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the area of parallelogram spanned by V1 and V2 pro-
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

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