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3. Interlude: A review of single variable integration

In order to understand what happened, we must ¬rst review the de¬nition of
the Riemann integral. In the usual de¬nition of the Riemann integral, the ¬rst step
b
is to divide the interval up into n evenly spaced subintervals. Thus, f (x)dx is
a
n
de¬ned to be the limit, as n ’ ∞, of f (xi )∆x, where {xi } are n evenly spaced
i=1
points in the interval [a, b], and ∆x = (b ’ a)/n. But what if the points {xi } are
n
not evenly spaced? We can still write down a reasonable sum: f (xi )∆xi , where
i=1
now ∆xi = xi+1 ’ xi . In order to make the integral well de¬ned, we can no longer
take the limit as n ’ ∞. Instead, we must let max{∆xi } ’ 0. It is a basic result
of analysis that if this limit converges, then it does not matter how we picked the
points {xi }; the limit will converge to the same number. It is this number that we
b
de¬ne to be the value of f (x)dx.
a


4. What went wrong?
1
We are now ready to ¬gure out what happened in section 2. Obviously, f (φ1 (a))da
’1
was not what we wanted. But let™s not give up on our general approach just yet: it
would still be great if we could use φ1 to ¬nd some function, that we can integrate on
[’1, 1], that will give us the same answer as the integral of f over M. For now, let™s
call this mystery function “?(a)”. We™ll ¬gure out what it has to be in a moment.
12 1. INTRODUCTION

f (φ(a2 ))
f
l3
l2
?(a2 ) M
L2 L3
l1
φ l4
? L1 L4

’1 a2 0 1
∆a
Figure 2. We want ?(a1 )∆a+?(a2 )∆a+?(a3 )∆a+?(a4 )∆a =
f (φ(a1 ))L1 + f (φ(a2 ))L2 + f (φ(a3 ))L3 + f (φ(a4 ))L4 .

1
Let™s look at the Riemann sum that we get for ?(a)da, when we divide the
’1
n
interval up into n pieces, each of width ∆a = 2/n. We get ?(ai )∆a, where ai =
i=1
’1+2/n. Examine Figure 2 to see what happens to the points, ai , under the function,
φ1 , for n = 4. Notice that the points {φ1 (ai )} are not evenly spaced along M. To use
these points to estimate the integral of f over M, we would have to use the approach
from the previous section. A Riemann sum for f over M would be:
4
f (φ1(ai ))li
i=1

= f (’1, 0)l1 + f (’1/2, 3/4)l2 + f (0, 1)l3 + f (1/2, 3/4)l4
= (0)l1 + (3/4)l2 + (0)l3 + (3/4)l4


The li represent the arc length, along M, between φ1 (ai ) and φ1 (ai+1 ). This is
a bit problematic, however, since arc-length is generally hard to calculate. Instead,
we can approximate li by substituting the length of the line segment which connects
φ1 (ai ) to φ1 (ai+1 ), which we shall denote as Li . Note that this approximation gets
better and better as we let n ’ ∞. Hence, when we take the limit, it does not
matter if we use li or Li .
So our goal is to ¬nd a function, ?(a), on the interval [’1, 1], so that the Riemann
4
sum, ?(ai )∆a equals (0)L1 + (3/4)L2 + (0)L3 + (3/4)L4 . In general, we want
i=1
4. WHAT WENT WRONG? 13

n n
f (φ1 (ai ))Li = ?(ai )∆a. So, we must have ?(ai )∆a = f (φ1 (ai ))Li . Solving, we
i=1 i=1
f (φ1 (ai ))Li
get ?(ai ) = .
∆a
What happens to this function as ∆a ’ 0? First, note that Li = |φ1 (ai+1 ) ’
φ1 (ai )|. Hence,

f (φ1(ai ))Li
lim ?(ai ) = lim
∆a
∆a’0 ∆a’0
f (φ1(ai ))|φ1 (ai+1 ) ’ φ1 (ai )|
= lim
∆a
∆a’0
|φ1 (ai+1 ) ’ φ1 (ai )|
= f (φ1 (ai )) lim
∆a
∆a’0
φ1 (ai+1 ) ’ φ1 (ai )
= f (φ1 (ai )) lim
∆a
∆a’0

But lim∆a’0 φ1 (ai+1 )’φ1 (ai ) is precisely the de¬nition of the derivative of φ1 at ai ,
∆a
Dai φ1 . Hence, we have lim∆a’0 ?(ai ) = f (φ1(ai ))|Dai φ1 |. Finally, this means that
1
f (φ1 (a))|Da φ1 |da, which should be a familiar
the integral we want to compute is
’1
integral from calculus.
1 π
f (φ1 (a))|Da φ1 |da = f (φ2 (t))|Dt φ2 |dt, using the func-
Exercise 1.1. Check that
’1 0
tion, f , de¬ned in section 2.

Recall that Da φ1 is a vector, based at the point φ(a), tangent to M. If we think
of a as a time parameter, then the length of Da φ1 tells us how fast φ1 (a) is moving
1
f (φ1 (a))|Da φ1 |da? Note that the
along M. How can we generalize the integral,
’1
bars | · | are a function which “eats” vectors, and “spits out” real numbers. So we can
generalize the integral by looking at other such functions. In other words, a more
1
general integral would be f (φ1 (a))ω(Da φ1 )da, where f is a function of points and
’1
ω is a function of vectors.
It is not the purpose of the present work to undertake a study of integrating with
all possible functions, ω. However, as with the study of functions of real variables,
a natural place to start is with linear functions. This is the study of di¬erential
forms. A di¬erential form is precisely a linear function which eats vectors, spits out
14 1. INTRODUCTION

numbers, and is used in integration. The strength of di¬erential forms lies in the fact
that their integrals do not depend on a choice of parameterization.

5. What about surfaces?

Let™s repeat the previous discussion (faster this time), bumping everything up a
dimension. Let f : R3 ’ R be given by f (x, y, z) = z 2 . Let M be the top half of the
sphere of radius 1, centered at the origin. We can parameterize M by the function,

φ, where φ(r, θ) = (r cos(θ), r sin(θ), 1 ’ r 2 ), 0 ¤ r ¤ 1, and 0 ¤ θ ¤ 2π. Again,
our goal is not to ¬gure out how to actually integrate f over M, but to use φ to set
up an equivalent integral over the rectangle, R = [0, 1] — [0, 2π].
Let {xi,j } be a lattice of evenly spaced points in R. Let ∆r = xi+1,j ’ xi,j , and
∆θ = xi,j+1 ’ xi,j . By de¬nition, the integral over R of a function, ?(x), is equal to
lim∆r,∆θ’0 ?(xi,j )∆r∆θ.
To use the mesh of points, φ(xi,j ), in M to set up a Riemann-Stiljes sum, we write
down the following sum: f (φ(xi,j ))Area(Li,j ), where Li,j is the rectangle spanned
by the vectors φ(xi+1,j ) ’ φ(xi,j ), and φ(xi,j+1) ’ φ(xi,j ). If we want our Riemann
f (φ(xi,j ))Area(Li,j )
sum over R to equal this sum, then we end up with ?(xi,j ) = .
∆r∆θ


R ‚φ
(x3,1 )
θ ‚r

2π φ
φ(x3,1 )

‚φ
(x3,1 )
‚θ


r
1
x3,1

)
Area(L
i,j
We now leave it as an exercise to show that as ∆r and ∆θ get small, ∆r∆θ con-
verges to the area of the parallelogram spanned by the vectors ‚φ (xi,j ), and ‚φ (xi,j ).
‚r ‚θ
The upshot of all this is that the integral we want to evaluate is the following:

‚φ ‚φ
f (φ(r, θ))Area (r, θ), (r, θ) drdθ
‚r ‚θ
R
5. WHAT ABOUT SURFACES? 15

Exercise 1.2. Compute the value of this integral for the function f (x, y, z) = z 2 .

The point of all this is not the speci¬c integral that we have arrived at, but the
form of the integral. We are integrating f —¦ φ (as in the previous section), times a
function which takes two vectors and returns a real number. Once again, we can
generalize this by using other such functions:

‚φ ‚φ
f (φ(r, θ))ω (r, θ), (r, θ) drdθ
‚r ‚θ
R
In particular, if we examine linear functions for ω, we arrive at a di¬erential form.
The moral is that if we want to perform an integral over a region parameterized by
R, as in the previous section, then we need to multiply by a function which takes a
vector and returns a number. If we want to integrate over something parameterized
by R2 , then we need to multiply by a function which takes two vectors and returns a
number. In general, an n-form is a linear function which takes n vectors, and returns
a real number. One integrates n-forms over regions that can be parameterized by
Rn .
CHAPTER 2


Forms

1. Coordinates for vectors

Before we begin to discuss functions on vectors we ¬rst need to learn how to
specify a vector. And before we can answer that we must ¬rst learn where vectors
live. In Figure 1 we see a curve, C, and a tangent line to that curve. The line can
be thought of as the set of all tangent vectors at the point, p. We denote that line
as Tp C, the tangent space to C at the point p.

T Cp

C


p




Figure 1. Tp C is the set of all vectors tangents to C at p.


What if C was actually a straight line? Would Tp C be the same line? To answer
this, let™s put down some coordinates. Suppose C were a straight line, with coordi-
nates, and p is the point corresponding to the number 5. Now, suppose you were to
draw a tangent vector to C, of length 2, which is tangent at p. Where would you
draw it? Would you put it™s base at 0 on C? Of course not...you™d put it™s base at
p = 5. So the origin for Tp C is in a di¬erent place as the origin for C. This is because
17
18 2. FORMS

we are thinking of C and Tp C as di¬erent lines, even though one may be right on
top of the other.
Let™s pause here for a moment to look at something a little more closely. What
did we really do when we chose coordinates for C? What are “coordinates” anyway?
They are a way of assigning a number (or, more generally, a set of numbers) to a
point in our space. In other words, coordinates are functions which take points of a
space and return (sets of) numbers. When we say that the x-coordinate of p is 5 we
really mean that we have a function, x : C ’ R, such that x(p) = 5.
What about points in the plane? Of course we need two numbers to specify such
a point, which means that we have two coordinate functions. Suppose we denote
the plane by P and x : P ’ R and y : P ’ R are our coordinate functions. Then
saying that the coordinates of a point, p, are (2, 3) is the same thing as saying that
x(p) = 2, and y(p) = 3. In other words, the coordinates of p are (x(p), y(p)).
So what do we use for coordinates in the tangent space? Well, ¬rst we need a
basis for the tangent space of P at p. In other words, we need to pick two vectors
which we can use to give the relative positions of all other points. Note that if
the coordinates of p are (x, y) then d(x+t,y) = 1, 0 , and d(x,y+t) = 0, 1 . We have
dt dt
changed to the notation “ ·, · ” to indicate that we are not talking about points of
P anymore, but rather vectors in Tp P . We take these two vectors to be a basis for
Tp P . In other words, any point of Tp P can be written as dx 0, 1 + dy 1, 0 , where
dx, dy ∈ R. Hence, “dx” and “dy” are coordinate functions for Tp P . Saying that
the coordinates of a vector V in Tp P are 2, 3 , for example, is the same thing as
saying that dx(V ) = 2 and dy(V ) = 3. In general we may refer to the coordinates of
an arbitrary vector in Tp P as dx, dy , just as we may refer to the coordinates of an
arbitrary point in P as (x, y).

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