Chapter 2: What Is Area? — The Sign Rule Hidden in Parallelepipeds

§2.0 Measuring Devices, Area, Volume, and Length

In Chapter 1, we defined $dx$, $dy$, and $dz$ as row vectors ($1$-forms) that eat one vector and return a scalar, and we reread $\int f\,dx$ as the limit of matrix actions. In this chapter we extend that framework and discover area-measuring devices ($2$-forms), which eat two vectors, and volume-measuring devices ($3$-forms), which eat three. Both are built from the same parts: $dx$, $dy$, and $dz$.

§2.1 The Limits of Elementary-School Area

Many readers know that “integration is, roughly speaking, a calculation for finding area.” But what is area, exactly? If we think about it again, we may realize that we have not really formalized the idea since elementary school.

So what was that elementary-school formalization? It was based on a very simple intuition: “how many $1\times 1$ squares (tiles) can be packed into the figure.”

For a figure drawn on two-dimensional paper (the $xy$ plane), this “tiling with squares” works well. For example, the area $S$ of the parallelogram spanned by two two-dimensional vectors $\mathbf{v}_1=(x_1,y_1)$ and $\mathbf{v}_2=(x_2,y_2)$ can be written as

However, we live in three-dimensional space. When we try to compute the area of a “tilted plane” in three-dimensional space, things get a bit painful.

Many readers were asked in high-school mathematics to find the area of the parallelogram spanned by three-dimensional vectors $\mathbf{v}_1=(x_1,y_1,z_1)$ and $\mathbf{v}_2=(x_2,y_2,z_2)$. Using high-school notation, we can write

If we expand in components, we fight through a calculation that fills a notebook page before we finally reach

The process is long enough that one would rather not do it more than a few times in life. Why does it become so painful? Because we originally defined area inside the two-dimensional world. In three-dimensional space, the tilt of the plane and the relation between the two vectors lying on it get entangled, and the formula becomes bulky once angles have to be handled too.

Note (terms defined later)

We use “angle” and “inner product” here, but at this stage the intuitive meanings from elementary school are enough. A formal definition of the inner product is given in Chapter 6, but it is not needed to follow the discussion here.

Here we change viewpoint completely. We set aside the elementary-school definition of area for now and redesign from scratch an algebraic measuring device that outputs area.

§2.2 The Three Rules an Area-Measuring Device Must Satisfy

As a first step in the design, we do not rely only on “$1\times 1$ squares.” We take as our base the parallelogram spanned by two vectors.

Note (why start with a parallelogram)

For readers used to “area = tiling with squares,” the motivation for suddenly starting with a parallelogram may not be obvious. Taking a parallelogram as the base relaxes the requirement that the sides be orthogonal, and the rules we are about to build become cleaner. A square appears naturally as a special case. When extending integrals to higher dimensions, this approach is also easier to handle.

What we want is a magic black box $S(\mathbf{v}_1,\mathbf{v}_2)$ that eats two vectors as input and spits out the area $S$ of the parallelogram they span.

This is somewhat sweeping, but for this black box to function properly as an “area-measuring device,” it is enough for it to satisfy the following three rules. Picture a parallelogram whose adjacent sides are two arrows emanating from the origin, and check each rule.

2.2.1 Rule 1: If one side stretches, the area stretches too (proportionality)

If one side vector $\mathbf{v}_1$ is doubled, the area of the parallelogram should naturally double:

Similarly, if we feed in the sum of two vectors, the area should split additively:

Together, these two properties are called linearity. The same “scalar multiple” and “additivity over sums” hold for the second argument $\mathbf{v}_2$ as well — in other words, $S$ is linear in both arguments (bilinear).

2.2.2 Rule 2: Overlap gives zero (alternation)

This is the most important rule. What if $\mathbf{v}_1$ and $\mathbf{v}_2$ are exactly the same vector? The parallelogram collapses to a line and the area is zero:

In fact, from this “same input gives zero” rule, a very strange property follows. Using Rule 1 (linearity), put $\mathbf{v}_1+\mathbf{v}_2$ into both arguments. By “same gives zero,”

Expanding the left-hand side by linearity gives

The first and fourth terms vanish by Rule 2, so what remains is $S(\mathbf{v}_1,\mathbf{v}_2)+S(\mathbf{v}_2,\mathbf{v}_1)=0$. Thus

If we swap the order of the input vectors, the sign of the area flips.

Note (signed area 1)

“Negative area? That sounds wrong.” Perhaps — but physicists do not ignore the “orientation” of nature. The direction of a current, the direction of a magnetic field — in all of these, a vector’s orientation is essential. So the idea that “area, too, has an orientation (signed area)” is rather natural for us physicists. This concept will later be indispensable when we distinguish the two sides of a surface, or compute magnetic flux or fluid flow through space. For now, do not worry too much; just keep moving.

Note (signed area 2)

Even in one variable, when $f(x)<0$ on an interval we can have $\displaystyle\int_a^b f(x)\,dx<0$ — signed “area” is already there. Signed area is that idea lifted to two dimensions.

2.2.3 Rule 3: Fixing the standard (normalization)

Finally, we must decide “what size counts as $1$.” Here we simply declare that the square formed by the basis vectors (standard basis vectors) $\hat{e}_x$ in the $x$ direction and $\hat{e}_y$ in the $y$ direction has area $1$. In this book’s convention, written as column vectors,

Note (basis vectors / standard basis)

One can think of the basis vectors (standard basis) as unit vectors along the axes of Cartesian coordinates. A more advanced definition is left to linear algebra.

Checkpoint so far

- An area-measuring device is a gadget that eats two vectors and returns a scalar. It must satisfy three rules: linearity, alternation, and normalization.

- From alternation (same vector twice gives zero) we get $S(\mathbf{v}_1,\mathbf{v}_2)=-S(\mathbf{v}_2,\mathbf{v}_1)$, so area becomes signed.

- In the next section we search concretely for an algebraic device that satisfies these three rules.

§2.3 An Area-Measuring Device Is an "Antisymmetric Matrix"

We have now set three rules while picturing in our minds the image of a figure called a “parallelogram.” However, once the rules are fixed, we no longer need to visualize the figure. “If we find an algebraic formula that satisfies these three rules, that becomes the definition of area.”

For simplicity, let us first consider two vectors whose $z$ component is always zero—that is, two vectors lying in the $xy$ plane.

Stating the conclusion first: under the constraint that the vectors lie in the $xy$ plane, the measuring device that outputs a quantity perfectly satisfying the three rules above settles into exactly one natural form in the world of linear algebra.

That is a computation of the form “row vector $\times$ matrix $\times$ column vector” with a certain $3 \times 3$ matrix sandwiched in the middle:

Let us actually substitute $\mathbf{v}_1$ and $\mathbf{v}_2$ and compute:

The quantity $x_1 y_2 - x_2 y_1$ we obtain here is precisely the content of the “two-dimensional area formula.”

Note (antisymmetric matrix and metric)

We have written $\mathbf{v}_1^T M \mathbf{v}_2$ (row vector $\times$ matrix $\times$ column vector). From a more advanced standpoint—that of general tensor analysis or general manifold theory—this way of laying vectors on their side without making the metric explicit is slightly bad manners. However, this book will proceed for the time being with standard Cartesian coordinates fixed. Within that scope there is no problem, so we will push forward with this notation here. After introducing the metric in Chapter 6, we will revisit this point.

2.3.1 Higher-Dimensional Extension of the Measuring Device

To repeat, we make the following standpoint clear here. Rather than putting first the naive intuition of the number of tiles we counted in elementary school, we now define the quantity obtained by feeding two vectors into this $3 \times 3$ antisymmetric matrix as the area component returned by this measuring device—the area-measuring device specialized to the $xy$ plane, or the reading along $dx \wedge dy$. It does not exhaust all the information about oriented area in three-dimensional space in a single number; it is merely a scalar cut out by one pair of glasses. The full picture will be given in §2.4.6 when we align all three kinds of measuring devices.

The $1$-form from the previous chapter was a “$1 \times 3$ matrix” that captures one vector. The matrix before us now has evolved into a “$3 \times 3$ antisymmetric matrix” that receives two vectors at once and computes area.

We will call this matrix (or operator) a $2$-form.

Note (quadratic form and $2$-form)

In linear algebra, “quadratic form” often refers to a quadratic polynomial. The $2$-form here is terminology from differential forms and is a different object.

Now the reader must be thinking: “I see, I understand that area comes out from a matrix. But this matrix has zeros in the third row and third column, so it completely ignores the $z$ component. This doesn’t solve that ‘painful three-dimensional calculation,’ does it?”

Exactly right. The matrix $\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ we now have in our hands is nothing more than one of the three basic measuring devices in three-dimensional space—the area-measuring device specialized to the $xy$ plane. This is like a “biased pair of glasses” that looks only at the $xy$ plane. In space, there also exist glasses that look at the $yz$ plane and glasses that look at the $zx$ plane.

Checkpoint so far

- An area-measuring device ($2$-form) is an algebraic device that measures the “pairing of two vectors,” not the parallelogram itself that is being measured.

- The true identity of an area-measuring device is an “antisymmetric matrix,” and the physical rules completely determine the form of the matrix.

- In three-dimensional space there exist three kinds of “biased glasses” (basis $2$-forms).

§2.4 Internal Structure of the Area-Measuring Device

2.4.1 Determining the Form of a "Device That Handles Two Inputs"

In §2.3, we said that the true identity of an area-measuring device ($2$-form) is the $3 \times 3$ antisymmetric matrix sandwiched in the middle. However, that matrix had been handed down from on high by the author.

Our next goal is this: “Can we combine the simple measuring devices $dx, dy, dz$ introduced in Chapter 1 as parts, and systematically build that complicated matrix?” If we can do this, then although we currently have only an area-measuring device specialized to the $xy$ plane, we will also be able to mass-produce area-measuring devices for the $yz$ plane and the $zx$ plane from the same parts by the same procedure.

To achieve this, let us first think about what the overall form of a “device that eats two vectors and spits out a scalar” ought to look like.

In Chapter 1, we saw that a device that eats one vector ($1$-form) can be represented as a "row vector" (row $\times$ column $=$ scalar). So how can we write a device that eats two vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$, at the same time, and moreover preserves the Rule 1 from §2.2 (proportionality and linearity) with respect to both?

The natural form is one in which a square table (matrix) is sandwiched between the two vectors—that is, the form “row $\times$ matrix $\times$ column” ($\mathbf{v}_1^T M \mathbf{v}_2$). Let us accept this as a requirement of linear algebra. What we must find is the contents of the matrix $M$ sandwiched in the middle.

2.4.2 Contraction—An Operation That Eliminates the Matrix

Here, let us give a name to the form "row $\times$ matrix $\times$ column" that we derived earlier.

Writing $\mathbf{v}_1^T M \mathbf{v}_2$ in components,

On the left-hand side we had the matrix $M$ ($3 \times 3 = 9$ numbers) and two vectors ($3+3=6$ numbers), but the right-hand side is just a single scalar. The indices $i$ and $j$ are "eliminated" by the sums, and 15 numbers collapse into 1. This is an example of an operation called contraction.

Contraction in the narrow sense is the operation of "taking a sum over one pair of matching indices." The simplest example is the $1$-form from Chapter 1. $dx(\mathbf{v}) = \sum_i (dx)_i v_i$ obtains a scalar by contracting once over the index $i$. Likewise, the matrix–vector product $\sum_j M_{ij} v_j$ is an operation that contracts once over the index $j$ to obtain a vector (the index $i$ remains).

Contraction in the broad sense is what you get by stacking this several times. The expression $\mathbf{v}_1^T M \mathbf{v}_2 = \sum_i \sum_j v_{1i} M_{ij} v_{2j}$ that we just saw creates a scalar from a matrix and two vectors by contracting over both indices $i$ and $j$—this is a double contraction. In the second half of this chapter (§2.5), the determinant appears as a contraction of three vectors with the Levi-Civita symbol $\epsilon_{ijk}$. Appendix A follows the entire process.

Note (the feel of contraction)

Indices disappear—this is the feel of contraction. A $3 \times 3$ matrix has two indices (row and column), and a vector has one index. In contraction, we make the index of the vector and an index of the matrix "the same letter" and take a sum, so that index vanishes from the expression. If two indices disappear, the number of remaining indices is zero—that is, we get a scalar. This operation of "eliminating indices" underlies every calculation in this book.

2.4.3 Filling in the Entries from the Desired Result

There is no need to search blindly. We already know the "answer" that ought to be output: the area formula for the $xy$ plane.

To produce this output, what numbers should we place in the entries of the matrix $M$? I want the reader to join in and solve the puzzle of filling in the entries while imagining the expansion of $\mathbf{v}_1^T M \mathbf{v}_2$.

1. To produce $x_1 y_2$: place $+1$ in the entry where the $x$ component of $\mathbf{v}_1$ ($x_1$) and the $y$ component of $\mathbf{v}_2$ ($y_2$) are multiplied—that is, row 1, column 2.
2. To produce $-x_2 y_1$: place $-1$ in the entry where the $y$ component of $\mathbf{v}_1$ ($y_1$) and the $x$ component of $\mathbf{v}_2$ ($x_2$) are multiplied—that is, row 2, column 1.
3. Everything else: terms like $x_1 x_2$ and $z$ components never appear in the area formula. Therefore all remaining entries are $0$.

In this way, that antisymmetric matrix that appeared from on high in §2.3 is uniquely determined in the following form:

2.4.4 Introduction of the Tensor Product

Now, if we look closely at the matrix $M$ of this area-measuring device, we see that it is built from the "subtraction" of two parts.

The first term on the right-hand side is "the matrix whose only nonzero entry is row 1, column 2," and it represents the operation of extracting the $x$ component from $\mathbf{v}_1$ and the $y$ component from $\mathbf{v}_2$ and multiplying them ($x_1 y_2$).

Let us recall the measuring devices from Chapter 1. The measuring device that extracts the $x$ component was $dx$, and the one that extracts the $y$ component was $dy$. Let us combine these two measuring devices into a new operator: it measures $\mathbf{v}_1$ with $dx$, measures $\mathbf{v}_2$ with $dy$, and multiplies the results. We call this the tensor product and write it as $dx \otimes dy$.

The tensor product $dx \otimes dy$ may sound intimidating, but there is nothing to fear. In the world of matrix representations, this is nothing more than the operation of “turning on the switch that sets only the entry in row 1, column 2 to $1$.”

Note (matrix representation of the tensor product)

For readers who like formulas: with $dx = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix}$ and $dy = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}$, the matrix representation of $dx \otimes dy$ is obtained as the outer product $dx^T dy$: turn the first row vector into a column, place it next to the second row vector, and multiply each column entry by each row entry to fill the matrix. You will probably find this construction in your linear algebra textbook as well.

2.4.5 Completion of the Wedge Product

The preparation is now complete. The area-measuring device matrix for the $xy$ plane that we derived was obtained by subtracting the "switch for $y_1 x_2$" from the "switch for $x_1 y_2$."

Translating this into the language of measuring devices ($1$-forms), we naturally arrive at a new operator: the wedge product $dx \wedge dy$.

The symbol $\wedge$ means "wedge" in English. Applying this definition to vectors gives:

Splendid — from combinations of the measuring devices ($1$-forms) of Chapter 1, we have constructed an area-measuring device ($2$-form) systematically. Because it is built from the subtraction of tensor products, please also notice that swapping the arguments flips the sign (alternating property)—Rule 2 from §2.2 is automatically "built in."

2.4.6 The Three Basis $2$-Forms

By the same method, we can also construct the remaining two "biased pairs of glasses."

Glasses that look at the $yz$ plane: $dy \wedge dz$

Glasses that look at the $zx$ plane: $dz \wedge dx$

That is everything!

Now let us apply each of the three pairs of glasses one by one to general vectors and see what each one measures.

For general three-dimensional vectors $\mathbf{v}_1 = \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} x_2 \\ y_2 \\ z_2 \end{pmatrix}$, we compute according to the definition in §2.4.5 (putting $\mathbf{v}_1$ in the first argument and $\mathbf{v}_2$ in the second):

Each computation is a single subtraction. Whatever the orientation of the vectors, the definition of the wedge product automatically returns the signed value measured by that pair of glasses.

2.4.7 The Meaning of the Three Numbers—Orthogonal Projection onto Each Coordinate Plane

Now, the reader surely expected the following: “If we feed in two vectors and add up the three readings from the three pairs of glasses, we get the area we learned in elementary school, right?”

That is,

However, it is not actually that simple.

When we designed the area-measuring device in §2.2, in the story confined to the $xy$ plane, a single number was indeed enough. But for a general parallelogram in three-dimensional space, orientation is not a simple $+/-$ choice. The readings obtained through the three basis $2$-forms (pairs of glasses),

do not represent the area of the parallelogram in space itself, but rather the signed areas obtained when it is orthogonally projected onto each coordinate plane.

Orthogonal projection onto each coordinate plane—this is the geometric meaning of the three readings. $A_{yz}$ is the signed area of the orthogonal projection of the parallelogram onto the $yz$ plane; $A_{zx}$ is the projection onto the $zx$ plane; and $A_{xy}$ is the projection onto the $xy$ plane.

Note (orthogonal projection)

Orthogonal projection means the "shadow" you get when you crush a figure flat onto a plane at right angles. It helps to picture a light source shining straight down onto the plane. For example, under orthogonal projection onto the $xy$ plane, you ignore the $z$ coordinate of the figure and take only the $x,y$ coordinates. From Chapter 3 onward, this book will sometimes call orthogonal projection simply a shadow.

So how do we extract the scalar area $S$ of the parallelogram itself from these projected areas? The answer has the same structure as the length of a vector.

The reader probably knows that to extract the "length (scalar)" from a vector $\mathbf{v} = x\hat{e}_x + y\hat{e}_y + z\hat{e}_z$, you compute the square root of the sum of squares of the components, $\sqrt{x^2+y^2+z^2}$. The operation that extracts the scalar area of a parallelogram from the three projected areas has exactly the same form.

This matches exactly the high-school area formula that we called "a painful calculation" in §2.1. We have at last returned safely from the design of an algebraic measuring device all the way to the "number of tiles" from elementary school.

Note (a general $2$-form)

The simple sum of the three readings, $(dy \wedge dz + dz \wedge dx + dx \wedge dy)(\mathbf{v}_1,\mathbf{v}_2)$, is also a perfectly good $2$-form that returns a single scalar. However, this value does not agree with the Euclidean area (the area we learned in elementary school). To obtain that, the nonlinear operation of "square root of the sum of squares" is required, and this nonlinearity foreshadows the appearance of the metric (inner product) that we will introduce in Chapter 6. It is a structure completely parallel to the fact that the length of a vector, $\sqrt{x^2+y^2+z^2}$, is the square root of the inner product $\mathbf{v}\cdot\mathbf{v}$.

2.4.8 One-Dimensional "Area" (Length) and Zero-Dimensional "Area" (a Point)

Let us lower our viewpoint a little. I said that the operation "square root of the sum of squares of the components" should look familiar—and in fact we already know the same pattern for one-dimensional figures (line segments).

The $1$-forms ($dx, dy, dz$) treated in Chapter 1 were measuring devices that "eat only one vector." The figure being measured is the vector itself ($\mathbf{v} = x\hat{e}_x + y\hat{e}_y + z\hat{e}_z$), which is data for a one-dimensional oriented measure—an oriented length. The operation that extracts the "scalar length" from this was $\sqrt{x^2+y^2+z^2}$.

So what is a $0$-form? Since a measuring device that eats $k$ vectors is a $k$-form, a $0$-form is a "measuring device that eats no vectors at all". It needs no input direction (no arrow); you simply place it on the spot and it returns a single scalar (temperature, density, and so on)—in other words, it is just a "scalar field (function)". The “zero-dimensional figure” in this analogy is simply a point; the value assigned there has only one component, and its magnitude is obtained by taking the absolute value. In standard mathematical language, a $0$-form is a function itself, but when this book says it "measures a point," that is a metaphor for lining up dimensions in a series: it returns a value when placed at that point without needing an input direction.

Point (0-dimensional), line segment (1-dimensional), parallelogram (2-dimensional). All of them were repetitions of the same structure: "eat $k$ vectors and return a scalar."

In §2.5 we will raise this one step further and apply the same structure to a three-dimensional parallelepiped (volume). To summarize:

• $k=0$: a $0$-form measures a point. One component; magnitude is absolute value
• $k=1$: a $1$-form measures a vector (oriented length). Three components; magnitude is $\sqrt{x^2+y^2+z^2}$
• $k=2$: a $2$-form measures the parallelogram spanned by two vectors (oriented area). Three components; magnitude is $\sqrt{A_{yz}^2+A_{zx}^2+A_{xy}^2}$
• $k=3$: a $3$-form measures the parallelepiped spanned by three vectors (oriented volume). One component; magnitude is absolute value

Because the stage of this book is three-dimensional space, we stop at $k=0,1,2,3$, but the principle that "a $k$-form eats $k$ vectors and returns a scalar" does not depend on $k$. This pattern carries forward into integration in Chapter 3 and the exterior derivative in Chapter 5.

2.4.9 Agreement with the Cross Product

Readers who learned the "vector cross product" in university linear algebra will recognize the following three components.

This is nothing other than the values of $dy \wedge dz$, $dz \wedge dx$, and $dx \wedge dy$ on $(\mathbf{v}_1,\mathbf{v}_2)$, listed in this order.

Note (for uneasy readers)

Even if this looks unfamiliar, there is no obstacle to understanding what follows. In this book we simply discovered the wedge product before the cross product.

Note ($2$-forms and the cross product)

Area as a $2$-form and the cross product as a three-component vector are related by what we will later call the Hodge dual. We will make clear in Chapter 6 why, in Cartesian coordinates, these must inevitably carry the same components; here we will leave it at "different ways of carrying information about oriented area."

Checkpoint so far

- The wedge product $dx \wedge dy := dx \otimes dy - dy \otimes dx$ automatically generates an antisymmetric matrix from the subtraction of tensor products. $dy \wedge dz$ and $dz \wedge dx$ are constructed similarly, and any area-measuring device can be assembled as a linear combination of the three basis $2$-forms.

- Feeding two vectors $\mathbf{v}_1,\mathbf{v}_2$ into the three basis $2$-forms yields the signed projected areas $A_{yz},A_{zx},A_{xy}$ onto the coordinate planes ($yz$, $zx$, $xy$).

- The scalar area (the positive magnitude corresponding to the number of tiles) is recovered by the square root of the sum of squares of the projected areas, $\sqrt{A_{yz}^2 + A_{zx}^2 + A_{xy}^2}$, returning to the formula of §2.1 (Lagrange's identity). It is the same kind of operation as the formula for the length of a vector. This nonlinearity foreshadows the metric we will introduce in Chapter 6.

§2.5 Volume-Measuring Devices and Determinants

In the previous section we combined two $1$-forms (measuring devices) $dx, dy$ via the wedge product and assembled the area-measuring device ($2$-form) $dx \wedge dy$. The area-measuring device was a device that “eats two vectors and returns the signed area of the parallelogram they span.”

Now at last we come to volume, and here the stance is the same. First we assign volume $1$ to the standard basis $\hat{e}_x,\hat{e}_y,\hat{e}_z$ lined up to form a unit cube (Rule 3), and we define the measuring device by alternativity and linearity. To return to the positive volume of elementary school—“how many unit cubes fit inside?”—we take the magnitude of the signed output. In three dimensions this will reduce to an absolute value, the one-component version of the same “square root of a sum of squares” structure we saw for area.

A natural question arises. “Can the device that measures the volume of the parallelepiped spanned by three vectors—the volume-measuring device ($3$-form)—also be assembled in the same way?”

2.5.1 Rules a Volume-Measuring Device Must Satisfy

Let us take exactly the same strategy as when we designed the area-measuring device in §2.2. First we decide “the rules that a volume-measuring device $V$ must satisfy,” and then we uniquely derive the form of the device from those rules.

A volume-measuring device $V(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3)$ is a function that eats three vectors and returns one scalar. The rules we impose on it are obtained by extending the three rules for the area-measuring device directly to three arguments:

Rule 1 (linear in each argument) If one vector doubles, the volume doubles; if one input is a sum of vectors, the output splits additively.

The same holds for the second and third arguments.

Rule 2 (alternativity — swapping any two arguments flips the sign)

The same for the other pairs. By the same argument as in §2.2, putting the same vector into two slots always gives zero—if $\mathbf{v}_1 = \mathbf{v}_2$, then $V(\mathbf{v}_1,\mathbf{v}_1,\mathbf{v}_3) = -V(\mathbf{v}_1,\mathbf{v}_1,\mathbf{v}_3)$, so $V=0$. Geometrically the parallelepiped collapses, but algebraically alternativity alone forces “collapse to zero.” This is the same structure as Rule 2 for the two-dimensional area-measuring device.

Rule 3 (normalization — the unit cube has volume 1)

2.5.2 Determining the Form from the Rules — Basis Expansion and Elimination

Let us follow the same procedure as when we derived the matrix for the area-measuring device in §2.3. Expand the three vectors in the basis:

By Rule 1 (multilinearity), $V(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3)$ expands into $3 \times 3 \times 3 = 27$ terms:

Twenty-seven terms is a lot, but Rule 2 (alternativity) eliminates most of them:

• Every term with a repeated index is zero (e.g., $V(\hat{e}_x, \hat{e}_x, \hat{e}_z) = 0$)
• Combinations without repetition are exactly those where $i, j, k$ are all distinct—that is, permutations of $(x,y,z)$

There are $3! = 6$ permutations of $(x, y, z)$ in all. By alternativity, even permutations have the same sign as $V(\hat{e}_x, \hat{e}_y, \hat{e}_z)$, and odd permutations have the opposite sign. Rule 3 fixed $V(\hat{e}_x, \hat{e}_y, \hat{e}_z) = 1$, so:

Therefore:

Readers who have studied linear algebra should recognize this as the determinant of the matrix formed by lining up the three vectors as columns:

Readers who do not know it yet may simply decide to call this six-term sum “the determinant” and move on—the essential fact is that it fell out of the rules above.

Here is the one point to keep in mind. The determinant is not merely a “procedure of multiplying and subtracting components”; it can also be read as an operator that returns a single number—the “signed volume” when three column vectors are fed into it in that order. The $1$-form of Chapter 1 is “one vector → scalar,” the $2$-form of §2.3 is “two vectors → scalar”—the determinant is “three vectors → scalar,” linking this series.

For the area-measuring device an antisymmetric matrix appeared; for the volume-measuring device the determinant appeared—but in both cases the rules we imposed first determined the algebraic form.

Note (the order of the six terms of the determinant) Some textbooks simply declare upfront that “the definition of the determinant is these six terms.” In the flow of this book, the order is rules of the measuring device → six terms.

2.5.3 Reinterpreting the $2$-form as a Determinant

In fact, a determinant was hiding in the wedge product of the $2$-form defined in §2.4 as well. Let us recall the definition:

The right-hand side is nothing other than the following $2 \times 2$ determinant:

In other words, the value of the wedge product is the determinant of the matrix with “which measuring device” across the rows and “which vector” down the columns.

2.5.4 Definition of $dx \wedge dy \wedge dz$ — Completing the Determinant Pattern

Extending this “determinant pattern” straightforwardly to $3 \times 3$ gives us the volume-measuring device.

Definition: For three vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$,

Each entry on the right-hand side is “the coordinate component that one of the $1$-forms $dx$, $dy$, or $dz$ extracts from vector $\mathbf{v}_i$,” so this is exactly:

This agrees completely with the $V$ derived in §2.5.2. The volume-measuring device can be constructed as the wedge product of the three measuring devices $dx$, $dy$, and $dz$.

Readers who learned determinants in linear algebra will easily see that this definition satisfies the three rules of §2.5.1:

• Multilinearity: the determinant is linear in each column, so this is automatic.
• Alternativity: swapping two columns of the determinant flips the sign.
• Normalization: inserting $\hat{e}_x, \hat{e}_y, \hat{e}_z$ gives the identity matrix, and $\det(I) = 1$.

2.5.5 Tensor-Product Representation of the Volume-Measuring Device — An Array of Three $3 \times 3$ Blocks Side by Side

In §2.4.4–2.4.5 we saw that the area-measuring device $dx \wedge dy$ can be built as the difference of tensor products $dx \otimes dy - dy \otimes dx$, and that the result is a $3 \times 3$ antisymmetric matrix. Can the same trick be applied to the volume-measuring device $dx \wedge dy \wedge dz$?

Why bring out a tensor array here at all? Because we want to show that the operation of eating three vectors and returning a determinant can be written with exactly the same idea of “contraction” as for the area-measuring device. As defined in §2.4.2, contraction is the operation of summing over a shared index and eliminating that index. $1$-forms, $2$-forms, and $3$-forms all run on this single principle.

The answer is yes. However, while tensor products of two measuring devices fit into $3 \times 3$ matrices, with three measuring devices one more index appears and we get a three-dimensional array of $3 \times 3 \times 3 = 27$ components. Such a multidimensional array with three indices is called a tensor in both mathematics and physics. A matrix (two indices) is a special case of a tensor; what appears here is a third-order tensor with three indices.

By the way, for tensors of third order and above, writing out every component makes the page explode, so even physicists rarely use array displays. Instead the usual practice is to write using an indexed representative such as the Levi-Civita epsilon component $\epsilon_{ijk}$, and leave the rest of the calculation to contraction. In this section too we follow that convention and, as needed, pull out individual components to carry the discussion forward.

Note (tensor notation) In tensor algebra, the notation itself $\epsilon_{ijk}$ often denotes “the entire third-order tensor” (the Levi-Civita epsilon). To avoid confusion, this book writes the entire third-order tensor as $\widehat{\epsilon}$ and a single component as $\epsilon_{ijk}$. That is, we distinguish matrix display (the $3\times3$ of §2.3) from “one component” in our notation. Keep this in mind when reading other books.

Note (upper and lower indices) An apology to detailed readers who are peeking ahead. In more advanced differential geometry and tensor analysis, the convention of distinguishing vectors from covectors and bases from dual bases by the position of upper vs. lower indices is often used. This book fixes the discussion to standard Cartesian coordinates, where the distinction between covector components and vector components does not surface, so we write everything with lower indices, as in $\epsilon_{ijk}$.

Now, antisymmetrizing this 27-component array (adding the six permutations with signs) produces an array where only 6 of the 27 entries are $\pm 1$ and the rest are zero. It is exactly the table of permutations derived in §2.5.2—the components are taken as follows (the $(i,j,k)$ component of $\widehat{\epsilon}$):

How to display these 27 numbers—there is a good way. Think of three $3 \times 3$ matrices lined up in a row:

The first index $i$ specifies “which slice from the left,” and inside each slice the contents are a $3 \times 3$ block indexed by $j,k$.

Eating three vectors and collapsing to a scalar — triple contraction

In formulas it is one line:

This is a triple contraction. We sum over all three indices $i, j, k$ and build a single scalar from the 27-component array and three vectors. It is the natural extension of $\mathbf{v}_1^T M \mathbf{v}_2$ (double contraction) from §2.4.2.

Viewing this triple contraction in two stages makes the structure even clearer:

• Step 1 (contract over index $i$ → matrix): Contract $\mathbf{v}_1$ and $\widehat{\epsilon}$ over index $i$. $\displaystyle M = \sum_{i=1}^3 v_{1i}\,\epsilon_{i,\cdot,\cdot}$ ($\epsilon_{i,\cdot,\cdot}$ is the $i$th $3 \times 3$ block of $\widehat{\epsilon}$ from the left). The result drops down to a single $3 \times 3$ matrix $M$. This $M$ is precisely the matrix of the area-measuring device constructed in §2.4.5–2.4.6:

• Step 2 (contract over indices $j,k$ → scalar): Sandwich the resulting matrix $M$ with $\mathbf{v}_2, \mathbf{v}_3$ and contract— $V = \mathbf{v}_2^T M \mathbf{v}_3$. An operation we are already familiar with from §2.4.

In Appendix A, we write out every component and confirm that this two-stage contraction reproduces all six terms of the determinant from §2.5.2.

2.5.6 Example: Signed Volume of a Parallelepiped

Let us verify by hand. Add a third vector to the two whose area we measured in §2.4.6:

Feed them into the volume-measuring device:

Computing according to the definition:

Note (negative volume?)

As with oriented area in §2.2, the order can yield a minus sign.

Let us try swapping $\mathbf{v}_1$ and $\mathbf{v}_2$:

2.5.7 Agreement with the Scalar Triple Product

Readers who learned the “scalar triple product” in vector analysis may know the following identity:

The right-hand side is exactly the $(dx \wedge dy \wedge dz)(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3)$ we defined.

Note (borrowing the dot product symbol early) The “$\cdot$” on the left is the symbol for the dot product; here we use it in the familiar coordinate-calculation sense of “sum of products of corresponding components.” In Chapter 6 we will give a formal definition using the metric $g$.

2.5.8 Output of the Volume-Measuring Device — Only One Independent Component

The $dx \wedge dy \wedge dz$ we assembled is a volume-measuring device ($3$-form) waiting for three vectors. Unlike the area-measuring device that eats two vectors, feeding three vectors into the volume-measuring device yields a single signed scalar $V$ (because, as we will see in §2.5.9, a $3$-form has only one independent component).

If we want the positive elementary-school volume—the scalar volume “how many unit cubes fit inside?”—we take the magnitude of the signed output. Since a $3$-form in three-dimensional space has only one independent component, this is simply $|V|=\sqrt{V^2}$. As with area, this nonlinear operation requires a metric; we will treat it again in Chapter 6.

2.5.9 The “Size” of a $3$-form — Why Is There Only One Independent Component?

An area-measuring device ($2$-form) could be represented by a $3 \times 3$ antisymmetric matrix, with 3 independent components (the coefficients of $dy \wedge dz$, $dz \wedge dx$, $dx \wedge dy$).

How many independent components does a volume-measuring device ($3$-form) have? Consider the candidates we can build as basis $3$-forms. The number of ways to choose three measuring devices $dx, dy, dz$ without repetition is … only one. Just $dx \wedge dy \wedge dz$ (changing the order only changes the sign; it does not produce a new basis).

Note (foreshadowing the Hodge dual) For area, feeding two vectors into the three basis $2$-forms gave three numbers, and obtaining scalar area required a nonlinear operation (square root of a sum of squares). By contrast, the output of the volume-measuring device $dx \wedge dy \wedge dz$ is a single scalar $V$ from the start. And this is no accident. A $0$-form (scalar field) also has 1 independent component, a $1$-form has 3, and a $2$-form has 3—that is, $1, 3, 3, 1$, mirrored from both ends. This sequence is a foreshadowing of the symmetry called the Hodge dual. We will go into detail in a later chapter.

2.5.10 Hierarchy of Measuring Devices — Organizing the Weapons Gained in Chapter 2

Here let us survey the overall picture of the “measuring devices” we have built up since Chapter 1.

From §2.4.8 onward, and especially through §2.5.8 and §2.5.9 (the number of independent components), we have seen that for each dimension $k$, the same pattern repeats: “a $k$-form eats $k$ vectors and returns a scalar.” Collecting the number of data components and the operation that extracts the ordinary scalar magnitude, we get:

Checkpoint so far

- The volume-measuring device $dx \wedge dy \wedge dz$ ($3$-form) is the wedge product of three measuring devices, and its value equals a $3 \times 3$ determinant.

- There is a $1, 3, 3, 1$ symmetry: $0$-form (1 component) → $1$-form (3 components) → $2$-form (3 components) → $3$-form (1 component).

- No matter the dimension, the principle is the same: a $k$-form eats $k$ vectors and returns a scalar. When we want the ordinary positive magnitude, we extract it by the appropriate “square root of a sum of squares” operation; in the one-component cases this reduces to an absolute value.

§2.6 Summary of This Chapter and Outlook Toward Chapter 3

Checkpoint so far — Chapter 2 as a whole

- Area is a $2$-form, an antisymmetric matrix that measures the “pairing of two vectors”; the physical rules completely determine the form of the matrix.

- The wedge product $\wedge$ is the operation that builds an antisymmetric matrix by “subtracting tensor products.” By attaching field coefficients to the three basis $2$-forms, we assemble a general $2$-form (the meaning of the coefficients comes in later chapters).

- The volume-measuring device $dx \wedge dy \wedge dz$ ($3$-form) is the wedge product of three measuring devices, and its value equals a $3 \times 3$ determinant. It can also be expressed by contraction with the antisymmetrized $\widehat{\epsilon}$ (components $\epsilon_{ijk}$, the Levi-Civita symbol) (verify all components in Appendix A).

In this chapter, starting from the measuring devices $dx$, $dy$, and $dz$ set up in Chapter 1, we have reached the point of building area ($2$-forms) and volume ($3$-forms) algebraically. The wedge product is not a tool for “drawing geometry directly”; it is a tool for mechanically generating formulas that satisfy the rules of measuring devices.

In the next chapter (Chapter 3), we will aggregate these forms over curves, surfaces, and regions—that is, rewrite line integrals, surface integrals, and volume integrals in a unified language of $1$-forms / $2$-forms / $3$-forms. Variable changes and pullback will also appear, and the Jacobian will arise naturally as the “calculation of shadows.”

Appendix A: Tensor-Product Representation of the Volume-Measuring Device — Full Component Calculation

In §2.5.5 we stated that contraction of $\widehat{\epsilon}$ (components $\epsilon_{ijk}$) with three vectors agrees with the determinant. Here we follow that entire process by hand calculation and verify it term by term.

A.1 Extension of the Tensor Product to Three Arguments

Let us review. The $(i,j)$ component of $dx \otimes dy$ was the product of the $i$th component of $dx$ and the $j$th component of $dy$:

Extension to three measuring devices is natural—we merely add one more index:

After substituting $dx=(1\ 0\ 0)$, $dy=(0\ 1\ 0)$, and $dz=(0\ 0\ 1)$, only one of the $3^3=27$ cells is nonzero: the entry at $(i,j,k)=(1,2,3)$ is $1$. The remaining 26 cells are $0$. Each of the other five permutations ($dx \otimes dz \otimes dy$, and so on) is likewise nonzero in exactly one place—for $dy \otimes dx \otimes dz$ the $1$ sits at $(2,1,3)$, and for $dz \otimes dy \otimes dx$ at $(3,2,1)$.

A.2 Antisymmetrization — Superposing Six Permutations with Signs

In §2.4.5, for area we had $dx \wedge dy = dx \otimes dy - dy \otimes dx$ (a signed sum of two terms). The three measuring devices are $dx, dy, dz$ in Cartesian coordinates. There are $3! = 6$ permutations, For each permutation $\sigma$, we arrange $(dx,dy,dz)$ in the permuted order, join them with tensor products $\otimes$, attach the sign $\mathrm{sgn}(\sigma)$, and sum over $S_3$—that is, the following six-line sum written in one line in the language of permutations.

Note (symmetric group and sign of a permutation)

$S_3$ is the symmetric group of degree 3 (the set of permutations). $\mathrm{sgn}(\sigma)$ is defined to be $+1$ for an even permutation and $-1$ for an odd permutation.

Written out:

Superposing these six “arrays with only one nonzero entry” with signs leaves only 6 of the 27 components at $\pm 1$; the other 21 are zero. Compare with the permutation table in §2.5.2—this is the construction principle that yields each component $\epsilon_{ijk}$ of $\widehat{\epsilon}$.

A.3 Writing Down the Matrix for Each Component

Let us verify cell by cell the three component matrices shown in §2.5.5.

Component 1 ($i=1$, $x$): among the six terms, those with $i = 1$ are $+1$ at $(1,2,3)$ and $-1$ at $(1,3,2)$.

Component 2 ($i=2$, $y$): $+1$ at $(2,3,1)$ and $-1$ at $(2,1,3)$.

Component 3 ($i=3$, $z$): $+1$ at $(3,1,2)$ and $-1$ at $(3,2,1)$.

To repeat: these are exactly the matrices of $dy \wedge dz$, $dz \wedge dx$, and $dx \wedge dy$, respectively.

A.4 Feeding Three Vectors — The Full Contraction Process

Step 1: weighted sum over index $i$ using the components of $\mathbf{v}_1$ (collapse index $i$)

Add the three area-measuring-device matrices, weighted by the components of $\mathbf{v}_1$. One vector is fed in and the $3$-dimensional array drops to a $3 \times 3$ matrix:

Adding component by component gives:

An area-measuring device specialized to $\mathbf{v}_1$ has appeared, built from the three area-measuring devices weighted by the components of $\mathbf{v}_1$. Note that it is an antisymmetric matrix—the antisymmetry in the indices of $\widehat{\epsilon}$ is inherited.

Step 2: sandwich the matrix with the remaining two vectors (collapse indices $j,k$)

What remains is the familiar “row $\times$ matrix $\times$ column” from §2.4:

First compute $\mathbf{v}_2^T \times M$ (row $\times$ matrix → row):

Finally, the sum of products of corresponding components of the resulting row vector and $\mathbf{v}_3$ (row $\times$ column → scalar):

Expanding and collecting terms gives:

This matches the determinant in §2.5.2 term for term—all six terms agree.