Let $M= \left(V, E, F \right)$ be an oriented triangulated surface without boundary and $p:V \rightarrow \mathbb{R}^3$ a realization. In earlier lectures we considered the space of piecewise linear functions on $M$:

\[W_{PL}:=\left\{ \tilde{f} :M\rightarrow \mathbb{R} \, \big \vert \,\left. \tilde{f}\right|_{\sigma} \mbox{ is affine for all } \sigma \in F \right\}.\]

A basis of $W_{PL}$ is given by : $\phi_i \in W_{PL}$ defined by $\phi_i(p_j) = \delta_{ij},$ an interpolation map is defined by:

\begin{align*} \mbox{int}_{PL} : \mathbb{R}^{V} & \rightarrow W_{PL}, \\

f & \mapsto \tilde{f} = \sum_{i \in V} f_i \phi_i.

\end{align*}

For some purposes piecewise constant function are more convenient then piecewise linear ones, therefore, we introduce a new function space on $M$ that consists of piecewise constant functions. In order to do that we assign a domain to each vertex:

For every edge $e_{ij} \in E$ the center is given by $S_{ij}= \frac{1}{2}\left( p_i + p_j \right)$ analog for every triangle $\sigma = \{i,j,k\} \in F$ the center of mass is defined as $S_{ijk} = \frac{1}{3} \left( p_i + p_j + p_k\right) $.

If we connect the center of the edges with the center of mass the triangle is divided into three parts with equal area. Now we can assign to each vertex $p_i$ the parts of the adjacent triangles that contain $p_i$ and denote it with $D_i$.

For the area of $D_i$ we obtain:

\[A_i := \frac{1}{3} \sum_{ \sigma = \{i,j,k\} \in F} A_{\sigma}. \]

The space of piecewise constant functions is defined in the following way:

\[W_{PC}:= \left\{\hat{f} :M \rightarrow \mathbb{R} \, \big \vert \, \left. \hat{f}\right|_{D_i} \mbox{ is constant for all } i \in V \right\}.\]

Also on $W_{PC}$ we have a canonical basis $\Psi_i \in W_{PC}$ defined by $\Psi_i(p_j) = \delta_{ij} $ and an interpolation map:

\begin{align*} \mbox{int}_{PC} : \mathbb{R}^{V} & \rightarrow W_{PC}, \\

f & \mapsto \hat{f} = \sum_{i \in V} f_i \Psi_i.

\end{align*}

There is a similar construction that assigns to each vertex an face and works purely combinatorial: For a triangulated $M$ surface one can define its discrete dual surface $M^{\ast}=\left( V^{\ast},E^{\ast},F^{\ast}\right)$. The faces of $M$ correspond to the vertices of $M^{\ast}$ and the vertices of $M$ to the faces of $M^{\ast}$. The set of edges stays the same. Two vertices in $M^{\ast}$ are connected by an edge if the corresponding faces in $M$ are adjacent. For more informations see here.

Since in general it is not easy to compute the area of the dual faces with the metric of the original one, we used the above construction of the $D_i$’s to define $W_{PC}$. On $W_{PL}$ we defined the usual scalar product $ \langle\!\langle \tilde{f}, \tilde{g} \rangle\!\rangle_{PL} := \int_M \tilde{f}\tilde{g}$ and computed the matrix $B$ representing it with respect to the basis $\left\{ \phi_1,\ldots \phi_N \right\},$ i.e. $b_{ij}:=\langle\!\langle\phi_i, \phi_j \rangle\!\rangle_{PL}.$ This gave us a scalar product on $\mathbb{R}^V$:

\[\langle\!\langle f, g \rangle\!\rangle_{PL} = f^tBg = \sum_{i,j \in V} f_i g_j b_{ij}.\]

On the same way we define an scalarproduct on $W_{PC}$:

\begin{align*}\langle\!\langle\cdot, \cdot \rangle\!\rangle_{PC} &: W_{PC}\times W_{PC} \rightarrow \mathbb{R}, \\ \langle\!\langle \hat{f}, \hat{g} \rangle\!\rangle_{PC} & := \int_M\hat{f}\hat{g}.\end{align*}

The matrix $A$ representing it with respect to the basis $\left\{ \Psi_1,\ldots \Psi_N \right\}$ is a diagonal matrix since:

\[a_{ij}:=\langle\!\langle\Psi_i, \Psi_j \rangle\!\rangle_{PC}\ = \int_M \Psi_i \Psi_j = A_i \delta_{ij}.\]

This gives us a new scalar product on $\mathbb{R}^V:$

\[\langle\!\langle f, g \rangle\!\rangle_{PC} := f^tAg = \sum_{i \in V} f_i g_i a_{ii}.\]

Note that the basis $\left\{ \Psi_1,\ldots \Psi_N \right\}$ is orthogonal with respect to $\langle\!\langle \cdot, \cdot \rangle\!\rangle_{PC}$, while the same is not true for $\left\{ \phi_1,\ldots \phi_N \right\}$ with respect to $\langle\!\langle \cdot, \cdot \rangle\!\rangle_{PL}$.

Now we have two scalar products on $\mathbb{R}^V$ one represented by a symmetric positive definite matrix $B$ and one by a positive definite diagonal one $A$.

Our aim is to define a discrete laplace operator $\Delta:\mathbb{R}^V \rightarrow \mathbb{R}^V$ analog to the smooth laplacian we discussed earlier. We will see that the discrtelaplacian depends on the choice of scalarproduct.

On a smooth surface $M$ we have:

\[\langle\!\langle \Delta f, g \rangle\!\rangle = \int_M \Delta f g =\, – \int_M \langle \mbox{grad}\, f , \mbox{grad}\, g \rangle.\]

This identity should also be true in the discrete case and will serve us as definition for the discrete laplacian. We already defined the bilinear operator

\begin{align} L: \mathbb{R}^V \times \mathbb{R}^V & \rightarrow \mathbb{R},\\ f,g & \mapsto \, – \int_M \langle \mbox{grad}\, f , \mbox{grad}\, g \rangle ,\end{align}

and discussed the relation between quadratic forms, symmetric bilinear maps and maps between dual spaces ( lecture: dirichlet energy 2 ).

\[\begin{array}{lcl} W_{PL} & \overset{E_D}{\longrightarrow} & W_{PL}^{\ast} \\ \uparrow \mbox{interpol.} & & \downarrow \mbox{interpol.}^{\ast} \\ \mathbb{R}^V & \overset{L}{\longrightarrow} & \mathbb{R}^{V\ast} \end{array} \]

In the finite case the quadratic form, bilinear form and map between the dual spaces can all be represented by the same matrix $L$ .

For $f,g \in \mathbb{R}^V$ we have:

\[f^tLg = \,- \int_M \langle \mbox{grad}\, f , \mbox{grad}\, g \rangle\overset{!}{=} \int_M \Delta f g = \langle\!\langle \Delta f, g \rangle\!\rangle .\]

This gives us one laplace operator for each scalarproduct on $\mathbb{R}^V$:

\begin{align} f^tLg = \left\{ \begin{array} {cc} \left(\Delta_{PL} f\right)^tBg, \\ \left(\Delta_{PC} f\right)^tAg,\end{array} \right. \\ \Rightarrow \left\{ \begin{array} {cc} \Delta_{PL} = \left(LB^{-1} \right)^t = B^{-1}L , \\ \Delta_{PC} = \left(LA^{-1} \right)^t = A^{-1}L. \end{array} \right. \end{align}

In general it is complicated and expensive to compute $B^{-1}$, such that it is much easier to use $\Delta_{PC}$ instead of $\Delta_{PL}$. Witch of the laplacians one has to choose depends on the nature of the problem.

One famous problem form physics and geometry is the *Poisson problem*:

Given $\rho \in \mathbb{R}^V$ find $u \in \mathbb{R}^V$ such that:

\[\Delta \,u = -\rho.\]

For Poisson problems both laplacians can be used since:

\[\Delta_{PL} u = \left(LB^{-1} \right)^t u = \,- \rho \Leftrightarrow Lu = -B\rho. \]

**Example: **Suppose $M$ consists of an uniform conducting material and $\rho: M \rightarrow \mathbb{R}$ is a charge density. For the electric field $E$ on $M$ there holds:

\[\mbox{div} E = \rho,\]

and for the potential $u: M \rightarrow \mathbb{R}$ we have

\[E=-\mbox{grad}\, u.\]Taking this together we have to solve:

\[\Delta \,u = -\rho,\]

in order to determine the potential for a given density.

In this setting piecewise constant functions are much more convenient to describe the charge density then piecewise linear ones. Therefore we use $\Delta_{PL} $ to solve this problem.