is the coordinate matrix for λ∗: V // V ∗ with respect to(v, θ).

Exercise 3.3.3. Let λ be an anti-symmetric 2-form on V . Let θ = (θ1, . . . , θn) be a basis for V ∗

and v = (v1, . . . , vn) be the corresponding dual basis of V . Then λ =∑

λijθi∧θj for some (unique)

anti-symmetric real matrix [λij ]. Show that [−λij ] is the coordinate matrix for λ∗: V // V ∗ withrespect to (v, θ), i.e., λ∗(vi) = −∑n

j=1 θjλji.

Let V , W be two-dimensional vector spaces, T : V //W a linear map. Let T ∗: W ∗ //V ∗ bethe dual of T , and ∧2T ∗ be the resulting endomorphism on ∧2V ∗: ∧2T ∗(α ∧ β) = T ∗(α) ∧ T ∗(β).Suppose that W = V , since dim∧2V ∗ = 1, then ∧2T ∗ must be the scalar multiplication by anumber.

Exercise 3.3.4. Show that this number is just detT .

3.3. DIGRESSION: LINEAR ALGEBRA 19

3.3.3 The area form on an oriented Euclidean plane

By definition, a Euclidean plane is just a pair (V, 〈, 〉) where V is a real two-dimensional vectorspace and 〈, 〉 is an inner product on V .

Observe that ∧2V ∗ is an one dimensional real vector space. By definition, an orientation onV is just a nonzero element of ∧2V ∗ with the understanding that two nonzero elements in ∧2V ∗

represent the same orientation if and only if they differ by the scalar multiplication of a POSITIVEnumber. It is clear that there are exactly two orientations on V .

Remark 3.3.2. This definition of orientation is intrinsic. The definition given in section 3.2 isextrinsic. The correspondence between these two definitions are fixed in this book as follows: LetV be a 2-dimensional vector subspace of R3, ω be a nonzero element in ∧2V ∗, n be a normal vectorof V in R3; then, ω and n represent the same orientation of V if there is a POSITIVE number λsuch that ω(v1, v2) = λ(v1 × v2) · n for any two vectors v1 and v2 in V .

An oriented Euclidean plane is just a triple (V, 〈, 〉, o) where V is a real two-dimensionalvector space, 〈, 〉 is an inner product on V and o is an orientation.

Note that there is an inner product on V ∗ which is naturally induced from the inner product onV : It is the one such that 〈, 〉∗: V //V ∗ is an isometry. It turns out that a basis is orthonormal ifand only if its dual basis is orthonormal. Then there is a natural inner product on T 2V ∗, too: it isthe (unique) inner product such that, when θ is an orthonormal basis for V ∗, { θi⊗θj√

2| 1 ≤ i, j ≤ n}

is an orthonormal minimal spanning set for T 2V ∗

Definition 3.3.1. Let (V, 〈, 〉, o) be an oriented Euclidean plane. The area form on V is the uniqueanti-symmetric unit two-form vol which represents orientation o.

Example 3.3.1. Let V = R2(= R2 × 0 ⊂ R3), with the standard inner product and standardorientation given. If we view V as a subset of R3, its standard orientation is more frequentlyrepresented by k as in section 3.2. Note that (i, j) is its standard orthonormal basis. Let (i, j) bethe corresponding dual basis for V ∗, then vol = i ∧ j . Note that vol is uniquely determined byvol(i, j) = 1

2! = 12!(i× j) · k .

Exercise 3.3.5. Let T be an endomorphism on R2, show that

vol(T (i), T (j)) =12!

(T (i)× T (j)) · k.

Exercise 3.3.6. Let a and b be vectors in R3. Show that

|a× b| =√

(a · a)(a · a)− (a · b)2.

Please be aware of our choice of the correspondence between the two different descriptionsof orientation: one given in this subsection (the intrinsic one) and one given in section 3.2 (theextrinsic one).


3.4 Fundamental Forms

3.4.1 The total differentials

Let Σ be a regular surface, p ∈ Σ, and f : Σ //R be a smooth map. Suppose that α: I //R3 isa parameterized smooth curve inside Σ with α(0) = p. Then f ◦ α: I // R is a smooth functionof one variable because f ◦ α = f ◦ α is a composition of two smooth maps, here f is a smoothextension of f . Therefore,

(f ◦ α)′(0) = (f ◦ α)′(0) = ∇f(p) · α′(0).

In view of exercise 2.3.1, we have shown that f induces a linear map

dfp : TpΣ // R (3.5)

which maps ~w to ∇f(p) · ~w. It is clear that dfp ∈ (TpΣ)∗ — the cotangent space of Σ at p. Weuse df to denote the map: p 7→ dfp. In the case f is a locally defined smooth function, df is alsodefined because dfp only depends on f locally. In any case, we call df the total differential of f .Note that df is in dependent of the choice of f because

dfp(α′(0)) = (f ◦ α)′(0) (3.6)

for any parameterized smooth curve inside Σ with α(0) = p.Let Σ be a regular surface, p ∈ Σ and φ: (R2, 0) ∼=loc (Σ, p) be a local parametrization. Write

φ−1 = (u, v), then u and v are locally defined smooth functions on Σ, so we have du and dv locallydefined on Σ. Note that (∂uφ(0), ∂vφ(0)) is a basis for TpΣ, and

dup(∂uφ(0)) = dup

(d

dt

∣∣∣∣t=0

φ(t, 0))

=d

dt

∣∣∣∣t=0

(u(φ(t, 0)) =d

dt

∣∣∣∣t=0

t = 1.

Similarly, dup(∂vφ(0)) = ddt

∣∣t=0

0 = 0. Likewise, dvp(∂vφ(0)) = 1 and dvp(∂uφ(0)) = 0. Therefore,we have shown that

Proposition 3.4.1. (dup, dvp) is a basis for (TpΣ)∗; in fact, it is the dual of (∂uφ(0), ∂vφ(0)).

Some Notations.

1. Let (Σ, p, φ) be as before, and f be a local smooth function or R3-valued local smooth functiondefined on the domain of φ. We write ∂uf(φ−1(q)) as fu(q) for any q in the domain of φ−1.In other word,

fu := ∂f∂u ◦ φ−1. (3.7)

We also write fuu as ∂2f∂u2 ◦ φ−1, fuvv as ∂3f

∂v∂v∂u ◦ φ−1, etc..

3.4. FUNDAMENTAL FORMS 21

2. Let F = (F 1, F 2, F 3) and G = (G1, G2, G3) be two locally defined smooth maps from Σ toR3. We write dF := (dF 1, dF 2, dF 3) and dG := (dG1, dG2, dG3) and

dF · dG := dF 1dG1 + dF 2dG2 + dF 3dG3. (3.8)

Therefore, if p is in the common domain of F and G, then (dF ·dG)|p is a symmetric two-formon TpΣ. Similarly, we write

dF ∧ dG := dF 1 ∧ dG1 + dF 2 ∧ dG2 + dF 3 ∧ dG3. (3.9)

Proposition 3.4.2. Let f : Σloc

//R be a locally defined smooth map, φ: (R2, 0)loc

// (Σ, p) be a

local diffeomorphism. Show that

df = (f ◦ φ)u du + (f ◦ φ)v dv (3.10)

on the common domain of f and φ.

Proof. Just need to verify that df(φu) = (f ◦φ)u and df(φv) = (f ◦φ)v. But the verification is justa computation. For example,

df(φu)|q =d

dt

∣∣∣∣t=0

f ◦ φ(φ−1(q) + (t, 0)

)= ∂u(f ◦ φ)(φ−1(q)) = (f ◦ φ)u|q.

Exercise 3.4.1. Let F : Σloc

//R3 be a locally defined smooth map. Then dF |p naturally defines

a linear map from TpΣ into R3: it maps X ∈ TpΣ to (dF 1p (X), dF 2

p (X), dF 3p (X)). In other word,

dF |p is a R3-valued 1-form on TpΣ. Show that

dFp(α′(0)) = (F ◦ α)′(0) (3.11)

for a parameterized smooth curve inside Σ with α(0) = p; consequently,

dF = (F ◦ φ)u du + (F ◦ φ)v dv. (3.12)

Let (Σ, p, φ) be as before, and ι: Σ //R3 be the inclusion map. Then ι is smooth because wecan take the identity map 1: R3 // R3 to be the smooth extension of ι. Since ι ◦ φ = φ, in viewof eq. (3.12) in the preceding exercise, locally we have

dι = φu du + φv dv. (3.13)


3.4.2 The first fundamental two-form

Now if we take V = TpΣ where Σ is a regular surface and p ∈ Σ. Then the dot product for vectorsin R3 defines a symmetric, positive-definite 2-form on TpΣ, denoted by Ip. I.e., for any two tangentvectors ~u, ~v in TpΣ ⊂ R3, we have

Ip(~u,~v) = ~u · ~v.

Clearly, Ip is just the inner product on TpΣ induced from the dot product of R3.

Exercise 3.4.2. In terms of notations introduced in the preceding subsection, show that1) I = dι · dι.2) Locally, we have I = E du2 + 2F du dv + Gdv2 where E = φu · φu, F = φu · φv, G = φv · φv.

3.4.3 The second fundamental two-form

Let n be the Gauss map. We have shown that Tnp is self adjoint with respect to inner product Ip,i.e.,

Ip(x, Tnp(y)) = Ip(Tnp(x), y)

for any x, y ∈ TpΣ. Define IIp(x, y) = −Ip(x, Tnp(y)) for any x, y ∈ TpΣ, then IIp is a symmetric2-form on TpΣ, called the second fundamental form of Σ at p. Why is IIp is symmetric? well,

IIp(x, y) = −Ip(x, Tnp(y)) = −Ip(Tnp(x), y) = −Ip(y, Tnp(x)) = IIp(y, x).

Exercise 3.4.3. In terms of notations introduced in the preceding subsection, show that1) II = −dn · dι.2) locally, we have II = e du2 + 2f du dv + g dv2 where e = n · φuu, f = n · φuv, g = n · φvv.3) (remembering the notation introduced in remark 3.3.1) the following triangle

(TpΣ)∗ TpΣooI∗p

TpΣ

(TpΣ)∗

−II∗p

²²

TpΣ

TpΣ

Tnp

ÂÂ???

????

????

is commutative. Consequently, we have

Kp = det(Tnp) = eg−f2

EG−F 2

∣∣∣p,

Hp = −12tr(Tnp) = 1

2eG+gE−2fF

EG−F 2

∣∣∣p.

(3.14)

Exercise 3.4.4. Let (Σ, p) be as before. Suppose that (Σ, p) is turned into (Σ′, p′) under the rigidtransformation R(x) = T (x) + a. show that TRp = T : TpΣ // Tp′Σ′ is compatible with the twofundamental forms. I.e., show that, for any x, y in TpΣ, we have

Ip(x, y) = I′p′(T (x), T (y)), IIp(x, y) = II′p′(T (x), T (y)). (3.15)

Exercise 3.4.5. In the local canonical form, we have TpΣ = R2 = R2 × 0 ⊂ R3. Compute Ip andIIp in this case.

3.5. EXERCISES 23

3.4.4 Area Form

Let Σ be an oriented regular surface and p a point on Σ. Note that TpΣ is an oriented Euclideanplane, so it has an area form volp. Then we have a map vol: p 7→ volp.

The basic objective here is to write down a local formula for vol. To do that, we let φ:(R2, 0)

loc// (Σ, p) be an orientation-preserving local parametrization, so (∂uφ, ∂vφ, n)|q form an

right-hand rule for all q in the domain of φ−1.In view of the example 3.3.1 and the subsequent exercises, we know that vol is uniquely deter-

mined by

vol(∂uφ, ∂vφ) =12(∂uφ× ∂vφ) · n, (3.16)

so

vol = (∂uφ× ∂vφ) · n du ∧ dv= |∂uφ× ∂vφ| du ∧ dv, (3.17)

i.e,

vol =√

EG− F 2 du ∧ dv. (3.18)

Note that, if we wish to indicate the dependence on Σ, we write vol as volΣ.Let n: Σ // S2 be the Gauss map. Then Tnp: TpΣ // TnpS2 is a linear map, then a linear

map∧2Tn∗p : ∧2(TnpS

2)∗ // ∧2 (TpΣ)∗

which maps volS2

npto a scalar multiple of volΣp . Note that, in the natural identification of TnpS2

with TpΣ, volS2

npgets identified with volΣp , therefore, in view of exercise 3.3.4 and the fact that

det Tnp = Kp, we have

n∗(volS2) = K volΣ

where K is the Gauss curvature, and notation n∗ is introduced: by definition, n∗(volS2)|p :=

∧2Tn∗p(volS2

np).

3.5 Exercises

Exercise 3.5.1. Compute the mean curvature and the Gauss curvature of the sphere with radiusr at any point. How about a general non-degenerate quadric?

Let Σ1 and Σ2 be two regular surfaces, p1 ∈ Σ1 and p2 ∈ Σ2. Let f : (Σ1, p1)loc

// (Σ2, p2) be

a local diffeomorphism. We say that f is an local isometry from (Σ1, p1) to (Σ2, p2) if Tfq is a(linear) isometry for each q in the domain of f .


Exercise 3.5.2. Let Σ be the cylindrical surface with base the unit circle, i.e.,

Σ = {(x, y, z) ∈ R3 |x2 + y2 = 1}.

Show that, 1) there is a local isometry from (Σ, p) to (R2, 0) for any point p ∈ Σ; 2) Σ and R2 havedifferent principal curvatures and different mean curvature, but the same Gauss curvature.

Exercise 3.5.3 (The third fundamental form). Let III = dn · dn. Show that IIIp(x, y) =Ip(dnp(x), dnp(y)) for any x, y in TpΣ. (Optional question 1: Locally we write III = Edu2 +Fdudv + Gdv2, can you work out the formulae for E , F and G in terms of E, F , G, e, f , g?)(Optional question 2: Show that dι ∧ dι = dn ∧ dι = dn ∧ dn = 0.)

Chapter 4

Intrinsic Theory

4.1 Vector Fields

Let Σ be a regular surface, X: Σ // R3 a smooth map. We say that X is a (smooth) vectorfield on Σ if X(p) ∈ TpΣ for any p ∈ Σ. In the case that X is only locally defined, X will be calleda local vector field.

Example 4.1.1. Let Σ be a regular surface and p ∈ Σ. Suppose that φ: (R2, 0)loc

// (Σ, p) is a

local parametrization, then

∂u : Σloc

// R3

q 7→ ∂φ

∂u

∣∣∣∣φ−1(q)

= φu(q). (4.1)

is a local vector field on Σ. Similarly, we have the local vector field ∂v. Note that (∂u, ∂v) is calleda local coordinate frame of Σ because, at each point where they are defined, it is a basis for thetangent plane of Σ at that point. Similarly, (du, dv) is called a local coordinate coframe of Σ;in fact, it is the dual of (∂u, ∂v). Note that ∂u = φu and ∂v = φv.

Remark 4.1.1. Let X be a local vector field and f be a local smooth function, both are locallydefined in an open set U . Then fX: p 7→ f(p)X(p) is a local vector field.

Exercise 4.1.1. Let φ, (∂u, ∂v) be as in the preceding example. Let X be a vector field, showthat, locally,

X = X1∂u + X2∂v

for some smooth local functions X1 and X2.

The following proposition is very basic and useful:

Proposition 4.1.1 (local existence of flow line). Let X be a local vector field defined on open setU of Σ. Then there is an open set W of U × R containing U × 0 and a smooth map Φ: W // Σsuch that, 1) Φ(q, 0) = q for any q ∈ U ; 2) ∂tΦ(q, t) = X(Φ(q, t)) for any (q, t) ∈ W . Moreover,1) any two such Φ’s must agree on the common domain of definition; 2) Φ(Φ(q, t), s) = Φ(q, t + s)when both sides make sense.

25

26 CHAPTER 4. INTRINSIC THEORY

The proof is just a direct consequence of local existence and uniqueness theorem for ODE, soit is omitted.

Remark 4.1.2. Map Φ in the proposition is called a flow map for X.

Example 4.1.2. Let Σ be a regular surface and p ∈ Σ. Suppose that φ: (R2, 0)loc

// (Σ, p) is a

local parametrization. LetΦ(q, t) = φ(φ−1(q) + (t, 0)).

Then Φ is a flow map for ∂u.

Example 4.1.3. Let S1 be the unit circle centered at the origin of R2, and X: S1 //R2 be thevector field whose value at (x, y) is (−y, x). Then

Φ((x, y), t) = (x cos t− y sin t, x sin t + y cos t)

is a flow map for X. Here W = S1 × R.

4.1.1 Lie derivatives of smooth functions

Let Σ be a regular surface, p ∈ Σ. Let X be a local vector field and f be a local smooth function,both are locally defined in an open neighborhood of p.

Definition 4.1.1. The Lie derivative of f with respect to X, denoted by LXf , is defined as follows:for any p in the common domain of f and X,

LXf(p) =d

dt

∣∣∣∣t=0

f(Φ(p, t)) ( = dfp(X(p)) )

where Φ is a flow map for X.

Note that LXf is a local smooth function, and LXf = 0 if either X = 0 or f is a constantfunction. Not also that LXf = df(X): p 7→ dfp(X(p)).

Exercise 4.1.2. Let X, Y be local vector fields, f and g be local smooth functions. Show that,on the common domain of X, f and g, we have

LX(f · g) = LXf · g + f · LXg.

LgXf = g · LXf.

LX(f + g) = LXf + LXg.

LX+Y f = LXf + LY f. (4.2)

Here the · between functions means the multiplication of functions.

4.2. DIRECTIONAL DERIVATIVES 27

4.1.2 Gradient vector fields

Let Σ be a regular surface and f a local smooth function on Σ. Note that I∗p: TpΣ // (TpΣ)∗ is anisomorphism. By definition, the gradient vector of f at p, denoted by ∇f |p, is the unique tangentvector which is mapped to dfp under I∗p. Then map ∇f : p 7→ ∇f |p is a local vector field. (Why is∇f smooth?)

Exercise 4.1.3. Let X be a local vector field, f is a local smooth function. Show that

I(X,∇f) = LXf = df(X).

4.1.3 Lie derivatives of Rm-valued smooth functions

Let Σ be a regular surface, p ∈ Σ. Let X be a local vector field and F be a local Rm-valued smoothfunction, both are locally defined in an open neighborhood of p.

Definition 4.1.2. The Lie derivative of F with respect to X, denoted by LXF , is defined as follows:for any p in the common domain of F and X,

LXF (p) =d

dt

∣∣∣∣t=0

F (Φ(p, t)) ( = dFp(X(p)) )

where Φ is a flow map for X.

Note that LXF is a local Rm-valued smooth function, and LXF = 0 if either X = 0 or F is aconstant function.

Exercise 4.1.4. Let X, Y be local vector fields, g be local smooth function, and F , G be localRm-valued smooth functions. Show that, on the common domain of X, Y , g, F and G, we have

LX(g · F ) = LXg · F + g · LXF.

LgXF = g · LXF.

LX(F + G) = LXF + LXG.

LX+Y F = LXF + LY F. (4.3)

Here the · means the scalar multiplication.

4.2 Directional Derivatives

4.2.1 Directional derivatives of Rm-valued smooth functions

Let Σ be a regular surface, p ∈ Σ. Let V ∈ TpΣ, F be a local Rm-valued smooth function. Let Xbe a local vector filed such that X(p) = V .


Definition 4.2.1. The directional derivative of F (as a Rm-valued smooth function) with respectto V , denoted by V (F ), is defined to be (LXF )(p).

Exercise 4.2.1. Use the results established in Ex. 4.1.4 to show that this definition makes sense,i.e., independent of choice of X. Moreover, show that, for local smooth function g and local smoothRm-valued smooth functions F and G, we have

V (g · F ) = V (g) · F (p) + g(p) · V (F ).

V (F + G) = V (F ) + V (G). (4.4)

Exercise 4.2.2. Let g be a smooth function on Σ, (∂u, ∂v) be a local coordinate frame. Then,locally

∂u(g) = (g ◦ φ)u, ∂v(g) = (g ◦ φ)v.

Exercise 4.2.3. Let V1 and V2 be two tangent vectors of Σ at p. Show that V1 = V2 ⇐⇒V1(g) = V2(g) for any smooth function g locally defined around p.

4.2.2 Covariant derivatives of vector fields

Let Σ be a regular surface, X, Y be local vector fields with a non-empty common domain. SinceY is a smooth map into R3, LXY is a well-defined R3-valued smooth functions on the commondomain of X and Y . However, LXY may not be a vector field anymore, but if we project out thethe component in the normal space, we get a vector field. More precisely, for any p ∈ Σ, if wedenote by πp the orthogonal projection of R3 onto TpΣ, and let ∇XY |p = πp (LXY (p)), then ∇XY :p 7→ ∇XY |p for p in the common domain of X and Y is a vector field — the covariant derivativeof Y with respect to X.

Exercise 4.2.4. 1) Show that

∇XY = LXY − (n · LXY )n.

2) Let X, Y , Z be local vector fields, g be local smooth function. Show that, on the commondomain of g, X, Y and Z, we have

∇X(g · Y ) = LXg · Y + g · ∇XY.

∇gXY = g · ∇XY.

∇X(Y + Z) = ∇XY +∇XZ.

∇X+Y Z = ∇XZ +∇Y Z. (4.5)

Here the · means the scalar multiplication. This implies that, locally, ∇ is completely determinedby ∇∂u∂u, ∇∂u∂v, ∇∂v∂u,∇∂v∂v.

We are now ready to make a definition.

4.2. DIRECTIONAL DERIVATIVES 29

Definition 4.2.2. Let Σ be a regular surface, p ∈ Σ. Let V ∈ TpΣ, Y be a local vector field. LetX be a local vector field such that X(p) = V . The directional derivative of Y (as a vector field)with respect to V , denoted by V (Y ), is defined to be (∇XY )|p.Exercise 4.2.5. Use the results established in Ex. 4.2.4 to show that this definition makes sense,

i.e., independent of choice of X. Moreover, show that, for local smooth function g and local vectorfield Y and Z, we have

V (g · Y ) = V (g) · Y (p) + g(p) · V (Y ).

V (Y + Z) = V (Y ) + V (Z). (4.6)

Let Σ be a regular surface and p ∈ Σ. Suppose that φ: (R2, 0)loc

//(Σ, p) is a local parametriza-

tion. Then ∂u and ∂v are local vector fields around p. Suppose that X is a local vector field aroundp, we wish to calculate ∇∂uX and ∇∂vX. First of all,

L∂uX|q =d

dt

∣∣∣∣t=0

X(φ(φ−1(q) + (t, 0))) =∂

∂u(X ◦ φ)|φ−1(q) = (X ◦ φ)u(q). (4.7)

In other word,

L∂uX = (X ◦ φ)u. (4.8)

Since ∂u ◦ φ = ∂uφ and ∂v ◦ φ = ∂vφ, we have

L∂u∂u = φuu,L∂u∂v = L∂v∂u = φuv,L∂v∂v = φvv. (4.9)

Therefore,

∇∂u∂u = φuu − (n · φuu)n,∇∂u∂v = ∇∂v∂u = φuv − (n · φuv)n,∇∂v∂v = φvv − (n · φvv)n. (4.10)

4.2.3 The covariant differentiation is non-commutative in general

Let Σ be a regular surface and p ∈ Σ. Suppose that φ: (R2, 0)loc

//(Σ, p) is a local parametrization.

The purpose here is to point out that, in general, the covariant differentiation is non-commutative.I.e., for vector field Z on regular surface Σ, in general,

R(∂u, ∂v)Z := ∇∂u(∇∂vZ)−∇∂v(∇∂uZ) 6= 0.

In fact, the failure of commutativity is measured by the Gauss curvature.First of all, we would like to mention that, in view of Ex. 4.2.4, by direct computations, we can

see that

R(∂u, ∂v)(g · Z) = g ·R(∂u, ∂v)Z (4.11)


for any local smooth function g and local vector field Z. Therefore,

R(∂u, ∂v)Z|pdepends only on the value of Z at p, so we have a map

F12|p : TpΣ // TpΣV 7→ R(∂u, ∂v)X|p (4.12)

where X is a local vector field whose value at p is V . It is not hard to see that F12|p is a linearmap, to determine it, we just need to compute F12(∂u) and F12(∂v) at p.

To simplify the computation, we may use the local canonical form, i.e., we may assume that,locally around p, Σ is the graph of

z = f(x, y) =12(κ1x

2 + κ2y2) + o(x2 + y2).

Thenφ(u, v) = (u, v, f(u, v)).

is a local parametrization. I will leave the rest to you as an exercise.

Exercise 4.2.6. Show that, at point p, we have

F12(∂u) = −K∂v

F12(∂v) = K∂u. (4.13)

where K = κ1κ2 is the Gauss curvature at p.

4.3 Riemann Curvature

Here we take a closer look at what we have done in the previous section, and that leads to the“discovery” of the Riemann curvature.

Let X, Y and Z be local vector fields, g be a local smooth function. On their common domain,do we have something like Eq. (4.11), i.e.,

(∇X∇Y −∇Y∇X)(gZ) = g(∇X∇Y −∇Y∇X)Z ?

Well, let us compute. Since

∇X∇Y (gZ) = ∇X(g∇Y Z + Y (g)Z)= g∇X∇Y Z + X(g)∇Y Z + Y (g)∇XZ + X(Y (g))Z, (4.14)

we have

(∇X∇Y −∇Y∇X)(gZ) = g(∇X∇Y −∇Y∇X)Z + (X(Y (g))− Y (X(g)))Z. (4.15)

Sorry, there is an extra term here because X(Y (g))− Y (X(g)) is not zero in general.

4.3. RIEMANN CURVATURE 31

4.3.1 Digression: Lie bracket

Locally, with a local parametrization φ chosen, we can write X = X1∂u + X2∂v and Y = Y 1∂u +Y 2∂v. Then

X(Y (g)) = X(Y 1∂u(g) + Y 2∂v(g))= X1∂u(Y 1∂u(g) + Y 2∂v(g)) + X2∂v(Y 1∂u(g) + Y 2∂v(g))= X1Y 1∂u(∂u(g)) + X1Y 2∂u(∂v(g)) + X1∂u(Y 1)∂u(g) + X1∂u(Y 2)∂v(g)+ X2Y 1∂v(∂u(g)) + X2Y 2∂v(∂v(g)) + X2∂v(Y 1)∂u(g) + X2∂v(Y 2)∂v(g)= X1Y 1∂u(∂u(g)) + X2Y 2∂v(∂v(g)) + X1Y 2∂u(∂v(g)) + X2Y 1∂v(∂u(g))

+X(Y 1)∂u(g) + X(Y 2)∂v(g) (4.16)

Note that ∂u(∂v(g)) = (g ◦ φ)vu = (g ◦ φ)uv = ∂v(∂u(g)), then we have

X(Y (g))− Y (X(g)) =[(X(Y 1)− Y (X1))∂u + (X(Y 2)− Y (X2))∂v

](g). (4.17)

Note that (X(Y 1) − Y (X1))∂u + (X(Y 2) − Y (X2))∂v is a local vector field, but its value at eachpoint of its domain is independent of φ because of Ex. 4.2.3 and the preceding equation. In otherword, there is a (global) vector field on Σ, denoted by [X, Y ], and is uniquely determined by thefollowing defining equation:

[X, Y ](g) := X(Y (g))− Y (X(g)). (4.18)

Locally, we have

[X, Y ] = (X(Y 1)− Y (X1))∂u + (X(Y 2)− Y (X2))∂v. (4.19)

Remark that [X, Y ] is called the Lie bracket of X and Y .

Exercise 4.3.1. Let X, Y and Z be local vector fields. On the common domain, show that

[X, Y ] = −[Y, X] skew symmetry[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0 Jacobi identity. (4.20)

Exercise 4.3.2. Let X, Y be local vector fields. On the common domain, show that

LXY − LY X = [X, Y ]. (4.21)

End of digression.

4.3.2 Curvarure Tensor

After the digression, we come back to Eq. 4.15 and rewrite it as follows:

(∇X∇Y −∇Y∇X)(gZ) = g(∇X∇Y −∇Y∇X)Z + [X, Y ](g)Z. (4.22)

Since∇[X,Y ](gZ) = [X, Y ](g)Z + g∇[X,Y ]Z,


we can rewrite the above equation as

(∇X∇Y −∇Y∇X −∇[X,Y ])(gZ) = g(∇X∇Y −∇Y∇X −∇[X,Y ])Z. (4.23)

In summary, if we let

R(X, Y ) = ∇X∇Y −∇Y∇X −∇[X,Y ], (4.24)

then

R(X, Y )(gZ) = gR(X, Y )Z. (4.25)

In fact, one can show thatR(fX, gY )(hZ) = (fgh)R(X, Y )Z;

consequently, at each point p, we have a multi-linear map

Rp : TpΣ× TpΣ× TpΣ // TpΣ(X, Y, Z) 7→ R(X, Y )Z|p (4.26)

where X, Y and Z are local vector fields whose values at p are X, Y and Z respectively. The mapR that assigns Rp to p ∈ Σ is called the Riemann curvature tensor.

Exercise 4.3.3. Let (X, Y ) be a basis of TpΣ. Show that1

Kp =〈Rp(X, Y, Y ), X〉〈X × Y, X × Y 〉 .

Therefore, if (X, Y ) is a local (tangent) frame, we have

K =〈R(X, Y )Y, X〉〈X × Y, X × Y 〉 . (4.27)

4.4 The covariant differentiation is intrinsic

The goal of these section is show that the covariant differentiation of vector fields on regular surfaceΣ is completely determined by the first fundamental form I on Σ.

To begin, we need to introduce some notations. Let

∇∂u∂u = Γ111∂u + Γ2

11∂v.∇∂u∂v = Γ1

12∂u + Γ212∂v.

∇∂v∂u = Γ121∂u + Γ2

21∂v.∇∂v∂v = Γ1

22∂u + Γ222∂v. (4.28)

Let g11 = E, g12 = F = g21 and g22 = G. In view of Eq. (4.10), after taking the inner product ofthe above equations with ∂u = φu and ∂v = φv, we have

12Eu = φuu · φu = Γ1

11g11 + Γ211g12 ≡ Γ111.

1Hint: use Ex. 4.2.6.

4.5. RIEMANN CURVATURE AND GAUSS CURVATURE ARE ALL INTRINSIC 33

Fu − 12Ev = φuu · φv = Γ1

11g12 + Γ211g22 ≡ Γ211.

12Ev = φuv · φu = Γ1

12g11 + Γ212g12 ≡ Γ112.

12Gu = φuv · φv = Γ1

12g12 + Γ212g22 ≡ Γ212.

φvu · φu = Γ121g11 + Γ2

21g12 ≡ Γ121.φvu · φv = Γ1

21g12 + Γ221g22 ≡ Γ221.

Fv − 12Gu = φvv · φu = Γ1

22g11 + Γ222g12 ≡ Γ122.

12Gv = φvv · φv = Γ1

22g12 + Γ222g22 ≡ Γ222.

It is clear then that Γabc’s are completely determined by the first fundamental form.Note that φuv = φvu, we have Γi

jk = Γikj and Γijk = Γikj . Since

(g11 g12

g21 g22

)=

(E FF G

)

is a non-singular matrix, we know that Γijk’s and Γabc’s determine each other. Consequently, Γi

jk’sand the covariant differentiation are completely determined by the first fundamental form.

4.5 Riemann curvature and Gauss curvature are all intrinsic

The last two sections imply that Riemann curvature and Gauss curvature are all intrinsic. Thatis because the covariant differentiation is intrinsic, Riemann curvature tensor is defined in termsof covariant differentiation and the Gauss curvature can be defined in terms of Riemann curvaturetensor and the first fundamental form (see Ex 4.3.3).

4.6 Exercises

Exercise 4.6.1. Let Σ be a regular surface, X, Y , Z be vector fields on Σ and ∇XY denote thecovariant derivative of Y with respect to X as defined before. Show that

1. ∇XY −∇Y X − [X, Y ] = 0 (torsion free);

2. LX(〈Y, Z〉) = 〈∇XY, Z〉+ 〈Y,∇XZ〉 (metric compatible ).

Show also that the two properties above plus the properties listed in part 2) of Exercise 4.2.4determine the covariant differentiation uniquely.

Exercise 4.6.2. Let Σ be a regular surface, X, Y , Z and W be vector fields on Σ and ∇XYdenote the covariant derivative of Y with respect to X as defined before. Show that

1. 〈R(X, Y )Z,W 〉 = −〈R(Y, X)Z, W 〉 = −〈R(X, Y )W,Z〉;


2. R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0;

3. 〈R(X, Y )Z,W 〉 = 〈R(Z,W )X, Y 〉;4. Let ∇W R be defines as follows:

(∇W R)(X, Y )Z := R(∇W X, Y )Z + R(X,∇W Y )Z + R(X, Y )(∇W Z)−∇W (R(X, Y )Z),

show that 1) (∇W R)(fX, gY )(hZ) = fgh(∇W R)(X, Y )Z and 2)

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

Chapter 5

Selected topics

5.1 Geodesics

Let Σ be a regular surface, P , Q be two points of Σ, α: [a, b] // Σ be a regular smooth map withα(a) = P and α(b) = Q, then the trace of α is a smooth path in Σ from P to Q. The length of thispath is

L[α] =∫ b

a|α′(t)| dt. (5.1)

We wish to find a length-minimizing path from P to Q. Note that for a smooth function f(x) toachieve minimal value at x0 we must have

∂f

∂xi

∣∣∣∣x0

= 0.

The same argument implies that, for L to achieve minimal value at α0 we must have

∂L

∂α(t)

∣∣∣∣α0

= 0.

The only difference is that while the index i in xi is discrete, the index t in α(t) is continuous.Note that in the case the index is continuous, people prefer to write ∂ as δ; thus, for L to achieveminimal value at α0 we must have

δL

δα(t)

∣∣∣∣α0

= 0. (5.2)

5.1.1 Digression

What is δLδα(t) then? Well, it is an infinite dimensional analogue of ∂f

∂xi , and is called the variationalderivative of L with respect to α. Note that

∆f =∑

i

∂f

∂xi∆xi + · · · .

35

36 CHAPTER 5. SELECTED TOPICS

Here · · · means terms in higher order of ∆xi’s. So, the infinite dimensional analogue will be

∆L =∫

dtδL

δα(t)∆α(t) + · · · .

Example 5.1.1. Let P , Q be points in R3. For each smooth map x: [t1, t2] //R3 with x(t1) = Pand x(t2) = Q, we let

S[x] =∫ t2

t1

12|x′(t)|2 dt.

What isδS

δx(t)

then? Well, we can compute it this way:

∆S =∫ t2

t1

12

(|(x + ∆x)′(t)|2 − |x′(t)|2) dt

=∫ t2

t1

12

(2x′(t) · (∆x)′(t) + |∆x(t)|2) dt

=∫ t2

t1

x′(t) · (∆x)′(t) dt + · · ·

=∫ t2

t1

[(x′ ·∆x)′(t)− x′′(t) ·∆x(t)] dt + · · ·

= x′(t) ·∆x(t)|t2t1 +∫ t2

t1

−x′′(t) ·∆x(t) dt + · · ·

=∫ t2

t1

−x′′(t) ·∆x(t) dt + · · · (5.3)

Therefore,δS

δx(t)= −x′′(t).

In reality, people compute δSδx(t) in a slightly more efficient way as follows:

δS =∫ t2

t1

x′(t) · δx′(t) dt

=∫ t2

t1

[(x′ · δx)′(t)− x′′(t) · δx(t)

]dt

= x′(t) · δx(t)|t2t1 +∫ t2

t1

−x′′(t) · δx(t) dt

=∫ t2

t1

−x′′(t) · δx(t) dt (5.4)

End of digression.

5.1. GEODESICS 37

After the above digression, we return to our question about length-minimizing path. Let φ bea local parametrization of Σ and write φ−1 ◦ α(t) as x(t) = (x1(t), x2(t)), then

α′(t) = (φ ◦ φ−1 ◦ α)′(t) = x1(t)φu(α(t)) + x2(t)φv(α(t)),

therefore, locally we have|α′|2 =

∑

i,j

hij(x)xixj

where

h11(x) = E(α) = (∂uφ · ∂uφ)(x)h12(x) = F (α) = (∂uφ · ∂vφ)(x)h22(x) = G(α) = (∂vφ · ∂vφ)(x). (5.5)

Instead of computing the variational derivative for L[α], let us first compute the variationalderivative for

S[α] =12

∫ b

a|α′(t)|2dt. (5.6)

Here it is:

δS[α] =12

∫ b

aδ(|α′(t)|2)dt

=12

∫ b

aδ(hij(x)xixj)dt

=12

∫ b

a(∂khij(x) δxkxixj + 2hij(x)xiδxj)dt

=12

∫ b

a(∂khij(x) δxkxixj − d

dt[2hij(x)xi]δxj)dt

=12

∫ b

a[∂khij(x)xixj − d

dt(2hik(x)xi)]δxkdt

=∫ b

a−[hik(x)xi +

12

(∂jhik(x) + ∂ihjk(x)− ∂khij(x)) xixj ]δxkdt (5.7)

Therefore, we have

− δS

δxl(t)= hil(x)xi + Γlij(x)xixj = hkl(x)(xk + Γk

ij(x)xixj).

So, for S to be minimal at x(t), x(t) must satisfy the geodesic equation:

xk + Γkij(x)xixj = 0. (5.8)

But how about the variational derivative of

L[α] =∫ b

adt |α′(t)|.

38 CHAPTER 5. SELECTED TOPICS

Well, you may do the computation directly, but it is bit com-plicated because of the appearanceof the square root. So I will do it differently. Let

S[α, e] =12

∫ b

a

[|α′(t)|2 1

e(t)+ e(t)

]dt. (5.9)

I claim that if L achieves a minimal at α(t) if and only if S achieves a minimal at (α(t), e(t)) forsome e(t).

Index

plane curve, 1

rigid motion, 3

smooth, 1

tangent line, 2

39

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Notes on Math 321