DDG2019

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*}

The graph of $\phi_i$

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 $D_i$.

The domain $D_i$ (gray colored) associated to the vertex $p_i$

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*}

Graph of $\Psi_i$

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.

A piece of an triangulated surface $M$ and its dual $M^{\ast}$. The red vertices and edges belong to $M$ and the blues ones to $M^{\ast}$.

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.

Let $M$ be a triangulated domain with the functionspace:
\[W_{PL}:=\bigl\{f:M\rightarrow\mathbb{R}\,\bigl\vert\bigr.\,\,\,\left. f\right|_{T_{\sigma}} \mbox{ is affine for all } \sigma \in \Sigma_2 \bigr\}.\]
On the interior of each triangle $T_{\sigma}$ in $M$ the gradient of a function $g \in W$ is well defined and constant. We will derive a nice formula to compute $\left. \mbox{grad}\;g \right|_{T_{\sigma}}$.

Theorem 1: Let $T \subset \mathbb{R}^2$ be a non degenerated triangle with vertices $p_i,p_j,p_k$ and edges $e_i :=p_k-p_j$, $e_j :=p_i-p_k$, $e_k :=p_j-p_i$. For an affine function $g:T\rightarrow \mathbb{R}$ we define $g_i:=g(p_i)$, $g_j:=g(p_j)$, $g_k:=g(p_k)$. Then there holds:
\[\mbox{grad}\, g = \frac{1}{2A}J\bigl( g_i e_i+g_j e_j+ g_k e_k\bigr),\]
where $A:=\frac{1}{2}\mbox{det}(e_i,e_j) = \frac{1}{2} \langle Je_i,e_j \rangle=\frac{1}{2}\langle Je_j,e_k \rangle=\frac{1}{2}\langle Je_k,e_i \rangle,$ denotes the area of $T$ and $J$ the usual $90$ degree rotation in the plan.

Proof: In order to compute $\mbox{grad}\, g$ consider the curve
\begin{align*} & \gamma : [0,1] \rightarrow \mathbb{R}^2,\\ & \gamma(t)  :=p_j + te_i. \end{align*}
The composition of $g$ and $\gamma$ is given by: $\left( g\circ \gamma\right)(t) = g_j + t\left( g_k – g_j\right)$. Taking the derivative direct and with respect to the product rule we get:
\[g_k -g_j = \left( g\circ \gamma\right)'(t) = g’\left(\gamma(t)\right) \cdot\gamma'(t) = \langle \mbox{grad} \, g, \gamma'(t) \rangle = \langle \mbox{grad} \, g,e_i \rangle. \]
On the other hand we have:
\begin{align*}\bigl\langle\frac{1}{2A}J\left( g_i e_i+g_j e_j+ g_k e_k\right),e_i \bigr\rangle & = \frac{1}{2A} \bigl(g_i \underbrace{\langle Je_i,e_i\rangle }_{=0} + g_j \underbrace{\langle Je_j,e_i\rangle}_{=-2A} + g_k \underbrace{\langle Je_k,e_i\rangle}_{= 2A}\bigr) \\ &= g_k -g_j.\end{align*}
And therfore:
\[\bigl\langle\frac{1}{2A}J\left( g_i e_i+g_j e_j+ g_k e_k\right),e_i \bigr\rangle = \bigl\langle \mbox{grad} \, g,e_i \bigr\rangle.\]
A cyclic permutation of $i,j,k$  gives us:
\[\bigl\langle\frac{1}{2A}J\left( g_i e_i+g_j e_j+ g_k e_k\right),e_j \bigr\rangle = \bigl\langle \mbox{grad} \, g,e_j \bigr\rangle,\]
and we obtain the desired result \[\mbox{grad}\, g = \frac{1}{2A}J\bigl( g_i e_i+g_j e_j+ g_k e_k\bigr).\]

$\square$

Definition: For $g \in W_{PL}$ the Dirichlet energy is defined as:
\[E_D(g) := \frac{1}{2} \int_M \left| \mbox{grad} \, g \right|^2.\]

Theorem 2: We use the same notation as in theorem 1. Additionally let $\alpha_i,\,\alpha_j,\, \alpha_k$ denote the angles of the triangle at the vertices $p_i, \, p_j, \, p_k$ respectively. Then there holds:
\[ \int_T \left| \mbox{grad} \, g \right|^2 = \frac{1}{2}\bigl( \cot (\alpha_i) (g_j-g_k) ^2 + \cot (\alpha_j) (g_k-g_i) ^2 + \cot (\alpha_k) (g_i-g_j) ^2\bigr).\]

Proof: \begin{multline}\int_T \left| \mbox{grad} \, g \right|^2 = \int_T \langle \frac{1}{2A}J\left( g_i e_i+g_j e_j+ g_k e_k\right), \frac{1}{2A}J\left( g_i e_i+g_j e_j+ g_k e_k\right) \rangle \\ =\frac{A}{4 A^2} \Bigl(\Bigr. g_i^2 |e_i|^2 +  g_j^2 |e_j|^2 + g_k^2 |e_k|^2 + 2g_ig_j \langle e_i,e_j \rangle + 2g_jg_j \langle e_j,e_k \rangle + 2g_kg_i \langle e_k,e_i \rangle \Bigl)\Bigr..  \end{multline}
On the other hand we have:
\begin{align}\frac{-1}{4A} \bigl(\bigr. \left( g_i -g_j\right)^2 & \langle e_i, e_j \rangle +\left( g_j -g_k\right)^2 \langle e_j, e_k \rangle +\left( g_k -g_i\right)^2 \langle e_k, e_i \rangle \bigl)\bigr. \\ &= \frac{-1}{4A} \bigl(\bigr. g_i^2 \langle e_i,\underbrace{e_j + e_k}_{= – e_i}\rangle -2g_ig_j\langle e_i,e_j \rangle + g_j^2 \langle e_j, \underbrace{e_i + e_k}_{= – e_j}\rangle \\ &\quad\quad\quad -2g_kg_j\langle e_k,e_j \rangle +g_k^2 \langle e_k, \underbrace{e_j + e_i}_{=- e_k}\rangle -2g_ig_k\langle e_i,e_k \rangle \bigl)\bigr. \\ &= \frac{1}{4 A} \bigl(\bigr. g_i^2 |e_i|^2 +  g_j^2 |e_j|^2 + g_k^2 |e_k|^2 \\ &\quad\quad\quad+ 2g_ig_j \langle e_i,e_j \rangle + 2g_jg_j \langle e_j,e_k \rangle + 2g_kg_i \langle e_k,e_i \rangle \bigl)\bigr. \\ & = \int_T \left| \mbox{grad} \, g \right|^2.\end{align}
With \begin{align*}\langle e_i,e_j \rangle & = -\cos (\alpha_k) |e_i||e_j|,\\2A = \frac{1}{2}\mbox{det}(e_i,e_j) & = \frac{1}{2} \langle Je_i,e_j \rangle
= \sin(\alpha_k) |e_i||e_j|.\end{align*}
We have:
\[\cot(\alpha_k) = \frac{-\langle e_i,e_j \rangle}{2A}.\]
And therefore we obtain for the Dirichlet energy of $g$ on $T$:
\[ \int_T \left| \mbox{grad} \, g \right|^2 = \frac{1}{2}\bigl( \cot (\alpha_i) (g_j-g_k) ^2 + \cot (\alpha_j) (g_k-g_i) ^2 + \cot (\alpha_k) (g_i-g_j) ^2\bigr).\]

$\square$

Corollary: The Dirichlet energy of $g \in W_{PL}$ is given by :
\[E_D(g) = \frac{1}{4} \sum_{\sigma = \{i,j,k\} \in \Sigma_2}  \cot (\alpha_i) (g_j-g_k) ^2 + \cot (\alpha_j) (g_k-g_i) ^2 + \cot (\alpha_k) (g_i-g_j) ^2.\]

If $E:= \left\{ \{i,j\} \in \Sigma\right\}$ denotes the set of edges and $\tilde{E}:= \left\{ (i,j) \in V \times V\, \big \vert \{i,j\} \in \Sigma\right\}$ the set of oriented edges, we choose $\hat{E} \subset \tilde{E}$ such that:

1)  $E = \left\{ \{i,j\} \subset V \big \vert (i,j) \in \hat{E}\right\}$,
2)  $(i,j) \in \hat{E} \rightarrow (j,i) \notin \hat{E}.$

I.e. for every edges we choose oneoriented one. For every $(i,j) \in \hat{E}$ let $\alpha_{ij}$ denote the angle of the opposite vertex to the left and $\beta_{ij}$ the one the opposite vertex to the right. Then we can write the Derichlet energy in the following form:
\[E_D(g) = \frac{1}{4} \sum_{(i,j) \in \hat{E}} \left( \cot(\alpha_{ij}) + \cot{\beta_{ij}}\right) \left(g_i-g_j \right)^2, \]
where we set the cotangent weights zero if the corresponding vertex does not exist because $(i,j)$ is an boundary edge.

Let $M \subset \mathbb{R}^2$ be a domain with smooth boundary $\partial M$ and outpointing normal vector field $N$. For a smooth function $f \in C^{\infty}(M,\mathbb{R})$ the gradient vector field $\mbox{grad} \, f :M \rightarrow \mathbb{R}^2$ is defined as :
\[ \mbox{grad} \, f := \left( \begin{array}{c} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{array} \right)  = \left( f’\right)^t. \]
For a vector field $v:M \rightarrow \mathbb{R}^2$ the divergence is given by:
\[\mbox{div} \, v := \frac{\partial v_1}{\partial x} +\frac{\partial v_2}{\partial y}.\]
Taking this together we can define the laplace operator $\Delta : C^{\infty}(M,\mathbb{R})\rightarrow C^{\infty}(M,\mathbb{R})$:
\[\Delta \,f := \mbox{div} \, \mbox{grad}\, f = \frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}.\]
The laplace operator plays an important role in physics and geometry. A typical problem from physics is the following called Dirichlet boundary problem: Given $g \in C^{\infty}(\partial M,\mathbb{R})$ find $f \in C^{\infty}(M,\mathbb{R})$ such that:
\[\left\{ \begin{array}{cc} \Delta \, f = 0 \\  \left. f\right|_{\partial M} = g.\end{array} \right.\]
If $f$ is a temperature distribution on $M$ and $g$ is the prescribed temperature on the boundary, then $f$ has to solve the heat equation: $\dot{f} = \Delta \, f.$ For a stationary temperature distribution there holds $\dot{f}=0$ and we obtain the above Dirichlet boundary problem.

Theorem:
1) Given $g \in C^{\infty}(\partial M,\mathbb{R})$ and $d \in C^{\infty}(M,\mathbb{R})$, then there is a unique $f \in C^{\infty}(M,\mathbb{R})$ such that: \[\left\{ \begin{array}{cc} \Delta \, f = d \\ \left. f\right|_{\partial M} = g.\end{array} \right.\]
2) If $d=0$ and $f$ is a solution of the corresponding Dirichlet boundary problem, then $f$ minimizes the Dirichlet energy \[ E_D(f):=\frac{1}{2}\int_M \left|\mbox{grad}\,f\right|^2,\] among all the functions $h \in C^{\infty}(M,\mathbb{R})$ with $\left. h\right|_{\partial M} = g.$

Proof:
1) The proof of the existence of $f$ is hard and can be found in some good books about PDE’s. We only proof the uniqueness of $f$.
Suppose there exists $f,\tilde{f} \in C^{\infty}(M,\mathbb{R})$ with: \( \Delta \, f =  \Delta \,\tilde{f} =0\) and \(\left. f\right|_{\partial M} =\left. \tilde{f}\right|_{\partial M} =g\), then $u:= \tilde{f} -f$ solves:
\[\left\{ \begin{array}{cc} \Delta \, u = 0 \\\left. u\right|_{\partial M} = 0.\end{array} \right.\]Now we consider the vector field $X:= u\, \mbox{grad} \,u$. For the divergence of $X$ we get:
\[ \mbox{div}\,X = \mbox{div} (u\, \mbox{grad} \,u) =\langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u\, \mbox{div}\, \mbox{grad} \, u =  \langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u \Delta \,u.\]
Using the fact that $X$ vanishes on the boundary and applying the divergence theorem we obtain:
\[0=\int_{\partial M}\langle X, N\rangle = \int_M \mbox{div}\,X = \int_M \langle \mbox{grad}\,u, \mbox{grad} \,u\rangle\ + u \Delta \,u = \int_M \left|\mbox{grad}\,u\right|^2.\]
Therefore, we have  $\mbox{grad} \,u=0$ on the whole of $M$ and $u$ is locally constant. Due to the fact that, $\left. u\right|_{\partial M} = 0$ $u$ is identically zero and therefore $f$ is unique.

2)Let $f$ be the unique solution of the Dirichlet boundary problem $\Delta \,f =0, \quad\left. f\right|_{\partial M} =g$ and $h$ an arbitrary smooth function with  $\left. h\right|_{\partial M} =g$. For $u:=h-f$ there holds $\left. u\right|_{\partial M} =0$ and we get:
\[\int_M \left|\mbox{grad},h\right|^2=\int_M \left|\mbox{grad}\,f+u\right|^2=\int_M \left|\mbox{grad}\,f\right|^2+ 2 \int_M \langle \mbox{grad}\,f ,\mbox{grad}\,u \rangle + \int_M \left|\mbox{grad}\,u\right|^2.\]
Using the usual product rule:
\[\mbox{div}(u\,\mbox{grad}\,f ) = \langle \mbox{grad}\,f ,\mbox{grad}\,u \rangle + u\, \Delta\,f, \] and the divergence theorem we obtain:
\begin{align*}2E_D(h)=\int_M \left|\mbox{grad}\,h\right|^2 &=\int_M \left|\mbox{grad}\,f\right|^2+ 2 \int_M \mbox{div}(u\,\mbox{grad}\,f ) -u\, \underbrace{\Delta\,f}_{=0}  + \int_M \underbrace{\left|\mbox{grad}\,u\right|^2}_{\geq 0}\\ &\geq \int_M \left|\mbox{grad}\,f\right|^2 + \int_{\partial M} \underbrace{u\langle \mbox{grad}\, f,N \rangle}_{=0}\\ &= \int_M \left|\mbox{grad}\,f\right|^2 = 2E_D(f).\end{align*}

$\square$

Let us now consider the space of smooth functions on $M$ that vanish at the boundary: \[V:= \left\{f \in  C^{\infty}(M,\mathbb{R}) \, \big\vert \, \left. f\right|_{\partial M}=0 \right\}\]
We can equip  $V$ with two  scalar products $\langle\!\langle \cdot, \cdot \rangle\!\rangle , \, \langle\!\langle\cdot, \cdot \rangle\!\rangle_D :V\times V \rightarrow \mathbb{R},$
\begin{align*}\langle\!\langle f, g\rangle\!\rangle &:=\int_M fg, \\ \langle\!\langle f, g\rangle\!\rangle_D & := \int_M \langle \mbox{grad}\,f ,\mbox{grad}\,g \rangle.\end{align*}
To see that $\langle\!\langle\cdot, \cdot \rangle\!\rangle_D$  is really a scalar product on $V$ one has to check that it is positive definite (symmetry and bi-linearity are obvious).
\begin{align*} & 0=\langle\!\langle f, f\rangle\!\rangle_D  = \int_M \left|\mbox{grad}\,f\right|^2 \\  \Leftrightarrow & \quad \left|\mbox{grad}\,f\right| = 0  \end{align*}
Therefore, $f$ is locally constant. With $\left. f\right|_{\partial M}=0$ one has $f=0$.

Propositon: On V the laplace operator is self adjoint, i.e. $\langle\!\langle \Delta \,f, g\rangle\!\rangle = \langle\!\langle f, \Delta \, g\rangle\!\rangle$ and  there holds: \[\langle\!\langle \Delta \,f, g\rangle\!\rangle =-\langle\!\langle f, g\rangle\!\rangle_D .\]

Proof: \begin{align*} \langle\!\langle \Delta \,f, g\rangle\!\rangle & = \int_M g\, \mbox{div} \, \mbox{grad}\, f  = \int_M  \mbox{div} (g \, \mbox{grad}\, f ) \,- \langle \mbox{grad} \,f ,\mbox{grad} \,g \rangle \\ & = \underbrace{\int_{\partial M} f \langle \mbox{grad} \,g , N \rangle}_{=0}  – \int_M \langle \mbox{grad} \,f ,\mbox{grad}\,g \rangle  \\ &=\,-  \langle\!\langle f, g\rangle\!\rangle_D \\ &= \,- \langle\!\langle g, f\rangle\!\rangle_D \\ & \,\, \vdots \\& =\langle\!\langle f, \Delta \, g\rangle\!\rangle.\end{align*}

$\square$

Let $M\subset \mathbb{R}^n$ be an open set and $f:M\to \mathbb{R}^k$ a smooth map. Then for each $p\in M$ the Jacobian $f'(p)$ is a $k\times n$ matrix. The derivative $d_pf$ of $f$ at $p$ then is the linear map given by matrix multiplication with $f'(p)$:

\begin{align*}d_pf:\mathbb{R}^n &\to \mathbb{R}^k \\\\ d_pf(X)&=f'(p)X.\end{align*}

For example, if $f$ describes a regular surface in $\mathbb{R}^3$ (so $n=2$ and $k=3$), the derivative $df$ assigns to each point $p\in M$ a linear map $d_pf:\mathbb{R}^2 \to \mathbb{R}^3$ and $f'(p)$ is a $3\times 2$ matrix:

$$f'(u,v)=\left(\begin{array}{cc}|&|\\ f_u&f_v\\ |&|\end{array}\right).$$

The image of $d_pf$ is a two-dimensional subspace of $\mathbb{R}^3$ and is called the tangent plane of $f$ at $p$.

Suppose now $k=1$. Then $\omega:=df$ is a so-called 1-form on $M$, which means that $\omega$ assigns to each $p\in M$ a linear form $\omega_p:\mathbb{R}^n\to\mathbb{R}$. Let us define for $j\in \{1,\ldots,n\}$ the $j$th coordinate function $x_j:M\to \mathbb{R}$ by

$$x_j(u_1,\ldots,u_n)=u_j.$$

Then for $p=(u_1\ldots,u_n)$ and

$$Y=\left(\begin{array}{c}y_1\\ \vdots \\y_n\end{array}\right).$$

we have

$$d_px_j(Y)= y_j.$$

Each 1-form $\omega$ on $M$ can be uniquely written as

$$\omega=a_1 \,dx_1 + \ldots + a_n \,dx_n.$$

with functions $a_1,\ldots a_n:M\to \mathbb{R}$ (they always will be assumed to be smooth). For example, if $f:M\to \mathbb{R}$ is a function then

$$df=\frac{\partial f}{\partial x_1} dx_1 +\ldots \frac{\partial f}{\partial x_n} dx_n.$$

A 2-form $\sigma$ on $M$ assigns to each $p\in M$ a skew-symmetric bilinear map

\begin{align*}\sigma_p:\mathbb{R}^n\times \mathbb{R}^n &\to \mathbb{R} \\\\ \sigma_p(X,Y)&=-\sigma_p(Y,X)\end{align*}

For two 1-forms $\omega$ and $\eta$ we define a 2-form $\omega \wedge \eta$ as

$$\omega \wedge \eta (X,Y)=\omega(X)\eta(Y)-\omega(Y)\eta(X).$$

For a 1-form $\omega$ given as above by functions $a_1,\ldots a_n$ we define the exterior derivative $d\omega$ as the 2-form

$$d\omega= da_1\wedge dx_1 +\ldots +da_n \wedge dx_n.$$

For example, for $n=2$ and $\omega=a\,du + b \,dv$ we have

$$d\omega= \left(\frac{\partial a}{\partial v}-\frac{\partial b}{\partial u}\right)du\wedge dv.$$

It is clear how to multiply a 1-form or a 2-form (from the left or from the right) by a function $f:M\to \mathbb{R}$. From these definitions we derive the following rules:

\begin{align*}\omega \wedge \eta&=-\eta \wedge \omega \\\\ d(f\omega)&=df\wedge \omega+f\,d\omega \\\\ d(\omega f)&=d\omega-\omega\wedge df.\end{align*}

If $\omega$ is a 1-form on $M$ and $\gamma:[a,b]\to \mathbb{R}^n$ is asmooth curve we define

$$\int_\gamma \omega := \int_a^b \omega(\gamma'(t))dt.$$

After defining $n$-forms we could give an $n$-dimensional version of the following definition, but for now let $M$ be a compact planar domain with smooth boundary $\partial M$ and let $\sigma=\rho\, du\wedge dv$ be a 2-form on $M$. Then we define

$$\int_M \sigma:= \int_M \rho\,\, du\,dv.$$

Now we can state a special case of the most important theorem that concerns differential forms.

Theorem of Stokes: Let $\omega$ be a 1-form on a compact planar domain$M$ with smooth boundary. Then

$$\int_M d\omega =\int_{\partial M} \omega.$$

Quaternions, $\mathbb{H}$ are a number system like real or complex numbers but with 4 dimensions. In particular, $\mathbb{H}$ is nothing but $\mathbb{R}^4$ together with a multiplication law. The identification of $\mathbb{H}$ and $\mathbb{R}^4$ is given by:
$$ \mathbb{H} = \lbrace x_0 + x_1 {\bf i} + x_2 {\bf j} + x_3 {\bf k} \,\vert \,x_0,x_1,x_2,x_3 \in \mathbb{R} \rbrace.$$
With the usual addition and scalar multiplication inherited from $\mathbb{R}^4$, $\mathbb{H}$ becomes a four dimensional  $\mathbb{R}$ vector space with canonical basis $\{ 1, {\bf i}, {\bf j}, {\bf k} \}.$ The quaternion multiplication is defined by:
\begin{align*}
&\left(x_0 + x_1 {\bf i} + x_2{\bf j} + x_3 {\bf k} \right) \cdot \left(y_0 + y_1{\bf i} + y_2{\bf j} + y_3  {\bf k} \right) \\:= & \left(x_0 y_0 – x_1 y_1 – x_2 y_2 – x_3 y_3 \right) \\
+ & \left(x_0 y_1 + x_1 y_0 + x_2 y_3 – x_3 y_2 \right){\bf i} \\
+ & \left(x_0 y_2 – x_1 y_3 + x_2 y_0 + x_3 y_1 \right){\bf j} \\
+ & \left(x_0 y_3 + x_1 y_2 – x_2 y_1 + x_3 y_0 \right) {\bf k}.
\end{align*}

A more simple way to define the quaternion multiplication is to say that the multiplication with $\alpha \in \mathbb{R}$ works as usually and that ${\bf i}, {\bf j}, {\bf k}$ satisfy the following  multiplication rules:

\[{\bf i}^2 = {\bf j}^2 = {\bf k}^2 = -1,\quad {\bf ij}=-{\bf ji}={\bf k}, \quad {\bf jk}=-{\bf kj}={\bf i}, \quad {\bf ki}=-{\bf ik}={\bf j}.\]

${\rm Re}(\mathbb{H}) := span\{1\}$ is called the real part of  $\mathbb{H}$ and ${\rm Im}(\mathbb{H}) := span\{{\bf i}, {\bf j}, {\bf k}\}$  the imaginary part and there holds:  $\mathbb{H} = {\rm Re}(\mathbb{H}) \oplus {\rm Im}(\mathbb{H})$. If we identify ${\rm Re}(\mathbb{H})$ with $\mathbb{R}$ and ${\rm Im}(\mathbb{H})$ with $\mathbb{R}^3$, we can write any ${\bf x} \in \mathbb{H} $ as: $\quad {\bf x} = \alpha + {\bf v},\quad $ for some  $\alpha \in \mathbb{R}$ and ${\bf v}\in \mathbb{R}^3.$ This notation gives us a new expression of the quaternion multiplication using the scalar and cross product of $\mathbb{R}^3$:
$${\bf p}{\bf q} = (\alpha + {\bf v}) (\beta + {\bf w}) = \alpha \beta -\langle {\bf v},{\bf w}\rangle + \alpha {\bf w}+ \beta {\bf v} + {\bf v} \times {\bf w}.$$

Due to the fact that the cross product is skew symmetric, we immediately see that the quaternion multiplication is not commutative, i.e. in general there holds ${\bf p}{\bf q} \neq {\bf q}{\bf p}$.

Proposition: The quaternion multiplication is associative i.e. for ${\bf p}, {\bf q}, {\bf r} \in \mathbb{H}$ there holds:
$$ ({\bf p}{\bf q}){\bf r}={\bf p}({\bf q}{\bf r})$$

Proof: Let \[ I :=  \left(\begin{array}{cc} 1 & 0 \\0 & 1\end{array} \right), \quad X := \left(\begin{array}{cc}{\bf i} & 0 \\ 0 & -{\bf i}\end{array}\right), \quad Y := \left(\begin{array}{cc} 0 & -1 \\ 1 & 0\end{array}\right), \quad Z := \left(\begin{array}{cc} 0 & -{\bf i} \\- {\bf i} & 0\end{array}\right). \]

Now we can define a linear map \begin{align*} F:\mathbb{H} &\rightarrow \mathbb{C}^{2\times 2} \\ x_0 + x_1 {\bf i} + x_2 {\bf j} + x_3 {\bf k} &\mapsto x_0 I + x_1 X + x_2 Y + x_3 Z \end{align*}
A straight forward calculation shows that $I,X,Y,Z$ satisfy the following product rules with respect to the matrix multiplication:
\begin{align*}  I^2=I , \quad &IX=XI=I, \quad XY=-YX = Z, \\ &IY=YI =Y , \quad YZ=-ZY=X, \\ &IZ=ZI=Z, \quad ZX=-XZ =Y.\end{align*}
Therfore, $F$ is an algebra isomorphism onto its image. Due to the fact that the matrix multiplication is associative we obtain that the quaternion multiplication is associative too.

The conjugate of an quaternion number $ {\bf x} = \alpha + {\bf v}$ is given by $ {\bf\bar{ x}} = \alpha – {\bf v}$.

Proposition: For $ {\bf p}, {\bf q} \in \mathbb{H}$ there holds : $ {\bf\overline{ pq}} = {\bf\bar{ q}}{\bf\bar{ p}}.$

Proof: \begin{align*} {\bf p}{\bf q} &= \alpha \beta -\langle {\bf v},{\bf w}\rangle + \alpha {\bf w}+ \beta {\bf v} + {\bf v} \times {\bf w} \\
\overline{ {\bf p}{\bf q}} &= \alpha \beta – \langle {\bf v},{\bf w}\rangle – \alpha {\bf w} – \beta {\bf v} + {\bf w} \times {\bf v}\\
&= {\bf \bar{q}}{\bf \bar{p}}\end{align*}

$\mathbb{H}$ also inherits the euclidean norm from $\mathbb{R}^4$: $|{\bf p}| = (\sum_{i=0}^3 p_i^2)^{\frac{1}{2}}$. Similar to the complex numbers we have $|{\bf p}|^2 = {\bf p}{\bf \bar{p}}={\bf \bar{p}}{\bf p}$. Additionally, there holds the following useful formula:

Proposition: For ${\bf p},{\bf q} \in \mathbb{H}$ there holds: $|{\bf p}{\bf q}|=|{\bf p}||{\bf q}|.$

Proof:\begin{align*} |{\bf p}{\bf q}| =(({\bf p}{\bf q}) (\overline{{\bf p}{\bf q}}))^{\frac{1}{2}} = ({\bf p}{\bf q} {\bf \bar{q}}{\bf \bar{p}})^{\frac{1}{2}}  = ({\bf p} |{\bf q}|^2{\bf \bar{p}})^{\frac{1}{2} } = (|{\bf p}|^2 |{\bf q}|^2)^{\frac{1}{2} } = |{\bf p}||{\bf q}|. \end{align*}

Note that for ${\bf q} \in \mathbb{H}$ the inverse element with respect to the quaternion multiplication is given by
$${\bf q} ^{-1} = \frac{{\bf \bar{q}} }{|{\bf q}|^2 }.$$
For quaternions with unit length this gives us immediately ${\bf q} ^{-1} = {\bf \bar{q}}.$
What makes quaternions so useful is the fact that one can describe rotation in $\mathbb{R}^3$ with them in a very elegant way. Therefore, we have to consider $\mathbb{R}^3$ as ${\rm Im}(\mathbb{H})$.

Theorem: Let ${\bf a} \in \mathbb{R}^3$ with $|{\bf a}|=1$ , $\alpha \in \mathbb{R}$ and ${\bf q} = \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}$. Then for all ${\bf y} \in \mathbb{R}^3$ we have:

(i)  ${\bf q}{\bf y} {\bf \bar{q}} \in {\rm Im} (\mathbb{H}) = \mathbb{R}^3$

(ii) The map $F:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, ${\bf y} \mapsto {\bf q}{\bf y}{\bf \bar{q}}$ is a rotation around ${\bf a}$ by the angle $\alpha$.

Proof: (i) First note that a quaternion ${\bf y}$ is purely imaginary if and only if ${\bf \bar{y}}= -{\bf y}$. Thus we have to show $\overline{{\bf q}{\bf y}{\bf \bar{q}}}= \,- {\bf q}{\bf y}{\bf \bar{q}}$.
$$\overline{{\bf q}{\bf y}{\bf \bar{q}}} = {\bf q}{\bf \bar{y}}{\bf \bar{q}} = \,- {\bf q}{\bf y}{\bf \bar{q}}.$$

(ii) $F$ is a linear map and therefore completely determined by its action on a basis. We extend ${\bf a}$ to an positive oriented  orthonormal basis $\{{\bf a}, {\bf b}, {\bf c} \}$ of $\mathbb{R}^3$.
\begin{align*}F({\bf a}) = {\bf q}{\bf a}{\bf \bar{q}} &= \left( \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}\right) {\bf a}\left(\cos\left(\frac{\alpha}{2}\right) – \sin\left(\frac{\alpha}{2}\right) {\bf a}\right) \\ &= \left( \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}\right)\left(\cos\left(\frac{\alpha}{2}\right){\bf a} + \sin\left(\frac{\alpha}{2}\right)\right ) \\ &= \left( \cos^2\left(\frac{\alpha}{2}\right) + \sin^2\left(\frac{\alpha}{2}\right)\right) {\bf a} + \cos\left(\frac{\alpha}{2}\right) \sin\left(\frac{\alpha}{2}\right) (1+{\bf a}^2) \\ & =  {\bf a}.\end{align*}

In the last step we used that ${\bf a}^2= \,-1$. This gives us that  the ${\bf a}$-axis is invariant under $F$. Now we consider the action of $F$ on the plane orthogonal to ${\bf a}$, i.e. the plane spanned by $ {\bf b}$ and ${\bf c}$. Since $ \langle {\bf a}, {\bf b} \rangle =0 $ we have:
$$ {\bf a}{\bf b} = {\bf a} \times {\bf b} = \,- {\bf b} \times {\bf a} = \,- {\bf b}{\bf a}.$$

Using this and the addition formulas for cosine and sine we get:
\begin{align*} F({\bf b}) & = \left( \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}\right) {\bf b} \left(\cos\left(\frac{\alpha}{2}\right) – \sin\left(\frac{\alpha}{2}\right) {\bf a}\right) \\ &= \left( \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}\right) \left(\cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a}\right)  {\bf b} \\ &=\left( \cos(\alpha) + \sin(\alpha) {\bf a}\right) {\bf b} = \cos(\alpha){\bf b} + \sin(\alpha)\left( {\bf a} \times {\bf b}\right) \\ &=  \cos(\alpha){\bf b} + \sin(\alpha) {\bf c}\end{align*}

Analog we obtain $F({\bf c}) =  \cos(\alpha){\bf b} – \sin(\alpha) {\bf b} $ and the matrix representation of $F$ with respect to the basis  $\{{\bf a}, {\bf b}, {\bf c} \}$ is given by:
\[ F =  \left(\begin{array}{ccc} 1 & 0 & 0 \\0 & \cos(\alpha) & -\sin(\alpha) \\ 0 & \sin(\alpha) & \cos(\alpha) \end{array} \right)\].

Now it is easy to see that $F$ describes a rotation around ${\bf a}$ by the angle $\alpha$.

Corollary: For every $ {\bf q} \in \mathbb{H}\backslash \{0\}$ the map $F:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, ${\bf y} \mapsto {\bf q}{\bf y}{\bf q}^{-1}$ is a rotation.

Proof: With ${\bf q} ^{-1} = \frac{{\bf \bar{q}} }{|{\bf q}|^2 }$ we get:
$$ {\bf q}{\bf y}{\bf q}^{-1} = \frac{{\bf q}}{|{\bf q}|}{\bf y}\overline{\left(\frac{{\bf q} }{|{\bf q}| }\right)}$$
Since $\frac{{\bf q}}{|{\bf q}|}$ has unit length there exists ${\bf a} \in \mathbb{R}^3$ with $|{\bf a}|=1$ and  $\alpha \in \mathbb{R}$  such that: $\frac{{\bf q}}{|{\bf q}|} = \cos\left(\frac{\alpha}{2}\right) + \sin\left(\frac{\alpha}{2}\right) {\bf a} $ and we can apply the theorem.

Note that the quaternion multiplication corresponds to the concatenation of rotations. Let ${\bf p}, {\bf q} \in \mathbb{H}$ with $|{\bf p}|= |{\bf q}| =1$ and $F,G:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ the corresponding rotations i.e. $F({\bf y}) = {\bf q}{\bf y}{\bf \bar{q}}$ and $G({\bf y}) = {\bf p}{\bf y}{\bf \bar{p}}$, then we get for the concatenation:

$$\left( F \circ G\right) (y) = {\bf q}\left( {\bf p}{\bf y}{\bf \bar{p}} \right) {\bf \bar{q}} = \left({\bf q} {\bf p}\right){\bf y}\overline{\left({\bf q} {\bf p}\right)}.$$