GEOMETRY OF CURVES AND SURFACES
ANDRÉ NEVES
Lecture 1
A parametrized curve is a smooth function
φ : [a, b] → Rn ,
( with n = 2 or n = 3)
where |φ0 (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);
φ3 : [0, 8π] → R3 ,
φ3 (θ) = (cos θ, sin θ, θ).
(3) helix
A non-example is
φ4 : [0, 8] → R2 ,
φ4 (t) = (|t|, t)
because the function is not smooth (it is not even C 1 ). Another non-example
is
φ5 : [−1, 1] → R2 , φ5 (t) = (0, t2 )
because |φ05 (0)| = |(0, 0)| = 0.
The requirement that φ : [a, b] → Rn has |φ0 (t)| 6= 0 for all a ≤ t ≤ b
serves two purposes. On one hand, we excluded maps like the constant map
from being a curve. On the other hand, every parametrized curve has a
tangent line passing through any given point. More precisely, the tangent
line to the parametrized curve φ that passes through φ(t0 ) is given by
L = {φ(t0 ) + φ0 (t0 )s : s ∈ R}.
For the purpose of geometry, what we are interested is in φ(I), the image of the parametrized curve, which we call a curve. The key thing to
understand is that a curve can have many different parametrizations. For
instance
φ6 : [0, 2] → R2 , φ6 (t) = (t3 /4 − 1, 3t3 /8 + 2)
parametrizes the same curve as φ1 (why?).
1
2
ANDRÉ NEVES
In general, if φ : [a, b] → Rn is a parametrized curve and f : [c, d] → [a, b]
is a smooth function with |f 0 (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
Z b
|φ0 (t)|dt.
length =
a
For instance, the length of the first curve is
Z 1
Z 1
Z
0
length =
|φ1 (t)|dt =
|(2, 3)|dt =
0
0
1√
13dt =
13.
0
The length of the third curve is
Z 8π
Z 8π
Z
length =
|φ03 (θ)|dθ =
|(− sin θ, cos θ, 1)|dθ =
0
√
0
8π
√
2 = 16π.
0
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 0 (s) 6= 0
for all s (why do I require this?). Then ψ = φ ◦ f : [c, d] → Rn is another
parametrization 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 the
same.
Let’s assume that n = 3 and, without loss of generality that f 0 > 0 (.i.e,
f is increasing). If φ(t) = (x(t), y(t), z(t)) we have
p
|ψ 0 (s)| = ((x ◦ f )0 (s))2 + ((y ◦ f )0 (s))2 + ((z ◦ f )0 (s))2
p
= (x0 (f (s))2 (f 0 (s))2 + (y 0 (f (s))2 (f 0 (s))2 + (z 0 (f (s))2 (f 0 (s))2
p
= f 0 (s) (x0 (f (s))2 + (y 0 (f (s))2 + (z 0 (f (s))2 = f 0 (s)|φ0 (f (s))|
and so, by the change of variable formula we obtain
Z d
Z d
0
length(ψ([c, d])) =
|ψ (s)|ds =
|φ0 (f (s))|f 0 (s)ds
c
c
Z
f (d)
=
f (c)
|φ0 (t)|dt =
Z
b
|φ0 (t)|dt = length(φ([a, b])).
a
Lecture 2
Given a parametrized curve φ : [a, b] → Rn we want to get rid of the fact
that there are infinitely many possible parametrizations of φ([a, b]) and so
we will choose a best possible one by requiring that |φ0 (t)| = 1 for all t. This
is called arc-length parametrization and it can always be done, as the next
lemma shows.
GEOMETRY OF CURVES AND SURFACES
3
Lemma 0.2. Given φ : [a, b] → Rn a parametrized curve, we can always
find ψ : [c, d] → Rn with φ([a, b]) = ψ([c, d])and |ψ 0 (s)| = 1 for all s.
Proof. The way to go is to guess what ψ should be and then find such ψ. We
know that we should have ψ = φ ◦ f for some increasing function f (why?).
Hence, with t = f (s) we want that
1 = |ψ 0 (s)| = f 0 (s)|φ0 (f (s))| =⇒ f 0 (s) = 1/|φ0 (f (s))| = 1/|φ0 (t)|.
Rt
So set h(t) = a |φ0 |(u)du. Then h0 (t) = |φ0 (t)| and we choose f to be the
inverse of h, which must exists and have f 0 (s) = 1/|φ0 (f (s))|. This should
do it.
A cool thing is that if φ : [a, b] → Rn is parametrized by arc-length then
the length of the curve is b − a (why?).
Consider φ : [0, L] → Rn parametrized curve that is parametrized by arclength. We define the geodesic curvature (or just curvature) as ~k = φ00 (t).
The first remark is that ~k(t) is always perpendicular to the tangent line
to the curve at φ(t). Indeed, we saw that the tangent line to the curve at
φ(t) is generated by φ0 (t) and thus, because
1 = φ0 (t).φ0 (t) =⇒ 0 =
d 0
φ (t).φ0 (t) = φ00 (t).φ0 (t)+φ0 (t).φ00 (t) = 2~k(t).φ0 (t)
dt
we obtain that ~k(t).φ0 (t) = 0.
The second remark is that ~k depends only on the curve φ([0, L)) and not
on the parametrization. The reason is that if φ and ψ are two parametrizations 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 φ0 (t) = ~b and φ00 (t) = 0, which means a straight line has curvature 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 on
the parametrization chosen. For instance, we can consider the following
parametrization of a straight line in the plane:
φ(t) = (0, et ),
φ00 (t)
0 ≤ t ≤ 1.
(0, et )
Then
=
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 means
that φ00 (t) = 0 for all t. Hence (assuming n = 3)
(x00 (t), y 00 (t), z 00 (t)) = 0
for all 0 ≤ t ≤ L
4
ANDRÉ NEVES
and thus one can easily see that
φ(t) = (x(t), y(t), z(t)) = φ0 (0)t + φ(0)
for all t.
Lecture 3
From now on we consider only plane curves, i.e., n = 2. The advantage
of this case is that the perpendicular vector to the curve becomes uniquely
determined by the tangent vector.
More precisely, if I have φ : [a, b] → R2 , φ(t) = (x(t), y(t)), then
N (t) = (−y 0 (t), x0 (t))/|φ0 (t)|
is a unit vector which is orthogonal to the tangent vector φ0 (t) = (x0 (t), y 0 (t)).
Indeed
1
φ0 (t).N (t) = 0
(−x0 (t)y 0 (t) + x0 (t)y 0 (t)) = 0 and |N (t)| = 1
|φ (t)|
for all a ≤ t ≤ b.
Because ~k is also orthogonal to φ0 (t) (regardless whether the parametrization is by arc-length or not – why?) we can instead of the vector ~k consider the scalar k = ~k.N , which we also denote by curvature. When φ
is parametrized by arc-length then, using the expression for N above, the
curvature k can be computed as
k = ~k.N (t) = φ00 (t).(−y 0 (t), x0 (t)) = x0 (t)y 00 (t) − y 0 (t)x00 (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 to
change it to
φ : [0, 2πr] → R2 ,
φ(θ) = ~c + (r cos(θ/r), r sin(θ/r)).
Then φ00 (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 circle
is 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 to
the curve. The normal vector N given above corresponds to rotating φ0 by
90 degrees on the counterclockwise direction. If we switch the orientation
of the curve, the new tangent vector becomes −φ0 and so, if we rotate it
counterclockwise by 90 degrees we obtain −N . As a result k changes by a
sign.
GEOMETRY OF CURVES AND SURFACES
5
Lecture 4
The example above is useful for the following geometric characterization
of 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 now
make 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 N
points 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]) does
not intersect the interior of Cr . Likewise, we say the circle Cr is too big if
there 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 big
if 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). If
η(t) > s2 then this means φ(t) lies outside Cs and if η(t) < s2 , this means
φ(t) lies inside Cs (why?).
We have η(0) = s2 and
d
((φ(t) − sN (0)).(φ(t) − sN (0)))
dt
= φ0 (t).(φ(t) − sN (0)) + (φ(t) − sN (0)).φ0 (t) = 2φ0 (t).(φ(t) − sN (0)).
η 0 (t) =
Thus η 0 (0) = 2φ0 (0).(−sN (0)) = 0 (why?) and
η 00 (0) = 2φ00 (0).(−sN (0)) + 2φ0 (0).φ0 (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 a
small neighborhood I of 0 so that η(t) > η(0) = s2 for all t ∈ I \ {0}. The
case s > r is handled similarly.
Exercise: Show that if γ is a planar curve with k = 1/r, then γ is
contained 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 which
k = 1/r (instead of k = −1/r).
Then
N.N = 1 =⇒ ∂t N.N = 0 =⇒ ∂t N = multiple of γ 0 .
Moreover,
∂t N.γ 0 = ∂t (N.γ 0 ) − N.γ 00 = 0 − N.~k = −k = −1/r
6
ANDRÉ NEVES
and so ∂t N = −γ 0 /r. Hence
c0 = γ 0 + rN 0 = γ 0 − γ 0 = 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. Show
that
k(t) = |φ0 (t)|−2 φ00 (t).N (t).
Solution: It suffices to see that the expression on the right hand side is
independent of the parametrization because, if the curve is parametrized by
arc length, then indeed the right hand side gives us k(t).
Suppose we have t = f (s), where f is a smooth strictly increasing function
with range [a, b], and consider ψ = φ ◦ f . Then
ψ 0 (s) = f 0 (s)φ0 (f (s)) ψ 00 (s) = f 00 (s)φ0 (f (s)) + (f 0 (s))2 φ00 (f (s)).
The vector f 00 (t)φ0 (f (s)) is tangent to the curve at ψ 0 (s) and so
f 00 (s)φ0 (f (s)).N (f (s)) = 0,
where N̄ (s) := N (f (s)) is the normal vector to ψ at ψ(s). Moreover,
|ψ 0 (s)| = f 0 (s)|φ0 (f (s))|. Hence
|ψ 0 (s)|−2 ψ 00 (s).N̄ (s) = |ψ 0 (s)|−2 ψ 00 (s).N (f (s)) = |ψ 0 (s)|−2 ψ 00 (s).N (f (s))
= |ψ 0 (s)|−2 (f 0 (t))2 φ00 (f (s)).N (f (s)) = |φ0 (f (s))|φ0 (f (s)).N (f (s))
= |φ0 (t)|−2 φ00 (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) = |c0 (t)|−1 c00 (t).N (t). = (1 + f 0 (t)2 )−1 (0, f 00 (t)).(−f 0 (t), 1)
=
f 00 (t)
1 + (f 0 (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 (0) = (1, 0) =⇒ N (0) = (0, 1).
From the previous exercise we can see that k(0) = 0 and we cannot find a
neighborhood U as described in the exercise for this curve γ. So the answer
GEOMETRY OF CURVES AND SURFACES
7
is NO.
Lecture 5
We continue with planar curves φ : [a, b] → R2 but we assume that the
curve 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 transversed
clockwise or counterclockwise. If γ is a curve centered at (0, 1) of radius less
than one, then r(γ, 0) = 0.
There is a nice way to compute the rotation number, which we usually
learn in complex analysis and is part of the Cauchy integral formula: the
formula is given by
Z
dz
1
.
r(γ, 0) =
i2π γ z
Writing the parametrization φ of γ as φ(t) = x(t) + iy(t), where a ≤ t ≤ b
we get
Z
Z b 0
Z b 0
1
dz
1
x (t) + y 0 (t)
1
(x + iy 0 )(x − iy)
=
dt =
dt
i2π γ z
i2π a x(t) + iy(t))
i2π a
x2 + y 2
Z b 0
1
x x + yy 0 + i(xy 0 − yx0 )
=
dt
i2π a
x2 + y 2
Z b 0
Z b 0
x x + yy 0
xy − yx0
1
1
=
dt
+
dt
i2π a x2 + y 2
2π a x2 + y 2
Z b
Z b 0
d p 2
1
xy − yx0
1
2
=
ln x + y dt +
dt
i2π a dt
2π a x2 + y 2
Z b 0
xy − yx0
1
=
dt.
2π a x2 + y 2
Thus we conclude
Z b
1
1
r(γ, 0) =
φ(t).(y 0 (t), −x0 (t))dt.
2π a |φ(t)|2
Given a closed curve γ parametrized by φ we define its winding number
w(γ) as the rotation number of the curve
F : [a, b] → R2 , F (t) = φ0 (t)
around the origin. In other words w(γ) = r(γ 0 , 0).
The winding number counts how many times F rotates around the origin
and so how many turns γ does, where turns counterclockwise are counted
positively and turns clockwise are counted negatively. Note that if we change
the orientation of the curve, then the winding number changes sign.
8
ANDRÉ NEVES
Theorem 0.4. If γ is a closed curve then
Z
1
w(γ) =
kds.
2π γ
Remark 0.5.
define
(1) Given a function f : γ → R defined on a curve γ we
Z
Z
f ds =
γ
b
f (φ(t))|φ0 (t)|dt,
a
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 also
show that the above definition is independent of the parametrization
chosen.
(2) The winding number of a curve is a topological invariant, i.e., if
we move the curve in a smooth way in the plane, then the number
of times γ rotates does not change. The technical lingo is to say
that w(γ) does not change under isotopies. On the other hand, the
curvature k is not invariant under smooth deformations. The content
of Theorem 0.4 is that if we integrate k, the value is a topological
invariant! This is a prototype for Gauss-Bonnet Theorem and that’s
why I have included it.
Proof. Assume the curve γ is parametrized by arc-length, i.e., |φ0 (t)| = 1.
Then F (t) = (x0 (t), y 0 (t)) and so, recalling the formula for N ,
F (t).(y 00 (t), −x00 (t)) = φ0 (t).(y 00 (t), −x00 (t)) = x0 (t)y 00 (t) − y 0 (t)x00 (t)
= (−y 0 (t), x0 (t)).~k(t) = N.~k = k(t).
Therefore we obtain that given a closed curve γ then
Z b
Z b
Z
1
1
1
w(γ) = r(F, 0) =
F (t).(y 00 (t), −x00 (t))dt =
k(t)dt =
kds.
2π a
2π a
2π γ
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 of
γ to be
Z
k 2 ds.
E(γ) = length(γ)
γ
This quantity is scale invariant, i.e., E(cγ) = E(γ) for every closed curve γ
and constant c > 0.
Corollary 0.6. If γ is a closed curve with w(γ) 6= 0 then E(γ) ≥ 4π 2 with
equality if and only if γ is a circle.
GEOMETRY OF CURVES AND SURFACES
9
The content of the above result is that E(γ) is a number that measures
how much a closed curve is curving. In particular, every closed curve curves
more than a circle and equality holds only for the circle. This type of results
are paramount in Geometry.
The assumption that w(γ) 6= 0 is not necessary and could be removed. I
will put that as an exercise later in the course.
Proof. Let’s orient the curve so that w(γ) ∈ N . From Holder’s inequality
and Theorem 0.4 we know that
s
Z
Z
p
2π ≤ 2πw(γ) = kds ≤ length(γ) k 2 ds = E(γ).
γ
γ
Moreover, if equality holds then w(γ) = 1 and thus
s
Z
Z
2π = kds = length(γ) k 2 ds.
γ
γ
Setting c0 = 2π/length(γ) we obtain
Z
Z
Z
Z
2
2
(k − c0 ) ds = k ds − 2c0 k + c20 ds
γ
γ
γ
γ
2
= 4π /length(γ) − 2(2π/length(γ))2π + (2π/length(γ))2 length(γ)
= 4π 2 /length(γ) − 8π 2 /length(γ) + 4π 2 /length(γ) = 0.
Therenfore k is a constant which we call 1/r. From Exercise 3 we deduced
that 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 be
a 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 we
know w(γ) = 1. Hence, indeed, γ is a circle or radius r.
Exercise: Give an example of a closed curve γ with no self-intersections
and E(γ) arbitrarily large. You can just draw the curve and explain in a
convincing 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 C 1 - why?) but we can makeRa tiny perturbation so that it becomes smooth. It’s length is 2π + 2L and γL k 2 ds = 2π,
which means E(γL ) = (2π + 2L)2π. So making L very large we can make
E(γL ) very large.
Lecture 7
In what follows Br (p) ⊂ R3 is the open ball of radius r centered at p.
10
ANDRÉ NEVES
The informal definition of surface is the following. We say that a set
S ⊂ R3 is a surface if locally looks just like a disc in R2 meaning that for all
p ∈ S we can find r small so that S ∩ Br (p) is diffeomorphic to a disc. For
examples 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 open
neighborhood 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
(q)
∂x
and
∂φ
(q)
∂y
are linearly independent vectors in R3 for all q ∈ U .
Conditions i) and ii) make sense because we want to say that “S ∩V looks
like an open set of R2 ” but one might wonderpwhy condition iii).
If we consider the cone S = {~x ∈ R3 : z = x2 + y 2 }, then the map
p
2
2
φ : R2 → R3 , φ(x, y) = e−1/(x +y ) (x, y, x2 + y 2 )
is smooth and injective (why?), has S = φ(R2 ), and thus, without condition
iii) (which fails for q = (0, 0)) the cone S would be a surface. The cone is
not 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 ,
satisfies
φ(x, y) = (x, y, f (x, y))
φ(R2 )
= S, φ is smooth and injective, and
∂φ
∂φ
= (1, 0, ∂x f (x, y)) and
= (0, 1, ∂y f (x, y))
∂x
∂y
are always linearly independent vectors (why?).
Another large class of examples comes from considering level sets of functions. 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 intersecting
along 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.
GEOMETRY OF CURVES AND SURFACES
11
Lecture 8
We start with two remarks regarding the last proposition. The first remark is that when F (x, y, z) = zy, then
∇F = (∂x F, ∂y F, ∂z F ) = (0, z, y)
and so ∇F (p) = (0, 0, 0) when (0, 0, 0) = p ∈ F −1 (0). In other words, the
condition about the gradient in Proposition cannot be removed.
The second remark is that we can now show the sphere
S 2 = {~x ∈ R3 : |~x| = 1}
is a surface. Indeed if we set F (x, y, z) = x2 + y 2 + z 2 − 1, then S 2 = 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 S 2 is a surface.
Note that if we were to do this using the definition of surface we would
have use more than one chart and so it would be not so straightforward as
in the graphical case (where we can get away with only one chart).
Exercise: Show that S 2 (see above) is a surface by finding the appropriate
charts.
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 and
L a smooth function in O. Assume that for some p = (x0 , y0 , z0 ) ∈ O we
have ∂z L(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 equation
L(x, y, z) = a for the variable z in a neighborhood V of p (i.e., express z in
terms of x and y) if ∂z L(p) 6= 0. This condition cannot be removed because
of examples like L(x, y, z) = z 2 − x2 + y 2 . Here you cannot uniquely solve
for 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 ∂z F (p) 6= 0. From the
implicit 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 locally
the graph of a function over some coordinate plane.
12
ANDRÉ NEVES
Exercise:With a > r consider the set
p
S = {(x, y, z) ∈ R3 : ( x2 + y 2 − a)2 + z 2 = r2 }.
Show that S is a surface and draw it.
Sketch of solution: Make sure Proposition 0.8 can be applied to
p
F (x, y, z) = ( x2 + y 2 − a)2 + z 2 − r2 .
Lecture 9
For the purpose of computations, it is useful to know that every surface
S is locally the graph of some function.
Lemma 0.10. Given a surface S and p ∈ S we can find an open neighborhood V of p so that S ∩ V can be written as the graph of some function
defined over one of the coordinate planes.
Proof. It suffices to find an open neighborhood A of p and F : A → R a
smooth 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 condition
above implies the existence of an open neighborhood V of p so that S ∩V can
be written as the graph of some function defined over one of the coordinate
planes.
Choose a chart φ : U ⊂ R2 → R3 so that φ(u0 , v0 ) = p. Set
P = span{∂u φ(u0 , v0 ), ∂v φ(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
∂T
∂T
= ∂u φ(u0 , v0 )
= ∂v φ(u0 , v0 )
= (0, 0, 1).
∂u
∂v
∂t
These 3 vectors are linearly independent and thus, from the Inverse Function
Theorem we have that T −1 is defined on an open set A ⊂ R3 containing
T (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∩A
then 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
Tp S = {α0 (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 the
charts. The disadvantage is that is impossible to compute things. For
instance if we want to compute the tangent plane to S = {z = x2 + y 2 }
GEOMETRY OF CURVES AND SURFACES
13
at p = (1, 1, 2) then it seems tricky. Thus the following lemma is clearly
helpful.
Lemma 0.11. Let φ : U → S ⊂ R3 a chart with φ(x0 , y0 ) = p. Then
∂φ
∂φ
Tp S = span
(x0 , y0 ),
(x0 , y0 ) .
∂x
∂y
In other words, the map dφ(x0 ,y0 ) : R2 → Tp S is bijective, where
dφ(x0 ,y0 ) (a, b) = a
∂φ
∂φ
(x0 , y0 ) + b (x0 , y0 ).
∂x
∂y
Note that the Jacobian dφ(x0 ,y0 ) can be represented by a 3 × 2 matrix,
where the first column is ∂x φ and the second column ∂y φ.
With this lemma we can compute the tangent plane to S = {z = x2 + y 2 }
at p = (1, 1, 2). Consider the chart φ(x, y) = (x, y, x2 + y 2 ). Then φ(1, 1) =
(1, 1, 2) and so
Tp S = span{∂x φ, ∂y φ} = 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) technical issues.
The charts used to define a surface are not unique. For instance if we have
a surface S and a chart φ : U ⊂ S ⊂ R3 , then for every smooth injective map
h ◦ U 0 ⊂ R2 → U with dh|q bijective for all q in U , we have that ψ = φ ◦ h
is also a chart (why?).
The converse is also true, i.e., if ψ : U 0 → R3 is also a chart and φ(U ) ∩
ψ(U 0 ) 6= ∅, then we can find open sets A, A0 such that V = φ(A) = ψ(A0 )
and so we can consider h = φ−1 ◦ ψ : A0 → A. The map h is called change
of parameters and naturally ψ = φ ◦ h.
There is technical issue that we address now. One needs to argue that h
is a differentiable map because of the following reason. What we learned in
calculus is that if L : O ⊂ Rn → O0 ⊂ Rn is differentiable, injective, and
DL|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 mentioned
cannot be applied, which means that φ−1 is continuous but not necessarily
differentiable. Thus, we cannot say that h is differentiable by saying is the
composition of two differentiable maps ψ and φ−1 .
The correct way of arguing that h is differentiable is the following. From
the proof of Lemma 0.10 we know that for every p ∈ V = φ(A) = ψ(A0 )
there is open neighborhood p ∈ O ⊂ R3 such that S ∩ O = F −1 (0), where
F : O → R is a function with ∇F =6= 0 at every point. Without loss of
generality assume that ∇F (p) = (0, 0, a) for some a 6= 0 and consider the
projection π : R3 → R2 , where π(x, y, z) = (x, y). Finally, set T = π ◦ φ and
G = π ◦ ψ. T is defined on an open set B such that φ(B) = S ∩ O and there
is q ∈ B so that p = φ(q).
14
ANDRÉ NEVES
We argue that dTp is injective. We have F (φ(x, y)) = 0 and so, differentiating with respect to x or y we obtain from the chain rule that
∂φ
∂φ
(q).∇F (p) = 0
(q).∇F (p) = 0.
∂x
∂y
Thus dTq is injective because if dTq (a, b) = 0 then a∂x φ(p) + b∂y φ(q) would
be collinear with ∇F (p) (why?) and the identities above show that is impossible.
As a result, the Inverse Function Theorem implies that T is invertible in
a 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. It
also 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
∂φ
∂φ
(x0 , y0 ),
(x0 , y0 ) ⊂ Tp S.
L = span
∂x
∂y
Let ~v = a∂x φ(x0 , y0 ) + b∂y φ(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
d
φ(x0 + at, y0 + bt)
dt |t=0
d
d
= (x0 + at)∂x φ(x0 , y0 ) + (y0 + bt)∂x φ(x0 , y0 ) = ~v .
dt
dt
Thus, by definition of Tp S we have ~v ∈ Tp S.
Choose α0 (0) ∈ Tp S. We want to conclude that α0 (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 we
have from the chain rule that
d(φ ◦ γ)
dx ∂φ
dy
∂φ
α0 (0) =
=
(0) (x0 , y0 ) + (0) (x0 , y0 ) ∈ L.
dt |t=0
dt
∂x
dt
∂y
If φ is not graphical, we know from Lemma 0.10 that there is a graphical 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 a
diffeomorphism (i.e., smooth, injective, with smooth inverse) and such that
φ = ψ ◦ h. In this case the chain rule implies that (why?)
∂ψ
∂ψ
∂ψ
∂ψ
span
(h(x0 , y0 )),
(h(x0 , y0 )) = span
(h(x0 , y0 )),
(h(x0 , y0 )) .
∂x
∂y
∂x
∂y
Because ψ is graphical we have
∂ψ
∂ψ
span
(h(x0 , y0 )),
(h(x0 , y0 )) = Tp S
∂x
∂y
and so the lemma is proven.
α0 (0) =
Another useful result to compute tangent planes is the following.
GEOMETRY OF CURVES AND SURFACES
15
Lemma 0.12. Let F : R3 → R so that ∇F 6= 0 on S = F −1 (0). Then for
any p ∈ S
Tp S = {~v : ∇F (p).~v = 0}.
Using this we can compute the tangent plane at every point of S 2 =
{|~x| = 1}. If F (x, y, z) = x2 + y 2 + z 2 − 1, then S 2 = F −1 (0) and so, if
p = (x0 , y0 , z0 ) ∈ S 2 , we have from the previous result that
Tp S 2 = {~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 a
function F points in the direction the function increases the most. Thus if
α0 (0) is a tangent vector, it means that F is constant along α and so α0 (0)
should be orthogonal to the gradient of F .
For the formal argument we set L = {~v : ∇F (p).~v = 0}. It suffices to
see that Tp S ⊂ L because dimL = dimTp S = 2 and so equality must hold.
Given α0 (0) ∈ Tp S we then have F ◦ α(t) = 0 because α(t) ⊂ S = F −1 (0).
Thus from the chain rule we have
d(F ◦ α)
∂F
∂F
∂F
0=
= x0 (0)
(p) + y 0 (0)
(p) + z 0 (0)
(p) = ~v .∇F (p)
dt
∂x
∂y
∂z
t=0
and so ~v ∈ L.
Lecture 11
There is one last set of definitions we need to get in order. Given a surface
S1 and a continuous map F : S1 → R3 , we say that F is smooth if for every
chart φ : U → S we have that F ◦ φ is smooth. Note that if φ and ψ are two
charts with the same image, then φ = ψ ◦ h for some diffeomorphism h and
thus, because h has a smooth inverse, we have that F ◦ φ is smooth if and
only if F ◦ ψ is smooth (why?).
Given a smooth map F : S → R3 we define its Jacobian at p ∈ S as the
linear map
d(F ◦ α)
(0).
dt
We need to make sure that the definition is consistent, which in this case
means that it does not depend on the curve α, only on α0 (0). Indeed, pick
p ∈ S and, using Lemma 0.10, pick a chart φ which is graphical, i.e., has
the 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 α10 (0) = α20 (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
dF|p : Tp S → R3 ,
dF|p (α0 (0)) =
d(F ◦ αi )
d(F ◦ φ(xi (t), yi (t)))
(0) =
(0)
dt
dt
∂(F ◦ φ)
∂(F ◦ φ)
= x0i (0)
(xi (0), yi (0)) + yi0 (0)
(xi (0), yi (0)).
∂x
∂y
16
ANDRÉ NEVES
The fact that α10 (0) = α20 (0) implies that (x01 (0), y10 (0)) = (x02 (0), y20 (0)) and
hence
d(F ◦ α1 )
d(F ◦ α2 )
(0) =
(0).
dt
dt
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 we
know that dφ(x0 ,y0 ) is map from R2 to Tp S which is bijective.
Lemma 0.13. If φ : U → S is a chart with φ(x0 , y0 ) = p, then
dF|p (∂x φ(x0 , y0 )) =
∂(F ◦ φ)
(x0 , y0 ),
∂x
dF|p (∂y φ(x0 , y0 )) =
∂(F ◦ φ)
(x0 , y0 ).
∂y
In other words d(F ◦ φ)|(x0 ,y0 ) = dF|p dφ|(x0 ,y0 ) .
Proof. Consider α(t) = φ(x0 + t, y0 ). Then
d(F ◦ α)
d(F ◦ φ)
(0) =
(x0 + t, y0 )
dt
dt |t=0
∂(F ◦ φ)
=
(x0 , y0 ).
∂x
The other derivative is computed similarly.
dF|p (∂x φ(x0 , y0 )) = dF|p (α0 (0)) =
Lecture 12
For instance, say that S = {z = x2 + y 2 } and F : S → R3 given by
F (x, y, z) = (cos(πz), xz, y + z 2 ). We already saw that with p = (1, 1, 2)
then Tp S = {z = 2x + 2y} = span{(1, 0, 2), (0, 1, 2)}. Let’s compute
dF|p (2, −1, 2).
We first compute d(F ◦ φ)|(1,1) , where φ(x, y) = (x, y, x2 + y 2 ). Then
F ◦ φ(x, y) = (cos(π(x2 + y 2 )), x(x2 + y 2 ), y + (x2 + y 2 )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) and
the second column (0, 2, 9). Now we noticed that (2, −1, 2) = 2(1, 0, 2) −
(0, 1, 2) = 2∂x φ(1, 1) − ∂y φ(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 a
smooth map with F (S1 ) ⊂ S2 , in which case we simply write F : S1 → S2 .
GEOMETRY OF CURVES AND SURFACES
17
From the definition of dF|p , we have that this linear map takes values into
TF (p) S2 , i.e., dF|p : Tp S1 → TF (p) S2 .
We now explain how to compute the normal vector N to a surface S at
a point p ∈ S. We have already seen that Tp S is a plane and so N (p) is by
definition a unit normal vector perpendicular to Tp S. In other words, N (p)
is a unit vector such that ~v .N (p) = 0 for all ~v ∈ Tp S.
If S = F −1 (0) is a surface and ∇F (p) 6= 0 for all p ∈ S, then we already
saw in Lemma 0.12 that ∇F (p) is perpendicular to Tp S 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 Lemma
0.11 that ∂x φ(x0 , y0 ), ∂x φ(x0 , y0 ) span Tp S and so we can determine the
normal vector as
∂x φ × ∂y φ
(x0 , y0 ).
N (p) =
|∂x φ × ∂y φ|
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 S 2 = {|~x| = 1}, we have that S 2 = F −1 (0) for F (x, y, z) = x2 +y 2 +z 2 −1
and so
∇F
2p
N (p) =
(p) =
= p.
|∇F |
|2p|
If S = {x2 + y 2 = 1} (infinite circle of radius one) then we have that
S = F −1 (0), where F (x, y, z) = x2 + y 2 − 1 and so
N (p) =
∇F
(2x, 2y, 0)
(x, y, 0)
=
=p
= (x, y, 0).
|∇F |
|(2x, 2y, 0)|
x2 + y 2
Finally, , when S = {z = x2 + y 2 } and p = (1, 1, 2) then using the chart
φ(x, y) = (x, y, x2 + y 2 ) we have
N (p) =
∂x φ × ∂y φ
(1, 0, 2) × (0, 1, 2)
(1, 1) =
|∂x φ × ∂y φ|
|(1, 0, 2) × (0, 1, 2)|
(−2, −2, 1)
=
= (−2/3, −2/3, 1/3).
|(−2, −2, 1)|
Lecture 13
Given a surface S and p ∈ S, there is always an ambiguity of choosing
N (p) or −N (p). When we can make such a choice unambiguously, we say the
surface 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 (Mobius
strip).
Exercise: Show that if the map N : S → R3 is continuous then it is actually
smooth.
18
ANDRÉ NEVES
Solution: Choose p ∈ S and a chart φ : U → R3 with φ(q) = p. Then
N (p) =
∂x φ × ∂y φ
(q)
|∂x φ × ∂y φ|
or
N (p) =
∂y φ × ∂y φ
(q).
|∂x φ × ∂y φ|
Say the second case case occurs. Then for each q 0 nearby q, because N is
continuous, we must necessarily have
N (φ(q 0 )) =
∂y φ × ∂y φ 0
(q ).
|∂x φ × ∂y φ|
The expression on the right-hand side is a smooth function of q 0 and so N ◦φ
is smooth.
Given an orientable surface S, the idea is to study the surface by looking
into the map N : S → R3 . This map is called the Gauss map.
Notice that N : S → S 2 = {|~x| = 1} ⊂ R3 because N (p) is always
a unit vector. Let’s compute it in some simple examples but note that if
N : S → S 2 is a Gauss map, then p 7→ −N (p) is also valid choice for the
Gauss map.
If S = {~x.v = 0} ⊂ R3 is a plane then N (p) = v for all p ∈ S and so the
Gauss map is constant.
If S = S 2 we already saw that N (p) = p for all p ∈ S 2 and so the Gauss
map N : S 2 → S 2 is the identity map in this case.
If S = {x2 + y 2 = 1}, we already saw that N : S → S 2 is given by
N (x, y, z) = (x, y, 0).
If S = {z = x2 + y 2 }, then for the chart φ(x, y) = (x, y, x2 + y 2 ) we have
∂φ ∂φ
×
= (1, 0, 2x) × (0, 1, 2y) = (−2x, −2y, 1)
∂x ∂y
and so the Gauss map is given by
(−2x, −2y, 1)
(−2x, −2y, 1)
√
N (x, y, z) = p
=
2
2
1 + 4z
1 + 4x + 4y
for all (x, y, z) ∈ S.
A crucial property of the Gauss map is that the Jacobian dN|p is a map
from Tp S to Tp S. Indeed dN|p : Tp S → TN (p) S 2 but Tq S 2 = {~x : ~x.q = 0}
for all q ∈ S 2 and so
TN (p) S 2 = {~x : ~x.N (p) = 0} = Tp S,
i.e., dN|p : Tp S → Tp S.
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 ∈ Tx S.
GEOMETRY OF CURVES AND SURFACES
19
If S = {|~x| = r} is the unit sphere of radius r then we have N : S → S 2
given by N (p) = p/r. Thus
dN|p (α0 (0)) =
d(N (α(t)))
1 d(α(t))
α0 (0)
(0) =
(0) =
,
dt
r dt
r
i.e., dN|p = 1r Id.
If S = {x2 + y 2 = r2 }, then Gauss map N : S → S 2 is given by
(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 map
is linear, we only need to determine it on a basis of T(x,y,z) S. If we set
e1 = (0, 0, 1) and e2 = (−y, x, 0)/r then we have
N (x, y, z) =
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 + t)
(0) = 0.
dt
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) (e1 ) = dN|(x,y,z) ((x, y, z + t)0 ) =
dN (r cos(θ0 + t/r), r sin(θ0 + t/r), z)
(0)
dt
1
1
1
= (− cos θ0 , sin θ0 , 0) = (−y/r, x/r, 0) = e2 .
r
r
r
The goal now is to study dN|p or equivalently the quadratic form
dN|(x,y,z) (e2 ) = dN|(x,y,z) (α0 (0)) =
A : Tp S × Tp S → R,
A(X, Y ) = −X.dN|p (Y ).
It is clear that knowing dN implies that we can compute A. The converse
is also true because if e1 , e2 is an orthonormal basis for Tp S 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 ∈ Tp S and
a, 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|p N (Z) = aA(X, Y ) + bA(X, Z).
20
ANDRÉ NEVES
To have a first idea of what A computes, consider a curve α : I → S with
α(0) = p. We have N (α(t)).α0 (t) = 0 for all t and so, from the definition of
Jacobian we have
A(α0 (0), α0 (0)) = −dNp (α0 (0)).α0 (0) = −
d(N ◦ α)
(0).α0 (0)
dt
d
(N ◦ α(t).α0 (t)) + N (p).α00 (0) = N (p).α00 (0).
dt |t=0
If |α0 (0)| = 1, then A(α0 (0), α0 (0)) = N (p).~k is the the projection of the
geodesic curvature of α on N . In other words, A(α0 (0), α0 (0)) computes
how much the curve α is curving because is lying inside S. If P is the
plane spanned by N (p) and α0 (0) that passes through the point p, then
A(α0 (0), α0 (0)) is the curvature of the planar curve S ∩ P ⊂ P at p.
Let S = {z = y 2 −z 2 } (a saddle) and consider the Gauss map N : S → S 2
with N (0, 0, 0) = (0, 0, 1). Then if e1 = (1, 0, 0) and e2 = (0, 1, 0) we have
from the interpretation given at the end of last lecture that Ap (e1 , e1 ) < 0
and Ap (e2 , e2 ) > 0, where p = (0, 0, 0). Drawing a picture, this should be
self-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 for
i, j = 1, 2
Aij := A(∂xi φ, ∂xj φ) = (∂xj xi φ).N ◦ φ.
Proof. From Lemma 4.1 of Lecture 9 we have for i = 1, 2,
∂(N ◦ φ)
dN (∂xi φ) =
∂xi
and thus, for i, j = 1, 2,
Aij = A(∂xi φ, ∂xj φ) = −∂xi φ.dN (∂xj φ) = −∂xi φ.
=
∂(N ◦ φ)
∂xi
∂
(∂xi φ.N ◦ φ) + (∂xj xi φ).N ◦ φ = (∂xj xi φ).N ◦ φ.
∂xi
From the identity above we see that in the same way that the tangent
plane 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 chart
φ(x1 , x2 ) = (x1 , x2 , f (x1 , x2 )) and so, using
∂x1 φ × ∂x2 φ = (1, 0, ∂x1 f ) × (0, 1, ∂x2 f ) = (−∂x1 f, −∂x2 f, 1)
we obtain
p
N ◦ φ(x1 , x2 ) = (−∂x1 f, −∂x2 f, 1)/ 1 + |∇f |2 .
Finally we have
∂2φ
=
∂xi ∂xj
∂2f
0, 0,
∂xi ∂xj
GEOMETRY OF CURVES AND SURFACES
21
and thus we have from Lemma 0.14 that
∂x2i xj f
Aij = p
,
1 + |∇f |2
i, j = 1, 2.
If S = {z = y 2 − 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 fundamental form of the saddle S. If we choose X = (2, 0, 0), Y = (1, −1, 0) both
in Tp S, then because
∂x1 φ(0, 0) = (1, 0, 0)
and ∂x2 φ(0, 0) = (0, 1, 0)
we have
A(X, Y ) = A(2∂x1 φ, ∂x1 φ − ∂x2 φ) = 2A11 − 2A12 = −4.
Lecture 16
Given a surface S the goal is to study the second fundamental form A :
Tp S × Tp S → 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 ∈
Tp S.
Proof. Pick a chart φ : U → S with φ(q) = p. From Lemma 0.14 we see
that Aij = Aij , which means the matrix (Aij )i,j=1,2 is symmetric. Thus if
we write Xi = α1i ∂x1 φ + α2i ∂x2 φ, with i = 1, 2 we have from the bilinearity
of A that
A(X1 , X2 ) =
2
X
αi1 αj2 A(∂xi φ, ∂xj φ) =
i,j=1
=
2
X
2
X
αi1 αj2 Aij
i,j=1
αi1 αj2 Aji =
i,j=1
2
X
αi1 αj2 A(∂xj φ, ∂xi φ) = A(X2 , X1 ).
i,j=1
It is a standard fact of Linear Algebra that bilinear forms on vector spaces
are diagonalizable. More precisely, we can find an orthonormal basis {e1 , e2 }
of Tp S so that A(ei , ei ) = λi , i = 1, 2 and A(e1 , e2 ) = A(e2 , e1 ) = 0. The
eigenvalues λ1 , λ2 are called the principal curvatures and the eigenvectors
e1 , e2 are called the principal directions.
Note that this is equivalent to say there there is an orthonormal basis
{e1 , e2 } of Tp S so that
dN|p (e1 ) = −λ1 e1
and dN|p (e2 ) = −λ2 e2 .
22
ANDRÉ NEVES
Indeed, if e1 , e2 are principal directions and λ1 , λ2 principal curvatures, then
because 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 = −λi ei ,
i = 1, 2.
Conversely, if dN|p (ei ) = −λi e1 for some orthonormal basis {e1 , e2 }, then
A(ei , ej ) = −dN|p (ei ).ej = λi (ei .ej ), i, j = 1, 2, and so e1 , e2 are principal
directions and λ1 , λ2 principal curvatures.
Exercise: If S is a surface, show that
λ1 (p) = min{A(X, X) : X ∈ Tp S, |X| = 1},
λ2 (p) = max{A(X, X) : X ∈ Tp S, |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
1
X.Y
dN|p = Id =⇒ A(X, Y ) = −
r
r
for all X, Y ∈ Tp S.
Thus the principal curvatures are λ1 = λ2 = − 1r and every unit vector in
Tp S is a principal direction. Convention dictates that we like the principal
curvatures of the sphere to be positive and so for Gauss map we should take
N (p) = −p/r (the interior pointing normal), which then makes A(X, Y ) =
1
X.Y
r and so we obtain λ1 = λ2 = r .
2
2
2
If S = {x + y = r }, 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 and
dN (e2 ) = er2 , which means the principal curvatures are λ1 = 0 and λ2 =
−1/r, and e1 , e2 are the principal directions. Again, the convention is that
we 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 and
λ2 = 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 ∈ Tp S.
Theorem 0.16. Say that S is a connected surface so that every point is
umbilical. Then S is contained in a plane or sphere.
In particular, if S has no boundary and is compact, then it must be a
sphere.
GEOMETRY OF CURVES AND SURFACES
23
Note that in Lecture 4 we saw an exercise where where we showed that
those 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 ∈ Tp S and
so for i = 1, 2
∂φ
∂(N ◦ φ)
= dN (
) = −(λ ◦ φ)∂xi φ.
∂xi
∂xi
Differentiating again, we have i, j = 1, 2
∂ 2 (N ◦ φ)
∂φ
∂2φ
= −∂xj (λ ◦ φ)
+ (λ ◦ φ)
∂xj ∂xi
∂xi
∂xj ∂xi
and thus, because
∂ 2 (N ◦ φ)
∂ 2 (N ◦ φ)
=
∂xi ∂xj
∂xj ∂xi
we obtain that
∂φ
∂φ
= ∂xi (λ ◦ φ)
.
∂xi
∂xj
Now ∂x1 φ, ∂x2 φ are linearly independent vectors and so the only way the
identity above can hold is if
∂xj (λ ◦ φ)
∂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 radius
1/λ0 .
With φ : U → S ⊂ R3 a chart we know that ∂xi (N ◦ φ) + λ−1
0 ∂xi φ = 0 for
−1
N
◦
φ
is
a
constant
c
.
Hence
||φ
−
c
||
i = 1, 2 and so φ + λ−1
0
0 = λ0 , which
0
means φ(U ) is contained in a sphere of center c0 and radius λ−1
0 .
We can now define the two most important concepts of this course. Given
a 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
1
λ1 (p) + λ2 (p)
H(p) = − tr dN|p =
.
2
2
Because K and H are functions, they are much easier to understand than
the second fundamental form A and already contain a lot of information
about the surface. The rest of the course is to understand what K and H
say about a surface.
Let’s compute this quantities when S = {z = (y 2 − x2 )/2}. In order to
do that, we need to do a bit of general theory.
24
ANDRÉ NEVES
Lemma 0.17. Given a chart φ : U ⊂ R2 → S, consider the matrix
g = (gij )i,j=1,2 ,
where
gij =
∂φ ∂φ
.
∂xi ∂xj
and the matrix A = (Aij )i,j=1,2 (see Lemma 0.14).
Then, with σ = g −1 A a 2 × 2 matrix, we have
K = detσ =
detA
detg
H=
trσ
σ11 + σ22
=
.
2
2
Note that a standard formula shows that
1
1
−1
−1
−1
g11
=
g22 , g12
= g21
=−
g12 ,
det g
det g
−1
g22
=
1
g11 .
det g
We use this lemma to compute K and H for S = {z = (y 2 − x2 )/2}. The
strategy is to find a chart φ , compute A = (Aij )i,j=1,2 , g = (gij )i,j=1,2 , and
−1
g −1 = (gij
)i,j=1,2 . Then we compute the matrix product σ = g −1 A and we
take the trace and determinant of this matrix.
We have
φ(x1 , x2 ) = (x1 , x2 , (x22 − x21 )/2)
and so
g11 = 1 + x21 ,
−1
g11
=
1 + x22
,
1 + x21 + x22
N=
g22 = 1 + x22 ,
g12 = −x1 x2 ,
−1
−1
g12
= g21
=
x1 x2
,
1 + x21 + x22
−1
g22
=
1 + x21
,
1 + x21 + x22
∂ x1 φ × ∂ x2 φ
(x1 , −x2 , 1)
=p
|∂x1 φ × ∂x2 φ|
1 + x21 + x22
and
1
,
A11 = − p
1 + x21 + x22
A12 = 0,
1
A22 = p
.
1 + x21 + x22
Thus
σ11 = −
1 + x22
,
(1 + x21 + x22 )3/2
σ12 =
x1 x2
,
(1 + x21 + x22 )3/2
σ21 = −
x1 x2
,
(1 + x21 + x22 )3/2
σ22 =
1 + x21
.
(1 + x21 + x22 )3/2
As a result we obtain
K = σ11 σ22 − σ12 σ21 = −
1 + x21 + x22
1
=−
(1 + x21 + x22 )3
(1 + x21 + x22 )2
and
x21 − x22
1
H = (σ11 + σ22 ) =
.
2
(1 + x21 + x22 )3
GEOMETRY OF CURVES AND SURFACES
25
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 : Tp S → Tp S with respect to the basis
∂x1 φ, ∂x2 φ. In other words, if
X = a1
∂φ
∂φ
+ a2
∂x1
∂x2
and dN|p (X) = −b1
∂φ
∂φ
− b2
,
∂x1
∂x2
then (b1 , b2 ) = σ(a1 , a2 )T . In particular
2
X
∂φ
∂φ
)=
σki
−dN|p (
.
∂xi
∂xk
k=1
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 (
2
2
X
X
∂φ
∂φ ∂φ
∂φ ∂φ
∂φ
)).
=(
σkj
).
=
σkj
.
∂xj ∂xi
∂xk ∂xi
∂xk ∂xi
k=1
k=1
=
2
X
k=1
σkj gki =
2
X
gik σkj .
k=1
and so, using matrix notation,
A = g.σ =⇒ σ = g −1 A.
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 surface
to 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 Tp S, i.e., if U is a small neighborhood of p then
S ∩ U \ {p} does not intersect p + Tp S.
Say that S is a surface with K(p) < 0. Then near p, S is in both sides
of Tp S, i.e., if U is a small neighborhood of p then S ∩ U \ {p} intersects
p + Tp S.
This proposition explains why K < 0 on the surface S = {z = (y 2 −
x2 )/2}.
The strategy is the following. Given p ∈ S we show that, nearby that
point, the surface is the graph of a function f which is defined over the
tangent plane Tp S and, moreover, Hess(f ) coincides with the matrix σ at
p. Thus if K(p) > 0, this means det σ = detHess(f ) > 0 and hence f has a
local 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
ANDRÉ NEVES
f has a saddle point, in which case the surface has points above and below
the 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 local
minimum. If detHess(f )(0) < 0, then the origin is neither a local maximum
nor a local minimum (i.e., it is a saddle point).
Proof. From Taylor’s formula we know that
1
f (x, y) = f (0, 0) + ∇f (0).(x, y) + (x, y)Hess(f )(0)(x, y)T + O(x3 + y 3 ),
2
where O(x3 + y 3 ) stands for a term whose absolute value is bounded from
above by C(x3 + y 3 ) for some constant C.
Because Hess(f ) is symmetric matrix, it means that is an orthonormal
basis {e1 , e2 } which diagonalizes Hess(f ) at (0, 0). Let’s represent the function 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
1
f˜(u, v) = f (0, 0) + ∇f˜(0).(uv) + (u2 λ1 + v 2 λ2 ) + O(u3 + v 3 ).
2
Now f˜(0) = (0, 0) (why?) and hence
1
f˜(u, v) = (u2 λ1 + v 2 λ2 ) + O(u3 + v 3 ).
2
It should be clear now that if λ1 λ2 > 0 then f˜ (and hence f ) has a local
max. or a local min. at the origin. if λ1 λ2 < 0 then f˜ (and hence f ) has a
saddle point at the origin.
Proof of Proposition 0.18. Let’s prove the Proposition in the following particular case first.
Assume p is the origin, Tp S is the xy-plane, and the surface S is graphical
with respect to this plane near the origin. What this means is that one can
find a coordinate chart of the form φ(x1 , x2 ) = (x1 , x2 , f (x1 , x2 )), where f
is a smooth function and defined in a neighborhood of the origin.
Because p is the origin we must have f (0, 0) = 0. Because Tp S is the
xy-plane, we must have N (p) = (0, 0, 1) and so ∇f (0, 0) = 0. In particular
∂x1 φ(0, 0) = (1, 0, 0) and ∂x2 φ(0, 0) = (0, 1, 0).
This means the matrix g = (gij )i,j=1,2 is the identity at the origin and
thus, using the computation we did for Aij in Lecture 13 for the graphical
case, we have at the origin
σij = Aij = ∂x2i xj f.
GEOMETRY OF CURVES AND SURFACES
27
In other words, the matrix σ at the origin is exactly the hessian of f and
thus K(p) = detHess(f )(0, 0).
Now we have that (0, 0) is a critical point for f and so the second derivative test says that if detHess(f )(0, 0) > 0 then the origin is strict local minimum or maximum. This proves the first statement because, near
the 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 a
saddle point (i.e., neither maximum nor minimum), and so there are points
near the origin where f (x1 , x2 ) < 0 and points where f (x1 , x2 ) > 0. This
proves the second statement.
In Lemma 0.10 we saw that S is graphical over one of the coordinate
planes near p but what we actually proved was that S is graphical near p
over any plane which does not contain N (p), the normal vector to Tp S. In
particular, S is graphical over Tp S is a neighborhood of p. Without loss of
generality 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 points
where the curvature is strictly positive.
Note that the example S = {z = (y 2 − x2 )/2} shows the condition that
the surface is compact is crucial.
Proposition 0.20. If S is a compact surface, then there is a point p in S
where K(p) > 0.
The condition that S is bounded is crucial because S = {z = (y 2 − x2 )/2}
has negative curvature everywhere. The condition that S is closed is crucial
because S = {x2 + y 2 < 1, z = 0} has curvature identical to zero. The
example S̄ = {x2 + y 2 ≤ 1, z = 0} does not count because it is not a surface
according to our definition (it has boundary). Note that if one were to define
what is a surface with boundary, as to include the previous example, the
correct statement in the theorem would be that “compact surfaces with no
boundary 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 contradiction
with the definition of r0 .
28
ANDRÉ NEVES
Choose p ∈ S ∩S0 . We must have Tp S = Tp S0 because otherwise we could
increase r0 a bit and still have S ∩ S0 6= ∅, a contradiction. The rigorous
way of checking this is the following.
For every path γ in S so that γ(0) = p, consider f (t) = |γ(t)|2 . Then
f (t) ≤ r02 = 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 this
implies f 0 (0) = 0. On the other hand, from the definition of f we see that
f 0 (0) = 2γ 0 (0).γ(0) = 2γ 0 (0).p. As a result we obtain that γ 0 (0) is orthogonal
to p. The arbitrariness of γ implies that p is perpendicular vector to Tp S
and so Tp S = Tp S0 .
Let’s also assume, without loss of generality, that p is the origin, Tp S is
the xy-plane and that the unit normal interior to S0 is N (p) = (0, 0, 1). By
the same reasoning as in the proof of Proposition 0.18 we have that S, S0 are
graphical near p over the xy-plane, which means there is in a neighborhood
of the origin so that S = {z = f (x, y)} and S0 = {z = f0 (x, y)}, where f
and 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 be
the eigenvalues of Hess(f ) (or Hess(f0 )) at the origin. In particular,
Hess(f0 )(v, v) =
|v|2
r0
for every vector v ∈ R2 ,
because both principal curvatures of S0 are the same and equal to r0−1 .
Because the interior unit normal to S0 is (0, 0, 1) we have that f0 ≤ f
near the origin or, equivalently, that h = f − f0 has a local minimum at the
origin. 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
|v|2
.
r0
Thus, the principal curvatures of S at the origin must be higher than 1/r0 ,
and so K(p) ≥ r0−2 > 0.
Hess(h)(v, v) ≥ 0 =⇒ Hess(f )(v, v) ≥ Hess(f0 )(v, v) =
Surfaces with H = 0 are called minimal surfaces and are ubiquitous in
science. They model from soap films to black holes and the previous proposition 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 a
local isometry if
dFp (X).dFp (Y ) = X.Y
for all p ∈ S1 , X, Y ∈ Tp S1 .
GEOMETRY OF CURVES AND SURFACES
29
In particular, |dFp (X)| = |X| for all p and X ∈ Tp S.
F being a local isometry also implies that dFp : Tp S1 → TF (p) S2 is bijective (why?). Hence, from the Inverse Function Theorem, we obtain that a
local 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 a
smooth 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 local
isometry we have
length(α) = length(F ◦ α).
Conversely, if F preserves the lengths of any given curve, then F is a
local isometry.
Proof. If β(t) = F (α(t)), then from the definition of Jacobian we have b0 (t) =
dFα(t) (α0 (t)). Thus
Z
Z
Z
0
0
length(F ◦ α) = |β (t)|dt = |dF (α (t))|dt = |α0 (t)|dt = length(α).
I
I
0
α (0)
I
Likewise, given p ∈ S and
∈ Tp S we have for all t small
Z t
Z t
|α0 (s)|ds = length(α[0,t] ) = length(F (α[0,t] )) =
|dF (α0 (s))|ds.
0
0
Differentiating both sides with respect to t we obtain |α0 (0)| = |dF (α0 (0))|
which means |dFp (V )| = |V | for all p ∈ S1 and V ∈ Tp S1 .
To obtain that dF (X).dF (Y ) = X.Y for all X, Y ∈ Tp S is now simple
trick. 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.B T = B T .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 surface
you get from applying the rigid motion to S. Naturally we expect S1 to be
isometric 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 ∈ Tp S we have dF (X) = BX and so
dF (X).dF (Y ) = (BX).(BY ) = (BX)T (BY ) = X T B T BY = X T Y = X.Y,
30
ANDRÉ NEVES
where U T 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 a
cylinder S2 = {(x, y, z) : x2 + y 2 = 1}. Hence, even if their shapes are
different, their intrinsic distances are the same.
Consider F : S1 → F (S1 ) ⊂ S 2 given by F (x, y, 0) = (cos x, sin x, y). F
is clearly smooth. Set e1 = (1, 0, 0) and e2 = (0, 1, 0). We have Tp S1 =
span{e1 , e2 } for all p ∈ S1 and
∂F
= (− sin x, cos x, 0),
∂x
∂F
dF (e2 ) =
= (0, 0, 1).
∂y
dF (e1 ) =
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 all X, Y ∈
Tp S1 .
The next lemma says that two surfaces are isometric if we can find charts
so that the matrix g = (gij )i,j=1,2 (defined in Lecture 15) are the same for
both 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 :=
∂ψ ∂ψ
∂φ ∂φ
.
=
.
∂xi ∂xj
∂xi ∂xj
on
U.
Conversely, if φ : U → S1 , ψ : U → S2 are charts so that
∂ψ ∂ψ
∂φ ∂φ
.
=
.
∂xi ∂xj
∂xi ∂xj
on
U
for all
i, j = 1, 2.
then φ(U ) is isometric to ψ(U ).
Proof. We have, for i = 1, 2,
∂φ
∂(F ◦ φ)
∂ψ
)=
=
∂xi
∂xi
∂xi
and so, because F is an isometry,
dF (
∂ψ ∂ψ
∂φ
∂φ
∂φ ∂φ
.
= dF (
).dF (
)=
.
.
∂xi ∂xj
∂xi
∂xj
∂xi ∂xj
To prove the second statement set F = ψ ◦ φ−1 : φ(U ) → ψ(U ). There is
the usual thing that φ−1 is not smooth and so technically speaking F is not
smooth but we already saw how to justify that so let’s just ignore it.
GEOMETRY OF CURVES AND SURFACES
31
F is bijective so we need to check that it preserves the dot product. Tp S1
is spanned by ∂x1 φ, ∂x2 φ and we have
dF (∂xi φ).dF (∂xj φ) = ∂xi (F ◦ φ).∂xj (F ◦ φ) = ∂xi ψ.∂xj ψ = ∂xi φ.∂xj φ
for all i, j = 1, 2. It is easy to see that this implies that dF (X).dF (Y ) = X.Y
for all X, Y ∈ Tp S1 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 intrinsic quantity, i.e., if F : S1 → S2 is a local isometry and K1 , K2 denote
the Gaussian curvature of S1 , S2 , then K2 ◦ F = K1 .
Note that the mean curvature H is not an intrinsic notion because, for
instance, 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 that
the trace of −dN is not intrinsic while the determinant of −dN is. Another
way of understanding how awesome this theorem is, is to note that K was
defined as the determinant of g −1 A, where g has intrinsic information only,
while A has information how the surfaces lies in space. Nonetheless, the
determinant only has intrinsic information, i.e., can be computed knowing
g only.
Proof of Theorema Egregium. The strategy is the following. Let S be a
surface with Gaussian curvature K, and φ : U → S a chart.
We want to show that K can be computed knowing only the matrix
g = (gij )i,j=1,2 , where gij = ∂xi φ.∂xj φ. More precisely, we would like to find
an expression for K which depends only on g, the first derivatives of g, and
the second derivatives of g.
If true, then for every other surface N which is locally isometric to S, we
can apply Lemma 0.23 to find some chart ψ into N so that the correspondent matrix hij = ∂xi ψ.∂xj ψ is identical to gij . As a result, the Gaussian
curvature of N , KN , can now be computed using the very same expression
mentioned in the previous paragraph and so it must be identical to K.
Claim 1: Let’s start by showing that
Kdet g = ∂x1 ([∂x22 x2 φ]T ).∂x1 φ − ∂x2 ([∂x21 x2 φ]T ).∂x1 φ,
where, given a vector X ∈ R3 , we denote by X T its tangential projection on
Tp S, i.e., X T = X − (X.N )N .
We have
[∂x22 x2 φ]T = ∂x22 x2 φ − (∂x22 x2 φ.N )N
[∂x21 x2 φ]T = ∂x21 x2 φ − (∂x21 x2 φ.N )N
32
ANDRÉ NEVES
and so differentiating both sides we obtain
∂x1 [∂x22 x2 φ]T = ∂x31 x2 x2 φ − ∂x1 (∂x22 x2 φ.N )N − (∂x22 x2 φ.N )∂x1 N
∂x2 [∂x21 x2 φ]T = ∂x32 x1 x2 φ − ∂x2 (∂x21 x2 φ.N )N − (∂x21 x2 φ.N )∂x2 N
which means
∂x1 [∂x22 x2 φ]T .∂x1 φ = ∂x31 x2 x2 φ.∂x1 φ − (∂x22 x2 φ.N )∂x1 N.∂x1 φ
∂x2 [∂x21 x2 φ]T .∂x1 φ = ∂x32 x1 x2 φ.∂x1 φ − (∂x21 x2 φ.N )∂x2 N.∂x1 φ
and so
∂x1 [∂x22 x2 φ]T .∂x1 φ − ∂x2 [∂x21 x2 φ]T .∂x1 φ
= −(∂x22 x2 φ.N )∂x1 N.∂x1 φ + (∂x21 x2 φ.N )∂x2 N.∂x1 φ.
Recalling Lemma 3.1 in Lecture 13 we can write this as
∂x1 [∂x22 x2 φ]T .∂x1 φ − ∂x2 [∂x21 x2 φ]T .∂x1 φ = A22 A11 − A212 = det A.
We know that K = det σ = det (g −1 A) = det A/det g and so det A = K det g
which means
Kdet g = ∂x1 ([∂x22 x2 φ]T ).∂x1 φ − ∂x2 ([∂x21 x2 φ]T ).∂x1 φ,
as desired.
Lecture 23
Claim 2: [∂x22 x2 φ]T can be computed in terms of g.
[∂x22 x2 φ]T lies in the tangent plane of S and so can be written as
[∂x22 x2 φ]T = Γ122 ∂x1 φ + Γ222 ∂x2 φ
for some Γ122 , Γ222 . We have
∂ x2
g22
∂x φ.∂x2 φ
= ∂ x2 2
= ∂x22 x2 φ.∂x2 φ = Γ122 g12 + Γ222 g22
2
2
and
g22
∂x φ.∂x2 φ
= ∂x2 (∂x2 φ.∂x1 φ) − ∂x1 2
2
2
= ∂x22 x2 φ.∂x1 φ + ∂x2 φ.∂x22 x1 φ − ∂x2 φ.∂x21 x2 = ∂x22 x2 φ.∂x1 φ
∂x2 g12 − ∂x1
= Γ122 g11 + Γ222 g21 .
In sum, using matrix notation,
1 ∂x2 g12 − ∂x1 g222
g11 g12
Γ22
=
∂x2 g222
g21 g22
Γ222
GEOMETRY OF CURVES AND SURFACES
33
and so we can solve it as
1 −1 Γ22
g11 g12
∂x2 g12 − ∂x1 g222 .
=
g21 g22
∂x2 g222
Γ222
1
g22 −g12
∂x2 g12 − ∂x1 g222 .
=
∂x2 g222
g11 g22 − (g12 )2 −g12 g11
Claim 3: ∂x1 [∂x22 x2 φ]T .∂x1 φ can be computed in terms of g.
From
[∂x22 x2 φ]T = Γ122 ∂x1 φ + Γ222 ∂x2 φ
we deduce that
∂x1 [∂x22 x2 φ]T .∂x1 φ
= ∂x1 Γ122 g11 + ∂x1 Γ222 g12 + Γ122 ∂x1 x1 φ.∂x1 φ + Γ222 ∂x2 x1 φ.∂x1 φ
g11
g11
= ∂x1 Γ122 g11 + ∂x1 Γ222 g12 + Γ122 ∂x1
+ Γ222 ∂x2
.
2
2
From Claim 2 we see that indeed ∂x1 [∂x22 x2 φ]T .∂x1 φ can be computed only
in terms of g.
Claim 4: ∂x2 [∂x21 x2 φ]T .∂x1 φ can be computed in terms of g.
This one is done like the previous cases and we obtain the following.
Setting
[∂x21 x2 φ]T = Γ112 ∂x1 φ + Γ212 ∂x2 φ
one can see that
1 1
g22 −g12
∂x2 g211
Γ12
=
∂x1 g222
Γ212
g11 g22 − (g12 )2 −g12 g11
and then
g11
g11
+ Γ212 ∂x2
.
2
2
Putting all these four claims together we see that we can find an expression
for K which depends only on the metric g and its derivatives.
For the sake of bookkeeping we obtain
∂x2 [∂x21 x2 φ]T .∂x1 φ = ∂x1 Γ112 g11 + ∂x1 Γ212 g12 + Γ112 ∂x1
K=
where
and
1
(∂x1 Γ122 − ∂x1 Γ112 )g11 + (∂x1 Γ222 − ∂x1 Γ212 )g12
g11 g22 − (g12 )2
g11
g11 + (Γ222 − Γ212 )∂x2
+(Γ122 − Γ112 )∂x1
2
2
Γ122
Γ222
1
=
g11 g22 − (g12 )2
Γ112
Γ212
g22 −g12
−g12 g11
1
=
g11 g22 − (g12 )2
g22 −g12
−g12 g11
∂x2 g12 − ∂x1 g222
∂x2 g222
∂x2 g211
∂x1 g222
34
ANDRÉ NEVES
Lecture 24
We now show the following theorem
Theorem 0.25 (Rigidity of sphere). Let S be a connected compact (closed
and bounded) surface with constant positive Gaussian curvature. Then S is
a 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 not
identical) such as the plane and the cylinder. So the above result is a specific
phenomena of positive Gaussian curvature.
The second one is that if we imagine a sphere being made of some inelastic
but flexible material (like paper) then we cannot deform it in any way apart
from rigid motions. Note that we can find compact isometric surfaces which
do not differ by rigid motions. It is an open problem to know whether
compact surfaces with no boundary are locally rigid.
The final remark is that S being compact is essential (see the exercise
below).
Exercise: Consider
S = {(φ(t) cos θ, φ(t) sin θ, ψ(t)) : 0 < θ < 2π, a < t < b}
a surface of revolution, where (φ0 (t))2 + (ψ 0 (t))2 = 1 for all t.
(1) Show that K = −φ00 /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 is
umbilical and then to use Theorem 4.2 of Lecture 14 to conclude that S
must be a sphere.
At every point x ∈ S we have principal curvatures λ1 (x) ≤ λ2 (x). Choose
a point p ∈ S so that
max λ2 (x) = λ2 (p).
x∈S
The fact that K = λ1 λ2 is constant implies that
min λ1 (x) = λ1 (p).
x∈S
Note that we necessarily have λ1 (p) ≤ λ2 (p). If we show that equality holds
then we must have λ1 (x) = λ2 (x) for all x ∈ S and so every point in S is
umbilical, which means we are done.
Claim: Suppose we have K > 0 and for some p ∈ S
λ1 (p) = min λ1 (x) ≤ max λ2 (x) = λ2 (p).
x∈S
x∈S
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
GEOMETRY OF CURVES AND SURFACES
35
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) = ∂x2i xj f (0). Moreover we can assume that the basis we choose for Tp S = {z = 0} diagonalizes
Hess(f )(0). Note that in this case the metric matrix is the identity and so
σij = Aij (recall Lecture 15), which means the eigenvalues of Hess(f )(0)
are λ1 , λ2 .
Consider the unit vectors
(1, 0, ∂x1 f )
(0, 1, ∂x2 f )
E1 = p
, E2 = p
1 + (∂x1 f )2
1 + (∂x2 f )2
which lie in the tangent plane of S and look at the functions
h1 (t) = A(E1 (0, t), E1 (0, t)) =
∂x21 x1 f
1
p
1 + (∂x1 f )2 1 + |∇f |2
∂x22 x2 f
1
p
.
1 + (∂x2 f )2 1 + |∇f |2
We have, from Exercise in Lecture 16, that
h2 (t) = A(E2 (t, 0), E2 (t, 0)) =
λ2 ≥ λ2 (t, 0) = max A(X, X) ≥ A(E2 (t, 0), E2 (t, 0)) = h2 (t),
{|X|=1}
which means h1 has a local minimum at the origin, h2 has a local maximum
at the origin and so h001 (0) − h002 (0) ≥ 0.
We will now compute h001 (0), h002 (0) and get a contradiction. We can either
just compute it or argue in the following longer but more intuitive way. Note
that by Taylor formula
f (x1 , x2 ) = λ1
x2
x21
+ λ2 2 + O(|x1 |3 + |x2 |3 )
2
2
and so
∇f (x1 , x2 ) = (λ1 x1 , λ2 x2 ) + O(|x1 |2 + |x2 |2 )
which means
(∂x1 f (0, t))2 = O(t3 ),
|∇f |2 (0, t) = λ22 t2 + O(t3 )
and hence
2
∂x21 x1 f (0, t)
3
2
2t
h1 (t) = p
+
O(t
)
=
∂
f
(0,
t)
1
−
λ
+ O(t3 ).
x1 x1
2
2
2
2
1 + λ2 t
Likewise we have
h2 (t) =
∂x22 x2 f (t, 0)
2
2t
+ O(t3 )
1 − λ1
2
and so
h001 (0) = ∂x42 x2 x1 x1 f (0, 0) − λ22 ∂x21 x1 f (0, 0) = ∂x42 x2 x1 x1 f (0, 0) − λ22 λ1
h002 (0) = ∂x41 x1 x2 x2 f (0, 0) − λ21 ∂x22 x2 f (0, 0) = ∂x41 x2 x2 x2 f (0, 0) − λ21 λ2 .
Therefore, h001 (0) − h002 (0) = λ1 λ2 (λ1 − λ2 ) < 0, a contradiction.
36
ANDRÉ NEVES
Lecture 25
Before we proceed and state the Gauss-Bonnet Theorem we need to be
able to integrate functions on surfaces.
Say that S is a surface with a chart φ : U ⊂ R2 → S. Given a compact
set D ⊂ U we define the area of φ(D) to be
Z
area(φ(D)) =
|∂x φ × ∂y φ| dxdy.
D
To see this makes sense recall that if ~u, ~v are two vectors in R3 , then they
span a parallelogram P in R3 with edges 0, ~u, ~v , ~u + ~v and the area of P is
exactly |~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
Z
area(P ) = area(φ(D)) =
|∂x φ × ∂y φ|dxdy
D
Z
|~u × ~v |dxdy = area(D)|~u × ~v | = |~u × ~v |
=
D
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 |∂x φ × ∂y φ| = detg and so
Z p
area(φ(D)) =
detg dxdy.
D
Proof. Given ~u, ~v two vectors we have |~u × ~v |2 = |~u|2 |~v |2 − (~u.~v )2 . This can
be seen at once from the fact that ~u and w
~ = ~v − (~u.~v )~u/|~u|2 are orthogonal
vectors 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
|∂x φ × ∂y φ|2 = |∂x φ|2 |∂y φ|2 − (∂x φ.∂y φ)2 = g11 g22 − (g12 )2 = detg.
Finally, given a function f defined on a surface S, if φ : U → S is a chart
then for every D compact set of U we define
Z
Z
Z
p
f dA =
f ◦ φ|∂x φ × ∂y φ|dxdy =
f ◦ φ detg dxdy.
φ(D)
D
D
We need to understand the case where we want to integrate a function
over 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 , . . . , Sk
of sets where
• S = S1 ∪ . . . ∪ Sk with Si ∩ Sj ⊂ ∂Si if i 6= j;
GEOMETRY OF CURVES AND SURFACES
37
• we have charts φi : Ui → S with Si = φi (Di ), where Di is a compact
subset of U .
Then we define
Z
k Z
X
f dA =
f ◦ φi |∂x φi × ∂y φi | dxdy.
S
i=1
Di
Of course for all this to make sense one needs to show that the value for the
integral does not depend on the decomposition made and one also needs to
make sure that the sets Si are somewhat “nice”. Let’s ignore this technical 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 S 2 .
The goal is still to be able to state the Gauss-Bonnet Theorem so we need
to 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 assume
orientable with a normal vector N . We assume the curve is parametrized
by 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 00 .
Consider E = N × γ 0 . Then N, γ 0 , E is an orthonormal basis for R3 and
so, because ~k is orthogonal to γ 0 , we can write it as
~k = (~k.N )N + (~k.E)E = kn N + kg E
kn is the normal curvature and we saw on Lecture 12 that A(γ 0 , γ 0 ) = ~k.N =
kn , i.e., kn is the amount γ is “forced” to cuve. kg is the geodesic curvature
and 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 seen
as a curve in R2 .
If S is the unit sphere {|~x| = 1} and γ(θ) = (cos θ, sin θ, 0), then choosing
N to be the interior unit normal
kn = A(γ 0 , γ 0 ) = γ 0 .γ 0 = 1
and
E = N × γ 0 = −γ × γ 0 = −(cos θ, sin θ, 0) × (− sin θ, cos θ, 0) = −(0, 0, 1),
which means kg = γ 00 .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 which
consists of three smooth curves, called the edges. The endpoints of these
curves are called the vertices of T (there are exactly three of them).
38
ANDRÉ NEVES
Given a compact surface Σ we say T = {T1 , . . . Tn } is a triangulation if
• Σ = ∪ni=1 Ti 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 a
triangulation.
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 distinct
edges, and V the number of distinct vertices. The great fact is that given any
two 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 triangulation
we used. For this reason we denoted it by χ(Σ).
If Σ, Σ0 are two surfaces which are homemorphic, i.e., we can find a a
continuos map F : Σ → Σ0 which is bijective, then χ(Σ) = χ(Σ0 ).
Let’s compute it in some examples. If Σ = {(x, y, 0) : x2 + y 2 ≤ 1} is a
disc, then χ(Σ) = 1. If Σ = {x1 + y 2 + z 2 = 1} is a sphere, then χ(Σ) = 2.
If Σ = {x2 + y 2 = 1, 0 ≤ z ≤ 1} is a cylinder, then χ(Σ) = 0. If Σ is a torus
then χ(Σ) = 0.
The general rule is the following. If Σ is a compact surface with no
boundary then χ(Σ) = 2 − 2g, where g is the number of handles the surface
has. 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 compact
surface Σ with boundary ∂Σ. Then
Z
Z
kg ds +
K dA = 2πχ(Σ).
Σ
∂Σ
The boundary of Σ is assumed to be oriented positively.
If ∂Σ = ∅, we have
Z
K dA = 2πχ(Σ).
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 normal
N on Σ. Given an arc-length parametrization γ of ∂Σ, we day that ∂Σ
is oriented positively if N × γ 0 points toward the inside of Σ. Note that
N × γ 0 is perpendicular to N and so lies in Tγ(t)Σ . Moreover, N × γ 0 is also
perpendicular to ∂Σ and so it either points away from Σ or into Σ.R
Note that such a condition is necessary because
R clearly χ(Σ) and Σ K dA
do not depend on how ∂Σ is oriented, while ∂Σ kg ds changes sign if we
GEOMETRY OF CURVES AND SURFACES
39
change the orientation of ∂Σ. Thus it is important that we have a canonical
way to choose the correct orientation.
For instance, if Σ = {(x, y, 0) : x2 + y 2 ≤ 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 0 (t) = (− sin t, cos t, 0),
N × γ 0 (t) = (− cos t, − sin t, 0), N = (0, 0, 1)
is a positively oriented basis. In this case kg = γ 00 .(N × γ 0 ) = 1 and indeed
Z
Z
Z
0 dA = 2π = 2πχ(Σ).
K dA = length(∂Σ) +
kg ds +
Σ
Σ
∂Σ
If ∂Σ = ∅, then the Gauss-Bonnet theorem
is the surface analog of the fact
R
that 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 topological
term (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 chart
φ : U ⊂ R2 → Σ so that
• φ is smooth in Ū , the closure of U , and Ū is diffeomorphic to a disc.
• Σ = φ(Ū ) with boundary ∂Σ = φ(∂U ).
Then
Z
Z
kg ds +
KdA = 2π,
∂Σ
Σ
where ∂Σ is positively oriented.
The basic idea is the following. In the setting of the Theorem above we
know from Lecture 25 that
Z
Z
KdA =
K ◦ φ|∂x1 φ × ∂x2 φ|dx1 dx2 .
Σ
U
The idea is to find functions M and L so that
K ◦ φ|∂x1 φ × ∂x2 φ| = ∂x1 M − ∂x2 L.
Hence from Green’s Theorem we have that (assuming the boundary of U is
oriented counterclockwise)
Z
Z
K ◦ φ|∂x1 φ × ∂x2 φ|dx1 dx2 =
Ldx1 + M dx2 .
U
∂U
We have gone from a double integral to a line integral and so the next thing
will be to show that
Z
Z
Ldx1 + M dx2 = 2π −
kg ds.
∂U
∂Σ
This identity will just follow from a simple computation plus a topological
fact.
40
ANDRÉ NEVES
Proof. The strategy is the following. Choose e1 : U → R3 , e2 : U → R3 so
that {e1 , e2 } is an orthonormal basis for Tφ(x) Σ for all x ∈ Ū . One way to
do this would be to choose
∂x1 φ
∂x2 φ − (∂x1 φ.e1 )e1
.
e1 =
, e2 =
|∂x1 φ|
|∂x2 φ − (∂x1 φ.e1 )e1 |
Moreover, we assume that N = e1 × e2 (switch the order if this does not
hold.)
Parametrize ∂Σ by a closed curve γ : [0, l] → ∂Σ with |γ 0 (t)| = 1. There is
σ a parametrization of ∂U so that γ = φ◦σ. We assume the parametrization
is positively oriented.
We can find a continuous function θ(t) so that
γ 0 (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.
0
0
(1) Show that k
R g =0 θ (t) − e1 (t).e2 (t).
(2) Show that σ θ (t)dt = 2π.
(3) Show that
Z l
Z
e1 .e02 dt =
K|∂x1 φ × ∂x2 φ|dx1 dx2 .
0
U
Putting (1), (2), and (3) together we obtain
Z
Z
Z l
Z
kg ds +
KdA =
kg dt +
K|∂x1 φ × ∂x2 φ|dx1 dx2
∂Σ
Σ
0
Z
=
l
θ0 dt −
U
Z
0
l
0
e1 .e02 dt +
Z
K|∂x1 φ × ∂x2 φ|dx1 dx2 = 2π.
U
Let’s prove (1).
We have
γ 0 = cos θe1 + sin θe2 =⇒ γ 00 = θ0 (− sin θe1 + cos θe2 ) + cos θe01 + sin θe02 .
Now e1 ×e2 = N and e1 , e2 being an orthonormal basis implies that e2 ×N =
e1 and N × e1 = e2 . Thus
N × γ 0 = N × (cos θe1 + sin θe2 ) = − sin θe1 + cos θe2 .
Therefore, using the fact that
e1 .e01 = e2 .e02 = 0 (why?) e1 .e02 = −e01 .e2 (why?)
we have
kg = γ 00 .(N × γ 0 ) = γ 00 .(− sin θe1 + cos θe2 )
= θ0 + (cos θe01 + sin θe02 ).(− sin θe1 + cos θe2 ) = θ0 + cos2 θe01 .e2 − sin2 θe1 .e02
= θ0 − (cos2 θ + sin2 θ)e1 .e02 = θ0 − e1 .e02 .
Let’s prove (3) and then (2).
GEOMETRY OF CURVES AND SURFACES
41
We will show that
∂x1 e1 .∂x2 e2 − ∂x1 e2 .∂x2 e1 = K
p
detg.
Assuming this for a moment, (3) follows because, writing σ(t) = (x1 (t), x2 (t))
we have from the chain rule
de2
e02 (t) =
(x1 (t), x2 (t)) = x01 ∂x1 e2 + x02 ∂x2 e2
dt
and so we obtain from Green’s Theorem that
Z l
Z l
0
e1 .e2 dt =
x01 e1 .∂x1 e2 + x02 e1 .∂x2 e2 dt
0
0
Z
Z
Z
∂x1 (e1 .∂x2 e2 )−∂x2 (e1 .∂x1 e2 )dx1 dx2
e1 .∂x2 e2 dx2 =
=
e1 .∂x1 e2 dx1 +
U
∂U
∂U
Z
Z
p
K detgdx1 dx2 .
(∂x1 e1 .∂x2 e2 − ∂x1 e2 .∂x2 e1 )dx1 dx2 =
U
U
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 some
αi and βi
∂xi e1 = αi e2 + (∂xi e1 .N )N = αi e2 − (e1 .∂xi N )N = αi e2 + A(e1 , ∂xi φ)N
and
∂xi e2 = βi e1 + (∂xi e2 .N )N = βi e1 + A(e2 , ∂xi φ)N
Thus
∂x1 e1 .∂x2 e2 − ∂x1 e2 .∂x2 e1
= A(e1 , ∂x1 φ)A(e2 , ∂x2 φ) − A(e2 , ∂x1 φ)A(e1 , ∂x2 φ)
Let’s write ∂xi φ = ai1 e1 + ai2 e2 , i = 1, 2. Because {e1 , e2 } is an orthonormal
basis we must have
p
detg = (∂x1 φ × ∂x2 φ).N = (a11 a22 − a12 a21 )N.N = (a11 a22 − a12 a21 )
Now, writing σij = A(ei , ej ) = −ei .dN (ej ), we have
= A(e1 , ∂x1 φ)A(e2 , ∂x2 φ) − A(e2 , ∂x1 φ)A(e1 , ∂x2 φ)
= (a11 σ11 + a12 σ12 )(a21 σ21 + a22 σ22 ) − (a11 σ21 + a12 σ22 )(a21 σ11 + a22 σ12 )
p
= (a11 a22 −a12 a21 )σ11 σ22 −(a11 a22 −a12 a21 )σ12 σ21 = detg(σ11 σ22 −σ12 σ21 )
p
= 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 that
γ = φ ◦ σ, where σ is a parametrization of ∂U . We proved in (1) that
42
ANDRÉ NEVES
kg = θ0 because e02 = 0. In this simplified case, the geodesic curvature kg
of γ coincides with the curvature k of the planar curve σ and so we obtain
that
Z
Z
Z l
0
kds = 2πw(σ) = 2π,
θ ds = kg ds =
0
γ
σ
where w(σ) (winding number) is one because σ is the boundary of a domain
U that is diffeomorphic to a disc.
0
0
R In0 the general case, if we write σ /|σ | = (cos β, sin β) 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.
R 0
Some
basic
results
in
algebraic
topology
imply
that
we
still
have
β dt =
R 0
θ dt even if the functions are different (they key word is that they are
homotopic!).
Now we want to prove Gauss-Bonnet for triangles. First we need to define
the 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 curves
so that γ1 (l) = γ2 (0). We assume the curves parametrized by arc length
and we assume that we have an orthonormal frame {e1 , e2 } which is defined
along γ1 ∪ γ2 and changes smoothly (even in the curve γ1 ∪ γ2 has one vertex
point). We set N = e1 × e2 . Finally we also assume that we have no cusp,
i.e., γ10 (l) 6= −γ20 (0).
We now define the exterior angle θ that γ10 (l) makes with γ20 (0). Loosely
speaking it is the counterclockwise angle in {e1 , e2 } that γ10 (l) makes with
γ20 (0). For the correct definition choose θ ∈ (−π, π) as the unique angle θ
so that cos θ = γ10 (l).γ20 (0) and θ has the positive sign if {γ10 (l), γ20 (0), N } is
a positive basis and the negative sign if {γ10 (l), γ20 (0), N } is a negative basis.
The only case not covered is if γ10 (l) and γ20 (0) are linearly dependent, in
which case γ10 (l) = γ20 (0) and so θ = 0. Note that if we switch e1 with e2 or
the 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. Consider
also a chart φ : U ⊂ R3 where T̄ ⊂ U and Σ = φ(T ). Then ∂Σ consists of 3
smooth 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 the
endpoint 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 }, and
similarly for θ2 and θ3 .
Lecture 29
We now state the triangular version of the Gauss-Bonnet Theorem.
GEOMETRY OF CURVES AND SURFACES
43
Theorem 0.29 (Triangular version of Gauss-Bonnet). Under the conditions
described at the end of the previous lecture we have
Z
3 Z
3
X
X
kg ds +
KdA = 2π.
θi +
γi
i=1
Σ
i=1
Proof. Like in the local version we have that for each γi , the geodesic curvature satisfies kg = θ0 − e1 .e02 . Moreover, step (3) carries over because Green’s
Theorem also holds for triangular domains, i.e.,
Z
Z
p
0
K ◦ φ detgdx1 dx2 .
e1 .e2 dt =
T
∂T
It is a topological theorem that
3 Z
X
i=1
θ0 dt +
γi
3
X
θi = 2π.
i=1
This should not come as a shock for the following reason. When ∂Σ is
smooth, i..e, θ1 = θ2 P
= 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 Z
X
i=1
kg ds +
γi
=
3
X
Z
θi
i=1
3 Z
X
i=1
+
KdA
Σ
kg ds +
γi
3
X
i=1
Z
θi −
e1 .e02 dt +
Z
∂T
KdA
Σ
=
3 Z
X
i=1
γi
kg ds +
3
X
θi = 2π.
i=1
The triangular version of Gauss Bonnet gives us the following classical
interpretation of Gaussian curvature. Let’s consider a triangle T inside a
surface Σ where the sides of T are geodesics, i.e., they have kg = 0. If Σ is a
plane, this is just saying that T is a proper triangle, i.e., the sides are straight
lines. We define the interior angles βi of the triangle to be βi = π − θi . Then
the theorem says that
Z
KdA = β1 + β2 + β3 − π.
T
When K = 0, this is saying that the sum of the interior angles of a triangle
is π (Thales’
Theorem), something we probably learned in high school. In
R
general, 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
ANDRÉ NEVES
is bigger than π and if the surface has negative curvature then the sum of
the 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 compact
surface Σ with boundary ∂Σ. Then
Z
Z
K dA = 2πχ(Σ).
kg ds +
Σ
∂Σ
The boundary of Σ is assumed to be oriented positively.
If ∂Σ = ∅, we have
Z
K dA = 2πχ(Σ).
Σ
Proof. Let’s assume that ∂Σ = ∅ and the case with boundary I leave it as
an exercise.
Consider a triangulation T = {Ti }Fi=1 of Σ and note that by making each
triangle very small, we can assume that for each T ∈ T there is a chart
φ : U → Σ so that T ⊂ φ(U ). Moreover, we can also assume that no T ∈ T
has cusps because if this happens for some triangle we can always bend the
edges 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 at
each vertex by θij , j = 1, 2, 3. Orient ∂Ti positively for all i. Then we have
F Z
X
i=1
Z
KdA +
Ti
kg ds +
∂Ti
F X
3
X
i=1 j=1
θij =
F
X
2π = 2πF.
i=1
Each edge of ∂Ti is also an edge of another triangle Tj with the opposite
orientation. Hence
Z
kg ds = 0.
∂Ti
Thus, denoting the internal angle by φij = π − θij we obtain
Z
KdA +
Σ
F
X
(π −
i=1
3
X
φij ) = 2πF
j=1
and so
Z
KdA + 3πF −
Σ
F X
3
X
φij = 2πF.
i=1 j=1
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 the
edge of two triangles, which means that each edge is being counted twice.
Thus the number of elements is also 2E.
GEOMETRY OF CURVES AND SURFACES
45
If we sum all the internal angles around a vertex we get 2π and so
F X
3
X
φij = 2πV.
i=1 j=1
Putting all these together
Z
KdA + 2πE − 2πV = 2πF
Σ
Z
KdA = 2π(F − E + V ) = 2πχ(Σ).
=⇒
Σ
Lecture 30
We now derive some basic consequences of Gauss-Bonnet.
Consequence 1: If Σ is a compact surface with no boundary which has
K ≥ 0, then it must homeomorphic to a sphere. Stating differently, every
surface which has genus g > 0 must have at least one point with negative
Gaussian curvature.
The reason is the following. From Gauss-Bonnet we know that
Z
0≤
KdA = 2πχ = 2π(2 − 2g).
Σ
Thus either g = 0, in which case Σ is topologically a sphere, or g = 1, in
which case Σ is topologically
R a torus. We now rule out this latter case. If
g = 1, then we would have Σ KdA = 0, which means K = 0 because K ≥ 0.
On the other hand, we know that every compact surface with no boundary
mast have a point p where K(p) > 0 and so this means that K = 0 cannot
happen.
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 intersect. The interpretation is the following: geodesics are Euclidean analogs of
straight lines and so what this result is saying is that under positive Gaussian curvature every two “straight lines” must intersect, i.e., Euclides fifth
axiom 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 so
that ∂S = γ1 ∪ γ2 (this is a standard result in algebraic topology). Applying
Gauss-Bonnet we have
Z
Z
Z
kg ds +
KdA = 2πχ(S) =⇒ 0 <
KdA = 2πχ(S).
∂S
S
S
46
ANDRÉ NEVES
We know that χ(S) = 2 − 2g − 2 ≤ 0 (S has two boundary components) and
this contradicts the formula above.
Consequence 3: Let S be a surface with K ≤ 0 and diffeomorphic to a
disc. 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]). Then
there 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 with
vertices p, q (because ∂Σ is smooth outside p and q, the other vertice will
have exterior angle 0 and so we do not worry about it). Hence we obtain
from the triangular version of Gauss-Bonnet that
Z
Z
Z
kg ds + θ1 + θ2 +
KdA = 2π =⇒
KdA = 2π − θ1 − θ2 .
a∪b
Σ
Σ
R
The angles θ1 and θ2 are strictly smaller than π and so Σ KdA > 0. This
contradicts K ≤ 0. The fact that θ1 , θ2 are strictly smaller than π is not
obvious but it happens is that were the case then a would be equal to b.