differential geometry notes

Contents

Content to be covered i

Chapter 1. Curve Theory 11. What is a curve? 12. Regular curves 33. Curvature and Torsion 54. Fundamental Theorem of Space Curves 145. Isoperimetric Inequality and Four Vertex Theorem 16

Chapter 2. Surfaces in R3 221. Surfaces: An introduction 222. Calculus on Surface 253. The First Fundamental Form 284. Local isometries and conformal maps 29

Chapter 3. Curvature of a surface 371. Second Fundamental Form 372. Gaussian, Mean and Principal Curvatures 39

Chapter 4. Geodesics and some fundamental results 481. Geodesics 492. Codazzi - Mainardi and Gauss equations 533. The Gauss-Bonnet Theorem 57

Content to be covered

Unit ISpace curves, planar curves, parameterization, curvature, torsion, signed curvature,Frenet-Serret equations, fundamental theorem of curve theory. Isoperimetric Inequal-ity, The Four Vertex Problem.

Unit IISurfaces: smooth surfaces, tangents, normals, first fundamental form, isometries ofsurfaces, conformal mappings of surfaces, surface area.

Unit IIISecond fundamental form, normal and principal curvature, Meunier’s theorem, Euler’stheorem, Gaussian and mean curvature, Gauss map.

Unit IVGauss equation, Christoffel symbols, Codazzi-Mainardi equations, Theorema Egregium,geodesics, local Gauss Bonnet theorem.

Reference Books(i) Andrew Pressly, Elementary Differential Geometry, SUMSeries, 2004.(ii) Goetz A., Introduction to Differential Geometry, Addison Wesley, Publ. Co., 1970.(iii) Weatherburn, C.E., Differential Geometry in Three Dimensions, Cambridge Uni-

versity Press, 1964.

i

Differential Geometry Notes

Author:Prakash A. Dabhi

CHAPTER 1

Curve Theory

1.1. What is a curve?Definition 1.1.1. A parametrized curve in Rn is a continuous function γ : I → Rn,where I is an interval in R.

Examples 1.1.2. Parametrization of Cartesian curves.

(i) A parametrization of a parabola y = x2 is γ(t) = (t, t2), t ∈ R. The curve γ1(t) =(t2, t4), t ∈ R is not a parametrization of y = x2. The curve γ(t) = (t3, t6), t ∈ R isa parametrization of y = x2.

(ii) The curve γ(t) = (a cos t, b sin t), t ∈ R is a parametrization of the ellipse x2

a2 + y2

b2 =1.

(iii) The curve γ(t) = (a sec t, b tan t), t ∈ (−π2 ,

π2 ) is a parametrization of the hyper-

bola x2

a2 − y2

b2 = 1.(iv) The curve γ(t) = (a cos3 t, a sin3 t), t ∈ (0, 2π] or t ∈ R is a parametrization of the

astroid x 23 + y 2

3 = a 23

Examples 1.1.3.

(i) Let γ : R → R2 be defined by γ(t) = (a cos t, b sin t), where a, b ∈ R\0. Thenthe image of γ is an ellipse in R2. Its Cartesian equation is x2

a2 + y2

b2 = 1. Inparticular when a = b, then the image of γ is a circle in R2.

(ii) Let γ : R→ R2 be defined by γ(t) = (t, t2). Then the image of γ is a parabola inR2. Its Cartesian equation is y = x2.

(iii) Let γ : R→ R2 be defined by γ(t) = (a cosh t, b sinh t), where a, b ∈ R\0. Thenthe image of γ is a part of the hyperbola x2

a2 − y2

b2 = 1.(iv) Let γ : R → R2 be defined by γ(t) = (et cos t, et sin t). Then the image of γ is a

logarithmic spiral in R2.(v) Let γ : R→ R3 be defined by γ(t) = (a+lt, b+mt, c+nt), where a, b, c, l,m, n ∈ R

and l2 +m2 +n2 6= 0. Then the image of γ is a line in R3 passing through (a, b, c)having direction (l,m, n). Its Cartesian equation is x−a

l = y−bm = z−c

n .1

2 1. CURVE THEORY

(vi) Let γ : R → R3 be defined by γ(t) = (a cos t, b sin t, ct), where a, b, c ∈ R\0.Then the image of γ is a helix in R3. It is a helix with the base ellipse x2

a2 + y2

b2 = 1.When a = b, it is called circular helix.

Definition 1.1.4. A parametrized curve γ in R3 is called planar if it is contained insome plane of R3. It is called non-planar or twisted if it is not planar.

Let γ : (a, b)→ Rn be a parametrized curve. Then γ = (γ1, γ2, . . . , γn), where eachγi is a mapping from (a, b) to R. The symbol γ is the derivative of γ and it means

γ = (γ ′1, γ ′2, . . . , γ ′n).Throughout it is assumed that all curves are smooth, i.e., all the derivatives of γ exist.

Examples 1.1.5.(i) The curve γ(t) = (t, t2, t3) is not planar.

Suppose that γ is planar. Then the exist a, b, c, d ∈ R, a2 +b2 + c2 6= 0, such thatat + bt2 + ct3 = d for all t. But then a = b = c = 0 (how???). This contradictsthe fact a2 + b2 + c2 6= 0.

(ii) The curve γ(t) = (cos t, sin t, 3 sin t + 4 cos t) is planar.One can see that the coordinates of γ satisfy the equation of the plane z = 4x+3y.Hence γ is planar.

(iii) The curve γ1(t) = 45 cos t, 1− sin t,−3

5 cos t) is a plane curve.

Definition 1.1.6. Let γ be a parametrized curve. Then the vector γ(t) is called thetangent vector to γ at the point γ(t).

If γ(t) 6= 0, then the equation of the tangent to γ at the point γ(t) is R − γ(t) =uγ(t), u ∈ R.

Exercise 1.1.7. Find the equation of the tangent to the following curves.(i) γ(t) = (a cos t, a sin t, bt), (a, 0, 2πb).(ii) γ(t) = (t, t2, t3), (1, 1, 1).(iii) γ(t) = (cos2 t, sin2 t), (1

2 ,12 ).

(iv) γ(t) = (et, t2), (1, 0).

Example 1.1.8. For a logarithmic spiral γ(t) = (et cos t, et sin t), show that the anglebetween γ(t) and γ(t) is independent of t.

Here γ(t) = (et cos t, et sin t) and γ(t) = (et cos t − et sin t, et sin t + et cos t). Letθ(t) be the angle between γ(t) and γ(t). Then

θ(t) = cos−1(

γ(t)γ(t)‖γ(t)‖‖γ(t)‖

)

2. Regular curves 3

= cos−1(

e2t√

2e2t

)= cos−1

(1√2

).

Therefore the angle between γ(t) and γ(t) is independent of t.

Proposition 1.1.9. Let γ be a parametrized curve in Rn. If γ = a, where a 6= 0, thenγ is a part of a line.

PROOF. Since γ = a, we have γ(t) = at + b. Hence γ is part of a line.

1.2. Regular curvesDefinition 1.2.1. The arc-length of a curve γ starting at the point γ(t0) is the functions(t) given by

s(t) =t∫

t0

‖γ(u)‖du.

Thus, s(t0) = 0 and s(t) is positive or negative according to whether t is larger orsmaller than t0.

Example 1.2.2. Compute the arc-length of the logarithmic spiral γ(t) = (ekt cos t, ekt sin t)starting at the point (1, 0).

We have γ(t) = (−ekt sin t + kekt cos t, ekt cos t + kekt sin t). Therefore

‖γ(t)‖ = ekt√

(− sin t + k cos t)2 + (cos t + k sin t)2 = ekt√

1 + k2.

Since γ(0) = (1, 0), the arc-length of γ starting at the point (1, 0) is

s(t) =t∫

0

eku√

1 + k2du =√

1 + k2

k (ekt − 1).

Definition 1.2.3. Let γ : (a, b)→ Rn be a parametrized curve. Then γ is called regularat γ(t) if γ(t) 6= 0 (or ‖γ(t)‖ > 0). A point γ(t) of the curve γ is called a singular pointif γ is not regular at that point. A curve γ is said to be regular if all its points areregular.

Definition 1.2.4. Let γ : (a, b)→ Rn be a parametrized curve. Then the scalar ‖γ(t)‖is called the speed of γ at the point γ(t). The curve γ is called a unit-speed curve if‖γ(t)‖ = 1 for all t.

It follows from the definitions that every unit-speed curve is regular. The converseis not true. For example γ(t) = (cos 2t, sin 2t) is regular but not unit-speed.

4 1. CURVE THEORY

Lemma 1.2.5. Let α : (a, b) → Rn be a parametrized curve. If ‖α(t)‖ = 1 for all t ,then α(t) and α(t) are perpendicular for all t. In particular, if γ is a unit-speedcurve, then γ(t) ⊥ γ(t) for all t.

PROOF. Since ‖α(t)‖ = 1 for all t , we have α(t)α(t) = 1. Hence α(t)α(t) = 0 for all t ,i.e., α(t) ⊥ α(t) for all t.

It follows that if γ is a unit-speed curve, then γ(t) ⊥ γ(t) for all t as ‖γ(t)‖ = 1 forall t.

Lemma 1.2.6. Let γ : (a, b) → Rn be a regular curve. Then the arc-length of γ,starting at any point of the curve, is a smooth map.

PROOF. We have s(t) =∫ tt0 ‖γ(u)‖du. Then ds

dt = ‖γ‖. Let γ = (γ1, γ2, . . . , γn). Since γ isregular, it follows that γ ′21 + γ ′22 + · · ·+ γ ′2n > 0. Define f : (0,∞)→ R and g : (a, b)→ Rby f (t) =

√t and g(t) = γ ′21 (t)+ · · ·+γ ′2n (t). Then both f and g are smooth maps. Since

the range of g is contained in (0,∞), the domain of f , it follows that the map f g is asmooth map. But f g = ds

dt . Therefore dsdt is a smooth map and hence s is a smooth

map.

Definition 1.2.7. A parametrized curve γ : (a, b)→ Rn is called a reparametrizationof γ : (a, b)→ Rn if there exits a bijective smooth map φ : (a, b)→ (a, b), whose inverseis also smooth, such that γ = γ φ. The map φ is called the reparametrization mapfor the above reparametrization.

If γ is a reparametrization of γ with the reparametrization map φ, then γ = γφ−1.Hence γ is a reparametrization of γ.

We note that two curves which are reparametrizations of each other have thesame image. Hence they have the same geometric properties.

Exercise 1.2.8. Let C be the collection of all parametrized curve in Rn. Let γ, β ∈ C.We say γ ∼ β if γ is a reparametrization of β. Show that the above relation on C is anequivalence relation.

Proposition 1.2.9. Any reparametrization of a regular curve is regular.

PROOF. Let γ : (a, b) → Rn be a reparametrization of a regular curve γ : (a, b) → Rn.Then there is a bijective smooth map φ : (a, b)→ (a, b), whose inverse is also smooth,such that γ = γ φ. Let ψ : (a, b)→ (a, b) be the inverse of φ. Then φ ψ(t) = t for allt ∈ (a, b). Therefore dφ

dtdψdt = 1 and hence dφ

dt (t) 6= 0 for any t ∈ (a, b). Since γ = γ φ,we have dγ

dt = dγdt

dφdt . Since γ is regular dγ

dt 6= 0. Since dφdt is never vanishing, it follows

that dγdt is never zero (vector), i.e., γ is regular.

3. Curvature and Torsion 5

Proposition 1.2.10. A parametrized curve has a unit-speed reparametrization iff itis regular.

PROOF. Let γ be a unit-speed reparametrization of the curve γ. Since γ is a reparametriza-tion of γ, γ is a reparametrization of γ. Since γ is regular (as it is unit-speed) and anyreparametrization of a regular curve is regular, it follows that γ is regular.

Conversely, assume that γ : (a, b) → Rn be regular. Let s be the arc-length ofγ, starting at the point γ(t0), i.e., s(t) =

∫ tt0 ‖γ(u)‖du. We note that s is a smooth

map. Since γ is regular, dsdt (t) = ‖γ(t)‖ > 0 for all t , and hence s is strictly increasing

and so it is one-one. Since dsdt (t) 6= 0 for all t , it follows from the inverse function

theorem that s−1 : (a, b) → (a, b) is smooth, where (a, b) is the range of s. Considerthe reparametrization γ : (a, b) → Rn given by γ = γ s−1, i.e., γ s = γ. Nowdγdt

dsdt = γ

dt . Then ‖dγdt ‖dsdt = ‖dγdt ‖ = ds

dt , i.e., ‖dγdt ‖ = 1. Hence γ has a unit-speedreparametrization.

Corollary 1.2.11. Let γ be a regular curve, and let γ be a unit-speed reparametriza-tion of γ given by γ u = γ, where u is a smooth map. Then u = ±s + c, wheres is the arc-length of γ, starting at any point, and c is a constant. Conversely, ifu = ±s + c, then γ is a unit-speed reparametrization of γ.

PROOF. Since γ u = γ, we have dγdt

dudt = dγ

dt . This implies ‖dγdt ‖∣∣dudt

∣∣ = ‖dγdt ‖. Therefore∣∣dudt

∣∣ = dsdt , as γ is unit-speed. This proves that u = ±s + c.

Conversely, assume that u = ±s+c. Clearly u is bijective, smooth and its inverseis also smooth. We have γ(±s + c) = γ. Hence ±dγ

dtdsdt = dγ

dt . This implies ‖dγdt ‖dsdt =

‖dγdt ‖ = dsdt . Therefore ‖dγdt ‖ = 1, i.e., γ is unit-speed reparametrization of γ.

1.3. Curvature and TorsionDefinition 1.3.1. Let γ : (a, b)→ R3 be a unit-speed curve. Then the curvature of κ(t)of γ at the point γ(t) is defined by κ(t) = ‖γ(t)‖.

Examples 1.3.2.(i) Let a, b ∈ R3, and let ‖a‖ = 1. Then the curvature of the unit-speed curve

γ(t) = ta + b is zero everywhere.(ii) Let (x0, y0) ∈ R2, and let R > 0. Then the curvature of the unit-speed curve

γ(t) = (x0 + R cos( tR ), y0 + R sin( tR )), which is a circle with center (x0, y0) andradius R, is 1

R everywhere.

6 1. CURVE THEORY

Proposition 1.3.3. If γ is a regular curve in R3, then its curvature κ is given by theformula

κ = ‖γ × γ‖‖γ‖3

.

PROOF. Let γ be a unit-speed reparametrization of γ given by γ s = γ. We notethat the curvature of γ(s(t)) is same as the curvature of γ at the point γ(t). We have(dsdt)2 = ‖γ‖2 = γ γ, and hence ds

dtd2sdt2 = γ γ. Let γ ′ denote the derivative of γ with

respect to s. Since γ s = γ, γ ′ dsdt = γ, i.e., γ ′ = γdsdt

. Differentiating it again, we have

γ ′′ =dsdt γ −

d2sdt2 γ(ds

dt)2

dtds =

(dsdt)2 γ − ds

dtd2sdt2 γ(ds

dt)4

= (γ γ)γ − (γ γ)γ‖γ‖4

= γ × (γ × γ)‖γ‖4

.

Hence κ = ‖γ ′′‖ = ‖γ × (γ × γ)‖‖γ‖4 = ‖γ‖‖(γ × γ)‖

‖γ‖4 = ‖γ × γ‖‖γ‖3 . The last step follows

because γ × γ ⊥ γ.

Example 1.3.4. Let γ(t) = (a cos t, a sin t, bt), where a, b > 0. Then computationsshows that κ(t) = a

a2+b2 .

Exercises 1.3.5. Compute the curvature of the following curves.(i) γ(t) = (ekt cos t, ekt sin t).(ii) γ(t) = (t, cosh t). Also determine a point having maximum curvature.(iii) γ(t) = (t, t2, t3).

Now we define a special type of curvature for plane curves only.

Definition 1.3.6. Let γ : (a, b) → R2 be a unit-speed curve. Let t be the unit tangentof γ, i.e., t = γ. The unit vector ns obtained by rotating the tangent vector t anti-clockwise by an angle π

2 is called the signed unit normal of γ. Since γ is unit-speed,γ is perpendicular to t and hence it is parallel to ns. Therefore there exists a mapκs : (a, b)→ R such that γ = κsns. The map κs is called the signed curvature of γ.

Since κ = ‖γ‖, it follows that κ = |κs|, i.e., the absolute value of the signed curva-ture is the curvature

Let Rθ : R2 → R2 be a rotation operator (it rotates a vector anti-clockwise byan angle θ). We note that Rθ is linear and Rθ(1, 0) = (cos θ, sin θ) and Rθ(0, 1) =


(− sin θ, cos θ). Let (x, y) ∈ R2. Then

Rθ(x, y) = Rθ(x(1, 0) + y(0, 1))= xRθ(1, 0) + yRθ(0, 1)= x(cos θ, sin θ) + y(− sin θ, cos θ).

HenceRθ(x, y) = (x cos θ − y sin θ, x sin θ + y cos θ).

In fact, Rθ is an isometry of R2, i.e., it satisfies

‖Rθ(x1, y1)− Rθ(x2, y2)‖ = ‖(x1, y1)− (x2, y2)‖ ((x1, y1), (x2, y2) ∈ R2).

In particular, when θ = π2 , R π

2(x, y) = (−y, x).

Fix a vector a ∈ R2. Define Ta : R2 → R2 by Ta(x) = x + a. Then Ta is called atranslation by a vector a (it translates given vector by a) and it is an isometry.

Definition 1.3.7. An isometry of R2 of the form Ta Rθ is called a direct isometry,where a ∈ R2 and θ ∈ [0, 2π).

Example 1.3.8. Compute the signed unit normal of the curve γ(t) = (ekt cos t, ekt sin t).The curve is not unit-speed. The unit tangent will be the unit vector in the direc-

tion of the tangent vector, i.e., t(t) = γ(t)‖γ(t)‖ . Now γ(t) = (−ekt sin t+kekt cos t, ekt cos t+

kekt sin t) and ‖γ(t)‖ = ekt√

1 + k2. Hence t(t) = 1√1+k2 (− sin t + k cos t, cos t + k sin t).

Since ns(t) = R π2(t(t)), we have

ns(t) = 1√1 + k2

(− cos t − k sin t,− sin t + k cos t)

Exercise 1.3.9. If γ : (a, b) → R2 is a unit-speed curve, then show that both signedunit normal and signed curvature are smooth maps.

Proposition 1.3.10. Let γ : (a, b)→ R2 be a unit-speed curve, and let s0 ∈ (a, b). Letφ0 ∈ R such that γ(s0) = (cosφ0, sinφ0). Then there exists a unique smooth mapφ : (a, b)→ R such that γ(s) = (cosφ(s), sinφ(s)) for all s ∈ (a, b) and φ(s0) = φ0.

PROOF. Let γ(s) = (f (s), g(s)) (s ∈ (a, b)). Since γ is smooth, both f and g are smoothmaps. Since γ is unit-speed, f2 + g2 = 1 and hence f f + gg = 0. Define φ : (a, b)→ Rby

φ(s) = φ0 +s∫

s0

(fg − fg).

8 1. CURVE THEORY

Since f and g are smooth, φ is a smooth map and φ(s0) = φ0.Let F = f cosφ + g sinφ and G = f sinφ − g cosφ. Then

F = f cosφ − f sinφφ + g sinφ + g cosφφ= (f + gφ) cosφ + (g − fφ) sinφ.

Now,

f + gφ = f + g(fg − fg) = f (1− g2) + fgg = f f2 + fgg = f (f f + gg) = 0.

By similar arguments it follows that g− fφ = 0. Hence F = 0, i.e., F is a constant map.Therefore F (s) = F (s0) for all s. Now

F (s0) = f (s0) cosφ(s0) + g(s0) sinφ(s0) = cos2 φ0 + sin2 φ0 = 1.

This shows that F = 1. Similarly, we get G = 0. Hence f cosφ + g sinφ = 1 andf sinφ−g cosφ−0. Solving these equations we get f = cosφ and g = sinφ. Thereforeγ(s) = (cosφ(s), sinφ(s)) for all s ∈ (a, b) and φ(s0) = φ0.

For the uniqueness, let ψ : (a, b) → R be a smooth map satisfying ψ(s0) = φ0and γ(s) = (cosψ(s), sinψ(s)) for all s. Then there is a map n : (a, b) → Z such thatφ(s)−ψ(s) = 2πn(s) for all s. Since both φ and ψ are smooth, n is smooth and hence itis continuous. Since (a, b) is connected, n is continuous and the non-empty connectedsubsets of Z are singletons only, it follows that n is a constant map, say, c. Thereforeφ(s)− ψ(s) = 2πc for all s. Since φ(s0) = ψ(s0) = φ0, we get c = 0. Hence ψ(s) = φ(s)for all s.

Definition 1.3.11. Let γ : (a, b) → R2 be a unit-speed curve. Then there is a smoothmap φ : (a, b) → R such that γ(s) = (cosφ(s), sinφ(s)) for all s, which is determineduniquely by the condition φ(s0) = φ0. The map φ is called the turning angle of γ.

Proposition 1.3.12. Let γ : (a, b) → R2 be a unit-speed curve, and let φ be theturning angle of γ. Then the signed curvature of γ can be determined by theformula κs = φ.

PROOF. Since φ is the turning angle of γ, we have γ(s) = (cosφ(s), sinφ(s)) for all s.Hence γ = (− sinφ, cosφ)φ. Now ns = R π

2(γ) = (− sinφ, cosφ). Therefore γ = φns

and hence the signed curvature κs of γ is φ, i.e., κs = φ.

It follows from the Proposition 1.3.12 that the signed curvature is the rate at whichthe tangent vector of the curve rotates.

Example 1.3.13. If γ : (a, b)→ R2 is a unit-speed curve, then its signed curvature andsigned unit normal are smooth maps.


The map ns : (a, b) → R2 is given by ns(t) = Rπ/2(t(t)). The map Rπ/2 : R2 → R2

defined by Rπ/2(x, y) = (−y, x) is a smooth map and the map t = γ is a smooth mapand hence the map Rπ/2 t = ns is a smooth map, i.e., the signed unit normal is asmooth map.

Now the map κs : (a, b) → R is given by γ(t) = κs(t)ns(t). Therefore κs(t) =γ(t)ns(t). Since both γ and ns are smooth maps, it follows that κs is a smooth map.

Theorem 1.3.14. Let k : (a, b) → R be a smooth map. Then there is a unit-speedcurve γ : (a, b)→ R2 whose signed curvature is k. Moreover, if γ : (a, b)→ R2 is anyother unit-speed curve whose signed curvature is k, then there is a direct isometryM of R2 such that γ = M γ.

PROOF. Let s0 ∈ (a, b). Define φ : (a, b) → R by φ(s) =∫ ss0k(u)du. Since k is smooth,

φ is a smooth map. Note that φ(s0) = 0 Define γ : (a, b)→ R2 by

γ(s) =

s∫

s0

cosφ(u)du,s∫

s0

sinφ(u)du

.

Then γ(s) = (cosφ(s), sinφ(s)) for all s. Therefore γ is unit-speed curve. It followsform the Proposition 1.3.10 that φ is the turning angle of γ satisfying φ(s0) = 0 andhence the signed curvature of γ is φ. But φ = k. Hence the signed curvature of γ isk.

Let γ : (a, b) → R2 be a unit-speed curve with signed curvature k. Let ψ be theturning angle of γ. Then ψ = k. But then ψ(s) =

∫ ss0k(u)du + ψ(s0) = φ(s) + θ, where

θ = ψ(s0) is a constant. So we have ψ = φ + θ. Since ψ is the turning angle of γ,˙γ(s) = (cosψ(s), sinψ(s)) for all s. Let a = γ(s0). Integrating the above equation, wehave

γ(s) =

s∫

s0

cosψdu,s∫

s0

sinψdu

+ a

= Ta

s∫

s0

cosψdu,s∫

s0

sinψdu

= Ta

s∫

s0

cos(φ + θ)du,s∫

s0

sin(φ + θ)du

= Ta

s∫

s0

(cosφ cos θ − sinφ sin θ)du,s∫

s0

(sinφ sin θ + cosφ cos θ)du

10 1. CURVE THEORY

= Ta

cos θs∫

s0

cosφdu − sin θs∫

s0

sinφdu, cos θs∫

s0

cosφdu + sin θs∫

s0

sinφdu

= Ta Rθ

s∫

s0

cosφdu,s∫

s0

sinφdu

= Ta Rθ(γ(s)).

Therefore γ(s) = M(γ(s)) for all s, where M = Ta Rθ is a direct isometry. Henceγ = M γ.

Example 1.3.15. Let γ : (a, b)→ R2 be a regular curve with constant signed curvature.Show that γ is (part of) a circle.

We may assume that γ is unit-speed. Let k be the curvature of γ. Let k < 0. Thenk = −` for some ` > 0. Let γ : (a, b) → R2 be defined by γ(t) = (1

` cos(`t),−1` sin(`t)).

Then γ is a unit-speed and t(t) = ˙γ(t) = (− sin(`t),− cos(`t)), ¨γ(t) = (−` cos(`t), sin(`t)).Hence ns(t) = (cos(`t),− sin(`t)). Therefore ¨γ(t) = −`ns(t). This implies that thesigned curvature of γ is −` = k. Since γ and γ have the same signed curvature, thereexists a direct isometry M of R2 such that γ = M γ. Since N γ is (part of) a circlefor any direct isometry N of R2, it follows that γ is also a circle.

Similarly one can show that γ is a circle when k > 0.

Examples 1.3.16.(i) Compute the signed curvature of γ(t) = (t, cosh t).

We have γ(t) = (1, sinh t). Let φ(t) be the angle between positive x-axis and γ(t).Then

cosφ(t) = (1, 0)(1, sinh t)√1 + sinh2 t

= 1cosh t .

Therefore sinφ(t) = tanh t and tanφ(t) = sinh t. Let s be the arc-length of γ,i.e., s(t) =

∫ t0 ‖γ(u)‖du = sinh t. Differentiating tanφ(t) = sinh t = s with respect

to s we have sec2 φφ = 1, i.e, (1 + tan2 φ)φ = 1. Therefore φ = 11+tan2 φ = 1

1+s2 =1

1+sinh2 t = 1cosh2 t . Hence κs(t) = 1

cosh2 t .(ii) Give an example of a plane curve having signed curvature 1

1+s2 .It is

γ(s) =

s∫

0

cos(tan−1 u)du,s∫

0

sin(tan−1 u)du

.

See the proof of Theorem 1.3.14.


Definition 1.3.17. Let γ : (a, b) → R3 be a unit-speed curve with nowhere vanishingcurvature. Then the function n = γ

κ is called the unit normal or principal normal ofγ.

It is clear that ‖n(t)‖ = 1 for all t. Since ‖t(t)‖ = 1 for all t , t t = 0, i.e., γ t = 0.This means t ⊥ n. The map b defined by b = t× n is called the unit binormal of γ.

Thus at each point γ(t) of a unit-speed curve γ, we have a right handed system oforthonormal basis t(t),n(t),b(t), i.e, t(t)×n(t) = b(t), n(t)×b(t) = t(t), b(t)×t(t) = n(t),t(t) t(t) = n(t) n(t) = b(t) b(t) = 1.

Since b(t) b(t) = 1 for all t , b b = 0, i.e, b ⊥ b. Also, b = t × n. Thereforeb = t × n + t × n = κ(n) × n + t × n = t × n. This implies that b ⊥ t. Hence b ‖ n.Therefore there is a map τ such that b = −τn. The map τ is called the torsion of γ.

Since n = b× t, we have n = b× t + b× t = −τ(n× t) + κ(b× n) = τb− κt.

Proposition 1.3.18. Let γ : (a, b) → R3 be a regular curve with nowhere vanishingcurvature. Then the torsion τ of γ is given by

τ = (γ × γ)...γ

‖γ × γ‖2.

PROOF. First assume that γ is unit-speed. Then γ = t, γ = κn and...γ = κn + κn = κn + κ(τb− κt). Now

(γ × γ)...γ = (t× κn)(κn + κ(τb− κt)) = κb(κn + κ(τb− κt)) = κ2τ,

and ‖γ × γ‖2 = ‖κb‖2 = κ2. Hence (γ×γ)...γ

‖γ × γ‖2 = κ2τκ2 = τ.

Let γ be a regular curve with nowhere vanishing curvature (not necessarily unit-speed). Let γ be its unit-speed reparametrization given by γ s = γ, where s is thearc-length of γ starting at any point of γ. Note that the torsion of γ at γ(s(t)) is sameas the torsion of γ at the point γ(t). Since γ s = γ, we have γ ′ dsdt = γ, i.e,. γ ′ = γ

dsdt

.Therefore γ ′′(dsdt )2+γ ′

d2sdt2 = γ and γ ′′′(dsdt )3+γ ′′2

dsdt

d2sdt2 +γ ′′ dsdt

d2sdt2 +γ ′ d3s

dt3 =...γ. It follows that

γ × γ = (γ ′ × γ ′′)(dsdt )3 and (γ × γ)...γ = (γ ′ × γ ′′)γ ′′′(dsdt )6. Also ‖γ × γ‖2 = ‖γ ′ × γ ′′‖2(dsdt )6.

Hence (γ×γ)...γ

‖γ × γ‖2 = (γ′×γ′′)γ′′′

‖γ ′ × γ ′′‖2 = τ , as γ is a unit-speed curve.

Exercises 1.3.19. Compute curvature and torsion of the following curves.

(i) γ(t) = (t, t2, t3).(ii) γ(t) = (a cos t, a sin t, bt)(iii) γ(t) = (et cos t, et sin t, et)

(iv) γ(t) = (45 cos t, 1− sin t, 3

5 cos t)(v) γ(t) = (1

3 (1 + t)3/2, 13 (1− t)3/2,

t√2 )

12 1. CURVE THEORY

Example 1.3.20.

(i) If γ : (a, b) → R3 is a unit-speed curve with nowhere vanishing curvature, thenthe unit normal, unit binormal, curvature and torsion are smooth maps.We know that n,b : (a, b)→ R3 and κ, τ : (a, b)→ R. By definition κ(t) = ‖γ(t)‖.Let γ = (γ1, γ2, γ3). Since κ is nowhere vanishing, γ ′′21 (t) + γ ′′22 (t) + γ ′′23 (t) > 0for all t. Define f : (0,∞) → R by f (t) =

√t and g : (a, b) → R by g(t) =

γ ′′21 (t) + γ ′′22 (t) + γ ′′23 (t). Then both f and g are smooth maps. Since g(t) > 0 forall t , f g is a smooth map. But f g = κ and hence κ is a smooth map.Since κ is a smooth map and κ(t) > 0 for all t , 1

κ is a smooth map. Thereforeγκ = n is a smooth map.By definition b = t × n and both t and n are smooth maps Therefore b is asmooth map.Now b = −τn. It means τ = −bn. Since both b (so is b) and n are smooth maps,τ is a smooth map.

(ii) Let γ be a regular curve in R3 with nowhere vanishing curvature. Then γ isplanar iff the torsion of γ is identically zero.We may assume that γ is unit-speed. Suppose that γ is planar. Then there exista unit vector a ∈ R3 and a constant d ∈ R such that γa = d. Differentiating, weget ta = 0. Differentiating it again, we have ta = 0, i.e., κ(na) = 0. This givesna = 0, as κ is nowhere vanishing. Therefore a ⊥ t and a ⊥ n. It means b = ±a.Since b is continuous, it follows that either b(t) = a for all t or b(t) = −a for allt. So, in either case, b = 0. Since b = −τn, it follows that τ = 0.Conversely, assume that τ = 0. Since b = −τn, we have b = 0, i.e, b is a constantunit vector. Now d

dt (γ b) = γ b = t b = 0. This means γ b is a constant function,say, d, i.e., γ b = d. Hence γ is lying on the plane R b = d, i.e., γ is planar.

(iii) Let γ be a regular curve in R3 with constant curvature and zero torsion. Then γis (part of) a line or (part of) a circle.Assume that the curvature κ = 0. Since γ = κn, γ = 0. Therefore γ is (part of)a line.Let κ > 0. Note that τ = 0. We see that d

dt (γ + 1κn) = t + 1

κ n = t + 1κ (τb− κt) = 0.

Therefore γ + 1κn = a for some constant vector a ∈ R3. But then ‖γ − a‖ = 1

k ,i.e., γ is lying on the sphere ‖R − a‖ = 1

κ . Since τ = 0, γ is lying on some plane.Therefore γ is lying on a sphere as well as a plane and hence it is (part of) acircle.

(iv) Let γ be a regular curve in R3 with nowhere vanishing curvature. Show that γis spherical iff τ

κ = ddt( κκ2τ).


We may assume that γ is unit-speed. Assume that γ is spherical. Then thereexist a constant vector a ∈ R3 and a constant R ∈ R such that (γ − a)(γ − a) =‖γ − a‖2 = R2. Differentiating it we get (γ − a)t = 0. Differentiating it again wehave (γ − a)n = − 1

κ . Again differentiation gives (γ − a)b = κτκ2 . Since t,n,b

is an orthonormal basis of R3, γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b.Therefore

γ − a = −1κn + κ

τκ2 b.

But then R2 = ‖γ − a‖2 = 1κ2 + κ2

τ2κ4 . Differentiating it we obtainτ2κ42κκ−κ2(2ττκ4+τ24κ2κ)

τ4κ8 − 2κκ3 = 0, i.e., κ2κτ−2κκ2τ−κ2κτ

κ4τ2 − τκ = 0, i.e., τ

κ = ddt( κκ2τ).

Conversely, assume that τκ = d

dt( κκ2τ). Let a = γ + 1

κn−κτκ2 b. Then

a = t− κκ2 n + 1

κ n− ddt

(κκ2τ

)b− κ

τκ2 b

= t− κκ2 n + 1

κ (τb− κt)− ddt

(κκ2τ

)b− κ

τκ2 (−τ)n

=(τκ −

ddt

(κκ2τ

))b = 0.

Hence a is a constant vector. Now ‖γ − a‖2 = 1κ2 + κ2

τ2κ4 . Again using τκ = d

dt( κκ2τ),

it follows that 1κ2 + κ2

τ2κ4 is a positive constant, say, R2. Therefore ‖γ − a‖2 = R2,and hence γ is spherical.

ORLet ρ = 1

κ and σ = 1τ . Assume that γ is spherical. Then there exist a constant

vector a ∈ R3 and a constant R > 0 such that (γ − a)(γ − a) = ‖γ − a‖2 = R2.Differentiating it we get (γ − a)t = 0. Differentiating it again we have (γ − a)n =− 1

κ = −ρ. One more differentiation gives (γ − a)b = −σρ. Now

γ − a = ((γ − a)t)t + ((γ − a)n)n + ((γ − a)b)b= −ρn− σρb.

Therefore R2 = ‖γ − a‖2 = ‖ − ρn− σρb‖2 = ρ2 + (σρ)2. Differentiation of thelast equation gives ρ

σ + (σρ)• = 0, which is required.Conversely, assume that ρ

σ + (σρ)• = 0. Then ρ2 + (σρ)2 = d for some constantd > 0(???). Let a = γ + ρn + σρb. Then a = t + ρn + ρ(τb − κt) + (σρ)•b −τσρn = ( ρσ + (σρ)•)b = 0. Therefore a is a constant vector. Now ‖γ − a‖2 =‖ − ρn− σρb‖2 = ρ2 + (σρ)2 = d. Hence γ is spherical.

(v) Let γ be a unit-speed curve in R3 with nowhere vanishing curvature. Show thatthe curve α = t is a regular curve and find the curvature and torsion of α.

14 1. CURVE THEORY

Here α = t. Therefore α = t = γ. Since the curvature κ = ‖γ‖ > 0, thecurve α is regular. Now α = t = κn, α = κn + κn = κn + κ(τb − κt) and...α = κn + κ(τb − κt) + κτb + κτb − κτ2n − 2κκt − κ3n. Now α × α = κ2τt + κ3b,(α × α)

...α = −κ3κτ + κ4τ , ‖α‖ = κ and ‖α× α‖ = κ2

√κ2 + τ2. Let κ1 be the

curvature of α and τ1 be the torsion of α. Then

κ1 = ‖α× α‖‖α‖3

= κ2√κ2 + τ2

κ3 =√κ2 + τ2

κand

τ1 = (α× α)...α

‖α× α‖2= κ(κτ − κτ)√

κ2 + τ2.

Theorem 1.3.21. Let γ be a unit-speed curve in R3. Let t, n and b be its unittangent, unit normal and unit binormal respectively. Let κ and τ be the curvatureand torsion of γ respectively. Then

t = κnn = −κt τbb = −τn

The equations in Theorem 1.3.21 are called Frenet-Serret equations or Serret-Frenet equations.

Example 1.3.22. Verify the Serret-Frenet equations for the curve

γ(a) =(a cos

(a√

a2 + b2

), a sin

(a√

a2 + b2

), b√

a2 + b2

).

1.4. Fundamental Theorem of Space CurvesTheorem 1.4.1. (Existence) Let k, t : (α, β) → R be smooth maps with κ(s) > 0 forall s. Then there exists a unit-speed curve γ : (α, β)→ R3 whose curvature is k andtorsion is t.

PROOF. Fix s0 ∈ (α, β). It follows from the theory of Differential Equations that theequations T = kN, N = −kT + tB and B = −tN have a unique solution T, N and B

such that T(s0),N(s0),B(s0) is a standard basis of R3. Since the matrix

0 k 0−k 0 t0 −t 0

expressing T, N and B in terms of T, N and B is skew-symmetric, it follows that T,N and B are orthogonal unit vectors. Since B is a smooth map orthogonal to both T

4. Fundamental Theorem of Space Curves 15

and N, there exists a smooth map λ : (α, β)→ R such that B = λT×N. Note that λ(s)is 1 or −1. Since λ is continuous, either λ(s) = 1 for all s or λ(s) = −1 for all s. SinceB(s0) = T(s0)×N(s0), we have λ(s) = 1 for all s, i.e., B = T×N. Define γ : (α, β)→ R3

by

γ(s) =s∫

s0

T(u)du.

Then γ = T. Therefore γ is a unit-speed curve and T is the unit tangent of γ. Nowγ = T = kN. Therefore ‖γ‖ = k, i.e., k is the curvature of γ. It follows fromγ = T = kN that N is the unit normal of γ. Since B = T × N, B is the unit binormalof γ. It follows from the equation B = −tN that t is the torsion of γ.

Theorem 1.4.2. (Uniqueness) If γ, γ : (α, β) → R3 are unit-speed curves with samecurvature and same torsion, then there is an isometry M of R3 such that γ = M γ.

PROOF. Let t, n and b be the unit tangent, unit normal and unit binormal of γ, and lett, n and b be those of γ. Fix s0 ∈ (α, β). Since t(s0),n(s0),b(s0) and t(s0), n(s0), b(s0)are both right handed orthonormal basis of R3, there is a rotation about origin thatmaps t(s0), n(s0) and b(s0) to t(s0), n(s0) and b(s0) respectively. Further, there is atranslation which takes γ(s0) to γ(s0) (and this has no effect on t, n and b). By applyingrotation followed by translation, we can therefore assume that γ(s0) = γ(s0), t(s0) =t(s0), n(s0) = n(s0) and b(s0) = b(s0). Hence to prove the theorem we need to proveγ = γ. Define A : (α, β)→ R by

A(s) = t(s)t(s) + n(s)n(s) + b(s)b(s).Then A is a smooth map. Note that A(s) ≤ 3 for all s and A(s0) = 3. Also observethat A(s) = 0 for all s (you need to verify this). Therefore A is a constant map. SinceA(s0) = 3, we have A(s) = 3 for all s. Hence it follows that t = t, n = n and b = b. Theequation t = t implies that γ = γ + c. Since γ(s0) = γ(s0), we get c = 0. Hence γ = γ.This proves the theorem.

Example 1.4.3. Describe all curves in R3 which have constant curvature κ > 0 andconstant torsion τ.

Let γ : (a, b) → R3 be a unit-speed curve with constant curvature κ > 0 andconstant torsion τ. Let a = κ

κ2+τ2 and b = τκ2+τ2 . Define γ : (a, b)→ R3 by

γ(t) =(a cos

(t√

a2 + b2

), a sin

(t√

a2 + b2

), b t√

a2 + b2

).

Then γ is a unit-speed curve. The curvature of γ is aa2+b2 = κ and its torsion is

ba2+b2 = τ. Therefore γ, γ : (a, b) → R3 are unit-speed curves with same curvature

16 1. CURVE THEORY

and same torsion. So, there is an isometry M of R3 such that γ = M γ. But γ is ahelix and hence M γ is a helix. This implies that γ is a helix. Hence any curve withconstant (non-zero) curvature and constant torsion is a helix.

1.5. Isoperimetric Inequality and Four Vertex TheoremDefinition 1.5.1. Let a ∈ R be a positive constant. A simple closed curve in R2 is a(regular) curve γ : R→ R2 such that γ(t) = γ(t ′) iff t ′ − t = ka for some k ∈ Z.

Thus, the point γ(t) returns to its starting point when t is increased by a, but notbefore that.

By Jordan Curve Theorem any simple closed curve in R2 has an ‘interior’ and‘exterior’. More precisely, the set of points in R2 that are not on the curve γ is thedisjoint union of two subsets of R2, denoted by int(γ) and ext(γ), with the followingproperties.

(i) int(γ) is bounded,(ii) ext(γ) is unbounded,(iii) both int(γ) and ext(γ) are connected, but any curve joining a point of int(γ) to a

point of ext(γ) crosses γ.

Proposition 1.5.2. Let γ be a regular curve with period a. Then its unit-speedreparametrization is `(γ)- periodic, where `(γ) is the length of γ.

PROOF. Since γ is a- periodic, the length, `(γ), of γ is

`(γ) =a∫

0

‖γ(u)‖du.

Let γ be a unit-speed reparametrization of γ given by γ s = γ, where s is the arc-length of γ starting at the point γ(0). Let t, t ′ ∈ R, and let t ≤ t ′. Then there isk ∈ N ∪ 0 such that t ′ − t = ka + ε , where 0 ≤ ε < a. Now

s(t ′)− s(t) =t ′∫

0

‖γ(u)‖du −t∫

0

‖γ(u)‖du =t ′∫

t

‖γ(u)‖du

=t+a∫

t

‖γ(u)‖du +t+2a∫

t+a

‖γ(u)‖du + · · ·+t+ka∫

t+(k−1)a

‖γ(u)‖du +t+ka+ε∫

t+ka

‖γ(u)‖du

= k`(γ) +t+ε∫

t

‖γ(u)‖du.

5. Isoperimetric Inequality and Four Vertex Theorem 17

Hence s(t ′) − s(t) = k`(γ) iff∫ t+εt ‖γ(u)‖du = 0 iff ε = 0 iff t ′ − t = ka. Now γ(s(t)) =

γ(s(t ′)) iff γ(t) = γ(t ′) iff t ′ − t = ka for some k ∈ Z iff s(t ′) − s(t) = k`(γ). Thereforeγ is `(γ)- periodic.

Let γ be a simple closed curve in R2. Let A(int(γ)) be the area of the interior ofγ. Then A(int(γ)) =

∫∫int(γ) dxdy.

Definition 1.5.3. A simple closed curve in R2 is called positively oriented if its unitsigned normal points towards the interior of the curve at ach point of the curve.

Theorem 1.5.4. (Green’s Theorem) Let γ be a simple closed curve in R2, and letγ be positively oriented. Let f and g be continuous on the closure of int(γ) andsmooth on int(γ). Then

∫∫

int(γ)

(∂g∂x −

∂f∂y

)dxdy =

∫

γ

[f (x, y)dx + g(x, y)dy].

Proposition 1.5.5. Let γ(t) = (x(t), y(t)) be a simple closed curve in R2 with perioda. Let γ be positively oriented. Then

A(int(γ)) = 12

a∫

0

(xy − yx)dt.

PROOF. Take f (x, y) = −12y and g(x, y) = 1

2x. Then both f and g are smooth functionson R2. It follows from the Green’s theorem that∫∫

int(γ)(12 − (−1

2 )dxdy =∫γ[

12xdy −

12ydx]. This implies

A(int(γ)) =∫γ[

12xdy −

12ydx] = 1

2∫ a

0 (xy − yx)dt.

Proposition 1.5.6 (Wirtinger’s Inequality). Let F : [0, π]→ R be a smooth map satis-fying F (0) = F (π) = 0. Then

π∫

0

F2dt ≥π∫

0

F2dt,

with equality holding iff F (t) = A sin t for some constant A.

PROOF. Define G : [0, π]→ R by G(t) = F (t)sin t . Note that G is a smooth map. Then

π∫

0

F2dt =π∫

0

(G sin t +G cos t)2dt

=π∫

0

G2 sin2 tdt + 2π∫

0

GG sin t cos tdt +π∫

0

G2 cos2 tdt

18 1. CURVE THEORY

=π∫

0

G2 sin2 tdt −π∫

0

G2(cos2 t − sin2 t)dt +π∫

0

G2 cos2 tdt

=π∫

0

G2 sin2 tdt +π∫

0

G2 sin2 tdt =π∫

0

F2dt +π∫

0

G2 sin2 tdt.

Henceπ∫

0

F2dt −π∫

0

F2dt =π∫

0

G2 sin2 tdt ≥ 0. (1.5.6.1)

By equation (1.5.6.1) it follows that∫ π

0 F2dt =∫ π

0 F2dt iff∫ π

0 G2 sin2 tdt = 0 iff G2 sin2 t =0 iff G sin t iff G = 0. But G = 0 implies G(t) = A for all t , for some constant A, i.e.,F (t) = A sin t for all t.

Example 1.5.7. The length and the are of a simple closed are invariant under isometry.Let γ : R → R2 be a simple closed curve with period ` , and let M : R2 → R2 be

an isometry1. Then there is an orthogonal matrix A and a ∈ R2 such that M(x) =A(x) + a (x ∈ R2). Also note that DM(x) = DA(x) = A (x ∈ R2) Let γ : R → R2 beγ = M γ. We need to prove that the lengths and the area of interiors of γ and γ aresame. Let t, t ′ ∈ R. Then γ(t ′) = γ(t) iff M(γ(t ′)) = M(γ(t)) iff γ(t ′) = γ(t) iff t ′ − t = k`for some k ∈ Z. Therefore γ is `-periodic. Now

`(γ) =`∫

0

‖ ˙γ(u)‖du =`∫

0

‖DM(γ(u))‖du

=`∫

0

‖DA(γ(u)) γ(u)‖du

=`∫

0

‖A(γ(u))‖du =`∫

0

‖γ(u)‖du = `(γ).

Since M is an isometry, there are a, b, c, d, e, f ∈ R2 such that a2 +b2 = c2 +d2 =a2 + c2 = b2 +d2 = 1, ac+bd = ab+ cd = 0 and M(x, y) = (ax+by + e, cx+dy + f ).

1We know that N : R2 → R2 is an isometry iff there are a, b, c, d, e, f ∈ R2 such that a2 +b2 = c2 +d2 =a2 + c2 = b2 + d2 = 1, ac + bd = ab + cd = 0 and N(x, y) = (ax + by, cx + dy) + (e, f ) OR N is anisometry of R2 if and only if there is a 2× 2 orthogonal matrix A and a ∈ R2 such that N(x) = Ax + a


Let γ(t) = (x(t), y(t)). Then γ(t) = (ax(t) + by(t) + e, cx(t) + dy(t) + f ). Now

A(int(γ)) = 12

`∫

0

((ax + by + e)(cx + dy + f )• − (ax + by + e)•(cx + dy + f ))dt

= 12

`∫

0

((ax + by + e)(cx + dy)− (ax + by)(cx + dy + f ))dt

= 12

(ad − bc)`∫

0

(xy − xy)dt + (ec − af )(x(`)− x(0)) + (ed − bf )(y(`)− y(0))

= 12

`∫

0

(xy − xy)dt =A(int(γ)),

as ad − bc = ±1, x(0) = x(`) and y(0) = y(`) (???).

Theorem 1.5.8 (The Isoperimetric Inequality). Let γ be a simple closed curve, let `be its length, and let A be the area of interior of γ. Then `2 ≥ 4πA, with equalityholding iff γ is a circle.

PROOF. We may assume that γ is unit-speed. Then the length of γ is its period, i.e.,γ is `- periodic. Consider the curve γ(t) = (x(t), y(t)) defined by γ(t) = γ

( t`π), i.e.,

γ = γ s or γ s−1 = γ, where s(t) = t`π . [Note that γ is a unit-speed reparametrization

of γ with the parametrization map s−1(t) = tπ` , i.e., s(t) = t`

π .] Then γ is a simple closedcurve with the same length ` and period π. Also the area of interior of γ is same asthe area of interior of γ. Since the length and the area of a curve are invariant underisometry, we may assume that γ(0) = γ(0) = (0, 0). Let x = r cos θ and y = r sin θ.Note that r, θ : [0, π] → R are smooth maps. Since γ is π- periodic, γ(0) = γ(π). Thisgives r(0) = r(π) as γ(0) = (0, 0). Now x2 + y2 = r2 + r2θ2 and xy − yx = r2θ. Sinceγ = γ s, we have ˙γ = γ ds

dt . But then r2 + r2θ2 = x2 + y2 = ‖ ˙γ‖2

= ‖γ‖2(dsdt )2 = `2π2 . We

also note that A =A(int(γ)) = 12∫ π

0 (xy − yx)dt = 12∫ π

0 r2θ. Now,

`2

4π −A = 14

π∫

0

(r2 + r2θ2)dt − 12

π∫

0

r2θdt

= 14

π∫

0

(r2 + r2θ2 − 2r2θ)dt

20 1. CURVE THEORY

= 14

π∫

0

r2(θ − 1)2dt + 14

π∫

0

(r2 − r2)dt ≥ 0.

It follows from the above equation that `24π−A = 0 iff 1

4∫ π

0 r2(θ−1)2dt+∫ π

0 (r2−r2)dt = 0iff∫ π

0 r2(θ − 1)2dt =∫ π

0 (r2 − r2)dt = 0 iff θ = 1 and r = A sin t for some A iff θ = t + αfor some constant α and r = A sin t for some A iff r = A sin(θ−α). But r = A sin(θ−α)is a circle. Hence `2 − 4πA = 0 iff γ is a circle.

Example 1.5.9. By applying the isoperimetric inequality to the ellipse x2

p2 + y2

q2 = 1(where p and q are positive constants), prove that

2π∫

0

√p2 sin2 t + q2 cos2 t dt ≥ 2π√pq,

with equality holding if and only if p = q.Consider γ(t) = (p cos t, q sin t). Then the trace of γ is the ellipse x2

p2 + y2

q2 = 1.Note the γ is 2π-periodic. Therefore length ` of γ is

` =2π∫

0

‖γ(t)‖dt =2π∫

0

√p2 sin2 t + q2 cos2 t dt.

Also, the area A of the interior of γ is

A = 12

2π∫

0

(xy − yx)dt = 12

2π∫

0

pq(cos t cos t + sin t sin t)dt = pq2 2π = pqπ.

It follows from the Isoperimetric inequality that

2π∫

0

√p2 sin2 t + q2 cos2 t dt

2

≥ 4πpqπ = 4π2pq,

i.e.,2π∫

0

√p2 sin2 t + q2 cos2 t dt ≥ 2π√pq.

Equality holds in above inequality iff γ is a circle iff p = q. [Because γ(t) =(p cos t, q sin t) is a circle iff p = q.]

Definition 1.5.10. A simple closed curve in R2 is called convex if its interior is aconvex subset of R2.


Definition 1.5.11. The vertex of a curve γ is a point of the curve where its signedcurvature κs has a stationary point, i.e, κs = 0 at that point.

Theorem 1.5.12 (Four Vertex Theorem). Every convex, simple closed curve in R2

has at least four vertices.

Example 1.5.13. Find the vertices of the curve γ(t) = (a cos t, b sin t), where a, b > 0and a 6= b.

Let s be the arc-length of γ starting at some point of γ. Denoting the derivativeof γ with respect to s by γ ′, we have γ ′(t) = (−a sin t, b cos t) dtds . Since ‖γ ′‖ = 1,dtds = 1√

a2 sin2 t+b2 cos2 t. Therefore

t(t) = γ ′(t) = 1√a2 sin2 t + b2 cos2 t

(−a sin t, b cos t).

Differentiating γ ′ with respect to s, we get

γ ′′(t) = ab(a2 sin2 t + b2 cos2 t)2

(−b cos t,−a sin t).

Now ns(t) = R π2(t(t)) = 1

(a2 sin2 t+b2 cos2 t)12(−b cos t,−a sin t). Hence

γ ′′ = ab(a2 sin2 t + b2 cos2 t) 3

2ns.

This implies that κs(t) = ab(a2 sin2 t+b2 cos2 t)

32. Now κs(t) = −3ab(a2−b2) sin t cos t

(a2 sin2 t+b2 cos2 t)52

. Since a, b > 0and a 6= b, we have κs(t) = 0 iff sin t cos t = 0 iff sin 2t = 0 iff t = nπ

2 for some n ∈ Z.Since γ is 2π- periodic, it follows that the vertices of γ are γ(0) = (a, 0), γ(π/2) = (0, b),γ(π) = (0,−a) and γ(3π/2) = (0,−b) only.

Example 1.5.14. The ellipse x2

a2 + y2

b2 = 1 (or the curve γ(t) = (a cos t, b sin t)) is aconvex curve.

Let (x, y), (z,w) ∈ int(γ), i.e., x2

a2 + y2

b2 < 1 and z2

a2 + w2

b2 < 1. Let X =(xa ,

yb)

andY =

( za ,

wb). Then ‖X‖ < 1 and ‖Y‖ < 1. Let t ∈ [0, 1]. Then ‖tX + (1− t)Y‖ ≤

t‖X‖+ (1− t)‖Y‖ < t + 1− t = 1. Now(tx + (1− t)z)2

a2 + (ty + (1− t)w)2b2 = ‖tX + (1− t)Y‖2 < 1.

Therefore t(x, y) + (1− t)(z,w) ∈ int(γ) and hence γ is convex.

CHAPTER 2

Surfaces in R3

2.1. Surfaces: An introductionDefinition 2.1.1. A subset S of R3 is called a surface if for every p ∈ S there is anopen subset W of R3 containing p and an open subset U of R2 such that S∩W and Uare homeomorphic.

Thus a surface S comes equipped with homeomorphisms φ : U → S ∩W , calledsurface patches or parametrizations of a surface. The collection of all these surfacepatches is called an atlas of the surface.

1.1. Examples.(i) The set S = (x, y, z) ∈ R2 : z = x2 + y2 is a surface.

Let W = R3 and U = R2. Then W is open in R3 and U is open in R2. Consider themap σ : U → S∩W defined by σ (u, v) = (u, v, u2 + v2). Then σ is a bijection andcontinuous. Since σ−1 is the restriction on S of the continuous map ψ : R3 → R2,ψ(x, y, z) = (x, y), the map σ−1 is continuous. Hence σ is a homeomorphismbetween S ∩W and U . This shows that S is a surface.In this example same W and same U work for all points of the surface. Theabove surface is an elliptical paraboloid.

(ii) The set S = (x, y, z) ∈ R3 : z = y2 − x2 is a surface.(iii) The set S = (x, y, z) ∈ R3 : z2 = x2 + y2, z ≥ 0 is a surface. The arguments to

show that both of above are surfaces are similar to the above examples.(iv) The set S = (x, y, z) ∈ R3 : x2 + y2 + z2 = 1 is a surface.

Let W1 = (x, y, z) ∈ R3 : x < 0 ∪ (x, y, z) ∈ R3 : y 6= 0, W2 = (x, y, z) ∈ R3 :x > 0 ∪ (x, y, z) ∈ R3 : z 6= 0 and U = (−π

2 ,π2 )× (0, 2π). Then both W1 and W2

are open in R3 and U is open in R2. For i = 1, 2, define σi : U → S ∩Wi by

σ1(u, v) = (cosu cos v, cosu sin v, sin v) and

σ2(u, v) = (− cosu cos v,− sin v,− cosu sin v).Then both σ1 and σ2 are homeomorphisms. Since σ1(U) ∪ σ2(U) = S, it followsthat S is a surface.

22

1. Surfaces: An introduction 23

We now see another parametrization of S. Let U = (u, v) ∈ R2 : x2 + y2 < 1.Let W1 = (x, y, z) ∈ R3 : z > 0, W2 = (x, y, z) ∈ R3 : z < 0, W3 = (x, y, z) ∈R3 : y > 0, W4 = (x, y, z) ∈ R3 : y < 0, W5 = (x, y, z) ∈ R3 : x > 0 andW6 = (x, y, z) ∈ R3 : x < 0. Then U is open in R2 and each Wi is open in R3.For i = 1, . . . , 6, define σi : U → S ∩Wi by

σ1(u, v) = (u, v,√

1− u2 − v2),

σ2(u, v) = (u, v,−√

1− u2 − v2),

σ3(u, v) = (u,√

1− u2 − v2, v),

σ4(u, v) = (u,−√

1− u2 − v2, v),

σ5(u, v) = (√

1− u2 − v2, u, v) and

σ6(u, v) = (−√

1− u2 − v2, u, v).Then each of σi is a homeomorphism. Since

⋃6i=1 σi(U) = S, the set S is a surface.

(v) The set S = (x, y, z) ∈ R3 : x2 + y2 = 1 is a surface.Let W1 = (x, y, z) ∈ R3 : x > 0 ∪ (x, y, z) ∈ R3 : y 6= 0, W2 = (x, y, z) ∈ R3 :x < 0 ∪ (x, y, z) ∈ R3 : z 6= 0, U1 = (0, 2π) × R and U2 = (−π, π) × R. Thenboth W1 and W2 are open in R3 and both U1 and U2 are open in R2. For i = 1, 2,define σi : Ui → S ∩Wi by

σi(u, v) = (cosu, sinu, v).

Then both σ1 and σ2 are homeomorphisms. Since σ1(U1) ∪ σ2(U2) = S, the set Sis a surface.Let U = (u, v) : 1

2 <√u2 + v2 < 3

2 and W = R3. Then U is open in R2 and Wis open in R3. Define σ : U → S ∩W by

σ (u, v) =(

u√u2 + v2

, v√u2 + v2

, tan(π√u2 + v2 − π)

).

Then σ is a homeomorphism. Hence S is a surface.(vi) Let a, b, c, d ∈ R and let a2 + b2 + c2 6= 0. Then the set S = (x, y, z) ∈ R3 :

ax + by + cz = d is a surface.Here S is a plane. We may assume that c 6= 0. Let U = R2 and W = R3. Then Uis open in R2 and W is open in R3. Define σ : U → S ∩W by

σ (u, v) = (u, v, 1c (d − au − bv)).

Then σ is a homeomorphism (what is its inverse?). Since σ (U) = S, it followsthat S is a surface.

24 2. SURFACES IN R3

(vii) The set S = (x, y, z) : x2 + y2 − z2 = 1 is a surface.Let U = (u, v) : 1

2 <√u2 + v2 < 3

2 and W = R3. Then U is open in R2 and Wis open in R3. Define σ : U → S ∩W by

σ (u, v) =(√

tan2(πr − π) + 1 ur ,√

tan2(πr − π) + 1 vr , tan(πr − π)),

where r =√u2 + v2. Then σ is a homeomorphism. Hence S is a surface.

(viii) The set S = (x, y, z) ∈ R3 : z2 = x2 + y2 is not a surface.Suppose that S is a surface. Then there exists an open subset W of R3 containing(0, 0, 0) such that S∩W is homeomorphic to some open subset U of R2. We mayassume that W is an open ball around origin (why?). Since S ∩W is connected,U is connected (why?). If we remove origin from S∩W , we get two components.While removal of any point from U gives connected subset. This is not possible(again why?). Hence our assumption that S is surface is false.

(ix) Let γ : (a, b) → R3 be a curve given by γ(v) = (0, f (v), g(v)), i.e, γ is a curve inyz- plane. We also assume that the curve does not intersect the z- axis and thecurve does not intersect itself. The surface obtained by rotating γ about the z-axis is called the surface of revolution. Let p = (x, y, z) be a point on this surfaceS. Then the distance of p from the z- axis is f (v) for some v ∈ (a, b). Let q bethe projection of p on the xy- plane. Let u be the angle between the segmentjoining q with the origin. Then the coordinates of q are (f (v) cosu, f (v) sinu, 0).Since the point p is at a distance g(v) from the xy- plane, the coordinates ofp are (f (v) cosu, f (v) sinu, g(v)). Now we prove that S is a surface. Let U1 :(0, 2π)× (a, b) and U2 = (−π, π)× (a, b), W1 = (x, y, z) ∈ R3 : x > 0∪ (x, y, z) ∈R3 : y 6= 0 and W2 = (x, y, z) ∈ R3 : x < 0 ∪ (x, y, z) ∈ R3 : z 6= 0. ThenU1, U2 are open in R2 and W1 and W2 are open in R3. For i = 1, 2, defineσi : Ui → S∩Wi by σi(u, v) = (f (v) cosu, f (v) sinu, g(v)). Then both σ1 and σ2 arehomeomorphisms and σ1(U1) ∪ σ2(U2) = S. Hence S is a surface.

(x) Let b > a > 0. Let C be the circle in xy-plane with centre (b, 0, 0) and radius a.The surface obtained by revolving C about z- axis is called the torus. We showthat torus is a surface.Let S be the torus. Let U1 = (0, 2π) × (0, 2π), U2 = (0, 2π) × (−π, π), U3 =(−π, π) × (0, 2π) and U4 = (−π, π) × (−π, π). The each Ui is open in R2. LetW1 = R3\

((x, y, z) : x2 + y2 = (a + b)2 ∪ (x, y, 0) : 0 ≤ x ≤ a + b

).

W2 = R3\((x, y, z) : x2 + y2 = (a − b)2 ∪ (x, y, 0) : 0 ≤ x ≤ a + b

).

W3 = R3\((x, y, z) : x2 + y2 = (a + b)2 ∪ (x, y, 0) : −(a + b) ≤ x ≤ 0

), and

W4 = R3\((x, y, z) : x2 + y2 = (a − b)2 ∪ (x, y, 0) : −(a + b) ≤ x ≤ 0

). Then

2. Calculus on Surface 25

each Wi is open R3. For i = 1, 2, 3, 4, define σi : Ui → S ∩Wi by

σi(u, v) = ((b + a cos v) cosu, (b + a cos v) sinu, a sin v).Then each σi is a homeomorphism and ∪4

i=1σi(Ui) = S. Therefore S is a surface.

2.2. Calculus on SurfaceIf σ : U → S ∩W ⊂ R3 is a surface patch of a surface S, then we denote the

partial derivative of σ with respect to the first variable and second variable by σu andσv respectively.

Definition 2.2.1. A surface patch σ : U → R3 is called a regular if σ is a smooth mapand σu and σv are linearly independent for all (u, v) in U .

A surface S is called smooth if it has an atlas consisting of regular patches.

Exercise 2.2.2. Check that the surfaces in examples (i), (ii), (iv), (v), (vi), (vii) in section1.1 are smooth surfaces.

Let S be a smooth surface, and let γ be a curve lying on a surface patch σ : U → R3

of S. Then γ has the form γ(t) = σ (u(t), v(t)), where u and v are smooth (real)functions.

Definition 2.2.3. Let p be a point on a surface S. A vector w ∈ R3 is called tangentvector to S at p if there is a curve in S passing through p whose tangent at p is w.

The tangent space at a point p of a surface S is the set of tangent vectors at p .We denote the tangent space of S at p ∈ S by TpS.

Proposition 2.2.4. Let σ : U → R3 be a patch of a surface S containing a point pof S. Let (u, v) be coordinates in U . Then the tangent space to S at p is a vectorsubspace of R3 spanned by the vectors σu and σv (the derivatives are evaluated atthe point (u0, v0) ∈ U such that σ (u0, v0) = p).

PROOF. Let γ be a curve in S passing through p. Then γ(t) = σ (u(t), v(t)) and γ(t0) =σ (u(t0), v(t0)) = σ (u0, v0) = p for some t0. Then the tangent to γ at p = γ(t0) will be

γ(t0) = σu(u(t0), v(t0))u(t0) + σv(u(t0), v(t0))v(t0)= σu(u0, v0)u(t0) + σv(u0, v0)v(t0).

Hence any tangent vector to S at p is a linear combination of σu(u0, v0) and σv(u0, v0),i.e., TpS ⊂ spanσu(u0, v0), σv(u0, v0).

Conversely, let a ∈ spanσu(u0, v0), σv(u0, v0). Then a = ξσu(u0, v0) + ησv(u0, v0)for some ξ, η ∈ R. Consider the curve γ(t) = σ (u0 + ξt, v0 + ηt). Then γ is lyingon the patch and σ and γ(0) = σ (u0, v0) = p. Now the tangent to γ at 0 will be


γ(0) = σu(u0, v0)ξ + σv(u0, v0)η = a. This shows that a is tangent to some curve in Spassing through p, i.e., a ∈ TpS. Therefore spanσu(u0, v0), σv(u0, v0) ⊂ TpS.

Definition 2.2.5. Let S be a surface, and let p be a point in the patch σ : U → R3 withσ (u, v) = p. Then the plane at p parallel to the plane generated by σu(u, v) and σv(u, v)is called the tangent plane to S at p. The vector

N(u, v) = σu(u, v)× σv(u, v)‖σu(u, v)× σv(u, v)‖

is called the unit normal to S at p.

Definition 2.2.6. Let S1 and S2 be smooth surfaces with surface patches σ1 : U1 → R3

and σ2 : U2 → R3.A map f : S1 → S2 is called a smooth map if the map σ−1

2 f σ1 : U1 → U2 is asmooth map. A map f : S1 → S2 is called a diffeomorphism if f is bijective and bothf and f−1 are smooth maps.

A map f : S1 → S2 is called a local diffeomorphism if for every p ∈ S1 there existsan open subset O of S1 containing p such that f|O : O → f (O) is a diffeomorphism.

Let f : S → S be a smooth map, and let p ∈ S. Let w ∈ TpS, i.e., w is a tangentvector to S at p. Then, by definition, there exists a curve γ in S through p suchthat γ(t0) = p and γ(t0) = w for some t0. Then γ = f γ is a curve in S passingthrough f (p) (γ(t0) = f (γ(t0)) = f (p)). Let w be the tangent to γ at the point f (p), i.e.,w = ˙γ(t0). (Hence we have a map which maps w ∈ TpS to w ∈ Tf (p)S.) The mapDpf : TpS → Tf (p)S given by Dpf (w) = w is called the derivative of f , Dpf , at p.

Note that the definition of Dpf (w) depends on f , p and w.Let σ : U → R3 be a surface patch containing p, say, p = σ (u0, v0), and let α, β

be smooth functions such that f (σ (u, v)) = σ (α(u, v), β(u, v)). Let w = λσu + µσvbe a tangent vector at p of a curve γ(t) = σ (u(t), v(t)), where u and v are smoothfunctions such that u(t0) = λ and v(t0) = µ. Since the corresponding curve on S isγ(t) = σ (α(u(t), v(t)), β(u(t), v(t))), we have

Dpf (w) = σu(uαu + vαv) + σv(uβu + vβv),

the derivatives of u and v are evaluated at t0. Thus

Dpf (w) = σu(λαu + µαv) + σv(λβu + µβv).

The righthand side depends only on p, f λ, µ (hence on w).

Proposition 2.2.7. Let f : S → S be a smooth map, and let p ∈ S. Then the derivativeDpf : TpS → Tf (p)S is a linear map.

2. Calculus on Surface 27

PROOF. Let w1 = λ1σu + µ1σv and w2 = λ2σu + µ2σv be in TpS, and let η ∈ R. ThenDpf (ηw1 +w2) = Dpf ((ηλ1 + λ2)σu + (ηµ1 + µ2)σv)

= [(ηλ1 + λ2)αu + (ηµ1 + µ2)αv]σu + [(ηλ1 + λ2)βu + (ηµ1 + µ2)βv]σv= η[(λ1αu + µ1αv)σu + (λ1βu + µ1βv)σv]

+(λ2αu + µ2αv)σu + (λ2βu + µ2βv)σv= ηDpf (w1) +Dpf (w2).

Hence the map Dpf is linear.

Note that the matrix of the linear map Dpf with respect to the basis σu, σv of

TpS and σu, σu of Tf (p)S is(αu αvβu βv

).

Proposition 2.2.8. Let S1, S2 and S3 be smooth surfaces.(i) If p ∈ S1, then the derivative at p of the identity map from S1 to itself is the

identity map from TpS1 to itself.(ii) If f : S1 → S2 and g : S2 → S3 are smooth maps, then for all p ∈ S1, Dp(g f ) =

Df (p)g Dpf .(iii) If f : S1 → S2 is a diffeomorphism, then for all p ∈ S1, the linear map Dpf :

TpS1 → Tf (p)S2 is invertible.PROOF. (i) Here I : S1 → S1 is the identity map, i.e., I(x) = x for all x ∈ S1. Let p be apoint on S1, and let w ∈ TpS. Let γ be a curve in S1 passing through p whose tangentat p is w. Since I is the identity map I γ = γ. Hence the tangent to I γ at p is witself. This shows that DpI(w) = w. Since w is an arbitrary point of TS, then map DpIis the identity map, i.e., DpI(w) = w for all w ∈ TpS1.

(ii) Let p be a point on S1, and let w ∈ TpS1. Let γ be a curve in S1 passing throughp whose tangent at p is w. Let γ1 = f γ. Then γ1 is a curve in S2 passing throughf (p). Let w1 be the tangent to γ1 at f (p). Then Dpf (w) = w1. Let γ2 = g γ1 = g f γ.Then γ2 is curve in S3 passing through g(f (p)). Let w2 be the tangent to γ2 at g(f (p)).Then Df (p)g(w1) = w2, i.e, Df (p)g(Dpf (w)) = w2, i.e., (Df (p)g Dpf )(w) = w2. Also, w2is the tangent to (g f ) γ at g f (p). Therefore Dp(g f )(w) = w2. This showsthat Dp(g f )(w) = (Df (p)g Dpf )(w). Since w is an arbitrary point of TpS1, we getDp(g f ) = Df (p)g Dpf .

(iii) Since f is a diffeomorphism, the map f−1 is smooth and the map f−1 f : S1 →S1 is the identity map. Let p ∈ S. Then I = Dp(f−1 f )

Proposition 2.2.9. Let S1 and S2 be surfaces and let f : S1 → S2 be a smooth map.Then, f is a local diffeomorphism if and only if, for all p ∈ S1, the linear mapDpf : TpS1 → Tf (p)S2 is invertible.


PROOF. Assume that f is a local diffeomorphism. Let p ∈ S1. Since f is a localdiffeomorphism, there exists an open subset O of S1 containing p such that f|O : O →f (O) is a diffeomorphism. It follows from Proposition 2.2.8 (ii) that Dpf is invertible.

The converse follows from the Inverse Function Theorem.

Example 2.2.10. Let f : S1 → S2 be a local diffeomorphism, and let γ be a regularcurve in S1. Show that f γ is a regular curve in S2.

We note that (f γ)•(t) is a tangent to the curve f γ at f γ(t). Hence bythe definition of the derivative Dγ(t)f (γ(t)) = (f γ)•(t) for all t. Since f is a localdiffeomorphism, Dpf is invertible for all t. In particular, Dγ(t)f is invertible for all t.Fix t. Since γ(t) 6= 0, and the linear map Dγ(t)f is invertible, we have (f γ)•(t) =Dγ(t)f (γ(t)) 6= 0. Since t is arbitrary, it follows that, ‖(f γ)•(t)‖ > 0 for all t , i.e., f γ isregular.

2.3. The First Fundamental FormDefinition 2.3.1. Let S be a surface, and let p ∈ S. Then the first fundamental formof S at p is a map Ip : TpS × TpS → R given by

Ip(v,w) = 〈v,w〉p = vw (v,w ∈ TpS).

Suppose that σ : U → R3 be a surface patch of a surface S. Fix p ∈ S. Then anytangent vector to S at p is a linear combination of σu and σv . Define du, dv : TpS → Rby du(w) = λ and dv(w) = µ if w = λσu + µσv . Then both du and dv are linear maps.Now

I(w,w) = 〈w,w〉 = λ2〈σu, σu〉+ 2λµ〈σu, σv〉+ µ2〈σv, σv〉.Let E = ‖σu‖2 = σuσu, F = σuσv and G = ‖σv‖2 = σvσv . The functions E, F,G : U → Rare called the first order magnitudes of the surface S. Hence I(w,w) = 〈w,w〉 =Edu(w)2 + 2Fdu(w)dv(w) +Gdv(w)2 or

〈·, ·〉 = Edu2 + 2Fdudv +Gdv2. (2.3.1.1)

Traditionally, the expression Edu2 + 2Fdudv + Gdv2 is called the first fundamentalform on the surface patch σ . Note that the first fundamental form on the surface doesnot depend on the choice of surface patch, but it depends on S and p.

Let γ : (α, β) → R be a curve lying on a surface patch σ (u, v) of a surface S.Then γ(t) = σ (u(t), v(t)). Therefore γ = uσu + vσv and hence ‖γ‖ = (〈γ, γ〉)1/2 =√Fu2 + 2Fuv +Gv2. If p = γ(t0) and q = γ(t1), then the length of the curve (on the

surface patch σ ) is∫ t1t0

√Fu2 + 2Fuv +Gv2dt.

Example 2.3.2. Compute the first fundamental form on the cylinder x2 + y2 = 1.

4. Local isometries and conformal maps 29

We know that the surface patch of the above surface is given by σ (u, v) =(cosu, sinu, v). Now σu = (− sinu, cosu, 0) and σv = (0, 0, 1). Therefore E = σuσu = 1,F = σuσv = 0 and G = σvσv = 1. Hence the first fundamental form on the surface isdu2 + dv2.

Exercise 2.3.3. Compute the first fundamental form on the following surfaces.(i) σ (u, v) = (f (u) cos v, f (u) sin v, g(u))(ii) σ (u, v) = (cosu cos v, cosu sin v, sinu)(iii) σ (u, v) = (sinhu sinh v, sinhu cosh v, sinhu)(iv) σ (u, v) = (u − v, u + v, u2 + v2)(v) σ (u, v) = (coshu, sinhu, v)(vi) σ (u, v) = (u, v, u2 + v2)

2.4. Local isometries and conformal mapsDefinition 2.4.1. Let S1 and S2 be surfaces. A smooth map f : S1 → S2 is called alocal isometry if it takes any curve in S1 to a curve of same length in S2.

A local isometry that is a diffeomorphism is called an isometry.

Example 2.4.2.(i) Let S1, S2 and S3 be surfaces. Let f : S1 → S2 and g : S2 → S3 be local isometry.

Then g f : S1 → S3 is a local isometry (i.e., composition of local isometries is alocal isometry).Let γ be a curve in S1. Since f is a local isometry f γ is a curve in S2 of thesame length. Since g is a local isometry, g f γ is a curve in S3 of the samelength. Hence if γ is a curve S1, then γ and g f γ have the same length, i.e.,g f is a local isometry.

(ii) The inverse of an isometry is an isometry.Let f : S1 → S2 be an isometry. Then by definition f−1 : S2 → S1 is a diffeomor-phism. Let γ be a curve in S2. Since f is a (local) isometry, the lengths of f−1 γand f f−1 γ = γ are same. Hence f−1 is a local isometry. Therefore f−1 is anisometry.

Definition 2.4.3. Let f : S1 → S2 be a smooth map, and let p ∈ S1. Define f∗ :TpS1 × TpS1 → R by

f∗〈v,w〉p = 〈Dpf (v), Dpf (w)〉f (p) (v,w ∈ TpS1).

Example 2.4.4. The map f∗ is a symmetric bilinear map.Let u, v,w ∈ TpS1 and λ, µ ∈ R. Then

f∗〈u, v〉p = 〈Dpf (u), Dpf (v)〉f (p) = Dpf (u)Dpf (v) = Dpf (v)Dpf (u)


= 〈Dpf (v), Dpf (u)〉f (p) = f∗〈v, u〉p.

Therefore f∗ is symmetric. Now

f∗〈λu + µv,w〉p = 〈Dpf (λu + µv), Dpf (w)〉f (p) = 〈λDpf (u) + µDpf (v), Dpf (w)〉f (p)

= λ〈Dpf (u), Dpf (w)〉f (p) + µ〈Dpf (v), Dpf (w)〉f (p)

= λf∗〈u,w〉p + µf∗〈v,w〉p.

Hence f∗ is bilinear.

Let f : S1 → S2 be a smooth map. Let γ be a curve in S1. Then γ(t0) is a tangent toS1 at the point γ(t0) = p. Now γ = f γ is a curve in S2 passing through f (γ(t0)) = f (p).Therefore by definition Dpf (γ(t0)) = ˙γ(t0). Since t0 arbitrary, we have Dpf (γ) = ˙γ

Example 2.4.5. Let T : Rn × Rn → R be a symmetric bilinear map. If T(x, x) = 0 forall x ∈ Rn, then T = 0.

Let x, y ∈ Rn. Then

0 = T(x + y, x + y) = T(x, x) + 2T(x, y) + T(y, y) = 2T(x, y).

Therefore T(x, y) = 0, i.e., T ≡ 0.It follows from the example that, if S and T are symmetric bilinear maps on Rn

and S(x, x) = T(x, x) for all x, then S = T .

Theorem 2.4.6. A smooth map f : S1 → S2 is a local isometry if and only if thesymmetric bilinear maps 〈·, ·〉p and f∗〈·, ·〉p on TpS1 are equal for every p ∈ S1.

PROOF. If γ is curve in S1, then the length of the part of γ with endpoints γ(t0) andγ(t1) is

∫ t1t0 〈γ, γ〉

1/2dt. Then γ = f γ is a curve in S2. The length of γ between thepoints γ(t0) and γ(t1) is

t1∫

t0

〈 ˙γ, ˙γ〉1/2γ dt =t1∫

t0

〈Dγf (γ), Dγf (γ)〉1/2γ dt =t1∫

t0

f∗〈γ, γ〉1/2γ dt (2.4.6.1)

Assume that the symmetric bilinear maps 〈·, ·〉p and f∗〈·, ·〉p are equal for every p ∈ S1.Then it follows from the equation (2.4.6.1) that any curve γ on S1 and the correspondingcurve γ = f γ have the same length. Therefore f is a local isometry.

Conversely, assume that f is a local isometry. Then for any curve γ and thecorresponding curve γ = f γ in S2 have the same length, i.e.,

t1∫

t0

〈γ, γ〉1/2γ dt =t1∫

t0

f∗〈γ, γ〉1/2γ dt.


Suppose that f∗〈·, ·〉p 6= 〈·, ·〉p for some p ∈ S1. Then there is v ∈ TpS1 such thatf∗〈v, v〉p 6= 〈v, v〉p. Let γ : (a, b → S1 ⊂ R3 such that γ(s) = 0 and γ(s) = v for somes ∈ (a, b). Then 〈γ(s), γ(s)〉

12γ(s) 6= f∗〈γ(s), γ(s)〉

12γ(s). We may assume that 〈γ(s), γ(s)〉

12γ(s)−

f∗〈γ(s), γ(s)〉12γ(s) > 0. By continuity of the map g(u) = 〈γ(u), γ(u)〉

12γ−f∗〈γ(u), γ(u)〉

12γ (u ∈

(a, b)) there exists ε > 0 and c > 0 such that 〈γ(u), γ(u)〉12γ(u) − f∗〈γ(u), γ(u)〉

12γ(u) > c for

all u ∈ (s − ε, s + ε). But then∫ s+εs−ε (〈γ(u), γ(u)〉

12γ(u) − f∗〈γ(u), γ(u)〉

12γ(u))du ≥ 2cε > 0.

i.e.,∫ s+εs−ε 〈γ(u), γ(u)〉

12γ(u)du >

∫ s+εs−ε f∗〈γ(u), γ(u)〉

12γ(u)du. It means the length of the curve

γ between the points γ(s − ε) to γ(s + ε) is not same as the length of curve γ = f γbetween the points γ(s − ε) to γ(s + ε), which is not possible as f is a local isometry.Hence 〈·, ·〉p = f∗〈·, ·〉p for every p ∈ S1.

Thus a smooth map f : S1 → S2 is a local isometry iff for every p ∈ S1,〈Dpf (v), Dpf (w)〉f (p) = 〈v,w〉p (v,w ∈ TpS1).

Exercise 2.4.7. LetH be a finite dimensional Hilbert space. A linear map T :H →His called an isometry if ‖Tx‖ = ‖x‖ for all x ∈ H.

Prove the following statements.(i) The map T is an isometry iff 〈Tx, Ty〉 = 〈x, y〉 for all x, y ∈ H.(ii) Every linear isometry on H is invertible.

Corollary 2.4.8. Let f : S1 → S2 be a smooth map. Then f is a local isometry iff forevery p ∈ S1, the linear map Dpf : TpS1 → Tf (p)S2 is an isometry.

Let σ : U → R3 be a surface patch. Fix (u0, v0) in U . Define γu0 and γv0 by γu0(v) =σ (u0, v) and γv0(u) = σ (u, v0). Then both γu0 and γv0 are curves on σ passing throughσ (u0, v0) (γu0(v0) = σ (u0, v0) = γv0(u0)). Now γu0(v0) = σv(u0, v0) and γv0(u0) = σu(u0, v0).Therefore σu(u0, v0) and σv(u0, v0) are tangent vectors at σ (u0, v0). The curves γu andγv are called parameter curves to the surface patch σ .

Let f : S1 → S2 be a smooth map. Let σ : U → R3 be a surface patch of S1. Letσ = f σ : U → R3. Let (u, v) ∈ U . Then

σu(u, v) = limh→0

f σ (u + h, v)− f σ (u, v)h

= limh→0

f γv(u + h)− f γv(u)h

= (f γv)•(u)Since γv(u) = σu(u, v), by definition of derivative Dpf of f at p = σ (u, v) we haveDpf (γv(u)) = (f γv)•(u), i.e., Dpf (σu(u, v)) = σu(u, v) or Dpf (σu) = σu. Similarly, we getDpf (σv) = σv .


Corollary 2.4.9. A local diffeomorphism f : S1 → S2 is a local isometry if and onlyif, for any surface patch σ of S1, the patches σ and f σ of S1 and S2, respectively,have the same first fundamental form.

PROOF. Let E, F and G be the first order magnitudes of σ and E1, F1 and G1 be thoseof f σ . Assume that f is a local isometry. Then to prove both σ and f σ have thesame fundamental form, we need to prove E = E1, F = F1 and G = G1. Since f is alocal isometry f∗〈v,w〉p = 〈v,w〉p for all p ∈ S1. In particular, f∗〈σu, σu〉p = 〈σu, σu〉p,f∗〈σu, σv〉p = 〈σu, σv〉p and f∗〈σv, σv〉p = 〈σv, σv〉p. Now,

E = 〈σu, σu〉p = f∗〈σu, σu〉p = 〈Dpf (σu), Dpf (σv)〉f (p) = 〈σu, σu〉f (p) = E1.

Similar arguments shows that F = F1 and G = G1. Hence the first fundamental formsof σ and f σ are same.

Conversely, assume that the first fundamental forms of σ and fσ are same. ThenE = E1, F = F1 and G = G1. Fix p = σ (u, v). Let w = λ1σu+µ1σv, z = λ2σu+µ2σv ∈ TpS.Then

f∗〈w, z〉p = 〈Dpf (w), Dpf (z)〉f (p) = 〈Dpf (λ1σu + µ1σv), Dpf (λ2σu + µ2σv)〉f (p)

= λ1λ2〈Dpf (σu), Dpf (σu)〉f (p) + (λ1µ2 + λ2µ1)〈Dpf (σu), Dpf (σv)〉f (p)

+µ1µ2〈Dpf (σv), Dpf (σv)〉f (p)

= λ1λ2〈σu, σu〉p + (λ1µ2 + λ2µ1)〈σu, σv〉p + µ1µ2〈σv, σv〉p= 〈λ1σu + µ1σv, λ2σu + µ2σv〉p = 〈w, z〉p.

Hence f∗〈·, ·〉p = 〈·, ·〉p. Since p ∈ S1 is arbitrary, f∗〈·, ·〉p = 〈·, ·〉p for every p ∈ S1, i.e.,f is a local isometry.

Definition 2.4.10. A local diffeomorphism f : S1 → S2 is called a conformal map if itpreserves angle between curves, i.e., if γ and γ1 are any two curves on S1 intersectingat a point p ∈ S1 and if γ2 and γ2 are their images under f , then the angle between γ1and γ1 at p is equal to the angle between γ2 and γ2 at f (p).

If σ : U → R3 is a surface patch, then σ may be viewed as a map from an opensubset of the plane (namely U) to the image of σ . We say that σ is a conformalparametrization or conformal surface patch if the map σ is conformal.

Exercise 2.4.11. Composition of two conformal maps is a conformal map.

Theorem 2.4.12. A local diffeomorphism f : S1 → S2 is conformal iff there is mapλ : S1 → R such that f∗〈v,w〉p = λ(p)〈v,w〉p for all p ∈ S1 and for all v,w ∈ TpS1.


PROOF. Let γ and γ be two curves on S1 that intersects at p ∈ S1. If θ is an anglebetween these two curves, then

cos θ = 〈γ,˙γ〉p

‖γ‖‖ ˙γ‖= 〈γ, ˙γ〉p〈γ, γ〉1/2p 〈 ˙γ, ˙γ〉1/2p

.

Let φ be the corresponding angle of intersection between the curves f γ and f γon S2 at f (p). Then

cosφ = 〈Dpf (γ), Dpf ( ˙γ)〉f (p)

‖Dpf (γ)‖‖Dpf ( ˙γ)‖

= 〈Dpf (γ), Dpf ( ˙γ)〉f (p)

〈Dpf (γ, Dpf (γ)〉1/2f (p)〈Dpf ( ˙γ), Dpf ( ˙γ)〉1/2f (p)

= f∗〈γ, ˙γ〉pf∗〈γ, γ〉1/2p f∗〈 ˙γ, ˙γ〉1/2p

.

Suppose that there is a map λ : TpS1 → R such that f∗〈v,w〉p = λ(p)〈v,w〉p for allp ∈ S1 and for all v,w ∈ TpS1. Then

cosφ = f∗〈γ, ˙γ〉pf∗〈γ, γ〉1/2p f∗〈 ˙γ, ˙γ〉1/2p

= λ(p)〈γ, ˙γ〉pλ(p)1/2〈γ, γ〉1/2p λ(p)1/2〈 ˙γ, ˙γ〉1/2p

= 〈γ, ˙γ〉p〈γ, γ〉1/2p 〈 ˙γ, ˙γ〉1/2p

= cos θ,

which proves that the map f is conformal.Conversely, assume that f is conformal. Then

f∗〈γ, ˙γ〉pf∗〈γ, γ〉1/2p f∗〈 ˙γ, ˙γ〉1/2p

= 〈γ, ˙γ〉p〈γ, γ〉1/2p 〈 ˙γ, ˙γ〉1/2p

(2.4.12.1)

Since every tangent vector to S1 is a tangent vector to a curve in S1, the equation(2.4.12.1) implies that

f∗〈v,w〉pf∗〈v, v〉1/2p f∗〈w,w〉1/2p

= 〈v,w〉p〈v, v〉1/2p 〈w,w〉1/2p

for all p ∈ S1 and v,w ∈ TpS1. Fix p ∈ S1. Let x, y be an orthonormal basis of TpS1.Let λ = f∗〈x, x〉, µ = f∗〈x, y〉 and ν = f∗〈y, y〉. Let w = cos θx + sin θy ∈ TpS1. Then w


makes an angle θ with x. Therefore

cos θ = 〈x,w〉p〈x, x〉1/2p 〈w,w〉1/2p

= λ cos θ + µ sin θ√λ(λ cos2 θ + 2µ sin θ cos θ + ν sin2 θ)

.

Take θ = π2 . Then µ = 0. Hence we get λ = λ cos2 θ + ν sin2 θ, which gives λ = µ, say,

λ(p). Let v = ζ1x + η1y,w = ζ2x + η2y ∈ TpS1. Thenf∗〈v,w〉p = f∗〈ζ1x + η1y, ζ2x + η2y〉p

= ζ1ζ2f∗〈x, x〉p + (ζ1η2 + ζ2η1)f∗〈x, y〉p + η1η2f∗〈y, y〉p= λ(p)(ζ1ζ2 + η1η2) = λ(p)(v ·w)= λ(p)〈v,w〉p.

Therefore f∗〈v,w〉p = λ(p)〈v,w〉p for all v,w ∈ TpS1. Since p ∈ S1 arbitrary, forevery p ∈ S1 there exists λ(p) ∈ R (hence there is a map λ : S1 → R) such thatf∗〈v,w〉p = λ(p)〈v,w〉p for all v,w ∈ TpS1. Hence the theorem.

Corollary 2.4.13. A local diffeomorphism f : S1 → S2 is conformal iff for any surfacepatch σ of S1, the first fundamental forms of patches σ of S1 and f σ of S2 areproportional.

In particular, a surface patch σ is conformal parametrization iff its first funda-mental form is λ(du2 + dv2) for some smooth map λ.

Example 2.4.14. Let Φ : U → V be a diffeomorphism between open subsets of R2.Write Φ(u, v) = (f (u, v), g(u, v)), where f and g are smooth functions on the uv-plane. Show that Φ is conformal if and only if either (fu = gv and fv = −gu) or(fu = −gv and fv = gu).

We have Φ(u, v) = (f (u, v), g(u, v)). Therefore Φu = (fu, gu) and Φv = (fv, gv). Thisgives E = f2

u + g2u, F = fufv + gugv and F = f2

v + g2v . Assume that Φ is conformally

parametrized. Then its first fundamental form is of the form λ(du2 +dv2), where λ isa smooth function of u and v. This gives F = fufv + gugv = 0 and f2

u + g2u = E = G =

f2v + g2

v . It means the vectors (fu, gu) and (fv, gv) are orthogonal and have the samenorm. Therefore (fv, gv) can be obtained by rotating (fu, gu) by an angle π/2 or 3π/2.Hence (fv, gv) = (gu,−fu) or (fv, gv) = (−gu, fu). Therefore (fu = −gv and fv = gu) or(fu = gv and fv = −gu).

Conversely, if (fu = −gv and fv = gu) or (fu = gv and fv = −gu), then F = 0 andE = G = f2

u + g2u. Therefore the first fundamental form of Φ is (f2

u + g2u)(du2 + dv2)

and hence Φ is conformally parametrized.


Definition 2.4.15. Let σ : U → R3 be a surface patch, and let R ⊂ U . Then the surfacearea, Aσ (R), of the part σ (R) of σ (U) is

Aσ (R) =∫∫

R

‖σu × σv‖ dudv.

Proposition 2.4.16. Let σ : U → R3 be a surface patch. Then ‖σu × σv‖ =√EG − F2.

PROOF. We know that if a , b, c, d ∈ R3, then (a×b)(c×d) = (a c)(b d)− (a d)(bc). Now‖σu × σv‖2 = (σuσu)(σvσv)− (σuσv)2 = EG − F2. Therefore ‖σu × σv‖ =

√EG − F2.

Example 2.4.17.(i) Compute the surface area of sphere of radius r.

We know that a parametrization of sphere of radius r is given by

σ (u, v) = (r cos v cosu, r cos v sinu, r sin v) (u ∈ (0, 2π), v ∈ (−π2 ,π2 )).

Now σu = (−r cos v sinu, r cos v cosu, 0), σv = (−r sin v cosu,−r sin v sinu, r cos v).Therefore E = σuσu = r2 cos2 v, F = σuσv = 0 and G = r2, and EG − F2 =r4 cos2 v. Now the surface area, A, of the sphere is

A =∫∫

(0,2π)×(− π2 ,

π2 )

r2 cos v dudv

= r2π/2∫

−π/2

2π∫

0

cos v dudv

= r2

π/2∫

−π/2

cos vdv

2π∫

0

du

= 4πr2.

(ii) Compute the area of (x, y, z) : x2 + y2 = 1, |z| ≤ 1.We know that the cylinder may be parametrized by σ (u, v) = (cosu, sinu, v),where u ∈ (0, 2π) and v ∈ R). We have E = G = 1 and F = 0. We want the areaof σ (R), where R = (u, v) : u ∈ (0, 2π), |v| ≤ 1. Therefore the area A of σ (R)is

A =∫∫

(0,2π)×[−1,1]

1dv =

2π∫

0

du

1∫

−1

dv

= 4π.

(iii) Let γ : (α, β) → R3 be a unit-speed curve in R3 whose curvature κ is strictly lessthan a−1. Consider the surface patch of the tube of radius a around γ given byσ (t, θ) = γ(t) + a(cos θn(t) + sin θb(t)), t ∈ (α, β), θ ∈ (0, 2π), where n and b are


the unit normal and unit binormal of γ respectively. We want to compute thesurface are of this tube. (Obviously the answer should be 2πa(β − α)).We have σt = t + a(cos θn + sin θb) = (1 − a cos θκ)t − a sin θτn + a cos θτb andσθ = −a sin θn + a cos θb. Therefore E = σtσt = (1− aκ cos θ)2 + a2τ , F = σtσθ =a2τ and G = σθσθ = a2. Now EG − F2 = ((1 − aκ cos θ)2 + a2τ)a2 − a4τ2 =a2(1− aκ cos θ)2. Now

Aσ =∫∫

(α,β)×(0×2π)

√EG − F2dtdθ

=β∫

α

2π∫

0

a(1− aκ cos θ)dtdθ

= 2πa(β − α).

CHAPTER 3

Curvature of a surface

3.1. Second Fundamental FormLet σ : U → R3 be a surface patch with the unit normal N. Fix (u, v) ∈ U . As

the parameter (u, v) changes to (u + δu, v + δv), the surface σ moves away from thetangent plane at σ (u, v) by a distance (σ (u+ δu, v + δv)− σ (u, v))N. Let δu and δv bevery small. By Taylor’s theorem

σ (u + δu, v + δv)− σ (u, v) = σuδu + σvδv + 12[σuu(δu)2 + 2σuvδuδv + σvv(δv)2] + ε

≈ σuδu + σvδv + 12[σuu(δu)2 + 2σuvδuδv + σvv(δv)2].

Hence (σ (u+δu, v+δv)−σ (u, v))N = 12 [L(δu)2+2M(δu)(δv)+N(δv)2], where L = σuuN,

M = σuvN and N = σvvN. Traditionally, the expression Ldu2 + 2Mdudv + Ndv2 iscalled the second fundamental form on a surface. The functions L,M and N arecalled the second order magnitudes.

Exercise 3.1.1. Calculate the second fundamental forms on sphere, cylinder, plane,torus, surface of revolution, hyperboloid with one sheets and two sheets, hyperboliccylinder.

Let S2 denote the unit sphere in R3, i.e., S2 = (x, y, z) : x2 + y2 + z2 = 1.

Definition 3.1.2. Let S be a surface. The map G : S → S2 defined by G(p) = Np,where Np is the unit normal at p ∈ S, is called the Gauss map.

An oriented surface is a smooth surface S together with a smooth choice of unitnormal N at each point, i.e., a smooth map N : S → R3 (It means that each of thethree components of N is a smooth function S → R) such that, for all p ∈ S, N(p)is a unit vector perpendicular to TpS. Al most all surfaces which we have seen areoriented and we will consider them only. There are surfaces which are not orientedfor example Mobius band is not oriented.

Let S be an oriented surface. The map G is a smooth map. Therefore, for p ∈ S,DpG : TpS → TG(p)S2 is a linear map. Note that TpS = TG(p)S2. Hence DpG : TpS → TpSis a linear map.

37

38 3. CURVATURE OF A SURFACE

Example 3.1.3. Find the image of the Gauss map for σ (u, v) = (u, v, u2 +v2), u, v ∈ R.We have σu = (1, 0, 2u) and σu = (0, 1, 2v). Therefore σu×σv = (−2u,−2v, 1). This

gives N = 1√1+4u2+4v2 (−2u,−2v, 1). Hence the image of the Gauss map is 1√

1+4u2+4v2 (−2u,−2v, 1) :u, v ∈ R.

Definition 3.1.4. Let p be a point on an oriented surface S. The map Wp : TpS → TpSdefined by Wp = −DpG is called the Weingarten map.

Since Wp = −DpG and DpG is a linear map, the map Wp : TpS → TpS is a linearmap.

The second fundamental form on a surface S at p ∈ S is a bilinear mapIIp : TpS × TpS → R defined by

IIp〈v,w〉p = 〈Wp(v), w〉p (v,w ∈ TpS).

Let σ be a surface patch of a surface. Fix a point p = σ (u, v) on the surface.Consider the parametric curves γv(t) = σ (t, v) and γu(t) = σ (u, t). Then γv(u) =σ (u, v) = p = γu(v). Now γv(u) = σu(u, v) and γu(v) = σv(u, v). Now G(γv)(t) =Nγv(t) = N(t, v). Hence tangent to the curveGγv atGσ (u, v) (or the tangent toGγvat G(p)) is (G γv)•(u) = Nu(u, v). Hence DpG(σu(u, v)) = Nu(u, v) or DpG(σu) = Nu.One may similarly show that DpG(σv) = Nv .

Lemma 3.1.5. Let σ be a surface patch of an oriented surface with the unit normalN. Then Nuσu = −L, Nuσv = −M = Nvσu and Nvσv = −N .

PROOF. Since σu and σv generates the tangent space and N is orthogonal to the tangentspace, we have Nσu = Nσv = 0. Differentiating Nσu = 0 partially with respect to u, weobtain Nuσu + Nσuu = 0. But Nσuu = L. Hence Nuσu = −L. By differentiating Nσu = 0partially with respect to v, we get Nvσu = −M . Similarly, by differentiating Nσv = 0partially with respect to u and v, we get the relations Nvσu = −M and Nvσv = −Nrespectively.

Proposition 3.1.6. Let σ be a surface patch of an oriented surface S containing apoint p of S, say, p = σ (u, v). Then for any v,w ∈ TpS,

IIp〈w,x〉 = Ldu(w)du(x) +M(du(w)dv(x) + du(x)dv(w)) +Ndv(w)dv(x),

where du, dv : TpS → R are defined by du(w) = λ and dv(w) = µ when w =λσu + µσv ∈ TpS.

PROOF. Since DpG(σu) = Nu and DpG(σv) = Nv , we have Wp(σu) = −Nu and Wp(σv) =−Nv . Let w = λ1σu + µ1σv and x = λ2σu + µ2σv be in TpS. Then

IIp〈w,x〉p = 〈Wp(w), x〉p

2. Gaussian, Mean and Principal Curvatures 39

= 〈Wp(λ1σu + µ1σv), λ2σu + µ2σv〉p= 〈λ1Wp(σu) + µ1Wp(σv), λ2σu + µ2σv〉p= 〈−λ1Nu − µ1Nv, λ2σu + µ2σv〉p= −λ1λ2Nuσu − λ1µ2Nuσv − λ2µ1Nvσu − µ1µ2Nvσv= Ldu(w)du(x) +M(du(w)dv(x) + du(x)dv(w)) +Ndv(w)dv(x).

This proves the result.

A bounded linear map T on a Hilbert space H is self-adjoint if 〈Tx, y〉 = 〈x, Ty〉for all x, y ∈ H. When the Hilbert space is real (i.e., when the field is R), self-adjointmap is often called symmetric.

Corollary 3.1.7. The second fundamental form at a point p ∈ S is a symmetricbilinear map. Equivalently, the Weingarten map is a self adjoint map.

PROOF. One can check (means you) that IIp is linear in both the variable (by fixingone), i.e., IIp is bilinear. It follows from the Proposition 3.1.6 that IIp〈w,x〉p = IIp〈x,w〉pfor all x,w ∈ TpS,i.e., IIp is symmetric. Hence the second fundamental form IIp is asymmetric bilinear map.

Let x,w ∈ TpS. Since IIp〈w,x〉p = IIp〈x,w〉p, we have 〈Wp(w), x〉p = 〈Wp(x), w〉p =〈w,Wp(x)〉p, i.e, 〈Wp(w), x〉p = 〈w,Wp(x)〉p. Hence Wp is symmetric.

3.2. Gaussian, Mean and Principal CurvaturesDefinition 3.2.1. Let Wp be the Weingarten map on an oriented surface S at p ∈ S.Then the Gaussian curvature, Kp, and the mean curvature, Hp, at p are defined by

Kp = det(Wp) Hp = 12 trace(Wp).

Proposition 3.2.2. Let σ : U → R3 be a surface patch of an oriented surface S, andlet p be in the image of σ , i.e., p = σ (u, v) for some (u, v) ∈ U . Then the matrix of

Wp with respect to the basis σu, σv of TpS is(E FF G

)−1( L MM N

).

PROOF. Since N is a unit vector, N N = 1. This will imply that NuN = 0 and NvN = 0.Hence Nu and Nv (and so −Nu and −Nv) are in the tangent space TpS. Thereforethere exist α, β, γ, δ ∈ R such that −Nu = ασu + γσv and −NV = βσu + δσv . SinceWp(σu) = −Nu andWp(σv) = −Nv , we haveWp(σu) = ασu+γσv andWp(σv) = βσu+δσv .

Therefore the matrix of Wp with respect to basis σu, σv of TpS is(α βγ δ

). Taking

dot product with σu and σv on both the sides of the equations −Nu = ασu + γσv and


−NV = βσu+δσv , we obtain L = αE+γF , M = αF+γG, M = βE+δF and N = βF+γG.The last four equations can be written as

(L MM N

)=(E FF G

)(α βγ δ

).

This gives (α βγ δ

)=(E FF G

)−1( L MM N

),

which proves the result.

We have(E FF G

)−1

= 1EG−F2

(G −F−F E

).

Corollary 3.2.3. Let σ be a surface patch of an oriented surface S containing pointp = σ (u, v). Then the Gaussian curvature Kp and the mean curvature Hp are

Kp = LN −M2

EG − F2 and Hp = LG − 2MF +NE2(EG − F2) .

PROOF. Since Kp = det(Wp), we have

Kp = det((

E FF G

)−1( L MM N

))

= det((

E FF G

)−1)det((

L MM N

))

= det((

E FF G

))−1

det((

L MM N

))

= LN −M2

EG − F2 .

Now(E FF G

)−1( L MM N

)= 1

EG−F2

(LG −MF MG −NFME − LF NE −MF

), and henceHp = 1

2 trace(Wp) =LG−2MF+NE

2(EG−F2) .

Exercise 3.2.4. Calculate the Gaussian curvature and mean curvature for sphere,cylinder, plane, torus, surface of revolution, hyperbolic paraboloid, . . ..

Gauss discovered a way to obtain the Gaussian curvature from the Gauss mapitself rather than its derivative, the Weingarten map. His result is an analogue of theresult which says that if γ is a unit-speed plane curve and if φ is the turning angle ofγ, then the signed curvature κs is κs = φ. It means the (signed) curvature is the rate


of change of direction of the tangent vector of γ per unit length. The ‘direction’ ofthe tangent plane of a surface is measured by the unit normal N, so we may expectthat a curvature of a surface is the ‘rate of change of N per unit area’. So to find thecurvature we need to consider the ratios Area(G(R))

Area(R) .

Theorem 3.2.5. Let σ : U → R3 be a surface patch, let (u0, v0) ∈ U , and let δ > 0be such that Rδ = (u, v) : (u − u0)2 + (v − v0)2 < δ2 ⊂ U . Then limδ→0

AN(Rδ )Aσ (Rδ ) = |K|,

where K is the Gaussian curvature of σ at σ (u0, v0).

PROOF. We have Nu ×Nv = LN−M2

EG−G2 (σu × σv) =K (σu × σv). Now

AN(Rδ)Aσ (Rδ)

=∫∫

Rδ ‖Nu ×Nv‖dudv∫∫Rδ ‖σu × σv‖dudv

=∫∫

Rδ |K|‖σu × σv‖dudv∫∫Rδ ‖σu × σv‖dudv

.

Let ε > 0. Since K is continuous, there exists η > 0 (η < δ) such that |K (u0, v0)| − ε <|K (u, v)| < |K (u0, v0)|+ ε for all (u, v) ∈⊂ Rη. Let 0 < β < η. Then

(|K (u0, v0)| − ε)∫∫

Rβ

‖σu × σv‖dudv ≤∫∫

Rβ

|K|‖σu × σv‖dudv

≤ (|K (u0, v0)|+ ε)∫∫

Rβ

‖σu × σv‖dudv.

This gives

|K (u0, v0)| − ε ≤∫∫

Rβ |K|‖σu × σv‖dudv∫∫Rβ ‖σu × σv‖dudv

≤ |K (u0, v0)|+ ε,

i.e.,|K (u0, v0)| − ε ≤

AN(Rβ)Aσ (Rβ)

≤ |K (u0, v0)|+ ε.

Hence ∣∣∣∣AN(Rβ)Aσ (Rβ)

− |K (u0, v0)|∣∣∣∣ < ε for every 0 < β < η,

i.e., limβ→0

AN(Rβ)Aσ (Rβ) = |K (u0, v0)|.

Example 3.2.6. For a plane, the image of the unit normal N is a point only. HenceAN(Rδ) = 0 for every δ > 0. Therefore |K| = 0, i.e., K = 0 at every point.

For a cylinder, the image of the unit normal N is a circle. Hence AN(Rδ) = 0 forevery δ > 0. Therefore |K| = 0, i.e., K = 0 at every point.


For a unit sphere, the unit normal is the identity map. Hence AN(Rδ) = Aσ (Rδ)for every δ > 0. Hence K = 1 at every point.

Proposition 3.2.7. Let p be a point on an oriented surface S. Then there existscalars κ1, κ2 and a basis t1, t2 of TpS such that Wp(ti) = κiti for i = 1, 2. Moreover,if κ1 6= κ2, then the vectors t1 and t2 are orthogonal.

PROOF. Since Wp is symmetric, both its eigen values are real, say, κ1 and κ2. Let t1and t2 be corresponding eigen vectors. Then Wp(ti) = κiti for i = 1, 2. If κ1 = κ2, thenWp = κ1I . But then every element of TpS is an eigen vector. So we may choose t1 andt2 so that they are linearly independent and hence a basis of TpS. Let κ1 6= κ2. Thenκ1〈t1, t2〉 = 〈κ1t1, t2〉 = 〈Wp(t1), t2〉 = 〈t1,Wp(t2)〉 = 〈t1, κ2t2〉 = κ2〈t1, t2〉. Since κ1 6= κ2,it follows that 〈t1, t2〉 = 0, i.e, t1 and t2 are orthogonal and hence t1, t2 is a basis ofTpS.

If x is an eigen vector of a linear map T , then any non-zero scalar multiple of xis also an eigen vector of T . Hence we may assume that the vectors t1 and t2 in aboveproposition are unit vectors.

Definition 3.2.8. Let p be a point on an oriented surface S. Then there exist scalarsκ1, κ2 and an orthonormal basis t1, t2 of TpS such that Wp(ti) = κiti for i = 1, 2.The scalars κ1 and κ2 are called the principal curvatures of S at p and the vectors(directions) t1 and t2 are called the principal vectors (directions) of S at p.

A point of a surface at which both the principal curvatures are equal is called anumbilical point.

It is known that every real symmetric matrix is diagonalizable, i.e., if A ∈ Mn(R)is symmetric, then there is an invertible matrix C ∈ Mn(R) such that C−1AC is adiagonal matrix and the entries on the main diagonal of the matrix are precisely theeigen values of A. Consider Wp as a matrix. Then Wp is symmetric as W ∗

p = Wp.

Therefore there is an invertible matrix C ∈M2(R) such that C−1WpC =(κ1 00 κ2

).

Hence it follows that κ1 = κ2 iff C−1WpC =(κ1 00 κ1

)= κ1I iff Wp = κ1I .

Hence we have the following result. A point p on a surface S is umbilic iff Wp isa scalar multiple of the Identity.

Proposition 3.2.9. Let p be a point on an oriented surface S. Let κ1 and κ2 beprincipal curvatures of S at p. Then the Gaussian curvature Kp and the meancurvature Hp at p are given by Kp = κ1κ2 and Hp = 1

2 (κ1 + κ2).


PROOF. Let t1 and t2 be the principal vectors. Then Wp(ti) = κiti for i = 1, 2. Then

the matrix of Wp with respect to basis t1, t2 of TpS is(κ1 00 κ2

). Hence Kp =

det(κ1 00 κ2

)= κ1κ2 and Hp = 1

2 trace(κ1 00 κ2

)= 1

2 (κ1 + κ2)

Proposition 3.2.10. The principal curvatures are the roots of the equation∣∣∣∣L − λE M − λFM − λF N − λG

∣∣∣∣ = 0,

and the corresponding principal vectors are t = ξσu + ησv where(L − λE M − λFM − λF N − λG

)(ξη

)=(

00

).

PROOF. We know that the principal curvatures at p are the eigen values of Wp, i.e., theprincipal curvatures are the roots of the characteristic equation of Wp. Since the ma-

trix of Wp with respect to basis σu, σv of TpS is(E FF G

)−1( L MM N

), the characteris-

tic equation will be

∣∣∣∣∣

(E FF G

)−1( L MM N

)− λ

(1 00 1

)∣∣∣∣∣ = 0, i.e.,∣∣∣∣L − λE M − λFM − λF N − λG

∣∣∣∣ = 0.

If w = ξσu + ησu is a principal direction, then it is an eigen vector of Wp corre-

sponding to some eigen value, say, λ. Therefore(E FF G

)−1( L MM N

)(ξη

)= λ

(ξη

),

i.e.,(L MM N

)(ξη

)= λ

(E FF G

)(ξη

), which will give the equation

(L − λE M − λFM − λF N − λG

)(ξη

)=(

00

).

Exercise 3.2.11. Compute the principal curvatures and principal vectors to sphere,cylinder, hyperbolic paraboloid, torus, surface of revolution, . . .

Let γ be a unit-speed curve on a surface S, then γ is a unit vector and by definitionit is a tangent vector to S. Therefore it is orthogonal to the unit normal Nγ . Since γis perpendicular to γ, γ is a linear combination of Nγ and Nγ × γ. Therefore thereexists functions κn and κg (of t) such that γ = κnNγ + κg (Nγ × γ).

Definition 3.2.12. Let γ be a unit-speed curve on an oriented surface S. Then thefunctions κn and κg (of t) satisfying γ = κnNγ + κg (Nγ × γ) are called the normalcurvature and the geodesic curvature of γ respectively.


Proposition 3.2.13. Let γ be a unit-speed curve on an oriented surface S. Thenκn = γNγ , κg = γ(Nγ × γ), κ2 = κ2

n + κ2g , κn = κ cosψ and κg = ±κ sinψ, where κ is the

curvature of γ and ψ is the angle between Nγ and the principal normal n of γ.

PROOF. Since Nγ and Nγ×γ are orthogonal vectors, by taking dot product with Nγ andNγ × γ both the sides of the equation γ = κnNγ + κg (Nγ × γ) we obtain κn = γNγ andκg = γ(Nγ × γ). Since Nγ and Nγ × γ are orthogonal unit vectors, κ2 = ‖γ‖2 = κ2

n + κ2g .

Since ψ is the angle between Nγ and n, cosψ = Nγn. Therefore κ cosψ = (κn)Nγ =γNγ = κn. Since κ2 = κ2

n + κ2g , we have κg = ±κ sinψ.

Definition 3.2.14. A normal section of a surface S is a curve γ on S such that theunit normal n of γ and the unit normal Nγ are parallel at each point of the curve γ.

Corollary 3.2.15. The curvature κ, the normal curvature κn and the geodesic cur-vature κg of a normal section of a surface are related by κn = ±κ and κg = 0.

Proposition 3.2.16. If γ is a unit-speed curve on an oriented surface S, then itsnormal curvature is given by the formula κn = IIγ〈γ, γ〉 = 〈Wγ(γ), γ〉γ . If σ is asurface patch of a surface S and if γ(t) = σ (u(t), v(t)) is a unit-speed curve on σ ,then κn = Lu2 + 2Muv +Nv2.

PROOF. Since γ is a tangent vector to S, Nγ(t)γ(t) = 0 for all t , i.e., (G γ(t))γ(t) = 0for all t. This gives (G γ(t))γ = −γ(G γ(t))• = −γDγG(γ(t)) = γWγ(γ(t)) for all t.Hence κn = Nγγ = Wγ(γ(t))γ = 〈Wγ(γ), γ〉.

Let γ(t) = σ (u(t), v(t)). Then γ = σuu + σv v. Now

κn = 〈Wγ(γ), γ〉 = 〈Wγ(σuu + σv v), σuu + σv v〉= 〈uWγ(σu) + vWγ(σv), σuu + σv v〉= 〈u(−Nu) + v(−Nv), σuu + σv v〉= u2(−Nuσu) + uv(−Nuσv) + uv(−Nvσu) + v2(−Nvσv)= Lu2 + 2Muv +Nv2.

This completes the proof

Let p be point on an oriented surface S, and let v be a unit vector in TpS. Thenthere is a unit-speed curve γ in S such that γ(t) = p and γ(t) = v. Then by above thenormal curvature of γ at the point γ(t) is 〈Wγ(t)(γ(t)), γ(t)〉 = 〈Wp(v), v〉 = IIp〈v, v〉.

Let t1 be a principal direction corresponding to the principal curvature κ1 at p.We may assume that t1 is a unit vector. Then the normal curvature of the surfaceat p in the unit direction t1 is κn = 〈Wp(t1), t1〉 = 〈κ1t1, t1〉 = κ1〈t1, t1〉 = κ1. Hence theprincipal curvatures are the normal curvatures.


Theorem 3.2.17. (Meusnier’s Theorem) Let p be a point on an oriented surface S,and let v be a unit tangent vector at p. Let Πθ be a plane containing a line throughp parallel to v and making an angle θ with the tangent plane at p and is not parallelto the tangent plane at p. Suppose that Πθ intersects S in a curve with curvatureκθ (at p). Then κθ sin θ is independent of θ.

PROOF. Let γθ be a unit-speed reparametrization of the curve of the intersection ofΠθ and the tangent plane at p. Then at p, γθ = ±v, so γθ is perpendicular to v andparallel to Πθ . Thus the unit normal n to the curve γθ at p makes an angle π

2 ± θ withthe unit normal N of the surface at p. Therefore κn = κθ cos(π2 − θ) = κθ sin θ. Sinceκn depends only on p and v, it follows that κn is independent of θ.

Theorem 3.2.18. (Euler’s Theorem) Let p be point on an oriented surface S, andlet κ1 and κ2 be the principal curvatures of S with principal vectors t1 and t2. Let vbe a unit vector in TpS. Then the normal curvature of S at p in the direction v isgiven by κn = κ1 cos2 θ + κ2 sin2 θ, where θ is the oriented angle t1 v.

PROOF. By replacing t2 by −t2 (if needed) we may assume that t1 t2 = π2 . Also we may

assume that both t1 and t2 are unit vectors. Since θ is an oriented angle between t1and v, v = cos θt1 + sin θt2. Now

κn = 〈Wp(v), v〉= 〈Wp(cos θt1 + sin θt2), cos θt1 + sin θt2〉= 〈κ1 cos θt1 + κ2 sin θt2, cos θt1 + sin θt2〉= κ1 cos2 θ + κ2 sin2 θ,

which proves the theorem.

Corollary 3.2.19. The principal curvatures at a point of a surface are maximumand minimum values of the normal curvatures. Moreover, the principal vectors(directions) are the directions giving these maximum and minimum values.

PROOF. Let κ1 and κ2 be the principal curvatures at a point p on the surface S. Wemay assume that κ1 ≤ κ2. Let t1 and t2 be the corresponding principal directions. Wemay assume that t1 t2 = π

2 . Let v be a unit tangent at p. Then the normal curvature atp in the direction v is κn = κ1 cos2 θ+κ2 sin2 θ, where θ is the oriented angle t1 v. Nowκ1 = κ1 cos2 θ+κ1 sin2 θ ≤ κ1 cos2 θ+κ2 sin2 θ = κn ≤ κ2 cos2 θ+κ2 sin2 θ = κ2. Hence κnlies between κ1 and κ2. Note that the normal curvature at p in the direction of t1 is κ1and the normal curvature at p in the direction of t2 is κ2. Hence the principal vectorsare the directions at p giving these maximum and minimum values.


Example 3.2.20. Let p be an oriented surface. Let t1 and t2 be principal directions atp, and let κ1 and κ2 be corresponding principal curvatures at p. If κn(θ) is a normalcurvature at p in the direction making an oriented angle θ with the principal direction,then show that 1

2π∫ 2π

0 κn(θ)dθ = Hp.We may assume that t1t2 = π

2 . It follows from the Euler’s theorem that κn(θ) =κ1 cos2 θ + κ2 sin2 θ. Now, the calculations give 1

2π∫ 2π

0 κn(θ)dθ = Hp.

Exercise 3.2.21. With the notations of the above example prove the following.(i) Hp = 1

2 (κn(θ) + κn(θ + π2 )) for any θ.

(ii) Letm ≥ 3. ThenHp = 1m

(κn(θ) + κn

(θ + 2π

m)

+ κn(θ + 4π

m)

+ · · ·+ κn(θ + 2π(m−1)

m

)).

Proposition 3.2.22. Let σ : U → R3 be a connected surface patch of an orientedsurface of which every point is umbilic. Then σ (U) is (part of) a plane or sphere.

PROOF. Since every point is an umbilical point, both the principal curvatures are same,say, κ and W = κI . Since Wp(σu) = −Nu and Wp(σv) = −Nv , we have Nu = −κσu andNv = −κσv . Now (κσu)v = −(Nu)v = −(Nv)u = (κσv)u. Therefore κvσu + κσuv =κuσv + κσuv , i.e, κvσu = κuσv . Since σu and σv are linearly independent, κu = κv = 0,i.e., κ is a constant map.

Suppose that κ = 0. Since Nu = −κσu and Nv = −κσv , we get Nu = Nv = 0, i.e., Nis a constant unit vector. Now (σN)u = σuN = 0 and (σN)v = σvN = 0. Hence σN = dfor some constant d ∈ R. This shows that σ is lying on a plane RN = d, i.e., S is (partof) a plane.

Suppose that κ 6= 0. Then (σ − 1κN)u = σu − 1

κNu = σu − 1κκσu = 0 and similarly

(σ − 1κN)v = 0. Hence σ − 1

κN is a constant vector, say, a. Therefore ‖σ − a‖ = 1|κ| and

hence σ is lying on the sphere ‖R − a‖ = 1|κ| , i.e., σ is (part of) a sphere.

Let σ be a surface patch of a surface S, and let p be in σ . By applying a rigidmotion to S we may assume that p = (0, 0, 0) and the principal vectors are t1 = (1, 0, 0)and t2 = (0, 1, 0). By Taylors theorem

σ (s, t) ≈ σ (0, 0) + sσu + tσv + 12[s2σuu + 2stσuv + t2σvv].

If σ (s, t) = (x, y, z), then z = 12 [s2σuu + 2stσuv + t2σvv]N = 1

2 (Ls2 + 2Mst + Nt2). Ift = sσu + tσv , then Ls2 + 2Mst + Nt2 = 〈W (t), t〉. Now t = xt1 + yt2. ThereforeW (t) = xW (t1) + yW (t2) = xκ1t1 + yκ2t2 = (κ1x, κ2y, 0). This gives Ls2 + 2Mst +Nt2 =(κ1x, κ2y, 0)(x, y, 0) = κ1x2 + κ2y2. Hence we obtain

z = 12(κ1x2 + κ2y2). (3.2.22.1)


(i) If both κ1 and κ2 are positive of both negative (i.e.,K > 0), then equation (3.2.22.1)represents an elliptic paraboloid. In this case p is called an elliptic point.

(ii) If both κ1 and κ2 are of opposite signs, both non-zero, (i.e., K < 0), then equation(3.2.22.1) represents a hyperbolic paraboloid. In this case p is called an hyperbolicpoint.

(iii) If one of κ1 and κ2 is zero and the other is non-zero, (i.e., K = 0), then equation(3.2.22.1) represents a parabolic cylinder. In this case p is called an parabolicpoint.

(iv) If both κ1 and κ2 are zero, (i.e.,K = 0), then equation (3.2.22.1) represents a plane.In this case p is called an flat point or a planar point.

Definition 3.2.23. A point p on an oriented surface S is called(i) an elliptic point if the Gaussian curvature at p is positive.(ii) a hyperbolic point if the Gaussian curvature at p is negative.(iii) a parabolic point if the Gaussian curvature at p is zero but one of the principal

curvature at p is non-zero.(iv) a flat point or a planar point if both the principal curvatures at p are zero.

Exercise 3.2.24. Determine the points on σ (u, v) = ((b+a cos v) cosu, (b+a cos v) sinu, a sin v)which are parabolic, elliptical and hyperbolic.

The set of ellptic points of σ is σ (u, v) : v ∈ (0, π2 ) ∪ (3π2 , 2π).

The set of parabolic points of σ is σ (u, v) : v ∈ π2 ,3π2 .

The set of hyperbolic points of σ is σ (u, v) : v ∈ (π2 ,3π2 ).

CHAPTER 4

Geodesics and some fundamental resultsDefinition 4.0.25. Let σ be a regular surface patch. Since σu, σv and N are linearlyindependent, there exist function Γk

ij such that

σuu = Γ111σu + Γ2

11σv + LNσuv = Γ1

12σu + Γ212σv + MN

σvv = Γ122σu + Γ2

22σv + NN

(4.0.25.1)

The functions Γkij are called the Christoffel symbols of second kind.

Proposition 4.0.26. Let σ be a regular surface patch. Then Γ111 = GEu−2FFu+FEv

2(EG−F2) , Γ211 =

2EFu−EEv−FEu2(EG−F2) , Γ1

12 = GEv−FGu2(EG−F2) , Γ2

12 = EGu−FEv2(EG−F2) , Γ1

22 = 2GFv−GGu−FGv2(EG−F2) and Γ2

22 = EGv−2FFv+FGu2(EG−F2) .

Definition 4.0.27. Let γ : (α, β) → R3 be a curve on a smooth surface S. A tangentvector field along γ is a smooth map from an interval (α, β) to R3 such that v(t) ∈ Tγ(t)Sfor all t ∈ (α, β).

Let v be a tangent vector field along a curve γ lying on a surface S. Then thecomponent of v along the unit normal Nγ is (vNγ)Nγ . Therefore the orthogonalprojection of v on the tangent space at TγS is ∇γ(v) = v − (vNγ)Nγ .

Definition 4.0.28. Let γ be a curve on a surface S, and let v be a tangent vector fieldalong γ. Then the covariant derivative of v along γ is the orthogonal projection∇γ(v) = v − (vNγ)Nγ of v on the tangent space TγS.

A tangent vector field v along a curve γ on a surface S is said to be parallel alongγ if ∇γ(v) = 0 at every point of γ.

Proposition 4.0.29. A tangent vector field v is parallel along a curve γ on a surfaceS iff v is perpendicular to the tangent space of S at every point of γ (i.e., v is parallelto the unit normal Nγ at every point of γ).

PROOF. Suppose that v is parallel along γ. Then ∇γ(v) = v − (vNγ)Nγ = 0, i.e., v =(vNγ)Nγ . Therefore v and Nγ are linearly dependent hence it is perpendicular to thetangent space of S at every point.

48

1. Geodesics 49

Conversely, assume that v is perpendicular to the tangent space of S at everypoint of γ. Then v = λNγ for some function λ. Now ∇γ(v) = v − (vNγ)Nγ = λNγ −(λNγ Nγ)Nγ = λNγ − λNγ = 0. Hence v is parallel along γ.

Proposition 4.0.30. Let γ(t) = σ (u(t), v(t)) be a curve on a surface patch σ , and letv(t) = α(t)σu(u(t), v(t)) + β(t)σv(u(t), v(t)) be a tangent vector field along a curve γ,where α and β are smooth maps of t. Then v is parallel along γ iff

α+ (Γ111u + Γ1

12v)α+ (Γ112u + Γ1

22v)β = 0β + (Γ2

11u + Γ212v)α+ (Γ2

12u + Γ222v)β = 0

(4.0.30.1)

(Note that these equations involve first order magnitudes of σ only.)

PROOF. Since v = ασu + βσv , we have

v = ασu + α(σuuu + σuv v) + βσv + β(σuvu + σvv v)= ασu + α((Γ1

11σu + Γ211σv + LNγ)u + (Γ1

12σu + Γ212σv +MNγ)v)

+βσv + β((Γ112σu + Γ2

12σv +MNγ)u + (Γ122σu + Γ2

22σv +NNγ)v)= (α+ (Γ1

11u + Γ112v)α+ (Γ1

12u + Γ122v)β)σu + (β + (Γ2

11u + Γ212v)α+ (Γ2

12u + Γ222v)β)σv

+(αuL + (αv + βv)M + βvN)Nγ.

Now v is parallel along γ iff v is parallel to Nγ iff the coefficients of σu and σv in theexpression of v are 0. Hence v is parallel along γ iff the equations in (4.1.2.1) hold.

4.1. GeodesicsDefinition 4.1.1. A curve γ on a smooth surface S is called a geodesic if γ(t) isperpendicular to the tangent space of the surface at γ(t) for all t (i.e., γ and Nγ arelinearly dependent at every point of γ).

Equivalently, γ is a geodesic iff γ is a tangent vector field parallel along γ.

Proposition 4.1.2. Let γ(t) = σ (u(t), v(t)) be a curve on a surface patch σ . Then γis a geodesic iff

u + (Γ111u + Γ1

12v)u + (Γ112u + Γ1

22v)v = 0v + (Γ2

11u + Γ212v)u + (Γ2

12u + Γ222v)v = 0

(4.1.2.1)

PROOF. Here γ(t) = σ (u(t), v(t)). Therefore γ = uσu + vσv . Then γ is a geodesic iff γis parallel along γ iff both the equations hold (See Proposition 4.0.30).

Proposition 4.1.3. Any geodesic has a constant speed.

50 4. GEODESICS AND SOME FUNDAMENTAL RESULTS

PROOF. Let γ be a geodesic on a surface S. We have ddt‖γ‖

2 = ddt (γγ) = 2γγ. Since γ

is a geodesic, γ is perpendicular to the tangent space and is therefore perpendicularto γ. So γγ = 0 and hence ‖γ‖ is constant (by continuity of ‖γ‖ ???).

Let γ be a geodesic on a surface S. Let λ = ‖γ‖. Take γ(t) = γ( tλ). Then γ is

a unit-speed curve also ¨γ = 1λ2 γ. Since γ is perpendicular to the tangent space, ¨γ is

perpendicular to the tangent space. Therefore γ is a geodesic. Hence we may nowassume that any geodesic is unit-speed.

Proposition 4.1.4. A unit-speed curve on a surface is a geodesic iff its geodesiccurvature is zero every where.

PROOF. Let γ be a unit-speed curve on a surface S. We know that a geodesic curvatureof a curve γ is given by κg = γ(Nγ × γ).

Assume that γ is a geodesic. Then γ and Nγ are linearly dependent. Henceκg = γ(Nγ × γ) = (γ ×Nγ)γ = 0.

Conversely, assume that κg = 0. Then γ is perpendicular to Nγ × γ. Since γ, Nγand γ × Nγ are mutually perpendicular and γ is perpendicular to γ, it follows that γand Nγ are linearly dependent. Therefore γ is a geodesic.

Proposition 4.1.5. Any (part of a) line on a surface is a geodesic.

PROOF. Let γ be (part of a) line lying on a surface. Then γ may be parametrized asγ(t) = t a + b. Then γ = 0. Hence γ is a geodesic.

Theorem 4.1.6. Let γ(t) = σ (u(t), v(t)) be a curve on a regular surface patch σ of asurface S. Then γ is a geodesic iff the following equations hold.

ddt (Eu + Fv) = 1

2 (Euu2 + 2Fuuv +Guv2)ddt (Fu +Gv) = 1

2 (Evu2 + 2Fvuv +Gv v2).

(4.1.6.1)

The equations in (4.1.6.1) are called geodesic equations.

PROOF. We know that σu, σv is a basis of the tangent space of a surface. Thereforeγ is a geodesic iff γ is perpendicular to both σu and σv , i.e., γσu = 0 and γσv = 0. Sinceγ(t) = σ (u(t), v(t)), γ = uσu + vσv and hence γ = d

dt (uσu + vσv). It follows that γ isa geodesic iff σu d

dt (uσu + vσv) = 0 and σv ddt (uσu + vσv) = 0. Now d

dt (σu(uσu + vσv)) =σu d

dt (uσu + vσv) + (uσu + vσv)(uσuu + vσuv). Therefore

σuddt (uσu + vσv) = d

dt (σu(uσu + vσv))− (uσu + vσv)(uσuu + vσuv)

= ddt (uE + vF )− (u2σuσuu + uv(σuσuv + σuuσv) + v2σuvσv)

1. Geodesics 51

= ddt (uE + vF )− (u2 1

2Eu + uvFu + v2 12Gu)

= ddt (uE + vF )− 1

2(u2Eu + uv2Fu + v2Gu).

Similarly, we get

σvddt (uσu + vσv) = d

dt (uF + vG)− 12(u2Ev + uv2Fv + v2Gv).

Hence γ is a geodesic iff both the equations in (4.1.6.1) are satisfied.

Example 4.1.7. Geodesics on the cylinder.We know that a parametrization of the cylinder S may be given as σ (u, v) =

(cosu, sinu, v). The first order magnitudes of the cylinder are E = 1, F = 0 andG = 1. Therefore Eu = Ev = Fu = Fv = Gu = Gv = 0. Let γ(t) = σ (u(t), v(t)) be aunit-speed curve on S. Then γ is a geodesic equation iff the geodesic equations hold,i.e., d

dt (u) = u = 0 and ddt (v) = v = 0. This implies that u(t) = at + b and v(t) = ct + d

for some constants a, b, c and d. Therefore

γ(t) = (cos(at + b), sin(at + b), ct + d).

Since γ is unit-speed curve, a2 + c2 = 1.If a 6= 0 and c = 0, then γ is (part of) a circle (in fact it is a part of the circle

which is the intersection of S with the plane z = d).If a = 0 and c 6= 0, then γ is (part of) a line.If a 6= 0 and c 6= 0, then γ is (part of) a helix.Hence the geodesics on S are (part of) lines, circles and helices on S.

Example 4.1.8. Determine the geodesics on a plane.(Ans. (part of) lines on the plane)

We may parametrize a plane by σ (u, v) = (u, v, au + bv), where a and b aresome constants. Then E = 1 + a2 F = ab and G = 1 + b2. Therefore Eu = Ev =Fu = Fv = Gu = Gv = 0. Let γ(t) = σ (u(t), v(t)) = (u(t), v(t), au(t) + bv(t)) be aunit-speed curve on the plane. Then γ is a geodesic iff d

dt ((1 + a2)u + abv) = 0 andddt (abu + (1 + b2)v) = 0 iff (1 + a2)u + abv = α and abu + (1 + b2)v = β for someconstants α and β iff u(t) = At + B and v(t) = Ct + D for some constants A and B.Hence γ(t) = (A,C,Aa +Cc)t + (B,D,Ba +Dc). Therefore γ is (part of) a line.

Example 4.1.9. Geodesics on the unit sphere S = (x, y, z) : x2 + y2 + z2 = 1.We know that the sphere S may be parameterized as

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


Then σu = (− sinu cos v, cosu cos v, 0), σv = (− cosu sin v,− sinu sin v, cos v). There-fore

E = cos2 v, F = 0 and G = 1,so Eu = Fu = Fv = Gu = Gv = 0 and Ev = −2 cos v sin v. Let γ(t) = σ (u(t), v(t)) be aunit-speed curve on S. Since γ is a unit-speed curve

1 = ‖σuu + σv v‖2 = u2 cos2 v + v2. (4.1.9.1)Suppose that γ is a geodesic. Then the geodesic equations imply that d

dt (u cos2 v) = 0and d

dt (v) = 12 (−2u2 cos v sin v). Therefore u cos2 v = Ω, for some constant Ω.

If Ω = 0, then u = 0, i,e, u is constant. Therefore γ is a part of meridian.Let Ω 6= 0. Since γ is a unit-speed curve, v2 = 1− u2 cos2 v = 1− Ω2

cos2 v . Now(dvdu

)2

=(v2

u2

)= cos4 v

Ω2 − cos2 v = cos2 v(Ω−2 cos2 v − 1

).

Therefore±du = dv

cos v√

Ω−2 cos2 v − 1.

Solving above differential equation we get

±(u − u0) = sin−1(

tan v√Ω−2 − 1

),

∴ tan v = ± sin(u − u0)√

Ω−2 − 1. (4.1.9.2)Take a = ∓ sinu0

√Ω−2 − 1 and b = ± cosu0

√Ω−2 − 1. Then

∓ sinu0√

Ω−2 − 1 cosu cos v ± cosu0√

Ω−2 − 1 sinu cos v

= ∓√

Ω−2 − 1 cos v(cosu sinu0 − cosu0 sinu)= ±

√Ω−2 − 1 cos v sin(u − u0)

= sin v. (∵ equation (4.1.9.2))Therefore every point on the curve γ is lying on the plane Ax + By = z. Hence γ ison the intersection of the sphere and the plane Ax+By = z. This shows that γ is partof a great circle of the sphere. Hence the geodesics on the sphere are great circles ofthe sphere. Since every great circle is a normal section, it follows that the geodesicson the sphere are precisely (part of) great circles of the sphere.

Example 4.1.10. Consider the surface of revolution σ (u, v) = (f (u) cos v, f (u) sin v, g(u)),where f > 0 and

( dfdu)2 +

(dgdu)2 = 1. Then

(i) Every meridian is a geodesic, i.e., the curves α(t) = σ (u(t), v0) is a geodesic.(ii) A parallel u = u0 (say) is a geodesic iff df

du (u0) = 0.

2. Codazzi - Mainardi and Gauss equations 53

Here σu = ( dfdu cos v, dfdu sin v, dgdu ) and σv = (−f sin v, f cos v, 0). Therefore E = 1,F = 0 and G = f2, and so Eu = Ev = Fu = Fv = Gv = 0 and Gu = 2f dfdu .

Let γ(t) = σ (u(t), v(t)) be a unit-speed curve on S. Since γ is unit-speed,

1 = ‖σuu + σv v‖2 = u2 + f2v2. (4.1.10.1)

Then γ is a geodesic iffddt (u) = f dfduv

2 and ddt (f

2v) = 0. (4.1.10.2)

(i) Let α(t) = σ (u(t), v0). Since α is unit-speed, u = ±1 [see (4.1.10.1)]. Hence theequations in (4.1.10.2) are satisfied, i.e., α is a geodesic.

(ii) Let β(t) = σ (u0, v(t)) be a parallel. Since β is unit-speed, v = ± 1f (u0) [see (4.1.10.1)].

Clearly the second equation in (4.1.10.2) holds and the first equation holds iff0 = d

dt (u) = f (u0) dfdu (u0), i.e., iff dfdu (u0) = 0 as f (u0) > 0. Hence β is a geodesic iff

dfdu (u0) = 0.

4.2. Codazzi - Mainardi and Gauss equationsProposition 4.2.1 (Codazzi - Mainardi Equations). Let σ be a surface patch of anoriented surface S. Then

Lv −Mu = LΓ112 +M(Γ2

12 − Γ111)−NΓ2

11

Mv −Nu = LΓ122 +M(Γ2

22 − Γ212)−NΓ2

12.

PROOF. Since σuuv = σuvu, we have

(Γ111σu + Γ2

11σv + LN)v = (Γ112σu + Γ2

12σv +MN)u.We first simplify both the sides of the above equation. Now

(σuu)v = (Γ111σu + Γ2

11σv + LN)v= (Γ1

11)vσu + Γ111σuv + (Γ2

11)vσv + Γ211σvv + LvN + LNv

= (Γ111)vσu + Γ1

11(Γ1

11σu + Γ211σv +MN

)+ (Γ2

11)vσv+Γ2

11(Γ1

22σu + Γ222σv +NN

)+ LvN + LNv

= ((Γ111)v + (Γ1

11)2 + Γ211Γ1

22)σu + (Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222)σv

+(MΓ111 +NΓ2

11 + Lv)N + LNv. (4.2.1.1)

Also

(σuv)u = (Γ112σu + Γ2

12σv +MN)u= (Γ1

12)uσu + Γ112σuu + (Γ2

12)uσv + Γ212σuv +MuN +MNu

= (Γ112)uσu + Γ1

12(Γ1

11σu + Γ211σv + LN

)+ (Γ2

12)uσv


+Γ212(Γ1

12σu + Γ212σv +MN

)+MuN +MNu

= ((Γ112)u + Γ1

12Γ111 + Γ1

12Γ212)σu + (Γ1

12Γ211 + (Γ2

12)u + (Γ212)2)σv

+(LΓ112 +MΓ2

12 +Mu)N +MNu. (4.2.1.2)

Taking dot product of equations (4.2.1.1) and (4.2.1.2) with N and comparing we getMΓ1

11+NΓ211+Lv = LΓ1

12+MΓ212+Mu. Therefore Lv−Mu = LΓ1

12+M(Γ212−Γ1

11)−NΓ211.

Since σuvv = σvvu, we have

(Γ112σu + Γ2

12σv +MN)v = (Γ122σu + Γ2

22σv +NN)u.Now

(σuv)v = (Γ112σu + Γ2

12σv +MN)v= (Γ1

12)vσu + Γ212σuv + (Γ2

12)uσv + Γ212σvv +MvN +MNv

= (Γ112)vσu + Γ2

12(Γ112σu + Γ2

12σv +MN) + (Γ212)uσv

+Γ212(Γ1

22σu + Γ222σv +NN) +MvN +MNv

= ((Γ112)v + Γ2

12Γ112 + Γ2

12Γ122)σu + ((Γ2

12)2 + (Γ212)u + Γ2

12Γ222)σv

+(MΓ212 +NΓ2

12 +Mv)N +MNv. (4.2.1.3)

Also

(σvv)u = (Γ122σu + Γ2

22σv +NN)u= ((Γ1

22)u + Γ122Γ1

11 + Γ222Γ1

12)σu + (Γ122Γ2

11 + (Γ222)u + Γ2

22Γ212)σv

+(LΓ122 +MΓ2

22 +Nu)N +NNu. (4.2.1.4)

Taking dot product of equations (4.2.1.3) and (4.2.1.4) with N and comparing we getMΓ2

12 + +NΓ212 +Mv = LΓ1

22 +MΓ222 +Nu. Therefore Mv −Nu = LΓ1

22 +M(Γ222−Γ2

12)−NΓ2

12.

Proposition 4.2.2. Let σ be a surface patch of an oriented surface S. Then

Nu = 1EG − F2 ((MF − LG)σu + (LF −ME)σv)

Nv = 1EG − F2 ((NF −MG)σu + (MF −NE)σv).

PROOF. We know that the matrix ofWp with respect to the basis σu, σv is(E FF G

)−1( L MM N

)=

1EG−F2

(LG −MF MG −NF−LF +ME −MF +NE

). Therefore Nu = −Wp(σu) = −

(LG−MFEG−F2 σu + −LF+ME

EG−F2 σv),

i.e., Nu = 1EG−F2 ((MF−LG)σu+(LF−ME)σv). Similarly one gets the desired expression

for Nv .

2. Codazzi - Mainardi and Gauss equations 55

Proposition 4.2.3 (Gauss Equations). Let σ be a surface patch of an oriented surfaceS, and let K be the Gaussian curvature of the surface. Then

EK = Γ111Γ2

11 + Γ211Γ2

22 − Γ211Γ1

12 − (Γ112)2 − (Γ2

12)u + (Γ211)v

FK = Γ111Γ1

12 + Γ112Γ2

12 − (Γ111)2 − Γ2

11Γ122 + (Γ2

12)u − (Γ111)v

FK = Γ112Γ2

12 + Γ212Γ2

22 − Γ211Γ1

22 − Γ212Γ2

22 − (Γ222)u + (Γ2

12)vGK = Γ1

11Γ122 + Γ1

12Γ222 − (Γ1

12)2 − Γ212Γ1

22 + (Γ122)u − (Γ1

12)v.

PROOF. We have

(σuu)v = ((Γ111)v + (Γ1

11)2 + Γ211Γ1

22)σu + (Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222)σv

+(MΓ111 +NΓ2

11 + Lv)N + LNv

= ((Γ111)v + (Γ1

11)2 + Γ211Γ1

22)σu + (Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222)σv

+(MΓ111 +NΓ2

11 + Lv)N + LEG − F2 ((NF −MG)σu + (MF −NE)σv)

=(

(Γ111)v + (Γ1

11)2 + Γ211Γ1

22 + L(NF −MG)EG − F2

)σu

+(

Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222 + L(MF −NE)

EG − F2

)σv

+(MΓ111 +NΓ2

11 + Lv)N. (4.2.3.1)

Also

(σuv)u = ((Γ112)u + Γ1

12Γ111 + Γ1

12Γ212)σu + (Γ1

12Γ211 + (Γ2

12)u + (Γ212)2)σv

+(LΓ112 +MΓ2

12 +Mu)N +MNu

= ((Γ112)u + Γ1

12Γ111 + Γ1

12Γ212)σu + (Γ1

12Γ211 + (Γ2

12)u + (Γ212)2)σv

+(LΓ112 +MΓ2

12 +Mu)N + MEG − F2 ((MF − LG)σu + (LF −ME)σv)

=(

(Γ112)u + Γ1

12Γ111 + Γ1

12Γ212 + M(MF − LG)

EG − F2

)σu

+(

Γ112Γ2

11 + (Γ212)u + (Γ2

12)2 + M(LF −ME)EG − F2

)σv

+(LΓ112 +MΓ2

12 +Mu)N. (4.2.3.2)

Since σuuv = σuvu and the vectors σu, σv and N are linearly independent, comparingthe coefficient of σu we get

(Γ111)v + (Γ1

11)2 + Γ211Γ1

22 + L(NF −MG)EG − F2 = (Γ1

12)u + Γ112Γ1

11 + Γ112Γ2

12 + M(MF − LG)EG − F2 .


This gives

FK = F LN −M2

EG − F2 = (Γ112)u + Γ1

12Γ111 + Γ1

12Γ212 − (Γ1

11)v − (Γ111)2 − Γ2

11Γ122.

Comparing the coefficients of σv , we have

Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222 + L(MF −NE)

EG − F2 = Γ112Γ2

11 + (Γ212)u + (Γ2

12)2 + M(LF −ME)EG − F2 .

This gives

EK = ELN −M2

EG − F2 = Γ111Γ2

11 + (Γ211)v + Γ2

11Γ222 − Γ1

12Γ211 − (Γ2

12)u − (Γ212)2.

The remaining equalities may be obtained by using σvvu = σuvv .

If σ is a regular surface patch, then EG − F2 > 0, i.e., at least one of E, F and Gis non-zero at any point. We also know that Γk

ij can be expressed by using the firstorder magnitudes E, F and G and their first order partial derivatives. Hence it followsfrom the Gauss equations that the Gaussian curvature K can be expressed by the firstorder magnitudes E, F and G and their partial derivatives.

Theorem 4.2.4 (Gauss’ Theorema Egregium). The Gaussian curvature of a surfaceis preserved under local isometries.

PROOF. Let f : S1 → S2 be a local isometry. Then the first fundamental forms (or thefirst order magnitudes) of S1 and S2 are same. It follows from the Gauss equationsthat the Gaussian curvature K1 of S1 can be expressed by the first order magnitudesE1, F1 and G1 and their partial derivatives. Since the first order magnitudes E2, F2 andG2 are same as E1, F1 and G1 respectively, the Gaussian curvature K2 of S2 is sameas the Gaussian curvature K1 of S1. Hence the Gaussian curvature of a surface ispreserved under local isometries.

The following result is analogous to the Fundamental Theorem of Space Curves.

Theorem 4.2.5 (Bonnet’s Theorem). Let σ : U → R3 and σ : U → R3 be surfacepatches with the same first and second fundamental forms. Then there is a directisometry M of R3 such that σ = M(σ ).

Moreover, let V be an open subset of R2 and let E, F , G, L, M and N be smoothfunctions on V . Assume that E > 0, G > 0, EG − F2 > 0 and that the Mainardi -Codazzi equations and Gauss equations hold, with K = LN−M2

EG−F2 . If (u0, v0) ∈ V , thenthere is an open set U contained in V and containing (u0, v0), and a surface patchσ : U → R3, such that Edu2 + 2Fdudv + Gdv2 and Ldu2 + 2Mdudv + Ndv2 are thefirst and second fundamental forms of σ , respectively.

3. The Gauss-Bonnet Theorem 57

4.3. The Gauss-Bonnet TheoremTheorem 4.3.1 (Gauss-Bonnet Theorem). Let γ be a positively oriented piecewisesmooth curve, consisting on n smooth arcs with exterior angles α1, α2, . . . , αn, ona smooth surface S with Gaussian curvature K . Let int (γ) be simply connected.Then

∫

γ

κg (s)ds +∫∫

int(γ)

K dudv +n∑

i=1

αi = 2π, (4.3.1.1)

where κg is the geodesic curvature of γ.

Example 4.3.2. The sum of interior angles of a regular n-gon on a plane is (n − 2)π.Let γ be the regular n-gon and that it is positively oriented. Since γ is made up

of lines of the plane (which are geodesics), κg = 0. We also note that the Gaussiancurvature K of the plane is 0. Let α1, α2, . . . , αn be the exterior angles of γ, and letβ1, β2, . . . , βn be the interior angles of γ. Then αi+βi = π (i = 1, 2, . . . , n). Clearly int(γ)is simply connected. Hence by the Gauss-Bonnet Theorem we have

∑ni=1 αi = 2π , i,e,∑n

i=1(π − βi) = 2π. This implies that∑n

i=1 βi = (n − 2)π.

Example 4.3.3. The sum of interior angles of a triangle on a sphere is > π.Let γ be the triangle and that it is positively oriented. Since γ is made up of

geodesics of the sphere (which are geodesics), κg = 0. We also note that the Gaussiancurvature K of the sphere of radius a is 1

a2 . Let α1, α2, α3 be the exterior anglesof γ, and let β1, β2, β3 be the interior angles of γ. Then αi + βi = π (i = 1, 2, 3).Clearly int(γ) is simply connected. Hence by the Gauss-Bonnet Theorem we have∫∫

int(γ)1a2dudv +

∑3i=1 αi = 2π. Therefore

∑3i=1 βi = π +

∫∫int(γ)

1a2dudv > π.


Some Applications of Differential GeometryBelow are some examples of how differential geometry is applied to other fields ofscience and mathematics.

(i) In physics, the following are some of applications.(a) Differential geometry is the language in which Einstein’s general theory of

relativity is expressed. According to the theory, the universe is a smoothmanifold equipped with a pseudo-Riemannian metric, which describes thecurvature of space-time. Understanding this curvature is essential for thepositioning of satellites into orbit around the earth. Differential geometry isalso indispensable in the study of gravitational lensing and black holes.

(b) Differential forms are used in the study of electromagnetism.(c) Differential geometry has applications to both Lagrangian mechanics and

Hamiltonian mechanics. Symplectic manifolds in particular can be used tostudy Hamiltonian systems.

(d) Riemannian geometry and contact geometry have been used to constructthe formalism of geometrothermodynamics which has found applicationsin classical equilibrium thermodynamics.

(ii) In economics, differential geometry has applications to the field of econometrics.(iii) Geometric modeling (including computer graphics) and computer-aided geomet-

ric design draw on ideas from differential geometry.(iv) In engineering, differential geometry can be applied to solve problems in digital

signal processing.(v) In probability, statistics, and information theory, one can interpret various struc-

tures as Riemannian manifolds, which yields the field of information geometry,particularly via the Fisher information metric.

(vi) In structural geology, differential geometry is used to analyze and describe geo-logic structures.

(vii) In computer vision, differential geometry is used to analyze shapes.(viii) In image processing, differential geometry is used to process and analyse data

on non-flat surfaces.(ix) Grigori Perelman’s proof of the Poincaré conjecture using the techniques of

Ricci flows demonstrated the power of the differential-geometric approach toquestions in topology and it highlighted the important role played by its analyticmethods.

(x) In wireless communications, Grassmanian manifold is used for beamformingtechniques in multiple antenna systems.

differential geometry notes

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