Multiplying k-forms: the wedge product

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.

🔗

Objectives

You should be able to:

Determine the wedge product of two $k$ -forms in $R^{3} .$
Relate the wedge product of two one-forms to the cross-product of the associated vector fields, and the wedge product of a one-form and a two-form to the dot product of the associated vector fields.
State and use general properties of the wedge product for $k$ -forms.

🔗

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

\land,

to which we now turn to.

🔗

Definition 4.2.1. The wedge product.

Consider a simple $k$ -form $ω$ and a simple $m$ -form $η$ on an open subset $U \subseteq R^{n}$ of the form:

ω = f d x_{i_{1}} \land \dots \land d x_{i_{k}}, η = g d x_{j_{1}} \land \dots \land d x_{j_{m}},

for smooth functions $f, g : U \to R .$ Then the wedge product $ω \land η$ is a $(k + m)$ -form defined by:

ω \land η = f g d x_{i_{1}} \land \dots \land d x_{i_{k}} \land d x_{j_{1}} \land \dots \land d x_{j_{m}} .

Note that the order is important here on the right, since we know that exchanging two $d x_{i}$ '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.

🔗

Example 4.2.2. The wedge product of two one-forms.

Consider the wedge product of the two one-forms $ω = x d x + x d y + e^{z} d z$ and $η = y d x + x d y + z d z$ on $R^{3} .$ By definition, the result is the two-form given by:

\begin{aligned} ω \land η = & (x d x + x d y + e^{z} d z) \land (y d x + x d y + z d z) \\ = & x y d x \land d x + x^{2} d x \land d y + x z d x \land d z + x y d y \land d x + x^{2} d y \land d y + x z d y \land d z \\ + y e^{z} d z \land d x + x e^{z} d z \land d y + z e^{z} d z \land d z \\ = & (x z - x e^{z}) d y \land d z + (- x z + y e^{z}) d z \land d x + (x^{2} - x y) d x \land d y, \end{aligned}

where in the last line we used the fact that $d x \land d x = d y \land d y = d z \land d z = 0$ and $d x \land d y = - d y \land d x,$ $d x \land d z = - d z \land d x,$ and $d y \land d z = - d z \land d y .$

🔗

Example 4.2.3. The wedge product of a one-form and a two-form.

Consider the wedge product of the one-form $ω = x y d x + y d y$ and the two-form $η = z d y \land d z$ on $R^{3} .$ The result is the three-form:

\begin{aligned} ω \land η = & (x y d x + y d y) \land (z d y \land d z) \\ = & x y z d x \land d y \land d z + y z d y \land d y \land d z \\ = & x y z d x \land d y \land d z, \end{aligned}

where we used the fact that $d y \land d y \land d z = 0$ since it has repeated factors.

🔗

Example 4.2.4. The wedge product of a zero-form and a $k$ -form.

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 $ω = y d y \land d z$ on $R^{3},$ the wedge product is the two-form:

f \land ω = x y d y \land d z .

We usually write $f ω,$ without the $\land$ 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 $R^{3} .$ Indeed, since we know that $k$ -forms with $k \geq 4$ necessarily vanish in $R^{3},$ this means that the only possible non-zero wedge products are:

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

ω

and

η .

ω \land η

equal to

η \land ω ?

The answer is no, not quite! The precise statement is the following lemma.

🔗

Lemma 4.2.6. Comparing $ω \land η$ to $η \land ω$ .

Let $ω$ be a $k$ -form and $η$ be an $m$ -form. Then

ω \land η = (- 1)^{k m} η \land ω .

In other words:

If either $k$ or $m$ is even, then $ω \land η = η \land ω,$ and the wedge product is commutative.
If both $k$ and $m$ are odd, then $ω \land η = - η \land ω,$ and the wedge product is anti-commutative.

In particular, if $ω$ is a $k$ -form with $k$ odd, then $ω \land ω = 0 .$

🔗

Proof.

This statement follows from the fact that exchanging two $d x_{i}$ 's in a basic form picks a sign. More precisely, let us first assume that $ω$ and $η$ take the simple forms:

ω = f d x_{i_{1}} \land \dots d x_{i_{k}}, η = g d x_{j_{1}} \land \dots \land d x_{j_{m}} .

Then

ω \land η = f g (d x_{i_{1}} \land \dots d x_{i_{k}}) \land (d x_{j_{1}} \land \dots \land d x_{j_{m}}),

while

η \land ω = f g (d x_{j_{1}} \land \dots \land d x_{j_{m}}) \land (d x_{i_{1}} \land \dots d x_{i_{k}}) .

To relate the second expression to the first, we need to move the $d x_{j}$ 's to the left of the $d x_{i}$ 's. We first move $d x_{i_{1}}$ to the left. Each time we pass a $d x_{j},$ we pick a sign. So in the end we get:

η \land ω = f g (- 1)^{m} d x_{i_{1}} \land (d x_{j_{1}} \land \dots \land d x_{j_{m}}) \land (d x_{i_{2}} \land \dots d x_{i_{k}}) .

We do the same thing with $d x_{i_{2}},$ moving it pass all the $d x_{j}$ 's, and so on all the way to $d x_{i_{k}} .$ The final result is

\begin{aligned} η \land ω = & f g (- 1)^{k m} d x_{i_{1}} \land \dots d x_{i_{k}} \land (d x_{j_{1}} \land \dots \land d x_{j_{m}}) \\ = & (- 1)^{k m} ω \land η . \end{aligned}

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

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

R^{3} :

The wedge product of a zero-form with a $k$ -form with $k \in {0, 1, 2, 3};$
A one-form with a one-form;
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

ω = f_{1} d x + f_{2} d y + f_{3} d z

and

η = g_{1} d x + g_{2} d y + g_{3} d z .

Then the wedge product is

ω \land η = (f_{2} g_{3} - f_{3} g_{2}) d y \land d z + (f_{3} g_{1} - f_{1} g_{3}) d z \land d x + (f_{1} g_{2} - f_{2} g_{1}) d x \land d y .

🔗

In terms of the associated vector fields

F = (f_{1}, f_{2}, f_{3}), G = (g_{1}, g_{2}, g_{3}),

🔗

what we are doing is constructing a new vector field, let's call it

H

for the time being, associated to

ω \land η,

with component functions given by (according to the dictionary established in Table 4.1.11 for relating a two-form to a vector field):

H = (f_{2} g_{3} - f_{3} g_{2}, f_{3} g_{1} - f_{1} g_{3}, f_{1} g_{2} - f_{2} g_{1}) .

🔗

What is this vector field? It is nothing but the cross-product of the vector fields

F

and

G!

Indeed,

F \times G = (f_{2} g_{3} - f_{3} g_{2}, f_{3} g_{1} - f_{1} g_{3}, f_{1} g_{2} - f_{2} g_{1}) .

🔗

Therefore, we end up with the following statement:

🔗

Lemma 4.2.7. The wedge product of two one-forms is the cross-product of the associated vector fields.

If $ω$ and $η$ are two one-forms on $U \subseteq R^{3},$ with associated vector fields $F$ and $G,$ then the vector field associated to the two-form $ω \land η$ via the dictionary in Table 4.1.11 is the cross-product $F \times G .$

🔗

Finally, we consider the wedge product of a one-form and a two-form. Let

ω = f_{1} d x + f_{2} d y + f_{3} d z

and

η = g_{1} d y \land d z + g_{2} d z \land d x + g_{3} d x \land d y .

Then the wedge product is the three-form:

ω \land η = (f_{1} g_{1} + f_{2} g_{2} + f_{3} g_{3}) d x \land d y \land d z .

🔗

According to the dictionary in Table 4.1.11, we thus conclude that the function associated to the three-form

ω \land η

is nothing but the dot product of the two vector fields

F

and

G

associated to

ω

and

η

respectively:

F \cdot G = f_{1} g_{1} + f_{2} g_{2} + f_{3} g_{3} .

🔗

Neat! So we get the following result:

🔗

Lemma 4.2.8. The wedge product of a one-form and a two-form is the dot product of the associated vector fields.

If $ω$ is a one-form and $η$ a two-form on $U \subseteq R^{3},$ with associated vector fields $F$ and $G$ via the dictionary in Table 4.1.11, then the function associated to the three-form $ω \land η$ is the dot product $F \cdot G .$

🔗

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

R^{3}

and vector calculus concepts

Differential form concept		Vector calculus concept
1-form	$ω$	vector field	$F$
1-form	$η$	vector field	$G$
2-form	$ω \land η$	vector field	$F \times G$

🔗

Table 4.2.10. Dictionary between the wedge product of a one-form and a two-form in

R^{3}

and vector calculus concepts

Differential form concept		Vector calculus concept
1-form	$ω$	vector field	$F$
2-form	$η$	vector field	$G$
3-form	$ω \land η$	vector field	$F \cdot G$

🔗

Exercises 4.2.3 Exercises

🔗

1.

Let $ω = e^{x} d x + y d z$ and $η = x y d x + z d y + y d z .$ Find $ω \land η,$ and write your result in standard form.

Solution.

We find (using the fact that $d x \land d x = d y \land d y = d z \land d z = 0$ ):

\begin{aligned} ω \land η = & (e^{x} d x + y d z) \land (x y d x + z d y + y d z) \\ = & z e^{x} d x \land d y + y e^{x} d x \land d z + x y^{2} d z \land d x + y z d z \land d y \\ = & - y z d y \land d z + (x y^{2} - y e^{x}) d z \land d x + z e^{x} d x \land d y . \end{aligned}

🔗

2.

Let $ω = x d y \land d z + y d z \land d x + z d x \land d y$ and $η = x d x + y d y + z d z .$ Find $ω \land η,$ and write your result in standard form.

Solution.

We find (using the fact that any wedge product with two repeated $d x,$ $d y$ or $d z$ is zero):

\begin{aligned} ω \land η = & (x d y \land d z + y d z \land d x + z d x \land d y) \land (x d x + y d y + z d z) \\ = & x^{2} d y \land d z \land d x + y^{2} d z \land d x \land d y + z^{2} d x \land d y \land d z \\ = & (x^{2} + y^{2} + z^{2}) d x \land d y \land d z, \end{aligned}

where we used the fact that $d y \land d z \land d x = d z \land d x \land d y = d x \land d y \land d z .$

🔗

3.

Let $ω = f_{1} d x + f_{2} d y$ and $η = g_{1} d x + g_{2} d y$ be one-forms on $R^{2},$ with associated vector fields $F = (f_{1}, f_{2})$ and $G = (g_{1}, g_{2}) .$ Show that

ω \land η = det (\begin{matrix} f_{1} & g_{1} \\ f_{2} & g_{2} \end{matrix}) d x \land d y .

Solution.

Let $ω = f_{1} d x + f_{2} d y,$ and $η = g_{1} d x + g_{2} d y .$ The associated vector fields are $F = (f_{1}, f_{2}),$ $G = (g_{1}, g_{2}) .$ We calculate the wedge product:

\begin{aligned} ω \land η = & (f_{1} d x + f_{2} d y) \land (g_{1} d x + g_{2} d y) \\ = & f_{1} g_{2} d x \land d y + f_{2} g_{1} d y \land d x \\ = & (f_{1} g_{2} - f_{2} g_{1}) d x \land d y \\ = & det (\begin{array}{c} f_{1} & g_{1} \\ f_{2} & g_{2} \end{array}) d x \land d y . \end{aligned}

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

🔗

4.

Let $ω = f d x + g d y + h d z$ be an arbitrary one-form on $R^{3} .$ By doing an explicit calculation, show that $ω \land ω = 0 .$

Solution.

We find:

\begin{aligned} ω \land ω = & (f d x + g d y + h d z) \land (f d x + g d y + h d z) \\ = & f g d x \land d y + f h d x \land d z + g f d y \land d x + g h d y \land d z + h f d z \land d x + h g d z \land d y \\ = & (g h - h g) d y \land d z + (h f - f h) d z \land d x + (f g - g f) d x \land d y . \end{aligned}

Since functions commute with each other, i.e. $g h = h g,$ $h f = f h,$ $f g = g f,$ we conclude that $ω \land ω = 0 .$

🔗

5.

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

Solution.

The one-forms associated to $F$ and $G$ are $ω = d x + d y$ and $η = 2 d y + 3 d z .$ We calculate the wedge product:

\begin{aligned} ω \land η = & (d x + d y) \land (2 d y + 3 d z) \\ = & 2 d x \land d y + 3 d x \land d z + 3 d y \land d z \\ = & 3 d y \land d z - 3 d z \land d x + 2 d x \land d y \end{aligned}

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

H = F \times G = (3, - 3, 2) .

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{aligned} F \times G = & det (\begin{array}{c} i & j & k \\ F_{1} & F_{2} & F_{3} \\ G_{1} & G_{2} & G_{3} \end{array}) \\ = & det (\begin{array}{c} i & j & k \\ 1 & 1 & 0 \\ 0 & 2 & 3 \end{array}) \\ = & 3 i - 3 j + 2 k, \end{aligned}

which, in component notation, reads $(3, - 3, 2) .$

🔗

6.

Let $A, B, C$ be vectors in $R^{3} .$ In your linear algebra course you may have seen that

A \cdot (B \times C) = B \cdot (C \times A) = C \cdot (A \times B) .

Prove this property by looking at the wedge product of the three one-forms $ω, η, λ$ associated to the vectors $A, B, C .$

Solution.

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

ω \land η \land λ = A \cdot (B \times C) d x \land d y \land d z,

η \land λ \land ω = B \cdot (C \times A) d x \land d y \land d z,

and

λ \land ω \land η = C \cdot (A \times B) d x \land d y \land d z .

But since $λ, ω, η$ are one-forms, we know that $ω \land η = - η \land ω,$ $η \land λ = - λ \land η,$ and $ω \land λ = - λ \land ω .$ Therefore,

ω \land η \land λ = η \land λ \land ω = λ \land ω \land η,

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

Section 4.2 Multiplying k-forms: the wedge product

Objectives

Subsection 4.2.1 Multiplying k-forms: the wedge product

Definition 4.2.1. The wedge product.

Example 4.2.2. The wedge product of two one-forms.

Example 4.2.3. The wedge product of a one-form and a two-form.

Example 4.2.4. The wedge product of a zero-form and a k-form.

Remark 4.2.5.

Lemma 4.2.6. Comparing ω∧η to η∧ω.

Proof.

Subsection 4.2.2 The wedge product and vector calculus

Lemma 4.2.7. The wedge product of two one-forms is the cross-product of the associated vector fields.

Lemma 4.2.8. The wedge product of a one-form and a two-form is the dot product of the associated vector fields.

Exercises 4.2.3 Exercises

1.

2.

3.

4.

5.

6.

Section 4.2 Multiplying $k$ -forms: the wedge product

Subsection 4.2.1 Multiplying $k$ -forms: the wedge product

Example 4.2.4. The wedge product of a zero-form and a $k$ -form.

Lemma 4.2.6. Comparing $ω \land η$ to $η \land ω$ .