5Smooth embedded surfaces (in ℝ3)
IB Geometry
5.4 Gaussian curvature
We will next consider the notion of curvature. Intuitively, Euclidean space is
“flat”, while the sphere is “curved”. In this section, we will define a quantity
known as the curvature that characterizes how curved a surface is.
The definition itself is not too intuitive. So what we will do is that we first
study the curvature of curves, which is something we already know from, say, IA
Vector Calculus. Afterwards, we will make an analogous definition for surfaces.
Definition (Curvature of curve). We let
η
: [0
, `
]
→ R
2
be a curve parametrized
with unit speed, i.e.
kη
0
k
= 1. The curvature
κ
at the point
η
(
s
) is determined
by
η
00
= κn,
where n is the unit normal, chosen so that κ is non-negative.
If
f
: [
c, d
]
→
[0
, `
] is a smooth function and
f
0
(
t
)
>
0 for all
t
, then we can
reparametrize our curve to get
γ(t) = η(f(t)).
We can then find
˙γ(t) =
df
dt
η
0
(f(t)).
So we have
k˙γk
2
=
df
dt
2
.
We also have by definition
η
00
(f(t)) = κn,
where κ is the curvature at γ(t).
On the other hand, Taylor’s theorem tells us
γ(t + ∆t) − γ(t) =
df
dt
η
0
(f(t))∆t
+
1
2
"
d
2
f
dt
2
η
0
(f(t)) +
df
dt
2
η
00
(f(t))
#
+ higher order terms.
Now we know by assumption that
η
0
· η
0
= 1.
Differentiating thus give s
η
0
· η
00
= 0.
Hence we get
η
0
· n = 0.
We now take the dot product of the Taylor expansion with n, killing off all the
η
0
terms. Then we get
(γ(t + ∆t) − γ(t)) · n =
1
2
κk˙γk
2
(∆t)
2
+ ··· , (∗)
where κ is the curvature. This is the distance denoted below:
γ(t)
γ(t + ∆t)
(γ(t + ∆t) − γ(t)) · n
We can also compute
kγ(t + ∆t) − γ(t)k
2
= k˙γk
2
(∆t)
2
. (†)
So we find that
1
2
κ
is the ratio of the leading (quadratic) terms of (
∗
) and (
†
),
and is independent of the choice of parametrization.
We now try to apply this thinking to embedded surfaces. We let
σ
:
V →
U ⊆ S
be a parametrization of a surface
S
(with
V ⊆ R
2
open). We apply
Taylor’s theorem to σ to get
σ(u + ∆u, v + ∆v) − σ(u, v) = σ
u
∆u + σ
v
∆v
+
1
2
(σ
uu
(∆u
2
) + 2σ
uv
∆u∆v + σ
vv
(∆v)
2
) + ··· .
We now measure the deviation from the tangent plane, i.e.
(σ(u + ∆u, v + ∆v) − σ(u, v)) · N =
1
2
(L(∆u)
2
+ 2M ∆u∆v + N(∆v)
2
) + ··· ,
where
L = σ
uu
· N,
M = σ
uv
· N,
N = σ
vv
· N.
Note that N and
N
are different things. N is the unit normal, while
N
is the
expression given above.
We can also compute
kσ(u + ∆u, v + ∆v) − σ(u, v)k
2
= E(∆u)
2
+ 2F ∆u∆v + G(∆v)
2
+ ··· .
We now define the second fundamental form as
Definition (Second fundamental form). The second fundamental form on
V
with σ : V → U ⊆ S for S is
L du
2
+ 2M du dv + N dv
2
,
where
L = σ
uu
· N
M = σ
uv
· N
N = σ
vv
· N.
Definition (Gaussian curvature). The Gaussian curvature
K
of a surface of
S
at P ∈ S is the ratio of the determinants of the two fundamental forms, i.e.
K =
LN − M
2
EG − F
2
.
This is valid since the first fundamental form is positive-definite and in particular
has non-zero derivative.
We can imagine that
K
is a capital
κ
, but it looks just like a normal capital
K.
Note that
K >
0 means the second fundamental form is definite (i.e. either
positive definite or negative definite). If
K <
0, then the second fundamental
form is indefinite. If
K
= 0, then the second fundamental form is semi-definite
(but not definite).
Example. Consider the unit sphere
S
2
⊆ R
3
. This has
K >
0 at each point.
We can compute this directly, or we can, for the moment, pretend that
M
= 0.
Then by symmetry, N = M . So K > 0.
On the other hand, we can imagine a Pringle crisp (also known as a hyperbolic
paraboloid), and this has
K <
0. More examples are left on the third example
sheet. For example we will see that the embedded torus in
R
3
has points at
which K > 0, some where K < 0, and others where K = 0.
It can be deduced, similar to the curves, that
K
is independent of parametriza-
tion.
Recall that around each point, we can get some nice coordinates where the
first fundamental form looks simple. We might expect the second fundamental
form to look simple as well. That is indeed true, but we need to do some
preparation first.
Proposition. We let
N =
σ
u
× σ
v
kσ
u
× σ
v
k
be our unit normal for a surface patch. Then at each point, we have
N
u
= aσ
u
+ bσ
v
,
N
v
= cσ
u
+ dσ
v
,
where
−
L M
M N
=
a b
c d
E F
F G
.
In particular,
K = ad − bc.
Proof. Note that
N · N = 1.
Differentiating gives
N · N
u
= 0 = N · N
v
.
Since
σ
u
, σ
v
and N for an orthogonal basis, at least there are some
a, b, c, d
such
that
N
u
= aσ
u
+ bσ
v
N
v
= cσ
u
+ dσ
v
.
By definition of σ
u
, we have
N · σ
u
= 0.
So differentiating gives
N
u
· σ
u
+ N · σ
uu
= 0.
So we know
N
u
· σ
u
= −L.
Similarly, we find
N
u
= σ
v
= −M = N
v
· σ
u
, N
v
· σ
v
= −N.
We dot our original definition of N
u
,
N
v
in terms of
a, b, c, d
with
σ
u
and
σ
v
to
obtain
−L = aE + bF −M = aF + bG
−M = cE + dF −N = cF + dG.
Taking determinants, we get the formula for the curvature.
If we have nice coordinates on
S
, then we get a nice formula for the Gaussian
curvature K.
Theorem. Suppose for a parametrization
σ
:
V → U ⊆ S ⊆ R
3
, the first
fundamental form is given by
du
2
+ G(u, v) dv
2
for some G ∈ C
∞
(V ). Then the Gaussian curvature is given by
K =
−(
√
G)
uu
√
G
.
In particular, we do not need to compute the second fundamental form of the
surface.
This is purely a technical result.
Proof. We set
e = σ
u
, f =
σ
v
√
G
.
Then e and f are unit and orthogonal. We also let N = e
×
f be a third unit
vector orthogonal to e and f so that they form a basis of R
3
.
Using the notation of the previous proposition, we have
N
u
× N
v
= (aσ
u
+ bσ
v
) × (cσ
u
+ dσ
v
)
= (ad − bc)σ
u
× σ
v
= Kσ
u
× σ
v
= K
√
Ge × f
= K
√
GN.
Thus we know
K
√
G = (N
u
× N
v
) · N
= (N
u
× N
v
) · (e × f)
= (N
u
· e)(N
v
· f ) − (N
u
· f )(N
v
· e).
Since N · e = 0, we know
N
u
· e + N · e
u
= 0.
Hence to evaluate the expression above, it suffices to compute N
·
e
u
instead of
N
u
· e.
Since e · e = 1, we know
e · e
u
= 0 = e · e
v
.
So we can write
e
u
= αf + λ
1
N
e
v
= βf + λ
2
N.
Similarly, we have
f
u
= −˜αe + µ
1
N
f
v
= −
˜
βe + µ
2
N.
Our objective now is to find the coefficients µ
i
, λ
i
, and then
K
√
G = λ
1
µ
2
− λ
2
µ
1
.
Since we know e · f = 0, differentiating gives
e
u
· f + e · f
u
= 0
e
v
· f + e · f
v
= 0.
Thus we get
˜α = α,
˜
β = β.
But we have
α = e
u
· f = σ
uu
·
σ
v
√
G
=
(σ
u
· σ
v
)
u
−
1
2
(σ
u
· σ
u
)
v
1
√
G
= 0,
since σ
u
· σ
v
= 0, σ
u
· σ
u
= 1. So α vanishes.
Also, we have
β = e
v
· f = σ
uv
·
σ
v
√
G
=
1
2
G
u
√
G
= (
√
G)
u
.
Finally, we can use our equations again to find
λ
1
µ
1
− λ
2
µ
1
= e
u
· f
v
− e
v
· f
u
= (e · f
v
)
u
− (e · f
u
)
v
= −
˜
β
u
− (−˜α)
u
= −(
√
G)
uu
.
So we have
K
√
G = −(
√
G)
uu
,
as required. Phew.
Observe, for
σ
as in the previous theorem,
K
depends only on the first
fundamental from, not on the second fundamental form. When Gauss discovered
this, he was so impressed that he called it the Theorema Egregium, which means
Corollary (Theorema Egregium). If
S
1
and
S
2
have locally isometric charts,
then K is locally the same.
Proof.
We know that this corollary is valid under the assumption of the previous
theorem, i.e. the existence of a parametrization
σ
of the surface
S
such that the
first fundamental form is
du
2
+ G(u, v) dv
2
.
Suitable
σ
includes, for each point
P ∈ S
, the geodesic polars (
ρ, θ
). However,
P
itself is not in the chart, i.e.
P 6∈ σ
(
U
), and there is no guarantee that there
will be some geodesic polar that covers
P
. To solve this problem, we notice that
K
is a
C
∞
function of
S
, and in particular continuous. So we can determine the
curvature at P as
K(P ) = lim
ρ→0
K(ρ, σ).
So done.
Note also that every surface of revolution has such a suitable parametrization,
as we have previously explicitly seen.