GEOMETRY OF CURVES AND SURFACES
ANDRÉ NEVES
1. Lecture 6
The informal definition of surface is the following. We say a set S ⊂ R3
is a surface if locally looks just like a disc in R2 that is, 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 1.1. 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 all the sense but one might wonder
why conp
dition iii). Well, 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.
A cheap 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
= (1, 0, ∂x f (x, y))
∂x
and
∂φ
= (0, 1, ∂y f (x, y))
∂y
are always linearly independent vectors (why?).
1
2
ANDRÉ NEVES
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}.
If we ask whether S is a surface, the answer is not so fast because 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 1.2. 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. 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 1.2 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 9: Show that S 2 (see above) is a surface by finding the appropriate
charts.
2. Lecture 7
To prove the proposition we need to recall the following theorem.
Theorem 2.1 (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.
This theorem should be proven on any decent multivariable calculus book.
The content 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.
GEOMETRY OF CURVES AND SURFACES
3
Proof of Proposition 1.2. 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.
Exercise 10: 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.
For the purpose of computations, it is useful to know that every surface
S is locally the graph of some function.
Lemma 2.2. Given a surface S and p ∈ S we can fins 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. Choose a chart φ : U ⊂ R2 → R3 so that φ(u0 , v0 ) = p. Set
P = span{∂u φ(u0 , v0 ), ∂v φ(u0 , v0 )} = dφ|(u0 ,v0 ) (R2 ).
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) and consider the projection
l(x, y, z) = (x, y).
We want to find a chart ψ : U 0 → S ⊂ R3 with ψ(U 0 ) a neighborhood of
p so that l ◦ ψ(x, y) = (x, y). This means that ψ(x, y) = (x, y, h(x, y)) and
so the surface is graphical over the xy-plane.
We have that l(P ) = R2 because otherwise P ∩ ker l = P ∩ span{(0, 0, 1)}
would be nonzero. Thus if we consider the map T = l ◦ φ : U → R2 we have
dT|(u0 ,v0 ) (R2 ) = l(P ) = R2 and so dT|(u0 ,v0 ) is invertible. From the inverse
function theorem we have that T −1 is defined on an open set U 0 ⊂ R2
containing T (u0 , v0 ) and T −1 (U 0 ) ⊂ U .
Set ψ = φ ◦ T −1 : U 0 → R3 . Then
l ◦ ψ(x, y) = l ◦ φ ◦ T −1 (x, y) = T ◦ T −1 (x, y) = (x, y)
and so we are done.
Exercise 11: Consider S 2 = {|~x| = 1}. Decompose S into open sets, so
that each is graphical over some coordinate plane and say what the defining
function is.
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ANDRÉ NEVES
3. Lecture 8
Before we finish talking about the definition of surface and move onto
more interesting stuff, there is one last issue we need to address. There
is nothing unique about the choice of charts φ that define a surface. 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 V =
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 an injective map with
dh|q bijective for all q ∈ A0 . The map h is called change of parameters and
naturally ψ = φ ◦ h.
There is technical issue that we address now. The fact that h is differentiable does not follow from general theory because 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 2.2 we know that for every q ∈ A0 , there is a neighborhood of h(q) in which T = l ◦ φ is invertible, where l is the orthogonal
projection in some coordinate plane. We have h = φ−1 ◦ ψ = T −1 ◦ l ◦ ψ and
written this way we see that h is the composition of two differentiable maps
(T −1 and l ◦ ψ) and hence differentiable.
Exercise 12: Show that dh|q is indeed bijective for all q ∈ A0 .
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 }
at p = (1, 1, 1) then it seems tricky. Thus the following lemma is clearly
helpful.
Lemma 3.1. Let φ : U → S ⊂ R3 a chart with φ(x0 , y0 ) = p. Then
∂φ
∂φ
(x0 , y0 ),
(x0 , y0 ) .
Tp S = span
∂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
GEOMETRY OF CURVES AND SURFACES
5
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}.
Let’s work out one more class of examples.
Lemma 3.2. 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}
4. Lecture 9
We start by proving the lemmas of last lecture.
Proof of Lemma 3.1. Set L = span{∂x φ(x0 , y0 ), ∂y φ(x0 , y0 )}. First we argue
that L ⊂ Tp S. Given ~v = a∂x φ + b∂y φ, 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. The idea is to
find a smooth curve γ in U with γ(0) = (x0 , y0 ) and α = ψ ◦ γ. 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
α0 (0) =
To find γ we want to say that γ = φ−1 ◦ α but we need to be careful
because φ−1 is not differentiable. The formal way to argue is the following.
Without loss of generality we can assume from Lemma 2.2 that we have
a chart ψ : U 0 → S of the form ψ(u, v) = (u, v, h(u, v)) for some function
h. If α(t) = (u(t), v(t), w(t)), then if we set σ(t) = (u(t), v(t)) we must
have ψ ◦ σ(t) = α(t). Moreover, by the change of parameters discussion,
there is a change of parameters h so that φ ◦ h = ψ in a neighborhood of
(u(0), v(0)). Thus if we set γ = h ◦ σ we have that γ is a smooth curve in
R2 with γ(0) = (x0 , y0 ) and φ ◦ γ = α.
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ANDRÉ NEVES
Proof of Lemma 3.2. 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=
(p) + y 0 (0)
(p) + z 0 (0)
(p) = ~v .∇F (p)
= x0 (0)
dt
∂x
∂y
∂z
t=0
and so ~v ∈ L.
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 change of parameters
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
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 3.1 we
know that dφ(x0 ,y0 ) is map from R2 to Tp S which is bijective.
dF|p : Tp S → R3 ,
dF|p (α0 (0)) =
Lemma 4.1. 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 ◦ φ)
d(F ◦ α)
(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)) =
5. Lecture 10
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).
GEOMETRY OF CURVES AND SURFACES
7
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 14: 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 ,
then from the definition of dF|p we have that this linear map takes valued
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 gave already seen that Tp S is a plane and so N (p) is by
definition a unit normal vector perpendicular to 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 3.2 that ∇F (p) is perpendicular to Tp S and so we can take
N (p) = ∇F (p)/|∇F (p)|.
For instance, if we choose 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
N (p) =
2p
∇F
(p) =
= p.
|∇F |
|2p|
If φ : U → S is a chart with φ(x0 , y0 ) = p, then we know from Lemma 3.1
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
N (p) =
(x0 , y0 ).
|∂x φ × ∂y φ|
For instance, 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)|
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ANDRÉ NEVES
In general, when we have a chart φ : U ⊂ R2 → S ⊂ R3 because we know
that ∂x φ(x, y), ∂y φ(x, y) form a basis for Tφ(x,y) S, then we can choose
N (φ(x, y)) =
∂x φ × ∂y φ
(x, y)
|∂x φ × ∂y φ|
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 15: Show that if the map N : S → R3 is continuous then it is
actually smooth.