5.1 Motivation

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5.1.1 Functions as transformations

I shall discuss maps f : I −! R2 and describe a geometric way of visualising

them, much used by topologists, which involves thinking about I = {x 2

R : 0 _ x _ 1} as if it were a piece of chewing gum1. We can use the same

way of thinking about functions f : R −! R. You are used to thinking

of such functions geometrically by visualising the graph, which is another

way of getting an intuitive grip on functions. Thinking of stretching and

deforming the domain and putting it in the codomain has the advantage

that it generalises to maps from R to Rn, from R2 to Rn and from R3 to

Rn for n = 1, 2 or 3. It is just a way of visualising what is going on, and

although Topologists do this, they don’t usually talk about it. So I may be

breaking the Topologist’s code of silence here.

Example 5.1.1. f(x) , 2x can be thought of as taking the real line, regarded

as a ruler made of chewing gum and stretching it uniformly and

moving it across to a second ruler made of, let’s say, wood. See figure 5.1 for

a rather bad drawing of this.

Example 5.1.2. f(x) , −x just turns the chewing gum ruler upside down

and doesn’t stretch it (in the conventional sense) at all. The stretch factor

1If, like the government of Singapore. you don’t like chewing gum, substitute putty

or plasticene. It just needs to be something that can be stretched and won’t spring back

when you let go

49

50 CHAPTER 5. GREEN’S THEOREM

Figure 5.1: The function f(x) , 2x thought of as a stretching.

is −1.

Example 5.1.3. f(x) , x+1 just shifts the chewing gum ruler up one unit

and again doesn’t do any stretching, or alternatively the stretch factor is 1.

Draw your own pictures.

Example 5.1.4. f(x) , |x| folds the chewing gume ruler about the origin

so that the negative half fits over the positive half; the stretch factor is 1

when x > 0 and −1 when x < 0 See figure 5.2 for a picture of this map in

chewing gum language.

Example 5.1.5. f(x) , x2 This function is more interesting because the

amount of stretch is not uniform. Near zero it is a compression: the interval

between 0 and 0.5 gets sent to the interval from 0 to 0.25 so the stretch is

one half over this interval. Whereas the interval between 1 and 2 is sent to

the interval between 1 and 4, so the stretch factor over the interval is three.

I won’t try to draw it, but it is not hard.

Remark 5.1.1. Now I want to show you, with a particular example, how

quite a lot of mathematical ideas are generated. It is reasonable to say of the

function f(x) , x2 that the amount of stretch depends on where you are. So

I would like to define the amount of stretch a function f : R −! R does at

a point. I shall call this S(f, a) where a 2 R.

5.1. MOTIVATION 51

Figure 5.2: The function f(x) , |x| in transformation terms.

It is plausible that this is a meaningful idea, and I can say it in English easily

enough. A mathematician is someone who believes that it has to be said in

Algebra before it really makes sense.

How then can we say it in algebra? We can agree that it is easy to define the

stretch factor for an interval, just divide the length after by the length before,

and take account of whether it has been turned upside down by putting in a

minus sign if necessary. To define it at a point, I just need to take the stretch

factor for small intervals around the point and take the limit as the intervals

get smaller.

Saying this in algebra:

S(f, a) , lim

_!0

f(a + _) − f(a)

_

Remark 5.1.2. This looks like a sensible definition. It may look familiar.

Remark 5.1.3. Suppose I have two maps done one after the other:

R g

−! R f

−! R

If S(g, a) = 2 and S(f, g(a)) = 3 then it is obvious that S(f _ g, a) = 6. In

general it is obvious that

S(f _ g, a) = (S(g, a))(S(f, g(a)))

52 CHAPTER 5. GREEN’S THEOREM

Figure 5.3: The change of variable formula.

Note that this takes account of the sign without our having to bother about

it explicitly.

Remark 5.1.4. You may recognise this as the chain rule and S(f, a) as

being the derivative of f at a. So you now have another way of thinking

about derivatives. The more ways you have of thinking about something the

better: some problems are very easy if you think about them the right way,

the hard part is finding it.

5.1.2 Change of Variables in Integration

Remark 5.1.5. This way of thinking makes sense of the change of variables

formula in integration, something which you may have merely memorised.

Suppose we have the problem of integrating some function f : U −! R

where U is an interval. I shall write [a, b] for the interval {x 2 R : a _ x _ b}

So the problem is to calculate

Z b

a

f(x) dx

Now suppose I have a function g : I −! [a, b] which is differentiable and, to

make things simpler, 1-1 and takes 0 to a and 1 to b. This is indicated in the

diagram figure 5.3

5.1. MOTIVATION 53

The integral is defined to be the limit of the sum of the areas of little boxes

sitting on the segment [a, b]. I have shown some of the boxes. We can

pull back the function f (expressed via its graph) to f _ g which is in a

sense the “same” function– well, it has got itself compressed, in general by

different amounts at different places, because g stretches I to the (longer in

the picture) interval [a, b].

Now the integral of f _ g over I is obviously related to the integral of f over

[a, b]. If we have a little box at t 2 I, the height of the function f _ g at t is

exactly the same as the height of f over g(t). But if the width of the box at

t is _t, it gets stretched by an amount which is approximately g0(t) in going

to [a, b]. So the area of the box on g(t), which is what g does to the box at t,

is approximately the area of the box at t multiplied by g0(t). And since this

holds for all the little boxes no matter how small, and the approximation

gets better as the boxes get thinner, we deduce that it holds for the integral:

Z 1

0

f _ g(t) g0(t) dt =

Z b

a

f(x) dx

This is the change of variable formula. It actually works even when g is not

1-1, since if g retraces its path and then goes forward again, the backward

bit is negative and cancels out the first forward bit. Of course, g has to be

differentiable or the formula makes no sense.2

When you do the integral

Z _/2

0

sin(t) cos(t) dt

by substituting x = sin(t), dx = cos(t) dt to get

Z 1

0

x dx

you are doing exactly this “stretch factor” trick. In this case g is the function

that takes t to x = sin(t); it takes the interval from 0 to _/2 to the interval

I (thus compressing it) and the function y = x pulls back to the function

y = sin(t) over [0, _/2]. The stretching factor is dx = cos(t) dt and is taken

care of by the differentials. We shall see an awful lot of this later on in the

course.

2g could fail to be differentiable at a finite number of points; we could cut the path

up into little bits over which the formula makes sense and works. At the points where it

doesn’t, well, we just ignore them, because after all, how much are they going to contribute

to the integral? Zero is the answer to that, so the hell with them.

54 CHAPTER 5. GREEN’S THEOREM

Figure 5.4: A vector field or 1-form with positive “twist”.

5.1.3 Spin Fields

Remark 5.1.6. I hope that you can see that thinking about functions as

stretching intervals and having an amount of stretch at a point is useful: it

helps us understand otherwise magical formulae. Now I am ready to use the

kind of thinking that we went through in defining the amount of stretch of a

function at a point. Instead, I shall be looking at differential 1-forms or R2

and looking at the amount of “twist” the 1-form may have at a point of R2.

The 1-form −ydx + xdy clearly has some, see figure 5.4

Remark 5.1.7. Think about a vector field or differential 1-form on R2 and

imagine it is the velocity field of a moving fluid. Now stick a tiny paddle

wheel in at a point so as to measure the “rotation” or “twist” or “spin” at a

point.

This idea, like the amount of stretch of a function f : R −! R is vague, but

we can try to make it precise by saying it in algebra.

Remark 5.1.8. If

V

_

x

y

_

, P (x, y) dx + Q(x, y) dy

is the 1-form, look first at Q(x, _ y) along the horizontal line through the point

a

b

_

5.1. MOTIVATION 55

Figure 5.5: The amount of rotation of a vector field

Figure 5.6: _Q _ @Q

@x _x

If 4x is small, the Q component to the right of

_

a

b

_

is

Q

_

a

b

_

+

@Q

@x

4x

and to the left is

Q

_

a

b

_

−

@Q

@x

4x

and the spin per unit length about

_

a

b

_

in the positive direction is

@Q

@x

Similarly there is a tendency to twist in the opposite direction given by

@P

@y

56 CHAPTER 5. GREEN’S THEOREM

Figure 5.7: Path integral around a small square

So the total spin can be defined as

@Q

@x

−

@P

@y

This is a function from R2 to R.

Example 5.1.6.

! , x2y dx + 3y2x dy

spin(!) = 3y2 − 2xy

Example 5.1.7.

! , −y dx + x dy

spin(!) = 2

Where 2 is the constant function.

Remark 5.1.9. Another way of making the idea of ‘twist’ at a point precise

would be to take the integral around a little square centred on the point and

divide by the area of the square.

If the square, figure 5.7 has side 24 we go around each side.

From

_

a +4

b −4

_

to

_

a +4

b +4

_

we need consider only the Q-component which

at the midpoint is approximately

5.2. GREEN’S THEOREM (CLASSICAL VERSION) 57

Q

_

a

b

_

+

@Q

@x

4

and so the path integral for this side is approximately

_

Q

_

a

b

_

+

_

@Q

@x

_

4

_

24.

The side from

_

a +4

b +4

_

to

_

a −4

b +4

_

is affected only by the P component

and the path integral for this part is approximately

_

P

_

a

b

_

+

@P

@y

4

_

(−24)

with the minus sign because it is going in the negative direction.

Adding up the contribution from the other two sides we get

_

@Q

@x

−

@P

@y

_

442

and dividing by the area (442) we get

spin(V) ,

_

@Q

@x

−

@P

@y

_

again.

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