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the kernel of a 1-form ω de¬nes an integrable plane ¬eld then ω § dω n = 0.
Our search for a plane ¬eld that is not integrable (i.e. not the tangent planes to
a foliation) has now been reduced to the search for a 1-form ω for which ω § dω = 0
somewhere. There are many such forms. An easy one is x dy + dz. We compute:

(x dy + dz) § d(x dy + dz) = (x dy + dz) § (dx § dy) = dz § dx § dy
Our answer is quite special. All we needed was a 1-form such that ω § dω = 0
somewhere. What we found was a 1-form for which ω § dω = 0 everywhere. This
means that there is not a single point of R3 which has a neighborhood in which the

planes given by Ker x dy + dz are tangent to a foliation. Such a plane ¬eld is called
a contact structure.
At this point you™re probably wondering, “What could Ker x dy + dz possibly
look like?!” It™s not so easy to visualize this, but we have tried to give you some
indication in Figure 11. A good exercise is to stare at this picture long enough to
convince yourself that the planes pictured cannot be the tangent planes to a foliation.



Figure 1. The plane ¬eld Ker x dy + dz.

We have just seen how we can use di¬erential forms to tell if a plane ¬eld is
integrable. But one may still wonder if there is more we can say about a 1-form,
assuming its kernel is integrable. Let™s go back to the expression ω § dω. Recall that
ω is a 1-form, which makes dω a 2-form, and hence ω § dω a 3-form.

1Figure drawn by Stephan Schoenenberger. Taken from Introductory Lectures on Contact Ge-
ometry by John B. Etnyre

A 3-form on Tp R3 measures the volume of the parallelepiped spanned by three
vectors, multiplied by a constant. For example, if ψ = ± § β § γ is a 3-form then the
constant it scales volume by is given by the volume of the parallelepiped spanned by
the vectors ± , β , and γ (where “ ± ” refers to the vector dual to the 1-form ±
introduced in Section 3 of Chapter 3). If it turns out that ψ is the zero 3-form then
the vector ± must be in the plane spanned by the vectors β and γ .
On R3 the results of Section 3 of Chapter 3 tell us that a 2-form such as dω can
always be written as ± § β, for some 1-forms ± and β. If ω is a 1-form with integrable
kernel then we have already seen that ω § dω = ω § ± § β = 0. But this tells us that
ω must be in the plane spanned by the vectors ± and β . Now we can invoke
Lemma 2.1 of Chapter 3, which says that we can rewrite dω as ω § ν, for some 1-form
If we start with a foliation and choose a 1-form ω whose kernel consists of planes
tangent to the foliation then the 1-form ν that we have just found is in no way
canonical. We made lots of choices to get to ν, and di¬erent choices will end up
with di¬erent 1-forms. But here™s the amazing fact: the integral of the 3-form ν § dν
does not depend on any of our choices! It is completely determined by the original
foliation. Whenever a mathematician runs into a situation like this they usually
throw up their hands and say, “Eureka! I™ve discovered an invariant.” The quantity
ν § dν is referred to as the Gobillion-Vey invariant of the foliation. It is a topic of
current research to identify exactly what information this number tells us about the
Two special cases are worth noting. First, it may turn out that ν § dν = 0
everywhere. This tells us that the plane ¬eld given by Ker ν is integrable, so we get
another foliation. The other interesting case is when ν § dν is nowhere zero. Then
we get a contact structure.

3. How not to visualize a di¬erential 1-form

There are several contemporary physics texts that attempt to give a visual in-
terpretation of di¬erential forms that seems quite di¬erent from the one presented
here. As this alternate interpretation is much simpler than anything described in
these notes, one may wonder why we have not taken this approach.

Let™s look again at the 1-form dx on R3 . Given a vector Vp at a point p the value
of dx(Vp ) is just the projection of Vp onto the dx axis in Tp R3 . Now, let C be some
parameterized curve in R3 for which the x-coordinate is always increasing. Then
dx is just the length of the projection of C onto the x-axis. To the nearest integer,
this is just the number of evenly spaced planes that C punctures that are parallel to
the y-z plane. So one way that you might visualize the form dx is by picturing these
This view is very appealing. After all, every 1-form ω, at every point p, projects
vectors onto some line lp . So can™t we integrate ω along a curve C (at least to the
nearest integer) by counting the number of surfaces punctured by C whose tangent
planes are perpendicular to the lines lp (see Figure 2)? If you™ve read the previous
section you might guess that the answer is a categorical NO!

z C





Figure 2. “Surfaces” of ω?

Recall that the planes perpendicular to the lines lp are precisely Ker ω. To say
that there are surfaces whose tangent planes are perpendicular to the lines lp is the
same thing as saying that Ker ω is an integrable plane ¬eld. But we have seen in
the previous section that there are 1-forms as simple as x dy + dz whose kernels are
nowhere integrable.

Can we at least use this interpretation for a 1-form whose kernel is integrable?
Unfortunately, the answer is still no. Let ω be the 1-form on the solid torus whose
kernel consists of the planes tangent to the foliation pictured in Figure 3 (This is
called the Reeb foliation of the solid torus). The surfaces of this foliation spiral
continually outward. So if we try to pick some number of “sample” surfaces then
they will “bunch up” near the boundary torus. This would seem to indicate that
if we wanted to integrate ω over any path that cut through the solid torus then we
should get an in¬nite answer, since such a path would intersect our “sample” surfaces
an in¬nite number of times. However, we can certainly ¬nd a 1-form ω for which
this is not the case.

Figure 3. The Reeb foliation of the solid torus.

We don™t want to end this section on such a down note. Although it is not in
general valid to visualize a 1-form as a sample collection of surfaces from a foliation,
we can visualize it as a plane ¬eld. For example, Figure 1 is a pretty good depiction of
the 1-form x dy + dz. All that we have pictured there is a few evenly spaced elements

of it™s kernel, but this is enough. To get a rough idea of the value of x dy + dz we
can just count the number of (transverse) intersections of the planes pictured with
C. So, for example, if C is a curve whose tangents are always contained in one of
these planes (a so called Legendrian curve) then x dy + dz will be zero. Inspection
of the picture reveals that examples of such curves are the lines parallel to the x-axis.

Exercise 6.3. Show that if C is a line parallel to the x-axis then x dy + dz = 0.


1. Forms on subsets of Rn

The goal of this chapter is to slowly work up to de¬ning forms in a much more
general setting than just on Rn . One reason for this is because Stokes™ Theorem
actually tells us that forms on Rn just aren™t very interesting. For example, let™s
examine how a 1-form, ω, on R2 , for which dω = 0 (i.e. ω is closed), integrates over
an 1-chain, C, such that ‚C = … (i.e. C is closed). It is a basic result of Topology
that any such 1-chain bounds a 2-chain, D. Hence, ω= dω = 0!!
Fortunately, there is no reason to restrict ourselves to di¬erential forms which
are de¬ned on all of Rn . Instead, we can simply consider forms which are de¬ned on
subsets, U, of Rn . For technical reasons, we will always assume such subsets are open
(i.e. for each p ∈ U, there is an « such that {q ∈ Rn |d(p, q) < «} ‚ U). In this case,
T Up = T Rn . Since a di¬erential n-form is nothing more than a choice of n-form on
T Rn , for each p (with some condition about di¬erentiability), it makes sense to talk
about a di¬erential form on U.

Example 7.1.
y x
ω0 = ’ dx + 2 dy
x2 + y 2 x + y2
is a di¬erential 1-form on R2 ’ (0, 0).

Exercise 7.1. Show that dω0 = 0.

Exercise 7.2. Let C be the unit circle, oriented counter-clockwise. Show that ω0 =
′ ′
2π. Hint: Let ω = ’y dx + x dy. Note that on C, ω0 = ω .

If C is any closed 1-chain in R2 ’ (0, 0), then the quantity ω0 is called the

winding number of C, since it computes the number of times C winds around the

Exercise 7.3. Let x+ denote the positive x-axis in R2 ’ (0, 0), and let C be any closed
1-chain. Suppose Vp is a basis vector of T Cp which agrees with the orientation of C at p.
A positive (resp. negative) intersection of C with x+ is one where Vp has a component
which points “up” (resp. “down”). Assume all intersections of C with x+ are either
positive or negative. Let P denote the number of positive ones, and N the number of
negative ones. Show that 2π ω0 = P ’N. Hint: Use the Generalized Stokes™ Theorem.

2. Forms on Parameterized Subsets

Recall that at each point a di¬erential from is simply an alternating, multilinear
map on a tangent plane. So all we need to de¬ne a di¬erential form on a more general
space is a well de¬ned tangent space. One case in which this happens is when we
have a parameterized subset of Rm . Let φ : U ‚ Rn ’ M ‚ Rm be a (one-to-one)
parameterization of M. Then recall that T Mp is de¬ned to be the span of the partial
derivatives of φ at φ’1 (p), and is an n-dimensional Euclidean space, regardless of the
point, p. Hence, we say the dimension of M is n.
A di¬erential k-form on M is simply an alternating, multilinear, real-valued func-
tion on T Mp , for each p ∈ M, which varies di¬erentiably with p. In other words, a
di¬erential k-form on M is a whole family of k-forms, each one acting on T Mp , for
di¬erent points, p. It is not so easy to say precisely what we mean when we say the
form varies in a di¬erentiable way with p. Fortunately, we have already introduced
the tools necessary to do this. Let™s say that ω is a family of k-forms, de¬ned on
T Mp , for each p ∈ M. Then φ— ω is a family of k-forms, de¬ned on T Rn’1 (p) , for
each p ∈ M. We say that ω is a di¬erentiable k-form on M, if φ— ω is a di¬erentiable
family on U.
This de¬nition illustrates an important technique which is used often when deal-
ing with di¬erential forms on manifolds. Rather than working in M directly we use
the map φ— to translate problems about forms on M into problems about forms on U.
These are nice because we already know how to work with forms which are de¬ned
on open subsets of Rn . We will have much more to say about this later.

Example 7.2. The in¬nitely long cylinder, L, of radius 1, centered along the

z-axis, is given by the parameterization, φ(a, b) = √a2a+b2 , √a2b+b2 , ln a2 + b2 ,
whose domain is R2 ’ (0, 0). We can use φ— to solve any problem about forms
on L, by translating it back to a problem about forms on U.

Exercise 7.4. Consider the 1-form, „ ′ = ’y dx + x dy, on R3 . In particular, this form
acts on vectors in T Lp , where L is the cylinder of the previous example, and p is any
point in L. Let „ be the restriction of „ ′ to vectors in T Lp . So, „ is a 1-form on L.
Compute φ— „ . What does this tell you that „ measures?

If ω is a k-form on M, then what do we mean by dω? Whatever the de¬nition,
we clearly want dφ— ω = φ— dω. So why don™t we use this to de¬ne dω? After all, we
know what dφ— ω is, since φ— ω is a form on Rn . Recall that Dφp is a map from T Rn p
to T Rm . However, if we restrict the range to T Mp , then Dφp is 1-1, so it makes sense
to refer to Dφ’1 . We now de¬ne

dω(Vp1 , ..., Vpk+1 ) = dφ— ω(Dφ’1(Vp1 ), ..., Dφ’1 (Vpk+1 ))
p p

Exercise 7.5. If „ ′ and „ are the 1-forms on R3 and L, respectively, de¬ned in the
previous section, compute d„ ′ and d„ . Answer: d„ ′ = 2 dx § dy and d„ = 0.

3. Forms on quotients of Rn (optional)


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