## Discrete Exterior Calculus

So far we’ve been exploring exterior calculus purely in the smooth setting. Unfortunately this theory was developed by some old-timers who did not know anything about computers, hence it cannot be used directly by machines that store only a finite amount of information. For instance, if we have a smooth vector field or a smooth 1-form we can’t possibly store the direction of every little “arrow” at each point — there are far too many of them! Instead, we need to keep track of a discrete (or really, finite) number of pieces of information that capture the essential behavior of the objects we’re working with; we call this scheme discrete exterior calculus (or DEC for short). The big secret about DEC is that it’s literally nothing more than the good-old fashioned (continuous) exterior calculus we’ve been learning about, except that we integrate differential forms over elements of our mesh.

Discrete Differential Forms

One way to encode a 1-form might be to store a finite collection of “arrows” associated with some subset of points. Instead, we’re going to do something a bit different: we’re going to integrate our 1-form over each edge of a mesh, and store the resulting numbers (remember that the integral of an $n$-form always spits out a single number) on the corresponding edges. In other words, if $\alpha$ is a 1-form and $e$ is an edge, then we’ll associate the number

$\hat{\alpha}_e := \int_e \alpha$

with $e$, where the use of the hat ($\ \hat{}\$) is supposed to suggest a discrete quantity (not to be confused with a unit-length vector).

Does this procedure seem a bit abstract to you? It shouldn’t! Think about what this integral represents: it tells us how strongly the 1-form $\alpha$ “flows along” the edge $e$ on average. More specifically, remember how integration of a 1-form works: at each point along the edge we take the vector tangent to the edge, stick it into the 1-form $\alpha$, and sum up the resulting values — each value tells us something about how well $\alpha$ “lines up” with the direction of the edge. For instance, we could approximate the integral via the sum

$\int_e \alpha \approx |e|\left(\frac{1}{N} \sum_{i=1}^N \alpha_{p_i}(\hat{e})\right),$

where $|e|$ denotes the length of the edge, $\{p_i\}$ is a sequence of points along the edge, and $\hat{e} := e/|e|$ is a unit vector tangent to the edge:

Of course, this quantity tells us absolutely nothing about the strength of the “flow” orthogonal to the edge: it could be zero, it could be enormous! We don’t really know, because we didn’t take any measurements along the orthogonal direction. However, the hope is that some of this information will still be captured by nearby edges (which are most likely not parallel to $e$).

More generally, a $k$-form that has been integrated over each $k$-dimensional cell (edges in 1D, faces in 2D, etc.) is called a discrete differential $k$-form. (If you ever find the distinction confusing, you might find it helpful to substitute the word “integrated” for the word “discrete.”) In practice, however, not every discrete differential form has to originate from a continuous one — for instance, a bunch of arbitrary values assigned to each edge of a mesh is a perfectly good discrete 1-form.

Orientation

One thing you may have noticed in all of our illustrations so far is that each edge is marked with a little arrow. Why? Well, one thing to remember is that direction matters when you integrate. For instance, the fundamental theorem of calculus (and common sense) tells us that the total change as you go from $a$ to $b$ is the opposite of the total change as you go from $b$ to $a$:

$\int_a^b \frac{\partial\phi}{\partial x} dx = \phi(b)-\phi(a) = -(\phi(a)-\phi(b)) = -\int_b^a \frac{\partial\phi}{\partial x} dx.$

Said in a much less fancy way: the elevation gain as you go from Pasadena to Altadena is 151 meters, so of the elevation “gain” in the other direction must be -151 meters! Just keeping track of the number 151 does you little good — you have to say what that quantity represents.

Therefore, when we store a discrete differential form it’s not enough to just store a number: we also have to specify a canonical orientation for each element of our mesh, corresponding to the orientation we used during integration. For an edge we’ve already seen that we can think about orientation as a little arrow pointing from one vertex to another — we could also just think of an edge as an ordered pair $(i,j)$, meaning that we always integrate from $i$ to $j$.

More generally, suppose that each element of our mesh is an oriented $k$-simplex $\sigma$, i.e., a collection of $k+1$ vertices $p_i \in \mathbb{R}^n$ given in some fixed order $(p_1, \ldots, p_{k+1})$. The geometry associated with $\sigma$ is the convex combination of these points:

$\left\{ \sum_{i=1}^{k+1} \lambda_i p_i \left| \sum_{i=1}^{k+1} \lambda_i = 1 \right. \right\} \subset \mathbb{R}^n$

(Convince yourself that a 0-simplex is a vertex, a 1-simplex is an edge, a 2-simplex is a triangle, and a 3-simplex is a tetrahedron.)

Two oriented $k$-simplices have the same orientation if and only if the vertices of one are an even permutation of the vertices of another. For instance, the triangles $(p_1, p_2, p_3)$ and $(p_2, p_3, p_1)$ have the same orientation; $(p_1, p_2, p_3)$ and $(p_2, p_1, p_3)$ have opposite orientation.

If a simplex $\sigma_1$ is a (not necessarily proper) subset of another simplex $\sigma_2$, then we say that $\sigma_1$ is a face of $\sigma_2$. For instance, every vertex, edge, and triangle of a tetrahedron $\sigma$ is a face of $\sigma$; as is $\sigma$ itself! Moreover, the orientation of a simplex agrees with the orientation of one of its faces as long as we see an even permutation on the shared vertices. For instance, the orientations of the edge $(p_2,p_1)$ and the triangle $(p_1,p_3,p_2)$ agree. Geometrically all we’re saying is that the two “point” in the same direction (as depicted above). To keep yourself sane while working with meshes, the most important thing is to pick and orientation and stick with it!

So in general, how do we integrate a $k$-form over an oriented $k$-simplex? Remember that a $k$-form is going to “eat” $k$ vectors at each point and spit out a number — a good canonical choice is to take the ordered collection of edge vectors $(p_2 – p_1, \ldots, p_{k+1}-p_1)$ and orthogonalize them (using, say the Gram-Schmidt algorithm) to get vectors $(u_1, \ldots, u_n)$. This way the sign of the integrand changes whenever the orientation changes. Numerically, we can then approximate the integral via a sum

$\int_\sigma \alpha \approx \frac{|\sigma|}{N} \sum_{i=1}^N \alpha_{p_i}(u_1, \ldots, u_n)$

where $\{p_i\}$ is a (usually carefully-chosen) collection of sample points. (Can you see why the orientation of $\sigma$ affects the sign of the integrand?) Sounds like a lot of work, but in practice one rarely constructs discrete differential forms via integration: more often, discrete forms are constructed via input data that is already discrete (e.g., vertex positions in a triangle mesh).

By the way, what’s a discrete 0-form? Give up? Well, it must be a 0-form (i.e., a function) that’s been integrated over every 0-simplex (i.e., vertex) of a mesh:

$\hat{\phi}_i = \int_{v_i} \phi$

By convention, the integral of a function over a zero-dimensional set is simply the value of the function at that point: $\hat{\phi}_i = \phi(v_i)$. In other words, in the case of 0-forms there is no difference between storing point samples and storing integrated quantities: the two coincide.

It’s also important to remember that differential forms don’t have to be real-valued. For instance, we can think of a map $f: M \rightarrow \mathbb{R}^3$ that encodes the geometry of a surface as an $\mathbb{R}^3$-valued 0-form; its differential $df$ is then an $\mathbb{R}^3$-valued 1-form, etc. Likewise, when we say that a discrete differential form is a number stored on every mesh element, the word “number” is used in a fairly loose sense: a number could be a real value, a vector, a complex number, a quaternion, etc. For instance, the collection of $(x,y,z)$ vertex coordinates of a mesh can be viewed as an $\mathbb{R}^3$-valued discrete 0-form (namely, one that discretizes the map $f$). The only requirement, of course, is that we store the same type of number on each mesh element.

The Discrete Exterior Derivative

One of the main advantages of working with integrated (i.e., “discrete”) differential forms instead of point samples is that we can easily take advantage of Stokes’ theorem. Remember that Stokes’ theorem says

$\int_\Omega d\alpha = \int_{\partial\Omega} \alpha,$

for any $k$-form $\alpha$ and $k+1$-dimensional domain $\Omega$. In other words, we can integrate the derivative of a differential form as long as we know its integral along the boundary. But that’s exactly the kind of information encoded by a discrete differential form! For instance, if $\hat{\alpha}$ is a discrete 1-form stored on the three edges of a triangle $\sigma$, then we have

$\int_\sigma d\alpha = \int_{\partial\sigma} \alpha = \sum_{i=1}^3 \int_{e_i} \alpha = \sum_{i=1}^3 \hat{\alpha}_i.$

In other words, we can exactly evaluate the integral on the left by just adding up three numbers. Pretty cool! In fact, the thing on the left is also a discrete differential form: it’s the 2-form $d\alpha$ integrated over the only triangle in our mesh. So for convenience, we’ll call this guy “$\hat{d}\hat{\alpha}$”, and we’ll call the operation $\hat{d}$ the discrete exterior derivative. (In the future we will drop the hats from our notation when the meaning is clear from context.) In other words, the discrete exterior derivative takes a $k$-form that has already been integrated over each $k$-simplex and applies Stokes’ theorem to get the integral of the derivative over each $k+1$-simplex.

In practice (i.e., in code) you can see how this operation might be implemented by simply taking local sums over the appropriate mesh elements. However, in the example above we made life particularly easy on ourselves by giving each edge an orientation that agrees with the orientation of the triangle. Unfortunately assigning a consistent orientation to every simplex is not always possible, and in general we need to be more careful about sign when adding up our piecewise integrals. For instance, in the example below we’d have

$(\hat{d}\hat{\alpha})_1 = \hat{\alpha}_1 + \hat{\alpha}_2 + \hat{\alpha}_3$

and

$(\hat{d}\hat{\alpha})_2 = \hat{\alpha}_4 + \hat{\alpha}_5 - \hat{\alpha}_2.$

Discrete Hodge Star

As hinted at above, a discrete $k$-form captures the behavior of a continuous $k$-form along $k$ directions, but not along the remaining $n-k$ directions — for instance, a discrete 1-form in 2D captures the flow along edges but not in the orthogonal direction. If you paid attention to our discussion of Hodge duality, this story starts to sound familiar! To capture Hodge duality in the discrete setting, we’ll need to define a dual mesh. In general, the dual of an $n$-dimensional simplicial mesh identifies every $k$-simplex in the primal (i.e., original) mesh with a unique $(n-k)$-cell in the dual mesh. In a two-dimensional simplicial mesh, for instance, primal vertices are identified with dual faces, primal edges are identified with dual edges, and primal faces are identified with dual vertices. Note, however, that the dual cells are not always simplices! (See above.)

So how do we talk about Hodge duality in discrete exterior calculus? Quite simply, the discrete Hodge dual of a (discrete) $k$-form on the primal mesh is an $(n-k)$-form on the dual mesh. Similarly, the Hodge dual of an $k$-form on the dual mesh is a $k$-form on the primal mesh. Discrete forms on the primal mesh are called primal forms and discrete forms on the dual mesh are called dual forms. Given a discrete form $\hat{\alpha}$ (whether primal or dual), its Hodge dual is typically written as $\hat{\star} \hat{\alpha}$.

Unlike continuous forms, discrete primal and dual forms live in different places (so for instance, discrete primal $k$-forms and dual $k$-forms cannot be added to each other). In fact, primal and dual forms often have different physical interpretations. For instance, a primal 1-form might represent the total circulation along edges of the primal mesh, whereas in the same context a dual 1-form might represent the total flux through the corresponding dual edges (see illustration above).

Of course, these two quantities (flux and circulation) are closely related, and naturally leads into one definition for a discrete Hodge star called the diagonal Hodge star. Consider a primal $k$-form $\alpha$. If $\hat{\alpha}_i$ is the value of $\hat{\alpha}$ on the $k$-simplex $\sigma_i$, then the diagonal Hodge star is defined by

$\hat{\star} \hat{\alpha}_i = \frac{|\sigma_i^\star|}{|\sigma_i|} \hat{\alpha}_i$

for all $i$, where $|\sigma|$ indicates the (unsigned) volume of $\sigma$ (which by convention equals one for a vertex!) and $|\sigma^\star|$ is the volume of the corresponding dual cell. In other words, to compute the dual form we simply multiply the scalar value stored on each cell by the ratio of corresponding dual and primal volumes.

If we remember that a discrete form can be thought of as a continuous form integrated over each cell, this definition for the Hodge star makes perfect sense: the primal and dual quantities should have the same density, but we need to account for the fact that they are integrated over cells of different volume. We therefore normalize by a ratio of volumes when mapping between primal and dual. This particular Hodge star is called diagonal since the $i$th element of the dual differential form depends only on the $i$th element of the primal differential form. It’s not hard to see, then, that Hodge star taking dual forms to primal forms (the dual Hodge star) is the inverse of the one that takes primal to dual (the primal Hodge star).

That’s All, Folks!

Hey, wait a minute, what about our other operations, like the wedge product ($\wedge$)? These operations can certainly be defined in the discrete setting, but we won’t go into detail here — the basic recipe is to integrate, integrate, integrate. Actually, even in continuous exterior calculus we omitted a couple operations like the Lie derivative ($\mathcal{L}_X$) and the interior product ($i_\alpha$). Coming up with a complete discrete calculus where the whole cast of characters $d$, $\wedge$, $\star$, $\mathcal{L}_X$, $i_\alpha$, etc., plays well together is an active and ongoing area of research, which may be of interest to aspiring young researchers like you (yes, you)!

## A Quick and Dirty Introduction to Exterior Calculus — Part V: Integration and Stokes’ Theorem

In the last set of notes we talked about how to differentiate $k$-forms using the exterior derivative $d$. We’d also like some way to integrate forms. Actually, there’s surprisingly little to say about integration given the setup we already have. Suppose we want to compute the total area $A_\Omega$ of a region $\Omega$ in the plane:

If you remember back to calculus class, the basic idea was to break up the domain into a bunch of little pieces that are easy to measure (like squares) and add up their areas:

$A_\Omega \approx \sum_i A_i.$

As these squares get smaller and smaller we get a better and better approximation, ultimately achieving the true area

$A_\Omega = \int_\Omega dA.$

Alternatively, we could write the individual areas using differential forms — in particular, $A_i = dx^1 \wedge dx^2(u,v)$. Therefore, the area element $dA$ is really nothing more than the standard volume form $dx^1 \wedge dx^2$ on $\mathbb{R}^2$. (Not too surprising, since the whole point of $k$-forms was to measure volume!)

To make things more interesting, let’s say that the contribution of each little square is weighted by some scalar function $\phi$. In this case we get the quantity

$\int_\Omega \phi\ dA = \int_\Omega \phi\ dx^1 \wedge dx^2.$

Again the integrand $\phi\ dx^1 \wedge dx^2$ can be thought of as a 2-form. In other words, you’ve been working with differential forms your whole life, even if you didn’t realize it! More generally, integrands on an $n$-dimensional space are always $n$-forms, since we need to “plug in” $n$ orthogonal vectors representing the local volume. For now, however, looking at surfaces (i.e., 2-manifolds) will give us all the intuition we need.

Integration on Surfaces

If you think back to our discussion of the Hodge star, you’ll remember the volume form

$\omega = \sqrt{\mathrm{det}(g)} dx^1 \wedge dx^2,$

which measures the area of little parallelograms on our surface. The factor $\sqrt{\mathrm{det}(g)}$ reminds us that we can’t simply measure the volume in the domain $M$ — we also have to take into account any “stretching” induced by the map $f: M \rightarrow \mathbb{R}^2$. Of course, when we integrate a function on a surface, we should also take this stretching into account. For instance, to integrate a function $\phi: M \rightarrow \mathbb{R}$, we would write

$\int_\Omega \phi \omega = \int_\Omega \phi \sqrt{\mathrm{det}(g)}\ dx^1 \wedge dx^2.$

In the case of a conformal parameterization things become even simpler — since $\sqrt{\mathrm{det}(g)} = a$ we have just

$\int_\Omega \phi a\ dx^1 \wedge dx^2,$

where $a: M \rightarrow \mathbb{R}$ is the scaling factor. In other words, we scale the value of $\phi$ up or down depending on the amount by which the surface locally “inflates” or “deflates.” In fact, this whole story gives a nice geometric interpretation to good old-fashioned integrals: you can imagine that $\int_\Omega \phi\ dA$ represents the area of some suitably deformed version of the initially planar region $\Omega$.

Stokes’ Theorem

The main reason for studying integration on manifolds is to take advantage of the world’s most powerful tool: Stokes’ theorem. Without further ado, Stokes’ theorem says that

$\int_\Omega d\alpha = \int_{\partial\Omega} \alpha,$

where $\alpha$ is any $n-1$-form on an $n$-dimensional domain $\Omega$. In other words, integrating a differential form over the boundary of a manifold is the same as integrating its derivative over the entire domain.

If this trick sounds familiar to you, it’s probably because you’ve seen it time and again in different contexts and under different names: the divergence theorem, Green’s theorem, the fundamental theorem of calculus, Cauchy’s integral formula, etc. Picking apart these special cases will really help us understand the more general meaning of Stokes’ theorem.

Divergence Theorem

Let’s start with the divergence theorem from vector calculus, which says that

$\int_\Omega \nabla \cdot X dA = \int_{\partial\Omega} N \cdot X d\ell,$

where $X$ is a vector field on $\Omega$ and $N$ represents the (outward-pointing) unit normals on the boundary of $\Omega$. A better name for this theorem might have been the “what goes in must come out theorem”, because if you think about $X$ as the flow of water throughout the domain $\Omega$ then it’s clear that the amount of water being pumped into $\Omega$ (via pipes in the ground) must be the same as the amount flowing out of its boundary at any moment in time:

Let’s try writing this theorem using exterior calculus. First, remember that we can write the divergence of $X$ as $\nabla \cdot X = \star d \star X^\flat$. It’s a bit harder to see how to write the right-hand side of the divergence theorem, but think about what integration does here: it takes tangents to the boundary and sticks them into a 1-form. For instance, $\int_\Omega X^\flat$ adds up the tangential components of $X$. To get the normal component we could rotate $X^\flat$ by a quarter turn, which conveniently enough is achieved by hitting it with the Hodge star. Overall we get

$\int_\Omega d \star X^\flat = \int_{\partial\Omega} \star X^\flat,$

which, as promised, is just a special case of Stokes’ theorem. Alternatively, we can use Stokes’ theorem to provide a more geometric interpretation of the divergence operator itself: when integrated over any region $\Omega$ — no matter how small — the divergence operator gives the total flux through the region boundary. In the discrete case we’ll see that this boundary flux interpretation is the only notion of divergence — in other words, there’s no concept of divergence at a single point.

By the way, why does $d \star X^\flat$ appear on the left-hand side instead of $\star d \star X^\flat$? The reason is that $\star d \star X^\flat$ is a 0-form, so we have to hit it with another Hodge star to turn it into an object that measures areas (i.e., a 2-form). Applying this transformation is no different from appending $dA$ to $\nabla \cdot X$ — we’re specifying how volume should be measured on our domain.

Fundamental Theorem of Calculus

The fundamental theorem of calculus is in fact so fundamental that you may not even remember what it is. It basically says that for a real-valued function $\phi: \mathbb{R} \rightarrow \mathbb{R}$ on the real line

$\int_a^b \frac{\partial \phi}{\partial x} dx = \phi(b) - \phi(a).$

In other words, the total change over an interval $[a,b]$ is (as you might expect) how much you end up with minus how much you started with. But soft, behold! All we’ve done is written Stokes’ theorem once again:

$\int_{[a,b]} d\phi = \int_{\partial[a,b]} \phi,$

since the boundary of the interval $[a,b]$ consists only of the two endpoints $a$ and $b$.

Hopefully these two examples give you a good feel for what Stokes’ theorem says. In the end, it reads almost like a Zen kōan: what happens on the outside is purely a function of the change within. (Perhaps it is Stokes’ that deserves the name, “fundamental theorem of calculus!”)

October 31, 2012 | Posted in: Notes | Comments Closed

## A Quick and Dirty Introduction to Exterior Calculus — Part IV: Differential Operators

Originally we set out to develop exterior calculus. The objects we’ve looked at so far — $k$-forms, the wedge product $\wedge$ and the Hodge star $\star$ — actually describe a more general structure called an exterior algebra. To turn our algebra into a calculus, we also need to know how quantities change, as well as how to measure quantities. In other words, we need some tools for differentiation and integration. Let’s start with differentiation.

In our discussion of surfaces we briefly looked at the differential $df$ of a surface $f: M \rightarrow \mathbb{R}^3$, which tells us something about the way tangent vectors get “stretched out” as we move from the domain $M$ to a curved surface sitting in $\mathbb{R}^3$. More generally $d$ is called the exterior derivative and is responsible for building up many of the differential operators in exterior calculus. The basic idea is that $d$ tells us how quickly a $k$-form changes along every possible direction. But how exactly is it defined? So far we’ve seen only a high-level geometric description.

Div, Grad, and Curl

Before jumping into the exterior derivative, it’s worth reviewing what the basic vector derivatives $\mathrm{div}$, $\mathrm{grad}$, and $\mathrm{curl}$ do, and more importantly, what they look like. The key player here is the operator $\nabla$ (pronounced “nabla”) which can be expressed in coordinates as the vector of all partial derivatives:

$\nabla := \left( \frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n} \right).$

For instance, applying $\nabla$ to a scalar function $\phi: \mathbb{R}^n \rightarrow \mathbb{R}$ yields the gradient

$\nabla\phi = \left( \frac{\partial f}{\partial x^1}, \ldots, \frac{\partial f}{\partial x^n} \right),$

which can be visualized as the direction of steepest ascent on some terrain:

We can also apply $\nabla$ to a vector field $X$ in two different ways. The dot product gives us the divergence

$\nabla \cdot X = \frac{\partial X^1}{\partial x^1} + \cdots + \frac{\partial X^n}{\partial x^n}$

which measures how quickly the vector field is “spreading out”, and on $\mathbb{R}^3$ the cross product gives us the curl

$\nabla \times X = \left( \frac{\partial X_3}{\partial x^2} - \frac{\partial X_2}{\partial x^3}, \frac{\partial X_1}{\partial x^3} - \frac{\partial X_3}{\partial x^1}, \frac{\partial X_2}{\partial x^1} - \frac{\partial X_1}{\partial x^2} \right),$

which indicates how much a vector field is “spinning around.” For instance, here’s a pair of vector fields with a lot of divergence and a lot of curl, respectively:

(Note that in this case one field is just a 90-degree rotation of the other!) On a typical day it’s a lot more useful to think of $\mathrm{div}$, $\mathrm{grad}$ and $\mathrm{curl}$ in terms of these kinds of pictures rather than the ugly expressions above.

Think Differential

Not surprisingly, we can express similar notions using exterior calculus. However, these notions will be a bit easier to generalize (for instance, what does “curl” mean for a vector field in $\mathbb{R}^4$, where no cross product is defined?). Let’s first take a look at the exterior derivative of 0-forms (i.e., functions), which is often just called the differential. To keep things simple, we’ll start with real-valued functions $\phi: \mathbb{R}^n \rightarrow \mathbb{R}$. In coordinates, the differential is defined as

$d\phi := \frac{\partial \phi}{\partial x^1} dx^1 + \cdots + \frac{\partial \phi}{\partial x^n} dx^n.$

It’s important to note that the terms $\frac{\partial \phi}{\partial x^i}$ actually correspond to partial derivatives of our function $\phi$, whereas the terms $dx^i$ simply denote an orthonormal basis for $\mathbb{R}^n$. In other words, you can think of $d\phi$ as just a list of all the partial derivatives of $\phi$. Of course, this object looks a lot like the gradient $\nabla \phi$ we saw just a moment ago. And indeed the two are closely related, except for the fact that $\nabla \phi$ is a vector field and $d\phi$ is a 1-form. More precisely,

$\nabla \phi = (d\phi)^\sharp.$

Directional Derivatives

Another way to investigate the behavior of the exterior derivative is to see what happens when we stick a vector $u$ into the 1-form $df$. In coordinates we get something that looks like a dot product between the gradient of $f$ and the vector $u$:

$df(u) = \frac{\partial f}{\partial x^1} u^1 + \cdots + \frac{\partial f}{\partial x^n} u^n.$

For instance, in $\mathbb{R}^2$ we could stick in the unit vector $u = (1,0)$ to get the partial derivative $\frac{\partial f}{\partial x^1}$ along the first coordinate axis:

(Compare this picture to the picture of the gradient we saw above.) In general, $df(u)$ represents the directional derivative of $f$ along the direction $u$. In other words, it tells us how quickly $f$ changes if we take a short walk in the direction $u$. Returning again to vector calculus notation, we have

$df(u) = u \cdot \nabla f.$

Properties of the Exterior Derivative

How do derivatives of arbitrary $k$-forms behave? For one thing, we expect $d$ to be linear — after all, a derivative is just the limit of a difference, and differences are certainly linear! What about the derivative of a wedge of two forms? Harkening back to good old-fashioned calculus, here’s a picture that explains the typical product rule $\frac{\partial}{\partial x}(f(x)g(x)) = f’(x)g(x) + f(x)g’(x)$:

The dark region represents the value of $fg$ at $x$; the light blue region represents the change in this value as we move $x$ some small distance $h$. As $h$ gets smaller and smaller, the contribution of the upper-right quadrant becomes negligible and we can write the derivative as the change in $f$ times $g$ plus the change in $g$ times $f$. (Can you make this argument more rigorous?) Since a $k$-form also measures a (signed) volume, this intuition also carries over to the exterior derivative of a wedge product. In particular, if $\alpha$ is a $k$-form then $d$ obeys the rule

$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{k}\alpha \wedge d\beta.$

which says that the rate of change of the overall volume can be expressed in terms of changes in the constituent volumes, exactly as in the picture above.

Exterior Derivative of 1-Forms

To be a little more concrete, let’s see what happens when we differentiate a 1-form on $\mathbb{R}^3$. Working things out in coordinates turns out to be a total mess, but in the end you may be pleasantly surprised with the simplicity of the outcome! (Later on we’ll see that these ideas can also be expressed quite nicely without coordinates using Stokes’ theorem, which paves the way to differentiation in the discrete setting.) Applying the linearity of $d$, we have

$\begin{array}{rcl} d\alpha &=& d(\alpha_1 dx^1 + \alpha_2 dx^2 + \alpha_3 dx^3) \\ &=& d(\alpha_1 dx^1) + d(\alpha_2 dx^2) + d(\alpha_3 dx^3). \end{array}$

Each term $\alpha_j dx^j$ can really be thought of a wedge product $\alpha_j \wedge dx^j$ between a 0-form $\alpha_j$ and the corresponding basis 1-form $dx^j$. Applying the exterior derivative to one of these terms we get

$d(\alpha_j \wedge dx^j) = (d\alpha_j) \wedge dx^j + \alpha_j \wedge \underbrace{(ddx^j)}_{=0} = \frac{\partial \alpha_j}{\partial x^i} dx^i \wedge dx^j.$

To keep things short we used the Einstein summation convention here, but let’s really write out all the terms:

$\begin{array}{rcccccc} d\alpha &=& \frac{\partial \alpha_1}{\partial x^1} dx^1 \wedge dx^1 &+& \frac{\partial \alpha_1}{\partial x^2} dx^2 \wedge dx^1 &+& \frac{\partial \alpha_1}{\partial x^3} dx^3 \wedge dx^1 \\ && \frac{\partial \alpha_2}{\partial x^1} dx^1 \wedge dx^2 &+& \frac{\partial \alpha_2}{\partial x^2} dx^2 \wedge dx^2 &+& \frac{\partial \alpha_2}{\partial x^3} dx^3 \wedge dx^2 \\ && \frac{\partial \alpha_3}{\partial x^1} dx^1 \wedge dx^3 &+& \frac{\partial \alpha_3}{\partial x^2} dx^2 \wedge dx^3 &+& \frac{\partial \alpha_3}{\partial x^3} dx^3 \wedge dx^3. \\ \end{array}$

Using the fact that $\alpha \wedge \beta = -\beta \wedge \alpha$, we get a much simpler expression

$\begin{array}{rcl} d\alpha &=& ( \frac{\partial \alpha_3}{\partial x^2} - \frac{\partial \alpha_2}{\partial x^3} ) dx^2 \wedge dx^3 \\ && ( \frac{\partial \alpha_1}{\partial x^3} - \frac{\partial \alpha_3}{\partial x^1} ) dx^3 \wedge dx^1 \\ && ( \frac{\partial \alpha_2}{\partial x^1} - \frac{\partial \alpha_1}{\partial x^2} ) dx^1 \wedge dx^2. \\ \end{array}$

Does this expression look familiar? If you look again at our review of vector derivatives, you’ll recognize that $d\alpha$ basically looks like the curl of $\alpha^\sharp$, except that it’s expressed as a 2-form instead of a vector field. Also remember (from our discussion of Hodge duality) that a 2-form and a 1-form are not so different here — geometrically they both specify some direction in $\mathbb{R}^3$. Therefore, we can express the curl of any vector field $X$ as

$\nabla \times X = \left( \star d X^\flat \right)^\sharp.$

It’s worth stepping through the sequence of operations here to check that everything makes sense: $\flat$ converts the vector field $X$ into a 1-form $X^\flat$; $d$ computes something that looks like the curl, but expressed as a 2-form $dX^\flat$; $\star$ turns this 2-form into a 1-form $\star d X^\flat$; and finally $\sharp$ converts this 1-form back into the vector field $\left( \star d X^\flat \right)^\sharp$. The take-home message here, though, is that the exterior derivative of a 1-form looks like the curl of a vector field.

So far we know how to express the gradient and the curl using $d$. What about our other favorite vector derivative, the divergence? Instead of grinding through another tedious derivation, let’s make a simple geometric observation: in $\mathbb{R}^2$ at least, we can determine the divergence of a vector field by rotating it by 90 degrees and computing its curl (consider the example we saw earlier). Moreover, in $\mathbb{R}^2$ the Hodge star $\star$ represents a rotation by 90 degrees, since it identifies any line with the direction orthogonal to that line:

Therefore, we might suspect that divergence can be computed by first applying the Hodge star, then applying the exterior derivative:

$\nabla \cdot X = \star d \star X^\flat.$

The leftmost Hodge star accounts for the fact that $d \star X^\flat$ is an $n$-form instead of a 0-form — in vector calculus divergence is viewed as a scalar quantity. Does this definition really work? Let’s give it a try in coordinates on $\mathbb{R}^3$. First, we have

$\begin{array}{rcl} \star X^\flat &=& \star( X_1 dx^1 + X_2 dx^2 + X_3 dx^3 ) \\ &=& X_1 dx^2 \wedge dx^3 + X_2 dx^3 \wedge dx^1 + X_3 dx^1 \wedge dx^2. \end{array}$

Differentiating we get

$\begin{array}{rcl} d \star X^\flat &=& \frac{\partial X_1}{\partial x^1} dx^1 \wedge dx^2 \wedge dx^3 + \\ && \frac{\partial X_2}{\partial x^2} dx^2 \wedge dx^3 \wedge dx^1 + \\ && \frac{\partial X_3}{\partial x^3} dx^3 \wedge dx^1 \wedge dx^2, \\ \end{array}$

but of course we can rearrange these wedge products to simply

$d \star X^\flat = \left( \frac{\partial X_1}{\partial x^1} + \frac{\partial X_2}{\partial x^2} + \frac{\partial X_3}{\partial x^3} \right) dx^1 \wedge dx^2 \wedge dx^3.$

A final application of the Hodge star gives us the divergence

$\star d \star X^\flat = \frac{\partial X_1}{\partial x^1} + \frac{\partial X_2}{\partial x^2} + \frac{\partial X_3}{\partial x^3}$

as desired.

In summary, for any scalar field $\phi$ and vector field $X$ we have

$\begin{array}{rcl} \nabla \phi &=& (d\phi)^\sharp \\ \nabla \times X &=& \left( \star d X^\flat \right)^\sharp \\ \nabla \cdot X &=& \star d \star X^\flat \end{array}$

One cute thing to notice here is that (in $\mathbb{R}^3$) $\mathrm{grad}$, $\mathrm{curl}$, and $\mathrm{div}$ are more or less just $d$ applied to a $0-$, $1-$ and $2-$ form, respectively.

The Laplacian

Another key differential operator from vector calculus is the scalar Laplacian which (confusingly!) is often denoted by $\Delta$ or $\nabla^2$, and is defined as

$\Delta := \nabla \cdot \nabla,$

i.e., the divergence of the gradient. Although the Laplacian may seem like just yet another in a long list of derivatives, it deserves your utmost respect: the Laplacian is central to the most fundamental of physical laws (any diffusion process and all forms of wave propagation, including the Schrödinger equation); its eigenvalues capture almost everything there is to know about a given piece of geometry (can you hear the shape of a drum?). Heavy tomes and entire lives have been devoted to the Laplacian, and in the discrete setting we’ll see that this one simple operator can be applied to a diverse array of tasks (surface parameterization, surface smoothing, vector field design and decomposition, distance computation, fluid simulation… you name it, we got it!).

Fortunately, now that we know how to write $\mathrm{div}$, $\mathrm{grad}$ and $\mathrm{curl}$ using exterior calculus, expressing the scalar Laplacian is straightforward: $\Delta = \star d \star d$. More generally, the $k$-form Laplacian is given by

$\Delta := \star d \star d + d \star d \star.$

The name “Laplace-Beltrami” is used merely to indicate that the domain may have some amount of curvature (encapsulated by the Hodge star). Some people like to define the operator $\delta := \star d \star$, called the codifferential, and write the Laplacian as $\Delta = \delta d + d \delta$.

One question you might ask is: why is the Laplacian for 0-forms different from the general $k$-form Laplacian? Actually, it’s not — consider what happens when we apply the term $d \star d \star$ to a 0-form $\phi$: $\star \phi$ is an $n$-form, and so $d \star \phi$ must be an $(n+1)$-form. But there are no $(n+1)$-forms on an $n$-dimensional space! So this term is often omitted when writing the scalar Laplacian.

| Posted in: Notes | Comments Closed

## 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.)

| Posted in: Notes | Comments Closed