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Section 2.4 The pullback of a one-form

All right, time to get serious! :-) In the previous section we introduced the notion of the pullback of a one-form with respect to a function \(\phi:V \to U\) with \(U,V \subseteq \mathbb{R}\) open subsets. But this notion of pullback can be generalized, and will become essential to develop our theory of integration (in fact, perhaps this class should be called “the power of the pullback”!). In this section we provide a more general definition of the pullback of a one-form.

Let \(\omega\) be a one-form on \(U \subseteq \mathbb{R}^n\text{,}\) where \(n \in \{1,2,3\}\text{,}\) i.e. we can be in one, two, or three dimensions. We now consider a smooth function \(\phi: V \to U\text{,}\) where \(V \subseteq \mathbb{R}^m\text{,}\) with again \(m \in \{1,2,3\}\text{.}\) Note that \(m\) and \(n\) don't have to be the same: we could have, say \(U \subseteq \mathbb{R}^3\text{,}\) and \(V \subseteq \mathbb{R}^2\text{.}\) Our goal is to define the pullback \(\phi^* \omega\text{,}\) which should be a one-form on \(V\text{.}\)

Just to be concrete: we could take, for instance, a one-form \(\omega\) on \(\mathbb{R}^3\text{,}\) and a smooth function \(\phi: \mathbb{R}^2 \to \mathbb{R}^3\text{.}\) The pullback \(\phi^* \omega\) should then be a one-form on \(\mathbb{R}^2\text{.}\)

Note that our notion of pullback should generalize the definition of pullback in Definition 2.3.4, which should consist in the case with \(m=n=1\text{.}\)

Subsection 2.4.1 The pullback of a function

Let us first define the pullback of a function in this context, generalizing Definition 2.3.3.

Definition 2.4.1. The pullback of a function.

Let \(U \subseteq \mathbb{R}^n\) and \(V \subseteq \mathbb{R}^m\) be open subsets, where \(m,n \in \{1,2,3\}\text{.}\) Let \(f: U \to \mathbb{R}\) and \(\phi: V \to U\) be smooth functions. The pullback of \(f\), which is denoted by \(\phi^* f\text{,}\) is the smooth function

\begin{equation*} \phi^* f = f \circ \phi: V \to \mathbb{R}. \end{equation*}

Explicitely, if we write \(\mathbf{t} \in V\) for an \(m\)-dimensional vector in \(V\text{,}\) then

\begin{equation*} \phi^*f (\mathbf{t}) = f(\phi(\mathbf{t})), \end{equation*}

where \(\phi(\mathbf{t})\) is an \(n\)-dimensional vector in \(U\text{.}\)

To make things more concrete, let us look at a specific example. Suppose that \(f\) is a smooth function on \(\mathbb{R}^3\text{,}\) that is \(f : \mathbb{R}^3 \to \mathbb{R}\text{.}\) Let \(\phi: \mathbb{R} \to \mathbb{R}^3\) be another smooth function, which takes a point in \(\mathbb{R}\) and maps it to a vector in \(\mathbb{R}^3\) (so it is a vector-valued function). We can write \(f\) explicitly as \(f=f(x,y,z)\text{.}\) As for the vector-valued function \(\phi\text{,}\) we write \(\phi(t) = (x(t),y(t), z(t) )\text{.}\) Then the pullback \(\phi^* f : \mathbb{R} \to \mathbb{R}\) is simply the composition:

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

For instance, if \(f(x,y,z) = x y + z\text{,}\) and \(\phi(t) = (t, t^2, 1)\text{,}\) then

\begin{equation*} \phi^* f(t) = f(t, t^2, 1) = t^3 + 1. \end{equation*}

We can do the same thing but pulling back to \(\mathbb{R}^2\) instead of \(\mathbb{R}\text{.}\) Suppose that \(f\) is a smooth function on \(\mathbb{R}^3\text{,}\) that is \(f : \mathbb{R}^3 \to \mathbb{R}\text{.}\) Let \(\phi: \mathbb{R}^2 \to \mathbb{R}^3\) be another smooth function, which takes a point in \(\mathbb{R}^2\) and maps it to a vector in \(\mathbb{R}^3\text{.}\) We can write \(f\) explicitly as \(f=f(x,y,z)\text{.}\) As for the vector-valued function \(\phi\text{,}\) we write \(\phi(t_1, t_2) = (x(t_1, t_2),y(t_1, t_2), z(t_1, t_2) )\text{.}\) Then the pullback \(\phi^* f : \mathbb{R} \to \mathbb{R}\) is simply the composition:

\begin{equation*} \phi^* f(t_1 t_2) = f(\phi(t_1, t_2)) = f(x(t_1, t_2), y(t_1, t_2), z(t_1, t_2) ). \end{equation*}

For instance, if \(f(x,y,z) = x y + z\text{,}\) and \(\phi(t_1, t_2) = (t_1, t_2^2, t_1+t_2)\text{,}\) then

\begin{equation*} \phi^* f(t_1, t_2) = f(t_1, t_2^2, t_1+t_2) = t_1 t_2^2 + t_1+t_2. \end{equation*}

Subsection 2.4.2 An axiomatic definition of the pullback of a one-form

We will take an axiomatic approach to the definition of the pullback of a one-form. Let us first recall three important properties of one-forms (from Subsection 2.1.1 and Definition 2.2.5):

  1. If \(\omega\) and \(\eta\) are one-forms on \(U\text{,}\) then \(\omega + \eta\) is a one-form on \(U\text{.}\)

  2. If \(\omega\) is a one-form on \(U\) and \(f\) a smooth function on \(U\text{,}\) then \(f \omega\) is a one-form on \(U\text{.}\)

  3. An exact one-form is a one-form \(\omega\) that can be written as the differential of a function \(f\) on \(U\text{:}\) \(\omega = d f\text{.}\)

We now want the pullback to be consistent with these properties. More precisely, we require that the pullback \(\phi^*\) satisfies the following properties:

  1. \(\phi^*(\omega + \eta) = \phi^* \omega + \phi^* \eta\text{.}\)

  2. \(\phi^* (f \omega) = (\phi^* f) (\phi^* \omega)\text{.}\)

  3. \(\phi^* ( d f ) = d (\phi^* f)\text{.}\)

It turns out that this is completely sufficient to fully determine the pullback of any one-form. Let us see why.

This follows from the third axiomatic property that we are imposing on the pullback. Recall from Remark 2.2.4 that we can think of \(dx\) as the differential \(d f\) of the function \(f(x,y,z) = x\text{.}\) By the third property of pullbacks, we want to impose that

\begin{equation*} \phi^*(dx) = \phi^*(df) = d(\phi^* f). \end{equation*}

From Definition 2.4.1, we can calculate \(\phi^* f\text{.}\) We get \(\phi^* f(\mathbf{t}) = x(\mathbf{t})\text{.}\) We thus obtain

\begin{equation*} \phi^*(dx) = d x(\mathbf{t}) = \sum_{i=1}^m \frac{\partial x}{\partial t_i} d t_i, \end{equation*}

where we use the definition of the differential of the function \(x(\mathbf{t})\text{.}\)

This result enables us to write down a general formula for the pullback of a one-form. For clarity, we will only write it down for a one-form on \(\mathbb{R}^3\text{,}\) but it is clear what the similar result should be for a one-form in \(\mathbb{R}^2\) or \(\mathbb{R}\text{.}\)

To prove this result, we use Lemma 2.4.4 (and the similar result for \(dy\) and \(dz\)), and the first and second axiomatic properties that we are imposing on the pullback. Using the first and second properties, we can write:

\begin{equation*} \phi^* \omega = \phi^*(f dx + g dy+ h dz) = (\phi^* f) \phi^*(dx) + (\phi^* g) \phi^*(dy) + (\phi^* h) \phi^*(dz). \end{equation*}

We then use Lemma 2.4.4 to evaluate \(\phi^*(dx), \phi^*(dy)\) and \(\phi^*(dz)\text{,}\) and from Definition 2.4.1 we know that \(\phi^* f(\mathbf{t}) = f(\phi(\mathbf{t}))\text{,}\) and similarly for \(g\) and \(h\text{.}\)

Suppose that \(\omega = f dx + g dy + h dz\) is a one-form on \(\mathbb{R}^3\text{.}\) Let \(\phi: \mathbb{R} \to \mathbb{R}^3\) be a smooth function, which takes a point in \(\mathbb{R}\) and maps it to a vector in \(\mathbb{R}^3\text{.}\) We write \(\phi(t) = (x(t),y(t), z(t) )\text{.}\) Then the pullback \(\phi^* \omega\) is a one-form on \(\mathbb{R}\) given by:

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

For instance, if \(\omega = x dx + xy dy + z^2 dz\text{,}\) and \(\phi(t) = (x(t), y(t), z(t)) = (t^2, t, 1) \text{,}\) then

\begin{align*} \phi^* \omega =\amp \left( x(t) \frac{dx}{dt} + x(t) y(t) \frac{dy}{dt} + z(t)^2 \frac{dz}{dt} \right) dt \\ =\amp \left( (t^2)(2 t) + (t^2)(t)(1) + (1)(0) \right) dt\\ =\amp 3 t^3 dt. \end{align*}

Suppose that \(\omega = f dx + g dy + h dz\) is a one-form on \(\mathbb{R}^3\text{.}\) Let \(\phi: \mathbb{R}^2 \to \mathbb{R}^3\) be a smooth function, which takes a point in \(\mathbb{R}^2\) and maps it to a vector in \(\mathbb{R}^3\text{.}\) We write \(\phi(\mathbf{t}) = (x(\mathbf{t}),y(\mathbf{t}), z(\mathbf{t}) )\text{,}\) with \(\mathbf{t} = (t_1, t_2)\text{.}\) Then the pullback \(\phi^* \omega\) is a one-form on \(\mathbb{R}^2\) given by:

\begin{align*} \phi^* \omega =\amp f(\phi(\mathbf{t})) \left(\frac{\partial x}{\partial t_1} d t_1 + \frac{\partial x}{\partial t_2} d t_2 \right) +g(\phi(\mathbf{t})) \left(\frac{\partial y}{\partial t_1} d t_1 + \frac{\partial y}{\partial t_2} d t_2 \right)\\ \amp+h(\phi(\mathbf{t})) \left(\frac{\partial z}{\partial t_1} d t_1 + \frac{\partial z}{\partial t_2} d t_2 \right) . \end{align*}

For instance, if \(\omega = x dx + xy dy + z^2 dz\text{,}\) and \(\phi(\mathbf{t}) = (x(\mathbf{t}), y(\mathbf{t}), z(\mathbf{t})) = (t_1 t_2, t_2, t_1+t_2) \text{,}\) then

\begin{align*} \phi^* \omega =\amp x(\phi(\mathbf{t})) \left(\frac{\partial x}{\partial t_1} d t_1 + \frac{\partial x}{\partial t_2} d t_2 \right) +x(\phi(\mathbf{t})) y(\phi(\mathbf{t})) \left(\frac{\partial y}{\partial t_1} d t_1 + \frac{\partial y}{\partial t_2} d t_2 \right)\\ \amp+z(\phi(\mathbf{t}))^2 \left(\frac{\partial z}{\partial t_1} d t_1 + \frac{\partial z}{\partial t_2} d t_2 \right) .\\ =\amp (t_1 t_2) (t_2 d t_1 + t_1 d t_2) + (t_1 t_2)(t_2)( dt_2) + (t_1+t_2)^2 (dt_1 + dt_2)\\ =\amp (t_1 t_2^2 + (t_1+t_2)^2 )d t_1 + (t_1^2 t_2 + t_1 t_2^2 + (t_1+t_2)^2) d t_2. \end{align*}

As a consistency check, we show that the pullback of a one-from from \(\mathbb{R}\) to \(\mathbb{R}\) reduces to Definition 2.3.4. Let \(\omega = f(x) dx\) on \(U \subseteq \mathbb{R}\text{,}\) and \(\phi: V \to U\) with \(V \subseteq \mathbb{R}\text{.}\) We write \(\phi(t) = x(t)\text{.}\) Then the pullback \(\phi^* \omega\) is the one-form on \(V\) given by:

\begin{equation*} \phi^* \omega =\left( f(x(t)) \frac{dx}{dt} \right) dt, \end{equation*}

which indeeds reproduces Definition 2.3.4 with our notation \(\phi(t) = x(t)\text{.}\)

We now have a very general definition of the pullback of a one-form. This will turn out to be very useful to define the integral of a one-form, which is what we now turn to.

Exercises 2.4.3 Exercises

1.

Consider the function \(f: \mathbb{R}^2 \to \mathbb{R}\) given by \(f(x,y) = e^{x+y}+ x + y\text{,}\) and the function \(\phi: \mathbb{R}^3 \to \mathbb{R}^2\) given by \(\phi(u,v,w) = (u + v, v+w)\text{.}\) Find the pullback \(\phi^* f\text{.}\) What is its domain?

Solution.

First, as \(f : \mathbb{R}^2 \to \mathbb{R}\) and \(\phi: \mathbb{R}^3 \to \mathbb{R}^2\text{,}\) we see that the composition \(\phi^* f = f \circ \phi: \mathbb{R}^3 \to \mathbb{R}^2 \to \mathbb{R}\text{,}\) i.e. the pullback \(\phi^* f\) is a function from \(\mathbb{R}^3 \) to \(\mathbb{R}\text{.}\) So its domain is \(\mathbb{R}^3\text{.}\)

We calculate its expression by composition:

\begin{align*} \phi^* f(u,v,w) =\amp f ( u+v, v+w) \\ =\amp e^{(u+v)+(v+w)} + (u+v) + (v+w) \\ =\amp e^{u+2v +w} + u + 2v + w. \end{align*}

2.

Consider the one-form \(\omega = x^2\ dx\) on \(\mathbb{R}\text{,}\) and the function \(\phi: \mathbb{R}^2 \to \mathbb{R}\) given by \(\phi(u,v) = u\text{,}\) which projects on the \(u\)-axis. Find the pullback one-form \(\phi^* \omega\) on \(\mathbb{R}^2\text{.}\) Interpret the result.

Solution.

Let us write \(\omega = f\ dx = x^2\ dx\text{.}\) By the definition of pullback, we get:

\begin{align*} \phi^* \omega =\amp f(\phi(u,v)) \left( \frac{\partial \phi}{\partial u}\ du + \frac{\partial \phi}{\partial v}\ dv \right)\\ =\amp u^2 \left( 1\ du + 0\ dv \right)\\ = \amp u^2\ du. \end{align*}

We see that the pullback one-form looks the same, but written in terms of \(u\) instead of \(x\text{.}\) However, \(\phi^* \omega\) is defined on \(\mathbb{R}^2\text{,}\) while \(\omega\) was defined on \(\mathbb{R}\text{.}\) Since the function \(\phi\) here simply projects on the \(u\)-axis, what the pullback does here is extend the one-form uniformly in the \(v\)-coordinate on the \(uv\)-plane; at any two points \((u, v_1)\) and \((u,v_2)\text{,}\) the one-form will be the same. Conceptually, this is what happens when we pullback using a “forgetful map”, i.e. a map that somehow “forgets” some information (in this case, the \(v\)-coordinate). The pullback then extends the object uniformly across the forgotten structure.

3.

Consider the one-form \(\omega = x^2\ dx + y^2\ dy\) on \(\mathbb{R}^2\text{,}\) and the map \(\Phi: \mathbb{R}^2 \to \mathbb{R}^2\) with \(\Phi(r,\theta) = (r \cos \theta, r \sin \theta)\text{,}\) which defines polar coordinates. Find the pullback \(\Phi^* \omega\text{.}\)

Solution.

We write \(\omega = f\ dx + g\ dy = x^2\ dx + y^2 \ dy\text{,}\) and \(\Phi(r,\theta) = (x(r,\theta), y(r, \theta) )\text{.}\) Then:

\begin{align*} \Phi^* \omega =\amp f(\Phi(r,\theta)) \left( \frac{\partial x}{\partial r}\ dr + \frac{\partial x}{\partial \theta}\ d \theta \right) + g(\Phi(r,\theta)) \left( \frac{\partial y}{\partial r}\ dr + \frac{\partial y}{\partial \theta}\ d \theta \right) \\ =\amp r^2 \cos^2 \theta \left( \cos \theta\ dr - r \sin \theta \ d\theta \right) + r^2 \sin^2\theta \left( \sin \theta\ dr + r \cos \theta\ d\theta \right)\\ =\amp r^2 (\cos^3 \theta + \sin^3 \theta)\ dr + r^3 (\sin^2 \theta \cos \theta - \cos^2 \theta \sin \theta )\ d \theta. \end{align*}

The notion of pullback allows us to easily calculate how one-forms change under changes of coordinates, such as going from Cartesian to polar coordinates in this case.

4.

Consider the one-form \(\omega = \frac{z^2\ dz}{\sqrt{x^2+y^2}}\) on \(U = \{(x,y,z) \in \mathbb{R}^3 \ | \ (x,y,z) \neq (0,0,z) \}\) (this is \(\mathbb{R}^3\) with the \(z\)-axis removed), and the function \(\phi: V \to U\) with \(\phi(r,\theta,\zeta) = r (\cos \theta, \sin \theta, \zeta)\text{,}\) and \(V = \{ (r, \theta, \zeta)\in \mathbb{R}^3\ | \ r \neq 0 \}\text{.}\) Determine the pullback one-form \(\phi^* \omega\text{.}\)

Solution.

By definition of the pullback, we get:

\begin{align*} \phi^* \omega =\amp \frac{1}{r} r^2 \zeta^2 \left( \frac{\partial}{\partial r} (r \zeta) \ dr + \frac{\partial}{\partial \theta}(r \zeta)\ d \theta + \frac{\partial}{\partial \zeta}(r \zeta) d \zeta \right)\\ =\amp r \zeta^2 (\zeta\ dr + r\ d \zeta). \end{align*}

5.

Let \(\omega\) be a one-form on \(\mathbb{R}^3\text{,}\) and \(Id: \mathbb{R}^3 \to \mathbb{R}^3\) the identity function defined by \(Id(x,y,z) = (x,y,z)\text{.}\) Show that \(Id^* \omega = \omega\text{.}\)

Solution.

Write \(\omega = f\ dx + g\ dy +h\ dz\text{.}\) By definition of the pullback, we get:

\begin{align*} Id^* \omega =\amp f(x,y,z) \left( \frac{\partial}{\partial x}(x)\ dx + \frac{\partial}{\partial y}(x)\ dy + \frac{\partial}{\partial z}(x)\ dz \right) \\ \amp+ g(x,y,z) \left( \frac{\partial}{\partial x}(y)\ dx + \frac{\partial}{\partial y}(y)\ dy + \frac{\partial}{\partial z}(y)\ dz \right)\\ \amp+h(x,y,z) \left( \frac{\partial}{\partial x}(z)\ dx + \frac{\partial}{\partial y}(z)\ dy + \frac{\partial}{\partial z}(z)\ dz \right) \\ =\amp f(x,y,z)\ dx + g(x,y,z)\ dy + h(x,y,z)\ dz\\ =\amp \omega, \end{align*}

which completes the proof.

6.

Let \(\omega\) be a one-form \(U \subseteq \mathbb{R}\text{,}\) and \(\phi:V \to U\) and \(\alpha:W \to V\) be smooth functions, with \(V,W \subseteq \mathbb{R}\) open subsets. Show that

\begin{equation*} (\phi \circ \alpha)^* \omega = \alpha^* (\phi^* \omega)\text{.} \end{equation*}

In other words, it doesn't matter whether we pullback in one or two steps through the chain of maps \(W \overset{\alpha}{\to} V \overset{\phi}{\to} U\text{.}\)

We note here that while the exercise is only asking you to prove it for open subsets \(U,V,W \subseteq \mathbb{R}\text{,}\) this property is true in general, not just in \(\mathbb{R}\text{.}\)

Solution.

Let us write \(\omega = f(x)\ dx\text{,}\) \(\phi=\phi(t)\text{,}\) and \(\alpha=\alpha(u)\text{.}\) On the one hand, we have:

\begin{equation*} (\phi \circ \alpha)^* \omega = f(\phi(\alpha(u))) \frac{d}{d u}( \phi(\alpha(u))) \ du. \end{equation*}

On the other hand, we have

\begin{equation*} \phi^* \omega = f(\phi(t)) \frac{d}{dt}(\phi(t))\ dt, \end{equation*}

and

\begin{equation*} \alpha^* (\phi^* \omega) = f(\phi(\alpha(u)) \left( \frac{d}{dt} \phi(t) \right)\Big |_{t=\alpha(u)} \frac{d}{du} \alpha(u)\ du. \end{equation*}

But

\begin{equation*} \frac{d}{d u}( \phi(\alpha(u))) = \left( \frac{d}{dt} \phi(t) \right)\Big |_{t=\alpha(u)} \frac{d}{du} \alpha(u) \end{equation*}

by the chain rule, and hence \((\phi \circ \alpha)^* \omega = \alpha^* (\phi^* \omega).\)