Disclaimer: Nothing in this note is original.

1. Some Conventions

We will use capital Latin letters such as $I,J,$ etc. to denote sets of indices, for example $a_ {I}$ is the shorthand notation for $a_ {i_ {1}\dots i_ {p}}$. To get rid of annoying factorial factors that frequently appear in differential form calculation, we require the indices to be arranged in increasing order, the indices are denoted $I_ {<}$,

\[I_ {<} = \left\{ i_ {1}<i_ {2}<\dots<i_ {p} \right\} .\]

We also adopt the so-called generalized Kronecker delta symbol,

\[\delta^{I}_ {J}:= \begin{cases} 1 & J \text{ is an even permutation of }I \\ -1 & J \text{ is an odd permutation of }I \\ 0 & \text{ otherwise} \end{cases}\]

for example, we have

\[\delta^{123}_ {132}=-1,\quad \delta^{123}_ {231}=1, \quad \delta^{123}_ {124}=0.\]

We can then define the familiar Levi-Civita symbol as

\[\epsilon_ {I} = \epsilon_ {i_ {1}\dots i_ {n}} =\epsilon^{I}= \delta^{I}_ {12\dots n}.\]

Note that it is called a symbol and it is not a tensor! So don’t use the metric to raise or lower the indices! Calculation-wise, the biggest difference between the Levi-Civita symbol and Levi-Civita tensor is that, for the symbol $\epsilon_ {I} = \epsilon^{I}$ by definition, however for the tensor what people usually do is to define $\epsilon_ {I}$ first, then use the metric to raise its indices, as for all tensors. It usually yields a factor involving $\det g$ and the signature of $g$, which is $-1$ for Minkowski metric.

In the above-mentioned notations, the wedge product of two differential forms $\alpha,\beta$ can be written as (in components)

\[(\alpha \wedge \beta)_ {I} = \delta_ {I}^{MN} \alpha_ {M_ {<}} \beta_ {N_ {<}}.\]

where repeated indices are supposed to be summed over. Equivalently

\[(\alpha \wedge \beta)(V_ {I}) = \delta^{MN}_ {I} \alpha(V_ {M_ {<}})\beta(V_ {N_ {<}}).\]

As an example of the last definition, take $\alpha,\beta$ to be two 1-forms acting on vectors $V_ {1},V_ {2}$, then by the definition

\[(\alpha \wedge \beta)(V_ {1}V_ {2}) = \delta^{MN}_ {12} \alpha(V_ {M_ {<}}) \beta(V_ {N_ {<}}) = \alpha(V_ {1}) \beta(V_ {2}) - \alpha(V_ {1})\beta(V_ {2}),\]

which reproduced our familiar definition for the wedge product.

As another example, let $\alpha$ be a 2-form and $\beta$ be a 1-form, then

\[(\alpha \wedge \beta)_ {523} = \delta_ {523}^{IJ}\; \alpha_ {I_ {<}}\beta_ {J_ {<}} = \delta^{253}_ {523}\alpha_ {25}\beta_ {3}+ \delta^{352}_ {523}\alpha_ {35}\beta_ {2} +\delta^{235}_ {523}\alpha_ {23}\beta_ {5}\]

which is

\[-\alpha_ {25}\beta_ {3} +\alpha_ {35}\beta_ {2} + \alpha_ {23}\beta_ {5}.\]

The notations we adopt here greatly simplifies the calculation.

The wedge product of three differential forms are also beautifully expressible in our notation,

\[(\alpha \wedge \beta \wedge \gamma)_ {I} = \delta^{ABC}_ {I} \alpha_ {A} \beta_ {B} \gamma_ {C},\quad A,B,C \text{ in increasing order}.\]

The contraction of indices also works in a simple way,

Lemma.

\[\sum_ {J_ {<}}\delta^{IJ}_ {M}\delta^{KL}_ {J} = \delta^{IKL}_ {J}.\]

2. A little bit more about differential forms

Although forms can be expanded in any bases, the geometric meaning of differential forms is most obvious in terms of Cartesian coordinates. By Cartesian coordinates we just mean that the metric is that of the Euclidean space $\mathbb{R}^{n}$. In this case, the bases for the tangent vector space is $\partial_ {i}$ and the dual bases are given by $dx^{i}$. We already know that $dx^{i}$ reads off the i-th component of a vector $\mathbf{v}$,

\[\left\langle dx^{i},\mathbf{v} \right\rangle = v^{i}.\]

Note that there will be problem if you think of $dx^{i}$ as an infinitesimal quantity, as is the case in the classical real analysis! Because then on the left hand side we have the inner product between an infinitesimal thing and one regular-sized thing, the final product should be infinitesimal of order one, but on the right hand side we clearly have a regular-sized thing, the i-th component of a finite-sized vector. In fact, $dx^{i}$ is a special case of $d\mathbf{v}$ for arbitrary vector $\mathbf{v}$, and $d\mathbf{v}$ should be regarded as a vector-valued 1-form! That is, something that takes an vector and spits out an vector. For more details, please refer to Frankel etc.

Let’s start with the 2-dimensional Euclidean space. Given two vectors $\mathbf{v,w}$ we have

\[dx^{i}\wedge dx^{j}(\mathbf{v,w}) = \begin{vmatrix} v^{i} & w^{j} \\ v^{j} & w^{i} \end{vmatrix} = \pm \text{ area projected on the } x^{i}x^{j} \text{ plane.}\]

We will show that exterior products yield a coordinate-free expression for the determinant! Consider an n-tuple of 1-forms $\tau^{1},\dots,\tau^{n}$, given in terms of bases

\[\tau^{i} = T^{i}_ {j} \sigma^{j}\]

where $\sigma^{i}$ are the bases. Then

\[\begin{align*} \tau^{1}\wedge \dots \wedge \tau^{n} &= T^{1}_ {i_ {1}}\sigma^{i_ {1}}\wedge \dots \wedge T^{n}_ {i_ {n} } \sigma^{i_ {n}} \\ &= T^{1}_ {i_ {1}}\dots T^{n}_ {i_ {n}} \sigma^{i_ {1}}\wedge \dots \wedge \sigma^{i_ {n}} \\ &= T^{1}_ {i_ {1}}\dots T^{n}_ {i_ {n}} \sigma^{1}\wedge \dots \wedge \sigma^{n} \epsilon^{i_ {1}\dots i_ {n}} \\ &= (\det T) \;\sigma^{1}\wedge \dots \wedge \sigma^{n} \end{align*}\]

For this reason the wedge product is very convenient for discussing linear dependence. For example, using form it is quite easy to tell if a set of forms are linearly independent, see the theorem below.

Theorem. The set of 1-forms $\tau^{i}$ are linearly dependent iff

\[\tau^{1} \wedge \tau^{2} \wedge \dots \wedge \tau^{n} = 0\]

Many formulas given in the form of vectors are more natural in terms of differential forms, here we shall briefly discuss the computations of differential forms and its relation with vector analysis.

Consider $\mathbb{R}^{3}$ as a 3-manifold with any (perhaps curvilinear) coordinate system $x^{1,2,3}$. Let $f$ be a 0-form, that is a function on $\mathbb{R}^{3}$. The 1-forms are written as

\[\alpha = \alpha_ {1}dx^{1} + a_ {2} dx^{2} +a_ {3}dx^{3}\]

and 2-forms are written as

\[\begin{align*} \gamma &= \gamma_ {12} dx^{1}\wedge dx^{2} + \gamma_ {23}\, dx^{2}\wedge dx^{3} + \gamma_ {13} dx^{1}\wedge dx^{3} \\ &:=\gamma_ {3} dx^{2}\wedge dx^{1} + \gamma_ {1}\, dx^{2}\wedge dx^{3} + \gamma_ {2} dx^{3}\wedge dx^{1} \end{align*}\]

where we used the convention that the indices are always arrange in increasing order in the first line, and indices are cyclic in the second line.

The 3-forms are the simplest for they only have one component,

\[\omega = \omega(x^{1},x^{2},x^{3})dx^{1}\wedge dx^{2}\wedge dx^{3}.\]

There are familiar expressions used in vector analysis in the case when the coordinates are Cartesian, involving line, surface, and volume integrals. Let’s look at some examples.

Given 1-forms $\alpha,\beta$, let $\mathbf{a,b}$ be the corresponding vectors defined components-wise, namely

\[\mathbf{a} = a^{i} \partial _ {i},\quad a^{i} \equiv \alpha_ {i}, \text{the same for }b \text{ and }\beta\]

then we have

\[\alpha \wedge \beta = \mathbf{a}\times \mathbf{b} \,dS^{i} \text{ component-wise}\]

where $ds^{1} \equiv dx^{2}\wedge dx^{3}$ and etc. Note that this correspondence in general does not hold in curvilinear coordinates.

There has already been a lot of discussions about the theoretical importance and philosophy of the exterior differential, I will not dive in them here, interested readers can refer to various textbooks and lecture notes. Here instead I’ll just focus on the calculation methods, starting with some examples in $\mathbb{R}^{3}$.

Given a function $f(x)$ which is by definition also a 0-form. The differential of $f$ is

\[df = \partial_ {x}f dx + \partial _ {y}f dy + \partial _ {z}f dz\]

If the coordinates are Cartesian, then the components are the components of the gradient of $f$,

\[\nabla_ {i} f = \partial _ {i}f,\quad df = \nabla f \cdot d\mathbf{x}\]

where $d\mathbf{x} = (dx^{1},dx^{2},dx^{3})$.

That is for 0-form. Given a 1-form $\alpha$ and the corresponding vector $a$ whose components are given by that by $\alpha$, we have

\[d \alpha = (\text{curl }a) \cdot d\mathbf{x} \text{ component wise.}\]

Finally, for a 2-form $\beta$ and corresponding vector $b$, namely $b_ {i} =\epsilon_ {ijk}\beta_ {jk}$, we have

\[d\beta = (\partial _ {1}b_ {1}+\partial _ {2}b_ {2}+\partial _ {3}b_ {3})dx\wedge dy\wedge dz,\]

in Cartesian coordinates $d\beta$ is the divergence of the vector $b$.

$d^{2}=0$ in any coordinate system; in Cartesian coordinates this yields the famous $\text{curl }\text{grad }=0$ and $\text{grad }\text{curl }=0$.

It is important to realize that it is no more difficult to compute d in a curvilinear coordinate system than in a Cartesian one. For example, in spherical coordinates, for 1-form

\[\alpha =\alpha_ {r} dr + \alpha_ {\theta}d\theta+ \alpha_ {\phi}d\phi\]

and the orientation is given by $dr\wedge d\theta \wedge d\phi$, then $d\alpha$ is simply

\[d\alpha = d\alpha_ {r}\wedge dr + d \alpha_ {\theta}\wedge d\theta + d\alpha_ {\phi} \wedge d\phi\]

where

\[d\alpha_ {r}\wedge dr = (\partial _ {\theta}\alpha_ {r}) d\theta \wedge dr + (\partial _ {\phi}\alpha _ {r}) d \phi \wedge dr, \text{ etc. }\]

In general, in situations where the “curl “ of a “vector” is required, the “vector” will most naturally appear in covariant form, i.e., it will be a 1-form.

Next we give the coordinate expression for $d$. Let $\alpha \in \Omega^{p}(\mathbb{R}^{n})$ and explicitly

\[\alpha = \alpha_ {I_ {<}} dx^{I}.\]

Then

\[d\alpha = (d \alpha_ {I_ {<}}) \wedge dx^{I} = (\partial _ {j}\alpha_ {I_ {<}}) dx^{j} \wedge dx^{I}.\]

In component form we have

\[(d\alpha)_ {I} = \delta_ {I}^{jK_ {<}} \;\partial _ {j}\alpha_ {K}.\]

3. Orientation and Pseudo-forms

Let $\mathbf{e}=(\mathbf{e_ {1},\dots,e_ {n}})$ and $\mathbf{f} = (\mathbf{f_ {1},\dots,f_ {n}})$ be two sets of bases, with transition matrix $P:\mathbf{e}\mapsto \mathbf{f}$ acting from the right, according to our convention, namely $\mathbf{f}=\mathbf{e}P$. They are said to be of the same orientation iff $\det P>0$. Thus orientation can be seen as equivalences classes of bases, with only two elements. We orient a vector space by declaring one of the two classes of bases to be positive. In our 3-space it is usual to declare the right-handed bases to be positively oriented, but we could just as well have the left-handed bases as positive.

Now consider a manifold $M$, at each point $p$ there is a tangent vector space $T_ {p}M$ on which we need to define an orientation. We shall define the orientation in a continuous manner. Given a parametrization $\mathbb{R}^{n}\to M$ of $M$ in terms of $x$, we have a set of natural bases given by

\[\left\{ \partial_ {1},\dots,\partial_ {n} \right\}\]

and we can simply define the orientation by saying that the order $\partial_ {1},\partial_ {2},\dots,\partial_ {n}$ is positive. Then any even permutation of it is still even, and odd permutation would be odd, simple as that.

We shall say that a manifold $M$ is orientable if we can cover $M$ by coordinate patches having positive Jacobians in each overlap. We can then declare the given coordinate bases to be positively oriented, and we then say that we have oriented the manifold. Briefly speaking, if a manifold is orientable it is then possible to pick out, in a continuous fashion, an orientation for each tangent space $T_ {x}M$ of $M$. Conversely, if it is possible to pick out continuously an orientation in each tangent space, we can assume that the coordinate frames in each coordinate patch have the chosen orientation and Mil must be orientable. An interesting fact is that complex manifolds are always orientable.

Next we introduce so-called pseudoforms. Consider ordinary $\mathbb{R}^{3}$ with its Euclidean metric. We would like to define the “volume 3-form” Vol$^{3}$ to be the form that assigns to any triple of vectors the volume of the parallelopiped spanned by the vectors; in particular $\text{Vol}(X, Y, Z)$ should be $1$ if $X, Y$, and $Z$ are orthonormal. But if $\text{vol}$ is to be a form we must then have $\text{vol}(Y, X, Z) = -1$, and yet $Y, X$, and $Z$ are orthonormal. We have asked too much of vol. In some books they get around this by taking absolute value $\left\lvert \text{Vol} \right\rvert$, but it this does great harm to the machinery of forms that we have labored to develop. What we could do is require that $\text{Vol}(X, Y, Z) = 1$ if the triple is an orthonormal right-handed system. This makes the volume form orientation-dependent. There is a serious drawback to this definition; what if we are dealing with a space that is not orientable? The physical space in which we live is, according to general relativity, curved and perhaps not orientable.

We compromise by defining a new type of form (called “form of odd kind” by its inventor Georges de Rham) differing from our usual forms (of “even kind”) in a way that will not seriously harm our machinery.

First note that the assignment of an orientation to a vector space $E$ is the same as the assignment of a function $o$ on bases of $E$ whose values are the two integers $\pm {1}$. If $(x)$ is a coordinate system, we should write $o(x)$ rather than $o(\partial_ {x})$.

Definition. A pseudo-p-form $\alpha$ on a vector space $E$ assigns,for each orientation $o(E)$, an exterior p-form $\alpha_ {o}$ such that if the orientation is reversed the exterior form is replaced by its negative,

\[\alpha_ {-o} = - \alpha_ {0}.\]

An example is the volume form of $\mathbb{R}^{3}$. Let $(x,y,z)$ be the Cartesian coordinate system in $\mathbb{R}^{3}$, we don’t endow it with an orientation yet, then

\[\text{Vol} := o(x,y,z) dx\wedge dy\wedge dz.\]

Note that the order in the orientation $o(x,y,z)$ is matched with the order in the differential form $dx,dy,dz$. If we adopt the familiar right-handed orientation then $\text{Vol}$ is $dx\wedge dy \wedge dz$, but if we adopt the left-handed orientation then $\text{Vol}$ would be $-dx \wedge dy \wedge dz$.

So roughly speaking

\[\text{pseudo-forms} = \text{orientation}\times \text{forms}.\]

Similarly we can define pseudovectors, pseudoscalars, and so on, pseudo always referring to a change of sign with a change of orientation.

The volume form in a Riemannian manifold is very similar to that in $\mathbb{R}^{3}$. The volume (pseudo)-n-form $\text{Vol}^{n}$ is by definition the unique n-form that 1) assigns to an orientation $o$ of the tangent space $M$ and 2) a positively oriented orthonormal bases $\mathbf{e}$ the value $+1$. Based on that, given an arbitrary coordinate system denoted by $(y)$, after some calculation we find that

\[\boxed{ \text{Vol}^{n} = o(y) \sqrt{ \left\lvert g \right\rvert }\; dy^{1} \wedge \dots dy^{n}. }\]

It is traditional to omit the orientation function $o(y)$, and we shall do so when no confusion can arise.

4. Interior Products and Vector Analysis

We know that the contraction between a contravariant index and a covariant index gives us a scalar, the two indices cancel each other. The interior product between a form and a vector is the application of that idea.

Let $v$ be a vector and $\alpha$ a p-form. Their interior product is a $(p-1)$-form $i_ {v}\alpha$ defined by

\[i_ {v}\alpha = 0 \text{ if } \alpha \text{ is a 0-form},\] \[i_ {v}\alpha = \alpha(v) \text{ if }\alpha \text{ is a }1\text{-form},\]

and

\[i_ {v}\alpha(w_ {2},\dots,w_ {p}) := \alpha(v,w_ {1},\dots,w_ {p}).\]

Sometimes we shall write $i(v)$.

Theorem. The interior product $i_ {v}: \Omega^{p} \to \Omega^{p-1}$ is an antiderivation, that is, for two forms $\alpha \in \Omega^{p}$ and $\beta \in \Omega^{q}$

\[i_ {v}(\alpha \wedge \beta) = (i_ {v}\alpha)\wedge \beta + (-1)^{p} \alpha \wedge i_ {v}\beta.\]

Pushing $i_ {v}$ through $\alpha$ gives us an extra factor $(-1)^{p}$, just like the exterior differentiation.

In components, we have

\[i_ {v}\alpha = \sum_ {I_ {<}}v^{j}\alpha_ {jI}\; dx^{I}.\]

Given a specific coordinate system, we can associate a 1-form to a vector by identifying the components. Of course this wouldn’t work under a change of coordinates, since a covariant vector (1-form) and a contravariant vector (regular vectors) have different transformation rules, but we can forget about the transformation rule for now. Writing the contraction as left-right angle brackets, i.e. for a vector $v$ and a 1-form $\alpha$

\[v(\alpha) = \alpha(v) =: \left\langle \alpha,v \right\rangle\]

where we adopted the convention that in $\left\langle -,- \right\rangle$ the vector appears on the right. The above mentioned associating to a vector a 1-form is just dualization, namely, given the 1-form $\nu$, which acts on any vector as $\left\langle \nu,- \right\rangle$, the associated vector is the unique vector $v$ that acts on any 1-forms as $\left\langle -,v \right\rangle$. We will write

\[\nu = \left\langle -,v \right\rangle .\]

For example,

\[\left\langle \nu, X \right\rangle \equiv \left\langle X,v \right\rangle,\]

where $\left\langle -,- \right\rangle$ is the inner product between two vectors on the right hand side. We use the same notation for all kind of “inner products”, may that be inner products between vectors, or between a vector and a 1-form.

In 3D Euclidean space, a vector can also be associated to a 2-form given by $v_ {1}dx^{2}\wedge dx^{3}+\dots$, so how does this work in a general coordinate system? For example, given a curved 3D Riemannian manifold, is there a universal way to associate a vector to a 2-form? To to that, we need the interior product we just introduced.

We claim that

\[\mathbf{v} \longleftrightarrow \tilde{\nu} := i_ {\mathbf{v}}\text{Vol}^{3}.\]

In curvilinear coordinates $(u)$, omitting the orientation $o(u)$, we have

\[i_ {v}\sqrt{ \left\lvert g \right\rvert }\; du^{1}\wedge du^{2}\wedge du^{3}=\sqrt{ \left\lvert g \right\rvert }(v^{1} du^{2}\wedge du^{3}+u^{2}du^{3}\wedge du^{1}+v^{3} du^{2}\wedge du^{1}).\]

Conversely, if $\beta$ is a 2-form, we can associate to it a vector $b$ with components

\[\mathbf{b} := \frac{1}{\sqrt{ \left\lvert g \right\rvert }}(b_ {23},b_ {31},b_ {12}) .\]

The same procedure will work with in any manifold $M^{n}$ having some distinguished volume form (not necessarily coming from a Riemannian metric). To the vector $v$ we may associate the (pseudo-)$(n-1)$-form $i_ {v}\text{Vol}^{n}$. It is pseudo since the volume is pseudo and $v$ is not.

Let $v,w$ be two vectors and $\nu,\omega$ be the corresponding 1-forms. What about the cross product of the vectors? We know that $v\times w$ is antisymmetric and so is $\nu \wedge\omega$ , thus we would guess that $\nu \wedge\omega$ is the 2-form associated to the so-called vector $v\times w$. It turns out to be correct. Recall that the 2-form associated to vector $v\times w$ is $i_ {v\times w}\text{Vol}^{3}$ , we have

\[\nu \wedge \omega=i_ {v\times w}\text{Vol}^{3}.\]

Although not usually mentioned in elementary books, the vector product is defined in $\mathbb{R}^{3}$ as follows: $v\times w$ is the unique pseudovector such that

\[\text{Vol}(v,w,c) = \left\langle v\times w, c \right\rangle \text{ for each vector }c.\]

Vector algebra in $\mathbb{R}^{3}$ is easily handled by use of interior and exterior products; the only question is, should one associate to a vector $\mathbf{B}$ its 1-form $\beta : = \left\langle -,\mathbf{B} \right\rangle$ or its 2-form $\beta^{(2)} = i_ {\mathbf{B}} \text{vol}^{3}$?

We already know that $df$ is associated to $\nabla f$.

We define the curl of $A$ by using $\mathbf{A}\leftrightarrow \alpha \in\Omega^{1}$ and then

\[\text{curl } \mathbf{A} = d\alpha,\quad d\alpha = i_ {\text{curl A}}\text{Vol}^{3}.\]

We define $\text{div }\mathbf{B}$ by using $\mathbf{B} \leftrightarrow \beta^{(2)}$ and

\[d \beta^{(2)} := \text{div }\mathbf{B} \;\text{Vol}^{3}.\]

Given in coordinates, we have

\[\text{div }\mathbf{B} = \frac{1}{\sqrt{ \left\lvert g \right\rvert }} \frac{ \partial }{ \partial u^{i} } \left( \sqrt{ \left\lvert g \right\rvert }b^{i} \right).\]

We define the Laplacian of a function f by

\[\nabla^{2}f = \Delta f := \text{div }\text{grad }f= \frac{1}{\sqrt{ \left\lvert g \right\rvert }} \frac{ \partial }{ \partial u^{i} } \left( \sqrt{ \left\lvert g \right\rvert } \partial ^{i}f \right),\quad \partial ^{i} = g^{ij} \partial _ {j}.\]

To continue with vector identities it is useful to associate a pseudo-3-form to each scalar $f$, namely

\[f \longleftrightarrow f(x)\,\text{Vol}^{n}.\]

Then we can use forms to calculate $\text{div }\mathbf{A}\times \mathbf{B}$. Recall that in $\mathbb{R}^{3}$ we have $(i_ {A}\text{Vol}^{3})\wedge \beta = \left\langle \mathbf{A,B} \right\rangle\text{Vol}^{3}$, we have

\[\begin{align*} \text{div }(\mathbf{A\times B }) \longleftrightarrow d(\alpha \wedge \beta) &= (d\alpha)\wedge \beta-\alpha \wedge d\beta \\ &= i_ {\text{curl} A}\text{Vol} \wedge \beta - \alpha \wedge i_ {\text{curl B}}\text{Vol} \\ &= \left\langle \text{curl }\mathbf{A},\mathbf{B} \right\rangle \text{Vol} - \left\langle \mathbf{A},\text{curl }\mathbf{B} \right\rangle\text{Vol} \\ \longleftrightarrow &\left\langle \text{curl }\mathbf{A},\mathbf{B} \right\rangle - \left\langle \mathbf{A},\text{curl }\mathbf{B} \right\rangle. \end{align*}\]

5. Dictionary

Let

\[\begin{align*} \text{Vol}^{3} &= dx\wedge dy\wedge dz = \text{ volume form} \\ 0\text{-form } F &= \text{functions f} \\ 1\text{-form } \alpha,\gamma &= \text{covariant expression for vectors } \mathbf{A},\mathbf{C} \\ 2\text{-form }\beta &=\text{vector }B \text{ through }i_ {B}\text{Vol}. \end{align*}\]

Recall that we use the interior product and the volume form to convert between vectors and 2-forms. Then we have the following rough, symbolic identifications.

\[\begin{align*} \mathbf{A}\times \mathbf{C}&\longleftrightarrow \alpha \wedge \gamma \text{ through }i_ {A\times C}\text{Vol} \\ \mathbf{A}\cdot \mathbf{B} &\longleftrightarrow \alpha \wedge \beta^{(2)} = A\cdot B\text{Vol} \\ \mathbf{A}\cdot \mathbf{C} &\longleftrightarrow i_ {\mathbf{C}}A \\ \text{grad }f &\longleftrightarrow df \\ \text{curl }\mathbf{A} &\longleftrightarrow d\alpha = i_ {\text{curl }A}\text{Vol} \\ \text{div }\mathbf{B}&\longleftrightarrow d\beta^{(2)}=\text{div }\mathbf{B}\text{Vol} \\ \nabla^{2}f &\longleftrightarrow d\,i_ {\text{grad }f}\text{Vol} = \nabla^{2}f \,\text{Vol} \end{align*}\]

6. Charge and Current in Classical Electromagnetism

Electromagnetic field is best described in the language of differential forms. However, electric field and magnetic field are given by forms of different ranks.

When it comes to counting the rank of forms, there are usually two simple ways to do it,

  1. differential forms are what we integrate on a manifold, the dimension of the form must be the same as the dimension of the manifold on which it is integrated. For example, a form which we integrate on $\mathbb{R}^{3}$ is a 3-form.
  2. Forms also are maps that turns vectors into numbers. We can count the rank of a form by counting how may vectors it acts on. For example, if a form maps $V\otimes V$ to real numbers, then it must be a 2-form.

In our case, it is easiest to adopt the first method. For example, given a test charge of charge $q$, if it moves along a curve $C$ in electric field $E$, the work done to the charge is given by

\[W = q \int_{C} E,\]

thus $E$ must be a 1-form. Similarly, the magnetic flux is given by the integral of magnetic field on some surface $S$

\[\Phi=\int_{S} \, B\]

thus $B$ must be a 2-form. Next we will dive into details.

We accept as a primitive notion the charge Q on a particle and we assume that there is a 3-form defined on the 3D space whose integral over a region gives us the total charge. Write

\[Q = \int \sigma,\quad \sigma = \rho(x) \text{Vol},\]

where $\rho$ is a charge density 0-form.

As for the spatial current 2-form, we have

\[\mathcal{j} = \sqrt{ \left\lvert g \right\rvert } \rho (v^{1}du^{2}\wedge du^{3}+\dots).\]

The electric and magnetic fields are defined operationally. In the following we shall use the euclidean metric and Cartesian coordinates of $\mathbb{R}^{3}$ (where there is no blatant distinction between covariant and contravariant vectors) and then we shall put the results in a form independent of the metric.

The electromagnetic force on a point mass of charge $q$ moving with velocity $v$ is given and measured by the Lorentz force law

\[\mathbf{F} = q(\mathbf{E}+\mathbf{v}\times \mathbf{B}).\]

You could say the fields $\mathbf{B,E}$ are defined by the Lorentz force law!

The 2-form $\mathcal{B}$ for magnetic field is defined from the pseudo vector $\mathbf{B}$ via

\[\mathcal{B} := i_ {B}\text{Vol}.\]

We then have for the Lorentz force covector \(f^{1} = q(\mathcal{E}+i_ {v}\mathcal{B})\)

where $\mathcal{E}$ is the electric field 1-form and $v$ the velocity of the test charge. This equation is independent of any metric or orientation.

In any coordinates we have

\[\mathcal{E} = E_ {1} dx^{1}+\dots,\quad \mathcal{B} = B_ {23} dx^{2}\wedge dx^{3}+\dots\]

If we introduce a metric, then we may consider the associated vector field $\mathbf{E}$ and the pseudovector $\mathbf{B}$. The pseudovector B has components $B_ {}1 = B_ {23} / \sqrt{ \left\lvert g \right\rvert }$, and so on.

Given a Riemannian metric, we can use it to raise or lower the indices, and we can define two new types of forms

\[\star \mathcal{E} := i_ {E}\text{Vol} = \sqrt{ \left\lvert g \right\rvert }\;(E^{1}dx^{2}\wedge dx^{3}+E^{2}dx^{3}\wedge dx^{1}+E^{3}dx^{1}\wedge dx^{2})\]

and

\[\star \mathcal{B} := \left\langle -,\mathbf{B} \right\rangle =B_ {1}dx^{1}+B_ {2}dx^{2}+B_ {3}dx^{3}.\]

For now, the star symbol can be simply regarded as a part of the name, later we will see that the star is a well-defined operation on forms.

Gauss’s Law. If $U$ is any compact 3D region,

\[\int _ {U} \, \star \mathcal{E} = 4\pi Q(U), \quad Q(u) \text{ is the total charge in U}.\]

We again conclude, when $\star \mathcal{E}$ is continuously differentiable, that

\[d \star \mathcal{E} = 4\pi \sigma \longleftrightarrow \text{div} \mathbf{E} = 4\pi\rho.\]

Faraday’s law in terms of forms reads

\[d\mathcal{E} = - \frac{d}{d t} \int_ {U} \, \mathcal{B},\quad U \text{ constant.}\]

Ampere-Maxwell law. If $M$ is a compact 2-sided surface with prescribed normal, then

\[\int _ {\partial M} \, \star \mathcal{B} = \int _ {M} \, 4\pi \mathbf{j} +\frac{ \partial \star \mathcal{E} }{ \partial t },\]

thus

\[d\star \mathcal{B} = 4\pi \mathbf{j} +\frac{ \partial \star \mathcal{E} }{ \partial t }.\]

(assuming $\mathcal{B}$ continuously differentiable) with vector expression $\text{curl }B=4\pi \mathbf{J}+\partial\mathbf{E} / \partial t$.

Note that the integral versions of Maxwell’s equations are more general than the partial differential equation versions since spatial derivatives do not appear in the equations. In particular, their continuity is of no concern !

However, that is what we have in 3D space. In 3D space the bases of forms are $dx,dy$ and $dz$, turning to relativistic 4D spacetime, we have an extra bases, i.e., $dt$. The expressions for electromagnetic field also needs some modifications.

Let’s start with the Lorentz force law. In non-relativistic case, the 3-dimensional Lorentz force is given by

\[\mathbf{f}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})\]

where the speed of light is set to $1$. We need to complete this spatial force vector to Minkowski 4-vector force. Lucky for us there is a trick to do that. The 3-force can be written as

\[\mathbf{f} = \frac{d\mathbf{p}}{dt} = \frac{d m_{0}\mathbf{u}}{dt}\]

where $\mathbf{u}$ is the three component of the four velocity $u$, $\mathbf{p}$ is the three momentum. As discussed before, a natural completion of the three momentum is the four momentum so in order to complete $\mathbf{f}$ we should replace $\mathbf{p}$ by the four momentum $P$, and we need to substitute $t$ for the Lorentz scalar $\tau$. Thus the four-force become

\[f^{\mu} = \frac{d}{d\tau}(m,\mathbf{p})=\left( \frac{dm}{dt},\gamma \frac{d\mathbf{p}}{dt} \right) = (f^{0},\gamma \mathbf{f}).\]

To obtain $f^{0}$, notice that the inner product of $P$ and itself is a constant,

\[P^{2} = \left\langle P,P \right\rangle =\text{const} = m_ {0}^{2}\]

and as a consequence, the four momentum $P$ is perpendicular to $dP / d\tau$. Since $P=m_{0} u$ and $m_{0}$ is a Lorentz scalar, $\mathbf{P}$ is parallel to $\mathbf{u}$, we have

\[\left\langle f,u \right\rangle =0= -\gamma f^{0} +\gamma^{2} \mathbf{f}\cdot \mathbf{v},\]

from where we can work out $f^{0}$,

\[f^{0} = \gamma \mathbf{f} \cdot \mathbf{v}.\]

Putting everything together, we have complete the three-force to the Lorentz contravariant four-force,

\[\boxed{ f = \gamma (\mathbf{f}\cdot \mathbf{v}, \mathbf{f}). }\]

Substitute the expression for $\mathbf{f}$ as the Lorentz force, in terms of classical vector form of electro-magnetic field, we have

\[f = \gamma q (\mathbf{E}\cdot \mathbf{v}, \mathbf{E}+\mathbf{v}\times \mathbf{B}).\]

On the other hand, the force itself is a 1-form since we can integrate it along some path $C$ to evaluate the work done,

\[W = \int_{C} \, f.\]

Coming back to the case of Lorentz force, to simplify the notations we write $E = E_{i}dx^{i}$ and $B= B_{i<j}\,dx^{i}\wedge dx^{j}$. The 1-form expression of the Lorentz force reads

\[f = -\gamma q\, i_{\mathbf{v}}E dt + \gamma q \, (E_{i}dx^{i}-i_{\mathbf{v}}B)\]

where $i_{\mathbf{v}}$ is the inner product with vector $\mathbf{v}$.

The Lorentz 1-form $f$ can be written as

\[f = -q i_{\mathbf{v}} F\]

where

\[\boxed{ F=E \wedge dt+B }\]

is the electromagnetic field strength 2-form. It was first introduced by Minkowski in 1907.

7. Maxwell’s Equations

In Minkowski space we define a 4-dimensional exterior differentiation $d$ in terms of the 3-dimensional exterior differentiation $\mathbf{d} = \sum_ {i}\partial_ {i}d^{i},i=x,y,z$ as

\[d = \mathbf{d}+dt\wedge \partial _{t}\]

then half of the Maxwell equations are given by

\[dF = 0\]

Following the Poincare’s lemma, if the spacetime manifold is simply connected,

\[F = dA,\quad A = -\phi dt+A_ {i}dx^{i}\]

for some 1-form $A$. Then we can further write down $E,B$ in terms of $A$, we would recover the familiar expressions.

To account for the other half of the Maxwell equation, we need to introduce the charged flux. Consider a charged fluid with local velocity $\mathbf{v}$, the current vector then is $\mathbf{j}=\rho \mathbf{v}$, $\rho$ is the density measured in the inertial system $x$. Let $\rho_{0}$ be the rest charge density, that is, the density measured by someone moving instantaneously with the fluid, then

\[\rho=\rho_{0} \gamma,\]

this is due to the contraction of the space. Thus

\[\mathbf{j} = \rho_{0}\gamma \mathbf{v}.\]

Since $\rho_{0}$ is independent of the observer by definition, we can construct the so-called current 4-vector

\[J:= \rho_{0} u = (\rho,\rho \mathbf{v}) := (\rho,\mathbf{J}).\]

We may then construct the associated current 3-form \(\begin{align*} \mathcal{J}&=i_{\mathbf{J}}\text{Vol}^{4} = \sigma^{(3)} - j^{(2)}\wedge dt \\ \sigma^{(3)}&= \rho dx\wedge dy\wedge dz, \\ j^{(2)}&= j_ {1}dy\wedge dz+j_ {2}dz\wedge dx+j_ {3}dx\wedge dy. \end{align*}\)

where $\text{Vol}^{4}$ is the 4-volume form, $\text{Vol}=dt\wedge dx \wedge dy \wedge dz$. Then the other half of the Maxwell’s equations can be written as

\[d\star F = 4\pi \mathcal{J}.\]

8. Hodge Operator

On a (pseudo-)Riemannian manifold $M$ we will first introduce a pointwise scalar product between $p$-forms, denoted by pointy brackets, then use it to define a global scalar product, denoted by parenthesis.

The local scalar product of two $p$-forms is defined to be

\[\left\langle \alpha,\beta \right\rangle := \alpha_ {I_ {<}} \beta^{I}\]

where again $I={i_ {1},\dots,i_ {p}}$ is the generalized index and $I_ {<}$ denotes that in the implied sum we have $i_ {1}<i_ {2}<\dots<i_ {p}$. We can denote the orthonormal bases of 1-forms by

\[\sigma^{1}, \dots,\sigma^{p}.\]

The global or Hilbert space scalar product is defined by

\[\left( \alpha,\beta \right) := \int _ {M} \, \left\langle \alpha,\beta \right\rangle \text{Vol} ^{n}.\]

whenever this makes sense. This will be the case when $M$ is compact, or, more generally, when the integrand has compact support.

Note that the space of smooth $p$-forms on a Riemannian $M$ that satisfy $\left( \alpha,\alpha \right)<\infty$ form only a pre-Hilbert space since it is not complete; a limit of square integrable smooth forms need not even be continuous. To get a Hilbert space we must “complete” this space. We shall not be concerned here with such matters, and we shall continue to use the inaccurate description “Hilbert space.” We shall even go a step further and use this denomination even in the pseudo-Riemannian case, where $\left( -,- \right)$ is not even positive definite.

If $\alpha \in\Omega^{1}(M)$, we may look at its contravariant version $A$ and define a $(n-1)$-form $i_ {A}\text{Vol}^{n}$. We can generalize this procedure, associate to each $p$-form a $(n-p)$-form $\star \alpha$, the Hodge-dual of $\alpha$, as follows. If

\[\alpha = \alpha_ {I_ {<}} dx^{I}\]

then

\[\boxed { \star \alpha = (\star\alpha) _ {J_ {<}} dx^{J}, \quad (\star\alpha)_ {J} = \sqrt{ \left\lvert g \right\rvert } \alpha^{K} \epsilon_ {K_ {<}J_ {<}.} }\]

and where the upper indices $K$ in $\alpha^{K}$ indicate that all of the covariant indices in $\alpha$ have been raised by the metric tensor. Note again that here $\epsilon_ {KJ}$ is not the Levi-Civita tensor but the Levi-Civita symbol, it simplifies the calculation but the price to pay is that $\alpha^{K}\epsilon_ {K_ {<}J_ {<}}$ can no longer be carelessly rewritten as $\alpha_ {K}\epsilon^{K_ {<}}_ {\;\;\;\; J_ {<}}$.

For an important special case, the 0-form that is the constant function $f=1$ has

\[\star 1 = \sqrt{ \left\lvert g \right\rvert } \epsilon_ {12..d} dx^{1}\wedge dx^{2}\dots \wedge dx^{d} = \text{Vol}^{d}.\]

We have

\[\alpha \wedge \star \beta=(\alpha \wedge \star \beta)_ {12\dots n} dx^{1}\wedge \dots \wedge dx^{n},\]

and then

\[(\alpha \wedge \star \beta)_ {12\dots n} = \alpha_ {A_ {<}}\beta_ {A} \sqrt{ \left\lvert g \right\rvert }dx^{1}\wedge \dots \wedge dx^{n}= \left\langle \alpha ,\beta \right\rangle\text{Vol}^{n} .\]

We have claimed that $\star$ generalized the interior product $\alpha\to i_ {A}\text{Vol}^{n}$. To see this, notice that

\[i_ {A} \text{Vol}^{n} = i_ {A}\sqrt{ \left\lvert g \right\rvert }\epsilon_ {I_ {<}}dx^{I} = \sqrt{ \left\lvert g \right\rvert } A^{i}\epsilon_ {iJ_ {<}}dx^{J} = \star \alpha,\quad \alpha_ {i}\equiv A^{i}.\]

Let $\mathbf{e} = (\mathbf{e}_ {1},\mathbf{e}_ {2},\dots,\mathbf{e}_ {n})$ be an orthonormal frame of vectors. Then $\sigma^{i}$ corresponding to $\mathbf{e}_ {i}$ are also orthonormal and

\[\sigma^{}\wedge \dots \wedge \sigma^{n} = \pm \text{Vol}^{n},\]

and

\[\star\sigma^{I} = \pm\sigma^{J}, \quad J\text{ is the complement of }I.\]

For example, look at the electromagnetic field in a perhaps curved space-time manifold $M^{4}$. Using the space-time metric, we have

\[\star F = \star(E\wedge dt)+\star B.\]

We know $E = E_ {i} dx^{1}$ so $\star (E\wedge dt)= \star (E_ {i}dx^{i}\wedge dt)= E_ {i}\star(dx^{i}\wedge dt)$ but what is, for example, $\star (dx^{1}\wedge dt)$? We usually don’t need to resort to the original definition which can be pretty cumbersome in calculations. Instead we notice that the Hodge dual is closed related to the inner product and volume form. In Minkowski space we have $\left\lvert g \right\rvert=1$ thus we can neglect it. Say, we want to calculate $\star(dx^{2}\wedge dx^{3})$, it has the property that

\[(dx^{2}\wedge dx^{3})\wedge \star (dx^{2}\wedge dx^{3}) = \left\langle dx^{2}\wedge dx^{3},dx^{2}\wedge dx^{3} \right\rangle\; \text{Vol}^{4},\]

which can be calculated using a relation which says that, given 1-forms $\alpha_ {i},\beta_ {j}$ we have

\[\left\langle \alpha_ {i_ {1}}\wedge \dots \wedge \alpha_ {i_ {d}}, \beta_ {j_ {1}} \wedge \dots \wedge \beta_ {j_ {d}}\right\rangle = \det \left\langle \alpha_ {i_ {m}},\beta_ {j_ {n}} \right\rangle\]

where the right hand side is a matrix with entry $(m,n)$ given by the inner product. Equipped with above relation and the fact that $dx$ are orthonormal, with convention $g=\text{diag} (1,-1,-1,-1)$ we have

\[\begin{align*} (dx^{2}\wedge dx^{3})\wedge \star (dx^{2}\wedge dx^{3}) &= \left\langle dx^{2}\wedge dx^{3},dx^{2}\wedge dx^{3} \right\rangle \text{Vol}^{4} = \left\langle dx^{2},dx^{2} \right\rangle \times \left\langle dx^{3},dx^{3} \right\rangle \text{Vol}^{4} \\ &=\text{Vol}^{4} = dt\wedge dx^{1}\wedge dx^{2}\wedge dx^{3} \end{align*}\]

thus we can read-off that

\[\star(dx^{2}\wedge dx^{3}) = dt\wedge dx^{1}.\]

likewise we have

\[\star(dt\wedge dx^{1}) = -dx^{2}\wedge dx^{3}.\]

To summarize, we have

\[\boxed { \begin{align*} \star(dx^{i}\wedge dx^{j} ) &= -\epsilon^{ijk} \, dx^{k}\wedge dt, \\ \star(dt \wedge dx^{i} ) &= -\frac{1}{2} \epsilon^{ijk}\, dx^{j}\wedge dx^{k}. \end{align*} }\]

It can be shown that, given $\alpha \in\Omega^{p}(M^{n})$,

\[\star^{2} = \text{det}(g) (-1)^{p(n-p)}\]

It is sufficient to verify these for terms of the form $\sigma^{I}$ and to assume these are orthonormal. Remember that $\star\sigma^{I_ {<}} = \pm \sigma^{J_ {<}}$ where $J$ is the compliment of $I$, and the sign dependent on the nature of the metric.

8.1. The Codifferential Operator $d^{\dagger}=d^{\star}=\delta$

The codifferential operator $d^{\dagger}$ is the dual of exterior differential $d$ in the sense that, in the global inner product,

\[(d \alpha^{p-1},\beta^{p} ) \equiv (\alpha^{p-1},d^{\dagger}\beta^{p-1}).\]

where the superscript denotes the dimension of the form. Thus $d^{\dagger}$ must send a $p$-form to a $(p-1)$-form.

Not recall that

\[(d \alpha^{p-1},\beta^{p} ) = \int_ {M} \, (d\alpha^{p-1})\wedge \star \beta^{p}=(-1)^{p-1}\int_ {M} \, \alpha^{p-1}\wedge d\star \beta^{p}\]

and

\[(\alpha^{p-1},d^{\dagger}\beta^{p-1}) = \int_ {M} \, \alpha^{p-1}\wedge \star d^{\dagger}\beta^{p}\]

then it can be shown that given $\beta \in\Omega^{p}(M)$ if we define

\[\boxed { d^{\dagger}\beta := \text{det}(g)\, (-1)^{n(p+1)+1}\star d\, \star \beta = (-1)^{p}\star^{-1}d \star \beta }\]

then we would have, given $\alpha \in\Omega^{p-1}(M)$,

\[(d\alpha,\beta) - (\alpha,d^{\dagger}\beta) = \int_ {M} \, d(\alpha \wedge \star \beta).\]

at least when $\alpha,\beta$ has compact support. If $M^{n}$ is closed then $d^{\dagger}$ is indeed the dual of $d$ in the pre-Hilbert space. If $M$ has a boundary, then the statement still holds if either of $\alpha$ and $\beta$ is zero on the boundary.

The operator $d^{^{\dagger}}$ is called the codifferential. The traditional notation is $\delta$ but we will not use it, since we want to keep $\delta$ for variation. Instead we will use $d^{\dagger}$.

A consequence of the definition for exterior derivative

\[d\alpha = (\partial _ {\mu}\alpha_ {I_ {<}})dx^{\mu}\wedge dx^{I}\]

is that, in a spacetime with a symmetric connection $\Gamma^{\alpha}_ {\;\mu \nu}=\Gamma^{\alpha}_ {\;\nu \mu}$, the partial derivative $\partial_ {\mu}$ can always be replaced by the covariant derivative $\nabla_ {\mu}$, as the readers can verify, the covariant derivatives introduce new terms concerning the connections, and these terms identically vanish due to the anti-symmetric nature of differential forms. Some calculation shows that a coordinate expression for the $(p-1)$-form $d^{\dagger}\beta$ is

\[(d^{\dagger}\beta)_ {K} = - \beta^{i}_ {\;K;j}.\]

8.2. Maxwell’s Equations in Curved Space-Time

We shall assume that the electromagnetic field is again described by an electromagnetic 2-form $F\in\Omega^{2}(M)$. In any local coordinates $(t,\mathbf{x})=(x^{0},x^{1},\dots,x^{4})$ we may decompose $F$ into part contain and doesn’t contain $dt$. The relation between $F$ and the electric 1-form $E\in\Omega^{1}(M)$ and magnetic 2-form $B\in\Omega^{2}(M)$ is given by

\[F = E\wedge dt + B.\]

Half of the Maxwell equation is given by $dF=0$. Define the charge density 3-form $\rho$ and current 2-form $j$, we have the 4D current 3-form

\[J = \rho - j \wedge dt\]

which is related to the 4-vector form of the current $J^{\mu}$ by

\[J = i_ {J^{\mu}}\text{Vol}^{4}.\]

where $\mu$ is just part of the name, not to be contracted with anything. Note that The current 4-vector depends on the metric! It is for this reason that the current 3-form is more basic than the 4-vector current.

The other half of the Maxwell equation can be written as

\[d\star F= 4\pi J^{(3)},\quad J^{(3)} \text{ is the current 3-form.}\]