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Section 4.3 Differentiating \(k\)-forms: the exterior derivative

We know how to add \(k\)-forms, and now also how to mutiply \(k\)-forms, thanks to the notion of wedge product. In this section we study how we can “differentiate” \(k\)-forms, using the notion of exterior derivative, which generalizes the differential of a function introduced in Definition 2.2.1.

Subsection 4.3.1 The exterior derivative

Let us start by recalling the definition of the differential of a function \(f:U \to \mathbb{R}\text{,}\) with \(U \subseteq \mathbb{R}^3\text{,}\) from Definition 2.2.1. The differential \(d f\) is the one-form on \(U\) given by

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

Since we now think of \(f\) as a zero-form, we see that the operator \(d\) takes a zero-form and outputs a one-form. Our goal is to generalize this operation, which we will now call the “exterior derivative”, to take a \(k\)-form and ouput a \((k+1)\)-form.

Definition 4.3.1. The exterior derivative of a \(k\)-form.

The exterior derivative of a zero-form \(f\) on \(U \subset \mathbb{R}^n\) is the one-form \(df\) on \(U\) given by:

\begin{equation*} df = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\ dx_i, \end{equation*}

which is the same thing as the differential introduced in Definition 2.2.1.

The exterior derivative of a \(k\)-form \(\displaystyle \omega = \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} f_{i_1 \cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k}\) on \(U \subset \mathbb{R}^n\) is the \((k+1)\)-form \(d \omega\) on \(U\) given by:

\begin{equation*} d\omega =\sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d (f_{i_1 \cdots i_k} ) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}, \end{equation*}

where \(d (f_{i_1 \cdots i_k})\) means the exterior derivative of the zero-form (function) \(f_{i_1 \cdots i_k}\text{.}\) In other words, we are applying the exterior derivative \(d\) to the component functions of \(\omega\text{.}\)

This definition may seem a little daunting because of the summations, so let us be more explicit for \(k\)-forms in \(\mathbb{R}^3\text{.}\)

We start with Definition 4.3.1 restricted to the case with \(n=3\text{.}\) For the exterior derivative of a zero-form, the statement is obvious. For the exterior derivatives of a one-form and a two-form, all we have to do is evaluate the exterior derivatives \(d(f), d(g)\) and \(d(h)\) of the zero-form \(f,g,h\text{,}\) and rearrange terms using Lemma 4.1.6.

Note that you certainly should not aim at learning these formulae by heart. The whole point is precisely that you don't need to learn these formulae! All you need to remember is that, to evaluate the exterior derivative of a \(k\)-form, you act with the exterior derivative on the component functions of the \(k\)-form. This may be clearer with examples.

We are already familiar with the calculation of the exterior derivative of a zero-form, since this is the same thing as the calculation of the differential of a function that we defined in Definition 2.2.1. But let us give an example here for completeness.

Let \(f(x,y,z) = y\ln(x) + z\) be a smooth function on \(U = \{ (x,y,z) \in \mathbb{R}^3\ | \ x>0\}\text{.}\) Its exterior derivative is the one-form \(d f\) on \(U\) given by:

\begin{align*} df =\amp \frac{\partial f}{\partial x}\ dx + \frac{\partial f}{\partial y}\ dy + \frac{\partial f}{\partial z}\ dz \\ =\amp \frac{y}{x} \ dx + \ln(x)\ dy + dz. \end{align*}

Let \(\omega = x y\ dx + (z+y)\ dy + x y z\ dz\) be a one-form on \(\mathbb{R}^3\text{.}\) Its exterior derivative is the two-form \(d \omega\) on \(\mathbb{R}^3\) given by:

\begin{align*} d\omega =\amp d(x y) \wedge dx + d(z+y) \wedge dy + d(x y z) \wedge dz\\ =\amp \left(y\ dx + x\ dy \right) \wedge dx + \left( dy + dz \right) \wedge dy + \left( y z \ dx+ x z\ dy+x y\ dz \right) \wedge dz \\ =\amp x\ dy \wedge dx + dz \wedge dy + y z\ dx \wedge dz + x z\ dy \wedge dz \\ =\amp (x z-1)\ dy \wedge dz - y z\ dz \wedge dx - x\ dx \wedge dy. \end{align*}

Let \(\omega = (x^2 +y^2)\ dy \wedge dz + \sin(z)\ dz \wedge dx + \cos(x y)\ dx \wedge dy\) be a two-form on \(\mathbb{R}^3\text{.}\) Its exterior derivative is the three-form \(d \omega\) on \(\mathbb{R}^3\) given by:

\begin{align*} d\omega =\amp d(x^2+y^2) \wedge dy \wedge dz + d(\sin(z))\wedge dz \wedge dx + d(\cos(x y))\wedge dx \wedge dy\\ =\amp (2 x\ dx + 2 y\ dy)\wedge dy \wedge dz + \left( \cos(z)\ dz \right)\wedge dz \wedge dx + \left(- y \sin( x y)\ dx - x \sin(x y)\ dy \right) \wedge dx \wedge dy \\ =\amp 2 x\ dx \wedge dy \wedge dz. \end{align*}

We see that going from the second line to the third line, most terms vanish, since anytime we take the wedge of \(dx\) with itself we get zero, and same for \(dy\) and \(dz\text{.}\)

To end this section, we note that the exterior derivative is linear. If \(\omega\) and \(\eta\) are two \(k\)-forms, and \(a,b \in \mathbb{R}\text{,}\) then

\begin{equation*} d(a \omega + b \eta) = a d \omega + b d \eta. \end{equation*}

This is proven in Exercise 4.3.4.3.

Subsection 4.3.2 The graded product rule

One of the most fundamental properties of the derivative is the product rule:

\begin{equation*} \frac{d}{dx} (f g) = \frac{df}{dx} g + f \frac{dg}{dx}. \end{equation*}

Now that we have defined the exterior derivative for differential forms, and that we know how to multiply differential forms using the wedge product, we could ask whether the exterior derivative satisfies a similar product rule with respect to the wedge product. It turns out that it does, but with a twist (or more precisely a sign). We call this the “graded product rule” for the exterior derivative.

First, we show that it is true for zero-forms. If \(\omega\) and \(\eta\) are zero-forms (that is \(k=l=0\)), that is functions \(\omega = f\) and \(\eta = g\text{,}\) then

\begin{align*} d (\omega \wedge \eta) =\amp d (f g)\\ =\amp \sum_{i=1}^n \frac{\partial}{\partial x_i} (f g)\ dx_i\\ =\amp \sum_{i=1}^n \left( \frac{\partial f}{\partial x_i} g + f \frac{\partial g}{\partial x_i} \right)\ dx_i\\ =\amp d(f) g + f d(g). \end{align*}

To prove the general statement we unfortunately need to use lots of summations. Let us introduce the following notation for the \(k\)-form \(\omega\) and the \(l\)-form \(\eta\text{:}\)

\begin{equation*} \omega = \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} w_{i_1 \cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k}, \qquad \eta = \sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} h_{j_1 \cdots j_l} dx_{j_1} \wedge \cdots \wedge dx_{j_l}. \end{equation*}

The exterior derivative of the wedge product is:

\begin{align*} d ( \omega \wedge \eta) =\amp \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n}\sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} d(w_{i_1 \cdots i_k}h_{j_1 \cdots j_l}) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_l}\\ =\amp\sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n}\sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} \left( d(w_{i_1 \cdots i_k})h_{j_1 \cdots j_l} +w_{i_1 \cdots i_k} d(h_{j_1 \cdots j_l}) \right) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_l}. \end{align*}

To go from the first to the second line, we used the fact that the product rule is satisfied for the exterior derivative of the product of zero-forms, as shown above.

Now let us study the terms on the right-hand-side of the equation we just obtained. First, we get:

\begin{align*} \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n}\amp \sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} d(w_{i_1 \cdots i_k})h_{j_1 \cdots j_l}\wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_l} \\ \amp= \left( \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d(w_{i_1 \cdots i_k})\wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}\right) \wedge \left(\sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} h_{j_1 \cdots j_l} dx_{j_1} \wedge \cdots \wedge dx_{j_l} \right)\\ \amp= d(\omega) \wedge \eta \end{align*}

All that we did to go from the first line to the second line is move the zero-form (or function) \(h_{j_1 \cdots j_l}\) to the right of the \(dx_i\)'s, which we can do since it is a zero-form and hence commutes with the \(dx_i\)'s by Lemma 4.2.6.

That takes care of the first set of terms. The remaining ones take the form

\begin{equation*} \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n}\sum_{1 \leq j_1 \lt \cdots \lt j_l \leq n} w_{i_1 \cdots i_k} d(h_{j_1 \cdots j_l}) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_l}. \end{equation*}

We would like to do the same and commute \(d(h_{j_1 \cdots j_l}) \) to the right of the \(dx_i\)'s, so that we can identify this term as \(\omega \wedge d(\eta)\text{.}\) However, \(d(h_{j_1 \cdots j_l}) \) is a one-form, and thus by Lemma 4.2.6 \(d(h_{j_1 \cdots j_l}) \wedge dx_i = - dx_i \wedge d(h_{j_1 \cdots j_l})\text{.}\) So every time we commute \(d(h_{j_1 \cdots j_l}) \) past a \(dx_i\text{,}\) we pick a sign. Since there are \(k\) \(dx_i\)'s in this expression, this tells us that it is equal to

\begin{equation*} (-1)^k \omega \wedge d(\eta). \end{equation*}

Putting all this together, we end up with the statement that

\begin{equation*} d (\omega \wedge \eta) = d(\omega) \wedge \eta + (-1)^k \omega \wedge d(\eta), \end{equation*}

which is the graded product rule stated in the lemma.

It is very important not to forget the sign in the graded product rule!

Let \(\omega = x y\ dz\) and \(\eta = (y+z)\ dx + 2 dy\) be two one-forms on \(\mathbb{R}^3\text{.}\) Suppose that we want to calculate the exterior derivative of the wedge product \(\omega \wedge \eta\text{.}\) There are two ways we can do that: we can first find an explicit expression for the wedge product, and then take the exterior derivative, or we can use the graded product rule. In this example we show that both calculations give the same result, as they should.

Let us first calculate the wedge product explicitly and take its exterior derivative. We have:

\begin{align*} \omega \wedge \eta =\amp (x y\ dz) \wedge ( (y+z) \ dx + 2\ dy) \\ =\amp - 2 x y dy \wedge dz + x y (y+z) dz \wedge dx. \end{align*}

We then calculate the exterior derivative, which gives the following three-form:

\begin{align*} d(\omega \wedge \eta) =\amp d(-2 x y) \wedge dy \wedge dz + d(x y (y+z) ) \wedge dz \wedge dx \\ =\amp \left(- 2 y\ dx - 2 x\ dy\right) \wedge dy \wedge dz + \left( y(y+z)\ dx + (2 x y + x z)\ dy + x y\ dz \right) \wedge dz \wedge dx \\ =\amp (- 2 y + (2 x y + x z) ) dx \wedge dy \wedge dz, \end{align*}

where we used the fact that \(d y \wedge dz \wedge dx = dx \wedge dy \wedge dz\text{.}\)

Let us now calculate the same exterior derivative but using the graded product rule. Since \(\omega\) is a one-form, we have:

\begin{equation*} d(\omega \wedge \eta) = d(\omega) \wedge \eta - \omega \wedge d(\eta). \end{equation*}

We calculate:

\begin{align*} d(\omega) =\amp d(xy) \wedge dz \\ =\amp y dx \wedge dz + x dy \wedge dz, \end{align*}

and

\begin{align*} d(\eta) =\amp d(y+z) \wedge dx + d(2) \wedge dy\\ =\amp dy \wedge dx + dz \wedge dx, \end{align*}

since \(d(2) = 0\text{.}\) Putting this together, we get:

\begin{align*} d(\omega \wedge \eta) =\amp (y dx \wedge dz + x dy \wedge dz) \wedge ((y+z)\ dx + 2 \ dy) - (x y\ dz) \wedge (dy \wedge dx + dz \wedge dx)\\ =\amp (-2 y + x(y+z)+ xy )dx \wedge dy \wedge dz, \end{align*}

which is indeed the same result as obtained above.

Remark 4.3.8.

In \(\mathbb{R}^3\text{,}\) the graded product rule can be split into the four following non-vanishing cases.

  1. If \(\omega = f\) is a zero-form (in which case we write \(f \wedge \eta = f \eta\) as usual when multiplying with a function) and \(\eta = g\) is a zero-form, then

    \begin{equation*} d(f g) = d(f) g + f d(g). \end{equation*}

  2. If \(\omega = f\) is a zero-form and \(\eta\) is a one-form, then

    \begin{equation*} d (f \eta) = d(f) \wedge \eta + f d(\eta). \end{equation*}

  3. If \(\omega = f\) is a zero-form and \(\eta\) is a two-form, then

    \begin{equation*} d( f\eta) = d(f) \wedge \eta + f d(\eta). \end{equation*}

  4. If \(\omega\) is a one-form and \(\eta\) is a one-form, then

    \begin{equation*} d(\omega \wedge \eta)= d(\omega)\wedge \eta - \omega \wedge d(\eta). \end{equation*}

Subsection 4.3.3 \(d^2=0\)

There's another fundamental property of the exterior derivative: if we apply the exterior derivative twice, we always get zero. This may seem surprising, as this is certainly not true for the ordinary derivative \(d/dx\text{,}\) but it is true for the exterior derivative because of antisymmetry of the wedge product. More precisely:

We write \(\omega = \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} w_{i_1 \cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k}\text{.}\) We have:

\begin{align*} d(d(\omega)) =\amp d\left( \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d(w_{i_1 \cdots i_k})\wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \right)\\ =\amp d\left( \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} \sum_{\alpha=1}^n \frac{\partial w_{i_1 \cdots i_k}}{\partial x_\alpha} dx_\alpha \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \right)\\ =\amp \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} \sum_{\alpha=1}^n d \left( \frac{\partial w_{i_1 \cdots i_k}}{\partial x_\alpha} \right) \wedge dx_\alpha \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ =\amp \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} \left( \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} dx_\beta \wedge dx_\alpha \right) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}. \end{align*}

If we can show that the term in brackets in the last line is zero, then clearly \(d(d(\omega)) =0 \text{.}\) We have:

\begin{align*} \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} dx_\beta \wedge dx_\alpha = \amp \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\alpha \partial x_\beta} dx_\beta \wedge dx_\alpha\\ = \amp - \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\alpha \partial x_\beta} dx_\alpha \wedge dx_\beta. \end{align*}

In the first, line, we used the fact that \(\frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} = \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\alpha \partial x_\beta}\) by the Clairaut-Schwarz theorem, since the coefficient functions \(w_{i_1 \cdots i_k}\) are assumed to be smooth. In the second line, we used the fact that \(dx_\beta \wedge dx_\alpha = - dx_\alpha \wedge d x_\beta\text{,}\) by Lemma 4.1.6. Finally, in the summation on the right-hand-side, we can simply rename \(\alpha\) to be \(\beta\text{,}\) and \(\beta\) to be \(\alpha\text{,}\) since those are indices that are summed over, and hence we can give them the name we want. We end up with the statement that

\begin{equation*} \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} dx_\beta \wedge dx_\alpha = - \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} dx_\beta \wedge dx_\alpha . \end{equation*}

But the only two-form that is equal to minus itself is the zero two-form, that is

\begin{equation*} \sum_{\alpha=1}^n \sum_{\beta=1}^n \frac{\partial^2 w_{i_1 \cdots i_k}}{\partial x_\beta \partial x_\alpha} dx_\beta \wedge dx_\alpha = 0. \end{equation*}

Therefore we conclude that \(d(d(\omega)) = 0\text{.}\)

Exercises 4.3.4 Exercises

1.

Let \(\omega = x y\ dx + \frac{y}{x}\ dy + \frac{z}{x}\ dz\) on \(U = \{(x,y,z) \in \mathbb{R}^3~|~x \neq 0 \}\text{.}\) Find the two-form \(d \omega\) and write your result in standard form. What is the vector field associated to \(d \omega\text{?}\)

Solution.

From the definition, we find:

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

Using the dictionary Table 4.1.11, the vector field associated to the two-form \(d \omega\) is

\begin{equation*} \mathbf{F}(x,y,z) = \left( 0, \frac{z}{x^2}, - x - \frac{y}{x^2} \right). \end{equation*}

2.

Find the three-form \(d \omega\) if \(\omega = x y z (dy \wedge dz + dz \wedge dx + dx \wedge dy)\text{,}\) and write your result in standard form.

Solution.

We find:

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

3.

Show that the exterior derivative is linear. That is, if \(\omega\) and \(\eta\) are two \(k\)-forms, and \(a,b \in \mathbb{R}\text{,}\) then
\begin{equation*} d(a \omega + b \eta) = a d \omega + b d \eta. \end{equation*}
Solution.

First, we show that the property holds if \(\omega\) and \(\eta\) are \(0\)-forms. Since those are simply functions, let us write them as \(f\) and \(g\text{.}\) Then:

\begin{align*} d(a f + b g) =\amp \sum_{i=1}^n \frac{\partial}{\partial x_i} (a f + b g)\ dx_i\\ =\amp \sum_{i=1}^n \left( a \frac{\partial f}{\partial x_i} + b \frac{\partial g}{\partial x_i} \right)\ dx_i\\ =\amp a \sum_{i=1}^n \frac{\partial f}{\partial x_i}\ dx_i + b \sum_{i=1}^n \frac{\partial g}{\partial x_i}\ dx_i\\ =\amp a d f + b dg. \end{align*}

For the second equality we used the fact that partial derivatives are linear.

Now we can prove the general case. Suppose that \(\omega\) and \(\eta\) are \(k\)-forms on \(U \subseteq \mathbb{R}^n\text{.}\) Let

\begin{equation*} \omega = \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} f_{i_1 \cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k} \end{equation*}

and

\begin{equation*} \eta = \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} g_{i_1 \cdots i_k} dx_{i_1} \wedge \cdots \wedge dx_{i_k}. \end{equation*}

Then

\begin{align*} d(a \omega + b \eta)= \amp \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d (a f_{i_1 \cdots i_k}+ b g_{i_1 \cdots i_k} ) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}\\ =\amp \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} \left( a\ d f_{i_1 \cdots i_k}+ b\ d g_{i_1 \cdots i_k} \right) \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}\\ =\amp a \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d f_{i_1 \cdots i_k}\wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}\\ \amp + b \sum_{1 \leq i_1 \lt \cdots \lt i_k \leq n} d g_{i_1 \cdots i_k}\wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k} \\ =\amp a d \omega + b d \eta. \end{align*}

In the second equality we used the calculation above that showed that linearity holds for \(0\)-forms.

4.

Let \(\omega = x e^y\ dx + z\ dy + y e^x\ dz\text{.}\) Show by explicit calculation that \(d^2 \omega = 0\text{.}\)

Solution.

Of course, we know that \(d^2 \omega =0\) since this is true for any differential form \(\omega\text{,}\) as proven in Lemma 4.3.9. But let us show that it is true by explicit calculation for this particular one-form \(\omega\text{.}\)

We first calculate the two-form \(d \omega\text{:}\)

\begin{align*} d \omega =\amp d( x e^y) \wedge dx + dz \wedge dy + d(y e^x) \wedge dz\\ =\amp x e^y\ dy \wedge dx + e^y\ dx \wedge dx + dz \wedge dy + y e^x\ dx \wedge dz + e^x\ dy \wedge dz\\ =\amp (e^x - 1)\ dy \wedge dz - y e^x\ dz \wedge dx - x e^y\ dx \wedge dy. \end{align*}

We then calculate the three-form \(d^2 \omega\text{:}\)

\begin{align*} d^2 \omega =\amp d(d \omega)\\ =\amp d(e^x-1) \wedge dy \wedge dz - d(y e^x)\wedge dz \wedge dx - d(x e^y) \wedge dx \wedge dy\\ =\amp e^x\ dx \wedge dy \wedge dz - y e^x\ dx \wedge dz \wedge dx - e^x\ dy \wedge dz \wedge dx - x e^y\ dy \wedge dx \wedge dy\\ \amp - e^y\ dx \wedge dx \wedge dy\\ =\amp (e^x - e^x) dx \wedge dy \wedge dz\\ =\amp 0, \end{align*}

as expected.

5.

Let \(\omega = (x^2-y^2)\ dx + y\ dz\) and \(\eta = (x^2+y^2)\ dy + y\ dz\text{.}\) By explicit calculation, show that

\begin{equation*} d(\omega \wedge \eta) = d \omega\wedge \eta - \omega \wedge d\eta, \end{equation*}

which is consistent with the graded product rule Lemma 4.3.6 since \(\omega\) is a one-form.

Solution.

We need to calculate \(d(\omega \wedge \eta)\text{,}\) \(d \omega\wedge \eta\text{,}\) and \(\omega \wedge d\eta\text{.}\) First, we calculate \(\omega \wedge \eta\text{:}\)

\begin{align*} \omega \wedge \eta =\amp ( (x^2-y^2)\ dx + y\ dz ) \wedge ( (x^2+y^2)\ dy + y\ dz) \\ =\amp (x^4-y^4) dx \wedge dy + y (x^2-y^2)\ dx \wedge dz + y (x^2+y^2)\ dz \wedge dy\\ =\amp - y(x^2+y^2)\ dy \wedge dz - y(x^2-y^2)\ dz \wedge dx + (x^4-y^4) dx \wedge dy. \end{align*}

Then

\begin{align*} d (\omega \wedge \eta) =\amp - d(y(x^2+y^2))\wedge dy \wedge dz - d(y(x^2-y^2))\wedge dz \wedge dx + d(x^4-y^4) \wedge dx \wedge dy\\ =\amp - 2 x y\ dx \wedge dy \wedge dz - (x^2 - 3 y^2)\ dy \wedge dz \wedge dx\\ =\amp (3 y^2 - 2 x y - x^2) dx \wedge dy \wedge dz. \end{align*}

Next, we calculate \(d\omega \wedge \eta\text{.}\) We have:

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

Then

\begin{align*} d \omega \wedge \eta =\amp (dy \wedge dz + 2 y dx \wedge dy) \wedge ((x^2+y^2)\ dy + y\ dz )\\ =\amp 2 y^2 dx \wedge dy \wedge dz. \end{align*}

Finally, we calculate \(\omega \wedge d \eta \text{.}\) We have:

\begin{align*} d \eta =\amp d(x^2+y^2) \wedge dy + dy \wedge dz\\ =\amp 2 x\ dx \wedge dy + dy \wedge dz\\ =\amp dy \wedge dz + 2 x\ dx \wedge dy. \end{align*}

Then

\begin{align*} \omega \wedge d\eta =\amp ((x^2-y^2)\ dx + y\ dz) \wedge (dy \wedge dz + 2 x\ dx \wedge dy)\\ =\amp (x^2-y^2) dx \wedge dy \wedge dz + 2 x y\ dz \wedge dx \wedge dy\\ =\amp (x^2 + 2 x y - y^2)\ dx \wedge dy \wedge dz. \end{align*}

Putting all this together, we conclude that

\begin{align*} d\omega \wedge \eta - \omega \wedge d\eta =\amp (3 y^2 - 2 x y -x^2) dx \wedge dy \wedge dz\\ =\amp d (\omega \wedge \eta), \end{align*}

as expected from the graded product rule Lemma 4.3.6.

6.

Let \(\omega\) be a \(k\)-form, \(\eta\) an \(m\)-form, and \(\lambda\) a \(\ell\)-form, all on \(U \subset \mathbb{R}^n\text{.}\) Show that

\begin{equation*} d(\omega \wedge \eta \wedge \lambda) = d\omega \wedge \eta \wedge \lambda + (-1)^k \omega \wedge d\eta \wedge \lambda + (-1)^{k+m} \omega \wedge \eta \wedge d\lambda. \end{equation*}
Solution.

First, using the graded product rule Lemma 4.3.6, we get that

\begin{equation*} d(\omega \wedge \eta \wedge \lambda) = d \omega \wedge (\eta \wedge \lambda) + (-1)^k \omega \wedge d(\eta \wedge \lambda). \end{equation*}

Next, again from the graded product rule we know that

\begin{equation*} d(\eta \wedge \lambda) = d \eta \wedge \lambda + (-1)^m \eta \wedge d\lambda. \end{equation*}

Putting this together, we get:

\begin{equation*} d(\omega \wedge \eta \wedge \lambda) = d\omega \wedge \eta \wedge \lambda + (-1)^k \omega \wedge d\eta \wedge \lambda + (-1)^{k+m} \omega \wedge \eta \wedge d\lambda. \end{equation*}

7.

Let \(\omega = (x+y)\ dx + x y\ dy\) be a one-form on \(\mathbb{R}^2\text{,}\) and let \(\phi: \mathbb{R}^2 \to \mathbb{R}^2\) be given by \(\phi(u,v) = (e^{u+v}, e^{u-v})\text{.}\) Find the two-form \(d(\phi^* \omega)\text{.}\)

Solution.

We first calculate the pullback one-form \(\phi^* \omega\text{.}\) We get:

\begin{align*} \phi^* \omega =\amp (e^{u+v} + e^{u-v}) \left( e^{u+v}\ du + e^{u+v}\ dv \right) + e^{u+v} e^{u-v}\left( e^{u-v}\ du - e^{u-v}\ dv \right) \\ =\amp \left( e^{2(u+v)} + e^{2u}+ e^{ 3u-v} \right)\ du + \left( e^{2(u+v)} + e^{2u} - e^{3u-v} \right)\ dv. \end{align*}

We can then calculate its exterior derivative. We get:

\begin{align*} d(\phi^* \omega) =\amp d\left( e^{2(u+v)} + e^{2u}+ e^{ 3u-v} \right)\wedge du + d\left( e^{2(u+v)} + e^{2u} - e^{3u-v} \right)\wedge dv\\ =\amp \left( 2 e^{2(u+v)} - e^{3u-v} \right) dv \wedge du + \left( 2 e^{2(u+v)} + 2 e^{2u} - 3 e^{3u-v} \right) du \wedge dv\\ =\amp 2 \left( e^{2 u} - e^{3u-v} \right) du \wedge dv. \end{align*}

8.

Let \(f,g: U \to \mathbb{R}\) be smooth functions with \(U \subseteq \mathbb{R}^n\text{,}\) and \(\alpha: [a,b] \to \mathbb{R}^n\) be a parametric curve whose image is in \(U\text{.}\) Using the product rule for the exterior derivative \(d(fg)\text{,}\) show that

\begin{equation*} \int_\alpha f dg = f(\alpha(b)) g(\alpha(b)) - f(\alpha(a)) g(\alpha(a)) - \int_\alpha g d f. \end{equation*}

One can think of this as the natural generalization of “integration by parts” to line integrals of one-forms. Indeed, if the curve is just an interval in \(\mathbb{R}\text{,}\) this reduces to the standard statement of integration by parts for definite integrals.

Solution.

From the graded product rule, we know that

\begin{equation*} d(fg) = f dg + g df, \end{equation*}

since \(f\) is a \(0\)-form (here we omit the wedge product symbol, since we are either multiplying two functions or a function with a one-form). By the Fundamental Theorem of line integrals, we know that

\begin{equation*} \int_\alpha d(f g) = f(\alpha(b)) g(\alpha(b)) - f(\alpha(a)) g(\alpha(a)), \end{equation*}

since \(d(fg)\) is an exact one-form. Using the graded product rule, we thus conclude that

\begin{equation*} \int_\alpha d(fg) = \int_\alpha f dg + \int_\alpha g df = f(\alpha(b)) g(\alpha(b)) - f(\alpha(a)) g(\alpha(a)). \end{equation*}

Solving for \(\int_\alpha f dg\text{,}\) we get the desired statement.

9.

Let \(\omega\) and \(\eta\) be one-forms that differ by the exterior derivative of a \(0\)-form, that is,

\begin{equation*} \eta = \omega + d f \end{equation*}

for some function \(f\text{.}\) Show that

\begin{equation*} d(\omega \wedge \eta) = d \omega \wedge d f. \end{equation*}
Solution.

We have:

\begin{equation*} d( \omega \wedge \eta) = d( \omega \wedge (\omega + d f) ) = d ( \omega \wedge \omega) + d( \omega \wedge d f). \end{equation*}

The first term on the right-hand-side is zero, since \(\omega \wedge \omega = 0\) for a one-form (see Lemma 4.2.6; the wedge product is anti-commutative for odd forms). As for the second term, we use the graded product rule and the fact that \(d^2 = 0\) to get:

\begin{equation*} d(\omega \wedge \eta) = d \omega \wedge d f - \omega \wedge d^2 f = d \omega \wedge d f. \end{equation*}