Lecture 1-30

GEOMETRY OF CURVES AND SURFACES

ANDRE NEVES

Lecture 1

A parametrized curve is a smooth function

φ : [a, b]→ Rn, ( with n = 2 or n = 3)

where |φ′(t)| 6= 0 for all t ∈ [a, b].Some examples are

(1) straight line

φ1 : [0, 1]→ R2, φ1(t) = (2t− 1, 3t+ 2);

(2) circle of radius r centred at the origin

φ2 : [0, 2π]→ R2, φ2(θ) = (r cos θ, r sin θ);

(3) helix

φ3 : [0, 8π]→ R3, φ3(θ) = (cos θ, sin θ, θ).

A non-example is

φ4 : [0, 8]→ R2, φ4(t) = (|t|, t)because the function is not smooth (it is not even C1). Another non-exampleis

φ5 : [−1, 1]→ R2, φ5(t) = (0, t2)

because |φ′5(0)| = |(0, 0)| = 0.

The requirement that φ : [a, b] → Rn has |φ′(t)| 6= 0 for all a ≤ t ≤ bserves two purposes. On one hand, we excluded maps like the constant mapfrom being a curve. On the other hand, every parametrized curve has atangent line passing through any given point. More precisely, the tangentline to the parametrized curve φ that passes through φ(t0) is given by

L = {φ(t0) + φ′(t0)s : s ∈ R}.

For the purpose of geometry, what we are interested is in φ(I), the im-age of the parametrized curve, which we call a curve. The key thing tounderstand is that a curve can have many different parametrizations. Forinstance

φ6 : [0, 2]→ R2, φ6(t) = (t3/4− 1, 3t3/8 + 2)

parametrizes the same curve as φ1 (why?).1

2 ANDRE NEVES

In general, if φ : [a, b]→ Rn is a parametrized curve and f : [c, d]→ [a, b]is a smooth function with |f ′(x)| 6= 0 for all x, then

φ ◦ f : [c, d]→ Rn, t 7→ φ(f(t))

is another parametrized curve with φ([a, b]) = φ ◦ f([c, d]).We are interested in properties of the curve which are invariant under

reparametrizations. A first example is the length of the curve given by

length =

∫ b

a|φ′(t)|dt.

For instance, the length of the first curve is

length =

∫ 1

0|φ′1(t)|dt =

∫ 1

0|(2, 3)|dt =

∫ 1

0

√13dt =

√13.

The length of the third curve is

length =

∫ 8π

0|φ′3(θ)|dθ =

∫ 8π

0|(− sin θ, cos θ, 1)|dθ =

∫ 8π

0

√2 = 16π.

Lemma 0.1. The length of a curve is invariant under reparametrizations.

Proof. Say that we have φ : [a, b]→ Rn and f : [c, d]→ [a, b] with f ′(s) 6= 0for all s (why do I require this?). Then ψ = φ ◦ f : [c, d] → Rn is anotherparametrization of φ([a, b]). We want to make sure that the length of φ([a, b])is the same as the length of ψ([c, d]) because the curve has remained thesame.

Let’s assume that n = 3 and, without loss of generality that f ′ > 0 (.i.e,f is increasing). If φ(t) = (x(t), y(t), z(t)) we have

|ψ′(s)| =√

((x ◦ f)′(s))2 + ((y ◦ f)′(s))2 + ((z ◦ f)′(s))2

=√

(x′(f(s))2(f ′(s))2 + (y′(f(s))2(f ′(s))2 + (z′(f(s))2(f ′(s))2

= f ′(s)√

(x′(f(s))2 + (y′(f(s))2 + (z′(f(s))2 = f ′(s)|φ′(f(s))|and so, by the change of variable formula we obtain

length(ψ([c, d])) =

∫ d

c|ψ′(s)|ds =

∫ d

c|φ′(f(s))|f ′(s)ds

=

∫ f(d)

f(c)|φ′(t)|dt =

∫ b

a|φ′(t)|dt = length(φ([a, b])).

�

Lecture 2

Given a parametrized curve φ : [a, b]→ Rn we want to get rid of the factthat there are infinitely many possible parametrizations of φ([a, b]) and sowe will choose a best possible one by requiring that |φ′(t)| = 1 for all t. Thisis called arc-length parametrization and it can always be done, as the nextlemma shows.

GEOMETRY OF CURVES AND SURFACES 3

Lemma 0.2. Given φ : [a, b] → Rn a parametrized curve, we can alwaysfind ψ : [c, d]→ Rn with φ([a, b]) = ψ([c, d])and |ψ′(s)| = 1 for all s.

Proof. The way to go is to guess what ψ should be and then find such ψ. Weknow that we should have ψ = φ ◦ f for some increasing function f (why?).Hence, with t = f(s) we want that

1 = |ψ′(s)| = f ′(s)|φ′(f(s))| =⇒ f ′(s) = 1/|φ′(f(s))| = 1/|φ′(t)|.

So set h(t) =∫ ta |φ

′|(u)du. Then h′(t) = |φ′(t)| and we choose f to be theinverse of h, which must exists and have f ′(s) = 1/|φ′(f(s))|. This shoulddo it. �

A cool thing is that if φ : [a, b] → Rn is parametrized by arc-length thenthe length of the curve is b− a (why?).

Consider φ : [0, L]→ Rn parametrized curve that is parametrized by arc-

length. We define the geodesic curvature (or just curvature) as ~k = φ′′(t).

The first remark is that ~k(t) is always perpendicular to the tangent lineto the curve at φ(t). Indeed, we saw that the tangent line to the curve atφ(t) is generated by φ′(t) and thus, because

1 = φ′(t).φ′(t) =⇒ 0 =d

dtφ′(t).φ′(t) = φ′′(t).φ′(t)+φ′(t).φ′′(t) = 2~k(t).φ′(t)

we obtain that ~k(t).φ′(t) = 0.

The second remark is that ~k depends only on the curve φ([0, L)) and noton the parametrization. The reason is that if φ and ψ are two parametriza-tions by arc length of the same curve then φ(t) = ψ(at + α) where α ∈ R,and a = 1 (orientation preserving )or a = −1 (orientation reversing) (why

?). Either way, they give the same ~k.

If we parametrize a straight line as φ(t) = ~a+ t~b, where ~b as length one,

then we have φ′(t) = ~b and φ′′(t) = 0, which means a straight line has cur-vature zero.

Note that the condition that the curve is parametrized by arc length

it is crucial because we want ~k to depend only on the curve and not onthe parametrization chosen. For instance, we can consider the followingparametrization of a straight line in the plane:

φ(t) = (0, et), 0 ≤ t ≤ 1.

Then φ′′(t) = (0, et) 6= 0, while, as we just saw in the previous example,~k = 0.

Exercise: Conversely, show that if φ : [0, L] → Rn has curvature zero,then it is a straight line.

Solution: Assume φ is parametrized by arc-length and so ~k = 0 meansthat φ′′(t) = 0 for all t. Hence (assuming n = 3)

(x′′(t), y′′(t), z′′(t)) = 0 for all 0 ≤ t ≤ L

4 ANDRE NEVES

and thus one can easily see that

φ(t) = (x(t), y(t), z(t)) = φ′(0)t+ φ(0) for all t.

Lecture 3

From now on we consider only plane curves, i.e., n = 2. The advantageof this case is that the perpendicular vector to the curve becomes uniquelydetermined by the tangent vector.

More precisely, if I have φ : [a, b]→ R2, φ(t) = (x(t), y(t)), then

N(t) = (−y′(t), x′(t))/|φ′(t)|

is a unit vector which is orthogonal to the tangent vector φ′(t) = (x′(t), y′(t)).Indeed

φ′(t).N(t) =1

|φ′(t)|(−x′(t)y′(t) + x′(t)y′(t)) = 0 and |N(t)| = 1

for all a ≤ t ≤ b.Because ~k is also orthogonal to φ′(t) (regardless whether the parametriza-

tion is by arc-length or not – why?) we can instead of the vector ~k con-

sider the scalar k = ~k.N , which we also denote by curvature. When φis parametrized by arc-length then, using the expression for N above, thecurvature k can be computed as

k = ~k.N(t) = φ′′(t).(−y′(t), x′(t)) = x′(t)y′′(t)− y′(t)x′′(t).

Let’s compute the curvature of a circle of radius r centered at the point~c. First we choose a parametrization, which could be

φ : [0, 2π]→ R2, φ(θ) = ~c+ (r cos θ, r sin θ).

The problem is that is not parametrized by arc-length and so we have tochange it to

φ : [0, 2πr]→ R2, φ(θ) = ~c+ (r cos(θ/r), r sin(θ/r)).

Then φ′′(θ) = −r−1(cos(θ/r), sin(θ/r)) and thus

k = ~k.N = −r−1(cos(θ/r), sin(θ/r)).(− cos(θ/r),− sin(θ/r)) = 1/r.

This makes sense, because the higher the radius, the less curved the circleis and the smaller the radius, the more curved the circle is.

Exercise : Explain why k depends on the parametrization of a curve(more precisely, on the orientation of the curve).

Solution: Given a parametrized curve φ, the vector ~k is intrinsic tothe curve. The normal vector N given above corresponds to rotating φ′ by90 degrees on the counterclockwise direction. If we switch the orientationof the curve, the new tangent vector becomes −φ′ and so, if we rotate itcounterclockwise by 90 degrees we obtain −N . As a result k changes by asign.


Lecture 4

The example above is useful for the following geometric characterizationof curvature.

Let γ be a curve on the plane and choose p in γ. Orient γ so that k(p) ≥ 0.Then the circle that ‘best’ approximates γ at p has radius 1/k(p). I nowmake this more precise.

Choose φ a parametrization by arc-length of γ and assume that φ(0) = p.For any r > 0, let Cr be then the unique circle of radius r that passes

though p, is tangential to the curve at p, and the unit normal vector Npoints towards the interior of the circle Cr.

We say the circle Cr is too small if there is ε > 0 so that Φ([ε, ε]) doesnot intersect the interior of Cr. Likewise, we say the circle Cr is too big ifthere is ε > 0 so that Φ([ε, ε]) does not intersect the exterior of Cr.

Lemma 0.3. Set r0 = k−1(p). Every Cr is too small if r < r0 and too bigif r > r0. Thus Cr0 is the circle that best approximates γ near p.

Proof. Without loss of generality assume φ(0) = p = (0, 0),and set η(t) =|φ(t)− sN(0)|2, where N(0) is the unit vector perpendicular to the curve at(0, 0). With this notation, Cs is the circle of radius s centered at sN(0). Ifη(t) > s2 then this means φ(t) lies outside Cs and if η(t) < s2, this meansφ(t) lies inside Cs (why?).

We have η(0) = s2 and

η′(t) =d

dt((φ(t)− sN(0)).(φ(t)− sN(0)))

= φ′(t).(φ(t)− sN(0)) + (φ(t)− sN(0)).φ′(t) = 2φ′(t).(φ(t)− sN(0)).

Thus η′(0) = 2φ′(0).(−sN(0)) = 0 (why?) and

η′′(0) = 2φ′′(0).(−sN(0)) + 2φ′(0).φ′(0)

= 2~k(0).(−sN(0)) + 2 = 2(1− sk(0)) = 2(1− s/r).

Therefore, if s < r then η(0) is a strict local minimum of η and so there is asmall neighborhood I of 0 so that η(t) > η(0) = s2 for all t ∈ I \ {0}. Thecase s > r is handled similarly. �

Exercise: Show that if γ is a planar curve with k = 1/r, then γ iscontained in a circle of radius r centered at some point.

Solution: Assume γ is parametrized by arc-length and set c(t) = γ(t) +rN(t), where N is the unit normal along the curve γ, with respect to whichk = 1/r (instead of k = −1/r).

Then

N.N = 1 =⇒ ∂tN.N = 0 =⇒ ∂tN = multiple of γ′.

Moreover,

∂tN.γ′ = ∂t(N.γ

′)−N.γ′′ = 0−N.~k = −k = −1/r

6 ANDRE NEVES

and so ∂tN = −γ′/r. Hence

c′ = γ′ + rN ′ = γ′ − γ′ = 0

and so c(t) = c(0) for all t, which means |γ − c(0)| = |rN | = r for all t, i.e.,γ is contained in a circle of radius r and center c(0).

Exercise: Consider φ : [a, b]→ R2 the parametrization of a curve. Showthat

k(t) = |φ′(t)|−2φ′′(t).N(t).

Solution: It suffices to see that the expression on the right hand side isindependent of the parametrization because, if the curve is parametrized byarc length, then indeed the right hand side gives us k(t).

Suppose we have t = f(s), where f is a smooth strictly increasing functionwith range [a, b], and consider ψ = φ ◦ f . Then

ψ′(s) = f ′(s)φ′(f(s)) ψ′′(s) = f ′′(s)φ′(f(s)) + (f ′(s))2φ′′(f(s)).

The vector f ′′(t)φ′(f(s)) is tangent to the curve at ψ′(s) and so

f ′′(s)φ′(f(s)).N(f(s)) = 0,

where N(s) := N(f(s)) is the normal vector to ψ at ψ(s). Moreover,|ψ′(s)| = f ′(s)|φ′(f(s))|. Hence

|ψ′(s)|−2ψ′′(s).N(s) = |ψ′(s)|−2ψ′′(s).N(f(s)) = |ψ′(s)|−2ψ′′(s).N(f(s))

= |ψ′(s)|−2(f ′(t))2φ′′(f(s)).N(f(s)) = |φ′(f(s))|φ′(f(s)).N(f(s))

= |φ′(t)|−2φ′′(t).N(t).

Exercise: Let f : R −→ R be a smooth function. Consider the curve

c : R −→ R, c(t) = (t, f(t)).

Compute the geodesic curvature k(t).

Solution: From the exercise above we obtain

k(t) = |c′(t)|−1c′′(t).N(t). = (1 + f ′(t)2)−1(0, f ′′(t)).(−f ′(t), 1)

=f ′′(t)

1 + (f ′(t))2

Exercise Say that γ is a planar curve which passes through p = (0, 0),has N(p) = (0, 1) and k(p) = 0. Can we find a neighborhood U of p so that

γ ∩ U ⊆ {(x, y)|y > 0} ∪ (0, 0)?

Given an example or explain why not.

Solution: Consider γ(t) = (t, t3). Then γ′(0) = (1, 0) =⇒ N(0) = (0, 1).From the previous exercise we can see that k(0) = 0 and we cannot find aneighborhood U as described in the exercise for this curve γ. So the answer


is NO.

Lecture 5

We continue with planar curves φ : [a, b] → R2 but we assume that thecurve is smooth and closed, i.e., φ(a) = φ(b) and, for any k-derivative,

φ(k)(a) = φ(k)(b). In what follows we consider the curve γ = φ([a, b]).Closed curves have a first nice invariant which is the rotation number

around 0, r(γ, 0). For instance if γ is a circle of radius centered at the origin,then r(γ, 0) is either 1 or −1, depending whether the curve is transversedclockwise or counterclockwise. If γ is a curve centered at (0, 1) of radius lessthan one, then r(γ, 0) = 0.

There is a nice way to compute the rotation number, which we usuallylearn in complex analysis and is part of the Cauchy integral formula: theformula is given by

r(γ, 0) =1

i2π

∫γ

dz

z.

Writing the parametrization φ of γ as φ(t) = x(t) + iy(t), where a ≤ t ≤ bwe get

1

i2π

∫γ

dz

z=

1

i2π

∫ b

a

x′(t) + y′(t)

x(t) + iy(t))dt =

1

i2π

∫ b

a

(x′ + iy′)(x− iy)

x2 + y2dt

=1

i2π

∫ b

a

x′x+ yy′ + i(xy′ − yx′)x2 + y2

dt

=1

i2π

∫ b

a

x′x+ yy′

x2 + y2dt+

1

2π

∫ b

a

xy′ − yx′

x2 + y2dt

=1

i2π

∫ b

a

d

dtln√x2 + y2dt+

1

2π

∫ b

a

xy′ − yx′

x2 + y2dt

=1

2π

∫ b

a

xy′ − yx′

x2 + y2dt.

Thus we conclude

r(γ, 0) =1

2π

∫ b

a

1

|φ(t)|2φ(t).(y′(t),−x′(t))dt.

Given a closed curve γ parametrized by φ we define its winding numberw(γ) as the rotation number of the curve

F : [a, b]→ R2, F (t) = φ′(t)

around the origin. In other words w(γ) = r(γ′, 0).The winding number counts how many times F rotates around the origin

and so how many turns γ does, where turns counterclockwise are countedpositively and turns clockwise are counted negatively. Note that if we changethe orientation of the curve, then the winding number changes sign.

8 ANDRE NEVES

Theorem 0.4. If γ is a closed curve then

w(γ) =1

2π

∫γkds.

Remark 0.5. (1) Given a function f : γ → R defined on a curve γ wedefine ∫

γfds =

∫ b

af(φ(t))|φ′(t)|dt,

where φ : [a, b]→ R2 is a parametrization of γ.Note that if f is identically one, then the above formula is simply

the length. In the same way we proved Lemma 0.1 one can alsoshow that the above definition is independent of the parametrizationchosen.

(2) The winding number of a curve is a topological invariant, i.e., ifwe move the curve in a smooth way in the plane, then the numberof times γ rotates does not change. The technical lingo is to saythat w(γ) does not change under isotopies. On the other hand, thecurvature k is not invariant under smooth deformations. The contentof Theorem 0.4 is that if we integrate k, the value is a topologicalinvariant! This is a prototype for Gauss-Bonnet Theorem and that’swhy I have included it.

Proof. Assume the curve γ is parametrized by arc-length, i.e., |φ′(t)| = 1.Then F (t) = (x′(t), y′(t)) and so, recalling the formula for N ,

F (t).(y′′(t),−x′′(t)) = φ′(t).(y′′(t),−x′′(t)) = x′(t)y′′(t)− y′(t)x′′(t)

= (−y′(t), x′(t)).~k(t) = N.~k = k(t).

Therefore we obtain that given a closed curve γ then

w(γ) = r(F, 0) =1

2π

∫ b

aF (t).(y′′(t),−x′′(t))dt =

1

2π

∫ b

ak(t)dt =

1

2π

∫γkds.

�

Lecture 6

The theorem above also has one nice consequence that I now explain.Given a closed curve γ with a parametrization φ, we consider the energy ofγ to be

E(γ) = length(γ)

∫γk2ds.

This quantity is scale invariant, i.e., E(cγ) = E(γ) for every closed curve γand constant c > 0.

Corollary 0.6. If γ is a closed curve with w(γ) 6= 0 then E(γ) ≥ 4π2 withequality if and only if γ is a circle.


The content of the above result is that E(γ) is a number that measureshow much a closed curve is curving. In particular, every closed curve curvesmore than a circle and equality holds only for the circle. This type of resultsare paramount in Geometry.

The assumption that w(γ) 6= 0 is not necessary and could be removed. Iwill put that as an exercise later in the course.

Proof. Let’s orient the curve so that w(γ) ∈ N . From Holder’s inequalityand Theorem 0.4 we know that

2π ≤ 2πw(γ) =

∫γkds ≤

√length(γ)

∫γk2ds =

√E(γ).

Moreover, if equality holds then w(γ) = 1 and thus

2π =

∫γkds =

√length(γ)

∫γk2ds.

Setting c0 = 2π/length(γ) we obtain∫γ(k − c0)2ds =

∫γk2ds− 2c0

∫γk +

∫γc2

0ds

= 4π2/length(γ)− 2(2π/length(γ))2π + (2π/length(γ))2length(γ)

= 4π2/length(γ)− 8π2/length(γ) + 4π2/length(γ) = 0.

Therenfore k is a constant which we call 1/r. From Exercise 3 we deducedthat γ is contained in a circle of radius r.

There is one last detail we have to take care. It is possible that γ could bea circle of radius r travelled k times, for instance φ(θ) = r(cos(kθ), sin(kθ)),where 0 ≤ θ ≤ 2π. This cannot be because in this case w(γ) = k and weknow w(γ) = 1. Hence, indeed, γ is a circle or radius r. �

Exercise: Give an example of a closed curve γ with no self-intersectionsand E(γ) arbitrarily large. You can just draw the curve and explain in aconvincing way why is the energy large.

Solution: Let γL be a curve that is the union of the following three pieces:{(x, 0) : 0 ≤ x ≤ L} ∪ {(x, 2) : 0 ≤ x ≤ L}, {x2 + (y − 1)2 = 1, x ≤ 0}, and{(x− L)2 + (y − 1)2 = 1, x ≥ L}.

This curve is not smooth (only C1- why?) but we can make a tiny pertur-bation so that it becomes smooth. It’s length is 2π+ 2L and

∫γLk2ds = 2π,

which means E(γL) = (2π + 2L)2π. So making L very large we can makeE(γL) very large.

Lecture 7

In what follows Br(p) ⊂ R3 is the open ball of radius r centered at p.

10 ANDRE NEVES

The informal definition of surface is the following. We say that a setS ⊂ R3 is a surface if locally looks just like a disc in R2 meaning that for allp ∈ S we can find r small so that S ∩ Br(p) is diffeomorphic to a disc. Forexamples you can see page 33 of Montiel-Ros book.

The correct definition is

Definition 0.7. A set S is a surface if for all p in S there is an openneighborhood V of p in R3, an open set U ⊂ R2 and a map φ : U → R3

(called chart) so that

i) φ(U) = S ∩ V ;ii) φ is smooth and injective;iii) for all q ∈ U , dφ|q : R2 → R3 is injective. In other words,

∂φ

∂x(q) and

∂φ

∂y(q)

are linearly independent vectors in R3 for all q ∈ U .

Conditions i) and ii) make sense because we want to say that “S∩V lookslike an open set of R2” but one might wonder why condition iii).

If we consider the cone S = {~x ∈ R3 : z =√x2 + y2}, then the map

φ : R2 → R3, φ(x, y) = e−1/(x2+y2)(x, y,√x2 + y2)

is smooth and injective (why?), has S = φ(R2), and thus, without conditioniii) (which fails for q = (0, 0)) the cone S would be a surface. The cone isnot smooth at the origin and so it should not be a surface.

An easy way of producing surfaces is to consider graphs. More precisely,for any smooth function f : R2 → R we can consider

S = graph(f) = {(x, y, f(x, y)) : (x, y) ∈ R2.}Then S is a surface because the map

φ : R2 → R3, φ(x, y) = (x, y, f(x, y))

satisfies φ(R2) = S, φ is smooth and injective, and

∂φ

∂x= (1, 0, ∂xf(x, y)) and

∂φ

∂y= (0, 1, ∂yf(x, y))

are always linearly independent vectors (why?).Another large class of examples comes from considering level sets of func-

tions. Let F : R3 → R be a smooth function and set

S = F−1(0) = {~x ∈ R3 : F (~x) = 0}.Note that S is not immediately a surface. For instance if F (x, y, z) = zy,then F−1(0) = {z = 0} ∪ {y = 0} is the union of two planes intersectingalong a line and hence not a surface. The correct statement is

Proposition 0.8. Assume that ∇F (p) 6= 0 for all p ∈ F−1(0). Then F−1(0)is a surface.

Before we prove this, let’s try to understand it.


Lecture 8

We start with two remarks regarding the last proposition. The first re-mark is that when F (x, y, z) = zy, then

∇F = (∂xF, ∂yF, ∂zF ) = (0, z, y)

and so ∇F (p) = (0, 0, 0) when (0, 0, 0) = p ∈ F−1(0). In other words, thecondition about the gradient in Proposition cannot be removed.

The second remark is that we can now show the sphere

S2 = {~x ∈ R3 : |~x| = 1}

is a surface. Indeed if we set F (x, y, z) = x2 +y2 +z2−1, then S2 = F−1(0).Moreover ∇F = (2x, 2y, 2z) and so for all p ∈ F−1(0) we have |∇F (p)| =2|p| = 2. Hence Proposition 0.8 implies that S2 is a surface.

Note that if we were to do this using the definition of surface we wouldhave use more than one chart and so it would be not so straightforward asin the graphical case (where we can get away with only one chart).

Exercise: Show that S2 (see above) is a surface by finding the appropriatecharts.

To prove the proposition we need to recall the following theorem.

Theorem 0.9 (Implicit function Theorem). Let O be an open set of R3 andL a smooth function in O. Assume that for some p = (x0, y0, z0) ∈ O wehave ∂zL(p) 6= 0. Then we can find a small neighborhood U ⊂ R2 of (x0, y0),a small neighborhood V ⊂ R3 of p, and a smooth function h on U so that

L(x, y, z) = L(p) ⇐⇒ z = h(x, y) for all points (x, y, z) ∈ V.

The content of the theorem is the following. You can solve the equationL(x, y, z) = a for the variable z in a neighborhood V of p (i.e., express z interms of x and y) if ∂zL(p) 6= 0. This condition cannot be removed becauseof examples like L(x, y, z) = z2 − x2 + y2. Here you cannot uniquely solvefor z in terms of x and y near the origin.

Proof of Proposition 0.8. Pick any (x0, y0, z0) = p ∈ F−1(0). Then

∇F (p) = (∂Fx(p), ∂Fy(p), ∂Fz(p)) 6= 0

and so without loss of generality we can assume ∂zF (p) 6= 0. From theimplicit function Theorem we can find U ⊂ R2 a neighborhood of (x0, y0),V ⊂ R3 a neighborhood of p, and a function h defined on U so that

F−1(0) ∩ V = {(x, y, h(x, y)) : (x, y) ∈ U}.

Thus φ : U → R3, φ(x, y) = (x, y, h(x, y)) is the chart we are looking for.The arbitrariness of p implies F−1(0) is a surface. �

Remark: Note that in the previous proof, we showed that F−1(0) is locallythe graph of a function over some coordinate plane.

12 ANDRE NEVES

Exercise:With a > r consider the set

S = {(x, y, z) ∈ R3 : (√x2 + y2 − a)2 + z2 = r2}.

Show that S is a surface and draw it.

Sketch of solution: Make sure Proposition 0.8 can be applied to

F (x, y, z) = (√x2 + y2 − a)2 + z2 − r2.

Lecture 9

For the purpose of computations, it is useful to know that every surfaceS is locally the graph of some function.

Lemma 0.10. Given a surface S and p ∈ S we can find an open neigh-borhood V of p so that S ∩ V can be written as the graph of some functiondefined over one of the coordinate planes.

Proof. It suffices to find an open neighborhood A of p and F : A → R asmooth function so that

S ∩A = F−1(0) and ∇F (q) 6= 0 for all q ∈ F−1(0).

This is because we saw in the proof of Proposition 0.8 that the conditionabove implies the existence of an open neighborhood V of p so that S∩V canbe written as the graph of some function defined over one of the coordinateplanes.

Choose a chart φ : U ⊂ R2 → R3 so that φ(u0, v0) = p. Set

P = span{∂uφ(u0, v0), ∂vφ(u0, v0)}For sure P does not contain one of the coordinate vectors (1, 0, 0), (0, 1, 0),or (0, 0, 1). Say that it does not contain (0, 0, 1).

Set T : U × R→ R3, T (u, v, t) = φ(u, v) + t(0, 0, 1).Then DT(u0,v0,0) is a 3× 3 matrix whose columns are

∂T

∂u= ∂uφ(u0, v0)

∂T

∂v= ∂vφ(u0, v0)

∂T

∂t= (0, 0, 1).

These 3 vectors are linearly independent and thus, from the Inverse FunctionTheorem we have that T−1 is defined on an open set A ⊂ R3 containingT (u0, v0, 0) = p and T−1(A) ⊂ U × R.

Denote the third component of T−1(x, y, z) by F (x, y, z). Then if q ∈ S∩Athen q = φ(u, v) for some (u, v) ∈ U and so T (u, v, 0) = q. Thus (u, v, 0) =T−1 ◦ T (u, v, 0) = T−1(q) and so F (q) = 0. In other words S ∩A = F−1(0).It is easy to see that ∇F (q) 6= 0 for all q ∈ F−1(0) (why?). �

Given a surface S ⊂ R3 and p ∈ S, we define its tangent plane at p as

TpS = {α′(0) : α : [−ε, ε]→ S is a smooth curve with α(0) = p}.The advantage of the definition above is that is does not depend on thecharts. The disadvantage is that is impossible to compute things. Forinstance if we want to compute the tangent plane to S = {z = x2 + y2}


at p = (1, 1, 2) then it seems tricky. Thus the following lemma is clearlyhelpful.

Lemma 0.11. Let φ : U → S ⊂ R3 a chart with φ(x0, y0) = p. Then

TpS = span

{∂φ

∂x(x0, y0),

∂φ

∂y(x0, y0)

}.

In other words, the map dφ(x0,y0) : R2 → TpS is bijective, where

dφ(x0,y0)(a, b) = a∂φ

∂x(x0, y0) + b

∂φ

∂y(x0, y0).

Note that the Jacobian dφ(x0,y0) can be represented by a 3 × 2 matrix,where the first column is ∂xφ and the second column ∂yφ.

With this lemma we can compute the tangent plane to S = {z = x2 +y2}at p = (1, 1, 2). Consider the chart φ(x, y) = (x, y, x2 + y2). Then φ(1, 1) =(1, 1, 2) and so

TpS = span{∂xφ, ∂yφ} = span{(1, 0, 2), (0, 1, 2)} = {(x, y, z) : z−2x−2y = 0}.

Lecture 10

In order to prove Lemma 0.11 we need to clarify some (important) tech-nical issues.

The charts used to define a surface are not unique. For instance if we havea surface S and a chart φ : U ⊂ S ⊂ R3, then for every smooth injective maph ◦ U ′ ⊂ R2 → U with dh|q bijective for all q in U , we have that ψ = φ ◦ his also a chart (why?).

The converse is also true, i.e., if ψ : U ′ → R3 is also a chart and φ(U) ∩ψ(U ′) 6= ∅, then we can find open sets A,A′ such that V = φ(A) = ψ(A′)and so we can consider h = φ−1 ◦ ψ : A′ → A. The map h is called changeof parameters and naturally ψ = φ ◦ h.

There is technical issue that we address now. One needs to argue that his a differentiable map because of the following reason. What we learned incalculus is that if L : O ⊂ Rn → O′ ⊂ Rn is differentiable, injective, andDL|x injective for all x ∈ O, then L−1 is also differentiable. In our case,

the chart φ is a map from R2 into R3 and so the result I just mentionedcannot be applied, which means that φ−1 is continuous but not necessarilydifferentiable. Thus, we cannot say that h is differentiable by saying is thecomposition of two differentiable maps ψ and φ−1.

The correct way of arguing that h is differentiable is the following. Fromthe proof of Lemma 0.10 we know that for every p ∈ V = φ(A) = ψ(A′)there is open neighborhood p ∈ O ⊂ R3 such that S ∩ O = F−1(0), whereF : O → R is a function with ∇F =6= 0 at every point. Without loss ofgenerality assume that ∇F (p) = (0, 0, a) for some a 6= 0 and consider theprojection π : R3 → R2, where π(x, y, z) = (x, y). Finally, set T = π ◦φ andG = π ◦ψ. T is defined on an open set B such that φ(B) = S ∩O and thereis q ∈ B so that p = φ(q).

14 ANDRE NEVES

We argue that dTp is injective. We have F (φ(x, y)) = 0 and so, differen-tiating with respect to x or y we obtain from the chain rule that

∂φ

∂x(q).∇F (p) = 0

∂φ

∂y(q).∇F (p) = 0.

Thus dTq is injective because if dTq(a, b) = 0 then a∂xφ(p) + b∂yφ(q) wouldbe collinear with ∇F (p) (why?) and the identities above show that is im-possible.

As a result, the Inverse Function Theorem implies that T is invertible ina neighborhood of p and that T−1 is smooth. Because h = T−1 ◦G (why?)we obtain that h is a composition of two smooth maps and thus smooth. Italso follows that dhq is injective for all q in its domain (why?).

Proof of Lemma 0.11. Let’s assume that φ is graphical meaning φ(x, y) =(x, y, f(x, y)) for some smooth function f . We show first that

L = span

{∂φ

∂x(x0, y0),

∂φ

∂y(x0, y0)

}⊂ TpS.

Let ~v = a∂xφ(x0, y0)+b∂yφ(x0, y0) for some a, b, and consider α(t) = φ(x0 +at, y0 + bt). Then α(t) ∈ S for all t, α(0) = p, and

α′(0) =d

dt |t=0φ(x0 + at, y0 + bt)

=d

dt(x0 + at)∂xφ(x0, y0) +

d

dt(y0 + bt)∂xφ(x0, y0) = ~v.

Thus, by definition of TpS we have ~v ∈ TpS.Choose α′(0) ∈ TpS. We want to conclude that α′(0) ∈ L. With α(t) =

(x(t), y(t), z(t)), consider γ(t) = (x(t), y(t)). Note that we must have α(t) =φ(γ(t)) (this is where we use the fact that φ is a graphical). In this case wehave from the chain rule that

α′(0) =d(φ ◦ γ)

dt |t=0=dx

dt(0)

∂φ

∂x(x0, y0) +

dy

dt(0)

∂φ

∂y(x0, y0) ∈ L.

If φ is not graphical, we know from Lemma 0.10 that there is a graphi-cal chart ψ. From the discussion we had at the beginning of this section,there is h a map from an open set of R2 into an open set of R2 which is adiffeomorphism (i.e., smooth, injective, with smooth inverse) and such thatφ = ψ ◦ h. In this case the chain rule implies that (why?)

span

{∂ψ

∂x(h(x0, y0)),

∂ψ

∂y(h(x0, y0))

}= span

{∂ψ

∂x(h(x0, y0)),

∂ψ

∂y(h(x0, y0))

}.

Because ψ is graphical we have

span

{∂ψ

∂x(h(x0, y0)),

∂ψ

∂y(h(x0, y0))

}= TpS

and so the lemma is proven. �

Another useful result to compute tangent planes is the following.


Lemma 0.12. Let F : R3 → R so that ∇F 6= 0 on S = F−1(0). Then forany p ∈ S

TpS = {~v : ∇F (p).~v = 0}.

Using this we can compute the tangent plane at every point of S2 ={|~x| = 1}. If F (x, y, z) = x2 + y2 + z2 − 1, then S2 = F−1(0) and so, ifp = (x0, y0, z0) ∈ S2, we have from the previous result that

TpS2 = {~x.∇F (p) = 0} = {~x.(2x0, 2y0, 2z0) = 0} = {~x ∈ R3 : ~x.p = 0}

Proof of Lemma 0.12. The reasoning is the following. The gradient of afunction F points in the direction the function increases the most. Thus ifα′(0) is a tangent vector, it means that F is constant along α and so α′(0)should be orthogonal to the gradient of F .

For the formal argument we set L = {~v : ∇F (p).~v = 0}. It suffices tosee that TpS ⊂ L because dimL = dimTpS = 2 and so equality must hold.Given α′(0) ∈ TpS we then have F ◦ α(t) = 0 because α(t) ⊂ S = F−1(0).Thus from the chain rule we have

0 =d(F ◦ α)

dt t=0= x′(0)

∂F

∂x(p) + y′(0)

∂F

∂y(p) + z′(0)

∂F

∂z(p) = ~v.∇F (p)

and so ~v ∈ L. �

Lecture 11

There is one last set of definitions we need to get in order. Given a surfaceS1 and a continuous map F : S1 → R3, we say that F is smooth if for everychart φ : U → S we have that F ◦φ is smooth. Note that if φ and ψ are twocharts with the same image, then φ = ψ ◦ h for some diffeomorphism h andthus, because h has a smooth inverse, we have that F ◦ φ is smooth if andonly if F ◦ ψ is smooth (why?).

Given a smooth map F : S → R3 we define its Jacobian at p ∈ S as thelinear map

dF|p : TpS → R3, dF|p(α′(0)) =

d(F ◦ α)

dt(0).

We need to make sure that the definition is consistent, which in this casemeans that it does not depend on the curve α, only on α′(0). Indeed, pickp ∈ S and, using Lemma 0.10, pick a chart φ which is graphical, i.e., hasthe form φ(x, y) = (x, y, f(x, y)) for some smooth function f . Pick α1,α2,two curves in S such that α1(0) = α2(0) = p and α′1(0) = α′2(0). Thus,with αi(t) = (xi(t), yi(t), zi(t)), i=1,2, we have that αi(t) = φ(xi(t), yi(t)).Moreover we obtain from the chain rule that

d(F ◦ αi)dt

(0) =d(F ◦ φ(xi(t), yi(t)))

dt(0)

= x′i(0)∂(F ◦ φ)

∂x(xi(0), yi(0)) + y′i(0)

∂(F ◦ φ)

∂y(xi(0), yi(0)).

16 ANDRE NEVES

The fact that α′1(0) = α′2(0) implies that (x′1(0), y′1(0)) = (x′2(0), y′2(0)) andhence

d(F ◦ α1)

dt(0) =

d(F ◦ α2)

dt(0).

Thus the definition is consistent.This definition is nice and elegant because does not use charts but for the

purpose of computations is not very useful. Note that from Lemma 0.11 weknow that dφ(x0,y0) is map from R2 to TpS which is bijective.

Lemma 0.13. If φ : U → S is a chart with φ(x0, y0) = p, then

dF|p(∂xφ(x0, y0)) =∂(F ◦ φ)

∂x(x0, y0), dF|p(∂yφ(x0, y0)) =

∂(F ◦ φ)

∂y(x0, y0).

In other words d(F ◦ φ)|(x0,y0) = dF|pdφ|(x0,y0).

Proof. Consider α(t) = φ(x0 + t, y0). Then

dF|p(∂xφ(x0, y0)) = dF|p(α′(0)) =

d(F ◦ α)

dt(0) =

d(F ◦ φ)

dt |t=0(x0 + t, y0)

=∂(F ◦ φ)

∂x(x0, y0).

The other derivative is computed similarly. �

Lecture 12

For instance, say that S = {z = x2 + y2} and F : S → R3 given byF (x, y, z) = (cos(πz), xz, y + z2). We already saw that with p = (1, 1, 2)then TpS = {z = 2x + 2y} = span{(1, 0, 2), (0, 1, 2)}. Let’s computedF|p(2,−1, 2).

We first compute d(F ◦ φ)|(1,1), where φ(x, y) = (x, y, x2 + y2). Then

F ◦ φ(x, y) = (cos(π(x2 + y2)), x(x2 + y2), y + (x2 + y2)2)

and we obtain from the previous lemma

d(F ◦ φ)|(1,1)(1, 0) = ∂x(F ◦ φ)(1, 1) = (0, 4, 8)

and

d(F ◦ φ)|(1,1)(0, 1) = ∂y(F ◦ φ)(1, 1) = (0, 2, 9).

Thus d(F ◦ φ)|(1,1) is a 3 × 2 matrix, where the first column is (0, 4, 8) andthe second column (0, 2, 9). Now we noticed that (2,−1, 2) = 2(1, 0, 2) −(0, 1, 2) = 2∂xφ(1, 1)− ∂yφ(1, 1) and so

dF|p(2,−1, 2) = d(F ◦ φ)|(1,1)(2,−1) = (0, 6, 7).

Exercise: Compute dF|p(2, 1, 2) using the chart ψ(x, z) = (x,√z − x2, z)

instead of φ given above.

One important remark is that if S1, S2 are surfaces and F : S1 → R3 is asmooth map with F (S1) ⊂ S2, in which case we simply write F : S1 → S2.


From the definition of dF|p, we have that this linear map takes values intoTF (p)S2, i.e., dF|p : TpS1 → TF (p)S2.

We now explain how to compute the normal vector N to a surface S ata point p ∈ S. We have already seen that TpS is a plane and so N(p) is bydefinition a unit normal vector perpendicular to TpS. In other words, N(p)is a unit vector such that ~v.N(p) = 0 for all ~v ∈ TpS.

If S = F−1(0) is a surface and ∇F (p) 6= 0 for all p ∈ S, then we alreadysaw in Lemma 0.12 that ∇F (p) is perpendicular to TpS and so we can take

N(p) =∇F (p)

|∇F (p)|.

If φ : U → S is a chart with φ(x0, y0) = p, then we know from Lemma0.11 that ∂xφ(x0, y0), ∂xφ(x0, y0) span TpS and so we can determine thenormal vector as

N(p) =∂xφ× ∂yφ|∂xφ× ∂yφ|

(x0, y0).

Let’s compute some examples.If S is a plane, for instance S = {z = 0}, then N(p) = (0, 0, 1) for all

p ∈ S.If S2 = {|~x| = 1}, we have that S2 = F−1(0) for F (x, y, z) = x2+y2+z2−1

and so

N(p) =∇F|∇F |

(p) =2p

|2p|= p.

If S = {x2 + y2 = 1} (infinite circle of radius one) then we have thatS = F−1(0), where F (x, y, z) = x2 + y2 − 1 and so

N(p) =∇F|∇F |

=(2x, 2y, 0)

|(2x, 2y, 0)|=

(x, y, 0)√x2 + y2

= (x, y, 0).

Finally, , when S = {z = x2 + y2} and p = (1, 1, 2) then using the chartφ(x, y) = (x, y, x2 + y2) we have


(1, 1) =(1, 0, 2)× (0, 1, 2)

|(1, 0, 2)× (0, 1, 2)|

=(−2,−2, 1)

|(−2,−2, 1)|= (−2/3,−2/3, 1/3).

Lecture 13

Given a surface S and p ∈ S, there is always an ambiguity of choosingN(p) or −N(p). When we can make such a choice unambiguously, we say thesurface is orientable. In these cases we have a continuous map N : S → R3.As we saw in the class there are surfaces which are non-orientable (Mobiusstrip).

Exercise: Show that if the map N : S → R3 is continuous then it is actuallysmooth.

18 ANDRE NEVES

Solution: Choose p ∈ S and a chart φ : U → R3 with φ(q) = p. Then


(q) or N(p) =∂yφ× ∂yφ|∂xφ× ∂yφ|

(q).

Say the second case case occurs. Then for each q′ nearby q, because N iscontinuous, we must necessarily have

N(φ(q′)) =∂yφ× ∂yφ|∂xφ× ∂yφ|

(q′).

The expression on the right-hand side is a smooth function of q′ and so N ◦φis smooth.

Given an orientable surface S, the idea is to study the surface by lookinginto the map N : S → R3. This map is called the Gauss map.

Notice that N : S → S2 = {|~x| = 1} ⊂ R3 because N(p) is alwaysa unit vector. Let’s compute it in some simple examples but note that ifN : S → S2 is a Gauss map, then p 7→ −N(p) is also valid choice for theGauss map.

If S = {~x.v = 0} ⊂ R3 is a plane then N(p) = v for all p ∈ S and so theGauss map is constant.

If S = S2 we already saw that N(p) = p for all p ∈ S2 and so the Gaussmap N : S2 → S2 is the identity map in this case.

If S = {x2 + y2 = 1}, we already saw that N : S → S2 is given byN(x, y, z) = (x, y, 0).

If S = {z = x2 + y2}, then for the chart φ(x, y) = (x, y, x2 + y2) we have

∂φ

∂x× ∂φ

∂y= (1, 0, 2x)× (0, 1, 2y) = (−2x,−2y, 1)

and so the Gauss map is given by

N(x, y, z) =(−2x,−2y, 1)√1 + 4x2 + 4y2

=(−2x,−2y, 1)√

1 + 4z

for all (x, y, z) ∈ S.A crucial property of the Gauss map is that the Jacobian dN|p is a map

from TpS to TpS. Indeed dN|p : TpS → TN(p)S2 but TqS

2 = {~x : ~x.q = 0}for all q ∈ S2 and so

TN(p)S2 = {~x : ~x.N(p) = 0} = TpS,

i.e., dN|p : TpS → TpS.

Lecture 14

Let’s compute the Jacobian of the Gauss map for simple examples.If S = {~x : ~x.q = 0} is a plane, then N(x) = q|q|−1 all x and so dN|x(~v) =

0 for all ~v ∈ TxS.


If S = {|~x| = r} is the unit sphere of radius r then we have N : S → S2

given by N(p) = p/r. Thus

dN|p(α′(0)) =

d(N(α(t)))

dt(0) =

1

r

d(α(t))

dt(0) =

α′(0)

r,

i.e., dN|p = 1r Id.

If S = {x2 + y2 = r2}, then Gauss map N : S → S2 is given by

N(x, y, z) =(x, y, 0)

r.

To compute dN(x,y,z) : T(x,y,z)S → T(x,y,z)S we note that, because the mapis linear, we only need to determine it on a basis of T(x,y,z)S. If we sete1 = (0, 0, 1) and e2 = (−y, x, 0)/r then we have

T(x,y,z)S = span{e1, e2}.The fastest way to see this is that e1 and e2 are both orthogonal to N(x, y, z).Moreover, {e1, e2} is an orthonormal basis for the tangent plane.

We need to compute dN|(x,y,z)(e1) and dN|(x,y,z)(e2). We have

dN|(x,y,z)(e1) = dN|(x,y,z)((x, y, z + t)′) =dN(x, y, z + t)

dt(0) = 0.

We know that (x, y) = (r cos θ0, y sin θ0) for some θ0 and so, setting α(t) =(r cos(θ0 + t/r), r sin(θ0 + t/r), z), we obtain

dN|(x,y,z)(e2) = dN|(x,y,z)(α′(0)) =

dN(r cos(θ0 + t/r), r sin(θ0 + t/r), z)

dt(0)

=1

r(− cos θ0, sin θ0, 0) =

1

r(−y/r, x/r, 0) =

1

re2.

The goal now is to study dN|p or equivalently the quadratic form

A : TpS × TpS → R, A(X,Y ) = −X.dN|p(Y ).

It is clear that knowing dN implies that we can compute A. The converseis also true because if e1, e2 is an orthonormal basis for TpS then

dN|p(X) = (dN|p(X).e1)e1 + (dN|p(X).e2)e2 = −A(e1, X)e1 −A(e2, X)e2.

A is called the second fundamental form.

Lecture 15

Exercise: Show that A is bilinear, i.e., A(aX+bY, Z) = aA(X,Z)+bA(Y,Z)and A(X, aY + bZ) = aA(X,Z) + bA(X,Y ) for all X,Y, Z ∈ TpS anda, b ∈ R.

Solution: We have from linearity of dN|p that

A(X, aY + bZ) = −X.dN|p(aY + bZ) = −X.dN|p(aY )−X.dN|p(bZ)

= −aX.dN|p(Y )− bX.d|pN(Z) = aA(X,Y ) + bA(X,Z).

20 ANDRE NEVES

To have a first idea of what A computes, consider a curve α : I → S withα(0) = p. We have N(α(t)).α′(t) = 0 for all t and so, from the definition ofJacobian we have

A(α′(0), α′(0)) = −dNp(α′(0)).α′(0) = −d(N ◦ α)

dt(0).α′(0)

= − d

dt |t=0(N ◦ α(t).α′(t)) +N(p).α′′(0) = N(p).α′′(0).

If |α′(0)| = 1, then A(α′(0), α′(0)) = N(p).~k is the the projection of thegeodesic curvature of α on N . In other words, A(α′(0), α′(0)) computeshow much the curve α is curving because is lying inside S. If P is theplane spanned by N(p) and α′(0) that passes through the point p, thenA(α′(0), α′(0)) is the curvature of the planar curve S ∩ P ⊂ P at p.

Let S = {z = y2−z2} (a saddle) and consider the Gauss map N : S → S2

with N(0, 0, 0) = (0, 0, 1). Then if e1 = (1, 0, 0) and e2 = (0, 1, 0) we havefrom the interpretation given at the end of last lecture that Ap(e1, e1) < 0and Ap(e2, e2) > 0, where p = (0, 0, 0). Drawing a picture, this should beself-evident.

More insight for A comes from the following lemma.

Lemma 0.14. Consider a surface S and a chart φ : R2 → S We have fori, j = 1, 2

Aij := A(∂xiφ, ∂xjφ) = (∂xjxiφ).N ◦ φ.Proof. From Lemma 4.1 of Lecture 9 we have for i = 1, 2,

dN(∂xiφ) =∂(N ◦ φ)

∂xiand thus, for i, j = 1, 2,

Aij = A(∂xiφ, ∂xjφ) = −∂xiφ.dN(∂xjφ) = −∂xiφ.∂(N ◦ φ)

∂xi

=∂

∂xi(∂xiφ.N ◦ φ) + (∂xjxiφ).N ◦ φ = (∂xjxiφ).N ◦ φ.

�

From the identity above we see that in the same way that the tangentplane is the “derivative” of a surface, then A is the “Hessian” of a surface.

For instance, if S = {z = f(x, y)} is graphical then we can take the chartφ(x1, x2) = (x1, x2, f(x1, x2)) and so, using

∂x1φ× ∂x2φ = (1, 0, ∂x1f)× (0, 1, ∂x2f) = (−∂x1f,−∂x2f, 1)

we obtain

N ◦ φ(x1, x2) = (−∂x1f,−∂x2f, 1)/√

1 + |∇f |2.Finally we have

∂2φ

∂xi∂xj=

(0, 0,

∂2f

∂xi∂xj

)


and thus we have from Lemma 0.14 that

Aij =∂2xixjf√

1 + |∇f |2, i, j = 1, 2.

If S = {z = y2 − x2} then at p = (0, 0, 0) we have

A11 = −2, A12 = A21 = A11 = 0, A22 = 2.

This agrees with the initial discussion we had regarding the second funda-mental form of the saddle S. If we choose X = (2, 0, 0), Y = (1,−1, 0) bothin TpS, then because

∂x1φ(0, 0) = (1, 0, 0) and ∂x2φ(0, 0) = (0, 1, 0)

we have

A(X,Y ) = A(2∂x1φ, ∂x1φ− ∂x2φ) = 2A11 − 2A12 = −4.

Lecture 16

Given a surface S the goal is to study the second fundamental form A :TpS × TpS → R.

One key fact is the following.

Lemma 0.15. A is symmetric, i.e., A(X1, X2) = A(X2, X1) for all X1, X2 ∈TpS.

Proof. Pick a chart φ : U → S with φ(q) = p. From Lemma 0.14 we seethat Aij = Aij , which means the matrix (Aij)i,j=1,2 is symmetric. Thus ifwe write Xi = α1i∂x1φ+ α2i∂x2φ, with i = 1, 2 we have from the bilinearityof A that

A(X1, X2) =2∑

i,j=1

αi1αj2A(∂xiφ, ∂xjφ) =2∑

i,j=1

αi1αj2Aij

=2∑

i,j=1

αi1αj2Aji =

2∑i,j=1

αi1αj2A(∂xjφ, ∂xiφ) = A(X2, X1).

�

It is a standard fact of Linear Algebra that bilinear forms on vector spacesare diagonalizable. More precisely, we can find an orthonormal basis {e1, e2}of TpS so that A(ei, ei) = λi, i = 1, 2 and A(e1, e2) = A(e2, e1) = 0. Theeigenvalues λ1, λ2 are called the principal curvatures and the eigenvectorse1, e2 are called the principal directions.

Note that this is equivalent to say there there is an orthonormal basis{e1, e2} of TpS so that

dN|p(e1) = −λ1e1 and dN|p(e2) = −λ2e2.

22 ANDRE NEVES

Indeed, if e1, e2 are principal directions and λ1, λ2 principal curvatures, thenbecause e1, e2 is an orthonormal basis we have

dN|p(ei) = (dN|p(ei).e1)e1 + (dN|p(ei).e2)e2

= −A(ei, e1)e1 −A(ei, e2)e2 = −λiei, i = 1, 2.

Conversely, if dN|p(ei) = −λie1 for some orthonormal basis {e1, e2}, thenA(ei, ej) = −dN|p(ei).ej = λi(ei.ej), i, j = 1, 2, and so e1, e2 are principaldirections and λ1, λ2 principal curvatures.Exercise: If S is a surface, show that

λ1(p) = min{A(X,X) : X ∈ TpS, |X| = 1},

λ2(p) = max{A(X,X) : X ∈ TpS, |X| = 1}.In particular, λ1 and λ2 have a definition which is independent of any chart.

Let’s work out our basic examples again.If S is a plane then the principal curvatures are λ1 = λ2 = 0.If S is a sphere of radius r then

dN|p =1

rId =⇒ A(X,Y ) = −X.Y

rfor all X,Y ∈ TpS.

Thus the principal curvatures are λ1 = λ2 = −1r and every unit vector in

TpS is a principal direction. Convention dictates that we like the principalcurvatures of the sphere to be positive and so for Gauss map we should takeN(p) = −p/r (the interior pointing normal), which then makes A(X,Y ) =X.Yr and so we obtain λ1 = λ2 = 1

r .

If S = {x2 + y2 = r2}, we already saw that if

e1 = (0, 0, 1) and e2 = (−y/r, x/r, 0)

then, with Gauss map N(x, y, z) = (x/r, y/r, 0), we have dN(e1) = 0 anddN(e2) = e2

r , which means the principal curvatures are λ1 = 0 and λ2 =−1/r, and e1, e2 are the principal directions. Again, the convention is thatwe should choose the interior pointing unit normal for the Gauss map, i.e.N(x, y, z) = −(x/r, y/r, 0) and so we obtain principal curvatures λ1 = 0 andλ2 = 1/r.Exercise:If S is a connected surface where λ1(p) = λ2(p) = 0 all p ∈ S,show that S is a subset of a plane.

Lecture 17

We say that p ∈ S is a umbilical point is λ1(p) = λ2(p), i.e., dN|p(X) =λ(p)X for all X ∈ TpS.

Theorem 0.16. Say that S is a connected surface so that every point isumbilical. Then S is contained in a plane or sphere.

In particular, if S has no boundary and is compact, then it must be asphere.


Note that in Lecture 4 we saw an exercise where where we showed thatthose curves with curvature k = 1/r are contained in a circle of radius r.The theorem above is the analogue of this statement for surfaces.

Proof. The first thing to show is that λ1(p) = λ2(p) = λ0 for all p ∈ S, i.e.,the principal curvatures are constant.

If we set λ(p) := λ1(p) = λ2(p), then dN(X) = −λ(p)X all X ∈ TpS andso for i = 1, 2

∂(N ◦ φ)

∂xi= dN(

∂φ

∂xi) = −(λ ◦ φ)∂xiφ.

Differentiating again, we have i, j = 1, 2

∂2(N ◦ φ)

∂xj∂xi= −∂xj (λ ◦ φ)

∂φ

∂xi+ (λ ◦ φ)

∂2φ

∂xj∂xi

and thus, because∂2(N ◦ φ)

∂xi∂xj=∂2(N ◦ φ)

∂xj∂xiwe obtain that

∂xj (λ ◦ φ)∂φ

∂xi= ∂xi(λ ◦ φ)

∂φ

∂xj.

Now ∂x1φ, ∂x2φ are linearly independent vectors and so the only way theidentity above can hold is if

∂x1(λ ◦ φ) = ∂x2(λ ◦ φ) = 0 =⇒ λ ◦ φ = constant.

This implies that λ(p) = λ0 for all p ∈ S.If λ0 = 0 then using an exercise from the previous lecture we see that

S has to be contained in a plane. If λ0 6= 0 we can assume λ0 is positive(why?) and so we have to show that S is contained in a sphere of radius1/λ0.

With φ : U → S ⊂ R3 a chart we know that ∂xi(N ◦φ) +λ−10 ∂xiφ = 0 for

i = 1, 2 and so φ+ λ−10 N ◦ φ is a constant c0. Hence ||φ− c0|| = λ−1

0 , which

means φ(U) is contained in a sphere of center c0 and radius λ−10 . �

We can now define the two most important concepts of this course. Givena surface S, we define its Gaussian curvature K to be

K(p) = det dN|p = λ1(p)λ2(p)

and its mean curvature H to be

H(p) = −1

2tr dN|p =

λ1(p) + λ2(p)

2.

Because K and H are functions, they are much easier to understand thanthe second fundamental form A and already contain a lot of informationabout the surface. The rest of the course is to understand what K and Hsay about a surface.

Let’s compute this quantities when S = {z = (y2 − x2)/2}. In order todo that, we need to do a bit of general theory.

24 ANDRE NEVES

Lemma 0.17. Given a chart φ : U ⊂ R2 → S, consider the matrix

g = (gij)i,j=1,2, where gij =∂φ

∂xi.∂φ

∂xj

and the matrix A = (Aij)i,j=1,2 (see Lemma 0.14).Then, with σ = g−1A a 2× 2 matrix, we have

K = detσ =detA

detgH =

trσ

2=σ11 + σ22

2.

Note that a standard formula shows that

g−111 =

1

det gg22, g−1

12 = g−121 = − 1

det gg12, g−1

22 =1

det gg11.

We use this lemma to compute K and H for S = {z = (y2− x2)/2}. Thestrategy is to find a chart φ , compute A = (Aij)i,j=1,2, g = (gij)i,j=1,2, and

g−1 = (g−1ij )i,j=1,2. Then we compute the matrix product σ = g−1A and we

take the trace and determinant of this matrix.We have

φ(x1, x2) = (x1, x2, (x22 − x2

1)/2)

and so

g11 = 1 + x21, g12 = −x1x2, g22 = 1 + x2

2,

g−111 =

1 + x22

1 + x21 + x2

2

, g−112 = g−1

21 =x1x2

1 + x21 + x2

2

, g−122 =

1 + x21

1 + x21 + x2

2

,

N =∂x1φ× ∂x2φ|∂x1φ× ∂x2φ|

=(x1,−x2, 1)√1 + x2

1 + x22

and

A11 = − 1√1 + x2

1 + x22

, A12 = 0, A22 =1√

1 + x21 + x2

2

.

Thus

σ11 = − 1 + x22

(1 + x21 + x2

2)3/2, σ12 =

x1x2

(1 + x21 + x2

2)3/2,

σ21 = − x1x2

(1 + x21 + x2

2)3/2, σ22 =

1 + x21

(1 + x21 + x2

2)3/2.

As a result we obtain

K = σ11σ22 − σ12σ21 = − 1 + x21 + x2

2

(1 + x21 + x2

2)3= − 1

(1 + x21 + x2

2)2

and

H =1

2(σ11 + σ22) =

x21 − x2

2

(1 + x21 + x2

2)3.


Lecture 18

Let’s prove Lemma 0.17.

Proof. Suppose we have a chart φ : U ⊂ R2 → S. Denote by σ = (σij)i,j=1,2

the matrix that represents −dN : TpS → TpS with respect to the basis∂x1φ, ∂x2φ. In other words, if

X = a1∂φ

∂x1+ a2

∂φ

∂x2and dN|p(X) = −b1

∂φ

∂x1− b2

∂φ

∂x2,

then (b1, b2) = σ(a1, a2)T . In particular

−dN|p(∂φ

∂xi) =

2∑k=1

σki∂φ

∂xk.

K is the determinant of σ and H is half the trace of σ.We need to find σ in terms of the matrix A = (Aij)i,j=1,2 and the matrix

g = (gij)i,j=1,2. From the symmetry of A and g we have

Aij = Aji = (−dN(∂φ

∂xj)).

∂φ

∂xi= (

2∑k=1

σkj∂φ

∂xk).∂φ

∂xi=

2∑k=1

σkj∂φ

∂xk.∂φ

∂xi

=

2∑k=1

σkjgki =

2∑k=1

gikσkj .

and so, using matrix notation,

A = g.σ =⇒ σ = g−1A.

σ is the matrix for which we need to compute the trace and the determinant.�

The next proposition gives a local picture of what is means for a surfaceto have positive and negative Gaussian curvature at the local level.

Proposition 0.18. Say that S is a surface with K(p) > 0. Then near p,S is all to one side of TpS, i.e., if U is a small neighborhood of p thenS ∩ U \ {p} does not intersect p+ TpS.

Say that S is a surface with K(p) < 0. Then near p, S is in both sidesof TpS, i.e., if U is a small neighborhood of p then S ∩ U \ {p} intersectsp+ TpS.

This proposition explains why K < 0 on the surface S = {z = (y2 −x2)/2}.

The strategy is the following. Given p ∈ S we show that, nearby thatpoint, the surface is the graph of a function f which is defined over thetangent plane TpS and, moreover, Hess(f) coincides with the matrix σ atp. Thus if K(p) > 0, this means det σ = detHess(f) > 0 and hence f has alocal minimum or a local max. In both cases, S nearby p is above (or below)its tangent plane. If K(p) < 0 then det σ = detHess(f) < 0, which means

26 ANDRE NEVES

f has a saddle point, in which case the surface has points above and belowthe tangent plane.

Let’s start by recalling the second derivative test.

Theorem 0.19. Let f be a function defined on the real plane with

f(0, 0) = 0, ∇f(0, 0) = (0, 0).

If detHess(f)(0) > 0, then the origin is either a local maximum or a localminimum. If detHess(f)(0) < 0, then the origin is neither a local maximumnor a local minimum (i.e., it is a saddle point).

Proof. From Taylor’s formula we know that

f(x, y) = f(0, 0) +∇f(0).(x, y) +1

2(x, y)Hess(f)(0)(x, y)T +O(x3 + y3),

where O(x3 + y3) stands for a term whose absolute value is bounded fromabove by C(x3 + y3) for some constant C.

Because Hess(f) is symmetric matrix, it means that is an orthonormalbasis {e1, e2} which diagonalizes Hess(f) at (0, 0). Let’s represent the func-

tion f with respect to this new basis, i.e., set f(u, v) = f(ue1 + ve2). Then

Hess(f)(0, 0) becomes a diagonal matrix with diagonal entries λ1, λ2, the

eigenvalues of Hess(f)(0, 0). The Taylor formula for f becomes

f(u, v) = f(0, 0) +∇f(0).(uv) +1

2(u2λ1 + v2λ2) +O(u3 + v3).

Now f(0) = (0, 0) (why?) and hence

f(u, v) =1

2(u2λ1 + v2λ2) +O(u3 + v3).

It should be clear now that if λ1λ2 > 0 then f (and hence f) has a local

max. or a local min. at the origin. if λ1λ2 < 0 then f (and hence f) has asaddle point at the origin. �

Proof of Proposition 0.18. Let’s prove the Proposition in the following par-ticular case first.

Assume p is the origin, TpS is the xy-plane, and the surface S is graphicalwith respect to this plane near the origin. What this means is that one canfind a coordinate chart of the form φ(x1, x2) = (x1, x2, f(x1, x2)), where fis a smooth function and defined in a neighborhood of the origin.

Because p is the origin we must have f(0, 0) = 0. Because TpS is thexy-plane, we must have N(p) = (0, 0, 1) and so ∇f(0, 0) = 0. In particular∂x1φ(0, 0) = (1, 0, 0) and ∂x2φ(0, 0) = (0, 1, 0).

This means the matrix g = (gij)i,j=1,2 is the identity at the origin andthus, using the computation we did for Aij in Lecture 13 for the graphicalcase, we have at the origin

σij = Aij = ∂2xixjf.


In other words, the matrix σ at the origin is exactly the hessian of f andthus K(p) = detHess(f)(0, 0).

Now we have that (0, 0) is a critical point for f and so the second de-rivative test says that if detHess(f)(0, 0) > 0 then the origin is strict lo-cal minimum or maximum. This proves the first statement because, nearthe origin, f(x1, x2) must be either strictly positive or strictly negative if(x1, x2) 6= (0, 0).

If detHess(f)(0, 0) < 0 the second derivative test says that (0, 0) is asaddle point (i.e., neither maximum nor minimum), and so there are pointsnear the origin where f(x1, x2) < 0 and points where f(x1, x2) > 0. Thisproves the second statement.

In Lemma 0.10 we saw that S is graphical over one of the coordinateplanes near p but what we actually proved was that S is graphical near pover any plane which does not contain N(p), the normal vector to TpS. Inparticular, S is graphical over TpS is a neighborhood of p. Without loss ofgenerality we can assume p is the origin. �

Lecture 19

Recall that compact sets of R3 are those that are closed and bounded.The next proposition says that compact surfaces must always have pointswhere the curvature is strictly positive.

Note that the example S = {z = (y2 − x2)/2} shows the condition thatthe surface is compact is crucial.

Proposition 0.20. If S is a compact surface, then there is a point p in Swhere K(p) > 0.

The condition that S is bounded is crucial because S = {z = (y2−x2)/2}has negative curvature everywhere. The condition that S is closed is crucialbecause S = {x2 + y2 < 1, z = 0} has curvature identical to zero. Theexample S = {x2 + y2 ≤ 1, z = 0} does not count because it is not a surfaceaccording to our definition (it has boundary). Note that if one were to definewhat is a surface with boundary, as to include the previous example, thecorrect statement in the theorem would be that “compact surfaces with noboundary are positively curved somewhere”.

Proof. S is bounded and so it is contained inside a large ball of radius R.Thus S ∩ {|~x| = R} = ∅. Choose

r0 = sup{r : S ∩ {|~x| = r} 6= ∅}.Denote by S0 = {|~x| = r0}.

The fact that S is closed implies that S ∩ S0 6= ∅. Indeed, if S ∩ S0 = ∅,the fact that S and S0 are closed and bounded implies that

inf{|x− y| : x ∈ S, y ∈ S0} > 0.

Thus we could decrease r0 a tiny bit and still have S∩S0 = ∅, a contradictionwith the definition of r0.

28 ANDRE NEVES

Choose p ∈ S∩S0. We must have TpS = TpS0 because otherwise we couldincrease r0 a bit and still have S ∩ S0 6= ∅, a contradiction. The rigorousway of checking this is the following.

For every path γ in S so that γ(0) = p, consider f(t) = |γ(t)|2. Thenf(t) ≤ r2

0 = f(0) for all t because S is inside S0 and p belongs to S0

(and so |p| = r0). Thus f has a local maximum at the origin and thisimplies f ′(0) = 0. On the other hand, from the definition of f we see thatf ′(0) = 2γ′(0).γ(0) = 2γ′(0).p. As a result we obtain that γ′(0) is orthogonalto p. The arbitrariness of γ implies that p is perpendicular vector to TpSand so TpS = TpS0.

Let’s also assume, without loss of generality, that p is the origin, TpS isthe xy-plane and that the unit normal interior to S0 is N(p) = (0, 0, 1). Bythe same reasoning as in the proof of Proposition 0.18 we have that S, S0 aregraphical near p over the xy-plane, which means there is in a neighborhoodof the origin so that S = {z = f(x, y)} and S0 = {z = f0(x, y)}, where fand f0 must satisfy

f(0, 0) = f0(0, 0) = 0 and ∇f(0, 0) = ∇f0(0, 0) = 0.

We also saw that the principal curvatures of S (or S0) at the origin must bethe eigenvalues of Hess(f) (or Hess(f0)) at the origin. In particular,

Hess(f0)(v, v) =|v|2

r0for every vector v ∈ R2,

because both principal curvatures of S0 are the same and equal to r−10 .

Because the interior unit normal to S0 is (0, 0, 1) we have that f0 ≤ fnear the origin or, equivalently, that h = f − f0 has a local minimum at theorigin. This is because S must be contained in the interior of S0. Hence,Hess(h) ≥ 0 at the origin, i.e., for every vector v we must have

Hess(h)(v, v) ≥ 0 =⇒ Hess(f)(v, v) ≥ Hess(f0)(v, v) =|v|2

r0.

Thus, the principal curvatures of S at the origin must be higher than 1/r0,and so K(p) ≥ r−2

0 > 0. �

Surfaces with H = 0 are called minimal surfaces and are ubiquitous inscience. They model from soap films to black holes and the previous propo-sition implies that there are no compact minimal surfaces.

Corollary 0.21. There are no compact minimal surfaces S in R3.

Proof. If S has H = 0, then λ1 = −λ2 and so K = −λ21 ≤ 0. By the previous

proposition, S cannot be compact. �

Lecture 20

Given two surfaces S1, S2 we say that a smooth map F : S1 → S2 is alocal isometry if

dFp(X).dFp(Y ) = X.Y for all p ∈ S1, X, Y ∈ TpS1.


In particular, |dFp(X)| = |X| for all p and X ∈ TpS.F being a local isometry also implies that dFp : TpS1 → TF (p)S2 is bijec-

tive (why?). Hence, from the Inverse Function Theorem, we obtain that alocal isometry is also a local diffeomorphism meaning that for every p ∈ S1,there are an opens set U, V ⊂ R3 containing p and F (p) respectively, and asmooth map G : S2 ∩ V → S1 ∩U so that F ◦G and G ◦ F are the identity.

When F is also bijective we say that F is an isometry.The map F being a local isometry means that it preserves the length of

curves.

Lemma 0.22. Let α : I → S1 be a curve. Then if F : S1 → S2 is a localisometry we have

length(α) = length(F ◦ α).

Conversely, if F preserves the lengths of any given curve, then F is alocal isometry.

Proof. If β(t) = F (α(t)), then from the definition of Jacobian we have b′(t) =dFα(t)(α

′(t)). Thus

length(F ◦ α) =

∫I|β′(t)|dt =

∫I|dF (α′(t))|dt =

∫I|α′(t)|dt = length(α).

Likewise, given p ∈ S and α′(0) ∈ TpS we have for all t small∫ t

0|α′(s)|ds = length(α[0,t]) = length(F (α[0,t])) =

∫ t

0|dF (α′(s))|ds.

Differentiating both sides with respect to t we obtain |α′(0)| = |dF (α′(0))|which means |dFp(V )| = |V | for all p ∈ S1 and V ∈ TpS1.

To obtain that dF (X).dF (Y ) = X.Y for all X,Y ∈ TpS is now simpletrick. We have

(1) dF (X − Y ).dF (X − Y ) = |X − Y |2

=⇒ |dF (X)|2 − 2dF (X).dF (Y ) + |dF (Y )|2 = |X|2 − 2X.Y + |Y |2

=⇒ dF (X).dF (Y ) = X.Y

�

Let’s work out some examples.Let B be a 3×3 matrix that lies in SO(3), i.e., B.BT = BT .B = Id. The

linear map x 7→ Bx corresponds to a rigid motion which fixes the origin,i.e., a reflection or rotation.

Given a surface S, we can consider the surface S1 = B(S), i.e., the surfaceyou get from applying the rigid motion to S. Naturally we expect S1 to beisometric to S because distances are preserved and that is indeed the case.

Consider the map F : S → S1, F (x) = Bx. The map F is bijective(why?). Given p ∈ S and X ∈ TpS we have dF (X) = BX and so

dF (X).dF (Y ) = (BX).(BY ) = (BX)T (BY ) = XTBTBY = XTY = X.Y,

30 ANDRE NEVES

where UT denotes the transpose of the matrix (or vector) U .

Lecture 21

We now argue that the plane S1 = {(x, y, 0) is locally isometric to acylinder S2 = {(x, y, z) : x2 + y2 = 1}. Hence, even if their shapes aredifferent, their intrinsic distances are the same.

Consider F : S1 → F (S1) ⊂ S2 given by F (x, y, 0) = (cosx, sinx, y). Fis clearly smooth. Set e1 = (1, 0, 0) and e2 = (0, 1, 0). We have TpS1 =span{e1, e2} for all p ∈ S1 and

dF (e1) =∂F

∂x= (− sinx, cosx, 0),

dF (e2) =∂F

∂y= (0, 0, 1).

Thus

dF (e1).dF (e1) = 1 = e1.e1, dF (e2).dF (e2) = 1 = e2.e2,

dF (e1).dF (e2) = 0 = e1.e2

and so it is straightforward to check that dF (X).dF (Y ) = X.Y for allX,Y ∈TpS1.

The next lemma says that two surfaces are isometric if we can find chartsso that the matrix g = (gij)i,j=1,2 (defined in Lecture 15) are the same forboth charts. The matrix g is called the metric.

Lemma 0.23. Consider S1, S2 surfaces and F : S1 → S2 an isometry.If φ : U → S1 is a chart, then ψ = F ◦ φ is a chart of S2 and

gij :=∂φ

∂xi.∂φ

∂xj=∂ψ

∂xi.∂ψ

∂xjon U.

Conversely, if φ : U → S1, ψ : U → S2 are charts so that

∂φ

∂xi.∂φ

∂xj=∂ψ

∂xi.∂ψ

∂xjon U for all i, j = 1, 2.

then φ(U) is isometric to ψ(U).

Proof. We have, for i = 1, 2,

dF (∂φ

∂xi) =

∂(F ◦ φ)

∂xi=∂ψ

∂xi

and so, because F is an isometry,

∂ψ

∂xi.∂ψ

∂xj= dF (

∂φ

∂xi).dF (

∂φ

∂xj) =

∂φ

∂xi.∂φ

∂xj.

To prove the second statement set F = ψ ◦ φ−1 : φ(U)→ ψ(U). There isthe usual thing that φ−1 is not smooth and so technically speaking F is notsmooth but we already saw how to justify that so let’s just ignore it.


F is bijective so we need to check that it preserves the dot product. TpS1

is spanned by ∂x1φ, ∂x2φ and we have

dF (∂xiφ).dF (∂xjφ) = ∂xi(F ◦ φ).∂xj (F ◦ φ) = ∂xiψ.∂xjψ = ∂xiφ.∂xjφ

for all i, j = 1, 2. It is easy to see that this implies that dF (X).dF (Y ) = X.Yfor all X,Y ∈ TpS1 and thus F is indeed an isometry. �

Lecture 22

Gauss in 1827 showed that the Gaussian curvature is an intrinsic quantity,i.e., it is invariant under isometries. This is a cornerstone in Geometry.

Theorem 0.24 (Theorema Egregium). The Gaussian curvature is an in-trinsic quantity, i.e., if F : S1 → S2 is a local isometry and K1, K2 denotethe Gaussian curvature of S1, S2, then K2 ◦ F = K1.

Note that the mean curvature H is not an intrinsic notion because, forinstance, the plane (with H = 0) is locally isometric to a cylinder of radius r(with H = 1/r) but the mean curvatures are different. It is remarkable thatthe trace of −dN is not intrinsic while the determinant of −dN is. Anotherway of understanding how awesome this theorem is, is to note that K wasdefined as the determinant of g−1A, where g has intrinsic information only,while A has information how the surfaces lies in space. Nonetheless, thedeterminant only has intrinsic information, i.e., can be computed knowingg only.

Proof of Theorema Egregium. The strategy is the following. Let S be asurface with Gaussian curvature K, and φ : U → S a chart.

We want to show that K can be computed knowing only the matrixg = (gij)i,j=1,2, where gij = ∂xiφ.∂xjφ. More precisely, we would like to findan expression for K which depends only on g, the first derivatives of g, andthe second derivatives of g.

If true, then for every other surface N which is locally isometric to S, wecan apply Lemma 0.23 to find some chart ψ into N so that the correspon-dent matrix hij = ∂xiψ.∂xjψ is identical to gij . As a result, the Gaussiancurvature of N , KN , can now be computed using the very same expressionmentioned in the previous paragraph and so it must be identical to K.Claim 1: Let’s start by showing that

Kdet g = ∂x1([∂2x2x2φ]T ).∂x1φ− ∂x2([∂2

x1x2φ]T ).∂x1φ,

where, given a vector X ∈ R3, we denote by XT its tangential projection onTpS, i.e., XT = X − (X.N)N .

We have

[∂2x2x2φ]T = ∂2

x2x2φ− (∂2x2x2φ.N)N

[∂2x1x2φ]T = ∂2

x1x2φ− (∂2x1x2φ.N)N

32 ANDRE NEVES

and so differentiating both sides we obtain

∂x1 [∂2x2x2φ]T = ∂3

x1x2x2φ− ∂x1(∂2x2x2φ.N)N − (∂2

x2x2φ.N)∂x1N

∂x2 [∂2x1x2φ]T = ∂3

x2x1x2φ− ∂x2(∂2x1x2φ.N)N − (∂2

x1x2φ.N)∂x2N

which means

∂x1 [∂2x2x2φ]T .∂x1φ = ∂3

x1x2x2φ.∂x1φ− (∂2x2x2φ.N)∂x1N.∂x1φ

∂x2 [∂2x1x2φ]T .∂x1φ = ∂3

x2x1x2φ.∂x1φ− (∂2x1x2φ.N)∂x2N.∂x1φ

and so

∂x1 [∂2x2x2φ]T .∂x1φ− ∂x2 [∂2

x1x2φ]T .∂x1φ

= −(∂2x2x2φ.N)∂x1N.∂x1φ+ (∂2

x1x2φ.N)∂x2N.∂x1φ.

Recalling Lemma 3.1 in Lecture 13 we can write this as

∂x1 [∂2x2x2φ]T .∂x1φ− ∂x2 [∂2

x1x2φ]T .∂x1φ = A22A11 −A212 = detA.

We know that K = det σ = det (g−1A) = detA/det g and so detA = K det gwhich means

Kdet g = ∂x1([∂2x2x2φ]T ).∂x1φ− ∂x2([∂2

x1x2φ]T ).∂x1φ,

as desired.

Lecture 23

Claim 2: [∂2x2x2φ]T can be computed in terms of g.

[∂2x2x2φ]T lies in the tangent plane of S and so can be written as

[∂2x2x2φ]T = Γ1

22∂x1φ+ Γ222∂x2φ

for some Γ122,Γ

222. We have

∂x2g22

2= ∂x2

∂x2φ.∂x2φ

2= ∂2

x2x2φ.∂x2φ = Γ122g12 + Γ2

22g22

and

∂x2g12 − ∂x1g22

2= ∂x2(∂x2φ.∂x1φ)− ∂x1

∂x2φ.∂x2φ

2

= ∂2x2x2φ.∂x1φ+ ∂x2φ.∂

2x2x1φ− ∂x2φ.∂

2x1x2 = ∂2

x2x2φ.∂x1φ

= Γ122g11 + Γ2

22g21.

In sum, using matrix notation,(∂x2g12 − ∂x1

g222

∂x2g222

)=

[g11 g12

g21 g22

](Γ1

22

Γ222

)


and so we can solve it as(Γ1

22

Γ222

)=

[g11 g12

g21 g22

]−1(∂x2g12 − ∂x1

g222 .

∂x2g222

)=

1

g11g22 − (g12)2

[g22 −g12

−g12 g11

](∂x2g12 − ∂x1

g222 .

∂x2g222

)Claim 3: ∂x1 [∂2

x2x2φ]T .∂x1φ can be computed in terms of g.From

[∂2x2x2φ]T = Γ1

22∂x1φ+ Γ222∂x2φ

we deduce that

∂x1 [∂2x2x2φ]T .∂x1φ

= ∂x1Γ122g11 + ∂x1Γ2

22g12 + Γ122∂x1x1φ.∂x1φ+ Γ2

22∂x2x1φ.∂x1φ

= ∂x1Γ122g11 + ∂x1Γ2

22g12 + Γ122∂x1

g11

2+ Γ2

22∂x2g11

2.

From Claim 2 we see that indeed ∂x1 [∂2x2x2φ]T .∂x1φ can be computed only

in terms of g.

Claim 4: ∂x2 [∂2x1x2φ]T .∂x1φ can be computed in terms of g.

This one is done like the previous cases and we obtain the following.Setting

[∂2x1x2φ]T = Γ1

12∂x1φ+ Γ212∂x2φ

one can see that(Γ1

12

Γ212

)=

1

g11g22 − (g12)2

[g22 −g12

−g12 g11

](∂x2

g112

∂x1g222

)and then

∂x2 [∂2x1x2φ]T .∂x1φ = ∂x1Γ1

12g11 + ∂x1Γ212g12 + Γ1

12∂x1g11

2+ Γ2

12∂x2g11

2.

Putting all these four claims together we see that we can find an expressionfor K which depends only on the metric g and its derivatives.

For the sake of bookkeeping we obtain

K =1

g11g22 − (g12)2

((∂x1Γ1

22 − ∂x1Γ112)g11 + (∂x1Γ2

22 − ∂x1Γ212)g12

+(Γ122 − Γ1

12)∂x1g11

2+ (Γ2

22 − Γ212)∂x2

g11

2

)where(

Γ122

Γ222

)=

1

g11g22 − (g12)2

[g22 −g12

−g12 g11

](∂x2g12 − ∂x1

g222

∂x2g222

)and (

Γ112

Γ212

)=

1

g11g22 − (g12)2

[g22 −g12

−g12 g11

](∂x2

g112

∂x1g222

)�

34 ANDRE NEVES

Lecture 24

We now show the following theorem

Theorem 0.25 (Rigidity of sphere). Let S be a connected compact (closedand bounded) surface with constant positive Gaussian curvature. Then S isa sphere.

Before we prove it we make several remarks.The first one is that, as we already saw, there are several surfaces with

zero Gaussian curvature which do not differ by rigid motions (i.e., are notidentical) such as the plane and the cylinder. So the above result is a specificphenomena of positive Gaussian curvature.

The second one is that if we imagine a sphere being made of some inelasticbut flexible material (like paper) then we cannot deform it in any way apartfrom rigid motions. Note that we can find compact isometric surfaces whichdo not differ by rigid motions. It is an open problem to know whethercompact surfaces with no boundary are locally rigid.

The final remark is that S being compact is essential (see the exercisebelow).

Exercise: Consider

S = {(φ(t) cos θ, φ(t) sin θ, ψ(t)) : 0 < θ < 2π, a < t < b}a surface of revolution, where (φ′(t))2 + (ψ′(t))2 = 1 for all t.

(1) Show that K = −φ′′/φ.(2) Find φ so that S has K = 1 but S is not contained in a sphere with

radius one.

Proof of Rigidity of sphere. The idea is to show that every point of S isumbilical and then to use Theorem 4.2 of Lecture 14 to conclude that Smust be a sphere.

At every point x ∈ S we have principal curvatures λ1(x) ≤ λ2(x). Choosea point p ∈ S so that

maxx∈S

λ2(x) = λ2(p).

The fact that K = λ1λ2 is constant implies that

minx∈S

λ1(x) = λ1(p).

Note that we necessarily have λ1(p) ≤ λ2(p). If we show that equality holdsthen we must have λ1(x) = λ2(x) for all x ∈ S and so every point in S isumbilical, which means we are done.Claim: Suppose we have K > 0 and for some p ∈ S

λ1(p) = minx∈S

λ1(x) ≤ maxx∈S

λ2(x) = λ2(p).

Then λ1(p) = λ2(p).Suppose by contradiction that λ1 := λ1(p) < λ2(p) := λ2. We can assume

without loss of generality that p = (0, 0, 0), N(p) = (0, 0, 1). Arguing like


in Lecture 17 we have that S near the origin can be written as S = {z =f(x1, x2)} where ∇f(0, 0) = 0, f(0, 0) = 0, and Aij(0) = ∂2

xixjf(0). More-

over we can assume that the basis we choose for TpS = {z = 0} diagonalizesHess(f)(0). Note that in this case the metric matrix is the identity and soσij = Aij (recall Lecture 15), which means the eigenvalues of Hess(f)(0)are λ1, λ2.

Consider the unit vectors

E1 =(1, 0, ∂x1f)√1 + (∂x1f)2

, E2 =(0, 1, ∂x2f)√1 + (∂x2f)2

which lie in the tangent plane of S and look at the functions

h1(t) = A(E1(0, t), E1(0, t)) =1

1 + (∂x1f)2

∂2x1x1f√

1 + |∇f |2

h2(t) = A(E2(t, 0), E2(t, 0)) =1

1 + (∂x2f)2

∂2x2x2f√

1 + |∇f |2.

We have, from Exercise in Lecture 16, that

λ2 ≥ λ2(t, 0) = max{|X|=1}

A(X,X) ≥ A(E2(t, 0), E2(t, 0)) = h2(t),

which means h1 has a local minimum at the origin, h2 has a local maximumat the origin and so h′′1(0)− h′′2(0) ≥ 0.

We will now compute h′′1(0), h′′2(0) and get a contradiction. We can eitherjust compute it or argue in the following longer but more intuitive way. Notethat by Taylor formula

f(x1, x2) = λ1x2

1

2+ λ2

x22

2+O(|x1|3 + |x2|3)

and so∇f(x1, x2) = (λ1x1, λ2x2) +O(|x1|2 + |x2|2)

which means

(∂x1f(0, t))2 = O(t3), |∇f |2(0, t) = λ22t

2 +O(t3)

and hence

h1(t) =∂2x1x1f(0, t)√

1 + λ22t

2+O(t3) = ∂2

x1x1f(0, t)

(1− λ2

2

t2

2

)+O(t3).

Likewise we have

h2(t) = ∂2x2x2f(t, 0)

(1− λ2

1

t2

2

)+O(t3)

and so

h′′1(0) = ∂4x2x2x1x1f(0, 0)− λ2

2∂2x1x1f(0, 0) = ∂4

x2x2x1x1f(0, 0)− λ22λ1

h′′2(0) = ∂4x1x1x2x2f(0, 0)− λ2

1∂2x2x2f(0, 0) = ∂4

x1x2x2x2f(0, 0)− λ21λ2.

Therefore, h′′1(0)− h′′2(0) = λ1λ2(λ1 − λ2) < 0, a contradiction. �

36 ANDRE NEVES

Lecture 25

Before we proceed and state the Gauss-Bonnet Theorem we need to beable to integrate functions on surfaces.

Say that S is a surface with a chart φ : U ⊂ R2 → S. Given a compactset D ⊂ U we define the area of φ(D) to be

area(φ(D)) =

∫D|∂xφ× ∂yφ| dxdy.

To see this makes sense recall that if ~u,~v are two vectors in R3, then theyspan a parallelogram P in R3 with edges 0, ~u,~v, ~u + ~v and the area of P isexactly |~u × ~v|. So at least if φ(x, y) = x~u + y~v and D = {0 ≤ x ≤ 1, 0 ≤y ≤ 1}, then using the definition we see that

area(P ) = area(φ(D)) =

∫D|∂xφ× ∂yφ|dxdy

=

∫D|~u× ~v|dxdy = area(D)|~u× ~v| = |~u× ~v|

and so we get the correct answer in case of a parallelogram.For the purpose of computations it is useful to have an expression in terms

of the matrix (gij)i,j=1,2.

Lemma 0.26. We have |∂xφ× ∂yφ| =√detg and so

area(φ(D)) =

∫D

√detg dxdy.

Proof. Given ~u,~v two vectors we have |~u× ~v|2 = |~u|2|~v|2 − (~u.~v)2. This canbe seen at once from the fact that ~u and ~w = ~v− (~u.~v)~u/|~u|2 are orthogonalvectors and so

|~u× ~v|2 = |~u× ~w|2 = |~u|2|~w|2 = |~u|2(|~v|2 − (~u.~v)2/|~u|2) = |~u|2|~v|2 − (~u.~v)2.

Thus

|∂xφ× ∂yφ|2 = |∂xφ|2|∂yφ|2 − (∂xφ.∂yφ)2 = g11g22 − (g12)2 = detg.

�

Finally, given a function f defined on a surface S, if φ : U → S is a chartthen for every D compact set of U we define∫

φ(D)fdA =

∫Df ◦ φ|∂xφ× ∂yφ|dxdy =

∫Df ◦ φ

√detg dxdy.

We need to understand the case where we want to integrate a functionover a compact surfaces S which cannot be all covered by a single chart.The way to do this is to break the surfaces S into a finite union S1, . . . , Skof sets where

• S = S1 ∪ . . . ∪ Sk with Si ∩ Sj ⊂ ∂Si if i 6= j;


• we have charts φi : Ui → S with Si = φi(Di), where Di is a compactsubset of U .

Then we define ∫Sf dA =

k∑i=1

∫Di

f ◦ φi|∂xφi × ∂yφi| dxdy.

Of course for all this to make sense one needs to show that the value for theintegral does not depend on the decomposition made and one also needs tomake sure that the sets Si are somewhat “nice”. Let’s ignore this techni-cal issues and proceed. We can see it well explained in Montiel and Ros book.

Exercise: Use the method above to compute the area of the unit sphere S2.

The goal is still to be able to state the Gauss-Bonnet Theorem so we needto define what is the geodesic curvature of a curve γ in a surfaces S.

Consider γ : [a, b] → S ⊂ R3 a curve in a surface S which we assumeorientable with a normal vector N . We assume the curve is parametrizedby arc-length. Because the curve γ is contained in S it has to “curve” if S

is curved and so we have the following decomposition for ~k = γ′′.Consider E = N × γ′. Then N, γ′, E is an orthonormal basis for R3 and

so, because ~k is orthogonal to γ′, we can write it as

~k = (~k.N)N + (~k.E)E = knN + kgE

kn is the normal curvature and we saw on Lecture 12 that A(γ′, γ′) = ~k.N =kn, i.e., kn is the amount γ is “forced” to cuve. kg is the geodesic curvatureand measures how much more γ is curving than the strictly necessary.

Let’s compute it on simple examples. If S = {z = 0} a plane and γ :[a, b]→ S, then kn is zero and kg is the curvature of the planar curve γ seenas a curve in R2.

If S is the unit sphere {|~x| = 1} and γ(θ) = (cos θ, sin θ, 0), then choosingN to be the interior unit normal

kn = A(γ′, γ′) = γ′.γ′ = 1

and

E = N × γ′ = −γ × γ′ = −(cos θ, sin θ, 0)× (− sin θ, cos θ, 0) = −(0, 0, 1),

which means kg = γ′′.E = 0.Curves with kg = 0, like the ones above, are called geodesics.

Lecture 26

We need to define what is the Euler characteristic of a surface.A region T ⊂ R3 is a triangle if T is homemorphic to a disc (i.e., there is

a bijective continuous map from a disc to T ) and ∂T is a closed curve whichconsists of three smooth curves, called the edges. The endpoints of thesecurves are called the vertices of T (there are exactly three of them).

38 ANDRE NEVES

Given a compact surface Σ we say T = {T1, . . . Tn} is a triangulation if

• Σ = ∪ni=1Ti and each Ti is a triangle.• If Ti ∩ Tj 6= ∅ and i 6= j, then Ti ∩ Tj is either a vertex or an edge.• For any edge, there are exactly only two triangles to which this edge

belongs, unless the edge belongs to ∂Σ.

It is a standard fact in topology that every compact surfaces admits atriangulation.

For each triangulation T we consider the Euler characteristic to be

χ = F − E + V,

where F is the number of faces (i.e. triangles), E is the number of distinctedges, and V the number of distinct vertices. The great fact is that given anytwo triangulations of the same surface, the Euler characteristic is the same,i.e., χ is an invariant of the surface which is independent of the triangulationwe used. For this reason we denoted it by χ(Σ).

If Σ,Σ′ are two surfaces which are homemorphic, i.e., we can find a acontinuos map F : Σ→ Σ′ which is bijective, then χ(Σ) = χ(Σ′).

Let’s compute it in some examples. If Σ = {(x, y, 0) : x2 + y2 ≤ 1} is adisc, then χ(Σ) = 1. If Σ = {x1 + y2 + z2 = 1} is a sphere, then χ(Σ) = 2.If Σ = {x2 + y2 = 1, 0 ≤ z ≤ 1} is a cylinder, then χ(Σ) = 0. If Σ is a torusthen χ(Σ) = 0.

The general rule is the following. If Σ is a compact surface with noboundary then χ(Σ) = 2− 2g, where g is the number of handles the surfacehas. If Σ is a compact surface where ∂Σ is a disjoint union of k closed curves,then χ(Σ) = 2− 2g − k, where g is the number of handles of Σ.

Finally, the goal now is to prove the following theorem

Theorem 0.27 (Gauss-Bonnet Theorem). Let Σ be an orientable compactsurface Σ with boundary ∂Σ. Then∫

∂Σkgds+

∫ΣK dA = 2πχ(Σ).

The boundary of Σ is assumed to be oriented positively.If ∂Σ = ∅, we have ∫

ΣK dA = 2πχ(Σ).

Various remarks are in order.The first one is to explain what is means for ∂Σ to be oriented positively.

Because the surface Σ is orientable we can choose a continuous unit normalN on Σ. Given an arc-length parametrization γ of ∂Σ, we day that ∂Σis oriented positively if N × γ′ points toward the inside of Σ. Note thatN × γ′ is perpendicular to N and so lies in Tγ(t)Σ. Moreover, N × γ′ is alsoperpendicular to ∂Σ and so it either points away from Σ or into Σ.

Note that such a condition is necessary because clearly χ(Σ) and∫

ΣK dAdo not depend on how ∂Σ is oriented, while

∫∂Σ kgds changes sign if we


change the orientation of ∂Σ. Thus it is important that we have a canonicalway to choose the correct orientation.

For instance, if Σ = {(x, y, 0) : x2 + y2 ≤ 1} and we choose N = (0, 0, 1),then if we orient the boundary of Σ counterclockwise γ(t) = (cos t, sin t, 0)we see that

γ′(t) = (− sin t, cos t, 0), N × γ′(t) = (− cos t,− sin t, 0), N = (0, 0, 1)

is a positively oriented basis. In this case kg = γ′′.(N × γ′) = 1 and indeed∫∂Σkgds+

∫ΣK dA = length(∂Σ) +

∫Σ

0 dA = 2π = 2πχ(Σ).

If ∂Σ = ∅, then the Gauss-Bonnet theorem is the surface analog of the factthat if γ ⊂ R2 is a closed curve, then

∫γ kds = 2πw(γ). In both identities

we integrate a geometric term (the left-hand side) to obtain a topologicalterm (the right-hand side).

Lecture 27

We will start proving a first version of Gauss-Bonnet theorem.

Theorem 0.28 (Local version of Gauss-Bonnet). Assume we have a chartφ : U ⊂ R2 → Σ so that

• φ is smooth in U , the closure of U , and U is diffeomorphic to a disc.• Σ = φ(U) with boundary ∂Σ = φ(∂U).

Then ∫∂Σkgds+

∫ΣKdA = 2π,

where ∂Σ is positively oriented.

The basic idea is the following. In the setting of the Theorem above weknow from Lecture 25 that∫

ΣKdA =

∫UK ◦ φ|∂x1φ× ∂x2φ|dx1dx2.

The idea is to find functions M and L so that

K ◦ φ|∂x1φ× ∂x2φ| = ∂x1M − ∂x2L.Hence from Green’s Theorem we have that (assuming the boundary of U isoriented counterclockwise)∫

UK ◦ φ|∂x1φ× ∂x2φ|dx1dx2 =

∫∂ULdx1 +Mdx2.

We have gone from a double integral to a line integral and so the next thingwill be to show that∫

∂ULdx1 +Mdx2 = 2π −

∫∂Σkgds.

This identity will just follow from a simple computation plus a topologicalfact.

40 ANDRE NEVES

Proof. The strategy is the following. Choose e1 : U → R3, e2 : U → R3 sothat {e1, e2} is an orthonormal basis for Tφ(x)Σ for all x ∈ U . One way todo this would be to choose

e1 =∂x1φ

|∂x1φ|, e2 =

∂x2φ− (∂x1φ.e1)e1

|∂x2φ− (∂x1φ.e1)e1|.

Moreover, we assume that N = e1 × e2 (switch the order if this does nothold.)

Parametrize ∂Σ by a closed curve γ : [0, l]→ ∂Σ with |γ′(t)| = 1. There isσ a parametrization of ∂U so that γ = φ◦σ. We assume the parametrizationis positively oriented.

We can find a continuous function θ(t) so that

γ′(t) = cos θ(t)e1 + sin θ(t)e2 ∈ Tγ(t)Σ,

where e1(t), e2(t) is short-hand notation for e1(σ(t)), e2(σ(t)).The proof will proceed in the following way.

(1) Show that kg = θ′(t)− e1(t).e′2(t).(2) Show that

∫σ θ′(t)dt = 2π.

(3) Show that∫ l

0e1.e

′2dt =

∫UK|∂x1φ× ∂x2φ|dx1dx2.

Putting (1), (2), and (3) together we obtain∫∂Σkgds+

∫ΣKdA =

∫ l

0kgdt+

∫UK|∂x1φ× ∂x2φ|dx1dx2

=

∫ l

0θ′dt−

∫ l

0e1.e

′2dt+

∫UK|∂x1φ× ∂x2φ|dx1dx2 = 2π.

Let’s prove (1).We have

γ′ = cos θe1 + sin θe2 =⇒ γ′′ = θ′(− sin θe1 + cos θe2) + cos θe′1 + sin θe′2.

Now e1×e2 = N and e1, e2 being an orthonormal basis implies that e2×N =e1 and N × e1 = e2. Thus

N × γ′ = N × (cos θe1 + sin θe2) = − sin θe1 + cos θe2.

Therefore, using the fact that

e1.e′1 = e2.e

′2 = 0 (why?) e1.e

′2 = −e′1.e2 (why?)

we have

kg = γ′′.(N × γ′) = γ′′.(− sin θe1 + cos θe2)

= θ′+ (cos θe′1 + sin θe′2).(− sin θe1 + cos θe2) = θ′+ cos2 θe′1.e2− sin2 θe1.e′2

= θ′ − (cos2 θ + sin2 θ)e1.e′2 = θ′ − e1.e

′2.

Let’s prove (3) and then (2).


We will show that

∂x1e1.∂x2e2 − ∂x1e2.∂x2e1 = K√detg.

Assuming this for a moment, (3) follows because, writing σ(t) = (x1(t), x2(t))we have from the chain rule

e′2(t) =de2

dt(x1(t), x2(t)) = x′1∂x1e2 + x′2∂x2e2

and so we obtain from Green’s Theorem that∫ l

0e1.e

′2dt =

∫ l

0x′1e1.∂x1e2 + x′2e1.∂x2e2dt

=

∫∂Ue1.∂x1e2dx1+

∫∂Ue1.∂x2e2dx2 =

∫U∂x1(e1.∂x2e2)−∂x2(e1.∂x1e2)dx1dx2∫

U(∂x1e1.∂x2e2 − ∂x1e2.∂x2e1)dx1dx2 =

∫UK√detgdx1dx2.

Lecture 28

Let’s finish the proof of (3). We need to show the identity that is missing.Recalling the definition of the second fundamental form A we have for someαi and βi

∂xie1 = αie2 + (∂xie1.N)N = αie2 − (e1.∂xiN)N = αie2 +A(e1, ∂xiφ)N

and∂xie2 = βie1 + (∂xie2.N)N = βie1 +A(e2, ∂xiφ)N

Thus

∂x1e1.∂x2e2 − ∂x1e2.∂x2e1

= A(e1, ∂x1φ)A(e2, ∂x2φ)−A(e2, ∂x1φ)A(e1, ∂x2φ)

Let’s write ∂xiφ = ai1e1 +ai2e2, i = 1, 2. Because {e1, e2} is an orthonormalbasis we must have√

detg = (∂x1φ× ∂x2φ).N = (a11a22 − a12a21)N.N = (a11a22 − a12a21)

Now, writing σij = A(ei, ej) = −ei.dN(ej), we have

= A(e1, ∂x1φ)A(e2, ∂x2φ)−A(e2, ∂x1φ)A(e1, ∂x2φ)

= (a11σ11 + a12σ12)(a21σ21 + a22σ22)− (a11σ21 + a12σ22)(a21σ11 + a22σ12)

= (a11a22−a12a21)σ11σ22−(a11a22−a12a21)σ12σ21 =√detg(σ11σ22−σ12σ21)

= K√detg.

This implies the desired claim.Let’s prove (2). I will be sketchy here because this is a result in topology

but I will explain what the issue is.Suppose that φ is just the map φ(x, y) = (x, y, 0) in which case Σ is just

U × {0} and we can take e1 = (1, 0, 0) e2 = (0, 1, 0) on U . Recall thatγ = φ ◦ σ, where σ is a parametrization of ∂U . We proved in (1) that

42 ANDRE NEVES

kg = θ′ because e′2 = 0. In this simplified case, the geodesic curvature kgof γ coincides with the curvature k of the planar curve σ and so we obtainthat ∫ l

0θ′ds =

∫γkgds =

∫σkds = 2πw(σ) = 2π,

where w(σ) (winding number) is one because σ is the boundary of a domainU that is diffeomorphic to a disc.

In the general case, if we write σ′/|σ′| = (cosβ, sinβ) then we still have∫σ β′dt = 2π (essentially by (1) again applied to U × {0}) but now θ(t)

is different from β(t) and so we cannot conclude the result immediately.Some basic results in algebraic topology imply that we still have

∫β′dt =∫

θ′dt even if the functions are different (they key word is that they arehomotopic!).

�

Now we want to prove Gauss-Bonnet for triangles. First we need to definethe exterior angle between two smooth arcs. The set-up is the following.

Suppose we have γ1 : (0, l] → R3, γ2 : [0, d) → R3 two smooth curvesso that γ1(l) = γ2(0). We assume the curves parametrized by arc lengthand we assume that we have an orthonormal frame {e1, e2} which is definedalong γ1∪γ2 and changes smoothly (even in the curve γ1∪γ2 has one vertexpoint). We set N = e1 × e2. Finally we also assume that we have no cusp,i.e., γ′1(l) 6= −γ′2(0).

We now define the exterior angle θ that γ′1(l) makes with γ′2(0). Looselyspeaking it is the counterclockwise angle in {e1, e2} that γ′1(l) makes withγ′2(0). For the correct definition choose θ ∈ (−π, π) as the unique angle θso that cos θ = γ′1(l).γ′2(0) and θ has the positive sign if {γ′1(l), γ′2(0), N} isa positive basis and the negative sign if {γ′1(l), γ′2(0), N} is a negative basis.The only case not covered is if γ′1(l) and γ′2(0) are linearly dependent, inwhich case γ′1(l) = γ′2(0) and so θ = 0. Note that if we switch e1 with e2 orthe orientations o γ1 and γ2, then θ changes sign.

Consider T ⊂ R2 a triangle, i.e., a region homeomorphic to a disc, where∂T has 3 smooth arcs with 3 vertices. We assume ∂T has no cusps. Consideralso a chart φ : U ⊂ R3 where T ⊂ U and Σ = φ(T ). Then ∂Σ consists of 3smooth curves γ1, γ2, γ3. The surface Σ is orientable with a normal N .

Orient positively the smooth arcs γi, i = 1, 2, 3, where we assume theendpoint of γ1 is the start point of γ2 and so on. We define the angle θ1

as the angle that γ1 makes with γ2 with respect to the basis {e1, e2}, andsimilarly for θ2 and θ3.

Lecture 29

We now state the triangular version of the Gauss-Bonnet Theorem.


Theorem 0.29 (Triangular version of Gauss-Bonnet). Under the conditionsdescribed at the end of the previous lecture we have

3∑i=1

∫γi

kgds+3∑i=1

θi +

∫ΣKdA = 2π.

Proof. Like in the local version we have that for each γi, the geodesic curva-ture satisfies kg = θ′−e1.e

′2. Moreover, step (3) carries over because Green’s

Theorem also holds for triangular domains, i.e.,∫∂Te1.e

′2dt =

∫TK ◦ φ

√detgdx1dx2.

It is a topological theorem that

3∑i=1

∫γi

θ′dt+3∑i=1

θi = 2π.

This should not come as a shock for the following reason. When ∂Σ issmooth, i..e, θ1 = θ2 = θ3 = 0, I motivated why indeed such formula is true.In the general case,

∑3i=1 θi is there to compensate the fact that ∂Σ is not

smooth.Putting all these facts together we obtain that

3∑i=1

∫γi

kgds+

3∑i=1

θi +

∫ΣKdA

=3∑i=1

∫γi

kgds+3∑i=1

θi −∫∂Te1.e

′2dt+

∫ΣKdA

=3∑i=1

∫γi

kgds+3∑i=1

θi = 2π.

�

The triangular version of Gauss Bonnet gives us the following classicalinterpretation of Gaussian curvature. Let’s consider a triangle T inside asurface Σ where the sides of T are geodesics, i.e., they have kg = 0. If Σ is aplane, this is just saying that T is a proper triangle, i.e., the sides are straightlines. We define the interior angles βi of the triangle to be βi = π−θi. Thenthe theorem says that ∫

TKdA = β1 + β2 + β3 − π.

When K = 0, this is saying that the sum of the interior angles of a triangleis π (Thales’ Theorem), something we probably learned in high school. Ingeneral,

∫T KdA measures the failure of Thales’ Theorem because if the

surface has positive curvature then the sum of the interior angles of a triangle

44 ANDRE NEVES

is bigger than π and if the surface has negative curvature then the sum ofthe interior angles of a triangle is smaller than π.

We can now prove the final version of Gauss-Bonnet Theorem.

Theorem 0.30 (Gauss-Bonnet Theorem). Let Σ be an orientable compactsurface Σ with boundary ∂Σ. Then∫

∂Σkgds+

∫ΣK dA = 2πχ(Σ).

The boundary of Σ is assumed to be oriented positively.If ∂Σ = ∅, we have ∫

ΣK dA = 2πχ(Σ).

Proof. Let’s assume that ∂Σ = ∅ and the case with boundary I leave it asan exercise.

Consider a triangulation T = {Ti}Fi=1 of Σ and note that by making eachtriangle very small, we can assume that for each T ∈ T there is a chartφ : U → Σ so that T ⊂ φ(U). Moreover, we can also assume that no T ∈ Thas cusps because if this happens for some triangle we can always bend theedges a tiny bit around the vertex with a cusp so that it disappears.

Denote the edges of each triangle Ti by Eij and the exterior angles ateach vertex by θij , j = 1, 2, 3. Orient ∂Ti positively for all i. Then we have

F∑i=1

∫Ti

KdA+

∫∂Ti

kgds+F∑i=1

3∑j=1

θij =F∑i=1

2π = 2πF.

Each edge of ∂Ti is also an edge of another triangle Tj with the oppositeorientation. Hence ∫

∂Ti

kgds = 0.

Thus, denoting the internal angle by φij = π − θij we obtain∫ΣKdA+

F∑i=1

(π −3∑j=1

φij) = 2πF

and so ∫ΣKdA+ 3πF −

F∑i=1

3∑j=1

φij = 2πF.

We now argue that 3F = 2E. Consider the set of edges

{Eij : i = 1, . . . , F, j = 1, 2, 3}.

This set has cardinality 3F (each triangle has three edges). Each edge is theedge of two triangles, which means that each edge is being counted twice.Thus the number of elements is also 2E.


If we sum all the internal angles around a vertex we get 2π and so

F∑i=1

3∑j=1

φij = 2πV.

Putting all these together∫ΣKdA+ 2πE − 2πV = 2πF

=⇒∫

ΣKdA = 2π(F − E + V ) = 2πχ(Σ).

�

Lecture 30

We now derive some basic consequences of Gauss-Bonnet.

Consequence 1: If Σ is a compact surface with no boundary which hasK ≥ 0, then it must homeomorphic to a sphere. Stating differently, everysurface which has genus g > 0 must have at least one point with negativeGaussian curvature.

The reason is the following. From Gauss-Bonnet we know that

0 ≤∫

ΣKdA = 2πχ = 2π(2− 2g).

Thus either g = 0, in which case Σ is topologically a sphere, or g = 1, inwhich case Σ is topologically a torus. We now rule out this latter case. Ifg = 1, then we would have

∫ΣKdA = 0, which means K = 0 because K ≥ 0.

On the other hand, we know that every compact surface with no boundarymast have a point p where K(p) > 0 and so this means that K = 0 cannothappen.

Consequence 2: If Σ is a compact surface with no boundary and K > 0,then every two closed simple (i.e., no self-intersections) geodesics must inter-sect. The interpretation is the following: geodesics are Euclidean analogs ofstraight lines and so what this result is saying is that under positive Gauss-ian curvature every two “straight lines” must intersect, i.e., Euclides fifthaxiom does not hold if the space positively curved.

The reason is the following: Suppose γ1, γ2 are two closed simple geodesics(i.e. kg = 0) which do not intersect. Then we can find a region S in Σ sothat ∂S = γ1∪γ2 (this is a standard result in algebraic topology). ApplyingGauss-Bonnet we have∫

∂Skgds+

∫SKdA = 2πχ(S) =⇒ 0 <

∫SKdA = 2πχ(S).

46 ANDRE NEVES

We know that χ(S) = 2−2g−2 ≤ 0 (S has two boundary components) andthis contradicts the formula above.

Consequence 3: Let S be a surface with K ≤ 0 and diffeomorphic to adisc. γ1, γ2 are two geodesics contained in S with γ1(0) = γ2(0) = p ∈ S.Then γ1 and γ2 only intersect at p.

Suppose γ1(l) = γ2(s) = q. Set a = γ1([0, l]) and b = γ2([0, s]). Thenthere is a region Σ diffeomorphic to a disc such that ∂Σ = a ∪ b. Let θ1

and θ2 be the exterior angles at p and q. Note that Σ is a triangle withvertices p, q (because ∂Σ is smooth outside p and q, the other vertice willhave exterior angle 0 and so we do not worry about it). Hence we obtainfrom the triangular version of Gauss-Bonnet that∫

a∪bkgds+ θ1 + θ2 +

∫ΣKdA = 2π =⇒

∫ΣKdA = 2π − θ1 − θ2.

The angles θ1 and θ2 are strictly smaller than π and so∫

ΣKdA > 0. Thiscontradicts K ≤ 0. The fact that θ1, θ2 are strictly smaller than π is notobvious but it happens is that were the case then a would be equal to b.

Lecture 1-30

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