jected onto the plane containing the vectors ω and

ν , and multiplied by the area of the parallelogram

spanned by ω and ν .

CAUTION: While every 1-form can be thought of as projected length not ev-

ery 2-form can be thought of as projected area. The only 2-forms for which this

interpretation is valid are those that are the product of 1-forms. See Exercise 2.15.

Let™s pause for a moment to look at a particularly simple 2-form on Tp R3 , dx§dy.

Suppose V1 = a1 , a2 , a3 and V2 = b1 , b2 , b3 . Then

a1 a2

dx § dy(V1, V2 ) =

b1 b2

This is precisely the (signed) area of the parallelogram spanned by V1 and V2 projected

onto the dx-dy plane.

Exercise 2.11. ω § ν( a1 , a2 , a3 , b1 , b2 , b3 ) = c1 dx § dy + c2 dx § dz + c3 dy § dz, for

some real numbers, c1 , c2 , and c3 .

26 2. FORMS

The preceding comments, and this last exercise, give the following geometric

interpretation of the action of a 2-form on the pair of vectors, (V1 , V2 ):

Every 2-form projects the parallelogram spanned by V1

and V2 onto each of the (2-dimensional) coordinate

planes, computes the resulting (signed) areas, multi-

plies each by some constant, and adds the results.

This interpretation holds in all dimensions. Hence, to specify a 2-form we need to

know as many constants as there are 2-dimensional coordinate planes. For example,

to give a 2-form in 4-dimensional Euclidean space we need to specify 6 numbers:

c1 dx § dy + c2 dx § dz + c3 dx § dw + c4 dy § dz + c5 dy § dw + c6 dz § dw

The skeptic may argue here. Exercise 2.11 only shows that a 2-form which is a

product of 1-forms can be thought of as a sum of projected, scaled areas. What about

an arbitrary 2-form? Well, to address this we need to know what an arbitrary 2-form

is! Up until now we have not given a complete de¬nition. Henceforth, we shall de¬ne a

2-form to be a bi-linear, skew-symmetric, real-valued function on Tp Rn —Tp Rn . That™s

a mouthful. This just means that it™s an operator which eats pairs of vectors, spits

out real numbers, and satis¬es the conclusions of Exercises 2.5 and 2.7. Since these

are the only ingredients necessary to do Exercise 2.11 our geometric interpretation

is valid for all 2-forms.

Exercise 2.12. If ω( dx, dy, dz ) = dx+5dy’dz, and ν( dx, dy, dz ) = 2dx’dy+dz,

compute

ω § ν( 1, 2, 3 , ’1, 4, ’2 )

Answer: ’127

Exercise 2.13. Let ω( dx, dy, dz ) = dx+5dy’dz and ν( dx, dy, dz ) = 2dx’dy+dz.

Find constants, c1 , c2 , and c3 , such that

ω § ν = c1 dx § dy + c2 dy § dz + c3 dx § dz

Answer: c1 = ’11, c2 = 4, and c3 = 3

Exercise 2.14. Express each of the following as the product of two 1-forms:

4. 2-FORMS ON Tp R3 (OPTIONAL) 27

(1) 3dx § dy + dy § dx

(2) dx § dy + dx § dz

(3) 3dx § dy + dy § dx + dx § dz

(4) dx § dy + 3dz § dy + 4dx § dz

4. 2-forms on Tp R3 (optional)

Exercise 2.15. Find a 2-form which is not the product of 1-forms.

In doing this exercise you may guess that in fact all 2-forms on Tp R3 can be

written as a product of 1-forms. Let™s see a proof of this fact that relies heavily on

the geometric interpretations we have developed.

Recall the correspondence introduced above between vectors and 1-forms. If

± = a1 dx + a2 dy + a3 dz then we let ± = a1 , a2 , a3 . If V is a vector then we let

V ’1 be the corresponding 1-form.

We now prove two lemmas:

Lemma 2.1. If ± and β are 1-forms on Tp R3 and V is a vector in the plane

spanned by ± and β then there is a vector, W , in this plane such that ± § β =

’1 ’1

§W

V .

Proof. The proof of the above lemma relies heavily on the fact that 2-forms

which are the product of 1-forms are very ¬‚exible. The 2-form ± § β takes pairs

of vectors, projects them onto the plane spanned by the vectors ± and β , and

computes the area of the resulting parallelogram times the area of the parallelogram

spanned by ± and β . Note that for every non-zero scalar c the area of the

parallelogram spanned by ± and β is the same as the area of the parallelogram

spanned by c ± and 1/c β . (This is the same thing as saying that ±§β = c±§1/cβ).

The important point here is that we can scale one of the 1-forms as much as we want

at the expense of the other and get the same 2-form as a product.

Another thing we can do is apply a rotation to the pair of vectors ± and β in

the plane which they determine. As the area of the parallelogram spanned by these

two vectors is unchanged by rotation, their product still determines the same 2-form.

In particular, suppose V is any vector in the plane spanned by ± and β . Then

we can rotate ± and β to ±′ and β ′ so that c ±′ = V , for some scalar c. We

28 2. FORMS

can then replace the pair ( ± , β ) with the pair (c ±′ , 1/c β ′ ) = (V, 1/c β ′ ). To

complete the proof, let W = 1/c β ′ .

Lemma 2.2. If ω1 = ±1 § β1 and ω2 = ±2 § β2 are 2-forms on Tp R3 then there

exists 1-forms, ±3 and β3 , such that ω1 + ω2 = ±3 § β3 .

Proof. Let™s examine the sum, ±1 § β1 + ±2 § β2 . Our ¬rst case is that the

plane spanned by the pair ( ±1 , β1 ) is the same as the plane spanned by the

pair, ( ±2 , β2 ). In this case it must be that ±1 § β1 = C±2 § β2 , and hence,

±1 § β1 + ±2 § β2 = (1 + C)±1 § β1 .

If these two planes are not the same then they intersect in a line. Let V be a

vector contained in this line. Then by the preceding lemma there are 1-forms γ and

γ ′ such that ±1 § β1 = V ’1 ’1

§ γ ′ . Hence,

§ γ and ±2 § β2 = V

’1 ’1

§ γ′ = V ’1

§ (γ + γ ′ )

±1 § β1 + ±2 § β2 = V §γ+ V

Now note that any 2-form is the sum of products of 1-forms. Hence, this last

lemma implies that any 2-form on Tp R3 is a product of 1-forms. In other words:

Every 2-form on Tp R3 projects pairs of vectors onto

some plane and returns the area of the resulting par-

allelogram, scaled by some constant.

This fact is precisely why all of classical vector calculus works. We explore this

in the next few exercises, and further in Section 4 of Chapter 5.

Exercise 2.16. Use the above geometric interpretation of the action of a 2-form on

Tp R3 to justify the following statement: For every 2-form ω on Tp R3 there are non-zero

vectors V1 and V2 such that V1 is not a multiple of V2 , but ω(V1 , V2 ) = 0.

Exercise 2.17. Does Exercise 2.16 generalize to higher dimensions?

Exercise 2.18. Show that if ω is a 2-form on Tp R3 then there is a line l in Tp R3 such

that if the plane spanned by V1 and V2 contains l then ω(V1 , V2 ) = 0.

Note that the conditions of Exercise 2.18 are satis¬ed when the vectors that are

perpendicular to both V1 and V2 are also perpendicular to l.

5. N-FORMS 29

Exercise 2.19. Show that if all you know about V1 and V2 is that they are vectors in

Tp R3 that span a parallelogram of area A, then the value of ω(V1 , V2 ) is maximized when

V1 and V2 are perpendicular to the line l of Exercise 2.18.

Note that the conditions of this exercise are satis¬ed when the vectors perpen-

dicular to V1 and V2 are parallel to l.

Exercise 2.20. Let N be a vector perpendicular to V1 and V2 in Tp R3 whose length

is precisely the area of the parallelogram spanned by these two vectors. Show that there

is a vector Vω in the line l of Exercise 2.18 such that the value of ω(V1 , V2 ) is precisely

Vω · N.

Remark. You may have learned that the vector N of the previous exercise is

precisely the cross product of V1 and V2 . Hence, the previous exercise implies that if

ω is a 2-form on Tp R3 then there is a vector Vω such that ω(V1, V2 ) = Vω · (V1 — V2 )

Exercise 2.21. Show that if ω = Fx dy § dz ’ Fy dx § dz + Fz dx § dy then Vω =

F1 , F2 , F3 .

5. n-forms

Let™s think a little more about our multiplication, §. If it™s really going to be

anything like multiplication we should be able to take three 1-forms, ω, ν, and ψ,

and form the product ω § ν § ψ. How can we de¬ne this? A ¬rst guess might be

to say that ω § ν § ψ = ω § (ν § ψ), but ν § ψ is a 2-form and we haven™t de¬ned

the product of a 2-form and a 1-form. We™re going to take a di¬erent approach and

de¬ne ω § ν § ψ directly.

This is completely analogous to the previous section. ω, ν, and ψ each act on

a vector, V , to give three numbers. In other words, they can be thought of as

coordinate functions. We say the coordinates of V are [ω(V ), ν(V ), ψ(V )]. Hence,

if we have three vectors, V1 , V2 , and V3 , we can compute the [ω, ν, ψ] coordinates of

each. This gives us three new vectors. The signed volume of the parallelepiped which

they span is what we de¬ne to be the value of ω § ν § ψ(V1 , V2 , V3 ).

There is no reason to stop at 3-dimensions. Suppose ω1 , ω2 , ..., ωn are 1-forms and

V1 , V2 , ..., Vn are vectors. Then we de¬ne the value of ω1 § ω2 § ... § ωn (V1 , V2 , ..., Vn )

30 2. FORMS

to be the signed (n-dimensional) volume of the parallelepiped spanned by the vectors

[ω1 (Vi ), ω2 (Vi ), ..., ωn (Vi )]. Algebraically,

ω1 (V1 ) ω2 (V1 ) ... ωn (V1 )

ω1 (V2 ) ω2 (V2 ) ... ωn (V2 )

ω1 § ω2 § ... § ωn (V1 , V2 , ..., Vn ) = . . .

. . .

. . .

ω1 (Vn ) ω2 (Vn ) ... ωn (Vn )

It follows from linear algebra that if we swap any two rows or columns of this

matrix the sign of the result ¬‚ips. Hence, if the n-tuple, V′ = (Vi1 , Vi2 , ..., Vin ) is

obtained from V = (V1 , V2 , ..., Vn ) by an even number of exchanges then the sign of

ω1 § ω2 § ... § ωn (V′) will be the same as the sign of ω1 § ω2 § ... § ωn (V). If the

number of exchanges were odd then the sign would be opposite. We sum this up by

saying that the n-form, ω1 § ω2 § ... § ωn is alternating.

The wedge product of 1-forms is also multilinear, in the following sense:

ω1 § ω2 § ... § ωn (V1 , ..., Vi + Vi′ , ..., Vn )

= ω1 § ω2 § ... § ωn (V1 , ..., Vi , ..., Vn ) + ω1 § ω2 § ... § ωn (V1 , ..., Vi′ , ..., Vn ),

and

ω1 § ω2 § ... § ωn (V1 , ..., cVi, ..., Vn ) = cω1 § ω2 § ... § ωn (V1 , ..., Vi, ..., Vn ),

for all i and any real number, c.

In general, we de¬ne an n-form to be any alternating, multilinear real-valued

function which acts on n-tuples of vectors.

Exercise 2.22. Prove the following geometric interpretation: Hint: All of the steps

are completely analogous to those in the last section.

An m-form on TpRn can be thought of as a function

which takes the parallelepiped spanned by m vectors,

projects it onto each of the m-dimensional coordinate

planes, computes the resulting areas, multiplies each

by some constant, and adds the results.

Exercise 2.23. How many numbers do you need to give to specify a 5-form on Tp R10 ?

5. N-FORMS 31

We turn now to the simple case of an n-form on Tp Rn . Notice that there is only

one n-dimensional coordinate plane in this space, namely, the space itself. Such

a form, evaluated on an n-tuple of vectors, must therefore give the n-dimensional

volume of the parallelepiped which it spans, multiplied by some constant. For this

reason such a form is called a volume form (in 2-dimensions, an area form).

Example 2.2. Consider the forms, ω = dx + 2dy ’ dz, ν = 3dx ’ dy + dz,

and ψ = ’dx ’ 3dy + dz, on Tp R3 . By the above argument ω § ν § ψ must

be a volume form. But which volume form is it? One way to tell is to compute

its value on a set of vectors which we know span a parallelepiped of volume 1,

namely 1, 0, 0 , 0, 1, 0 , and 0, 0, 1 . This will tell us how much the form scales

volume.

3 ’1

1

’1 ’3 = 4

ω § ν § ψ( 1, 0, 0 , 0, 1, 0 , 0, 0, 1 ) = 2

’1 1 1

So, ω § ν § ψ must be the same as the form 4dx § dy § dz.

Exercise 2.24. Let ω( dx, dy, dz ) = dx + 5dy ’ dz, ν( dx, dy, dz ) = 2dx ’ dy + dz,

and γ( dx, dy, dz) = ’dx + dy + 2dz.

(1) If V1 = 1, 0, 2 , V2 = 1, 1, 2 , and V3 = 0, 2, 3 , compute ω §ν §γ(V1 , V2 , V3 ).

Answer: ’87

(2) Find a constant, c, such that ω § ν § γ = c dx § dy § dz.

Answer: c = ’29

(3) Let ± = 3dx § dy + 2dy § dz ’ dx § dz. Find a constant, c, such that

± § γ = c dx § dy § dz.

Answer: c = 5

Exercise 2.25. Simplify:

dx § dy § dz + dx § dz § dy + dy § dz § dx + dy § dx § dy

Exercise 2.26. Let ω be an n-form and ν an m-form. Show that

ω § ν = (’1)nm ν § ω

CHAPTER 3

Di¬erential Forms

1. Families of forms

Let™s now go back to the example in Chapter 1. In the last section of that chapter

we showed that the integral of a function, f : R3 ’ R, over a surface parameterized

by φ : R ‚ R2 ’ R3 is

‚φ ‚φ

f (φ(r, θ))Area (r, θ), (r, θ) drdθ

‚r ‚θ

R

This was one of the motivations for studying di¬erential forms. We wanted to

generalize this integral by considering functions other than “Area(·, ·)” which eat

pairs of vectors and return numbers. But in this integral the point at which such a