3 Surfaces
3.1 Regular Parametrized Surfaces
We approach surfaces in E
3
similarly to how we considered curves; a parametrized surface is a
function x : U → E
3
where U is some open subset of the plane R
2
. Our main purpose is to develop
and measure the curvature of a surface in terms of the parametrizing function x.
Our primary definition should mostly be familiar from elementary multivariable calculus.
Definition 3.1. A (smooth local) surface is the range S = x(U) of a smooth function x : U → E
3
, where
U is a connected open subset of R
2
.
Given co-ordinates u, v on U, the co-ordinate tangent vector fields are the partial derivatives x
u
=
∂x
∂u
and x
v
=
∂x
∂v
.
The exterior derivative or differential of the surface is the vector-valued 1-form dx = x
u
du + x
v
dv.
A surface is regular at P = x(p) if the tangent vectors x
u
(p) and x
v
(p) are linearly independent: other-
wise said, at P, the surface has a well-defined
Tangent plane T
P
S = Span
x
u
(p), x
v
(p)
(a 2-dim subspace of T
P
E
3
), and
Unit normal vector n(p) =
x
u
(p) × x
v
(p)
||
x
u
(p) × x
v
(p)
||
∈ T
P
E
3
S is regular if it is regular everywhere. An orientation is a smooth choice of unit normal vector field n.
The M¨obius strip (Exercise 9) shows that not every surface is orientable!
For brevity, we will often refer to the parametrizing function x as the surface, though many different
parametrizations will exist! A general surface typically needs to be parametrized by several overlap-
ping functions x
1
, x
2
, . . .. Our definition is local since there is only one x.
∂
∂v
p
∂
∂u
p
p
U
x
The partial derivatives x
u
(p), x
v
(p) are tangent to the surface at x(p): if p = (u
0
, v
0
) then the curve
y(t) := x(t, v
0
) lies in the surface and passes through P = x(p); its tangent vector at P is then
y
′
( u
0
) = lim
h→0
y(u
0
+ h) −y(u
0
)
h
= lim
h→0
x(u
0
+ h, v
0
) − x(p)
h
= x
u
(p)
47
To help distinguish between domain and codomain, we standardize notation.
Domain U ⊆ R
2
: Points are written lower case or as row vectors: e.g., p = (u
0
, v
0
) ∈ U. Typically we’ll
use u, v as co-ordinates unless it is more natural to use angles such as ϕ, θ.
Tangent vectors/fields are written with an arrow in our new notation: e.g.,
w
p
=
∂
∂u
p
∈ T
p
R
2
.
Codomain E
3
: Points are written upper case or as row vectors, e.g., P = (3, 4, 8) ∈ E
3
. Co-ordinates on
E
3
will typically be x, y, z.
Vectors are written bold-face as either row or column vectors: e.g., x(u, v) =
u, v, u
2
+ v
2
.
Tangent vectors/fields use the old notation:
16
e.g., if P = x(p), then x
u
(p) =
∂x
∂u
p
∈ T
P
E
3
.
Example 3.2. Consider the sphere of radius a parametrized using spherical polar co-ordinates:
x(θ, ϕ) = a
cos θ cos ϕ
sin θ cos ϕ
sin ϕ
, dx = x
θ
dθ + x
ϕ
dϕ = a cos ϕ
−sin θ
cos θ
0
dθ + a
−cos θ sin ϕ
−sin θ sin ϕ
cos ϕ
dϕ
The unit normal field is simply n =
1
a
x. The domain U = (0, 2π) × (−
π
2
,
π
2
) is an open rectangle
whose image S = x(U) is the sphere minus the (dashed) semicircle x(0, ϕ). While we could extend θ
to wrap round the equator, we cannot extend to the north or south poles without sacrificing regularity:
x
θ
= a cos ϕ
−sin θ
cos θ
0
= 0 when ϕ = ±
π
2
ϕ
θ
∂
∂ϕ
p
∂
∂θ
p
π
2
3π
2
2ππ
0
π
2
−
π
2
x
This illustrates the term local: indeed the famous hairy ball theorem from topology says that it is im-
possible to find a regular parametrization of the entire sphere by a single function.
Also observe how the tangent vectors
∂
∂ϕ
p
,
∂
∂θ
p
∈ T
p
R
2
are mapped by dx to tangent vectors
dx
dϕ
p
= dx
∂
∂ϕ
p
!
,
dx
dθ
p
= dx
∂
∂θ
p
!
∈ T
x(p)
S
16
To use our new notation in E
3
would require a subtle redefinition of dx: if
w is a vector field on U, then dx(
w) is the
vector field on S such that
dx(
w)
[ f ] =
w
f ◦x
for all f : S → R. In co-ordinates this benefits from tensor notation:
x(u
1
, u
2
) =
x
1
(u
1
, u
2
), x
2
(u
1
, u
2
), x
3
(u
1
, u
2
)
=⇒ dx =
∑
i,j
∂x
j
∂u
i
∂
∂x
j
⊗du
i
In more general situations this approach is necessary, but it is overkill for our purposes!
48
Theorem 3.3. Let S = x( U) be a smooth surface containing the point P = x(p):
1. The differential at p is a linear map dx : T
p
R
2
→ T
P
E
3
mapping tangent vectors in R
2
to vectors
tangent to S.
2. S is regular at P if and only if dx is injective (1–1) at p. In such a case we can view it as an
invertible linear map dx : T
p
R
2
→ T
P
S.
Proof. 1. The differential at p is linear since the co-ordinate 1-forms du, dv are linear: indeed
dx
a
∂
∂u
p
+ b
∂
∂v
p
!
= x
u
(p) du
a
∂
∂u
p
+ b
∂
∂v
p
!
+ x
v
(p) dv
a
∂
∂u
p
+ b
∂
∂v
p
!
= ax
u
(p) + bx
v
(p) = a dx
∂
∂u
p
!
+ b dx
∂
∂v
p
!
This expression is moreover tangent to S at x(p): if this last assertion is unconvincing, see
Exercise 8.
2. The range of dx at p is plainly Span{x
u
(p), x
v
(p)}. This is 2-dimensional (and thus defines the
tangent plane) if and only if rank dx = 2 ⇐⇒ dx is 1–1.
It is worth reiterating two crucially important properties of dx:
• At a regular point, dx : T
p
R
2
→ T
P
S is an invertible linear map. We shall shortly use this to
pull-back calculations from S to U.
• The differential is co-ordinate independent and thus does not depend on the parametrization
of S. This follows since dx is the unique 1-form satisfying dx(
w) =
w[x] for all vector fields
w
on U; a description that does not depend on co-ordinates.
Aside: change of co-ordinates To more clearly spell this out, suppose we choose a new
parametrization y(s, t) = x
F(s, t)
where F(s, t) = (u, v) is a change of co-ordinates on U. By
the chain rule,
y
s
y
t
=
∂u
∂s
∂v
∂s
∂u
∂t
∂v
∂t
x
u
x
v
and (du dv) = (ds dt)
∂u
∂s
∂v
∂s
∂u
∂t
∂v
∂t
from which
dy = (ds dt)
y
s
y
t
= (du dv)
∂u
∂s
∂v
∂s
∂u
∂t
∂v
∂t
−1
∂u
∂s
∂v
∂s
∂u
∂t
∂v
∂t
x
u
x
v
= (du dv)
x
u
x
v
= dx
The matrix of partial derivatives is the Jacobian of the co-ordinate change.
To be completely strict, dx and dy are not identical since they feed on tangent vectors with respect to
different co-ordinates. Formally
y = x ◦ F =⇒ dy = dx ◦ dF
where dF maps tangent vectors in Span{
∂
∂s
,
∂
∂t
} to those in Span{
∂
∂u
,
∂
∂v
}: in matrix language, dF is
precisely the above Jacobian!
49
Common Surfaces
You should have met many of these families/examples in multi-variable calculus.
Graphs If f (x, y) is a smooth function, its graph may be parametrized by x(u, v) =
u, v, f (u, v)
.
Its differential and unit normal field are
dx =
1
0
f
u
du +
0
1
f
v
dv n =
1
p
1 + f
2
u
+ f
2
v
−f
u
−f
v
1
This is regular at all points, regardless of f .
Examples 3.4. 1. The standard circular paraboloid may be parametrized x(u, v) =
u, v, u
2
+ v
2
.
2. The upper half of the unit sphere is the graph of z = f (x, y) =
p
1 − x
2
−y
2
where x
2
+ y
2
< 1.
3. A plane has equation ax + by + cz = d where a, b, c, d are constant. Since at least one of a, b, c
must be non-zero, this may be written as a function and graphed. For instance, if b = 0 we
have y = f (x, z) =
1
b
( d − ax −cz) and n =
1
√
a
2
+b
2
+c
2
a, b, c
.
Surfaces of Revolution If a smooth positive function x = f (z) is rotated around the z-axis, we
obtain a parametrization
x(θ, v) =
f (v) cos θ, f (v) sin θ, v
, (θ, v) ∈ (0, 2π) ×dom( f )
with differential and unit normal field
dx =
−f (v) sin θ
f (v) cos θ
0
dθ +
f
′
( v) cos θ
f
′
( v) sin θ
1
dv n =
1
p
1 + f
′
( v)
2
cos θ
sin θ
−f
′
( v)
Examples 3.5. 1. The simplest example ( f (z) ≡ 1) is the right circular cylinder of radius 1.
2. We may rotate around any axis! For instance, if we
rotate the curve z = 2 + cos x around the x-axis, the
resulting surface may be parametrized
x(θ, v) = (2 + cos v)
0
cos θ
sin θ
+
v
0
0
This time v measures distance along the x-axis and θ
the angle of rotation around it.
The differential and unit normal field are
dx = (2 + cos v)
0
−sin θ
cos θ
dθ +
1
−sin v cos θ
−sin v sin θ
dv n =
1
p
1 + sin
2
v
sin v
cos θ
sin θ
Note the orientation of the surface: the unit normal field points outward, away from the x-axis.
50
Ruled Surfaces Given functions y( u), z(u), define
x(u, v) = y(u) + vz(u)
Through each point P = x(u
0
, v
0
) passes a line t 7→ x(u
0
, t) = y(u
0
) + tx(u
0
) lying in the surface. The
surface can be visualized as moving a ruler through space. Ruled surfaces are common in engineering
applications since they may be constructed using straight beams.
Definition 3.6. The tangent developable of a smooth curve y is the special case when z = y
′
.
Examples 3.7. 1. Every plane is a ruled surface! Let y be a line in the plane and z any other tan-
gent direction. For instance, the plane passing through (1, 0, 9) and spanned by (2, −3, −5) ad
(1, 2, 3) may be parametrized
x(u, v) =
1, 0, 9
+
2, −3, −5
u
| {z }
y(u)
+
1, 2, 3
| {z }
z(u)
v
2. A helicoid is built by joining each point of a helix to its axis of rotation. From the standard helix,
we obtain the helicoid x(u, v) =
v cos u, v sin u, u
for v > 0.
3. The hyperboloid of one sheet is a doubly ruled surface: through each point there are two lines lying
on the surface. It may be parametrized as a ruled surface by
x(u, v) =
1
0
0
+ u
0
1
1
+ v
2u
u
2
−1
u
2
+ 1
though convincing yourself there are two lines through each point takes a little more work. . .
Helicoid Hyperboloid
51
Implicitly Defined Surfaces
Definition 3.8. A regular implicitly defined surface is the zero set of a smooth function f : E
3
→ R for
which d f = 0 (equivalently ∇f = 0).
Recall that the directional derivative of f in the direction v is D
v
f (P) = v · ∇f (P). This is zero if
and only if v is orthogonal to ∇f (P) . In particular, this says that ∇f provides a normal field to an
implicitly defined surface.
Examples 3.9. 1. Let a, b, c, d be constants. The function f (x, y, z) = ax + by + cz −d has
d f = a dx + b dy + c dz
which is non-zero provided at least one of a, b, c are non-zero. This defines a plane with unit
normal field n =
1
||
∇f
||
∇f =
1
√
a
2
+b
2
+c
2
a, b, c
.
2. The sphere of radius a is the zero set of f (x, y, z) = x
2
+ y
2
+ z
2
− a
2
. It has unit normal field
n =
1
||
∇f
||
∇f =
1
a
x, y, z
The sphere is everywhere regular since at least one of x, y, z is non-zero at all points of the
sphere. Contrast this with our earlier example of the parametrized sphere which could not be
made regular at the north and south poles. The lack of regularity in this case is an aspect of the
parametrization, not the surface itself.
3. The function f (x, y, z) = x
2
+ y
2
−z
2
−c has
d f = 2(x dx + y dy −z dz)
which is non-zero away from (x, y, z) = (0, 0, 0). Depending on the sign of c, the zero set is a
hyperboloid or a cone; visualize the horizontal cross-sectional circles to determine which.
c > 0 Hyperboloid of 1-sheet: x
2
+ y
2
= z
2
+ c > 0 for all z
c = 0 Cone: x
2
+ y
2
= z
2
contains a non-regular point (0, 0, 0)
c < 0 Hyperboloid of 2-sheets: x
2
+ y
2
= z
2
−
|
c
|
≥ 0 only when
|
z
|
≥
p
|
c
|
52
Our next result, a corollary of the famous implicit function theorem, ties together the notions of regular-
ity. In particular, it says that we can always assume the existence of local co-ordinates.
Theorem 3.10. A regular implicitly defined surface f (x, y, z) = 0 is (locally) the image of a regular
local surface.
Proof. Suppose P = (x
0
, y
0
, z
0
) lies on the surface and ∇f (P) = 0. At least one of the partial deriva-
tives of f is non-zero; suppose WLOG that f
z
(P) = 0. By the implicit function theorem, there exists
U ⊆ R
2
and a function g : U → R for which g(x
0
, y
0
) = z
0
and f
x, y, g(x, y)
= 0. The surface is
then (locally) the graph of z = g(x, y).
x : U → E
3
: (u, v) 7→
u, v, g(u, v)
Example 3.11. The zero set of f (x, y, z) = x
2
+ y
2
−z
2
−6 is a hyperboloid of one sheet. It has unit
normal vector field
n(x, y, z) =
1
||
∇f
||
∇f =
1
p
x
2
+ y
2
+ z
2
x
y
−z
=
1
√
6 + 2z
2
x
y
−z
whenever (x, y, z) is a point on the hyperboloid. For instance, at P = (3, 1, 2) the unit normal is
n(P) =
1
√
14
(3, 1, 2) and the tangent plane has equation
3x + y −2z = 6
Alternatively, the hyperboloid can be parametrized in several ways.
(a) In the language of the proof, near P = (3, 1, 2) it is the graph of z = g(x, y) =
p
x
2
+ y
2
−6.
This results in a (local) regular parametrization
x(u, v) =
u, v,
p
u
2
+ v
2
−6
(b) The hyperboloid is a surface of revolution around the z-axis:
x(θ, v) =
√
6 + v
2
cos θ
√
6 + v
2
sin θ
v
For this parametrization, the differential and normal field are
dx =
p
6 + v
2
−sin θ
cos θ
0
dθ +
1
√
6 + v
2
v cos θ
v sin θ
√
6 + v
2
dv
n =
x
θ
×x
v
||
x
θ
×x
v
||
=
1
√
6 + 2v
2
√
6 + v
2
cos θ
√
6 + v
2
sin θ
−v
which is precisely what we obtained above.
Yet another expression could be obtained using a parametrization as a ruled surface (e.g., page 51).
53
Exercises 3.1. 1. Show that parametrization x(r, θ) =
r cos θ, r sin θ,
√
1 −r
2
of the upper hemi-
sphere is non-regular at r = 0.
2. Explain why the parametrization in Example 3.11(a) is local: what is left out?
3. (a) Compute dx and n for the paraboloid x(u, v) =
u, v, u
2
+ v
2
.
(b) Repeat for the polar co-ordinate parametrization y(r, θ) =
r cos θ, r sin θ, r
2
. Is this
parametrization everywhere regular?
(c) Using x = r cos θ, etc., write dx in terms of r, θ, dr, dθ. What do you observe?
(d) By viewing the paraboloid as the zero set of f (x, y, z) = z − x
2
− y
2
, find another expres-
sion for the unit normal field.
4. (a) Find a parametrization for the tangent developable of the helix
y(u) =
cos u, sin u, u
. Compute dy and the unit normal field n.
(The picture covers v ∈ (−3, 6) with the original curve y(u) in green)
(b) If y is a unit speed biregular curve, prove that its tangent devel-
opable x(u, v) = y(u) + vy
′
( u) is a regular surface except when
v = 0. Express the differential and unit normal field in terms of the
Frenet frame of y.
5. Let f (x, y, z) = z
2
. Show that the zero set of f has a regular parametriza-
tion despite the gradient of f vanishing at z = 0.
6. Let a, b, c be positive constants and define x(θ, ϕ) =
a cos θ cos ϕ
b sin θ cos ϕ
c sin ϕ
, (θ, ϕ) ∈ (0, 2π) × (−
π
2
,
π
2
)
(a) Show that x parametrizes the ellipsoid
x
2
a
2
+
y
2
b
2
+
z
2
c
2
= 1. What part(s) of the ellipsoid are
‘missing’ from the parametrization?
(b) Describe geometrically the curves θ = constant and ϕ = constant on the ellipsoid.
(c) Calculate the differential of x and show that dx is 1–1 for each p ∈ U.
7. The tube of radius a > 0 centered on a curve y(t) may be parametrized
in terms of the Frenet frame of y:
x(ϕ, t) = y(t) + a cos ϕ N(t) + a sin ϕ B(t)
(a) Briefly explain why the normal field is n = cos ϕ N(t) + sin ϕ B(t) .
(b) Suppose y is unit speed. Prove that x is everywhere regular if and
only if κ(t) <
1
a
at all points of the generating curve.
8. Let c(t) : (−ϵ, ϵ) → U be a curve and y(t) = x
c(t)
the corresponding
curve in the surface x : U → E
3
. Prove that dx
c
′
(0)
= y
′
(0).
(Hint: Recall how to write c
′
( t) as a vector field)
9. (M
¨
obius strip) Show that x(u, v) =
(2+v cos
u
2
) cos u
(2+v cos
u
2
) sin u
v sin
u
2
!
is regular and ori-
entable whenever 0 < u < 2π and −1 < v < 1. By computing n(0, 0)
and n(2π, 0), explain what happens if we try to extend u to [0, 2π].
54
3.2 The Fundamental Forms
Our immediate goal is to use differentials to describe the shape of a surface. Before making the main
definition, we need another product of 1-forms.
Definition 3.12. Given 1-forms α, β on U, define the symmetric 2-form αβ by
αβ(
v,
w) =
1
2
α(
v)β(
w) + α(
w)β(
v)
where
v,
w are vector fields on U. Note that α
2
(
v,
w) := αα(
v,
w) = α(
v)α(
w).
Symmetric 2-forms behave the way you (hopefully!) think they should.
Lemma 3.13. On each tangent space, αβ : T
p
R
n
× T
p
R
n
→ R is a symmetric and bilinear.
Moreover αβ = βα, and the product is linear in each slot:
α(β + γ) = αβ + αγ and ( α + β)γ = αγ + βγ (∗)
Take care when using co-ordinate 1-forms; convention dictates that dx
2
= (dx)
2
is a symmetric 2-
form, not the exterior derivative (1-form) d(x
2
) = 2x dx.
Example 3.14. Let
v = a
∂
∂x
+ b
∂
∂y
and
w = c
∂
∂x
+ d
∂
∂y
. Then
dx
2
(
v,
w) = ac, dy
2
(
v,
w) = bd, dx dy(
v,
w) =
1
2
(ad + bc)
In particular,
dx
2
+ dy
2
(
v,
w) = ac + bd is the dot product in disguise.
To evaluate symmetric 2-forms with respect to co-ordinates, linearity and distributivity (∗) are all
you need. For instance, if α = x dx −dy and β = xy dy, then αβ = x
2
y dxdy − xy dy
2
.
If α, β take values in E
n
, we use the dot product for multiplication of the resulting vectors α(
v), etc.
( α · β)(
v,
w) :=
1
2
α(
v) · β(
w) + α(
w) · β(
v)
Definition 3.15. The first and second fundamental forms of a regular local surface x : U → E
3
are
I = dx ·dx, I = −dx ·dn
where dn is the differential of the unit normal field (I requires that the surface be oriented). The first
fundamental form is also commonly denoted ds
2
(see Example 3.17 and Theorem 3.20 for why).
Example 3.16. If x(u, v) =
u, uv, 1 + u
, then
dx =
1
v
1
du +
0
u
0
dv, n =
1
√
2
1
0
−1
, dn = 0
from which I = (2 + v
2
) du
2
+ 2uv du dv + u
2
dv
2
and I = 0.
55
Why should we care about I & I?
Basic interpretation I(
v,
w) = dx(
v) ·dx(
w) pulls back the dot product from T
P
S to T
p
R
2
. The length
of and angle between tangent vectors to the surface S at P may now be computed in T
p
R
2
.
I(
v,
w) = −
1
2
dx(
v) · dn(
w) + dx(
w) · dn(
v)
describes how the normal field n changes over
the surface. In the example, I ≡ 0 encapsulates the constancy of the normal field: the surface is
(part of) the plane x ·(1, 0, −1) = −1.
Co-ordinate invariance Since dx is independent of co-ordinates, so also is I. The unit normal field is
independent of oriented co-ordinate changes. More formally, if y(s, t) = x(u, v) parametrize the
same surface, then
17
I
y
= I
x
and I
y
=
(
I
x
if the orientations are identical
−I
x
if the orientations are reversed
The upshot is that the fundamental forms provide a co-ordinate independent way to compute infor-
mation about a surface from within the parametrization space U.
Example 3.17. For the sphere of radius a in spherical polar co-ordinates, recall Example 3.2:
x(θ, ϕ) = a
cos θ cos ϕ
sin θ cos ϕ
sin ϕ
=⇒ dx = a cos ϕ
−sin θ
cos θ
0
dθ + a
−cos θ sin ϕ
−sin θ sin ϕ
cos ϕ
dϕ
=⇒ I = a
2
cos
2
ϕ dθ
2
+ dϕ
2
If you revisit the pictures in Example 3.2, the effect of I is easy to visualize:
• I(
∂
∂θ
,
∂
∂θ
) =
||
x
θ
||
2
= a
2
cos
2
ϕ: the tangent vector x
θ
is shorter near the poles, where cos ϕ → 0.
• I(
∂
∂ϕ
,
∂
∂ϕ
) =
x
ϕ
2
= a
2
: the tangent vector x
ϕ
always has the same length.
• I(
∂
∂θ
,
∂
∂ϕ
) = x
θ
·x
ϕ
= 0: the co-ordinate tangent vectors are always orthogonal.
At a point P = x(p) on the sphere, if we increase the co-ordinates by tiny quantities ∆p =
∆θ, ∆ϕ),
then the distance ∆s travelled along the surface approximately satisfies
( ∆s)
2
≈
||
x(p + ∆p) − x(p)
||
≈
x
θ
∆θ + x
ϕ
∆ϕ
2
= a
2
cos
2
ϕ (∆θ)
2
+ a
2
( ∆ϕ)
2
with equality in the limit ∆θ, ∆ϕ → 0. Near the poles, a change in longitude ∆θ corresponds to a
smaller distance on the sphere. This is analogous to how a standard map of the Earth works, with
distances appearing distorted near the poles. We’ll return to this idea shortly. . .
Computing I is very easy for the sphere, since n =
1
a
x is merely the scaled position vector:
I = −dx ·dn = −
1
a
dx ·dx = −
1
a
I = −a
cos
2
ϕ dθ
2
+ dϕ
2
17
As in the Aside on page 49, we strictly have I
y
= I
x
◦dF, etc., where y(s, t) = x
F( s, t)
= x(u, v). The ±-sign in the
expressions for I is that of the determinant of the Jacobian dF.
56
The fundamental forms I, I may be computed directly in terms of co-ordinates u, v.
Theorem 3.18. If x : U → E
3
is a regular (oriented) surface, then
I = E du
2
+ 2F du dv + G dv
2
and I = l du
2
+ 2m du dv + n dv
2
where the smooth functions E, F, G, l, m, n : U → R are defined by
E = x
u
·x
u
F = x
u
·x
v
G = x
v
·x
v
l = x
uu
·n = −x
u
·n
u
m = x
uv
·n = −x
u
·n
v
= −x
v
·n
u
n = x
vv
·n = −x
v
·n
v
The expressions for I come from differentiating x
u
· n = 0 = x
v
· n, and are particularly helpful
because they avoid computing derivatives of n (which likely contains a square-root).
Example 3.19. Parametrize the graph of z = f (x, y) by x(u, v) =
u, v, f (u, v)
to obtain,
x
u
=
1
0
f
u
, x
v
=
0
1
f
v
=⇒ E = 1 + f
2
u
, F = f
u
f
v
, G = 1 + f
2
v
=⇒ I = (1 + f
2
u
) du
2
+ 2 f
u
f
v
du dv + (1 + f
2
v
) dv
2
x
uu
=
0
0
f
uu
, x
uv
=
0
0
f
uv
, x
vv
=
0
0
f
vv
, n =
1
p
1 + f
2
u
+ f
2
v
−f
u
−f
v
1
=⇒ l =
f
uu
p
1 + f
2
u
+ f
2
v
, m =
f
uv
p
1 + f
2
u
+ f
2
v
, n =
f
vv
p
1 + f
2
u
+ f
2
v
=⇒ I =
1
p
1 + f
2
u
+ f
2
v
f
uu
du
2
+ 2 f
uv
du dv + f
vv
dv
2
As a particular example, the circular paraboloid z = x
2
+ y
2
has fundamental forms
I = (1 + 4u
2
) du
2
+ 8uv du dv + (1 + 4v
2
) dv
2
= du
2
+ dv
2
+ 4
u du + v dv
2
I =
2
√
1 + 4u
2
+ 4v
2
du
2
+ dv
2
As a sanity check, compare with the parametrization of the same paraboloid in polar co-ordinates
y(r, θ) =
r cos θ, r sin θ, r
2
(Exercise 3.1.3). By computing the partial derivatives y
r
, y
θ
, y
rr
, y
rθ
, y
θθ
directly, one easily verifies that
I = (1 + 4r
2
) dr
2
+ r
2
dθ
2
, I =
2
√
1 + 4r
2
dr
2
+ r
2
dθ
2
These expressions are identical to the originals (same orientation!) since
(
du = cos θ dr −r sin θ dθ
dv = sin θ dr + r cos θ dθ
=⇒
(
du
2
+ dv
2
= dr
2
+ r
2
dθ
2
u du + v dv
2
= r
2
dr
2
57
Curves in Surfaces: interpreting I and I
Given a regular (oriented) surface x : U → E
3
and a curve c(t) in U, we may transfer this curve to
the surface y(t) = x
c(t)
. Its tangent vector (Exercise 3.1.8) and speed are then
y
′
( t) = dx
c
′
( t)
=⇒
y
′
( t)
=
q
dx
c
′
( t)
·dx
c
′
( t)
=
q
I
c
′
( t), c
′
( t)
We can do something similar for the second fundamental form.
Theorem 3.20. Let y(t) = x
c(t)
parametrize a curve in a surface x with unit normal field n.
1. If a < b, then the arc-length of y between y(a) and y(b) is
R
b
a
q
I
c
′
( t), c
′
( t)
dt.
2. The normal acceleration of the curve is y
′′
( t) ·n = I(c
′
, c
′
).
This puts some flesh on our earlier observations (page 56). I measures infinitesimal squared-distance
on the surface, while I measures how the surface bends away from the normal field: recall how
force/acceleration motivated the curvature κ of a curve (Definition 1.15).
Proof. 1. Arc-length is the integral of the speed
||
y
′
( t)
||
=
q
I
c
′
( t), c
′
( t)
.
2. Since y
′
lies in the tangent plane, we have y
′
·n ≡ 0. Differentiate to obtain
0 =
d
dt
( y
′
·n) = y
′′
·n + y
′
·
d
dt
n
c(t)
= y
′′
·n + dx(c
′
) · dn(c
′
) = y
′′
·n −I(c
′
, c
′
)
Example (3.17, cont). Consider the curve c(t) = (θ(t), ϕ(t)) =
2t, t
where 0 ≤ t ≤
π
4
. This has
tangent field c
′
( t) = 2
∂
∂θ
+
∂
∂ϕ
. Translated to the unit sphere, the resulting curve has arc-length
Z
π
4
0
q
I
c
′
, c
′
) dt =
Z
π
4
0
p
4 cos
2
t + 1 dt ≈ 1.619
In the parametrization space U, c(t) is a straight line. The shortest path between the endpoints of the
curve on the sphere is the great circle arc with length
2π
4
=
π
2
≈ 1.571; its pre-image in U appears longer
but isn’t due to the cos
2
ϕ factor in the first fundamental form. By spending more time at northerly
latitudes, I is smaller for more of the great circle arc and the resulting arc-length is shorter.
ϕ
θ
0
π
2
0
π
4
58
If a map of the Earth covers a small latitude range (almost constant ϕ ≈ ϕ
0
), the first fundamental
form is almost similar to a standard dot product I ≈ (a cos ϕ
0
dθ)
2
+ (a dϕ)
2
. If not, say when we
travel by plane, the distortion becomes much more apparent.
The picture shows the shortest path from Irvine (California) to Irvine (Scotland), as flown by an
aircraft in ideal conditions. The straight line on the map corresponds to a longer real-world path.
If we travel at constant speed, it can be checked that great circles are precisely those curves whose
acceleration is entirely normal to the surface. This observation, and its relation to geodesics (paths
minimizing distance), is a matter for another course.
Example 3.21. A skater descends into a paraboloidal bowl z =
1
2
r
2
following the path described
by c(t) = (r(t), θ(t)) = (1 − t, 4t
2
) in polar co-ordinates. If we parametrize the bowl in polar co-
ordinates x(r, θ) =
r cos θ, r sin θ,
1
2
r
2
, the fundamental forms are seen to be
I = (1 + r
2
) dr
2
+ r
2
dθ
2
I =
1
√
1 + r
2
(dr
2
+ r
2
dθ
2
)
For the skater’s path, c
′
( t) = −
∂
∂r
+ 8t
∂
∂θ
, whence
I(c
′
, c
′
) = (1 + (1 −t)
2
) + 64t
2
(1 − t)
2
The path therefore has arc-length
Z
1
0
q
I(z
′
, z
′
) dt =
Z
1
0
q
1 + (64t
2
+ 1)(1 − t)
2
dt ≈ 1.82
and normal acceleration
y
′′
·n = I(c
′
, c
′
) =
1
p
1 + (1 −t)
2
1 + 64t
2
(1 − t)
2
0
5
y
′′
· n
0 1
t
By Newton’s second law, this is proportional to the component of the force experienced by the skater
pushing perpendicularly out from the surface.
59
Exercises 3.2. 1. Verify the final details of Example 3.19: that is, compute I, I directly using the polar
co-ordinate parametrization y(r, θ) =
r cos θ, r sin θ, r
2
.
2. Find the fundamental forms for the surface of revolution x(θ, v) =
f (v) cos θ, f (v) sin θ, v
.
3. Compute the first fundamental forms of each parametrized surface wherever they are regular
(a, b, c are non-zero constants). Where does each parametrization fail to be regular?
(a) Ellipsoid x(θ, ϕ) = (a cos θ cos ϕ, b sin θ cos ϕ, c sin ϕ)
(b) Elliptic paraboloid x(r, θ) = (ar cos θ, br sin θ, r
2
)
(c) Hyperboloid of two sheets x(u, v) = (a sinh u cos v, b sinh u sin v, c cosh u)
4. Calculate the fundamental forms of Enneper’s surface
x(u, v) =
u −
1
3
u
3
+ uv
2
, v −
1
3
v
3
+ vu
2
, u
2
−v
2
5. Compute dy for the parametrization y(r, θ) =
r cos θ, r sin θ,
√
1 −r
2
of the upper unit hemi-
sphere. Verify that the first fundamental form is the same as in Example 3.17.
6. Let x be the tangent developable of a unit speed biregular curve y (Exercise 3.1.4).
(a) Compute the fundamental forms of x in terms of the curvature and torsion of y.
(b) If y(u) =
cos
u
√
2
, sin
u
√
2
,
u
√
2
is the unit speed helix, show that
I =
1 +
v
2
4
du
2
+ 2 dudv + dv
2
, I = −
v
4
du
2
7. Prove that I ≡ 0 if and only if x is (part of) a plane.
8. Parametrize the great circle in Example 3.17 (cont) by z(t) =
cos t,
1
√
2
sin t,
1
√
2
sin t
, 0 ≤ t ≤
π
2
.
Verify that the arc has length
π
2
and that the acceleration of z is entirely normal; z
′′
= (z
′′
·n)n.
9. Equip the upper half plane y > 0 with the abstract first fundamental form I =
1
y
2
dx
2
+ dy
2
.
Compare the arc-length between the points ( 1, 1) and (−1, 1):
(a) Over the circular arc c(t) =
√
2
cos t, sin t
centered at the origin.
(b) Over the ‘straight’ line y = 1.
This is the Poincar´e half-plane model of hyperbolic space. There is neither a surface x : U → E
3
nor a
second fundamental form I!
10. (Hard) The torus obtained by rotating the unit circle in the x, z-plane centered at (2, 0, 0) around
the z-axis may be parametrized
x(u, v) =
(2 + cos ϕ) cos θ, (2 + cos ϕ) sin θ, sin ϕ
, (θ, ϕ) ∈ R
2
Let k = 0 be constant and consider the curve y(t) = x(kt, t) on the torus.
(a) Prove that y(t) has a self-intersection (∃s = t such that y(t) = y(s)) if and only if k ∈ Q.
(b) If k ∈ Q, show that the curve is periodic in that there exists a minimum positive T for which
y(t + T) = y(t) for all t. Find T in terms of k and write down (don’t evaluate!) the integral
for the arc-length of the curve over one period.
60
3.3 Principal, Gauss & Mean Curvatures
Since I and I are symmetric bilinear forms on each tangent space T
p
R
2
, they may be expressed in
matrix form: their matrices with respect to linearly independent vector fields
s,
t are
[I] :=
I(
s,
s) I(
s,
t)
I(
s,
t) I(
t,
t)
and [I] :=
I(
s,
s) I(
s,
t)
I(
s,
t) I(
t,
t)
Otherwise said
I
f
s + g
t, h
s + k
t
=
f g
[I]
h
k
and similarly for I. Matters are simplest when these matrices are diagonal. . .
Definition 3.22. Linearly independent vector fields
s,
t are said to be orthogonal if I(
s,
t) = 0. They
additionally describe curvature directions if I(
s,
t) = 0.
Co-ordinates u, v are orthogonal/curvature-line if the above apply to the the co-ordinate fields
∂
∂u
,
∂
∂v
.
In the language of Theorem 3.18, the matrices of the fundamental forms with respect to
∂
∂u
,
∂
∂v
are
A :=
E F
F G
and B :=
l m
m n
(∗)
Co-ordinates are orthogonal iff F = x
u
· x
v
≡ 0 (I has no du dv term), and are curvature-line iff I is
also diagonal:
I = E du
2
+ G dv
2
and I = l du
2
+ n dv
2
While the meaning of orthogonal is clear, the reason for the term curvature-line will take a little work.
Examples 3.23. 1. Since the sphere of radius a has I = −
1
a
I, any orthogonal co-ordinates on the
sphere are curvature-line! E.g., spherical polar co-ordinates: I = a
2
cos
2
ϕ dθ
2
+ dϕ
2
.
2. (Example 3.2.3.19) Standard polar co-ordinates are curvature-line for the paraboloid z = r
2
:
I = (1 + 4r
2
) dr
2
+ r
2
dθ
2
, I =
2
√
1 + 4r
2
dr
2
+ r
2
dθ
2
3. In curvature-line co-ordinates n
u
= −
l
E
x
u
and n
v
= −
n
G
x
v
(see Exercise 11).
A Little Linear Algebra The existence of curvature directions is equivalent to the simultaneous diag-
onalization of both matrices (∗). This requires an extension of the concepts of eigenvalues/vectors.
Definition 3.24. Let A, B be square matrices of the same dimension. A non-zero vector
v is an
eigenvector of B with respect to A with eigenvalue λ if
(B −λA)
v =
0
If A = I is the identity matrix, these are standard eigenvalues/vectors. We compute in the usual
manner: solve the characteristic polynomial and find
v ∈ N(B − λA) in the nullspace. . .
61
Example 3.25. Let A =
2 3
3 5
and B =
0 1
1 3
.
det(B − λA) =
−2λ 1 −3λ
1 −3λ 3 −5λ
= λ
2
−1 = 0 ⇐⇒ λ = ±1
v
1
∈ N(B − A) = N
−2 −2
−2 −2
= Span
1
−1
,
v
2
∈ N(B + A) = N =
2 4
4 8
= Span
2
−1
Note that {
v
1
,
v
2
} =
1
−1
,
2
−1
is a basis of R
2
consisting of eigenvectors of B with respect to A.
Theorem 3.26. Let A, B be symmetric matrices of the same dimension, with A positive-definite.
18
1. There exists a basis of eigenvectors of B with respect to A. Moreover, all eigenvalues are real.
2. If
s,
t are eigenvectors corresponding to distinct eigenvalues, then
s
T
A
t = 0 =
s
T
B
t.
Proof. 1. This follows from the famous spectral theorem in linear algebra.
19
2. Assume B
s = k
1
A
s and B
t = k
2
A
t where k
1
= k
2
, and apply the symmetry of A and B,
s
T
B
t =
s
T
( k
2
A
t) = k
2
s
T
A
t
∥
t
T
B
s =
t
T
( k
1
A
s) = k
1
t
T
A
s = k
1
s
T
A
t
=⇒ (k
2
−k
1
)
s
T
A
t = 0 =⇒
s
T
A
t = 0
Application to Regular Surfaces With respect to independent vector fields, the matrices A, B of I, I
are symmetric. Moreover, the regularity of x guarantees the positive-definiteness of A:
∀
w =
0 =⇒ I(
w,
w) = dx(
w) · dx(
w) =
||
dx(
w)
||
2
> 0
We may therefore apply Theorem 3.26 to the matrices of the fundamental forms.
Definition 3.27. The principal curvatures k
1
, k
2
: U → R of an oriented surface x : U → E
3
are the
eigenvalues of I with respect to I. Corresponding eigenvectors are curvature directions.
The Gauss and mean curvatures are, respectively, K := k
1
k
2
and H :=
1
2
( k
1
+ k
2
).
A point x(p) is umbilic if k
1
(p) = k
2
(p).
18
For all non-zero vectors,
v
T
A
v > 0. Equivalently, all eigenvalues of A are positive. This means that
⟨
v,
w
⟩
:=
v
T
A
w
defines an inner product on R
n
. In Example 3.25, A has eigenvalues
1
2
(7 ±
√
45) > 0.
19
In case you’re interested: A has an orthogonal eigenbasis {
x
1
, . . . ,
x
n
} by the spectral theorem. Since its eigenvalues
µ
i
are positive, we may scale such that
||
x
i
||
2
=
1
µ
i
. Let X = (
x
1
···
x
n
) so that X
T
AX = I is the identity matrix. But then,
det(B −λA) = det(X
T
)
−1
det
X
T
BX −λI
det(X
−1
) = 0 ⇐⇒ det(X
T
BX −λI) = 0
Since X
T
BX is symmetric (spectral theorem again), it has an orthogonal eigenbasis {
w
1
, . . . ,
w
n
} and real eigenvalues λ
k
.
Each
v
k
:= X
w
k
is an eigenvector of B with respect to A with eigenvalue λ
k
. Since X is invertible, {
v
1
, . . . ,
v
n
} is a basis.
62
The curvatures are independent of oriented co-ordinate changes. If we reverse orientation, then k
1
, k
2
and H change sign, while K = k
1
k
2
is unchanged.
At non-umbilic points, Theorem 3.26 says that curvature directions diagonalize both fundamental
forms, in line with Definition 3.22.
At umbilic points, I = kI and all directions are curvature directions; any orthogonal directions nec-
essarily diagonalize both fundamental forms.
Example 3.28. Here are two totally umbilic surfaces where the curvatures are constant.
1. A plane: I ≡ 0 =⇒ all curvatures are zero.
2. A sphere of radius a: I = −
1
a
I =⇒ k
1
= k
2
= −
1
a
, K =
1
a
2
and H = −
1
a
.
In fact these comprise all totally umbilic surfaces (see Exercise 12).
Theorem 3.29. 1. In co-ordinates, the Gauss and mean curvatures are given by
K =
ln − m
2
EG − F
2
=
det B
det A
= det(A
−1
B) and H =
lG + nE −2mF
2(EG − F
2
)
=
1
2
tr A
−1
B
2. At non-umbilic points, the curvatures k
1
, k
2
, K, H are smooth functions and the curvature direc-
tions may be described locally by (smooth) vector fields.
Proof. 1. The principal curvatures are the solutions to the quadratic equation
det
l m
m n
−λ
E F
F G
= (EG − F
2
) λ
2
−(lG + nE −2mF)λ + (ln − m
2
)
of which K and H are the product and half the sum of the roots.
2. The roots
−b±
√
b
2
−4ac
2a
of a quadratic are smooth functions of the coefficients unless b
2
−4ac = 0,
in which case we have a repeated root (k
1
= k
2
). At non-umbilic points, each eigenspace is
one-dimensional, so there is no obstruction to choosing smooth eigenvectors.
20
Examples 3.30. 1. (Example 3.19) For the paraboloid x(r, θ) =
r cos θ, r sin θ, r
2
, standard polar
co-ordinates are curvature-line:
A = [I] =
1 + 4r
2
0
0 r
2
B = [I] =
2
√
1+4r
2
0
0
2r
2
√
1+4r
2
!
The curvatures are therefore
k
1
=
2
(1 + 4r
2
)
3/2
, k
2
=
2
√
1 + 4r
2
, K =
4
(1 + 4r
2
)
2
, H =
2 + 4r
2
(1 + 4r
2
)
3/2
The curvatures make sense at the single umbilic point (r = 0), but the co-ordinates are not
curvature-line there since the parametrization fails to be regular (x
θ
(0, θ) = 0).
20
At an umbilic point x(p), the eigenspace is 2-dimensional so lim
q→p
v(q) need not exist and
v need not be continuous.
63
2. Parametrize a graph z = f (x, y) in the usual manner x(u, v) =
u, v, f (u, v)
. Then
A = [I] =
1 + f
2
u
f
u
f
v
f
u
f
v
1 + f
2
v
B = [I] =
1
p
1 + f
2
u
+ f
2
v
f
uu
f
uv
f
uv
f
vv
Theorem 3.29 tells us that
K =
f
uu
f
vv
− f
2
uv
(1 + f
2
u
+ f
2
v
)
2
H =
f
vv
(1 + f
2
u
) + f
uu
(1 + f
2
v
) − 2 f
u
f
v
f
uv
2( 1 + f
2
u
+ f
2
v
)
3/2
In the abstract, solving for the curvatures and directions is disgusting. As a sanity check, you
should verify that f (u, v) = u
2
+ v
2
recovers exactly the curvatures in the previous example!
3. (Exercise 3.2.6) The tangent developable of the unit-speed helix has
A = [I] =
1 +
v
2
4
1
1 1
B = [I] =
−
v
4
0
0 0
Now solve for the curvatures:
−
v
4
−λ
1 +
v
2
4
−λ
−λ −λ
=
v
2
4
λ
2
+
v
4
λ = 0 =⇒ k
1
= 0, k
2
= −
1
v
, K = 0, H = −
1
2v
In this case an explicit computation of the curvature directions is not difficult:
k
1
= 0 =⇒ N(B −k
1
A) = N
−
v
4
0
0 0
= Span
0
1
⇝
s =
∂
∂v
k
2
= −
1
v
=⇒ N(B −k
2
A) = N
1
v
1
v
1
v
1
v
= Span
1
−1
⇝
t =
∂
∂u
−
∂
∂v
where we made the natural choice of vector fields
s,
t. As a sanity check, here are the matrices
of the fundamental forms with respect to
s,
t:
I(
s,
s) =
0 1
A
0
1
= 1 . . . =⇒ [I] =
I(
s,
s) I(
s,
t)
I(
s,
t) I(
t,
t)
=
1 0
0
v
2
4
[I] =
0 0
0 −
v
4
in which the principal curvatures are clearly visible: 1k
1
= 0,
v
2
4
k
2
= −
v
4
Constant Gauss & Mean Curvature Surfaces
Minimal Surfaces H ≡ 0: Among all surfaces whose boundary is a given closed curve, a surface with
minimal surface area has H ≡ 0. This is the shape made by a soap film whose boundary is the
curve: it minimizes the ‘total tension’ of the soap film. More gen-
erally, constant mean curvature (CMC) surfaces model soap bubbles.
Constant Gauss Curvature Surfaces: We’ve see that planes, cones and
cylinders have K = 0, and that spheres have constant positive
Gauss curvature. A pseudosphere with constant K = −1 is shown
in the picture.
64
Existence of (Curvature-Line) Co-ordinates
At non-umbilic points, Theorems 3.26 and 3.29 tell us how to find curvature directions as vector fields
s,
t. Unfortunately, being able to compute explicit curvature co-ordinates is exceptionally unlikely.
Example (3.30.3 cont). Recall that we chose curvature direction fields
s =
∂
∂v
and
t =
∂
∂u
−
∂
∂v
. By
inspection, the functions s = u + v and t = u satisfy the required equations:
s[s] = 1 =
t[t],
s[t] = 0 =
t[s] (∗)
It follows that
s =
∂
∂s
and
t =
∂
∂t
for curvature-line co-ordinates s, t, as you can easily verify using the
chain rule. Indeed
I =
v
2
4
du
2
+ d(u + v)
2
= ds
2
+
v
2
4
dt
2
, I = 0 ds
2
−
v
4
dt
2
so that the co-ordinates really do diagonalize both fundamental forms.
The simple reason the example is so unlikely is that mixed partial derivatives must commute: if
s =
∂
∂s
and
t =
∂
∂t
are co-ordinate fields (∃s, t : U → R), then their Lie bracket (Exercise 2.3.10) vanishes:
[
s,
t] =
∂
∂s
,
∂
∂t
=
∂
∂s
◦
∂
∂t
−
∂
∂t
◦
∂
∂s
= 0
The astonishing fact is that this simple condition is locally sufficient.
Theorem 3.31 (Co-ordinate fields). Let
s,
t be linearly independent vector fields on U ⊆ R
2
.
1. If there exist functions s, t : U → R such that
s =
∂
∂s
,
t =
∂
∂t
, then [
s,
t] = 0.
2. Suppose [
s,
t] = 0 and let p ∈ U. Then there exists a neighborhood V of p and co-ordinate
functions s, t : V → R for which
s =
∂
∂s
,
t =
∂
∂t
.
Examples 3.32. 1. The fields
s =
∂
∂x
+ y
∂
∂y
and
t =
∂
∂y
do not arise simultaneously from co-ordinates:
[
s,
t] =
∂
2
∂x∂y
+ y
∂
2
∂y
2
−
∂
2
∂y∂x
−
∂
∂y
−y
∂
2
∂y
2
= −
∂
∂y
= 0
2. The cylindrical paraboloid x(u, v) = (u, v,
1
2
u
2
+ v) has curvatures and curvature directions
k
1
= 0,
s =
∂
∂v
, k
2
=
2
[2 + u
2
]
3/2
,
t = 2
∂
∂u
−u
∂
∂v
The Lie bracket condition [
s,
t] = 0 is satisfied, so co-ordinates s, t corresponding to these fields
must exist. You can try to find such by inspection, though simultaneously solving (∗) is messy.
Alternatively, following the proof of part 2 (Exercise 13), observe that the dual 1-forms are
α =
1
2
u du + dv, β =
1
2
du (α(
s) = β(
t) = 1, α(
t) = β(
s) = 0)
These forms are exact: α = d(
1
4
u
2
+ v) and β = d(
1
2
u). We therefore conclude that s =
1
4
u
2
+ v
and t =
1
2
u are suitable curvature-line co-ordinates.
65
The Lie bracket condition says that explicit co-ordinates corresponding to given vector fields are very
unlikely to exist. This is no matter: we typically only require co-ordinates s, t whose fields
∂
∂s
,
∂
∂t
are
parallel to
s,
t: that is
∂
∂s
= f
s and
∂
∂t
= g
t for some functions f , g (equivalently
s[t] = 0 =
t[s])
Such co-ordinates indeed exist, though only locally, as shown by one of the most important founda-
tional results in differential geometry.
Theorem 3.33 (Frobenius). Let
s,
t be linearly independent vector fields on a domain U. Then there
exist local co-ordinates s, t whose co-ordinate fields are parallel to
s,
t.
In particular, if x(p) is a non-umbilic point on a surface x : U → E
3
, then there exists a neighborhood
V of p and curvature-line co-ordinates s, t on V.
Frobenius’ theorem comes in many guises and generalizes to higher dimensions, taking the place
of Picard’s ODE existence/uniqueness theorem (1.39) for particular classes of PDE. Its proof is too
involved for us, though the informal idea is to search for functions f , g such that [ f
s, g
t] = 0, a
lengthy process that indeed depends on Picard’s theorem.
Exercises 3.3. 1. Find the eigenvalues of B =
−1 −1
−1 1
with respect to A =
1 1
1 2
. If
s,
t are
corresponding eigenvectors, verify that
s
T
A
t = 0 =
s
T
B
t.
2. Parametrize the graph of x = z
2
; compute I, I and the principal, Gauss and mean curvatures.
3. Use Theorem 3.29 to find the Gauss and mean curvatures of the graph of y = x
2
−z
2
.
4. Show that Enneper’s surface (Exercise 3.2.4) is minimal.
5. Let x(u, v) = y(u) + vy
′
( u) be the tangent developable of a unit speed biregular curve y.
(a) Find the principal curvatures, Gauss and mean curvatures of x.
(b) Compute the curvature directions and find curvature line co-ordinates.
(This is very similar to Example 3.30.3 - keep track of the changes!)
6. With respect to some co-ordinates u, v, suppose that a surface has fundamental forms
I = u
2
du
2
+ v
2
dv
2
, I = u
2
du
2
+ 2uv du dv + v
2
dv
2
(a) Show that the principal curvatures are constant: k
1
= 0 and k
2
= 2.
(b) Show that
s = v
∂
∂u
−u
∂
∂v
and
t = v
∂
∂u
+ u
∂
∂v
are curvature directions.
(c) Compute the Lie bracket [
s,
t] to show that these are not vector fields with respect to some
curvature-line co-ordinates s, t.
(d) Find explicit curvature-line co-ordinates for the surface; functions s, t such that
∂
∂s
,
∂
∂t
are
parallel to
s,
t and express I, I with respect to s, t.
(Hint: try to guess solutions to
s[t] = 0 =
t[s])
66
7. Rotate y = f (x) around the x-axis and parametrize the surface via
x(ϕ, v) =
v, f (v) cos ϕ, f (v) sin ϕ
(a) Verify that the co-ordinates ϕ, v are curvature-line, compute the principal curvatures, and
show that the Gauss and mean curvatures are
K = −
f
′′
( v)
f (v)
1 + f
′
( v)
2
2
H =
f (v) f
′′
( v) −1 − f
′
( v)
2
2 f (v)
1 + f
′
( v)
2
3/2
(b) Demonstrate the following by choosing suitable f (v):
i. A cylinder has K = 0;
ii. A cone has K = 0;
iii. A sphere of radius a has K =
1
a
2
.
iv. A catenoid f (v) = a
−1
cosh(av −c) is a minimal surface.
(c) Suppose x is a minimal surface H ≡ 0. By writing g(v) = 1 +
f
′
( v)
2
, show that
1 + f
′2
= g = a
2
f
2
for some constant a
By substituting f (v) = a
−1
cosh
ah(v)
, show that the surface is a catenoid.
(d) Plainly K ≡ 0 if and only if f
′′
( v) ≡ 0. What are these surfaces? More generally, if the
surface has constant non-zero Gauss curvature K, show that f satisfies a non-linear ODE
K f
2
= (1 + f
′2
)
−1
+ c for some constant c
(Solving for f requires an elliptic integral when c = 0, so don’t try!)
8. The tractrix is parametrized by y(t) =
sinh
−1
t −t(1 + t
2
)
−1/2
, (1 + t
2
)
−1/2
. By revolving this
curve around the x-axis, show that the resulting surface is a pseudosphere with K ≡ −1.
9. We know that the Gauss and mean curvature are defined in terms of the principal curvatures.
By writing down a suitable quadratic polynomial, prove that knowing of H, K is sufficient to
recover the principal curvatures.
10. The graph of a function z = f (x, y) is parametrized by x(u, v) =
u, v, f (u, v)
. What can you
say about the surface if (u, v) are curvature-line co-ordinates?
(Hint: recall Example 3.19)
11. Suppose u, v are curvature-line co-ordinates for a surface x. Explain why n
u
= −k
1
x
u
and
n
v
= −k
2
x
v
.
12. Suppose that a surface x : U → E
3
is totally umbilic I = kI for some function k : U → R.
(a) Use Exercise 11 and n
uv
= n
vu
to prove that k is constant.
(b) Prove that x is (part of) a plane or a sphere (recall Example 3.28).
(Hint: If k = 0 consider c := x +
1
k
n. . . )
13. We prove part 2 of Theorem 3.31. Given the assumptions, define the dual 1-forms to
s,
t:
α(
s) = 1 = β(
t) and α(
t) = 0 = β(
s)
Use Exercise 2.3.10 to prove that dα = 0 = dβ. Hence conclude (footnote, page 43) that (locally)
α = ds and β = dt for some functions s, t.
67
3.4 Power Series Expansions and Euler’s Theorem
In this section we intersect a surface with certain planes and consider the resulting curves. The
curvatures provide data about these curves and thus tell us something about the local shape of the
surface. The key is to see how curvatures describe a quadratic approximation to a surface.
At a regular point P on a surface S, choose axes such that P is
the origin and the (x, y)-plane is tangent
21
to S. By Theorem
3.10, S is locally the graph of a function z = f (x, y), which we
may parametrize in the usual manner
x(u, v) =
u, v, f (u, v)
The unit normal vector n
P
= k is therefore the standard vertical basis vector. Since the tangent plane
at P is the (x, y)-plane, we see that f
u
(0, 0) = 0 = f
v
(0, 0); substituting into Example 3.19 yields the
fundamental forms at P:
I
P
= du
2
+ dv
2
I
P
= f
uu
du
2
+ 2 f
uv
dudv + f
uv
dv
2
[I]
P
=
1 0
0 1
[I]
P
= Hess f =
f
uu
f
uv
f
uv
f
vv
The last matrix is the Hessian of f , and the Gauss and mean curvatures at P are
K(P) = det Hess f (0, 0) and H(P) =
1
2
tr Hess f (0, 0)
It bears repeating that these expressions are only valid at the origin O ∈ U (equivalently P ∈ S).
Although the co-ordinates u, v will extend nearby on the surface, the first fundamental form need
not be diagonal anywhere except at the origin.
Now suppose we rotate the (x, y)-plane so that the axes point in the principal directions. Then the
Hessian is also diagonal ( f
uv
(0, 0) = 0) and the principal curvatures at P are
k
1
= f
uu
(0, 0) and k
2
= f
vv
(0, 0)
Theorem 3.34. If the graph of z = f (x, y) is tangent to the (x, y)-plane at the origin O so that the
axes are the curvature directions, then the Maclaurin approximation of the function f (x, y) is
f (x, y) ≈ f (O) + (x y) ∇f
|
O
+
1
2
(x y) Hess f (O)
x
y
+ higher order terms
=
1
2
k
1
(O)x
2
+
1
2
k
2
(O)y
2
+ higher order terms
Example 3.35. Let f (x, y) = x
2
−y
2
(above picture). At the origin, x(u, v) =
u, v, u
2
−v
2
has
I = du
2
+ dv
2
, I = 2(du
2
−dv
2
), k
1
= 2, k
2
= −2, K = −4, H = 0
In this case the Maclaurin approximation is exact!
1
2
k
1
x
2
+
1
2
k
2
y
2
= x
2
−y
2
= f (x, y)
21
This amounts to applying a rigid motion (direct isometry) to the surface, which does nothing to the fundamental forms.
68
Level Curves: intersections with planes parallel to the tangent plane
If c is small, then the intersection of S with a plane cn
P
+ T
P
S parallel to the tangent plane is a level
curve; in our analysis, they correspond to level curves f (x, y) = constant. Theorem 3.34 tells us how
level curves depend on the curvatures. For instance, if k
1
, k
2
have opposite signs, then for small c,
k
1
x
2
+ k
2
y
2
≈ 2c
is approximately a hyperbola.
Definition 3.36. Suppose k
1
, k
2
, K, H are the curvatures of a surface S at a point P. We say that P is:
Elliptic ⇐⇒ K > 0 ⇐⇒ k
1
, k
2
= 0 and have the same sign.
Level curves near P are approximately ellipses.
Hyperbolic ⇐⇒ K < 0 ⇐⇒ k
1
, k
2
= 0 and have opposite signs.
Level curves near P are approximately hyperbolæ.
Parabolic ⇐⇒ K = 0 and H = 0 ⇐⇒ exactly one of k
1
, k
2
is zero.
Level curves near P are approximately a pair of parallel lines, e.g. x = ±c.
Planar ⇐⇒ K = H = 0 ⇐⇒ k
1
= k
2
= 0.
The curvatures provide no data as to the level curves near P.
Example (3.35, mk. II). For the graph of z = x
2
−y
2
, the level curve x
2
−y
2
= c = 0 is a hyperbola.
In fact this is true everywhere on this surface: under the usual
parametrization x(u, v) =
u, v, u
2
−v
2
, we have
K = −
4
(1 + 4u
2
+ 4v
2
)
2
and H =
4(v
2
−u
2
)
(1 + 4u
2
+ 4v
2
)
3/2
Since K < 0 everywhere, all points are hyperbolic.
In the picture, shifted tangent planes cn
P
+ T
P
S and their intersections with the surface are drawn
for two points. In both cases the level curves are genuine hyperbolæ.
Normal Curvature: intersections with planes containing the normal vector
Theorem 3.34 is the surface analogy of Exercise 1.6.5: a regular curve in E
2
passing through the origin
horizontally at t = 0 has its graph given locally by
y =
1
2
κ(0)x
2
+ higher order terms (∗)
We put this to work by considering the curvature of curves passing through a point on a surface.
Definition 3.37. Let S be a surface and v
P
∈ T
P
S a non-zero tangent vector.
The normal curvature ν(v
P
) is the curvature at P of the curve
22
defined by the intersection of the
surface S and the normal plane Span{v
P
, n
P
}.
We say that v
P
is asymptotic if ν(v
P
) = 0.
22
Strictly, the curve is the connected component of S ∩Span{v
P
, n
P
} containing P.
69
Example (3.35, mk. III). Consider the hyperbolic paraboloid z = x
2
− y
2
at the origin P = O.
Fix an angle ψ and let v
P
= ( cos ψ, sin ψ). The intersection curve y ⊆ S ∩ Span{v
O
, n
O
} may be
parametrized using polar co-ordinates:
y(r) =
r cos ψ, r sin ψ, r
2
(cos
2
ψ − sin
2
ψ)
which amounts to the graph of the function g(r) = r
2
cos 2ψ.
The normal curvature is the curvature at r = 0 of this curve:
ν(v
O
) = κ(0) =
g
′′
(0)
[1 + g
′
(0)
2
]
3/2
= 2 cos 2ψ
Think about how the this corresponds to the picture and observe that
v
O
is asymptotic ⇐⇒ cos 2ψ = 0 ⇐⇒ ψ = ±
π
4
Our next result generalizes the method in the example.
Theorem 3.38 (Euler). Suppose v
P
makes angle ψ with the first principal curvature direction. Then
ν(v
P
) = k
1
cos
2
ψ + k
2
sin
2
ψ
In particular, the principal curvatures are the extremes of normal curvature: if k
1
≤ k
2
, then
k
1
≤ ν(v
P
) ≤ k
2
where the bounds are realized precisely when v
P
points in a curvature direction.
Proof. Choose axes so the curvature directions at P are i, j, and n
P
= k. The surface is locally a graph
z = f (x, y). If (r, ψ) are polar co-ordinates in the (x, y)-plane, Theorem 3.34 says that
z = f (x, y) ≈
1
2
k
1
(r cos ψ)
2
+
1
2
k
2
(r sin ψ)
2
+ ··· =
1
2
( k
1
cos
2
ψ + k
2
sin
2
ψ)r
2
+ ···
Fix ψ and let v
P
=
cos ψ
sin ψ
(assume unit length since only the direction matters). Our curve of interest
y ⊆ S ∩Span{v
P
, n
P
} may be parametrized
y(r) = rv
P
+ f (r cos ψ, r sin ψ) n
P
=
r cos ψ
r sin ψ
f ( r cos ψ, r sin ψ)
=
r cos ψ
r sin ψ
1
2
νr
2
+ ···
The last equality used observation (∗), where ν is the normal curvature. For the first result, simply
compare the z-expressions in the displayed equations. For the final observation, note that
ν(v
P
) = k
1
(1 − sin
2
ψ) + k
2
sin
2
ψ = k
1
+ (k
2
−k
1
) sin
2
ψ ∈ [k
1
, k
2
]
and that the bounds are achieved precisely when ψ = 0,
π
2
, when v
P
is a curvature direction.
70
Examples 3.39. 1. If P is a planar point (k
1
= k
2
= 0), all normal curvatures are zero and all
directions are asymptotic.
2. (Example 3.30.1) All points of the paraboloid z = r
2
are elliptic (everywhere k
1
, k
2
> 0). The
surface has no asymptotic directions at any point, indeed the normal curvature in the direction
v
P
= (cos ψ, sin ψ) at P = (r cos θ, r sin θ, r
2
) is
ν(v
P
) = k
1
cos
2
ψ + k
2
sin
2
ψ =
2
(1 + 4r
2
)
3/2
cos
2
ψ + (1 + 4r
2
) sin
2
ψ
> 0
3. If k
2
= 0, then v
P
=
cos ψ
sin ψ
is asymptotic ⇐⇒ tan ψ = ±
q
−
k
1
k
2
.
The Second Fundamental Form and the Local Shape of a Surface
Our standard approach is to transfer calculations about surfaces back to the parametrization space.
With this in mind, we consider special tangent vectors with respect to the second fundamental form.
Definition 3.40. Let x : U → E
3
be an oriented surface and
w
p
∈ T
p
R
2
.
1. A tangent vector
w
p
=
0 is asymptotic if I(
w
p
,
w
p
) = 0.
2. The Dupin indicatrix at p ∈ U is the set of tangent vectors
w
p
such that I(
w
p
,
w
p
) = ±1.
Theorem 3.41. The notions of asymptotic in Definitions 3.37 & 3.40 coincide:
v
P
= dx(
w
p
) ∈ T
P
S is asymptotic ⇐⇒
w
p
∈ T
p
R
2
is asymptotic
The proof is an exercise. Recalling Theorem 3.20, a direction is asymptotic if and only if the normal
acceleration in said direction is zero.
The Dupin indicatrix turns out to precisely describe level curves near a point. To see this, write
w
p
= a
s
p
+ b
t
p
where
s
p
,
t
p
are orthonormal curvature directions;
23
the indicatrix at p has equation
I(
w
p
,
w
p
) =
a b
k
1
0
0 k
2
a
b
= k
1
a
2
+ k
2
b
2
= ±1
This defines a conic in the tangent space T
p
R
2
whose type depends on the signs of the principal cur-
vatures. In essence, the Dupin indicatrix indicates the level curve obtained by taking the intersection
S ∩ (cn
P
+ T
P
S) for infinitesimal c. We summarize all possibilities in a table using the point-types
introduced in Definition 3.36:
type of point # asymptotic directions Dupin indicatrix
elliptic 0 ellipse
hyperbolic 2 two hyperbolæ
parabolic 1 two parallel lines
planar ∞ empty
23
With respect to
s
p
,
t
p
, the matrices of the fundamental forms at p are [I
p
] =
1 0
0 1
and [I
p
] =
k
1
0
0 k
2
71
Examples 3.42. For a parametrized surface x at a given point p = (u
0
, v
0
), write
w
p
= a
∂
∂u
p
+ b
∂
∂v
p
.
1. (Exercise 3.2.6) The tangent developable of the unit-speed helix has
I(
w
p
,
w
p
) = −
v
0
4
du
2
(
w
p
,
w
p
) = −
v
0
4
a
2
The single asymptotic direction is
w
p
=
∂
∂v
p
. The Dupin indicatrix is a pair of parallel lines
−
v
0
4
a
2
= ±1 =⇒
w
p
= ±
2
p
|
v
0
|
∂
∂u
p
+ b
∂
∂v
p
(b is arbitrary!)
2. In its usual parametrization, the surface z = x
2
y has
I =
2
√
1 + 4u
2
v
2
+ u
4
v du
2
+ 2u dudv
At p = (−1, 2) (i.e., x(p) = (−1, 2, 2)) we see that
I(
w
p
,
w
p
) =
2
√
18
2a
2
−2ab
=
2
√
2
3
a(a −b)
b
a
T
p
R
2
The point is hyperbolic with asymptotic directions
∂
∂v
p
and
∂
∂u
p
−
∂
∂v
p
(a = 0 and a − b = 0). The indicatrix comprises two hyperbolæ a(a − b) = ±
3
2
√
2
.
Exercises 3.4. 1. Consider the graph of the function z = x
2
−3y
2
+ 7xy
3
+ 9y
4
.
(a) Find the Gauss and mean curvatures at the origin.
(Hint: use Theorem 3.34)
(b) Find the normal curvature at the origin for the curve in the surface described by x = y.
2. As in Example 3.35, mk. III (page 70), find the asymptotic directions at the origin for the surface
z = y
2
−3x
2
.
3. For the elliptic paraboloid z = x
2
+ y
2
, let P = (1, 2, 5) be a fixed point.
(a) Find the maximum and minimum values for the normal curvature at P.
(b) Find the Dupin indicatrix at P.
4. For the hyperbolic paraboloid z = x
2
− y
2
, let p = (u
0
, v
0
) and P =
u
0
, v
0
, u
2
0
− v
2
0
. If c = 0,
prove that the intersection of the parallel plane cn
P
+ T
P
S and the paraboloid may be expressed
(x −u
0
)
2
−(y −v
0
)
2
= constant, z = x
2
−y
2
That is, the level curves really are hyperbolæ.
5. Consider the graph of the surface z = x
2
+ y
4
.
(a) Compute the Gauss curvature and classify all points according to Definition 3.36.
(b) Sketch the level curves z = 1,
1
100
and
1
10000
and compare to the Dupin indicatrix at (0, 0).
6. Prove Theorem 3.41 by considering the normal acceleration of the curve S ∩Span{v
P
, n
P
}.
72
3.5 Adaptive Frames & Gauss’ Remarkable Theorem
In this section we repurpose the idea of a moving frame first encountered when studying curves.
Definition 3.43. Let x : U → E
3
parametrize a surface S. A moving frame for S is a triple of smooth
functions e
1
, e
2
, e
3
on U such that, for each p ∈ U,
e
1
(p), e
2
(p), e
3
(p)
is a positively oriented orthonormal basis of T
x(p)
E
3
When S is oriented, we say that a moving frame is adaptive if e
3
= n is the unit normal field.
For an adaptive frame, the tangent plane at each point is T
x(p)
S = Span{e
1
(p), e
2
(p)}.
We will often refer to the matrix-valued function E =
e
1
e
2
e
3
: U → SO
3
(R ) as the frame.
Examples 3.44. We’ll repeatedly analyze three examples through this section.
1. The parabolic cylinder x(u, v) =
u, v,
1
2
u
2
has an adaptive frame
e
1
=
1
√
1 + u
2
1
0
u
e
2
=
0
1
0
e
3
=
1
√
1 + u
2
−u
0
1
2. The sphere of radius R in spherical polar co-ordinates x(ψ, ϕ) has an adaptive frame
e
1
=
−sin ψ
cos ψ
0
e
2
=
−cos ψ sin ϕ
−sin ψ sin ϕ
cos ϕ
e
3
= x =
cos ψ cos ϕ
sin ψ cos ϕ
sin ϕ
We use ψ instead of θ since we’ll need the latter for something else momentarily. . .
3. The paraboloid x(r, ψ) =
r cos ψ, r sin ψ,
1
2
r
2
has an adaptive frame
e
1
=
1
√
1 + r
2
cos ψ
sin ψ
r
e
2
=
−sin ψ
cos ψ
0
e
3
=
1
√
1 + r
2
−r cos ψ
−r sin ψ
1
In the pictures we’ve reduced the lengths of the frame vectors for clarity.
In each case e
1
, e
2
were obtained by differentiating with respect to the co-ordinates (and normalizing
if necessary). This works because the co-ordinate systems for all three examples are orthogonal.
73
As with the Frenet frame approach to curves, our strategy is to analyse a surface x : U → E
3
two
stages:
1. Describe how x moves with respect to the frame E.
2. Describe how the frame E moves (with respect to itself).
We describe infinitesimal changes using 1-forms, following an approach pioneered by
´
Elie Cartan
around 1899.
Definition 3.45. Let x : U → E
3
be a smooth map and E = (e
1
e
2
e
3
) a moving frame. The metric
forms θ
j
and connection forms ω
jk
are the 1-forms on U defined by
θ
j
:= e
j
·dx, ω
jk
= e
j
·de
k
where j, k ∈ {1, 2, 3}.
Since e
1
, e
2
, e
3
are orthonormal, these forms are nothing more than the co-ordinates of dx, de
1
, de
2
and de
3
with respect to the moving frame:
dx =
3
∑
j=1
( e
j
·dx)e
j
= e
1
θ
1
+ e
2
θ
2
+ e
3
θ
3
, de
k
=
3
∑
j=1
e
j
ω
jk
(∗)
The frame is adaptive if and only if θ
3
= 0. Moreover, as the next result shows, for any frame there
are only three independent connection forms (compare this with Theorem 1.29).
Lemma 3.46. For all j, k, we have ω
jk
= −ω
kj
. In particular ω
jj
= 0.
Proof. Take the exterior derivative of the identity e
j
·e
k
= 0 or 1, to obtain
0 = de
j
·e
k
+ e
j
·de
k
= ω
kj
+ ω
jk
If (∗) are arranged in matrix format, the subscripts follow the usual row/column convention:
dx =
e
1
e
2
e
3
θ
1
θ
2
θ
3
= EΘ, dE =
e
1
e
2
e
3
0 ω
12
ω
13
−ω
12
0 ω
23
−ω
13
−ω
23
0
= Eω
The second expression should remind you of the Frenet–Serret equations for a curve! The metric
forms get their name because they measure small changes on the surface. The connection forms tell
us how nearby frames are related (connected): abusing notation a little, if
s
p
∈ T
p
R
2
, then
E(p +
s
p
) − E(p) ≈ dE(
s
p
) = E(p)ω(
s
p
)
The fundamental forms of x can be written in terms of Θ and ω; in an adaptive frame this is partic-
ularly simple.
Lemma 3.47. In an adaptive frame
I = dx ·dx = θ
2
1
+ θ
2
2
and I = −dx ·de
3
= −θ
1
ω
13
−θ
2
ω
23
74
Examples (3.44, mk. II). You needn’t compute all exterior derivatives de
k
: use the skew-symmetry
of ω to help; also consider which frame fields are easier to differentiate! The expressions for the
fundamental forms should be a sanity check since we know how to compute them already.
1. The parabolic cylinder has
dx =
1
0
u
du +
0
1
0
dv =
p
1 + u
2
e
1
du + e
2
dv =⇒ θ
1
=
p
1 + u
2
du, θ
2
= dv
=⇒ I = (1 + u
2
) du
2
+ dv
2
Since e
2
is constant, we have de
2
= 0 from which
ω
12
= e
1
·de
2
= 0, ω
23
= −ω
32
= −e
3
·de
2
= 0
The final connection form requires a derivative:
ω
13
= e
1
·de
3
=
1
√
1 + u
2
1
0
u
·
−u
(1 + u
2
)
3/2
−u
0
1
+
1
√
1 + u
2
−1
0
0
=
−1
1 + u
2
du
Putting it together, we have
ω =
1
1 + u
2
0 0 −1
0 0 0
1 0 0
du and I = −θ
1
ω
13
−θ
2
ω
23
= du
2
2. For the sphere of radius R, dx = R cos ϕe
1
dψ + Re
2
dϕ, whence
θ
1
= R cos ϕ dψ, θ
2
= Rdϕ =⇒ I = R
2
cos
2
ϕ dψ
2
+ dϕ
2
de
1
=
−cos ψ
−sin ψ
0
dψ =⇒
(
ω
12
= −e
2
·de
1
= −sin ϕ dψ
ω
13
= −e
3
·de
1
= cos ϕ dψ
ω
23
= e
2
·de
3
=
−cos ψ sin ϕ
−sin ψ sin ϕ
cos ϕ
·
−sin ψ cos ϕ
cos ψ cos ϕ
0
dψ +
−cos ψ sin ϕ
−sin ψ sin ϕ
cos ϕ
dϕ
= dϕ
=⇒ I = −θ
1
ω
13
−θ
2
ω
23
= −R
cos
2
ϕ dψ
2
+ dϕ
2
3. For the paraboloid,
dx =
cos ψ
sin ψ
r
dr + r
−sin ψ
cos ϕ
0
dψ =
p
1 + r
2
e
1
dr + re
2
dψ
=⇒ θ
1
=
p
1 + r
2
dr, θ
2
= r dψ =⇒ I = (1 + r
2
)dr
2
+ r
2
dψ
2
The connection forms are comparatively ugly. The low-hanging fruit is de
2
=
−cos ψ
−sin ψ
0
dψ,
which quickly yields two of them:
ω
12
= e
1
·de
2
= −
dψ
√
1 + r
2
, ω
23
= −ω
32
= −e
3
·de
2
=
−r dψ
√
1 + r
2
75
The last connection form requires a nastier differentiation, though only one of the three terms
in de
3
provides a non-zero result when dotted with e
1
:
ω
13
= e
1
·de
3
=
1
√
1 + r
2
cos ψ
sin ψ
r
·
··· +
1
√
1 + r
2
−cos ψ
−sin ψ
0
dr
=
−dr
1 + r
2
We therefore obtain the connection form matrix
ω =
1
√
1 + r
2
0 −dψ
−1
√
1+r
2
dr
dψ 0 −r dψ
1
√
1+r
2
dr r dψ 0
and second fundamental form
I = −
p
1 + r
2
dr
−dr
1 + r
2
−r dψ
−r dψ
√
1 + r
2
=
1
√
1 + r
2
dr
2
+ r
2
dψ
2
The Structure Equations for a Moving Frame
The metric and connection forms satisfy matrix equations dx = EΘ and dE = Eω. Since d
2
= 0,
something nice happens when we take the exterior derivatives of these expressions:
0 = d
2
x = d(dx) = d(E Θ) = dE ∧Θ + E dΘ = E(ω ∧Θ + dΘ)
0 = d
2
E = d(dE) = d(Eω) = dE ∧ ω + Edω = E
ω ∧ω + dω
The notation ω ∧Θ means matrix multiplication using the wedge product of forms to evaluate each
entry.
24
Since each E(p) is an invertible matrix, we conclude two identities.
Theorem 3.48. The metric and connection forms satisfy the structure equations; each amounts to
three separate equations after multiplying out the matrix expressions.
1. dΘ + ω ∧Θ = 0, equivalently dθ
j
+
∑
k=j
ω
jk
∧θ
k
= 0 for each j = 1, 2, 3
2. dω + ω ∧ ω = 0, equivalently dω
jk
+ ω
ji
∧ω
ik
= 0 where i, j, k are distinct.
These are easy to remember if you pay attention to the indices! In an adaptive frame (θ
3
= 0), things
are a little simpler and some of the equations get special names:
First structure equations
dθ
1
+ ω
12
∧θ
2
= 0
dθ
2
+ ω
21
∧θ
1
= 0
Symmetry equation ω
31
∧θ
1
+ ω
32
∧θ
2
= 0
Gauss equation dω
12
+ ω
13
∧ω
32
= 0
Codazzi equations
dω
13
+ ω
12
∧ω
23
= 0
dω
23
+ ω
21
∧ω
13
= 0
24
Be careful not to reverse the order: Θ ∧ ω makes no sense since the dimensions of the matrices are incompatible!
Similarly, ω ∧ ω is unlikely to be zero. . .
76
Examples (3.44, mk. III). 1. For the parabolic cylinder, Θ =
√
1+u
2
du
dv
0
and ω =
1
1+u
2
0 0 1
0 0 0
−1 0 0
du,
so all the structure equations are trivial:
dΘ = 0 = −ω ∧Θ, dω = 0 = −ω ∧ ω
2. For the sphere, Θ = R
cos ϕ dψ
dϕ
0
and ω =
0 sin ϕ dψ −cos ϕ dψ
−sin ϕ dψ 0 dϕ
cos ϕ dψ −dϕ 0
, from which
dΘ = R
−sin ϕ
0
0
dϕ ∧dψ = −ω ∧Θ
dω =
0 cos ϕ sin ϕ
−cos ϕ 0 0
−sin ϕ 0 0
dϕ ∧dψ = −ω ∧ω
3. For the paraboloid, Θ =
√
1+r
2
dr
r dψ
0
and ω =
1
√
1+r
2
0 −dψ −
dr
√
1+r
2
dψ 0 −r dψ
dr
√
1+r
2
r dψ 0
.
The first equations aren’t too bad to check:
dΘ =
0
1
0
dr ∧dψ = −ω ∧ Θ
The second are a little nastier: you should check that
dω =
1
(1 + r
2
)
3/2
0 r 0
−r 0 −1
0 1 0
dr ∧dψ = −ω ∧ ω
Gauss’ Remarkable Theorem
Suppose we have an adaptive frame for an oriented local surface x. If θ
1
, θ
2
were linearly dependent
at p, then the differential dx = e
1
θ
1
+ e
2
θ
2
: T
p
R
2
→ T
x(p)
S = Span{e
1
(p), e
2
(p)} would have
rank ≤ 1 and thus not be a bijection. We conclude that {θ
1
, θ
2
} forms a basis of the space of 1-forms
at p, and that any other 1-form may be written as a linear combination thereof. . .
Lemma 3.49. There exist unique functions a, b, c such that
ω
13
= aθ
1
+ bθ
2
, ω
23
= bθ
1
+ cθ
2
With respect to these functions, the second fundamental form, Gauss and mean curvatures are
I = −aθ
2
1
−2bθ
1
θ
2
−cθ
2
1
, K = ac −b
2
, H = −
1
2
(a + c)
77
Proof. That ω
13
= aθ
1
+ bθ
2
and ω
23
=
ˆ
bθ
1
+ cθ
2
are linear combinations of θ
1
, θ
2
is the above discus-
sion. By the symmetry equation and the fact that θ
1
∧θ
2
= 0,
0 = ω
13
∧θ
1
+ ω
23
∧θ
2
= (−b +
ˆ
b) θ
1
∧θ
2
=⇒
ˆ
b = b
The formula for I follows from Lemma 3.47.
Moreover, if
w
1
and
w
2
are the dual vector fields to θ
1
, θ
2
, then the matrices of I, I with respect to
these fields are the identity matrix
25
and B =
−a −b
−b −c
. The Gauss and mean curvatures are the
determinant and half the trace of B (Theorem 3.29).
Now consider the final connection form ω
12
. Since θ
1
, θ
2
form a basis at each point, we may write
ω
12
= f θ
1
+ gθ
2
for some functions f , g : U → R. Applying the 1
st
structure equations,
dθ
1
= −ω
12
∧θ
2
= −f θ
1
∧θ
2
dθ
2
= −ω
21
∧θ
1
= −θ
1
∧ω
12
= −gθ
1
∧θ
2
whence f , g (and ω
12
) are determined by θ
1
, θ
2
. This brings us to the capstone result of these notes.
Theorem 3.50 (Gauss’ Theorem Egregium). The Gauss curvature depends only on the first funda-
mental form.
Proof. By the above discussion, ω
12
(and thus dω
12
) depends only on θ
1
, θ
2
, which may be recovered
from I = θ
2
1
+ θ
2
by writing it as a sum of squares. But now the Gauss equation reads
dω
12
= ω
13
∧ω
23
= (aθ
1
+ bθ
2
) ∧ (bθ
1
+ cθ
2
) = (ac −b
2
) θ
1
∧θ
2
= Kθ
1
∧θ
2
An explicit formula for K as a function of the coefficients E, F, G of I can be found; see Exercise 9.
Egregium (Latin for remarkable/outstanding) is the (modest!) term Gauss applied after proving his
result in 1827. Why did he consider it so remarkable? The original definition of K relied on the nor-
mal field; an object outside the surface which helps describe its position/orientation in E
3
. Gauss’
Theorem, however, says that K is intrinsic to the surface: it depends only on the metric (first fun-
damental form) which may be understood by an occupant of the surface with no ability to escape
(travel outside the surface) in to view its shape. By contrast, the second fundamental form and the
mean curvature depend on how a surface is embedded; these are extrinsic quantities.
As a nice side-effect, the result provides what is often a faster method for calculating K.
1. Compute the first fundamental form I = dx · dx and express it as a sum of squares I = θ
2
1
+ θ
2
2
.
2. Write ω
12
= f θ
1
+ gθ
2
and compute f , g using the 1
st
structure equations.
3. Use the Gauss equation to find K.
We need only calculate 1-forms θ
1
, θ
2
, ω
12
that are related to the tangent part of the moving frame.
The unit normal e
3
needn’t be considered or calculated.
25
θ
j
(
w
k
) = δ
jk
=
(
1 j = k
0 j = k
implies that dx(
w
1
) = e
1
and dx(
w
2
) = e
2
are orthonormal.
78
Examples (3.44, mk. IV). We return to our examples one last time. Even though we’ve already
calculated the connection forms, the goal is to see that ω
12
= f θ
1
+ gθ
2
and thus K may be found
directly from I.
1. The parabolic cylinder has I = (1 + u
2
) du
2
+ dv
2
so the natural choice is
θ
1
=
p
1 + u
2
du and θ
2
= dv
Since dθ
1
= 0 = dθ
2
we see that f = g = 0. We conclude that
ω
12
= 0 =⇒ dω
12
= 0 =⇒ K = 0
2. For the sphere I = R
2
(cos
2
ϕ dψ
2
+ dϕ
2
so we choose θ
1
= R cos ϕ dψ and θ
2
= R dϕ. Certainly
0 = dθ
2
= −gθ
1
∧θ
2
=⇒ g = 0. Moreover,
dθ
1
= −f θ
1
∧θ
2
=⇒ R sin ϕ dψ ∧dϕ = −f R
2
cos ϕ dψ ∧ dϕ =⇒ f = −R
−1
tan ϕ
We conclude that ω
12
= −R
−1
tan ϕ θ
1
= −sin ϕ dψ, from which
dω
12
= cos ϕ dψ ∧dϕ =
1
R
2
θ
1
∧θ
2
=⇒ K =
1
R
2
3. For the paraboloid, I = (1 + r
2
)dr
2
+ r
2
dψ
2
so we choose θ
1
=
√
1 + r
2
dr and θ
2
= r dψ. This
time dθ
1
= 0 =⇒ f = 0 and
dθ
2
= −gθ
1
∧θ
2
=⇒ dr ∧dψ = −gr
p
1 + r
2
dr ∧dψ =⇒ g = −
1
r
√
1 + r
2
We conclude that ω
12
= −
1
r
√
1+r
2
θ
2
= −
1
√
1+r
2
dψ, from which
dω
12
=
r
(1 + r
2
)
3/2
dr ∧dψ =
1
(1 + r
2
)
2
θ
1
∧θ
2
=⇒ K =
1
(1 + r
2
)
2
Since K depends only on the metric, it is invariant under isometric transformations of the surface. This
helps explain why the Gauss curvature of a cylinder and a cone are both zero: both may constructed
by rolling up a flat plane without other distortion.
The contrapositive of Gauss’ Theorem is also important: surfaces with distinct Gauss curvatures
cannot be isometric. Since the metric I determines how we measure angle and length, this explains
why a perfect flat map (K = 0) of any part of the Earth (K =
1
R
2
) is impossible to achieve. The holy
grail of map-making would be a map free of direction, angle and length/area distortion:
1. Straight lines on the map should correspond to paths of shortest distance on the Earth.
2. Angles on the map should equal corresponding angles on the Earth’s surface.
3. Areas on the map and the Earth should be in constant ratio.
Gauss’ Theorem implies that you cannot have all these properties in one map. In fact, at most one of
these properties is possible in a single map.
79
Riemannian Geometry
We can even employ the method when there is no surface! The idea is to equip a domain with an
abstract first fundamental form and use it to compute lengths, angles, area, geodesics, curvature, etc.
Example 3.51. The Poincar
´
e disk model of hyperbolic space is the disk
(x, y) ∈ R
2
: x
2
+ y
2
< 1
equipped with the metric (first fundamental form)
I =
4( dx
2
+ dy
2
)
(1 − x
2
−y
2
)
2
=
4( dr
2
+ r
2
dψ
2
)
(1 − r
2
)
2
As one approaches the boundary of the disk, the idea is that measured distance gets larger: the
boundary circle is in fact infinitely far from any point inside the disk. To express I as a sum of squares,
a natural choice is θ
1
=
2dr
1−r
2
and θ
2
=
2r dψ
1−r
2
, from which dθ
1
= 0 =⇒ f = 0 and
dθ
2
= −gθ
1
∧θ
2
=⇒
2( 1 + r
2
)
(1 − r
2
)
2
dr ∧dψ = −
4gr
(1 − r
2
)
2
dr ∧dψ =⇒ g = −
1 + r
2
2r
from which
dω
12
= d(gθ
2
) = −d
1 + r
2
1 −r
2
∧dψ =
−4r
(1 − r
2
)
2
dr ∧dψ = −θ
1
∧θ
2
=⇒ K = −1
Hyperbolic space is the canonical example of a negatively curved geometry. There is no surface here, no
second fundamental form, and no mean curvature! Since there is no surface, it is harder to visualize
what K means in this context (e.g. Section 3.4).
26
The Gauss curvature of a surface is the simplest avatar of a more general object called the Riemann
curvature tensor. As an example of how this is applied, in general relativity
27
mass is construed as
changing the metric of spacetime (i.e. I); it can be seen that this metric is compatible with unique
connection (essentially ω) from which the curvature (dω + ω ∧ ω) may be computed. When a
physicist says spacetime is curved, this is what they mean: there is no exterior to spacetime from which
we can measure curvature, so everything is computed intrinsically.
The Fundamental Theorem of Surfaces
Recall the equivalence of spacecurves up to rigid motions (Theorem 1.38) and the Fundamental The-
orem of Biregular Spacecurves (Corollary 1.42). A similar discussion is available for surfaces once we
replace curvature and torsion with the fundamental forms I, I.
The equivalence problem is almost identical. Suppose x : U → E
3
is an oriented surface, and that
A ∈ O
3
(R ) and b = E
3
are constants. Then y := Ax + b is a new surface, the result of applying an
isometry to x. A moving frame for x is transformed to a frame for y via
E
y
=
Ae
1
Ae
2
± Ae
3
where ±1 = det A
26
The most famous consequence concerns angle-sums of geodesic triangles: A + B + C = π +
R
△
K. If K < 0, the angle-
sum of a geodesic triangle is less than 180°. When K > 0 (e.g., a sphere), the angle sum is greater than 180°. This topic, the
related Gauss–Bonnet Theorem, and other consequences, are a matter for another course.
27
Really this is pseudo-Riemannian geometry, since I is not positive-definite.
80
The upshot is that n
y
= (det A)An
x
, and I, I transform exactly as κ, τ:
I
y
= dy ·dy = (Adx) · (Adx) = dx ·dx = I
x
I
y
= −dy ·dn
y
= −(det A) (Adx) · (Adn
x
) = (det A)I
x
As with curves, we may ask the question in reverse. If we know the fundamental forms, can we also
recover the surface up to a rigid motion? The answer is yes, though with a caveat: unlike κ, τ for
spacecurves, the fundamental forms cannot be chosen independently.
Theorem 3.52 (Bonnet). Suppose I and I are symmetric bilinear forms where I is positive-definite.
Provided the Gauss–Codazzi equations are satisfied, there exists a local parametrized surface with
these fundamental forms, which is moreover unique up to rigid motions.
Everything ultimately depends on a generalization of the existence/uniqueness theorem for ODE
(another version of the Frobenius Theorem (3.33)). Here is a rough sketch of how the process works.
1. Suppose we are given I, I on U, and initial conditions at some p ∈ U (for the surface x(p) = x
0
and frame E(p) = E
0
).
2. Since I is positive-definite, it may be written I = θ
2
1
+ θ
2
2
.
3. The first structure equations determine ω
12
and I determines ω
13
and ω
23
(Lemma 3.49).
4. The Frobenius Theorem shows that the initial value problem
dE = Eω E(p) = E
0
(∗)
has a unique local solution provided the Gauss–Codazzi equations dω + ω ∧ ω = 0 are sat-
isfied. The solution E is SO
3
(R )-valued and supplies an adapted frame (compare Corollary
1.41).
5. To find the surface, solve a second initial value problem
dx = EΘ x(p) = x
0
Frobenius says this has a unique solution provided dΘ + ω ∧ Θ = 0. Since this is precisely
what we used to determine ω in step 2, we don’t need to check this condition.
6. Any different choice of metric forms in step 2 merely rotates E around n = e
3
and does not
affect the resulting surface.
It is a little easier to understand the integrability condition when written in co-ordinates: (∗) is a
linear system of eighteen PDE in nine unknowns
(
∂E
∂u
= EP
∂E
∂v
= EQ
where P = ω
∂
∂u
, Q = ω
∂
∂v
are skew-symmetric matrix functions
The Gauss–Codazzi equations are essentially the fact that mixed partial derivatives commute:
28
0 = E
uv
−E
vu
= E
v
P + EP
v
−E
u
Q + EQ
u
= E
P
v
− Q
u
−[P, Q]
P
v
− Q
u
−[P, Q] =
∂
∂v
ω
∂
∂u
−
∂
∂u
ω
∂
∂v
−
ω
∂
∂u
, ω
∂
∂v
=
dω + ω ∧ω
∂
∂v
,
∂
∂u
28
[P, Q] = PQ −QP and dω is evaluated as in Exercise 2.3.10.
81
The part that requires some proof is that the integrability condition (P
v
−Q
u
= [P, Q]) is sufficient for
a solution. This is not as hard as it sounds; here is another sketch:
1. If p = (u
0
, v
0
), use Picard’s ODE existence/uniqueness theorem to
solve an initial value problem on the horizontal line v = v
0
:
d
e
E
du
=
e
EP(u, v
0
),
e
E(u
0
, v
0
) = E
O
2. For each u
1
, apply the ODE theorem to solve another IVP on the
vertical line u = u
1
:
dE
dv
= EQ(u
1
, v), E(u
1
, v
0
) =
e
E(u
1
, v
0
)
v
u
p
u
0
v
0
3. Finally, one shows that the resulting E is differentiable with respect to u, and uses the integra-
bility condition to check that E
u
= EP as required.
The first two steps may be accomplished approximately using a numerical method to desired accu-
racy, so this amounts to an algorithm for the approximation of E. The same approach can then be
followed to approximate the surface.
The Gauss–Codazzi equations in curvature-line co-ordinates Suppose (u, v) are curvature-line
co-ordinates. Then the fundamental forms are
I = E du
2
+ G dv
2
, I = k
1
E du
2
+ k
2
Gdv
2
(†)
where E, G are positive functions and k
1
, k
2
are the principal curvatures. We therefore choose metric
forms θ
1
=
√
E du and θ
2
=
√
G dv. In the language of Lemma 3.49,
a = −k
1
, b = 0, c = −k
2
, ω
13
= −k
1
√
E du, ω
23
= −k
2
√
G dv
The first structure equations determine
ω
12
=
1
2
√
EG
(
E
v
du −G
u
dv
)
(see Exercise 9). Moreover, the Gauss–Codazzi equations are equivalent to
dω
12
+ ω
21
∧ω
13
= 0 ⇐⇒
G
u
√
EG
u
+
E
v
√
EG
v
= −2k
1
k
2
√
EG
dω
13
+ ω
12
∧ω
23
= 0 ⇐⇒ 2(k
1
)
v
E = (k
2
−k
1
)E
v
dω
23
+ ω
21
∧ω
13
= 0 ⇐⇒ 2(k
2
)
u
G = (k
1
−k
2
)G
u
These equations show the relationship between I and I: we cannot independently choose the metric
(E, G) and the curvatures (k
1
, k
2
). However, if E, G, k
1
, k
2
satisfy these equations, Bonnet’s theorem
guarantees the existence of a surface with fundamental forms (†), unique up to rigid motions.
While I, I cannot be chosen independently, Bonnet’s result is considered the best description of the
minimal data for a surface. You might suspect/hope that knowledge of K, H would be enough to
determine a surface up to rigid motions, but Exercise 10 shows such to be vain!
82
Exercises 3.5. 1. The unit cylinder x(ϕ, v) =
cos ϕ, sin ϕ, v
has adaptive frame
e
1
=
−sin ϕ
cos ϕ
0
, e
2
=
0
0
1
, e
3
= e
1
×e
2
=
cos ϕ
sin ϕ
0
(a) Directly compute the metric forms θ
j
and connection forms ω
jk
.
(b) That the six structure equations are satisfied should be obvious from your answers to (a):
why?
(c) Why is it completely obvious from your answer to (a) that K ≡ 0?
2. For a general regular surface, explain why we cannot, in general, find co-ordinates u, v for
which I = du
2
+ dv
2
.
3. For the paraboloid example (3.44.3) verify the Gauss–Codazzi equations dω + ω ∧ ω = 0.
(Hint: this is easier if you treat the three equations separately!)
4. Verify that the metric I =
dx
2
+dy
2
y
2
on the upper half-plane y > 0 has curvature K = −1.
(Hint: Recall Example 3.51 and Exercise 3.2.9)
5. Consider the catenoid x(u, v) =
cos u cosh v, sin u cosh v, v
obtained by revolving the cate-
nary x = cosh z around the z-axis.
(a) Show that there exists a moving frame for which the metric forms are
θ
1
= cosh v du, θ
2
= cosh v dv
(b) Show that ω
12
= tanh v du =
sinh v
cosh v
du and use it to prove that the Gaussian curvature of
the catenoid is
K = −
1
cosh
4
v
6. We re-prove Exercise 3.3.12 using our new language.
(a) Suppose a surface x is totally umbilic: I = λI, where λ is some function. Explain why
ω
13
= −λθ
1
and ω
23
= −λθ
2
.
(b) Use the 1
st
structure equations and the Codazzi equations to prove that dλ = 0.
(c) If a = 0, what is x?
(d) If a = 0, define c := x −
1
a
e
3
. Prove that dc = 0 and hence conclude that the surface is
(part of a) round sphere.
7. Suppose E =
e
1
e
2
e
3
is an adaptive frame for a surface. Any other adaptive frame (with the
same orientation) is obtained by rotating around e
3
: that is
ˆ
E =
ˆ
e
1
ˆ
e
2
e
3
where
ˆ
e
1
= cos φ e
1
+ sin φ e
2
,
ˆ
e
2
= −sin φ e
1
+ cos φ e
2
for some smooth function φ : U → R.
(a) Compute θ
1
, θ
2
in terms of
ˆ
θ
1
,
ˆ
θ
2
and conclude that
ˆ
θ
1
∧
ˆ
θ
2
= θ
1
∧θ
2
.
(b) Use Definition 3.45 to compute
ˆ
ω
12
in terms of ω
12
and φ. Verify that d
ˆ
ω
12
= dω
12
so that
the Gauss equation is identical for the new moving frame.
83
8. Suppose I is the 1
st
fundamental form of a surface. Suppose I = θ
2
1
+ θ
2
2
for some 1-forms θ
1
, θ
2
.
Prove that there exists a moving frame E = (e
1
e
2
e
3
) for which dx = e
1
θ
1
+ e
2
θ
2
.
(Hint: consider the dual vector fields to θ
1
, θ
2
)
9. Suppose u, v are orthogonal co-ordinates so that θ
1
=
√
E du and θ
2
=
√
G dv.
(a) Use the structure equations to prove that
ω
12
=
1
2
√
EG
(
E
v
du −G
u
dv
)
(b) Hence deduce an explicit formula for the Gauss curvature in terms of the coefficients of
the 1
st
fundamental form:
K = −
1
2
√
EG
∂
∂u
G
u
√
EG
+
∂
∂v
E
v
√
EG
This can be multiplied out to remove the square roots, though you’ll get more terms. A nastier
expression (the Brioshi formula) may be found for general co-ordinates with F = 0.
10. In Exercise 3.3.5 we saw that the tangent developable x(u, v) = y(u) + vy
′
( u) of a unit-speed
curve has curvatures K = 0, H = −
τ
2vκ
. Use this to describe two surfaces with the same curva-
ture functions which are not related by a direct isometry.
11. Show that the surfaces parametrized by
x(u, v) =
u cos ϕ , u sin ϕ, ln u
, y(u, v) =
u cos ϕ , u sin ϕ, ϕ
have the same Gauss curvature but distinct first fundamental forms I
x
= I
y
. To do this properly,
you should argue that there is no reparametrization of y so that K
x
= K
y
and I
x
= I
y
.
(Gauss’ Theorem isn’t biconditional: surfaces can have the same K without being locally isometric)
12. Consider the family of surfaces
x
t
( u, v) = cos t
sin u sinh v
−cos u sinh v
u
+ sin t
cos u cosh v
sin u cosh v
v
, t ∈ [0,
π
2
]
When t = 0 this is a helicoid. When t =
π
2
this is the catenoid from Exercise 5.
(a) Compute the first fundamental form of x
t
and show that it is independent of t (the family
x
t
is therefore isometric).
(b) Show that the unit normal of x
t
is also independent of t:
n
t
=
1
cosh v
cos u
sin u
−sinh v
Hence compute the second fundamental form of x
t
for each t.
(c) Find the Gauss and mean curvatures of all surfaces x
t
. What is special about this family?
Relate this to Gauss’ Theorem.
84
