Differential forms revisited: an algebraic approach

Section 4.1 Differential forms revisited: an algebraic approach

Now that we are familiar with one-forms, we take a step back, and revisit the definition. We introduce a more algebraic approach to one-forms, which will allow us to generalize it and introduce the concept of

k

-forms.

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Objectives

You should be able to:

Define basic one-forms as linear maps, and basic two- and three-forms as multilinear maps.
Define $k$ -forms in $R^{3} .$
Prove the antisymmetry propeties of basic $k$ -forms.
Relate two-forms to vector fields.

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Subsection 4.1.1 An algebraic approach to one-forms

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When we introduced one-forms in Definition 2.1.1, we said that the objects

d x, d y

and

d z

could be understood a placeholders. We paused briefly in Remark 2.2.4 to relate these objects to the differentials of “projection functions” on

R^{3},

but this wasn't entirely satisfactory. Now we go back and give a more rigorous definition of what these objects stand for.

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Definition 4.1.1. The basic one-forms.

The basic one-form $d x_{i}$ , with $i \in {1, \dots, n},$ is a linear map $d x_{i} : R^{n} \to R$ which takes a vector $u = (u_{1}, \dots, u_{n}) \in R^{n}$ and projects it onto the $x_{i}$ -axis:

d x_{i} (u_{1}, \dots, u_{n}) = u_{i} .

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Remark 4.1.2.

When we are working on $R^{3},$ it is customary to write $(x, y, z)$ for $(x_{1}, x_{2}, x_{3}),$ and we write the basic one-forms as $d x, d y,$ and $d z$ (instead of $d x_{1}, d x_{2}$ and $d x_{3}$ ).

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This gives a rigorous meaning of these placeholders. Using this definition, we can write a general linear map

M : R^{3} \to R

M = A d x + B d y + C d z,

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where

A, B, C \in R

are just constants. In other words, it is an arbitrary linear combination of the three projection operators. In general, given an abstract vector space

V,

the set of linear maps

M : V \to R

forms a vector space itself, which is called the “dual vector space” and denoted by

V^{*}

(see for instance https://en.wikipedia.org/wiki/Dual_space).

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Let us now look back at the definition of one-forms in Definition 2.1.1, focusing on

R^{3}

for simplicity. A (differential) one-form was defined as a linear combination of basic one-forms with coefficients that are smooth functions on an open subset

U \subseteq R^{3} .

That is, we can write a one-form as:

ω = f d x + g d y + h d z,

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for smooth functions

f, g, h : U \to R .

With our new understanding of the placeholders

d x, d y, d z,

we can make sense of this object. For any point

P \in U,

the one-form

ω

defines a linear map

R^{3} \to R

(or equivalently an element of the vector space dual to

R^{3}

). This is the dual concept to vector fields: a vector field is a rule that assigns to all points on

U

a vector in

R^{3},

while a one-form is a rule that assigns to all points on

U

a linear map

R^{3} \to R

(that is, a “dual vector”). Nice!

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Subsection 4.1.2 Basic $k$ -forms

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This algebraic understanding of the basic one-forms as linear maps allows us to define a natural generalization. Instead of looking only at linear maps

R^{n} \to R,

we now also define multilinear maps

R^{n} \times R^{n} \to R,

and

R^{n} \times R^{n} \times R^{n} \to R,

and so on. The definition relies on the determinant.

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Definition 4.1.3. Basic two-forms.

The basic two-form $d x_{i} \land d x_{j}$ , for $i, j \in {1, \dots, n}$ is a multilinear map¹ $d x_{i} \land d x_{j} : R^{n} \times R^{n} \to R$ which takes two vectors $(u, v) \in R^{n} \times R^{n},$ with $u = (u_{1}, \dots, u_{n})$ and $v = (v_{1}, \dots, v_{n}),$ and maps them to the following determinant:

d x_{i} \land d x_{j} (u, v) = det (\begin{matrix} u_{i} & v_{i} \\ u_{j} & v_{j} \end{matrix}) .

A multilinear map is a function of several variables that is linear separately in each variable.

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In the same way we can define the notion of basic three-forms.

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Definition 4.1.4. Basic three-forms.

The basic three-form $d x_{i} \land d x_{j} \land d x_{k}$ , for $i, j, k \in {1, \dots, n}$ is a multilinear map $d x_{i} \land d x_{j} \land d x_{k} : R^{n} \times R^{n} \times R^{n} \to R$ which takes three vectors $(u, v, w) \in R^{n} \times R^{n} \times R^{n},$ with $u = (u_{1}, \dots, u_{n})$ and $v = (v_{1}, \dots, v_{n}),$ and $w = (w_{1}, \dots, w_{n}),$ and maps them to the following determinant:

d x_{i} \land d x_{j} \land d x_{k} (u, v, w) = det (\begin{matrix} u_{i} & v_{i} & w_{i} \\ u_{j} & v_{j} & w_{j} \\ u_{k} & v_{k} & w_{k} \end{matrix}) .

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We wrote the definition of basic one-, two-, and three-forms explicitly for clarity, but in fact they are just special cases of the more general definition of basic

k

-forms. We now define basic

k

-forms, for completeness, but don't worry if the notation is confusing you: as we will see, in

R^{3}

all basic

k

-forms with

k \geq 4

automatically vanish, so the above definitions are sufficient.

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Definition 4.1.5. Basic $k$ -forms.

The basic $k$ -form $d x_{i_{1}} \land \dots \land d x_{i_{k}}$ , with $i_{1}, \dots, i_{k} \in {1, \dots, n},$ is a multilinear map $d x_{i_{1}} \land \dots \land d x_{i_{k}} : (R^{n})^{k} \to R$ which takes $k$ vectors $(u^{1}, \dots, u^{k}) \in (R^{n})^{k},$ with $u^{j} = (u_{1}^{j}, \dots, u_{n}^{j}),$ and maps them to the following determinant:

d x_{i_{1}} \land \dots \land d x_{i_{k}} (u^{1}, \dots, u^{k}) = det (\begin{matrix} u_{i_{1}}^{1} & \dots & u_{i_{1}}^{k} \\ ⋮ & ⋱ & ⋮ \\ u_{i_{k}}^{1} & \dots & u_{i_{k}}^{k} \end{matrix}) .

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It looks like there are many possibilities here, but many of them either vanish or are related to each other. More precisely, basic

k

-forms satisfy the following antisymmetry relations, which significantly reduce the number of non-zero basic forms

R^{3} .

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Lemma 4.1.6. Antisymmetry of basic $k$ -forms.

The basic two-forms satisfy the following properties. For any $i, j \in {1, \dots, n},$

d x_{i} \land d x_{j} = - d x_{j} \land d x_{i} .

In particular,

d x_{i} \land d x_{i} = 0, i \in {1, \dots, n} .

Similarly, the basic $k$ -forms pick a sign whenever we exchange the order of two of the $d x_{i}$ 's, and vanish if two of the $d x_{i}$ 's are the same.

It follows that the only non-vanishing basic $k$ -forms in $R^{n}$ are for $1 \leq k \leq n .$ In particular, in $R^{3}$ we only have basic one-, two-, and three-forms. The independent basic two-forms in $R^{3}$ are (using the $x, y, z$ notation for $R^{3}$ ):

d y \land d z, d z \land d x, d x \land d y,

and the only independent basic three-form in $R^{3}$ is

d x \land d y \land d z .

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Proof.

This follows directly from the property of the determinant. If we exchange a $d x_{i}$ with $d x_{j},$ we exchange two rows in the matrix that we take the determinant of, and hence the determinant picks a sign. Similarly, if two of the $d x_{i}$ 's are the same, then the matrix that we take the determinant of has two equal rows, and hence the determinant is zero.

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Remark 4.1.7.

We note here that the basic $k$ -forms can be given a geometric interpretation. We focus on $R^{3}$ for simplicity. It's easier to start with the basic three-form in $R^{3} .$ The three vectors $u, v$ and $w$ in $R^{3}$ span a three-dimensional parallelepiped. Then the basic three-form $d x \land d y \land d z (u, v, w)$ calculates its oriented volume, since this is what the determinant calculates.

As for the basic two-forms, the two vector $u$ and $v$ in $R^{3}$ span a parallelogram. The basic two-form $d x \land d y (u, v)$ calculates the oriented area of its projection on the $x y$ -plane, while the two-forms $d x \land d z (u, v)$ and $d y \land d z (u, v)$ calculate the oriented area of its projection on the $x z$ - and $y z$ -planes respectively.

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Subsection 4.1.3 $k$ -forms in $R^{3}$

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We are now ready to introduce the concept of

k

-forms, with

k \in {0, 1, 2, 3},

R^{3},

which naturally generalizes the one-forms introduced in Definition 2.1.1.

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Definition 4.1.8. $k$ -forms in $R^{3}$ .

Let $U \subseteq R^{3}$ be an open subset.

A zero-form is a smooth function $f : U \to R .$
A one-form is an expression of the form
$f d x + g d y + h d z,$
for smooth functions $f, g, h : U \to R .$
A two-form is an expression of the form
$f d y \land d z + g d z \land d x + h d x \land d y,$
for smooth functions $f, g, h : U \to R .$
A three-form is an expression of the form
$f d x \land d y \land d z,$
for a smooth function $f : U \to R .$

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Of course there is no point in defining

k

-forms with

k \geq 4

R^{3},

as those would necessarily vanish, by Lemma 4.1.6. But we note that this definition of

k

-forms can naturally be generalized to

R^{n}

using the general definition of basic

k

-forms in Definition 4.1.5.

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Remark 4.1.9.

Our definition of two- and three-forms involves a specific choice of basic two- and three-forms. For instance, we used $d x \land d y \land d z$ instead of $d z \land d x \land d y .$ When we express the differential forms with the choice of basic form in Definition 4.1.8, we say that the $k$ -forms are in standard form. We generally want to present differential forms in standard form, to simplify things.

With this being said, you may wonder why we chose this particular choice of ordering for the basic two-forms, i.e. why we wrote

f d y \land d z + g d z \land d x + h d x \land d y

instead of, say,

f d x \land d y + g d x \land d z + h d y \land d z .

The reason behind this choice will become clear in the next section, when we relate the wedge product of differential forms to the cross-product of vector fields.

An easy way to remember this choice of ordering is to rename in your head $(x, y, z) \mapsto (x_{1}, x_{2}, x_{3}),$ and to denote the component functions by $f_{1}, f_{2}, f_{3} .$ Then the proper choice of ordering is

f_{1} d x_{2} \land d x_{3} + f_{2} d x_{3} \land d x_{1} + f_{3} d x_{1} \land d x_{2},

which runs through the three cyclic permutations of $(1, 2, 3)),$ namely $(1, 2, 3), (2, 3, 1)$ and $(3, 1, 2) .$

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Remark 4.1.10.

Using the algebraic interpretation of the basic one-, two-, and three-forms in the previous subsections, we can give a geometric meaning to $k$ -forms: a $0$ -form assigns a number to all points in $U,$ while a $k$ -form (with $k \geq 1$ ) assigns a multilinear map $(R^{3})^{k} \to R$ to all points in $U .$ In other words, if we act on a given set of vectors, a $k$ -form assigns a notion of $k$ -dimensional oriented volume for the corresponding projection of the $k$ -dimensional parallepiped generated by the $k$ vectors.

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Subsection 4.1.4 $k$ -forms and vector calculus

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As has become customary, we end this section by relating our construction in the world of differential forms to the traditional concepts in vector calculus. In Principle 2.1.3, we saw that we can naturally associate to a one-form a corresponding vector field. This correspondence can be generalized to

k

-forms in

R^{3} .

We get the following table, which provides a dictionary between differentials forms in

R^{3}

and vector calculus concepts. Note that to establish the dictionary, we write the

k

-forms on the left-hand-side of the table in standard form (see Remark 4.1.9).

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Table 4.1.11. Dictionary between

k

-forms in

R^{3}

and vector calculus concepts

Differential form concept		Vector calculus concept
0-form	$f$	function	$f$
1-form	$f d x + g d y + h d z$	vector field	$(f, g, h)$
2-form	$f d y \land d z + g d z \land d x + h d x \land d y$	vector field	$(f, g, h)$
3-form	$f d x \land d y \land d z$	function	$f$

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Exercises 4.1.5 Exercises

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1.

Show that $d z \land d x \land d y = d x \land d y \land d z .$

Solution.

We know that $d z \land d x = - d x \land d z,$ and $d z \land d y = - d y \land d z .$ So we can write

d z \land d x \land d y = - d x \land d z \land d y = d x \land d y \land d z .

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2.

List the independent non-vanishing basic $k$ -forms in $R^{4} .$

Solution.

Because of anti-symmetry, we know that the only non-vanishing basic $k$ -forms in $R^{4}$ are for $k \leq 4 .$ Let us write the basic one-forms by

d x_{1}, d x_{2}, d x_{3}, d x_{4} .

Then the basic two-forms are obtained by pairing those two-by-two, up to anti-symmetry. We get the basic two-forms:

d x_{1} \land d x_{2}, d x_{1} \land d x_{3}, d x_{1} \land d x_{4}, d x_{2} \land d x_{3}, d x_{2} \land d x_{4}, d x_{3} \land d x_{4} .

For the basic three-forms, we pair the one-forms three-by-three, without repeated factors (otherwise they would vanish). We get:

d x_{1} \land d x_{2} \land d x_{3}, d x_{1} \land d x_{2} \land d x_{4}, d x_{1} \land d x_{3} \land d x_{4}, d x_{2} \land d x_{3} \land d x_{4} .

Finally, there is only one independent basic four-form, since there cannot be repeated $d x_{i}$ factors. (There is always only one independent basic “top-form”, i.e. basic $n$ -form in $R^{n}$ ). It reads:

d x_{1} \land d x_{2} \land d x_{3} \land d x_{4} .

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3.

Write down the vector field $F : R^{3} \to R^{3}$ associated to the two-form

ω = x y d x \land d y + x y z d x \land d z + x^{2} d y \land d z .

Solution.

Before we extract the vector field we need to make sure that we write the one-form in the correct form according to the dictionary Table 4.1.11. We have:

ω = x^{2} d y \land d z - x y z d z \land d x + x y d x \land d y .

Then the associated vector field is:

F (x, y, z) = (x^{2}, - x y z, x y) .

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4.

Let $ω = d x \land d z$ be a basic two-form on $R^{3},$ and $u = (1, 2, 3)$ and $v = (3, 2, 1)$ be vectors in $R^{3} .$ Evaluate

ω (u, v) .

Solution.

By definition of a basic two-form (and recalling that we use the standard notation here that $d x \land d z = d x_{1} \land d x_{3}$ ), we have:

d x \land d z (u, v) = det (\begin{matrix} u_{1} & v_{1} \\ u_{3} & v_{3} \end{matrix}) .

Substituting the entries for the vectors $u$ and $v,$ we get:

d x \land d z (u, v) = det (\begin{matrix} 1 & 3 \\ 3 & 1 \end{matrix}) = 1 - 9 = - 8.

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5.

Let $ω = d x \land d y \land d z$ be the basic three-form on $R^{3} .$

Show that $ω (u, v, w) = 1,$ with $u = (1, 0, 0),$ $v = (0, 1, 0)$ and $w = (0, 0, 1)$ basis vectors in $R^{3} .$
Show that $ω (v, u, w) = - 1 .$
In general, show that there are three choices of ordering of the basis vectors for which $ω$ evaluates to $1,$ and three choices for which it evaluates to $- 1 .$

Solution.

(a) By definition of a basic three-form, we get:

d x \land d y \land d z (u, v, w) = det (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) = 1.

(b) We changed the ordering of the basis vectors here. We get:

d x \land d y \land d z (v, u, w) = det (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) = - 1.

d x \land d y \land d z (u, v, w) = d x \land d y \land d z (w, u, v) = d x \land d y \land d z (v, w, u) = 1,

while the following three orderings give $- 1 :$

d x \land d y \land d z (v, u, w) = d x \land d y \land d z (w, v, u) = d x \land d y \land d z (u, w, v) = - 1.

The reason is that whenever we permute two basis vectors, we exchange two columns in the matrix that we are taking the determinant of. But we know from properties of the determinant that swapping two columns of matrix changes its detereminant by $- 1 .$ So we conclude that doing an even number of two-by-two swaps of basis vectors does not change the determinant, while doing an odd numbers of two-by-two swaps changes the determinant by $- 1 .$

FYI: in the language of group theory, the group of permutations of three objects is called the “symmetric group” $S_{3},$ whose elements are the permutations. We call a permutation that is a swap of two objects a “transposition”. All permutations can be obtained by doing a finite number of transpositions (i.e. by swapping objects two-by-two a finite number of time). We define the “sign” of a permutation as being positive if the permutation can be obtained by an even number of transpositions, and negative if it requires an odd number of transpositions. The statement above would then be that the determinant is unchanged if the basis vectors are related by a positive permutation, and picks a sign if they are related by a negative permutation.

Section 4.1 Differential forms revisited: an algebraic approach

Objectives

Subsection 4.1.1 An algebraic approach to one-forms

Definition 4.1.1. The basic one-forms.

Remark 4.1.2.

Subsection 4.1.2 Basic k-forms

Definition 4.1.3. Basic two-forms.

Definition 4.1.4. Basic three-forms.

Definition 4.1.5. Basic k-forms.

Lemma 4.1.6. Antisymmetry of basic k-forms.

Proof.

Remark 4.1.7.

Subsection 4.1.3 k-forms in R3

Definition 4.1.8. k-forms in R3.

Remark 4.1.9.

Remark 4.1.10.

Subsection 4.1.4 k-forms and vector calculus

Exercises 4.1.5 Exercises

1.

2.

3.

4.

5.

Subsection 4.1.2 Basic $k$ -forms

Definition 4.1.5. Basic $k$ -forms.

Lemma 4.1.6. Antisymmetry of basic $k$ -forms.

Subsection 4.1.3 $k$ -forms in $R^{3}$

Definition 4.1.8. $k$ -forms in $R^{3}$ .

Subsection 4.1.4 $k$ -forms and vector calculus