Hodge star

Section 4.8 Hodge star

In this section we introduce one last operator on differential forms, called the “Hodge star”. While we will not really use it in this course, it is an integral part of the theory of differential forms, so it is worth being introduced to it briefly. The Hodge star can be used to recover the Laplacian operator in the language of vector calculus.

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Objectives

You should be able to:

Define the Hodge star operator in $R^{n} .$
Determine the Hodge star of zero-, one-, two-, and three-forms in $R^{3} .$
Relate to the Laplacian operator in vector calculus.

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Subsection 4.8.1 The Hodge star

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The Hodge star is an operator that provides some sort of duality between

k

-forms and

(n - k)

-forms in

R^{n} .

It is easiest to define it in terms of the basic

k

-forms from Definition 4.1.5, and then extend to general differential forms by applying it to each summand.

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Definition 4.8.1. The Hodge star dual of a $k$ -form in $R^{n}$ .

Let $ω = d x_{i_{1}} \land \dots \land d x_{i_{k}}$ be a basic $k$ -form on $R^{n} .$ Then the Hodge star dual of $ω$ , which is denoted by $⋆ ω,$ ¹ is the unique basic $(n - k)$ -form with the property:

ω \land ⋆ ω = d x_{1} \land \dots \land d x_{n} .

This standard notation should not be confused with the pullback of a differential form; those are very different things.

To define the Hodge star dual of a general $k$ -form on $U \subseteq R^{n},$ we apply the Hodge star to each summand. More precisely, if

η = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}

is a $k$ -form on $U,$ then its Hodge star dual $* η$ is the $(n - k)$ -form given by

⋆ η = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} ⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}}) .

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To make sense of this definition, let us look at the Hodge star action on the basic

k

-forms for low-dimensional space.

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Example 4.8.2. The action of the Hodge star in $R$ .

There are only two basic $k$ -forms in $R,$ namely the zero-form $1$ and the one-form $d x .$ From the definition, we want $1 \land ⋆ 1 = d x$ and $d x \land ⋆ d x = d x,$ from which we conclude that:

⋆ 1 = d x, ⋆ d x = 1.

The Hodge star thus provides a duality between zero-forms and one-forms in $R .$

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Example 4.8.3. The action of the Hodge star in $R^{2}$ .

It becomes a little more interesting in $R^{2} .$ The basic forms are the zero-form $1,$ the one-forms $d x$ and $d y,$ and the two-form $d x \land d y .$ From the definition, we want $1 \land ⋆ 1 = d x \land d y,$ $d x \land ⋆ d x = d x \land d y,$ $d y \land ⋆ d y = d x \land d y,$ and $(d x \land d y) \land ⋆ (d x \land d y) = d x \land d y .$ We conclude that

\begin{matrix} ⋆ 1 = d x \land d y, ⋆ (d x \land d y) = 1, \\ ⋆ d x = d y, ⋆ d y = - d x . \end{matrix}

It thus provides a duality between zero-forms and two-forms in $R^{2},$ and a “self-duality” for one-forms. Note that the sign is important here for the action on the basic one-forms.

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Example 4.8.4. The action of the Hodge star in $R^{3}$ .

Things become even more interesting in $R^{3} .$ The basic forms are the zero-form $1,$ the one-forms $d x, d y, d z,$ the two-forms $d y \land d z, d z \land d x, d x \land d y,$ and the three-form $d x \land d y \land d z .$ From the definition, we get that:

\begin{matrix} ⋆ 1 = d x \land d y \land d z, ⋆ (d x \land d y \land d z) = 1, \\ ⋆ d x = d y \land d z, ⋆ d y = d z \land d x, ⋆ d z = d x \land d y, \\ ⋆ (d y \land d z) = d x, ⋆ (d z \land d x) = d y, ⋆ (d x \land d y) = d z . \end{matrix}

Thus, in $R^{3},$ it provides a duality between zero-forms and three-forms, and between one-forms and two-forms.

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Example 4.8.5. An example of the Hodge star action in $R^{3}$ .

Consider the two-form $ω = x y z d y \land d z + e^{x} d x \land d y .$ Its Hodge star dual is the one-form:

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

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The action of the Hodge star in

R^{3}

naturally justifies our dictionary to translate between

k

-forms in

R^{3}

and vector calculus objects in Table 4.1.11. Indeed, let

ω = f d x + g d y + h d z

be a one-form on

R^{3} .

Then its Hodge dual is the two-form

⋆ ω = f d y \land d z + g d z \land d x + h d x \land d y .

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This is why we used this particular choice for the basic two-forms in Table 4.1.11; it's because this is what one gets through Hodge duality, which identifies one-forms and two-forms in

R^{3} .

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In fact, what this means is that we really only needed the first two lines in Table 4.1.11. Indeed, we can always transform a two-form into a one-form by taking its Hodge dual, and a three-form into a zero-form. So, in the end, all that we need to establish a dictionary between differential forms and vector calculus objects is to say that zero-forms are functions, and one-forms correspond to vector fields.

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For instance, if

F

is the vector field associated to a one-form

ω,

we could have defined the curl

\nabla \times F

to be the vector field associated to the one-form

⋆ d ω .

That is,

ω \leftrightarrow F ⋆ d ω \leftrightarrow \nabla \times F .

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Similarly, we could have defined the divergence

\nabla \cdot F

to be the function given by

⋆ d ⋆ ω .

That is,

ω \leftrightarrow F ⋆ d ⋆ ω \leftrightarrow \nabla \cdot F .

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We could translate all vector calculus identities in Section 4.4 using the Hodge star, but in the end, as far as we are concerned in this course, this is just a fancier way of saying the same thing. :-)

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Example 4.8.6. Maxwell's equations using differential forms (optional).

Recall from Example 4.4.7 the statement of Maxwell's equations, which form the foundations of electromagnetism. They can be written in terms of the electric vector field $E$ and the magnetic vector field $B$ on $R^{3}$ as follows (I am now using units with $c = 1$ as is standard in modern physics, and I have rescaled the electric charge $ρ$ and the electric current density $J$ to absorb the factor of $4 π$ ):

\begin{aligned} \nabla \cdot E = & ρ, \\ \nabla \cdot B = & 0, \\ \nabla \times E + \frac{\partial B}{\partial t} = & 0, \\ \nabla \times B - \frac{\partial E}{\partial t} = & J . \end{aligned}

It turns out that there is a very nice way of rewriting Maxwell's equations using differential forms, which makes them manifestly relativistic (i.e. consistent with special relativity). Moreover, this reformulation works in any number of dimensions! It defines the natural generalization of Maxwell's equations to higher-dimensional spacetimes, which is useful in physics theories like string theory.

To write Maxwell's equations in this form, we need to consider them as living on spacetime, i.e. $R^{4},$ with coordinates $(t, x, y, z) .$ We first construct a two-form $F$ on $R^{4}$ that combines the electric field $E$ and the magnetic field $B$ as follows:

F = B_{x} d y \land d z + B_{y} d z \land d x + B_{z} d x \land d y + E_{x} d x \land d t + E_{y} d y \land d t + E_{z} d z \land d t .

We also construct a three-form which combines the electric current $J$ and the electric charge $ρ$ as:

J = ρ d x \land d y \land d z - j_{x} d t \land d y \land d z - j_{y} d t \land d z \land d x - j_{z} d t \land d x \land d y .

Using these definitions, and the definition of the Hodge star on $R^{4},$ ² we can rewrite Maxwell's equations neatly as the following two equations (you can check this!):

\begin{aligned} d F = & 0, \\ d ⋆ F = & J . \end{aligned}

Isn't that neat? The first equation is simply saying that the two-form $F$ is closed. The second equation is the “source equation”; if there is no source (i.e. $J = 0$ ), it is simply saying that the two-form $⋆ F$ is also closed. Moreover, not only is this formulation nice and clean, but it is manifestly Lorentz invariant (as it is formulated in four-dimensional space time).

To be precise, we need to modify the definition of the Hodge star operator a little bit here. The reason is that spacetime is

R^{4},

but with a different definition of length as for standard Euclidean space. It is called Minkowski spacetime; the definition of length is given by a metric, and in this case the metric is Lorentzian, which is a bit different from the standard Euclidean metric (it has an extra sign). The reason is that the fourth dimension of time behaves somewhat differently from the three space dimensions, which is reflected in this choice of Lorentzian metric. In the end, the correct definition of the Hodge star operator on Minkowski spacetime

R^{4}

is the operator

⋆

that acts on basic

k

-form as

ω \land ⋆ ω = \pm d t \land d x \land d y \land d z,

with a plus sign whenever

ω

does not contain

d t,

and a minus sign whenever

ω

contains

d t .

Furthermore, this formulation of Maxwell's equation naturally generalizes to any number of dimensions. In $R^{n},$ $F$ remains a two-form, but $J$ becomes an $(n - 1)$ -form. Then $⋆ F$ is an $(n - 2)$ -form, and the two equations make sense in $R^{n} .$ As mentioned above, this higher-dimensional generalization is useful in modern physical theories such as string theory. This is an example of the power of the formalism of differential forms!

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Subsection 4.8.2 The Hodge star and the Laplacian

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We can combine the Hodge star with the exterior derivative to define a new operation on differential forms, called the “Laplace-Beltrami operator”.

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Definition 4.8.7. The codifferential and the Laplace-Beltrami operator.

Let $ω$ be a $k$ -form on $U \subseteq R^{n} .$ We define the codifferential, which is denoted by $δ,$ as the operator acting on $ω$ as:

δ ω = (- 1)^{k} ⋆ d ⋆ ω .

We define the Laplace-Beltrami operator, denoted by $Δ,$ as the operator acting on $ω$ as:

Δ ω = - (d δ + δ d) ω,

where $d$ is the exterior derivative.

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This looks very fancy, but it is just the natural generalization of the Laplacian of a function to differential forms, as we now see.

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Lemma 4.8.8. The Laplace-Beltrami operator and the Laplacian of a function.

Let $f$ be a zero-form (a function) on $U \subseteq R^{3} .$ Then

Δ f = \nabla^{2} f = \nabla \cdot \nabla f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}},

where $\nabla^{2} f$ is called the Laplacian of $f$ .

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Proof.

First, we notice that if $f$ is a function,

δ f = ⋆ d ⋆ f = 0,

since $⋆ f$ is a three-form, and hence its exterior derivative vanishes in $R^{3} .$ Thus

Δ f = - δ d f = ⋆ d ⋆ d f .

We now translate into the language of vector calculus. The vector field associated to the one-form $d f$ is the gradient $\nabla f .$ Its Hodge dual is a two-form, and taking its exterior derivative means that we take the divergence of this vector field. So the result is that

Δ f = \nabla \cdot \nabla f .

Expanding in coordinates $(x, y, z),$ we get:

Δ f = \nabla \cdot \nabla f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}},

which we write as the Laplacian $\nabla^{2} f$ of the function $f .$

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The Laplace-Beltrami operator applied to a one-form in

R^{3}

gives rise to another operation in vector calculus, called the “Laplacian of a vector field”.

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Lemma 4.8.9. The Laplace-Beltrami operator and the Laplacian of a vector field.

Let $ω$ be a one-form on $U \subseteq R^{3},$ and $F$ be its associated vector field. Then the vector field associated to the one-form $Δ ω$ is the Laplacian of the vector field $F$ , which is denoted by $\nabla^{2} F .$ It acts on the vector field as

\nabla^{2} F = \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} + \frac{\partial^{2} F}{\partial z^{2}},

and satisfies the identity

\nabla^{2} F = \nabla (\nabla \cdot F) - \nabla \times (\nabla \times F) .

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Proof.

If $ω$ is a one-form, by definition

Δ ω = - d δ ω - δ d ω = d ⋆ d ⋆ ω - ⋆ d ⋆ d ω,

which is also a one-form. Let us now extract its associated vector field. Let $F$ be the vector field associated to $ω .$

We look at the first term on the right-hand-side. $⋆ ω$ is a two-form associated to $F .$ $d ⋆ ω$ then takes the divergence $\nabla \cdot F .$ $⋆ d ⋆ ω$ maps this to a zero-form, and then $d ⋆ d ⋆ ω$ take the gradient of the resuling zero-form. The result is that the vector field associated to the one-form $d ⋆ d ⋆ ω$ is

\nabla (\nabla \cdot F) .

Let us now look at the second term on the right-hand-side. $d ω$ takes the curl $\nabla \times F .$ $⋆ d ω$ then maps it to a one-form associated to the vector field $\nabla \times F .$ $d ⋆ d ω$ then takes the curl again, $\nabla \times (\nabla \times F),$ and finaly $⋆ d ⋆ d ω$ maps it back to a one-form. The result is that the vector field associated to the one-form $⋆ d ⋆ d ω$ is

\nabla \times (\nabla \times F) .

We conclude that the vector field associated to the one-form $Δ ω$ is

\nabla (\nabla \cdot F) - \nabla \times (\nabla \times F),

which we can take as the definition of the Laplacian of the vector field $F .$ To show that it takes the form

\nabla^{2} F = \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} + \frac{\partial^{2} F}{\partial z^{2}},

one only needs to do an explicit calculation in $R^{3},$ see Exercise 4.8.4.5.

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Subsection 4.8.3 Two more vector calculus identities

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To end this section, let us prove two more vector calculus identities, this time involving the Laplacian of a function.

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Lemma 4.8.10. Vector calculus identities, part 5.

Let $f$ and $g$ be smooth functions on $U \subseteq R^{3} .$ Then it satisfies the identities:

$\nabla^{2} (f g) = f \nabla^{2} g + 2 \nabla f \cdot \nabla g + g \nabla^{2} f,$
$\nabla \cdot (f \nabla g - g \nabla f) = f \nabla^{2} g - g \nabla^{2} f .$

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Proof.

These two identities follow from the graded product rule for the exterior derivative.

\begin{aligned} d F = & 0, \\ d ⋆ F = & J . \end{aligned}

We start with the first one:

\begin{aligned} \nabla^{2} (f g) = & Δ (f g) \\ = & ⋆ d ⋆ d (f g) \\ = & ⋆ d ⋆ (g d f + f d g) \\ = & ⋆ d (g ⋆ d f + f ⋆ d g) \\ = & ⋆ (d g \land ⋆ d f + g d ⋆ d f + d f \land ⋆ d g + f d ⋆ d g) \\ = & \nabla g \cdot \nabla f + g \nabla^{2} f + \nabla f \cdot \nabla g + f \nabla^{2} g \\ = & f \nabla^{2} g + 2 \nabla f \cdot \nabla g + g \nabla^{2} f . \end{aligned}

In the proof we used the fact that $\nabla^{2} f = ⋆ d ⋆ d f$ as in the proof of Lemma 4.8.8.

For the second identity, we get the following:

\begin{aligned} \nabla \cdot (f \nabla g - g \nabla f) = & ⋆ d ⋆ (f d g - g d f) \\ = & ⋆ d (f ⋆ d g - g ⋆ d f) \\ = & ⋆ (d f \land ⋆ d g + f d ⋆ d g - d g \land ⋆ d f - g d ⋆ d f) \end{aligned}

To proceed we need to use a result which we haven't proved. For any two $k$ -forms $ω$ and $η$ on $R^{3},$ there's a general result that says that

ω \land ⋆ η = ⋆ ω \land η .

Note that it is important that $ω$ and $η$ are both $k$ -forms (same $k$ ), otherwise it wouldn't apply. It it not difficult to prove this statement, but since we do not need it anywhere else, we leave the proof as an exercise (see Exercise 4.8.4.4).

Now in our previous expression we had the terms $d f \land ⋆ d g$ and $- d g \land ⋆ d f .$ Since $d f$ and $d g$ are both one-forms,

d f \land ⋆ d g = ⋆ d f \land d g,

and these two terms cancel out. Thus

\begin{aligned} \nabla \cdot (f \nabla g - g \nabla f) = & f (⋆ d ⋆ d g) - g (⋆ d ⋆ d f) \\ = & f \nabla^{2} g - g \nabla^{2} f . \end{aligned}

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Exercises 4.8.4 Exercises

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1.

Let $ω$ be the two-form

ω = x y d y \land d z + x y z d z \land d x + y d x \land d y

on $R^{3} .$ Find $⋆ ω .$

Solution.

We calculate the one-form $⋆ ω$ using the action of the Hodge star on basic two-forms in $R^{3} :$

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

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2.

For any $k$ -form $ω$ on $R^{n},$ show that

⋆ ⋆ ω = (- 1)^{k (n - k)} ω .

Solution.

We only need to prove the statement for basic $k$ -forms as by definition of the action of the Hodge star operator in Definition 4.8.1 it will then follow for all $k$ -forms.

Let $α$ be a basic $k$ -form on $R^{n} .$ By definition of the Hodge star, we know that $⋆ α$ is the unique basic $(n - k)$ -form such that

α \land ⋆ α = d x_{1} \land \dots \land d x_{n} .

Now consider the basic $(n - k)$ -form $⋆ α .$ By definition of the Hodge star, $⋆ ⋆ α$ will be the unique basic $k$ -form such that

⋆ α \land ⋆ ⋆ α = d x_{1} \land \dots \land d x_{n} .

As the right-hand-side for both equations is the same, we get

α \land ⋆ α = ⋆ α \land ⋆ ⋆ α .

Using graded commutativity of the wedge product as in Lemma 4.2.6, we can rewrite the right-hand-side as:

α \land ⋆ α = (- 1)^{k (n - k)} ⋆ ⋆ α \land ⋆ α .

Finally, given $⋆ α,$ we know that the left-hand-side uniquely defines $α$ (by definition of the Hodge star), while the right-hand-side uniquely deefines $⋆ ⋆ α$ (again by definition of the Hodge star), and therefore we must have

α = (- 1)^{k (n - k)} ⋆ ⋆ α .

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3.

Let $ω$ and $η$ be one-forms on $R^{n},$ and $F$ and $G$ be the associated vector fields. Show that

⋆ (ω \land ⋆ η) = F \cdot G .

Solution.

We can write the one-forms $ω$ and $η$ as:

\begin{aligned} ω = & \sum_{i = 1}^{n} f_{i} d x_{i}, \\ η = & \sum_{i = 1}^{n} g_{i} d x_{i}, \end{aligned}

for smooth functions $f_{i}, g_{i} : R^{n} \to R .$ Since $⋆ η$ is an $(n - 1)$ -form, $ω \land ⋆ η$ is an $n$ -form on $R^{n} .$ It then follows that the only non-vanishing terms in $ω \land ⋆ η$ are those of the form $d x_{i} \land ⋆ d x_{i},$ as all other “cross-terms”, i.e. terms of the form $d x_{i} \land ⋆ d x_{j}$ with $i \neq j,$ will necessarily vanish since $⋆ d x_{j}$ necessarily contains a $d x_{i} .$ Thus we get:

\begin{aligned} ω \land ⋆ η = & \sum_{i = 1}^{n} f_{i} g_{i} d x_{i} \land ⋆ d x_{i} \\ = & (\sum_{i = 1}^{n} f_{i} g_{i}) d x_{1} \land \dots \land d x_{n}, \end{aligned}

where we used the definition of the Hodge star for basic one-forms. Finally, since $⋆ (d x_{1} \land \dots d x_{n}) = 1,$ we get:

⋆ (ω \land ⋆ η) = \sum_{i = 1}^{n} f_{i} g_{i} ⋆ (d x_{1} \land \dots \land d x_{n}) = \sum_{i = 1}^{n} f_{i} g_{i} = F \cdot G,

where the last equality is for the associated vector fields $F = (f_{1}, \dots, f_{n})$ and $G = (g_{1}, \dots, g_{n}) .$

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4.

Let $ω$ and $η$ be $k$ -forms on $R^{n} .$ Show that

ω \land ⋆ η = (- 1)^{k (n - k)} ⋆ ω \land η .

Note that it is important that $ω$ and $η$ are both $k$ -forms (same $k$ ), otherwise this property wouldn't apply.

Solution.

The proof is similar in spirit to the solution of the previous problem. Since $ω$ and $η$ are both $k$ -forms on $R^{n},$ we can write both as linear combinations of basic $k$ -forms in $R^{n} :$

\begin{aligned} ω = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}, \\ η = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} g_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}}, \end{aligned}

for smooth functions $f_{i_{1} \dots i_{k}}, g_{i_{1} \dots i_{k}} : R^{n} \to R .$ Then

⋆ η = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} g_{i_{1} \dots i_{k}} ⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}}) .

But $⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}})$ is an $(n - k)$ -form on $R^{n},$ and thus $ω \land ⋆ η$ is an $n$ -form in $R^{n} .$ It follows that the only terms in $ω \land ⋆ η$ that will be non-vanishing are those involving the wedge product of a basic $k$ -form with its own Hodge star dual, as all other “cross-terms” will necessarily involve the wedge products of repeated $d x$ 's which trivially vanish. So we get:

ω \land ⋆ η = \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} g_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}} \land ⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}}) .

Similarly, we have:

\begin{aligned} ⋆ ω \land η = & \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} g_{i_{1} \dots i_{k}} ⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}}) \land d x_{i_{1}} \land \dots \land d x_{i_{k}} \\ = & (- 1)^{k (n - k)} \sum_{1 \leq i_{1} < \dots < i_{k} \leq n} f_{i_{1} \dots i_{k}} g_{i_{1} \dots i_{k}} d x_{i_{1}} \land \dots \land d x_{i_{k}} \land ⋆ (d x_{i_{1}} \land \dots \land d x_{i_{k}}), \end{aligned}

where we used graded commutativity of the wedge product, Lemma 4.2.6. We thus conclude that

ω \land ⋆ η = (- 1)^{k (n - k)} ⋆ ω \land η .

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5.

Let $F = (f_{1}, f_{2}, f_{3})$ be a smooth vector field in $R^{3} .$ Show that

\nabla (\nabla \cdot F) - \nabla \times (\nabla \times F) = \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} + \frac{\partial^{2} F}{\partial z^{2}} .

This is the definition of the Laplacian of the vector field $\nabla^{2} F$ as in Lemma 4.8.9.

Solution.

This is just an explicit and rather painful calculation. Let us do it step-by-step. First,

\nabla \cdot F = \frac{\partial f_{1}}{\partial x} + \frac{\partial f_{2}}{\partial y} + \frac{\partial f_{3}}{\partial z} .

Thus

\nabla (\nabla \cdot F) = (\frac{\partial^{2} f_{1}}{\partial x^{2}} + \frac{\partial^{2} f_{2}}{\partial x \partial y} + \frac{\partial^{2} f_{3}}{\partial x \partial z}, \frac{\partial^{2} f_{1}}{\partial y \partial x} + \frac{\partial^{2} f_{2}}{\partial y^{2}} + \frac{\partial^{2} f_{3}}{\partial y \partial z}, \frac{\partial^{2} f_{1}}{\partial z \partial x} + \frac{\partial^{2} f_{2}}{\partial z \partial y} + \frac{\partial^{2} f_{3}}{\partial z^{2}}) .

Next, we move on to the curl. First, we have

\nabla \times F = (\frac{\partial f_{3}}{\partial y} - \frac{\partial f_{2}}{\partial z}, \frac{\partial f_{1}}{\partial z} - \frac{\partial f_{3}}{\partial x}, \frac{\partial f_{2}}{\partial x} - \frac{\partial f_{1}}{\partial y}) .

Taking the curl again, we get:

\begin{aligned} \nabla \times (\nabla \times F) = & (\frac{\partial^{2} f_{2}}{\partial y \partial x} - \frac{\partial^{2} f_{1}}{\partial y^{2}} - \frac{\partial^{2} f_{1}}{\partial z^{2}} + \frac{\partial^{2} f_{3}}{\partial z \partial x}, \frac{\partial^{2} f_{3}}{\partial z \partial y} - \frac{\partial^{2} f_{2}}{\partial z^{2}} - \frac{\partial^{2} f_{2}}{\partial x^{2}} + \frac{\partial^{2} f_{1}}{\partial x \partial y}, \\ \frac{\partial^{2} f_{1}}{\partial x \partial z} - \frac{\partial^{2} f_{3}}{\partial x^{2}} - \frac{\partial^{2} f_{3}}{\partial y^{2}} + \frac{\partial^{2} f_{2}}{\partial y \partial z}) \end{aligned}

Putting these two calculations together, and using the fact that partial derivatives commute by Clairaut's theorem (since the vector fields are assumed to be smooth), we get:

\begin{aligned} \nabla & (\nabla \cdot F) - \nabla \times (\nabla \times F) \\ = & (\frac{\partial^{2} f_{1}}{\partial x^{2}} + \frac{\partial^{2} f_{1}}{\partial y^{2}} + \frac{\partial^{2} f_{1}}{\partial z^{2}}, \frac{\partial^{2} f_{2}}{\partial x^{2}} + \frac{\partial^{2} f_{2}}{\partial y^{2}} + \frac{\partial^{2} f_{2}}{\partial z^{2}}, \frac{\partial^{2} f_{3}}{\partial x^{2}} + \frac{\partial^{2} f_{3}}{\partial y^{2}} + \frac{\partial^{2} f_{3}}{\partial z^{2}}) \\ = & \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} + \frac{\partial^{2} F}{\partial z^{2}} . \end{aligned}

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6.

In this problem we prove the statement in Example 4.8.6 about Maxwell's equations.

Write down the action of the Hodge star operator on basic $k$ -forms in Minkowksi $R^{4}$ (see Footnote 4.8.2 ).
Using your result in part (a), show that the two equations
$\begin{aligned} d F = & 0, \\ d ⋆ F = & J, \end{aligned}$
with $F$ and $J$ defined in Example 4.8.6, reproduce Maxwell's equations.

Section 4.8 Hodge star

Objectives

Subsection 4.8.1 The Hodge star

Definition 4.8.1. The Hodge star dual of a k-form in Rn.

Example 4.8.2. The action of the Hodge star in R.

Example 4.8.3. The action of the Hodge star in R2.

Example 4.8.4. The action of the Hodge star in R3.

Example 4.8.5. An example of the Hodge star action in R3.

Example 4.8.6. Maxwell's equations using differential forms (optional).

Subsection 4.8.2 The Hodge star and the Laplacian

Definition 4.8.7. The codifferential and the Laplace-Beltrami operator.

Lemma 4.8.8. The Laplace-Beltrami operator and the Laplacian of a function.

Proof.

Lemma 4.8.9. The Laplace-Beltrami operator and the Laplacian of a vector field.

Proof.

Subsection 4.8.3 Two more vector calculus identities

Lemma 4.8.10. Vector calculus identities, part 5.

Proof.

Exercises 4.8.4 Exercises

1.

2.

3.

4.

5.

6.

Definition 4.8.1. The Hodge star dual of a $k$ -form in $R^{n}$ .

Example 4.8.2. The action of the Hodge star in $R$ .

Example 4.8.3. The action of the Hodge star in $R^{2}$ .

Example 4.8.4. The action of the Hodge star in $R^{3}$ .

Example 4.8.5. An example of the Hodge star action in $R^{3}$ .