Differential Geometry
Differential Geometry
Differential Geometry
Chapter 1
Graphs and Level Sets 1
Chapter 2
Vector Fields 6
~apter 3
The Tangent Space 13
~hapter 4
Surfaces 16
Chapter 5 -
Vector Fields on Surfaces; Orientation 23
Chapter 6
The Gauss Map. 31
ytmpter 7
Geodesics 38
Chapter 8
Parallel Transport 45
xi
xii Contents
Chapter 9
?'he Weingarten Map 53
Chapter 10
~urvature of Plane Curves 62
Chapter 11
~rc Length and Line Integrals 68
Chapter 12
~urvature of Surfaces 82
Chapter 13
Convex Surfaces 95
Chapter 14
/parametrized Surfaces 108
Chapter 15
Aocal Equivalence of Surfaces and Parametrized
Surfaces 121
Chapter 16
Focal Points 132
Chapter 17
Surface Area and Volume 139
Chapter 18
~inimal Surfaces 156
Chapter 19
The Exponential Map 163
Chapter 20
Surfaces with Boundary 177
Chapter 21
The Gauss-Bonnet Theorem 190
Chapter 22
Rigid Motions and Congruence 210
Contents xiii
Chapter 23
Isometries 220
Chapter 24
Riemannian Metrics 231
Bibliography 245
Associated with each real valued function of several real variables is a collec-
tion of sets, called level sets, which are uSerulin'studying qualitative proper-
ties of the function. Given a functionf: U -. R, where U c:. R" + I, its level sets
are the sets f-I(C) defined, for each real number c, by
f-I(C) == {(Xb ... , X,,+l) e U:f(xl:; ... , X.+l) == c}.
The number c is caned the height ofthe level set, andf-I(c) is called the level
set at height c. Since f-l(C) is the solution set of the equation
f(Xb ... , xa+ •.)-,c, the level set f-:-l(C) is often described as "the set
f(Xb ... , X.+l)= c."
The "level set" and "height" termmotops arise from the relation be-
tween' the level sets or a function ·-.t.4$ismpQ., The .graph of a function
f: U -. R is the subset of R" + 2 defined by
graph of
I(xl) = xi
(a): n = 0 (b):n=l
Figure 1.1 The level sets f-l(C) (c> 0) for the function
f(x., ... , x n +.) = xi + ... + X;+l'
Figure 1.2 Level sets for the functionf(x., ... , x n +.) = xi + ... + X;+l'
For n = 1, level sets are (at least for non-constant differentiable functions)
generally curves in 1R2. These curves play the same roles as contour lines on a
topographic map. If we think of the graph off as a land, with local maxima
representing mountain peaks and local minima representing valley bottoms,
then we can construct a topographic map of this land by projecting orthog-
onally onto 1R2. Then all points on any given level curve f-l(C) correspond
to points on the land which are at exactly height c above" sea level"
(X3 = 0).
Just as contour maps provide an accurate picture of the topography of a
land, so does a knowledge of the level sets and ,their heights accurately
portray the graph of a function. For functions f: 1R2 --+ IR, study of the level
curves can facilitate the sketching of the graph off. For functionsf: 1R3 --+ IR,
1 Grapbs and Level Sets 3
the graph lies in 1R4 , prohibiting sketches and leaving the level sets as the best
tools for studying the behavior of the function.
One· way of visualizing the graph of a function f: U-+ IR, U C 1R2, given
its level sets, is as follows. Think of a plan", parallel to the (Xh x2)-plane,
moving vertically. %en it reaches height c thi&plane, X3 = C, cuts the graph
off in the translate to this plane of the level set f - 1(c). As ~he plane moves,
these sets generate . the graph off (see .Figure 1.3).
X3 = 2
<a>
<h}
FitUre 1.3 Level Set5.and'sraphs of ftmctiotts f: .R Z ..... R. The label on each level set
indicates its'heisbt. (a~f (x bXZ)- - xi +xi ~ (b) A fanction with two local minima.
The same princip1ecan be used to help visualize level sets ofi functions
f: .U ~ R, . where U c: Rl .. Each plane Xi = constant will cut the level set
f~l(cl(c fixed} in some subset, usually 8carve.Letiing the plane move, by
changing the selected value of the xt-coordinate,. these subsets will generate
Xl = 0
Xl
XI=t,X~+X~ =~
Xl = i,x~ +X~ = t
X I = 0, X~ + X~ = 1
xl = - t,x~ +X~ = t
(a)
Xl = - 2
{ X3 =x~-4
(b)
Figure 1.4 Level sets in ~3, as generated by intersections with the planes
Xl = constant. (a) XI + x~ + x~ = 1. (b) XI - x~ + X3 = o.
EXERCISES
In Exercises 1.1-1.4 sketch typical level curves and the graph of each
function.
1.1. f(Xh X2) = Xl.
1 Graphs and Level Sets 5
•
Vector Fields
The tool which will allow us to study the geometry of level sets is the
calculus of vector fields. In this chapter we develop some of the basic ideas.
A vector at a point p E IRn+ 1 is a pair v = (p, v) where v E IRn+ 1. Geomet-
rically, think of v as the vector v translated so that its tail is at p rather than
at the origin (Figure 2.1). The vectors at p form a vector space IR;+ 1 of
v I,V)
p
6
2 Vector Fields 7
J;?V+W
P
(p, w) E R:, p e R3, the. cross product is also defined, using the standard
crossproduct on,R\by (P,v)x(P,'w) (p,V)(w). =
Usitt. ttieldfjt~,,~yle1tgtff,t'fd!iaveetor t:::(p, v) at .p ,and the
anglel1betweetf tWo Vect6l1~,===~'Q>~v) andw == (p, w) are defined by
i.
1,.ti\'.ti~,:('f7-V)1/2
cos (J ==V • "/~vtI 11"11
A~or~AfXODCJ .. ~R"+ a function whidl~igns t~each point
of U. a vector at :diatptfb:tt. Thus
X<P)·~··tp,.i(P»:·'·
····'·:/;~1!~~ft~~~ ,'I:~' ;~'::~:\i' ',r·····\·ZI..a2· .';.,.w+ 1 f: '1'
for .o~~~~!·,~.~,R"·" ..~:.. ecwr-:~·:OIl~< are 0 ten most easl y
d~:,.,,:~lthiI~«lfunct~ X. Three typical vector fields
onR·are,~·itt :f'iJute23.· ,
~ ( -" .:,'~' .
- ,\ t / /
---- --
-- ,'.~. /
-- -
-,....-
----.~
----.. -......
'_--"
.. - " ,
.~-----
...-.... -7+"', .-....
/' I ~ '"
Il\ \
(a): X(P) == (1,01 (b): X(p). p
t R.ecall that U c:: R" + 1 is open if for each p e U there is an t > 0 such that q e U
whenever IIq - pil < t.
8 2 Vector Fields
~ (t)
for each t the position at time t of a particle moving in [Rn+ 1 then rl(t)
represents the velocity of this particle at time t.
A parametrized curve rl: I -+ [Rn+ 1 is said to be an· integral curve of the
vector field X on the open set U in [Rn+ 1 if rl(t) E U and ~(t) = X(rl(t)) for
all tEl. Thus rl has the ,property that its velocity vector at each point of
the curve coincides with the value of the vector field at that point (see
Figure 2.5).
Theorem. Let X be a smooth vector field on an open set U C [Rn+. and let
p E U. Then there exists an open interval I containing 0 and an integral curve
rl: 1-+ U of X such that
(i) rl(O) = p.
2 Vector Fields 9
• dXl
-;jt(t) = Xt(Xt(t), ... , x,,+t(t))
(E)
10 2 Vector Fields
EXAMPLE. Let X be the vector field X(p) = (p, X(p)) where X(Xl' x 2 ) =
(-X2' xd (Figure 2.3.(c)). A parametrized curve (X(t) = (x 1 (t), X2(t)) is an
integral curve of X if and only if the functions x 1 (t) and X2(t) satisfy the
differential equations
dX2
dt = Xl·
The general solution of this pair of equations is
x 1 (t)= C 1 cos t + C 2 sin t
x 2(t) = C 1 sin t - C 2 cos t.
Thus the integral curve of X through the point (1,0) (with Xl(O) = 1 and
X2(0) = 0) is
(X(t) = (cos t, sin t),
whereas the integral curve through an arbitrary point (a, b) (with x 1 (0) =a
and X2(0) = b) is
P(t) = (a cos t - b sin t, a sin t + b cos t)
(see Figure 2.6).
Figure 2.6 Integral curves of the vector field X(Xh X2) = (Xl, X2, -X2, Xl).
EXERCISES
2.1. Sketch the following vector fields on R2: X(p) = (p, X(p» where
(a) X(p) = (0, 1) (d) X(Xh X2) = (X2' xd
(b) X(p) = -p (e) X(Xh X2);::: (-2X2' txt>.
(c) X(Xh X2) = (X2, ,-Xl)
2.2. Find and sketch the gradient field of each of the following functions:
(a) f(Xh X2) +
= Xl X2
(b) f(xtt X2) = xi + x~
(c) f(Xh X2) = Xl - x~
(d) f(Xh X2) = (xf- xH/4.
2.3. The divergence of af}smooth vector field X on U c R"+I,
X(p) = (p, X I (p), ... , XII + l(P» for p e U,
is the function div X: U -+ R defined by div X = ~).! f (oXi/ox,). Find the
divergence of each of the vector fields in Exercises 2.1 and 2.2.
2.4. Explain why an integral curve of a vector field cannot cross itself as does the
parametrized curve in· Figure 2.4.
2.5. Find the integral curve through p = (1, 1) of each of the vector fields in Exer-
cise 2.1.
2.6. Find the integral curve through p = (a, b) of each of the vector fields in Exer-
cise 2.1.
2.7. A smooth vector field X on an open set U of R"+ I is said to be complete if for
• each p e U the maximal integral curve of X through p has domain ,qual to IR.
Determine which of the following vector fields are complete:
(a) X(Xh X2) = (Xh X2, 1,0), U = R2.
(b) X(Xh X2) ;::: (Xh X2, 1,0), U = 1R2 - {(O, O)}.
(c) X(Xh X2) = (Xh X2, -X2' XI~ U = 1R2 - {(O, O)}.
(d) X(Xh X2) ;::: (Xh X2, 1 + xi, 0), U = R2.
12 2 Vector Fields
2.8. Let U be an open set in IRn+ 1, let p E U, and let X be a smooth vector field on
U. Let rt: 1-+ U be the maximal integral curve of X through p. Show that if
p: 1 -+ U is any integral curve of X, with P(t o) = p for some to E 1, then P(t) =
rt(t - to) for all t E 1. [Hint: Verify that if Pis defined by P(t) = P(t + to) then P
is an integral curve of X with P(O) = p.]
2.9. Let U be an open set in IRn+ 1 and let X be a smooth vector field on U. Suppose
rt: I -+ U is an integral curve of X with rt(O) = rt(to) for some to E I, to =1= O.
Show that rt is periodic; i.e., show that rt(t + to) = rt(t) for all t such that both t
and t + to E I. [lfint: See Exercise 2.8.]
2.10. Consider the vector field X(Xh X2) = (Xh X2, 1,0) on 1R2. For t E IR and
p E 1R2, let CPt(p) = rtp(t) where rtp is the maximal integral curve of X through p.
(a) Show that, for each t, CPt is a one to one transformation from 1R2 onto itself.
Geometrically, what does this transformation do?
(b) Show that
CPo = identity
CPt1 +t2 = CPt1 CPt2 for all t l> t2 E IR
0
-} (c)
(a): n ~ 1 (b): n = 2
13
14 3 The Tangent Space
IfVf(p) = 0, this lemma says nothing. But ifVf(p) =1= 0, it says that the set
of all vectors tangent to f-l(C) at p is contained in the n-dimensional vector
subspace [Vf (p)]1. of ~~ + 1 consisting of all vectors orthogonal to Vf (p). A
point P E ~n+ 1 such that Vf(p) =1= 0 is called a regular point off.
The vector field Y has domain the open subset of U where Vf =1= O. Since p is
a regular point of f, p is in the domain of Y. Moreover, since X(p) = V E
[Vf(p)]l., Y(p) = X(p). Thus we have obtained a smooth vector field Y such
that Y(q) 1- Vf(q) for all q E domain (Y), and Y(p) = v.
Now let (X be an integral curve of Y through p. Then (X(O) = p,
a(O) = Y((X(O)) = Y(p) = X(p) = v and
Thus we see that at each regular point p on a level setf-l(c) ofa smooth
function there is a well defined tangent space consisting of all velocity vectors
at p of all parametrized curves inf--l(c) passing through p, and this tangent
space is precisely [Vf(p)]1. (see Figure 3.2).
EXERCISES
3.1. Sketch the level sets f-l( -1), f- 1 (0), and f-l(l) for f(x., ... , x n + 1) = xi +
., . + x~ - X~+ 1; n = 1, 2. Which points p of these level sets fail to have tangent
spaces equal to [Vf(p)JL?
3 The Tangent Space 15
tangent space
at p
(a): n == 1 (b): n = 2
Figure 3.2 Tangent space at a typical point of the level set f-l(l~ where
f(Xh ... , xlI+d = xf + ... + X;+l ..
EXAMPLE 2. For 0 =1= (aI' ... , an+1) E IRn + 1 and b E IR, the n-plane
a1x1 + ... + an+1xn+1 = b is the level setf-1(b) wheref(x h ... , x n+1) =
16
4 Surfaces 17
(a): n = 1
and Vg(Xh ... , Xn+1) = (Xh ... , Xn+h -Of/OXh"" -of/oxn, 1) is never
zero.
Xl Xl
O-sphere I-sphere
in 111 (unit circle)
in JR.2
(a): n = 0 (b): n = I
Figure 4.2 The cylinder g-1(1) over the n-sphere: g(X1' ... , Xn + d = xi + ... + x;.
EXAMPLE 5. Let C be a curve in 1R2 which lies above the x I-axis. Thus
C = f-1(C) for some f: U ~ IR with Vf(p) =1= 0 for all p E C, where U is
contained in the upper half plane X2 > O. Define S = g- l(C) where
g: U x IR ~ IR by g(Xh X2' X3) = f(Xh (x~ + X~)1/2). Then S is a 2-surface
(Exercise 4.7). Each point p = (a, b) E C generates a circle of points of S,
namely the circle in the plane Xl = a consisting of those points
(Xh X2, X3) E 1R3 such that Xl = a, x~ + x~ = b2. S is called the surface of
revolution obtained by rotating the curve C about the XI-axis (see Figure
4.3).
4 SurfaceS t9
Figure 4.3 The surface of revolution S obtained by rotating the curve C about the
xl· axis.
for all v E S" and s~ Vg(p) = lVf(P) for some A., as required. o
Remark. IfS is compact (closed and boundedt) then every smooth function
g: U -+ IR attains a maximum on S and a minimum on S. The above theorem .
can then be used to locate candidates for these extreme points. If S is not
compact, there may be no extrema.
and
t S is closed if R"+ 1 - S is open; S is bounded if there exists MeR such that np~ < M
for all peS.
20 4 Surfaces
Figure 4.4 Level curves of the function g(Xl' Xl) = axf + 2bxIXl + cxi
(ac - bl > 0), The four points where these curves are tangent to the unit circle S
are the extreme points of g on S '.
is an eigenvector of (b ~) then
EXERCISES
4.1. For what values of e is the level set f-l(e) an n-surface, where
(a) f(x., ... , X,,+l) = xi + ... + X:+l
(b) f(x., ... , X,,+l) = xi + ... + x: - X:+l
(c) f(x., ... , X,,+l) = Xl Xl ... X,,+l + 1
4.2. Show that the cylinder xl + xi = 1 in -Il 3 can be represented as a level set of
each of the following functions:.
(a) f(x., Xl, X3) = xi + x~
(b) f(x., Xl, X3) == -xl - xi
(c) f(x., Xl, X3) = 2xi + 2xi + sin(xt + xU.
4.3. Show that if an n-surface S is represented both asf-l(e) and as g-l(d) where
Vf(p) =F 0 and Vg(p) =F 0 for all peS, then for each peS, Vf(p) = AVg(p) for
some real number A =F O.
4.4. Sketch the graph of the function /: Rl .... R given by I(x., Xl) - X~ - 3xi Xl'
[Hint: First find the level setf-l(O~ In what region of the plane isf> O? Where
is f < O?] The 2-surface graph (f) is called a monkey saddle. (Why?)
4.5. Sketch the cylinders f -1 (0) where
(a) f(x., Xl) == Xl
(b) f(x., Xl, X3) = Xl - X~
(c) f(x., Xl, X3) = (xt/4) + (xi/9) - 1
4.6. Sketch the cylinder e)Ver the graph off(x) = sin x.
4.7. Verify that a surface of revolution (Example S) is a 2-surface.
4.8. Sketch the surface of revolution obtained by rotating C about the X I-axis,
where C is the curve
(a) Xl = 1 (cylinder)
(b) -xi + xi = 1, xl> 0 (I-sheeted hyperboloid)
(c) xl + (Xl - 2)1 = 1 (torus)
4.9. Show that the set S of all unit vectors at all points of Rl forms a 3-surface in 1Il4.
[Hint: (Xh Xl, X3, X4) e S if and only if xJ + xl- 1.]
4.10. Let S = f-l(e) be a 2-surface in 1Il3 which lies in the half space X3 > O. Find a
function g: U .... III (U open in 1Il4) such that g-l(e) is the 3-surface obtained by
rotating the 2-surface S about the (x h xl)-plane.
4.11. Let a, b, e e III be such that ae - b l > O. Show that the maximum and mini-
mum values of the function g(x h Xl) = xi +' X~ on the ellipse axi + 2bx 1 Xl +
ex! = 1 are ofthe form l/Al and l/Al where Al and Al are the eigenvalres ofthe
matrix (: ~).
4.12. Show that the maximum and minimum values of the function
g(Xh ... , X,,+l) = ~lj~l ajjxjxJ on the unit n-sphere xi + ... + X:+l = 1,
where (au) is a symmetric n x n matrix of real numbers, are eigenvalues of the
matrix (ajJ).
22 4 Surfaces
p=(~ ! n?
Vector Fields on Surfaces;
Orientation
5 .
(a) (b)
Figure 5.1 Vector fields on the I-sphere: (a) a tangent vector field, (b) a normal
vector field.
23
24 5 Vector Fields on Surfaces; Orientation
andf IX(O) = f(p) = c, sof IX(t) = c for all tEl. Conditions (i) and (ii) are
0 0
clearly satisfied, and condition (iii) is satisfied because any p: 1-+ S satisfy-
ing (i) and (ii) is also an integral curve of the vector field Y on W so the
theorem of Chapter 2 applies. 0
X to S. Then Pis also an integral curve oCx, sending 0 to !X( t 1 ~ as is the curve
& defined by &(t) = (X(t + tl)' By uniqueness of integral curves, (X(t) =
&(t - t 1 ) = P(t - t 1 ) E S for all t such that t -'tl is in the common domain
of & and p. But this contradicts the facrthat !X(t)¢ S for values of t arbitrarily
close to t 1• Hence (X(t) E S for all tel with t> to. The proof for t < to is
~~ 0
clearly has the required properties, as does the vector field N 2 defined by
N 2 (p) = -N 1 (p) for all pES.
To show that these are the only two such vector fields, suppose N3 were
26 5 Vector Fields on Surfaces; Orientation
another. Then, for each PES, N 3 (p) must be a multiple ofN 1 (p) since both
lie in the I-dimensional subspace S; c IR;+ 1. Thus
N 3 (p) = g(p)N 1 (p)
where g: S -+ IR is a smooth function on S (g(p) = N 3 (p) • N 1 (p) for pES).
Since N 1 (p) and N 3 (p) are both unit vectors, g(p) = ± 1 for each pES.
Finally, since g is smooth and S is connected, g must be constant on S (see
Exercise 5.2). Thus either N3 = N1 or N3 = N 2 • 0
seen by picking a unit normal vector at some point on the central circle and
trying to extend it continuously to a unit normal vector field along this
circle. After going around the circle once, the normal vector is pointing in
the opposite direction! Since there is no smooth unit normal vector field on
B, B cannot be expressed as a level set f - 1 (c) of some smooth function
f: U -+ IR with Vf(p) =1= 0 for all PES, and hence B is not a 2-surface accord-
5 Vector Fields on Surfaces; Orientation 27
(a) (b)
Figure 5.4 Orientation on a plane curve: (a) the chosen normal direction at each
point determines (b) a choice of tangent direction at each point.
On a 2..surface in 1R 3 , an orientation can be used to define a directibn of
rotation in th~ tangent space at each point of the surface. Given 0 E IR, the
positive O-rotation at the point p of the oriented 2..surface S is the line~
transformation ~: S" .... S" defined by ~(v) = (cos O)v + (sin O)N(p) x v
where N(p) is th~ orientation normal direction atp. Ro is usually descri~d
as the" right..handed rotation about N(P) thro~gh the angle 0" (see Figure
5.5).
Figure 5.5 Orientation on the 2-sphere: at each point the chosen normal direction
determines a sense of positive rotation in the tangent space. The satellite figure is an
, enlarged view of one tangenf space.
28 5 Vector Fields on Surfaces; Orientation
det(jL)
is positive, where N(p) = (p, N(p)) is the orientation normal direction at p
and ei = (p, ei) for i E {I, 2, 3}; the basis is left-handed if the determinant is
negative.
On an n-surface in IRn + 1 (n arbitrary), an orientation can be used to
partition the collection of all ordered bases for each tangent space into two
subsets, those consistent with the orientation and those inconsistent with the
orientation. An ordered basis {Vh ... , .vn} (not necessarily orthonormal) for
the tangent space Sp at the point p of the oriented n-surface S is said to be
consistent with the orientation N on S if the determinant
det( Vl )
~~)
is positive; the basis is inconsistent with N if the determinant is negative.
Here, as usual, Vi = (p, Vi) and N(p) = (p, N(p)).
EXERCISES
5.1. Show that the unit n-sphere XI + ... + x;+ 1 = 1 is connected if n > 1.
5.2. Show that if S is a connected n-surface in IR n+ 1 and g: S -+ IR is smooth and
takes on only the values + 1 and -1, then g is constant. [Hint: Let pES. For
q E S, let IX: [a, b] -+ S be continuous and such that lX(a) = p, lX(b) = q. Use the
intermediate value theorem on the composition g IX.] 0
5.3. Show by example that if S is not connected then Theorem 2 ofthis section fails.
5.4. Show that the two orientations on the n-sphere XI + ... + x;+ 1 = r2 of radius
r> 0 are given by N 1 (p) = (p, plr) and N 2 (p) = (p, -plr).
5.5. IR n may be viewed as the n-surface x n + 1 = 0 in IR n+ 1. Let N be the orientation
on IR n c IR n+ 1 defined by N(p) = (p,O, ... ,0,1) for each p E IRn. (This N is
called the natural orientation on IRn.) Show that, given this orientation for each
n,
(a) the positive tangent direction at p E 1R1 is the direction (p, 1,0),
(b) the positive 8-rotation in IR~, p E 1R2, is counterclockwise rotation through
the angle 8, and
5 Vector Fields on Surfaces; Orientation 29
(c) the ordered orthonormal basis {(P, 1,0, 0, O~ (p,O, 1, 0, O~ (p, 0,0, 1,0)}
for R:, p E R 3, is right-handed.
5.6. Let C be an oriented plane curve and let v be a nonzero vector tangent to Cat
p E C. Show that the basis {v} for C" is consistent with the orientation on C if
and only if the positive tangent direction at p is v/llvil. [Hint: Let 6 denote the
angle measured counterclockwise from (p, 1,0) to the orientation direction
N(p~ so that N(p) = (p, cos 6, sin 6). Express both v and the positive tangent
direction at p in terms of 6.]
5.7. Recall that the cross product v x l ' of two vectors v = (p, V .. V2, V3) and
w = (p, WI. W2, W3) in R: (p E R 3 ) is defined by
(a) Show that v x l ' is orthogonal to both v and " and that Ilv x I'll ==
Ilvllllwll sin 6, where 6 == cos- 1 (v ••/lIvll~wll) wthe angle between v and w.
(b) Show that if u = (p, u., U2, U3) then
u'(VX'W)==vo(wxa)=-wo(axv)==I:: ::
Wl W2
::1.
W3
(c) Show that the only,vector x in R: such that 'a' x is equal to the determi-
nant above (part b) for all • E R: is x == v x w.
5.8. Let S be an orie.nted 2-surface in R' and let {v, w} be an ordered basis for the
tangent spaceS, to S at p E S.Show t~at the consistency of {v, w} with the
orientation N of S is equivalent with each of the following conditions:
(a) N(P)· (v x 1'» 0
(b) w/llw~ == R,(v/llvU for some 6 with 0 < 8 < 1t, where R, is the positive
6-rotation in S".
5.9. Let S be an oriented 3-surface in R4 and letp e S.
(a) Show that, given vectors v = (p, v) and l' == (p, w) in S", there is a unique
vector v )( l ' e S" such that
a • (v x 1') == det( : ) (
N(P)
for all a == (p, u) E S", where N(P) == (p, N(P» is the orientation direction
at' p.This vector v x l ' is the cross product of v and w.
(b) Check that the cross ,product in S" has the following properties:
(i) (v + 1') X X = V X X + l ' X x
(ii) v x (1' + x) = v x l ' + V X x
(iii) (cv) x l ' == c(v x 1')
(iv) v x (cw) == c(v x 1')
(v) V X l ' = -1' X V ~
(vi) a' (v x 1') = V • (1' X a) = l ' • (a .x v)
(vii) v x l ' is orthogonal to both v and l '
30 5 Vector Fields on Surfaces; Orientation
------s
31
32 6 The Gauss Map
(a): n = 1
s
(b): n = 2
PROOF. The idea ofthe proof is as follows. Given v e S', consider the n-plane
If. By moving this n-plane far enough in the v-direction, it will have null
intersection with S. Bringing it back in until it just touches S at some point p,
it will be tangent there (see Figure 6.3). Hence at this point, N(p) = ±v. If
that there exist tl and t2 between a and b with! /X(td > c and! /X(t2) < c.
0 0
But then, by the intermediate value theorem, there would exist t 3 between t 1
and t2 such that! /X(t3) = c, contradicting (ii).
0
Image (Xl
Image (X2
Figure 6.5 Given a compact connected n-surface S =f-l(C~ the nearby level sets
f-l(C - a) andf-l(c + a) are slightly inside and slightly outside S.
(Possibly,f- 1(c + e) might also contain some points far away from S but we
can ignore such points in the present argument.) Similarly, for small enough
e > 0, the level set f - 1 (c - e) will be an n-suIface S _ on the other side of S,
gotten by pushing each point of S a short distance out along - VI Denoting
by V the set of points between S_ and S+, by V+ the set of points in
/R"+ 1 - V which lie on the same side of SasS + , and by V_ the set ofpoin~
in IR" + 1 - V which lie on the other side of S, we can define 1: IR" + 1 ~ R by
I
f(P) for P E V
1(P) = c + e for P E V+
C - e for P E V_ .
EXERCISES
Figure 6.6 The curve 0: must cross the compact n-surface S an even number of times.
6 The Gauss Map 37
Geodesics are curves in n-surfaces which play the same role as do straight
lines in IRn. Before formulating a precise definition, we must introduce the
process of differentiation of vector fields and functions defined along pa-
rametrized curves. In order to allow the possibility that such vector fields and
functions may take on different values at a point where a parametrized curve
crosses itself, it is convenient to regard these fields and functions to be
defined on the parameter interval rather than on the image of the curve.
A vector field X along the parametrized curve oc f ~ IR n+ 1 is a function
which assigns to each t E f a vector X{t) at (X{t); i.e., X{t) E IR~(~) 1 for all t E f.
A function f along (X is simply a function f: f ~ IR. Thus, for example, the
velocity IX of the parametrized curve (X: f ~ IR n + 1 is a vector field along (X
(Figure 7.1); its length II IX II : f ~ IR, defined by II IX II (t) = IIIX{t)II for all t E f, is a
function along (x. II IX II is called the speed of (x.
Figure 7.1 The velocity field along a parametrized curve rl. Note that rl(t 1 ) = rl(t2)
does not imply that ~(td = ~(t2)'
38
7 Geodesics 39
X(t) measures the rate of change of the vector part (Xt(t), ... , X,,+t(t)) of
X(t) along ex. Thus, for example, the acceleration eX of a parametrized curve ex
is the vector field along ex obtained by differentiating the velocity field eX
[eX = (&)] (see Figure 7.2).
a(to)
Figure~7.2 The acceleration Ci(to) is the derivative at to ofthe velocity vector field~.
Figure 7.3 Geodesics ~(t) = (cos(at + b), sin(at + b), ct + d) in the cylinder
xi + x~ = 1. (a) a = 0, (b) c = 0, (~) a 1= 0, c 1= 0.
EXAMPLE 3. For each pair of orthogonal unit vectors {e 1 , e2} in 1R3 and each
a E IR, the great circle (or point if a = 0) a(t) = (cos at)e 1 + (sin at)e2 is a
geodesic in the 2-sphere xi + x~ + x~ = 1 in II~P, because &(t) = (a(t),
- a2a(t)) = ±a2N(a(t)) for all t E IR (see Figure 7.4).
Intuitively, it seems clear that given any point p in an n-surface S and any
initial velocity v at p (v ESp) there should be a geodesic in S passing through
7 Geodesics 41
Theorem. ut S be an n-surface in !R"+ 1, let pES, and let v e Sp' Then there
exists an open .interval I containing 0 and a geodesic a: 1-+ S such that
(i) a(O) = p and eX(O) = v.
(ii) If p: 1-+ S is any other geodesic in S with P(O) = p and /J(O) = v, then
1 cJ and P(t) = a(t)for all t e'1.
.Remark. The geodesic a is called the maximal geodesic in S passing through
p with initial velocity v.
PIlOOF. S = f - ~ (c) for some c E IR and some smooth function f: U -+ IR (U
+
open in 1R"+ 1 ) with Vf(p) 0 for all pES. Since Vf(p) 0 for all pin some +
open set containing S, we may assume (by shrinking U if necessary) that
+
Vf(p) 0 for all p E U. Set N = VflIIVfll.
By definition, a parametrized curve a: I -+ S is a geodesic of S if and only
if its acceleration is everywhere perpendicular to S; that is, if and only if a( t)
is a multiple of N(a(t» for all tEl:
ti(t) = g(t)N(a(t»
for all tEl, where g: 1-+ IR. Taking the dot product of both sides of this
equation with N(a(t» we find
g = Ci • N 0 a= (a • N 0 a)' - a • N. ~ a
= -a . N ~ ex,
since eX • N oa = O. Thus a: I -+ S is a geodesic if and only if it satisfies the
differential equation
(G) ji + (a • N ~ a)(N a) = O. 0
42 7 Geodesics
(a . N 0 (X)' = &• N 0 (X + a. N ; (X =0
by (G), so a . N 0 (X is constant along (x, and
(a • N 0 (X)(O) = v . N(p) = 0
since v ESp and N (p) .1 Sp' It follows then that
(f (X)'(t) = Vf((X(t)) • a(t) = IIVf((X(t))IIN((X(t)) . a(t) = 0
0
for all tEl so f (X is constant, and f((X(O)) = f(p) = c so f((X(t)) = c for all
0
It follows from the theorem just proved that each maximal geodesic on
the unit 2-sphere in 1R·3 (Example 3) is either a great circle (parametrized by a
constant speed parametrization) or is constant ((X(t) = P for all t, some p)
since such a curve can be found through each point p with any given initial
velocity. Similarly, each maximal geodesic on the cylinder xi + x~ = 1 in 1R3
(Example 2) is either a vertical line, a horizontal circle, a helix (spiral), or is
constant.
t See e.g. W. Hurewicz, Lectures on Ordinary Differential Equations, Cambridge, Mass.: MIT
Press (1958), pp. 32-33. See also Exercise 9.15.
7 Geodesics 43
EXERCISES
7.1. Find the velocity, the acceleration, and the speed of each of the following
parametrized curves:
(a) IX(t) = (t, t 2)
(b) IX(t) = (cos t, sin t)
(c) IX(t) = (cos 3t, sin 3t)
(d) IX(t) = (cos t, sin t, t)
(e) a(t) = (cos t, sin t, 2 cos t, 2 sin t).
7.2. Show that if IX: 1-+ 1R"+ 1 is a parametrized curve with cons~~nt speed then
Ci(t) .1 a(t) for all tel.
, ~
7.3. Let IX: 1-+ 1R"+ 1 be a parametrized curve with a(t) =1= 0 for all tel. Show that
there exists a unit speed reparametrization Pof IX; i.e., show that there exists an
interval J and a smooth function h: J -+ I (onto) such that h' > 0 and such that
P= IX 0 h has unit speed. [Hint: Set h = S-1 where s(t) = J~o Ila(t)11 dt for some
to E I.]
7.4. Let X and Y be smooth vector fields along the parametrized curve IX: 1-+ 1R"+ 1
and let I: I -+ IR be a smooth function along IX. Verify that
+
(a) (X Y) = X + Y
(b) (Ix) = !'X + IX
(c) (X, Y)' = X . Y + X . Y.
7.5. Let S denote ihe cylinder xf + xi = r2 of radius r > 0 in R3. Show that IX is a
geodesic of S if and only if IX is of the form
IX(t) = (r cos (at + b), r sin(at + b), ct + d)
for some a, b, c, d E R.
7.6. Show that a parametrized curve IX in the unit n-sphere xf + ... + x:+ 1 = 1 is a
geodesic if and only if it is of the form
IX(t) = (cos,at)el + (sin at)e2
for some orthogonal pair of unit vectors {el' e2} in 1R"+ 1 and some a E R. (For
a =1= 0, these curVes are" great circles" on the n-sphere.)
7.7. Show that if IX: I -+ S is a geodesic in an n-surface S and if P= IX 0 h is a repar-
ametrization of IX(h: i -+ I) then Pis a geodesic in S if and only if there exist
a, b E R with a> 0 such that h(t) = at + b for all t E 1.
7.8. Let C be a plane curve in the upper half plane X2 > 0 and let S be the surface of
revolution obtained by rotating C about the xl-axis (see Example 5, Chapter 4).
Let IX: 1-+ C, IX(t) = (Xl(t), X2(t)), be a constant speed parametrized c~rve in C.
F or each 8 E IR, define CXs: I -+ S by I
The curves ex6 are called meridians of S, and the circles Pt are parallels of S (see
Figure 7.5).
(a) Show that meridians and parallels always meet orthogonally; i.e.,
~(t) . Pt(8) = 0 for all tEl, 8 E R
(b) Show that each meridian ClfI is a geodesic of S. [Hint: Note that {~(t), Pt(8)}
spans SI" where p = ex6(t). Hence it suffices to check that a6(t) is perpendicu-
lar to both ~(t) and Pt(8).]
(c) Show that a parallel Pt is a geodesic of S if and only ifthe slope X2(t)/X'I(t)
of the tangent line to C at ex(t) is zero.
7.9. Let S be an n-surface in /Rn + I, let v ESp, PES, and let ex: I -+ S be the maximal
geodesic in S with initial velocity v. Show that the maximal geodesic Pin S with
initial velocity cv (c E /R) is given by the forqlula P(t) = ex(ct).
7.10. Let S be an n-surface in /Rn+l, let PES, let v ESp, and let ex: I -+S be the
maximal geodesic in S passing through p with velocity v. Show that if p: 1-+ S
is any geodesic in S with P(to) = p and P(to) = v for some to E 1 then P(t) =
ex(t - to) for all tEl.
7.11. Let S be an n-surface in /Rn + I and let p: I -+ S be a geodesic in S with P( to) =
P(O) and P(to) = P(O) for some to E I, to =1= O. Show that P is periodic by show-
ing that P(t + to) = P(t) for all t such that both t and t + to E I. [Hint: Use
Exercise 7.10.]
7.12. An n-surface S in /Rn+ I is said to be geodesically complete if every maximal
geodesic in S has domain IR. Which of the following n-surfaces are geodesicalll
complete?
(a) The n-sphere xi + ... + x;+ I = 1.
(b) The n-sphere with the north pole deleted: xi + ... + X;+I = 1, Xn+1 =1= 1.
(c) The cone xi + x~ - x~ = 0, X3 > 0 in /R 3 •
(d)
(e)
The
The
cylinder xi + x~ = 1 in /R 3 •
,e.
cylinder in /R3 with a straight line deleted: xi + x~ = 1, XI =1= 1.
Parallel Transport
X(t)
N(oc(t)) /
1P\/s.(t)
L
_ X'(~)
Figure 8.1 The covariant derivative X'(t) is the orthogonal projection onto the
tangent space of the ordinary derivative X(t).
projecting onto the tangent space to S defines an operation with the same
properties as differentiation, except that now differentiation of vector fields
tangent to S yields vector fields tangent to S. This operation is called covar-
iant differentiation.
Let S be an n-surface in·lRn + 1, let ex: 1-+ S be a.parametrizedcul-ve in S,
and let X be a smooth vector field tangent to S along ex. The covariant
derivative of X is the vector field X' tangent to S along ex defined by
I
p
!
(a) (b)
Figure 8.2 Euclidean parallelism in IR 2 : (a) parallel vectors; (b) a parallel vector
field.
parametrized curve ex: I -+ IRn + 1 is Euclidean parallel if X(t.) = X(t 2) for all •
t 1 , t2 E I, where X(t) = (ex(t), X(t)) for tEl (see Figure 8.2(b)). Thus X is
Euclidean parallel along ex if and only if X = o.
Given an n-surface S in IRn + 1 and a parametrized curve ex: I -+ S, a
smooth vector field X tangent to S along ex is said to be Levi-Civita parallel,
or simply parallel, if X' = O. Intuitively, X is parallel along ex if X is a constant
vector field along ex, as seen from S.
8 Parallel Transport 47
~
dt
IIXI1 2 = ~ (X • X) = 2X' • X = O.
dt
(ii) If X and Yare two parallel vector fields along ex, then X • Y is constant
along ex, since
(X • V)' = X' • Y + X • Y' = O.
(iii) If X and Yare parallel along ex, then the angle cos-I(X • Y/IIXIIIIYII)
between X and Y is constant along ex, since X • Y, IIXII, and IIYII are each
constant along ex.
(iv) If X and Yare parallel along ex then so are X + Y and eX, for all e E IR.
(v) The velocity vector field along a parametrized curve ex in S is parallel if
and only if ex is a geodesic.
= V + (V • N ~ ex)N ex 0
where the N j UE {I, ... , n + I}) are the components ofN.) By the existence
and uniqueness theorem for solutions of first order differential eqljlations,
there exists a unique vector field V along ~ satisfying equation (P) tbgether
with the initial condition V(t o) = v (that is, satisfying Yt(t o) = Vi for
i E {I, ... , n + I}, where v = (ex(to), Vb ... , vn + d). The existence and
uniqueness theorem does not guarantee, however, that V is tangent to S
along ex.
48 8 Parallel Transport
= [- (V • N ~ ex)N 0 ex] • N 0 ex + V • N ~ ex
= - V • N ~ ex + V . N ~ ex = 0,
so V • N ex is constant along ex and, since (V • N ex)(to) = v • N(ex(to)) = 0,
0 0
this constant must be zero. Finally, this vector field V, tangent to S along ex,
is parallel because it satisfies equation (P). 0
Remark. We have implicitly assumed in the above proof that the solution
V of (P) satisfying V(t o) = v is defined on the whole interval] and not just on
some smaller interval containing to . That this is indeed the case can be seen
from the following argument. Suppose 1 c ] is the maximal interval on
which there exists a solution V of (P) satisfying V(t o) = v. If 1 =1= ], there
exists an endpoint b of 1 with bE]. Let {t i} be a sequence in 1 with
limi .... oo ti = b. Since IIVII is constant on 1, IIV(ti)11 ~ Ilvll for all i, so the
sequence {V(ti)} of vector parts of {V(ti)} takes values in a compact set, the
sphere of radius Ilvll about the origin in IRn + 1. It follows that {V(ti)} must
have a convergent subsequence {V(tik)}' Let w = limk .... oo V(t ik ), and let W be
a solution of (P), on some interval J containing b, with W(b) = (ex(b), w).
Then W - V is also a solution of (P), on 1 n J, and in particular IIW - VII is
constant on 1 n J. But
lim II W(t ik ) - V(t ik )II = Ilw - wll = 0
k .... oo
Figure 8.3 Levi-Civita parallel vector fields along geodesics in the 2-sphere.
8 Parallel Transport 49
PROOF. The "only if" statement is immediate from properties (i) and (iii)
above. So suppose both IIXliand the angle (J between X and ~ are constant
along cx, Let to E I and let v E S«(to) be a unit vector orthogonal to ~(to). Let V
be the unique parallel vector field along a such that V(to ) = v. Then IIVII = 1
°
and V • ~ = along cx so {~(t), V(t)} is an orthogonal hasis for Srz(t) , for each
tel. In particular, 'there exist smooth functions f, g: 1-+ IR such that
X ::;:: ff!. + gV. Since
Prz/l(V) = Vo(x)
= (cos (J)(q, -cos (J, -sin (J, 0) - (sin (J)(q, -sin (J, cos (J,O)
= - (q, cos 2(J, sin 2(J,0).
50 8 Parallel Transport
Image 0(0
Image a n l4
Image a ni2
Pal4 (v)
Note that parallel transport from p to q is path dependent; that is, if ex and
fJ are two parametrized curves in S from p to q and v ESp, then, in genera~
PCl(v) =1= Pp(v).
Tangent vectors v ESp, PES, may also be transported along piecewise
smooth curves in S. A piecewise smooth parametrized curve ex in S is a contin-
uous map ex: [a, b] -+ S such that the restriction of ex to [tb ti+ d is smooth
for each i E {O, 1, ... , k}, where a = to < tl < ... < tk + 1 = b (see Figure 8.5).
along ex is a vector space isomorphism which preserves dot products; that is,
(i) PIX is a linear map
(ii) P« is one to one and onto
(iii) P«(v) • P«(w) = v • w for all v, WE Sp.
PROOF. Property (i) is an immediate consequence of the fact that if V and W
are parallel vector fields along a parametrized curve in S, then so are V + W
and cV, for all c E IR. Similarly, property (iii) follows from the fact that if V
and Ware parallel then V • W is constant. Finally, the kernel (null space) of
PIX is zero because IIP«(v)/I = 0 implies IIvll = 0, by (iii~ so PIX is a one to one
linear map from one n-dimensional vector space to another. But all such
maps are onto. 0
EXERCISES .
8.1. Let S be an n-surface in.R"+ 1, let~: I ..... S be a parametrized curve, and let X and
Y be vector fields tangent to S along ~. Verify that
(a) '(X + Y)' = X' + Y', and
(b) (IX)' =!'X +IX',
for all smooth functions I along ~.
8.2. Let S be an n-plane alXl + ... + a;'+lx.+l == b in R"+I, let p, q E S, and let
v = (p, v) ESp. Show that if~ is any parametrized curve in S from p to q then
PII(v) = (q, v). Conclude that, in an n-plane, parallel transport is path
independent. .
8.3. Let ~: [0, x] ..... 52 be the half great circle in 52, running from the north pole
p = (0, 0, 1) to the south pole q = (0, 0, -1~ defined by ~(t) = (sin t,O, cos t).
Show that, for v == (p, Vb Vl' 0) E 5~, PII(l') = (q, -Vb V2, 0). [Hint: Check this
first when v = (p, 1, 0, .0) and when v = (p, 0, 1, 0); then use the linearity of PII']
8.4. Let p be a point in the 2-sphere 52 and let v and wE 5~, be such that ~l'lI == Ilwll·
Show that there is a piecewise smooth parametrized curve~: [a, b] ..... 52, with
~(a) = ~(b) = p, such that p.(l') = w. [Hint: Consider closed curves ~, with
&(a) ,,;,. v, which form geodesic triangles with ~(t) .l p for t in the "middle
segment" of [a, b].]
8.5. Let~: I ..... R"+ 1 be a parametrized curve with ~(t) E SIn S 2 for all tel, where
S 1 and S2 are two n-surfaces in R"+ 1. Suppose X is a vector field along ~ which is
tangent both to SI and to S2along~.
(a) Show by example that X may be parallel along ~ viewed as a curve in S 1 but
not parallel along ~ viewed as a curve in S 2' I
(b) Show that if S 1 is tangent to S 2 along. ~ (that is,(S 1)11(0 = (S 2)1I(t) for all tel)
then X is parallel along ~ in S 1 if and only if X is parallel along ~ in S 2 .
(c) Show that, if S 1 and S i. aren-surfaces which are tangent along a par-
ametrized curve ~: I ..... S 1· n S 2, then ~ is a geodesic in S 1 if and only if ~ is a
geodesic in S2 .
52 8 Parallel Transport
8.6. Let 8 be an n-surface and let ct: 1-+8 be a parametrized curve in 8. Let p: 1-+ 8
be defined by p = ct h where h: 1-+ I is a smooth function with h'(t) =1= 0 for all
0
8.7. Let 8 be an n-surface in [Rn+l, let p E 8, and let Gp denote the group of non-
singular linear transformations from 8 p to itself. Let
Hp = {T E Gp: T= P(I. for some piecewise smooth ct: [a, b] -+ 8 with
ct(a) = ct(b) = pl.
Show that H p is a subgroup of Gp by showing that
(i) for each pair of piecewise smooth curves ct and pin 8 from p to p there is a
piecewise smooth curve '}' from p to p such that
Py = PpoP(I.' and
(ii) for each ct in 8 from p to p there is a p in 8 from p to p such that P p = P; 1.
(The subgroup Hp is called the holonomy group of 8 at p.)
8.8. Let ct: I -+ 8 be a unit speed curve in an n-surface 8, and let X be a smooth vector
field, tangent to 8 along ct, which is everywhere orthogonal to ct (X(t)) . &(t) = 0
for all t E 1). Define the Fermi derivative X' of X by
X'(t) = X'(t) - [X'(t) • &(t)]&(t).
(a) Show that if X and Yare smooth vector fields along ct which are tangent to 8
and orthogonal to ct then
(i) (X + V)' = X' + Y'
(ii) (fX)' =!'X + fX' for all smooth functions f along ct, and
(iii) (X . V)' = X' . Y + X . Y'.
(b) Show that if a E I and v E 8(1.(a) is orthogonal to ~(a) then there exists a
unique vector field V along ct, tangent to 8 and orthogonal to ct, such that
V' = 0 and V(a) = v. (V is said to be Fermi parallel along ct.)
(c) For ct: [a, b] -+ 8 a parametrized curve in 8 and v E 8(1.(a), with v 1. &(0), let
F(I.(v) = V(b) where V is as in part b). Show that F(I. is a vector space isomor-
phism from ~(a)l. onto ~(b)\ where &(t)l. is the orthogonal complement of
&(t) in 8(1.(t). Also show that F(I. preserves dot products. (F(I.(v) is the Fermi
transport of v along ct to ct(b).)
The Weingarten Map
9
Vv+./= Vv / + V.I
53
54 9 The Weingarten Map
where (X: 1-+ S is any parametrized curve in S with ti(to ) = v. Note that the
value of Vv f is independent of the curve (X in S passing through p with
velocity v, since
Vv f = (J (X)'(t o ) = VJ((X(t o )) • ti(to ) = VJ(p) • v
0
for all smooth vector fields X and Y on U (or on S) and all smooth functions
f: U -+ !R (or f: S -+ !R). Here, the sum X + Y of two vector fields X and Y is
the vector field defined by (X + Y)(q) = X(q) + Y(q), the product of a func-
tionf and a vector field X is the vector field defined by (fX)(q) = j;(q)X(q),
and the dot product of vector fields X and Y is the function defined by
(X· Y)(q) = X(q) • Y(q), for all q E U (or for all q E S). Moreover, for each
smooth vector field X, the function which sends v into VvX is a linear map,
9 The Weingarten Map 55
from IR:+ 1 into IR:+ 1 if X is a vector field on an open set U, and from Sp into
IR:+ 1 if X is a vector field on an n-surface S.
Note that the derivative VvX of a tangent vector field X on an n-surface S
with respect to a vector v tangent to S at PES will not in general be tangent
to S. In later chapters we will find it useful to consider the tangential com-
ponentDvX of VvX:
DvX = VvX - (VvX • N(p))N(p),
where N is an orientation on S. DvX is called the covariant derivative of the
tangent vector field X with respect to v E Sp. Note that DvX = (X 0 a)'(to )
where a: I ..... S is any parametrized curve in S with a(to) = v. Covariant
differentiation has the same properties ((i)-(iii) above, with V replaced by D)
as ordinary differentiation (see Exercise 9.5). Moreover, for each smooth
tangent vector field X on S, the function which sends v into DvX is a linear
map from Sp into Sp.
We are now ready to study the rate of change of the normal direction N
on an oriented n-surface S in jR" + 1. Note that, for PES and v ESp, the
derivative VvN is tangent to S (i.e., VvN 1.. N(P)) since
0= Vv(1) = Vv(N • N)= (VvN) • N(p) + N(p)· (VvN)
= 2(Vv N) • N(p).
The linear map Lp: S p-+ S p defined by.
Lp(v) = -VvN
is called the Weingarten map of S at p. The geometric meaning of Lp can be
seen from the formula
VvN = -(N ; a)(to )
where a: I ..... S is any parametrized curve in S with ~(to) = v: Lp(v) measures
(up to sign) the rate of change of N (i.e., the turning of N since N has
constant length) as one passes through palong any such curve a. Since the
tangent space S(J(t) to Sat a(t) is just [N(a(t))J1., the tangent space turns as the
normal N turns and so Lp(v) can be interpreted as a measure of the turning
of the tangent space as one passes through p along a (see Figure 9.1). Thus
Lp contains information about the shape of S; for this reason Lp is
sometimes called the shape operator of S at p.
For computational purposes, it is important to note that Lp(v) can be
obtained from the formula
Lp(v) = -VvN = -(p, VvN h ..• , VvN,.+d
= -(p, VN 1 (p)· v, ... , VN,.+l(P) • v),
where N is any smooth vector field defined on an open set U containing S
with ~(q) = N(q) for all q E S. Note that ~(q) need not be a unit vector for
56 9 The Weingarten Map
Figure 9.1 The Weingarten map. Lp(v) = - (N ~ 1X)(to) measures the turning of the
normal, hence the turning of the tangent space, as one passes through p along the
curve IX.
q ¢ S. When f: U -+ IR is such that S = f-l(C) for some IR and N(q) = CE
Vf(q)/IIVf(q)11 for all q E S, it is natural to take N = VJ7IIVfll. Sometimes,
however, another choice of N is more convenient, as the following example
shows.
r----~
s
Lp(')·'
t_______~_
Figure 9.2 All parametrized curves in S passing through p with the same velocity
will necessarily have. the same normal component of acceleration at p. In the figure,
a(tt} = P(tl) = a(h) = v; fJ is a geodesic.
0= [ti • (N 0 (X)]'(t o )
= (i(t o ) • (N 0 (X)(t o ) + ti(to ) • (N ~ (X)(t o )
= (i(t o ) • N((X(t o )) + v· VvN
= (i(t o ) • N(p) - v . Lp(v)
so (i(t o ) • N(p) = Lp(v) • v as claimed. o
Theorem 2. The Weingarten map Lp is self-adjoint; that is,
Lp(v) • W = v • Lp(w)
for all v, WE Sp.
1 n+ I 02f
IIVf(p)11 i. ~ I OXi OX j (P)ViWj'
Since 02f10Xi OXj = 02f10Xj OXi for all (i, j), we can finally conclude that
1 1I+1 02j
Lp(v) • W = - IIVj(p)11 i'~1 OXi OXj (P)viWj
1 n+1 02j
IIVj(p)11 i, ~ 1 OXj OXi (P)viWj
1 n+l 02/
IIVj( )11 . ~ ~ (P)wjVi = Lp(w) • V. 0
P l,j=1 UXj uXi
EXERCISES
9.1. Compute Vvf wheref: R"+l-+R and v e R;+l, p e R"+ I , are given by
(a) f(Xh X2) = 2x~ + 3x~, v = (1,0, 2, 1) (n = 1)
(b) = x~ - x~, v = (1, 1, cos 8, sin 8)
f(Xh X2) (n = 1)
(c) f(Xh X2, X3) = Xl X2 xL v = (1, 1, 1, a, b, c) (n = 2)
(d) f(q) = q • q, v = (p, v) (arbitrary n).
9.2. Let U be an open set in R"+ 1 and letf: U -+ R be a smooth function. Show that
if e, = (p,0, ... , 1, ... , 0) where p eU and the 1 is in the (i + 1)ih ~pot (i spots
after the p), then V.I f = (bflox,)(P).
9.3. Compute VvX where v e R;+l, p e R"+ I , and X are given by
(a) X(Xh (Xh X2, Xl X2, X~), V = (1, 0, 0, 1)
X2) = (n = 1)
(b) X(Xh (Xl', X2, -X2, Xl), v = (cos 8, sin 8, -sin 8, cos 8)
X2) = (n = 1)
(c) X(q) = (q, 2q), v = (0, ... ,0, 1, ... , 1) , (arbitrary n).
9.4. Verify that differentiation of vector fields has the properties (i)-(iii) as stated on
page 54.
9.5. Show that covariant differentiation of vector fields has the following proper-
ties: for peS and v e S"
(i) Dv(X + V) = DvX + DvV ,
(ii) Dv(fX) = (Vv f)X(p) + f(p)DvX
(iii) Vv(X • V) = (DvX) ~ Y(p) + X(p) • (Dv V)
for all smooth tangent vector fields X and V on S and all smooth functions
f:S-+R.
9.6. Suppose X is a smooth unit vector field on an n-surface S in R"+ I; i.e.,
IIX(q)1I = 1 for all q e S. Show that VvX.J.. X(p) for all ve S" peS. Show
further that, if X is a unit tangent vector field on S, then DvX .J.. X(p ~
9.7. A smooth tangent vector field X on an n-surface S is said to be a geodesic vector
field, or geodesic flow, if all integral curves of X are geodesics of S.
(a) Show that a smooth tangent vector field X on S is a geodesic field if and
only if Dx(;~X = 0 for all peS.
(b) Describe a geodesic flow on a 2-surface of revolution in R3.
60 9 The Weingarten ~ap
9.9. Show that if S is an n-surface and N is a unit normal vector field on S, then the
Weingarten map of S oriented by - N is the negative of the Weingarten
map of S oriented by N.
9.10. Let V be a finite dimensional vector space with inner product (dot product).
Let L: V -+ V be a linear map.
(a) Show that there exists a unique linear map L*: V -+ V such that
L*(v) • W = v • L(w) for all v, WE V. [Hint: Choose an orthonormal basis
{eb ... , en} for Vand compute L*(ej) for each i]. (L* is the adjoint of L.)
(b) Show that the matrix for L* relative to an orthonormal basis for V is the
transpose of the matrix for L relative to that basis. Conclude that L is
self-adjoint (L* = L) if and only if the matrix for L relative to any orthonor-
mal basis for V is symmetric.
9.12. Let S be an n-surface in ~n+ 1, oriented by the unit normal vector field N.
Suppose X and Yare smooth tangent vector fields on S.
(a) Show that
(VX(p)Y) . N(p) = (Vy(p)X) . N(p)
for all pES. [Hint: Show that both sides are equal to Lp(X(p)) • Y(p).]
(b) Conclude that the vector field [X, Y] defined on S by [X, Y](p)
= VX(p)Y - VY(p)X is everywhere tangent to S. ([X, Y] is called the Lie
bracket of the vector fields X and Y.)
d1x, + L N· oNJtE:J.dxt = 0
dt" 'ox" dtdt
(in n + 1 variables Xi) to the first order differential system
dXi
fit=u,.
I
(in 2n + 2 variables Xi and u,). This first order system ,of differential equations isjust
the ditTerential equation for the integral curves of X in U x RIJ + 1 c: R"r+". The
vector field X is called a geodesic spray.
Curvature of Plane Curves
10
Let C = f-l(C), where f: U --+ IR, be a plane curve i~ the open set U c 1R2,
oriented by N = VflIIVfll. Then, for each P E C, the Weingarten map Lp is a
linear transformation on the I-dimensional space C p' Since every linear
transformation from a I-dimensional space to itself is mUltiplication by a
real number, there exists, for each P E C, a real number K(p) such that
Lp(v) = K(p)V for all v E C p.
K(p) is called the curvature of C at p.
If v is any non-zero vector tangent to the plane curve C at p E C then
Lp(v) • v= K(p)llvI1 2
so the curvature of C at p is given by the formula
K(p) = Lp(v) • v/llvl12.
In particular, if IX: I --+ C is any parametrized curve in C with ci(t) =1= 0 for all
tEl then, by Theorem 1 of Chapter 9,
= Lp(ci(t)) • ci(t) = cx(t) • N(IX(t))
( ( ))
K IX t 11&(t)112 11&(t)112
If IX is a unit speed parametrized curve in C, this formula reduces to
, K(IX(t)) = Ci(t) • N(IX(t)).
Thus the curvature of C at p E C measures the normal component of acceler-
ation of any unit speed parametrized curve in C passing through p.
Note in particular the significance of the sign of K(P): if K(p) > 0 then the
curve at p is turning toward its normal N(p), and if K(p) < 0 the curve is
turning away from N(p) (see Figure 10.1).
62
10 Curvature of ~lane Curves 63
Figure 10.1 The curvature of C is positive at points where C is bending toward its
normal and is negative where C is bending away from its normal.
One way to compute the curvature of a plane curve is to use the formula
K IX = (tX· N 1X)/lltiI1 2 (or the equivalent formula in Exercise 10.1), where IX
0 0
for all t E 1.
Local parametrizations of plane curves are unique up to reparametriza-
tion: given any parametrization 13:' I-+- C of a segment of C containing p,
there exists a smooth function h: I-+- IR, withh'(t) > O.for all tEl, such that
f3(t) = lX(h(t)) for aU tEl, where IX is the unit speed local parametrization
constructed above. Indeed, since {X(f3(t))} is a basis for the I-dimensional
vector space CJJ(t) , /3(t) is necessarily a multiple of X(f3(t)~ In fact,
/3(t) = 1I/3(t)IIX(f3(t)) since IIXII = 1 and since {/3(t)} and {X(f3(t))} are both
64 10 Curvature of Plane Curves
f IIP(T)II dT,
t'
h(t) =
to
IX(t) for all t E domain 13 h- 1 • In other words, f3(t) = IX(h(t)) for all tEl, as
0
claimed. Note in particular that if 13: I-+- C is a unit speed local parametriza-
tion of C with f3(t o) = P then h(t) = t - to and f3(t) = IX(t - to)for all t E 1.
EXAMPLE. Let C be the circle f-l(r 2), where f(Xl' X2) = (Xl - a)2 +
(X2 - b)2, oriented by the outward normal VflIIVfll. Since Vf(p) =
(p, 2(Xl - a), 2(X2 - b)) for p = (Xl' X2) E 1R2, the integral curves of
X(p) = (p, 2(X2 - b), -2(Xl - a)) will be local parametrizations of C. The
integral curve through (a + r, b) gives the global parametrization IX(t) =
(a + r cos 2t, b - r sin 2t). Hence
ii(t) • N(IX(t)) ii(t) Vf(IX(t))
K(IX(t)) = 11~(t)112 = Ilti(t)112 • IIVf(IX(t))11
_ (-4r cos 2t, 4r sin 2t) • (2r cos 2t, -2r sin 2t)
- 11(-2r sin 2t, -2r cos 2t)11211(2r cos 2t, -2r sin 2t)11
-8r2 1
= (4r2)(2r) = - r
If C had been oriented by the inward normal, the curvature would have been
+ l/r at each point.
For C an arbitrary oriented plane curve and p E C such that K(p) =1= 0,
there exists a unique oriented circle 0, called the circle of curvature of C at p
(see Figure 10.2), which
(i) is tangent to C at p (i.e., C p = Op)
(ii) is oriented consistently with C (i.e., N(p) = N l(P) where Nand N 1 are
respectively -the orientation normals of C and 0), and
(iii) whose normal turns at the same rate at p as does the normal to C (i.e.,
VvN = VvN 1 for all v E C p = Op).
10 Curvature of Plane Curves 6S
N(q)
Figure 10.2 The circle of curvature at two points of an oriented plane curve C.
This circle of curvature is the circle which hugs the curve C closest among
all circles containing p (see Exercises 10.8 and 10.9~ Condition (i) says that
the center of 0 is on the nonnalline to C at p, condition (iii) says that its
radius r satisfies the equation l/r = IK(P) I, where K(P) is the curvature of C
at p, and condition (ii) says that N(p) points toward the center of 0 if
K(P) > 0 and away from the center of 0 if K(P) < O. The radius r = 1/ IK(P) I
of the circle of curvature is called the radius of curvature of C at p; its center
is the. center of curvature of C at p.
EXERCISES
10.1. Let rx(t) == (x(t~y(t» (te I) be a local parametrization of the oriented plane
curve C. Show that
'K orx == (x'y" - y'x")/(X,2 + y'2)3/2.
10.2. Let g: 1- R be a smooth function and let ~ denote the graph of g. Show that
the curvature of C at the point (t, g(t» is g"(t)f(1 + (g'(t»2)3/2, for an appro-
priate choice of orientation.
10.3. Find global parametrizations of cachof the following plane curves, oriented
by VJi'~V/l where/is the function defined by the left side of each equation.
(a) axl + bX2 == c, (a, b)::/= (0, O~
(b) xi/a 2 + X~/b2 == 1, a::/= 0, b::/= O.
(c) X2 - axi == c, a::/= O.
(d) xi - x~ == 1, Xl > O.
10.4. Find the curvature 'K of each of the oriented plane curves in Exerci~ 10.3.
10.5. Let C be an oriented plane curve. Let pee and let N(P) == (p, N(p» denote
the orientation unit normal vector-at p. Show that if rx: I - C is any unit speed
local parametrization of C with «(to) == p, and h(t) == (rx(t) - p) • N(P) (see
Figure 10.3~ then h(to) == h'(to) = 0 and h"(to) == 'K(P~
66 10 Curvature of Plane Curves
------------~--p~~------------cp
Figure 10.3 h(t) is the projection of ~(t) - P along N(p). h(t) may be thought of as
the "height of ~(t) above the line tangent to S at p."
10.6. Let C be a plane curve oriented by the unit normal vector field N. Let
~:1-+ C be a unit speed local parametrization of C. For tEl, let T(t) = ~(t).
Show that
These formulas are called the Frenet formulas for a plane curve.
10.7. Let ~: 1-+ 1R3 be a unit speed parametrized curve in 1R3 such that
~(t) x iX(t) =f= 0 for all tEl. Let T, N, and B denote the vector fields along ~
defined by T(t) = ~(t), N(t) = iX(t)/lliX(t)ll, and B(t) = T(t) x N(t) for all tEl.
(a) Show that {T(t), N(t), B(t)} is orthonormal for each tEl.
(b) Show that there exist smooth functions K: I -+ IR and 't': I -+ IR such that
t=KN
N= -KT+'t'B
B= -'t'N.
These formulas are called the Frenet formulas for parametrized curves in
1R3. The vector fields Nand B are respectively the principal normal and the
binormal vector fields along ~. The functions K and 't' are called the curva-
ture and torsion of ~.
10.8. Show that the circle of curvature 0 at a point p of an oriented plane curve C,
where K(p) #- 0, has second order contact with Cat p; i.e., show that if ~ and p
are unit speed local parametrizations of C and of 0, respectively, with ~(O) =
P(O) = p, then ci(O) = p(O) and Ci(O) = p(O).
10.9. Let C be an oriented plane curve, let pEe, and let ~: 1-+ C be a unit speed
local parametrization of C with ~(O) = p. Assume K(p) =f= O. For q E 1R2 and
r > 0, define f: 1-+ IR by f(t) = 11~(t) - ql12 - r2. Show that q is the center
of curvature and r the radius of curvature of C at p if and only if
f(O) = f'(0) = f"(O) = O.
10 Curvature of Plane Curves 67
10.10. Let rx: I ~ C be a unit speed local parametrization of the oriented plane curve
C. Suppose (J: I ~ R is smooth and-is such that
a(t) == (rx(t~ cos (J(t~ sin (J(t))
for all tel (see Figure 10.4). (We shall, in the next chapter, be able to prove
that such a function (J exists; see Exercise 11.15.) Show that d(J/dt == " 0 IX.
Arc Length and Line
11 Integrals
Before analyzing the Weingarten map for n-surfaces (n > 1) we shall pause
to see how parametrizations of plane curves can be used to evaluate integrals
over the curve.
The length l(IX) of a parametrized curve IX: I -+ ~n + 1 is defined to be the
integral of the speed of IX:
f Ila(t)11 dt,
b
I(IX) =
a
where a and b are the endpoints of I (possibly ± (0). Note that I(IX) may be
± 00. Also note that the length of a parametrized curve is the total" distance
travelled." Whenever IX retraces itself then that portion of the image which is
covered more than once is counted more than once.
Note that if fJ: 1-+ ~n+ 1 is a reparametrization of IX, then l(fJ) = I(IX).
Indeed, if fJ = IX h where h: 1-+ I is such that h'(t) > 0 for all tEl, then
0
f IIP(t)11 dt = f 11&(h(t))llh'(t) dt
d d
l(fJ) =
c c
= f 11&(u)11 du = I(IX),
b
f f 1 dt = t2 -
t2 t2
11&(t)11 dt = t1
t1 t1
68
11 Arc Length and Line Integrals 69
where 81" >-0 and 82 > 0 are chosen small enough so that A is contained in
the domain of/and so that Vf(q) • (q, v) > 0 for aD q E A (see Figure 11.1~
That this last condition can ,be satisfied is a consequence of,the continuity
of (Of!OXh Of!OX2)·~ v and the fact that «Of!OXh Of!OX2)· v)(po) =
«Of!OX1)(PO))2 + «OPOX2)(PO))2 > O. The condition Vf(q) • (q, v) > 0 for aU
q E A guarantees that, forlrl < 8h the function gr(s) = !(Po + ru 1+ sv) is
strictly increasing on the interval - 82 <. S < 82 (g;.(S) > 0) and hence that for
each r with Irl < 81 there is at most ones with ~I < 82 such that
gr(s) =!(Po + ru + sv) =c. In other words, for each r with Irl < 81 there is at
most ones with lsi < 82 such that Po + ru + sve C. Nowy(t) =- Po + hl(t)u +
70 11 Arc Length and Line Integrals
Image ~r
Figure 11.1 The rectangle A about Po is chosen so that each line segment ~r(S) =
(Po + ru) + sv (r fixed, - E2 < s < E2) meets C at most once. In the shaded rectangle
B, each such line segment meets C exactly once.
h2(t)v, where h1(t) = (y(t) - Po)· u/llul1 2 and h2(t) = (y(t) - Po)· vlllvl1 2 are
smooth functions of t with h1 (0) = h2(0) = O. Using the continuity ofy and of
h'1' together with the fact that
h'1 (0) = Y(O) • (Po, u/llul1 2 ) = X(Po) . X(Po)/IIX(Po)11 2 = 1,
we can choose t1 < 0 and t2 > 0 in the domain ofy so that both y(t) E A and
h'1 (t) > 0 for all t E (t1' t 2) (Figure 11.1). Setting r1 = h1 (td and r2 = h1 (t 2) it
follows that for each r E (r1' r2) there is exactly one t E (tb t 2) with h1 (t) = r
(since h1(t) is continuous and strictly increasing) and that s = h2(t) for this t
is an s (and therefore the unique s) with Is I < e2 such that Po + ru + sv E C.
In other words, if B is the rectangle
B = {Po + ru + sv: r1 < r < r2, Is I < e2},
then Po + ru + sv E B n C if and only if r = h1(t) and s = h2(t) for some
t E (tl' t 2); i.e., if and only if Po + ru + sv E Image y. Thus C nBc
Image y, as required. 0
unit tangen t vector field on C constructed as in the previous chapter and let IX
be the maximal integral curve of X with IX(O) = P(ttl = P(t 2). Then, since Pis
also an integral curve of X, uniqueness of integral curves implies that
P(t) = IX(t - i1 )
and, at the same time,
P(t) = IX(t - t 2)
for all tEl. Setting t = t2 - t 1 , we have
P(t) = IX(t - ttl = IX((tt) - t 2) = P(t + t)
for all t such that both t and t + tEl. Thus if P is not one to one then it is
periodic.
If Pis periodic then C must be compact because C is the image under the
continuous map P of a closed interval [to, to + t]. On the other hand, if Pis
not periodic, then Pmust be one to one so C cannot be compact because the
function P- 1: C ~ ~ is continuous on C yet attains no maximum value. The
continuity of p- 1 can be checked as follows. Given to E I and e > 0, set
y(t) = P(t + to) for t such that rt I < B and t + to E I, and choose an open
rectangle B about Po = P( to) = y(O) as in the proof of Theorem 1. Then
C nBc Image y so IP- 1 (p) - to I = Iy - 1 (P) I < e whenever p E C n B, as
required for continuity. 0
Recall that the period of a periodic function P is the smallest t such that
P( t + t) = P( t) for all t such that both t an~ t +. t are in the domain of p. If t
is the period of P then any subset of the domain of P of the form [to, to + t)
is called afundamental domain of p. Note that the restriction of any periodic
global parametrization Pof a compact plane curve to a fundamental domain
maps that fundamental domain one to one onto Image p. Hence, if we allow
half-open intervals as well as open intervals as domains of parametrized curves,
every connected oriented plane curve admits a one to one unit speed global
parametrization. Moreover, any two such parametrizations IX: I ~ C and
p: I ~ C have parameter intervals I and I of the same length. Indeed, IX and
P are related by P(t) = IX(t - to) for some to E ~ so I is simply a translate of
I. Hence we can define the length of a connected oriented plane curve C to be
the length of I where IX: I ~ C is anyone to one unit speed global parametri-
zation of C. ,
Since the Jength of P is the same as the length of IX where IX is any
reparametrization of p, it follows that the length of a connected oriented
plane curve C3Jl be computed from the formula
EXAMPLE. Let C denote the circle (Xl - a)2 + (X2 - b)2 = r2 oriented by
its outward normal. Then tX: I ~ C, defined by tX(t) = (a + r cos 2t,
b - r sin 2t) is a global parametrization of C, as we saw in the previous
chapter. tX is periodic with period n so the restriction of tX to the interval [0, n)
is a one to one global parametrization of C. Hence
I(C) = f1t IltX(t)11 dt = f1t 11(-2r sin 2t, -2r cos 2t)11 dt = 2nr.
o 0
EXAMPLE 3. For each i E {1, ... , n + 1}, let Xi: U ~ ~ (U C ~n+ l) be defined
by
Note that the sum of two smooth I-forms is smooth and that the product of
a smooth function and a smooth I-form is smooth.
Given a "i-form w and a vector field X on U c IR"+ 1, we can define a
function w(X): U ~ IR by
(w(X»)(P) = w(X(P)).
Note that if w and X are both smooth then so is w(X~
This formula shows that thefunGtionsjj, if they exist, are unique,d also
that they are· smooth if w is smooth. On the other hand, if we defule fune-
tionsfj by the above formula, then the I-forms w and Li':!l Ii dx i have the
same value on each of the basis vectors Xi(P) for ~ + 1 and hence by linearity
they have the same value on all vectors in IR;+ 1, P E U, so ill = L7:! I Ii dXi'
Clearly w is smooth if each Ii is smooth. 0
74 11 Arc Length and Line Integrals
f = f w(tX(t)) dt.
b
W
a a
f f = f w(ci(h(t)h'(t))) dt
d d
W =
w(/J(t)) dt
p e e
d b
= f w(ci(h(t))h'(t)) dt = f w(tX(u)) du = f w.
c a a
fe w f w,=
a
f a
W = .L
1=0
fw
ai
f df = f df(ex(t)) dt = f '(1
ez
b
II.
b
II
0 tX)'(t) dt == f(tX(b)) - f(tX(a)~
-. X2 dx Xl dx
,,- - x~ + x~ 1 + x~ + x~ 2
and tet C·denote the ellipse (xi/0 2) + (xJ/b2) = 1, oriented by its inward
normaL The. parapletrized curve tX: [0, 2nl~ C defined by tX(t) = (a cos t,
b sin· t) restricts to a one to one global parametrization of C on the interval
[0, 2n) so
2#
Jc ,,== I" == f
CI 0
,,(<«t»dt
== t2#00
0 2 008 2 t +b1 sin 2 t dt =4 t
.. n/2 (b/a) sec2 t
1 + (b/a)2 tan 2 t dt
== 4 foo . dU 2 == 2n.
o 1 +u
Since its integral over the compact curve C is n()t zero, the 1-form" of
Example 2 cannot be exact. However, its restriction to V (or, more precisely,
to V x 1R2~.where V is the complement in 1R2 of any ray through the 0rgin,
is exact. Indeed, for v any llD;it vector in 1R2 and .
V == 1R2 - {rv: r ~ O},
" == d8y where 8y : V ~ IR is defin'ed as follows. Let 8.,denote the Unique r~al
number with 0 ~ 8~ < 2n such that v = (cos 8." sin 8.,). Then, for each
76 11 Arc Length and Line Integrals
(x, y) E V, define Ov(x, y) to be the unique real number with Ov :::;; Ov(x, y) <
Ov + 2n such that .
((x 2 +Xy2 )'/2' (x 2 : y2 )'/2 ) = (cos /ly(X, y), sin /ly(x, y))
(see Figure 11.2). In order to verify that dO v = '1 Iv simply note that
tan Ov(x, y) = y/x and cot Ov(x, y) = x/y so in each sufficiently small open
set we can solve one or the other of these equations for Ov and compute
dO v
ao ao
= ~v dx 1 + ~v dx 2 = - 2
X2
2 dx 1 + 2
Xl
2 dX2 .
uXl uX2 Xl +X2 Xl +X2
Figure 11.2 Oy(x, y) is the inclination angle of the line segment from the origin
to (x, y), Ov :$; Oy < Ov + 2n.
where at is the restriction of a to the interval [a, t] and qJ(a) is chosen so that
a(a)/lIa(a)1I = (cos qJ(a), sin qJ(a)).
11 Arc Length and Line Integrals
We claim that
(*) ex(t)/llex(t)11 = (cos q>(t~ sin q>(t))
for all t E[a, b]. Indeed, let to denote the least upper bound of the set
{t E [a, b]: (*) holds for a ::; t ::; t}. By continuity, (*) must hold at t = to.
Setting v =- ex(to)/llex(to)11 and. defining 8y as above we find that
(cos 8y(ex(to)~ sin 8y(ex(to»)) = ex(to)/Il ex(to) II = (cos q>(to), sin q>(to)) so
q>(to) - 8y(ex(to» = 21tm for some integer m. Choosing ~ > 0 so that ex(t) E V
for all t E [a, b] with It - to I < ~ we find further that, for t E [a, b],
It - to I < ~, and t =1= ti (t i a point where ex fails to be smooth),
so q>(t) - 8y{ex(t)) = 2n:m for all t E [a, b] with It - to I < ~. But then
(cos q>(t), sin q>(t)) = (cos 8v(ex(t)~ sin 8y(ex(t))) == ex(t)/lIex(t)1I
for aU such t, which is possible only if to = b. Thus to == b md (*) holds for
all t E [a, b], as claimed.
Finally, sjnce ex(a) = ex(b~ (*) implies that
(cos q>(a~ sin q>(a)) = (cos q>(b), sin q>(b)~
so q>(b) - q>~a) = 21tk for some integer k, and
. .
o
The integer k(ex) = (1/2n) J« " is called the winding number of ex since it
counts the number of times the closed curve ,ex winds around the origin.
EXERCISFS
11.7. f(Xh X2) = xi - x~, U = {(Xh X2): 0 < Xl < 2}, c = 1. [Set up, but do not
evaluate, the integra1.]
11.8. f(Xh X2) = -9xi + 4x~, U = {(Xh X2): Xl > 0, 0 < X2 < 3}, c = O. [Hint:
Note that there is a parametrization ex(t) = (Xl(t), X2(t)) of f- 1 (0) with
X2(t) = t.]
11.9. Show that if C is a connected oriented plane curve and C is the same curve
with the opposite orientation, then I(C) = I(C).
11.10. Let C be a connected oriented plane curve, let ex: 1- C be a one to one unit
speed parametrization of C, and let ,,: C - IR denote the curvature of C.
(a) Show that S~ I" ex(t) I dt, where a and b are the endpoints of I, is
0
of C.
U~ I" 0 ex(t) I dt is called the total curvature of C.]
11.11. Let f and g be smooth functions on the open set U c IRn + 1. Show that
(a) d(f + g) = df + dg.
(b) d(fg) = gdf + fdg.
(c) If h: IR - IR is smooth then d(h 0 f) = (h' 0 f) df
11.12. Compute the following line integrals.
(a) (X2 dXl - Xl dX2) where ex(t) = (2 cos t, 2 sin t), 0 ~ t ~ 2n.
S<x
(b) Se (-X2 dx 1 + Xl dX2) where C is the ellipse (xi/a 2 ) + (xVb 2) = 1,
oriented by its inward norma1.
(c) S<x ~J~ I Xj dXj where ex: [0, 1] - IRn + 1 is such that ex(O) = (0, 0, ... , 0)
and ex(l) = (1, 1, ... , 1). [Hint: Find an f: IRn+ 1 - IR such that df =
L7~ I Xj dXj.]
11.13. Let w = L7~1 Ii dXj be a smooth I-form on IRn+ 1 and let ex(t) =
(xl(t1 ... , x n + l(t)), where the Xi are smooth real valued functions on [a, b].
Show that
b n+ 1 dx·
<x
W
f f
= II j~l (Ii 9-'ex) dt. Tt
11.14. Let C =f-l(C) be a compact plane curve oriented by VflIIVfll, let X be the
unit vector field on U = domain (f) obtained by rotating Vfl I Vf I through
the angle -nI2, and let Wx be the I-form on U dual to X. Show that
Se Wx = I(C).
11.15. Let ex: 1- 1R2 be a unit speed curve, let to E I, and let (Jo E IR be such that
&(to) = (ex(to), cos (Jo, sin (Jo). Show that there exists a unique smooth function
(J: 1- IR with (J = (Jo such that &(t) = (ex(t), cos (J(t), sin (J(t)) for all t E 1.
[Hint: Set (J(t) = (Jo + S:o '1(/3(-r)) d-r where '1 is the I-form of Theorem 3 and
f3 = dexldt.]
11.16. Let ex: [a, b] - 1R2 - {O} be closed piecewise smooth parametrized curve.
Show that the winding number of ex is the same as the winding number offex
where.!: [a, b] - IR is any piecewise smooth function along ex withf(a) = f(b)
11 Arc Length and Line Integrals 79
andf(t) > 0 for all t E [a, b]. Conclude that ex and ex/llexll have the same wind-
ing number.
11.17. Let a = to < t1 < ... < tA;+ 1 = b and let qJ: [a, b] x [0, 1] ~ 1R2 - to} be a
continuous map such that, for each u E [0, 1], the map qJu: [a, b] - IR defined
by qJu(t) = qJ(t, u) is smooth on each interval [ti' ti+~]. Assume that qJu(a) =
qJu(b) for all u E [0, 1]. Show that the winding number k(qJu) is a continuous
function ofu and hence that k(qJu) is constant and, in particular, k(qJo) = k(qJd·
[The map qJ is called a homotopy between qJo and qJ1']
11.18. The whIding number of dex/dt (which is the same, by Exercise 11.16, as the
winding number of dex/dt/lldex/dtll), where ex: [a, b] - 1R2 is a regular (&(t) =1=
for all t) parametrized curve with &(a) = &(b~ is called the rotation index of ex.
°
(a) Show by example that for each integer k there is an ex with rotation index
k.
(b) Show ~hat if ex is the restriction to [a, b] of a periodic regular parametrized
curve with period t = b - a and if ex is one to one on [a, b) then the
rotation index of ex is ± 1. [Hint: See Figure 11.3. Let u E 1R2, U =1= 0, and
choose to so that h(t) = ex(t) • u has an absolute minimum at to. For
Figure 11.3 To show that the rotation index of ex is ± 1, to is chosen so that Image ex
lies completely on one side ofthe tangent line at ex(to), t/! is the normalized secant map
t/!(t h t2) = (ex(t2) - ex(t1»/llex(t2) - ex(t 1)11 extended continuously to the closed
triangle T, andt/>: [to, to + t] x [0, 1] - T is the homotopy which maps horizontal
line segments to piecewise smooth curves in T, as indicated.
80 11 Arc Length and Line Integrals
~~ (t,) ~I ~: (t,) I
- ~: (to) ~I ~~ (to) I if t1 = to and
+
(d) Show that if/is any polynomial withJ(z) 0 for all z E C with Iz I ~ 1
then k(J) = n. [Apply Exercise 11.17 to qJ: [0, 21t] x [0, 1] -+ 1R2 - to},
qJ(t, u) =
u'1(! (cos t + i sin t))
u
'if u +0
1a,.(cos t + i sin tr if u = 0,
where a,. is the leading coefficient ofJ(z).]
+
(e) Conclude that if J is a polynomial with J(z) 0 for all z E C then the
degree ofJmust be zero. (This exercise proves theJundamental theorem oj
algebra: every non-constant polynomial with complex coefficients must
have a root in C.)
11.21. Let ex: [a, b] -+1R2 - to} be smooth and such that ex(a) = ex(b~ Suppose ex hits
the positive x l-axis only finitely many times. Show that the winding number
k(ex) is equal to the algebraic number of crossings of the positive xl-axis by ex,
where each upward crossing is counted positively and each downward cross-
ing is counted negatively. (Hint: Let tl < t2 < ... < t", be the set of all
t E [a, b] such that ex(t) lies on the positive xl-axis. Let V and 8y be asin our
discussion of the winding number, with v = (1, O~ Then
k(ex) = f lim
'=0 ,"'0
(1+1-'
,,+£
d8~(<«t)) dt
where to = a and t", + 1 = b. (If t 1 = a, the sum will range only from 1 to
m-1.)] .
11.22. Let ex: 1 -+1R2 be a piecewise smooth closed parametrized curve. For
p = (a, b) E H2 - Image ex,define
k ()
J1 ex =
f
II
-(X2 - b) dx + l a)dx2
(Xl"';'
(Xl -a)2 +(xz -:" b)2. .
(a) Show that k,(ex) is an integer. [Hint: Show that k,(ex) is the winding
number of fJ: 1 -+ Hl - to} where fJ(t) = ex(t) - p.] .
(b) Show that if p and q E IR z - Image ex can bejoined by a continuous curve
in'H2 -Image ex thenk,(ex) = kf(lX~
(The integer k,(cx) is the winding number of ex about p.)