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首页哥本哈根大学:曲线与曲面几何1笔记 - 参数化方法与曲率研究
"Curves and Surfaces - Lecture Notes for Geometry 1是由Henrik Schlichtkrull教授编写的教材,出自于哥本哈根大学数学系。这是一份关于微分几何的讲义,微分几何是通过运用微积分技术,特别是函数的导数来研究几何对象的学科。主要讨论的对象是二维和三维空间中的曲线与曲面,这些对象的形状特性主要通过参数化方法进行深入研究。 在这份讲义中,关键概念包括曲线和曲面的切线以及它们的切空间。对于曲线,主要关注其切线的概念,这是通过对参数化函数的一阶导数进行定义的,而曲面的切线空间则涉及二阶导数。曲率是另一个核心概念,它描述了这些几何体在某一点弯曲的程度,通过对参数化函数的更高阶导数进行分析得出。曲率的计算涉及到对几何对象局部性质的深入理解,这对于理解曲面尤为重要。 对于曲面,尤其是曲率的理解,占据了讲义的一大篇幅。著名的高斯定理在此发挥了重要作用,它指出,一个曲面的所谓高斯曲率可以直接由曲面上的量来计算,无需依赖外部的物理测量。这个定理展示了微分几何如何将抽象的理论与实际的几何特征联系起来,对于深入理解几何形态和性质具有重要意义。 此外,讲义还可能涵盖其他相关主题,如曲线和曲面的嵌入、正交坐标系的选择、曲面的等参变换,以及如何应用这些概念到物理学、工程学和计算机图形学等领域。通过这些内容的学习,读者能够掌握基本的微分几何工具,为后续更高级的几何学研究打下坚实基础。" 这些笔记不仅适合数学专业学生,也对需要处理空间形状和运动的工程师、物理学家以及计算机科学家有着实际的应用价值。通过深入学习这些笔记,学生将能够理解和操作复杂的几何结构,解决实际问题中的曲面分析问题。
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12 Chapter 1
Corollary 1.6. Let f: Ω → R be a smooth function, where Ω ⊂ R
3
is open.
Let
S = {(x, y, z) ∈ Ω | f(x, y, z) = c}
and let p = (x
0
, y
0
, z
0
) ∈ S. Assume that p is not a critical point.
Then there exist an open interval W ⊂ Ω around p, such that S∩W is the
graph of a smooth function h, considered either as z = h(x, y), as y = h(x, z)
or a s x = h(y, z).
In particular, it follows that the l evel set can be parametrized as a smooth
surface in a neighborhood of each non-critical point.
Proof. By assumption at least one of the partial derivatives
∂f
∂x
,
∂f
∂y
and
∂f
∂z
is not zero at p. Interchanging z with x or y if necessary, we may assume
that it is
∂f
∂z
. The conclusion then follows from the previous theorem.
Example 1.6 .1. The eq uat ion for a plane in R
3
, ax + by + cz = d, where
(a, b, c) 6= (0, 0, 0), satisfies the assumption of the preceding corollary. If c 6=
0, the plane is the graph z = h(x, y) of the function h(x, y) = (d−ax−by)/c.
If c = 0 the plane is vertical, and we cannot exhibit it as a gra ph of the
form z = h(x, y), but then we can exhibit it as a graph over one of the other
coordinate planes.
Example 1.6.2. The surfaces introduced in Examples 1.2.2-1.2.4, sphere,
cylinder and cone, are level sets for, respectively, f(x, y, z) = x
2
+ y
2
+ z
2
,
f(x, y, z) = x
2
+ y
2
and f(x, y, z) = x
2
+ y
2
−z
2
. These functions all satisfy
the assumption of the corollary above, except for the vertex (0, 0, 0) of the
cone.
Example 1.6.3. Let f(x, y, z) = ze
x
+ yz and consider the equation
f(x, y, z) = 1 in a neighborhood of the point p = (0, 0, 1) (which solves
the equation). The partial derivative
∂f
∂z
= e
x
+ y is 1 at this point, so by
the implicit function theorem the solution z exists as a function o f (x, y)
in a neighborhood of (0, 0, 1). In fact, the equation has the solution z =
1/(e
x
+ y). However, if the equation is replaced by for example f(x, y, z) =
ze
x
+ sin(yz) = 1, then it is impossible to write down a solution to the
equation in terms of known functions, but the conclusion from the implicit
function theorem remains the same since we still have
∂f
∂z
= 1 at p.
1.7 The implicit function theorem, more equations
We have seen that an equation f(x, y) = c in R
2
defines a plane curve,
and that an equation f(x, y, z) = c i n R
3
defines a surface (under suitable
circumstances). In order to define a curve in R
3
we need two equations. For
example, the x-axis is the set defined by equations y = 0 and z = 0. Consider
two equations of the form
f
1
(x, y, z) = c
1
,
f
2
(x, y, z) = c
2
,
![](https://csdnimg.cn/release/download_crawler_static/16837647/bg11.jpg)
Parametrized curves and surfaces 13
where f
1
, f
2
maps an o pen set Ω ⊂ R
3
into R. We say that we have two
equations in three variables. It is actually more convenient to wri te the
equations in the form
f(x, y, z) = c
where f = (f
1
, f
2
) maps Ω into R
2
, a nd where c = (c
1
, c
2
). We want to
generalize Theorem 1.6 in order to deal with this situation.
In fact, we will generalize even further, to functions f : Ω ⊂ R
n
→ R
m
, that
is, to the case of m equatio ns in n variables, the only condition being that
n > m. For the application to space curves, only n = 3 and m = 2 is needed,
and the reader is encouraged to specialize to this case at first reading.
It is convenient for the comparison with Theorem 1.6 to write elements in
R
n
as (x, y) where x = (x
1
, . . . , x
n−m
) ∈ R
n−m
and y = (y
1
, . . . , y
m
) ∈ R
m
.
The goa l is to obtain y as a function of x.
Theorem 1.7. Let f: Ω → R
m
be a smooth functio n, where Ω ⊂ R
n
is
open. Let c ∈ R
m
be fixed . Let
C = {(x, y) ∈ Ω | f (x, y) = c}
and let p = (x
0
, y
0
) ∈ C. Assume that the determinant of the m × m matrix
A =
∂f
i
∂y
j
(p),
consis ting of the last m columns of the Jacobian Df(p), is non-zero.
Then there exist open intervals I ⊂ R
n−m
and J ⊂ R
m
around x
0
and y
0
,
respectively, such that W = I × J ⊂ Ω, and a smooth map h: I → J such
that
C ∩ W = {(x, h(x)) | x ∈ I},
that is, in the neighborh ood W of p, C is the graph of h: R
n−m
→ R
m
.
Proof. The most common technique for solv ing several equations in several
variables is elimination of variables. That is, we use one of the equations to
express a particular variable in terms of the others, and insert this expression
in t he remaining equations. The chosen variable has then been eliminated,
and the number of equations is reduced by one. This will also be our strategy
in the present proof.
We prove the theorem by induction on m. The case m = 1 was already
treated in Theorem 1.6. Thus, we a ssume that m ≥ 2 and that the conclusion
of the theorem is valid for functions into R
m−1
. We can safely assume that
c = 0, since this can be arranged by subtraction of t he constant from f .
Since det A is non-zero, A is invertible. We want to replace f : R
n
→ R
m
by the function A
−1
◦f : R
n
→ R
m
, obtained by multiplying all image vectors
f(x, y) ∈ R
m
with t he constant matrix A
−1
. Since multiplication by A
−1
![](https://csdnimg.cn/release/download_crawler_static/16837647/bg12.jpg)
14 Chapter 1
is a bijection, the equations f(x, y) = 0 and A
−1
f(x, y) = 0 are equival ent.
The Jacobian of the linear map, multi plication by A
−1
, is the matrix A
−1
itself (see Example B.1), and hence it follows from the chain rule that
D(A
−1
◦f )(p) = A
−1
· Df(p).
We see that the last m columns of D(A
−1
◦ f)(p) comprise a unit matrix
δ
kj
. By the replacement of f with A
−1
◦ f we thus obtain a function whose
Jacobian matrix at p has a unit matrix in its last m col umns, and for which
the solution set C is unaltered. From now on we assume this replacement has
been carried out, that is, we assume ∂f
k
/∂y
j
= δ
kj
.
In particular, for the function f
m
whose derivatives are in the last row
of Df, we have that ∂f
m
/∂y
m
(p) = 1. We will apply Theorem 1.6 to the
equation f
m
(x, y) = 0. The effect of the theorem is that the last variable,
y
m
, can be written as a smooth function of the remaining variables. We
write the remaining variables as (x, y
′
) ∈ R
n−1
where y
′
= (y
1
, . . . , y
m−1
).
More precisely, it then follows that there exists an interval neighborhood
W
1
= I
1
×J
1
around p, where I
1
⊂ R
n−1
and J
1
⊂ R, and a smooth function
h: I
1
→ J
1
such that for (x, y) = ( x, y
′
, y
m
) ∈ W
1
we have f
m
(x, y) = 0 if
and only if
y
m
= h(x, y
′
).
Let the function F : I
1
⊂ R
n−1
→ R
m−1
be defined by
F
k
(x, y
′
) = f
k
(x, y
′
, h(x, y
′
)) (3)
for k = 1, . . . , m − 1, where as before y
′
= (y
1
, . . . , y
m−1
). T he partial
derivatives of F
k
are obtained by applying the chain rule to (3):
∂F
k
∂y
j
=
∂f
k
∂y
j
+
∂f
k
∂y
m
∂h
∂y
j
(j = 1, . . . , m − 1),
and at p we thus have
∂F
k
∂y
j
=
∂f
k
∂y
j
= δ
kj
(because
∂f
k
∂y
m
= 0). The determinant
of t his mat rix being non-zero, we can apply our induction hypothesis to
F , and we obtain the existence of an interval neighborhood W
2
= I
2
× J
2
around (x
0
, y
0
′
), where I
2
⊂ R
n−m
and J
2
⊂ R
m−1
, and a smooth function
g: I
2
→ J
2
such that the solution set for the equation F (x, y
′
) = 0 in W
2
is
the gra ph of g, that is, F (x, y
′
) = 0 if and only if y
′
= g(x).
Let the interval W ⊂ R
n
be defined by
W = W
1
∩ {(x, y) | (x, y
′
) ∈ W
2
}.
We now see that for (x, y) in this set we have
(x, y) ∈ C
![](https://csdnimg.cn/release/download_crawler_static/16837647/bg13.jpg)
Parametrized curves and surfaces 15
if a nd only if
f
k
(x, y) = 0, k = 1, . . . , m
if a nd only if
f
k
(x, y) = 0, k = 1, . . . , m − 1 and y
m
= h(x, y
′
)
if a nd only if
F (x, y
′
) = 0 and y
m
= h(x, y
′
)
if a nd only if
y
′
= g(x) and y
m
= h(x, y
′
)
if a nd only if
y = (g(x), h(x, g(x))).
The function x 7→ (g(x), h(x, g(x))) is thus seen to be the desired function
whose graph is C in a neighborhood of p.
Corollary 1.7. Let f: Ω → R
2
be smooth, where Ω ⊂ R
3
is open. Let c ∈ R
2
and
p ∈ C = {(x, y, z) ∈ Ω | f(x, y, z) = c}.
Assume the rows of Df(p) (a 2 ×3 matrix) are linearly independent.
Then the re exist an open interval W ⊂ Ω around p, such that C ∩ W can
be parametrized as a s mooth curve in the form of a graph, considered either
as (y, z) = h(x), as (x, z) = h(y) or as (x, y) = h(z).
Proof. At l east one of the three 2 × 2 submatrices of Df(p) has non-zero
determinant. With suitable reorganization of variables the theorem can be
applied.
Example 1. 7 . The set of equations
x + xy + z
2
= 3 ∧ x
3
+ 2xz − y
2
z
2
= 2 (4)
has the form f (x, y, z) = c wi th
f(x, y, z) =
x + xy + z
2
x
3
+ 2xz − y
2
z
2
, c =
3
2
.
The Jacobian matrix is
Df(x, y, z) =
1 + y x 2z
3x
2
+ 2z −2yz
2
2x −2y
2
z
.
In the point (1, 1, 1) the equations (4) are satisfied and the Jacobian is
2 1 2
5 −2 0
.
![](https://csdnimg.cn/release/download_crawler_static/16837647/bg14.jpg)
16 Chapter 1
The determinant of the last two columns is det A = 4. Since this determinant
is not zero, the implicit function theorem assures that the equations can
be solved for (y, z) as function of x, in a neighborhood of (1,1 ,1). In this
neighborho od the set of soluti ons can thus be parametrized as a smooth curve
in the form of a 3-dimensional graph (t, h(t)) where h(t) = (y(t), z(t)) ∈ R
2
.
This example demonstrates the theoretical power o f the implicit function
theorem, since the explicit solving of (4) for y and z as functions of x is
clearly a difficult task.
1.8 Exercises
1 Let γ: I → R
n
be a parametrized curve with γ
′′
(t) = 0 for al l t. Show that
it is a line or a constant curve.
2 The following parametrized curve is called the cycloid
γ(t) = (t − sin t, 1 − cos t), (t ∈ R).
It is constructed by a circle of radius 1 rolling wi thout slipping along the
positive x-axis. The curve is the path o f a point on the circumference of
the circle. Explain the formula above from this construction.
x
y
3 Determine a parametri zed surface with image {(x, y, z) | x + 2y −2z = 1}.
4 Consider the equation x
3
+xy
2
−2ay
2
= 0 in R
2
, where a > 0 is a constant.
Show that the parametrized curve
γ(t) = (
2at
2
1 + t
2
,
2at
3
1 + t
2
),
where t ∈ R, is bijective onto the set of solutions. This curve is called the
cissoid of Diocles. Draw a sketch of it (say for a = 1).
5 Let S ⊂ R
3
denote the set of solutions to the equation x
2
+ y − z
2
= 1.
a. Show that the map σ: R
2
→ R
3
given by
σ(u, v) = (u + v, 1 − 4uv, u −v)
is an injective parametrized surface with image S.
b. Show that S contains no critical points for the function f(x, y, z) =
x
2
+ y −z
2
, and determine a parametrization of S as a graph of a smooth
function h.
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