# Lecture 18—Laplace Beltrami

In the next two lectures we’ll take a deep dive into one of the most important objects not only in discrete differential geometry, but in differential geometry at large (not to mention physics!): the Laplace-Beltrami operator. This operator generalizes the familiar Laplacian you may have studied in vector calculus, which just gives the sum of second partial derivatives: $\Delta \phi = \sum_i \partial^2 \phi_i / \partial x_i^2$. We’ll motivate Laplace-Beltrami from several points of view, talk about how to discretize it, and show how from a computational point of view it really is the —Swiss army knife— of geometry processing algorithms, essentially replacing the discrete Fourier transform from classical signal processing.

You can find a video lecture for these slides, from a talk given by Etienne Vouga, here.

## 6 thoughts on “Lecture 18—Laplace Beltrami”

1. Lqq88888 says:

Do we have video lectures for these slides?

1. Keenan says:

See this page for an updated discussion of the smooth setting, including a video. There’s also a video that covers the original slides, including the discrete side of things, recorded by one of my colleagues (Etienne Vouga), which you can find here.

2. Lqq88888 says:

For $\int_M{f\nabla f }dA$, should I realize it as a quadratic form?
If so, should I take f as $(a_0, a_1, a_2, …)$ with the Laplacian spectrum as its basis?
Or should I take f as $(f(p_1), f(p_2), …)$ with indicator functions ${\{\mathbb I_{\{p\}}: p \in M}\}$? as its basis?

1. Keenan says:

Great question.

First of all, you have to be a careful with these little triangles: $\nabla$ denotes the gradient, whereas $\Delta$ denotes the Laplacian. You can type the former in TeX as \nabla, and the latter as \Delta. The Dirichlet energy of a function $u$ on a closed surface $M$ can then be expressed as

$E_D(u) := \int_M u \Delta u\ dA.$

Second, this energy is indeed a quadratic form—and it’s important to realize that you do not need a basis to define a quadratic form. Avoiding discussion of a basis allows you to easily define (and understand) quadratic forms even in the “infinite dimensional” case, i.e., even for functions like $E_D$ whose domain is a continuous function, rather than a finite-dimensional vector. In particular, the Dirichlet energy is a quadratic form in the sense that

$E_D(cu) = c^2 u$

for any constant $c \in \mathbb{R}$, and the expression

$E_D(u+v) – E_D(u) – E_D(v)$

is linear in each of $u$ and $v$, i.e., it is bilinear. Nothing more is needed.

Hope that helps, and let us know if you have any other questions.

1. Lqq88888 says:

Thanks! I typoed on $\Delta$. Actually I asked that question because I was wondering how can I intuitively imagine $\Delta$ as a super large matrix. I mean since a baby quadratic form is like $x^{T}Ax$, maybe it’s also possible to imagine $\int_{\Omega}{f\Delta f}dA$ as something similar?

1. Keenan says:

Yeah, if you really want to think of Dirichlet energy in terms of a matrix then there are two good ways to go. One is to think of it in terms of the discrete Laplacian, or more precisely, the cotan matrix $A_{ij}$ (which does not include the mass matrix, i.e., it captures $d \ast d$ rather than $\ast d \ast d$). Then the Dirichlet energy is

$E_D(u) = \frac{1}{2} u^T A u,$

where $u \in \mathbb{R}^{|V|}$ is a vector storing one value per vertex. A different way is to indeed consider, in the smooth setting, eigenfunctions of the Laplace-Beltrami operator, i.e., functions $\phi_i$ satisfying

$\Delta \phi_i = \lambda_i u_i$

for associated eigenvalues $\lambda_i \in \mathbb{R}$. These eigenfunctions form an orthonormal basis for $L^2(M;\mathbb{R})$, i.e., any square-integrable function $u: M \to \mathbb{R}$ can be expressed as an infinite sum

$u = \sum_{i=1}^\infty \underbrace{\langle\!\langle u, \phi_i \rangle\!\rangle}_{c_i \in \mathbb{R}} \phi_i,$

where $\langle\!\langle \cdot, \cdot \rangle\!\rangle$ denotes the $L^2$ inner product

$\langle\!\langle u, v \rangle\!\rangle := \int_M u(x) v(x) dx.$

The values $c_i$ are the “coordinates” of the function $u$ in the infinite spectral basis $\{\phi_1, \phi_2, \ldots\}$.

It’s actually kind of fun to write out the Dirichlet energy in this setting. Assuming $M$ has no boundary, we have
$\begin{array}{rcl} E_D(u) &=& \int_M |\nabla u|^2\ dA = -\int_M u \Delta u\ dA \\ &=& -\int_M u \Delta \sum_{i=1}^\infty c_i \phi_i\ dA \\ &=& -\int_M u \sum_{i=1}^\infty c_i \Delta \phi_i\ dA \\ &=& -\int_M u \sum_{i=1}^\infty c_i \lambda_i \phi_i\ dA, \\ \end{array}$
i.e., if we expand $u$ in the eigenbasis, then all $\Delta$ does is multiply each component by the associated eigenvalue. We can take this calculation a step further by expanding the other instance of $u$ in the eigenbasis. Since the basis is orthonormal, we get just
$E_D(u) = \sum_{i=1}^\infty \lambda c_i^2.$
So, in the eigenbasis, the Dirichlet energy is nothing more than the squared norm of the function, but with each component weighted by the eigenvalues of $\Delta$. Since higher frequencies will have larger eigenvalues, that means that the Dirichlet energy will penalize rapid oscillation much more than low-frequency variation. Finally, if you want to think of Dirichlet energy as a matrix (which I think was your original goal), you can now think of it in terms of a diagonal “matrix”
$A = \left[ \begin{array}{ccc} \lambda_1 & & \\ & \lambda_2 & \\ & & \ddots \end{array} \right]$
that goes on forever. The Dirichlet energy itself then looks (conceptually) like
$c^T A c,$
where $c$ is the (infinite) vector of all the coefficients $c_i$. A common thing to do in practice is to truncate this expansion, and work only with the first $k$ eigenvectors of the discrete Laplace matrix, rather than the infinite set of eigenfunctions of the Laplace-Beltrami operator. The nice thing about this treatment is that it allows one to control how much computation is done, by selecting the cutoff $k$. This idea is the basis of spectral geometry processing, which I mention in the lecture.