The title and word choice are a bit out-of-date. Surface fairing refers to mesh smoothing and denoising. The ansatz of the paper is to cast a given mesh as a signal that can be smoothed with a low-pass filter.
The ansatz is motivated and explained in the 1D case. Curves. A discrete laplacian definition is given without proof or background.
\[\Delta x_i = \frac{1}{2}\Big(x_{i+1} - x_{i}\Big) + \frac{1}{2}\Big(x_{i-1} - x_{i}\Big)\]The difference of differences around $x_i$ definition given is an approximation of the second derivative, and when viewing the polygon vertices as a discrete sequence, it is essentially the second finite difference of the sequence ${x_i}$,
\[\begin{align} \Delta^2 x_i &= x^{i+1} - 2 x_{i} + x^{i-1} \\ &= \Big(x_{i+1} - x_{i}\Big) + \Big(x_{i-1} - x_{i}\Big) \\ \end{align}\]When written as a matrix the Laplacian of the sequence is the circulant matrix, \(K = -\frac{1}{2}\begin{bmatrix} -2 & 1 & 0 & \cdots & 0 & 1\\ 1 & -2 & 1 & 0 & \cdots & 0 \\ 0 & 1 & \ddots & \ddots & \ddots & \vdots \\ 0 & 0 & \ddots & \ddots & 1 & 0 \\ \vdots & \vdots & \ddots & 1 & -2 & 1 \\ 1 & 0 & \cdots & 0 & 1 & -2 \\ \end{bmatrix}.\)
By the Spectral Theorem $K$ symmetric implies that it can be diagonalized by some real orthogonal matrix, $\Lambda = O K O^T$. Therefore, $K$ has real eigenvalues. Though, orthogonally diagonalizing $K$ is overkill.
Alternate proof:
Suppose $0 \neq \lambda \in \mathbb{C}$ is an eigenvalue of $K$, for some $v \in \mathbb{C}^n$. Also define the inner product on $\mathbb{C}^n$ to be $v \cdot w := v^T \bar{w}$. The product $Kv \cdot v$ can be computed two different ways:
Then, one must conclude $\bar{\lambda} = \lambda \ \Longrightarrow \lambda \in \mathbb{R}$. $\Box$
By a similar argument, it can be shown that $v \neq w$ eigenvectors of $K$, which do not span each other, must be orthogonal, if they have different eigenvalues. Just compute $Kv \cdot w$ two different ways. Assuming $\lambda_v \neq \lambda_w$ then
Since $\lambda_v \neq \lambda_w \ \Longrightarrow \ v \cdot w = 0$.
Eigenvectors of $K$ belonging to the same eigenspace are also orthogonal. For a proof checkout these notes.
$K$ is a circulant matrix so we can calculate eigenvalues. In my notes on circulant matrices, I did just that. In general the eigenvalues have the following form.
\(\begin{equation} \lambda_C = c_0 + c_1 e^{i\frac{2\pi k}{n}} + c_2 e^{i\frac{4\pi k}{n}} +\cdots + c_{n-1}e^{i\frac{2\pi k(n-1)}{n}} \end{equation}\) For $\Delta \approx K$, all coefficients $c_i$ are zero except for $c_0 = -2$, $c_1 = 1$, and $c_{n-1} = 1$. Substituting those specific coefficients into the eigenvalue formula and simplifying gives,
\[\begin{align} \lambda_k &= -2 + e^{i\frac{2\pi k}{n}} + e^{i\frac{2\pi k(n-1)}{n}} \\ &= -2 + e^{i\frac{2\pi k}{n}} + \cancelto{1}{e^{i2\pi k}} + e^{-i\frac{2\pi k}{n}} \\ &= -1 + 2\cos\big(\frac{2\pi k}{n}\big) \end{align}\]Based on the previous section, we expected the eigenvalues of $K$ to be real, and they are indeed real.