Math 497C
Mar 3, 2004
1
Curves and Surfaces
Fall 2004, PSU
Lecture Notes 10
2.3
Meaning of Gaussian Curvature
In the previous lecture we gave a formal definition for Gaussian curvature
K
in terms of the differential of the gauss map, and also derived explicit
formulas for
K
in local coordinates. In this lecture we explore the geometric
meaning of
K
.
2.3.1
A measure for local convexity
Let
M
⊂
R
3
be a regular embedded surface,
p
∈
M
, and
H
p
be hyperplane
passing through
p
which is parallel to
T
p
M
. We say that
M
is
locally convex
at
p
if there exists an open neighborhood
V
of
p
in
M
such that
V
lies on
one side of
H
p
. In this section we prove:
Theorem 1.
If
K
(
p
)
>
0
then
M
is locally convex at
p
, and if
k
(
p
)
<
0
then
M
is not locally convex at
p
.
When
K
(
p
) = 0, we cannot in general draw a conclusion with regard to
the local convexity of
M
at
p
as the following two exercises demonstrate:
Exercise 2.
Show that there exists a surface
M
and a point
p
∈
M
such
that
M
is strictly locally convex at
p
; however,
K
(
p
) = 0 (
Hint:
Let
M
be
the graph of the equation
z
= (
x
2
+
y
2
)
2
. Then may be covered by the Monge
patch
X
(
u
1
, u
2
) := (
u
1
, u
2
,
((
u
1
)
2
+(
u
2
))
2
). Use the Monge Ampere equation
derived in the previous lecture to compute the curvature at
X
(0
,
0).).
Exercise 3.
Let
M
be the
Monkey saddle
, i.e., the graph of the equation
z
=
y
3
−
3
yx
2
, and
p
:= (0
,
0
,
0). Show that
K
(
p
) = 0, but
M
is not locally
convex at
p
.
1
Last revised: October 31, 2004
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
After a rigid motion we may assume that
p
= (0
,
0
,
0) and
T
p
M
is the
xy
plane. Then, using the inverse function theorem , it is easy to show that
there exists a Monge Patch (
U, X
) centered at
p
, as the follwing exercise
demonstrates:
Exercise 4.
Define
π
:
M
→
R
2
by
π
(
q
) := (
q
1
, q
2
,
0). Show that
dπ
p
is
locally onetoone. Then, by the inverse function theorem, it follows that
π
is a local diffeomorphism. So there exists a neighborhood
U
of (0
,
0) such
that
π
−
1
:
U
→
M
is onetoone and smooth. Let
f
(
u
1
, y
2
) denote the
z

coordinate of
π
−
1
(
u
1
, u
2
), and set
X
(
u
1
, u
2
) := (
u
1
, u
2
, f
(
u
1
, u
2
)). Show that
(
U, X
) is a proper regular patch.
The previous exercisle shows that local convexity of
M
at
p
depends on
whether or not
f
changes sign in a neighborhood of the origin. To examine
this we need to recall the Taylor’s formula for functions of two variables:
f
(
u
1
, u
2
) =
f
(0
,
0) +
2
i
=1
D
i
f
(0
,
0)
u
i
+
1
2
2
i,j
=1
D
ij
(
ξ
1
, ξ
2
)
u
i
u
j
,
where (
ξ
1
, ξ
2
) is a point on the line connecting (
u
1
, u
2
) to (0
,
0).
Exercise 5.
Prove the Taylor’s formula given above.
(
Hints
: First re
call Taylor’s formula for functions of one variable:
g
(
t
) =
g
(0) +
g
(0)
t
+
(1
/
2)
g
(
s
)
t
2
, where
s
∈
[0
, t
]. Then define
γ
(
t
) := (
tu
1
, tu
2
), set
g
(
t
) :=
f
(
γ
(
t
)), and apply Taylor’s formula to
g
. Then chain rule will yield the
desired result.)
Next note that, by construction,
f
(0
,
0) = 0. Further
D
1
f
(0
,
0) = 0 =
D
2
f
(0
,
0) as well. Thus
f
(
u
1
, u
2
) =
1
2
2
i,j
=1
D
ij
(
ξ
1
, ξ
2
)
u
i
u
j
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Math, Geometry, Tp M, di N

Click to edit the document details