## 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.)

## Homework 2: Normals of Discrete Surfaces

[Homework 2 has been released (below) and is due next Friday by 5pm. We suspect there will be a lot of questions about C++, libDDG, etc., so please do not hesitate to ask questions either in the comments (preferred) or via email. Good luck!]

For a smooth surface in $$\mathbb{R}^3$$, the normal direction is easy to define: it is the unique direction orthogonal to all tangent vectors — in other words, it’s the direction sticking “straight out” of the surface. For discrete surfaces the story is not so simple. If a mesh has planar faces (all vertices lie in a common plane) then of course the normal is well-defined: it is simply the normal of the plane. But if the polygon is nonplanar, or if we ask for the normal at a vertex, then it is not as clear how the normal should be defined.

In practice there are a number of different possibilities, which arise from different ways of looking at the smooth geometry. But before jumping in, let’s establish a few basic geometric facts.

Area of a Polygon

Here’s a simple question: how do you compute the area of a polygon in the plane? Suppose our polygon has vertices $$p_1, p_2, \ldots, p_n$$. One way to compute the area is to stick another point $$q$$ in the middle and sum up the areas of triangles $$q,p_i,p_{i+1}$$ as done on the left:

A cute fact about life is that if we place $$q$$ anywhere and sum up the signed triangle areas, we still recover the polygon area! (Signed area just means negative if our vertices are oriented clockwise; positive if they’re counter-clockwise.) You can get an idea of why this happens just by looking at the picture: positive triangles that cover “too much” area get accounted for by negative triangles.

The proof is an application of Stokes’ theorem — consider a different expression for the area $$A$$ of a planar polygon $$P$$:

$A = \int_P dx \wedge dy.$

Noting that $$dx \wedge dy = d(xdy) = -d(ydx)$$, we can also express the area as

$A = \frac{1}{2} \int_P d(xdy) - d(ydx) = \frac{1}{2} \int_{\partial P} xdy - ydx,$

where we’ve applied Stokes’ theorem in the final step to convert our integral over the entire surface into an integral over just the boundary. Now suppose that our polygon vertices have coordinates $$p_i = (x_i,y_i)$$. From here we can explicitly work out the boundary integral by summing up the integrals over each edge $$e_{ij}$$:

$\int_{\partial P} xdy - ydx = \sum \int_{e_{ij}} xdy - ydx.$

Since the coordinate functions $$x$$ and $$y$$ are linear along each edge (and their differentials $$dx$$ and $$dy$$ are therefore constant), we can write these integrals as

$\begin{array}{rcl} \sum \int_{e_{ij}} xdy - ydx &=& \sum \frac{x_i+x_j}{2}(y_j-y_i) - \frac{y_i+y_j}{2}(x_j-x_i) \\ &=& \frac{1}{2} \sum (p_i + p_j) \times (p_j - p_i) \\ &=& \frac{1}{2} \sum p_i \times p_j - p_i \times p_i - p_j \times p_j - p_j \times p_i \\ &=& \sum p_i \times p_j. \end{array}$

In short, we’ve shown that the area of a polygon can be written as simply

$A = \frac{1}{2} \sum_i p_i \times p_j.$

Exercise 2.1 Complete the proof by showing that for any point $$q$$ the signed areas of triangles $$(q,p_i,p_{i+1})$$ sum to precisely the expression above.

Vector Area

A more general version of the situation we just looked at with polygon areas is the vector area of a surface patch $$f: M \rightarrow \mathbb{R}^3$$, which is defined as the integral of the surface normal over the entire domain:

$N_\mathcal{V} := \int_M N dA.$

A very nice property of the vector area is that it depends only on the shape of the boundary $$\partial M$$ (as you will demonstrate in the next exercise). As a result, two surfaces that look very different (such as the ones above) can still have the same vector area — the physical intuition here is that the vector area measures the total flux through the boundary curve.

For a flat region the normal is constant over the surface and we get just the usual area times the unit normal vector. Things get more interesting when the surface is not flat — for instance, the vector area of a circular band is zero since opposing normals cancel each-other out:

Exercise 2.2 Using Stokes’ theorem, show that the vector area can be written as

$N_\mathcal{V} = \frac{1}{2} \int_{\partial M} f \wedge df,$

where the product of two vectors in $$\mathbb{R}^3$$ is given by the usual cross product $$\times$$.

Here’s another fairly basic question: consider a triangle sitting in $$\mathbb{R}^3$$, and imagine that we’re allowed to pull on one of its vertices $$p$$. What’s the quickest way to increase its area $$A$$? In other words, what’s the gradient of $$A$$ with respect to $$p$$?

Exercise 2.3 Show that the area gradient is given by

$\nabla_p A_\sigma = \frac{1}{2} \mathbf{u}^{\perp}$

where $$\mathbf{u^\perp}$$ is the edge vector across from $$p$$ rotated by an angle $$\pi/2$$ in the plane of the triangle (such that it points toward $$p$$). You should require only a few very simple geometric arguments — there’s no need to write things out in coordinates, etc.

Ok, with these facts out of the way let’s take a look at some different ways to define vertex normals. There are essentially only two definitions that arise naturally from the smooth picture: the area gradient and the volume gradient; we’ll start with the former.

The area gradient asks, “which direction should we `push’ the surface in order to increase its total area $$A$$ as quickly as possible?” Sliding all points tangentially along the surface clearly doesn’t change anything: we just end up with the same surface. In fact, the only thing we can do to increase surface area is move the surface in the normal direction. The idea, then, is to define the vertex normal as the gradient of area with respect to a given vertex.

Since we already know how to express the area gradient for a single triangle $$\sigma$$, we can easily express the area gradient for the entire surface:

$\nabla_p A = \sum_\sigma \nabla A_\sigma.$

Of course, a given vertex $$p$$ influences only the areas of the triangles touching $$p$$. So we can just sum up the area gradients over this small collection of triangles.

Exercise 2.4 Show that the gradient of surface area with respect to vertex $$p_i$$ can be expressed as

$\nabla_p A = \frac{1}{2}\sum_j (\cot\alpha_j + \cot\beta_j)( p_j - p_i )$

where $$p_j$$ is the coordinate of the $$j$$th neighbor of $$p_i$$ and $$\alpha_j$$ and $$\beta_j$$ are the angles across from edge $$(p_i,p_j)$$.

Mean Curvature Vector

The expression for the area gradient derived in the last exercise shows up all over discrete differential geometry, and is often referred to as the cotan formula. Interestingly enough this same expression appears when taking a completely different approach to defining vertex normals, by way of the mean curvature vector $$HN$$. In particular, for a smooth surface $$f: M \rightarrow \mathbb{R}^3$$ we have

$\Delta f = 2HN$

where $$\Delta$$ is the Laplace-Beltrami operator, $$H$$ is the mean curvature, and $$N$$ is the unit surface normal (which we’d like to compute). Therefore, another way to define vertex normals for a discrete surface is to simply apply a discrete Laplace operator to the vertex positions and normalize the resulting vector.

The question now becomes, “how do you discretize the Laplacian?” We’ll take a closer look at this question in the future, but the remarkable fact is that the most straightforward discretization of $$\Delta$$ leads us right back to the cotan formula! In other words, the vertex normals we get from the mean curvature vector are precisely the same as the ones we get from the area gradient.

This whole story also helps us get better intuition for $$\Delta$$ itself. Earlier, while studying exterior calculus, we saw that the (0-form) Laplacian can be expressed as $$\Delta = \star d \star d$$. But a different way to think about the Laplacian is simply as the sum of second partial derivatives, i.e.,

$\Delta \phi = d(d\phi(X))(X) + d(d\phi(Y))(Y),$

where $$X$$ and $$Y$$ are (unit) orthogonal directions.

What’s the geometric meaning here? Remember that for a good old-fashioned function $$\phi: \mathbb{R} \rightarrow \mathbb{R}$$ in 1D, second derivatives basically tell us about the curvature of a function, e.g., is it concave or convex?

Well, since $$\Delta$$ is a sum of second derivatives, it’s no surprise that it tells us something about the mean curvature!

Exercise 2.5 Show that the relationship $$\Delta f = 2HN$$ holds.

An alternative way to come up with normals is to look at the volume gradient. Suppose that our surface encloses some region of space with total volume $$\mathcal{V}$$. As before, we know that sliding the surface along itself tangentially doesn’t really change anything: we end up with the same surface, which encloses the same region of space. Therefore, the quickest way to increase $$\mathcal{V}$$ is to again move the surface in the normal direction. A somewhat surprising fact is that, in the discrete case, the volume gradient actually yields a different definition for vertex normals than the one we got from the area gradient. To express this gradient, we’ll use three-dimensional versions of of our “basic facts” from above.

First, much like we broke the area of a polygon into triangles, we’re going to decompose the volume enclosed by our surface into a collection of tetrahedra. Each tetrahedron includes exactly one face of our discrete surface, along with a new point $$q$$. For instance, here’s what the volume might look like in the vicinity of a vertex $$p$$:

Just as in the polygon case the location of $$q$$ makes no difference, as long as we work with the signed volume of the tetrahedra. (Can you prove it?)

Next, what’s the volume gradient for a single tetrahedron? One way to write the volume of a tet is as

$\mathcal{V} = \frac{1}{3} A h,$

where $$A$$ is the area of the base triangle and $$h$$ is the height. Then using the same kind of geometric reasoning as in the triangle case, we know that

$\nabla_p \mathcal{V} = \frac{1}{3} A N,$

where $$N$$ is the unit normal to the base.

To express the gradient of the enclosed volume with respect to a given vertex $$p$$, we simply sum up the gradients for the tetrahedra containing $$p$$:

$\nabla_p \mathcal{V} = \sum_i \mathcal{V}_i = \frac{1}{3} \sum_i A_i N_i.$

At first glance this sum does not lead to a nice expression for $$\Delta_p \mathcal{V}$$ — for instance, it uses the normals $$N_i$$ of faces that have little to do with our surface geometry. However, remember that we can place $$q$$ anywhere we please and still get the same expression for volume. In particular, if we put $$q$$ directly on top of $$p$$, then the $$N_i$$ and $$A_i$$ coincide with the normals and areas (respectively) of the faces containing $$p$$ from our original surface:

Exercise 2.6 Show that the volume gradient points in the same direction as the vector area $$N_\mathcal{V}$$ (i.e., show that they are the same up to a constant factor).

Other Definitions

So far we’ve only looked at definitions for vertex normals that arise from some smooth definition. This way of thinking captures the essential spirit of discrete differential geometry: relationships from the smooth setting should persist unperturbed in the discrete setting (e.g., $$\Delta f = 2HN$$ should be true independent of whether $$\Delta$$, $$H$$, and $$N$$ are smooth objects or discrete ones). Nonetheless, there are a number of common definitions for vertex normals that do not have a known origin in the smooth world. (Perhaps you can find one?)

Uniform Weighting

Perhaps the simplest way to get vertex normals is to just add up the neighboring face normals:

$N_U := \sum_i N_i$

The main drawback to this approach is that two different tessellations of the same geometry can produce very different vertex normals, as illustrated above.

Tip-Angle Weights

A simple way to reduce dependence on the tessellation is to weigh face normals by their corresponding tip angles $$\theta$$, i.e., the interior angles incident on the vertex of interest:

$N_\theta := \sum_i \theta_i N_i$

Sphere-Inscribed Polytope

Here’s another interesting approach to vertex normals: consider the sphere $$S^2$$ consisting of all points unit distance from the origin in $$\mathbb{R}^3$$. A nice fact about the sphere is that the unit normal $$N$$ at a point $$x \in S^2$$ is simply the point itself! I.e., $$N(x) = x$$. So if we start out with a polytope whose vertices all sit on the sphere, one reasonable way to define vertex normals is to simply use the vertex positions.

In fact, it’s not too hard to show that the direction of the normal at a vertex $$p_i$$ can be expressed purely in terms of the edge vectors $$e_j = p_j – p_i$$, where $$p_j$$ are the immediate neighbors of $$p_i$$. In particular, we have

$N_S = \frac{1}{c} \sum_{j=0}^{n-1} \frac{e_j \times e_{j+1}}{|e_j|^2 |e_{j+1}|^2}$

where the constant $$c \in \mathbb{R}$$ can be ignored since we’re only interested in the direction of the normal. (For a detailed derivation of this expression, see Max, “Weights for Computing Vertex Normals from Facet Normals.”) Of course, since this expression depends only on the edge vectors it can be evaluated on any mesh (not just those inscribed in a sphere).

Coding Assignment

Implement the following methods in libDDG:

• Vertex::normalEquallyWeighted()
Purpose: returns unit vertex normal using uniform weights $$N_U$$
• Vertex::normalAreaWeighted()
Purpose: returns unit vertex normal using face area weights $$N_\mathcal{V}$$
• Vertex::normalAngleWeighted()
Purpose: returns unit vertex normal using tip angle weights $$N_\theta$$
• Vertex::normalMeanCurvature()
Purpose: returns unit mean curvature normal $$\Delta f$$
• Vertex::normalSphereInscribed()
Purpose: returns unit sphere-inscribed normal $$N_S$$

(The definitions for these methods can be found in libddg/src/Vertex.cpp.)

Once you’ve successfully implemented these methods, test them out on the provided meshes. For convenience you can flip through the different methods using the keys [1]-[5]. You can also see what the mesh looks like by viewing it in “wireframe” mode. Do you notice that some definitions work better than others? When? Why? The only thing you need to submit for the coding assignment (via email) is your modified version of Vertex.cpp. (Of course, if you modify other files you should submit these too! For instance, you might find it convenient to add a method HalfEdge::cotan() that computes the cotangent of the angle across from a given half edge.)

Example meshes can be found here: ddg_hw2_meshes.zip

## 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.

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.

November 1, 2011 | Comments Closed

## 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!”)

November 5, 2011 | Comments Closed

## 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)!

November 9, 2011 | Comments Closed

## Homework 3: The Discrete Laplacian

In the course notes we mentioned that the Laplace-Beltrami operator (more commonly known as just the Laplacian) plays a fundamental role in a variety of geometric and physical equations. In this homework we’ll put the Laplacian to work by coming up with a discrete version for triangulated surfaces. Similar to the homework on vertex normals, we’ll see that the same discrete expression for the Laplacian (via the cotan formula) arises from two very different ways of looking at the problem: using test functions (often known as Galerkin projection), or by integrating differential forms (often called discrete exterior calculus).

Poisson Equations

Before we start talking about discretization, let’s establish a few basic facts about the Laplace operator $$\Delta$$ and the standard Poisson problem

$\Delta \phi = \rho.$

Poisson equations show up all over the place — for instance, in physics $$\rho$$ might represent a mass density in which case the solution $$\phi$$ would (up to suitable constants) give the corresponding gravitational potential. Similarly, if $$\rho$$ describes an charge density then $$\phi$$ gives the corresponding electric potential (you’ll get to play around with these equations in the code portion of this assignment). In geometry processing a surprising number of things can be done by solving a Poisson equation (e.g., smoothing a surface, computing a vector field with prescribed singularities, or even computing the geodesic distance on a surface).

Often we’ll be interested in solving Poisson equations on a compact surface $$M$$ without boundary.

Exercise 3.1: A twice-differentiable function $$\phi: M \rightarrow \mathbb{R}$$ is called harmonic if it sits in the kernel of the Laplacian, i.e., $$\Delta \phi = 0$$. Argue that the only harmonic functions on a compact domain without boundary are the constant functions. (This argument does not have to be incredibly formal — there are a just couple simple observations that capture the main idea.)

This fact is quite important because it means we can add a constant to any solution to a Poisson equation. In other words, if $$\phi$$ satisfies $$\Delta \phi = \rho$$, then so does $$\phi+c$$ since $$\Delta(\phi+c) = \Delta\phi + \Delta c = \Delta\phi + 0 = \rho$$.

Exercise 3.2: A separate fact is that on a compact domain without boundary, constant functions are not in the image of $$\Delta$$. In other words, there is no function $$\phi$$ such that $$\Delta \phi = c$$. Why?

This fact is also important because it tells us when a given Poisson equation admits a solution. In particular, if $$\rho$$ has a constant component then the problem is not well-posed. In some situations, however, it may make sense to simply remove the constant component. I.e., instead of trying to solve $$\Delta \phi = \rho$$ one can solve $$\Delta \phi = \rho – \bar{\rho}$$, where $$\bar{\rho} := \int_M \rho\ dV / |M|$$ and $$|M|$$ is the total volume of $$M$$. However, you must be certain that this trick makes sense in the context of your particular problem!

When working with PDEs like the Poisson equation, it’s often useful to have an inner product between functions. An extremely common inner product is the $$L^2$$ inner product $$\langle \cdot, \cdot \rangle$$, which takes the integral of the pointwise product of two functions over the entire domain $$\Omega$$:

$\langle f, g \rangle := \int_\Omega f(x) g(x) dx.$

In spirit, this operation is similar to the usual dot product on $$\mathbb{R}^n$$: it measures the degree to which two functions “line up.” For instance, the top two functions have a large inner product; the bottom two have a smaller inner product (as indicated by the dark blue regions):

Similarly, for two vector fields $$X$$ and $$Y$$ we can define an $$L^2$$ inner product

$\langle X, Y \rangle := \int_\Omega X(x) \cdot Y(x) dx$

which measures how much the two fields “line up” at each point.

Using the $$L^2$$ inner product we can express an important relationship known as Green’s first identity. Green’s identity says that for any sufficiently differentiable functions $$f$$ and $$g$$

$\langle \Delta f, g \rangle = -\langle \nabla f, \nabla g \rangle + \langle N \cdot \nabla f, g \rangle_\partial,$

where $$\langle \cdot, \cdot \rangle_\partial$$ denotes the inner product on the boundary and $$N$$ is the outward unit normal.

Exercise 3.3: Using exterior calculus, show that Green’s identity holds. Hint: apply Stokes’ theorem to the 1-form $$g df$$.

One last key fact about the Laplacian is that it is positive-semidefinite, i.e., $$\Delta$$ satisfies

$\langle \Delta \phi, \phi \rangle \geq 0$

for all functions $$\phi$$. (By the way, why isn’t this quantity strictly greater than zero?) Words cannot express the importance of positive-(semi)definiteness. Let’s think about a very simple example: functions of the form $$\phi(x,y) = ax^2 + bxy + cy^2$$ in the plane. Any such function can also be expressed in matrix form:

$\phi(x,y) = \underbrace{\left[ \begin{array}{cc} x & y \end{array} \right]}_{\mathbf{x}^T} \underbrace{\left[ \begin{array}{cc} a & b/2 \\ b/2 & c \end{array} \right]}_{A} \underbrace{\left[ \begin{array}{c} x \\ y \end{array} \right]}_{\mathbf{x}} = ax^2 + bxy + cy^2,$

and we can likewise define positive-semidefiniteness for $$A$$. But what does it actually look like? It looks like this — positive-definite matrices ($$\mathbf{x}^T A \mathbf{x} > 0$$) look like a bowl, positive-semidefinite matrices ($$\mathbf{x}^T A \mathbf{x} \geq 0$$) look like a half-cylinder, and indefinite matrices ($$\mathbf{x}^T A \mathbf{x}$$ might be positive or negative depending on $$\mathbf{x}$$) look like a saddle:

Now suppose you’re a back country skiier riding down this kind of terrain in the middle of a howling blizzard. You’re cold and exhausted, and you know you parked your truck in a place where the ground levels out, but where exactly is it? The snow is blowing hard and visibility is low — all you can do is keep your fingers crossed and follow the slope of the mountain as you make your descent. (Trust me: this is really how one feels when doing numerical optimization!) If you were smart and went skiing in Pos Def Valley then you can just keep heading down and will soon arrive safely back at the truck. But maybe you were feeling a bit more adventurous that day and took a trip to Semi Def Valley. In that case you’ll still get to the bottom, but may have to hike back and forth along the length of the valley before you find your car. Finally, if your motto is “safety second” then you threw caution to the wind and took a wild ride in Indef Valley. In this case you may never make it home!

In short: positive-definite matrices are nice because it’s easy to find the minimum of the quadratic functions they describe — many tools in numerical linear algebra are based on this idea. Same goes for positive definite linear operators like the Laplacian, which can often be thought of as sort of infinite-dimensional matrices (if you took some time to read about the spectral theorem, you’ll know that this analogy runs even deeper, especially on compact domains). Given the ubiquity of Poisson equations in geometry and physics, it’s a damn good thing $$\Delta$$ is positive-semidefinite!

Exercise 3.4: Using Green’s first identity, show that $$\Delta$$ is negative-semidefinite on any compact surface $$M$$ without boundary. From a practical perspective, why are negative semi-definite operators just as good as positive semi-definite ones?

Test Functions

The solution to a geometric or physical problem is often described by a function: the temperature at each point on the Earth, the curvature at each point on a surface, the amount of light hitting each point of your retina, etc. Yet the space of all possible functions is mind-bogglingly large — too large to be represented on a computer. The basic idea behind the finite element method is to pick a smaller set of functions and try to find the best possible solution from within this set. More specifically, if $$u$$ is the true solution to a problem and $$\{\phi_i\}$$ is a collection of basis functions, then we seek the linear combination of these functions

$\tilde{u} = \sum_i x_i \phi_i,\ x_i \in \mathbb{R}$

such that the difference $$||\tilde{u}-u||$$ is as small as possible with respect to some norm. (Above we see a detailed curve $$u$$ and its best approximation $$\tilde{u}$$ by a collection of bump-like basis functions $$\phi_i$$.)

Let’s start out with a very simple question: suppose we have a vector $$v \in \mathbb{R}^3$$, and want to find the best approximation $$\tilde{v}$$ within a plane spanned by two basis vectors $$e_1, e_2 \in \mathbb{R}^3$$:

Since $$\tilde{v}$$ is forced to stay in the plane, the best we can do is make sure there’s error only in the normal direction. In other words, we want the error $$\tilde{v} – v$$ to be orthogonal to both basis vectors $$e_1$$ and $$e_2$$:

$\begin{array}{rcl} (\tilde{v} - v) \cdot e_1 &=& 0, \\ (\tilde{v} - v) \cdot e_2 &=& 0. \\ \end{array}$

In this case we get a system of two linear equations for two unknowns, and can easily compute the optimal vector $$\tilde{v}$$.

Now a harder question: suppose we want to solve a standard Poisson problem

$\Delta u = f.$

How do we check whether a given function $$\tilde{u}$$ is the best possible solution? The basic picture still applies, except that our bases are now functions $$\phi$$ instead of finite-dimensional vectors $$e_i$$, and the simple vector dot product $$\cdot$$ gets replaced by the $$L^2$$ inner product. Unfortunately, when trying to solve a Poisson equation we don’t know what the correct solution $$u$$ looks like (otherwise we’d be done already!). So instead of the error $$\tilde{u} – u$$, we’ll have to look at the residual $$\Delta \tilde{u} – f$$, which measures how closely $$\tilde{u}$$ satisfies our original equation. In particular, we want to “test” that the residual vanishes along each basis direction $$\phi_j$$:

$\langle \Delta \tilde{u} - f, \phi_j \rangle = 0,$

again resulting in a system of linear equations. This condition ensures that the solution behaves just as the true solution would over a large collection of possible “measurements.”

Next, let’s work out the details of this system for a triangulated surface. The most natural choice of basis functions are the piecewise linear hat functions $$\phi_i$$, which equal one at their associated vertex and zero at all other vertices:

At this point you might object: if all our functions are piecewise linear, and $$\Delta$$ is a second derivative, aren’t we just going to get zero every time we evaluate $$\Delta u$$? Fortunately we’re saved by Green’s identity — let’s see what happens if we apply this identity to our triangle mesh, by breaking up the integral into a sum over individual triangles $$\sigma$$:

$\begin{array}{rcl} \langle \Delta u, \phi_j \rangle &=& \sum_k \langle \Delta u, \phi_j \rangle_{\sigma_k} \\ &=& \sum_k \langle \nabla u, \nabla \phi_j \rangle_{\sigma_k} + \sum_k \langle N \cdot \nabla u, \phi_j \rangle_{\partial\sigma_k}. \\ \end{array}$

If the mesh has no boundary then this final sum will disappear since the normals of adjacent triangles are oppositely oriented, hence the boundary integrals along shared edges cancel each-other out:

In this case, we’re left with simply

$\langle \nabla u, \nabla \phi_j \rangle$

in each triangle $$\sigma_k$$. In other words, we can “test” $$\Delta u$$ as long as we can compute the gradients of both the candidate solution $$u$$ and each basis function $$\phi_j$$. But remember that $$u$$ is itself a linear combination of the bases $$\phi_i$$, so we have

$\langle \nabla u, \nabla \phi_j \rangle = \left\langle \nabla \sum_i x_i \phi_i, \nabla \phi_j \right\rangle = \sum_i x_i \langle \nabla \phi_i, \nabla \phi_j \rangle.$

The critical thing now becomes the inner product between the gradients of the basis functions in each triangle. If we can compute these, then we can simply build the matrix

$A_{ij} := \langle \nabla \phi_i, \nabla \phi_j \rangle$

and solve the problem

$A x = b$

for the coefficients $$x$$, where the entries on the right-hand side are given by $$b_i = \langle f, \phi_i \rangle$$ (i.e., we just take the same “measurements” on the right-hand side).

Exercise 3.5: Show that the aspect ratio of a triangle can be expressed as the sum of the cotangents of the interior angles at its base, i.e.,

$\frac{w}{h} = \cot{\alpha} + \cot{\beta}.$

Exercise 3.6: Let $$e$$ be the edge vector along the base of a triangle. Show that the gradient of the hat function $$\phi$$ on the opposite vertex is given by

$\nabla \phi = \frac{e^\perp}{2A},$

where $$e^\perp$$ is the vector $$e$$ rotated by a quarter turn in the counter-clockwise direction and $$A$$ is the area of the triangle.

Exercise 3.7: Show that for any hat function $$\phi$$ associated with a given vertex

$\langle \nabla \phi, \nabla \phi \rangle = \frac{1}{2}( \cot\alpha + \cot\beta )$

within a given triangle, where $$\alpha$$ and $$\beta$$ are the interior angles at the remaining two vertices.

Exercise 3.8: Show that for the hat functions $$\phi_i$$ and $$\phi_j$$ associated with vertices $$i$$ and $$j$$ (respectively) of the same triangle, we have

$\langle \nabla \phi_i, \nabla \phi_j \rangle = -\frac{1}{2} \cot \theta,$

where $$\theta$$ is the angle between the opposing edge vectors.

Putting all these facts together, we find that we can express the Laplacian of $$u$$ at vertex $$i$$ via the infamous cotan formula

$(\Delta u)_i = \frac{1}{2} \sum_j (\cot \alpha_j + \cot \beta_j )( u_j - u_i ),$

where we sum over the immediate neighbors of vertex $$i$$.

Differential Forms

The “Galerkin” approach taken above reflects a fairly standard way to discretize partial differential equations. But let’s try a different approach, based on exterior calculus. Again we want to solve the Poisson equation $$\Delta u = f$$, which (if you remember our discussion of differential operators) can also be expressed as

$\star d \star d u = f.$

We already outlined how to discretize this kind of expression in the notes on discrete exterior calculus, but let’s walk through it step by step. We start out with a 0-form $$u$$, which is specified as a number $$u_i$$ at each vertex:

Next, we compute the discrete exterior derivative $$d u$$, which just means that we want to integrate the derivative along each edge:

$(du)_{ij} = \int_{e_{ij}}\!du = \int_{\partial e_{ij}}\!u = u_j - u_i.$

(Note that the boundary $$\partial e_{ij}$$ of the edge is simply its two endpoints $$v_i$$ and $$v_j$$.) The Hodge star converts a circulation along the edge $$e_{ij}$$ into the flux through the corresponding dual edge $$e^\star_{ij}$$. In particular, we take the total circulation along the primal edge, divide it by the edge length to get the average pointwise circulation, then multiply by the dual edge length to get the total flux through the dual edge:

$(\star du)_{ij} = \frac{|e^\star_{ij}|}{e_{ij}}(u_j - u_i).$

Here $$|e_{ij}|$$ and $$e^\star_{ij}$$ denote the length of the primal and dual edges, respectively. Next, we take the derivative of $$\star d u$$ and integrate over the whole dual cell:

$(d \star d u)_i = \int_{C_i} d \star d u = \int_{\partial C_i} \star d u = \sum_j \frac{|e^\star_{ij}|}{|e_{ij}|}( u_j - u_i ).$

The final Hodge star simply divides this quantity by the area of $$C_i$$ to get the average value over the cell, and we end up with a system of linear equations

$(\star d \star d u)_i = \frac{1}{|C_i|} \sum_j \frac{|e^\star_{ij}|}{|e_{ij}|}( u_j - u_i ) = f_i$

where $$f_i$$ is the value of the right-hand side at vertex $$i$$. In practice, however, it’s often preferable to move the area factor $$|C_i|$$ to the right hand side, since the resulting system

$(\star d \star d u)_i = \sum_j \frac{|e^\star_{ij}|}{|e_{ij}|}( u_j - u_i ) = |C_i| f_i$

can be represented by a symmetric matrix. (Symmetric matrices are often easier to deal with numerically and lead to more efficient algorithms.) Another way of looking at this transformation is to imagine that we discretized the system

$d \star d u = \star f.$

In other words, we converted an equation in terms of 0-forms into an equation in terms of $$n$$-forms. When working with surfaces, the operator $$d \star d$$ is sometimes referred to as the conformal Laplacian, because it does not change when we subject our surface to a conformal transformation. Alternatively, we can think of $$d \star d$$ as simply an operator that gives us the value of the Laplacian integrated over each dual cell of the mesh (instead of the pointwise value).

Exercise 3.9: Consider a simplicial surface and suppose we place the vertices of the dual mesh at the circumcenters of the triangles (i.e., the center of the unique circle containing all three vertices):

Demonstrate that the dual edge $$e^\star$$ (i.e., the line between the two circumcenters) bisects the primal edge orthogonally, and use this fact to show that

$\frac{|e^\star_{ij}|}{|e_{ij}|} = \frac{1}{2}(\cot \alpha_j + \cot \beta_j).$

Hence the DEC discretization yields precisely the same “cotan-Laplace” operator as the Galerkin discretization.

Meshes and Matrices

So far we’ve been giving a sort of “algorithmic” description of operators like Laplace. For instance, we determined that the Laplacian of a scalar function $$u$$ at a vertex $$i$$ can be approximated as

$(\Delta u)_i = \frac{1}{2} \sum_j (\cot\alpha_j+\cot\beta_j)(u_j - u_i),$

where the sum is taken over the immediate neighbors $$j$$ of $$i$$. In code, this sum could easily be expressed as a loop and evaluated at any vertex. However, a key aspect of working with discrete differential operators is building their matrix representation. The motivation for encoding an operator as a matrix is so that we can solve systems like

$\Delta u = f$

using a standard numerical linear algebra package. (To make matters even more complicated, some linear solvers are perfectly happy to work with algorithmic representations of operators called callback functions — in general, however, we’ll need a matrix.)

In the case of the Poisson equation, we want to construct a matrix $$L \in \mathbb{R}^{|V| \times |V|}$$ (where $$|V|$$ is the number of mesh vertices) such that for any vector $$u \in \mathbb{R}^{|V|}$$ of values at vertices, the expression $$Lu$$ effectively evaluates the formula above. But let’s start with something simpler — consider an operator $$B$$ that computes the sum of all neighboring values:

$(Bu)_i = \sum_j u_j$

How do we build the matrix representation of this operator? Think of $$B$$ a machine that takes a vector $$u$$ of input values at each vertex and spits out another vector $$Bu$$ of output values. In order for this story to make sense, we need to know which values correspond to which vertices. In other words, we need to assign a unique index to each vertex of our mesh, in the range $$1, \ldots, |V|$$:

It doesn’t matter which numbers we assign to which vertices, so long as there’s one number for every vertex and one vertex for every number. This mesh has twelve vertices and vertex 1 is next to vertices 2, 3, 4, 5, and 9. So we could compute the sum of the neighboring values as

$(Bu)_1 = \left[ \begin{array}{cccccccccccc} 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{array} \right]\left[ \begin{array}{c} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \\ u_7 \\ u_8 \\ u_9 \\ u_{10} \\ u_{11} \\ u_{12} \end{array} \right].$

Here we’ve put a “1″ in the $$j$$th place whenever vertex $$j$$ is a neighbor of vertex $$i$$ and a “0″ otherwise. Since this row gives the “output” value at the first vertex, we’ll make it the first row of the matrix $$B$$. The entire matrix looks like this:

$B = \left[ \begin{array}{cccccccccccc} 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ \end{array} \right]$

(You could verify that this matrix is correct, or you could go outside and play in the sunshine. Your choice.) In practice, fortunately, we don’t have to build matrices “by hand” — we can simply start with a matrix of zeros and fill in the nonzero entries by looping over local neighborhoods of our mesh.

Finally, one important thing to notice here is that many of the entries of $$B$$ are zero. In fact, for most discrete operators the number of zeros far outnumbers the number of nonzeros. For this reason, it’s usually a good idea to use a sparse matrix, i.e., a data structure that stores only the location and value of nonzero entries (rather than explicitly storing every single entry of the matrix). The design of sparse matrix data structures is an interesting question all on its own, but conceptually you can imagine that a sparse matrix is simply a list of triples $$(i,j,x)$$ where $$i,j \in \mathbb{N}$$ specify the row and column index of a nonzero entry and $$x \in \mathbb{R}$$ gives its value.

Scalar Poisson Equation

In the first part of the coding assignment you’ll build the cotan-Laplace operator and use it to solve the scalar Poisson equation

$\Delta \phi = \rho$

on a triangle mesh, where $$\rho$$ can be thought of as a (mass or charge) density and $$\phi$$ can be thought of as a (gravitational or electric) potential. Once you’ve implemented the methods below, you can visualize the results via the libDDG GUI. (If you want to play with the density function $$\rho$$, take a look at the method Viewer::updatePotential.)

Coding 3.1: Implement the method Mesh::indexVertices() which assigns a unique ID to each vertex in the range $$0, \ldots, |V|-1$$.

Coding 3.2: Derive an expression for the cotangent of a given angle purely in terms of the two incident edge vectors and the standard Euclidean dot product $$(\cdot)$$ and cross product $$(\times)$$. Implement the method HalfEdge::cotan(), which computes the cotangent of the angle across from a given halfedge.

Coding 3.3: Implement the methods Face::area() and Vertex::dualArea(). For the dual area of a vertex you can simply use one-third the area of the incident faces — you do not need to compute the area of the circumcentric dual cell. (This choice of area will not affect the order of convergence.)

Coding 3.4: Using the methods you’ve written so far, implement the method Mesh::buildLaplacian() which builds a sparse matrix representing the cotan-Laplace operator. (Remember to initialize the matrix to the correct size!)

Coding 3.5: Implement the method Mesh::solveScalarPoissonProblem() which solves the problem $$\Delta\phi = \rho$$ where $$\rho$$ is a scalar density on vertices (stored in Vertex::rho). You can use the method solve from SparseMatrix.h; $$\rho$$ and $$\phi$$ should be represented by instances of DenseMatrix of the appropriate size. Be careful about appropriately incorporating dual areas into your computations; also remember that the right-hand side cannot have a constant component!

You should verify that your code produces results that look something like these two images (density on the left; corresponding potential on the right):

(Take a look at the new menu items in the GUI.)

Implicit Mean Curvature Flow

Next, you’ll use nearly identical code to smooth out geometric detail on a surface mesh (also known as fairing or curvature flow). The basic idea is captured by the heat equation, which describes the way heat diffuses over a domain. For instance, if $$u$$ is a scalar function describing the temperature at every point on the real line, then the heat equation is given by

$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}.$

Geometrically this equation simply says that concave bumps get pushed down and convex bumps get pushed up — after a long time the heat distribution becomes completely flat. We also could have written this equation using the Laplacian: $$\frac{\partial u}{\partial t} = \Delta u$$. In fact, this equation is exactly the one we’ll use to smooth out a surface, except that instead of considering the evolution of temperature, we consider the flow of the surface $$f: M \rightarrow \mathbb{R}^3$$ itself:

$\frac{\partial f}{\partial t} = \Delta f.$

Remember from our discussion of vertex normals that $$\Delta f = 2HN,$$ i.e., the Laplacian of position yields (twice) the mean curvature times the unit normal. Therefore, the equation above reads, “move the surface in the direction of the normal, with strength proportional to mean curvature.” In other words, it describes a mean curvature flow.

So how do we compute this flow? We already know how to discretize the term $$\Delta f$$ — just use the cotangent discretization of Laplace. But what about the time derivative $$\frac{\partial f}{\partial t}$$? There are all sorts of interesting things to say about discretizing time, but for now let’s use a very simple idea: the change over time can be approximated by the difference of two consecutive states:

$\frac{\partial f}{\partial t} \approx \frac{f_h - f_0}{h},$

where $$f_0$$ is the initial state of our system (here the initial configuration of our mesh) and $$f_h$$ is the configuration after a mean curvature flow of some duration $$h > 0$$. Our discrete mean curvature flow then becomes

$\frac{f_h - f_0}{h} = \Delta f.$

The only remaining question is: which values of $$f$$ do we use on the right-hand side? One idea is to use $$f_0$$, which results in the system

$f_h = f_0 + h\Delta f_0.$

This scheme, called forward Euler, is attractive because it can be evaluated directly using the known data $$f_0$$ — we don’t have to solve a linear system. Unfortunately, forward Euler is not numerically stable, which means we can take only very small time steps $$h$$. One attractive alternative is to use $$f_h$$ as the argument to $$\Delta$$, leading to the system

$\underbrace{(I - h \Delta)}_A f_h = f_0,$

where $$I$$ is the identity matrix (try the derivation yourself!) This scheme, called backward Euler, is far more stable, though we now have to solve a linear system $$A f_h = f_0$$. Fortunately this system is highly sparse, which means it’s not too expensive to solve in practice. (Note that this system is not much different from the Poisson system.)

Coding 3.6: Implement the method Mesh::buildFlowOperator(), which should look very similar to Mesh::buildLaplacian.

Coding 3.7: Implement the method Mesh::computeImplicitMeanCurvatureFlow(). Note that you can treat each of the components of $$f$$ ($$x$$, $$y$$, and $$z$$) as separate scalar quantities.

You should verify that your code produces results that look something like these two images (original mesh on the left; smoothed mesh on the right):

Skeleton Code and Handin

Skeleton code for this assignment can be found here:

ddg_poisson.zip

(You may want to use your Makefile from the last project!) You can easily locate all the methods that need to be implemented by searching for the string “TODO.” Please submit the entire project this time as a single zip file (i.e., not just the files you modify).