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Section 3.3 The pullback of a one-form along a curve

We are moving closer to our goal of introducing the integral of one-forms along curves in \(\mathbb{R}^2\) and \(\mathbb{R}^3\text{.}\) The key insight will be to define integration using parametric curves. It is not so easy to deal with a curve as a subset of \(\mathbb{R}^2\) or \(\mathbb{R}^3\text{.}\) However, given a parametrization, we have a map from an interval to the curve. This map gives us a way to do calculus on a curve, by pulling back to the interval, where we know how to do calculus. This is how we will define integration in the next section. But to do this, we need to understand how to pull back a one-form along a curve.

Subsection 3.3.1 The pullback of a one-form along a curve

To be able to do integrate over curves, we will use a parametrization to pull back the integrand (a one-form) to an interval in \(\mathbb{R}\text{.}\) To do this we need to generalize the notion of pullback of one-forms in Definition 2.3.4 to parametrizations.

Definition 3.3.1. The pullback of a one-form along a curve in \(\mathbb{R}^2\) or \(\mathbb{R}^3\).

Let \(\alpha: [a,b] \to \mathbb{R}^2\) be a parametric curve as in Definition 3.2.2, with \(\alpha(t) = (x(t), y(t))\text{,}\) and let \(\omega = f(x,y)\ dx + g(x,y)\ dy\) be a one-form on an open subset \(U \subseteq \mathbb{R}^2\) containing \(C = \text{Im}(\alpha)\text{.}\) The pullback of \(\omega\), denoted by \(\alpha^* \omega\text{,}\) is a one-form on an open subset \(V \subseteq \mathbb{R}\) containing \([a,b]\) defined by

\begin{equation*} \alpha^* \omega = \left(f(x(t), y(t)) \frac{dx}{dt} + g(x(t),y(t)) \frac{dy}{dt} \right)\ dt. \end{equation*}

Similarly, given a parametric curve \(\alpha: [a,b] \to \mathbb{R}^3\text{,}\) with \(\alpha(t) = (x(t), y(t), z(t))\text{,}\) and a one-form \(\omega = f(x,y,z)\ dx + g(x,y,z)\ dy + h(x,y,z)\ dz \) on \(U \subseteq \mathbb{R}^3\text{,}\) the pullback \(\alpha^* \omega\) is given by:

\begin{equation*} \alpha^* \omega = \left( f(x(t), y(t), z(t)) \frac{dx}{dt} + g(x(t), y(t), z(t)) \frac{dy}{dt} + h(x(t), y(t), z(t)) \frac{dz}{dt} \right) \ dt. \end{equation*}

What is important to note here is that \(\alpha^* \omega\) is a one-form on \(\mathbb{R}\text{:}\) as you can see, it is written as a function of the single variable \(t\text{,}\) times \(dt\text{.}\) We start with a one-form on \(\mathbb{R}^2\) or \(\mathbb{R}^3\text{,}\) but we use the parametric curve to pull it back to \(\mathbb{R}\text{.}\)

Consider the counterclockwise parametrization of the unit circle introduced in Example 3.2.3, given by the function \(\alpha:[0,2\pi] \to \mathbb{R}^2\) with \(\alpha(\theta) = (\cos \theta, \sin \theta).\) Let \(\omega = x^2 y\ dx + e^y\ dy\) be a one-form on \(\mathbb{R}^2\text{.}\) We can pull it back along the parametrized unit circle to get a new one-form on \(\mathbb{R}\text{:}\)

\begin{align*} \alpha^* \omega =\amp \left( (\cos \theta)^2 \sin \theta \frac{d}{d \theta} \cos \theta + e^{\sin \theta} \frac{d}{d \theta} \sin \theta \right) \ d\theta \\ =\amp \left( - \cos^2 \theta \sin^2 \theta + \cos \theta e^{\sin \theta} \right)\ d \theta. \end{align*}

Subsection 3.3.2 Reformulation in terms of the associated vector field

As is becoming customary, we can translate between the language of differential forms and the language of vector fields.

This is simply a translation of Definition 3.3.1 in terms of the associated vector fields.

Exercises 3.3.3 Exercises

1.

Consider the one-form \(\omega = x y\ dx + z^2\ dy + z\ dz\) on \(\mathbb{R}^3\text{.}\) Find its pullback along the helix centered around the \(z\)-axis and with tangent vector \(\mathbf{T}(t) = (-3 \sin t, 3 \cos t, 4)\text{,}\) \(0 \leq t \leq 2 \pi\text{,}\) and initial position \(\mathbf{r}(0) = (3,0,1)\text{.}\)

Solution.

First, we find a parametrization for the helix. Integrating the tangent vector, we know that the parametrization must be given by \(\alpha:[0,2 \pi] \to \mathbb{R}^3\) with \(\alpha(t) = (3 \cos t + A, 3 \sin t + B, 4 t + C)\) for some constants \(A,B,C\text{.}\) Using the fact that \(\alpha(0) = (3,0,1)\text{,}\) we conclude that \(A=0\text{,}\) \(B=0\text{,}\) and \(C = 1\text{.}\) Thus the parametrization is \(\alpha(t) = (3 \cos t, 3 \sin t, 4 t + 1)\text{.}\)

We can then calculate the pullback \(\alpha^* \omega\text{.}\) We get:

\begin{align*} \alpha^* \omega =\amp \left( (3 \cos t)(3 \sin t) (-3 \sin t) + (4t+1)^2 (3 \cos t) + (4 t + 1) (4) \right) \ dt\\ =\amp \left( - 27 \cos t \sin^2 t + 3 (4t+1)^2 \cos t + 4 (4t+1) \right)\ dt. \end{align*}

2.

Consider the one-form \(\omega = \frac{x}{x^2+y^2}\ dx + \frac{y}{x^2+y^2}\ dy + z\ dz\text{.}\) Explain why you cannot pull it back along the parametric curve \(\alpha:[0,2] \to \mathbb{R}^3\) with \(\alpha(t) = (t-1, t-1, t^2)\text{.}\)

Solution.

The key here is to be careful about the domain of definition of the one-form \(\omega\text{.}\) The largest open subset \(U \subseteq \mathbb{R}^3\) over which \(\omega\) is defined is all points in \(\mathbb{R}^3\) such that \(x^2+y^2 \neq 0\text{,}\) so that the denominators in the component functions do not vanish. But \(x^2 + y^2 = 0\) if and only if \((x,y) = (0,0)\text{,}\) so the domain of definition of \(\omega\) is \(U = \{ (x,y,z) \in \mathbb{R}^3~|~ (x,y) \neq (0,0) \}\text{.}\) In other words, it consists of all points in \(\mathbb{R}^3\) except the points on the \(z\)-axis.

To be able to pull back our one-form along the parametric curve \(\alpha\text{,}\) we must make sure that its image \(C = \text{Im}(\alpha)\) lies within \(U\text{.}\) Unfortunately, we see that at \(t=1\text{,}\) \(\alpha(1) = (0,0,1)\text{,}\) so the parametric curve intersects the \(z\)-axis, which is not in \(U\text{!}\) Thus we can't pull back along \(\alpha\text{.}\)

Just for fun, let's see what would happen if we had naively tried to pull back along \(\alpha\text{.}\) What we would get is:

\begin{equation*} \alpha^* \omega = \left( \frac{t-1}{2(t-1)^2} + \frac{t-1}{2(t-1)^2} + t^2 (2t) \right) \ dt = \left(\frac{1}{t-1} + 2 t^3 \right)\ dt. \end{equation*}

The problem is that this is not defined at \(t=1\text{,}\) which is part of the interval \([0,2]\) for the parametric curve. In other words, the result of the pullback is not actually a well defined one-form on an open subset containing \([0,2]\text{,}\) since it is not defined at \(t=1\text{.}\)

3.

Suppose that \(\omega= d f\) is an exact one-form on some open subset \(U \subseteq \mathbb{R}^2\text{,}\) and that \(\alpha:[a,b]\to \mathbb{R}^2\) is a parametric curve whose image lies within \(U\text{.}\) Show that

\begin{equation*} \alpha^*(d f) = d(\alpha^* f). \end{equation*}

In other words, pullback commutes with exterior derivative, as we already saw in Exercise 2.3.3.3 for the pullback of a one-form in \(\mathbb{R}\text{.}\)

Solution.

The pullback of a function is just composition. If we write \(\alpha(t) = (x(t), y(t))\text{,}\) then

\begin{equation*} \alpha^* f (t) = f(x(t), y(t) ).\text{.} \end{equation*}

It then follows that

\begin{equation*} d(\alpha^* f) = \frac{d f(x(t),y(t))}{d t} \ dt. \end{equation*}

Using the chain rule for functions of two variables, we can rewrite this as follows:

\begin{align*} d(\alpha^* f) =\amp \left( \frac{\partial f}{\partial x} (x(t), y(t)) \frac{dx}{dt} + \frac{\partial f}{\partial y}(x(t), y(t)) \frac{dy}{dt} \right) \ dt \\ =\alpha^* (d f), \end{align*}

where we used the definition of the pullback of a one-form.