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Section 4.2 Multiplying \(k\)-forms: the wedge product

With the introduction of \(k\)-forms we can now introduce a new operation: we can multiply differential forms. This operation is called the “wedge product”, which is what we study in this section.

Subsection 4.2.1 Multiplying \(k\)-forms: the wedge product

When we introduced one-forms in Section 2.1, we explained how we can add two one-forms to get another one-form, and how we can multiply a one-form by a function to get another one-form. Let us first state that the addition property obviously holds for two- and three-forms as well: the sum of two two-forms is a two-form, and the sum of two three-forms is a three-form. Note that it doesn't really make sense to add a one-form with a two-form, etc.

When we introduced one-forms we did not however talk about multiplying two one-forms together. To do this, we need the full theory of \(k\)-forms, as multiplying two one-forms will give us a two-form. The operation of multiplying \(k\)-forms is called the “wedge product”, and denoted by \(\wedge\text{,}\) to which we now turn to.

Definition 4.2.1. The wedge product.

Consider a simple \(k\)-form \(\omega\) and a simple \(m\)-form \(\eta\) on an open subset \(U \subseteq \mathbb{R}^n\) of the form:

\begin{equation*} \omega = f\ dx_{i_1} \wedge \ldots \wedge dx_{i_k}, \qquad \eta = g\ dx_{j_1} \wedge \ldots \wedge dx_{j_m}, \end{equation*}

for smooth functions \(f,g: U \to \mathbb{R}\text{.}\) Then the wedge product \(\omega \wedge \eta\) is a \((k+m)\)-form defined by:

\begin{equation*} \omega \wedge \eta = f g\ dx_{i_1} \wedge \ldots \wedge dx_{i_k} \wedge dx_{j_1} \wedge \ldots \wedge dx_{j_m}. \end{equation*}

Note that the order is important here on the right, since we know that exchanging two \(dx_i\)'s will pick a sign, by the definition of basic forms.

The wedge product of arbitrary \(k\)- and \(m\)-forms is then defined by linearity, i.e. by distributing the wedge product term by term in the expression of the forms as sums of terms.

This definition looks complicated, but in the end it is fairly simple. It is easier to understand it by working through some examples.

Consider the wedge product of the two one-forms \(\omega = x\ dx + x\ dy + e^z\ dz\) and \(\eta = y\ dx + x\ dy + z\ dz\) on \(\mathbb{R}^3\text{.}\) By definition, the result is the two-form given by:

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

where in the last line we used the fact that \(dx \wedge dx = dy \wedge dy = dz \wedge dz = 0\) and \(dx \wedge dy = - dy \wedge dx\text{,}\) \(dx \wedge dz = - dz \wedge dx\text{,}\) and \(dy \wedge dz = -dz \wedge dy\text{.}\)

Consider the wedge product of the one-form \(\omega = xy\ dx + y\ dy\) and the two-form \(\eta = z\ dy \wedge dz\) on \(\mathbb{R}^3\text{.}\) The result is the three-form:

\begin{align*} \omega \wedge \eta =\amp \left( xy\ dx + y\ dy \right) \wedge \left( z\ dy \wedge dz \right) \\ =\amp xyz\ dx \wedge dy \wedge dz + yz\ dy \wedge dy \wedge dz\\ =\amp xyz\ dx \wedge dy \wedge dz, \end{align*}

where we used the fact that \(dy \wedge dy \wedge dz = 0\) since it has repeated factors.

The wedge product with a zero-form is just a standard product, since a zero-form is just a function. For instance, given the zero-form \(f = x\) and the two-form \(\omega = y dy \wedge dz\) on \(\mathbb{R}^3\text{,}\) the wedge product is the two-form:

\begin{equation*} f \wedge \omega = xy\ dy \wedge dz. \end{equation*}

We usually write \(f \omega\text{,}\) without the \(\wedge\) symbol, when we multiply with a zero-form, since it is just a function.

Note that this generalizes the statement that a one-form mutiplied by a function is another one-form; this is also true for a two-form and a three-form.

Remark 4.2.5.

These examples pretty much exhaust the possible non-zero wedge products in \(\mathbb{R}^3\text{.}\) Indeed, since we know that \(k\)-forms with \(k \geq 4\) necessarily vanish in \(\mathbb{R}^3\text{,}\) this means that the only possible non-zero wedge products are:

  • A zero-form (i.e. a function) with a \(k\)-form, with \(k \in \{0,1,2,3\}\text{;}\)

  • A one-form with a one-form, which gives a two-form;

  • A one-form with a two-form, which gives a three-form.

All other wedge products will necessarily vanish.

Now that we are familiar with the wedge product, a natural question is whether it is “commutative”. Pick two differential forms \(\omega\) and \(\eta\text{.}\) Is \(\omega \wedge \eta\) equal to \(\eta \wedge \omega\text{?}\) The answer is no, not quite! The precise statement is the following lemma.

This statement follows from the fact that exchanging two \(dx_i\)'s in a basic form picks a sign. More precisely, let us first assume that \(\omega\) and \(\eta\) take the simple forms:

\begin{equation*} \omega = f\ dx_{i_1} \wedge \ldots dx_{i_k}, \qquad \eta = g\ dx_{j_1} \wedge \ldots \wedge d x_{j_m}. \end{equation*}

Then

\begin{equation*} \omega \wedge \eta = f g \ \left( dx_{i_1} \wedge \ldots dx_{i_k} \right) \wedge \left( dx_{j_1} \wedge \ldots \wedge d x_{j_m} \right), \end{equation*}

while

\begin{equation*} \eta \wedge \omega = f g\ \left( dx_{j_1} \wedge \ldots \wedge d x_{j_m} \right) \wedge \left( dx_{i_1} \wedge \ldots dx_{i_k} \right). \end{equation*}

To relate the second expression to the first, we need to move the \(dx_{j}\)'s to the left of the \(dx_i\)'s. We first move \(dx_{i_1}\) to the left. Each time we pass a \(dx_j\text{,}\) we pick a sign. So in the end we get:

\begin{equation*} \eta \wedge \omega = f g (-1)^m \ dx_{i_1} \wedge \left( dx_{j_1} \wedge \ldots \wedge d x_{j_m} \right) \wedge \left( dx_{i_2} \wedge \ldots dx_{i_k} \right). \end{equation*}

We do the same thing with \(dx_{i_2}\text{,}\) moving it pass all the \(dx_j\)'s, and so on all the way to \(dx_{i_k}\text{.}\) The final result is

\begin{align*} \eta \wedge \omega =\amp f g (-1)^{ km }\ dx_{i_1} \wedge \ldots dx_{i_k} \wedge \left( dx_{j_1} \wedge \ldots \wedge d x_{j_m} \right) \\ =\amp (-1)^{k m} \omega \wedge \eta. \end{align*}

Finally, the statement is proved for general differential forms by doing this manipulation term by term after distributing the wedge product.

Subsection 4.2.2 The wedge product and vector calculus

In Table 4.1.11 we established a dictionary between differential forms in \(\mathbb{R}^3\) and vector calculus concepts. We can extend this dictionary to understand the concept of wedge product in vector calculus.

As we saw, there are really only three types of non-zero wedge products in \(\mathbb{R}^3\text{:}\)

  1. The wedge product of a zero-form with a \(k\)-form with \(k \in \{0,1,2,3 \}\text{;}\)

  2. A one-form with a one-form;

  3. A one-form with a two-form.

Let us now translate each of those operations in the language of vector calculus.

First, multiplying a zero-form by a \(k\)-form is just multiplying the \(k\)-form by a function. So the same is true in vector calculus: we multiply the corresponding vector field or function by a function. No big deal.

Multiplying two one-forms is interesting though. Let \(\omega = f_1 dx + f_2\ dy + f_3\ dz\) and \(\eta = g_1\ dx + g_2\ dy + g_3\ dz\text{.}\) Then the wedge product is

\begin{equation*} \omega \wedge \eta = (f_2 g_3 - f_3 g_2) dy \wedge dz + (f_3 g_1- f_1 g_3) dz \wedge dx+ (f_1 g_2 - f_2 g_1) dx \wedge dy . \end{equation*}

In terms of the associated vector fields

\begin{equation*} \mathbf{F} = (f_1, f_2, f_3), \qquad \mathbf{G} = (g_1, g_2, g_3), \end{equation*}

what we are doing is constructing a new vector field, let's call it \(\mathbf{H}\) for the time being, associated to \(\omega \wedge \eta\text{,}\) with component functions given by (according to the dictionary established in Table 4.1.11 for relating a two-form to a vector field):

\begin{equation*} \mathbf{H} = \left( f_2 g_3 - f_3 g_2, f_3 g_1 - f_1 g_3, f_1 g_2 - f_2 g_1 \right). \end{equation*}

What is this vector field? It is nothing but the cross-product of the vector fields \(\mathbf{F}\) and \(\mathbf{G}\text{!}\) Indeed,

\begin{equation*} \mathbf{F} \times \mathbf{G} = \left( f_2 g_3 - f_3 g_2, f_3 g_1 - f_1 g_3, f_1 g_2 - f_2 g_1 \right). \end{equation*}

Therefore, we end up with the following statement:

Finally, we consider the wedge product of a one-form and a two-form. Let \(\omega = f_1\ dx + f_2\ dy + f_3\ dz\) and \(\eta = g_1\ dy \wedge dz+ g_2\ dz \wedge dx + g_3\ dx \wedge dy \text{.}\) Then the wedge product is the three-form:

\begin{equation*} \omega \wedge \eta = (f_1 g_1 + f_2 g_2 + f_3 g_3 ) \ dx \wedge dy \wedge dz. \end{equation*}

According to the dictionary in Table 4.1.11, we thus conclude that the function associated to the three-form \(\omega \wedge \eta\) is nothing but the dot product of the two vector fields \(\mathbf{F}\) and \(\mathbf{G}\) associated to \(\omega\) and \(\eta\) respectively:

\begin{equation*} \mathbf{F} \cdot \mathbf{G} = f_1 g_1 + f_2 g_2 + f_3 g_3. \end{equation*}

Neat! So we get the following result:

These two lemmas justify the ordering in the definition of two-form mentioned in Remark 4.1.9.

We can summarize the dictionary between the wedge product and vector products in the following two tables:

Table 4.2.9. Dictionary between the wedge product of two one-forms in \(\mathbb{R}^3\) and vector calculus concepts
Differential form concept Vector calculus concept
1-form \(\omega\) vector field \(\mathbf{F}\)
1-form \(\eta\) vector field \(\mathbf{G}\)
2-form \(\omega \wedge \eta\) vector field \(\mathbf{F} \times \mathbf{G}\)
Table 4.2.10. Dictionary between the wedge product of a one-form and a two-form in \(\mathbb{R}^3\) and vector calculus concepts
Differential form concept Vector calculus concept
1-form \(\omega\) vector field \(\mathbf{F}\)
2-form \(\eta\) vector field \(\mathbf{G}\)
3-form \(\omega \wedge \eta\) vector field \(\mathbf{F} \cdot \mathbf{G}\)

Exercises 4.2.3 Exercises

1.

Let \(\omega = e^x \ dx + y\ dz\) and \(\eta = x y\ dx + z\ dy + y\ dz\text{.}\) Find \(\omega \wedge \eta\text{,}\) and write your result in standard form.

Solution.

We find (using the fact that \(dx \wedge dx = dy \wedge dy = dz \wedge dz = 0 \)):

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

2.

Let \(\omega = x\ dy \wedge dz + y\ dz \wedge dx + z\ dx \wedge dy\) and \(\eta = x\ dx + y\ dy + z\ dz\text{.}\) Find \(\omega \wedge \eta\text{,}\) and write your result in standard form.

Solution.

We find (using the fact that any wedge product with two repeated \(dx\text{,}\) \(dy\) or \(dz\) is zero):

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

where we used the fact that \(dy \wedge dz \wedge dx = dz \wedge dx \wedge dy = dx \wedge dy \wedge dz. \)

3.

Let \(\omega = f_1\ dx + f_2\ dy\) and \(\eta=g_1\ dx + g_2\ dy\) be one-forms on \(\mathbb{R}^2\text{,}\) with associated vector fields \(\mathbf{F}=(f_1, f_2)\) and \(\mathbf{G}=(g_1, g_2)\text{.}\) Show that

\begin{equation*} \omega \wedge \eta = \det \begin{pmatrix} f_1 \amp g_1 \\ f_2 \amp g_2 \end{pmatrix} dx \wedge dy. \end{equation*}
Solution.

Let \(\omega = f_1 \ dx + f_2\ dy\text{,}\) and \(\eta = g_1\ dx + g_2\ dy\text{.}\) The associated vector fields are \(\mathbf{F} = (f_1, f_2)\text{,}\) \(\mathbf{G} = (g_1, g_2)\text{.}\) We calculate the wedge product:

\begin{align*} \omega \wedge \eta =\amp (f_1 \ dx + f_2\ dy) \wedge (g_1\ dx + g_2\ dy) \\ =\amp f_1 g_2 \ dx \wedge dy + f_2 g_1\ dy \wedge dx\\ =\amp (f_1 g_2 - f_2 g_1) dx \wedge dy\\ =\amp \det \begin{pmatrix} f_1 \amp g_1 \\ f_2 \amp g_2 \end{pmatrix} dx \wedge dy. \end{align*}

We note that this determinant could be taken as a “definition” of what “cross-product” of vector fields means in \(\mathbb{R}^2\text{.}\) Note that the result however is a function, not a vector field. From the point of view of differential forms, we could define the “cross-product” of vector fields in \(\mathbb{R}^n\) as follows: take two one-forms \(\omega\) and \(\eta\) on \(\mathbb{R}^n\text{,}\) with associated vector fields \(\mathbf{F}\) and \(\mathbf{G}\text{.}\) We would define the “cross-product” of the vector fields as being given by the two-form \(\omega \wedge \eta\) in \(\mathbb{R}^n\text{.}\) Note that only on \(\mathbb{R}^3\) can we associate to the result a new vector field (this is because of Hodge duality between one-forms and two-forms in \(\mathbb{R}^3\text{,}\) see Section 4.8). For instance, in \(\mathbb{R}^4\) a two-form has 6 component functions (there are 6 independent non-vanishing basic two-forms in \(\mathbb{R}^4\)), so it cannot be associated to a vector field.

4.

Let \(\omega = f\ dx + g\ dy + h\ dz\) be an arbitrary one-form on \(\mathbb{R}^3\text{.}\) By doing an explicit calculation, show that \(\omega \wedge \omega = 0\text{.}\)

Solution.

We find:

\begin{align*} \omega \wedge \omega =\amp (f\ dx + g\ dy + h\ dz) \wedge (f\ dx + g\ dy + h\ dz)\\ =\amp f g\ dx \wedge dy + f h \ dx \wedge dz + g f\ dy \wedge dx + g h \ dy \wedge dz + h f\ dz \wedge dx + h g\ dz \wedge dy\\ =\amp (g h - h g)\ dy \wedge dz + (h f - f h)\ dz \wedge dx + (f g - g f)\ dx \wedge dy. \end{align*}

Since functions commute with each other, i.e. \(g h = hg\text{,}\) \(h f = f h\text{,}\) \(f g = g f\text{,}\) we conclude that \(\omega \wedge \omega = 0\text{.}\)

5.

Find the cross-product of the vectors \(\mathbf{F}=(1,1,0)\) and \(\mathbf{G}=(0,2,3)\) in \(\mathbb{R}^3\) by computing the wedge product of the associated one-forms.

Solution.

The one-forms associated to \(\mathbf{F}\) and \(\mathbf{G}\) are \(\omega = dx + dy\) and \(\eta = 2 dy + 3 dz\text{.}\) We calculate the wedge product:

\begin{align*} \omega \wedge \eta =\amp (dx + dy) \wedge (2 dy+ 3 dz) \\ =\amp 2 dx \wedge dy + 3 dx \wedge dz + 3 dy \wedge dz\\ =\amp 3 dy \wedge dz - 3 dz \wedge dx + 2 dx \wedge dy \end{align*}

According to the dictionary Table 4.1.11, the vector field associated to this two-form is

\begin{equation*} \mathbf{H} = \mathbf{F} \times \mathbf{G} = (3 ,-3, 2). \end{equation*}

This is indeed the cross-product, as you can calculate using standard formulae from linear algebra. For instance, you may have seen the formula:

\begin{align*} \mathbf{F} \times \mathbf{G} =\amp \det \begin{pmatrix} \mathbf{i} \amp \mathbf{j} \amp \mathbf{k} \\ F_1 \amp F_2 \amp F_3 \\ G_1 \amp G_2 \amp G_3 \end{pmatrix}\\ =\amp \det \begin{pmatrix} \mathbf{i} \amp \mathbf{j} \amp \mathbf{k} \\ 1 \amp 1 \amp 0 \\ 0 \amp 2 \amp 3 \end{pmatrix}\\ =\amp 3 \mathbf{i} - 3 \mathbf{j} + 2 \mathbf{k}, \end{align*}

which, in component notation, reads \((3,-3,2)\text{.}\)

6.

Let \(\mathbf{A}, \mathbf{B}, \mathbf{C}\) be vectors in \(\mathbb{R}^3\text{.}\) In your linear algebra course you may have seen that

\begin{equation*} \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}) = \mathbf{C} \cdot ( \mathbf{A} \times \mathbf{B} ). \end{equation*}

Prove this property by looking at the wedge product of the three one-forms \(\omega, \eta, \lambda\) associated to the vectors \(\mathbf{A}, \mathbf{B}, \mathbf{C}\text{.}\)

Solution.

As discussed in Lemma 4.2.7 and Lemma 4.2.8, we know that

\begin{equation*} \omega \wedge \eta \wedge \lambda = \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) dx \wedge dy \wedge dz, \end{equation*}
\begin{equation*} \eta \wedge \lambda \wedge \omega = \mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}) dx \wedge dy \wedge dz, \end{equation*}

and

\begin{equation*} \lambda \wedge \omega \wedge \eta = \mathbf{C} \cdot ( \mathbf{A} \times \mathbf{B} ) dx \wedge dy \wedge dz. \end{equation*}

But since \(\lambda, \omega, \eta\) are one-forms, we know that \(\omega \wedge \eta = - \eta \wedge \omega\text{,}\) \(\eta \wedge \lambda = - \lambda \wedge \eta\text{,}\) and \(\omega \wedge \lambda = - \lambda \wedge \omega\text{.}\) Therefore,

\begin{equation*} \omega \wedge \eta \wedge \lambda = \eta \wedge \lambda \wedge \omega = \lambda \wedge \omega \wedge \eta, \end{equation*}

and the statement is proved. In other words, it follows directly from anti-commutativity of the wedge product of one-forms.