Exact and closed k-forms

Section 4.6 Exact and closed $k$ -forms

We define exact and closed

k

-forms using the exterior derivative. We show that exact forms are always closed, and determine when closed forms are exact. We rephrase these statements in the language of vector calculus.

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Objectives

You should be able to:

Define closed and exact $k$ -forms using the exterior derivative.
Determine when a $k$ -form is closed or exact, focusing on one-, two-, and three-forms in $R^{3} .$
Show that the general definition of closeness for one-forms reduces to our previous definition in terms of partial derivatives.
Show that exact $k$ -forms are always closed.
Recall the statement of Poincare's lemma.
Rephrase and use these statements in the language of vector calculus.

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Subsection 4.6.1 Exact and closed $k$ -forms

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We introduced the notion of exact one-forms in Definition 2.2.5, using the concept of differential. We also introduced closed one-forms in

R^{2}

in Definition 2.2.9 and in

R^{3}

in Definition 2.2.14, but our definition was rather ad hoc. These concepts are much more natural now that we have introduced the exterior derivative.

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Definition 4.6.1. Exact and closed $k$ -forms.

Let $ω$ be a $k$ -form on $U \subseteq R^{n} .$ We say that $ω$ is closed if $d ω = 0 .$ We say that it is exact if there exists a $(k - 1)$ -form $η$ on $U$ such that $ω = d η .$

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Example 4.6.2. Exact and closed one-forms in $R^{3}$ .

Before we move on, let us show that this reproduces the definitions that we used in Definition 2.2.5 and Definition 2.2.14 for one-forms in $R^{3} .$

First, if $ω$ is a one-form, according to Definition 4.6.1 is it exact if there exists a zero-form $f$ such that $ω = d f .$ As we know that the exterior derivative of a zero-form is the same thing as the differential of a function introduced in Definition 2.2.1, this is precisely the definition of an exact one-form that we gave in Definition 2.2.5.

As for closeness, according to Definition 4.6.1 a one-form $ω$ is closed if $d ω = 0 .$ If $ω$ is a one-form on $U \subset R^{3},$ we can write $ω = f d x + g d y + h d z .$ Then

d ω = (\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z}) d y \land d z + (\frac{\partial f}{\partial z} - \frac{\partial h}{\partial x}) d z \land d x + (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) d x \land d y .

Thus $d ω = 0$ if and only if

\frac{\partial h}{\partial y} = \frac{\partial g}{\partial z}, \frac{\partial f}{\partial z} = \frac{\partial h}{\partial x}, \frac{\partial g}{\partial x} = \frac{\partial f}{\partial y},

which is precisely the condition stated in Definition 2.2.14. In fact, this explains where this strange condition comes from!

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As for one-forms, it is easy to show that exact forms are always closed: it follows directly from the fact that

d^{2} = 0,

which is a key property of the exterior derivative proved in Lemma 4.3.9.

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Lemma 4.6.3. Exact $k$ -forms are closed.

If a $k$ -form $ω$ on $U \subseteq R^{n}$ is exact, then it is closed.

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Proof.

This is a direct consequence of the fact that $d^{2} = 0 .$ If $ω$ is exact, then there exists a $(k - 1)$ -form $η$ such that $ω = d η .$ But then

d ω = d (d η) = 0

by Lemma 4.3.9. Therefore $ω$ is closed.

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As for one-forms, the converse statement is much more subtle. When are closed

k

-forms exact? The answer depends on the domain of definition of the

k

-form. However, there is a simple case for which closed

k

-forms are always exact, as in Theorem 3.6.1. This is called “Poincare's lemma” for

k

-forms. We will state the result here without proof.

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Theorem 4.6.4. Poincare's lemma for $k$ -forms, version 1.

Let $ω$ be a $k$ -form defined on all of $R^{n} .$ Then $ω$ is exact if and only if it is closed.

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In fact, the theorem can be generalized slightly, as in Theorem 3.6.4. What matters in the proof is not so much that

ω

is defined on all of

R^{n},

but rather that it is defined on a domain

U

that is contractible, which intuitively means that it can be continuously shrunk to a point within

U .

A more precise statement of Poincare's lemma could then be formulated as follows.

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Theorem 4.6.5. Poincare's lemma for $k$ -forms, version II.

Let $ω$ be a $k$ -form defined on an open ball $U \subseteq R^{n} .$ Then $ω$ is exact if and only if it is closed.

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It is important to note however that if

U

is not an open ball or the whole of

R^{n},

then closed forms may not necessarily be exact.

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Remark 4.6.6.

We should note that contrary to Poincare's lemma for one-forms in Theorem 3.6.4, for $k \geq 2$ it is not true that closed $k$ -forms on simply connected open subsets $U \subseteq R^{n}$ are necessarily exact. This is only true for one-forms; simple-connectedness is not sufficient for $k \geq 2 .$ What we need is a higher-dimensional analog of simple-connectedness; this is why it was replaced by the statement that $U$ is an open ball in Theorem 4.6.5. (Slightly more generally, one could say that $U$ must be “contractible”. Every contractible space is simply connected, but not the other way around. This is the kind of thing that is studied in cohomology and homology, see Remark 4.6.7.)

Just to highlight the subtlety here, consider the two-form

ω = \frac{1}{(x^{2} + y^{2} + z^{2})^{3 / 2}} (x d y \land d z + y d z \land d x + z d x \land d y),

which is defined on $U = R^{3} ∖ {(0, 0, 0)},$ which is simply connected. One can check that $ω$ is closed, but not exact. Does that contradict Poincare's lemma? Fortunately it doesn't, as $U$ is not an open ball in $R^{3}$ (as it does not contain the origin). But it shows that there exists $k$ -forms with $k \geq 2$ defined on simply connected open sets that are closed but not exact.

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Remark 4.6.7. The world of cohomology (this is just for fun and beyond the scope of this class!).

In fact, the relation between closed and exact forms is quite deep. As we have seen, it is closely connected to the existence of “holes” in a space, which is the subject of topology. In fact, studying when closed forms are not exact gives rise to the topic of cohomology, which is an important branch of geometry and topology. Believe me, cohomology is all over the place. You wouldn't believe it, but it is even used to describe gauge theories in physics!

While describing cohomology is way beyond the scope of this course, let me explain in a few words what it is about, just for fun, in the context of differential forms. Suppose that $ω$ and $η$ are closed $k$ -forms on $U .$ If they differ from each other by an exact form, that is $ω - η = d ρ$ for some $(k - 1)$ -form $ρ,$ then we say that $ω$ and $η$ are “equivalent” (or “cohomologous”). In this way, we construct equivalence classes of closed $k$ -forms that differ by an exact form. Those equivalence classes (which are called “cohomology classes”) form a vector space, which is called the “de Rham cohomology space” $H^{k} (U) .$ ¹

Mathematically, what we are doing here is constructing a quotient vector space. If we write

Z^{k} (U)

for the vector space of closed

k

-forms on

U,

and

B^{k} (U)

for the vector space of exact

k

-forms on

U,

then the de Rham cohomology space is constructed as the quotient space

H^{k} (U) = Z^{k} (U) / B^{k} (U) .

If $U$ is the whole of $R^{n},$ or an open ball in $R^{n},$ then all closed $k$ -forms are exact, and hence they are all equivalent to the zero $k$ -form. The de Rham cohomology spaces $H^{k} (U)$ (with $k \geq 1$ ) are then all trivial (the zero vector space). So the de Rham cohomology spaces $H^{k} (U)$ with $k \geq 1$ control, in a sense, how topologically non-trivial $U$ is.

For instance, if we consider $U = R^{2} ∖ {(0, 0)},$ one can show that

H^{1} (U) = R, H^{2} (U) = 0.

The fact that $H^{2} (U) = 0$ says that all (closed) two-forms on $U$ are exact. However, the interesting fact here is that $H^{1} (U) = R,$ which says that not all closed one-forms are exact; indeed, we saw one example of such a one-form in Example 2.2.13, which was closed but not exact. Since $H^{1} (U)$ is one-dimensional, this means that all closed one-forms that are not exact differ from the one we saw in that example by an exact form (they are in the same cohomology class), up to overall rescaling.

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Subsection 4.6.2 Translation into vector calculus for $R^{3}$

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Focusing on

R^{3},

we can rephrase the notions of exactness and closeness in terms of vector calculus objects, using Table 4.1.11. In

R^{3},

we focus on zero-, one-, two-, and three-forms. In fact, the interesting statements are for one-forms and two-forms.

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We already translated the exactness and closeness statements for one-forms in the language of vector calculus in Section 2.2, but let summarize the statements here.

A one-form $ω$ on $R^{3}$ is exact if and only if its associated vector field $F$ is conservative, which is the statement that
$F = \nabla f$
for some potential function $f .$
A one-form $ω$ on $R^{3}$ is closed if and only if its associated vector field $F$ is curl-free, that is
$\nabla \times F = 0$
(see Remark 4.4.5).
The fact that exact one-forms are closed is the statement that if $F$ is conservative, then it is curl-free, namely $\nabla \times F = 0 .$ We call this the “screening test for conservative vector fields”. However, while conservative vector fields are curl-free, curl-free vector fields are not necessarily conservative.
Poincare's lemma translates into the statement that if $F$ is defined and has continuous first order partial derivatives on all of $R^{3}$ (or an open simply connected subset therein), then $F$ is conservative if and only if it is curl-free.

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We can do a similar translation for two-forms. The result is the following four statements.

A two-form $ω$ on $R^{3}$ is exact if and only if its associated vector field $F$ has a vector potential, which is a vector field $G$ such that
$F = \nabla \times G .$
A two-form $ω$ on $R^{3}$ is closed if and only if its associated vector field $F$ is divergence-free, that is
$\nabla \cdot F = 0 .$
The fact that exact two-forms are closed is the statement that if there exists a vector potential for $F,$ then $F$ is divergence-free, namely $\nabla \cdot F = 0 .$ We call this the “screening test for vector potentials”. However, while vector fields with a vector potential are divergence-free, divergence-free vector fields do not necessarily have a vector potential.
Poincare's lemma translates into the statement that if $F$ is defined and has continuous first order partial derivatives on all of $R^{3}$ (or an open ball therein), then $F$ has a vector potential if and only if it is divergence-free.

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Remark 4.6.8.

Suppose that a vector field $F$ on $R^{3}$ has a vector potential $G .$ This means that there exists a vector field $G$ such that

F = \nabla \times G .

Finding $G$ is not always obvious, as one would need in principle to integrate fairly complicated partial differential equations. Moreover, $G$ is far from unique.

It turns out that there is a nice result that drastically simplifies calculations. One can show that, if $F$ has a vector potential $G,$ then we can always choose $G$ to have vanishing $z$ -component function. In other words, if $F$ has a vector potential, then there always exists a $G = (g_{1}, g_{2}, 0)$ such that

F = \nabla \times G .

This is very helpful when trying to find a vector potential.

In the language of differential forms, this corresponds to the statement that if $ω$ is an exact two-form on $R^{3},$ then there exists a one-form $η = f d x + g d y$ (with vanishing third component function) such that $ω = d η .$ We prove this statement in Exercise 4.6.3.6.

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Exercises 4.6.3 Exercises

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1.

Let $ω$ be the following two-form on $R^{3} :$

ω = (e^{x} + y) d y \land d z + (y e^{x} + e^{z}) d z \land d x + (e^{x + y} - 2 z e^{x}) d x \land d y .

Determine whether $ω$ is exact. If it is, find a one-form $η$ such that $d η = ω .$

Solution.

First, we notice that $ω$ is defined on all of $R^{3},$ so Poincare's lemma applies. So we know that it is exact if and only if it is closed. Let us first determine whether it is closed.

We calculate:

\begin{aligned} d ω = & \frac{\partial}{\partial x} (e^{x} + y) d x \land d y \land d z + \frac{\partial}{\partial y} (y e^{x} + e^{z}) d y \land d z \land d x \\ + \frac{\partial}{\partial z} (e^{x + y} - 2 z e^{x}) d z \land d x \land d y \\ = & (e^{x} + e^{x} - 2 e^{x}) d x \land d y \land d z \\ = & 0. \end{aligned}

Thus $ω$ is closed, and hence, by Poincare's lemma, it is also exact.

We now want to find a one-form $η$ such that $d η = ω .$ We assume that $η$ takes the form

η = f d x + g d y .

Its exterior derivative is

d η = - \frac{\partial g}{\partial z} d y \land d z + \frac{\partial f}{\partial z} d z \land d x + (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) d x \land d y .

So we need to solve the three equations:

- \frac{\partial g}{\partial z} = e^{x} + y, \frac{\partial f}{\partial z} = y e^{x} + e^{z}, \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} = e^{x + y} - 2 z e^{x} .

Integrating the first one, we get:

g (x, y, z) = - z e^{x} - y z + α (x, y)

for some function $α (x, y) .$ Integrating the second one, we get

f (x, y, z) = y z e^{x} + e^{z} + β (x, y)

for some function $β (x, y) .$ Substituting these expressions in the third equation, we get:

\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} = - z e^{x} + \frac{\partial α (x, y)}{\partial x} - z e^{x} - \frac{\partial β (x, y)}{\partial y} = e^{x + y} - 2 z e^{x},

so we must have

\frac{\partial α (x, y)}{\partial x} - \frac{\partial β (x, y)}{\partial y} = e^{x + y} .

We can choose any function $α$ and $β$ that work. So why not choose $β = 0,$ and $α (x, y) = e^{x + y} .$ Then we get that the one-form

η = (y z e^{x} + e^{z}) d x + (- z e^{x} - y z + e^{x + y}) d y

is such that $d η = ω .$

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2.

Let $ω$ be the following two-form on $R^{3} :$

ω = x y d y \land d z + y z d z \land d x + z x d x \land d y .

Determine whether $ω$ is exact. If it is, find a one-form $η$ such that $d η = ω .$

Solution.

$ω$ is defined on all of $R^{3},$ so Poincare's lemma applies. Thus $ω$ is exact if and only if it is closed. Let us determine whether it is closed.

We calculate:

\begin{aligned} d ω = & \frac{\partial}{\partial x} (x y) d x \land d y \land d z + \frac{\partial}{\partial y} (y z) d y \land d z \land d x + \frac{\partial}{\partial z} (z x) d z \land d x \land d y \\ = & (y + z + x) d x \land d y \land d z . \end{aligned}

As this is non-zero, $ω$ is not closed. We then conclude that $ω$ is not exact.

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3.

Let $F$ be the following vector field on $R^{3} :$

F = (y z, x z, x y) .

Determine whether $F$ has a vector potential $G .$ If it does, find such a vector potential.

Solution.

Since the component functions of $F$ are smooth on all of $R^{3},$ Poincare's lemma applies. So $F$ will have a vector potential if and only if it is divergence-free. We calculate its divergence:

\begin{aligned} \nabla \cdot F = & \frac{\partial}{\partial x} (y z) + \frac{\partial}{\partial y} (x z) + \frac{\partial}{\partial z} (x y) \\ = & 0, \end{aligned}

and thus we conclude that there exists a vector potential $G$ such that $\nabla \times G = F .$

Now we need to find $G .$ We assume that $G = (g_{1}, g_{2}, 0) .$ Then the condition that $\nabla \times G = F$ is

\nabla \times G = (- \frac{\partial g_{2}}{\partial z}, \frac{\partial g_{1}}{\partial z}, \frac{\partial g_{2}}{\partial x} - \frac{\partial g_{1}}{\partial y}) = (y z, x z, x y) .

So we get three equations to solve. We integrate the first one (equality of the $x$ -component functions) to get:

g_{2} (x, y, z) = - \frac{y z^{2}}{2} + α (x, y) .

We integrate the second one (equality of the $y$ -component functions) to get:

g_{1} (x, y, z) = \frac{x z^{2}}{2} + β (x, y) .

Substituting in the third one (equality of the $z$ -component functions), we get:

\frac{\partial g_{2}}{\partial x} - \frac{\partial g_{1}}{\partial y} = \frac{\partial α (x, y)}{\partial x} - \frac{\partial β (x, y)}{\partial y} = x y .

We need to find any two functions $α$ and $β$ that satisfy this condition. We choose $β = 0,$ $α = \frac{x^{2} y}{2} .$ As a result, we have found a vector potential

G (x, y, z) = (\frac{x z^{2}}{2}, \frac{y}{2} (x^{2} - z^{2}), 0) .

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4.

Consider the two-form

ω = \frac{1}{(x^{2} + y^{2} + z^{2})^{3 / 2}} (x d y \land d z + y d z \land d x + z d x \land d y),

which is defined on $U = R^{3} ∖ {(0, 0, 0)} .$ Show that $ω$ is closed.

Solution.

We want to show that $d ω = 0 .$ First, we calculate

\begin{aligned} \frac{\partial}{\partial x} \frac{x}{(x^{2} + y^{2} + z^{2})^{3 / 2}} = & \frac{(x^{2} + y^{2} + z^{2})^{3 / 2} - 3 x^{2} (x^{2} + y^{2} + z^{2})^{1 / 2}}{(x^{2} + y^{2} + z^{2})^{3}} \\ = & \frac{(x^{2} + y^{2} + z^{2}) - 3 x^{2}}{(x^{2} + y^{2} + z^{2})^{5 / 2}} . \end{aligned}

Similarly,

\frac{\partial}{\partial y} \frac{y}{(x^{2} + y^{2} + z^{2})^{3 / 2}} = \frac{(x^{2} + y^{2} + z^{2}) - 3 y^{2}}{(x^{2} + y^{2} + z^{2})^{5 / 2}}

and

\frac{\partial}{\partial z} \frac{z}{(x^{2} + y^{2} + z^{2})^{3 / 2}} = \frac{(x^{2} + y^{2} + z^{2}) - 3 z^{2}}{(x^{2} + y^{2} + z^{2})^{5 / 2}} .

Now, we get:

\begin{aligned} d ω = & \frac{\partial}{\partial x} \frac{x}{(x^{2} + y^{2} + z^{2})^{3 / 2}} d x \land d y \land d z + \frac{\partial}{\partial y} \frac{y}{(x^{2} + y^{2} + z^{2})^{3 / 2}} d y \land d z \land d x \\ + \frac{\partial}{\partial z} \frac{z}{(x^{2} + y^{2} + z^{2})^{3 / 2}} d z \land d x \land d y \\ = & \frac{3 (x^{2} + y^{2} + z^{2}) - 3 x^{2} - 3 y^{2} - 3 z^{2}}{(x^{2} + y^{2} + z^{2})^{5 / 2}} d x \land d y \land d z \\ = & 0. \end{aligned}

We conclude that $ω$ is closed.

What is interesting with this example is that $ω$ is closed, but one can show that it is not exact. This doesn't contradict Poincare's lemma, as $ω$ is not defined on all of $R^{3}$ (or an open ball thereof). But it is interesting since $ω$ is defined on a simply connected subset of $R^{3},$ so it shows that there exists $k$ -forms with $k \geq 2$ defined on simply connected subsets that are closed but not exact.

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5.

Consider the vector field

F (x, y, z) = (- \frac{y}{x^{2} + y^{2}}, \frac{x}{x^{2} + y^{2}}, z) .

Find the domain of definition of $F .$ Is it path connected? Simply connected?
Determine the divergence of $F .$
Determine the curl of $F .$
Does $F$ have a vector potential? Justify your answer. If it does, find such a vector potential.
Is $F$ conservative? Justify your answer. If it is, find a potential function.

Solution.

$F$ is defined (and in fact, is smooth) wherever the denominator is non-zero. This is wherever $(x, y) \neq (0, 0) .$ So the domain of definition of $F$ is

$U = {(x, y, z) \in R^{3} | (x, y) \neq (0, 0)} .$

This is $R^{3}$ minus the $z$ -axis. It is path connected, since any two points in $U$ can be connected by a path. It is however not simply connected, since a closed curve around the $z$ -axis cannot be continuously contracted to a point within $U$ (it would hit the $z$ -axis, which is not in $U$ ).
We calculate the divergence of $F :$

$\begin{aligned} \nabla \cdot F = & \frac{\partial}{\partial x} (- \frac{y}{x^{2} + y^{2}}) + \frac{\partial}{\partial y} (\frac{x}{x^{2} + y^{2}}) + \frac{\partial}{\partial z} (z) \\ = & \frac{2 x y}{(x^{2} + y^{2})^{2}} - \frac{2 x y}{(x^{2} + y^{2})^{2}} + 1 \\ = & 1. \end{aligned}$
We calculate the curl of $F :$

$\begin{aligned} \nabla \times F = & (\frac{\partial}{\partial y} (z) - \frac{\partial}{\partial z} (\frac{x}{x^{2} + y^{2}}), \frac{\partial}{\partial z} (- \frac{y}{x^{2} + y^{2}}) - \frac{\partial}{\partial x} (z), \\ \frac{\partial}{\partial x} (\frac{x}{x^{2} + y^{2}}) - \frac{\partial}{\partial y} (- \frac{y}{x^{2} + y^{2}})) \\ = & (0, 0, \frac{x^{2} + y^{2} - 2 x^{2}}{(x^{2} + y^{2})^{2}} + \frac{x^{2} + y^{2} - 2 y^{2}}{(x^{2} + y^{2})^{2}}) \\ = & (0, 0, 0) . \end{aligned}$

So $F$ is curl-free.
We found in (b) that the divergence of $F$ is non-zero. This means that $F$ cannot have a vector potential, since vector fields that have a vector potential are divergence-free.
We found in (c) that $F$ is curl-free, so it passes the screening test for conservative vector fields. However, since its domain of definition $U$ is not simply connected, Poincare's lemma does not apply. We thus cannot conclude whether $F$ is conservative from the statement that it is curl-free.

In fact, we can show that it is not conservative by showing that its integral along a closed loop is non-zero, as in Exercise 3.4.3.2. Let $ω$ be the one-form associated to $F :$

$ω = - \frac{y}{x^{2} + y^{2}} d x + \frac{x}{x^{2} + y^{2}} d y + z d z .$

Consider the parametric curve $α : [0, 2 π] \to R^{3}$ with $α (t) = (\cos (t), \sin (t), 0),$ which is the unit circle (counterclockwise) around the origin in the $x y$ -plane. The pullback of $ω$ is:

$\begin{aligned} α^{*} ω = & (- \frac{\sin (t)}{\cos^{2} (t) + \sin^{2} (t)} (- \sin (t)) + \frac{\cos (t)}{\cos^{2} (t) + \sin^{2} (t)} \cos (t)) d t \\ = & d t . \end{aligned}$

The line integral of $ω$ along $α$ is thus:

$\int_{α} ω = \int_{0}^{2 π} α^{*} ω = \int_{0}^{2 π} d t = 2 π .$

Since this is non-zero, by Corollary 3.4.3 (or, in other words, the Fundamental Theorem for line integrals), $ω$ cannot be exact, since the line integrals of exact one-forms along closed curves vanish. Equivalently, the vector field $F$ is not conservative.

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6.

Let $ω$ be an exact two-form on $R^{3} .$ Show that there exists a one-form $η$ of the form

η = f d x + g d y

(i.e. with a vanishing $z$ -component function) such that $d η = ω .$

In other words, in the language of vector calculus, if a vector field $F$ has a vector potential $G,$ then $G$ can always be chosen to take the form $G = (g_{1}, g_{2}, 0) .$

Solution.

Since $ω$ is exact, we know that there exists a one-form $β$ such that $d β = ω .$ Suppose that we find such a $β :$

β = b_{1} d x + b_{2} d y + b_{3} d z,

for some component functions $b_{1}, b_{2}, b_{3} .$ $β$ is certainly not unique; there are many one-forms such that their exterior derivative equals $ω .$ In fact, since $d^{2} = 0,$ we can add to $β$ any exact one-form $d F$ for a function $F,$ and we get another one-form whose exterior derivative is $ω .$ That is, if we define

\tilde{β} = β - d F

for any function $F,$ then $d \tilde{β} = d β - d^{2} F = ω .$

In particular, since

d F = \frac{\partial F}{\partial x} d x + \frac{\partial F}{\partial y} d y + \frac{\partial F}{\partial z} d z,

if we can choose $F$ such that

\frac{\partial F}{\partial z} = b_{3},

then we see that

\tilde{β} = (b_{1} - \frac{\partial F}{\partial x}) d x + (b_{2} - \frac{\partial F}{\partial y}) d y,

and hence it is of the form that we are looking for (no $z$ -component function). But this is easy to do; simply pick

F (x, y, z) = \int b_{3} (x, y, z) d z,

i.e. any antiderivative in the $z$ -variable will do. So we conclude that we can always find a one-form $\tilde{β}$ with no $z$ -component and such that $d \tilde{β} = ω .$

Section 4.6 Exact and closed k-forms

Objectives

Subsection 4.6.1 Exact and closed k-forms

Definition 4.6.1. Exact and closed k-forms.

Example 4.6.2. Exact and closed one-forms in R3.

Lemma 4.6.3. Exact k-forms are closed.

Proof.

Theorem 4.6.4. Poincare's lemma for k-forms, version 1.

Theorem 4.6.5. Poincare's lemma for k-forms, version II.

Remark 4.6.6.

Remark 4.6.7. The world of cohomology (this is just for fun and beyond the scope of this class!).

Subsection 4.6.2 Translation into vector calculus for R3

Remark 4.6.8.

Exercises 4.6.3 Exercises

1.

2.

3.

4.

5.

6.

Section 4.6 Exact and closed $k$ -forms

Subsection 4.6.1 Exact and closed $k$ -forms

Definition 4.6.1. Exact and closed $k$ -forms.

Example 4.6.2. Exact and closed one-forms in $R^{3}$ .

Lemma 4.6.3. Exact $k$ -forms are closed.

Theorem 4.6.4. Poincare's lemma for $k$ -forms, version 1.

Theorem 4.6.5. Poincare's lemma for $k$ -forms, version II.

Subsection 4.6.2 Translation into vector calculus for $R^{3}$