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Section 4.4 The exterior derivative and vector calculus

In this section we continue developing our dictionary between differential forms and standard vector calculus concepts. We introduce the vector calculus operations corresponding to the exterior derivative.

Subsection 4.4.1 Grad, div and curl

In \(\mathbb{R}^3\text{,}\) we saw that there are three possibilities to get a non-zero differential forms as a result of acting with the exterior derivative: either we take the exterior derivative of a zero-form, a one-form, or a two-form. All three of these operations are given separate names and notation in the standard vector calculus language.

Definition 4.4.1. The gradient of a function.

Let \(f\) be a zero-form (a function) on \(U \subseteq \mathbb{R}^3\text{.}\) Its exterior derivative \(df\) is the one-form

\begin{equation*} d f = \frac{\partial f}{\partial x}\ dx + \frac{\partial f}{\partial y}\ dy + \frac{\partial f}{\partial z}\ dz. \end{equation*}

We define the gradient of \(f\), and denote it by \(\boldsymbol{\nabla} f\text{,}\) to be the vector field associated to the one-form \(df\) according to Table 4.1.11:

\begin{equation*} \boldsymbol{\nabla f} = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right). \end{equation*}

Note that the input of the gradient is a function, and the output is a vector field.

Of course, we already knew the definition of the gradient of a function, and how it is the vector field associated to the exterior derivative of a zero-form: this was already stated in Fact 2.2.2. We include the statement here for completeness. From our point of view, we could take this as the definition of the gradient of a function: it is the vector field associated to the exterior derivative of a zero-form.

Definition 4.4.2. The curl of a vector field.

Let \(\omega = f_1\ dx + f_2\ dy+ f_3 \ dz\) be a one-form on \(U \subseteq \mathbb{R}^3\text{,}\) with its associated vector field \(\mathbf{F} = (f_1,f_2,f_3)\text{.}\) Its exterior derivative \(d \omega\) is the two-form

\begin{equation*} d \omega = \left( \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} \right) dy \wedge dz + \left( \frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x} \right) dz \wedge dx + \left( \frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} \right) dx \wedge dy. \end{equation*}

We define the curl of \(\mathbf{F}\), and denote it by \(\boldsymbol{\nabla} \times \mathbf{F}\text{,}\) to be the vector field associated to the two-form \(d \omega\) according to Table 4.1.11:

\begin{equation*} \boldsymbol{\nabla} \times \mathbf{F} = \left( \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} , \frac{\partial f_1}{\partial z} - \frac{\partial f_3}{\partial x}, \frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} \right) . \end{equation*}

Note that input of the curl is a vector field, and the output is also a vector field.

Finally, we can apply the exterior derivative to a two-form to get a three-form.

Definition 4.4.3. The divergence of a vector field.

Let \(\eta = f_1\ dy \wedge dz + f_2 \ dz \wedge dx + f_3\ dx \wedge dy\) be a two-form on \(U \subseteq \mathbb{R}^3\text{,}\) with its associated vector field \(\mathbf{F} = (f_1,f_2,f_3)\text{.}\) Its exterior derivative \(d \eta\) is the three-form

\begin{equation*} d \eta = \left( \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z} \right )dx \wedge dy \wedge dz. \end{equation*}

We define the divergence of \(\mathbf{F}\), and denote it by \(\boldsymbol{\nabla} \cdot \mathbf{F}\text{,}\) to be the function associated to the three-form \(d \eta\) according to Table 4.1.11:

\begin{equation*} \boldsymbol{\nabla}\cdot \mathbf{F} = \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} + \frac{\partial f_3}{\partial z}. \end{equation*}

Note that the input of the diveregence is a vector field, and the output is a function.

Remark 4.4.4.

Now you can start to see the power of developing the framework of differential forms. These three operators, namely grad, div, and curl, which appear as independent operators in vector calculus, are just the action of the same operator, namely the exterior derivative, but on zero-, one-, and two-forms respectively. Moreover, we don't need to remember these definitions by heart: all we need to remember is how to act with the exterior derivative on \(k\)-forms, which simply amounts to acting with the exterior derivative on the component functions. So much simpler!

Even more powerful is the fact that the framework of differential forms naturally extend to any dimension, not only \(\mathbb{R}^3\text{.}\) However, the defintions of grad, curl, div, the cross-product, etc. rely on the geometry of \(\mathbb{R}^3\text{.}\) The natural generalization to higher dimensions is just the action of the exterior derivative as we defined it.

Remark 4.4.5.

The introduction of the curl of a vector field allows us to rephrase the screening test for conservative vector fields in \(\mathbb{R}^3\) in a nicer way. Looking at the screening test in Lemma 2.2.16, it is clear that the screening test for a vector field \(\mathbf{F}\) is satisfied if and only if

\begin{equation*} \boldsymbol{\nabla} \times \mathbf{F} = 0. \end{equation*}

In other words, the screening test was simply saying that the curl of the vector field vanishes.

Remark 4.4.6.

Just as for the cross product of two vectors, in standard vector calculus textbooks a determinant formula is usually given to remember how to calculate the curl of a vector field \(\mathbf{F}=(f_1, f_2, f_3)\) in \(\mathbb{R}^3\text{:}\)

\begin{equation*} \boldsymbol{\nabla} \times \mathbf{F} = \det \begin{pmatrix} \mathbf{i} \amp \mathbf{j} \amp \mathbf{k} \\ \frac{\partial}{\partial x} \amp \frac{\partial}{\partial y} \amp \frac{\partial}{\partial z} \\ f_1 \amp f_2 \amp f_3 \end{pmatrix}, \end{equation*}

where \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors in the \(x,y,z\) directions. You can use this formula if you want. Or you can remember that the curl is obtained by taking the exterior derivative of the one-form associated to \(\mathbf{F}\text{.}\)

Maxwell's equations form the foundations of electromagnetism. It turns out that they are written in terms of the divergence and the curl. More precisely, if \(\mathbf{E}\) is the electric vector field, and \(\mathbf{B}\) is the magnetic vector field, both defined on \(\mathbb{R}^3\) (our space), Maxwell's equations state that

\begin{align*} \boldsymbol{\nabla} \cdot \mathbf{E} =\amp 4 \pi \rho,\\ \boldsymbol{\nabla} \cdot \mathbf{B} =\amp 0,\\ \boldsymbol{\nabla} \times \mathbf{E} + \frac{1}{c} \frac{\partial \mathbf{B}}{\partial t} =\amp 0,\\ \boldsymbol{\nabla} \times \mathbf{B} - \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} =\amp \frac{4 \pi}{c} \mathbf{J}, \end{align*}

where \(c\) is the speed of light, \(\rho\) is the total electric charge density, and \(\mathbf{J}\) is the total electric current density (which is a vector field). In particular, the equations simplify when there is no charge or current (such as in vacuum), with \(\rho = \mathbf{J} = 0\text{.}\)

Note that we are abusing notation a little bit here. Those equations are the “time-dependent” Maxwell's equations. What this means is that we think of \(\mathbf{E}\) and \(\mathbf{B}\) as vector fields in \(\mathbb{R}^3\) (in space), but that also depend on another variable \(t\) corresponding to time. This is why the equations above include partial derivatives of \(\mathbf{E}\) and \(\mathbf{B}\) with respect to \(t\text{.}\) The “time-independent” Maxwell's equations, in which \(\mathbf{E}\) and \(\mathbf{B}\) are true vector fields on \(\mathbb{R}^3\) (with no time dependence), would correspond to setting the two terms involving partial derivatives with respect to \(t\) to zero.

We can summarize the dictionary between the exterior derivative in \(\mathbb{R}^3\) and vector calculus operations in the following table.

Table 4.4.8. Dictionary between the exterior derivative in \(\mathbb{R}^3\) and vector calculus concepts
Differential form concept Vector calculus concept
\(d\) of a 0-form \(d f\) gradient \(\boldsymbol{\nabla} f\)
\(d\) of a 1-form \(d \omega\) curl \(\boldsymbol{\nabla} \times \mathbf{F}\)
\(d\) of a 2-form \(d \eta\) divergence \(\boldsymbol{\nabla} \cdot \mathbf{F}\)

Subsection 4.4.2 The graded product rule and vector calculus identities

The power of the formalism of differential forms is further highlighted by the following lemma. We saw above that the standard concepts of grad, curl, and div, are just reformulations of the exterior derivative. We showed in Lemma 4.3.6 that the exterior derivative satisfies a graded product rule. In \(\mathbb{R}^3\text{,}\) this graded product rule can be split into two cases, as in Remark 4.3.8, depending on whether \(\omega\) is a zero- or a one-form. Using the definition of grad, curl, and div in Definition 4.4.1, Definition 4.4.2 and Definition 4.4.3, those statements can be translated into vector calculus identities. The result is the following lemma.

This is just the reformulation in terms of vector fields of the four different non-vanishing cases of the graded product rule for differential forms in \(\mathbb{R}^3\) (see Remark 4.3.8).

Now you may be starting to like this. Learning this kind of vector calculus identities by heart is frustrating. But these are just reformulations of the one and only graded product rule for the exterior derivative Lemma 4.3.6, which is all that you have to remember (sure, there is an annoying sign in the graded product rule, but it's much better than learning vector calculus identities!).

Subsection 4.4.3 \(d^2=0\) and vector calculus identities

Another key property of the exterior derivative is that \(d^2=0\text{,}\) see Lemma 4.3.9. In \(\mathbb{R}^3\text{,}\) this corresponds to two separate statements, namely that \(d(d(f)) =0 \) with \(f\) a zero-form (a function), and \(d(d(\omega))=0\) with \(\omega\) a one-form. The corresponding vector calculus identities are the following:

This is just the reformulation of the statement that \(d^2=0\) for a zero-form and a one-form in \(\mathbb{R}^3\text{.}\)

The game of translating easy statements for the exterior derivative into complicated statements for grad, curl, and div is fun, isn't it? Let's prove one more vector calculus identity for now, which follows by combining the statement that \(d^2=0\) with the graded product rule.

Consider the action of the exterior derivative on the two-form \(f d g \wedge dh\text{:}\)

\begin{equation*} d (f dg \wedge dh ) = d f \wedge (dg \wedge dh) + f d( dg \wedge dh), \end{equation*}

where we used the graded product rule and the fact that \(f\) is a zero-form. Using the graded product rule again, we know that

\begin{align*} d(dg \wedge dh) =\amp d(dg) \wedge dh - dg \wedge d(dh)\\ =\amp 0, \end{align*}

since \(d(dg) = d(dh) = 0\text{.}\) Thus

\begin{equation*} d (f dg \wedge dh ) = d f \wedge (dg \wedge dh). \end{equation*}

The translation for the associated vector fields is:

\begin{equation*} \boldsymbol{\nabla} \cdot ( f (\boldsymbol{\nabla g} \times \boldsymbol{\nabla} h) ) = \boldsymbol{\nabla} f \cdot (\boldsymbol{\nabla} g \times \boldsymbol{\nabla} h ), \end{equation*}

as claimed.

Subsection 4.4.4 Two more vector calculus identities

We end this section by noting that there are two more vector calculus identities involving grad, curl and div. We will present the identities without proof here. To get them from differentials forms, we would need to introduce the concept of Lie derivatives, which is beyond the scope of this course.

Keep in mind that you certainly do not need to learn these identities by heart! We are presenting them here just so that you are aware of them.

Finally, there are a few more vector calculus identities that involve the Laplacian operator, which in the language of differential forms requires the introduction of the Hodge star operator. We will come back to this in Section 4.8.

Exercises 4.4.5 Exercises


Let \(\mathbf{F} = (x y, y z, x z)\) be a vector field on \(\mathbb{R}^3\text{.}\) Find its curl \(\boldsymbol{\nabla} \times \mathbf{F}\) and divergence \(\boldsymbol{\nabla}\cdot \mathbf{F}\text{.}\)


The curl of the vector field is given by:

\begin{align*} \boldsymbol{\nabla} \times \mathbf{F} =\amp \det \begin{pmatrix} \mathbf{i} \amp \mathbf{j} \amp \mathbf{k} \\ \frac{\partial}{\partial x} \amp \frac{\partial}{\partial y} \amp \frac{\partial}{\partial z} \\ x y \amp y z \amp x z \end{pmatrix}\\ =\amp - y \mathbf{i} - z \mathbf{j} - x \mathbf{k}. \end{align*}

In component notation, this reads \(\boldsymbol{\nabla} \times \mathbf{F}= (-y, -z, -x).\)

As for the div, we get:

\begin{align*} \boldsymbol{\nabla} \cdot \mathbf{F} =\amp \frac{\partial}{\partial x}(x y) + \frac{\partial}{\partial y}(y z) + \frac{\partial}{\partial z}(x z) \\ =\amp y + z + x. \end{align*}

Note that we could have done these calculations using differential forms. To get the curl, we associate to \(\mathbf{F}\) a one-form \(\omega = x y\ dx + y z\ dy+ x z\ dz\) and calculate its exterior derivative:

\begin{align*} d \omega =\amp x dy \wedge dx + y dz \wedge dy + z dx \wedge dz\\ =\amp - y dy \wedge dz - z dz \wedge dx - x dx \wedge dy \end{align*}

The curl \(\boldsymbol{\nabla} \times \mathbf{F}\) is then the vector field associated to this two-form, that is \(\boldsymbol{\nabla} \times \mathbf{F} = (-y, -z, -x)\text{.}\)

To calculate the divergence \(\boldsymbol{\nabla} \cdot \mathbf{F}\text{,}\) we associate to \(\mathbf{F}\) a two-form \(\eta = x y \ dy \wedge dz + y z \ dz \wedge dx + x z \ dx \wedge dy\) and calculate its exterior derivative:

\begin{align*} d \eta =\amp y dx \wedge dy \wedge dz + z dy \wedge dz \wedge dx + x dz \wedge dx \wedge dy \\ =\amp (x+y+z) dx \wedge dy \wedge dz. \end{align*}

Therefore \(\boldsymbol{\nabla} \cdot \mathbf{F} = x + y + z\text{.}\)


For the following two vector fields, find their curl and divergence:

  1. \(\displaystyle \mathbf{F}(x,y,z) = \frac{1}{\sqrt{x^2+y^2}} (x, y, 0).\)

  2. \(\displaystyle \mathbf{G}(x,y,z) = \frac{1}{\sqrt{x^2+y^2}} (- y , x , 0).\)


Let's solve this one using differential forms. You can do it directly using the formulae for curl and div as well.

(a) To find the curl, we associated a one-form \(\omega\) to \(\mathbf{F}\text{:}\)

\begin{equation*} \omega = \frac{1}{\sqrt{x^2+y^2}} (x \ dx + y\ dy). \end{equation*}

We calculate its exterior derivative:

\begin{align*} d \omega =\amp -\frac{x y}{(x^2+y^2)^{3/2}} dy \wedge dx - \frac{x y}{(x^2+y^2)^{3/2}} dx \wedge dy \\ =\amp 0. \end{align*}

Thus we conclude that \(\boldsymbol{\nabla} \times \mathbf{F} = 0\text{.}\)

To find the divergence, we associate a two-form \(\eta\) to \(\mathbf{F}\text{:}\)

\begin{equation*} \eta = \frac{1}{\sqrt{x^2+y^2}} (x \ dy \wedge dz + y\ dz \wedge dx). \end{equation*}

We calulate its exterior derivative:

\begin{align*} d \eta =\amp \left( (x^2+y^2)^{-1/2} - x^2(x^2+y^2)^{-3/2} \right) dx \wedge dy \wedge dz \\ \amp+ \left((x^2+y^2)^{-1/2} - y^2(x^2+y^2)^{-3/2} \right) dy \wedge dz \wedge dx\\ =\amp \left( \frac{2}{\sqrt{x^2+y^2}} - \frac{x^2+y^2}{(x^2+y^2)^{3/2}} \right) dx \wedge dy \wedge dz\\ =\amp \frac{1}{\sqrt{x^2+y^2}} dx \wedge dy \wedge dz. \end{align*}

We conclude that

\begin{equation*} \boldsymbol{\nabla} \cdot \mathbf{F} = \frac{1}{\sqrt{x^2+y^2}}. \end{equation*}

(b) To find the curl, we associate a one-form \(\omega\) to \(\mathbf{G}\text{:}\)

\begin{equation*} \omega = \frac{1}{\sqrt{x^2+y^2}} (-y \ dx + x \ dy). \end{equation*}

We calculate its exterior derivative:

\begin{align*} d \omega =\amp \left( - (x^2+y^2)^{-1/2} + y^2 (x^2+y^2)^{-3/2} \right) dy \wedge dx \\ \amp + \left( (x^2+y^2)^{-1/2} - x^2 (x^2+y^2)^{-3/2} \right) dx \wedge dy\\ =\amp \left(\frac{2}{\sqrt{x^2+y^2}} - \frac{x^2+y^2}{(x^2+y^2)^{3/2}} \right) dx \wedge dy \\ =\amp \frac{1}{\sqrt{x^2+y^2}} dx \wedge dy. \end{align*}

Thus we conclude that

\begin{equation*} \boldsymbol{\nabla} \times \mathbf{G} = \left(0, 0, \frac{1}{\sqrt{x^2+y^2}} \right). \end{equation*}

To find the divergence, we associate to \(\mathbf{G}\) the two-form:

\begin{equation*} \eta = \frac{1}{\sqrt{x^2+y^2}}( -y\ dy \wedge dz + x\ dz \wedge dx ). \end{equation*}

We calculate its exterior derivative:

\begin{align*} d \eta =\amp \frac{x y}{(x^2+y^2)^{3/2}} dx \wedge dy \wedge dz - \frac{x y}{(x^2+y^2)^{3/2}} dy \wedge dz \wedge dx\\ =\amp 0 \end{align*}

Therefore \(\boldsymbol{\nabla} \cdot \mathbf{G} = 0\text{.}\)


Find a vector field \(\mathbf{F} = (0, f_2, f_3)\) such that \(\boldsymbol{\nabla} \times \mathbf{F} = ( 0 ,z ,y)\text{.}\)


Since \(\mathbf{F}=(0,f_2, f_3)\text{,}\) and using the definition of the curl, we know that

\begin{equation*} \boldsymbol{\nabla} \times \mathbf{F} = \left( \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z}, - \frac{\partial f_3}{\partial x}, \frac{\partial f_2}{\partial x} \right). \end{equation*}

Thus we need to solve the equations

\begin{equation*} \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z}=0, \qquad - \frac{\partial f_3}{\partial x} = z, \qquad \frac{\partial f_2}{\partial x} = y. \end{equation*}

Integrating the last two equations, we get:

\begin{equation*} f_2 = x y + g(y,z), \qquad f_3 = - x z + h(y,z), \end{equation*}

for some functions \(g(y,z), h(y,z)\text{.}\) The first condition then imposes that

\begin{equation*} \frac{\partial}{\partial y} h(y,z) = \frac{\partial}{\partial z} g(y,z). \end{equation*}

There are many possible choices, but the simplest would be \(g(y,z) = h(y,z) = 0\text{.}\) We would then conlude that

\begin{equation*} \mathbf{F}=(0, x y, -x z) \end{equation*}

is a vector field such that \(\boldsymbol{\nabla} \times \mathbf{F} = (0, z, y).\)


Is there a vector field \(\mathbf{F}\) on \(\mathbb{R}^3\) such that \(\boldsymbol{\nabla} \times \mathbf{F} = \left( x, y+ x z, z \right)\text{?}\) Justify your answer.


We know that \(\boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \times \mathbf{F}) = 0\) for any vector field \(\mathbf{F}\text{.}\) So if there is a vector field \(\mathbf{F}\) such that \(\boldsymbol{\nabla} \times \mathbf{F} = \left( x, y+ x z, z \right)\text{,}\) then the divergence of the vector on the right-hand-side (let's call it \(\mathbf{G}\)) must be zero. But

\begin{equation*} \boldsymbol{\nabla} \cdot \mathbf{G} = 1 + 1 + 1 = 3, \end{equation*}

which is obviously non-zero. Therefore, we conclude that there does not exist a vector field \(\mathbf{F}\) such that \(\boldsymbol{\nabla} \times \mathbf{F} = \left( x, y+ x z, z \right)\text{.}\)


Let \(\mathbf{F} = (xy, y^2, xy+z)\) and \(\mathbf{G}=( xyz, yz, z^2)\) be smooth vector fields on \(\mathbb{R}^3\text{,}\) and \(\alpha:[0,2 \pi] \to \mathbb{R}^3\) be the parametric curve given by \(\alpha(t) =(\sin(t), \cos(t), t(t- 2 \pi) ) \text{.}\) Show that the line integral of the vector field

\begin{equation*} \mathbf{F} \times (\boldsymbol{\nabla} \times \mathbf{G}) + (\boldsymbol{\nabla} \times \mathbf{F}) \times \mathbf{G} + (\mathbf{G} \cdot \boldsymbol{\nabla}) \mathbf{F} + (\mathbf{F} \cdot \boldsymbol{\nabla}) \mathbf{G} \end{equation*}

along \(\alpha\) is zero.


Well, you certainly do not want to evaluate this line integral, it would be painful!

First, we notice that the parametric curve \(\alpha\) is closed, since

\begin{equation*} \alpha(0) = (0,1,0) = \alpha(2 \pi). \end{equation*}

Next, we notice that we can use the vector calculus identity 1 from Lemma 4.4.12, which states that

\begin{equation*} \boldsymbol{\nabla} (\mathbf{F} \cdot \mathbf{G}) = \mathbf{F} \times (\boldsymbol{\nabla} \times \mathbf{G}) + (\boldsymbol{\nabla} \times \mathbf{F}) \times \mathbf{G} + (\mathbf{G} \cdot \boldsymbol{\nabla}) \mathbf{F} + (\mathbf{F} \cdot \boldsymbol{\nabla}) \mathbf{G}. \end{equation*}

So the vector field that we want to evaluate the line integral of is \(\boldsymbol{\nabla} (\mathbf{F} \cdot \mathbf{G})\text{.}\) As this is the gradient of a function, this means that the vector field is conservative. Therefore, its integral along any closed curve vanishes. We conclude that the integral along \(\alpha\) is zero!


Suppose that you study a vector field \(\mathbf{F}\) in a lab. You measure that

\begin{equation*} \mathbf{F}(x,y,z) = (x z + y z + x^2 y, \alpha (y z + x^2 z), \beta (x y z + y) ), \end{equation*}

for some constants \(\alpha, \beta\) that you are not able to determine experimentally. However, from theoretical considerations you know that \(\mathbf{F}\) must be divergence-free (i.e., its divergence is zero). Find the values of \(\alpha\) and \(\beta\text{.}\)


We know that

\begin{align*} \boldsymbol{\nabla}\cdot \mathbf{F} =\amp \frac{\partial}{\partial x} (x z +yz+ x^2 y) + \alpha \frac{\partial}{\partial y}(y z + x^2 z) + \beta \frac{\partial}{\partial z}(x y z + y) \\ =\amp z + 2 x y + \alpha z + \beta x y. \end{align*}

Since we know that \(\boldsymbol{\nabla} \cdot \mathbf{F} = 0\text{,}\) and that \(\alpha\) and \(\beta\) are constants (i.e. do not depend on \(x,y,z\)), we conclude that we must have

\begin{equation*} \alpha = -1, \qquad \beta = - 2. \end{equation*}


Let \(\mathbf{F}(x,y,z) = (f(x), g(y), h(z))\) for some smooth functions \(f(x), g(y), h(z)\) on \(\mathbb{R}\) (note that those are functions of a single variable), and let \(q(x,y,z)\) be an arbitrary smooth function on \(\mathbb{R}^3\text{.}\) Show that

\begin{equation*} \boldsymbol{\nabla} \cdot (\mathbf{F} \times \boldsymbol{\nabla} q ) = 0. \end{equation*}

Let us first solve it using vector calculus identities, and then provide an alternative but equivalent solutions using differential forms. Identity 4 of Lemma 4.4.9 states that

\begin{equation*} \boldsymbol{\nabla} \cdot (\mathbf{F} \times \mathbf{G} ) = (\boldsymbol{\nabla} \times \mathbf{F}) \cdot \mathbf{G} - \mathbf{F} \cdot (\boldsymbol{\nabla} \times \mathbf{G}). \end{equation*}

Applying this to the case at hand, we get:

\begin{equation*} \boldsymbol{\nabla} \cdot (\mathbf{F} \times \boldsymbol{\nabla} q ) = (\boldsymbol{\nabla} \times \mathbf{F}) \cdot \boldsymbol{\nabla} q - \mathbf{F} \cdot (\boldsymbol{\nabla} \times \boldsymbol{\nabla} q ). \end{equation*}

We then calculate the curl of \(\mathbf{F}\text{.}\) We get:

\begin{align*} \boldsymbol{\nabla} \times \mathbf{F} =\amp \left( \frac{\partial h(z)}{\partial y} - \frac{\partial g(y)}{\partial z}, \frac{\partial f(x)}{\partial z} - \frac{\partial h(z)}{\partial x}, \frac{\partial g(y)}{\partial x} - \frac{\partial f(x)}{\partial y} \right)\\ =\amp 0. \end{align*}

Moreover, from Identity 1 of Lemma 4.4.10, we know that

\begin{equation*} \boldsymbol{\nabla} \times \boldsymbol{\nabla} q = 0. \end{equation*}


\begin{equation*} \boldsymbol{\nabla} \cdot (\mathbf{F} \times \boldsymbol{\nabla} q ) = 0. \end{equation*}

Let us now solve the question using differential forms. Let \(\omega = f(x)\ dx + g(y)\ dy + h(z)\ dz\) be the one-form associated to \(\mathbf{F}\text{.}\) Then \(\boldsymbol{\nabla} \cdot (\mathbf{F} \times \boldsymbol{\nabla} q )\) is the function associated to \(d( \omega \wedge d q)\text{.}\) So we want to show that

\begin{equation*} d(\omega \wedge d q) = 0. \end{equation*}

Using the graded product rule, we have:

\begin{align*} d(\omega \wedge dq) =\amp d \omega \wedge dq - \omega \wedge d^2 q \\ =\amp d \omega \wedge dq, \end{align*}

where we used the fact that \(d^2 =0 \text{.}\) But

\begin{align*} d \omega =\amp df(x) \wedge dx + dg(y) \wedge dy + dh(z) \wedge dz \\ =\amp 0, \end{align*}

since by evaluating the differentials we only get terms involving the vanishing basic two-forms \(dx \wedge dx = dy \wedge dy = dz \wedge dz = 0\text{.}\) We thus conclude that

\begin{equation*} d(\omega \wedge d q) = 0. \end{equation*}