2 Regular Surfaces

In this half of the course we approach surfaces in E3 in a similar way to which we considered curves.A parameterized surface will be a function1 x : U → E3 where U is some open subset of the planeR2. Our purpose is twofold:

1. To be able to measure quantities such as length (of curves), angle (between curves on a surface),area using the parameterization space U. This requires us to create some method of taking tangentvectors to the surface and ‘pulling-back’ to U where we will perform our calculations.2

2. We want to find ways of defining and measuring the curvature of a surface.

Before starting, we recall some of the important background terms and concepts from other classes.

Notation

Surfaces being functions x : U ⊆ R2 → E3, we will preserve some of the notational differencesbetween R2 and E3. Thus:

• Co-ordinate points in the parameterization space U ⊂ R2 will be written as lower case letters or,more commonly, row vectors: for example p = (u, v) ∈ R2.

• Points in E3 will be written using capital letters and row vectors: for example P = (3, 4, 8) ∈ E3.

• Vectors in E3 will be written bold-face or as column-vectors: for example v =

(2−1π2

).

Sometimes it will be convenient to abuse notation and add a vector v to a point P, the result will bea new point P + v.

Open Sets in Rn

The domains of our parameterized functions will always be opensets in R2. These are a little harder to describe than open intervalsin R: the definitions are here for reference.

Definition 2.1. Let p ∈ Rn. The open ball of radius r around p is theset of all points whose distance to p is less than r: i.e.

B(p, r) = {x ∈ Rn : |x− p| < r}

A set U ⊆ Rn is open if for all points p ∈ U there is some radiusrp > 0 such that B(p, rp) ⊆ U.

Otherwise said, around any point of an open set, there exists anopen ball which is contained entirely in that set.

prp

U

B(p, rp)

qB(q, rq)

1For example x(u, v) =( u

vu2+v2

)describes a paraboloid.

2One of the great difficulties of differential geometry is that it is hard to see at this stage why this process is beneficial:why not measure all these things directly within E3? Part of the reason is that we want to be able to generalize: we want tobe able to deal with surfaces (and eventually, though not in this class) with higher-dimensional objects (manifolds) withoutreference to some larger space in which they lie. As an example, suppose that a physicist says, ‘space-time is curved.’ Theyare not imagining an observer measuring its bendiness from the outside. On the contrary, the curvature is detected withinspacetime: in general relativity gravity is precisely the curvature of space-time.

1

Examples

1. In R, B(p, r) is just the open interval (p− r, p + r).

2. (0, 1) ⊆ R is open: Let p ∈ (0, 1) and choose any rp with 0 < rp ≤ min(p, 1 − p), thenB(p, rp) ⊆ (0, 1).

3. [0, 1) ⊆ R is not open: for p = 0 it is impossible to find a radius r > 0 such that B(p, r) = (−r, r)is contained entirely in [0, 1).

4. In R2, B(p, r) is the interior (without edge) of the circle of radius r centered at p. In R3 it is theinterior of the sphere of radius r centered at p.

We use the concept of an open set in Rn similarly to how we used open intervals when describingcurves: a curve was usually defined as a function x : I → En where I was some open interval;surfaces will be defined on open subsets U ⊂ R2 similarly, e.g. x : U → E3. Openness is importantbecause edges of sets cause difficulties for differentiation. Unless otherwise said, you should assumethat the domain of any parameterized surface is open.

Partial Differentiation and the Gradient Vector

A multivariable scalar function is an assignment f : U ⊆ Rn → R, i.e. f (x1, x2, x3) = x21 − x3 cos x2 is

a function f : R3 → R. Note that we could write f (x) = x21 − x3 cos x2, where x = (x1, x2, x3).

Supposing it exists, the partial derivative of f with respect to xi is the function

∂ f∂xi

= limh→0

f (x1, . . . , xi−1, xi + h, xi+1, . . . , xn)− f (x1, . . . , xn)

h

The function f is differentiable if ∂ f∂xi

exists for all i = 1, . . . , n. It is smooth if partial derivatives of all

orders and all combinations exist: for example, if f : R3 → R is smooth then ∂4 f∂x1∂x2

2∂x3is defined and

continuous. Essentially all our surfaces will be smooth. Some notations for functions:

• C(U, V) is the set of continuous functions f : U → V.

• C1(U, V) is the set of continuously differentiable functions f : U → V with (differentiable func-tions with continuous derivatives).

• C∞(U, V) is the set of smooth functions.

It is common to say, for instance, that “ f is C1” rather than writing f ∈ C1(U, V). Sets C2, C3, etc., aredefined similarly.

Example If f (x) = x21 − x3 cos x2, then it is a smooth function f ∈ C∞(R3, R). Moreover,

∂ f∂x1

= 2x1,∂ f∂x2

= x3 sin x2,∂ f∂x3

= − cos x2

Definition 2.2. The gradient of a differentiable function f : U ⊆ Rn → R is the function∇ f : U → Rn

defined by

∇ f (x1, . . . , xn) =

(∂ f∂x1

, . . . ,∂ f∂xn

)If f is as above, we have ∇ f = (2x1, x3 sin x2,− cos x3).

2

2.1 Vector Fields and 1-forms

We are almost ready to make the critical definition where we turn differential operators into vectors.The first step involves thinking about the gradient of a scalar function f : En → R: in such a situationwe would write ∇ f as a column vector. Moreover, as the next definition facilitates, the gradientencodes both the direction in which the value of f changes maximally, and the magnitude of that rateof change.

Definition 2.3. If v = (v1, . . . , vn)T is a vector based at a point P ∈ U ⊆ En and f : U → R is afunction, then the directional derivative of f at P in the direction v is the scalar

Dv f (P) := v · ∇ f (P) =n

∑k=1

vk∂ f∂xk

∣∣∣∣P

The directional derivative allows one to compute the rate of change of f as one travels in any direc-tion. Indeed:

Theorem 2.4. 1. If h is a small number, then f (P + hv) ≈ f (P) + Dv f (P)h.

2. If x(t) is a curve passing through P at t = 0 and with tangent vector x′(0) = v, then

Dv f (x(0)) =ddt

∣∣∣∣t=0

f (x(t))

is the rate of change of f at P as you travel along the curve.

3. If v is a unit vector, then Dv f (P) is maximal when v points in the same direction as the gradient vector.

4. ∇ f points in the direction of greatest increase of f at P. If v is the unit vector in that direction, then|∇ f (P)| = Dv f (P).

Example In E3, let P = (x1, x2, x3) and suppose that f (P) = x21 + x2x3 + x3 − x1. Then

∇ f =

2x1 − 1x3

x2 + 1

Now let v =

( 12−1

). Then the directional derivative in the direction of v is

Dv f = v · ∇ f =3

∑k=1

vk∂ f∂xk

= 2x1 − 1 + 2x3 − (x2 + 1)

Be careful with notation! Dv f is a continuous function E3 → R. If we evaluate at a specific pointP = (0, 1, 2), say, then Dv f (P) = −1 + 4− 2 = 1 is a number.

3

The Directional Derivative Operator

One of the reasons for putting the function f at the end of the directional derivative expression is thatit tempts us to write the directional derivative at a point P as an operator:

Dv|P =n

∑k=1

vk∂

∂xk

∣∣∣∣P

This is a map (function) from the space of continuously differentiable functions U ⊆ En → R to thereal numbers. It is similarly tempting to drop any reference to the point P and define a more generaloperator

Dv =n

∑k=1

vk∂

∂xk

This is a map from the space of differentiable functions U ⊆ En → R to the space of continuousfunctions U → R: feed a C1 function f to Dv and the result is a continuous function Dv f .It is worth noting that v determines the operator Dv and vice versa.3

One of the principal ideas of vector fields in differential geometry is to view them as directionalderivative operators or, as we’ve seen above, linear first-order partial differential operators. This soundsartificially complicated but the rational is simple: we really only care about vectors in terms of howfunctions/surfaces change in said direction.

We now come to one of the fundamental definitions of differential geometry.

Definition 2.5. Let U ⊆ Rn be open and let p ∈ U. The tangent space to at p, denoted TpU is the set ofall directional derivative operators Dv|p at p. A vector field v on U is a (smooth) choice for each p ∈ Uof an element of TpU.

By smooth, we mean that the entries of v are infinitely differentiable functions. Note that we currentlyuse a lower-case v ∈ Rn. We will often write v|p for the particular tangent vector4 which is the valueof the vector field at p.

Example v = 3x1∂

∂x1+ 2x1x3

∂∂x2− x1

∂∂x3

is a vector field on R3. If R3 is Euclidean space E3, then thiscorresponds to the vector-valued function

v

x1x2x3

=

3x12x1x3−x1

v is an infinitely differentiable function E3 → E3. It should be clear that the operators ∂

∂x1, ∂

∂x2, ∂

∂x3form a basis of the tangent space at each point.

3Given a linear partial differential operator Dv, observe that the co-ordinate xk : En → R is a C1 function. Thusvk = Dvxk, which recovers v, entry by entry.

4The prefix tangent helps us discriminate between these and position vectors. To reiterate: in differential geometry almostevery vector is a tangent vector: a directional derivative operator.

4

Definition 2.6. If v is a vector field on U ⊆ Rn, and f : U → R is a function, then we write v[ f ] forthe result of applying the vector field v to the function f . I.e.,

v =n

∑k=1

vk∂

∂xk=⇒ v[ f ] =

n

∑k=1

vk∂ f∂xk

If f is also smooth, then v[ f ] is itself a smooth function U → R.

Example Let f (x) = x21x2 be a function on R3, and v be the vector field in the previous example.

Then

v[ f ] = 3x1∂ f∂x1

+ 2x1x3∂ f∂x2− x1

∂ f∂x3

= 6x21x2 + 2x3

1x3

Theorem 2.7. Let v, w be vector fields on U ⊆ Rn and let f , g : U → R be smooth. Then

1. Each tangent space is a vector space: indeed f v and v + w are vector fields, where

( f v)[g] = f v[g], (v + w)[ f ] = v[ f ] + w[ f ]

2. Vector fields act linearly on smooth functions: if a, b ∈ R are constant, then

v[a f + bg] = av[ f ] + bv[g]

3. Vector fields satisfy a product rule known as the Leibniz rule:

v[ f g] = f v[g] + gv[ f ]

All these results are straightforward from the definition of a vector field as a linear first order partialdifferential operator.

Notation warning! Much of the difficulty of working with vector fields (and the associated 1-formswhich are to come) is of becoming confused with notation. Remember that v[ f ] is a function Rn → R,whose value at p ∈ Rn is v[ f ](p) = v|p [ f ]. The previous Theorem written at the point p, produces atotal mess! In practice, extra brackets are very helpful. . .

( f v)[g](p) = f (p)v[g](p), (v + w)[ f ](p) = v[ f ](p) + w[ f ](p),v[a f + bg](p) = av[ f ](p) + bv[g](p),v[ f g](p) = f (p)v[g](p) + g(p)v[ f ](p)

Note in particular that f v denotes the vector field obtained by multiplying each vector in the vectorfield by the value of the function f at the corresponding point. It does not mean apply the function f tothe vector v, which makes no sense.For example, if v = x2

2∂

∂x1− x1

∂∂x2

is a vector field on R2, f (x1, x2) = x21x2 and p = (2,−1), then

f v = x21x3

2∂

∂x1− x3

1x2∂

∂x2is a vector field.

v[ f ] = x22

∂∂x1

(x21x2)− x1

∂∂x2

(x21x2) = 2x1x3

2 − x31 is a function.

( f v)(p) = f (p) v|p = −4 ∂∂x1

+ 8 ∂∂x2

is a tangent vector.

(v[ f ])(p) = −4− 8 = −12 is a number.

5

1-forms

Make sure you understand what a tangent vector is before moving on to this section, as 1-forms aredefined in terms of tangent vectors.

Definition 2.8. A 1-form at p ∈ U ⊆ Rn (sometimes called a covector), is a linear map

α|p : TpU → R

The set of 1-forms at p is the cotangent space T∗p U.If v|p is a tangent vector at p, then we write α|p (v|p) for the value of this function.

Example Let p = (1, 0,−2) ∈ R3. Then α|p defined by

α|p

(a

∂

∂x1

∣∣∣∣p+ b

∂

∂x2

∣∣∣∣p+ c

∂

∂x3

∣∣∣∣p

)= 7a + 3b + c

is a 1-form. Indeed

α|p

(−2

∂

∂x1

∣∣∣∣p+ 3

∂

∂x2

∣∣∣∣p− 4

∂

∂x3

∣∣∣∣p

)= −2 · 7 + 3 · 3− 4 · 1 = −9

Definition 2.9. Write dxk for the 1-form at p defined by

dxk

(∂

∂xj

∣∣∣∣p

)= δjk =

{1 j = k0 j 6= k

(†)

The 1-forms dxk (where k = 1, . . . , n) form a basis5 of the cotangent space T∗p U, hence the general1-form at p has the form

α|p =n

∑k=1

ak dxk

for unique constants ak.

Definition 2.10. A 1-form on U ⊆ Rn is an assignment

α =n

∑k=1

ak dxk

where each ak : U → R is a smooth function. We also write dxk for the 1-form on U whose restrictionto a point p is dxk.If v is a vector field on U, we write α(v) for the function U → R obtained by mapping p 7→ α|p (v|p).

5The tangent space and the cotangent space at p both have the same dimension n. The ∗ denotes ‘dual space’ in linearalgebra: if V is a vector space over a field F, then V∗ is the vector space of linear maps V → F. Equation (†) says that the

vectors ∂∂xk

∣∣∣p

and the covectors dxk are dual bases to each other.

6

Example In R2, let α = 2x2 dx1 − 3 dx2 and v = 3x21

∂∂x1

+ ∂∂x2

. Then α(v) = 6x21x2 − 3 is a function

R2 → R.

Theorem 2.11. Let f : U → R be smooth. Then there exists a unique 1-form denoted d f with the propertythat d f (v) = v[ f ] for all vector fields v on U.

Definition 2.12. d f is the exterior derivative of f .If a 1-form α = dg is the exterior derivative of a function g, we say that it is exact.

Proof. A linear map is defined uniquely by what it does to a basis, thus, at each p ∈ U, it is enoughto define d f |p by

d f |p

(∂

∂xi

∣∣∣∣p

)=

∂ f∂xi

∣∣∣∣p

d f is then the assignment p 7→ d f |p. Otherwise said,

d f =∂ f∂x1

dx1 + · · ·+∂ f∂xn

dxn (∗)

This is smooth since all derivatives of f are smooth.

Example If f (x1, x2) = 2x21x2 − 1, then d f = 4x1x2 dx1 + 2x2

1 dx2.

Proposition 2.13. If f , g are smooth functions, then

1. d( f + g) = d f + dg

2. d( f g) = f dg + g d f

3. d f = 0 ⇐⇒ f is a constant function

Proof. 1. and 2. follow immediately from the definition of d f and the relevant parts of Theorem 2.7.For 3.,

d f = 0 ⇐⇒ d f [v] = 0 for all vector fields v

⇐⇒ d f(

∂

∂xi

)= 0 for all i = 1, . . . , n

⇐⇒ ∂ f∂xi

= 0 for all i

⇐⇒ f is constant

Example: Polar Co-ordinates

Co-ordinate functions xi : U → R have played a large role in our discussion of vector fields and1-forms: at each point p ∈ U the tangent and contangent spaces have bases generated by thesefunctions:

TpU = Span

{∂

∂x1

∣∣∣∣p

, . . . ,∂

∂xn

∣∣∣∣p

}T∗p U = Span {dx1, . . . , dxn}

7

As an example of this, we consider the change of co-ordinates in U = R2 \ {(0, 0)} from rectangularto polar. Recall:{

x = r cos θ

y = r sin θ!

{r =

√x2 + y2

tan θ = yx

The chain rule tells us:

∂

∂x=

∂r∂x

∂

∂r+

∂θ

∂x∂

∂θ=

x√x2 + y2

∂

∂r− y

x2 + y2∂

∂θ= cos θ

∂

∂r− sin θ

r∂

∂θ

Similarly

∂

∂y=

∂r∂y

∂

∂r+

∂θ

∂y∂

∂θ=

y√x2 + y2

∂

∂r+

xx2 + y2

∂

∂θ= sin θ

∂

∂r+

cos θ

r∂

∂θ

We can also compute the 1-forms: by (∗),

dx = d(r cos θ) = cos θ dr− r sin θ dθ, dy = sin θ dr + r cos θ dθ

We can check that the dual basis relations hold:

dx(

∂

∂x

)= (cos θ dr− r sin θ dθ)

(cos θ

∂

∂r− sin θ

r∂

∂θ

)= cos2 θ dr

(∂

∂r

)− cos θ sin θ

rdr(

∂

∂θ

)− r sin θ cos θ dθ

(∂

∂r

)+ sin2 θ dθ

(∂

∂θ

)= 1

Similarly

dy(

∂

∂y

)= 1 dx

(∂

∂y

)= dy

(∂

∂x

)= 0

One can also compute ∂∂r , ∂

∂θ , dr, dθ in terms of ∂∂x , ∂

∂y , dx, dy, either by pure linear algebra from theabove, or directly using the chain rule.

Relation to elementary calculus

When n = 1 we are used to writing d fdx for the derivative of a function f : R → R. Contrary to what

you might expect, d fdx is not the quotient of two 1-forms, rather it is the result of the application of the

vector field ddx to the function f , or equivalently the application of the 1-form d f to the vector field

ddx :

d fdx

=d

dx[ f ] = d f

(d

dx

)Vector fields in R are written with a straight d rather than partial ∂ because there is only one directionin which to differentiate. The notion of tangent vector is at best useless and at worst misleading in R!

8

Line integrals

Where you have seen 1-forms before is in integration.∫ 1

0 g(x)dx is the integral of the 1-form g dxover the interval [0, 1]. Indeed we integrate 1-forms over oriented curves.

Definition 2.14. If x : [a, b]→ U ⊆ Rn is a curve, and α is a 1-form on U, we define∫x

α :=∫ b

aα(

x′(t))

dt

In order to feed the 1-form a tangent vector, we are obliged think of the derivative of the curve x(t)as a family of tangent vectors along the curve,6 i.e.

x(t) =(

f1(t)f2(t)

)=⇒ x′(t) = f ′1(t)

∂

∂x1+ f ′2(t)

∂

∂x2

Examples

1. Integrate α = x1dx2 over the unit-circle counter-clockwise. Here x(t) =

(cos tsin t

), so that

x′(t) = − sin t ∂∂x1

+ cos t ∂∂x2

. Then α(x′(t)) = x1 cos t which, restricted to the curve x(t), iscos2 t. Around the unit circle is t = 0 to 2π, thus,∫

xα =

∫ 2π

0cos2 t dt =

12

∫ 2π

01 + cos 2t dt = π

2. Integrate α = x22dx1 − x2

1dx2 over the curve x(t) = (t, t2) between (0, 0) and (1, 1).∫x

α =∫ 1

0α(x′(t))dt =

∫ 1

0(t2)2 − 2t3dt =

15− 1

2= − 3

10

Theorem 2.15. The integral of a 1-form along a curve is independent of the choice of (orientation-preserving)parameterization.

Proof. Suppose we are integrating α over a curve parameterized by x(t) where t runs from a to b.Now suppose that t is a function of s (t(p) = a, t(q) = b) where s′(t) > 0 (orientation-preserving)and integrate using s as a parameter. The critical observation is that d

ds x(t(s)) = x′(t(s)) dtds by the

chain rule.∫x

α =∫ q

pα

(dds

x(t(s)))

ds =∫ q

pα

(dtds

x′(t(s)))

ds

=∫ q

pα(x′(t(s)))

dtds

ds =∫ b

aα(x′(t))dt

=∫

xα (integrating with respect to t)

6This should make sense: the velocity x′ indicates magnitude and direction or travel, not position, so it should bethought of as a tangent vector rather than a position vector.

9

If we change the orientation of the curve then the order of the limits is reversed and∫

α becomes−∫

α.

As the next theorem shows, integration of 1-forms that are the exterior derivative of a function isindependent of the path you choose. This is essentially the fundamental theorem of calculus forcurves.

Theorem 2.16 (Fundamental Theorem of Line Integrals). Let x(t) be a curve for a ≤ t ≤ b. If f is afunction on Rn, then,∫

xd f = f (x(b))− f (x(a))

The integral of d f over any curve between two points depends only on the values of f at those points.

Proof.∫x

d f =∫ b

ad f (x′)dt =

∫ b

ax′[ f ]dt

However,

x′[ f ] = x′1∂ f∂x1

+ · · ·+ x′n∂ f∂xn

=dx1

dt∂ f∂x1

+ · · ·+ dxn

dt∂ f∂xn

=ddt

( f (x(t)))

Hence∫x

d f =∫ b

a

ddt

( f (x(t)))dt = f (x(b))− f (x(a))

Example If α = cos(x1x2)(x2dx1 + x1dx2), find the integral of α over any curve between the points(π, 1

3

)and

( 12 , π

).

Observe that α = d sin(x1x2). Thus if γ is any curve between (1, 2) and (3, 4), we have

∫γ

α = sin(x1x2)∣∣∣( 1

2 ,π)

(π, 13 )

= sinπ

2− sin

π

3= 1−

√3

2

Summary

There should be two take-aways from this discussion:

• Vector fields and tangent vectors allow one to express the notion that vectors encode a directionin which something can change (i.e. a derivative).

• Vector fields and 1-forms essentially break standard derivatives into two pieces: the result is analternative, and ultimately more flexible, language for describing familiar results from multi-variable calculus.

The real pay-off is hard to see until we apply our new language to surfaces. Thus. . .

10

2.2 Surfaces

A surface can be thought of in two ways:

1. A two-dimensional subset of E3 with certain properties.

2. The image of a function x : U → E3 defined on an open set U ⊆ R2.

The second notion is easier to calculate with since we can describe points on the surface in termsof co-ordinates on U. It is, however, less general and arguably less geometric. In practice, a sur-face is typically defined as being the union of the images of several functions xi with suitable rulesgoverning what happens on overlaps.7

Definition 2.17. 1. A (local) parameterized surface S is the image of a map x : U → E3 where Uis an open subset of R2. I.e. S = x(U). In what follows, we will often abuse language and referto the parameterization x as the surface.

2. Let u, v be co-ordinates on U. The co-ordinate tangent vector fields to x are

xu =∂x∂u

xv =∂x∂v

3. The surface x (with co-ordinates u, v) is regular at a point p ∈ U if xu(p), xv(p) are linearlyindependent.

4. If x is regular at p ∈ U, then the tangent plane to the surface at x(p) is Span{xu(p), xv(p)}.

5. A surface is regular if it is regular at all points p ∈ U.

6. If x is regular, then the unit normal vector field to the surface is

U =xu × xv

|xu × xv|

A choice of one of the two unit normal directions is called choosing an orientation of the surface.

Throughout we will assume that a surface is smooth and regular: i.e. all derivatives of x exist at allpoints of U and the tangent plane is well defined at each point of x(U).

We now make use of some of the language of the previous section and obtain an important result.

Definition 2.18. Suppose that x : U → E3 is a smooth surface. The differential of x is the vector-valued1-form dx.

Suppose that u, v are co-ordinates on U ⊆ R2 and that x is a smooth surface. Then the differential ofx can be written

dx =∂x∂u

du +∂x∂v

dv = xudu + xvdv

where du, dv are the co-ordinate 1-forms on U.7See the aside below. . .

11

Theorem 2.19. If x : U → E3 is a surface and y ∈ TpU is a tangent vector to U at p ∈ U, then dx(y) istangent to the surface at x(p). Indeed at each point p ∈ U we have a linear map

dx|p : TpU → Tx(p)S

Proof. Write y = a ∂∂u

∣∣∣p+ b ∂

∂v

∣∣∣p. Then dx(y) = axu + bxv is clearly tangent to the surface at x(p).

We say that dx is a 1-form with values in E3: it maps tangent vectors to U to genuine tangent vectorsto the surface. This indeed is one reason for calling linear 1st-order differential operators at p ∈ Utangent vectors.

Examples

1. Sphere of radius a. Use spherical polar co-ordinates:

x(φ, θ) = a

sin φ cos θsin φ sin θ

cos φ

Here x : U → E3, where U = (0, π)× (0, 2π). Note that the image of x is the sphere minusthe (dashed) semicircle defined by θ = 0, so that our domain is open. We could argue forextending θ modulo 2π so that co-ordinates coontinue round the sphere. In contrast, however,the co-ordinates cannot be extended to the north or south poles without losing regularity: in thisparameterization the sphere is not regular at the poles. This explains the term local: we usuallyneed more than one parameterization to cover a complicated surface.

θ

φ

∂∂θ

∂∂φ

0 π2 π

0

π2

π

3π2

2π

x

Observe in the picture how the tangent vectors ∂∂φ , ∂

∂θ to U are mapped by the differential dx tothe tangent vectors

dxdφ

= dx(

∂

∂φ

),

dxdθ

= dx(

∂

∂θ

)12

2. Graph of a smooth function z = f (x, y). Parameterize by x(u, v) =

uv

f (u, v)

. This has

dx =

10fu

du +

01fv

dv U =1√

1 + f 2u + f 2

v

− fu− fv

1

where fu = ∂ f

∂u , etc. This is clearly regular at all points, regardless of the function f .

For example, the paraboloid z = x2 + y2 may be parameterized x(u, v) =

uv

u2 + v2

.

3. Surfaces of revolution. For instance, a smooth curve z = f (x) can be rotated around the x-axisto obtain

x(φ, u) =

uf (u) cos φf (u) sin φ

(φ, u) ∈ (0, 2π)× dom( f )

This has differential and unit normal field

dx =

0− f (u) sin φf (u) cos φ

dφ +

1f ′(u) cos φf ′(u) sin φ

du

U =1√

1 + f ′(u)2

− f ′(u)cos φsin φ

The surface is regular everywhere.

Just as with the sphere, we should ignore thecurve φ = 0, but it is not crucial. The pic-ture is the surface of revolution defined by thecurve z = f (x) = 2 + cos x for x ∈ (0, 2π).The co-ordinate φ essentially measures the an-gle around the axis of rotation.

4. Ruled surfaces. Given two curves y(u), z(u), define

x(u, v) = y(u) + vz(u)

In general, the surface is ruled by the lines parameterized by v: that is, through each point onthe surface, there exists a line parameterized which is contained entirely within the surface.The particular case where z(u) = y′(u) is called the tangent developable of y(u).

Ruled surfaces are very useful in engineering applications as they are very simple to construct:for example corrugated iron is a ruled surface built on a curve which is approximately a sinewave. Two famous examples of ruled surfaces are shown below:

13

• The helicoid is built from a helix by attaching to each point a horizontal line pointing to the

z-axis. A particular example is given by x(u, v) =

v cos uv sin u

u

for v > 0.

• The hyperboloid of one sheet is in fact a doubly ruled surface: through each point there are twolines lying on the surface. It may be parameterized as a ruled surfaces by

x(u, v) =

100

+ u

011

+ v

2uu2 − 1u2 + 1

though convincing yourself of the double ruling might take a little more work. . .

5. Tube of radius a > 0 about a curve y(u). Parameterize as

x(u, φ) = y(u) + a cos φN(u) + a sin φB(u)

where N, B are the normal and binormal vectors of the Frenetframe of y.

A tube around a helix is drawn.

14

Implicitly Defined Surfaces

You are probably more used to surfaces being defined implicitly—as the level sets of some function—rather than being parameterized. This method of defining surfaces is extremely useful. Not havingan explicit parameterization means that calculations are harder, however you gain by having a moreglobal view of a surface.

Definition 2.20. An implicitly defined surface is the zero set of a smooth function f : E3 → R. We saythat it is regular if ∇ f (equivalently d f ) does not vanish on the surface.

Examples

1. The sphere of radius a is implicitly defined by f (x, y, z) = x2 + y2 + z2 − a2. Here

d f = 2(xdx + ydy + zdz)

is never zero since at least one of x, y, z is non-zero at all points on the surface f = 0. The sphereis thus regular. Note the contrast with our earlier example of the parameterized sphere whichwe could not make regular at the north and south pole. The lack of regularity in this case is afunction of the parameterization, not the surface itself.

2. The function f (x, y, z) = x2 + y2 − z2 − a has

d f = 2(xdx + ydy− zdz),

which is non-zero away from (x, y, z) = (0, 0, 0). Depending on the sign of a, the set of pointswith f = 0 is a hyperboloid or a cone.

a > 0 Hyperboloid of 1-sheet: x2 + y2 = z2 + a > 0 for all z

a = 0 Cone: (0, 0, 0) is not a regular point of the cone

a > 0 Hyperboloid of 2-sheets: x2 + y2 = z2 − a ≥ 0 only when |z| ≥√

a

The next theorem ties together the currently distinct notions of regular. More importantly, it says thatwe can always assume the existence of local co-ordinates.

Theorem 2.21. A regular implicitly defined surface is (locally) the image of a regular local surface.

15

We omit the proof, as it is essentially the famous (and difficult!) Implicit Function Theorem8 frommulti-variable calculus in disguise.In order to calculate with implicitly defined surfaces, you need some of the ingredients of parame-terized surfaces. In particular: if f (x, y, z) = 0 describes a regular surface S and s ∈ S, then ∇ f (s) isorthogonal to the surface at s and so the tangent plane to S at s is

TsS = {x ∈ E3 : (x− s) · ∇ f (s) = 0}

Example The hyperboloid of one sheet defined implicitly by the equation x2 + y2 − z2 = 12 hasunit normal vector

U(x, y, z) =1√

x2 + y2 + z2

xy−z

=1√

12 + 2z2

xy−z

whenever (x, y, z) is a point on the hyperboloid. Thus at the point (3, 2, 1), the unit normal is

1√14(3, 2,−1), and the tangent plane has equation

3x + 2y− z = 12

Of course, the hyperboloid could have been parameterized: it is a surface of revolution (around thez-axis) so we could choose

x(u, θ) =

√

12 + u2 cos θ√12 + u2 sin θ

u

Then the differential is

dx =

u(12 + u2)−1/2 cos θ

u(12 + u2)−1/2 sin θ1

du +

−√

12 + u2 sin θ√12 + u2 cos θ

0

dv

with resulting normal field

U =xu × xθ

|xu × xθ |=

1√12 + 2u2

−√

12 + u2 cos θ

−√

12 + u2 sin θu

Notice that this is precisely the negative of what we obtained earlier, but no matter. Yet anotherexpression could be obtained using our parameterization as a ruled surface (page 14).

8If f (x, y, z) = 0 is the implicit surface and ∇ f 6= 0, then, at any given P = (x0, y0, z0), at least one of the partial

derivatives of f is non-zero: suppose WLOG that ∂ f∂z

∣∣∣P6= 0. The implicit function theorem says that there exists an open

set U ⊆ R2 and a unique function g : U → R such that g(x0, y0) = z0 and f(

x, y, g(x, y))= 0. The surface may then

(locally) be parameterized as a graph:

x : U → E3 : (u, v) 7→

uv

g(u, v)

16

Aside: Global surfaces We have only rigorously defined a local parameterized surface. In general asurface S is any subset of E3 which locally ‘looks like’ a parameterized surface. That is, for any points ∈ S, there exists a subset V ⊆ S containing s, an open subset U of R2 and a smooth local surfacex : U → V.The upshot is that we may require more than one co-ordinate chart (subsets x(U) ⊆ S) to coveran entire surface. For example, the sphere described using spherical polar co-ordinates covers thewhole sphere in a regular fashion, except for a semi-circle (or, pushing things with periodicity in θ,everywhere except two points). To cover the entire sphere with co-ordinate charts we need to chooseanother point to be the center of a new chart. Indeed a famous topological result (the Hairy DogTheorem) says that it is impossible to find global co-ordinates on the sphere: the best you can do is tohave the entire sphere except for one point on a single chart (stereographic projection does the job, ifyou know what this is. . . ).

When we start thinking about surfaces globally (such as the entire sphere at once), we see that it ispossible to construct non-orientable surfaces such as the Mobius strip. A Mobius strip is certainly asurface: near any point it looks exactly like a bit of R2, and so we can find local co-ordinates. How-ever if you try to look at the surface globally you find you cannot orient it. Consider walking aroundthe strip once. An orientation at a point is a choice of unit normal vector, a choice of which way is‘up’. On circuiting the strip you find that ‘up’ is now ‘down’. It is thus impossible to choose a globalregular parameterization.

Change of co-ordinates Suppose that y(s, t) = x(

F(s, t))

where F(s, t) = (u, v) is a change of co-ordinates on the parameterization space U. The differential takes care of everything. By the chainrule the differentials dx and dy may be thought of as related by a matrix multiplication:(

∂y∂s∂y∂t

)=

(∂u∂s

∂v∂s

∂u∂t

∂v∂t

)(∂x∂u∂x∂v

)(du dv) = (ds dt)

(∂u∂s

∂v∂s

∂u∂t

∂v∂t

)=⇒ (ds dt) = (du dv)

(∂u∂s

∂v∂s

∂u∂t

∂v∂t

)−1

from which

dy = (ds dt)

(∂y∂s∂y∂t

)= (du dv)

(∂u∂s

∂v∂s

∂u∂t

∂v∂t

)−1 ( ∂u∂s

∂v∂s

∂u∂t

∂v∂t

)(∂x∂u∂x∂v

)= (du dv)

(∂x∂u∂x∂v

)= dx

The matrix of partial derivatives is the Jacobian of the co-ordinate change, as you should have met inmulti-variable calculus. Strictly speaking, dx and dy are not identical as they feed on tangent vectorswith respect to different co-ordinates. We should properly write dy = dx ◦ dF as a composition oflinear functions, where dF maps tangent vectors in Span{ ∂

∂s , ∂∂t} to those in Span{ ∂

∂u , ∂∂v}.

17

2.3 The Fundamental Forms

Now that we have differentials, we can start using them to describe surfaces. The required objects abased on a little more algebra.

Definition 2.22. Given 1-forms α, β at p ∈ Rn, define a symmetric bilinear form αβ of two vectorsX, Y ∈ TpRn by,

αβ(X, Y) =12(α(X)β(Y) + α(Y)β(X))

Lemma 2.23. αβ is indeed a symmetric bilinear form on each tangent space TpRn.

Proof. αβ(X, Y) = αβ(Y, X) is clear from the formula, hence the product is symmetric. Moreover,fixing X, we see that αβ(X,−) : Y → αβ(X, Y) is linear since both α, β are linear on TpRn. Bysymmetry we have bilinearity.

Not only is αβ a symmetric bilinear form on each tangent space, it is symmetric itself: αβ = βα. Wewrite α2 = αα. Clearly

α2(X, Y) = α(X)α(Y)

Examples

1. In R2, let α = x1 dx1 − dx2, β = x1x2 dx2, X = ∂∂x1− x2

∂∂x2

, Y = ∂∂x2

. Then,

αβ(X, Y) =12((x1 + x2)x1x2 + x1x2

2)=

12(x2

1x2 + 2x1x22)

2. The dot product of tangent vectors in TpRn is given by9

(ds)2 = (dx1)2 + · · ·+ (dxn)

2

I.e. if X = 3 ∂∂x1− ∂

∂x2and Y = 2 ∂

∂x1+ 5 ∂

∂x2, then

(ds)2(X, Y) = dx21(X, Y) + dx2

2(X, Y) = dx1(X)dx1(Y) + dx2(X)dx2(Y) = 6− 5 = 1

That is,(

3−1

)·(

25

)= 1.

If 1-forms α, β are vector-valued (e.g. α(X) ∈ E3) then we can still define symmetric products, butwe must specify how the vectors are to be multiplied. For example

(α · β)(X, Y) =12(α(X) · β(Y) + α(Y) · β(X))

requires multiplication of vectors by the dot product, as in the following definitions.

9s =√

x21 + · · ·+ x2

n is the distance of a point from the origin.

18

Definition 2.24. Given a local parameterized surface x : U → E3, we define the first I and second I

fundamental forms

I = dx · dx, I = −dx · dU

Here dU is the differential of the unit normal field. The fundamental forms are symmetric bilinearforms on each tangent space TpU.

By the discussion in the previous aside, if y(s, t) = x((u(s, t), v(s, t))

), then dy · dy = dx · dx so that

the first fundamental form is independent of co-ordinates. Similarly the second fundamental form isindependent of co-ordinates up to a change of sign if the orientation is reversed.

Before worrying about the meanings of the fundamental forms, we compute a few.

Examples

1. The sphere of radius a.

dx = a

cos φ cos θcos φ sin θ− sin φ

dφ + a

− sin φ sin θsin φ cos θ

0

dθ

In this case I = a2 dφ2 + a2 sin2 φ dθ2. Computing I is very easy here, because the normal vectoris simply the scaled position vector:

U =1a

x =⇒ dU =1a

dx =⇒ I = −1a

I = −a dφ2 − a sin2 φ dθ2

2. If x is a surface of revolution around the z-axis, recall that

dx =

f ′(u) cos φf ′(u) sin φ

1

du + f (u)

− sin φcos φ

0

dφ

Thus I = (1 + f ′(u))du2 + f (u)2dφ2. The unit normal vector is

U =xu × xφ∣∣xu × xφ

∣∣ = 1√f ′(u)2 + 1

− cos φ− sin φ

f ′(u)

and so

I = −dx · dU

= −

f ′(u) cos φf ′(u) sin φ

1

du + f (u)

− sin φcos φ

0

dφ

· 1√

f ′(u)2 + 1

00

f ′′(u)

du +

sin φ− cos φ

0

dφ

=

1√f ′(u)2 + 1

(− f ′′(u)du2 + f (u)dφ2)

19

The first fundamental form I transfers the dot product from E3 to U. Indeed if dx(zi) = xi, i = 1, 2,then I(z1, z2) = x1 · x2. This is in the spirit of transferring calculations from E3 back to U. Thefirst fundamental form I encodes ‘infinitessimal distances’ on a surface. For example, the distance smeasured between two very close points on the surface of the sphere of radius a is given by,

s2 ≈ a2(∆φ)2 + a2 sin2 φ(∆θ)2

where ∆φ, ∆θ are the small changes in the co-ordinates. If all you want to do on a surface is tomeasure distances, then all you need to know is the first fundamental form—you don’t even need toknow the surface!The second fundamental form measures how the unit normal vector of a surface is changing in com-parison to the surface itself. Since U has unit length, dU|p (X) is orthogonal to U for any X ∈ TpUand is thus a vector in the tangent space to the surface at x(p). The second fundamental formI = −dx · dU thus measures how the unit normal changes relative to the surface.

Curves in surfaces

One of the best ways of understanding what the fundamental forms I, I are measuring is to considercurves lying in a surface.Let x : U → E3 be a local surface. If z(t) = (u(t), v(t)) is a curve in U then the compositiont 7→ x(z(t)) describes a curve in E3 lying in the surface. It is often useful to think of z(t) as beinga curve in the surface, even though it is only a curve in U. We will abuse notation by thinking ofx′ = d

dt x(z(t)) as the velocity vector of the curve in E3. In a similar way, define U′ = ddt U(z(t)).

Example Consider the curve z(t) = (φ(t), θ(t)) = (2t, 5t) on the surface of the unit sphere. In thiscase,

x(t) =

sin 2t cos 5tsin 2t sin 5t

cos 2t

Lemma 2.25. Considering the fact that z′(t) = u′(t) ∂

∂u + v′(t) ∂∂v , we have

x′ = dx(z′), U′ = dU(z′)

Proof.ddt

x(z(t)) = xuu′ + xvv′ = dx(

u′(t)∂

∂u+ v′(t)

∂

∂v

)= dx(z′)

If x(z(t)) is a curve in the surface x, then the speed of the curve is

v(t) =√

x′ · x′ =√

dx(z′) · dx(z′) =√

I(z′, z′)

We’ve therefore proved the following:

Theorem 2.26. Let x(z(t)) be a curve in a surface x : U → E3 with first fundamental form I. Then thearc-length of x(z(t)) between x(z(a)) and x(z(b)) is given by∫ b

a

√I(z′, z′)dt

20

Examples

1. Consider the unit sphere with the curve given above z(t) = (2t, 5t). Here z′(t) = 2 ∂∂φ + 5 ∂

∂θ .

Recalling that the first fundamental form of the unit sphere is I = dφ2 + sin2 φ dθ2, we have

I(z′, z′) = 4 + 25 sin2 5t

The arc-length of the curve over a ≤ t ≤ b is therefore given by10

∫ b

a

√4 + 25 sin2 5t dt

Generally for a curve z(t) = (φ(t), θ(t)) on a sphere, the arc-length between t = a, b is given by,∫ b

a

√(φ′)2 + sin2 φ(θ′)2 dt

2. Suppose that a surface satisfying y > 0 has first fundamental form I = dx2+dy2

y2 . Find the arc-length over the circular path z(t) = (cos t, sin t) between t0 and t1.

Here z′(t) = − sin t ∂∂x + cos t ∂

∂y so that I(z′, z′) = 1sin2 t

. Thus the arc-length is

∫ t1

t0

1sin t

dt =[− ln

1 + cos tsin t

]t1

t0

= lnsin t1(1 + cos t0)

sin t0(1 + cos t1)

Notice that as t0 → 0+ or as t1 → π− the arc-length becomes infinite.11

The examples show the advantages of our approach:

• If you know I, then you can change the curve lying in a given surface and still compute the arc-length with minimal effort. Before this, you would have had to go back to the start: computeddt x(z(t)) again, then it’s length. . . Since applications often involve studying all curves simulta-neously, this approach is essential for being able to quickly compare the outcomes.

• You don’t need to know the surface explicitly, all you need is I. Example 2 is a description ofthe Poincare half-plane model of hyperbolic space: it is simply a domain U (here the upperhalf-plane) with an abstract first fundamental form. There is no surface x : U → E3 and so noI to worry about.

Interpreting I Suppose you were skateboarding in a bowl: how much effort are your legs goingto have to make in order to keep you standing up? The second fundamental form along a curvemeasures exactly this.

Theorem 2.27. For a curve x(z(t)) lying in a surface x with unit normal U and second fundamental form I,

U · d2

dt2 x(z(t)) = I(z′, z′).

10Don’t try to compute this elliptic integral explicitly!11You may see this example again as hyperbolic space.

21

Proof. Write x′ = ddt x(z(t)) as before. Since the tangent vector to the curve is always perpendicular

to the unit normal, we have

0 =ddt

(x′ ·U) = x′′ ·U + x′ ·U′ = x′′ ·U + dx(z′) · dU(z′)

= x′′ ·U− I(z′, z′)

Theorem 2.27 says that the second fundamental form along a curve measures the acceleration in thenormal direction to the surface that is required in order to keep the curve lying in the surface.

I, I in co-ordinates

Suppose that u, v are co-ordinates on U ⊆ R2. We can write the fundamental forms of a surface x interms of easily calculated derivatives.

Proposition 2.28. Define functions E, F, G and l, m, n on x : U → E3 by

E = xu · xu F = xu · xv G = xv · xv

l = xuu ·U = −xu ·Uu m = xuv ·U = −xu ·Uv n = xvv ·U = −xv ·Uv

Then

I = Edu2 + 2Fdudv + Gdv2 and I = ldu2 + 2mdudv + ndv2

These should be clear: I is just the formula dx · dx, while the expressions for I come from differenti-ating xu ·U = 0 = xv ·U.

Examples

1. Consider the graph of a function z = f (x, y). Parameterizing in the usual way gives us,

xu =

10fu

, xv =

01fv

, U =1√

1 + f 2u + f 2

v

− fu− fv

1

,

E = 1 + f 2u , F = fu fv, G = 1 + f 2

v ,

l =fuu√

1 + f 2u + f 2

v, m =

fuv√1 + f 2

u + f 2v

, n =fvv√

1 + f 2u + f 2

v

Combining these gives us,

I = (1 + f 2u)du2 + 2 fu fv dudv + (1 + f 2

v )dv2,

I =1√

1 + f 2u + f 2

v

{fuu du2 + 2 fuv dudv + fvv dv2

}2. A skater follows a path parameterized by

z(t) = (r(t), θ(t)) = (1 − t, 4t2) in theparaboloidal bowl z = 1

2 r2. Compute:

(a) The fundamental forms of the surface:

(b) The speed of the skater at time t.

(c) The normal acceleration experienced bythe skater at time t.

22

The surface can be parameterized x(r, θ) =

r cos θr sin θ

12 r2

which yields

xr =

cos θsin θ

r

, xθ =

−r sin θr cos θ

0

, U =1√

1 + r2

−r cos θ−r sin θ

1

,

E = 1 + r2, F = 0, G = r2,

l =1√

1 + r2, m = 0, n =

r2√

1 + r2,

I = (1 + r2)dr2 + r2 dθ2, I =1√

1 + r2(dr2 + r2 dθ2)

For the skater’s path, we have z′(t) = − ∂∂r + 8t ∂

∂θ . Therefore

I(z′, z′) = (1 + (1− t)2) + 64t2(1− t)2

The skater’s speed is therefore

v(t) =√

I(z′, z′) =√

1 + (64t2 + 1)(1− t)2

The second fundamental form along the path is

I(z′, z′) =1√

1 + (1− t)2(1 + 64t2(1− t)2)

Theorem 2.27 says that this is the normal acceleration of the skater and thus proportional (viaNewton’s second law F = ma) to the force experienced by the skater.

2.4 Principal Curvatures

The idea of the next section is to think about finding directions on a surface with special properties:this amounts to finding co-ordinates with respect to which the fundamental forms look particularlysimple. The most obvious way to may the forms simple is to insist that there be no dudv term.

Definition 2.29. Co-ordinates u, v on U are orthogonal for a surface x : U → E3 if the first fundamentalform is diagonal: that is

I = E du2 + G dv2

where there is no dudv term. Otherwise said, xu · xv = 0 so the tangent vectors generated by theco-ordinate vector fields ∂

∂u , ∂∂v are always orthogonal.

In addition, u, v are curvature-line co-ordinates if the second fundamental form is diagonal:

I = l du2 + n dv2

While the meaning of orthogonal is clear, to understand why we use the term curvature-line will needa little work.

23

Examples

1. Recall the sphere using spherical polar co-ordinates: since both I, I are diagonal, these are cur-vature line co-ordinates.12

2. The graph of a function z = f (x, y) has its usual parameterization (u, v) = (x, y) in orthogonalco-ordinates if and only if f (x, y) is a function only of one of the variables ( fx = 0 or fy = 0).In this case the co-ordinates are also curvature-line. The surface is necessarily ruled, with astraight line in the surface parallel to either the x or the y axis.

3. A surface of revolution parameterized using (u, φ) where u is the distance along the axis ofrotation and φ the angle round the axis, has

I = I = (1 + f ′(u))du2 + f (u)2dφ2, I =1√

1 + f ′(u)2(− f ′′(u)du2 + f (u)dφ2)

These are curvature-line co-ordinates.

Finding Curvature Directions: A Little Linear Algebra

Finding curvature-line co-ordinates is difficult in general. It starts with finding curvature directionsat a point, which amounts to finding directions with respect to which both I and I are diagonal. Firstwe need some background linear algebra to extend the idea of eigenvalues and eigenvectors.

Definition 2.30. Let A, B be two n× n matrices. The eigenvalues of B with respect to A are the rootsof the polynomial

det(B− λA) = 0

A vector v is an eigenvector of B with respect to A and with eigenvalue λ if (B− λA)v = 0

Theorem 2.31. Suppose that A is an n× n symmetric, positive definite13 matrix and that B is n× n sym-metric. Then the eigenvalues of B with respect to A are real, and there exists a basis v1, . . . , vn of Rn such thateach vk is an eigenvector of B with respect to A.

You should skip the proof if you haven’t studied the spectral theorem in linear algebra.

Proof. The spectral theorem states that there exists an orthogonal basis x1, . . . , xn of Rn of eigenvectorsof A. In particular, since A is positive definite, all of the eigenvalues are positive and we may choosex1, . . . , xn to be orthonormal with respect to A: i.e.

xTi Axj =

{1 i = j0 i 6= j

With respect to the basis x1, . . . , xn the matrix of A is the identity matrix.Now let X = (x1, . . . , xn) be the matrix with column vectors x1, . . . , xn. Then XT AX = I, whence

A = (XT)−1X−1

12This is something of a cheat: every pair of orthogonal co-ordinates on the sphere is curvature-line (exercise).13vT Av ≥ 0 for all v, with equality if and only if v = 0. This says that (v, w) := vT Aw is an inner product.

24

It follows that

det(B− λA) = det((XT)−1(XTBX− λI)X−1) = 0 ⇐⇒ det(XTBX− λI) = 0

XTBX is a symmetric matrix and so has real eigenvalues λ1, . . . , λn, and n distinct orthogonal eigen-vectors w1, . . . , wn. Define vk = Xwk: these form a basis of Rn such that vk is an eigenvector of Bwith respect to A with eigenvalue λi.

Example Let A =

(2 33 5

)and B =

(0 11 3

). Then the eigenvalues of A are 7±

√48 so that A is

positive definite. Now

det(B− λA) =

∣∣∣∣ −2λ 1− 3λ1− 3λ 3− 5λ

∣∣∣∣ = λ2 − 1 = 0 ⇐⇒ λ = ±1

The eigenvalues of B with respect to A are thus ±1. When λ = 1 we have

(B− A)v1 =

(−2 −2−2 −2

)v1 = 0 ⇐⇒ v1 =

(1−1

)

When λ = −1 we obtain v2 =

(2−1

).

The eigenvectors of B with respect to A are(

1−1

)and

(2−1

)with eigenvalues 1,−1 respectively.

Diagonalizing the Fundamental Forms Suppose we have co-ordinates u, v on U, then for each p ∈U we have a basis ∂

∂u

∣∣∣p

, ∂∂v

∣∣∣p, with respect to which I, I are symmetric bilinear forms with matrices

(E FF G

),

(l mm n

)respectively. Think about what this means: if Y = y1

∂∂u

∣∣∣p+ y2

∂∂v

∣∣∣p

and Z = z1∂

∂u

∣∣∣p+ z2

∂∂v

∣∣∣p

are

tangent vectors at p, then

I(Y, Z) = (y1 y2)

(E FF G

)(z1z2

), I(Y, Z) = (y1 y2)

(l mm n

)(z1z2

)Indeed the matrix of I is positive definite, since

I(Y, Y) = dx(Y) · dx(Y) = |dx(Y)|2 ≥ 0

According to Theorem 2.31, there is a basis

X1|p = X(1)1

∂

∂u

∣∣∣∣p+ X(2)

1∂

∂v

∣∣∣∣p

, X2|p = X(1)2

∂

∂u

∣∣∣∣p+ X(2)

2∂

∂v

∣∣∣∣p

25

of TpU and scalars k1, k2 such that Xi|p is an eigenvector of I with respect to I with eigenvalue ki.Otherwise said,((

l mm n

)− ki

(E FF G

))(X(1)

i

X(2)i

)= 0, for i = 1, 2

Since the fundamental forms depend smoothly on the position p ∈ U, there is nothing stopping usfrom considering k1, k2 as smooth functions U → R and X1, X2 as vector fields on U.

Definition 2.32. If x : U → E3 is a surface, the functions k1, k2 : U → R are the principal curvatures ofx. The vector fields X1, X2 are the principal curvature directions.A point x(p) is umbilic if k1(p) = k2(p). At an umbilic point, all tangent directions are principalcurvature directions.

Corollary 2.33. Suppose that X1, X2 are the principal curvature directions on x. Then, at any non-umbilicpoint, I(X1, X2) = 0 = I(X1, X2).

Proof. Suppose that x(p) is non-umbilic and fix a basis of TpU so that the matrices of I, I are the

symmetric matrices A, B in Theorem 2.31. Viewing X1 =

(X(1)

1

X(2)1

), X2 =

(X(1)

2

X(2)2

)as column vectors

with respect to the basis of TpU, we see that BXi = ki AXi for each i. It is then straightforward tocompute:

I(X1, X2) = XT1 BX2 = XT

1 (k2AX2) = k2I(X1, X2)

I(X2, X1) = XT2 BX1 = XT

2 (k1AX1) = k1I(X1, X2)

Since k1 6= k2 it follows that I(X1, X2) = 0 = I(X1, X2).

Examples

1. The sphere of radius a has fundamental forms related by I = − 1a I. It follows that k1(p) =

k2(p) = − 1a at all points: every point on the sphere is umbilic!

2. Suppose that a surface14 has fundamental forms

I = u2 du2 + v2 dv2, I = u2 du2 + 2uv dudv + v2 dv2

Then the principal curvatures are the solutions λ of the equation

det((

u2 uvuv v2

)− λ

(u2 00 v2

))= 0

Multiplying this out, we obtain

u2v2(1− λ)2 − u2v2 = 0 ⇐⇒ (1− λ)2 = 1 ⇐⇒ λ = 0, 2

whence the principal curvatures are k1 = 0, k2 = 2. We can also compute the eigenvectors:

14For example x(u, v) = 12

cos( u2+v2√

2)

sin( u2+v2√2

)

u2−v2√2

though it is unimportant.

26

k1 = 0(

u2 uvuv v2

)(X(1)

1

X(2)1

)=

(00

)=⇒

(X(1)

1

X(2)1

)=

(v−u

)

k2 = 2(−u2 uvuv −v2

)(X(1)

2

X(2)2

)=

(00

)=⇒

(X(1)

2

X(2)2

)=

(vu

)We therefore have curvature directions

X1 = v∂

∂u− u

∂

∂v, X2 = v

∂

∂u+ u

∂

∂v

You can easily checked that I(X1, X1) = 2u2v2, etc., and that the matrices of I and I with respectto this basis are

[I] = 2u2v2(

1 00 1

)[I] = 2u2v2

(0 00 2

)in which the principal curvatures k1 = 0 and k2 = 2 are clearly visible.

In fact we can get more than just curvature directions, we can always locally assume the existence ofcurvature co-ordinates.

Theorem 2.34 (Frobenius: easy version). There exist curvature-line co-ordinates on a neighborhood of anynon-umbilic point.

A more thorough discussion is given below in the aside. Two things are important for us:

• We can only guarantee the existence of curvature-line co-ordinates locally.

• If s1, s2 are curvature-line co-ordinates for x, and X1, X2 are curvature directions, then X1 =f1

∂∂s1

and X2 = f2∂

∂s2for some functions f1, f2. That is, the vector fields ∂

∂siand Xi are parallel but

not necessarily equal!

Example, mark II In the above example, s = u2− v2 and t = u2 + v2 are curvature line co-ordinates.You should check the following.

• The co-ordinate vector fields are

∂

∂s=

1u

∂

∂u− 1

v∂

∂v=

1uv

X1 and∂

∂t=

1u

∂

∂u+

1v

∂

∂v=

1uv

X2

which are parallel but not equal to X1, X2.

• The fundamental forms really are diagonalized by s, t:

I = 2ds2 + 2dt2 I = dt2

We finish by considering totally umbilic surfaces: the proof is left to the homework.

Theorem 2.35. Suppose that all points of x : U → E3 are umbilic. Then x is part of a plane or a sphere.

27

Aside: The Frobenius Theorem and Curvature-line co-ordinates

As seen above, we can find curvature directions X1, X2, using linear algebra. What we’d really likeare curvature co-ordinates: functions s, t : U → R such that ∂

∂s = X1 and ∂∂t = X2. Unfortunately

this is usually too much too ask! A necessary condition is immediate: since mixed partial derivativesshould be equal, we’d need

X1X2 =∂2

∂s∂t=

∂2

∂t∂s= X2X1

That this condition is locally sufficient, and the consequent existence of curvature co-ordinates isgiven by the following result, which provides a deep link between differential geometry and thetheory of partial differential equations.

Theorem 2.36 (Frobenius: hard version). 1. Suppose that X, Y are linearly independent vector fields onan open set U containing p.

(a) If the Lie bracket [X, Y] = XY − YX of the vector fields is zero, then there exists a neighborhoodV ⊆ U of p and functions s, t : V → R such that

X =∂

∂s, Y =

∂

∂t

(b) There exists a neighborhood W ⊆ U of p and functions f , g : W → R such that [ f X, gY] = 0 onW. Part (a) then says there exists V ⊆ W and co-ordinates s, t : W → R such that ∂

∂s = f X and∂∂t = gY are parallel to X, Y.

2. If x(p) is a non-umbilic point of a surface x : U → E3, then there exist curvature-line co-ordinates for xon some neighborhood of p. Indeed if X, Y are curvature directions such that [X, Y] = 0, let α, β be thedual 1-forms to X, Y, then (locally) α = ds and β = dt.

The proof is a little too technical for this course: at its heart is the usual existence and uniquenesstheorem for differential equations and a bit more technology on forms than we’ve yet developed. Itis moreover completely generalizable to arbitrary dimensions.

Example, mark III Let’s revisit our example again! Frobenius’ Theorem actually tells us how tofind curvature co-ordinates. We give a sketch, though in practice this involves invoking the existenceof solutions to certain differential equations!

1. Hunt for functions f , g such that the vector fields X1 = f X1, X2 = gX2 satisfy the Lie bracketcondition:

[X1, X2] = 0

Multiply this out (it’s messy) to see that it is equivalent to being able to solve{v ∂ f

∂u + u ∂ f∂v = − f u2+v2

uv

v ∂g∂u − u ∂g

∂v = −g u2−v2

uv

(∗)

That this system can be solved is essentially the theorem. Indeed you can check that f = g = 1uv

does the job!

28

2. The co-ordinate direction vector fields

X1 =1u

∂

∂u− 1

v∂

∂vand X2 =

1u

∂

∂u+

1v

∂

∂v

satisfy the Lie bracket condition. The dual basis 1-forms to these fields are easy to find usinglinear algebra:

α =12(u du− v dv) and β =

12(u du + v dv)

3. Observe that α = ds and β = dt where s = u2− v2 and t = u2 + v2. We therefore have curvatureline co-ordinates s and t.

There are sometimes short-cuts to this procedure if you are feeling lucky, but in the abstract thisis what is needed. For instance, we could have gone straight to α, β by noting that we have to haveα(X2) = 0 = β(X1): by luck we would observe that α, β are the exterior derivatives of some functionsand we’re done! In practice, all this trick does is transform the problem to an equivalent systemof PDE. In general, given independent vector fields X1, X2 on U ⊆ R2, we know that there existfunctions a, b such that

[X1, X2] = aX1 + bX2

Hunting for functions f , g such that [ f X1, gX2] = 0 is equivalent to f , g satisfying the PDEs

X2[ f ] = a f X1[g] = −bg

This is the system (∗) in our example, and the problem of finding curvature-line co-ordinates essen-tially boils down to solving this. The theorem assures us that solutions exist, so we can always assumethe existence of local curvature co-ordinates. Finding them explicitly may be essentially impossible!

2.5 Gauss and mean curvature

The principal curvatures of a surface are usually rearranged in two combinations which have adeeper geometric meaning.

Definition 2.37. Suppose that x : U → E3 has principal curvatures k1, k2. The mean curvature H andGauss curvature K of x are defined respectively by

H = 12 (k1 + k2) K = k1k2

Example The sphere of radius a (with outward pointing normal) has k1 = k2 = − 1a . Thus H = − 1

aand K = 1

a2 .

Proposition 2.38. In co-ordinates, the mean and Gauss curvatures are given by,

H =lG + nE− 2mF

2(EG− F2), K =

ln−m2

EG− F2

29

Proof. In the basis ∂∂u , ∂

∂v , I and I have matrices(

E FF G)

and ( l mm n ) respectively. Now expand the deter-

minant,

det((

l mm n

)− λ

(E FF G

))= 0

to obtain

(EG− F2)λ2 − (lG + nE− 2mF)λ + (ln−m2) = 0

K is the product and H half the sum of the roots of this equation.

Examples

1. A surface of revolution found by rotating the curve x = f (z) around the z-axis, and parameter-ized in the usual way, has principal curvatures

k1 = − f ′′(u)(1 + f ′(u)2)3/2 , k2 =

1f (u)(1 + f ′(u)2)1/2

whence

H =1 + f ′(u)2 − f (u) f ′′(u)

2 f (u)(1 + f ′(u)2)3/2 , K = − f ′′(u)f (u)(1 + f ′(u)2)2

The simplest example would be a cylinder, where the radius f (u) is constant. We immediatelysee that a cylinder has zero Gauss curvature and mean curvature half the radius.

2. If z = f (x, y) is the graph of a function then, parameterizing in the usual way, we have

H =fvv(1 + f 2

u) + fuu(1 + f 2v )− 2 fu fv fuv

2(1 + f 2u + f 2

v )3/2 , K =

fuu fvv − f 2uv

(1 + f 2u + f 2

v )2

The Gauss and mean curvatures are extremely important objects, the Gauss curvature especially so.In particular, note that these are invariants of a surface:

• The Gauss curvature is independent of parameterization, while the mean curvature is invariantunder orientation-preserving reparameterizations.

Many of the applications of these curvatures will have to wait until another course: here are a few towhet your appetite.

Definition 2.39. Constant mean curvature (CMC) surfaces are surfaces for which H is constant. Aminimal surface is a surface for which H is identically zero.

CMC surfaces have many useful applications and appear throughout mathematics. Minimal surfacesare so-called because they minimize areas. For instance, suppose that y is a closed curve in E3. Imag-ine filling in the curve so that you have some sort of warped disc surface whose edge is y. There areinfinitely many choices, but the one with minimal area will have H = 0. The same holds for surfacesjoining a pair of closed curves. As such, minimal surfaces are often the solutions to many practicalquestions: for example, the shape assumed by a soap film is typically a minimal surface as this willbe the surface which minimizes the total tension in the film.

It can be shown that constant Gauss curvature surfaces are spheres if K > 0, and planes, cones orcylinders when K = 0.

30

Definition 2.40. A pseudosphere is a surface with constant negative Gauss curvature.

It is an exercise to show that the surface of revolution defined by a tractrix15 defines a pseudospherewith K = −1, and that the surface obtained by rotating a catenary f (u) = a cosh(a−1u + c) is a mini-mal surface.

Perhaps the most important fact regarding Gauss curvature is that it depends only on the first funda-mental form. This is Gauss’ famous Theorem Egregium, whose proof will also have to wait for alittle more technology. It is therefore entirely detectable to any inhabitant of a surface who can onlymeasure distance and angle, but is incapable of understanding the normal direction. This conceptis important when extended to several dimensions and the theory of relativity and spacetime: wecannot travel ‘normal’ to the Universe in order to measure curvature! Indeed the Gauss curvature isthe 2-dimensional version of the more general Riemann curvature tensor which rules the discussion ofrelativity.

2.6 Power series expansion and Euler’s theorem

We can analyze the properties of a surface at a fixed point in terms of the curvatures described in theprevious sections. Essentially we view a surface as a graph of a function nearby a given point. Theanalysis will tell us what happens at that point and approximately nearby, but not globally.

Theorem 2.41. Let s be a point on a regular surface S. Choose axes such that s is the origin and the (x1, x2)-plane is tangent to the surface. The surface is then locally the graph of a function x3 = f (x1, x2). Parameter-izing in the usual way we have that at the origin,

I(0,0) = du2 + dv2

I(0,0) = fuu(0, 0)du2 + 2 fuv(0, 0)dudv + fuv(0, 0)dv2

Proof. That the surface is locally a graph is essentially Theorem 2.21 after rotation and translation of

the surface to the origin in such a way that U|s becomes the vertical basis vector( 0

01

). Parameterizing

in the usual way, we take (u, v) to be the standard orthonormal co-ordinates on the tangent plane TsSsuch that (0, 0) corresponds to s and obtain a parameterization x(u, v) =

( uv

f (u,v)

). Section 2.3 shows

the fundamental forms for such a surface:

I = (1 + fu)du2 + 2 fu fv dudv + (1 + f 2v )dv2

I =1√

1 + f 2u + f 2

v

{fuu du2 + 2 fuv dudv + fvv dv2

}from which evaluation at (0, 0) is immediate.

Note that we’re only computing the fundamental forms at the origin: in particular, although the co-ordinates u, v will extend at least nearby on the surface, the first fundamental form need not bediagonal anywhere except at the origin.

15Recall the definition from our discussion of curves.

31

Definition 2.42. The matrix of I(0, 0) with respect to the co-ordinates (u, v) is the Hessian of f at(0, 0):given by

Hess f (0, 0) =(

fuu fuvfuv fvv

)∣∣∣∣(0,0)

It is immediate that the Gauss and mean curvatures satisfy

K(s) = det Hess f (0, 0) and H(s) =12

tr Hess f (0, 0)

Suppose now that we rotate the (x1, x2)-plane such that the axes point in the principal directions at(0, 0). Then I(0, 0) is also diagonal ( fuv(0, 0) = 0). We have the following result.

Theorem 2.43. Suppose that x(u, v) =( u

vf (u,v)

)is a local surface defined as a graph z = f (x, y), such that

the (x, y)-plane is tangent at x(0, 0) and both I(0, 0) and I(0, 0) are diagonal. Then the principal curvaturesat (0, 0) are given by

k1 = fuu(0, 0) and k2 = fvv(0, 0)

Furthermore, the Taylor expansion of x3 = f (x1, x2) about the origin is

Tx3 =k1

2x2

1 +k2

2x2

2 + higher order terms

Proof. The principal curvatures are clear since both fundamental forms are diagonal. For the latter,recall Taylor series for multi-valued functions:

Tx3 = f (0, 0) + (x1 x2) ∇ f |(0,0) +12(x1 x2) Hess f |(0,0)

(x1x2

)+ higher order terms

= f (0, 0) + (x1 x2)

(00

)+

12(x1 x2)

(fuu fuvfuv fvv

)∣∣∣∣(0,0)

(x1x2

)+ higher order terms

=12(x1 x2)

(k1 00 k2

)(x1x2

)+ higher order terms

=k1

2x2

1 +k2

2x2

2 + higher order terms

Again note that we are not claiming that u, v are curvature co-ordinates on the surface: we onlyhave to have diagonal fundamental forms at the origin. Nearby there is no requirement that u, v becurvature-line.

The Theorem is the surface analogy of a result we saw earlier regarding curves: a regular curve in E2

passing through the origin horizontally at t = 0 has its graph given locally by

y =κ(0)

2x2 + higher order terms ∗

We can put this correspondence to work by considering the curvature of certain curves passingthrough a point on a surface.

32

Definition 2.44. Let v be a tangent vector at a point s to a surface S in E3. Let Πs(v) be the planethrough s spanned by v and U|s. Its intersection Πs(v) ∩ S with the surface contains a connectedcomponent which describes a curve lying in the surface passing through s. The normal curvature of Sin the direction v is the curvature at s of this curve.

The normal curvature is surprisingly easy to compute.

Theorem 2.45 (Euler). Suppose that v is a tangent vector to S making angle θ with the first principal curva-ture direction. Then the normal curvature of S in the direction v is

κ = k1 cos2 θ + k2 sin2 θ

Proof. Choose axes such that the principal curvature directions at s are(

100

)and

( 010

), and the unit

normal is U|s =( 0

01

). Consider the position vector

x =

t cos θt sin θ

z

We can think about this in two ways:

1. Let θ be the angle made by v and the principal direction(

100

). Then x(t, z) parameterizes the

plane Πs(v). By observation (∗), the plane curve in the intersection Πs(v) ∩ S has the form

y(t) =

t cos θt sin θ

12 κt2 + higher order terms

where κ is the normal curvature.

2. By Theorem 2.41, the surface is locally a graph z = f (t, θ), where t, θ are polar co-ordinates inthe horizontal plane. Theorem 2.43 says that

z = 12 k1(t cos θ)2 + 1

2 k2(t sin θ)2 + higher order terms

=12(k1 cos2 θ + k2 sin2 θ)t2 + higher order terms

Comparing the two expressions for z yields the result.

A straightforward consequence is that the normal curvature always lies between the two principalvalues k1 and k2. The principal curvature are therefore the extremes of curvature at any given point.

33

2.7 Elliptic and hyperbolic points: the Dupin Indicatrix

Theorem 2.43 provides a description of a surface near a point s.

Definition 2.46. Suppose that k1, k2, K, H are the principal, Gauss and mean curvatures of a surfaceS at a given point s. We say that s is:

1. Elliptic ⇐⇒ K > 0 ⇐⇒ k1, k2 are non-zero and have the same signIn this case, H has the same sign as the principal curvatures: it is positive if and only if thesurface bends towards the unit normal.

2. Hyperbolic ⇐⇒ K < 0 ⇐⇒ k1, k2 are non-zero and have opposite signs

3. Parabolic ⇐⇒ K = 0 and H 6= 0 ⇐⇒ exactly one of k1, k2 are zero

4. Planar ⇐⇒ K = H = 0 ⇐⇒ k1 = k2 = 0

Recalling Theorem 2.43, we see that a surface may be written locally as

z ≈ 12

k1x2 +12

k2y2

Near an elliptic point, level curves (z =constant) of the surface are approximately ellipses. Near ahyperbolic point they are approximately hyperbolæ. The correspondence doesn’t extend to the othertypes of point. Near a parabolic point a surface looks approximately like a parabolic cylinder, whichnear a planar point a surface looks approximately like part of a plane.

Examples

1. The (elliptic) paraboloid x(u, v) =( u

vu2+v2

)has

I = (1 + 4u2)du2 + 8uv dudv + (1 + 4v2)dv2 I =2√

1 + 4u2 + 4v2(du2 + dv2)

from which we can calculate16

K =4

(1 + 4u2 + 4v2)2 and H =1 + 2u2 + 2v2

(1 + 4u2 + 4v2)3/2

Both are positive everywhere, and so all points are elliptic with the surface always bendingtowards the unit normal.

2. The (hyperboloic) paraboloid x(u, v) =( u

vu2−v2

)has

K = − 4(1 + 4u2 + 4v2)2 and H =

4(v2 − u2)

(1 + 4u2 + 4v2)3/2

so that all points are hyperbolic.

16Revisit Proposition 2.38 and the examples following to do this quickly.

34

3. The surface x(u, v) =( u

vu2+v4

)has

K =24v2

(1 + 4u2 + 16v6)2 and H =6v2(1 + 4u2) + 1 + 16v6

(1 + 4u2 + 16v6)3/2

Away from v = 0, x is entirely elliptic and curves towards the unit normal vector. The pointsx(u, 0) are parabolic. This doesn’t mean that the surface contains a straight line, indeed allpoints x(u, v) where (u, v) 6= (u0, 0) lie on the same side of the tangent plane at (u0, 0). To seethis, note that

U(u0, 0) =1

(1 + 4u20)

1/2

−2u001

and so

(x(u, v)− x(u0, 0)) ·U(u0, 0) =1

(1 + 4u20)

1/2

u− u0v

u2 + v2 − u20

·−2u0

01

=

1(1 + 4u2

0)1/2

((u− u0)2 + v2) > 0

for all (u, v) 6= (u0, 0). As this example shows, a parabolic point can still be convex (or indeeda saddle).

Definition 2.47. Recall from Euler’s theorem that the normal curvature takes values at each pointbetween the two principal values. A tangent vector v at a point on a surface is asymptotic if thenormal curvature defined by v is zero.

Examples

1. The elliptic paraboloid has no asymptotic directions at any point: the principal curvatures areboth positive at all points, whence the normal curvature at an point is k1 cos2 θ + k2 sin2 θ > 0.

2. The hyperbolic paraboloid has asymptotic directions at every point. The principal curvatureshave opposite signs, whence there are two angles θ ∈ (−π

2 , π2 ) such that k1 cos2 θ + k2 sin2 θ = 0:

i.e.

tan2 θ = − k1

k2

Computing this more specifically is awkward, as the principal curvatures themselves are ugly.At the origin, however, things are easier: k1 = 2 = −k2, so the asymptotic directions satisfytan2 θ = 1 ⇐⇒ θ = ±π

4 . In this case the asymptotic directions point half way between theprincipal curvature directions.

3. in the third example, there are only asymptotic directions at the parabolic points when v = 0.These point parallel to the y-axis.

In keeping with our theme of transferring calculations back to the parameterization space U, weprove theorem.

35

Theorem 2.48. Suppose that x : U → E3 is a local surface and let v ∈ TpU be a non-zero tangent vector.Then dx(v) is asymptotic iff I(v, v) = 0.

Because of the theorem we tend to call the tangent vector v asymptotic in its own right.

Proof. Let v ∈ TpU be a non-zero tangent vector and let Π be the plane at x(p) spanned by U(p) anddx(v). Let y be the curve defined by the intersection of the surface and Π, parameterized such thaty(0) = x(p). Then the curvature of y at x(p) is precisely the normal curvature of the surface in thedirection dx(v). Now recall Theorem 2.27, from which we see that

y′′(0) ·U = I(v, v)

But y′′(0) · U is zero iff the curvature of y is zero at t = 0 which is iff the normal curvature in thedirection dx(v) vanishes.

Definition 2.49. Let x : U → E3 be a parameterized surface. The Dupin indicatrix at p ∈ U is the setof tangent vectors v ∈ TpU such that I(v, v) = ±1.

We can easily derive an equation for the Dupin indicatrix. Suppose that

v = x1X1 + x2X2

where X1, X2 are unit-length principal curvature vectors.17 Then

I(v, v) = x21I(X1, X1) + 2x1x2I(X1, X2) + x2

2I(X2, X2)

= k1x21 + k2x2

2 = ±1

defines a curve in the tangent space TpU. This curve depends on the signs of k1, k2.

1. If k1, k2 have the same sign, then the Dupin indicatrix is an ellipse (with respect to the basisX1, X2). This motivates the name elliptic point. In the special case of an umbilic point k1 = k2the indicatrix is a circle (with respect to X1, X2).

2. If k1, k2 are non-zero with opposite signs, then the indicatrix is a pair of hyperbolae, hencehyperbolic point.

3. If one of k1, k2 is zero (a parabolic point) the indicatrix is a pair of lines.

4. If both curvatures are zero, the indicatrix is empty.

When we say that the indicatrix is an ellipse or a circle, etc., with respect to the basis X1, X2, whatwe really mean is with respect to the first fundamental form. Remember that TpU, although a vectorspace, has no inherent notion of length or angle (it is not a Euclidean space). It is the first fundamen-tal form that defines what is meant by length and angle in TpU.

The Dupin indicatrix (at least for elliptic and hyperbolic points) gives a visualization of what thesurface looks like near a point. Near an elliptic point, planar slices through the surface orthogonal tothe unit normal are approximately ellipses. Near a hyperbolic point planar slices through the surface

17I.e. I(X1, X1) = 1 = I(X2, X2) and I(X1, X2) = 0.

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orthogonal to the unit normal are approximately hyperbolae. Typical indicatrices for an elliptic anda hyperbolic point are shown below, with the asymptotic directions indicated by dashed arrows forthe hyperbolic point. The first is based on an umbilic point k1 = k2 = 0.9 (this is a circle with respectto the chosen principal curvature directions X1, X2), while the second has k1 = −k2 = 0.3.

X1

X2

X1

X2

Corollary 2.50. A point s on a surface has

012∞

asymptotic directions if s is

ellipticparabolichyperbolicplanar

.

Proof. The equation for asymptotic directions v is I(v, v) = 0. This requires solving the singular conic

I(v, v) = k1x21 + k2x2

2 = 0

The options are:

• Elliptic point: (x1, x2) = (0, 0) if k1, k2 are the same sign. We obtain no asymptotic directionsv = x1X1 + x2X2.

• Hyperbolic point: if k1, k2 have opposite signs, the conic factors as two distinct lines x2 =±√−k1/k2x1 and we obtain two distinct asymptotic directions.

• Parabolic point: if k1 = 0, then the conic is the x1-axis (x2 = 0). We have a single asymptoticdirection along this axis. (k2 = 0 is similar.)

• Planar point: if k1 = k2 = 0 then all vectors v satisfy I(v, v) = 0 whence any direction isasymptotic.

There are thus at most two asymptotic directions at each non-planar point on a surface.

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