# A Quick and Dirty Introduction to Exterior Calculus — Part III: Hodge Duality

Previously we saw that a $$k$$-form measures the (signed) projected volume of a $$k$$-dimensional parallelpiped. For instance, a 2-form measures the area of a parallelogram projected onto some plane, as depicted above. But here’s a nice observation: a plane in $$\mathbb{R}^3$$ can be described either by a pair of basis directions $$(\alpha,\beta)$$, or by a normal direction $$\gamma$$. So rather than measuring projected area, we could instead measure how well the normal of a parallelogram $$(u,v)$$ lines up with the normal of our plane. In other words, we could look for a 1-form $$\gamma$$ such that

$\gamma(u \times v) = \alpha \wedge \beta(u,v).$

This observation captures the idea behind Hodge duality: a $$k$$-dimensional volume in an $$n$$-dimensional space can be specified either by $$k$$ directions or by a complementary set of $$(n-k)$$ directions. There should therefore be some kind of natural correspondence between $$k$$-forms and $$(n-k)$$-forms.

The Hodge Star

Let’s investigate this idea further by constructing an explicit basis for the space of 0-forms, 1-forms, 2-forms, etc. — to keep things manageable we’ll work with $$\mathbb{R}^3$$ and its standard coordinate system $$(x^1, x^2, x^3)$$. 0-forms are easy: any 0-form can be thought of as some function times the constant 0-from, which we’ll denote “$$1$$.” We’ve already seen the 1-form basis $$dx^1, dx^2, dx^3$$, which looks like the standard orthonormal basis of a vector space:

What about 2-forms? Well, consider that any 2-form can be expressed as the wedge of two 1-forms:

$\alpha \wedge \beta = (\alpha_i dx^i) \wedge (\beta_j dx^j) = \alpha_i \beta_j dx^i \wedge dx^j.$

In other words, any 2-form looks like some linear combination of the basis 2-forms $$dx^i \wedge dx^j$$. How many of these bases are there? Initially it looks like there are a bunch of possibilities:

$\begin{array}{ccc} dx^1 \wedge dx^1 & dx^1 \wedge dx^2 & dx^1 \wedge dx^3 \\ dx^2 \wedge dx^1 & dx^2 \wedge dx^2 & dx^2 \wedge dx^3 \\ dx^3 \wedge dx^1 & dx^3 \wedge dx^2 & dx^3 \wedge dx^3 \\ \end{array}$

But of course, not all of these guys are distinct: remember that the wedge product is antisymmetric ($$\alpha \wedge \beta = -\beta \wedge \alpha$$), which has the important consequence $$\alpha \wedge \alpha = 0$$. So really our table looks more like this:

$\begin{array}{ccc} 0 & dx^1 \wedge dx^2 & -dx^3 \wedge dx^1 \\ -dx^1 \wedge dx^2 & 0 & dx^2 \wedge dx^3 \\ dx^3 \wedge dx^1 & -dx^2 \wedge dx^3 & 0 \\ \end{array}$

and we’re left with only three distinct bases: $$dx^2 \wedge dx^3$$, $$dx^3 \wedge dx^1$$, and $$dx^1 \wedge dx^2$$. Geometrically all we’ve said is that there are three linearly-independent “planes” in $$\mathbb{R}^3$$:

How about 3-form bases? We certainly have at least one:

$dx^1 \wedge dx^2 \wedge dx^3.$

Are there any others? Again the antisymmetry of $$\wedge$$ comes into play: many potential bases are just permutations of this first one:

$dx^2 \wedge dx^3 \wedge dx^1 = -dx^2 \wedge dx^1 \wedge dx^3 = dx^1 \wedge dx^2 \wedge dx^3,$

and the rest vanish due to the appearance of repeated 1-forms:

$dx^2 \wedge dx^1 \wedge dx^2 = -dx^2 \wedge dx^2 \wedge dx^1 = 0 \wedge dx^1 = 0.$

In general there is only one basis $$n$$-form $$dx^1 \wedge \cdots \wedge dx^n$$, which measures the usual Euclidean volume of a parallelpiped:

Finally, what about 4-forms on $$\mathbb{R}^3$$? At this point it’s probably pretty easy to see that there are none, since we’d need to pick four distinct 1-form bases from a collection of only three. Geometrically: there are no four-dimensional volumes contained in $$\mathbb{R}^3$$! (Or volumes of any greater dimension, for that matter.) The complete list of $$k$$-form bases on $$\mathbb{R}^3$$ is then

• 0-form bases: 1
• 1-form bases: $$dx^1$$, $$dx^2$$, $$dx^3$$
• 2-form bases: $$dx^2 \wedge dx^3$$, $$dx^3 \wedge dx^1$$, $$dx^1 \wedge dx^2$$
• 3-form bases: $$dx^1 \wedge dx^2 \wedge dx^3$$,

which means the number of bases is 1, 3, 3, 1. In fact you may see a more general pattern here: the number of $$k$$-form bases on an $$n$$-dimensional space is given by the binomial coefficient

$\left( \begin{array}{c} n \\ k \end{array} \right) = \frac{n!}{k!(n-k)!}$

(i.e., “$$n$$ choose $$k$$”), since we want to pick $$k$$ distinct 1-form bases and don’t care about the order. An important identity here is

$\left( \begin{array}{c} n \\ k \end{array} \right) = \left( \begin{array}{c} n \\ n-k \end{array} \right),$

which, as anticipated, means that we have a one-to-one relationship between $$k$$-forms and $$(n-k)$$-forms. In particular, we can identify any $$k$$-form with its complement. For example, on $$\mathbb{R}^3$$ we have

$\begin{array}{rcl} \star\ 1 &=& dx^1 \wedge dx^2 \wedge dx^3 \\ \star\ dx^1 &=& dx^2 \wedge dx^3 \\ \star\ dx^2 &=& dx^3 \wedge dx^1 \\ \star\ dx^3 &=& dx^1 \wedge dx^2 \\ \star\ dx^1 \wedge dx^2 &=& dx^3 \\ \star\ dx^2 \wedge dx^3 &=& dx^1 \\ \star\ dx^3 \wedge dx^1 &=& dx^2 \\ \star\ dx^1 \wedge dx^2 \wedge dx^2 &=& 1 \end{array}$

The map $$\star$$ (pronounced “star”) is called the Hodge star and captures this idea that planes can be identified with their normals and so forth. More generally, on any flat space we have

$\star\ dx^{i_1} \wedge dx^{i_2} \wedge \cdots \wedge dx^{i_k} = dx^{i_{k+1}} \wedge dx^{i_{k+2}} \wedge \cdots \wedge dx^{i_n},$

where $$(i_1, i_2, \ldots, i_n)$$ is any even permutation of $$(1, 2, \ldots, n)$$.

The Volume Form

So far we’ve been talking about measuring volumes in flat spaces like $$\mathbb{R}^n$$. But how do we take measurements in a curved space? Let’s think about our usual example of a surface $$f: \mathbb{R}^2 \supset M \rightarrow \mathbb{R}^3$$. If we specify a region on our surface via a pair of unit orthogonal vectors $$u, v \in \mathbb{R}^2$$, it’s clear that we don’t want the area $$dx^1 \wedge dx^2(u,v)=1$$ since that just gives us the area in the plane. Instead, we want to know what a unit area looks like after it’s been “stretched-out” by the map $$f$$. In particular, we said that the length of a vector $$df(u)$$ can be expressed in terms of the metric $$g$$:

$|df(u)| = \sqrt{df(u) \cdot df(u)} = \sqrt{g(u,u)}.$

So the area we’re really interested in is the product of the lengths $$|df(u)||df(v)| = \sqrt{g(u,u)g(v,v)}$$. When $$u$$ and $$v$$ are orthonormal the quantity $$\det(g) := g(u,u)g(v,v)-2g(u,v)$$ is called the determinant of the metric, and can be used to define a 2-form $$\sqrt{\det(g)} dx^1 \wedge dx^2$$ that measures any area on our surface. More generally, the $$n$$-form

$\omega := \sqrt{\det(g)} dx^1 \wedge \cdots \wedge dx^n$

is called the volume form, and will play a key role when we talk about integration.

On curved spaces, we’d also like the Hodge star to capture the fact that volumes have been stretched out. For instance, it makes a certain amount of sense to identify the constant function $$1$$ with the volume form $$\omega$$, since $$\omega$$ really represents unit volume on the curved space:

$\star 1 = \omega$

The Inner Product on $$k$$-Forms

More generally we’ll ask that any $$n$$-form constructed from a pair of $$k$$-forms $$\alpha$$ and $$\beta$$ satisfies

$\alpha \wedge \star \beta = \langle\langle \alpha, \beta \rangle\rangle \omega,$

where $$\langle\langle \alpha, \beta \rangle\rangle = \sum_i \alpha_i \beta_i$$ is the inner product on $$k$$-forms. In fact, some authors use this relationship as the definition of the wedge product — in other words, they’ll start with something like, “the wedge product is the unique binary operation on $$k$$-forms such that $$\alpha \wedge \star \beta = \langle\langle \alpha, \beta \rangle\rangle \omega$$,” and from there derive all the properties we’ve established above. This treatment is a bit abstract, and makes it far too easy to forget that the wedge product has an extraordinarily concrete geometric meaning. (It’s certainly not the way Hermann Grassmann thought about it when he invented exterior algebra!). In practice, however, this identity is quite useful. For instance, if $$u$$ and $$v$$ are vectors in $$\mathbb{R}^3$$, then we can write

$u \cdot v = \star\left(u^\flat \wedge \star v^\flat\right),$

i.e., on a flat space we can express the usual Euclidean inner product via the wedge product. Is it clear geometrically that this identity is true? Think about what it says: the Hodge star turns $$v$$ into a plane with $$v$$ as a normal. We then build a volume by extruding this plane along the direction $$u$$. If $$u$$ and $$v$$ are nearly parallel the volume will be fairly large; if they’re nearly orthogonal the volume will be quite shallow. (But to be sure we really got it right, you should try verifying this identity in coordinates!) Similarly, we can express the Euclidean cross product as just

$u \times v = \star(u^\flat \wedge v^\flat)^\sharp,$

i.e., we can create a plane with normal $$u \times v$$ by wedging together the two basis vectors $$u$$ and $$v$$. (Again, working this fact out in coordinates may help soothe your paranoia.)

### One Response to “A Quick and Dirty Introduction to Exterior Calculus — Part III: Hodge Duality”

1. Keenan says:

I’ve added a short blurb on the dot product and the cross product to the notes on Hodge duality — this information will be useful in the upcoming assignment!