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It will be helpful in the future to be able to distinguish between the vector 2, 3
in Tp P and the vector 2, 3 in Tq P , where p = q. We will do this by writing 2, 3 p
for the former and 2, 3 q for the latter.
Let™s pause for a moment to address something that may have been bothering
you since your ¬rst term of calculus. Let™s look at the tangent line to the graph of
y = x2 at the point (1, 1). We are no longer thinking of this tangent line as lying
in the same plane that the graph does. Rather, it lies in T(1,1) R2 . The horizontal
2. 1-FORMS 19

y

dy
l



1 dx




x
1


Figure 2. The line, l, lies in T(1,1) R2 . Its equation is dy = 2dx.


axis for T(1,1) R2 is the “dx” axis and the vertical axis is the “dy” axis (see Fig. 2).
Hence, we can write the equation of the tangent line as dy = 2dx. We can rewrite
dy
this as dx = 2. Look familiar? This is one explanation of why we use the notation
dy
in calculus to denote the derivative.
dx


Exercise 2.1.

(1) Draw a vector with dx = 1, dy = 2, in the tangent space T(1,’1) R2 .
(2) Draw ’3, 1 (0,1) .


2. 1-forms

Recall from the previous chapter that a 1-form is a linear function which acts
on vectors and returns numbers. For the moment let™s just look at 1-forms on Tp R2
for some ¬xed point, p. Recall that a linear function, ω, is just one whose graph is
a plane through the origin. Hence, we want to write down an equation of a plane
though the origin in Tp R2 — R, where one axis is labelled dx, another dy, and the
third, ω (see Fig. 3). This is easy: ω = a dx + b dy. Hence, to specify a 1-form on
Tp R2 we only need to know two numbers: a and b.
20 2. FORMS

ω




dy




dx

Figure 3. The graph of ω is a plane though the origin.


Here™s a quick example: Suppose ω( dx, dy ) = 2dx + 3dy then

ω( ’1, 2 ) = 2 · ’1 + 3 · 2 = 4.

The alert reader may see something familiar here: the dot product. That is, ω( ’1, 2 ) =
2, 3 · ’1, 2 . Recall the geometric interpretation of the dot product; you project

’1, 2 onto 2, 3 and then multiply by | 2, 3 | = 13. In other words

Evaluating a 1-form on a vector is the same as pro-
jecting onto some line and then multiplying by some
constant.

In fact, we can even interpret the act of multiplying by a constant geometrically.
Suppose ω is given by a dx + b dy. Then the value of ω(V1 ) is the length of the
a,b
projection of V1 onto the line, l, where | a,b |2 is a basis vector for l.
This interpretation has a huge advantage... it™s coordinate free. Recall from the
previous section that we can think of the plane, P , as existing independent of our
choice of coordinates. We only pick coordinates so that we can communicate to
someone else the location of a point. Forms are similar. They are objects that exist
2. 1-FORMS 21

independent of our choice of coordinates. This is one of the keys as to why they are
so useful outside of mathematics.
There is still another geometric interpretation of 1-forms. Let™s ¬rst look at the
simple example ω( dx, dy ) = dx. This 1-form simply returns the ¬rst coordinate of
whatever vector you feed into it. This is also a projection; it™s the projection of the
input vector onto the dx-axis. This immediately gives us a new interpretation of the
action of a general 1-form, ω = a dx + b dy.


Evaluating a 1-form on a vector is the same as pro-
jecting onto each coordinate axis, scaling each by some
constant, and adding the results.

Although this interpretation is a little more cumbersome it™s the one that will
generalize better when we get to n-forms.
Let™s move on now to 1-forms in n dimensions. If p ∈ Rn then we can write p in co-
ordinates as (x1 , x2 , ..., xn ). The coordinates for a vector in Tp Rn are dx1 , dx2 , ..., dxn .
A 1-form is a linear function, ω, whose graph (in Tp Rn — R) is a plane through the
origin. Hence, we can write it as ω = a1 dx1 + a2 dx2 + ... + an dxn . Again, this can be
thought of as either projection onto the vector a1 , a2 , ..., an and then multiplying
by | a1 , a2 , ..., an | or as projecting onto each coordinate axis, multiplying by ai , and
then adding.

Exercise 2.2. Let ω( dx, dy ) = ’dx + 4dy.

(1) Compute ω( 1, 0 ), ω( 0, 1 ), and ω( 2, 3 ).
(2) What line does ω project vectors onto?

Exercise 2.3. Find a 1-form which

(1) projects vectors onto the line dy = 2dx and scales by a factor of 2.
(2) projects vectors onto the line dy = 1 dx and scales by a factor of 5 .
1
3
(3) projects vectors onto the dx-axis and scales by a factor of 3.
1
(4) projects vectors onto the dy-axis and scales by a factor of 2 .
(5) does both of the two preceding operations and adds the result.
22 2. FORMS

3. Multiplying 1-forms

In this section we would like to explore a method of multiplying 1-forms. You may
think, “What™s the big deal? If ω and ν are 1-forms can™t we just de¬ne ω · ν(V ) =
ω(V ) · ν(V )?” Well, of course we can, but then ω · ν isn™t a linear function, so we
have left the world of forms.
The trick is to de¬ne the product of ω and ν to be a 2-form. So as not to
confuse this with the product just mentioned we will use the symbol “§” (pronounced
“wedge”) to denote multiplication. So how can we possibly de¬ne ω § ν to be a 2-
form? To do this we have to say how it acts on a pair of vectors, (V1 , V2 ).
Note ¬rst that there are four ways to combine all the ingredients:

ω(V1 ) ν(V1 ) ω(V2) ν(V2 )

The ¬rst two of these are associated with V1 and the second two with V2 . In other
words, ω and ν together give a way of taking each vector and returning a pair of
numbers. And how do we visualize pairs of numbers? In the plane, of course! Let™s
de¬ne a new plane with one axis being the ω-axis and the other the ν-axis. So,
the coordinates of V1 in this plane are [ω(V1), ν(V1 )] and the coordinates of V2 are
[ω(V2 ), ν(V2 )]. Note that we have switched to the notation “[·, ·]” to indicate that we
are describing points in a new plane. This may seem a little confusing at ¬rst. Just
keep in mind that when we write something like (1, 2) we are describing the location
of a point in the x-y plane, whereas 1, 2 describes a vector in the dx-dy plane and
[1, 2] is a vector in the ω-ν plane.
Let™s not forget our goal now. We wanted to use ω and ν to take the pair of
vectors, (V1 , V2 ), and return a number. So far all we have done is to take this pair of
vectors and return another pair of vectors. But do we know of a way to take these
vectors and get a number? Actually, we know several, but the most useful one turns
out to be the area of the parallelogram that they span. This is precisely what we
de¬ne to be the value of ω § ν(V1 , V2 ) (see Fig. 4).



Example 2.1. Let ω = 2dx ’ 3dy + dz and ν = dx + 2dy ’ dz be two 1-
forms on Tp R3 for some ¬xed p ∈ R3 . Let™s evaluate ω § ν on the pair of
3. MULTIPLYING 1-FORMS 23

z ν
V1
V2

ν(V1 )
ω
ω(V1)
x y

Figure 4. The product of ω and ν.

vectors, ( 1, 3, 1 , (2, ’1, 3 ). First we compute the [ω, ν] coordinates of the
vector 1, 3, 1 .


[ω( 1, 3, 1 ), ν( 1, 3, 1 )] = [2 · 1 ’ 3 · 3 + 1 · 1, 1 · 1 + 2 · 3 ’ 1 · 1]
= [’6, 6]
Similarly we compute [ω( 2, ’1, 3 ), ν( 2, ’1, 3 )] = [10, ’3]. Finally, the area
of the parallelogram spanned by [’6, 6] and [10, ’3] is
’6 6
= 18 ’ 60 = ’42
10 ’3


Should we have taken the absolute value? Not if we want to de¬ne a linear
operator. The result of ω § ν isn™t just an area, it™s a signed area. It can either be
positive or negative. We™ll see a geometric interpretation of this soon. For now we
de¬ne:

ω(V1 ) ν(V1 )
ω § ν(V1 , V2 ) =
ω(V2 ) ν(V2 )
Exercise 2.4. Let ω and ν be the following 1-forms:
ω( dx, dy ) = 2dx ’ 3dy
ν( dx, dy ) = dx + dy
(1) Let V1 = ’1, 2 and V2 = 1, 1 . Compute ω(V1), ν(V1 ), ω(V2 ) and ν(V2 ).
(2) Use your answers to the previous question to compute ω § ν(V1 , V2 ).
24 2. FORMS

(3) Find a constant c such that ω § ν = c dx § dy.

Exercise 2.5. ω § ν(V1 , V2 ) = ’ω § ν(V2 , V1 ) (ω § ν is skew-symmetric).

Exercise 2.6. ω § ν(V, V ) = 0. (This follows immediately from the previous exercise.
It should also be clear from the geometric interpretation).

Exercise 2.7. ω § ν(V1 + V2 , V3 ) = ω § ν(V1 , V3 ) + ω § ν(V2 , V3 ) and ω § ν(cV1 , V2 ) =
ω § ν(V1 , cV2 ) = cω § ν(V1 , V2 ), where c is any real number (ω § ν is bilinear).

Exercise 2.8. ω § ν(V1 , V2 ) = ’ν § ω(V1, V2 ).

It™s interesting to compare Exercises 2.5 and 2.8. Exercise 2.5 says that the 2-
form, ω § ν, is a skew-symmetric operator on pairs of vectors. Exercise 2.8 says that
§ can be thought of as a skew-symmetric operator on 1-forms.

Exercise 2.9. ω § ω(V1 , V2 ) = 0.

Exercise 2.10. (ω + ν) § ψ = ω § ψ + ν § ψ (§ is distributive).

There is another way to interpret the action of ω§ν which is much more geometric,
although it will take us some time to develop. Suppose ω = a dx + b dy + c dz. Then
we will denote the vector a, b, c as ω . From the previous section we know that if
V is any vector then ω(V ) = ω · V , and that this is just the projection of V onto
the line containing ω , times | ω |.
Now suppose ν is some other 1-form. Choose a scalar x so that ν ’ xω is
perpendicular to ω . Let νω = ν ’ xω. Note that ω § νω = ω § (ν ’ xω) =
ω § ν ’ xω § ω = ω § ν. Hence, any geometric interpretation we ¬nd for the action
of ω § νω is also a geometric interpretation of the action of ω § ν.
ω νω
Finally, we let ω = | ω | and νω = | νω | . Note that these are 1-forms such
that ω and νω are perpendicular unit vectors. We will now present a geometric
interpretation of the action of ω § νω on a pair of vectors, (V1 , V2 ).
First, note that since ω is a unit vector then ω(V1 ) is just the projection of V1
onto the line containing ω . Similarly, νω (V1 ) is given by projecting V1 onto the
line containing νω . As ω and νω are perpendicular, we can thus think of the
quantity
3. MULTIPLYING 1-FORMS 25


ω(V1 ) νω (V1 )
ω § νω (V1 , V2 ) =
ω(V2 ) νω (V2 )
as being the area of parallelogram spanned by V1 and V2 , projected onto the plane
containing the vectors ω and νω . This is the same plane as the one which contains
the vectors ω and ν .
Now observe the following:

ω νω 1
ω § νω = § ω § νω
=
| ω | | νω | | ω || νω |
Hence,


ω § ν = ω § νω = | ω || νω |ω § νω
Finally, note that since ω and νω are perpendicular the quantity | ω || νω |
is just the area of the rectangle spanned by these two vectors. Furthermore, the
parallelogram spanned by the vectors ω and ν is obtained from this rectangle by
skewing. Hence, they have the same area. We conclude

Evaluating ω § ν on the pair of vectors (V1, V2) gives

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