Integrating a one-form over an interval in \mathbb{R}

Section 3.1 Integrating a one-form over an interval in $R$

Our goal is to define the integral of a one-form along a curve. But as a starting point, we define how to integrate one-forms over a closed interval in

R .

We attach particular importance to our guiding principles of reparametrization-invariance and orientability.

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Objectives

You should be able to:

Define the integral of a one-form over an interval in $R .$
Determine how the integral changes when reversing the orientation of the interval.
Relate the transformation property of one-forms to the substitution formula for definte integrals.

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Subsection 3.1.1 The integral of a one-form over an interval

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Consider a one-form

ω = f (x) d x

over an open subset

U \subseteq R

that contains the interval

[a, b] .

We define the integral of

ω

over

[a, b] .

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Definition 3.1.1. The integral of a one-form over $[a, b]$ .

We define the integral of $ω$ over $[a, b]$ , with $a \leq b,$ as follows:

\int_{[a, b]} ω = \int_{a}^{b} f (x) d x,

where on the right-hand-side we use the standard definition of definite integrals from calculus.

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Well, that was simple! This is just the standard notion of definite integrals from calculus. All we are doing is introducing some fancy notation for it. Great!

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Example 3.1.2. An example of an integral of a one-form over an interval.

Consider the one-form $ω = x^{3} d x$ on $R .$ Suppose that you want to integrate it over the interval $[0, 1] .$ Then the integral is

\int_{[0, 1]} ω = \int_{0}^{1} x^{3} d x = \frac{1}{4},

where we used the Fundamental Theorem of Calculus to evaluate the integral as usual, since we are back in the realm of the definite integrals that we know and love.

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What is interesting however is to study what our guiding principles of reparametrization-invariance and orientability become in this simple context. Let us start with orientability.

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Subsection 3.1.2 Integrals of one-forms over intervals are oriented

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If we look at Definition 3.1.1, there is something a bit peculiar. On the left-hand-side, we are integrating over an interval

[a, b],

so by definition we must have

a \leq b .

However, on the right-hand-side, we could exchange the limits of integration, and instead of integrating from

a

b,

we could integrate from

b

a .

But what would that correspond to on the left-hand-side?

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What is going on here is that the definition involves an implicit choice of orientation for the interval. Indeed, when we write

\int_{a}^{b} f (x) d x,

we say that we integrate “from

a

b

”: this is a choice of direction, of orientation. We think of the interval

[a, b]

as being implicitly given a choice of direction of increasing real numbers, i.e. from

a

b .

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But we could have decided to consider the interval

[a, b],

but with the opposite choice of orientation, i.e. going from

b

a,

in the direction of decreasing real numbers. That would be another choice of orientation for the interval. Let us be a little more precise.

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Definition 3.1.3. The orientation of an interval.

We define the orientation of an interval in $R$ to be a choice of direction. There are two choices: either in the direction of increasing real numbers, or decreasing real numbers.

Let $a \leq b .$ By $[a, b]_{+},$ we mean the interval $[a, b]$ with the orientation of increasing real numbers, i.e. “from $a$ to $b$ ”. By $[a, b]_{-},$ we denote the same interval but with the orientation of decreasing real numbers (from $b$ to $a$ ).

We define the canonical orientation to be the orientation of increasing real numbers. When we write $[a, b]$ without specifying the orientation, we always mean the interval with its canonical orientation.

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With this clarification, we can extend Definition 3.1.1 to intervals with the opposite choice of orientation.

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Definition 3.1.4. The oriented integral of a one-form.

We define the integral of $ω$ over the oriented interval $[a, b]_{\pm}$ , with $a \leq b,$ as follows:

\int_{[a, b]_{\pm}} ω = \pm \int_{a}^{b} f (x) d x .

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When the orientation is canonical, that is

[a, b]_{+} = [a, b],

we recover our original definition Definition 3.1.1. But when the interval is oriented in the direction of decreasing real numbers, that is

[a, b]_{-},

we get:

\int_{[a, b]_{-}} ω = - \int_{a}^{b} f (x) d x = \int_{b}^{a} f (x) d x,

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where we used properties of definite integrals to exchange the limits of integration. In other words, we can think of the integral on the right-hand-side as “going from

b

a

”, which is consistent with the orientation of decreasing real numbers as

a \leq b .

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Subsection 3.1.3 Orientation-preserving reparametrizations

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So we know that integrals of one-forms over intervals are oriented. What about reparametrization-invariance?

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Suppose that

ω = f (x) d x

is a one-form over an open subset

U \subseteq R

that contains the interval

[a, b] .

What happens to the integral if we do a change of variables

x = ϕ (t) ?

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We know how to study this question, as we know how one-forms transform under changes of variables. The key: the pullback. Let us be a bit more precise.

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Suppose that we are given a function

ϕ : [c, d] \to [a, b],

and assume that

ϕ

can be extended to a

C^{1}

-function¹ on an open set

V \subseteq R

that contains

[c, d]

(we call

V

and “open neighborhood of

[c, d]

”). We know how the one-form changes under this change of variables: as in Section 2.3, we can pullback

ω

to get a new one-form

ϕ^{*} ω

which can be integrated over

[c, d] .

The question is: does the integral change when we do such a change of variables? In other words, is the integral of

ϕ^{*} ω

over

[c, d]

equal to the integral of

ω

over

[a, b] ?

C^{1}

-function is a differentiable function whose derivative is continuous.

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To answer this question we need to be a little more precise. We impose a further requirement on

ϕ :

we require that

ϕ (c) = a

and

ϕ (d) = b,

that is, it maps the left endpoint of the interval

[c, d]

to the left endpoint of the interval

[a, b],

and similarly for the right endpoints. This means that

ϕ

“preserves the orientation” of the intervals: it maps the smallest real number to the smallest one, and the largest one to the largest one. In other words, it preserves the direction of increasing real numbers.

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Note that we can think of

ϕ

as a “reparametrization”, in the sense that we can think of the interval

[a, b] \subset R

as a “curve” in

R,

and

ϕ

as a parametrization of the curve.

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Lemma 3.1.5. Integrals of one-forms over intervals are invariant under orientation-preserving reparametrizations.

Let $ω$ be a one-form on $U \subseteq R$ with $[a, b] \subset U,$ and $ϕ : [c, d] \to [a, b]$ be a function that can be extended to a $C^{1}$ -function on an open neighborhood of $[c, d]$ and such that $ϕ (c) = a$ and $ϕ (d) = b .$ Then

\int_{[c, d]} ϕ^{*} ω = \int_{[a, b]} ω .

Explicitly, $ω = f (x) d x,$ and $ϕ^{*} ω = f (ϕ (t)) \frac{d ϕ}{d t} d t,$ thus the statement becomes

\int_{c}^{d} f (ϕ (t)) \frac{d ϕ}{d t} d t = \int_{a}^{b} f (x) d x .

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Proof.

The proof is clear from the explicit statement

\int_{c}^{d} f (ϕ (t)) \frac{d ϕ}{d t} d t = \int_{a}^{b} f (x) d x,

since this is nothing else but the substitution formula for definite integrals that you proved in Calculus I! Indeed, since $a = ϕ (c)$ and $b = ϕ (d),$ we can rewrite this equation as

\int_{c}^{d} f (ϕ (t)) \frac{d ϕ}{d t} d t = \int_{ϕ (c)}^{ϕ (d)} f (x) d x,

which is the substitution formula if we do the change of variables $x = ϕ (t) .$

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What this means is that we can think of the integral

\int_{C} ω,

where we think of

C = [a, b]

as a “curve” in

R,

as being defined intrinsically in terms of the geometry of the curve and a choice of orientation. It does not matter how we parametrize the curve, as long as we preserve the orientation: the integral is the same.

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Moreover, as stated in the proof of the lemma, reparametrization-invariance is nothing else but the substitution formula for definite integrals. Isn't that cool? What this means is that the substitution formula for definite integrals is simply the statement that integrals of one-forms over intervals are invariant under orientation-preserving reparametrizations, or equivalently that they are invariant under pullback. Neat! The fact that we need to “transform the differential

d x

” when we do a substitution is now clear: it comes from the transformation property for one-forms under changes of variables studied in Section 2.3, formulated mathematically in terms of pullback.

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Subsection 3.1.4 Orientation-reversing reparametrizations

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In the previous lemma we considered functions

ϕ

that preserve the orientation. What happens if instead we consider a function such that

ϕ (c) = b

and

ϕ (d) = a,

i.e. that maps the left endpoint of the interval

[c, d]

to the right endpoint of the interval

[a, b],

and vice-versa? Such a

ϕ

reverses the orientation from increasing real numbers to decreasing real numbers.

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Lemma 3.1.6. Integrals of one-forms over intervals pick a sign under orientation-reversing reparametrizations.

\int_{[c, d]} ϕ^{*} ω = \int_{[a, b]_{-}} ω = - \int_{[a, b]} ω .

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Proof.

From the substitution formula for definite integrals, we know that

\int_{c}^{d} f (ϕ (t)) \frac{d ϕ}{d t} d t = \int_{ϕ (c)}^{ϕ (d)} f (x) d x = \int_{b}^{a} f (x) d x = - \int_{a}^{b} f (x) d x .

We recognize the right-hand-side as $- \int_{[a, b]} ω,$ which completes the proof.

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It is thus not true that the integral is invariant under all reparametrizations: it is only invariant under reparametrizations that perserve the orientation of the interval. Under reparametrizations that reverse the orientation, the integral changes sign, as expected.

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We have thus shown that the integral of a one-form over an interval is both oriented and reparametrization-invariant, our two guiding principles. In the context of definite integrals, orientability reduces to the statement that definite integrals pick a sign when we exchange limits of integration, and reparametrization-invariance to the substitution formula. Neat!

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Exercises 3.1.5 Exercises

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1.

Consider the one-form $ω = x e^{x^{2}} d x .$ Compute the integral of $ω$ over the interval $[1, 2],$ with both choices of orientation.

Solution.

We first calculate $\int_{[1, 2]} ω$ with the canonical choice of orientation in the direction of increasing real numbers, i.e. from $1$ to $2 .$ We get:

\int_{[1, 2]_{+}} ω = \int_{1}^{2} x e^{x^{2}} d x = \frac{1}{2} \int_{1}^{4} e^{u} d u = \frac{1}{2} (e^{4} - e),

where we used the substitution $u = x^{2} .$ As for the other choice of orientation in the direction of decreasing real numbers, i.e. from $2$ to $1,$ we get:

\int_{[1, 2]_{-}} ω = \int_{2}^{1} x e^{x^{2}} d x = \frac{1}{2} \int_{4}^{1} e^{u} d u = - \frac{1}{2} (e^{4} - e),

which is minus the other integral as expected.

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2.

Let $ω = \frac{1}{x^{2}} d x$ be a one-form on $R_{> 0},$ and $ϕ : R \to R_{> 0}$ be given by $ϕ (t) = e^{t} .$ Show that we can write

\int_{[1, e^{3}]} ω = \int_{[0, 3]} e^{- t} d t .

Solution.

The map $ϕ (t) = e^{t}$ can be restricted to the interval $[0, 3] \subset R .$ We see that its image is the interval $[1, e^{3}],$ and that the map is injective. Moreover, $ϕ (0) = 1$ and $ϕ (3) = e^{3},$ so it preserves orientation. Thus we know that

\int_{[0, 3]} ϕ^{*} ω = \int_{[1, e^{3}]} ω .

We calculate the pullback one-form:

ϕ^{*} ω = \frac{1}{(ϕ (t))^{2}} ϕ^{'} (t) d t = \frac{1}{e^{2 t}} e^{t} d t = e^{- t} d t .

Therefore, we conclude that

\int_{[1, e^{3}]} ω = \int_{[0, 3]} e^{- t} d t .

Note that this is just a fancy way of doing a substitution. Indeed, we could write the original integral as follows:

\int_{[1, e^{3}]} ω = \int_{1}^{e^{3}} \frac{1}{x^{2}} d x .

We can do the substitution $x = e^{t},$ and the integral becomes

\int_{[1, e^{3}]} ω = \int_{1}^{e^{3}} \frac{1}{x^{2}} d x = \int_{0}^{3} e^{- t} d t = \int_{[0, 3]} e^{- t} d t,

as claimed. Indeed, as we have seen, orienting-preserving reparametrizations of the integral is just the substitution formula for definite integrals.

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3.

Let $ω = \sin (x^{2}) d x$ be a one-form on $R .$ TRUE or FALSE:

\int_{[1, 4]} ω = 2 \int_{[- 2, - 1]} t \sin (t^{4}) d t .

Solution.

To go from the expression in $x$ to the expression in $t$ we need to do a change of variables. More precisely, we consider the smooth function $ϕ : R \to R_{\geq 0}$ with $ϕ (t) = t^{2} .$ We see that $ϕ : [- 2, - 1] \to [1, 4],$ with $ϕ (- 2) = (- 2)^{2} = 4$ and $ϕ (- 1) = (- 1)^{2} = 1 .$ This means that it changes the orientation on the interval. So we should get

\int_{[- 2, - 1]} ϕ^{*} ω = - \int_{[1, 4]} ω .

By definition of pullback, we get

ϕ^{*} ω = \sin (t^{4}) \frac{d ϕ}{d t} d t = 2 t \sin (t^{4}) d t,

so we conclude that

\int_{[1, 4]} ω = - 2 \int_{[- 2, - 1]} t \sin (t^{4}) d t .

Therefore the statement is FALSE.

For fun, let us check that this is consistent with what we expect from the substitution formula. Recall that the substitution formula tells us that

\int_{c}^{d} f (ϕ (t)) \frac{d ϕ}{d t} d t = \int_{ϕ (c)}^{ϕ (d)} f (x) d x .

In our case, this means that

2 \int_{- 2}^{- 1} t \sin (t^{4}) d t = \int_{4}^{1} \sin (x^{2}) d x .

We recognize the left-hand-side as

\int_{[- 2, - 1]} ϕ^{*} ω,

and the right-hand-side as

- \int_{[1, 4]} ω,

which is consistent with what we wrote above.

Section 3.1 Integrating a one-form over an interval in R

Objectives

Subsection 3.1.1 The integral of a one-form over an interval

Definition 3.1.1. The integral of a one-form over [a,b].

Example 3.1.2. An example of an integral of a one-form over an interval.

Subsection 3.1.2 Integrals of one-forms over intervals are oriented

Definition 3.1.3. The orientation of an interval.

Definition 3.1.4. The oriented integral of a one-form.

Subsection 3.1.3 Orientation-preserving reparametrizations

Lemma 3.1.5. Integrals of one-forms over intervals are invariant under orientation-preserving reparametrizations.

Proof.

Subsection 3.1.4 Orientation-reversing reparametrizations

Lemma 3.1.6. Integrals of one-forms over intervals pick a sign under orientation-reversing reparametrizations.

Proof.

Exercises 3.1.5 Exercises

1.

2.

3.

Section 3.1 Integrating a one-form over an interval in $R$

Definition 3.1.1. The integral of a one-form over $[a, b]$ .